4th Iberoamerican Mathematical Olympiad Problems 1989

A1.  Find all real solutions to: x + y - z = -1; x² - y² + z² = 1, -x³ + y³ + z³ = -1.

A2.  Given positive real numbers x, y, z each less than π/2, show that π/2 + 2 sin x cos y + 2 sin y cos z > sin 2x + sin 2y + sin 2z.

A3.  If a, b, c, are the sides of a triangle, show that (a - b)/(a + b) + (b - c)/(b + c) + (c - a)/(a + c) < 1/16.

B1.  The incircle of the triangle ABC touches AC at M and BC at N and has center O. AO meets MN at P and BO meets MN at Q. Show that MP·OA = BC·OQ.

B2.  The function f on the positive integers satisfies f(1) = 1, f(2n + 1) = f(2n) + 1 and f(2n) = 3 f(n). Find the set of all m such that m = f(n) for some n.

B3.  Show that there are infinitely many solutions in positive integers to 2a² - 3a + 1= 3b² + b. 

Solutions

Problem A1

Find all real solutions to: x + y - z = -1; x² - y² + z² = 1, -x³ + y³ + z³ = -1.

Solution

Answer: (x, y, z) = (1, -1, 1) or (-1, -1, -1).

From the first equation x = z - y - 1. Substituting in the second equation: 2z² - 2yz + 2y - 2z = 0, so (z - 1)(z - y) = 0. Hence z = 1 or y = z. If z = 1, then from the first equation x + y = 0, and hence from the last equation, x = 1, y = -1. If y = z, then x = -1, and hence from the last equation y = z = -1.

Many thanks to Allan Lacy Mora for pointing out that the original solution was wrong.

Problem A2

Given positive real numbers x, y, z each less than π/2, show that π/2 + 2 sin x cos y + 2 sin y cos z > sin 2x + sin 2y + sin 2z.

Solution

Solution by David Krumm

We have sin 2x + sin 2y + sin 2z - 2 sin x cos y - 2 sin y cos z = 2 sin x(cos x - cos y) + 2 sin y(cos y - cos z) + 2 sin z cos z, so we wish to show that sin x(cos x - cos y) + sin y(cos y - cos z) + sin z cos z < π/2 (*).

We have to consider six cases: (1) x ≤ y ≤ z; (2) x ≤ z ≤ y; (3) y ≤ x ≤ z; (4) y ≤ z ≤ x; (5) z ≤ x ≤ y; (6) z ≤ y ≤ x. The first case is obvious from the diagram, because the lhs represents the shaded area, and the rhs represents the whole quarter circle.

In cases (2) and (5) the second term is negative, and - sin y < - sin x, so the sum of the first two terms is less than sin x (cos x - cos y) + sin x (cos y - cos z) = sin x (cos x - cos z). But by the same argument as the first case the two rectangles represented by sin x( cos x - cos z) and sin z cos z are disjoint and fit inside the quarter circle. So we have proved (2) and (5).

In cases (3) and (4), the first term is negative. The remaining two terms represent disjoint rectangles lying inside the quarter circle, so again the inequality holds.

In case (6) the first two terms are negative. The last term is ½ sin 2z ≤ ½ < π/2, so the inequality certainly holds.

Problem A3

If a, b, c, are the sides of a triangle, show that (a - b)/(a + b) + (b - c)/(b + c) + (c - a)/(a + c) < 1/16.

Solution

Put f(a, b, c) = (a - b)/(a + b) + (b - c)/(b + c) + (c - a)/(a + c). Let A, B, C be the permutation of a, b, c, with A <= B <= C. If (A, B, C) = (b, a, c), (a, c, b) or (c, b, a), then f(a, b, c) = X, where X = (B - A)/(B + A) + (C - B)/(C + B) - (C - A)/(A + C). If (A, B, C) = (a, b, c), (b, c, a) or (c, a, b), then f(a, b, c) = -X.

Put B = A + h, C = B + k = A + h + k, where h, k ≥ 0. Since A, B, C are the sides of a triangle, we also have A + B > C or A > k. So X = h/(2A + h) + k/(2A + 2h + k) - (h + k)/(2A + h + k) = hk(h + k)/( (2A + h)(2A + h + k)(2A + 2h + k) ). This is obviously non-negative. We claim also that it is < 1/20. That is equivalent to: 20h²k + 20hk² < (2A + h)(2A + h + k)(2A + 2h + k). Since k < A it is sufficient to show that 20h²k + 20hk² ≤ (2k + h)(2k + h + k)(2k + 2h + k) = 18k³ + 27hk² + 13h²k + 2h³ or 18k³ + 7hk² - 7h²k + 2h³ ≥ 0. But 7k² - 7hk + 2h² = 7(k - h/2)² + h²/4 ≥ 0 and h and k are non-negative, so 18k³ + h(7k² - 7hk + 2h²) ≥ 0.

Thus we have established that 0 <= X < 1/20, which shows that f(a, b, c) < 1/20, which is slightly stronger than the required result. 

Problem B1

The incircle of the triangle ABC touches AC at M and BC at N and has center O. AO meets MN at P and BO meets MN at Q. Show that MP.OA = BC.OQ.

Solution

The key to getting started is to notice that angle AQB = 90^o.

Angle BAQ = 90^o - B/2, so angle OAQ = 90^o - B/2 - A/2 = C/2. So OQ = AO sin C/2. Thus we have to show that MP = BC sin C/2.

Let the incircle touch AB at L and let Y be the midpoint of ML (also the intersection of ML with AO). Angle NMC = 90^o - C/2. It is also A/2 + angle MPY, so angle MPY = 90 - C/2 - A/2 = B/2. Hence MP = MY/sin B/2. We have MY = MO sin MOA = r cos A/2 (where r is the inradius, as usual). So MP = (r cos A/2)/sin B/2. We have BC = BN + NC = r (cot B/2 + cot C/2), so MP/BC = (cos A/2)/( sin B/2 (cot B/2 + cot C/2) ). Hence MP/(BC sin C/2) = ( cos A/2 )/( cos B/2 sin C/2 + sin B/2 cos C/2) = cos A/2 /sin(B/2 + C/2) = 1.

 Problem B2

The function f on the positive integers satisfies f(1) = 1, f(2n + 1) = f(2n) + 1 and f(2n) = 3 f(n). Find the set of all m such that m = f(n) for some n.

Solution

We show that to obtain f(n), one writes n in base 2 and then reads it in base 3. For example 12 = 1100₂, so f(12) = 1100₃ = 36. Let g(n) be defined in this way. Then certainly g(1) = 1. Now 2n+1 has the same binary expansion as 2n except for a final 1, so g(2n+1) = g(2n) + 1. Similarly, 2n has the same binary expansion as n with the addition of a final zero. Hence g(2n) = 3 g(n). So g is the same as f. Hence the set of all m such that m = f(n) for some n is the the set of all m which can be written in base 3 without a digit 2. 

Problem B3

Show that there are infinitely many solutions in positive integers to 2a² - 3a + 1= 3b² + b.

Solution

Put A = a - 1 and the equation becomes A(2A + 1) = b(3b + 1). Let d be the greatest common divisor of A and b. Put A = dx, b = dy. Then x(2dx + 1) = y(3dy + 1). Since x and y are coprime, x must divide 3dy + 1. So put 3dy + 1 = nx. Then 2dx + 1 = ny. Solving for x and y in terms of n and d we get x = (n + 3d)/(n² - 6d²), y = (n + 2d)/(n² - 6d²).

So we would certainly be home if we could show that there were infinitely many solutions to n² - 6d² = 1. It is not hard to find the first few: 1² - 6.0² = 1, 5² - 6.2² = 1, 49² - 6.20² = 1. We notice that 49² = 2.5² - 1, so we wonder whether n = 2.49² - 1 might be another solution and indeed we find it gives d = 1960 = 2.49.20. This suggests we try (2 n² - 1)² - 6(2nd)² = 4n⁴ - 4n² + 1 - 24n²d² = 4n²(n² - 6d² - 1) + 1 = 1. So there are indeed infinitely many solutions to n² - 6d² = 1 and we are done.