HMMT 十一月 2019 · 冲刺赛 · 第 24 题

HMMT November 2019 — Guts Round — Problem 24

[12] Let P be a point inside regular pentagon ABCDE such that ∠ P AB = 48 and ∠ P DC = 42 . Find ∠ BP C , in degrees. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2019, November 9, 2019 — GUTS ROUND Organization Team Team ID#

解析

[12] Let P be a point inside regular pentagon ABCDE such that ∠ P AB = 48 and ∠ P DC = 42 . Find ∠ BP C , in degrees. Proposed by: Dylan Liu ◦ Answer: 84 ◦ ◦ ◦ Since a regular pentagon has interior angles 108 , we can compute ∠ P DE = 66 , ∠ P AE = 60 , and ◦ ◦ ∠ AP D = 360 − ∠ AED − ∠ P DE − ∠ P AE = 126 . Now observe that drawing P E divides quadrilateral ◦ P AED into equilateral triangle P AE and isosceles triangle P ED , where ∠ DP E = ∠ EDP = 66 . That is, we get P A = P E = s , where s is the side length of the pentagon. ◦ ◦ ◦ Now triangles P AB and P ED are congruent (with angles 48 − 66 − 66 ), so P D = P B and ∠ P DC = ◦ ∠ P BC = 42 . This means that triangles P DC and P BC are congruent (side-angle-side), so ∠ BP C = ∠ DP C . ◦ ◦ Finally, we compute ∠ BP C + ∠ DP C = 2 ∠ BP C = 360 − ∠ AP B − ∠ EP A − ∠ DP E = 168 , meaning ◦ ∠ BP C = 84 .