PUMaC 2012 · 团队赛 · 第 3 题

PUMaC 2012 — Team Round — Problem 3

(3 digits) Suppose you draw 5 vertices of a convex pentagon (but not the sides!). Let N be the number of ways you can draw at least 0 straight line segments between the vertices so that no two line segments intersect in the interior of the pentagon. What is N − 64? (Note what the question is asking for! You have been warned!) 2012

解析

Problem: (3 digits) Suppose you draw 5 vertices of a convex pentagon (but not the sides!). Let N be the number of ways you can draw at least 0 straight line segments between the vertices so that no two line segments intersect in the interior of the pentagon. What is N − 64? (Note what the question is asking for! You have been warned!) Answer: 288 Solution: Note that the five outer edges don’t really matter. As in we don’t have to worry about whether they intersect anything else. Case work! Let the five vertices be A, B, C, D, E in that order. If AC is drawn other interior segments we could have are { CE } , { AD } , ∅ , so 3 in this case. If AC is not drawn, we could have ∅ , { AD } , { AD, BD } { BD } , { BE } , { BD, BE } , { BE, CE } , { CE } , so 8 in this case. Thus, 11 total for the interior segments. 5 Multiply by 2 for the outer edges, so there are 11 × 32 = 352 ways. The answer is then 352 − 64 = 288 . Author: Alan 2012