IMO 1978 SL 16

Origin: YUG

Problem

Determine all the triples (a, b, c) of positive real numbers such that the system ax + by −cz = 0, a 1 −x2 + b 1 −y2 −c 1 −z2 = 0, is compatible in the set of real numbers, and then find all its real solutions.

Solution

Clearly |x| \leq1. As x runs over [−1, 1], the vector u = (ax, a \sqrt 1 −x2) runs over all vectors of length a in the plane having a nonnegative vertical component. Putting v = (by, b

1 −y2), w = (cz, c \sqrt 1 −z2), the system becomes u+v = w, with vectors u, v, w of lengths a, b, c respectively in the upper half-plane. Then a, b, c are sides of a (possibly degenerate) triangle; i.e, |a −b| \leqc \leqa + b is a necessary condition. Conversely, if a, b, c satisfy this condition, one constructs a triangle OMN with OM = a, ON = b, MN = c. If the vectors −−\to OM, −−\to ON have a positive nonnegative component, then so does their sum. For every such triangle, putting u = −−\to OM, v = −−\to ON, and w = −−\to OM +−−\to ON gives a solution, and every solution is given by one such triangle. This triangle is uniquely determined up to congruence: \alpha = \angleMON = \angle(u, v) and \beta = \angle(u, w). Therefore, all solutions of the system are x = cos t, y = cos(t + \alpha), z = y = cos(t + \beta), t \in[0, \pi −\alpha] or x = cos t, y = cos(t −\alpha), z = y = cos(t −\beta), t \in[\alpha, \pi].