1965 IMO Problems/Problem 3
Problem
Given the tetrahedron whose edges and have lengths and respectively. The distance between the skew lines and is , and the angle between them is . Tetrahedron is divided into two solids by plane , parallel to lines and . The ratio of the distances of from and is equal to . Compute the ratio of the volumes of the two solids obtained.
Solution
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 = k3(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 k2(k+1)2 times smaller. Its height is 1/(k+1) times the height of A above ABCD, so vol prism = 3 k2(k+1)3 vol ABCD. Thus vol ABWXYZ = (k3 + 3k2)/(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) k2(k+3)/(3k+1).
See Also
1965 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |