HMMT 十一月 2009 · 冲刺赛 · 第 24 题

HMMT November 2009 — Guts Round — Problem 24

[ 12 ] Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nd 2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND 4 3 2

解析

[ 12 ] Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points? 1 Answer: Label the vertices of the cube A , B , C , D , E , F , G , H , such that ABCD is the top 2 face of the cube, E is directly below A , F is directly below B , G is directly below C , and H is directly 1 below D . We can obtain a volume of by taking the vertices A , B , C , F , and H . To compute the 2 volume of ACBF H , we will instead compute the volume of the parts of the cube that are not part of 1 ABCF H . This is just the three tetrahedrons CF GH , AEF H , and ACDH , which each have volume 6 1 (by using the bh formula for the area of a pyramid). Therefore, the volume not contained in ACBF H 3 1 1 1 1 is 3 · = , so the volume contained in ACBF H is 1 − = . 6 2 2 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nd 2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND 4 3 2