PUMaC 2011 · 组合（A 组） · 第 3 题

PUMaC 2011 — Combinatorics (Division A) — Problem 3

[ 4 ] Two points are chosen uniformly at random on the sides of a square with side length 1. If p is the probability that the distance between them is greater than 1, what is b 100 p c ? (Note: b x c denotes the greatest integer less than or equal to x .)

解析

Fix one edge for the first point to lie on. If the second point lies on the opposite edge, it will be of distance greater than one (with 1 / 4 probability), and if it lies on the same edge, then it will be of distance less than one (again 1 / 4 probability). Suppose then that the second point lies on one of the other two adjacent edges. If the first point is distance x from this edge, then √ 2 the other point must lie farther than 1 − x from the vertex shared by the edges the two points lie on. Therefore, this reduces to a geometric probability problem where we are finding the area of a region outside of a quarter circle in the unit square (we could also view this as ∫ √ ( ) 1 1 1 π 6 − π 2 the integral 1 − x dx , although this is a bit more work). Thus, p = + 1 − = , 0 4 2 4 8 and b 100 p c = 35 .