PUMaC 2009 · 几何（A 组） · 第 7 题

PUMaC 2009 — Geometry (Division A) — Problem 7

You are given a convex pentagon ABCDE with AB = BC , CD = DE , ∠ ABC = 150 , ◦ ◦ ∠ BCD = 165 , ∠ CDE = 30 , BD = 6. Find the area of this pentagon. Round your answer to the nearest integer if necessary. ◦ ◦

解析

You are given a convex pentagon ABCDE with AB = BC , CD = DE , ∠ ABC = 150 , ◦ ◦ ∠ BCD = 165 , ∠ CDE = 30 , BD = 6. Find the area of this pentagon. Round your answer to the nearest integer if necessary. ◦ Solution. 9. The condition ∠ BCD = 165 is not necessary. The following proof works with ◦ any other angle x instead of 165 . Denote AB = BC = a , CD = DE = b , AC = p and CE = q . We first compute pq : apply Cosine Rule in trianlge ABC and CDE respectively, we get 2 2 2 2 2 ◦ 2 ◦ p = a + a − 2 a cos ∠ ABC = 2 a (1 − cos 150 ) = 2 a (1 + cos 30 ) 2 2 2 2 2 ◦ q = b + b − 2 b cos ∠ CDE = 2 b (1 − cos 30 ) Then 2 2 2 2 2 ◦ 2 2 p q = 4 a b (1 − cos 30 ) = a b = ⇒ pq = ab 3 By Sine Rule, the area of the pentagon is 1 1 1 2 ◦ 2 ◦ ◦ ◦ [ ABC ] + [ CDE ] + [ ACE ] = a sin 150 + b sin 30 + pq sin( x − 15 − 75 ) 2 2 2 1 2 2 ◦ = ( a + b − 2 ab sin(90 − x )) 4 1 2 2 = ( a + b − 2 ab cos x ) 4 1 2 The last expression is exactly BD = 9 by applying Cosine Rule to triangle BCD . 4 ◦ ◦