25th International Mathematical Olympiad 1984 Problems & Solutions

A1.  Prove that 0 ≤ yz + zx + xy - 2xyz ≤ 7/27, where x, y and z are non-negative real numbers satisfying x + y + z = 1.

A2.  Find one pair of positive integers a, b such that ab(a+b) is not divisible by 7, but (a+b)⁷ - a⁷ - b⁷ is divisible by 7⁷.

A3.  Given points O and A in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point X in the plane, the circle C(X) has center O and radius OX + ∠AOX/OX, where ∠AOX is measured in radians in the range [0, 2π). Prove that we can find a point X, not on OA, such that its color appears on the circumference of the circle C(X).

B1.  Let ABCD be a convex quadrilateral with the line CD tangent to the circle on diameter AB. Prove that the line AB is tangent to the circle on diameter CD if and only if BC and AD are parallel.

B2.  Let d be the sum of the lengths of all the diagonals of a plane convex polygon with n > 3 vertices. Let p be its perimeter. Prove that:     n - 3 < 2d/p < [n/2] [(n+1)/2] - 2, where [x] denotes the greatest integer not exceeding x.

B3.  Let a, b, c, d be odd integers such that 0 < a < b < c < d and ad = bc. Prove that if a + d = 2^k and b + c = 2^m for some integers k and m, then a = 1. 

Solutions

Problem A1

Prove that 0 ≤ yz + zx + xy - 2xyz ≤ 7/27, where x, y and z are non-negative real numbers satisfying x + y + z = 1.

Solution

(1 - 2x)(1 - 2y)(1 - 2z) = 1 - 2(x + y + z) + 4(yz + zx + xy) - 8xyz = 4(yz + zx + xy) - 8xyz - 1. Hence yz + zx + xy - 2xyz = 1/4 (1 - 2x)(1 - 2y)(1 - 2z) + 1/4. By the arithmetic/geometric mean theorem (1 - 2x)(1 - 2y)(1 - 2z) ≤ ((1 - 2x + 1 - 2y + 1 - 2z)/3)³ = 1/27. So yz + zx + xy - 2xyz ≤ 1/4 28/27 = 7/27. 

Problem A2

Find one pair of positive integers a, b such that ab(a+b) is not divisible by 7, but (a+b)⁷ - a⁷ - b⁷ is divisible by 7⁷.

Solution

We find that (a + b)⁷ - a⁷ - b⁷ = 7ab(a + b)(a² + ab + b²)². So we must find a, b such that a² + ab + b² is divisible by 7³.

At this point I found a = 18, b = 1 by trial and error.

A more systematic argument turns on noticing that a² + ab + b² = (a³ - b³)/(a - b), so we are looking for a, b with a³ = b³ (mod 7³). We now remember that a^φ(m) = 1 (mod m). But φ(7³) = 2·3·49, so a³ = 1 (mod 343) if a = n⁹⁸. We find 2⁹⁸ = 18 (343), which gives the solution 18, 1.

This approach does not give a flood of solutions. n⁹⁸ = 0, 1, 18, or 324. So the only solutions we get are 1, 18; 18, 324; 1, 324. 

Problem A3

Given points O and A in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point X in the plane, the circle C(X) has center O and radius OX + (∠AOX)/OX, where ∠AOX is measured in radians in the range [0, 2π). Prove that we can find a point X, not on OA, such that its color appears on the circumference of the circle C(X).

Solution

Suppose the result is false. Let C¹ be any circle center O. Then the locus of points X such that C(X) = C₁ is a spiral from O to the point of intersection of OA and C₁. Every point of this spiral must be a different color from all points of the circle C₁. Hence every circle center O with radius smaller than C₁ must include a point of different color to those on C₁. Suppose there are n colors. Then by taking successively smaller circles C₂, C₃, ... , C_n+1 we reach a contradiction, since each circle includes a point of different color to those on any of the larger circles. 

Problem B1

Let ABCD be a convex quadrilateral with the line CD tangent to the circle on diameter AB. Prove that the line AB is tangent to the circle on diameter CD if and only if BC and AD are parallel.

Solution

If AB and CD are parallel, then AB is tangent to the circle on diameter CD if and only if AB = CD and hence if and only if ABCD is a parallelogram. So the result is true.

Suppose then that AB and DC meet at O. Let M be the midpoint of AB and N the midpoint of CD. Let S be the foot of the perpendicular from N to AB, and T the foot of the perpendicular fromM to CD. We are given that MT = MA. OMT, ONS are similar, so OM/MT = ON/NS and hence OB/OA = (ON - NS)/(ON + NS). So AB is tangent to the circle on diameter CD if and only if OB/OA = OC/OD which is the condition for BC to be parallel to AD. 

Problem B2

Let d be the sum of the lengths of all the diagonals of a plane convex polygon with n > 3 vertices. Let p be its perimeter. Prove that:

    n - 3 < 2d/p < [n/2] [(n+1)/2] - 2, where [x] denotes the greatest integer not exceeding x.

Solution

Given any diagonal AX, let B be the next vertex counterclockwise from A, and Y the next vertex counterclockwise from X. Then the diagonals AX and BY intersect at K. AK + KB > AB and XK + KY > XY, so AX + BY > AB + XY. Keeping A fixed and summing over X gives n - 3 cases. So if we then sum over A we get every diagonal appearing four times on the lhs and every side appearing 2(n-3) times on the rhs, giving 4d > 2(n-3)p.

Denote the vertices as A₀, ... , A_n-1 and take subscripts mod n. The ends of a diagonal AX are connected by r sides and n-r sides. The idea of the upper limit is that its length is less than the sum of the shorter number of sides. Evaluating it is slightly awkward.

We consider n odd and n even separately. Let n = 2m+1. For the diagonal A_iA_i+r with r ≤ m, we have A_iA_i+r ≤ A_iA_i+2 + ... + A_iA_i+r. Summing from r = 2 to m gives for the rhs (m-1)A_iA_i+1 + (m-1)A_i+1A_i+2 + (m-2)A_i+2A_i+3 + (m-3)A_i+3A_i+4 + ... + 1.A_i+m-1A_i+m. Now summing over i gives d for the lhs and p( (m-1) + (1 + 2 + ... + m-1) ) = p( (m² + m - 2)/2 ) for the rhs. So we get 2d/p ≤ m² + m - 2 = [n/2] [(n+1)/2] - 2.

Let n = 2m. As before we have A_iA_i+r <= A_iA_i+2 + ... + A_iA_i+r for 2 ≤ r ≤ m-1. We may also take A_iA_i+m ≤ p/2. Summing as in the even case we get 2d/p = m² - 2 = [n/2] [(n+1)/2] - 2. 

Problem B3

Let a, b, c, d be odd integers such that 0 < a < b < c < d and ad = bc. Prove that if a + d = 2^k and b + c = 2^m for some integers k and m, then a = 1.

Solution

a < c, so a(d - c) < c (d - c) and hence bc - ac < c(d - c). So b - a < d - c, or a + d > b + c, so k > m.

bc = ad, so b(2^m - b) = a(2^k - a). Hence b² - a² = 2^m(b - 2^k-ma). But b² - a² = (b + a)(b - a), and (b + a) and (b - a) cannot both be divisible by 4 (since a and b are odd), so 2^m-1 must divide b + a or b - a. But if it divides b - a, then b - a ≥ 2^m-1, so b and c > 2^m-1 and b + c > 2^m. Contradiction. Hence 2^m-1 divides b + a. If b + a ≥ 2^m = b + c, then a ≥ c. Contradiction. Hence b + a = 2^m-1.

So we have b = 2^m-1 - a, c = 2^m-1 + a, d = 2^k - a. Now using bc = ad gives: 2^ka = 2^2m-2. But a is odd, so a = 1.

Read more

Solutions are also available in     Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.