HMMT 十一月 2009 · 冲刺赛 · 第 24 题
HMMT November 2009 — Guts Round — Problem 24
题目详情
- [ 12 ] Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figurewhose vertices are the 5 chosen points?
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd
4 3 2
英文原题
[ 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?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 nd ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND
解析
英文解析
- [ 12 ] Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figurewhose vertices are the 5 chosen points?
Answer: Label the vertices of the cube A , B , C , D , E , F , G , H , such that ABCD is the top 1
face of the cube, E is directly below A , F is directly below B , G is directly below C , and H is directly 2
below D . We can obtain a volume of by taking the vertices A , B , C , F , and H . To compute the 1
volume of ACBF H , we will instead compute the volume of the parts of the cube that are not part of 2
ABCF H . This is just the three tetrahedrons CF GH , AEF H , and ACDH , which each have volume 1
(by using the bh formula for the area of a pyramid). Therefore, the volume not contained in ACBF H16
1 1 1 13
is 3 · = , so the volume contained in ACBF H is 1 − = .
6 2 2 2
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd
4 3 2