PUMaC 2013 · 几何（B 组） · 第 7 题

PUMaC 2013 — Geometry (Division B) — Problem 7

[ 7 ] A tetrahedron ABCD satisfies AB = 6, CD = 8, and BC = DA = 5. Let V be the 4 n m maximum volume of ABCD possible. If we can write V = 2 3 for some integers n and m , find mn .

解析

[ 7 ] A tetrahedron ABCD satisfies AB = 6, CD = 8, and BC = DA = 5. Let V be the 4 n m maximum volume of ABCD possible. If we can write V = 2 3 for some integers n and m , find mn . 1 Solution Let h be the distance between line AB and CD . Suppose that the four vertices 0 0 0 0 A, B, C, D are projected vertically along h to vertices A , B , C , D respectively. It is known 0 0 0 0 that if we denote ✓ be the angle between A B and C D , the volume is given by 1 V = AB · CD · h · | sin ✓ | . 6 AS AB and CD are fixed, we should maximize the value of h sin ✓ . Meanwhile we can observe p p 0 0 0 0 0 0 0 0 2 2 2 2 A B = AB, C D = CD and B C = BC h , D A = DA h . By the law of cosines p p ! ! 0 0 0 0 0 0 2 0 0 2 0 0 0 0 | A B C D | = A B + C D 2 A B · C D cos ✓ = 100 96 cos ✓ . but also by triangle inequality p p ! ! ! ! 0 0 0 0 0 0 0 0 0 0 0 0 2 2 | A B C D | = | C B + D A |  | B C | + | D A | = 2 25 h = 100 4 h , p so it should give h  24 cos ✓ . Now considering h sin ✓ can be bounded by following use of AM-GM inequality p 2 2 p p sin ✓ sin ✓ 2 2 1 / 4 h sin ✓ = 24 cos ✓ sin ✓ = 48(cos ✓ · · ) 2 2 2 2 sin ✓ sin ✓ 2 p cos ✓ + + 2 2 3 / 4 1 / 4  4 3( ) = 4 · 3 , 3 1 / 4 5 1 / 4 this gives maximum of V to be 1 / 6 · 6 · 8 · (4 · 3 ) = 2 · 3 . Thus ( n, m ) = (20 , 1). Thus -20