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Page 1: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

BULGARIAN NATIONAL OLYMPIAD IN MATHEMATICS

Third round

����

Problem �� Let p and q be positive numbers such that the parabola y � x� � �px� q hasno common point with the x�axis� Prove that there exist points A and B on the parabola suchthat the segment AB is parallel to the x�axis and � AOB � ��� O is the coordinate origin if

and only if p� � q � �

�� Find the values of p and q for which the points A and B are de ned in

an unique way�Solution� Since the parabola has no common point with the x�axis� then the roots of the

equation x�� �px� q � � are not real and hence p� � q� Let the points Ax�� y� and Bx�� y�Figure � be with the required properties� Then x� and x� are the roots of the equationx� � �px � q � y� � � and y� � q � p�� because the vertex of the parabola has coordinatesp� q� p�� On the other hand OA� � x�� � y�� � OB

� � x�� � y�� � AB� � x� � x�

� and it followsfrom the Pythagorean theorem that y�� � x�x� � �� But x�x� � q� y� and thus y�� � y� � q � ��Consequently the existence of the points A and B is equivalent to the assertion that the equationfy � y� � y � q � � has a solution y� � q � p�� A and B are de ned in an unique way ifthis is the only solution� A necessary condition is that the discriminant of the equation is not

negative� i�e� q � �

�� The last condition is su�cient because fq � p� � q � p� � p� � � and

��

�� q � p�� The corresponding solution y� is unique i� q �

��

Figure �� Figure ��

Page 2: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Let A�A�A�A�A�A�A�� B�B�B�B�B�B�B�� C�C�C�C�C�C�C� be regularheptagons with areas SA� SB and SC � respectively� Let A�A� � B�B� � C�C�� Prove that

��

SB � SC

SA� ��

p��

Solution� Let A�A� � a� A�A� � b� A�A� � c Figure �� By the Ptolomeus theorem for the

quadrangle A�A�A�A� it follows that ab�ac � bc� i�e�a

b�a

c� �� Since �A�A�A�

�� �B�B�B��

thenB�B�

B�B�

�a

band hence B�B� �

a�

b� Analogously C�C� �

a�

c� Therefore

SB � SC

SA�

a�

b��a�

c��

Thena�

b�a�

c�

�a

b�a

c� �

�equality is not possible because

a

b�� a

c� On the other hand

a�

b��a�

c��

�a

b�a

c

��� �a�

bc� �� �a�

bc� �

By the sine theorem we geta�

bc�

sin��

sin��

�sin

��

��

� cos��

�� � cos

��

�� Since cos

��

��

cos�

��

p�

�� then

a�

bc�

p�

�� �

p�

�p���� From here and from � we get the right hand

side inequality of the problem�Problem �� Let n � � be an integer� Find the number of the permutations a�� a�� � � � � an of

the numbers �� �� � � � � n with the following property� there exists only one index i � f�� �� � � � � n��g such that ai � ai��

Solution� Denote by pn the number of the permutations with the given properties� Obviously�p� � � and p� � �� Let n � �� The number of the permutations with an � n is equal to pn���Consider all the permutations a�� a�� � � � � an with ai � n� where � � i � n � � is xed� Their

number is

�n � �

i� �

�� Consequently

pn � pn�� �n��Xi�

�n � �

i� �

�� pn�� � �n�� � ��

From here

pn � �n�� � � � �n�� � � � � � �� �� �

� �n � n � ��

Problem �� Let n � � and � � xi � � for i � �� �� � � � � n� Prove the inequality

x� � x� � � � �� xn� x�x� � x�x� � � � �� xn��xn � xnx� ��n

��

When is there an equality�Solution� Denote by Sx�� x�� � � � � xn the left hand side of the inequality� This function is

linear with respect to each of the variables xi� Particularly�

Sx�� x�� � � � � xn � max S�� x�� � � � � xn� S�� x�� � � � � xn �

Page 3: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

From here it follows by induction that it is enough to prove the inequality when all xi areequal to � or �� On the other hand for arbitrary xi we have

�Sx�� x�� � � � � xn � n � �� x��� x�� �� x��� x�� � � ����� xn�� x�� x�x� � x�x� � � � � � xnx�

i�e� Sx�� x�� � � � � xn � n

�� when xi � ��� ��� In the case when xi are equal to � or �� the left

hand side of the last inequality is an integer� Consequently Sx�� x�� � � � � xn ��n

�� It follows

from that when n is even� the equality is satis ed i� x�� x�� � � � � xn � �� �� �� �� � � � � �� �� when n is odd� the equality is satis ed i� x�� x�� � � � � xn � x� �� �� �� �� � � � � �� �� where

x � ��� �� is arbitrary�Problem �� The points A�� B�� C� lie on the sides BC� CA� AB of the triangle ABC

respectively and the lines AA�� BB�� CC� have a common point M � Prove that if the point Mis center of gravity of �A�B�C�� then M is the center of gravity of �ABC�

Solution� Let M be the center of gravity of �A�B�C�� Let A� be a point on MA� suchthat B�A�C�A� is a parallelogram� The points B� and C� are constructed analogously� SinceA�C�kA�B�kC�B�� then the points A�� C�� B� are colinear and C� is the midpoint of A�B��The same is true for the points A�� B�� C� and C�� A�� B�� We shall prove that A� � A� B� � B

and C� � C� which will solve the problem�Assume that A� �� A and let A be between A� and M � Then C� is between C and M � B is

between B� and M and consequently A� is between A and M � which is a contradiction�

Problem �� Find all pairs of positive integers x� y for whichx� � y�

x� yis an integer which

is a divisor of �����Solution� It is enough to nd all pairs x� y for which x � y and x� � y� � kx� y� where

k divides ���� � ��������� We shall use the following well�known fact� if p is a prime numberof the type �q � � and if it divides x� � y�� then p divides x and y� For p � �� �� �� this canbe proved directly� If k is divisible by � then x and y are divisible by � too� Simplifying by� we get an equality x�� � x�� � k�x� � y�� where k� divides ������� Considering � and �� inan analogous way we get an equality a� � b� � �a � b it is not possible to get an equalitya�� b� � a� b� where a � b� From here �a� ��� �b� �� � ��� i�e� a � �� b � � or a � ��b � ��

The above considerations imply that the pairs we are looking for are of the type �c� c��c� c� c� �c� c� �c� where c � �� �� �� ��� ���� ����� ����� �������

Page 4: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

BULGARIAN NATIONAL OLYMPIAD IN MATHEMATICS

Fourth round

����

Problem �� Find the number of all integers n � �� for which the number a���a is divisibleby n for every integer a�

Solution� Let n be with the required property� Then p� �p prime� does not divide n sincep� does not divide p�� � p� Hence n is a product of pairwise di�erent prime numbers� On theother hand ��� � � �������������� �� But n is not divisible by �� and � � because �� � ��mod ��� and �� � � �mod � ��� The Fermat theorem implies that a�� � a �mod p�

when p �� � �� �� �� Thus n should be equal to the divisor of � � � � � � � �� di�erent from ��Therefore the number we are looking for is �� � � ��

Problem �� A triangle ABC with semiperimeter p is given� Points E and F lie on theline AB and CE CF p� Prove that the excircle k� of �ABC to the side AB touches thecircumcircle k of �EFC�

Solution� Let P and Q be the tangent points of k� with the lines CA and CB� respectively�Since CP CQ p� then the points E� P � Q and F lie on the circle with center C and radius p�We denote by i the inversion de�ned by this circle� Since i�P � P � i�Q� Q� then i�k�� k��On the other hand i�E� E and i�F � F � Hence i�k� is the line AB� But k� touches AB andthus k touches k��

Problem �� Two players A and B take stones one after the other from a heap with n � �stones� A begins the game and takes at least � stone but no more then n � � stones� Eachplayer on his turn must take at least � stone but no more than the other player has taken beforehim� The player who takes the last stone is the winner� Find who of the players has a winningstrategy�

Solution� Consider the pair �m� l�� where m is the number of the stones in the heap and l

is the maximal number of stones that could be taken by the player on turn� We must �nd forwhich n the position �n� n � �� is winning �i�e� A wins� and for which n it is losing �B wins��We shall apply the following assertion several times� If �m� l� is a losing position and l� � l�then �m� l�� is losing too�

Now we shall prove that �n� n� �� is a losing position i� n is a power of ��Su�ciency� Let n �k� k � �� If k � then B wins on his �rst move� Assume that

��k� �k� �� is a losing position and let consider the position ��k��� �k��� ��� If A takes at least�k stones on his �rst move� then B wins at ones� Let A take l stones� where � � l � �k� Bythe inductive assumption B could play in such a way that he could win the game ��k� l� sincel � �k��� the last move will be the move of B� After this move we get the position ��k� m� withm � l� which is losing for A� according to the inductive assumption�

Necessity� It is enough to prove that if n is not a power of �� then �n� n � �� is a winningposition� Let n �k � r� where � � r � �k � �� On his �rst move A takes r stones and B isfaced to the position ��k� r�� which is losing for B�

Page 5: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� The points C�� A� and B� lie on the sides AB� BC and CA of the equilateraltriangle ABC respectively in such a way that the inradii of the triangles C�AB�� B�CA�� A�BC�

and A�B�C� are equal� Prove that A�� B� and C� are the midpoints of the corresponding sides�Solution� We shall prove that BA� CB� AC� �Figure Figure ��

��� Assume the contrary and let BA� � CB� � AC�� Let �

be the rotation at ���� which center coincides with the in centerof the incircle of �ABC� This rotation transforms the incirclesof the triangles C�BA�� A�CB� and B�AC� to the incircles ofthe triangles A�CB�� B�AC� and C�BA�� respectively� Let A� ��A��� B� ��B�� and C� ��C��� It follows that BB� � BC�

and BC� � BA�� But the incircles of the triangles BC�A� andBC�B� have equal radii �because ���AC�B�� �BC�B��� whichis a contradiction�

Let r be the radius of the incircles of the triangles C�AB��B�CA�� A�BC� and A�B�C�� From the triangle B�AC� we have

r ��B�C�

��

p

� and from �A�B�C� which is equilateral we

have r B�C��

p

�� From here B�C�

�and consequently A�� B�� C� are midpoints of the

corresponding sides�Problem �� Let A f�� �� � � � � m � ng� where m and n are positive integers and let the

function f � A� A be de�ned by the equations�

f�i� i� � for i �� �� � � � � m� �� m� �� � � � � m� n � �

f�m� � and f�m� n� m� ��

a� Prove that if m and n are odd then there exists a function g � A� A such that g�g�a�� f�a� for all a � A�

b� Prove that if m is even then m n i� there exists a function g � A � A such thatg�g�a�� f�a� for all a � A�

Solution� a� Let m �p��� n �q�� and g�i� p� i�� for i �� �� � � � � p� g�i� q� i��for i m � �� m� �� � � � � m� q� g��p� �� p � �� g�p� �� �� g�m� �q � �� m � q � ��g�m� q � �� m� �� It is easy to check that g�g�a�� f�a� for all a � A�

b� Let m n and g�i� m� i for i �� �� � � � � m� g�m� i� i � � for i �� �� � � � � m� ��g��m� ��

For the converse let M f�� �� � � � � mg� It follows by the de�nition of f that the elementsof M remain in M after applying the powers of f with respect to superposition� Moreover�these powers scoop out the whole M � The same is true for the set A nM � The function f isbijective in A and if there exists g verifying the condition� then g is bijective too� We shall provethat g�M�

TM � It follows from the contrary that there exists i � M such that g�i� � M �

Consider the sequence i� g�i�� g��i�� � � � and the subsequence i� f�i�� f��i�� � � �� It is easy to see thatg�M� M � We deduce that there exists a permutation a�� a�� � � � � am of elements of M � suchthat g�ai� ai�� for i �� �� � � � � m� �� g�am� a� and f�a�i��� a�i�� for i �� �� � � � � s� ��f�a�s��� a�� where m �s� The last contradicts to the properties of f which were mentionedalready� Thus g�M�

TM � Analogously g�A nM� A nM � if g�i� � A nM for i � A nM �

At last let us observe that when starting from an element of M and applying g we go to A nM �but when applying g for a second time we go back to M � The same is true for the set A nM �

Page 6: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

From here and from the bijectivity of g it follows that M and A nM have one and the samenumber of elements� i�e� n m�

Problem �� Let x and y be di�erent real numbers such thatxn � yn

x� yis an integer for some

four consecutive positive integers n� Prove thatxn � yn

x� yis integer for all positive integers n�

Solution� Let tn xn � yn

x� y� Then tn�� � b�tn�� � c�tn � for b ��x� y�� c xy� where

t� �� t� �� We shall show that b� c � Z� Let tn � Zfor n m�m� �� m� �� m� � Sincecn �xy�n t�n�� � tn�tn�� �Zwhen n m�m� �� then cm� cm�� �Z� Therefore c is rationaland from cm�� �Zit follows that c �Z� On the other hand

b tmtm�� � tm��tm��

cm�

i�e� b is rational� From the recurrence equation it follows by induction that tn could be rep�resented in the following way tn fn���b�� where fn���X� is a monic polynomial with integercoe�cients and degfn�� n � �� Since b is a root of the equation fm�X� tm��� then b � Z�Now from the recurrence equation it follows that tn �Zfor all n�

Page 7: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

BULGARIAN NATIONAL OLYMPIAD IN MATHEMATICS

Third round

����

Problem �� Prove that for all positive integers n � � there exist an odd positive integersxn and yn� such that

�x�n � y�n � �n�

Solution� If n � � we have x� � y� � �Suppose that for an integer n � � there are odd positive integers xn� yn� such that �x�n�y

�n �

�n� We shall prove that for each pair�X �

xn � yn

�� Y �

j�xn � ynj

�and

�X �

jxn � ynj

�� Y �

�xn � yn

we have �X� � Y � � �n��� Indeed�

�xn � yn

��

��xn � yn

��

� ���x�n � y�n

�� � � �n � �n���

Since xn and yn are odd� i�e� xn � �k � and yn � �l � k� l are integers�� thenxn � yn

�� k � l � and

jxn � ynj

�� jk � lj� which shows that one of the numbers

xn � yn

andjxn � ynj

�is odd� Thus� for n � there are odd natural numbers xn�� and yn�� with the

required property�Problem �� The circles k� and k� with centers O� and O� respectively are externally tangent

at the point C� while the circle k with center O is externally tangent to k� and k�� Let � bethe common tangent of k� and k� at the point C and let AB be the diameter of k� which isperpendicular to �� the points A and O� lie in one and the same semiplane with respect to theline �� Prove that the lines AO�� BO� and � have a common point�

Figure � Figure ��

Page 8: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� Denote by r� r� and r� the radii of k� k� and k�� by M and N the tangent pointsof k with k� and k�� respectively and by P the common point of � and AB Figure ��

It follows from O�O� � � and AB � � that �BON � �CO�N � Then � CNO� � � ONB�

and consequently the points C� N and B are colinear� Also�BN

CN�

BO

CO�

�r

r�� Analogously�

A� M and C are colinear andAM

MC�

r

r��

The lines AN � BM and � have a common point H � which is the altitude center of �ABC�

By Ceva�s theorem we have AP

PB�BN

NC�CM

MA�AP

PB�r

r��r�

r� � from where

r�

PB�

r�

AP� �

Let now D� and D� be the common points of the line � with the lines BO� and AO��

respectively� Obviously� �O�CD� � �BPD� Figure ��� from whereCD�

D�P�

r�

PB� Analogously�

CD�

D�P�

r�

APand according to � we have

CD�

D�P�CD�

D�P� which shows that D� D�� Thus� the

lines AO�� BO� and � have a common point�Problem �� a� Find the maximal value of the function y �

���x� � �x�� in the interval ��� ��

b�Let a� b and c be real numbers and M be the maximal value of the function y ����x� � ax� � bx� c�� in the interval ��� �� Prove that M � � For which a� b� c is the equality

reached�Solution� a� Using that

��x� � �x

��� �x� � � �Figure ��

� ��� � ��

r r r r

�x�

��x�

�� we �nd that the function

y �����x� � �x

���

has a local maximums when x � �

�� Then its maximal

value in the interval ��� � is the biggest among the

numbers y��� y�� y

��

�and y

�Figure ���

But y�� � y

��

�� y

�� y� � � thus the

maximal value is equal to �b� Let fx� � �x� � ax� � bx � c� Assume that there exist numbers a� b� c� for which

the maximal value M of the function y � jfx�j in ��� � is less than � i�e� M � � Then� � fx� � for all x ��� ��

Consider the function gx� � fx����x� � �x

�� ax� � b� ��x� c� We have g�� � ��

g

��

�� �� g

�� � and g� � �� Consequently gx� changes its sign at least � times�

which means that the quadratic equation ax� � b � ��x� c � � has at least � di�erent roots�This is possible only if a � b� � � c � �� i�e� if fx� � �x� � �x� According to a� the maximalvalue of y �

���x� � �x�� is � �

The equality M � is reached only when a � �� b � �� and c � ��Problem �� The real numbers a�� a�� � � � � an n � �� form an arithmetic progression� There

exists a permutation ai� � ai�� � � � � ain of the same numbers� which is a geometric progression�

Page 9: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Find the numbers a�� a�� � � � � an� if they are pairwise di�erent and the biggest among them isequal to ����

Solution� Let a� � a� � � � �� an � ��� and q be the quotient of the geometric progressionai� � ai�� � � � � ain � We have q �� � and q �� � The numbers ain � ain�� � � � � � ai� form also a geometric

progression which quotient is

q� Thus� we can assume that jqj � � i�e� q � or q � �� Then

jai� j � jai� j � � � �� jain j� from where ai �� � for all i�More exactly� either all numbers are positive q � � and then ai� � ai� � � � � � ain � which

together with a� � a� � � � � � an shows that aik � ak � i�e� the numbers a�� a�� � � � � an forman arithmetic as well as a geometric progression� or the numbers ai� � ai� � � � � � ain change theirsigns alternatively q � �� and then the positive ones form an increasing geometric progressionwith quotient q�� and the order is the same as in the arithmetic progression� The numbersa�� a�� � � � � an could not be all negative� because an � ��� � ���

Assume now that � among the numbers a�� a�� � � � � an are positive� Then � � an�� � an�� �

an and they form a geometric as well as an arithmetic progression� Therefore �an�� � an���anand a�

n�� � an��an� From here an�� � an�� � an� which is a contradiction�Thus at most two among the numbers are positive� Analogously� at most two among the

numbers are negative� Consequently� n � ��Let n � �� Then a� � a� � � � a� � a� and �a� � a� � a�� �a� � a� � a�� But q � � and

the geometric progression is either a�� a�� a�� a� or a�� a�� a�� a�� Let it be a�� a�� a�� a�� Thena� � a�q� a� � a�q

� and a� � a�q�� Thus� �a�q � a�q

� � a� and �a� � a�q � a�q�� From here

q � � which contradicts to q � ��So n � �� There are two possibilities I� a� � a� � � � a� � ���� Then the geometric progression is a�� a� � a�q� a� � a�q

�� Itfollows from �a� � a��a� that �a� � a�q

��a�q� i�e� q��q�� � �� Thus� q � ��� ��a� � ����a� � ���� and the numbers are ����������� �����

II� a� � � � a� � a� � ���� Now the geometric progression is a�� a� � a�q� a� � a�q��

From �a� � a� � a� we obtain �a� � a�q � a�q�� i�e� again q � ��� Therefore� a� � �a� � ���

and a� � ���� The numbers are ����� ���� �����Problem �� A convex quadrilateral ABCD� for which � ABC � � BCD � ���� is given�

The common point of the lines AB and CD is E� Prove that � ABC � � ADC if and only if

AC� � CD � CE �AB �AE�

Solution� Let � ABC � �� � ADC � �� � BAC � and � CAD � Figure ��� The point Ais between E and B and the point D is between E and C� Also� � AEC � � � � � ����

Applying the sine theorem to the trian� Figure ��gles ACD�ACE�ABC and again to�ACE�we obtain

CD

AC�

sin

sin ��

CE

AC� �

sin

sin � � � ��

AB

AC�

sin � � �

sin ��

AE

AC� �

sin � � �

sin � � � ��

Page 10: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

From here the equation AC� � CD �CE � AB �AE is equivalent to the equations

CD

AC�CE

AC�AB

AC�AE

AC� � ��

sin � � � sin � � � sin � � sin sin sin �� sin � sin � sin � � �� � ��

cos � � � � � �� cos � � � � � �� sin �� cos � �� cos � �� sin �� cos � � ��� cos � � ��� sin � � �� � �

sin � � � �� sin �� � � � ��� sin �� � � � � �� sin � � � � � ��

sin � � �� cos � � � � � sin � � �� cos � � � � � ��

sin � � �� sin � � � sin � �

But sin �� � and sin � � � �� �� Consequently� sin � � �� � � and � � ��Problem �� A rectangle m n m � � n � � is divided into mn squares with lines�

parallel to its sides� In how many ways could two of the squares be canceled and the remainingpart be covered with dominoes � �

Solution� Denote by F m�n� the number we are looking for� Since every domino coversexactly two squares� then F m�n� � � if m and n are odd�

Let at least one of the numbers m and n be even� We color theFigure ��squares in two colors � white and black in such a way that everytwo neighbor squares with common side� are of di�erent colorsFigure ��� The number S� of the white squares is equal to the

number S� of the black ones and S� � S� �mn

��

Each domino covers one white and one black square� If twowhite or two black squares are canceled� then it is impossible tocover by dominoes the remaining part of the rectangle� Now weshall show that if one white and one black squares are canceled

then the remaining part can be covered by dominoes�Since mn is even� then mn � �t� where t � �� We make induction with respect to t� The case

t � � is obvious� Let t� � � and the proposition is true for all � � t � t�� Let mn � � t� � ��Denote by T� the rectangle consisted of the �rst two rows of the considered rectangle� and by T�� the rectangle consisted of the remaining rows� If the two canceled squares are in T� or in T��then we can cover each of the rectangles by dominoes and consequently we can cover the givenrectangle�

Let one of the canceled squares be in T�� and the other � in T�� We place a domino insuch a way that it covers one square from T� and one square from T�� It is also possible thatthe canceled square from T� and the covered square from T� are of di�erent colors� Thus� thecanceled square from T� and the covered square from T� are of di�erent colors� According tothe inductive assumption the remaining part of T� and the remaining part of T� can be covered�Thus� the rectangle can be covered too�

Finally one white and one black squares can be chosen in S� �S� �

�mn

��ways� Therefore�

in this case F m�n� �m�n�

��

Page 11: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

BULGARIAN NATIONAL OLYMPIAD IN MATHEMATICS

Fourth round

����

Problem �� Find all primary numbers p and q� for which��p � �p���q � �q�

pqis an integer�

Solution� Let p be prime number and pj ��p � �p�� It follows by the Fermat theorem that�p � �p � � �mod p�� Consequently p ��

Let now p and q be such prime numbers that��p � �p���q � �q�

pqis an integer� If pj ��p � �p��

then p �� Since ��� �� � � � � �� then either qj ��q � �q�� i�e� q �� or q �� Therefore thepairs ������ ������ ����� satisfy the problem condition� It remains the case when p � �� q � ��Now pj ��q � �q� and qj ��p � �p�� We can assume that p � q� It is clear that �p� q � � andconsequently� there are positive integers a and b� for which ap � b�q � � �Bezou theorem��Since �q� �� �q� �� � it follows by the Fermat theorem that �q�� � �q�� �mod q�� From�p � �p� �mod q� we deduce that �ap � �ap �mod q� and therefore� �b�q����� � �b�q�����

�mod q�� But �b�q����� � � �mod q� and �b�q����� � � �mod q�� Thus� q �� which is acontradiction� Finally� �p� q� ��� ��� ��� ��� ��� ���

Problem �� Find the side length of the smallest equilateral triangle in which three diskswith radii �� � and � without common inner points can be placed�

Solution� Let in a equilateral �ABC two disks with radii � and � without common innerpoints be placed� It is clear that a line � exists� which separates them� i�e� the disks are indi�erent semiplanes with respect to � �Figure ��

Figure � Figure ��

This line divides the triangle into a triangle and a quadrilateral or into two triangles� Inboth cases the disks can be replaced in the gure in a way that each of them is tangent to twoof the sides of �ABC� It is clear that the new disks have no common inner point� Let the disksbe inscribed in � A and � B of �ABC� respectively� We translate the side BC parallelly to itself

Page 12: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

towards the point A� till the disk which is inscribed in � B touches the disk which is inscribed in� A �Figure ��� Thus� we get an equilateral �A�B�C� with a smaller sides� in which two diskswith radii � and � and without common inner points are placed�

Let A�B� x� I be the incenter of �A�B�C�� while O�Figure ��and O� be the centers of the two disks� Then A�I B�I xp�� A�O� �� B�O� �� Since the disk with radius � is

inside the �A�B�C�� then O� � IB�� Thus� B�O� � B�I �

i�e� x � �p�� On the other hand O�I

xp���� O�I

xp��

�� O�O� � and by the cosine theorem for �O�O�I we nd

that

�xp�� �

���

�xp�� �

���

�xp�� �

��xp�� �

� ���

But x � �p�� and from here x

p�� Consequently� AB �

p�� On the other hand in the equilateral �ABC with

side length p� three disks with radii �� � and � �without

common inner points� can be placed inscribing circles withthese radii in the angles of the triangle �Figure ���

Note that the disks wit radii � and � are tangent to each other� It follows from the aboveconsiderations that the solution of the problem is

p��

Problem �� The quadratic functions f�x� and g�x� are with real coe�cients and have thefollowing property� if the number g�x� is integer for a positive x� then the number f�x� is integertoo� Prove that there are such integers m and n� that f�x� mg�x� � n for all real x�

Solution� Let g�x� px��qx�r� We can assume that p � �� Since g�x� p�x�q

�p���r� q�

�p

after the variable change of x by x �q

�pwe reduce the problem for the following quadratic

functions f�x� ax��bx�c and g�x� px��s� p � �� Let k be such an integer that k � s andsk � s

p�

q

�p� Since g

�sk � s

p

� k is integer� then f

�sk � s

p

a�k � s�

p� b

sk � s

p� c

is an integer too� Consequently� the number

f

�sk � � s

p

�� f

�sk � s

p

bpp�

pk � � s �

pk � s

�a

p��

is an integer for all k which are su�ciently big� It follows from here thata

pis an integer� Indeed�

suppose thata

pis not an integer� If b � �� we chose k in a way that

bpp�

pk � � s�

pk �

�a

p

�� � a

p�

and if b � �� we chose k in a way that

bpp�

pk � � s�

pk �

�a

p

�� a

p�

In both cases there is a contradiction with the fact that �� is an integer� Now� it followsthat

bpp�

pk � � s�

pk � s

Page 13: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

for all k� which are su�ciently big� This is possible only when b ��

Leta

p m� Then� f

�sk � s

p

� m�k� s� � c is an integer �when k is su�ciently big�� i�e�

c�ms is an integer� Let n c�ms� Now it is clear that f�x� mg�x� � n for all x�Problem �� The sequence fang�n�� is de ned by

a� � an�� an

n�

n

an� n � �

Prove that ba�nc n when n � � �it is denoted by bxc the integer part of the number x��Solution� Let f�x�

x

n�

n

x� Since f�a� � f�b�

�a� b��ab� n��

abn� it follows that the

function f�x� is decreasing in the interval ��� n��

Firstly� by induction we shall prove thatpn � an � np

n� when n � �� We have

a� � a� � and a� �� i�e�p� � a� � �p

�� Let

pn � an � np

n� for an integer n � ��

Then� an�� f�an� � f�pn�

n� pn

and an�� f�an� � f

�npn�

npn�

�pn �

and thus the induction nishes�Since an � p

n� it remains to prove� that an �pn� � We have an�� f�an� �

f

�npn �

npn�

when n � �� Consequently� an � n � pn � �

when n � �� Then�

an�� f�an� � f

�n � pn� �

�n� �� � n��n� ��

�n� �npn� �

�pn � �

when n � �� �The last inequality is equivalent to �n��n � �� � �n � � ��� Therefore�pn � an �

pn� when n � �� i�e� ba�nc n�

Problem �� The quadrilateral ABCD is inscribed in a circle� The lines AB and CD meeteach other in the point E� while the diagonals AC and BD � in the point F � The circumcirclesof the triangles AFD and BFC have a second common point� which is denoted by H � Provethat � EHF ����

Solution� Let O be the circumcenter of ABCD� We shall prove that O is the second commonpoint of the circumcircles of �AHB and �CHD� �Since AB and CD are not parallel� thenO � H �� After that we shall prove that the points E� H and O are colinear and � OHF ����

We shall consider the possible positions of H �Let G be the common point of AD and BC �these lines are not parallel because the cir�

cumcircles of �AFD and �BFC are not tangent�� It is clear that H is in the interior of� AGB�

� H is in �CGD� Then� � CHD � CHF � � DHF ��� � � CBF � ��� � � DAF ���� � dCD � ��� �from dCD � dAB and dAB � dCD � ������ which is impossible�

�� H is in �CFD� Then� � CHD ���� � � CHF � � DHF � CBF � � DAF dCD � COD and � AHB � AHF � � BHF � ADF � � BCF dAB � AOB�

�� H is in �ABF � Analogously� � CHD � COD and � AHB � AOB��� H is on AB� Again � CHD � COD and ��� � AHB � AOB� Consequently� O is

the midpoint of AB��� H is not in �ABG� Then � CHD � COD and � AHB ���� � � AOB�

Page 14: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Note that in the cases ��� �� and �� the points O and H are in one and the same semiplanewith respect to the line AB and with respect to the line CD� Consequently� the points A�B�Hand O are concyclic� We have the same for the points C�D�H and O�

Now we shall prove that the line EH passes through the point O� Let this line meet thecircumcircles of �AHB and �CHD at the points O� and O�� respectively� Since these pointslie on the ray EH�� it follows from the equalities EH�EO� EA�EB EC�ED EH�EO�

that O� � O� � O�The points O and H are in one and the same semiplane with respect to CD and we can

assume that the quadrilateral COHD is convex� Let H be in �CFD� Then� � OHF � FHC�� OHC ��� � � FBC � � ODC ��� � �

� �� COD � ���� � �

� �� COD� ���� The other

possibility for H is to be inside � AGB and outside the non�convex quadrilateral AFBG�Hence � OHF � OHC � � FHC � ODC � � FBC ����Problem �� A square table of size � � with the four corner squares deleted is given�

a� What is the smallest number of squares which need to be colored blackFigure ��so that a ��square entirely uncolored Greek cross �Figure �� cannot be foundon the table�

b� Prove that it is possible to writte integers in each square in a way thatthe sum of the integers in each Greek cross is negative while the sum of allintegers in the square table is positive�

Solution� Denote the square in row i and column j by �i� j�� Note thata cross is uniquely determined by its central cell� The cross with central cell

�i� j� is denoted by Cij � � � i� j � �� The number of all crosses is ���a� The squares �� i�� �i� �� ��� i�� �i� ��� i �� � � � � �� are included in exactly one cross� the

squares ������ ������ ������ ����� are included in exactly � crosses� the squares ��� i�� �i� ��� ��� i���i� ��� i �� �� � � in exactly � crosses and nally� the squares �i� j�� � � i� j � �� are includedin exactly � crosses� Since C�� contains a colored square� then at least one of ����� ����� ������������ ����� should be colored� Similarly� at least one of ������ ����� ������ ������ ������ at leastone of ������ ����� ������ ������ ����� and at least one of ������ ������ ������ ������ ����� are colored�Any of them is contained in at most four crosses �the rst two in each quintuple in cross� thethird one in � crosses and the remaining two � in � crosses�� Denote by x the number of thecolored squares among �i� j�� �� i� j � �� The number of crosses containing a colored square isnot greater than ��� � �x� whence

� � �x � ��� ��

Thus� x � � and the number of the colored cells is at least ��Suppose that the number of the colored squares is �� Then x �� Moreover� acording ��

there exists at most one Greek cross containing more than one colored square� If two squareswith a common side or a common vertex are colored� then it is easy to check that there are twocrosses with at least two colored squares each� Therefore� the squares ������ ������ ������ ����������� should be uncolored� With no loss of generality� let the two colored squares be ����� and������ But now the crosses C�� and C� do not contain colored squares�

To prove that the minimal number of colored squares is �� color for example the squares������ ������ ������ ������ ������ ������ ������

b� Let us write �� in each of the colored squares from a� and in the remaining squares�Since every Greek cross contains a colored square� the sum of the numbers in its squares does notexceed ����� � � �� The sum of all numbers in the table is �������������� � � �and we are done�

Page 15: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLV National Mathematics

Olympiad� �rd round� April

����

Problem �� Find all natural numbers a� b and c such that theroots of the equation

x� � �ax� b � �

x� � �bx� c � �

x� � �cx� a � �

are natural numbers�

Solution� Let fx�� x�g� fx�� x�g and fx�� x�g be the roots of the�rst� second and third equation respectively� and let all of them benatural numbers�

Assume that xi � � for all i � � �� � � � � � Then �a � x� � x� �x�x� � b� �b � x� � x� � x�x� � c and �c � x� � x� � x�x� � a�Thus ��a � b � c� � a � b � c� which is impossible� since a� b� c arenatural numbers�

Page 16: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Therefore at least one of the numbers xi equals � Without lossof generality suppose x� � � so � �a� b � ��

If xi � � for i � � �� �� � then

��b� c� � �x� � x�� � �x� � x�� � x�x� � x�x� � c� a�

whence ���a��c� � c�a� c � �� a� which is impossible whena� b� c are natural numbers�

So at least one of x�� x�� x�� x� equals � Let x� � � Now ��b�c � �� Assuming that x� � � and x� � �� we get �c � x� � x� � a�

so ���b� � � b�

�� Thus �b � �� a contradiction�

Therefore at least one of the numbers x�� x� is and it followsthat � �c� a � �� Further

� � � � �a� b� � �� �b� c� � �� �c � a� � � �a� b� c�

and since a� b� c are natural numbers� it follows that a � b � c � �

Direct veri�cation shows that a � b � c � satisfy the conditionsof the problem�

Problem �� Given a convex quadrilateral ABCD which can beinscribed in a circle� Let F be the intersecting point of diagonalsAC and BD and E be the intersecting point of the lines AD andBC� If M and N are the midpoints of AB and CD� prove that

MN

EF�

������ABCD � CD

AB

���� �

Page 17: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� Let � AEB � �� EC � c� ED � d���i �

c� ��EC and

��j �

d� ��ED� Since ABCD is an inscribed quadrilateral�

AB

CD�

AE

CE�

BE

DE� k� Therefore

��EA � kc

��j and

��EB � kd

��i � Since

F � AC and F � BD there exist x and y such that

��EF � x

��EA� � � x�

��EC � xkc

��j � �� x�c

��i

and ��EF � y

��EB � � � y�

��ED � ykd

��i � �� y�d

��j �

Comparing the coe�cients of��i and

��j in these equalities gives

xkc � � � y�d and ykd � � � x�c� This implies x �kd� c

�k� � �c�

Therefore

��EF �

k

k� �

��kd� c�

��j � �kc� d�

��i�

and thus

EF � �

�k

k� �

�� ��kd � c�� � �kc� d�� � ��kd � c��kc� d� cos �

��

On the other hand

���MN �

�����AD �

��BC

��

�����ED ���EA�

��EC ���EB

����d� kc�

��j � �c� kd�

��i�

and it follows that

MN� �

����d � kc�� � �c� kd�� � ��d � kc��c� kd� cos �

��

Page 18: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

ThereforeMN�

EF ��

�k� �

k

��

�k �

k

��and so

MN

EF�

������ABCD � CD

AB

����Problem �� Prove that the equation

x� � y� � z� � �x� y � z� � � � �

has no solution in rational numbers�

Solution� It is easy to see that the equation is equivalent to

��x� �� � ��y � �� � ��z � �� � ��

It has a solution in rational numbers if and only if there existinteger numbers a� b� c and a natural number m such that

��� a� � b� � c� � �m��

Suppose such numbers exist and give m its smallest possiblevalue� There are two cases�

�I� m � �n is an even number� Now a� � b� � c� is divisible by ��This implies that all numbers a� b� c are even ones� so a � �a��b � �b�� c � c�� It follows now that a�� � b�� � c�� � �n�� whichcontradicts the way m have been chosen�

�II� m � �n � is an odd number� Now m� �mod �� andtherefore a�� b�� c� � �mod ��� which is impossible whena� b� c are integer�

Page 19: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Find all continuous functions f�x� de�ned in the setof real numbers and such that

f�x� � f

�x� �

for all real x�

Solution� Let f�x� be a function satisfying the conditions of theproblem� Obviously f�x� is an even function�

Let x� � �� There are two cases�

�I� � � x� �

�� Consider the sequence

�� x�� x�� � � � � xn� � � �

de�ned by the equalities xn�� � x�n�

��

It is easy to see by induction that � � xn �

�for all n� Moreover

xn�� � xn � x�n� xn �

���xn �

��� ��

which implies that �� is a monotone increasing function� Since it isbounded� it follows that it is a convergent function� Let lim

n��xn � ��

Now �� � ��

�� �� so � �

��

On the other hand� since f�x� is a continuous function�

limn��

f �xn� � f

��

Page 20: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

But

f�xn��� � f

�x�n�

�� f�xn�

for all n� Thus f�x�� � f�x�� � � � � which means that f�x� � f

for all x� ����

�II� x� �

�� Consider the following sequence�

��� x�� x�� � � � � xn� � � �

de�ned by xn�� �

sxn �

��

As in the previous case� we show that ��� is a convergent function

and limn��

xn �

�� Further lim

n��f �xn� � f

�and since

f�xn��� � f

�x�n�� �

�� f�xn�

for all n we get that f�x� � f

��

Therefore f�x� is a constant function in the interval ������ andsince it is even� it is a constant for all x�

Conversely� any constant function satis�es the conditions of theproblem�

Problem �� Two squares K� and K� with centres M and N andsides of length are placed in the plane in a way that MN � �� twoof the sides of K� are parallel to MN and one of the diagonals of

Page 21: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

K� lies on the line MN � Find the locus of midpoints of segmentsXY where X is an interior point for K� and Y is an interior pointfor K��

Solution� The locus is the interior of a regular hexagon centred at

the midpoint of the segment MN and with a side length of

��

To prove this� proceed as follows�

Fix the point Y in the interior of K�� When X varies in theinterior of K� the locus of the midpoints of XY is a square K �

� which

is homothetic to K� by homothecy with centre Y and coe�cient

��

Obviously the side of this square is

�and its centre is the midpoint

of MY � When Y varies in the interior of K� then the locus ofthe midpoints of MY is a square K �

� which is homothetic to K� by

homothecy with centreM and coe�cient

�� The side of this square

equals

�and its centre Q is the midpoint of MN � Finally� when

the centres of K �� vary in the interior of K �

�� the squares vary in theinterior of a regular hexagon centred at Q and with a side length

of

��

Page 22: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem � Find the number of non�empty sets of Sn � f� �� � � � � ngsuch that there are no two consecutive numbers in one and the sameset�

Solution� Denote the required number by fn� It is easy to see thatf� � � f� � �� f� � ��

Divide the subsets of Sn having no two consecutive numbers intotwo groups�those that do not contain the element n and thosethat do� Obviously the number of subsets in the �rst group is fn���

Let T be a set of the second group� Therefore either T � fng orT � fa�� � � � � ak��� ng� where k � � It is clear that ak�� �� n � �so fa�� � � � ak��g � Sn��� whence the number of sets in the secondgroup is fn�� � � Therefore

fn � fn�� � fn�� � �

After substituting un � fn � we get

u� � �� u� � � un � un�� � un���

Therefore the sequence fung�n� coincides with the Fibonacci se�quence from its third number onwards� Thus we obtain

fn �p�

�� �

p�

�n��

���p�

�n���A � �

Page 23: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLV National Mathematics

Olympiad� �th round� May

����

Problem �� Consider the polynomial

Pn�x� �

�n

��

�n

�x�

�n

�x� � � � � �

�n

k � �

�xk�

where n � � is a natural number and k ��n � �

�a� Prove that Pn���x� � Pn���x�� Pn���x� � �x� ��Pn�x��

�b� Find all integer numbers a such that Pn�a�� is divisible by

�n��

�� for all n � �

Solution� �a� Compare the coe cients in front of xm� � � m ��n� �

� It su ces to show that

�n �

m � �

��

�n� �

m� �

��

�n� �

m� �

��

�n

m � �

��

�n

m� �

��

Page 24: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Using the identity

�a� �

b

��

�a

b

��

�a

b� �

�we get that

��n�

m� �

���n� �

m� �

��� �

��n � �

m � �

���n� �

m� �

���

��n� �

m� �

���

n

m� �

����

n

m� �

��

�n� �

m� �

�� �

�n � �

m � �

��

�n

m� �

���

n

m� �

��

��n� �

m� �

���n� �

m� �

���

���

n� �

m� �

���

n

m� �

����

n

m� �

��

�n� �

m

���n

m

���

n

m� �

��

� ��

�b� Suppose a satis�es the condition of �b� Then P��a�� ��� � a� is divisible by � and so a � �� �mod � On the contrary�let a � �� �mod � Now a� � � � � �mod �� Since P��a�� ��� P��a�� � and P��a�� � � it follows by induction from �a� that

Pn�a�� is divisible by �n��

�� for any n Therefore the required values

of a are all integer numbers congruent to � modulo

Problem �� Let M be the centroid of �ABC Prove the inequal�ity

sin � CAM � sin � CBM � �p

Page 25: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�a� if the circumscribed circle of�AMC is tangent to the lineAB�

�b� for any �ABC

Solution� We use the standard notation for the elements of�ABCLet G be the midpoint of AB

�a� It follows from the conditions of the problem that

�c

��� GA� � GM �GC �

m�

c ��

����a� � �b� � c���

and therefore a� � b� � �c� Using the median formula we get ma �p

�b and mb �

p

�a Further

A � sin � CAM � sin � CBM � S

��

bma

��

amb

��

�a� � b�� sin �pab

From the Cosine Law a�� b���ab cos � � c� �a� � b�

�� so a�� b� �

�ab cos � Therefore A ��psin �� � �p

�b� There are two circles through C and M tangent to the lineAB Denote the contact points by A� and B� and let A� � GA��B� � GB� Since G is the midpoint of A�B� and CM �MG � � � ��M must be the centroid of �A�B�C Furthermore it is clear that� CAM � � CA�M and � CBM � � CB�M Suppose � CA�M � ���

and � CB�M � ��� Now sin � CAM � sin � CBM � sin � CA�M �

sin � CB�M � �p

Page 26: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

It remains to consider the case � CA�M � ���� � CB�M � ���

�the above angles could not both be obtuse� It follows from�CA�M

that CM� � CA�� �A�M

�� so

���b�� � �a�� � c��� � b�� �

���b�� � �c�� � a���

�a�� b�� c� are sides of�A�B�C� We know from �a� that a���b�� � �c��

and the above inequality becomes a�� � �b�� Again from �a� weobtain

sin � CB�M �b� sin ��

a�p

�b�

a�p

vuut� ��a�� � b���a�b�

��

Substitutingb��a��

� x we get that

sin � CB�M ��

�p

p��x � x� � � �

�p

s�� �

��� � �

since x ��

� Therefore

sin � CAM � sin � CBM � � � sin � CB�M � � ��

��

�p�

Note� The inequality holds only for �ABC with angles � ������� � � ������� � � ��� or � � ������� � � ������ � � ���

Problem �� Let n and m be natural numbers such that m � i �aib

�i for i � �� �� � � � � n� where ai and bi are natural numbers and ai

is not divisible by a square of a prime number Find all n for whichthere exists an m such that a� � a� � � � �� an � ��

Page 27: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� It is clear that n � �� Since ai � � if and only ifm�i isa perfect square� at most three of the numbers ai equal � �prove it��It follows now from a� � a� � � � �� an � �� that n � �

We show now that the numbers ai are pairwise distinct Assumethe contrary and let m� i � ab�i and m� j � ab�j for some � � i �

j � n Therefore � � n � � � �m � j� � �m � i� � a�b�j � b�i � Itis easy to see that the former is true only if �bi� bj� a� � ��� �� �� or��� � �� and in either case a� � a� � � � �� an � ��

All possible values of ai are �� �� � �� �� �� �� and �� There arethree possibilities for n� n � � and fa�� a�g � f�� ��g� f�� ��g� f�� �g�n � and fa�� a�� a�g � f�� �� �g� f�� � �g� n � � and fa�� a�� a�� a�g �f�� �� � �g Suppose n � � and fa�� a�� a�� a�g � f�� �� � �g Now��b�b�b�b��� � �m� ���m����m� ��m��� � �m� ��m� ���� ��which is impossible Therefore n � � or n �

If n � and �a�� a�� a�� � ��� �� ��� then m � has the requiredproperty� and if n � � and �a�� a�� � ���� ��� then m � �� has therequired property �It is not di cult to see that the remaining casesare not feasible�

Problem �� Let a� b and c be positive numbers such that abc � �Prove the inequality

� � a� b�

� � b� c�

� � c� a� �

� � a�

� � b�

� � c�

Solution� Let x � a � b � c and y ��

a�

b�

c� ab � bc � ca

�abc � �� It follows from Cauchy�s Inequality that x � and y � Since both sides of the given inequality are symmetric functions of

Page 28: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

a� b and c� we transform the expression as a function of x� y andabc � � After simple calculations we get

� �x� y � x�

�x� y � x� � xy� �� � �x� y

� � �x � �y�

which is equivalent to

x�y � xy� � �xy � �x� � y� � ��x� y � �� � ��

Write the last inequality in the form

��

x�y � �x�� � �

xy�

� y�� � �

xy�

� y� � �

x�y � ��x� �

� �xy�

� x� � �xy � �x� � �xy � ��� � ��

When x � � y � � all terms in the left hand side are non�negativeand the inequality is true Equality holds when x � � y � � whichimplies a � b � c � �

Problem �� Given a �ABC with bisectors BM and CN �M �AC� N � AB� The rayMN� intersects the circumcircle of �ABCat point D Prove that

BD�

AD�

CD�

Solution� Let A�� B� and C� be the orthogonal projections of Don the lines BC� CA and AB� respectively It follows from �DAB�

and the Sine Law that DB� � DA � sin � DAB� � DA � sin � DAC �DA �DC

�R�R is the circumradius of �ABC� Analogously DA� �

Page 29: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

DB �DC�R

and DC� �DA �DB

�R Our equality is now equivalent to

AD �CD � BD � CD �AD �BD and so it su ces to prove that

��� DB� � DA� �DC�

Denote by m the distance from M to AB and BC and by n the

distance from N to AC and BC LetDM

MN� x �x � �� Further�

DB�

n� x�

DC�

m� x� � and

DA� �m

n �m� x Therefore DB� � nx�

DC� � m�x� �� and DA� � nx�m�x� �� � DB� �DC� and ���holds

Problem � Let X be a set of n � � elements� n � � Orderedn�tuples �a�� a�� � � � � an� and �b�� b�� � � � � bn� formed from distinct ele�ments of X are called �disjoint� if distinct indices i and j exist suchthat ai � bj Find the maximal number of n�tuples any two of whichare �disjoint�

Solution� For n � � denote by A�n� �� the maximum number ofordered n�tuples such that any two of them are �disjoint� Also letS�X� be a set of such n�tuples for which jS�X�j � A�n � �� It isclear that for any � � X the following holds�

jf�a�� a�� � � � � an� � S�X�ja� � �gj � A�n��

Thus A�n� �� � �n� ��A�n� Therefore

A�n� �� � �n� ��n � � � A���

Direct veri�cation shows that A�� � � so A�n� �� � �n� ���

Page 30: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

We prove now that A�n� �� ��n� ���

�by constructing a set of

�n� ���

�ordered n�tuples� any two of which are �disjoint� We may

assume that X � f�� �� � � � � n � �g Consider a set E of all evenpermutations of �� �� � � � � n� � �A permutation �a�� a�� � � � � an��� iscalled even if the number of pairs �i� j� such that i � j and ai � ajis an even number� The set

f�a�� a�� � � � � an�j�a�� a�� � � � � an� an��� � Eg

has�n� ���

�ordered n�tuples� any two of which are �disjoint�

Page 31: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLVII National

Mathematics Olympiad�

�rd round� ����� April �

Problem �� Find the least positive integer number n �n � ��with the following property� for any colouring of n di�erent pointsA�� A�� � � � � An on a line and such that A�A� � A�A� � � � � �An��An in two colours� there are three points Ai� Aj� A�j�i� � i �

j � i � n� which have the same colour�

Solution� Assume the two colours are white and black� Consider� points coloured as follows� A�� A�� A�� A� �white�� A�� A�� A�� A�

�black�� Obviously no three points Ai� Aj� A�j�i � � i � j� i � n�have the same colour and therefore n � �

If we can show that n � has the required property� we will bedone� Suppose there are points coloured black or white and nothree points Ai� Aj� A�j�i � � i � j� i � n� have the same colour�

First assume that for i � �� i � � or i � � points Ai and Ai�

have the same colour �say white�� Then the points Ai��� Ai�� Ai�

Page 32: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

should be black �note that i � � and i � � � �� which is acontradiction�

Suppose now that for i � �� �� � the points Ai and Ai� havedi�erent colours� Without loss of generality assume A� is a whitepoint� Then A� and A� are black� Because of the symmetry we maysuppose that A� is white and A� is black� Consiquently A� is white�A� is black � � � � � �� and A is white �� � � � ��� ThereforeA� should be both white � � � � � � and black � � � � ���which is again a contradiction�

Consequently the assumption is not true and so n � �

Problem �� Let ABCD be a quadrilateral such that AD � CD

and � DAB � � ABC � ��� The line passing through D and themidpoint of the segment BC intersects the line AB at the point E�Prove that � BEC � � DAC�

Solution� Let M be the midpoint of BC and let AD and BC

meet at point N and AN and EC meet at point P � It follows fromMenelaus� Theorem aplied to �DMN and �DEN that DP �NC �

ME � PN � CM � ED and DA � NB � ME � AN � BM � ED�Combining the above equalities with AN � BN � BE � CE andAD � CD� we get DP �NC � DC � PN � Therefore CP is bisectorof � DCN �

Consequently � ACP � � ACD� � DCP � �

��� NDC� � DCN� �

� NAB so � DCP � � CAB and we obtain � BEC � � ABC �� BCE � � BAD � � DCP � � DAC� Q� E� D�

Page 33: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Note� The assertion is also true if the condition � ABC � ��

is left out�

Problem �� Let R be the set of all positive real numbers� Provethat there is no function f � R � R

such that

�f�x��� � f�x � y��f�x� � y�

for arbitrary positive real numbers x and y�

Solution� Suppose there exists a function satisfying the conditionsof the problem� Write the initial equality in the form

f�x�� f�x� y� �f�x�y

f�x� � y�

First we prove that f�x� � f�x � � � �

�for x � �� Obviously

f is a �strictly� monotone non�increasing function� Fix x � � andchoose a natural number n� such that n � F �x � � � � Whenk � �� � � � � � n� � we obtain that

f�x�k

n�� f�x�

k �

n� �

f�x� k

n� �n

f�x� k

n� � �

n

n�

Adding the above inequalities gives f�x�� f�x� � � �

��

Let the natural number m be such that m � f�x�� Therefore

f�x�� f�x�m� �m��X

i��

�f�x� i�� f�x � i� �� �m

� f�x��

and so f�x�m� � �� But this contradicts the fact that f is strictlypositive�

Page 34: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Let f�x� � x�� �x�� Find the number of di�erentreal solutions of the equation f�f�x�� � ��

Solution� Since f ��x� � ��x���x�� it follows that f is strictlymonotone non�decreasing in the intervals ������ and ���� andstrictly monotone non�increasing in the interval ��� �� Moreoverlim

x���f�x� � ��� f��� � �� f�� � �� f��� � � � and so

the equation f�x� � � has three distinct roots x�� x�� x� such thatx� � � � x� � � x� � �� Therefore f�x� � x� has only one realroot �which is less than �� and f�x� � x� and f�x� � x� have threedistinct real roots each �one in each of the intervals ������� ��� �and ������ Since the roots of f�f�x�� � � are exactly the rootsof these three equations� we conclude that it has seven distinct realroots�

Problem �� The convex pentagon ABCDE is inscribed in a circlewith radius R� The inradii of the triangles ABC� ABD� AEC andAED are denoted by rABC� rABD� rAEC and rAED� Prove that

a�� cos � CAB � cos � ABC � cos � BCA � �rABC

R�

b�� If rABC � rAED and rABD � rAEC � then �ABC � �AED�

Solution� a�� Using the standard notation for the elements of�ABC� we obtain that

r

R�

S

pR�

�S�

pabc�

��p � a��p� b��p � c�

abc�

Page 35: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

abc�a�b� � c� � a�� � b�c� � a� � b�� � c�a� � b� � c��� abc� �

� cos � CAB � cos � ABC � cos � BCA� �

b�� Applying continuously the equality from a�� and using thefact that ABCDE is inscribed in a circle convex pentagon� it is easyto see that

rABC � rAEC � rEDC � rAED � rABD � rBCD�

The condition of the problem and the above equality imply rBCD �rEDC� Since �BCD and �EDC have a side in common and equalcircumradiuses we get that �BCD � �EDC �prove it using a����In particular BC � ED and using rABC � rAED again� we get�ABC � �AED�

Problem � Show that the equation

x�y� � z��z� � x� � y��

has no solution in positive integer numbers�

Solution� Suppose x� y� z is a solution for whichxy

z�which is a

natural number�why�� has its minimum value�

Write x� y� z in the form x � dx�� y � dy�� z � dz�� where d ��x� y� z�� Our equation is equivalent to x��y

�� � z���z

�� � x�� � y���� Let

u � �x�� z��� v � �y�� z��� x� � ut� y� � vw� Since z� divides x�y��we get z� � uv� A substitution in the last equality gives

Page 36: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�u� � w���v� � t�� � u�v�� Further� since �x�� y�� z�� � � itfollows that �u�w� � � �v� t� � � Therefore

u� � w� � v�� v� � t� � u�

oru� � w� � v�� v� � t� � u��

Without loss of generality we may assume that the �rst pair ofequalities hold� It is easily seen that v and u are odd integer numbers�It follows now from u� � w� � v� that u � m� � n�� w � mn� v �m� � n�� where m and n are coprime natural numbers �of distinctparity�� Substitution in v� � t� � u� shows that t� � �mn�� ��m� � n��� and so t � p� � q��mn � pq�m� � n� � p� � q� for somenatural numbers p and q� Therefore

p�q� � m��m� � p� � q���

which shows that p� q� m is a solution of the original equation� It re�

mains to be seen thatpq

m� n � d�p� � q��mn �

xy

z� which contra�

dicts the way we have chosen x� y� z�

Page 37: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLVII National

Mathematics Olympiad� �th

round� ����� May �

Problem �� Let n be a natural number� Find the least naturalnumber k for which there exist k sequences of ��s and ��s of length�n � � with the following property� any sequence of ��s and ��s oflength �n�� coincides in at least n�� positions with some of thesek sequences�

Solution� We shall prove that k � Assume that k � � and letthe respective sequences be ai�� a

i�� � � � � a

i�n�� for i �� � � � � k� Since

k � � there is a sequence b�� b�� � � � � b�n�� such that �b�l��� b�l�� ��ai�l��� a

i�l�� for l �� � � � � � n and i �� � � � � k� This is a contra�

diction� For k it is easily seen that the sequences ��� � � � �� ��� � � � ����� � � � �� ��� � � � � have the required property�

Problem �� The polynomials Pn�x� y � n �� �� � � � are de�ned by

P��x� y �� Pn���x� y �x�y�� �y�� Pn�x� y�� ��y�y� Pn�x� y �

Page 38: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Prove that Pn�x� y Pn�y� x for all x� y and n�

Solution� We know that P��x� y � and P��x� y xy � x �y � �� Assume that Pn���x� y and Pn�x� y � �n � � are symmetricpolynomials� Then

Pn���x� y� � �x� y � ���y � ��Pn�x� y � �� � �y � y��Pn�x� y�

� �x� y � ���y � ��Pn�y � �� x� � �y � y��Pn�y� x�

� �x� y � ���y � ��

��x� y � ���x� ��Pn���y � �� x� ��� �x� x��Pn���y � �� x�

� �y � y��

��y � x� ���x� ��Pn���y� x� ��� �x� x��Pn���y� x�

� �x� y � ���y � ���x� y � ���x� ��Pn���y � �� x� ��

� �y � y���x� x��Pn���y� x�

� �x� y � ���y � ���x� x��Pn���y � �� x�

� �y � y���x� y � ���x� ��Pn���x� �� y��

and by induction it follows that all the polynomials are symmetric�

Problem �� On the sides of a non�obtuse triangle ABC a square�a regular n�gon and a regular m�gon �n�m � � are constructedexternally� so that their centres are vertices of a regular triangle�Prove that m n � and �nd the angles of ABC�

Solution� Let the square� the n�gon and the m�gon be constructedon the sides AB� BC and CA� respectively� Denote their centresby O�� O� and O�� denote by A�� B� and C� the centres of theequilateral triangles constructed externally on BC� CA and AB�

Page 39: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

The lines O�C�� O�A� and O�B� intersect at the circumcentre O of�ABC� Since �A�A�A� is equilateral� it follows straightforwardlythat �O�O�O� is equilateral if and only if C�A�jjO�O�� A�B�jjO�O�

and B�C�jjO�O�� This is equivalent to

OC�

C�O�

OA�

A�O�

OB�

B�O�

k�

On the other hand�

OC�

C�O�

cotC � tan ���

cot �� � tan ����

OA�

A�O�

cotA� tan ���

cot ����

n� tan ���

OB�

B�O�

cotB � tan ���

cot ����

m� tan ���

Set cot����

n x and cot

����

m y� The above identities imply that

cotA kx� k � �p�� cotB ky � k � �p

�� cotC k � k � �p

��

From the identity cotA cotB � cotB cotC �cotC cotA � we get

k ��x� y � � �p� p

�xy � �p� � � �x� y �

p�� �

cotC x� y � xy � �� �

p�p

�xy � �p� � � �x� y �

p�� �

Since m�n � � it follows that x� y � p��i� e� xy � p

��x � y � ��The inequality cotC � � implies x � y � xy � � � �

p� � �� and

therefore x� y� �� �p� � xy � p��x� y � � �i� e� x� y � �

p��

This shows that x y p�� i� e�� m n �� Hence cotC

�� cotA cotB �� so � C ���� � A � B ���

Page 40: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Let a�� a�� � � � � an be real numbers� not all of themzero� Prove that the equation

p� � a�x�

p� � a�x� � � ��

p� � anx n

has at most one nonzero real root�

Solution� The given equation is equivalent to xnXi��

aip� � aix

n�

Since the functionaip

� � aix�ai � � is strictly decreasing� it follows

that this equation has at most one nonzero root�

Problem �� Let m and n be natural numbers such that A ��m�� n � � ��m is integer� Prove that A is an odd integer number�

Solution� Assume that A is an even integer number� i� e�� �m �� n � � �km� Then m is an even integer number� Moreovermj�n��� which shows that m �t�� and n is odd� Let m ��m��where � � � and m� is odd� Then ��j�n � � and therefore � � �

� Since m�j�n � �� it follows that m�ja� � �� where a �n��

� � It iswell�known that in this case m� �t���� Since m ����t��� hasthe form �t�� and � � � � � we see that � �� Then m ��t���and from �m � � n � � �km it follows that j�n � �� which isimpossible�

Problem � The sides and the diagonals of a regular n�gon X arecoloured in k colours so that�

Page 41: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�i for each colour a and any two vertices A and B of X� thesegment AB is coloured in colour a or there is a vertex C suchthat AC and BC are coloured in colour a�

�ii the sides of any triangle with vertices among the vertices of Xare coloured in at most two colours�

Prove that k � ��

Solution� Assume that the colouring involves at least three di�er�ent colours a� b� c� We shall construct an in�nite subset of verticesof X� which will imply a contradiction�

Let Z � X and A� is a vertex such that the colour of A�Zis a� From �i it follows that there is a vertex B�� such that thecolour of B�Z and B�A� is b� Analogously there is a vertex C�

such that the colour of C�Z and C�B� is c� Considering the tri�angles C�A�Z and C�A�B� we see �using condition �ii that thecolour of C�A� is c� Let A� � X be such that the colour of A�C�

and A�Z is a� It is easily seen that A� � A� and the colourA�A� and A�B� is a� Now we shall proceed by induction� Let thevertices A�� B�� C�� � � � � Ak��� Bk��� Ck�� be such that the colour ofAiAj� AiBj� AiCj is a� the colour of BiAj� BiBj� BiCj is b and thecolour CiAj� CiBj� CiCj is c� �� � j � i � k �

Take a vertex Ak such that the colour of AkCk�� and AkZ is a�Then considering the triangles ZAkBj and Ck��AkBj �j � k � ZAkCj

and Bj��AkCj �j � � � k � AkAjBj and AkAjBj we see that thecolour of AkBj� AkCj and AkCj is a� The vertices Bk and Ck areconstructed in a similar way�

Page 42: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLVIII National

Mathematics Olympiad�

�rd round� ����� April �

Problem �� Find all triples �x� y� z� of natural numbers such thaty is a prime number� y and � do not divide z� and x� � y� � z��

Nikolay Nikolov

Solution� Since �x� y���x� y����xy�� � x�� y� � z�� it followsfrom the conditions of the problem that x�y and �x�y����xy arerelatively prime� Therefore x� y � u� and x� � xy � y� � v�� Thus�y� � �v � x � y��v � x � y� and since y is a prime number�there are three cases to consider

�� v � x � y � y� v � x � y � �y� Now x � �� which isimpossible�

� v � x � y � �� v � x � y � �y�� Now �y� � � � �x �y� � �u� � �y� and it follows that � divides u� � �� which isimpossible�

Page 43: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�� v � x� y � �� v � x � y � y�� Now y� � � � �x � y� ��u���y� and it follows that �y����� �u�� � �� Thereforey � � u � �� x � �� z � ��� Direct veri�cation shows that��� � ��� is a solution of the problem�

Problem �� A convex quadrilateral of area S is inscribed in acircle whose centre is a point interior to the quadrilateral� Provethat the area of the quadrilateral whose vertices are the projectionsof the point of intersection of the diagonals on the sides does not

exceedS

� Christo Lesov

Solution� Let ABCD be a quadrilateral inscribed in a circle withcentre O and radius R� Denote by E the point of intersection ofAC and BD� Denote further by M � N � P � Q and F the projectionsof E on AB� BC� CD� DA and MN � respectively� We know that

MN � BE sin � ABC �BE �AC

R� Also�

EF � EM sin � EMN �AE �BE sin � AEB

ABsin � CBE�

Since BE sin � CBE � CE sin � BCE � CEAB

Rand AE � CE �

R� �OE�� it follows that EF �R� �OE�

Rsin � AEB� Therefore

S�MEN �MN � EF

�AC �BE sin � AEB�R� �OE��

�R��

Page 44: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Similar equalities hold for S�NEP � S�PEQ and S�QEM � By combin�ing the above we obtain

SMNPQ �AC �BD sin � AEB�R� �OE��

�R�� SABCD

Q� E� D�

Problem �� In a competition � judges marked the contestants byyes or no� It is known that for any two contestants� two judges gaveboth a yes� two judges gave the �rst one a yes and the second onea no� two judges gave the �rst one a no and the second one a yes�and �nally� two judges gave both a no� What is the greatest possiblenumber of contestants� Emil Kolev

Solution� Denote the number of contestants by n� Consider atable with � rows and n columns such that the cell in the ith row andjth column contains � ��� if the ith judge gave the jth contestanta no �a yes�� The conditions of the problem now imply that thetable formed by any two columns contains among its rows each ofthe pairs ��� ��� �� and �� twice� We shall prove that � columnshaving this property do not exist� Assume the opposite� It is easilyseen that if in any column all �s are replaced by �s and vice versa�the above property is retained� Therefore without loss of generalitysuppose that the �rst row consists of �s� Denote the number of�s in the ith row by ai� It is clear that the total number of �s is� � � � �� Further� the number of occurrences of �� is

��

�� � ���

On the other hand the same number is�Xi��

�ai

�� Since a� � �� it

Page 45: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

follows that�Xi��

ai � �� It is easy to prove now that�Xi��

�ai

�� ���

Therefore �� ��Xi��

�ai

�� ��� which is false�

The diagram on the right showsthat it is possible to have exactly contestants

� � � � � � �� � � � � � �� � � � � � �� � � � � � �� � � � � � �� � � � � � �� � � � � � �� � � � � � �

Problem �� Find all pairs �x� y� of integer numbers such that x� �y� � y� � �� Nikolay Nikolov and Emil Kolev

Solution� It is obvious that x � y� On the other hand x � y �� �� �y � ��� � y� � y� � � �� y�y � �� � �� Therefore ify � � or y � �� the problem has no solution� Direct veri�cationyields all pairs �x� y� which satisfy the equality x� � y� � y� � ��namely ������� ����� and ��� ���

Problem �� Let B� and C� be points on the sides AC and AB

of �ABC� The straight lines BB� and CC� intersect at point D�Prove that the quadrilateral AB�DC� is circumscribed if and only ifthe incircles of �ABD and �ACD are tangent�

Rumen Kozarev and Nikolay Nikolov

Page 46: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� Note that the incircles of �ABD and �ACD are tan�gent if and only if AB�AD�BD � AC�AD�CD� so AB�CD �AC �BD�

Suppose AB�DC� is circumscribed and the incircle touches AB��B�D� DC�� C�A in the points M�N�P�Q� respectively� ThereforeAB � CD � AQ� BQ� CP �DP � AM � BN � CM �DN �AC �BD�

Conversely� let the incircles of �ABD and �ACD be tangent�Denote the point of intersection of the tangent through C �di�erentfrom CA� with the incircle of �ABB� by D�� It follows from theabove that BD� �CD� � AB �AC � BD�CD� so DD� � jCD�CD�j� Therefore D� � D� which completes the proof�

Problem � Each interior point of an equilateral triangle of side �

lies in one of six circles of the same radius r� Prove that r �p�

���

Nikolay Nikolov and Emil Kolev

Solution� Divide each side of the triangle into �ve equal partsand draw lines parallel to the sides through these points� Thus the

triangle is divided into � equilateral triangles of side�

�� The total

number of vertices is � � � � �� Therefore there exist � points whichare interior to one and the same circle� It is easy to see now that

r �p�

��� which solves the problem�

Page 47: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLVIII National

Mathematics Olympiad� �th

round� ����� May ����

Problem �� The faces of an orthogonal parallelepiped whose di�mensions are natural numbers are painted green� The parallelepipedis partitioned into unit cubes by planes parallel to its faces� Findthe dimensions of the parallelepiped if the number of cubes havingno green face is one third of the total number of cubes�

Sava Grozdev

Solution� Let x � y � z be the dimensions of the parallelepiped�It follows from the conditions of the problem that x � � and �x ����y � ���z � �� �

xyz

�� Since

x� �

x� y � �

y� z � �

z when x �

we obtain that�x� ���y � ���z � ��

xyz� �

�� �

�� Therefore x �

and thus x � � x � � x � � or x � �

�� If x � � then �y � ���z � �� � yz which is impossible�

Page 48: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�� If x � � then ��y � ���z � �� ��yz

� so �y � ��z � � � ���

In this case the only solutions are ��� � ��� ��� �� ��� ��� �� ���and ��� ��� ����

�� If x � � then ��y� ���z� �� ��yz

� so ��y� ����z� �� � ���

Therefore the solutions are ��� �� �� ��� � ��� and ��� � ���

�� If x � then ��y � ���z � �� � �yz so �y � ���z � �� � ��Thus there is an unique solution � � � ���

Answer� The problem has � solutions���� � ��� ��� �� ��� ��� �� ������ ��� ��� ��� �� �� ��� � ��� ��� � �� and � � � ���

Problem �� Let fang�n�� be a sequence of integer numbers suchthat

�n � ��an�� � �n � ��an � ��n� ��

for any n � �� If ���� divides a���� �nd the smallest n � � suchthat ���� divides an� Oleg Mushkarov� Nikolai Nikolov

Solution� It is obvious that a� � � and an�� �n� �

n� �an � � when

n � �� Therefore the sequence is uniquely determined by its secondterm� Furthermore the sequence an � �n � ���cn � �� �where c �a�

�� � is an arbitrary real number� satis�es the equality from the

conditions of the problem� We conclude now that all sequences whichsatisfy this equality are of the kind given above� Since all terms areinteger numbers and ���� divides a���� it is easy to see that c isinteger and c � ����m � �� Therefore ���� divides an if and only if

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���� divides �n����n���� Thus n � �k�� and k�k��� is divisibleby ��� � �� � �� Since k and k � � are relatively prime we get thatthe smallest n � � equals � � ��� � � � ����

Problem �� The vertices of a triangle have integer coordinatesand one of its sides is of length

pn where n is a square�free natural

number� Prove that the ratio of the circumradius and the inradiusis an irrational number� Oleg Mushkarov� Nikolai Nikolov

Solution� Suppose thatR

r� q where q is a rational number�

Without loss of generality assume that one of the ends of the sideof length

pn is at the origin of the coordinate system� Let the

remaining two vertices have coordinates �x� y� and �z� t� where x� y� zand t are integers� The sides of our triangle have lengths a �

pA� b �p

B and c �pC where n � A � x� � y�� B � z� � t� and C �

�x�z����y� t��� It follows from the conditions of the problem that

q �R

r�abc

�S� pS

�abc�a� b� c�

�S��

where S is the area of the triangle� Since S is rational �prove it�� itfollows that

pABC�

pA�

pB �

pC� � �S�q is a rational number�

Therefore ApBC �B

pAC � �S�q�C

pAB and after squaring we

obtain thatpAB is a rational number� Thus AB is a perfect square�

By analogy both BC and CA are perfect squares� Let AB � E�BC � F � and CA � G� where E F and G are integer� Write A�Band C in the following form� A � a�a

��� B � b�b

�� and C � c�c

��

where a�� b� and c� are square�free integers� So a�b��a�b��� � m�

and therefore a�b� is a perfect square whence a� � b�� By analogy

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a� � c�� Thus A � ma��� B � mb�

�� C � mc�

� where m is square�free�

It follows from ma�� � n that m � n� a� � � and we obtain�������

x� � y� � n

z� � t� � nb���x� z�� � �y � t�� � nc��

Since both b� and c� are integer it follows from the Triangle Inequal�ity that ��b� � c� and ��c� � b� whence b� � c�� It is easy to deter�mine now that x��y� � ��xz�yt� and consequently ��xz�yt� � n�Let ��xt � yz� � k� Then n� � k� � ��x� � y���z� � t�� � �n�b��so k� � n���b�� � ��� Therefore �b�� � � is a perfect square which isimpossible�

Problem �� Find the number of all natural numbers n� � � n ����� such that their binary representations do not contain threeconsecutive equal digits� Emil Kolev

Solution� Denote by an� n � � the number of sequences of zeroesand ones of length n which begin with � and do not contain threeconsecutive equal digits� Also for any a� b � f�� �g denote by xn

ab

the number of sequences of zeroes and ones of length n which beginwith � and do not contain three consecutive equal digits such thatthe last two terms are respectively a and b� It is easy to see that forn � �

xn��� xn��

���xn

��� xn��

��� xn��

���xn

��� xn��

��� xn��

���xn

��� xn��

���

Adding up the above equalities we obtain an � an���xn���� �xn���� �

an���an��� Since a� � � and a� � � it follows that a � � a � ��a� � �� a� � �� a� � �� and a�� � ���

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Since the required number is equal to a��a�� � � ��a�� it followsfrom the above that the answer is ����

Problem �� The vertices A B and C of an acute triangle ABClie on the sides B�C�� C�A� and A�B� of �A�B�C� and � ABC �� A�B�C� � BCA � � B�C�A� � CAB � � C�A�B�� Prove that theorthocentres of �ABC and �A�B�C� are equaly remote from thecircumcentre of �ABC� Nikolai Nikolov

Solution� Denote byH the orthocentre of�ABC� Since � CHB ����� � � CAB � ���� � � C�A�B� we have that A� lies on thecircumcircle k� of �BHC� Similarly B� and C� lie on circum�circles k� and k� of �CHA and �AHB� Therefore � B�HC� �� B�HA � � C�HA � � B�CA � � C�BA � �� B�A�C� and likewise� C�HA� � � � C�B�A� and � A�HB� � �� A�C�B� so H is the cir�cumcentre of �A�B�C��

Let us draw straight lines passing through the vertices of �ABC

and parallel to the corresponding sides and denote their points of in�tersection byA�B� and C�� Since � A�B�C� � � A�B�C�� � B�C�A� �� B�C�A� and � C�A�B� � � C�B�A� it follows from the above thatthe segments A�H�B�H and C�H are of equal length and are diam�eters of k�� k� and k�� It is clear now that there exists a compositionof a rotation and a homothecy both centred at H such that theimage of �A�B�C� is �A�B�C�� Therefore the image of ortho�centre H� of �A�B�C� is the orthocentre H� of �A�B�C�� Thus� HH�H� � � HA�A� � ��� and to solve the problem we have toshow that the circumcentre O of �ABC is the midpoint of HH��

Indeed the image of �ABC by a homothecy centred at the cen�

Page 52: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

troid of �ABC and with coe�cient �� is �A�B�C�� Therefore��

MH�� �� ��

MH and since��

MH� �� ��

MO we obtain��

OH�� � ��

OH �

Problem � Prove that the equation

x� � y� � z� � t� � ����

has in�nitely many integer solutions� Grigor Grigorov

Solution� Since ��� � ��� � ������ �� � ���� we are looking forsolutions of the form x � ��� k� y � ��� k� z � ��� l� t � l wherek and l are integer� After simple calculations we obtain that ourequation is equivalent to l�l��� � ��k� whence ��l�������k� � ��The latter is Pell�s equality� Since l � �� k � � is a solution allsolutions are of the form �ln� kn� where �ln � � � kn

p�� � �� �p

���n� n � �� �� � � �� Therefore the original equation has in�nitelymany integer solutions�

Page 53: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Union of Bulgarian Mathematicians

Sava Grozdev Emil Kolev

BULGARIAN

MATHEMATICAL COMPETITIONS

����

So�a� ����

Page 54: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Winter Mathematical Competition

Russe� ��� February ����

Problem ��� Given the inequality �n�� ��x � ��n�� �n��n���where n is an integer

a� Factorize the expression ��n� � �n� � n� ��

b� Find all n� for which the inequality holds true for any positivenumber x�

Solution a� ��n� � �n� � n � � �n� ������ �n��

b� Since ��x � � is not true for any x it follows that n � ��� Ifn � �� then the inequality is equivalent to �n���x � �n�������n��If n � then ��x � �� which is not true for any x� Let n � ��

If n � � � �� then�n � ����� �n�

n � �� � and the inequality is not

true for any positive x� If n � � � �� then the inequality becomes

x ��n� ���� � �n�

n� �and n � is the only solution If n � �� then

�n� ���� � �n�

n� �� � and there exists x � �� which is not a solution

Therefore the inequality has no solution when n � ���

Problem ��� In an isosceles �ABC�AC BC� the points A�� B�

and C� are midpoints of BC�AC and AB respectively Points A�

and B� are symmetric points of A� and B� with respect to AB� LetM be the intersecting point of CA� and A�C�� and let N be theintersecting point of CB� and B�C�� The intersecting point of ANand BM is denoted by P� Prove that AP BP�

Page 55: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution Since CC� k A�A� and CC� A�A�� we have thatCC�A�A� is a parallelogram Thus� A�M C�M� But A�B�C�B isalso a parallelogram and therefore the intersecting point of BM andAC is B�� Hence P lies on the median BB�� Analogously P lies onthe median AA�� In the isosceles�ABC the medians AA� and BB�

are of the same length Therefore AP �

�AA�

�BB� BP�

Problem �� Find all pairs of prime numbers p and q� such thatp� � �pq � q� is

a� a perfect square�

b� a power of �

Solution a� Let p� � �pq � q� r�� where p and q are primenumbers If p � �� q � �� then p� � �pq � q� � ��mod �� andr� � ��mod ��� a contradiction Without loss of generality p �and we get that q� � �q � � r� and �q� � ��q � �� ��r���Therefore ��q��r�����q��r��� ��� We may assume that r � �and so �q��r�� �� or �q��r�� ��� In the �rst case q�r ��which is impossible and in the second case solving the system

����� q � r ���q � �r � � ��

we �nd q �� Because of the symmetry the only solutions are p �� q � and p �� q ��

b� Let p� � �pq � q� �n� where n is a natural number Sincep � �� q � �� we have p� � �pq � q� � �� and so n � �� It followsnow that ����p� � �pq � q�� and ���p� � �pq � q�� �p� q��� �pq�Thus� ���p � q�� and ����p � q��� Therefore ����pq� showing thatp � or q �� But if p � then q � �and vice versa� Weobtain p� � �pq � q� ��� ��� The only solution of the problemis p q ��

Page 56: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �a�� Given the equation�

jx� �j �

jx� ��aj � where a is

a parameter

a� Solve the equation

b� If a is the square of a prime number prove that the equationhas a solution which is a compose integer

Solution After squaring and simple calculations we obtain ���a���x ���a� �����a � ���

a� If a �

��then every x � � is a solution If a � �

��then the

only solution is x ��a� ��

b� Let a p� where p is a prime number If p � then x ���is not a prime If p � � then p �k � � and x ����k � ��� � � �A� ��� which is divisible by ��

Problem �a�� The quadrilateral ABCD is inscribed in a circle withdiameter BD� Let M be the symmetric point of A with respect toBD and let N be the intersecting point of the straight lines AM andBD� The line passing through N� which is parallel to AC� intersectsCD and BC in P and Q respectively Prove that the points P�C�Qand M are vertices of a rectangle

Solution It follows from the condition of the problem that M lies

on the circumcircle of ABCD� Since �� MAC �� MBC �

MC� and

�� MNQ �� MAC ie �� MNQ �� MBC� we get that the pointsM�N�B and Q lie on a circle Since�� MNB ��� we conclude that�� BQM ���� Also� since �BDC is a right angle triangle we have

thatMQ k PC� From the other hand �� MDC �� MAC �

MC�� and

therefore �� MDC �� MNQ� so the points N�P�M and D lie on acircle Thus� �� MPD �� MND ��� and MP k CQ� ThereforeP�C�Q and M are vertices of a rectangle

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Problem �a� See Problem ��

Problem �b�� Given the system �����������

x� y� x a� �

x

x� y a� ��

where a is a real parameter

a� Solve the system if a ��

b� Find all values of a� such that the system has an unique solu�tion

c� If a � ��� �� and �x� y� is a solution of the system �nd all values

of a such that the expressionx

y�y

xtakes its minimal value

Solution It follows easily that x and�

x� yare roots of the equation

t� � �a� ��t� a� � �� There are two cases to be considered

���

���������

x �

x� y a� �

���������

x �

y � � a

a� ��

when a � �

and

���

���������

x a� �

x� y �

��������x a� �

y � � a�

a� When a � we have �x� y� �����

�� or �x� y� ���� ���

Page 58: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

b� The sytem has an unique solution ��� �� when a � and ��� ��when a ��

c� If �x� y� is a solution and a � ��� �� thenx

yand

y

xare positive

Furtherx

y�y

x� �� and equality occures when

x

y

y

x� It follows

from ��� and ��� thatx

ya� �

� � a� Using the equality

a� �

�� a

� � a

a� �

we �nd the only value of a �

��

Problem �b�� Given an acute �ABC� The bisector of �� ACBintersects AB at point L� The feet of the perpendiculars from L toAC and BC are denoted by M and N respectively Let P be theintersecting point of AN and BM� Prove that CP AB�

Solution Let l be the line through C which is parallel to AB� LetF and E be respectively the intersecting points of AN and BM withl� The intersecting point of CP and AB is denoted by D� We obtainAD

CF

PD

PC

BD

CE� so

AD

BD

CF

CE� From the other hand

AM

CM

AB

CEand

BN

CN

AB

CF� But CM CN and we get

AM

BN

CF

CE�

ThereforeAD

BDAM

BN� which implies

���AM

ADBN

BD

Further if CH AB�H � AB� then �ALM � �AHC and soAL

AC

AM

AH� In the same manner

BL

BC

BN

BH� But CL is a bisec�

tor and thereforeAL

AC

BL

BC� so

AM

AH

BN

BH� The last equation

combined with ��� gives D � H which implies CP AB�

Problem �b� Prove that the digit of the hundreds of ������������

Page 59: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

����� is even

Solution Write the number ����������������� in the form ��������� � �� ���������� �������������� ����������������� Since �� ���� ��� ���� and ��� ������ we have that the last two digits of��� coincide with the last two digits of ���� so the last two digits of��� are ���Moreover the last two digits of ����� are also ��� Thereforethe last two digits of the given number are the last two digits of theproduct ����������� which are � and �� Since ����� � ����� � ����� isdivisible by � and it ends by ��� the digit of the hundreds is even

Problem ���� Find all values of the real parameter a such that thenonnegative solutions of the equation ��a��� sin x����a� sin �x sin �x form an in�nite arithmetic progression

Solution Since sin �x � sin x cos x and sin �x sinx�� cos� x����we may write the equation in the formsinx �� cos� x� ��� a� cos x� a� �� Thus sinx �� cos x � or

cos x �a�� The nonnegative solutions of the equations sinx �

and cosx � are x k� and x �k�� k �� �� �� � � � respectively

Let jaj � �� In this case the equation cosx �a�has no solution

and therefore the nonnegative solutions of the initial equation are�� �� ��� � � � � which form an arithmetic progression Let now jaj �� and let x� be the only solution of the equation cos x �a

�in

the interval ��� ��� In this case the nonnegative solutions of the lastequation are x x� � �k� and x �� � x� � �k�� It is clear nowthat the nonnegative solutions form an arithmetic progression only

when x� �� x� �

�and x� �� so giving a ��� a � and a ��

The values of a are a � and jaj � ��

Problem ���� Let O� I and H be respectively the circumcenter�incenter and orthocenter for an acute nonequilateral �ABC� Prove

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that if the circumcircle of �OIH passes through one of the verticesof �ABC then it passes through another vertex of �ABC�Solution Assume thatO� I�H and C lie on a circle It is well knownthat CI is the bisector of �� HCO� Thus �� IHO �� ICO �� ICH�� HOI and it follows from �IHO that IH IO t�

We shall prove that if �� BAC � ��� then O� I�H and A lie ona circle Denote by M and N the projection points of I respec�tively on AO and AH� Let O� and O� be such that IO� IO� tand O� lies between A and M� and M lies between O� and O��Analogously let H� and H� be such that IH� IH� t and H�

lies between A and N� and N lies between H� and H�� If O �O��H � H� or O � O��H � H�� then �AIO � �AIH and there�fore AO AH� But AH �AO cos �� BAC and so �� BAC ����If O � O��H � H� or O � O��H � H�� then it follows from�� IO�O� �� IO�O� �� IH�H� �� IH�H� that AOIH is inscribed

Suppose now that A and B do not lie on the circumcircle of�OIH� In this case �� BAC �� ABC ��� and therefore �ABCis equilateral which is a contradiction

Problem ��� In each of the cells of a � � table is written a realnumber The element in the i�th row and j�th column equals tothe modulus of the di�erence of the sum of the elements from thei�th row and the sum of the elements from the j�th column Provethat every element of the table equals either to the sum or to thedi�erence of two other elements of the table

Solution Let p�� p�� p� and q�� q�� q� be the sum of the elements inthe �rst� second and third row and in the �rst� second and thirdcolumn respectively It is clear that p� � p� � p� q� � q� � q��Therefore the element in the �rst row and the �rst column equals tojp� � q�j� From the other hand jp� � q�j jp� � p� � q� � q�j whichimplies jp��q�j ��jp��q�j���jp��q�j where ��� �� � ������ Since

Page 61: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

jp� � q�j � � it is clear that �� �� �� is impossible Thereforejp� � q�j is either the sum or the di�erence of jp� � q�j and jp� � q�j�By analogy every element of the table is the sum or the di�erenceof two other elements

Problem ����� Prove that for every positive number a the se�

quence fxng�n��� such that x� �� x� a� xn�� �

qx�n��xn� n � ��

is convergent and �nd its limit

Solution It follows by induction that the terms of the sequencefxng�n�� can be expressed as xn a�n� where fng�n��� is a siquence

de�ned by � �� � �� n�� �n�� � n

�� n � �� Thus� n�� �

n�� ��

��n�� � n� and therefore n�� � n��

���

�n�� �

�� ���

�n� Adding the equalities k���k��

���

�kfor k

�� �� � � � � n after simple calculations we obtain n���� ���

��

���

��

� � � � ����

�n

� ����

�n��� �

���

� �

��� �

���

�n��

�� Since

� � and limn��

���

�n � it follows that lim

n��n

�� This shows

that the sequence fxng�n�� is convergent and limn��

xn a�

� �

Problem ���� Given a convex quadrilateral ABCD where M isthe intersecting point of its diagonals It is known that DB �DM�AM MC�

a� Express BC and CD by the sides of �ABD�b� Prove that if � �� ADB� �� ABD ����� then �� DBC

� �� BDC�

Solution a� Denote AB c�BC p�CD q�DA b and

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DB a� It follows from the condition of the problem that DM �

�a�MB

�a� The formula for the median in a triangle applied for

�ABC and �ACD gives

�����

�a� �p� � �c� � �AM��

�a� �q� � �b� � �AM�

Let �� AMB � From the Law of Cosines for �AMB and �AMDwe obtain

c� AM� ��a�

�� �

�AM�a cosb� AM� �

a�

��

�AM�a cos

Thus �b� � c� �AM� ��

�a� and a substitution in ��� gives

��� p� �a� � ��b� � �c�

�q�

�b� � �c� � �a�

b� It follows from the condition of the problem that �� ADB ���� and so c � a� Let D� be a point on AB� such that D�B DB�

Further �� AD�B �� ABD ������ �� ABD

����� �� ABD

�� ADB� Therefore �AD�D � �ADB and thus b� c�c � a�� A

substitution in ��� implies p �c � �a

�and q�

�c� � �ac� �a�

��

Hence q� p�p � a�� Let B� be a point on BC �B lies between C

and B��� such that BB� BD a� Sinceq

pp� a

qit follows that

�DBC � �B�DC� Therefore �� DBC �� BB�D� �� BDB� � �� BB�D � �� BDC�

Problem ��� See problem ���

��

Page 63: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Spring Mathematical Tournament

Jambol� ����� March ����

Problem ���� Let f�x� be a linear function such that f��� ��and f�f���� ���� Find all values of m� for which the set of thesolutions of the inequality f�x��f�m�x� � � is an interval of length��

Solution Let f�x� ax�b� It follows from f��� �� that b ���and from f�f���� ��� that a �� Therefore the function is f�x� �x � �� Consider the inequality ��x � �����m � x� � �� � � ���x� ����m� �� �x� � �� The solution of the last inequality is an

interval with end points�

�and

�m� �

�� Therefore

������ ��m� �

���� ��

so j��mj �� Finally� we obtain that m � and m ��

Problem ��� Given an isosceles right angle triangle ABC with�� ACB ���� Point P lies on BC�M is the midpoint of AB andlet L and N be points from the segment AP such that CN APand AL CN�

a� Find �� LMN �

b� If the area of �ABC is � times greater than the area of�LMN �nd �� CAP�

Solution a� Let �� CAP � We �nd �� ACN ��� � and�� MCN �� ACN���� ���� �� LAM� Since AM CM andAL CN it follows that �AML � �CMN� Therefore �� AML �� CMN and so �� LMN ���� �� AML� �� CMN ����

b� It follows from�AML � �CMN that LM MN and since

��

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�� LMN ��� we get SLMN MN�

�� Also SABC

AC�

�� Now

�MN�

�AC�

�applies that MN

AC

�� Denote by Q the midpoint

of AC� Thus QM QN AC

� MN� Therefore �QMN is

equilateral and �� QNM ���� which implies�� QNA ������� ���� But AQ QN and so �� CAP �� QNA ����

Problem �� There are ���� white balls in a box There are alsosu�ciently many white� green and red balls The following opera�tions are allowed

�� Replacement of two white balls with a green ball�

�� Replacement of two red balls with a green ball�

�� Replacement of two green balls with a white bal and a redball�

�� Replacement of a white ball and a green ball with a red ball�

�� Replacement of a green ball and a red ball with a white ball�

a� After �nitely many of the above opperations there are threeballs left in the box Prove that at least one of them is a green ball

b� Is it possible after �nitely many opperations to have only oneball left in the box�

Solution Consider a box with x white� y green and z red ballsDirect veri�cation shows that after applying any of the allowed op�erations the sum x� �y � �z does not change modulo � Since theinitial values are x ����� y z �� we obtain that this sum iscongruent to � modulo �

a� There are � balls in the box and therefore x � y � z ��Moreover x��y��z � ��mod��� If a green ball is not in the box theny � and so x�z � and x��z � x�����x� � ���x � ��mod���

��

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which is impossible

b� Suppose that there is only one ball left in the box Thereforex� y � z � and x� �y � �z � ��mod��� which is impossible

Problem �a�� Find all values of m such that the equation��

x�m�

m

x�m� �m

m� � x�

��jx�mj �m� �

has exactly one nonnegative root

Solution The equation�

x�m�

m

x�m� �m

m� � x� � is equivalent

to �when x � �m� to �m���x �m�m���� When m �� it hasin�nitely many roots � all numbers x � ��� If m � �� we obtainx �m� The equation jx �mj �m � has two roots x � andx �m when m � �� an unique root x � when m � and hasno roots when m � ��

Let m � � and m � ��� In this case the equation has an uniqueroot x �m and it is nonnegative Let m �� Then the equation

becomes�

xjxj �� which obviously has no roots Let m � �� The

equation has three roots x �m�x �� x �m and two of themare nonnegative

Thus� the desired values of m are m � � and m � ���Problem �a�� Given an acute�angled triangle ABC and let � and � be its angles respectively to A�B and C� For an arbitraryinterior point M denote by A�� B� and C� respectively the feet ofthe perpendiculars from M to BC�CA and AB� Find the locus ofM for which the triangle A�B�C� is a right angle triangle

Solution Let M be a point such that �� A�C�B� ���� Since thequadrilateral AC�MB� is inscribed we have �� B�C�M �� B�AM x� Analogously �� A�C�M �� A�BM y� Denote z �� AMB� Itfollows now from the quadrilateral AMBC that x� � � y� ����� �

��

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z� ����� Since x � y �� A�C�B� ��� we get z ��� � ��Therefore �� AMB ��� � �� Thus� the locus of M such that�� A�C�B� ��� is an arc G� in the interior of �ABC for whichthe segment AB is seen under angle ��� � �� Analogously one canprove that the locus of M such that �� C�B�A� ��� is an arc G�

in the interior of �ABC� for which AC is seen under angle ��� � �Further� the locus of M such that �� B�A�C� ��� is an arc G� inthe interior of �ABC for which BC is seen under angle ��� � �The desired locus is the union of the three arcs G� �G� � G��

Problem �a� See Problem ��

Problem �b�� The real numbers x and y are such that x��xy�y� �� If F x�y � xy��

a� prove that F � ���b� �nd the greatest possible value of F�

Solution a� It follows from the condition of the problem that �x�y�� ��xy and therefore xy � ��� From the other hand x�� y� �� xy� Thus� xy � �� so z xy � ���� ��� Hence F xy�x�� y�� z�� � z� and the inequality F � �� is equivalent to the inequalityz� � z � � � �� The latter one is true for any z � ���� ���

b� Assume that the greatest value of F exists and denote it byA� This implies that the system

���

����� xy�x� � y�� Axy � �x� � y�� �

has a solution It is clear that t� xy and t� x� � y� areroots of the quadratic equation t� � t � A �� But t� � t� �

��x � y�� �

��x � y�� � � and the equality is impossible since if

so then x y � which is a contradiction to x� � xy � y� ��

Therefore t� � t� and so D � � �A � �� ie A ��

�� The system

��

Page 67: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

��� is equivalent to

����� xy t�x� � y� t��

and the last is equivalent to����� �x� y�� t� � �t��x� y�� t� � �t��

The latter system has a solution i� the in�

equalities t�� �t� � � and t���t� � � hold true It su�ces to prove

that when A ��

�the roots t� � t� of the equation g�t� t�� t�A

satisfy the inequalities t� � �t� � � and t� � �t� � �� If A �then t� � and t� � and the inequalities hold true It is clear

that we can consider only the case A � ����

��� Now t� � �� t� � �

and the inequality t� � �t� � � holds true From the other hahd

t� � t� � and the inequality t� � �t� � � is equivalent to t� � �

��

The same inequality is equivalent to t� � �

�� For the roots t� and

t� we obtain t� � �

�t� � �

�� which is equivalent to

����� g���� � �

g���� � ��

Therefore A � �

�� Conversely� if A

�then x y �

p�

�satisfy

the condition of the problem

Problem �b�� A line l is drown through the orthocenter of anacute�angled triangle ABC� Prove that the lines symmetric to l withrespect to the sides of the triangle intersect in a point

Solution Let the intersecting points of l with the sides AC andBC be Q and P respectively and let the intersecting point of l withthe extention of AB be R� Without loss of generality A lies betweenR and B� Denote the symmetric points of H with respect to AC�BCand AB by B�� A� and C� respectively It is well known that A�� B�

and C� lie on the circumcircle k of �ABC� It is clear also that thesymmetric lines of l are B�Q�A�P and C�R� Let B�Q�A�P S andB�Q�C�R T�We obtain �� CB�Q� �� CA�P �� CHQ� �� CHP ���� and therefore SA�CB� is inscribed Thus� S k�B�Q� From

��

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the other hand �� RC�A �� RHA �� AB�T and therefore B�ATC�

is inscribed Hence� T k � B�Q� This implies that T � S whichshows that the three lines intersect in a point

Problem �b� See Problem ��

Problem ���� Solve the equationpx� �

px� � �

px� ���

Solution It is clear that x � �� Note that x � is a root ofthe equation We shall prove that there are no other roots Ras�ing the equation to the forth power we get x� � ��

px�� �px� � �

��px��� �

px� ��� � ��

px�� �px� ��� � � �

px� ��

x� ���

Let f�x� ��px�� �px� � � ��

px��� �

px� ���

���px�� �px� ����� �

px� ��� Obviously f�x� is an increasing func�

tion If x � � then it follows from the inequalities x� � x� f�x� �f��� �� that x� � f�x� � x � ��� a contradiction If x � �then it follows from the inequalities x� � x� f�x� � f��� �� thatx� � f�x� � x� ��� a contradiction This completes the prove

Problem ���� The incircle of an isosceles �ABC touches the legsAC and BC at points M and N respectively A tangent t is drawn

to the smaller of the arcs�

MN and let t intersects NC and MC atpoints P and Q respectively Let T be the intersecting point of thelines AP and BQ�

a� Prove that T lies on the segment MN �

b� Prove that the sum of the areas of triangles ATQ and BTPis the smallest possible when t is parallel to AB�

Solution a� Let the incircle touches AB and PQ at points R andS respectively Let MN and SR intersect QB in points T� andT� respectively Since �� T�MQ �� T�NC ����� �� T�NB and�� MT�D �� BT�N it follows from the Law of Sines for�MT�D and

��

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�BT�N thatDT�MD

BT�BN

� soQT�BT�

MQ

BN� By analogy

QT�BT�

SQ

BR�

It follows now from MQ SQ and BN BR thatQT�BT�

QT�BT�

so T� T�� In the same manner one can prove that AP passesthrough the intersecting point of MN and SR� This implies thatAP�BQ�MN and SR intersect in T�

b� We have SATQ�SBPT SABQ�SABP ��SABT � Since�ABCis isosceles we get MN k AB and therefore SABT is constant Thus�SATQ�SBPT is minimal exactly when SABQ�SABP is minimal The

latter sum equals toAQ�AB sin�BP�AB sin

AQ�BP �AB sin��

� where is the angle to the base of the triangleTherefore it su�ces to �nd the minimum of AQ � BP� It is easilyseen that AQ � BP AM � BN � PQ and therefore we have to�nd when PQ is minimal Let r be the inradius and O be the centerof the incircle of �ABC� Using that PQ r�cotg�� cotg � where� �� OQP� �� OPQ we obtain from ABPQ that ���� ��

����� Therefore �� ���� � � Thus PQ �r sin

cos��� � � cos�

It is clear now that PQ is minimal exactly when cos��� � �� so� PQ k AB�Problem ��� There are n � � points in the plane such that thedistance between any two of them is an integer Prove that at least�

�from the distances between them are divisible by �

Solution We show �rst that the assertion from the problem is truefor n � ie for � points with integer distances between them atleast one distance is divisible by � Denote the points by A�B�C andD �it is easy to be seen that WLOG �� BAD �� BAC� �� CAD�By the Law of Cosines for �ABC��ACD��ABD we obtain

BC� AB� �AC� � ��AB�AC cos

��

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CD� AD� �AC� � ��AD�AC cos

BD� AB� �AD� � ��AB�AD cos �

where �� BAC� �� CAD� � �� BAD � �

Suppose that all distances are integers not divisible by � There�fore AB� � AC� � AD� � BC� � CD� � BD� � ��mod �� andso ��AB�AC cos � ��AD�AC cos � ��AB�AD cos � � ��mod ���Thus� ��AB�AC cos���AD�AC cos � ��AC��AB�AD� cos cos� AC��AB�AD� cos cos � ��mod ���

Note that cos� cos and cos � are rational numbers Moreover�

if cos p

q� cos

r

s� where p� q and r� s are relatively prime

then p� q� r and s are not divisible by � Hence p� � q� � r� �s� � ��mod �� and cos � cos cos � sin sin cos cos �pq� � p�

q

ps� � r�

s� Therefore ��AC��AB�AD� sin sin is divisible

by � and after multiplying ��AB�AD cos � � ��mod �� by AC� weobtain ��AC��AB�AD� cos cos � ��mod ��� a contradiction toAC��AB�AD� cos cos � ��mod ��� Therefore at least one of thedistances is divisible by �

Let n � �� Since there exist

�n

�sets with four elements each

there exist at least

�n

�distances �counted more than once� divisible

by � Each such distance is counted exactly

�n� �

�times and we

get that the desired number is at least

�n

��n���

� �

�n

��

Problem ���� Let f�x� x� � �x � �

x� � �x� ���

��

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a� Find the greatest value of f�x��

b� Find the greatest value of the function

�x� � �x� ��

x� � �x� ��

�fx��

Solution a� We shall prove that the greatest value of f�x� equals� Since x� � �x� �� � ���x� we obtain f�x� � � �x� ��� � ��and an equality occurs only when x ���

b� Let g�x� x� � �x� ��

x� � �x� ��� Since x���x��� � ���x� we obtain

g�x� � � �x � ��� � �� and equality occurs only when x ���Since x���x��� � ���x� we have that the function h�x� g�x�fx�

is correctly de�ned Further� when f�x� � �� ie x � �������� thenh�x� � �� �� From the other hand g�x� � � x � �� and soh�x� � � when f�x� � �� Therefore the greatest value of h�x� equals� and h�x� � when x ���Problem ���� A point A� is chosen on the side BC of a triangleABC such that the inradii of �ABA� and �ACA� are equal De�note the diameters of the incircles of �ABA� and �ACA� by da�In the same manner de�ne db dc� If BC a�CA b�AB c� p a� b� c

�and ha� hb� hc are the altitudes of the triangle ABC and d

is the diameter of the incircle of �ABC prove that

a� da �

qp�p � a�

ad ha�

b� da � db � dc � p � ha � hb � hc�

Solution a� LetO� and O� be the incenters of�ABA� and�ACA�

and p� and p� be semiperimeters of the same triangles We haveSABC SABA�

�SACA� ra�p�� ra�p� rc�p�� p�� rc��p�CC���

where ra da�� Therefore ra�p � AA�� SABC � If P and Q are

the touching points of the incircles of �ABA� and �ACA� with

��

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BC then the quadrilateral O�PQO� is rectangle and so O�O� PQ PA� � QA� p� �AB � p� �AC p� � p� �AC �BC p�AA� �AC �BC� It follows from the similarity of �IO�O� and

�IBC� where I is the incenter of �ABC thatO�O�

BC

r � rar

AA� � p�AC �BC

BCr � rcr

� We obtain the system������ra�p�AA�� SAA� �AB � p

c

r � rar

Thus� AA� qp�p � a� and ra

rp

p �qp�p � a�

rpp�pp �pp � a�

a

ha�� r

qp�p � a�

a� which

implies a�

b� By a� the inequality is equivalent to

pp � a

a�

pp� b

b�

pp� c

c�pp

d� Further

pp � a

a�

pp� b

b�

pp� c

c

pp � a

�p � b� � �p � c��

pp� b

�p � a� � �p � c��

pp� c

�p� a� � �p� b��

pp � a

�pp� b

pp� c

pp � b

�pp� a

pp � c

pp� c

�pp� a

pp � b

pq�p� a��p� b��p� c�

pp

d�

Problem ��� See problem ���

��

Page 73: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

XLIX National Mathematical Olympiad

Third Round� ���� April ����

Problem �� Find all value of the real parameter a such that theequation

�t � �a�t � �� a� �

has an unique root in the interval ��� ���

Solution After the substitutions x �t and f�x� x� � �ax �� � a� the problem is equivalent to �nd all values of a such thatthe equation f�x� � has an unique root in the interval ��� ��� Itis easily seen that if f��� � or f��� �� then a �� a �� ora ��� are not solutions of the problem Therefore the equationf�x� � has an unique root in the interval ��� �� when f����f��� � �or Df �� �a � ��� ��� It follows now that f����f��� � � � �a�����a����a���� � � � a � �������� or �a��� �� �a � ��� ���

and so a �p�

��

Problem �� In �ABC�CH�H � AB� is altitude and CM andCN�M�N � AB� are bisectors respectively of �� ACH and �� BCH�The circumcenter of �CMN coincides with the incenter of �ABC�Prove that S�ABC

AN�BM

��

Solution Let I be the incenter of ABC� Denote by P�Q and R thecommon points of the incircle of ABC respectively with AB�BCand CA� It follows from IP IQ IR and IC IM IN that�IMP��INP��ICQ and �ICR are congruent Let �� MIP

�� NIP �� QIC �� CIR �� We have that �� MCN �

��� MIN

�� Since CM and CN are bisectors we get �� ACB � �� MCN ��� It follows now from the quadrilateral IQCR that �� QIR�

��

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�� QCR ���� � �� ���� � � ���� Therefore �ABC isa right angle triangle which implies �� BCM �� BMC �� BAC ��� ABC

�and�� ANC �� ACN �� ABC�

�� BAC

�� ThereforeBC

BM and AC AN and so SABC AC�BC

�AN�BM

��

Problem � Let fang�n�� be a sequence such that a� ��� a� ���� an�� �an � an�� for n � �� Prove that

a� an and an�� are relatively prime for all n�

b� for every natural numberm there exist in�nitely many naturalnumbers n� such that an � � and an�� � � both are divisible by m�

Solution a� Suppose that there exist natural numbers n andm � ��such that m divides both an and an��� It follows from an�� an����an that m divides an��� By induction m divides both a� and a��which is impossible since a� and a� are relatively prime

b� Consider the sequence fang de�ned by an�� an�� � �an fornegative indices Compute a� a� � �a� ��� a�� a� � �a� �� a�� a� � �a� �� a�� a�� � �a�� � and so on We �nd asequences fang��� of integers such that an�� �an � an�� for everyn� Since the pairs �p�mod m�� q�mod m�� are �nitely many we getthat there exist integers r and s � r such that ar � as�mod m�and ar�� � as���mod m�� It is clear now that ar�i � as�i�mod m�for every i� Therefore the sequence fang��� is periodic Since a�� �a�� � ��mod m�� there exist in�nitely many natural numbers n suchthat both an � � and an�� � � are divisible by m�

Problem �� Given a convex quadrilateralABCD� such that�� BCD�� CDA� The bisector of �� ABC intersects the segment CD inpoint E� Prove that �� AEB ��� if and only if AB AD �BC�

��

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Solution Let �� AEB ���� Since �� CEB � ��� there exists apoint F on the sideAB such that �� BEF �� BEC�Thus�BCE ��BFE� which implies BF BC and �� BFE �� BCE� From theother hand

��� AE

sin �� AFE

AF

sin �� AEF�

AE

sin �� ADE

AD

sin �� AED�

Since �� AED �� AEF and �� AFE� �� ADE ����� we obtainAF AD and so AB AD �BC�

Conversely� let AB AD � BC� There exists a point F onthe segment AB such that AF AD and BF BC� Therefore�BFE � �BCE� so �� BFE �� BCE and �� BEF �� BEC� Itfollows from ��� and AF AD that sin �� AED sin �� AEF� Since�� AED� �� AEF � ����� we have �� AED �� AEF and therefore�� AEB ����

Problem � Prove that for any two real numbers a and b thereexists a real number c � ��� ��� such that

����ac� b��

c� �

���� � �

���

Solution Consider f�x� ax��

x� �and let m and M be respec�

tively the minimum and the maximumvalue of the function f in the

interval ��� ��� Since f��� �� f��� a� �� and f ��x� a� �

�x� ���

there are four cases for the di�erence M �m�

�� a � �

�� Thus f ��x� � � for x � ��� �� and soM�m f����f���

�� a � �

��

��

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�� a � �� Thus f ��x� � � for x � ��� �� and so M�m f����f��� a� �

�� �

��

��

�� a � �� Thus d

�pa� � � ��� ��� f ��x� � � for x � ��� d� and

f ��x� � � for x � �d� ���

���

�� a � �

�� Since f��� � f���� we have M � m f��� �

f�d� �� �pa�� � ���p�

����

���

�� a � �� Since f��� � f���� we have M � m f��� �

f�d� �

���pa� ��� � �

��p� � ����

In all four cases M �m ��

��� which impliesM � b �

��or m� b �

��� The assertion of the problem follows now by continuity

Problem �� Find all sets S of four points in the plane such that forany two circles k� and k�� having diameters with endpoints � pointsfrom S there exists a point A � S � k� � k��Solution Let S fA�B�C�Dg� Consider the circles k� and k� withdiameters respectively AB and CD� It follows that at least one ofthe angles ACB�ADB�CAD and CBD is right angle Without lossof generality �� ACB ���� Let k� and k be circles with diametersrespectively AC and BD� Since �� ABC � ���� we obtain that thecommon point of k� and k is one of the points A�C or D�

� Let A � k� � k� Now �� BAD ��� and therefore D � l��l� AB�A � l�� It is easily seen that �� BDC � ��� and since�� BAC � ���� �� ABD � ���� it follows that the common point of

��

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the circles with diameters AD and BC is the point C and �� ACD ���� Therefore D l� �BC�

� Let C � k� � k� Now �� BCD ��� and therefore D lies onthe line AC� Since �� BDC � ���� �� BAC � ��� and �� ACD � ����we obtain that the common point of the circles with diameters ADand BC is B and �� ABD ���� Therefore D l� � AC� wherel� AB�B � l��

� Let D � k� � k� Now �� ADC ��� and therefore D � k��Since �� BAC � ��� and �� ACD � ���� we obtain that the commonpoint of the circles with diametersAD and BC is B or D� In the �rstcase �� ABD ��� and thus D � l�� from the other hand l��k� ��which is a contradiction In the second case �� BDC ���� and weget that D is the orthogonal projection of C on AB�

In conclusion� all sets S� satisfying the condition of the problemare those consisting of the vertices of a right triangle and the foot ofthe altitude to the hypothenusis

��

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XLIX National Mathematical Olympiad

Fourth Round� ���� May ����

Problem �� In an orthogonal coordinate system xOy a set consisit�ing of ���� points Mi�xi� yi�� is called �good� if � � xi � ��� � �yi � � i �� �� � � � � ���� and xi � xj for i � j� Find all naturalnumbers n with the following properties a� For any �good� setsome n of its points lie in a square of side length ��

b� There exists a �good� set such that no n � � of its points liein a square of side length ��

�A point on a side of a square lies in the square�

Solution We shall prove that n �� is the only solution of theproblem We show �rst that for a �good� set some �� points liein a square of side length � All points lie in the rectangle � �x � ��� � � y � �� Divide this rectangle to �� squares of sidelength �� If some �� points lie in one of these rectangles then weare done Conversely� in every square there are at least �� or atmost �� points We prove now that there exists a square with atleast �� points and there exists a square with at most �� pointsIndeed� if a square with �� points does not exist then the points areat most ����� ���� � ����� If there is no square with less than�� points then the points are ����� � �� � ����� Further� move thesquare with more than �� points towards the square with less that�� points Since the number of points in this square changes at mostby one we get the assertion of the problem

To prove b� let x� �� xi xi�� ���

����and yi � for i

�� �� �� � � � � ���� while yi � for i �� �� � � � � ����� Let XY ZT be anunit square WLOG we assume that it intersects the lines

��

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y � and y � in the points P�Q and R�S respectively Then

PQ�QR�RS�SP � PY�Y Q�QZ�ZR�RT�TS�SX�XP ��

But SP�QR � � and we get PQ�RS � �� This implies that in thesquare XY ZT there are no more than �� points

We found that n �� is a solution of the problem It is unique

since �� �����

��� ���

Problem �� Given an acute �ABC� Prove that there exist uniquepoints A�� B� and C� on BC�AC and AB respectively with the fol�lowing property Each of the points is the midpoint of the segmentwith ends the orthogonal projections of the other two points on thecorresponding side Prove that �A�B�C� is similar to the triangleformed by the medians of �ABC�

Solution It is easy to be seen that the perpendicular through C� toAB intersects A�B� at its midpoint The same is applied for A� andB�� Therefore the tree perpendiculars intersect at the medcenter of�A�B�C�� the point P�

��

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Let T be the midpoint ofA�B�� �� PB�A� x �� PCA���� PA�B� y �� PCB�� The Law of Sines for �B�TP and �A�TP gives

sin

sinxB�T

TPA�T

AP

sin

sin y�

wheresin x

sin y

sin

sin� Further� if CN is the median in�ABC� similar

arguments imply thatsin �� ACG

sin �� BCG

sin

sin� Since x� y

�� ACG� �� BCG � and � is acute it follows that x �� ACG� y �� BCG� It is clear now that CP is symmetric to the median in�ABC through C with respect to the bisector of angle �� The sameis true for AP and BP� Therefore the point P is unique �and there�fore A�� B� and C� are unique� Further� �� B�C�A� �� B�C�P��� A�C�P �� B�AP� �� A�BP �� BAG� �� ABG �� BGK� whereK is the symmetric point of G with respect to the midpoint of BC�By analogy �� C�A�B� �� GBK and �� A�B�C� �� GKB� There�fore �A�B�C� is similar to the triangle formed by the medians of�A�B�C��

��

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Problem � Let p � � be a prime number and a�� a�� � � � � ap�� bea sequence of natural numbers such that p does not divide both akand akk� � for all k �� �� � � � � p� �� Prove that the product of someelements of the sequence is congruent to � modulo p�

Solution Consider the sequence �� a�� a�� � � � � ap��� We shallprove by induction that for any i �� �� � � � � p � � there exist in�tegers b�� b�� � � � � bi each of which is a product of some elements fromthe above sequence and bm �� bn�mod p� for m � n� Indeed� fori � we can choose b� �� b� a��a� �� ��mod p��� Suppose wehave chosen b�� b�� � � � � bi such that bm �� bn�mod p� for m � n� Con�sider b�ai� b�ai� � � � � biai� It is easily seen that any two of them are notcongruent modulo p� Further� if bjai for any j is congruent to bl forsome l we get that b�ai� b�ai� � � � � biai modulo p is a permutation ofb�� b�� � � � � bi modulo p� Thus�

�b�ai���b�ai�� � � � ��biai� � b��b�� � � � �bi�mod p�

and therefore aii � ��mod p�� a contradiction

It follows now that for any s �� �� � � � � p � � there exist fewelements of the sequence from the condition of the problem suchthat their product is congruent to s modulo p�

Problem �� Find all polynomials P �x� with real coe�cients suchthat

P �x��P �x� �� P �x��

for any real x�

Solution We show �rst that for any natural number n there existsat most one polynomial P �x� of degree n such that P �x��P �x���

��

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P �x��� Indeed� if

P �x� anxn � an��x

n�� � � � �� a�x� a��

comparing the coe�cients in front of the same degrees of x in

P �x��P �x � �� and P �x��� we get an � and an�� �n�� It is

easily seen that each of the subsequent coe�cients is a solution of anequation of �rst degree of the form �x� b �� where b is a functionof already chosen coe�cients Therefore� if a polynomial of degreen with the desired property exists it is unique one

We prove now that a polynomial P �x� of odd degree such thatP �x��P �x��� P �x�� does not exist Let P �x� xm�x� ��n�Q�x��m or n could be zero�� where Q�x� is a polynomial such that Q��� �� and Q��� � �� A substitution gives

�x� ��m�xn�Q�x��Q�x� �� xm��x� ��n�Q�x���

If m � n the substitution x � leeds to a contradiction Thereforem n and so Q�x��Q�x� �� Q�x��� where Q�x� is of odd degreeTherefore there exists a real number x� � �� �� such that Q�x�� ��It is clear that x� �� since otherwise the substitution x ��shows that Q��� � which is impossible A substitution x x�gives that Q�x��� �� thus Q�x�� � and so on Since x� � ��� � itfollows that there are in�nitely many distinct terms in the sequencex�� x

��� � � � � x

�k� � which is impossible

Direct veri�cation shows that for any even natural number n �k� the polynomial P �x� xk�x���k is of degree n and is a solutionof the problem

Therefore all polynomials such that P �x��P �x � �� P �x�� areP �x� xk�x� ��k�

Problem � Let D be the midpoint of the base AB of an isoscelesacute �ABC�A point E is chosen on AB and O is the circumcenter

��

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of �ACE� Prove that the line through D perpendicular to DO� theline through E� perpendicular to BC and the line through B� parallelto AC intersect in a point

Solution Let �� ABC �� BAC and let G be the circumcenterof �ABC� Consider points F � and F �� on the line through B parallelto AC such that OD DF � and BC EF ��� Denote by H � and H ��

respectively the projection points of F � and F �� on the lineAB� SinceO is inner point for �� ADC and �� ACB � ���� we have that F � andF �� lie in the interior of �� BAC� It su�ces to prove that F � � F ���ie F �H � F ��H ��� Let O� and G� be respectively the projectionpoints of O on AB and of G on OO�� Since �DH �F � � �OO�D� itfollows that

DH �

F �H �

OO�

DO�� Also� �� GOG� and so

B�H �

F �H �OG�

GG��

It follows from GG� DO�� O�G� DG and �� DBG � � ���

that F �H � BD�O�D�

GD �tg��O�D� Denote BC �EF �� I� Since

�� CBF �� ���� � � and BE �O�D� it is clear that F ��H ��

BF �� sin BI sin

cos����� � �� �BE sin cos

cos � �O�Dtg�

F �H �� This completes the prove

Problem �� Let A be the set of all binary sequences of length nand let � � A be the sequence with zero elements The sequence c c�� c�� � � � � cn is called sum of a a�� a�� � � � � an and b b�� b�� � � � � bnif ci � when ai bi and ci � when ai � bi� Let f A � A bea function such that f��� � and if the sequences a and b di�er inexactly k terms then the sequences f�a� and f�b� di�er also exactlyin k terms Prove that if a� b and c are sequences from A such thata� b� c �� then f�a� � f�b� � f�c� ��

Solution Consider the sequence e� �� �� � � � � �� �� e� �� �� � � � � ��� � � � en�� �� �� � � � � �� �� en �� �� � � � � �� �� It follows from the condi�

��

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tion of the problem that �since f��� �� for all p� � � p � n thereexists q� � � q � n� such that f�ep� eq�

It is clear also that f�ep� � f�eq� for � � p� q � n� p � q�Therefore

��� ff�e��� f�e��� � � � � f�en�g fe�� e�� � � � � eng�

Consider an arbitrary sequence a a�� a�� � � � � an with t ones Iff�ep� eq� and ap �� then the q�th term of the sequence f�a� isalso � �otherwise ep and a di�er at t � � terms whereas f�ep� eqand f�a� di�er at t� � terms� By analogy if ap �� then the q�thterm of f�a� is also �

Finaly� consider the sequences a a�� a�� � � � � an� b b�� b�� � � � � bnand c c�� c�� � � � � cn such that a � b � c �� This means thatfor every i� � � i � n the sum ai � bi � ci is even number Fixi� � � i � n and let f�ei� ej� It follows now that the j�th terms ofthe sequences f�a�� f�b�� f�c� coincide with ai� bi� ci respectively andusing ��� we obtain that f�a� � f�b� � f�c� ��

��

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Union of Bulgarian Mathematicians

Sava Grozdev Emil Kolev

BULGARIAN

MATHEMATICAL COMPETITIONS

����

So�a� ����

Page 86: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Winter mathematics competition

Bourgas� ��� February ����

Problem ���� a� Draw all points in the plane with coordinates�x� y� such that

�j�x� yj � ���j�x� yj � �� � ��

b� Find all x and y for which��������j�x� yj � ���j�x� yj � �� � �

y � fxg � �� � x �

�For a real number x we denote the unique number in the interval��� � for which x� fxg is an integer by fxg��Solution a� It is easy to see that these are the points on the fourlines l� �x�y � �� l� �x�y � ��� l� �x�y � � and l� �x�y ����

b� There are two types of solutions

� Solutions of

�������y � ��x� �y � fxg� � x �

�� Solutions of

�������y � �x � �y � fxg� � x �

Denote the integer part of x by �x�� i�e� x � �x� � fxg� Inthe �rst case we have ��x�� � fxg � x� �x�� i�e �x � �� �x��

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Since the right hand side is an integer all posiible values of x are

���

����

�� � � � ��� Direct veri�cation shows that only x �

�� y �

��

x ��

�� y �

��x � � y � � are solutions�

In the second case �x� � � fxg � x � �x�� i�e� x � � � �x��It follows that x is an integer� i�e� x � �� � or � Direct veri�cationshows that solution is only x � �� y � ��

Thus� there are four solutions

x ��

�� y �

��x �

�� y �

��x � � y � ��x � �� y � ��

Problem ���� Points A�� B� and C� are chosen on the sides BC�CAand AB of a triangle ABC� Point G is the centroid of �ABC� andGa� Gb and Gc are centroids of �AB�C���BA�C� and �CA�B�

respectively� The centroids of �A�B�C� and �GaGbGc are denotedby G� and G� respectively� Prove that

a� the points G�G� and G� lie on a straight line�

b� lines AGa� BGb and CGc intersect in a point if and only ifAA�� BB� and CC� intersect in a point�

Solution a� Let O be an arbitrary point and

�OA� � � �OB � � � �� �OC � �OB� � � �OA� � � �� �OC�

�OC� � � �OA � �� �� �OB�

where �� �� � � ��� �� The existence of �� � and � follows fromthe fact that A�� B� and C� lie on the sides BC�CA and AB of�ABC� Then we have

�OG �

��OA � �OB � �OC

��

�� �OA � � �OB � � �OC

��

�OG� ���

��OA� � �OB� � �OC�

��

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h�� � �� �OA� ��� � � � �OB � ��� �� �� �OC

i�

By analogy �OGa �

h� � � � �� �OA � �� �� �OB � � � �� �OC

i�

�OGb �

h� �OA � �� � �� �� �OB � �� �� �OC

i�

�OGc �

h� �OA� � �OB � ��� �� �� �OC

i�

Thus� �OG� �

��OA� � �OB� � �OC�

��

h� � �� � ��� �OA � �� � �� � ��� �OB � �� � ��� ��� �OC

i�

Since �GG� � �OG� � �OG we have

�GG� �

h�� � � � � �OA � ��� �� �OB � � � �� �� �OC

i�

Using the same arguments we obtain �GG� � �OG� � �OG �

h��� � �� � �� �OA � ��� � ��� �OB � �� � �� � ��� �OC

i� It follows

from the last two equalities that �GG� ��

��GG� and we are done�

b� Since the lines AA�� BB� and CC� intersect in a point Ceva�s

theorem givesAC�

C�B� BA�

A�C� CB�

B�A� �

Denote the intersecting points of AGa� BGb and CGc with the sidesBC�CA and AB by A�� B� and C� respectively� A necessary andsu�cient condition for lines AGa� BGb and CGc to intersect in a

point isCA�

A�B� BC�

C�A� AB�

B�C� �

Denote the midpoint of B�C� by A�� Let h� and h� be the alti�

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tudes of �AA�C and AA�B� from A� and A�� We have

SAA�CSAA�B�

�h��AC

h��AB��

AA��AC

AA��AB��

By analogySAA�BSAA�C�

�AA��AB

AA��AC��

Dividing the above equalities and using that SAA�C� � SAA�B�and

SAA�CSAA�B

�CA�

BA�we obtain

CA�

A�B�AC

AB� AC�

AB��

By analogy

BC�

C�A�CB

CA� CB�

CA��AB�

B�C�

BA

BC� BA�

BC��

Multiplying the above equalities gives

CA�

A�B

BC�

C�A

AB�

B�C�AC

AB

AC�

AB�� BCAC

CB�

CA�� BABC

BA�

BC��

AC�

C�B

BA�

A�C

CB�

B�A�

Therefore the lines AGa� BGb and CGc intersect in a point i� thelines AA�� BB� and CC� interect in a point�

Problem ��� Let An be the number of sequences from ��s and �sof length n� such that no four consequtive elements equal ��� Findthe parity of A�����

Solution Denote the number of sequences of length n� the lastthree terms of which are ijk� where i� j� k � f�� g and no four con�sequtive elements equal to �� by anijk�

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Obviously an��ijk � an�ij � an�ij when ijk �� � and an��

��� � an����Adding the above equalities for all values of ijk we obtainAn�� � �An � an����

Therefore An�� an��� an����� � an����� an����� � an����� � an����� an����� � an����� an����� � an����� � an����� � an����� an����� � an����� � an����� �an����� � an����� � an����� � an����� � an����� � An���mod ���

Thus� Ak��m Ak�mod �� for all k and m�

Therefore A���� A� � � ��mod ���

Problem ��a��� Find all pairs �a� b� of integers such that the system����� x� � �ax� �a� � �y� � �by � x � �

has exactly three real solutions�

Solution Let �a� b� satisfy the condition of the problem� Since thereare exactly three real solutions the �rst equation has two distinctroots x� � x�� which implies D� � �a� � �a � � �� The discrim�inant of the second equation in respect to y equals D� � �b� � x�and therefore there are exactly three real solutions i� x� � x� � b��

These conditions are satis�ed i� a and b are integers such thata� � �a � � and b� � �a � p

a� � �a� � It follows from thelatter equality that a� � �a � � c�� where c is a positive integer�Therefore the discriminant of a���a��c� � � is a perfect square�i�e� � � � � c�� � d�� where d is nonnegative integer� The lastequality can be written in the form �d��c��d��c� � � which impliesthat d� �c � � d� �c � �� i�e� d � �� c � � Hence� a�� �a� � with roots a � � and a � ��� Respectively b � � or b � ��� Directveri�cation shows that all pairs �a� b� � ��� �� ������ ���� �� and������� satisfy the condition of the problem�

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Problem ��a��� The tangential point of a circle k through thevertex C of a �ABC and the line AB is the vertex B� The circle kintersects for a second time the side AC snd the median of �ABCthrough C at points D and E respectively� Prove that if the inter�secting point of the tangents to k through C and E lies on the lineBD then �� ABC � ����

Solution Let F be the intersecting point of the tangents tC andtE to k at C and E�G � BD CE and ACBH is parallelogram�

We have that �� ABC � ��� �� tC k AB �� FD

FB�

CD

CA�

From the other handCD

CA�

CD

BH�GD

GB� i�e� it su�ces to show that

FD

FB�

GD

GB� �� Since �FBE � �FED and �FBC � �FCD�

we obtainFB

FE�

BE

EDand

FB

FC�

BC

CD� Further� FC� � FE� �

FB�FD and soFD

FB�

CD�ED

CB�EB� Now the equality from � � follows

fromCD�ED

CB�EB�SCEDSCEB

�GD

GB� which completes the proof�

Problem ��a�� Ivan and Peter alternatively write down � or until each of them has written ��� digits� Peter is a winner if thenumber� which binary representation has been obtained� cannot beexpressed as a sum of two perfect squares� Prove that Peter has awinning strategy�

Solution First we prove that if the binary representation of a pos�itive integer ends with two ones and even number of zeroes thenthis integer cannot be represented as sum of two squares� Indeed�such a number is of the form k�s � �� and if we suppose thatx�� y� � k�s� ��� then using the fact that x�� y� equals � mod�ulo i� x and y are even we get that there exist integers p� q forwhich p� � q� � s� �� which is a contradiction�

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The winning strategy of Peter could be

If one of Ivan�s digits is then Peter simply repeats all digitswritten by Ivan� The �nal number is of the form k�s � �� andcannot be written as x� � y�� If all Ivan�s digits are zeroes thenthe �rst three digits of Peter are � � after which he writes onlyzeroes� The �nal number is ����� � � � ��� � ���� and cannotbe represented as p� � q� since � cannot be written in that form�

Problem ��b��� Find all values of the real parameter a such thatthe equation

logx �x� � x� a�� �

has unique solution�

Solution The equation is equivalent to �x� � x � a�� � x�� x �� x �� � Further� �x� � x � a�� � x� �� �x � a���x� � x �

a� � � with roots x� � �a and x��� �� �p� �a

� provided

� �a � �� Since��p� �a

� �� we obtain that only x� �

�a� x� � � �p� �a

can be roots of the equation from the prob�

lem� If a � � then x� � � and x� � �� which implies that theequation has no roots� If a � � then x� � and x� �� Notethat x� � x� implies a � � which is a contradiction� Therefore ifthe equation has unique solution then it is necessary to have a � �and one of the two roots equals � Thus� x� � � a � � andx� � � a � ���

Problem ��b��� On each side of a right isosceles triangle withlegs of length is chosen a point such that the triangle formed fromthese three points is a right triangle� What is the least value of thehypotenuse of this triangle�

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Solution Consider a right isosceles triangle ABC with right angleat C� Let A� � BC�B� � CA and C� � AB be such that �A�B�C�

is right triangle�

� Suppose �� B�A�C� � ���� Assume that the circle k with di�ameter B�C� intersects BC at point X �� A�� It is easy to be seenthat X is an interior point for the line segment BC� Draw a tangentl to k which is parallel to BC and the tangential point Y belongs

to the smaller of the arcs�

A�X � Denote the intersecting points of lwith AB and AC by C� and B� respectively� Consider a homothetyof center A such that the image of B��C�� is C�B�� It is clear thatthe image of �B�C�Y is inscribed in �ABC and its hypotenuseis less than B�C�� Therefore wlog we may assume that BC is tan�gent to k� Thus� if �� C�B�A� � �� then �� C�A�B � �� From the

Sine Low for �BC�A� we obtain BA� �B�C� sin� sin�� � ���

sin ���

B�C� sin��sin� � cos��� Hence

�� � BA� �A�C � B�C� cos� sin��B�C� sin��sin�� cos��

Therefore � B�C��sin� � � sin ��� �� B�C�� � � sin �� �

cos ��� � �� From the other hand � sin �� � cos �� �p�

��p�sin �� � p

�cos ��

��p� sin��� � � �

p�� where is

such that sin �p�and cos �

�p�� Note that the equality holds

when �� � � ��� and since � ��� we have that � � ����

Further B�C�� �p�� � B�C�� � � sin �� � cos ��� � � and

so B�C� �p� �

�� It is easily seen that if B�C� �

p��

�and

�� � � ��� then �� holds true and therefore there exists triangle

with B�C� �

p��

��

�� Suppose �� A�C�B� � ���� Denote the projection of A� and

Page 94: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

B� on AB by A� and B� respectively� Obviously AB� � B�B�

and BA� � A�A�� Let P be the midpoint of A�B� and Q � themidpoint of A�B�� Thus� A�B� � �PC� � �PQ � A�A��B�B� andA�B� � A�B� � AB �AB� � BA� � AB �B�B� � A�A�� Adding

the above equalities gives �A�B� � AB �p� and so A�B� �

p�

��

Since

p�

p� �

�we obtain that the least possible value equalsp

��

��

Problem ��b�� An element x is chosen from the setA � f� �� � � � � �ng� n � �� Questions of the type Does x belongto B � A where the sum of the elements of B equals �n����n � �are allowed� Prove that one can �nd x with exactly n questionsstated in advance�

Solution

Lemma There exist n sets each with �n�� elements� the sumof the elements of each set equals �n����n � � with the followingproperty the elements from the set f� �� � � � � �ng get as answersdistinct n�tuples from �yes� and �no��

Proof Induction by n � �� For n � � the sum of the elementsof B is � and we use the sets B� � f� �� �� �g� B� � f� �� �� �g andB� � f� � �� �g� The table shows that the elements from A get asanswers distinct triples�

� � � � � �B� � f� �� �� �g � � � � � � � �B� � f� �� �� �g � � � � � � � �B� � f� � �� �g � � � � � � � �

� means �yes�� � means �no��

Page 95: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

The number of elements in each set is �� � �����

Suppose the assertion is true for some m� Therefore there existsetsB�� B�� � � � � Bm each with �m�� elements� the sum of the elementsof each set equals �m����m�� and every element from f� �� � � � � �mggets distinct m�tuple�

Consider the set f� �� � � � � �m��g� For any i� � i � m if Bi �fa�i� a�i� � � � � a�m��ig� let

Di � fa�i� a�i� � � � � a�m��i� a�i � �m� a�i � �m� � � � � a�m��i � �mg�It is clear that each set Di� � i � m has exactly �m elementsand the sum of the elements of Di is equal to ��a�i � a�i � � � � �a�m��i���m����m � ���m����m�����m�� � ��m����m�����m�� ��m����m�� ��� It is easily seen that only the elements t and t��m

for � t � �m get equal m�tuples� Let P and Q be nonintersectingsets such that jP j � jQj � �m�� and P�Q � f� �� � � � � �mg� Considera set �Q obtained from Q by adding �m to each of its elements� Theset Dm�� � P � �Q has �m elements and the sum of its elementsis equal to

Ps�P s �

Ps�Q s � �m����m � �m����m � � � ��m�� �

�m����m�� � �� It is clear that exactly one element of each pair�t� t � �m� for � t � �m belongs to Dm��� Hence� we have foundthe desired sets D��D�� � � � �Dm���

It is obvious now that the sets� given by the Lemma solve theproblem�

Problem ����� A sequence a�� a�� � � � � an� � � � is de�ned by

a� � k� a� � �k � � and an�� � �an�� � �an� n � �

where k is a real number�

a� Find all values of k� such that the sequence fang�n�� is conver�gent�

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b� Prove that if k � then

an�� �

��a�n�� � �anan��

� an � an��

�� n � �

where �x� denotes the integer part of x�

Solution a� Write the given reccurent relation in the form an�� �an�� � ��an�� � an� and consider the sequence cn � an�� � an�It is obvious that if the sequence fang�n�� is convergent then thesequence fcng�n�� is also convergent� From the other hand it followsfrom c� � a��a� � k�� and cn�� � �cn that the sequence fcng�n��

is an arithmetic progression and cn � �k � ����n�� for any positiveinteger n� Thus� if k�� �� � then the sequence fcng�n�� is unbounded

and therefore is not convergent one� Therefore k�� � �� i�e� k �

��

For this value of k all terms of the sequence fang�n�� are equal to

�and therefore the sequence is convergent�

b� It easily follows by induction that an � �n�� From the otherhand the equality from the condition of the problem is satis�ed forn � � �� Suppose it is true for n � m and n � m� � Then�

�a�m�� � �amam��

� am � am��

��

����m�� � �� � ���m � ���m�� � �

� �m � � �m�� �

��

�����m�� � �m��

���m � �

����m �

�� �m�� � � am���

which completes the proof�

Problem ����� On each side of a triangle with angles ���� ��� and��� and hypotenuse is chosen a point such that the triangle formedfrom these three points is a right triangle� What is the least valueof the hypotenuse of this triangle�

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Solution Consider triangle ABC with angles �� � and �� LetA� � BC�B� � CA and C� � AB be such that �A�B�C� is rightwith �� A�C�B� � ���� Suppose that the circle k of diameter A�B�

intersects AB at point X �� C�� It is easy to see that if � � ���

and � � ��� then X is an interior point for line segment AB� Drawa tangent l to k which is parallel to AB and the tangential point

Y belongs to the smaller of the arcs�

C�X � Denote the intersectingpoints of l and lines AC and BC by B� and A� respectively� Con�sider homothety of center C for which the image of B��A�� is thepoint A�B�� It is obvious that the image of �B�A�Y is inscribedin �ABC and its hypotenuse is less than B�A�� Therefore wlog wemay suppose that AB is tangent to k� In this case if �� B�A�C� � then �� B�C�A � �

The Sine Low for �AC�B� and �BC�A� gives

AC� �B�A� sin sin��� �

sin��BC� �

B�A� cos sin���� � � ��

sin��

Since AB � AC� � C�B we obtain

�� AB �B�A� sin sin��� �

sin��B�A� cos sin���� � � ��

sin �

This equality is equivalent to

�AB � A�B��cotg� � cotg� � � sin � � �cotg�� cotg�� cos ���

From the other hand

� sin � � �cotg�� cotg�� cos � �q � �cotg�� cotg��� sin�� � �� �

q � �cotg� � cotg����

where

cos � ��q

� �cotg� � cotg���� sin� �

cotg�� cotg�q � �cotg�� cotg���

Page 98: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

It is easily seen that such an angle always exists�

Finally� the least value is

B�A� ��AB

cotg�� cotg� �q � �cotg�� cotg���

It is clear that if B�A� ��AB

cotg�� cotg� �q � �cotg�� cotg���

and � � � � ��� then �� holds true and therefore there exists a

triangle with B�A� ��AB

cotg�� cotg� �q � �cotg�� cotg���

It remains to compute the above expression for the three possibleways to inscribe a right triangle in triangle of angles ���� ��� and ���

and hypotenuse � We obtain

p���p��

p� � �

p�

� The least

value is

p���p��

which is the answer of the problem�

Problem ���� The plane is divided into unit squares by linesparallel to coordinate axes of an orthogonal coordinate system� Findthe number of paths of length n from the point with coordinates��� �� to the point with coordinates �a� b� moving along the sides ofthe unit squares�

Solution Divide the path into unit paths� i�e� paths between twoneighbouring points of integer coordinates� Denote the number ofmoves up by x�� down by y�� right by x� and left by y�� The conditionof the problem gives x� � x� � y� � y� � n� x� � y� � b� x� � y� � a�

Thus� y� � x� � b�x� �n � a� b

�� x�� y� �

n� a� b

�� x�� Since

jaj � jbj � jx� � y�j � jx� � y�j � x� � x� � y� � y� � n we obtainthat it is necessary to have a � b � n and a � b n�mod ��� Let

Page 99: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

us �rst �x the moves up and rigth� This can be done by

�n

n�a�b�

ways� After that in these already �xedn � a� b

�positions �x the

moves up and �nally� in the remainingn� a� b

�positions �x the

moves left� We obtain

�n

n�a�b�

� n�a�b

�Xi�b

�n�a�b

i

��n�a�b

�n�a�b

� � i

��

The sum

n�a�b

�Xi�b

�n�a�b

i

��n�a�b

�n�a�b

�� i

�equals to the coe�cient in front

of xn�a�b

� in the expansion of � � x�n�a�b

� � � x�n�a�b

� � � � x�n�

Therefore the sum equals

�n

n�a�b�

�� The number of paths is

�n

n�a�b�

��n

n�a�b�

��

Answer If jaj � jbj n or a � b � n�mod ��� the number of

paths is �� otherwise the number of paths is

�n

n�a�b�

��n

n�a�b�

��

Page 100: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Spring mathematics tournament

Kazanlak� �� March � � April ����

Problem ���� Let a be a real parameter such that � � a � � Provethat the solutions of the inequality

jxj� jax�

�j �

form an interval of length greater or equal to �

Solution If a � � then all numbers from the interval�

���

of

length are solutions of the problem� If a � then the inequality

becomes jxj � jx �

�j � � The solutions of this inequality are all

numbers from the interval���

� which is of length �

Further� assume that � � a � and let x � �� In this case

ax�

� � and therefore the solutions are x �

�� � a�� Let x � ��

If ax�

�� �� i�e� x � �

�� the solutions satisfy x � �

��� a�� If

ax�

�� �� i�e� x � �

�a� the solutions satisfy x � �

�� � a��

Let a �

�� Then �

��� a�� �

�aand so all numbers x �

�� � a�are solutions of the inequality� Since �

�� � a� �

�ait has no other solutions� Therefore all solutions form an interval��

�� � a��

�� � a�

�� of length

�� � a��

�� � a��

� a� �

Let a

�� Then � �

�� � a�� �

�aand the solutions are x �

Page 101: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

� �

�� � a�

x � �

�aor x �

�� �

�� � a���

�a

��� Therefore all so�

lutions form an interval

�� �

�� � a��

�� � a�

�� of length

� a� �

Problem ���� Given a square ABCD of side length � PointM � BC and point N � CD are such that the perimeter of �MCNis ��

a� Find �� MAN �

b� If P is the foot of the perpendicular from A to MN� �nd thelocus of the point P�

Solution a� Let K be a point on the extension of CB such thatBK � DN� Then MK � MB � BK � MB � DN � � CM ��CN � �� �CM �CN� �MN �using that CM �CN �MN ���� Since �ABK �� �ADN we have that AK � AN� Therefore�AMN �� �AMK and it follows that �� MAN ��� MAK� i�e�

�� MAN �

��� KAN� Furthermore �� KAB ��� NAD� which im�

plies that �� KAN ��� BAD � ���� Therefore �� MAN � ���

b� It follows from �AMN �� �AMK that �� AMN ��� AMK�Thus� �APM �� �ABM and AP � AB � � Therefore P lies on acircle of center A and radius � Finally� the locus of P is an arc of acircle with center A and radius excluding points B and D�

Problem ��� a� Prove that

� � �� � � � �� n� �n�n� ���n � �

holds true for any positive integer n�

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b� Find the least integer n� n for which

� � �� � � � �� n�

nis a perfect square�

Solution

a� Direct veri�cation shows that

n�n � ���n � �

�� �n� �� �

�n� ��n� ����n � ��

�for any integer n� Using this equality the proof is easily done byinduction�

b� It follows from a� that n�n � ���n � � � �m�� Since �n � is odd then n � is even� i�e� n is odd� Let n � �k � � Thenk�k � � � �m�� Therefore ��k or ��k � i�e� ��k or ��k � � Letk � �l� Then l��l � � � m�� Since �l� �l � � � � we have l � n�

and �l� � v�� The latter equality is impossible �contradiction bothmodulo and modulo ��� Let k � �l�� Then ��l���l�� � m��Since ��l � � l � � � we get that �l � � u�� l � � v�� u � v � Verifying consequently for v � �� �� �� �� � �� � � � we obtainl � �� �� �� ��� ��� �� ��� � � � � Thus� �l� � �� �� ��� �� �� ��� ����rst perfect square�� So� the least n is �k � � where k � �l � ����� � � �� and therefore n � ����

Problem ���� Let f�x� � x���ax� a where a is a real parameter�

a� Find all values of a for which the equation f�x� � � has atleast one real root�

b� If x� and x� are the real roots of f�x� � � �not necessarilydistinct� �nd the least value of the expression

A ��a� a�

� � x��� � x��� ��a� �

�� �a� x��� � �a� x���

Page 103: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution a� The discriminant of f�x� is D � ��a��a�� Thereforethe equation f�x� � � has at least one real root i� �a� � a � �� so

giving a �����

� �������

b� Since f�x� � �x � x���x � x�� we obtain f��� � �� �x����� x�� � � � x��� � x�� � ����� �a���� a � � �a andf���a� � ���a�x�����a�x�� � ���a����a���a��a � � �a� Therefore the denominators of the two fractions of A areequal to � �a and

A ����a� � a� � �a�

� �a�

�� �a���a� � �a� �

� �a� �a���a��

The quadratic function g�a� � �a� � �a � attains its minimal

value for a � �

�������

� ������� Therefore the minimal

value of A equals to the smallest of the numbers g��� and g�

��

i�e� this value is g�

�� ���

��

Problem ���� Given a convex quadrilateral ABCD such that

OA �OB�OD

OC �OD� where O is the intersecting point of its diago�

nals� The circumcircle of �ABC intersects the line BD at point Q�Prove that CQ is the bisector of �� DCA�

Solution Let CQ�� Q� � BD be the bisector of �� DCO� Therefore

DQ�

Q�O�DC

CO

This equality� combined with the condition of the problem gives

OA�OC �OD� � OB�OD �� OAQ�O �DQ�

Q�OCD � OB�OD

��

Page 104: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�� OA�CO�DO

Q�O� OB�OD �� OA�CO � Q�O�OB�

Therefore the quadrilateral ABCQ� is cyclic� Thus� Q� Q�

Problem ��� Prove that there exist eight consecutive positiveintegers such that non of them can be written in the formj�x� � �xy � �y�j� where x and y are integers�

Solution Denote f�x� y� � �x� � �xy � �y�� Since f��� �� � ��f��� � � �� f�� �� � �� f�� � � and f��� �� � ��� �rst possiblesequence of eight positive integers is �� �� � � � � �� We shall provethat non of these integers can be written in the form j�x���xy��y�j�where x and y are integers�

Let f�x� y� � �k� where x and y are integers� It su�ces to provethat f�x� y� � �k has no solutions for k � f�� �� � � � � �g�

Suppose k is even� Then x and y are also even� If x � �x� andy � �y� we get the equality f�x�� y�� � �k which implies that k isdivisible by � Thus� k �� and k �� �� Let k � � and considerthe equation f�x�� y�� � �� which is equivalent to f�x�� y�� � ��As above we conclude that x� and y� are both even and let x� � �x�and y� � �y�� Therefore f�x� y� � �� By analogy if k � � we getthe equation f�x� y� � ���

Multiply the equation f�x� y� � �k by �� and write it in theform

�x� �y�� � ��y� � ���k�Since �� � ��� it is appropriate to consider modules � and ��

Denote t � x � �y and consider all possibilities for k� i�e�k � f� �� �� �� �� �g�

� If k � � then t� ����� �mod �� and so t� �� �mod��� Raising this congruence to ��th power gives t�� ���� �

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�mod �� which is a contradiction to Fermat�s theorem� Thereforek �� ��

�� If k � � then t� ����� �� �mod ��� Raising in ��thpower gives t�� ���� � �mod ��� a contradiction�

The cases k � �� �� � � are treated similarly�

Problem ��a��� Let a and b be positive numbers such that bothof the equations �a� b�x�� � a� b and �ab��x�� � ab� havetwo distinct real roots� Prove that if the two bigger roots are equalthen the two smaller roots are also equal�

Solution Let a and b satisfy the condition of the problem� Bothequations have two distinct roots i� a b and ab � Since a �

it follows that a� ab � i�e� a � Therefore a b

aand

a � It follows from the condition of the problem that a � b �pa� b � ab� �

pab� � i�e�

pa� b � �a� ��b� � �

pab� �

If a b � thenpa� b � p

ab� �� �a � ��b � � �� �� b � � Therefore b � � Conversely if

a� b � then

pa� b � p

ab� �� �a� ��b� � � � �� b � � i�e� b � �Thus� the two bigger roots are equal i� b � and a � In this casethe two smaller roots are also equal�

Problem ��a��� Let A� and B� be points respectively on the sidesBC and AC of �ABC�D � AA�BB� and E � A�B�CD� Provethat if �� A�EC � ��� and the points A�B�A�� E lie on a circle� thenAA� � BA��

Solution Let F � AE BC� We prove that EA� is the bisec�tor of �� BEF� which solves the problem� Indeed� then we have�� BAA� ��� BEA� ��� FEA� ��� ABA�� i�e� AA� � BA��

��

Page 106: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Using Ceva�s and Menelaus�s theorems for �AA�C we obtainAD

A�D� A�F

CF� CB�

AB�� �

AD

A�D� A�B

CB� CB�

AB�� and so

��A�F

A�B�

CF

CB

Let B� be a point on the ray A�B� such that �� B �EA� ��� A�EF�

Then EA� is the bisector of �� B�EF� and since �� A�EC � ��� itfollows that EC is the external bisector of the same angle� ThereforeA�F

A�B� �CF

CB� and it follows from �� thatA�B

CB�A�B

CB� � i�e� B � B�

which completes the proof�

Problem ��a�� Find all positive integers x and y such that

x� � y� � x�y�

�x� y��

is a nonnegative integer�

Solution Let x and y be positive integers such that

z �x� � y� � x�y�

�x� y��

is a nonnegative integer� Substitute a � x� y and b � xy and writethe expression in the form b���ab�a��a�z� � �� The discriminantof this quadratic equation a��a � � � z� is a perfect square� so�a � � � z� � ��t � ��� Thus� a � t� � t � z � � and from theequation for b we obtain that b � a�t � �� Since t � � we have�x � y�� � a� � a�t � � � �a � ��t � ���� From the other handa � t� and therefore a� � a�t � � � �a � ��t � � � ���� Sincea�� a�t� � �� �a� ��t� �� �� �the two numbers are of di�erentparity� it follows that �x� y�� � a�� a�t� � � �a� ��t� �� ����

��

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Thus� a � t� i�e� t � z � � and so t � �� z � �� which impliesa � b � � Therefore x � y � � and these are the only positiveintegers satisfying the condition of the problem�

Problem ��b��� Solve the equation

�log��cosx�sinx � �

� � �log��cosx�sinx �p��

Solution Note that the admissible values are those x for whichcos x � sinx �� cos x � sinx �� After simple calculations theequation becomes

�p�� � cos x� �

p� � � sin x �

p��

This equation is equivalent top� �p�

cosx�

p� �

p�

sinx �

��

Since cos �� �p��p�

�and sin �� �

p��p�

�we have

sin�x� ��� �

��

From the roots of the later equation only x � �� � k���� are ad�missible� Therefore the roots are x � �� � k�����

Problem ��b��� Given a triangle ABC� Let M be such an interiorpoint of �� BAC that �� MAB ��� MCA and �� MAC ��� MBA�Analogously� let N be such an interior point of �� ABC that�� NBA ��� NCB and �� NBC ��� NAB� and let P be such an inte�rior point of �� ACB that �� PCA ��� PBC and �� PCB ��� PAC�Prove that linesAM�BN and CP intersect in a point on circumcircleof �MNP�

Page 108: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution Suppose �� C � ���� The case �� C ��� is treated simi�larly� Denote the intersecting point of line CP and line segment ABby C�� Since PC� is the bisector of �� APB��� APC� ��� BPC� �

� ��� ACB� we getAC�

BC��

AP

BP� Triangles APC and BPC are sim�

ilar which givesAP

CP�

AC

BCand

BP

CP�BC

AC�� AP

BP�

AC�

BC��

b�

a��

By analogy� if A� and B� are the intersecting points of AM and

BN with BC and AC respectively thenBA�

A�C�c�

b�and

CB�

B�A�a�

c��

Thus�AC��BA��CA�

C�B�A�C�B�A� � i�e� according to Ceva�s theorem lines

AM�BN and CP intersect in a point� Dnote this point by K� Since�� APB ��� AOB � ��� where O is the circumcenter of �ABC we

get that A�P�O and B lie on a circle and PC� intersects the arc�

ABat its midpoint C�� Therefore OC� is a diameter of this circle and�� OPC� ��� OPC � ���� It easily follows now that M�N and P lieon a circle of diameter OK�

Problem ��b�� Consider a set P of six four�letter words over analphabet of two letters a and b� Denote by QP the set of all wordsover the same alphabet which do not contain as subwords the wordsfrom P � Prove that

a� if QP is �nite then it does not contain words of length � �

b� there exists a set P such that QP is �nite and it contains wordof length ��

Solution a� Suppose that QP contains a word of length � Wewill show that QP contains words of any length� Let be a word oflength � There are � subwords of of length �� Since there are �distinct words of length � it follows that there exists a word � thatappears as subword of twise� Let � � ������ where �i � fa� bg�

��

Page 109: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Consider the subword of obtained after the second appearance of�� i�e� consider the word

� � � �������� � � � ������

Write �� after ��� Obviously the new word does not contain subwordsfrom P�

� � � �������� � � � ������

By analogy write �� and so on� Thus� we �nd words of any lengthwithout subwords from P� This contradiction shows that there areno words of length � in QP �

b� A direct veri�cation shows that the set

P � f����� ���� ��� ��� �� � g

is such that QP is �nite and ������ � QP �

Problem ����� Prove that there exist unique numbers � and �such that cos� � ��� �tg� � and � � � � � � �

Solution The function f�x� � cos x � x� is continuous in the in�terval ��� �� f��� � � and f�� � cos � � �� Therefore thereexists � such that cos� � ��� In the interval ��� � the function cosxis decreasing one� whereas the function x� is increasing� Thereforethere exists unique � � ��� � such that cos� � ���

We show now that in the interval ��� � there exists a unique� such that g��� � � where g�x� � xtgx � � The function g�x� isincreasing one because xtgx is increasing as product of two increasing

functions� Moreover tg�

��

� � �� g�� � tg � � tg �

tg�

� and the uniquennes of � follows� Since sinx � x for positive

��

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x it follows that g��� � �tg� � �sin�

�� � �� Further� g�x� is

increasing in the interval ��� � and so � � ��

Problem ����� Let AA� and BB� be the altituds of obtuse non�isosceles �ABC� and O and O� are circumcenters of �ABC and�A�B�C respectively� A line through C intersects the line segmentsAB and A�B� at points D and D� respectively and E is point on

the line OO� such that �� ECD � ���� Prove thatEO�

EO�CD�

CD�

Solution Let F and F� be the feet of the perpendiculars from Oand O� to CD� Since CO� � AB we have��EO�

EO�CF�

CF�CO� cos �� O�CF�

CO cos �� OCF�

CO� sin �� BDC

CO sin��� DCB� �� BAC��

From the other hand the Low of Sine�s for �A�B�C��ABC��A�D�C and �BDC gives

��� CO� �A�C

� sin �� A�B�C�

A�C

� sin �� ABC

��� CO �BC

� sin �� BAC

��

CD� � CA�sin �� B�A�C

sin��� B�A�C� �� D�CA���

CA� sin �� BAC

sin��� BAC� �� DCB�

��� CD �BC sin �� ABC

sin �� BDC

��

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It follows now from ������������� and ��� that

EO�

EO�A�C

BC

sin �� BAC sin �� BDC

sin �� ABC sin��� BAC� �� DCB��CD�

CD�

which completes the proof�

Problem ���� There are ��� towns in a country every one ofwhich is connected with at least ��� towns by direct bus line� Findthe largest n for which there exist n towns any two of which areconnected by direct bus line�

Solution Let S� snd S� be two towns connected by direct bus line�If k is the number of towns connected to both S� and S� by bus linethen �����k�������k��k � ��� which implies that k � ���Therefore there exists town S� connected to both S� and S�� Further�let k be the number of towns connected to all S�� S� and S��Therefore��� � k� � ���� � k� � k � ���� so giving k � ���� Thereforethere exists town S� connected to all S�� S� and S�� By analogy let kbe the number of towns connected to all S�� S�� S� and S�� We have����� k�� ����� k�� k � ���� which implies k � ���� Thereforethere exists S� connected to all S�� S�� S� and S�� For the number nfrom the condition of the problem we obtain n � �� We show thatn � �� For� number the towns by S�� S�� � � � � S���� and connect Skand Sm with direct line for all k and m for which k � m�mod���

Since����

� ��� we have that each town is connested to ��� or

�� other towns� i�e� the condition of the problem is satis�ed�

For arbitrary � towns the numbers of at least two are equal mod�ulo � and therefore they are not connected to each other� Thereforen � � and so n � ��

��

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L National Mathematics Olympiad

�rd round� ��� April ����

Problem �� For which values of the real parameter a the equation

lg�x����a��x��a���x�����a�x��a� � lg�x����a��x�a��has exactly one root�

Solution Write the equation in the form

lg�x� � ��a� �x� �a�� �x� � ��a� �x� �a�

��

lg�x� � ��a� �x� a�� �x� � ��a� �x� a�

��

Since lg t�t

�is an increasing function this equality is equivalent

to

����� x� � ��a� �x� �a� � x� � ��a� �x� a�

x� � ��a� �x� a� �

The equation in the above system is equivalent to f�x� � � wheref�x� � x� � ��a � �x � �a�� Substitution x� � �ax � x � �a� inthe inequality gives x � �a�� Therefore we have to �nd those a forwhich the equation f�x� � x�� ��a� �x��a� � � has exactly oneroot such that x � �a�� There are three cases to be considered

� f�x� � � has two roots and �a� is between the roots� In thiscase f��a�� � �� a��a� �� � �� which is impossible�

�� f�x� � � has two roots one of which equals �a� and thesecond one is less than �a�� in this case f��a�� � � and so a � �

��

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or a � �� When a � � we obtain x� � ��� x� � �� which impliesthat a � � is a solution� If a � � we get x� � �� x� � �� whichimplies that a � � is also a solution�

�� f�x� � � has one root which is less than �a�� Since D � � we

get that �a� � a � � �� and so a��� ���p�

�� It is easy to

check that for both values of a the corresponding root is less than�a��

Therefore the solutions are a � �� a � �� a ���p�

��

Problem �� Diagonals AC and BD of a cyclic quadrilateral ABCDintersect in a point E� Prove that if �� BAD � ��� and AE � �CEthen the sum of two of the sides of the quadrilateral equals the sumof the other two�

Solution Set �� ABD � x and �� CBD � y� From the Low ofSine�s we obtain

AE

AB�

sin x

sin���� � y � x��CE

BC�

sin y

sin���� � x� y�

andAB

BC�

sin���� � x�

sin���� � y�� Therefore � �

AE

CE�

sinx� sin���� � x�

sin y� sin���� � y��

Hence ��cos��y� ����� cos ���� � cos��x� ����� cos ���� i�e� � cos��x � ���� � �� � cos��y � ����� and so sin��x � ���� �� sin��y������ Therefore sin�x����� cos ��� � � cos ��� sin�y������i�e� sinx � sin���� � x� � ��sin y � sin���� � y��� Again fom theLow of Sine�s we get AD �AB � ��CD �BC�� i�e�

AD �BC � AB � CD or AD � CD � AB �BC�

��

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Problem � Find the least positive integer n such that there existsa group of n people such that

� There is no group of four every two of which are friends�

�� For any choice of k � people among which there are nofriends there exists a group of three among the remaining n � kevery two of which are friends�

Solution Consider a group of � people A�� A�� � � � � A�� such thatAi� i � � � � � � � � is not friend only with Ai�� and Ai�� �we set thatA � A� and A� � A��� It is easily seen that there are no four everytwo of which are friends� Also� for any choice of k � people �inthis case k is or �� every two of which are not friends there existsa group of three among remaining � � k people every two of whichare friends� Therefore k � ��

We prove that for any group of � which satisfy the condition �it is possible to choose a group of k � every two of which are notfriends such that among remaining �� k a group of three every twoof which are friends does not exist�

Denote the people byA�� A�� � � �A�� If some of them is friend withthe other � �wlog suppose this is A�� it is clear that there are no threeamong A�� A�� � � �A� every two of which are friends� Therefore thechoice of A� solves the problem�

Suppose one of then is friend with exactly four others �wlog as�sume A� is friend with A�� A�� A� and A�� Then the choice of A� andA� solves the problem�

Therefore each person has �� � � or � friends� It is obviuos thatthere exists a group of three any two of which are friends �otherwisethe problem is trivial�� Assume that the group A�� A� and A� hasthis property� Wlog A� and A� are not friends� Therefore amongA�� A�� A� and A� there is a group of three any two of which are

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friends� If this is A� and A� together with one of A� or A� then thereexists a person who is friend with at least four others� a contradiction�Therefore wlog suppose that this group is A�� A� and A�� Since A�

and A� are not friends and since among the others there is a group ofthree friends we obtain that either A� or A� has at least four friendswhich is a contradiction�

Therefore k � ��

Problem �� Given a right triangle ABC with hypotenuse AB�A point D distinct from A and C is chosen on the ray AC� suchthat the line through incenter of �ABC parallel to the bisector of�� ADB is tangent to the incircle of �BCD� Prove that AD � BD�

Solution First we show that C lies on the line segment AD� For�suppose the contrary� i�e� �� ADB ���� Denote the tangentialpoint of incircle k�I� r� of �BCD and the side BD by P and thetangential point of k and the line through the center J parallel tobisector l of �� ADB by T � Since l � DI we get T � DI� Thus�

�� IJT ��� CBD

���� IBP and since IT � r � IP we have that

�IJT and �IBP are congruent� In particular IJ � IB whichimplies that �� BJC � ���� a contradiction� Let A� be such point onthe ray DA� that DA� � DB and E is the midpoint of A�B� Denotethe tangential point of incircle of �A�BC by F and its incenter byJ �� Then we have

EF � jA�E �A�F j �

�jA�B � �A�B �A�C �BC�j �

�jBC �A�Cj �

�jBC � �A�D � CD�j �

��jBC � CD �BDj � r�

Therefore the line J �F parallel to l is tangent to k� This impliesthat J � � JF� From the other hand J � � JC and since JC is not

��

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parallel to JF ��� ACJ � ��

��� ADB ��� ADI�� we get J � J ��

Thus� �� ABJ ��� CBJ ��� A�BJ� i�e� A � A� which completes theproof�

Problem � Find all triples of positive integers �a� b� c� such thata� � b� � c� is divisible by a�b� b�c and c�a�

Solution If d � gcd�a� b� c� it is easy to see that

�a

d�b

d�c

d

�is

also a solution of the problem� Thus� it su�ces to �nd all triples�a� b� c� such that gcd�a� b� c� � � Let gcd�a� b� � s and supposes � If p is a prime divisor of s then p divides a� p divides b andp divides a� � b� � c�� Hence p divides c� and so p divides c� Thisis a contradiction to gcd�a� b� c� � � Therefore gcd�a� b� � andby analogy gcd�a� c� � gcd�b� c� � � It follows from a���a� � b� �c��� b���a�� b�� c��� c���a�� b�� c�� and gcd�a�� b�� � gcd�a�� c�� �gcd�b�� c�� � that a� � b� � c� is divisible by a�b�c�� In particulara� � b� � c� � a�b�c�� Wlog we may assume that a � b � c� Then

�c� � a� � b� � c� � a�b�c� � c � a�b�

�� Suppose a � Then

a�b�

� b� � b� � a�a� b� � a� � ab� b� and the inequality ��� c

a� � ab � b� holds� From the other hand since b � a � � we havethat b� � �b � a� b and since c b� we obtain ���� c a� b� Now

�� and ��� give�a� � ab� b���a� b�

c�� �� a� � b�

c�� � which

is impossible since a��b�

c�is an integer� Thus� a � � In this case we

have that �b��c� is divisible by b�c�� Consider the following cases

� if b � c it is easy to see that b � c � � Indeed� �� � � is asolution of the problem�

�� if b � we obtain the same solution�

�� if b � � we obtain c � �� Triple �� �� �� is a solution of the

��

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problem�

Supose now that c b � �� Since � b� � c� � b�c�� it follows

that �c� �b��c� and so �c b� or c b�

�� It follows from �c b�

that �c b� � b � � i�e� ����b� � b�

c� �� From the other hand

when b � � the inequalitiesc

b�

b � � ��

b�

c�

�hold�

Multiplying ��� and �� givesb� �

c�� � which is impossible since

this number is an integer� Direct veri�cstion shows that when b � �or b � we get no new solutions� Therefore all solutions of theproblem are triples �k� k� k� and �k� �k� �k� �and its permutations�for arbitrary positive integer k�

Problem �� Given a pack of �� cards� The following opperationsare allowed

� Swap the �rst two cards�

�� Put the �rst card on the last place�

Prove that using these opperations one can order the cards inarbitrary manner�

Solution First we show that it is possible to change any two neigh�bouring cards� Indeed� using �� we can move the two cards on the�rst two positions� After that apply � and again �� to put all re�maining cards on their initial positions�

Consider two arbitrary cards ai and ai�j� We can change thesetwo cards by changing ai and ai�� then ai and ai�� and so on� up toai and ai�j� After that we change ai�j and ai�j�� and so on� up toai�j and ai���

Since we can change any two cards we can order the cards inarbitrary manner�

Page 118: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

L National Mathematics Olympiad

�th round� ���� May ����

Problem �� Consider the sequence fang such that a� � � a� � ��and an��an���an�� � � for n � �� Prove that there exist sequences

fxng and fyng of positive integers such that an �y�n � �

xn � ynfor any

n � ��

Solution Let xn �an � an��

�� x� � � and yn �

an � an���

� y� � �

Then xn � �xn���yn�� and yn � �xn����yn��� Since an � xn�yn�

it su�ces to prove that xn � yn �y�n � �

xn � yn� i�e� x�n � �y�n � ��

We prove this by induction� The assertion is obvious for n � ��Suppose that x�n�� � �y�n�� � �� Writing this equality in the form��xn�� � yn���� � ���yn�� � �xn���� � � gives x�n � �y�n � �� whichcompletes the proof�

Problem �� Given nonisosceles triangle ABC� Denote the tan�gential points of the inscribed circle k of center O with the sidesAB� BC and CA by C�� A� and B� respectively� Let AA� k �A�� BB� k � B� and let A�A�� B�B� be bisectors in triangleA�B�C� �A� � B�C�� B� � A�C��� Prove that

a� A�A� is bisector of �� B�A�C��

b� if P and Q are the intersecting points of circumcircles of tri�angle A�A�A� and triangle B�B�B� then the point O lies on the linePQ�

Solution a� From the Low of Sine�s we obtain

AB�

AA��

sin �� B�A�A�

sin ���AC�

AA��

sin �� C�A�A�

sin���

��

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where �� ��� A�B�C�� �� � A�C�B�� Since AB� � AC� we get

A�B�

A�C��

sin �� B�A�A�

sin �� C�A�A��

sin ��sin ��

�A�B�

A�C��A�B�

A�C��

which implies that A�A� is the bisector of �� B�A�C��

b� LetM � AA�BB�� It follows fromMA� �MA� � MB� �MB�

that M lies on the line PQ� Therefore it su�ces to prove thatOM � O�O�� where O� and O� are circumcenters of �A�A�A� and�B�B�B�� It follows from a� that the diametrically opposite pointof A� in k� � the circumcircle of �A�A�A�� lies on the line B�C��

Therefore O� � B�C�� Moreover �� B�B�A� ��� CA�A� � � ���

��

It easily follows now that O� coinsides with the intersecting pointof B�C� and BC� Let OO� A�A� � N and OO� B�B� � K� Itfollows from �OA�O� that ON �OO� � OA�

� � r� and by analogyOK �OO� � r� where r is the radius of k� Since O� N � M and K lieon the circle k� of diameter OM we have that the line O�O� is theimage of k� by inversion of center O and degree r�� i�e� OO� � OM �

Problem � For a permutation a�� a�� � � � an of the numbers � � � � � nit is allowed to change the places of any two consecutive blocks� i�e�from

a�� � � � � ai� ai��� ai�� � � � � ai�p� z �A

� ai�p��� ai�p�� � � � � ai�q� z �B

� ai�q��� � � � � an

by replacing A and B one can obtain

a�� � � � � ai� ai�p��� ai�p�� � � � � ai�q� z �B

� ai��� ai�� � � � � ai�p� z �A

� ai�q��� � � � � an�

Find the least number of such changes after which from n� n �� � � � one can obtain � �� � � � n�

Solution Call the change of two blocks a move� We shall provethat the least number of moves such that from n� n�� � � � one can

��

Page 120: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

obtain � �� � � � n is�n �

��

Consider the number of pairs ai� ai�� such that ai � ai��� Thisnumber is � in the initial permutation n� n � � � � � �� and is n � in the �nal permutation � �� � � � � n � � n�

First we show that a move changes the number of pairs ai� ai��

such that ai � ai�� at most by two� For� consider a move of theblocks a � � � � b and c� � � � d of the permutation

� � � p� a � � � � b� c� � � � d� q � � �

As a result we obtain

� � � p� c � � � � d� a� � � � b� q � � �

It is clear that at most three pairs can change the ordering� Supposethe elements in all three pairs change the ordering� i�e� from p a�b c and d q we get p � c� d � a and b � q� Adding the�rst three inequalities gives p � b � d a � c � q and adding thelast three implies p � b� d � a � c � q� a contradiction� Thereforea move changes the number of pairs ai� ai�� for which ai � ai��

at most by two� It is easily seen that the �rst and the last moveschange this number by one� Therefore if x is desired number we have

�� ��x� �� � n� which implies x ��n�

�� It remaines to �nd

a sequence of�n�

�moves such that from n� n� � � � � �� we get

� �� � � � � n � � n� Let n be even number� i�e� n � �k� Number thepositions from right to left by � � � � � n� First change the places ofthe blocks from positions � � � � � � k � and k� k � � Next� changethe places of the blocks from positions �� � � � � � k and k � � k � ��Third� change the blocks �� � � � � k� and k��� k�� and so on� Onthe k�th step change the blocks from positions k� k� � � � � �k�� and�k� � �k� As a result we obtain k�� k��� � � � � �k� � �� � � � � k� The

��

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last change is � �� � � � � k and k� � k�� � � � � �k� When n is odd� i�e�n � �k� the �rst change is of the blocks from positions � �� � � � � kand k � � k �� after that �� �� � � � � k � and k � �� k � � and so on�The last k � change is � � � � � k and k � � � � � � �k and we obtain� �� � � � � �k � �

Problem �� Let n � � be �xed integer� At any point with integercoordinates �i� j� we write i � j modulo n� Find all pairs �a� b� ofpositive integers such that the rectangle with vertices ��� ��� �a� ����a� b�� ��� b� has the following properties

� the remainders �� � � � � � n � written in its interior pointsappear equal number of times�

�� the remainders �� � � � � � n� written on its boundary appearequal number of times�

Solution Let pi and si for i � �� � � � � � n � be the number ofresidues i on the sides and in the interior of the rectangle respectively�If a� � a � kn for an integer k then the corresponding numbers for�a�� b� are p�i � pi � �k and s�i � si � k�b� ��

Therefore if �a� b� is a solution then �a� kn� b� and by analogy�a� b � ln� are also solutions� Thus� wlog we may suppose that � a� b � n�

If n � � then all possible values of �a� b� are

�a� b� � �� �� �� ��� ��� �� ��� ���

Pair ��� �� is not a solution because there is unique interior point�s� � � s� � �� for the rectangle� The remaining three cases givesolutions�

Therefore for n � � solutions are all pairs �a� b� where either aor b is an odd number�

Let n � and suppose �a� b� is a solution of the problem for

��

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which � a� b � n� In the vertices ��� �� and �a� b� are written �and a� b modulo n respectively� whereas all residues � �� � � � � a�b � appear on the boundary even number of times� �once on theboundary �� ��� � � � � �a� ��� �a� �� � � � � �a� b� � and once on theboundary ��� �� � � � � ��� b�� �� b�� � � � � �a � � b��� If n does notdivide a � b then � and the residue of a � b appear odd number oftimes whereas at least one of the remaining residues �n �� appearseven number of times� Therefore n divides a � b and so a � b � nor a� b � �n� When a� b � �n we have that a � b � n� This pairis not a solution since the number of interior points is �n� �� andis not divisible by n� It remains to consider the case a � b � n� Ifa and b then the rectangle has interior points� For any suchpoint �i� j� we have � � i � a� � � j � b� � � i� j � a� b � n andtherefore the residue modulo n is not �� Therefore a � � b � n � �or a � n � � b � � which is a solution� Therefore for n � allsolutions are a � �kn� b � n��ln and a � n��kn� b � �ln�where k� l � �� � �� � � ��

Problem � Find all real numbers t for which there exist realnumbers x� y� z such that

�x� � �xz � z� � � �y� � �yz � z� � � x� � xy � y� � t�

Solution� We shall prove that the answer is t ����p�

��

��

Let x� y� t� � satisfy the conditions of the problem� Consider fourpoints A�B�C�O in the plane such that AO � x�BO � y�CO �zp�and �� AOB � ���� �� BOC ��� COA � ���� By the cosine

theorem it follows that t �� AB �pt� BC �

�p�� CA �

p��

Since �� ACB ��� AOB � ��� and �� BAC ��� BOC � ���� we

��

Page 123: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

have

��

��p�

����

�p�

�� � �pt��

� �p�

�p�

� cos �� ACB

��

����pt�� �

��p�

�� � ��p�

���p� �p

� cos �� BAC �p�

��

These inequalities are equivalent to t � and t �pt � �� i�e�

t ��� �p�

��

��

Conversely� let t ����p�

��

�� Then we can construct a tri�

angle ABC such that AB �pt� BC �

�p�� CA �

p�� By �� and

��� it follows that �� ACB � ��� and �� ABC ��� BAC � ���� LetkA be the circle through B and C such that the arc BC� lying inthe half�plane �with respect to BC� containing the point A� is equalto ���� Denote by kB the analogous circle trough A and C� Then itis easy to see that the second common point O of kA and kB liesin the interior of �ABC� �Indeed� assume the contrary� If O lies in�� ACB� then �� AOB ��� AOC� �� BOC � ��� � ��� � ����

which is impossible� If O lies in the opposite angle of �� ACB�then �� AOB ��� AOC� �� BOC � ��� � ��� � ��� which con�tardicts to the inequalities �� AOB ��� ACB � ���� The other casesfor the position of O can be rejected in the same manner�� Hence�� AOB � ����� �� AOC� �� BOC � ���� � ��� � ��� � ����Set x � AO� y � BO� z � CO

p�� Applying the cosine theorem for

triangles AOB�BOC and COA it follows that x� y� z� t satisfy theconditions of the problem�

Page 124: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Given the equation

�p� ��x� � �p� �y� � px� �p � ��y � �

where p is �xed prime number of the form k � �� Prove that

a� If �x�� y�� is a solution of the equation where x� and y� are positiveintegers� then p divides x��

b� The given equation has in�nitely many solutions �x�� y�� wherex� and y� are positive integers�

Solution a� Set y � � z and write the equation in the form

��� x� � �z � x���p� ��z � x� � p��

If z � x and �p � ��z � x� � p are relatively prime then they areperfect squares which is impossible since the second one is of theform k��� Let q be a common divisor of these numbers� It followsfrom ��� that q�x and so q�z� Since q��p���z�x�� p we have q�pgiving q � p which completes the proof�

b� It su�ces to prove that ��� has in�nitely many solutions inpositive integers� Let x � px� and z � pz�� Then x�� � �z��x����p���z� � x�� � � and therefore there exist positive integers a and bsuch that z� � x� � a�� x� � ab and �p � ��z� � x�� � � b�� Itfollows now that

�� �p � ��b� � �p � ��a� b�� � �

Letq

p� � �qp �

��k��

� mk

qp � � � nk

qp � for any k �

�� � � � � � where mk and nk are positive integers� It is obvious that

�qp � ��

qp� ��k�� � mk

qp� � � nk

qp �

and after multiplyingwe obtain that �p���m�k��p��n�k � � i�e� b �

mk and a�b � nk are solutions of ��� Hence� x � pmk�nk�mk� and

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z � pnk�nk�mk� are solutions of ���� The assertion of b� follows fromthe fact that both sequences m��m�� � � � and n�� n�� � � � are strictlyincreasing�

Page 126: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

SPRING MATHEMATICAL COMPETITION

����

Grade ��

Problem �� Find all values of a� for which the system�������x� �jyj � jxj

jyj� jx� aj � �

has exactly two solutions�Solution� Let �x� y be a solution with x � � Then y � and jx� aj � �� i�e� x � a� �� It

is obvious that when a � � the system has two solutions with x � � namely �a��� � �a��� �When �� � a � � the system has only one solution �a � �� with x � � Let x � � Then

jyj � �x�and consequently jx � aj � � �

x

�� Since jx � aj � � then � �

x

�� or x � ���

We have x � a � � �x

�and x � ��a � �� x � a � �� � x

�� From here x �

��a � �� From

�� � ��a � � � we get�� � a � �� and from �� � �

��a � � � we get �� � a � ��

Obviously if �x� y is a solution of the system with x � � then y �� and therefore �x��yis a solution too� Thus� when a � �� the system has no solution� When a � �� we get

x � ��a� � � �� and x ��

��a� � � ��� i�e� the system has exactly two solutions �������

If � � a � ��� then ��a�� �� �

��a� � and the two values of x give solutions� thus we get four

solutions with x � �Therefore the system has two solutions only when a � �� namely �a� �� and �a��� and

when a � ��� namely �������Problem �� Let M be the midpoint of the side BC of the parallelogram ABCD� N be the

common point of AM and BD� while P be the common point of AD and CN � Prove thata AP � AD b CP � BD i� AB � AC�Solution� a Consider �ABC� The lines AM and BD are medians and thus N is the center

of gravity� If Q is the intersection point of CP and AB� then Q is the midpoint of AB� It�s easyto see that �APQ �� �BCQ �Figure� �� from where AP � BC � AD�

b Let AB � AC� Then �ABC is isosceles and AM is a median in it� It is easy tosee that �NBC is isosceles and BN � CN � Analogously NA is a median and an altitudein �PND� thus PN � DN � i�e� PC � PN � CN � DN � BN � BD �Figure �� LetCP � BD� Trough B we draw a line BF � parallel to CP � Obviously PFBC is a parallelogramand therefore � CPD � � BFP � Since CP � BF � then �DBF is isosceles� Consequently

Page 127: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

� NDA � � BFP � � CPD and DN � NP � A is the midpoint of PD and NA is perpendicularto AD and BC� We get that AM is perpendicular to BC and we deduce from here that �ABCis isosceles� i�e� AB � AC�

Figure �� Figure ��

Problem �� A convex polygon with n sides� n � �� is given� No four vertices of it lie onone and the same circle�

a Prove that there exists a circle through � vertices of the polygon which contains theremaining vertices in its interior�

b Prove that there exists a circle through � consecutive vertices of the polygon which containsthe remaining vertices in its interior�

Solution� a Let AB be a side of the polygon� All the vertices lie in one of the half�planeswith respect to AB� The segment AB is seen under di�erent angles from the vertices which aredi�erent from A and B� Let C be the vertex from which AB is seen under the smallest angle�The circle we are looking for is de�ned by A� B and C�

b We shall use the following two lemmas�Lemma �� Let the segment AB be seen from the point X under the angle � and from the

point Y under the angle �� where � � � � � ��� Then the radius of the circumcircle of�ABX is greater than the radius of the circumcircle of �ABY �

Lemma �� For each convex polygon there exist a side and a vertex from which this side isseen under an acute angle�

Proof� Let Ai� Ai�� and Ai�� be three consecutive vertices of the polygon �Figure �� Atleast one of the angles �Ai��AiAi�� and �AiAi��Ai�� is acute� This is enough for the proof�

Figure �� Figure ��

Let now M be the set of all pairs� that are consisted of a side and a vertex from which thisside is seen under an acute angle� Let �AB� C be an element ofM � Consider the circumcircle of�ABC and let N be the set of all such circles� We denote by k the circle in N with the biggestradius� Let k be the circumcircle of �A�A�Am� where A�A� is a side of the polygon �Figure

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�� We shall show that k is the circle we are looking for� Assume that there exists a vertex Ap

which is outside k� Then � A�ApA� � � A�AmA� � �� and the circumcircle of �A�A�Ap isfrom N and its radius is greater than the radius of k �Lemma �� We get a contradiction�

Consider one of the acute angles in �A�A�Am� Let � � � A�A�Am be acute� We shall provethat Am � A�� which means that k passes through � consecutive vertices of the polygon� Assumethat Am �� A�� Then A� is situated in the way which is shown in the Figure �� If � � � A�A�Am�then � � � � ���� i�e� ��� � � � � � ��� Thus � A�AmA� � ��� which means that �A�A��Am is an element ofM � Let c be the circumcircle of�A�A�Am and X be the intersection pointof c and the segment bisector of A�Am� Since � A�XAm � ��� � � � � � � A�A�Am� then theradius of c is greater than the radius of k �Lemma �� This is a contradiction�

Grade ��

Problem �� Let M be an arbitrary point on the side AB � � of the equilateral triangleABC� The points P and Q are orthogonal projections of M on AC and BC� while P� and Q�

are orthogonal projections of P and Q on AB�

a Prove that P�Q� ��

��

b Find the position of M for which the segment PQ is with the smallest length�Solution� a We have SABC � SACM � SBCM � Figure ��

��AC�MP �BC�MQ �

��MP �MQ �Figure �� On

the other hand SABC �

p�

�� Thus MP �MQ �

p�

��

Now from the rectangular triangles P�MP and MQ�Q

we evaluate P�M �

p�

�MP and MQ� �

p�

�MQ� From

here P�Q� � P�M �MQ� �

p�

��MP �MQ �

��

b The orthogonal projection of the segment PQ onAB is the segment P�Q� and thus PQ � P�Q�� There�fore PQ is with minimal length when it is parallel toP�Q�� The last is true exactly when AP � BQ� We getthat�AMP �� �BMQ and hence AM � BM � So� PQis minimal when M is the midpoint of AB�

Problem �� The quadratic function f�x � �x� � �px� p� � is given� Let S be the areaof the triangle with vertices at the intersection points of the parabola y � f�x with the x�axisand the vertex of the same parabola� Find all rational p� for which S is an integer�

Solution� The discriminant of f�x isD � ���p��p�� andD � for all real p� Consequentlyf�x has two real roots x� and x�� i�e� f�x intersects the x�axis in two di�erent points � A

and B� The vertex C of the parabola has coordinates �p and h � f��p � �p� � p� � � � Wehave

AB � jx� � x�j �q�x� � x�� �

q�x� � x�� � �x�x� � �

q�p� � p� ��

Now we �nd S � SABC �AB�h

�� ��p��p��

� � Denote q � �p��p��� Since q is rational

and q� � S� is an integer� then q is an integer too� ThenS

qis rational and

�S

q

��� q is integer�

Page 129: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

thusS

qis integer too� Therefore q � n�� where n is a positive integer� i�e� �p� � p� �� n� � �

The quadratic equation �with respect to p has a rational root exactly when its discriminant��n� � �� is a square of a rational number� Consequently ��n� � �� � m�� and we can considerm to be a positive integer� From the equality ��n � m��n � m � �� we get �n � m � ���n�m � �� or �n�m � �� �n�m � �� From here n � �� m � � or n � �� m � �� The rational

numbers we are looking for are � ���

����

��

Problem �� Let n be a positive integer and X be a set with n elements� Prove thata The number of all subsets of X �X and included is equal to �n�b There exist �n�� subsets of X each pair of which is with common element�c There do not exist �n�� � � subsets of X � each pair of which is with common element�Solution� a We use induction with respect to n� The base of the induction is obvious�

Assume that the assertion is true for a set with n � � elements and let X be with n elements�We can assume that X � f�� �� � � � � ng� Let Y � f�� �� � � � � n� �g� All subsets of X are dividedinto two groups� I group � those which do not contain n and II group � those� which containn� Both groups have one and the same number of elements because each set of the II group isobtained from exactly one set of the I group by annexing n� Thus the number of the elementsof X is twice greater than the number of the subsets of the I group� But the subsets of the Igroup are exactly the subsets of Y and according to the inductive assumption their number is�n��� Thus the number of the subsets of X is �n�

b According to a the number of the subsets of X from the II group is �n�� and each pairof them has a common element � the number n�

c If A X � let A � X nA� All subsets of X are divided into pairs fA�Ag and the number ofthese pairs is �n��� Now if we have �n���� arbitrary subsets of X � according to the pigeonholeprinciple� it is not possible that they are in di�erent pairs of the type fA�Ag� Consequentlyamong the given �n � � subsets there exist two pairs of the type fA�Ag� which obviously haveno common element�

Grade ���

Problem �� Find all values of the real parameters p and q� for which the roots of theequations x� � px � � � and x� � qx � � � form �in a suitable order an arithmeticprogression with four members�

Solution� Denote by x�� x� the roots of the equation x� � px � � � and by y�� y� theroots of x� � qx � � � � It is clear that for all p and q the numbers x�� x�� y�� y� are realand x�x� � y�y� � ��� Assume that x� � � x� and y� � � y�� If four numbers a� b� cand d form an arithmetic progression then the numbers d� c� b� a form an arithmetic progressiontoo� So� we can assume that x�� x�� y�� y� in a suitable order form an increasing arithmeticprogression� If x� � y� �the case y� � x� is analogous then there are two possibilities�

I� The arithmetic progression is x�� y�� y�� x�� Then x� � x� � y� � y�� from where p � q�i�e� x� � y�� x� � y� which is impossible�

II� The arithmetic progression is x�� y�� x�� y�� Then x� � x� � y� � y�� from wherepp� � � �

pq� � �� i�e� p� � q� and since p �� q� then p � �q �� �

In the same way we have y� �x� � x�

��

p

�and therefore

p�

�� pq

�� � � � Since p � �q�

then p� ��

�and p � � �p

�� Hence �p� q �

��p��� �p

��

�� �p

���p�

��

Page 130: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Triangle ABC with AB � ��� BC � ��� CA � �� is given�a If M is the center of gravity of �ABC� prove that AM� � CM� � BM��b Find the locus of points P from the plane of �ABC� for which AP � � CP � � BP ��c Find the minimal and maximal values of BP � if AP � � CP � � BP ��Solution� a We have Figure ��

AM ��

�ma �

�� ��

p�b� � �c� � a� �

p���

CM ��

�mc �

�� ��

p�a� � �b� � c� � ��

BM ��

�mb �

�� ��

p�a� � �c� � b� � ���

Hence AM� � CM� � �� � �� � ��� � BM��b Let E be the midpoint of AC and D be symmetric

to B with respect to E �Figure �� We shall prove that thelocus we are looking for is a circle k with center D and radius ��� For an arbitrary pointP we have �PE� � �PA� � �PC� � AC� and �PE� � �PB� � �PD� � BD�� from where��PA��PC��PB� � �PD�� �BD��AC�� But BD � �BE � ��� i�e� PA��PC��PB� �PD������� It follows from here that the equality PA��PC� � PB� is equivalent to PD � ���

c Let the circle k intersects the line BD at the points M and N � It follows from b thatBP is minimal when P coincides with M and then BP � BM � ��� BP is maximal when P

coincides with N � which gives BN � ���Problem �� Find the smallest positive integer n� for which there exist n di�erent positive

integers a�� a�� � � � � an satisfying the conditions�a the smallest common multiple of a�� a�� � � � � an is ���� b for each i� j � f�� �� � � � � ng the numbers ai and aj have a common divisor �� � c the product a�a� � � �an is a perfect square and is divisible by ����Find all n�ples �a�� a�� � � � � an� satisfying a� b and c�Solution� Since ���� � �������� and ai������ then for all i

ai � ��i��i��i���i �

where the numbers �i� �i� i� i are equal to or �� There is no ai which is divisible by �� Wehave a�a� � � �an � k�� where k is a positive integer and since ��� � �� divides k�� then �� dividesk� and since among the numbers � there is no one which is divisible by �� then n � �� Thenumbers � � which are divisible by � are

� ��� ��� ���� ����� ������ ������ �������� �

They are � in number� Let n � �� Then the numbers a�� a�� � � � � an are among � � It iseasy to see that the product of any � numbers from � is not a perfect square� Thus n � ��Let n � �� It is not possible that all the numbers a�� a�� � � � � an are divisible by �� because insuch a case ���k� and �� �� k�� Therefore � numbers from a�� a�� � � � � an are divisible by �� i�e�they are from � and at least one of them �for example a� is not divisible by �� It followsfrom a that among a�� a�� � � � � a� there is at least one which is divisible by �� at least one whichis divisible by � and at least one which is divisible by ��� Since each pair of these numbershas a common divisor� and � �� a�� then a� must be divisible by ������� i�e� a� � ������� At

Page 131: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

last note that the product of all numbers � is equal to ���������� The only possibility isa�a� � � � a� � ���������� From here a�a� � � �a� � ������������ and the only possibility is

fa�� a�� � � � � a�g � f� � �� � � �� � � ��� � � � � �� � � � � ��� � � � � ��g�

Therefore n � � and

fa�� a�� a�� a� a�� a�� a�g � f� � �� � � �� � � ��� � � � � �� � � � � ��� � � � � �� � � � � ��g�

Grade ���

Problem �� Let an �n� �

�n��

���

��

��

�� � � �� �n

n

�� n � �� �� �� � � �� Prove that

a an�� � an for all n � � b the sequence fang�n� is convergent and �nd its limit�Solution� a We have

an�� �n � �

�n��

���

��

��

�� � � �� �n��

n� �

��

n� �

��n� ��an � ��

From here

an�� � an�� ��n� ���an�� � an� �an � �

��n� ��n� �� n � �� �� � � � � �

Since an � for all n� then if an�� � an � � we have an�� � an�� � � But a� ��

�and

a ��

�� i�e� a � a� � � Hence a� � a � � a� � a� � � � � �� an�� � an � �

b The sequence a�� a�� a�� � � � is decreasing when n �Figure ��� and it is bounded �an � for all n� Therefore thissequence is convergent� Let lim

n��an � a� From the

equality an�� �n� �

��n� ��an�� after passing to in�nity

we get a ��

��a� �� i�e� a � ��

Problem �� The quadrilateral ABCD is inscribedin a circle with center O� The diagonals AC and BDintersect each other in the point M � M �� O� The linethrough M which is perpendicular to OM intersects thesides AB and CD of the quadrilateral ABCD in thepoints X and Y � respectively� Prove that AB � CD i�BX � CY �

Solution� If AB � CD� then ABCD is isoscelestrapezoid� Hence OM�AD and OM�BC� from where

XY kBC and BX � CY �Let BX � CY � Denote by P and Q the midpoints of AB and CD� respectively� The quadri�

laterals OPXM and OQYM are inscribed� thus � OPM � � OXM and � OQM � � OYM �We shall prove that � OPM � � OQM � The triangles ABM and DCM are congruent and

Page 132: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

MP and MQ are medians in them� Therefore �MPB � �MQC and � MPB � � MQC�Then � OPM � �� � � MPB � �� � � MQC � � OQM� �Or � OPM � � MPB � �� �� MQC � �� � � OQM �

Hence � OXM � � OYM and OX � OY � Therefore�OXB �� �OY C� From here � OBA �� OCD and the isosceles triangles ABO and DCO are equal� i�e� AB � CD

Remark� If X � P and Y � Q� then PB � QC ��

�AB �

�CD�

Problem �� Let n be a positive integer and let

f�x � xn � �k � �xn�� � ��k � �xn�� � � � �� ��n� �k� �x� nk � ��

a Prove that f��� k � n� ��b Prove that if n � � and k is an integer �k �� � then the equation f�x � has no integer

solution�Solution� a Since

xn�� � �xn�� � � � �� �n� �x� n

� �xn�� � � � �� x� � � �xn�� � � � �� x� � � � � �� �x� � x� � � �x� � � �

�xn � �

x� ��xn�� � �

x � �� � � �� x� � �

x� ��x� � �

x � ��x� �

x� ��

then

xn�� � �xn�� � � � �� �n� �x� n ��

x� �

�xn�� � �

x� �� �n� �

�xn�� � �n� �x� n

�x� ��

when x �� �� Thus

f�x � xn � xn�� � � � �� x� � � k�xn�� � �xn�� � � � �� n

�xn�� � �

x� �� k

xn�� � �n� �x� n

�x� ���

i�e�

f�x �xn�� � �k � �xn�� � �k�n� � � ��x� kn � �

�x� ��

when x �� ��From here

f��� k ���k�n� � � ����� k � kn � �

k�� n � ��

when �� k �� �� If �� k � �� i�e� k � � then f�� � n � ��b Let n � � and k be an integer �k �� � Assume that f�a � � Obviously a �� � We have

an � an�� � � � �� a� � � �k�an�� � �an�� � � � �� �n� �a� n�

If a � ��� the left hand side of the equation is equal to or ��� while the right hand side isneither nor ��� because jn� �n� � � �n � �� � � � j � � when n � � �k �� � Thus a �� ���The equation f�a � can be written in the following way�

��a � k � ��an�� � �an�� � � � �� �n� �a� n � n � ��

Page 133: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Hence rn�a � an�� � �an�� � � � �� �n� �a� n divides n� �� If a � �� then rn�a � n� �when n � � and this is impossible� From a �� and a �� �� it follows that a � ��� We shallprove that the inequality jrn�tj � n � � is satis�ed for all integers t � �� and for all n � �

except t � ��� n � � and t � ��� n � �� Since rn��� � ���n�� � �n � �

�� then r���� � ��

r��� � �� and r���� � �� i�e� jr����j � �� From r��t � t���t�� it follows that jr��tj � �when t � ���

Now we shall use induction� If jrn�tj � n � � for t � �� and n � �� then rn���t �t�rn�t�n�� and jrn���tj � jtj�jrn�tj��n�� � ��n���n�� � n��� i�e� jrn���tj � n���Hence jrn�tj � n� � when n � � and t � �� except the cases n � �� t � �� and n � �� t � ���Thus rn�a does not divide n � � when n � � and a � ��� because r���� � �� r��� � ���For all others n � � and a � �� we have jrn�aj � n� ��

Grade ���

Problem �� The function f�x �p�� x �x � � is given� Let F �x � f�f�x�

a Solve the equations f�x � x and F �x � x�b Solve the inequality F �x � x�c If a� � �� �� prove that the sequence fang�n�� determined by an � f�an�� for n �

�� �� � � � � is convergent and �nd its limit�Solution� The functions f�x� F �x and F �x� x are de�ned for all x � �� ���

a The equation f�x � x� i�e�p�� x � x has only one root � �

�� �p�

�� It is clear that

� � �� � and the roots of the equation F �x � x are x� � � x� � �� x� � ��b In �� �� the function F �x� x has a constant sign� The contrary would imply that there

is � � �� � such that F �� � � and this contradicts the result from a� Analogously in ��� ��

the function F �x � x has a constant sign� On the other hand�

�� �� ��

�� ��� � and

F

��

�� �

�� � F

��

�� �

�� � From here F �x � x i� x �

���� �p

��

c Let a� � �� It follows from a that an � � for n � � �� �� � � � and hence the sequence is

convergent and its limit is �� Let now a� � �� Since f ��x � � �

�p�� x

� for all x � �� ��

then f�x is decreasing� From here frac�a� � frac�� � �� i�e� a� � �� By inductiona�n � �� � and a�n�� � ��� � for all n � � �� �� � � �� On the other hand it follows from theresult of b that for all x � �� � we have F �x � x� while for x � ��� � we have F �x � x�respectively� Also F ��x � f ��f�x�f ��x � � i�e� F �x is increasing and hence F �x � �x� � ifx � �� � and F �x � ��� x if x � ��� �� By induction we get

a� � a� � � � � � a�n � � � �� ��a� � a� � � � � � a�n�� � � � � � ��

Both sequences are convergent and let their limits be �� and ��� respectively� We haveF �a�n � a�n�� and F �a�n�� � a�n�� for n � � �� �� � � �� The function F �x is continuous andthus F ��� � �� and F ��� � ��� We get �� � �� � � because the only solution of F �x � x

in �� � is �� Therefore the sequence fang�n� is convergent and its limit is �� The case � � a�is analogous�

Page 134: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� The sides AC and BC of the triangle ABC are diameters of two circles� eachof which touches internally a circle k� which is concentric to the incircle of �ABC�

a Prove that AC � BC�

b If cos � BAC ��

�� �nd the ratio of the radii of k and the incircle of �ABC�

Solution� a Let I be the center of the incircle of �ABC� N be the common point ofthis circle with AC and M be the midpoint of AC� Let r� be the radius of k and r be the

radius of the incircle of �ABC� From the condition it follows that IM �b

�� r� and from the

rectangular �INM we get MN� �

�b

�� r�

��� r�� On the other hand AN �

b� c� a

�� i�e�

MN � jAM � AN j � jc� aj�

� Therefore

�c� a� � �b� �r�� � �r�� ��

Analogously�c� b� � �a� �r�

� � �r�� ��

Assume that a �� b� From �� and �� we get �b� a��c� a� b � �b� a�a� b� �r�� i�e��c� a� b � a� b� �r�� From here b� �r� � c� a and from �� it follows that r � � which isimpossible� Thus a � b�

b Letr�r

� t and � BAC � �� Then c � �b cos� ��

�b and

r �c

�tan

�� b cos� tan

�� b cos�

s�� cos�

� � cos��

�b

�p��

Since b � �r�� from �� it follows that t �b�p�c� b� � �r�

�rand substituting c and r we

get t ��

��p�� ��

Problem �� n points �n � �� no three of which are colinear are given in the plane� Morethan n triangles are constructed with vertices among these points� Prove that at least twotriangles have exactly one common vertex�

Solution� Assume the contrary and let k be the smallest number for which the assertion isnot true� This means that there are constructed at least �k � � triangles using k points� Itfollows from the pigeonhole principle that there exists a point A which is a vertex of at least �triangles� Let ABC be the �rst triangle� At least one of the points B and C is a vertex of thesecond triangle� which we denote by ABD� If ACX is the third triangle then X � D� Thus theforth triangle must contain B or C� which is impossible� Therefore if A and B are vertices of twotriangles then they are vertices of all the four triangles� Let A be a vertex of t triangles� t � ��These triangles are of the kind ABA�� ABA�� � � �� ABAt� where all the points A�� A�� � � � � At

are pairwise di�erent� Obviously it is not possible to exist a triangle of the type BXY � whereX and Y are points which are di�erent from A�� A�� � � � � At� Triangles BAiAj and AiAjAm donot exist too� Hence the points A� B� A�� A�� � � �� At are vertices only of the triangles ABA��ABA�� � � �� ABAt� In such a way we use t� � points and get t triangles� It is not possible thatt� � � k� because all triangles are t � k� The number of remaining points is k� � k� t� � andby them there are constructed at least k � �� t � k� triangles� such that no two of them have

Page 135: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

exactly one common vertex� The triangles are more than the points k�� and thus k� � �� Wehave found a number k� � k for which the assertion is not true� This contradicts the choice ofk�

Page 136: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

SPRING MATHEMATICAL COMPETITION

����

Grade �

Problem �� Prove that for all real a � ��� �� the area of the �gure encountered by the

graphs of the functions y � �� jx� �j and y � j�x� aj is less than �

Solution Firstly we shall �nd the common points of the given functions For this purposewe solve the equation

j�x� aj � �� j�� xj� ���

Since � � a � � thena

�� � and we shall consider the cases� x � a

�a

�� x � � and x � �

We have�� When x � a

�the equation ��� takes the form a � �x � x Then x �

a

� which satis�es

��� becausea

��a

� Whena

�� x � � the equation ��� takes the form �x� a � x ie x � a which does not

satisfy ��� because a � �

� When x � � the equation ��� takes the form �x � a � � � x Then x �a � �

� which

satis�es ��� becausea� �

�� � when a � �

Thus the graphs of the two functions have two common Figure �points �Figure �� the �rst of which �denoted by A� has coor

dinates xA �a

� yA �

a

� while the second one �denoted by B�

has coordinates xB �a� �

� yB � �� a� �

��

�� a

� Denote

by C D and E the points with coordinates xC � � yC � ��

xD �a

� yD � � and xE � � yE � � respectively

Then the �gure encountered by the two graphs is thequadrilateral ACBD and its area S is obtained by subtractingthe areas of the triangles ODA and BDE from the area of thetriangle OEC Therefore

S � SOEC � SODA � SBDE

� �� �

��OD � yA � �

���DE � yB

� �� �

�� a�� a�� �

����� a

�� �� a

Page 137: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

� �� �

��� a� � �

��� ��� a��

��

����a� � �a� �

��

����� �a� ���

��

��

Problem �� The altitude CD of the rectangle triangleFigure �ABC � � ACB � ���� is a diameter of the circle k which meetsthe sides AC and BC in E and F respectively The intersec tion point of the line BE and the circle k which is di�erentfromE is denoted byM Let the intersection point of the linesAC and MF be K and the intersection point of the lines EFand BK be P

a� Prove that the points B F M and P are concyclic�b� If the points D M and P are colinear �nd the angles

A and B of the triangle ABCSolution a� Since � ECF � ��� then EF is a diameter of

the circle k and consequently � EMF � ��� � � BMK ThenBC and KM are altitudes in �BEK �Figure �� which means

that the point F is the altitude center of this triangle It follows from here that EP � BK ie� BPF � ���

Thus � BPF � � BMF � ��� and consequently the points B F M and P lie on a circlewith diameter BF

b� We have�Figure �

� BDM � ��� � � MDC � ��� � � MEC

� � CBE � � FBM � � FPM�

If the points D M and P are colinear �Figure �� then theequality between the angles � BDM and � FPM implies thatthe lines AB and EF are parallel ie EF and CD are di ameters of the circle k which are perpendicular to each otherThis means that CD is the bisector of � ACB ConsequentlyAC � BC and � BAC � � ABC � ���

Problem �� In a state every town is connected with thenearest town by a straight way The distances between the

pairs of towns are pairwise di�erent Prove thata� no two ways have common points�b� every town is connected by ways with at most � other towns�c� there is no closed piecewise line consisted of waysSolution a� Suppose that the ways AC and BD meet each other �Figure �� and let C be

the nearest town to A while D be the nearest town to B Then AC � AD and BD � BCfrom where AC �BD � AD �BC

On the other hand if O is the common point of AC and BD then AO � OD � AD andBO �OC � BC from where AC �BD � AD �BC This is a contradiction

b� Let the town X be connected by ways with the towns A and B �Figure �� Then AB isthe longest side of �XAB Indeed if we assume that for example AX is the longest side then

Page 138: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

A should not be the nearest town to X and X should not be the nearest town to A as wellConsequently the way AX should not exist

Figure � Figure � Figure �

Therefore � AXB is the biggest angle in �XAB from where � AXB � ��� Now if weassume that the town X is connected with at least � towns then the sum of the angles at Xwould be greater than � � ��� � ���� which is impossible Thus every town is connected withat most � other towns

c� Assume that there exists a closed piecewise line A�A� � � �An consisted of ways �Figure�� The distances A�An and A�A� are di�erent Let A�An � A�A� Then A� is not the nearesttown to A� and consequently �because the way A�A� exists� A� is the nearest town to A� Itfollows from here that A�A� � A�A�

Proceeding in this way we obtain the chain� A�An � A�A� � A�A� � � � � � AnA� whichleads to contradiction

Grade �

Problem �� Find the values of the real parameter b for which the di�erence between themaximal and the minimal values of the function f�x� � x���bx�� in the interval ��� �� is equalto �

b � �

Figure �

b� �

Figure �

b� �

Figure �

Solution It is clear that the minimal value of the quadratic function f�x� is obtained whenx � b We shall consider the following three cases�

�� Let b � � In this case the function f�x� is increasing in the interval ��� �� �Figure ��and the maximal value is f��� � �� �b while the smallest one is f��� � � From the condition

f���� f��� � �� �b � � we �nd b � ��

� which is a solution of the problem

Page 139: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�� Let b � � Now the function f�x� is decreasing in the interval ��� �� �Figure �� andthe maximal value is f��� � � while the minimal one is f��� � � � �b From the condition

f���� f��� � �b� � � � we �nd b ��

� which is a solution of the problem

�� Let � � b � � �Figure �� The minimal value of f�x� in the interval ��� �� is f�b� � �� b�while the the maximal one is f��� or f��� From f���� f�b� � b� � � we �nd b � �� which arenot solutions �because in this case � � b � �� From f���� f�b� � ��� b�� � � we �nd b � � orb � �� which are not solutions too

Finally the answers are� b � ��

�and b �

Problem �� The quadrilateral ABCD is inscribed in aFigure ��circle with radius � a circle can be inscribed in it and AB �AD Prove that�

a� the area of the quadrilateral ABCD does not exceed ��b� the inradius of the quadrilateral ABCD does not exceedp

Solution We have �Figure ��� AB �CD � AD�BC andfrom AB � AD we get� BC � CD Let AB � AD � x BC �CD � y The triangles ACB and ACD are equal �by SSS�from where � B � � D But ABCD is inscribed in a circle andconsequently � B � � D � ���� Therefore � B � � D � ���

Then AC is diameter of the circumcircle of the quadrilateral and particularly AC � � andx� � y� � AC� � �

Let S p and r be the area of the quadrilateral its semiperimeter and inradius respectivelya� We have S � SABC � SACD � �SACB � xy and

xy �qx�y� � x� � y�

��AC�

�� �

Thus S � � �and S � � only if x � y and then the quadrilateral ABCD is a square�b� We shall make use of the formula S � pr which is true for all polygons that can be

inscribed in a circle We have�

r� �S�

p��

S�

�x� y��� fracS�x� � y� � �xy �

S�

� � �S�

Now

r �p�

� r� � �

� S�

� � �S� �

S� � S � � � � �S � ���S � �� � �

S � �

�because S � �� The last inequality is true according to a� and consequently the inequality

r �p�

�is true too

Problem �� This problem is the same as problem � grade �

Page 140: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Grade ��

Problem �� Find in the plane the locus of points with coordinates �x� y� for which thereexists exactly one real number z satisfying the equality�

xz� � yz� � � �x� jyj� z� � yz � x � ��

Solution Let �x� y� be the coordinates of a point from the locus we are looking for Thismeans that there exists exactly one real number z for which

xz� � yz� � � �x� jyj� z� � yz � x � ��

If x � � the above equality takes the form

z�yz� � �jyjz � y

�� ��

This equality is satis�ed for at least two di�erent values of z �for example z � � and z � �if y � � and z � � and z � �� if y � �� which shows that the condition of the problem is notveri�ed Consequently the points from the y axis do not belong to the locus

Let x �� � Thus if z satis�es the given equality then z �� � It is easy to see that in this

case the number�

zsatis�es the same equality and consequently z �

z From here z� � � ie

z � ��Case �� Let z � � Then x � y � � �x� jyj� � y � x � � from where y � jyj � � This

shows that y � � If y � � and then the given equality takes the form x�z� � �

��� � and

consequently it is satis�ed also by z � �� Thus we can assume that y � � Then

xz� � yz� � ��x� y�z� � yz � x � ��

from where�z � ���

�xz� � �y � �x�z � x

�� ��

We have one of the following possibilities��i� The number z � � is the only to satisfy the equality xz���y��x�z�x � � This is true

if x� y� �x� x � � and D � �y� �x��� �x� � y�y� �x� � � from where y � ��x and y � ��ii� There is no real z which satisfy the equality xz� � �y � �x�z � x � � Then D �

�y � �x�� � �x� � y�y � �x� � � from where y � ��x and y � �Therefore every time when y � ��x and y � � the point �x� y� belongs to the locusCase �� Let z � �� Analogously to the previous case we �nd that y � � and then the

given equality takes the form

�z � ����xz� � �y � �x�z � x

�� ��

We have the following possibilities��i� the number z � �� is double root of the equation

xz� � �y � �x�z � x � �� ���

�ii� the equation ��� has no real rootBy computing we deduce that here y � �x and y � �

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Finally the locus �Figure ��� consists of the internal as well as of the boundary points of theangle which is de�ned by the graphs of the linear functions y � �x and y � ��x when x � �without the points from the negative part of the x � axis

Problem �� In the triangle ABC ha and hb are the altitudes from A and B respectively�c is the internal bisector of � ACB while O I and H are the circumcenter the incenter and

the altitude center respectively Prove that if�c

ha�

�c

hb� � then OI � IH

Solution Let BC � a AC � b � ACB � � CL � �c Figure ��

�Figure ��� Since SABC � SALC � SBLC then�

�� �ca sin �

��

�� �cb sin �

��

�� ab sin� from where �c �

�ab cos�

�a� b

But

ha � b sin � and hb � a sin � Then�c

ha�

�c

hb�

�a cos�

��a� b� sin�

�b cos�

��a� b� sin �

��

sin�

�a

a� b�

b

a� b

��

sin�

which

shows that sin�

��

� ie

�� ��� because

�� ��� Conse

quently � � ���Let CL meet the circumcircle of �ABC in M Then

OM � AB But CH � AB and therefore � OMC � � MCH On the other hand OM � OC �R �R is the circumradius� Consequently � OMC � � OCM � � MCH Let H� be the foot of

the altitude from B From �HCH� we have CH �CH�

sin��because � H�HC � ��� � � ACH

� � CAB � �� But CH� � a cos� ie CH �a

sin�� cos� � �R cos � Since � � ��� then

CH � RWe consider �COI and �CHI Since the point I lies on CM then � OCI � � HCI Also

CO � R � CH and consequently �COI � �CHI Thus OI � IH Problem �� Let A� A� � � � An �n � �� be n points in the plane no � of which are colineara� Prove that there is at most one point As such that all triangles AsAiAj �i� j � �� �� � � � � n�

are acuteb� Let among A� A� � � � An be a point which is a vertex of an acute triangles only We

consider the angles de�ned by the given points Denote by Nk the number of the acute angles� AiAkAj �i� j � �� �� � � � � n� for which the point Ak is their vertex Find the minimal value ofNk

Solution a� Assume that there is more than one point with the given property and let Aand B be such two points while X and Y be any two of the remaining points There are twopossibilities for the points A B X and Y �

�i� A B X and Y de�ne a convex quadrilateral Because the sum of its internal angles is���� then at least one of these angles is not acute Consequently at least one of the triangleswith vertex A or B is not acute

�ii� One of the points A B X and Y is inside the triangle de�ned by the other three Thenthe sum of the three angles with a vertex inside is ���� Consequently in this case a trianglewith vertex in A or B is not acute again

The contradictions show that it is not possible to exist more than one point which is a vertex

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of acute triangles onlyb� Let A� be the point which is a vertex of acute triangles only We consider the angles

� AkA�As �Figure ��� Let A�A�An be the biggest one Because all angles with vertex A� areacute the points A� A� � � � An�� are inside the acute angle � A�A�An We can assume thatthe points are enumerate in such a way that � A�A�Ak � � A�A�Ak�� for k � �� �� � � � � n

Consider � AiAkAj where � � i � k � j � n Figure ��We assume that Ak is internal for �A�AiAj Then

� A�AkAi � � A�AkAj � ����

and at least one of the angles � A�AkAi and � A�AkAj will not beacute which is a contradiction

Therefore the points A� Ai Ak Aj de�ne a convex quadrilat eral If we assume that � AiAkAj � ��� then there exists an angleof this quadrilateral with vertex A� Ai or Aj which is � ��� Thisis impossible

Therefore � AiAkAj � ��� and it is clear that � AiAkAj � ��� when i� j � k or i� j � kParticularly it follows from here that no of the angles AiAkAj is right The number of all

angles with vertex Ak is equal to�n� ���n� ��

� If Tk is the number of the obtuse angles with

vertex Ak then Nk ��n� ���n� ��

�� Tk Thus Nk is minimal when Tk is maximal

It is easy to see that T� � T� � Tn � � Let � � k � n � � It is clear that the number ofthe points Ai for which � � i � k is k� � and the number of the points Aj for which k � j � n

is n� k Then Tk � �k � ���n� k� for k � �� �� � � � � n� �

But �k � ���n� k� � �n� ���

� and the equality is reached when k � � � n � k ie when

k �n� �

� We have two possibilities�

�i� The number n is even Thenn� �

�is integer and consequently the maximal value of Tk

is Tn��

��n� ���

� Therefore the minimal value of Nk in this case is Nn��

��n� ���n� �

��

�n� ���

��n�n � ��

�ii� The number n is odd Then the nearest integers ton � �

�are

n� �

�and

n � �

� It is

easy to see that now the maximal value of Tk is Tn��

� Tn��

�n� �

�� �

��n � n � �

��

�n� ���n� ��

� Consequently the minimal value of Nk in this case is

Nn��

� Nn��

��n� ���n� ��

�� �n� ���n� ��

��

�n� ���

��

Grade ��

Problem �� Find the values of the real parameter a for which the inequality x� � �x� ���x� � ax� � ��x� � �x� � � � is satis�ed for all real x

Page 143: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution When x � � the given inequality is satis�ed for all a Thus it is enough to �nd

such a that

�x� �

x�

�� �

�x� �

x�

�� ��

�x�

x

�� a � � for all x � � and

�x� �

x�

��

�x� �

x�

�� ��

�x�

x

��a � � for all x � �

Denote t � x ��

x It is clear that x � � t � � and x � � t � �� But

x� ��

x�� t� � � and x� �

x�� t� � �t

We consider the function f�t� ��t� � �t

�� �

�t� � �

�� ��t� � t� � �t� � �t � �� � a

The problem is reduced to �nd such a that f�t� � � for all t � � and f�t� � � for all t � ��simultaneously

But f ��t� � �t� � ��t � � � ��t � ���t � �� from where it isFigure ��easy to obtain that �Figure ��� f�t� � � for all t � � i� f��� � �and f�t� � � for all t � �� i� f���� � � Since f��� � a � ��and f���� � a � �� we �nd for a� a � ��� and a � �� Finally��� � a � ��

Problem �� The point D lies on the arc�

BC of the circumcircleof �ABC which does not contain the point A and D �� B D �� C

On the raysBD� and CD� there are taken pointsE and F such thatBE � AC and CF � ABLet M be the midpoint of the segment EF

a� Prove that � BMC is right

b� Find the locus of the points M when D describes the arc�

BC

Figure �� Figure ��

Solution Denote by � � the angles corresponding to the vertexes A B C of �ABCand by a b c the lengths of the sides BC CA AB respectively Let � BCD � It follows

from D � �

BC that � � � � ie � ��� ��

a� We have �Figure ������BM �

�����BF �

��BE

��

�����CF �

��BE �

��BC

�and

���CM �

�����CF �

��CE

��

�����CF �

��BE � ��BC

� Then we �nd for the scalar product

���BM � ���CM �

���BM � ���CM �

������

CF ���BE

�����BC

��

Page 144: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

��

�����CF

���BE

� � � ��CF � ��BE � a��

��

���c� � b� � �bc cos�� � ��� a�

��

���c� � b� � �bc cos� � a�

�� ��

�We have used that jBEj � jACj � b jCF j � jABj � c and � ���BE�

��CF � � � BDC � � � ��

Consequently���BM � ���CM ie the angle � BMC is right

b� It is clear that the points M and A are in di�erent semiplanes with respect to the line BCAccording to a� � BMC is a right angle and consequently if the point M is from the locus thenM lies on a semicircle k with diameter BC k and A are in di�erent semiplanes with respect to

the line BC Let � BCM � � Then CM � a cos� �because � BMC ��

�� and

���CM � ��CB � a�a cos�� � cos� � �a cos����

On the other hand by the sine theorem we �nd�

���CM � ��CB �

�����CB �

��BE �

��CF

�� ��CB

��

���a� � ab cos �� � ��� �� � ac cos

��

�a���� b

acos��� � �

c

acos

��

�a��� �

sin � cos� sin cos��� �

sin�

��

�a��� �

sin�� � �� sin� � � � �

� sin�

��

�a� �� � cos� � �� �

�a cos

���

Consequently cos�� � cos� �

� But

��

� �

��

�and therefore � �

� ie

� BCM � �

�and � BCM �

�� � �

� In addition we have � CBM �

�� � BCM �

�� � �

�and � CBM �

��

���� �

Let K and N be points from the semicircle k for which � BCK �

�and � CBN �

� Then

angleBCN ��

�� �

����

�and � CBK �

��

��� � �

It follows from the above considerations that when the point D describes the arc�

BCthen CM� describes the interior of � �CK�� CN�� while BM� describes the interior of

� �BK�� BN�� Consequently the locus we are looking for is the arc�

NK from the semi circle k �Figure ���

Problem �� is the same as problem � grade ��

Page 145: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Spring mathematics

tournament�����

Problem ���� Given the equation jx� aj��� � �jx��j� where ais a real parameter

a� Prove that for any value of a the equation has exactly twodistinct roots x� and x�

b� Prove that jx� � x�j � � and �nd all values of a for whichjx� � x�j � �

Solution� a� When x � ��� the equation is equivalent to jx �aj � ��x � � � which has a solution only if ��x � � � �� i e� if

x � ��

� Considering the cases of both x � a and x � a shows that

if x � ��� the given equation has a unique root x�� and if x � ���it has a unique root x��

x� �

�������

a� �

� a � ��

�a� �

�� a � ��

x� �

�������

� � a

�� a � �

�a� �

� a �

Page 146: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

b� It follows from a� that

jx� � x�j �

���������������

��� � ��a

��� a � ��

�� ��

�� a � �

���a� ���

��� a �

It remains to be seen that��� � ��a

��� � when a � ��

�and

��a� ���

��� � when a �

� Therefore jx� � x�j � � and equal�

ity obtains only when a � ���

���

��

Problem ���� Let O be the intersecting point of the diagonalsof the convex quadrilateral ABCD and let � DAC � � DBC Themidpoints of AB and CD are respectively M and N and P and Q

are points on AD and BC respectively such that OP � AD andOQ � BC Prove that MN � PQ

Solution� Denote by E and F the midpoints of AO and BO Then

�PEM �� �MFQ since MF ��

�AO � PE� QF �

�OB � ME

and � MEP � � MEO � � OEP � � MEO � �� DAC � � MFO ��� DBC � � MFQ MFOE is a parallelogram� Therefore MP �MQ The case when E and F are interior points for � PMQ istreated similarly prove that there are no other possibilities� Byanalogy we conclude that NP � NQ Hence M and N lie on theaxis of symmetry of PQ and so MN � PQ

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Problem ����� Find all natural numbers n such that there existsan integer number x for which ������ n � �� � x��x

Solution� Let n be a solution of the problem Then �x � ��� ����� ��� n � ��� If n � � we get �x � ��� � ��� � and so�x � � � ��� Therefore x � ��� and x � ���� satisfy theconditions of the problem Let n � � Now �x � ��� is divisibleby ��� � which is a prime number� and so �x � ��� is divisible by��� � But this is impossible� since ������� n���� is not divisibleby ��� � when n � � The only solution is n � �

Problem ���� Find all values of the real parameter m such thatthe equation x�� �mx� �m�����x�� �x� �mm����� � � hasexactly three distinct roots

Solution� Suppose m satis�es the conditions of the problem andthe equations

x� � �mx� �m� � �� � �

x� � �x� �mm� � �� � �

share the root x� After subtracting we get �m � ��x� � �m ���m���� and so x� � m��� note that ifm � �� the two equationscoincide� Substituting x� in any of the equations gives the equationm� � ��m� � �m � �� � � with roots m � �� and m � � Directveri�cation shows that the condition is satis�ed only for m � �

Let now �� and �� share no roots Since D� � � � �m� � ��� always has two distinct roots and therefore �� should have equal

Page 148: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

roots Thus D� � � � �mm� � �� � � and so m � �� But thiscase has already been considered Thus we determine that m � �

Problem ���� The area of the equilateral triangleABC is PointsM and N are chosen respectively on AB and AC so that AN � BM Denote byO the intersecting point of the straight linesBN and CM The area of BOC is �

a� Prove that MB � AB � � � � or MB � AB � � � �

b� Find � AOB

Solution� a� DenoteMB

AB� x Therefore SABN � x � SBMC

and so SBOM � x � � and SAMON � SBOC � � Further SCON �

� � � � � x � �� � � � x� SANO �x

� � x SCNO �

x� � x�

� � x�

SAMO ��� x

x SBOM �

�� x

x x � �� It follows from SAMON �

SANO � SAMO that � �x�� x�

�� x�

� � x

x x� ��� and thus �x� �

�x�� � �� The roots of the above equation are x� ��

�and x� �

b� Since �ABN �� �BMC� we get � BOM � � BCM �� CBO � � MBO� � CBO � ��� Further � MAN � � MON � ����

and therefore the quadrilateral AMON is inscribed in a circle LetMB

AB�

�� i e� AM � �BM � �AN Denote by Q the mid�

point of AM Triangle AQN is isosceles and has an angle equal to���� so it is equilateral Therefore Q is the circumcentre of AMON

and � AOM � � ANM � ��� Thus � AOB � ���� Similarly� if

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MB

AB�

�� i e� �AM �MB � AN � we get � AMN � � AON � ����

so � AOB � ���

Problem ���� Given n points� n � �� in the plane such thatno three lie on a line John and Peter play the following game�On his turn each of them draws a segment between any two pointswhich are not connected The winner is the one after whose moveevery point is an end of at least one segment If John is to begin thegame� �nd the values of n for which he can always win no matterhow Peter plays

Solution� Call a point isolated if it is not an end of a segmentJohn wins exactly when there are � or � isolated points before hislast move Peter is forced to reach the above only if before his movethere are exactly � isolated points and any of the remaining n � �points are connected by a segment Indeed� if there are at least �isolated points he could connect one of them with a non�isolatedpoint If the isolated points are � but not all of the remainig n � �points are connected he could draw a missing segment Since the

number of segments with ends in n � � points isn� ��n� ��

��

we determine that John wins only whenn� ��n� ��

�is an odd

integer number This is true when n is of the form �k� � or �k ��

Problem ��� Let fx� � x���ax�a�� �

�where a is real param�

eter Find all values of a such that the inequality jfx�j � � holdsfor any x in the interval ��� ��

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Solution� Let M and m be the maximum and minimum valuesof fx� in the interval ��� �� Then the condition of the problem isequivalent toM � � and m � �� There are three cases to consider

Case �� a � ��� �� Then m � fa� � ��a�� �

�and M � f�� �

�a�� �

�or M � f�� � �a� � �a�

� It follows from m � �� and

M � � that a � ��p�

��

p�

�� and a � �����

�� � ���

���� In this

case the solution is a � ���

p�

��

Case �� a � � Now m � f�� and M � f�� From m � ��and M � � we get that a � ���

���

��� Therefore a � ���

�� ��

Case �� a � � Now m � f�� and M � f�� It follows from

m � �� and M � � that a � ���

���

�� which is a contradiction with

a � �

Thus the solution is a � ���

��

p�

��

Problem ��� Let I and G be the incentre and the centre of�ABC with sides AB � c� BC � a� CA � b

a� Prove that if a � b� the area of CIG equalsa� b�r

�where r

is the inradius of ABC�

b� If a � c�� and b � c��� prove that the segment IG is parallelto AB and �nd its length

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Solution� a� We shall use the usual notation for a triangle LetCL and CM be respectively the bisector and the median from C

It follows from CG ��

�CM that SCIG �

�SCIM Thus SCIM �

SCLM � SILM �LM hc

�� LM r

��

LM

�hc � r� We �nd from

AL �BL � c andAL

BL�

b

athat AL �

bc

a� band so LM � AM �

AL �c

�� bc

a� b�

ca� b�

�a� b� Also� hc � r �

�S

c� r �

�pr

c� r �

r�p � c�

c�

ra� b�

c Therefore SCIG �

�SCIM �

� ca� b�

�a� b�

ra� b�

c�

a� b�r

b� The distances from I and G to AB are respectively r andhc

� Hence r �

S

p�

chc

�p�

chc

�c�

hc

�and so IGkAB Therefore

the altitude from C of triangle CIG equals�

�hc � �r Thus SCIG �

IG r On the other hand SCIG �a� b�r

��r

�and so IG �

Problem ��� Let n � n be an even number and A be a subsetof f�� �� � � � � ng Consider the sums of the form ��x� � ��x� � ��x��where x�� x�� x� are integer numbers in A not necessarily distinct����� ��� �� at least one of which is not �� belong to f��� �� �g andnone of the elements of A appears with coe�cients � and �� in anyof the sums Call A a �free� set if n divides none of the above sums

a� Construct a �free� set having �n

�� elements �x� is the least in�

teger number less than or equal to x�

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b� Prove that no set of �n

�� � � elements is �free�

Solution� a� The set A � f�� �� � � � � ��n�� � �g is �free� and has

�n

�� elements prove it��

b� Let n � �m and suppose that A f�� �� � � � � ng is a �free� sethaving �

n

��� � � m�� elements Without loss of generality assume

that A f�� �� � � � � �mg since we can replace x � A by n� x� Weshall show that there exist two elements of A whose sum is equalto another element of A Indeed� let a� � a� � � am�� be theelements of A and consider the set B � fa� � ai � i � �� �� � � � �m��g There are �m � � integer numbers a�� a�� � � � � am��� �a�� a� �a�� � � � � a��am�� fromA�B and they lie in the interval �a�� a��am����which contains exactly am�� � � � �m � � integer numbers Thisgives ai � a� � aj for some i� j But then a� � aj � ai � �� whichis impossible The case n � �m � � is settled in a similar fashionNotice that �m� � cannot be an element of a �free� set

Problem ���� Find the least natural number a such that the

equation cos� �a�x��� cos�a�x��cos��x

�acos�x

�a��

���� � �

has a root

Solution� The roots of the our equations are the common roots of

cos �x� a� � � and cos��x

�a cos�x

�a��

�� � � � � The roots of

the �rst one are x � a� �n� n � ������ � � � and the roots of the

second one are x � �ak� �

��� k � ���� � � � � Therefore a �

�n

�k � �

Page 153: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

for some integer numbers n and k It is easy to see now that theleast natural number with the required property is a � �

Problem ���� Point F lies on the base AB of a trapezoid ABCDand is such that DF � CF Let E be the intersecting point of ACand BD and O� and O� are circumcentres of ADF and FBC respec�tively Prove that the straight lines FE and O�O� are orthogonal

Solution� Let k� and k� be circles with centres O� and O� and letthe intersecting points of the two circles be points P and Q It is wellknown that PQ � O�O� On the other hand� if L is an arbitrarypoint and two lines through L intersect k� and k� in points A� Band C� D respectively Then L � PQ if and only if LA LB �LC LD Let k� and k� be the circumscribed circles of �AFDand �FBC and let G be the intersecting point of FE with CDDenote by C� and D� those points on DC for which AD�jjCF andBC�jjDF � i e� such that the quadrilaterals AFCD� and BFDC�

are parallelograms Using that FD � FC we get � CFB � � FCD �� FDC � ���� � � BC�C This means that F � B� C and C� lie ona circle and so the line DC intersects k� in C and C� By analogy� AFD � � FDC � � FCD � � AD�D and so line DC meets k�in points D and D� In accordance with the initial notes FE isperpendicular to OO� if and only if GC GC� � GD GD� It followsfrom �GCE � �FAE� �GDE � �FBE and �DCE � �BAEthat

GC

AF�

CE

EA�

DC

ABand

GD

BF�

DE

EB�

DC

AB Thus GC �

DC AFAB

and GD �BF DC

AB On the other hand GC� � jDC��

DGj � jBF�DGj � BF j��DGBF

j � BF j��DCAB

j � BF

ABjAB�DCj�

Page 154: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

GD� � jCD��CGj � jAF �CGj � AF j�� CG

AFj � AF j��DC

ABj �

AF

ABjAB�DCj ThereforeGC GC� �

DC AF BFAB�

jAB�DCj �GD GD�

Problem ���� Find all natural numbers n for which a convex n�gon can be partitioned into triangles through its diagonals in such away that there is an even number of diagonals from each vertex Ifthere is a vertex with no digonals through it� assume that there isan even number zero� of diagonals from this vertex�

Solution� It is easy to see by induction that if an n�gon is par�titioned into triangles through d non�intersecting diagonals thenn � d � � Let n be a natural number and A�A� An is a convexn�gon which can be partitioned into triangles through d diagonalsin a way that there is an even number of diagonals through eachvertex Since n � � is a solution we may assume that n � �

It is clear that at least one side of each triangle is a diagonal of then�gon We say that a triangle is of type tk� k � �� �� �� if exactly k ofits sides are diagonals Denote by xk the number of triangles of typetk It is easy to see that �x��x� � n � d�� abd x���x���x� � �dIt follows now that x� � x� � �� so x� � � Therefore there existsa triangle two of whose sides are sides of the n�gon Let that beAj��AjAj�� Diagonal Aj��Aj�� is a side of another triangle�e g�Aj��Aj��As Assume that Aj��As or Aj��As is a side of the n�gonIf it is Aj��As then s � j � � It follows now that there are nodiagonals from Aj�� distinct from Aj��Aj�� because such a diagonalintersects Aj��As This contradicts the premise that there is an even

��

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number of diagonals from each vertex Therefore both Aj��As andAj��As are diagonals so Aj��Aj��As is of type t� Hence there is atriangle of type t� adjacent to each triangle of type t� If distincttriangles of type t� are adjacent to distinct triangles of type t� thenx� � x� � x� � � � x�� a contradiction Therefore there are atleast two triangles of type t� adjacent to one and the same triangleof type t� Without loss of generality assume these are the trianglesA�AnAn�� and An��An��An�� Consider the polygon A�A� An��Obviously the diagonals partition this polygon into triangles andthere is an even number of diagonals through each vertex

Conversely� if the polygon A�A� An�� can be partitioned in therequired way� then adding the verticesAn��� An��� An and diagonalsAn��An�� and A�An�� shows that the same is true for the polygonA�A� An

Therefore a natural number n � � is a solution if and only ifn� � is a solution It is easy to see that n � � is a solution� whereasn � � and n � � are not Thus all natural numbers satisfying theconditions of the problem are n � �k� k � �� �� � � �

Problem ����� For any real number b denote by fb� the maximal

value of j sinx� �

� � sinx� bj Find the minimal value of fb�

Solution� Substitute t � sinx and gt� � t ��

� � t� b Since

gt� is an increasing function in the interval ���� ��� it follows thatfb� � maxjg���j� jg��j� � maxjbj� jb� �

�j� Now from the graph

of the function fb� we conclude that minfb� � f��

�� �

��

Page 156: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem ����� A convex quadrilateralABCD is such that � DAB �� ABC � � BCD Let H and O be respectively the orthocentre andthe circumcentre of �ABC Prove that H� O and D lie on a line

Solution� Let � CAB � �� � ABC � � � BCA � Note that� � and � There are three cases to consider� � ���� � ��� and � ��� Suppose �rst that � ��� Then O andH are interior points for �ABC and � ACO � � CAO � � HCB �� HAB � ��� � Therefore O is an interior point for �HAC and� HAO � � � � ACD� � HCO � � � � � CAD� � HAD �� HCD � � � ��� It follows from the Sine Theorem for �AHD�

�CHD and �ACD thatsin � AHD

sin � HAD�

AD

HD�sin � HCD

sin � CHD�

HD

CD�

sin � CAD

sin � ACD�CD

AD By multiplying the above equalities we get

sin � AHDsin � HCOsin � CAO � sin � CHDsin � HAOsin � ACO�

It follows now from Ceva�s Theorem that AO� CO and HD intersectin a point and so H� O and D lie on a line In the case of � ���

we obtain that H � B� O is the midpoint of AC and AHCD is arectangle Therefore H� O and D lie on a line Finally� let � ���In this case B and O are interior points for �AHC and �ADCrespectively Similarly to the case � ��� we get that the points H�O and D lie on a line

Problem ����� For any natural number n � � denote by mn�the maximum number of points which can be placed inside or onthe outline of a regular n�gon with side � in a way that the distancebetween any two of them is greater than � Find all n for whichmn� � n � �

��

Page 157: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� We prove �rst that mn� � n � � for n � �� �� � Let nbe one of the above numbers and let B�� B�� � Bn be points satis�fying the conditions of the problem for the regular n�gon A�A� An

of side � and centre O It is obvious that OBi � � and thereforeno three points O� Bi� Bj� � � i �� j � n lie on a line Further�

more at least one of the angles OBiBj is less than��

n� ��� and

it follows from the Cosine Theorem that BiB�

j � OB�

i � OB�

j ��OBiOBj cos

��

n� B �

iB�j

�where B �

i and B �j are points on the seg�

ments OA� and OA� such that OB�i � OBi� B �

j � OBj Whenn � �� �� � the greatest side in �OA�A� is A�A� � � and thereforeBiBj � B�

iB�j � �� which is a contradiction Thus mn� � n � �

if n � �� �� � It is easily seen that for these n there exist n � �points on the outline of a regular n�gon A�A� An with the re�quired property Therefore mn� � n � � if n � �� �� � We shallprove now that if n � then mn� � n Let Bi � AiAi�� be points

such that AiBi � �i���� � � i � n � � where � � � ��

�n��is arbi�

trary chosen From n � we obtain cos � A�A�A� � � cos��

n� ��

��

It follows now that B�B�� � ����������������� � ����� � ��

Similarly BiBi�� � � when � � i � n � � Further� it is clearthat B�Bn�� � A�An � � Since OA� � OA� � � OAn � ��AiAj � � when ji � jj � � and Bi � Ai when � � �� it followsthat we can make � so small that OBi � � when � � i � n� � andBiBj � � when ji � jj � � Then the points B�� B�� � Bn��� O

satisfy the conditions of the problem and so mn� � n when n � Since it is obvious that m�� � �� we come to the conclusion thatmn� � n � � only when n � �� �� �

��

Page 158: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Spring mathematics

tournament � ����

Problem ����� Find all values of the real parameter a such that

the inequalities jx � �j � j� � xj � a and�a� �

�x� �a� �� ��

�are

equivalent�

Solution� We begin with the rst inequality� jx��j� j��xj � a�

If x � �� it is equivalent to �x� �� x� � � a or x �� � a

� i� e�

�� a

�� x � �� which has a solution if

� � a

�� �� i� e� if a � ��

If x � ��� �� then jx � �j � j� � xj � x � � � � � x � � and theabove equation has a solution only when a � �� Finally if x � �

then x � � � x � � � a or x �a� �

� i� e� � � x �

a� �

� which

has a solution when a � �� Therefore when a � � the inequalityhas no solution and when a � � the solutions form the interval��� a

� � � a

��

Page 159: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Let us rewrite the second inequality in the form�x � �a� ��

�x� �a� ��

�� Its solutions form either the interval���� �a

� �a� �

�or the

interval��a� �

� �� � �a

� depending on which of the two numbers

�a� �

�and

�� � �a

�is greater�

Therefore the two inequalities are equivalent if

� � a

��

�� � �a

��

� � a

��

�a� �

or�� a

��

�a� �

��

� � a

��

�� � �a

��

The rst pair is satised by a � � the second one by a � �� But ifa � � the rst inequality has no solution whereas the second onedoes� Thus the only solution is a � ��

Problem ����� Let M and N be the midpoints of the sides BCand AC and BH� �H � BC� the altitude in �ABC� The straightline perpendicular to the bisector of � HMN intersects the line AC

in point P such that HP ��

��AB �BC� and � HMN � ��deg�

a�� Prove that �ABC is isosceles�

b�� Find the area of �ABC if HM � ��

Solution� a� Since HM is a median to the hypothenuse BC ofthe right triangle �BHC and MN is a middle segment in �ABCit follows that HM �

�BC �if H � C again HM �

�BC� and

Page 160: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

MN ��

�AB� Therefore HP � HM �MN � Denote � HMN � ��

There are two cases to consider for the points H�N and P � �� N

lies between H and P �� H lies between N and P �

�� Let us nd a point K

on the extension of HM suchthat MK � MN � ThenHP � HK and � HKP �

��deg���� The statement

of the problem implies thatMP is the bisector of � NMK

�external to �HMN� i� e�� NMP � � KMP � There�fore �PNM �� �PKM

and � PNM � � PKM �

��deg���� On the other hand

� PNM � ��deg �� i� e�

�� deg���� �� deg ��

whence � � ��deg and � HNM � ��� deg� Also � MHB � ��degand since HM � MB it follows that � HMB � ��deg� Thus� BMN � ��� deg � � ANM i� e� ABMN is a isosceles trapezoidAN � BM � Therefore AC � BC�

Page 161: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�� Let H lie between N

and P and let K be a pointsuch that MK � MN andHP � HK� It follows fromthe isoscelesness of �KHP

that � HPK � � HKP ��

��

Since MP is the bisectorof � NMK in the isosceles�NMK it follows thatMP is the axis of symmetryof NK� Thus NP � KP

and �NMP �� �KMP �

Therefore � PNM ��

�and

since �NMP it follows that

� ��

�� �� deg � ��� deg�

and hence � � ��deg� Therefore MH � AC and since BH � ACwe get that H coincides with C and � BCA � ��deg � ABC �� BAC � ��deg and so AC � BC�

b� �� When � � ��deg � � ACB � ��deg HM � HB � �

AC � BC � �� Then SABC ��

�AC BH � ��

�� When � � ��deg � AC � BC AC � BC � �BH � ��

Then SABC ��

�AC BC � ��

Problem ���� Problem for the UBM award�

Is it possible to nd ��� straight lines in the plane such that there

Page 162: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

are exactly ���� intersecting points�

Solution� Consider �� lines such that �� of them are parallel andthe remaining �� lines pass through a single point and intersect all�� parallel lines� Then the total number of intersecting points is�� �� � � � ����� Choose the last line in such a way that itintersects all lines and does not pass through any of the points� Nowthere are ���� � �� � ���� intersecting points�

Problem ���� The graph of a linear function is parallel to the

graph of y ��

�x�

��

� passing through M��� ���� and intersects

the coordinate axes Ox and Oy in A and B correspondingly�

�a� Find the coordinates of A and B�

�b� Consider the unity grid in the plane� Find the number ofsquares containing points of AB �in their interiors��

Solution� �a� The graph of a linear function is parallel to the

graph of y ��

�x �

��

�when the linear function is of the form y �

�x�b� From the condition thatM belongs to its graph we determine

��� � ��

�� b whence b � ���

�� The coordiantes of A and B are

respectively the solutions of the systems

������y �

�x� ��

�y � �

and

������y �

�x� ��

�x � �

Page 163: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

whence we get A��� �� and B�� ���

��

�b� The coordinates of the points on AB satisfy the conditions

���

������y �

�x� ��

�� � x � ��

Find the number of points with integer coordinates lying on thesegment AB i� e� the number of integer solutions of ���� It follows

from y ��

�x � ��

�that y � x � �� �

x� �

�and if x and y are

integer thenx� �

�� t is integer� Conversely if t is integer then

x � �t � � and y � �t � �� are integer� Since � � x � �� we nd

� � �t�� � �� i� e� ��

�� t � �� This observation implies that the

points with integer coordinates lying on AB satisfy the conditionsx � �t � � y � �t� �� for t � �� �� �� �� �� Therefore there are �such points�

Further� the segment AB exits a square and enters another oneonly when it intersects a line of the grid� The number of such inter�sections is �� � �� � ��� But when AB passes through a knot �i� e�through a point with integer coordinates� the passage from a squareto another one involves crossing two lines� This happens � times �notcounting A�� Therefore the required number is �� � � � � � ���

Problem ���� Let l� and l� be the loci of the centroid G and theincentre I of the right triangle ABC whose hypotenuse AB is a givensegment of length c�

�a� Find l� and l��

Page 164: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�b� Find the area of �ABC when the length of GI is minimal�

Solution� �a� The vertex C of �ABC could be placed in eithersemiplane with respect to AB� The centroid G lies on a circle k�of radius

c

�and centred at the midpoint O of segment AB� The

incentre I lies on one of the two arcs which are the locus of thepoints X such that � AXB � ��� deg� One of the arcs is part of acircle k�

�with centre Q� and radius R and the other one is part of a

circle k���with centre Q�� and the same radius and Q� and Q�� lie in

di�erent semiplanes with respect to AB�

Conversely� Let G be a point on k� distinct from the intersectingpoints M and N of k� and AB� Then the vertex C of �ABC is

uniquely determined by CO �c

�� Since AO � BO � CO it follows

that �ABC is a right triangle� Choose that of the points Q� and Q��

which does not lie in the same semiplane as C does with respect toAB� Without loss of generality this is Q�� Denote the intersectingpoint of the bisector of � BAC and k�

�by I� We shall prove that

I is the incentre �ABC� Since � ABI � ��deg�� BAI we get� � ABI � ��deg�� � BAI � � ABC�

Therefore BI is the bisector of � ABC�

Let I be a point of k��such that � AIB � ��� deg� Then the

vertex C is uniquely determined by� AI is the bisector of � BACBI is the bisector of � ABC and C lies in one and the same semiplanewith I with respect to AB� Since � BAI� � ABI � ��deg it follows

that �ABC is a right triangle� Therefore CO �c

�and hence the

intersecting point G of CO and k� is the centroid of �ABC�These observations imply that l� is the circle k� without M and

Page 165: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

N and l� consists of two arcs k��and k��

�such that if X � k�

�or

X � k��� then � AXB � ��� deg�

�b� We shall show that R ��

�c and it will follow �how�� that

k� lies in the interior of both k��and k��

�� Let P be the intersecting

point of OQ� and k� which lies between O and Q��

Since PQ� �c

��

c

��

c

��

OP � ON and �PNO it fol�lows that OP � ON � PN

and so PQ� � PN � Hence� PNQ� � � PQ�N � Therefore� PNQ� � ��deg � � PQ�N ���deg so � ANQ� � � AQ�N �

It follows for �AQ�N that AQ� � AN i� e�R ��

�c� Let G � l� and

I � l� be such that Q� O G and I lie in a line in this order� Weshall show that GI is the least possible segment�

Case �� G� � l� I� � l� andO lie on a line� Then G�I� �

GI� Indeed it follows from�Q�I�O that OQ� � OI� �

Q�I� i� e� OQ��OG��G�I� �

R � OQ� � OG � GI and soG�I� � GI�

Page 166: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Case �� G� � l� I� � l�and O are not on a line� LetOI� intersect k� in a point G��Then G�I� � G�I�� Indeedit follows from �OG�I� thatOG��G�I� � OI�� But OI� �OG��G�I� � OG��G�I� andtherefore G�I� � G�I�� Butfrom case � we get that G�I� �

GI and so G�I� � GI�

Thus �ABC having the required property is such that Q� O Gand I lie on a straight line� SimilarlyQ�� determines a triangle equalto the rst one� Since OQ� � AB �ABC is an isosceles triangle

and so SABC �c�

��

Problem ��� Problem for the UBM award�

Given n points on a circle such that no three chords with endsin the given points intersect in a point� Prove that there exists n

such that there aren� � �n� �

�chords with ends in the given points

partitioning the interior of the circle into ���� regions�

Solution� First we shall prove the following Lemma� The numberof regions into which the interior of a circle is divided by drawing all�n

�chords with ends in n given points provided no three chords

intersect in a point is

�n

��

�n

�� ��

Page 167: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Use induction by n� It is easy to see that the above formula holdsfor n � �� Suppose it is true for some n� To obtain the result wecount how many new regions are added when a new point appears onthe circle� It is easily seen that if a chord intersects t other chordsthen it �adds� t� � new regions� Therefore the new regions are�

n��Xk��

�k�n� k � �� � ���

Given that ���� �n�� ��n� ��n

�and ������ ��n���� �

�n� ��n��n � ��

� we easily obtain that the above sum equals

�n � �

��

�n� �

�� � �

��n

��

�n

�� �

��

which completes the proof�

We show now that if n � �� it is possible to drawn� � �n � �

��

��� chords such that there are ���� regions� Let us draw all chordswith ends in �� of the given points �there are ��� such chords��

It follows then that the interior of the circle is divided into

���

���

��

�� � � ���� regions� Draw a chord connecting the ��th point

with one of the rst �� in a way that there are � and � points on thetwo sides of the drawn chord� This chord intersects � � � �� chordsand therefore there are �� new regions� Therefore the total numberof regions is ���� � �� � �����

��

Page 168: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem ���� Find all parameters a such that the inequality jax���x� �j � �� �x holds for any x � ��� ���

Solution� Observe that the inequality is equivalent to the system����� ax� � � � �

ax� � �x � � ��

It follows from the second inequality that a � � because if a � �then ax� � �x � � � is not true for x � �� Further the second

inequality gives x� � �

a� Since this inequality is true for x � ��� ��

we get � � �

a� a � �� Let D � � � a be the discriminant of

ax� � �x� �� There are two cases to consider�

�� D � �� The solution of the second inequality isx � ��� x�� �x� ��� where x� � x� are the roots of ax���x�� ��� Therefore ��� �� � ��� x�� or ��� �� � �x� ����

���� ��� �� � ��� x��� It follows from the above that

��������a����� � ����� � � �a �� � � � � � �x� � x�

�� �

��

���������

a ��a ��

a� �

���������a ��a �a � �

which is impossible�

���� ��� �� � �x� ���� Therefore

��������a����� � ����� � � �a��� � ��� � � �x� � x�

�� ��

��

���������

a ��a ��

a� ��

��

Page 169: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

which is also impossible�

�� D � � � the inequality ax�� �x� � � holds for anyreal value of x� It then follows that ��a � � �� a �� Thereforethe solution is a � ��

Problem ���� The quadrangle ABCD is inscribed in a circle� Thetangents to the circle passing through A and C intersect at thepoint P � If PA� � PD PB and P does not lie on DB provethat the intersecting point of AC and BD is the midpoint of AC�

Solution�

Let E be the second in�tersecting point of PB andthe circle� Then PA� �PE PB � PD � PE�Hence � APD � � EPC� �IfO is the centre of the cir�cle the above follows fromthe similarity of �ODPand �OEP �� Therefore�ADP � �PCB be�cause � APD � � BPC andAP

BP�

DP

CP so

AD

BC�

AP

BP� ���

Also �APB � �DCP because � APB � � DPC � � APC �� APD and

AP

DP�BP

CPand so

AB

DC�BP

CP� ���

��

Page 170: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

From ��� and ��� we getAB ADBC DC �

AP BPBP CP � �� On the

other hand

SABD

SCBD�

�AB AD sin � DAB

�BC DC sin � DCB

�AB ADBC DC � �

�� DAB � ��� deg� � DCB��Therefore SABD � SCBD i� e� the diagonal BD halves AC�

Problem ��� Problem for the UBM award�

See problem ����

Problem ����� Find all values of the real parameter a such thatthe inequality x� � ax� � �a � ��x� � ax � � � � holds for all realvalues of x�

Solution� If x � � the inequality holds for any a� Suppose x �� ��Dividing both sides of the inequality by x� gives x� � ax� a� � �a

x�

x�� � or

�x �

x

��

� a�x �

x

�� a � � � �� Substitute

t � x��

x� If x � � it is true that t � and if x � � it is true that

t � ��� Therefore we want to nd all a such that the inequalityt� � at � a � � � � holds for any t � ��� ��� �� ���� Denotef�t� � t� � at� a� �� The discriminant of f�t� is D � a� � �a� ��

Let D � � i� e� a � �� � �p� � � �

p��� Then the inequality

f�t� � � holds for any real t in particular for t � ��� ��� �� ����

Let D � i� e� a � ��� � � p�� �� �p� ���� Then the

inequality f�t� � � holds for any t � ��� ��� �� ��� if and

��

Page 171: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

only if ���������� � �a

�� �

f��� � �f���� � �

The solutions of this system when a � ��� ��p�� ���p� ���are

a ����

� �� �

p�i� So a �

���

� � � �

p���

Problem ����� A quadrangle with perpendicular diagonals ACand BD is inscribed in a circle with centre O and radius ��

a�� Calculate the sum of the squares of the sides of the quadrangle�

b�� Find the area of ABCD if a circle with centre I is inscribed in

it and OI ��p��

Solution� a�� Denote � BAC � � � CAD � �� It follows fromthe Sine Theorem for �ABC and �ABD that BC � � sin� AD �� sin � ABD � � sin��� deg��� � � cos �� Analogously CD � � sin�AB � � cos �� Therefore AB� � BC� � CD� � DA� � � sin� � �� cos� �� � sin� � � � cos� � � ��

b�� The statement of the problem implies that AB � CD �AD�BC or � cos ��� sin� � � cos ��� sin� i� e� sin�����deg� �sin�����deg�� Therefore � � � or � � ��deg��� We shall consideronly the case � � � �the case � � ��deg�� is settled in a similarfashion�� NowAD � AB CD � BC and � ABC � � ADC � ��deg�The points O and I lie on AC and when � �� deg I lies on

��

Page 172: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

the segment AO� In �ACD �DI is a bisector� determine AI �� cos �

sin� � cos�� Consequently OI � AO �AI �

sin�� cos�

sin� � cos��

�p��

Raising both sides to the �nd power gives� � sin ��

� � sin ���

�� Hence

sin �� ��

�� It then follows that the area of ABCD is equal to

S � �SABC � AB BC � � cos� sin� � � sin �� � ��

Problem ���� Problem for the Atanas Radev award�

Find all natural numbers n such that� If a and b are naturalnumbers and n divides a�b� � then n divides a� � b�

Solution� Obviously the condition holds for n � � and supposen �� Let a be a natural number coprime to n� It follows fromBezout�s Theorem that there exists a natural number b such thata�b�� is divisible by n� Further n divides a�� b and since a�� � �a��a� � b�� �a�b� �� it follows that n divides a� � ��

Let n � ��k where k is an odd number and � �� Supposek �� Then n and k � � are coprime and therefore ��k divides�k � ��� � � so k divides ��� Hence n is of the form n � �� �� ��where � �� � � �� � �� Consequently n divides ��� � � and weconclude that � � ��

It is easy to see now that n divides �� � � � ����

Conversely let n be a divisor of ���� Then n satises the con�dition of the problem� Indeed if � divides a�b � � then � dividesboth a� � � and a� � � and therefore � divides a� � b� Similarly if� divides a�b � � then � divides a� � b� If we can show that the

��

Page 173: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

same property holds for �� �� � and �� we will be done� Assume �k

divides a�b� � where � � k � �� Then a is an odd integer numberand therefore a�� � is divisible by �� Consequently b�� is divisibleby �k and thus a� � b is divisible by �k as well� Assume �� dividesa�b��� Then a is an odd integer number and it is easy to verify thata� is congruent to � or � modulo ��� Further b should be congruentto �� or � modulo �� correspondingly and again a� � b is divisibleby ���

The required numbers are all divisors of ����

Problem �����

a�� Let p be a positive real parameter� Find the least values of

the functions f�x� � x�p

xand g�x� � x �

p

x�in the interval

�� ����

b�� Let a�� a�� a� be positive real numbers� Prove that ��a� �pa�a� � �

pa�a�a�� � ��a� � a� � a���

Solution� a�� Since f ��x� � �� p

x��

�x�pp��x�pp�x�

we get

that the function f�x� decreases in the interval �� pp� and increases

in the interval �pp ���� Therefore the minimal value of f�x� in

�� ��� equals f�pp� � �

pp� Analogously it follows from g��x� �

���p

x�that the least value of g�x� in �� ��� equals g� �

q�p� �

� �p�p

��

b�� After substituting x � a�� y �pa�a�� z � �

pa�a�a� our

inequality becomes � � �

�x�

y�

x� y�

z�

y�� z� It follows from a��

��

Page 174: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

that�

�y � �

�x �

y�

xand

�y �

z�

y� z� Therefore

�x �

y�

x�

y ��

z�

y�� z �

�y �

z�

y�� z � which completes the proof� Note

that equality occurs when x � �y � �z i� e� a� � �a� � ��a�� Notefurther that b�� could also be solved by the following inequalities�

a� �pa�a� � �

pa�a�a� � a� �

pa��a� �

��pa��a���a� �

� a� ��

��a� � �a�� �

���a� � �a� � ��a�� �

��a� � a� � a���

Problem ����� Let I and r are the incentre and inradius of�ABCand N is the midpoint of the median through C� Prove that ifr � CN � IN then AC � BC or � ACB � ��deg �

Solution� Use the standart notation for the elements of �ABCand apply the formula for a median in a triangle� Since IN andIM are medians in �CIM and �AIB respectively we get IN� ��

���CI���MI��CM�� � �

���CI��AI��BI�� �

�AB��CM���

Therefore

CN� � IN� ��

���CM� �

�AB� � �CI� �AI� �BI�� �

��a��b����p�c����r���p�a���r���p�b���r�� � �p � c�c

��r��

It follows from the statement of the problem that IN� � �CN �r�� so CN� � IN� � r� � �CN r� It follows from what we haveproved above that �CM r � �p � c�c� Taking the square of both

��

Page 175: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

sides of this equality and using the formul� �CM� � �a� � �b� � c�

and r� ��p� a��p� b��p� c�

p we get ��a���b��c���p�a��p�b� �

p�p�c�c�� After some simple calculations the above equality becomes�a� � b� � c���a � b�� � �� Therefore a � b i� e� AC � BC ora� � b� � c� i� e� � ACB � ��deg�

Problem ���� Problem for the Atanas Radev award�

See problem �����

��

Page 176: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Spring mathematics

tournament�Kazanl�ak�

�� March�� April �

Problem ���� Given an inequality jx� �j � ax� where a is a realparameter�

a� Solve the inequality�

b� Find all values of a such that the inequality has exactly twointeger solutions�

Chavdar Lozanov� Kiril Bankov� Teodosi Vitanov

Solution� a� I� Let x � �� Then the inequality is equivalent tox� � � ax �� ��� a�x � ��

�� � � a � �� a � � � x ��

�� a�

� � a� � �� a � ��

Therefore � � a � �� � � x ��

� � a�

Page 177: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

� �� a � � a � � ��x � �� Therefore a �� x � ��

�� �� a � � � � � a � x � �� Therefore � � a� x � ��

II� Let x � �� Then the inequality is equivalent to � � x �ax �� � � �a� ��x�

�� a�� � � �� a � ��� �

a� �� x � ��

a� �� � �� a � ��

Therefore a � ���

a� �� x � �� when �� � a � � no solution

exists�

� a� � � �� a �� � � � � � x� so no solution exists�

�� a � � � � �� a � �� � x ��

a� �� �� Therefore

a � ��� x � �

a� ��

So when a � ��� then x � �a��

for �� � a � � no solution

exists when � � a � �� then �a�� � x � �

��a when � � a� then� � x�

b� It follows from a� that the inequality could have two integer

solutions only if � � a � �� Since in this case � ��

a� �� � �

�� a�

we �nd that there are exactly two integer solutions if and only if

��

�� a� ��

Therefore the answer is�

� a �

��

Page 178: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem ���� Let M be the midpoint of the side BC of �ABCand � CAB ��� � ABC ����

a� Find � AMC�

b� Prove that AM AB �BCAC

Chavdar Lozanov

Solution� a� Draw CH AB� Now � ACH ��� � CAH and� HCB ���� For �ACH it is true that AH HC� Further it

follows from �CHB that CH �

CB HM � Therefore AH

HM � so � MAH � AMH ���� � � AHM

� Note that � CHM

��� and � AHM ��� � ��� ����� Therefore � AMH ���� Weobtain that � AMC � HMC � � AMH ��� � ��� ����

b� Let S be the area of�ABC� We know that S AB � CH

AB � CB�

� Since AM is a median� it follows that SAMC S

� If

CP AM � thenS

AM � CP

� But � CAM � CAB� � MAB

��� � ��� ���� and thus CP AC

� Therefore

S

AM �AC�

NowAB � CB

�AM �AC

� and we obtain AM

AB �BCAC

Problem ���� Consider all points in the plane whose coordinates�x� y� in an orthogonal coordinate system are integer numbers and� � x � ��� � � y � �� Each point is painted green� red or blue�

Page 179: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Prove that there exists a rectangle with sides parallel to the coordi�nate axes whose vertices are all of the same colour�

Kiril Bankov

Solution� Since the number of coloured points is � � �� �� andthere are three di�erent colours� it follows that there are at least �points of the same colour �say blue�� Denote by p�� p�� � � � � p�� thelines parallel to the ordinate axis and passing through the points��� ��� �� ��� � � � � ���� ��� respectively� Let n�� n�� � � � � n�� be the num�ber of blue points on the lines p�� p�� � � � � p��� It is clear that � �ni � � for i �� � � � � � ��� Without loss of generality assumen� � n� � � � � � n��� Since n� � n� � � � � � n�� �� we obtainthat n� � �

�� Let n� �� Then the remaining blue points lie on �� linesand so n� � � Therefore there are two blue points on each ofthe lines p� and p� all having the same ordinates� These pointsform the required rectangle�

� Let n� �� Then the remaining � blue points lie on �� linesand therefore n� � �

a� If n� �� then there is a blue rectangle with vertices onp� and p��

b� Let n� � Then n� n� n� n� � The numberof ways of choosing two blue points on p� is � and thecorresponding number for each of p�� p�� � � � � p� is �� Thetotal number of blue pairs is ��� �� �� which is greaterthat ��the number of ways of choosing two horizontallines out of � lines� Therefore there exists a blue rectangle�

Page 180: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�� Let n� � Then n� n� � � � n � Since � � �� weapply the same reasoning as in b��

Problem ���� Find all rational numbers a such that j�a � j � �

and A �a� �

�a�is integer� Ivan Tonov

Solution� It follows from j�a� j � � that�

�� a � �

�� Also� it is

clear that when a � �

�� then A � � and A � only if a

�� Let

k be a positive integer such that A k� Then �a� � �al � l ��

where l �

k� Multiply the above equality by � and write it in the

following way�

��a� � ��a� � � � ��a� � �al � �l � � � ��

��a� � ��� � ��a� ��� � �a�l � �� � ��l � �� ��

so ��a� � ��� � ��a � ��� � ��l � ���� � �a� �� Therefore ��l �������a� � �� which is possible �recall a �

�� only if l � � or k � ��

But since k is a positive integer� it follows that k � and a �

��

Therefore the required values are a �

�and a

��

Problem ���� Given a �ABC� Let M be the midpoint of AB�� CAB ��� and � ABC ����

a� Find � ACM �

Page 181: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

b� Prove that CM AB �BCAC

Chavdar Lozanov

Solution� a� Let AH BC� Then � HAB ��� and AH AB

HM � It follows from � HAC � HAB � � CAB ���

that AH HC� Thus HM HC and � HCM ���� � � MHC

But � MHC � AHB � � AHM ��� � ��� ���� Therefore� HCM ���� so � ACM � HCM � � HCA ��� � ��� ����

b� Let S be the area of�ABC� We know that S BC �AH

BC �AB�

� Since CM is a median� we obtain SACM S

� If MP

AC� thenS

AC � PM

� It follows from �PMC that PM

MC�� PCM ����� Therefore

S

AC �MC

�� so

AC �MC

BC �AB�

� which impliesMC AB �BCAC

Problem ���� Given n points on a circle denoted consecutivelyby A�� A�� � � � � An �n � ��� Initially � is written at A� and � at allremaining points� The following operation is allowed� choose a pointAi where a � is written and replace the numbers a� b and c writtenat the points Ai��� Ai and Ai�� by ��a� ��b and ��c� respectively��Here A means An and An�� means A���

a� If n ����� is it possible to have a � in all points after per�forming the described operation a �nite number of times�

Page 182: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

b� Find all values of n such that it is not possible to have a � inall points after �nite number of operations�

Kiril Bankov

Solution� a� After performing the transformation from the con�ditions of the problem consecutively for the pointsA�� A�� A�� � � � � An���we have the following distribution of �s and �s�

���A� A� A� � � � An�� An�� An�� An

� � � � � � � � � �

If n ����� then arrange the obtained ���� ones in ��� groups ofthree �s and then perform the operation on each of the groups� Weobtain a zero in every point�

b� If n �k � � we can repeat the steps from a� and we getonly zeroes in the points�

If n �k�� then starting from ��� and performing the operationwith An�� one obtains�

A� A� A� � � � An�� An�� An

� � � � � � � � �

There are �k ones which can be arranged in k groups of � ones eachand again to obtain anly zeroes�

We shall prove now that if n �k� it is not possible to have onlyzeroes after a �nite number of operations� Assume the opposite� i� e��that after a �nite number of operations we have only zeroes� Denotethe number of operations performed with the point Ai by ai� Sinceeach operation changes the number of ones by an odd number ��or ��� it follows that the sum S a�� a�� � � �� an of all operations

Page 183: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

is an odd number� On the other hand S �a� � a� � a�� � �a� �a�� a��� � � �� �a�k��� a�k��� a�k�� Note that a�� a�� a� is equalto the number of changes �from � to � or vice versa� of the numberwritten at A�� Since at the beginning there is a � written at A� anda � again at the end� it follows that a� � a� � a� is an even number�The same applies for a� � a� � a� and so on� Therefore S is a sumof even numbers� a contradiction to the fact that S is odd� Answerto b�� all numbers divisible by ��

Problem ��� It is known that if the real parameter a equals anyof the numbers p � q � r� then at least one of the remaining two isa root of the equation

x� � �� a�x� a� � a� � ��

Prove that a� p � ��

� b� p � ��� Sava Grozdev

Solution� a� It follows from the conditions of the problem that

D � � a�� � ��a� � a � �� � �� so ��

�� a � � and therefore

p � ��

�� When a ��

�� then the equation has an unique root

x �

�� and when a

�� the roots are x ��

�and x

�� Thus in

the case of p ��

�three distinct numbers p� q and r satisfying the

condition do not exist� Therefore p � ��

��

b� Suppose that when a p� then x q is a root� The casex r is treated in the same fashion� Since a and x are symmetric in

Page 184: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

the equality x�� ��a�x�a��a�� �� we get that when a q�then x p is a root� When a r� at least one of the numbers p andq is a root� Let that be p� Now when a p� the equation has x ras a root �because of the symmetry�� We obtain that when a p�the roots are q and r� We conclude now that in all cases p�q�r �It is clear that the roots of x�� ��p�x�p��p�� � are greaterthan p and therefore p� � � � p�p � p� � p � � � �� which gives

p � �� or p ��

�� But �p � p � q � r and p �

�� Therefore

p � ���

Problem ��� Through an interior point K of the non�equilateral�A�A�A� lines Q�P� k A�A�� Q�P� k A�A� and Q�P� k A�A� aredrawn� �Q�� Q�� Q� lie on A�A�� A�A� and A�A�� respectively��Points P�� P�� P�� Q�� Q�� Q� lie on a circle k� Prove that�

a� �P�P�P� �Q�Q�Q� �A�A�A�

b� point K� the centre of k and the circumcentre�A�A�A� lie ona line�

Rumen Kozarev

Solution� a� It follows from the conditions of the problem thatthe arcs �P�Q�� �P�Q� and �P�Q� are equal� Therefore � P�P�P� P�Q�P�

�P�Q�

�Q�P�

�P�Q�

�Q�P�

� Since the quadrilat�

eral A�P�KQ� is a parallelogram we get � A�A�A� � Q�KP� �Q�P�

�Q�P�

� so � P�P�P� � A�A�A�� Similarly� one can show

Page 185: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

that � Q�Q�Q� � A�A�A�� and the same equalities for the remain�ing pairs of angles� Thus �P�P�P� �Q�Q�Q� �A�A�A� andsince the �rst two triangles have the same circumcircle� they areidentical�

b� Since A�P�KQ� is a parallelogram� we obtain � A�P�Q�

� KQ�P� �P�Q�

�P�Q�

� P�P�P� � A�A�A�� Similarly � A�Q�P�

� A�A�A��

Let O be the circumcentre of �A�A�A�� It is easy to see thatOA� P�Q��OA� P�Q� and OA� P�Q�� Denote the midpointof OK by S� If R�� R� and R� are the midpoints of P�Q�� P�Q� andP�Q� then SR� k OA� SR� k OA� SR� k OA� �SR� is a middlesegment in �OKA� � SR� k OA��� Therefore S lies on the axesof symmetry of P�Q� P�Q� and P�Q�� so S is the centre of the circlethrough the points P�� P�� P�� Q�� Q�� Q��

Problem ��� Find all polynomials f�x� xn � an��xn�� � � � ��

a�x � a� a � � with integer coe�cients such that f�ai� �� i �� � � � � n� �� Sava Grozdev

Solution� It is clear that n � �� Since f�x� �x � a��x �a�� � � � �x � an���� it follows from f��� a ����na � � � an�� thatjaij � for i �� � � � � � n� ��

First case� jaj �� Now f�x� �x� ��p�x� ��q� p� q n � ��We know that �x� ��p�x���q �xp� pxp��� � � ���xq� qxq��� � � ��and comparing the coe�cients in front of xn�� and xn�� we obtain����� q � p an�� ��

q�q���� � p�p���

� � pq an�� �� �

��

Page 186: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

It is easy to �nd now that p � q � and so p �� q orp � q �� In the �rst case we get that f�x� x� � x� � x � ��which is a solution and in the second one f�x� x� � x� � x � ��which is not�

Second case� jaj � � Now � f�a� jan � an��an�� � � � � �

a�a�aj � jajn�jajn���� � � jaj��jaj�jaj ja�j�ja�j����ja�jn�����ja�j�� �

�� Therefore jaj � Moreover an��an�� � � � � � a�a� a have the same

negativity�the opposite to those of an �

We conclude that a �� n is an even number and ai ����i��

for i �� � � � � � n � �� If n � � then a� �� and � f���� ����n � ����n����n�� � � � � � ��������� � �n� � �� whichis impossible� Therefore n and f�x� x� � x � � which is asolution�

Answer�

f�x� x� � x� and f�x� x� � x� � x� ��

Problem ���� Prove that the inequality

x� � �x � �x� � x�

x� � x� �� � �

holds for any real number x� Rumen Kozarev

Solution� �� Let x � �� Then � �x � � We shall show that�x� � x�

x� � x� �� � The last inequality is equivalent to �x� � x� �

x� �x� �� x�x� �� � � �� x ��� �����

��

�and

therefore it holds for x � ��

��

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� Let x � �� We prove that�x� � x�

x� � x� �� � � �x� Assume

the opposite� i� e���x� � x�

x� � x� �� � �x � �x� � x�

x� � x� �� � �

�� x�x� �� � � �� x ��� �� � ��

���� Since x � ��

we obtain that x ��

��

�� Therefore

�x� � x�

x� � x� �� � �x �

� � �

� � � �� �x� � x � � �x� � �x � �� x� � x � � �� �� x ��� � �

p� � �� �

p ��� � x �� �

p ����

because x � �

� Thus

�x� � x�

x� � x� �� � �x � � ���

p� � � �� ���

Since obviously�x� � x�

x� � x� ��

�x� � �x � �

x� � x� � � for any x � �� we

get a contradiction�

Problem ���� Let M be an interior point in the square ABCD�Denote the second points of intersection of the lines AM � BM � CM �DM with the circumcircle of ABCD by A�� B�� C�� D�� respectively�Prove that

A�B� � C�D� A�D� �B�C��

Emil Kolev

Solution� Since�ABM �A�B�M �BCM �B�C�M �CDM �C�D�M �DAM �D�A�M � we obtain

AB

A�B�

BM

A�M BC

B�C�

BM

C�M CD

C�D�

DM

C�M DA

D�A�

DM

A�M�

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It is easy to see now that

AB

A�B�� CD

C�D�

BM �DMA�M � C�M

BC

B�C�� DA

D�A��

Using that AB BC CD DA it follows from the above thatA�B� � C�D� A�D� �B�C�� Q� E� D�

Problem ���� Consider n points in the plane such that no threelie on a line� What is the least number of segments having their endsin the given points such that for any two points A and B there existsa point C connected to both A and B� Emil Kolev

Solution� Denote the points by A�� A�� � � � � An� Draw segmentsconnecting A� with all remaining points� Also� draw segments A�A��A�A�� � � � � An��An when n is odd and A�A�� A�A�� � � � � An��An���A�An when n is even� It is easy to see that the condition of the

problem is met and that there are��n � �

�segments �dMe denotes

the least natural number which is greater or equal to M��

Suppose it is possible to draw less than��n � �

�segments and

to meet the condition of the problem� Obviously each point is con�nected by a segment with another one� If each point is connected toat least three others� we will have that the number of segments is

at least�n

� which is greater than

��n� �

�� Therefore there exists

a point �let that be A�� connected with at most two others� If A� isconnected to exactly one point �let that be A��� a point connected toboth A� and A� does not exist� Therefore A� is connected to exactlytwo points �let them be A� and A��� It is easy to see that A� and A�

��

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are connected by a segment� Consider the pairs A� and Ai for anyi � �� It is clear that the point connected to both A� and Ai couldbe either A� or A�� In both cases Ai is connected to A� or A�� Sincethere are at least two segments from each point Ai� i � � then thenumber of segments from Ai� i � � is at least �n���� Further� sinceat least n� � from these points connect some point of Ai� i � � withA� or A� �and therefore they are counted once� the total number of

drawn segments is at least � � n� � ��n � �

���n� �

��

This is a contradiction with the number of drawn segments�

Therefore the answer is��n� �

��

Problem ����� Given a function f�x� de�ned for any real x andf�tgx� sin x for x ���

� ��

� �� Find the minimum and the maxi�mum of the function f�sin� x� � f�cos� x��

Oleg Mushkarov� Nikolai Nikolov

Solution� Let t tgx� Then sin x t

� � t�and it follows from

the conditions of the problem that f�t� t

� � t�for any t� Therefore

f�sin� x� � f�cos� x� � sin� x � cos� x�� � sin� x��� � cos� x�

��sinx � cos x��� ��sin x � cos x�� � �sin x � cos x��

��

Page 190: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Let u sinx�cosx �

sin x� Then u ���

��

� and we have to �nd

the minimumand the maximumof the function g�u� �u�

� �u� � u�

in the interval ���

��

�� We obtain

g��u� �u���� u���u� � u� � ��

� � �u� � u���� �

when u ���

��

� and so g�u� is an increasing function in the in�

terval ���

��

�� It follows now that max

u� � �

�� ���g�u� g�

��and

minu� � �

�� ���g�u� g���

� ��

���

Problem ����� A circle is tangent to the circumcircle of �ABCand to the rays

��AB and

��AC at points M and N � respectively� Prove

that the excentre to side BC of �ABC lies on the segment MN �Oleg Mushkarov� Nikolai Nikolov

Solution� Let O be the circumcentre of �ABC and L be the cen�tre of the circle tangent to the circumcircle of �ABC� First we

shall �nd the radius � of this circle� For �OAL we get AL �

sin A�

AO R� OL R� �� � OAL jB � Cj

� From the Cosine Law we

obtain �R � ��� R� ���

sin� A

� R� cos B�C�

sin A

� After simpli�cation

��

Page 191: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

the above becomes

� cos�A

R sin

A

�sin

A

� cos

B � C

� R sin

A

cos

B

cos

C

Since sinA

s�p� b��p� c�

bc� cos

B

sp�p � b�

ac� cos

C

sp�p � c�

aband abc �RS� it follows from the previous equality that � ra

cos� A

� where ra is the exradius to side BC of �ABC� Let I

AL �MN and let T be the projection of I on the line AB� Since

AL MN � it follows that IT AI � IMAM

AI �AM � LMAM �AL

AI � LMAL

LM cos�A

ra� Since I lies on the bisector of � A� we

conclude that I is the excentre to side BC of �ABC�

Problem ����� Given an orthogonal coordinate system with ori�gin O in the plane� Distinct real numbers are written at the pointswith integer coordinates� Let A be a nonempty �nite set of integerpoints which is central�symmetric regarding O and O � A� Provethat there exists an integer point X such that if AX is the image of Aunder translation de�ned by �OX � then at least half of the numberswritten at the points of AX are greater than the number writtenat X� Avgustin Marinov

Solution� Let us denote the number of points in A� which is obvi�ously even� by s� Connect all integer points X with the points fromAX by arrows so that the arrow points to the smaller number� Sup�pose no point X with the required property exists� Then there areat least s�� arrows pointing out of any point X� For every natural

��

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number n denote the square with vertices �n� n�� ��n� n�� ��n��n�and �n��n� by Kn�

Since A is a �nite set� there exists a natural number d such thatA � Kd� For every n denote the number of arrows within the squareKn by Sn� Since there are at least s�� arrows pointing out of everyinteger point of Kn �and these arrows are within the square Kn�d��it follows that �n�����s��� � Sn�d� On the other hand� since AX

is a central�symmetric set� there are at most s�� arrows pointing toevery integer point of Kn�d� Therefore Sn�d � �n�d�����s� ���Thus �n � ����s � �� � �n � d � ����s � ��� so s � � � �� �d

n � ����s� �� for any n� When n�� one obtains s�� � s� ��

a contradiction� Therefore a point X with the required propertydoes exist�

��

Page 193: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

WINTER MATHEMATICAL COMPETITION

����

Grade � � First Group�

Problem �� Prove that for every positive integer n the following proposition holds��The number � is a divisor of �n � n� if and only if � is a divisor of �n�n� � ��Solution� If � is a divisor of n then � is neither a divisor of �n�n� nor a divisor of �n�n����Let � be not a divisor of n� In this case � divides n� � � � �n� � � �n� � � and since � is a

prime number then � divides either n� � � or n� � �� Now the above proposition follows fromthe equalities�

�n�n� � � � �n� � � ��n � � � �n� � �n

and�n�n� � � � �n� � � ��n � � � �n� � �n �

Problem �� Let ABCDE be a convex pentagon and let M P N Q be the midpoints ofthe segments AB BC CD DE respectively� If K and L are the midpoints of the segmentsMN and PQ respectively and the segment AE is of length a �nd the length of the segmentKL�

Figure �� Figure ��

Solution� Let F be the midpoint of the segment AD �Figure � � Then the quadrilateralMPNF is a parallelogram� Hence K is midpoint of the segment FP � It follows from here that

KL ��

��FQ� On the other hand FQ �

��a �because F and Q are midpoints of ED and AD

respectively �

Therefore KL ��

��a�

Page 194: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem �� Every point in the plane is colored either in black or in white� Prove thatthere exists a right angled triangle with hypotenuse of length � and an acute angle of ��� whichvertices are colored in one and the same colour�

Solution� First we shall show that there exist two points which are colored in one and thesame colour and the distance between them is �� Indeed let ABC be a equilateral triangle ofside �� Obviously two of its vertices �say A and B are colored in one and the same colour �e�g�white � Let AXYBZT be a regular hexagon with a big diagonal AB �Figure � �

If one of the vertices X Y Z T �e�g� X is white then the vertices of the triangle ABX�� AXB � ��� � XAB � ��� are colored in one and the same colour� Otherwise the verticesof the triangle XY T � � Y XT � ��� � XY T � ��� are colored in one and the same colour�

Grade ��

Problem �� Let A ��p

�x� � �x� �and B �

�x� �px� � �x� �

� Find all integer values of x

for which the number C ��A�B

�is an integer�

Solution� We have

A ��p

��x� � ��

j�x� �j � B ���x� � p�x� � �

���x� �

jx� �j �

and

C ��

���

j�x� �j �x� �

jx� �j��

�� Let x � �� Then

C ��

���

�x� �� �

��

��x� �

���x� � � �

and

C � � ���x� �

���x� � � � �

�� �x

���x� � � ��

Hence � � C � � i�e� C is not an integer for any x � ��

�� Let ��

�� x � �� Then x � � �because x is an integer and C � �� Thus x � � is a

solution of the problem�

�� Let x � ��

�� Then x � �� �because x is an integer � It is clear that

C ��

�� �

�x� �� �

�� � � �x� �

���x� � � �

and

C � � ��

�� �

�x� �� �

�� � � �� ��x� �

���x� � �

�x� �

� ��x� � � ��

Hence �� � C � � i�e� C � � and x � ���Finally only x � � and x � �� are solutions of the given problem�Problem �� Let M and N be the midpoints of the sides BC and AC of the triangle ABC

�AB �� AC AB �� BC and G be the intersection point of the lines AM and BN � The anglebisectors of � BAC and � ABC intersect BC and AC in the points D and E respectively� Prove

Page 195: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

that the quadrilateral DEMN is inscribed in a circle if and only if there exists a circle inscribedin the quadrilateral CMGN �

Figure �� Figure ��

Solution� Without loss of generality we can suppose that N is between A and E� There aretwo possibilities for the points D and E which are shown in the Figure � and Figure ��

The quadrilateral DEMN is inscribed in a circle if and only if � CNM � � CDE i�e�� BAN � � BDE � ���� �because MNkAB and � CNM � � BAN � Thus the quadrilateralDEMN is inscribed in a circle i� the quadrilateral ABDE is inscribed in a circle� This isequivalent to � DAE � � DBE i�e� to AC � BC�

Therefore we should prove that there exists a circle inscribed in the quadrilateral CMGNif and only if AC � BC�

Let AC � BC� Then CM � CN and since G is the cen� Figure ��

ter of gravity of the triangle ABC we have GM ��

�AM �

�BN � GN � Hence CM �GN � CN �GM i�e� there ex�

ists a circle inscribed in the quadrilateral CMGN �Figure� �

Conversely if there exists a circle inscribed in the quadri�

lateral CMGN then CM � GN � CN �GM and�

�BC �

�BN �

�AC �

�AM � Hence AM � BN �

��BC � AC �

Let K and L be the points of contact of the circle and thesides CM and CN respectively� Obviously CK � CL�

On the other hand this circle is inscribed in the triangles

ACM and BCN � Hence CK ��

��AC �CM �AM CL �

��BC�CN�BN � Thus AM �BN �

��AC�BC � It follows from here that

��BC�AC �

��AC � BC i�e� AC � BC�

Problem �� Thirty points are given in the plane� Some of them are connected with segmentsas it is shown in the Figure �� The points are labeled with di�erent positive integers�

If a is a segment and p and q are the numbers corresponding to its endpoints we denote� �a � jp� qj�

a Construct an example of labeling of the points with the integers �� �� � � ��� in which there

Page 196: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

exists exactly one segment a with � �a � ��b Prove that for every labeling there exists at least one segment a with � �a � ��

Figure �� Figure ��

Solution� a A possible example is shown in the Figure ��b Let the points be labeled with the positive integers m � m� � m� � � � � � m�� � M � It

is clear that M � m� ���Let A and B be the points labeled with m andM respectively and let a� � AC� a� � C�C�

� � � ak�� � Ck��Ck�� ak � Ck��B be the shortest path of segments connecting A and B�It is not di�cult to see that k � �� If we assume that � �ai � � for i � �� �� � � �k thenm� �� �M � m� �k � m� �� which is a contradiction�

Grade ��

Problem �� Let m be a real number such that the roots x� and x� of the equation

f�x � x� � �m� � x�m� � �m� � � �

are real numbers�a Find all values of m for which x�� � x�� � ��b Prove that

� �mx���� x�

�mx���� x�

� � � ���

��

Solution� a Since x� and x� are real numbers then

D �f � �m� � � � ��m� � �m� �

�� ��m� � �m� � � ��

Hence ��

�� m � �� On the other hand

� � x�� � x�� � �x� � x� � � �x�x� � �m� � �m� ���

We obtain from here that m � �� �p�� But

���p� � ��

�� �� �

p� � �

Page 197: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

and therefore only m �p�� � is a solution of the given problem�

b We have

mx���� x�

�mx���� x�

�m�x����� x� � x����� x� �

f��

�x�� � x�� � x�x��x� � x�

m� �

�m� � �m� � ��m� �

m � �� m� � �m� �

Thus if F �mx���� x�

�mx���� x�

� � then F � �m� � � and

���

��

���

�� �

��

� F � ��� � � � ��

Problem �� The point D lies inside the acute triangle ABC� Three of the circumscribedcircles of the triangles ABC ABD BCD and CAD have equal radii� Prove that the fourthcircle has the same radius�

Solution� There are two cases��� The radii of the circumscribed circles of the Figure ��

triangles ABD BCD CAD are equal�Let O� O� and O� be the centers of these cir�

cles �Figure � � Obviously the quadrilaterals O�CO�DO�AO�D and O�BO�D are rhombuses�

Hence O�CkO�DkAO� and O�C � AO�� Thus thequadrilateral CAO�O� is a parallelogram� It followsfrom here that ACkO�O� and since O�O� � DB thenBD � AC� Analogously CD � AB and AD � BC�Therefore D is the altitude center of �ABC�

Now it is easy to see that � BDC � ����� � BACwhich implies that the circumscribed circles of�BCDand �ABC are symmetric according to the line BC�

Therefore their radii are equal��� The radii of the circumscribed circles of the triangles ABC ACD and BCD are equal�In this case the circumscribed circles of �ABC and �BCD are symmetric according the

line BC� Hence � BDC � ���� � � BAC� Analogously � ADC � ���� � � ABC� Therefore� ADB � ����� � BDC � � ADC � ����� � ACB and the circumscribed circles of �ABD and�ABC are symmetric according to the line AB� Thus they have equal radii�

Problem �� Let A be a set with � elements� Find the maximal number of ��element subsetsof A such that the intersection of any two of them is not a ��element set�

Solution� Let B�� B�� � � � � Bn be subsets of A such that jBij � � jBiBj j �� � �i� j � �� � � � � n �Assume that there exists an element a A which belongs to four of the subsets B�� B�� � � � � Bn

�e�g� a B�� B�� B�� B� � Then jBi Bj j � � �i� j � �� � � � � � � But Bi �� Bj if i �� j i�e�jBi Bj j �� �� Thus jBi Bj j � � �i� j � �� �� �� � � It follows from here that jAj � � � ��� � �

Page 198: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

which is a contradiction� Therefore every element of A belongs to at most three of the subsetsB�� B�� � � � � Bn� Then �n � ��� i�e� n � ��

If A � fa�� a�� � � � � a�g then the subsets

B� � fa�� a�� a�g� B� � fa�� a�� a�g� B� � fa�� a�� ag� B� � fa�� a�� a�g�B� � fa�� a�� a�g� B� � fa�� a�� ag� B � fa�� a�� a�g� B� � fa�� a�� ag

provide an example of exactly eight ��element subsets of A such that jBi Bj j �� ��Therefore the searched number is n � ��

Grade ���

Problem �� Find all positive roots of the equation

logx�a����

x� �� loga ��

where a � � is a real number�

Solution� It is clear that if a � � and x � � then x � a � � � � and�

x� �� �� Hence in

this case logx�a����

x� �is well de�ned� Since

logx�a����

x� ��

loga�

x� �loga �x� a� �

then the given equation is equivalent to

x� �� �logax�a���� ��

The function�

x� �is strictly decreasing in the interval ����� � The function loga�x�a��

is strictly increasing in the interval ����� and obviously the same is true for the function�logax�a���� Therefore the equation �� has no more than one root in the interval ����� �

On the other hand it is easy to check that x � � is a root of this equation�Problem �� A circle k with center O and diameter AB is given� The points C and D are

moving along the arc�

AB so that C is between B and D and if � BOC � �� and � AOD � ��

then tan� � tan� ��

�� Prove that the lines which are perpendicular to CD and divide CD in

ratio � � � measured from C pass through a �xed point of the given circle�

Solution� Let E be such a point on the arc�

AB not containing C and D that if � BOE � �then tan � � �Figure � �

We shall show that the point E satis�es the problem�s conditions�It is enough to prove that if F is the foot of the perpendicular from E to CD then the point

F is between C and D andCF

FD�

��

We havetan� � ECD � tan�� ECA� � ACD � tan�

�� � �

Page 199: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�tan

��

�� tan�

�� tan

��

�� tan�

�� tan�

�� tan�

�� � � tan�

�� tan�

��� �

�tan� �

���tan� �

� �� � � tan��

�� tan�

�� � � tan�

�� � tan��

tan�� EDC � tan�� EDB � � BDC � tan� � � �tan � tan �

�� tan � tan��

�� tan�

�� � tan��

Thus Figure ��

tan�� ECD � � tan�� EDC �

It follows from the last equality that either

� ECD �

�and � EDC �

or� ECD �

�and � EDC �

��

But these angles belong to the triangle EDC�Hence they are acute angles� Therefore the pointF is between the points C and D� Since

tan� � ECD �EF

FCand tan�� EDC �

EF

FD�

thenEF

FC� �

EF

FD i�e� FD � ��FC�

Problem �� Find all prime numbers p for which the number p��p�� � � is a k�th power�k � � of a positive integer�

Solution� Let p��p�� � � � xk �x � � is an integer � It is clear that p �� � i�e� p � �q � � isan odd number� Since p�x then x � p�y �y is a positive integer and ��q � � ��q � � � pk��yk�At least one of the numbers �q � � and �q � � is a k�th power of an integer because they arerelatively prime numbers�

�� Let �q � � � zk i�e� �q � zk � �� If k is even then zk � � is not divisible by � � Henceq � � p � � and p��p�� � � � ���

If k � �l� � then �q � �z � � �z�l � z�l�� � � � � � z � � i�e� z � � � �� where � � � � q�On the other hand

�q � ��� � � �l�� � � � ����A� ����l� � �

�A is an integer � The last equality contradicts with � � q��� Let �q � � � zk i�e� �q � zk � �� If k is odd we obtain a contradiction as in the previous

case�If k � �l then �zl � � �zl � � � �q and since GCD�zl � �� zl � � � � we have zl � � � �

i�e� q � � p � � p��p�� � � � ���� � ����The sought numbers are p � � and p � ��

Page 200: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Grade ���

Problem �� Find all the values of the real parameter p for which the range of the function

f�x ����� p � cos x

p� sin� x�

contains the interval ��� ���Solution� Let y � cos x� The problem is to �nd all values of p such that for every k ��� ��

the equation���� p � y

p� � � y�� k

has at least one root y� ���� �� i�e� we should �nd all values of q � �� p such that for everyk ��� �� the equation

ky� � y � q�k � � � �

has at least one root y�� �� q y� ���� ���If y�� � q then �y� � �q � � y� � ��q and �q� � q i�e� q� � � or q� �

��

If q � � the roots of the equation ky� � y � � are y� � � y� ��

k� But y� �

k ���� �� and

y�� �� q � � for every k ��� ��� Thus q � � satis�es the problem�s conditions�

If q ��

� then the roots of the equation ky� � y � �

��k � � � � are y� � ��

� y� �

��

k�

But y� ���� �� only if k � �� Thus q ��

�doesn�t satisfy the problems conditions�

Let q �� � and q �� �

�� The equation ky��y�q�k�� � � has real roots i�D � ���k�k�� q �

� i�e� i� q � � �

�k�k � � for every k ��� ��� Hence q � � �

��� The vertex of the parabola

g�y � ky� � y � q�k � � has as its �rst coordinate y� ��

�k ��� �� � ���� ��� Therefore the

equation g �y � � has at least one root y� ���� �� i� at least one of the inequalities g��� � �

and g�� � � holds� It is easy to obtain from here that q � k � �

k � �for every k ��� ��� Hence

q � �

� i�e� q

�� �

����

�and q �� �

��

Finally p ��

����

��

� p �� �

��

Problem �� The point O is on the edge AB of the tetrahedron ABCD� The circumscribedsphere of the tetrahedron AOCD intersects the edges BC and BD in the points M and N

�M �� C N �� D respectively� The circumscribed sphere of the tetrahedron BOCD intersectsthe edges AC and AD in the points P and Q �P �� C Q �� D respectively� Prove that thetriangles OMN and OQP are similar�

Solution� Since the points A C M and O are in one and the same plane and are lying ona sphere then the quadrilateral ACMO is inscribed in a circle �Figure �� � Then �BOM �BCA and OM

CA�BM

BA�Figure �� � Analogously from �ABD�

OQ

BD�AQ

AB� Hence

OM

OQ�BM

AQ� ACBD

� ��

Page 201: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

From �BCD we haveMN

CD�BM

BD and from �ACD�

PQ

CD�AQ

AC� Thus

Figure ��� Figure ���

MN

PQ�BM

AQ� ACBD

� ��

It follows from �� and �� thatOM

OQ�MN

PQ� Similarly

ON

OP�MN

PQ�

Therefore �OMN �OQP �Problem �� Solve in positive integers the equation�

� � �x � �y � �z��t�

Solution� If y � � and z � � then the right side of the given equation is divisible by �� But� � �x � � �mod � and hence min�y� z � �� On the other hand �y � � �mod � � Thus y isdivisible by � �� is the index of � modulo � � It follows from here that y � � z � min�y� z � �and the equation is � � �x � ��y� � ���t where y � �y� �y� is a positive integer �

If t � � then �x���y� � �� Using congruence modulo � we obtain that x is even i�e� x � �x��Hence ��x� � ��y� ��x� � ��y� � � from where x� � y� � � i�e� x � � and y � ��

Let t � �� Then ��y� � � �mod �� and y� is divisible by � �� is the index of � modulo �� �Hence y� � �y� and � � �x � ���y� � ���t� On the other hand ��� � � �mod �� and �x � ���t

�mod �� � Obviously x � t� Thus �x�t � � �mod �� which is not true�Therefore the given equation has a unique solution �x� y� z� t � ��� �� �� � �

Grade ���

Problem �� For every real number x we denote by f�x the maximal value of the functionpt� � �t� � in the interval �x� �� x��a Prove that f�x is an even function and �nd its minimal value�b Prove that the function f�x is not di�erentiable for x � ��c Prove that the sequence an � ff�n g n � �� �� � � � is convergent and �nd its limit��For every real number a we denote with fag the unique real number in the interval ��� �

for which the number a� fag is an integer�

Page 202: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� It is clear that the maximal value of the functionpt� � �t� � in the interval

�x��� x� is reached at the endpoints of this interval� Since the inequality �x�� ����x�� ���x� � �x� � is equivalent to x � � then

f�x �

��

px� � �x� � if x � �

px� � �x� � if x � ��

a Obviously f�x � f��x i�e� f�x is an even function and its minimal value is f �� � ��b Since

limx��

x��

f�x � f��

x� �� lim

x��

x��

px� � �x� �� p�

x� �

p�

and

limx��

x��

f�x � f��

x� �� lim

x��

x��

px� � �x� ��p�

x�

p�

��

then f ��� doesn�t exist�

c It is clear that an �np

n� � �n� �o� On the other hand n� � �

pn� � �n� � � n� �

i�e� an �pn� � �n� �� �n� � �

Therefore

limn��

an � limn��

�pn� � �n� � � n� �

� ��

Problem �� The points A B C and D lie onFigure ���a straight line in the given order� A circle k passesthrough the points B and C and AM AN DK andDL are tangents to k�

a Prove that the points P � MN BC and Q �KL BC don�t depend on the circle k�

b � If AD � a BC � b �a � b and the segmentBC is moving along AD �nd the minimal length ofthe segment PQ�

Solution� a From the Steward�s formula for thetriangle AMN and the segment AP �Figure �� wehave�

AP ��MN � AM��NP � AN��MP �MN�MP�NP

� �AM� �MP�NP �MN

�here AM � AN because they are tangents to a circle �Hence

AP � � AM� �MP�NP � AB�AC �BP�CP

� AB�AC � �AC �AP �AP � AB

� ��AB�AC � AP �AB �AC �AP ��

��

Page 203: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

i�e� AP ��AB�AC

AB �AC� Analogously DQ �

�DB�DC

DC �DB� These equalities show that the position

of the points P and Q doesn�t depend on the circle k�b Let us denote AB � x BC � y and CD � z� It follows from a that

PQ � AD �AP �DQ � x� y � z � �x�x� y

y � �x� �z�y � z

y � �z

� �x� y

��� �x

y � �x

�� z

��� ��y � z

y � �z

�y��x� y � z

�y � �x �y � �z �

Having in mind that y � b x� y � z � a we obtain

PQ �b�a

�a� x� z �a� z � x �

b�a

a� � �x� z �� b�

a�

Therefore the minimal length of the segment PQ isb�

aand this length is reached i� AB �

CD � x � z �a� b

��

Problem �� Find all prime numbers p and q such that the number �p � �q is divisible byp�q�

Solution�Lemma� If k � � then k doesn�t divide �k�� � ��Proof� Assume that k divides �k�� � �� Obviously k is odd� Let k � p��� p��� � � � p�rr where

p�� p�� � � � � pr are odd prime numbers �pi �� pj if i �� j ��� ��� � � � � �r are positive integers andr � �� Let pi � � � �mi �ti where ti are odd integers � i � �� �� � � � � r � Let m� be the smallestnumber in the sequence m�� m�� � � � � mr� It follows from pi � � �mod�pi � � that pi � ��mod �mi and p�ii � � �mod �mi i � �� �� � � � � r� Hence k � � � �m� �u �u is an integer � If

�k�� � �� �mod k then ��m� �u � �� �mod k and �p�����u � �� �mod p� because t� is

odd� But �p��� � � �mod p� � a contradiction�Let �p � �q is divisible by p�q� We have three cases��� p and q are odd prime numbers� Then �p � �q � � �mod p and since �p � � �mod p

then �q � ���modp and �pq � ��� p � �� �mod p � Similarly �pq � �� �mod q � Thus�pq�� � �� �mod pq which is a contradiction with the lemma�

�� p � � q � �� Then � � �q � ��modq and it follows from �q � � �mod q that� � ��modq and q � �� It is clear that �� � �� � �� � � �mod � � � �

�� p � q � �� Then �� � �� � � � � �mod � � � �Therefore the sought numbers are� p � q � �� p � � q � �� p � � q � ��

��

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WINTER MATHEMATICAL COMPETITION

����

Grade �

Problem �� For which integer values of the parameter a the equation j�x� �j� jx� �j � a

has an integer solutions�Solution� We shall consider the following cases�I� x � �� Then �x� � � x � � � and the equation is equivalent to �x� � � x � � � a�

Thus x ��� a

� which is a solution when

� � a

�� � i�e� when a � ��

II� ��

�� x � �� Then �x�� � x�� � and the equation has the form �x��� x��� � a�

Thus x � a � � which is a solution when ��

�� a� � � � i�e�

�� a � �� In this interval the

integers are a � �� �� � and respectively we get x � � �� � which are integer solutions�

III� x � ��

�� Then �x�� � x�� � and the equation is equivalent to� �x���� x��� �

a or ��x � a� �� Thus x ��� a

� which is a solution when

�� a

�� ��

� i�e� when a �

��

According to the considered cases the given equation has solution when a ��

�� When a � �

the only integer solution is x � � When a � � there are two integer solutions x � � and x � ���When a � � x � � is the only integer solution� When a � � the equation has two solutions

x� ��� a

�and x� �

�� a

�� For these solutions we have� if a � �k � � then x� is integer only�

if a � �k � � then x� is integer only� if a � �k there is no integer solution�As a result we get integer solutions when a � � or a � �k� � where k is a positive integer

greater than ��Problem �� The bisector AD D � BC� of the acute isosceles triangle ABC divides it into

two isosceles triangles� Let O and I be the incenter and the circumcenter of�ABC respectively�AO meets BC in point E while F is the intersection point of the lines BI and DO� Prove that�

a� the quadrilaterals ABEF and ADCF are rhombi with equal side lengths�b� If H is the altitude center of �ABE then the points A D E F H are concyclic�Solution� Firstly let us justify the position of AD� If it is a bisector of the angle between the

two equal sides then it is perpendicular to BC �ADB and �ADC are isosceles� ConsequentlyAD � BD � CD and � BAD � � CAD � ��� i�e� � BAC � �� which contradicts to thecondition that�ABC is acute� It follows that AD is bisector of the angle belonging to the baseAB while �ABD and �ACD are isosceles Figure ���

a� If � BAD � � CAD � � then � ABC � ��� Since � ADB � � CAD � � BAD the onlypossibility is � ADB � � ABC � �� and AD � AB� But � ACD � � ADB� � CAD � ���� �

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� i�e� � ACD � � CAD and CD � AD� From the equality ������� � ��� we de�ne � � ����Thus � ACB � ��� � BAC � � ABC � ���� The points O and I lie on the bisector of � ACBwhich is a segment bisector of AB� We have OA � OC and � OAC � � OCA � ��� � � OADwhile � OAB � ��� � ��� � ��� � AEB � ��� � ��� � ��� � ��� � � BAE� It follows thatBE � AB and the bisector BI of � ABC is the segment bisector of AE� In the triangle ACDwhich is isosceles we have that AO and CO are bisectors and it follows that DO is the bisectorof � ADC and the segment bisector of AC� Then F is the intersection point of the segmentbisectors of AE and AC i�e� F is the circumcenter of �ACE the radius of the circumcircle isAF � EF � CF � We have � AFE � �� ACE � ��� while � AFB � � BFE � ��� � � ABF �� EBF and consequently AF � AB � BE � EF i�e� ABEF is a rhombus� In the same wayCF � AF � AB � AD � CD and ADCF is a rhombus which side is AD � AB�

b� It is clear that � AEF � � AEB � ��� and Figure ��

� ADF ��

�� ADC � frac�� ��� � ���� � ���

the segment AF is seen from the points D and E

under ���� We draw a line throughA perpendicularto BC which intersects BI in the point H � thealtitude center of �ABE because BI�AE� But in�ABD the altitude AH is a bisector of � BAD ����� Thus � BAH � ��� � AHF � ��� � ��� ���� � � AEF � � AOF � Note that the points D EH are in one and the same semiplane with respectto AF � Consequently D EH together with A andF are concyclic�

Problem �� Every day a student preparinghimself for the Winter competition in mathemat�ics has been solving problems during a period of �weeks� He has been solving at least one problem daily but no more than � problems weekly�

a� Prove that during some consecutive days the student has solved �� problems exactly�b� If � � n � �� is a natural number prove that during some consecutive days the student

has solved n problems exactly�Remark� Every week begins on Monday and ends on Sunday�Solution� Since a� is a particular case of b� we shall solve b� only�According to the condition the student has been solving problems during � � � �� days and

has solved at most � � � � problems� Let xi be the number of the problems solved duringthe i�th day i � �� �� � � � � ����

Let � � n � �� be a �xed natural number� We want to prove that there exist such k � lthat xk�� � � xl � n� Denote Xi � x� � � xi� Obviously

� � X� � X� � � � � � X�� � �

and the problem is to prove the existence of such k � l that Xl �Xk � n�Case �� � � n � ��� We consider the numbers

X� � X� � � � � � X��� X� � n � X� � n � � � � � X�� � n� ��

which are integers and their number is �� Obviously they are in the interval ��� � � n� inwhich there are � � n � � � �� � � integers� Consequently among the numbers �� there

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are at least two equal� The �rst �� of them as well as the next �� are di�erent from each other�Therefore there exist such k and l for which Xl � Xk � n i�e� for which Xl �Xk � n�

Case �� � � n � ��� Firstly we shall prove the followingLemma� If the integers z�� z�� � � � � zm belong to the interval ��� �n� and if m � n then among

the numbers z�� z�� � � � � zm there are two the di�erence of which is equal to n exactly�Proof� With the numbers from the interval ��� �n� we construct the following pairs�

�� n� ��� �� n� ��� � � � � n� �n��

The number of these pairs is n and the di�erence of the numbers in each pair is equal ton� Since m � n at least two of the numbers z�� z�� � � � � zm belong to one and the same pair�Therefore their di�erence is equal to n�

Let us �nish now the solution of the problem�If n � �� then �n � � and thus all the numbers X�� X�� � � � � X�� are in the interval

��� �n�� On the other hand n � �� and according to the lemma there are two numbers amongX�� X�� � � � � X�� which di�erence is n�

If � � n � �� we represent the interval ��� �� as an union of the intervals ��� �n� ��n��� ���In the second one there are � � �n � �� � � � � � �n integers� Then the number of theintegers among X�� X�� � � � � X�� which belong to the interval ��� �n� is at least ��� �� �n� ��n� �� � �� �� � �� � n� Consequently we can apply the lemma again�

Remark� The case � can be solved by the lemma proved above�

Grade �

Problem �� Let f x� � x�� p���x�� � p� �� p� ��x��p�� ��p��� where p is a realparameter�

a� Prove that f �� p� � �b� Find all values of p for which two of the roots of the equation f x� � are lengths of the

cathetuses of a rectangle triangle which hypotenuse is equal to �p��

Solution� a� We have f x� � x� p� �� x�� � p� ��x� � p� ����b� The roots of f x� � are x��� � p � � � pp� � �p� �� and x� � � � p� If p � �

then x� � x� � and x� � � The equation x�� � x�� � �� ���p���

gives p � �� which is

impossible� If p � � then two of the roots are equal to and this case gives no solution� If p � �then x� � x� � and x� � � Therefore �� � x��� x�� � p� ��

pp� � �p� ����� �� p���

This equation is equivalent to

p� �� � p� �� � �qp� � �p� ��� � �

If p � �� we get �pp� � �p� �� � � � � p� � from where �p� � ��p � �� � and

p ���� �

p�

�� Since

�� � �p�

�� � and

��� �p�

�� � we �nd p� � �� and p� �

��� �p�

��

Problem �� The incenter of the quadrilateral ABCD is O� The lines lA � OA lB � OBlC � OC and lD � OD are drawn through the points A B C and D respectively� The lines lAand lB meet each other in the point K lB and lC � in L lC and lD � in M lD and lA � inN �

a� Prove that the lines KM and LN meet each other in the point O�

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b� If the lengths of the segments OK OL and OM are p q and r respectively �nd thelength of the segment ON �

Solution� a� We shall prove that the points N O and L are colinear� Denote � ABC � � B� BCD � � C � CDA � � D and � DAC � � A� Since O is incenter the segments OAOBOCand OD are the bisectors of the corresponding angles of the quadrilateral ABCD� Note thateach of the quadrilaterals AKBO BLCO CMDO and DNAO is inscribed�

Consequently� � NOK � � KOL � � �Figure ��� ONA � � OKA � � � � OKB � � OLB ��� � ADO� � ABO��� � BAO� � BCO �

�� � � D

��

� B

��

� A

��

� C

�� � �� � � � ��

It follows from here that the points N O andL are colinear� Analogously the points KOandM are colinear� Therefore O is the inter�section point of the diagonals of the quadri�lateral KLMN �

b� Firstly we shall prove that the quadri�lateral KLMN is inscribed� Indeed

� NKL� � NML

� � AKO � � OKB � � DMO � � OMC

�� B

��

� A

��

� C

��

� D

�� ���

Thus OK�OM � OL�ON from where

ON �OK�OM

OL�p�r

q�

Problem �� A square with side length � is divided into unit squares by parallel to its sideslines� Let A be the set of the vertexes of the unit squares which are not on the sides of the givensquare� How many points from A can be chosen at most in a way that no three of them arevertexes of isosceles rectangle triangle�

Solution� We shall prove that the maximal number is �� Let us enumerate the points in theway shown on the table ��

It is easy to be seen that no � of the points ������ and �� are vertexes of a isoscelesrectangle triangle� Assume that there exists a set of � points with the desired property� Notethat if � points form a square then at most � of them can be among the already chosen ones�The points ���� and ��� ���� and �� ����� and � form squares�

Consequently at most � of the chosen points lie on the contour� It�� �� �� ��� � �� ��� � � �� � � �

Table ��

follows from here that at least one of the points ��� and �� is from thechosen ones� Due to the symmetry we may assume that this is the point ��Since the points ��� and �� �� and �� form isosceles rectangle trianglesthen at most two of the points ���� and �� are from the chosen ones�The points ���� and �� form a square and therefore at most one of thepoints ��� and �� is from the chosen ones� It follows from here that at

least � points are chosen from �������� and ��� By the pigeonhole principle we deduce thatat least two points are chosen in one of the sets ����� and ������ It is easy to see that ifthe two points are in the �rst set then we have two possibilities � � and �� or � and � in both

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cases it is not possible to chose more points on the square which encounters ��� Analogously ifthe two points are in the second set then the possibilities are two again � � and �� or � and� in both cases it is not possible to chose more points on the square which encounters ��� Thecontradiction shows that the maximal number of points which can be chosen is equal to ��

Grade ��

Problem �� Let p and q be such integers that the roots x� and x� of the quadratic equationx� � px � q � are real numbers� Prove that if the numbers � x� x� in some order� form ageometric progression then the number q is a perfect cube�

Solution� There are two possibilities for the order of the numbers in the geometric progres�sion� x�� �� x� and �� x�� x�� In the �rst case we get q � x�x� � �� i�e� q � � is a perfectcube� Let now x� � x��� We have �p � x� � x� � x� � x�� i�e� x� satis�es the equationx��� x�� p � � On the other hand x� satis�es also x

��� px�� q � � From these two equations

we get p � ��x� � q � p� � � If p � � then x� is rational and q � x�x� � x�x�� � x�� is the

cube of a rational number� Since q is integer then it is a perfect cube� If p � � then q � p andthe quadratic equation becomes x� � x� � � � The last equation has no real root�

Problem �� A triangle ABC with a radius R of Figure ��the circumcircle is given� Let R� and R� be the radii ofthe circles k� and k� respectively which pass throughC and are tangent to the line AB in A and B respec�tively�

a� Prove that the numbers R� R and R� form ageometric progression�

b� Find the angles of �ABC if the radius of thecircle which is tangent to k� k� and the line AB is

equal toR

��

Solution� a� Let AB � c BC � a and CA � b�Denote by O� and O� the centers of the circles k� and k� Figure ��� Then O� is the intersectionpoint of the perpendicular from A to AB and the segment bisector of AC� Since � MAO� �

j��� � Aj M is the midpoint of AC� then R� � AO� �AM

cos j�� � � Aj �AC

� sin � Aand by the

sine theorem it follows that

R� � R ba� ��

Analogously

R� � R ab� ��

From here R�R� � R� and the proposition is proved�b� Let O be the center of the circle which touches k� k� and the line AB Figure ��� Denote

by T the tangent point of this circle with AB and by r its radius� From the rectangle trapezoidATOO� it follows that AT �

p R� � r�� � R� � r�� � �

prR�� Analogously BT � �

prR��

Then c � AT � TB � �prR� � �

prR� and from here we �nd r �

c�

� pR� �

pR���

� Using ��

and �� we get

r �abc�

�R a� b��� ��

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Sinceab

a� b��� �

�and c� � �R� it follows from �� that r � R

�� The equation is reached

when a � b and c � �R� Now it follows that � A � � B � ��� and � C � ���Problem �� A positive integer n and a Figure ��

real number � are given in a way that cos� ��

n� Find all positive integers k for which the

number cosk� is an integer�Solution� Case �� n � �� Then cos� � �

and � � �m� m � ���� � � ��� For all k � Nthe number cosk� is an integer�

Case �� n � �� Then cos� ��

�and

� � ���� �m� m � ���� � � ��� It is clear

that for all k � N which are divisible by �the number cosk� is an integer�

Case �� n � �� We shall prove that for

all k � N the number cosk� is not an integer� Let n be odd� We have cos� ��

n cos �� �

� cos� �� � ��� n�

n�and �� n�� n� � �� We shall prove by induction that cos k� �

a

nk where

a� n� � �� Assume that the assertion is true for all integers from � to k� We shall check it fork � �� From

cos k � ���� cos k � ��� � � cosk� cos�

it follows that cos k� ��� ��

n ank

� b

nk���

�a� bn�

nk�� where cos k� �

a

nkand cos k � ��� �

b

nk��� We have a� n� � � and b� n� � � according to the inductive assumption� It is clear that

�a � bn�� n� � � and this ends the proof� The case when n is even is analogous� Now cosk�is expressed by a fraction which denominator is equal to �pk with n � �p while the nominatorhas no common divisor with p�

Answer� if n � � �k � N�if n � � k � �q where q � N�if n � � there is no solution�

Grade ��

Problem �� Find the values of the real parameter a for which the function

f x� � x� � �x� jx� �� aj � jx� �j� �

has nonnegative values for all real x�Solution� Firstly let � � a � � i�e� a � �� Then

f x� �

�����

x� � �� a� x � � � ax� � �x� � � a� � � a � x � �x� � �x� � � a� x � ��

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If � � a � � i�e� a � � we �nd that

f x� �

�����

x� � �� a� x � �x� � �x� �� a� � � x � � � ax� � �x� � � a� x � � � a�

Hence the smallest value of f x� is reached in one of the points � � � � a�We have f � � � � ja� �j f �� � � � jaj f �� � �� j�� aj f � � a� � a� � ja� �j� ��

These four numbers must be nonnegative� We �nd that a � ���� ��� Then ja � �j � � � a andf � � a� � a� � a� � � for all a�

Let now a � ���� ��� If a � �� then x� � �x� �� a � when x � �� Analogously if a � �then x� � �� a � when x � �

Finally a � ���� ���Problem �� The point O is circumcenter of the acute triangle ABC� The points P and Q

lie on the sides AB and AC respectively� Prove that O lies on the line PQ if and only if

sin �� �PB

PAsin � �

QC

QAsin ���

where � � � BAC � � ABC � � � ACB�Solution� Firstly let O be on PQ� Denote Figure ��

x � � AOP Figure ��� By the sine theorem forthe triangles AOP BOP AOQ and COQ we �ndthat

PA

sin x�

PO

sin �� � ���

PB

sin �� � x��

PO

sin �� � ���

QA

sin ��� � x��

QO

sin �� � ��

QC

sin � � ��� � x��

QO

sin �� � ��

Then

PB

PAsin � �

QC

QAsin �� �

sin �� � x�

sin xsin � � sin � � x�

sin xsin �� �

sin x sin �� cos x� sin x cos ��� sin� � sin � cosx� sin x cos �� sin ���

� � sin � cos �� � sin �� cos � � � sin � � �� � sin ���

Conversely let the given equality be satis�ed� Denote by Q� the intersection point of OP

and AC� It follows from the above that sin �� �PB

PAsin � �

Q�C

Q�Asin ��� Then

Q�C

Q�A�QC

QA�

Since the points Q� and Q lie on the segment AC then Q� � Q�Problem �� Find all functions f x� with integer values and de�ned in the set of the integers

such that�f x�� �f f x�� � x

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for all integers x�Solution� The function f x� � x satis�es the condition of the problem�Let f x� be a function which satis�es the condition� Let g x� � f x��x� The condition can

be written in the form�f f x��� �f x� � f x�� x�

which is equivalent tog x� � �g f x���

From here we obtain

g x� � �g f x�� � ��g f f x��� � ��g f f f x���� � ��g f f f f x����� � � � � �

Since the numbers g f f � � � f x�� � � �� are integer then g x� is divisible by �n for all integersx and all natural numbers n� This is possible only if g x� � � Thus f x� � x is the onlysolution of the problem�

Page 212: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Winter mathematics

competition�Burgas� ����

Problem ���� Let F be the set of points with coordinates �x� y�such that jjxj � jyjj� jxj� jyj � ��

�a� Draw F �

�b� Find the number of points in F such that �y � j�x� �j � �

Solution� �a� If jxj � jyj then jjxj � jyjj� jxj� jyj � jxj � jyj�jxj�jyj � �jxj � � thus jxj � � and therefore � � jyj so �� � y � ��We conclude that the segments �� � y � � on the lines x � � andx � �� belong to F �

If jxj � jyj then jjxj � jyjj� jxj � jyj � �jxj� jyj� jxj� jyj ��jyj � � thus jyj � � and therefore � � jxj so �� � x � �� Weconclude that the segments �� � x � � on the lines y � � andy � �� also belong to F �

Thus we have determined that F consists of the sides of a squarewith vertices A�������� B������� C��� ���D���� ���

Page 213: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�b� We �nd the number of solutions of �y � j�x��j� on eachof the segments AB�BC�CD�DA�

The segment CD consists of all points �x� y� such that �� � x �� y � �� The equation � � j�x � �j � has no solution x when�� � x � �� Therefore �y � j�x� �j � has no solution on CD�

The segment AB consists of all points �x� y� such that �� � x �� y � ��� The equation �� � j�x� �j � has two solutions� x � and x � �� Therefore �y � j�x� �j � has two solutions on AB�

As above we get that �y � j�x � �j � has a unique solutionon AD� �x� y� � ���� � and a unique solution on BC� �x� y� �������� Note that the last one has already been obtained as a pointon AB� Thus there are three solutions of �y � j�x � �j � in F ��x� y� � ���� �� � ����� �������

Problem ���� Let H be the orthocentre of an acute triangle ABC�Prove that the midpoints of AB and CH and the intersecting pointof the internal bisectors of � CAH and � CBH lie on a line�

Page 214: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� Denote by M P and N the midpoints of AB and CHand the intersecting point of the internal bisectors of � CAH and� CBH ��g� ���

Let AA� �A� � BC� and BB� �B� � AC� be altitudes in�ABC�We show �rst thatM N and P lie on the axis of symmetry l ofA�B��From � CA�H � � CB�H � � � we get PA� � PB� �

CH

�� Similarly

from � AA�B � � AB�B � � � we getMA� � MB� �AB

�� Therefore

M � l and P � l� We prove now that �NMA��� �NMB��

Now � BAA� � � � � � and � ABB� � � � � � thus � NBB� �� NBA� � �

�� A�BB� � ��� � �

�� By analogy � NAA� � ��� � �

so � ANB � �� � � � NAB � � NBA � � �� Therefore �ABN is aright triangle and MN � MA� � MB� �

AB

��

Further � NMA� � � NMB � � A�MB � � � � �� By analogy� NMB� � � � � � and therefore the considered triangles are iden�tical� It follows now that NA� � NB� so N � l�

Problem ���� The n points A�� A�� � � � � An�� lie a circle in thisorder and divide it into equal arcs� Find an orderingB�� B�� � � � � Bn��

Page 215: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

of the same points such that the length of B�B� � � �Bn�� is maximal�

Solution� Let �rst n � �k��� Clearly a chord AiAj is of maximallength if ji� jj � k or k � �� Cosider the following points�

A�� Ak� A�k� Ak��� A�k��� Ak��� A�k��� � � � � A�� Ak���

Since each segment is of maximal length it follows that the lengthof A�AkA�kAk��A�k��Ak��A�k�� � � �A�Ak�� is maximal�

Let now n � �k� A chord AiAj is of maximal length if ji� jj �k� There are k such segments� A�Ak� A�Ak��� � � � � Ak��A�k��� Thesecond longest chord AiAj is obtained when ji� jj � k� � or k���Cosider the following points�

A�� Ak� A�k��� Ak��� A�k��� Ak��� � � � � Ak��� A��

It is easy to see that there are k segments of maximal length andk � � segments of the second greatest length� Trivially this is therequired ordering�

Problem ���� Let � �� � be the roots of the equation x��px�q � � For any natural number n denote�

an ��n � �n

�� ��

�a� Find p and q such that for any natural number n the followingequality holds�

an��an�� � anan�� � ����n�

Page 216: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�b� Prove that for these p and q it is true that

an � an�� � an��

for any natural number n�

�c� Prove that for any natural number n an is integer and if divides n then an is even�

Solution� �a� Since � and � are the roots of x�� px� q � weknow that �� � � �p �� � q and therefore�

����n ��n�� � �n��

� � � �

n�� � �n��

�� �� �n � �n

� � � �

n�� � �n��

�� �

��

��� �������� ������n�� � ��� � �������n�

��

p� � �q�pqn�� � p�p� � q�qn�

�qn

p� � �q��p� � �pq� � �pqn�

Thus

��� pqn � ����n���It follows from ��� for n � � and n � � that pq � � and pq� � ��and so p � �� q � ��� Direct veri�cation shows that p � �� andq � �� satisfy ��� for any n� Also � �� ��

�b� Since � and � are the roots of x��x�� � we know that�� � �� � and �� � � � �� Therefore

an � an�� ��n � �n

�� ���n�� � �n��

� � ���n�� � �� � �n�� � ��

�� �

Page 217: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

��n��� � �n���

�� ���n�� � �n��

� � �� an���

which completes the proof of �b��

�c� Since a� ��� �

�� �� � and a� �

�� � ��

�� �� � � � � � it

follows by induction from

��� an � an�� � an��

that an is integer for any n� From ��� we obtain�

an�� � an�� � an�� � an�� � an � an�� � �an�� � an�

Observe that a� � a� � a� � � is an even number� It is easy to seenow �again by induction� that an is even for n � k�

Problem ���� A pentagon ABCDE is inscribed in a circle� Let Pbe the intersecting point of AC andBD and letQ be the intersectingpoint of AD and CE� Prove that if the triangles ABP AEQ CDP CDQ and APQ have the same area then ABCDE is a regularpentagon�

Solution� It su�ces to prove that the sides of ABCDE are equal�Since�ABP and�CDP have equal areas so do the triangles ACDand ADB� Therefore the quadrilateral ABCD ��g� �� is a trapezoidinscribed in a circle so AB � CD� By analogy from SAEQ � SCDQ

we get ACkDE and AE � CD� Now � PCQ � � ACE �

AE

��

AB

�� � ADB � � PDQ� Therefore CDQP is inscribed in a circle�

Page 218: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

On the other hand it follows from SCDP � SCDQ that CDQP is a

trapezoid� Thus � CDP � � DCQ �

BC��

DE and therefore BC �DE� It remains to show that AB � BC� Consider the trianglesABP and APQ whose areas are equal� They have AP as a commonside and � AQP � � APQ � � ACD � � ABD � � ABP � It is easy tosee now that �ABP �� �APQ� Since � APB � � ADP � � PAQ �� PAQ

it follows that � APB � � APQ � � AQP � � ABP � Therefore�

AB ��

CD��

AE ��

DE so�

AB��

DE��

BC or AB � BC whichcompletes the proof�

Problem ���� Given a rectangular table of � rows and ����columns� The table is �lled with zeroes and ones in such a waythat there are at least �� ones in any column� Prove that it ispossible to remove �� rows in such a way that there is at most one

Page 219: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

column consisting of zeroes in the remaining table �� rows and ����columns��

Solution� We show �rst that there is a row with at least ����ones� Assume the contrary� Denote by ai the number of ones in thei�th row �i � �� �� � � � � � � and by bi the number of ones in the i�th

column �i � �� �� � � � � ������ Now����Xi��

bi ����Xi��

ai� Note that the sum

on the left�hand side is at least ���� �� whereas the sum on theright�hand side is at most ���� � a contradiction�

Without loss of generality assume that the �rst row begins by ����ones� Consider the table formed by the last ��� columns� As abovewe prove that there is a row of this table �not necessarily distinctfrom the �rst one� with at least �� ones� Let that be the secondrow �if it is not the �rst one� and let it begin with �� ones in thenew table� Now consider the table formed by the last ��� columns�Analogously there exists a row having at least � ones� Let that bethe third row �if it is not the �rst or the second one� and let it beginwith �� ones in the new table� Consider next the table formed bythe last � columns and note that there exists a row that originallyhas �� ones� Finally consider the table formed by the last � rowsand note that there exists a row that originally has � ones�

We now have � rows �if there are fewer of them we add arbitraryrows�� Remove the remaining �� rows of the original table� Since�������������� � ���� there is at most one column consistingof zeroes�

Page 220: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem ���� Find all real numbers x such that tan��

��� x

tan�

��and tan

��

��� x

�form �in some order� a geometric progres�

sion�

Solution� Denote a � tan�

��and y � tan x� There are three cases

to consider�

�� tan��

��� x

�tan

��

��� x

�� tan�

��� Now

a� y

� � ay a� y

�� ay�

a�� Therefore a� � y� � a��� � a�y�� and so �a � ��y� � �Since a �� �� we get that y � so tan x � � Obviouslyall numbers x of the kind x � k� k � Zare solutions to theproblem�

�� tan�

��tan

��

��� x

�� tan�

��

��� x

�� We obtain

aa� y

�� ay�

�a� y

� � ay

��

� �a����y�ay�� �a�� ��y�a� � �

The case of y � is settled in �� Let y� and y� be the roots ofthe equation ay�� �a� � ��y � a � � Since a � tan �� deg �tan��� deg� deg� � ��p we get y� � y� �

p so tanx �p

� Obviously all x of the kind x ��

�k� k � Zare solutions

of the problem�

� tan�

��tan

��

��� x

�� tan�

��

��� x

�� The substitution z �

�x transforms this case to the previous one� Therefore x �

��� k� k � Z�

Page 221: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

The required numbers are x � k� and x � ��� k� k � Z�

Problem ���� Two points C and M are given in the plane� LetH be the orthocentre of �ABC such that M is a midpoint of AB�

�a� Prove that CH CD � jAM� � CM�j where D � AB andCD � AB�

�b� Find the locus of points H when AB is of given length c�

Solution� �a� It is easy to see that the equality holds if �ABCis a right triangle� If �ABC is not a right triangle then �BDHand �ADC exist and �BDH � �ADC� Therefore

DH

AD�

BD

CDand so

��� CD DH � AD BD

There are three cases�

Case �� �ABC is an acute triangle ��g� ���

Case �� �ABC is obtuse triangle and � C � � � ��g� ���

Case � �ABC is obtuse triangle and � A � � � or � B � � �

��g� ��

In cases � and � it follows from ��� that

CDDH � �AM DM��AM�DM� � AM��DM� � AM���CM��CD���

Page 222: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

so

��� AM� � CM� � CD�DH � CD��

In case � we have CD � CH �DH and from ��� we get

AM� � CM� � CD�DH � CH �DH� � �CH CD�

In case � we have CD � DH � CH and from ��� we get

AM� � CM� � CD�DH �DH � CH� � CH CD�

In case it follows from ��� that CD DH � �DM AM��DM�AM� � DM� �AM� � CM� � CD� �AM� and therefore CM� �AM� � CD�DH � CD�� Now DH � CH � CD and so CM� �AM� � CD�CH � CD � CD� � CH CD� Summarising all threecases considered �and the case of a right triangle �ABC� we get

CH CD � jAM� � CM�j�

��

Page 223: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

�b� It follows from the conditions of the problem that A and Bare diametrically opposite in the circle �k� of centre M and radiusc

�� When the diameterAB varies in �k� we obtain all triangles ABC

with the �xed vertex C� There are three cases�

Case ��c

�� jCM j ��g� ��� Case ��

c

�� jCM j ��g� ��� Case �

c

� jCM j ��g� ���In case � C lies on �k� and triangles ABC are right triangles and

their orthocentres coincide with C� In this case the locus consists ofpoint C�

��

Page 224: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Consider now case �� Let AB be a diameter in �k� perpendicularto CM � The orthocentre H of �ABC lies on line CM � MoreoverH is an external point for �k� because �ABH is acute� The lastfollows from � AHB � �� deg�� ACB and � ACB � � deg�

Let A�B� be a arbitrary diameter of �k� such that C � A�B��Denote by H� the orthocentre of A�B�C and by D� the foot of thealtitude from the vertex C� Since �A�B�C is obtuse we get thatH� �� C� Consider �MD�C and �CHH� and use a�� We are in

case � of �a�� For �ABC we obtain that CH CM �c�

��CM� and

for �A�B�C that CH� CD� �c�

��CM�� Therefore

CH

CH��CD�

CM

so �MD�C � �CHH�� Since �MD�C is a right triangle weknow that H�H�CM and so H� lies on a line l through H which isperpendicular to CM � Let H� �� H be an arbitrary chosen point ofl and let A�B� be a diameter of �k� such that AB � CH��

Since H is an external point for �k� and l�CM through H itfollows that all points of l are external for �k�� In particular H� �� Cand therefore A�B� exists� The orthocentre of �A�B�C lies on CH�

Page 225: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

and it follows from the above that it lies on l� Since H� is theintersecting point of these two lines it is obvious that H� is theorthocentre of �A�B�C�

Therefore the locus is a line l perpendicular to CM through a

point H on the ray opposite to CM� and of distancec�

�jCM j � jCM jfrom point C�

Consider case � Let AB be a diameter of �k� perpendicular toCM � The orthocentre H of �ABC lies on CM but now H is aninternal point for �k�� Consider an arbitrary diameter A�B� of �k�such that H � A�B�� Let H� be the orthocentre of �A�B�C andCD�� the altitude� Consider �MD�C and �CHH�� It is essentialto show that �CHH� is uniquely determined �for �MD�C it isobvious�� It su�ces to prove that H� �� C� If the contrary is truethen �A�B�C is a right triangle with its right angle at C and sinceA�B� is a diameter C must lie on �k� a contradiction� Since�ABCis acute we apply case � of �a� so CH CM � CM� � c�

�� There

are three cases for �A�B�C�

�� if it is an acute triangle we apply case � of �a� and therefore

CH� CD� � CM� � c�

��

�� if it is an obtuse triangle we apply case of �a� and we get thesame equality �note that �A�B�C is not obtuse at C becauseC is an external point for �k���

� if it is a right triangle �say at B�� then D� � B� � H� and

therefore CH� CD� � CB�� � CM� � c�

��

��

Page 226: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

The further considerations follow those from case �� We concludethat the locus is a line perpendicular to CM and passing through

H � CM� of distance jCM j � c�

�jCM j from C�

In particular it follows that PQ�CM a well�known property�

Problem ���� How many natural numbers a�a� � � � a�n exist suchthat�

�a� none of the digits ai is zero�

�b� the sum a�a� � a�a � � � �� a�n��a�n is an even number�

Solution� Denote the required number byAn� The product a�i��a�iis even if at least one of the digits a�i�� and a�i is even� Thereforethere are � � � � � � � � � �� choices for a�i�� and a�i such thata�i��a�i is even� Similarly a�i��a�i is odd when both a�i�� and a�iare odd� There are � � � �� choices for a�i�� and a�i such thata�i��a�i is odd� The number of a�a� � � � a�n such that i of the items

a�i��a�i are odd is�n

i

���i��n�i� Therefore An �

Pi

�n

�i

����i��n��i�

Let Bn �Pi

�n

�i��

����i����n��i��� Obviously

An �Bn �nXi��

�n

i

���i��n�i � ��� � ���n � ��n�

An �Bn �nXi��

����i�n

i

���i��n�i � ��� � ���n � �n�

Thus �An � ��n � �n and An � ���n � �n���

��

Page 227: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem ����� The sequence fxng�n�� is de�ned as�

x� � � xn�� � x�n � xn � �� n � �� �� � � � �

�a� Prove that fxng�n�� is monotone increasing and unbounded�

�b� Prove that the sequence fyng�n�� de�ned as yn ��

x� � ��

x� � �� � �

xn � � n � �� �� � � � � is convergent and �nd

its limit�

Solution� �a� xn�� � xn � �xn � ��� � and so fxng�n�� is amonotone increasing function� We now prove by induction that xn �n � �� Obviously this equality holds for n � �� Suppose it is truefor n � k � �� Then

xk�� � xk�xk � � � � � �k � ���k � �� � � � k � �

Therefore xn � n � � when n � �� �� � � � � and so the sequence isunbounded�

�b� It follows from the recursive de�nition of our sequence thatxk�� � � � �xk � ���xk � ��� Hence

xk�� � ��

�xk � ���xk � ���

xk � �� �

xk � ��

so�

xk � ��

xk � �� �

xk�� � ��

By adding the above equalities for k � �� �� � � � � n we get

yn ��

x� � �� �

xn�� � �� � � �

xn�� � ��

��

Page 228: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Since � �

xn�� � �� �

nit follows that lim

n��

xn�� � �� so

limn��

yn � ��

Problem ����� Given �ABC such that � ABC � � deg and� BAC � � deg� Let BL �L � AC� be the internal bisector of� ABC and AH�H � BC be the altitude from A� Find � AHL if� BLC � � AHL�

Solution� Denote � � � AHL � � � BAC � � � ABC and � �

� ACB� From the Sine Law for �AHL and �CHL we getsin �

AL�

sin � ALH

AHand

sin�� deg���CL

�sin � CLH

CH� Since

CL

AL�

BC

BA�

sin�

sin �and

AH

CH� tan � �from � ACB � deg it follows that H lies

on a ray CB�� we get

���cos �

sin ��

sin�

cos ��

It follows from the conditions of the problem that � � � ��

� so

���� � �� �� Thus cos � � � cos������ and ��� is equivalent tocos � cos��� � �� � sin � sin� � � cos��� � �� � cos��� � �� �cos�� � �� � cos�� � �� � � � sin �� sin�� � �� � cos�� ��� cos �� � � cos ���� sin �� sin�� � �� � cos�� � ��� � �There are two cases to consider�

�� cos �� � � From � � deg it follows that � � ��deg��Working backwards we get that �� deg � � BLC � � AHL

��

Page 229: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

for any �ABC such that � ��

�� �� deg��

�� cos �� �� � It follows from � � � ��

�that

��� � sin �� � cotg�

��

But � deg � � �� deg and so

�� cotg�

� cotg deg �

p�

On the other hand also from the conditions of the problem we

get �� deg � � � � ��

�� � deg so deg � � � deg�

Therefore

��� � sin �� � � sin � deg �p�

The inequalities �� and ��� show that ��� holds only if � � deg� In this case � � � � � � � deg� �It is obviously truethat � deg � � BLC � � AHL for any equilateral �ABC��

Problem ����� Find all integer numbers m�n � � such that

� �m�n �m���n

n

is integer�

��

Page 230: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� Letm and n satisfy the conditions of the problem� Sincen is an odd integer number we get �m�n� � � and n � �� Whenn � all m � � such that m � ��mod � are solutions becauseif m � ���mod � then � �m�n �m���n � � � � � � � ��mod ��Let now n � � It follows that m�n �� ��mod n� because otherwise� � m�n � m���n � �mod n� i� e� n� On the other hand � �

m�n � m���n �m�n�� � �

m�n � �and therefore m�n�� � ��mod n�� Let k

be the least natural number such that mk � ��mod n�� Furtherkn�� and k� n so k � n��� Let �n� be Euler�s function� From�m�n� � � it follows that m�n� � ��mod n� so k � �n�� Thereforen�� � �n� � n� � which is impossible�

The required numbers are� n � and all m � � such thatm � ��mod ��

��

Page 231: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Winter mathematics

competition�Pleven� ���

February ����

Dedicated to the One Hundredth Anniversary of the UBM

Problem ���� Let three numbers a� b and c be chosen so thata

b�b

c�c

a�

Page 232: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

a�� Prove that a � b � c�

b�� Find the sum x � y ifx

�y�

y

x� y�

�x � �y

xand the

expression ��x� � ��y � has its maximum value�

Solution�

a�� It is obvious that a �� �� b �� �� c �� �� The �rst equalitygives b� � ac� whence by multiplying both sides by b we getb� � abc� Similarly a� � abc and c� � abc� Hence a� � b� � c�

and therefore a � b � c�

b�� By multiplying both the numerator and the denominator ofthe second fraction by � and using the result of a�� we obtainx � �y� Thus ��x� � ��y � � ����y� � �y � �� � � ����y� ������ and its maximum value is � when y� � � ��Therefore y � �

�and x � �

�� i� e�� x� y � �

Problem ���� In the acute triangle �ABC with � BAC � �deg�BE �E � AC� and CF �F � AB� are altitudes� Let H� M andK be the orthocentre of ABC and the midpoints of BC and AH�respectively�

a�� Prove that the quadrangle MEKF is a square�

b�� Prove that the diagonals of the quadrangle MEKF intersectat the midpoint of OH� whereO is the circumcentre of�ABC�

c�� Find the length of EF when the circumradius of �ABC is ��

Page 233: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution�

a�� The segments EM and FM are medians to the hypotenusesof �BCE and �BCF and therefore EM � FM � �

�BC�

Similarly� for �AHE and �AHF we get EK � FK � �

�AH�

Since � BAC � �deg� we �nd that �AEB and �CEH areisosceles� Hence AE � BE and EC � EH� i� e�� �AHE ���BCE� Therefore EK � EM � Thus MEKF is a rhombus�Furthermore�

� MEK � � MEB � � HEK � � CBE � � HEK

� � EAH � � HEK � � EAH � � AHE � ��deg�

i� e�� the quadrangle is a square�

b�� It follows from a�� that the intersecting point S of the diagonalsof the quadrangle MEKF is the midpoint of both diagonals�Since�AEB is isosceles� E lies on the axis of symmetry of thesegment AB and therefore EO�AB� i� e�� EOkHF � SimilarlyFOkEH� Thus the quadrangle EOFH is a parallelogram�From the above we conclude that S is the midpoint of OH�

Page 234: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

c�� a�� implies that in the acute triangle �ABC with orthocentreH and � BAC � �deg it is true that AH � BC� �AFEis of the same type and therefore EF � AO � �� �It followsfrom b�� that O is orthocentre of this triangle��

Problem ���� Let ��� points be chosen on the plane so that out ofany �� it is possible to choose �� that lie inside a circle of diameter ��Find the smallest number of circles of diameter su�cient to coverall ��� points�

�We say that a circle covers a certain number of points if allpoints lie inside the circle or on its outline��

Solution� Consider a regular hexagon with a side of length ��Choose ��� points as follows� the � vertices of the hexagon and��� points inside a circle of diameter � centred at the centre of thehexagon� It is clear that the above ��� points satisfy the conditionof the problem� Moreover any circle of radius � covers at most one ofthe vertices of the hexagon� Therefore the required number is no less

than � �in our case� � circles for each vertex and a single circle forthe remaining points��

Now we shall prove that the required number is no greater than ��Arbitrarily choose points and add other �� for a total of ��� It isclear that there is a circle of diameter � covering at least �� of these�� points� At most � points lie outside the circle and thereforeat least of the initially chosen points lie inside the circle� Thedistance between these two points is no greater than �� We haveproved that among any points there always exist such that thedistance between them is no greater than ��

Page 235: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Now choose a circle of radius � centred in one of the points� Ifthe remaining points lie inside the circle� the required number is �and thus no greater than �� If this is not the case� take anotherpoint outside the �rst circle� If all points lie in the two circles� thenthe required number is and thus no greater than �� Continuing inthis way we either obtain no more than � circles covering all pointsor have � circles and a point that lies outside all circles� Considerthis point and the centres of the chosen circles� There exist pointsamong these such that the distance between them is no greaterthan �� But this is impossible because of the way we chose ourpoints�

Together the two parts of the proof demonstrate that the requirednumber is ��

Problem ���� Find all quadratic functions f�x� � x��ax�b withinteger coe�cients such that there exist distinct integer numbers m�n� p in the interval ��� �� for which jf�m�j � jf�n�j � jf�p�j � ��

Solution� Let f�x� be a function satisfying the conditions of theproblem� Such a function cannot take one and the same valuefor three di�erent arguments �otherwise we would have a quadraticequation having three distinct roots�� Therefore two of the numbersf�m�� f�n� and f�p� equal � �or ��� and the third one equals ���or ���

Case �� f�m� � f�n� � �� f�p� � ��� Without loss ofgenerality we may assume that m � n� Since m� n are roots ofx� � ax � b � � � �� we obtain that a � m � n� b � mn � ��

Page 236: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Subtracting the two equalities

m� � am� b � �p� � ap� b � ���

we �nd

�� � m� � p� � a�m� p� � �m� p��m � p� a� � �m� p��p � n��

Thus the numbers m� p and p� n are either both positive or bothnegative and since m � n� they are positive� Moreover they areinteger and therefore are equal to � and �� or to and �� But sincem�n� p � ��� �� it follows that neither m � p nor p � n is ��� Thereare two cases to consider� m � p � � p � n � � and m � p � ��p� n � � i� e�� either m � p�� n � p� � or m � p��� n � p� �It is obvious that in both cases at least one of m�n� p lies outside theinterval ��� ���

Case �� f�m� � f�n� � ��� f�p� � �� As in Case � we geta � m � n� b � mn � � and �m � p��p � n� � ���� Using similararguments we obtain that either m � p � � p � n � �� or m �p � �� p � n � �� �Without loss of generality we suppose thatjm � pj � jp � nj�� Therefore the two options are m � p � �n � p � � and m � p � � n � p � �� Simple calculations show thatall triples �m�n� p� satisfying the conditions are ��� � ��� ��� �� ����� �� � and ��� � ��� So the functions are f�x� � x� � ��x � ���f�x� � x� � ��x � �� f�x� � x� � �x� � and f�x� � x� � �x � ��

Problem ���� Three points A�� B� and C� lie on the sides BC�CA and AB of �ABC so that AB� � C�B� and BA� � C�A�� LetD be the re�exion of C� in A�B� �D �� C�� Prove that the line CDis perpendicular to the straight line through the circumcentres of�ABC and �A�B�C�

Page 237: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� It su�ces to prove that D is the second intersectingpoint of the two circumcircles� We know that

� A�DB� � � A�C�B� � � � deg�� BC�A� � � AC�B�

� � � deg�� C�BA� � � C�AB� � � A�CB��

On the other hand� A�D � A�C� � A�B and B�D � B�C� �B�A� which shows that A� and B� are the circumcentres of �BC�D

and �AC�D� Therefore � ADB � � ADC� � � BDC� ��

�� AB�C� �

�� BA�C� � ��deg� � C�AB����deg�� C�BA� � � ACB� Since C

and D lie in one and the same semiplane in regard to both A�B� andAB� it follows from � A�DB� � � A�CB� and � ADB � � ACB thatD is the second intersecting point of the circumcircles of �A�B�C

and �ABC� This completes the proof�

Problem ���� All natural numbers from � to ��� inclusive arewritten � times �so that there are � ones� � twos and so on� in the

Page 238: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

cells of a rectangular table with � rows and ��� collumns� so thatthe di�erence between any two elements lying in one and the samecolumn is no greater than �� Find the maximum possible value ofthe smallest sum amongst all ��� sums of the elements lying in oneand the same column�

Solution� Consider the placement of the ones in the columns� Ifthey are all in a single column� then the minimum sum of elementslying in one column is �� Let all ones lie in exactly columns�Therefore there are at least ones in a single column and thus theminimal sum is no greater that � � � � � � � �� If all ones areplaced in exactly � columns� then the sum of all numbers in thesethree columns is at most � � �� � � � � � � � � �� Hence the minimalsum is at most � � � � �� If all ones are placed in exactly �columns� then the sum of all numbers in these columns is at most� � � � � � � � � � � � � � � �� and therefore the minimal sum isat most �� � �� i� e�� � It is impossible to have ones in more than �columns� because in that case the total number of s� �s and �s doesnot su�ce to �ll the remaining cells� Therefore the required sum isat most ��

The following exampleshows that this sum canbe �� consequently theanswer is ��

� � � � � � � ��� � � � � � � � ��� � � � � � � � ��� � � � � � � � ��� � � � � � � � ��� � � � � � � � ��� � � � � � � � � � � ��� � � � � � � � � � � ��� � � � � � � � � � � ���

Page 239: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Problem ���� Find all values of the real parameter a for whichthe equation x���x���a���x�a� � � has three distinct roots x��

x� and x� such that sin��

�x�

�� sin

��

�x�

�and sin

��

�x�

�form

�in some order� an aritmetic progression�

Solution� Since x�� �x�� �a���x� a� � �x� ���x�� x� a���in order for there to be three distinct real roots it is necessary thatD � � � a� � �� Therefore a� � � and thus � � p

� � a� � �� Theroots of our equation are x� � �� x� � ��

p� � a�� x� � ��p�� a��

It follows now that x� � x� � and � x� � � and � � x� � ��

There are two cases to consider�

�� The second term of the progression is sin��

�x�

�� Then

sin��

�x�

�� sin

��

�x�

�� sin

��

sin��

�x� � x�

��cos

��

�x� � x�

��� sin

��

cos��

��x� � x��

�� ��

But�

�jx� � x�j �

p� � a� �

�� and hence

��x� � x�� �

���

���

��� Therefore cos

��

��x� � x��

�� � when x� � x�� which is

impossible� since the roots are distinct�

� The �rst or the third term of the progression is sin��

�x�

��

Then

sin��

�� sin

��

�xi

�� sin

��

�� � xi�

Page 240: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

for i � or �� Hence

sin�

�� sin

��

�xi

�� sin

��

�cos

��

�xi

�� cos

��

�sin

��

�xi

��

After simple calculations we get cos��

�xi

�� ��

� From the re�

strictions for x� and x� we obtain xi � � or xi � � In the �rst casea� � �� which is impossible� and in the second case x� � � x� � �and a� � ��

Thus a has a unique value and it is a � ��

Problem ���� A point C lies on the periphery of a circle� Two pointsA and B are chosen anticlockwise away from C such that if � CAB �� and � CBA � �� the following equality holds�

cos��

� �

�� sin

��

� �

��

Prove that the bisectors of � CAB pass through a �xed point�

Solution� It is easy to see that � � ��deg� � � �deg and tan�

� � � ��deg satisfy the condition for � and �� Therefore the

required point is the midpoint of OE�F ��g� ���

From the premises of the problem we obtain

cos�

cos� � sin

sin� � sin

cos� � cos

sin�

��

Page 241: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

and after dividing by cos�

cos � �cos

�� � �why��� and if cos� � ��

i� e�� � � ���� we have one of the two cases already considered�we obtain

tan�

�� � tan �� � � tan �

It follows in particular that if � is �xed� then � is uniquely deter�mined�

Suppose that tan�

� � Thus

� �� and therefore � � ����

If tan � � �� we get � � ���� which is impossible� since ��� � � ���If tan � � �� we get � tan � � �� � tan �� i� e�� tan � � �� whichis a contradiction�

Therefore tan�

and B lies on CED where � GCF � � FCD

and tan � GCF � ��g� ���

Fix the point B such that � � ���� Let T be the midpoint of thearc CB and let A� be the intersecting point of TF and the circle�

We shall show that A� A� We obtain � OA�F ��

� ���� � �� �

���

���� and � A�FO � ����

����� � ���

� It follows from

��

Page 242: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

the Sine Theorem for �A�FO thatsin

�� � �

�sin

�� � �

�� ���

� � � which is

equivalent to cos��

�� �

�� sin

��

�� �

�� Therefore A� A�

The case of � � ��� can be dealt with by analogy� The conditionfor B to lie on CED shows that A� lies between C and B ��g� ���

Problem ���� Let n be a natural number� Find the number ofsequences a�a� � � � a�n� where ai � �� or ai � �� for i � �� � � � � n�such that ������

�lXi��k��

ai

������

for all k and l for which � k l n�

Solution� It is clear that a sequence having a�k�� � a�k � � for� k n satis�es the condition of the problem� because any sumof the form

P�l

i��k�� ai equals zero� There are n such sequences�Let us determine the number of sequences such that there existsa k for which a�k�� � a�k �� �� Let k�� k�� � � � � ks be all k with theabove property� It is easily seen that if a�ki�� � a�ki � ���� thena�ki���� � a�ki�� � � ��� Therefore all sums a�ki�� � a�ki �and soalso a�ki��a�ki� are uniquely determined by a�k��� � a�k� �there aretwo possibilities for a�k���a�k��� There are two possibilities for anyof the remaining n � s pairs �for which a�t�� � a�t � ��� Thereforethere are

n��n���n

���n��

�n

��� � ���n�k

�n

k

��� � ���

�n

n� �

���

�n

n

��

Page 243: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

sequences with the required property� By adding and subtracting n

to and from the above expression we get�

�n�n

��n��

�n

��n��

�n

�� � � ��

�n

n� �

��

�n

n

���n���n�n

Thus there are � �n � n sequences�

Problem ����� Consider the function f�x� �px �

px� � �p

x� � �px� �� x � ��

a�� Find limx�� f�x��

b�� Prove that f�x� is an increasing function�

c�� Find the number of real roots of the equation f�x� � aq

x��x�

where a is a real parameter�

Solution� a�� By grouping the �rst and third radicals and thesecond and fourth radicals and rationalising we get that when x � ��f�x� � �p

x�px�� � �p

x���px�� � Therefore limx�� f�x� � ��

b�� When x � ��

f ��x� ��

px� �

px� �

� �

px� �

��

px� �

��

px��px���px�� �px��� �

� �

pxpx���px�px��� � ��

��

Page 244: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Therefore f�x� is an increasing function if x � ��

c�� It follows from a�� and b�� that f�x� � � when x � �� i� e��the equation could have a solution only if a � �� Let a � �� The

function g�x� � aq

x��x

� aq�� �

xis decreasing and continuous and

limx�� g�x� � a � �� Since f�x� is an increasing and continuous func�

tion and limx�� f�x� � �� in accordance with the Bolzano�Weierstra�

Theorem the equation f�x� � g�x� has a solution �and it is a uniqueone� exactly when f��� g���� i� e�� if �� �p�� a � ��

Problem ����� The convex quadrangle ABCD is inscribed in acircle with centre O� Let E be the intersecting point of AC and BD�Prove that if the midpoints of AD� BC and OE lie on a straight line�then AB � CD or � AEB � ��deg�

Solution� It su�ces to prove that if � AEB �� ��deg� then AB �CD� Let � AEB �� ��deg� If O E� then ABCD is a rectangleand therefore AB � CD� Suppose O � E� Let M � N � P � Q be themidpoints of AD� BC� AC� BD� respectively� and R be the inter�secting point of the straight lines through P and Q perpendicularto BD and AC� respectively� It is clear that MPNQ and OPRQ

are parallelograms� Therefore the midpoints of MN and OR coin�cide with the midpoint of PQ� and since the midpoint of OE lieon MN � we get that REkMN � On the other hand R is the ortho�centre of �PQE and therefore RE�PQ� Hence MN�PQ� i� e��the parallelogram MPNQ is a rhombus� It is easy to see now thatAB � PN � NQ � CD� which solves the problem�

��

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Note� The above solution shows that if O is the intersectingpoint of the axes of symmetry of AC and BD� then the assertion ofthe problem and its opposite are true for a quadrangle that is notinscribed in a circle� This could be demonstrated by using complexnumbers or trigonometry�

Problem ����� Let famg�m�� be a sequence of integer numberssuch that their decimal representations consist of even digits �a� � �a� � �� a� � ��� � � �� Find all integer numbersm such that am � �m�

Page 246: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� Let m be an integer number such that m � b�� b� � �� � �� bn � n� Denote f�m� � b��b� � ��� � � ��bn � ��n� It is clearthat ff�m� j m � Ng is the set of integer numbers with only evendigits in their decimal representation� Since f�m�� � f�m�� ��m� � m�� it follows that am � f�m� for any m� Therefore it su�cesto �nd all m such that

��b� � b� � � � � �� bn � n� � b� � b� � �� � � � �� bn � ��n�

i� e��

��� ��b� � b� � � � � �� bn � n� � b� � b� � �� � � � �� bn � ��n�

Since b��b���� � ��bn�n n���� and b��b������ � ��bn���n � ��n�it follows from ��� that ��n�� � �� � ��n� i� e�� � � n�� � ��n� Thusn � �� and therefore n �� If n � �� we get from ��� thatb� � �b� � ��b� � �b� � ��b� � ��� which is imposible� In thesame way it is easy to show that n � �� i� e�� n � �� In this caseb���b����b� � �b�� Obviously b� � � and b� � b�� because b��b� isdivisible by � As a result we have the equation b��b� � ��� and itssolutions are b� � � b� � � and b� � �� b� � �� Therefore all integernumbersm with the required property are m � � ��� ���� ��� and m � � � � � � � � � � � � ��

��

Page 247: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Winter mathematics

competition�Varna� ����

Problem ���� Find all natural numbers x and y such that�

a��

x� �

y�

��

b��

x�

y�

��

xy

Solution� a� The equation is equivalent to �y � �x � xy so

x ��y

y � ��

�y � �� �

y � �� �� �

y � � Therefore y � � � � and thus

y � � Hence there is a unique solution x � � y � �

b� Let x � y be a solution of the problem Now�

��

x�

y�

xy�

y� �

xy�

y � �

xy�

y

xy�

x giving x � � When x � � x �

or x � � no solution exists When x � � it follows that y � � andx � � implies y � � If y � x we apply the same reasoning The

Page 248: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

problem has four solutions�

x � �� y � ��x � �� y � ��x � �� y � ��x � �� y � ��

Problem ���� Given an acute �ABC with centroid G and bisec�tors AM�M � BC�� BN�N � AC�� CK�K � AB� Prove that oneof the altitudes of �ABC equals the sum of the remaining two ifand only if G lies on one of the sides of �MNK

Solution� We shall repeatedly usethe following property� A segment con�necting a vertex of a triangle with apoint on the opposite side divides thetriangle into two triangles such that theratio of their areas equals the ratio ofthe parts into which the point dividesthe side

Let G � MN and G� G� G� be the projections of G on BC ACand AB respectively Further denote the projections of N on BCand AB byN� and N� and those ofM on AC and AB byM� andM�We shall prove that GG� � GG��GG� and from the above propertyit will follow straightforwardly that the altitude from C is equal tothe sum of the remaining two altitudes We obtain

GG�

NN�

�CM �GG�

CM �NN�

�SGMC

SNMC

�GM

NM�

By analogyGG�

MM�

�GN

NM implying that

GG�

NN�

�GG�

MM�

� � so

��� GG� �MM� �GG� �NN� �MM� �NN��

Page 249: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Let M� and G� be the projections of M and G on NN� It easilyfollows now that

NN� �GG�

NN� �MM�

�GN

NM�

GG�

MM�

Further using that NN� � NN� and MM� � MM� we obtainGG� � MM� � GG� � MM� � GG� � MM� and therefore GG� �GG� �GG�

Conversely let the altitude through C be the sum of the remain�ing two Now GG� � GG� � GG� If G� � GG� �MN then itfollows straightforwardly that the sum of the distances from G� toAC and BC equals to G�G� It is easy to check now that G� �� G

Problem ���� Let n be a natural number Find all integer valuesof m such that k � m�� is integer and A � ����k � � is a sumof the squares of n integers �not necessarily distinct and di�erentfrom zero�

Solution� It is su�cient to consider only nonnegative values of m

�� n � � and k �m

is integer only if m � �p and m � �p�

If m � �p then A � � � ���� � ���p � � and it follows by inductionthat A is of the form A � �a � � so A is congruent to � modulo �We conclude that A is not a perfect square If m � �p � thenA � � � � �� � ���p�� � � and it follows by induction that A is ofthe form A � �a� � Therefore A is not a perfect square because� divides A but � does not

� n � and k � m is integer for any m Now A � ����m � �When m � � we get A � � which is not a sum of two squares

Page 250: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

When m � � we obtain A � ��� � ��� � ��� If m � thenA � � � ��� � ��m � � and as above A is congruent to � modulo �On the other hand the sum of two perfect squares is congruent to� � or modulo � and so no solution exists in this case

�� n � and k � m is integer for any m Now A � �� � �� ����m�� which can be written in the form A � �a�� Therefore Ais congruent to � modulo � whereas a sum of three perfect squaresis congruent to � � � � � or � modulo � Thus no solution existsin this case

�� n � � and k is integer for any m Now A � �����m�n�� �� � � � �� � �� and if a� � ����m�n�� � a� � � a� � a� � �� a� � a� �� � � � an � � then A � a�� � a�� � � � �� a�n

Answer� if n � � no solution exists�if n � there is an unique solution m � ��if n � � no solution exists�if n � � any m � � is a solution

Problem ���� Let p be a real parameter such that the equationx� � �px � p � � has real and distinct roots x� and x�

a� Prove that �px� � x�� � p � �

b� Find the least possible value of

A �p�

�px� � x�� � �p�

�px� � x�� � �p

p��

When does equality obtain�

Page 251: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

Solution� a� It follows from the equation that x�� � �px��p andso �px� � x�� � p � �p�x� � x�� � �p� � � The inequality is strictbecause otherwise x� � x� � �

b� As in a� we obtain �px� � x�� � �p � �px� � x�� � �p ��p� � �p � � �the last inequality follows from the conditions of theproblem x� and x� to be distinct and real giving p � �� Therefore

A �p�

�p� � �p�

�p� � �p

p��

�from the Arithmetic�Geometric Mean Inequality� and equality ob�tains when �p� � �p � p� i e when p � ���

Problem ���� Given an acute�ABC such that AC � BC letMbe the midpoint of AB and let CD AP and BQ be the altitudesDenote the circumcircle of �PQC by k� and the circumcircle of�DRP by k� where R is the point of intersection of AB and PQProve that�

a� MP is tangent to both k� and k�

b� RH CM where H is the orthocentre of �ABC

Solution� a� Note that H � k� Since � APM � � PAM ���� � � ABC � � BCD we obtain that MP is a tangent to k�On the other hand � MPD � � MPB� � DPB � � MPB� � DPB �� MBP�� BPR ��ABC � �DBP � so � ARP � � MBP�� BPR �� MBP � � QPC � � MBP � � BAC��ABC � �PQC� Therefore� MPD � � MBP and thus MP is a tangent to k�

Page 252: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

b� Let L � CM � k� It follows froma� thatML �MC �MP � �MD �MRWe conclude that L lies on the circum�cicrle of �DRC and therefore RL CM Further HL CM since HC isa diameter of k� Hence RH CM

Problem ���� A square table �lled with nonnegative �not neces�sarily distinct� integer numbers is said to be a magic square with summ if the sum of the numbers in each row and each column equals mProve that the number of magic squares � � � of sum m such thatthe minimal element among the elements on the main diagonal lies

in the centre is

�m� �

Solution� It is evident that knowing the elements of main diagonaland the element in the cell ��� � �see �g �� one can determine allelements in the table Indeed there is a unique choice for all remain�ing cells �see �g � Therefore it su�ces to see when all elementsare nonnegative and b is the minimal element among the elementson the main diagonal

a d

b

c

a d m� a� d

m� c� a� b� d b a� d � c

b� d� c m� b� d c

Fig � Fig

It is clear from �g that the following inequalities hold�

��� a� d � m�� � b� d � m�

Page 253: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

��� c � a� d���� c � b� d���� a� b� d � c � m

The conditions of the problem imply b � a and b � c It isclear now that ��� follows from ��� and � � and ��� follow from ���Therefore we can consider only ��� and ���

Consider the following chain of inequalities

b � b� d� c � a� b� d� c � a� d � m

�the �rst follows from ��� the second from b � a the third fromb � c and the fourth is equivalent to ���� It is easy to see thatknowing the quadruple �b� b�d�c� a�b�d�c� a�d� we can uniquelydetermine a� b� c and d and so �nd a magic square Therefore therequired number equals the number of quadruples which is

�m��

Problem ���� Find all values of the real positive parameter asuch that the inequality acos�x � a� sin

� x � holds for any real x

Solution� We know that acos�x � a� sin� x � a��� sin

� x � a� sin� x �

a

a� sin� x

� a� sin� x Substitute t � a� sin

� x Since � � sin� x � �

we obtain that t is between � and a� Our inequality now becomesa

t� t � � t� � t � a � � Since it holds true for any x

�i e for any t between � and a�� it follows that the roots of f�t� �t�� t� a � � lie outside the open interval determined by � and a�Therefore f��� � � and f�a�� � � The �rst inequality gives a � �and the second one implies a� � a� � a � � � a� � a � � �� � �a����a��a��� � � Since a � � we obtain a��a�� � �

Page 254: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

The solution of this inequality is a �����p�

��� �p�

� So we

obtain a ������ �p�

Problem ���� Let CH and CM be an altitude and a median ina non�obtuse �ABC Let the bisector of angle BAC meet CH andCM at points P and Q respectively If � ABP � � PBQ � � QBCprove that�

a� �ABC is a right triangle�

b� BP � CH

Solution� a� Let R � BC�AP T � AC � BP and S � AC �BQ Denote AB � c BC � aCA � b It is easy to see thatP lies between A and Q �otherwise� ABP � � PBQ� It follows fromCeva�s Theorem for point P that�

���AH

HB� BRRC

� CTTA

� � � b cos�

a cos �� cb� SBTCSABT

� �

� b cos�

a cos �� cb� BT � a � sin

��

BT � c sin �

� � � cos�

cos �� sin

��

sin �

� �

� cos�

cos�� sin

�cos beta

sin �

� � � cos�

cos��

cos �

Page 255: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

It follows from Ceva�s Theorem for point Q that�

� �

AM

MB� BRRC

� CSSA

� � � c

b� SBSCSABS

� � � c

b� BS � a � sin

BS � c sin ��

� �

� c

b� a sin

c sin ��

� � � a sin �

b sin �

�cos �

� � � a

b� cos

��

Now ��� and � � implycos�

cos��

b

a From the Sine Law we obtain

sin�

sin��

b

a so

cos�

cos��

sin �

sin�� sin � � sin � If � � � then

the triangle is isosceles and therefore P � Q implying that � ABP �� PBQ � � QBC � �� which is impossible Thus � � � � ��� andtherefore � ACB � ���

b� It follows from �BCS that cos�

��

a

BS Combining the

above with � � gives b � BS Note that �ABC � �BHC whichimplies

BP

CH�BS

AC� Therefore BP � CH

Problem ���� LetA be a set of natural numbers with no zeroes intheir decimal representation It is known that if a � a�a� � � � ak � Athen b � b�b� � � � bk where bj� � � j � k is the remainder of �ajmodulo �� belongs to A and the sum of the digits of b equals thesum of the digits of a

a� Prove that the sum of the digits of a k�digit number in A equals�k

Page 256: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

b� Find the smallest k�digit number which could be an elementof A

Solution� a� Let a � a�a� � � � ak be a k�digit number from A thesum of whose digits is S

Consider the following numbers� b � b�b� � � � bk� c � c�c� � � � ckand d � d�d� � � � dk where bj� � � j � k is the remainder of �ajmodulo �� cj� � � j � k is the remainder of �bj modulo �� anddj � � � j � k is the remainder of �cj modulo ��

By the conditions of the problem all b c and d belong to AFurther

��� S �kX

j��

aj �kX

j��

bj �kX

j��

cj �kX

j��

dj�

Direct veri�cation shows that for �xed j the sum aj � bj � cj � djis equal to � �e g if aj � � then bj � �� cj � �� dj � � andtherefore aj � bj � cj � dj � �� It follows now from ��� that �S �Pk

j�� aj�Pk

j�� bj�Pk

j�� cj�Pk

j�� dj �Pk

j���aj�bj�cj�dj� � �kTherefore S � �k Q� E� D

b� We shall prove that the required number is a � a�a� � � � a�twhere a� � �� a� � �� � � � � at � �� at�� � �� at�� � �� � � � � a�t � � ifk � t and b � b�b� � � � b�t�� where b� � �� b� � �� � � � � bt � �� bt�� ��� bt�� � �� � � � � b�t�� � � if k � t� � It is easy to see that a and bcould be elements of a set having the required property

Let k � t and suppose there exists c � c�c� � � � c�t � A suchthat c � a Since there are no zeroes among the digits of c weobtain c� � c� � � � � � ct � � But it follows from a� that thesum of the digits of c is �k � ��t The last is possible only if

��

Page 257: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

ct�� � ct�� � � � � � c�t � � Hence c � a a contradiction with thechoice of c

Similarly suppose k � t� � and there exists c � c�c� � � � c�t�� �A such that c � b Since there are no zeroes among the digits of cwe obtain c� � c� � � � � � ct � � But it follows from a� that thesum of the digits of c is �k � ��t � � The latter is possible only ifct�� � � and since c � b it follows that ct�� � � It is easy to seenow that ct�� � ct�� � � � � � c�t � � Hence c � b a contradictionwith the choice of c

Problem ����� Given the sequence an � n�apn� � �� n � �� � � � �

where a is a real number�

a� Find the values of a such that the sequence fang�n�� is conver�gent

b� Find the values of a such that the sequence fang�n�� is mono�tone increasing

Solution� a� If a � �� the sequence fang�n�� is convergent be�

cause an � n �pn� � � �

��n �

pn� � �

� ��

n

� � �

n�

� � when

n�� Conversely let the sequence fang�n�� be convergent Sincean � n � p

n� � � � �a � ��pn� � � we get that the sequence

�a���pn� � � is also convergent Since

pn� � ��� when n��

it follows that a� � � � so a � ��b� Let fang�n�� be a monotone increasing sequence i e an�� �

��

Page 258: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

an for any n This inequality is equivalent to

��a� n� ��q

�n � ��� � � �pn� � �

� ���

Since

limn��

n � �q�n� ��� � � �

pn� � �

� limn��

� �

nq�� � �

n�� � �

n��q� � �

n�

� �

it follows from �� that a � ��Conversely let a � �� It follows from n� �q

�n� ��� � � �pn� � �

n � �

n� � � n� � that �� holds true so the sequence fang�n�� is increas�

ing The required values of a are a � �������

Problem ����� Given a �ABC with circumcentreO and circum�radius R The incircle of �ABC is of radius r and touches the sidesAB�BC and CA in the points C�� A� and B� Let the lines deter�mined by the midpoints of the segments AB� and AC� BA� andBC� CA� and CB� meet at points C�� A� and B� Prove that the

circumcircle of �A�B�C� is of centre O and radius R�r

Solution� We show �rst that the projection B� of B� on AC is themidpoint of AC Let Ba and Bc be the midpoints of AB� and CB�We shall use the standard notation for the elements of �ABC

Page 259: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

We obtainBaB�

B�B�

� tg�

r

p � a

andBcB�

B�B�

� tg

r

p� c so

BaB�

BcB�

�p � c

p � a Since BaBc �

b

it follows that BaB� �p � c

CBc

and BcB� �

p � a

ABa

which gives AB� � CB� There�fore B�O � B�B� � B�O ��p � c��p � a�

r�R cos �

We shall show that the above expression equals R�r

and by analogy

A�O � C�O � R �r

which will complete the proof

We obtain that�p � c��p� a�

r�R cos � �

r

�R � S

�p � b��

S

p� R�� � cos �� � Sb

p�p � b�� R sin�

� r

p � b�

�R

bsin�

� tg

sin� �

sin�� sin� � sin

cos

which is

a true equality

Problem ����� Find the smallest natural number n such that thesum of the squares of its divisors �including � and n� equals �n����

Solution� It is clear that n has at least three divisors and let � �d� � d� � � � � � dk � n be those di�erent from � and n The

��

Page 260: BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS · BULGARIAN NA TIONAL OL YMPIAD IN MA THEMA TICS Third round Problem Let p and q b e p ositiv en um b ers suc h that the parab ola

conditions of the problem imply

�� d�� � d�� � � � �� d�k � �n � ��

Let n � p� where p is a prime number It follows now from �� thatp� � p� � � � � � p���� � �p� � � so pn� and therefore p � Theabove equality implies � � p� � p� � � � �� p���� � �p��� � whichis impossible

Therefore k � �� �� � because otherwise the number of divisorsof n equals � � � i e n � p� n � p� or n � p� where p is aprime number Suppose that k � � Since didk�i � n it followsfrom �� that �dk�� � d��� � �dk�� � d��� � �dk�� � d��� � � Thelast inequality is impossible since the numbers dk�� � d�� dk�� � d�and dk��� d� are distinct �if for example dk��� d� � dk��� d� � Athen d��A � d�� � d��A � d�� so d� � d�� We conclude now thatk � or k � �

Assume k � � Then n has � divisors and thus n is of the formn � p � q� where p and q are distinct prime numbers �n is not ofthe form n � p�� It follows from �� that

�� p� � q� � q� � p�q� � �pq� � ��

If q � � then q� � p�q� � pq� � ��pq� � �pq � � and thereforeq � or q � � Direct veri�cation shows that inequality �� isimpossible Thus k � and hence n � pq where p and q aredistinct prime numbers such that

p� � q� � �pq � ��

Since qnp� � � it is easy to see that if p � �� then p � � q � ��and n � �� Since ��� � �� � �� we conclude that the smallestn with the required property is n � ��

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