HMMT 十一月 2009 · 冲刺赛 · 第 27 题

HMMT November 2009 — Guts Round — Problem 27

[ 14 ] ABCD is a regular tetrahedron of volume 1. Maria glues regular tetrahedra A BCD , AB CD , ′ ′ ′ ′ ′ ′ ABC D , and ABCD to the faces of ABCD . What is the volume of the tetrahedron A B C D ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nd 2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND

解析

[ 14 ] ABCD is a regular tetrahedron of volume 1. Maria glues regular tetrahedra A BCD , AB CD , ′ ′ ′ ′ ′ ′ ABC D , and ABCD to the faces of ABCD . What is the volume of the tetrahedron A B C D ? 125 Answer: Consider the tetrahedron with vertices at W = (1 , 0 , 0), X = (0 , 1 , 0), Y = (0 , 0 , 1), 27 1 1 1 and Z = (1 , 1 , 1). This tetrahedron is similar to ABCD . It has center O = ( , , ). We can 2 2 2 ′ ′ ′ ′ ′ ′ ′ ′ ′ construct a tetrahedron W X Y Z in the same way that A B C D was constructed by letting W ′ 1 1 1 ′ be the reflection of W across XY Z and so forth. Then we see that Z = ( − , − , − ), so OZ has 3 3 3 √ √ 5 5 1 5 ′ ′ ′ ′ 6 length 3, whereas OZ has length 3. We thus see that W X Y Z has a side length = that 1 6 2 3 2 ′ ′ ′ ′ of W XY Z , so by similarity the same is true of A B C D and ABCD . In particular, the volume of ( ) 3 5 125 ′ ′ ′ ′ A B C D is that of ABCD , so it is . 3 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nd 2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUND