HMMT 十一月 2017 · THM 赛 · 第 8 题
HMMT November 2017 — THM Round — Problem 8
题目详情
- U ndecillion years ago in a galaxy far, far away, there were four space stations in the three-dimensional space, each pair spaced 1 light year away from each other. Admiral Ackbar wanted to establish a base somewhere in space such that the sum of squares of the distances from the base to each of the stations does not exceed 15 square light years. (The sizes of the space stations and the base are negligible.) Determine the volume, in cubic light years, of the set of all possible locations for the Admiral’s base.
解析
- U ndecillion years ago in a galaxy far, far away, there were four space stations in the three-dimensional space, each pair spaced 1 light year away from each other. Admiral Ackbar wanted to establish a base somewhere in space such that the sum of squares of the distances from the base to each of the stations does not exceed 15 square light years. (The sizes of the space stations and the base are negligible.) Determine the volume, in cubic light years, of the set of all possible locations for the Admiral’s base. Proposed by: Yuan Yao √ 27 6 Answer: π 8 1 1 1 √ √ √ Solution 1: Set up a coordinate system where the coordinates of the stations are ( , , ) , 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 √ √ √ √ √ √ √ √ √ ( − , − , ) , ( , − , − ) , and ( − , , − ). The sum of squares of the distances 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 is then ∑ 1 1 3 3 2 2 2 2 2 2 2( x − √ ) + 2( x + √ ) = 4( x + y + z ) + = 4 r + , 2 2 2 2 2 2 sym √ 3 3 6 2 where r is the distance from the center of the tetrahedron. It follows from 4 r + ≤ 15 that r ≤ , 2 4 √ √ 3 6 4 π 3 27 6 so the set is a ball with radius R = , and the volume is R = π . 4 3 8 Solution 2: Let P be the location of the base; S , S , S , S be the stations; and G be the center of 1 2 3 4 the tetrahedron. We have: 4 4 ∑ ∑ 2 ~ ~ P S = P S · P S i i i i =1 i =1 4 4 ( ) ( ) ∑ ∑ 2 ~ ~ ~ ~ P S = P G + GS · P G + GS i i i i =1 i =1 4 4 ( ) ∑ ∑ 2 ~ ~ ~ ~ ~ ~ P S = P G · P G + 2 P G · GS + GS · GS i i i i i =1 i =1 Since GS = GS = GS = GS , we have: 1 2 3 4 4 4 ( ) ∑ ∑ 2 2 2 ~ ~ P S = 4 P G + 4 GS + 2 P G · GS i i 1 i =1 i =1 ( ) 4 4 ∑ ∑ 2 2 2 ~ ~ P S = 4 P G + 4 GS + 2 P G · GS i i 1 i =1 i =1 ~ ~ ~ ~ ~ Since G is the center of the tetrahedron, GS + GS + GS + GS = 0 . Thus: 1 2 3 4 4 ∑ 2 2 2 P S = 4 P G + 4 GS i 1 i =1 3 27 2 2 Since GS = , we have P G ≤ . Thus, the locus of all good points is a ball centered at G with 1 8 8 √ √ 27 4 27 6 π 3 radius r = . Then, the volume is V = πr = . 8 3 8