IMO 1978 SL 7

Origin: FRA

Problem

We consider three distinct half-lines Ox, Oy, Oz in a plane. Prove the existence and uniqueness of three points A \inOx, B \inOy, C \inOz such that the perimeters of the triangles OAB, OBC, OCA are all equal to a given number 2p > 0.

Solution

Let x = OA, y = OB, z = OC, \alpha = \angleBOC, \beta = \angleCOA, \gamma = \angleAOB. The conditions yield the equation x + y +

x2 + y2 −2xy cos \gamma = 2p, which transforms to (2p −x −y)2 = x2 + y2 −2xy cos \gamma, i.e. (p −x)(p −y) = xy(1 −cos \gamma). Thus p −x x \cdot p −y y = 1 −cos \gamma, and analogously p−y y \cdot p−z z = 1 −cos \alpha, p−z z \cdot p−x x = 1 −cos \beta. Setting u = p−x x , v = p−y y , w = p−z z , the above system becomes uv = 1 −cos \gamma, vw = 1 −cos \alpha, wu = 1 −cos \beta. This system has a unique solution in positive real numbers u, v, w: u =

(1−cos \beta)(1−cos \gamma) 1−cos \alpha , etc. Finally, the values of x, y, z are uniquely de- termined from u, v, w. Remark. It is not necessary that the three lines be in the same plane. Also, there could be any odd number of lines instead of three.