IMO 1977 LL GBR21

Origin: GBR

Problem

Given that x1+x2+x3 = y1+y2+y3 = x1y1+x2y2+x3y3 = 0, prove that x2 x2 1 + x2 2 + x2 + y2 y2 1 + y2 2 + y2 = 2 3.

Solution

Let us consider the vectors v1 = (x1, x2, x3), v2 = (y1, y2, y3), v3 = (1, 1, 1) in space. The given equalities express the condition that these three vec- tors are mutually perpendicular. Also, x2 x2 1+x2 2+x2 3 , y2 y2 1+y2 2+y2 3 , and 1/3 are the squares of the projections of the vector (1, 0, 0) onto the directions of v1, v2, v3, respectively. The result follows from the fact that the sum of squares of projections of a unit vector on three mutually perpendicular directions is 1.