IMO 1984 SL 9

Origin: POL

Problem

Let a, b, c be positive numbers with \sqrta+ \sqrt b+\sqrtc = \sqrt 2 . Prove that the system of equations \sqrty −a + \sqrtz −a = 1, \sqrt z −b + \sqrt x −b = 1, \sqrtx −c + \sqrty −c = 1, has exactly one solution (x, y, z) in real numbers.

Solution

Let us show first that the system has at most one solution. Suppose that (x, y, z) and (x′, y′, z′) are two distinct solutions and that w.l.o.g. x < x′. Then the second and third equation imply that y > y′ and z > z′, but then \sqrty −a + \sqrtz −a > \sqrty′ −a + \sqrt z′ −a, which is a contradiction. We shall now prove the existence of at least one solution. Let P be an arbitrary point in the plane and K, L, M points such that PK = \sqrta, PL = \sqrt b, PM = \sqrtc, and \angleKPL = \angleLPM = \angleMPK = 120◦. The lines through K, L, M perpendicular respectively to PK, PL, PM form an equilateral triangle ABC, where K \inBC, L \inAC, and M \inAB. Since its area equals AB2\sqrt 3/4 = S\triangleBPC + S\triangleAPC + S\triangleAPB = AB \sqrta + \sqrt b + \sqrtc

/2, it follows that AB = 1. Therefore x = PA2, y = PB2, and z = PC2 is a solution of the system (indeed, \sqrty −a + \sqrtz −a = \sqrt PB2 −PK2 + \sqrt PC2 −PK2 = BK + CK = 1, etc.).