PUMaC 2014 · 代数（A 组） · 第 1 题

PUMaC 2014 — Algebra (Division A) — Problem 1

[ 3 ] On the number line, consider the point x that corresponds to the value 10. Consider 24 distinct integer points y , y , ..., y on the number line such that for all k such that 1 ≤ k ≤ 12, we have 1 2 24 that y is the reflection of y across x . Find the minimum possible value of 2 k − 1 2 k ∑ 24 ( | y − 1 | + | y + 1 | ) n n n =1

解析

[ 3 ] On the number line, consider the point x that corresponds to the value 10. Consider 24 distinct integer points y , y , ..., y on the number line such that for all k such that 1 ≤ k ≤ 12, 1 2 24 we have that y is the reflection of y across x . Find the minimum possible value of 2 k − 1 2 k ∑ 24 ( | y − 1 | + | y + 1 | ) n n n =1 Solution: Let k be the coordinate of a given point (different from variable used in problem statement). Then, the reflection of k across x is 20 − k . WLOG let k < 10. For nonzero integers m , | m − 1 | + | m + 1 | = | 2 m | . If 1 ≤ k < 10, then since 20 − k ≥ 1, it follows that | k − 1 | + | k + 1 | + | 20 − k − 1 | + | 20 − k + 1 | = 2 k + 2(20 − k ) = 40. If k = 0, then | k − 1 | + | k + 1 | + | 20 − k − 1 | + | 20 − k + 1 | = 2 + 40 = 42. If k ≤ − 1, then 20 − k ≥ 1, so | k − 1 | + | k + 1 | + | 20 − k − 1 | + | 20 − k + 1 | = − 2 k + 2(20 − k ) = 40 − 4 k ≥ 44. So, to attain the minimum, we pick the 12 smallest such k . This gets us 9 · 40 + 42 + 44 + 48 = 494 .