HMMT 十一月 2013 · THM 赛 · 第 5 题
HMMT November 2013 — THM Round — Problem 5
题目详情
英文原题
- [ 7 ] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Company XYZ wants to locate theirbase at the point P in the plane minimizing the total distance to their workers, who are located atvertices A , B , and C . There are 1, 5, and 4 workers at A , B , and C , respectively. Find the minimumpossible 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 ,
3 2 1 010
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 basen n − 1 0 b
∑
b ), then it is equal to a b .inii =0
It turns out that for every positive integer b > 1, every positive integer k has a unique base b representation. That is, for every positive integer k , there exists a unique n and digits 0 ≤ a , . . . , a < b
0 nsuch 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 aunique 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
英文解析
- [ 7 ] Let ABC be a triangle with AB = 13, BC = 14, CA = 15. Company XYZ wants to locate theirbase at the point P in the plane minimizing the total distance to their workers, who are located atvertices A , B , and C . There are 1, 5, and 4 workers at A , B , and C , respectively. Find the minimumpossible 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 .