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Auckland Mathematical Olympiad 2002

Division 1

Questions

1. Nine identical books cost less than $10 in total and ten such bookscost more than $11. How much does one book cost?

2. Two squares, whose areas are A and B, are placed in a semicircle asshown below. Find the ratio A/B.

B

A

3. The Land of Oz is blessed by many things but not efficient structure oftheir military forces. Their armored brigade consists of two armoredbatallions and one infantry batallion. Their infantry brigade consistsof two infantry batallions and one armored batallion. Any armoredbatallion consists of two armored squadrons and one infantry squadronand any infantry batallion consists of two infantry squadrons and onearmored squadron. Any armored squadron consists of two armoredsquads and one infantry squad and any infantry squadron consistsof two infantry squads and one armored squad. Any armored squadconsists of two armored soldiers and one infantry soldier and any in-fantry squad consists of two infantry soldiers and one armored soldier.What is the least number of soldiers that must be retrained in orderto transform an infantry brigade into an armored brigade?

4. Find all integer solutions of the equation xy + 3x 5y = +1.5. Given that there exists a triangle with lengths of its sides a,b,c. prove

that there exists a triangle with lengths of its sides

a,

b,

c.

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Auckland Mathematical Olympiad 2002

Division 2

Questions

6. Solve the system of equations

x2 + y2 + z2 = 14x2 + y2 + t2 = 21x2 + t2 + z2 = 26t2 + y2 + z2 = 29

7.

The number 11 . . . 1 m

is divisible by 37. Prove that the number of unitsm is divisible by 3.

8. Cut a rectangle 1.5 4 into two pieces with which it is possible tocover the surface of a unit cube.

9. How many negative roots does the following equation have

2x5 + x4 5x3 + 4x2 13x + 4 = 0?

10. Three circles c1, c2, c3 pass through the common point I. Let P, Q,R be the points of intersection of c1 and c2, c2 and c3, c3 and c1,

respectively, different from I. Let A be a point on c1 lying outside c2and c3 as shown in the picture below.

c3

c2 c1

C

BI

Q

R

P

A

Let B be the point of intersection of AP with c2, C be the point ofintersection of BQ with c3. Prove that CR will pass through A.

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Solutions

1. Let x be the price of one book. Then 9x < 10 and 10x > 11 fromwhich we have 1.1 < x < 1.1111... The only possibility is x = 1.11, i.e.one dollar and 11 cents.

2. Let us denote the lengths of the sides of these squares as a and b,respectively.

a

bO

M

N

We have (a/2)2 + a2 = ON2 = OM2 = (a/2 + b)2 + b2 from which forthe ratio x = a/b we find x2x2 = 0. Hence a/b = 2 and A/B = 4.

3. Answer: one soldier. Indeed, one soldier should be retrained to trans-form an infantry squad into an armored squad, hence one soldier shouldbe retrained to convert an infantru squadron into an armored squadronand an infantry batallion into an armored batallion, and finally an in-fantry brigade into an armored brigade.

4. Since we cannot have any solutions with y = 3, we can write ourequation as

x =5y + 1

y + 3= 5 14

y + 3.

A solution is possible only when y + 3 is a divisor of 14, i.e y + 3 =1,2,7,14. Respectively, we have eight solutions: (9,2);(19,4); (2,1); (12,5); (3, 4);(7,10); (4, 11); (6,17).

5. Suppose that a

b

c. Then the existence of a triangle with sides

a,b,c is equivalent to a + b c. Then, since a + b = (a)2 + (b)2 (

a +

b)2, we obtain

a + b a +

b

c

a + b a +

b,

which, together with

a

b c, gives us the existence of atriangle with sides

a,

b,

c.

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6. Adding all equations together we get 3(x2 + y2 + z2 + t2) = 90 or

(x2 + y2 + z2 + t2) = 30. Subtracting each equation in turn from thistotal, we get x2 = 1, y2 = 4, z2 = 9, t2 = 16. Thus x = 1, y = 2,z = 9, t = 4

7. Let N = 11 . . . 1 m

be a multiple of 37. Suppose m = 3k + 2, then

N = 11 . . . 1 m

= 11 . . . 1 3k

00 + 11

Since 111 = 373, we see that 11 . . . 1

3k

00 is divisible by 111 and hence

by 37. But then 11 = N 11 . . . 1 3k

00 must be divisible by 37, which is

impossible. Similarly we consider the case m = 3k + 1.

8. We cut it as shown below on the left picture and join the two parts asshown on the right one.

9. The equation

2x5 + x4 5x3 + 4x2 13x + 4 = 0,

may be written as x4 + 4x2 + 4 = 2x5 + 5x3 + 13x or (x2 + 2)2 =2x5 + 5x3 + 13x. This equation cannot have negative roots becausewhen x is negative, the right-hand-side is negative whereas the left-hand-side is always positive.

10. Instead of proving that CR passes through A, we will draw the seg-ments CR and AR and prove that CRA = 180. Let us draw thesegments IP,IQ,IR. Since the quadrilateral APIR is cyclic, we ob-tain that ARI = 180 AP I and therefore ARI = BP I.Similarly IRC = BQI. But BP I + BQI = 180. Hence

ARC = ARI +IRC = BP I + BQI = 180.

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