IMO 1972 SL 7

Origin: GBR

Problem

(a) A plane \pi passes through the vertex O of the regular tetrahedron OPQR. We define p, q, r to be the signed distances of P, Q, R from \pi measured along a directed normal to \pi. Prove that p2 + q2 + r2 + (q −r)2 + (r −p)2 + (p −q)2 = 2a2, where a is the length of an edge of a tetrahedron. (b) Given four parallel planes not all of which are coincident, show that a regular tetrahedron exists with a vertex on each plane.

Solution

(i) Consider the circumscribing cube OQ1PR1O1QP1R (that is, the cube in which the edges of the tetrahedron are small diagonals), of side b = a \sqrt 2/2. The left-hand side is the sum of squares of the projections of the edges of the tetrahedron onto a perpendicular l to \pi. On the other hand, if l O Q1 P R1 O1 Q P1 R forms angles ϕ1, ϕ2, ϕ3 with OO1, OQ1, OR1 respectively, then the projections of OP and QR onto l have lengths b(cosϕ2 + cos ϕ3) and b| cos ϕ2 −cos ϕ3|. Summing up all these expressions, we obtain 4b2(cos2 ϕ1 + cos2 ϕ2 + cos2 ϕ3) = 4b2 = 2a2. (ii) We construct a required tetrahedron of edge length a given in (i). Take O arbitrarily on \pi0, and let p, q, r be the distances of O from \pi1, \pi2, \pi3. Since a > p, q, r, |p−q|, we can choose P on \pi1 anywhere at distance a from O, and Q at one of the two points on \pi2 at distance a from both O and P. Consider the fourth vertex of the tetrahedron: its distance from \pi0 will satisfy the equation from (i); i.e., there are two values for this distance; clearly, one of them is r, putting R on \pi3.