Olympiad Corner~Oi1i~.~.fiJ c=)

!1RsJM

+ ;\i!t*C!f;f , JIM:!: ill T -fft*~~~~tJi:(Euler) 0 f&~~':(£=+~~!f;fB~§~aJI ' {g.~gj]!c:£*~~~~ ' ~ftRM 0 ~~J:~Ij)~

:@. ..i}A:fD.1J;l;t.i)~fm~:fJ1. ..majj~ ~ ajj (JIg 0 ~ ~ :fJ1. .:(£ ~f/ if?d (JIg ~ ~

tJi:,fj (Euler Line)- -~.~=-~M~OO.m:~(JIgI5(J1g~iV&o

1997 Chinese Mathematical °.zympiad:

Part I (8:00-12:30, January 13, 1997)

Problem 1~ Let ;XI, X2, ..., X1997 be realnumbers &atisfying th~ following twoconditions: -

1 -

(1) -~~Xi ~J3 (i= 1, 2, ..., 1997);-v3

(2) XI +X2+ ...+X1997 = -318J3~

Find the maximum value of

12 12 12XI +xz +"'+XI997'

stfl.ii ' AH:fa A 'O51.8U ~ffi {~~~%ABC;f[]A'B'C'~J;J~~ ' mPj,

AH:A'0=BC:B'C'=2:1

f'tiJ:5t1!? ' :l;.JL\ G -tl?reJt:l~ AA' 9f.fjX,

AG:A'G=2:1

1E11I:t. ' =- ~ % HAG;f[] OA'G ffi ~ 0

8311: t.:Ift ~~-ff~~.:=.~~ ABC, ;;:m~E~.:=. ..iif1'9 ~~ (mid-poinJS),A" B' ~Dc' $t;8tl1~ m~.:=. ~jj~.¥ iR Sf, $t ~(perpendicular bisectors) 0 ~1r}~rn :

~.:=...¥iRSf,:i1-~mx~~~I!i'llPIII-~15 0 0 ~15 0 ~~.:=.~~ABC ~'J}~IIJL\ (circumcentre)' rn}J.mfg**B~;;Ho

A

Problem 2. Let AIBI C1D1 be anarbitrary convex quadrilateral. Let P bea point inside the quadrilateral such thatthe segments from P to each vertex formacute angles with the two sides throughthe vertex. Recursively define Ak.Bk. Ckand Dk as the points symmetric to P withrespect to the lines Ak-1Bk-l. Bk-lCk-l.Ck-'IDk-l and Dk-lAk-l. respectively(k=2.3 ).

LHGA = LOGA'

mJ.-.1 0 ' G' H JjJ(.-~~ ' mm~lli~, ~ROG:GH= 1:2 0

A

B C, (III.=.)", r'I B~..."

~m~OH~~l5m1'i¥-}L,'E~:I=::g~ fL IS iii (Nine-point circle) ~IiI/L' 0 m~1L!!i1il~1~-OO;;m~=.

:;llj ~ ABC ~ =.~ ~~!!i A', B',c', =.~~.¥.ff:. D' E' FPJ&=.m~;fO.¥/L,r~ ~ ~ IS K' L' M ~ III(liJlZ9) 0

(continued onpag(! 4)

/ ' '0-,~ '"~'-+-:~. ~

L-r "~;,-~B / ," :A" c (II-)

.~-1iOO ~ .=-;:I1!J7f;:;i A'B'C';fDABC:;;p{gffifJ'), , ffij1i.f;JLI!.if~1'J 0 ~{I.if

~./j\~~rI"J.=-;:I1!J7f;:;iA'B'C' fim.=-;:I1!J 7f;:;i ABC rI"J ~ I!i .=- ;:I1!J 7f;:;i (medialtriangle) 0 E rI"J .=- f~;@j (altitudes) I;n

tlfi,t~.:t£ OA' , DB' :fa OC'L ' I!I JJ:t

0 l!it!?ffi~7 ~1!i.=-;:I1!J7f;:;i A'B'C' rI"J¥;JL' (orthocentre);:I1!J@ (111=) 0

([lJIZY)

:fiI ~ ~fL 15 m ~ ~ # 11/ a';J::7f; ~ m aJj~~¥~W.~tl (Poncelet) ~ 1821~§Jt.t(tIili a';J , f&~ A', B', C',K, L, M ~155}J3X".R:fiIm:m:lZYl!im1i!i ' ~ f& m aJj ~ 'fIm m 1i!i a';J IZY I!i # iii, ~;fIJffl~.=. 'fImm-g-a';Jm:m:11maJj

~.=. 'fIm fil. 'N L ~ ~ -'fIm fil ' JWi: 1:&maJj D , E, F-&:(£~IIIL 0 .~'frJ~~f&a';Jm~:

Mathematical Excalib!!! '-- Vat. No.1 J@-Feb, 97 ~age 2

~ Bijj ~ IX tt'J ~ W (.=>(continued from page 1) .J2 :;I;.J.I.a9~1I1I~

:wm*~~~*.X5!li

:t£=.:fiJ ~ ABH J:t:1 ' c' f[) L:St.8tl ~.if AB f[) HB (I/;J J:t:11!i ' 1!l1l:t. C'L SJZ fTAH 0 rm:tJ. ' :t£ =. :fiJ ~ ACH J:t:1 'B'M SJZfT AH 0 ffl Pj, C'L SJZfT B'M 0W ~ 1.:1. =.::fiJ ~ ABC f[] HBC , flJ mrm fl (I/;J J:t:11!i ~:tJ. ' DJ:;I;:D B'C' SJZ fjML f[) CB 0 EI3 ~ AD ~ ~ BC, 1!l1l:t.B'C'LM~OO;;§~ 0 ;;g~(l/;Jml!i'M"~

#[;]0

r 1lofiiJml3J3 .J2 ~_$II:IVE? J

r B~ji71'?ff~ I ~ .J2=m/n , "iiJ"iI mfa n 1'~~fMII: 0 Ei:I* m2 = 2n2, m ~\~fM:fi ' ~f~ 2k, ~U 4t2 = 2n2, 2k2 =n2 , n:Ic n /J)' ~fMI'i. ' ~ J§ I J

J::jzI!;~aj3 , Rmitl~.f~'r:I:.~ ' *~B~"5Jfi~ Q 52.m.:t~~ (ARISTOTLE)f£1}ftBfj 330 ~~;fJre'E (PJ.~fPJ~~) a~*, m{~&~~~~~'"5J~f£~~1I~~3i@J$B~~ffl~~T Q ~

., ~E!:j~~~JJ}3~:fl[]IltMiJ? ' m~fx:~!J!.*~f§{~~~:;;r\"~~@J*~~~~~ ' [ffi

ma~ff~~~~A'C'KMfDA'B'KL ~ r *f!fLJJ}3 J ~m~ Q"CJ~~~fr~-tt:?~~§~ ' lZlJ!:t/if"Ztl#iii Q f.§.~-=-flW~~~:::fj#~ti~ f£~OO~JJ}3~, 2 ?~~trM~3U ' ~T~ ' RnJ;t~liJ(circumclrcle)~~r~ ~g-1I~:fI.' rP].~,~,~17J"5J~m,(III-'-) Q 9-~¥J~~{~'r:I:.~~W:.~ ' .:@miU~/'\ [!J-1-PfE-.)tm'r:I:.~ Q :jIj::ftf.IIi/j/WF ' ~~

~ijg~~JJ}3::e: PI ' ..., P s ~ s 1I~rP]~

~:fI. ' ~U ~-:::-P; ~;4!\1;;~. Q [!JIlt, ::e:

H:;;r\"~:J\:=;~~1J' ~u.JH~;4!\I;;~fx:Q~.' :fl[]*l!j.~.~:i:lm~f!1-1-PfE-.)tm'r:I:.~' ~~g-1I~JJ}31h$~' ~P~.-:fI. m2 = Hn2 m'~~~.[!J-1-te~~-*fx:'-~-1M'~J§I

A

B n c: (rilE)

~* ' ~f&.r"I:Jitl'Ea"J~fPJ~~ ' jiOO

~fPJ~~'~~~~~=T~~~~.fi.~*jj:m~jj~a"Jti~ ~fD.iI~ ~PJ1}~.a"J~~~! ~PJ1}~., ~1~~ff;(£-1}#~. ' ~Jj~~fD.iI~~~~1}#~.a"J~T~fi.ffi'~~~~,.J2 ~ ~~~fi. 0 (PJ. ~~,ZI!;. , ~;j:t~H. EVES a"J::I=f~ "ANINTRODUcrIONTO THE ffiSTORY OF MA11ffiMATICS"a"J~ 3 .' 3rd edition, 1969 0 ) ;(£~

IIII=P~AP~~15~a"Jf;j~~ACfa.ilAB a"J1}#~:m ' RP~AC=jAP fa AB= kAP 0 .f~ BICI ~ BICI.¥;lR~AC,~ff CB = CBI 0 ~.~m BCI = BICI= ABI ' rtlJJ:t.

ACI=AB -ABI = AB -(AC-AB)= ME -AC= (2k- flAP,

ABI = AC -AB = (j -k)AP 0

?!~ : ACI fD ABI ~-OO~/J\a"J~jj~a"Jf;j~~fD.iI ' ~~OO~/J\a"J~jj~a"J-iIABI/J\~}g:~jj~a"J-iI AB a"J-~ 0 ~JJ:t.7Ji~m~~~ ' ~\~itl-OOff:.~/j\a"J~jj:Jf:;i; , 'Ea"J-iI ABt IJ\~ AP, f'§' ABtW17Jf&~ AP a"J~T~fi.ffi ' ~;jF~~I (~@fi.¥~*~mE~~.fi.~*~{£IIJj::J.JJ:tjj~~~~PJ 1}~:m~ ~, ;f:§"¥;~{£IIJjj:~~*mf~~5tfi.~OO

~a"J~~oPJm"¥;~a"Jfi.~*~~m~'~~~r*~~~~';(£E~.fi.~

~::I= (EUCLID'SELEMENTS) a"J~wr~f:tcmPJ~ I )

A

B D C (111:;;"\)

:;r\"~tJ"1fi1:;r\"DJ~g::'Fif#~~f~ ' ~!i:tA', B', C', K, L, M~m#iI 0 ~-1500 ' LA'DK~~~(III[g) , iliJA'K~MjlI!;~miltJ"1~f~ ' ~!i:tD{t!. :(£!i:t ~ I!i 1111: 0 ~ ~, E:fO F {t!.:(£!i:t~~fil1: ' fifrP)'fL~#fiI 0

.ft~*~m=1I~SJ3 0 ~.JH = m/n, "PI'i" m fD n ~/;}#I!IT 0 EB~ m2 =

Hn2 = n(Hn) , n ~\~~ 1 ~ -1 ,~p~

NH~1I%~~$' ~~ I ~1ImSJ3mFJUm1lmSJ3:fi -l!i~.f§ /PJ ' ~tgmIJ::¥.~-m.'it:A/;J'~R ' mSJ3T :E=:fi~fIl.~f\:fIl.~fIl. ' ~U~~\~~fIl. 0 (f\:fIl.~fIl.~t~

~-~fIl.*fIl.~~.J:\$m 1'+CN-lx""-l+ ...+c.x+co=O A/;Jm ' f9lJllD.JH ~ ;- -H =0

A/;Jmo~~~~~fim~~@$~o)

fLl5li] :fa ~ :tV. ~ ::Iff ~)if 11m f* ?

*~~tfj*IBJL\.tt~mOO rn 1S~:(£fLI!i III 1:.. ~ .=. f!!J % A 'B'C' :fa KIM (iii-t;) 0 E8~ KA' , LB':fa MC' ~fLlS

Ii] at ~ 1~ ' lE11l:t.=. f!!J % KIM m. fL

15 iii ~ iii JL\ ~" 1800 "6J ~ .=. ffJ ]f;::;A 'B'C' 0 .=..ffJ]f;::; ABC ~ ~:tV.~ OH m

Hiffl1~ tSlE $t 8tl ~ .=. f!!J ]f;::; A 'B'C' :faKIM~~JL\ ("6J~~IiI=&IiIIZY) ,

lE11l:t~~~.=.f!!J]f;::; A'B'C':fO KIM ~

JJL1!I5'E'fr:IJ~I=PI5:§tt.~fLl5li]~iii JL\ 0

c

pV'"A D

~?-1g : AC1IAB1 = (2k -j)/(j -k) , ffij"m = j -k :IE~t-*.Wr~~ ~m/J\mt~ma)3

~ ~ m 0 [!l:m AC1IAB1 =.J2 ' f£ ~

(2k -fllm =.J2 ' ~p~ m.J2 = 2k- j ~

~f;;T o"lifJ(;1MT~t-*.Wr~~ma)3~~~~~W~~ ' ~~~p)re~m~~m

:;"\1Wma)3 : ;:S:.J2 = jlk ~MlM~ :StfJI:~

, ~u~.J2 = (2k -fll(j -k) (:Ii~[!l:m

j.J2 -k.J2 = 2k -j) , f.§. k < j < 2k

(~:m 1

~

Solution: CHEUNG Tak Fai (ValtortllCollege, Form 6) and Gary NG KaWing (STFA Leung Kau Kui College,Form 4).

Problem 48. Squares ABDE and BCFGare drawn outside of triangle ABC.Prove that triangle ABC is isosceles ifDG is parallel to AC.

Problem CornerWe welcome readers to submit solutionsto the problems posed below forpublication consideration. Solutionsshould be preceded by the solver's name,address; school affiliation and gradelevel. Please send submissions to Dr.Kin-Yin Li, Dept of Mathematics, HongKong University of Science andTechnology, Clear Water Bay, Kowloon.The deadline tor submitting solutions isApr. 5, 1997.

Solution: Henry NG Ka Man (S1FALeung Kau Kui College, Form 6), GaryNG Ka Wing {S1FA Leung Kau KuiCollege, Form 4) and YUNG Fai(CUHK).

From B, draw a perpendicular line to AC(and hence also perpendicular to DG.)Let it intersect AC at X and DG at Y.Since LABX = 900 -LDBY = LBDYand AB = BD, the right triangles ABXand BDY are congruent and AX = BY.Similarly, the right triangles CBX andBGYare congruent and BY = CX. So AX= CX, which impliesAB = CR.

Suppose

X13 + x + 90= (~ -x + a)q(x),

where q(x) is a polynomial with integercoefficients. Taking x = -1, 0, 1, we get

88 = (2+a)q(-1),90 = aq(O)

and 92 = aq(l).

Since a divides 90, 92 and a+2divides 88, a can only be 2 or -1.. NowX2 -X -1 has a positive root, butX13 + x + 90 cannot have a positive root.So a can only be 2. We can check bylong division that ~ -x + 2 dividesX13 + x + 90 or observe that if w is any ofthe two roots of x2 -x + 2, then..J-=w-2, W4 = -3w+ 2, W8 = -3w- 14,W12 = 45w-46 andw13 + w+ 90 = O.

Problem 51. Is there a positive integer nr--r--

Problem 52. Let a, b, c be distinct realnumbers such that a3 = 3(b2+c1 -25,b3 = 3(c2+a1 -25, C3 = 3(a2+b1 -25.Find the value of abc.

Problem 53. For MBC, define A' onBC so that AB + BA' = AC + CA' andsimilarly define B' on CA and C' on AB.Show thatAA', BB', CC' are concurrent.(The point of concurrency is called theNagel point of MBC.)

Other commended solvers: CHANMingChiu (La Salle College, Form 6), CHANWing Sum (HKUST) and WilliamCHEUNG Pok-man (S.T.F.A. LeungKau Kui College, Fonn 6).

Problem 47. If x, y, z are real numberssuch that x2 + y2 + i = 2, then show thatx + y + z ~ xyz + 2.

Problem 54. Let R be the set of realnumbers. Find all func~ons f: R -+ Rsuch that

f(j{x+y» = .f{x+y) + .f{x)fl.y) -xy

for all x. Y E R. (Source: 1995 Byelorussian

Mathematical Olympiad (Final Round»

Solution: CHAN Ming Chiu (La SalleCollege, Form 6).

Comments: This was a problem on the1988 Leningrad MathematicalOlympiad. Most solvers gave solutionsusing pure geometry or a bit oftrigonometry. The editor will like topoint out there is also a simple vectorsolution. Set the origin 0 at the

-+ -+midpoint of AC. Let OC = m, OB = nand k be the unit vector perpendicular to

-+ -+the plane. Then AB = n + m, CB = n -m,

-+ -+BD= -(n + m) x k, BG= (n -m) x k

-+ -+ -+and DG= BG- BD = 2n x k. IfDG isparallel to AC, then n x k is a multiple of

-+ -+m and so m = OC and n = OB areperpendicular. Therefore, triangle ABCis isosceles.Problem 55. In the beginning, 65

beetles are placed at different squares ofa 9 x 9 square board. In each move,every beetle creeps to a horizontal orvertical adjacent square. If no beetlemakes either two horizontal moves ortwo vertical moves in succession, showthat after some moves, there will be atleast two beetles in the same square.(Source: 1995 Byelorussian Mathematical

Olympiad (Fmal Round»

If one of x, y, z is nonpositive, say z, then

l+xyz-x- y-z = (l-x-y)-z(l-xy) ~O

because Other commended solvers: CHANWing Chiu (La Salle College, Form 4),Calvin CHEUNG Cheuk Lon (S.T.F.A.Leung Kau Kui College, Form 5),William CHEUNG Pok-man (S.T.F.A.Leung Kau Kui College, Form 6), YvesCHEUNG Yui Ho (S.T.F.A. Leung KauKui College, Form 5), CHING WaiHung (S.T.F.A. Leung Kau Kui College,Form 5), Alan LEUNG Wing Lun(STF A Leung Kau Kui College, Form5), .OR Fook Sing & WAN Tsz Kit(Valtorta College, Form 6), TSANG SaiWing (Valtorta College, Form 6),WONG Hau Lun (STFA Leung KauKui College, Form 5), Sam YUEN ManLong (STFA Leung Kau Kui Coll~ge,Form 4).

x+y~ ~2(X2+y2)~2

andxy ~ (X2 + Y1/2 ~ 1

*****************Solutions

*****************

So we may assume x, y, z are positive,sayO 1, then

(x + y) + z ~~2(x+ y)2 +Z2)

= 2j";i+i ~ xy + 2 ~ xyz + 2.

Problem 46. For what integer a doesX2 -x + a divide X13 + x + 90? (Source:1963 Putnam Exam.)

Comments: This was an unused problemin the 1987 IMO and later appeared as aproblem on the 1991 PolishMathematical Olympiad. (continued on page 4)

such that ..jn-l +..jn+l is a rationalnumber?

Mathematical Excalibur, Vol. No.1, lan-Feb. 97 PaJr,e 4

c = 2(y -,. 2z + 3w -2x),d= 2(z-2w+ 3y -2z).

From that point on, a, b, c, d will alwaysbe even, so Ibc-adl, lac-bdl, lab-cdl

will always be divisible by 4.

Part II (8:00-12:30. January 14, 1997)

Problem 4. Let quadrilateral ABCD beinscribed in a circle. Suppose lines ABand DC intersect atP and lines AD andBC intersect at Q. From Q, construct thetwo tangents QE and QF to the circlewhere E and F are the points oftangency. Prove that the three points P,E, F are collinear.

Problem Corner(continued from page 3)

Problem 49. Let Ul, U2, U3~ ...be asequence of integers such that Ul = 29,Uz = 45 and Un+2 = Un+i2 -Un for n= 1,2,

3, Show that 1996 divides infini~lymany terms of this sequence. (Source:1986 Canadian Mathematical Olympiadwith modification)

Solution 2: Official Solution.

Solution: William CHEUNG Pok-man(STFA LeungKau Kui College, Form 6)and YUNG Fai (CUHK).

17}

After n ?; 1 steps, the sum of the integerswill be O. So d = -a -b -c. Then

bc -ad = bc +a(a + b + c)= (a + b) (a + c).

Similarly,

Problem 5. Let A = {I, 2, 3, ...,For a mapping f: A -+ A, denote

fllJ(X) =f(x) ,flk+lJ(X) = fV1kJ(X)) (k = 1,2,3,

Consider one-to-one mappings f from Ato A satisfying the condition: there existsa natural number M such that

(1) for m < M, 1 $ i $ 16,

f[m](i+ 1)- f[m](i) ~:tl (mod 17),

f[m](I)- f[m](17) ~:tI (mod 17);

Let Un be the remainder of Un upon

division by 1996, i.e.,

Un = Un (mod 1996).

Consider the sequence of pairs (Un,Un+V.There are at most 19962 distinct pairs.So let (Up,Up+V = (Uq,Uq+J be the first

repetition with p < q. If p > 1, then the

recurrence relation implies (UP"'-l,Up} =

(Uq-l,Uq) resulting in an earlierrepetition. So p= 1 and the sequence of

pairs (Un,Un+V is periodic with period q-1. Since U3 = 1996, we have 0 = U3 =

U3+.1:(q:--l) and so 1996divid.es U3+.1:(q-l) forevery positive integer k. --

(2) tor 1 ~ i ~ 16;

f[M](i+.1)-f[M](i) =1oc-1 (moo 17),

f[M](1)-f[M]