AMC 8 2007 Problem 22
AMC8 2007 年第 22 题
四边形与多边形★★★☆☆
A lemming sits at a corner of a square with side length 10 meters. The lemming runs 6.2 meters along a diagonal toward the opposite corner. It stops, makes a right turn and runs 2 more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
一只旅鼠坐在一个边长为 10 米的正方形的某个角落里。旅鼠沿着对角线朝对面的角落跑了 6.2 米，然后停下来，右转 90 度，又继续跑了 2 米。一位科学家测量了旅鼠和正方形各边之间的最短距离。这四段距离的平均值是多少米？
- A.2
- B.4.5
- C.5
- D.6.2
- E.7
正确答案 · Correct Answer
C
解析 · Solution
解法1
先理清题意：正方形边长 10，旅鼠从一个角出发，沿对角线跑 6.2 米、停下右转 、再跑 2 米，落在某点 P。求 P 到四条边的垂直距离的平均值。
关键一句话：这串跑动其实是幌子——平均值跟 P 落在哪、跑了多远都无关，恒等于 5。
把正方形放进坐标系：左下角 (0,0)、右上角 (10,10)，设 P=(x,y)。P 到四条边的垂直距离就是：到左边 、到右边 10-x、到下边 、到上边 10-y（见下图）。
四个距离相加：x+(10-x)+y+(10-y)=20，平均 。 和 自动消掉了——所以无论 P 在正方形内哪里，答案都是 5。题里的 6.2 和 2 只决定 P 的位置，对平均值毫无影响。
易错点：别去算 、右转后的坐标这些——那是出题人放的烟雾；一看到「到四边距离的平均」，就该想到相对两边的距离之和 = 边长。
不设坐标也行：对正方形内任意一点，它到「左边 + 右边」的距离之和，正好横穿整个正方形 =10；到「上边 + 下边」的距离之和也 =10。所以四条距离之和恒为 10+10=20，平均 =5。一句话搞定。
真把 P 算出来也行，正好用来确认它是幌子：沿对角线 (1,1) 方向跑 6.2，到 ；右转 变成 (1,-1) 方向再跑 2，到 。四条距离 ，相加正好 =20，平均 5。绕这么一大圈，和解法2 的一句话得到同一个答案——正好印证「跑动是幌子」。
前置（按需补）：点到一条与坐标轴平行的边的垂直距离 = 坐标差；正方形内一点到相对两边的距离之和 = 边长。抓住这条，这类题都能秒杀。
Solution 1
*(Elementary — set up coordinates.)* Put the square with corners (0,0) and (10,10), and let the lemming land at P=(x,y). Its perpendicular distances to the four sides are to the left, 10-x to the right, to the bottom, 10-y to the top (see figure).
Adding them: x+(10-x)+y+(10-y)=20, so the average is . The and cancel — so wherever lands inside the square, the answer is 5. The 6.2 and 2 only fix where is; they do not affect the average.
Trap: don't bother computing or the turn — that is a smokescreen. On seeing "average distance to the four sides," think: opposite sides' distances sum to the side length.
*(Slick — pair opposite sides.)* No coordinates needed: for any interior point, its distance to the left side plus its distance to the right side spans the whole square =10; likewise top plus bottom =10. So the four distances always sum to 10+10=20, and the average is 5. One line.
*(The long way — confirming the smokescreen.)* You can actually locate : running 6.2 along the (1,1) diagonal reaches ; a right turn to the (1,-1) direction for 2 more gives . The four distances sum to exactly 20, average 5 — the same as the one-line Solution 2, confirming the motion was a red herring.
Prerequisite: the perpendicular distance from a point to an axis-parallel side is a coordinate difference; in a square, the distances to a pair of opposite sides sum to the side length.
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