AIME 1993 · 第 2 题

AIME 1993 — Problem 2

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\text{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\text{th}}_{}$ day?

解析

Solution

On the first day, the candidate moves $[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$ , and so on. The east/west displacement is thus $\frac{1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2}{2} = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|$ . Applying difference of squares, we see that

$\begin{aligned} \left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. \end{aligned}$ Similarly, the north/south displacement is

$\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.$ Since $\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45$ , the two distances evaluate to $8(45) + 10\cdot 4 = 400$ and $8(45) + 10\cdot 6 = 420$ . By the Pythagorean Theorem, the answer is $\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}$ .