PUMaC 2016 · 几何（A 组） · 第 3 题

PUMaC 2016 — Geometry (Division A) — Problem 3

Let C be a right circular cone with apex A . Let P , P , P , P and P be points placed evenly 1 2 3 4 5 along the circular base in that order, so that P P P P P is a regular pentagon. Suppose 1 2 3 4 5 that the shortest path from P to P along the curved surface of the cone passes through the 1 3 midpoint of AP . Let h be the height of C , and r be the radius of the circular base of C . If 2 ( ) 2 h a can be written in simplest form as , find a + b . r b

解析

The net of the curved surface of C is a sector of a disc. Denote by P the point on the disc corresponding to the point P on C . Then the shortest path along the curved surface of ′ ′ ′ ′ ′ ′ C between P and P corresponds to the line segment P P . By symmetry, P P ⊥ A P . 1 3 1 3 1 3 2 ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ Thus P lies on the perpendicular bisector of A P , so P P = P A = P A and 4 A P P is 1 2 1 2 1 2 1 2 ′ ′ ′ equilateral ( A is the center of the disc). Then if s = AP = A P is the slanted height of C , 1 1 ( ) 2 2 2 2 5 6 r h h s − r 11 we get 2 πr = 2 πs ⇒ s = . Then = = = so a + b = 11 + 25 = 36 . 2 2 6 5 r r r 25 Problem written by Mel Shu. 1