IMO 1988 SL 6

Origin: CZS

Problem

In a given tetrahedron ABCD let K and L be the centers of edges AB and CD respectively. Prove that every plane that contains the line KL divides the tetrahedron into two parts of equal volume.

Solution

Let \omega be the plane through AB, parallel to CD. Define the point trans- formation f : X "\toX′ in space as follows. If X \inKL, then X′ = X; otherwise, let \omegaX be the plane through X parallel to \omega: then X′ is the point symmetric to X with respect to the intersection point of KL with \omegaX. Clearly, f(A) = B, f(B) = A, f(C) = D, f(D) = C; hence f maps the tetrahedron onto itself. We shall show that f preserves volumes. Let s : X "\toX′′ denote the symmetry with respect to KL, and g the transformation mapping X′′ into X′; then f = g ◦s. If points X′′ 1 = s(X1) and X′′ 2 = s(X2) have the property that X′′ 1 X′′ 2 is parallel to KL, then the segments X′′ 1 X′′ 2 and X′ 1X′ 2 have the same length and lie on the same line. Then by Cavalieri’s principle g preserves volume, and so does f. Now, if \alpha is any plane containing the line KL, the two parts of the tetra- hedron on which it is partitioned by \alpha are transformed into each other by f, and therefore have the same volumes. Second solution. Suppose w.l.o.g. that the plane \alpha through KL meets the interiors of edges AC and BD at X and Y . Let −−\to AX = \lambda−\to AC and −−\to BY = µ−−\to BD, for 0 \leq\lambda, µ \leq1. Then the vectors −−\to KX = \lambda−\to AC −−−\to AB/2, −−\to KY = µ−−\to BD+−−\to AB/2, −−\to KL = −\to AC/2+ −−\to BD/2 are coplanar; i.e., there ex- ist real numbers a, b, c, not all zero, such that A B C D K L X Y −\to0 = a−−\to KX + b−−\to KY + c−−\to KL = (\lambdaa + c/2)−\to AC + (µb + c/2)−−\to BD + b −a −−\to AB. Since −\to AC, −−\to BD, −−\to AB are linearly independent, we must have a = b and \lambda = µ. We need to prove that the volume of the polyhedron KXLY BC, which is one of the parts of the tetrahedron ABCD partitioned by \alpha, equals half of the volume V of ABCD. Indeed, we obtain VKXLY BC = VKXLC + VKBY LC = 1 4(1 −\lambda)V + 1 4(1 + µ)V = 1 2V.