Solutions

Summary

Twenty-seventh International Olympiad, 1986

1986/1

It will be enough to prove that at least one of the numbers 2d − 1, 5d − 1, 13d − 1 is not a perfect square. Assume, by the way of contradiction, that 2d = x² + 1, 5d = y² + 1, 13d = z² + 1 for some integers x, y, z. By the first equation, x must be odd: x = 2t + 1. Hence also the number d = (x² + 1)/2 = 2t² + 2t + 1 is odd, and consequently y and z are even: y = 2u, z = 2v.

From z² − y² = 8d we get (v + u)(v − u) = 2d. The number 2d is divisible by 2 but not by 4. This gives the desired contradiction because the numbers v + u and v − u are of the same parity.

1986/2

First Solution. Consider the transformation of the plane defined as the composition f = r₃ ∘ r₂ ∘ r₁, where r_j is the rotation about A_j through 120∘ clockwise. The mapping f preserves the length and direction of every vector; hence it is the translation by a certain vector v. By the conditions of the problem, f(P₀) = P₃, and by periodicity, f ⁿ(P₀) = P_3n; the symbol f ⁿ denotes the n-fold composition (iterate) f ∘ f ∘ · · · ∘ f. Since f is the translation by v, f ⁿ is the translation by the vector nv. For n = 662 we get P_3n = P₁₉₈₆ = P₀. Thus v is the zero vector, which means that f is the identity map. Denote the point r₂(A₁) by B. Then A₁ = f (A₁) = r₃(r₂(r₁(A₁))) = r₃(r₂(A₁))) = r₃(B).

The isosceles triangles A₁A₂B and BA₃A₁ do not coincide; they have equal angles (∠A₂ = ∠ A₃ = 120°) and a common base A₁B, hence they are congruent. Consequently A₁A₂BA₃ is a rhombus (with angles 60° and 120°), and hence A₁A₂A₃ is an equilateral triangle.

Second Solution. The same reasoning can be neatly written down in terms of the complex plane. The given points A_k, P_k (k = 0, 1, 2, …) are represented by the complex numbers a_k, p_k.