IMO 1986 SL 3

Origin: USA

Problem

Let A, B, and C be three points on the edge of a circular chord such that B is due west of C and ABC is an equilateral triangle whose side is 86 meters long. A boy swam from A directly toward B. After covering a distance of x meters, he turned and swam westward, reaching the shore after covering a distance of y meters. If x and y are both positive integers, determine y.

Solution

Let E be the point where the boy turned westward, reaching the shore at D. Let the ray DE cut AC at F and the shore again at G. Then EF = AE = x (because AEF is an equilateral triangle) and FG = DE = y. From AE \cdot EB = DE \cdot EG we obtain x(86 −x) = y(x + y). If x is odd, then x(86 −x) is odd, while y(x + y) is even. Hence x is even, and so y must also be even. Let y = 2y1. The above equation can be rewritten as (x + y1 −43)2 + (2y1)2 = (43 −y1)2. Since y1 < 43, we have (2y1, 43−y1) = 1, and thus (|x+y1 −43|, 2y1, 43− y1) is a primitive Pythagorean triple. Consequently there exist integers a > b > 0 such that y1 = ab and 43 −y1 = a2 + b2. We obtain that a2 + b2 + ab = 43, which has the unique solution a = 6, b = 1. Hence y = 12 and x = 2 or x = 72. Remark. The Diophantine equation x(86−x) = y(x+y) can be also solved directly. Namely, we have that x(344 −3x) = (2y + x)2 is a square, and since x is even, we have (x, 344 −3x) = 2 or 4. Consequently x, 344 −3x are either both squares or both two times squares. The rest is easy.