IMO 1965 Problem A3 The tetrahedron ABCD is divided into two parts by a plane parallel to AB and CD. The distance of the plane from AB is k times its distance from CD. Find the ratio of the volumes of the two parts.
Let the plane meet AD at X, BD at Y, BC at Z and AC at W. Take plane parallel to BCD through WX and let it meet AB in P. Since the distance of AB from WXYZ is k times the distance of CD, we have that AX = k·XD and hence that AX/AD = k/(k+1). Similarly AP/AB = AW/AC = AX/AD. XY is parallel to AB, so also AX/AD = BY/BD = BZ/BC. vol ABWXYZ = vol APWX + vol WXPBYZ. APWX is similar to the tetrahedron ABCD. The sides are k/(k+1) times smaller, so vol APWX = k 3 (k+1) 3 vol ABCD. The base of the prism WXPBYZ is BYZ which is similar to BCD with sides k/(k+1) times smaller and hence area k 2 (k+1) 2 times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k 2 (k+1) 3 vol ABCD. Thus vol ABWXYZ = (k 3 + 3k 2 )/(k+1) 3 vol ABCD. We get the vol of the other piece as vol ABCD - vol ABWXYZ and hence the ratio is (after a little manipulation) k 2 (k+3)/(3k+1). s are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 7th IMO 1965 © John Scholes jscholes@kalva.demon.co.uk 27 Sep 1998 Last corrected/updated 26 Sep 2003