PUMaC 2015 · 几何（A 组） · 第 2 题

PUMaC 2015 — Geometry (Division A) — Problem 2

[ 3 ] Terry the Tiger lives on a cube-shaped world with edge length 2. Thus he walks on the outer surface. He is tied, with a leash of length 2, to a post located at the center of one of the faces of the cube. The surface area of the region that Terry can roam on the cube can ” pπ be represented as + a b + c for integers a, b, c, p, q where no integer square greater than 1 q divides b , p and q are coprime, and q > 0. What is p + q + a + b + c ? (Terry can be at a location if the shortest distance along the surface of the cube between that point and the post is less than or equal to 2.) ◦

解析

[ 3 ] Terry the Tiger lives on a cube-shaped world with edge length 2. Thus he walks on the outer surface. He is tied, with a leash of length 2, to a post located at the center of one of the faces of the cube. The surface area of the region that Terry can roam on the cube can ” pπ be represented as + a b + c for integers a, b, c, p, q where no integer square greater than 1 q divides b , p and q are coprime, and q > 0. What is p + q + a + b + c ? (Terry can be at a location if the shortest distance along the surface of the cube between that point and the post is less than or equal to 2.) Solution: Figure 1: Figures 1 and 2 First note that Terry cannot get to the face opposite the face that the post is on. So if we unfold the remaining 5 faces in a cross shape as in the figure 1, we can see that the region that Terry can get to is the intersection of this cross with a circle of radius 2 centered at P , where the post is. We divide up the area into 4 sectors and 8 triangles, as done in the figure 2. ” ◦ We can see that each sector’s central angle is 60 and each triangle has a base of 3 − 1 and a height of 1. Therefore the total area is: ” ” ( 3 − 1 ) ⋅ 1 4 8 π 2 ⋅ π ⋅ 2 + 8 ⋅ = + 4 3 − 4 6 2 3 1 So our answer is 8 + 3 + 4 + 3 − 4 = 14 . ◦