IMO 1990 SL 19

Origin: POL

Problem

Let P be a point inside a regular tetrahedron T of unit volume. The four planes passing through P and parallel to the faces of T partition T into 14 pieces. Let f(P) be the joint volume of those pieces that are neither a tetrahedron nor a parallelepiped (i.e., pieces adjacent to an edge but not to a vertex). Find the exact bounds for f(P) as P varies over T .

Solution

Let d1, d2, d3, d4 be the distances of the point P to the tetrahedron. Let d be the height of the regular tetrahedron. Let xi = di/d. Clearly, x1 + x2 + x3 + x4 = 1, and given this condition, the parameters vary freely as we vary P within the tetrahedron. The four tetrahedra have volumes x3 1, x3 2, x3 3, and x3 4, and the four parallelepipeds have volumes of 6x2x3x4, 6x1x3x4, 6x1x2x4, and 6x1x2x3. Hence, using x1 + x2 + x3 + x4 = 1 and setting g(x) = x2(1 −x), we directly verify that f(P) = f(x1, x2, x3, x4) = 1 −  i=1 x3 i −6  1\leqi<j<k\leq4 xixjxk = 3(g(x1) + g(x2) + g(x3) + g(x4)) . We note that g(0) = 0 and g(1) = 0. Hence, as x1 tends to 1 and other variables tend to 0, f(x1, x2, x3, x4) = 0. Thus f(P) is sharply bounded downwards at 0. We now find an upper bound. We note that g(xi + xj) = (xi + xj)2(1 −x1 −x2) = g(xi) + g(xj) + 2xixj  1 −3 2(xi + xj)  ; thus for xi + xj \leq2/3 and xi, xj > 0 we have g(xi + xj) + g(0) \geq g(xi) + g(xj). Equality holds only when xi + xj = 2/3. Assuming without loss of generality x1 \geqx2 \geqx3 \geqx4, we have g(x1) + g(x2)+g(x3)+g(x4) < g(x1)+g(x2)+g(x3+x4). Assuming y1+y2+y3 = 1 and y1 \geqy2 \geqy3, we have g(y1) + g(y2) + g(y3) \leqg(y1) + g(y2 + y3). Hence g(x1) + g(x2) + g(x3) + g(x4) < g(x) + g(1 −x) for some x. We also have g(x) + g(1 −x) = x(1 −x) \leq1/4. Hence f(P) \leq3/4. Equality holds for x1 = x2 = 1/2, x3 = x4 = 0 (corresponding to the midpoint of an edge), and as the variables converge to these values, f(P) converges to 3/4. Hence the bounds for f(P) are 0 < f(P) < 3 4.