HMMT 十一月 2013 · THM 赛 · 第 5 题
HMMT November 2013 — THM Round — Problem 5
题目详情
- [ 7 ] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Company XYZ wants to locate their base at the point P in the plane minimizing the total distance to their workers, who are located at vertices A , B , and C . There are 1, 5, and 4 workers at A , B , and C , respectively. Find the minimum possible total distance Company XYZ’s workers have to travel to get to P . Bases Many of you may be familiar with the decimal (or base 10) system. For example, when we say 2013 , 10 3 2 1 0 we really mean 2 · 10 +0 · 10 +1 · 10 +3 · 10 . Similarly, there is the binary (base 2) system. For example, 10 9 8 7 6 5 4 3 2 1 0 11111011101 = 1 · 2 + 1 · 2 + 1 · 2 + 1 · 2 + 1 · 2 + 0 · 2 + 1 · 2 + 1 · 2 + 1 · 2 + 0 · 2 + 1 · 2 = 2013 . 2 10 In general, if we are given a string ( a a . . . a ) in base b (the subscript b means that we are in base n n − 1 0 b ∑ n i b ), then it is equal to a b . i i =0 It turns out that for every positive integer b > 1, every positive integer k has a unique base b repre- sentation. That is, for every positive integer k , there exists a unique n and digits 0 ≤ a , . . . , a < b 0 n such that ( a a . . . a ) = k . n n − 1 0 b We can adapt this to bases b < − 1. It actually turns out that if b < − 1, every nonzero integer has a unique base b representation. That is, for every nonzero integer k , there exists a unique n and digits 0 ≤ a , . . . , a < | b | such that ( a a . . . a ) = k . The next five problems involve base − 4. 0 n n n − 1 0 b Note: Unless otherwise stated, express your answers in base 10.
解析
- [ 7 ] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Company XYZ wants to locate their base at the point P in the plane minimizing the total distance to their workers, who are located at vertices A , B , and C . There are 1, 5, and 4 workers at A , B , and C , respectively. Find the minimum possible total distance Company XYZ’s workers have to travel to get to P . Answer: 69 We want to minimize 1 · P A + 5 · P B + 4 · P C . By the triangle inequality, ( P A + P B ) + 4( P B + P C ) ≥ AB + 4 BC = 13 + 56 = 69, with equality precisely when P = [ AB ] ∩ [ BC ] = B .