HMMT 二月 2021 · 冲刺赛 · 第 32 题
HMMT February 2021 — Guts Round — Problem 32
题目详情
- [18] Let acute triangle ABC have circumcenter O , and let M be the midpoint of BC . Let P be the ◦ unique point such that ∠ BAP = ∠ CAM , ∠ CAP = ∠ BAM , and ∠ AP O = 90 . If AO = 53 , OM = 28 , and AM = 75, compute the perimeter of 4 BP C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT Spring 2021, March 06, 2021 — GUTS ROUND Organization Team Team ID#
解析
- [18] Let acute triangle ABC have circumcenter O , and let M be the midpoint of BC . Let P be the ◦ unique point such that ∠ BAP = ∠ CAM , ∠ CAP = ∠ BAM , and ∠ AP O = 90 . If AO = 53 , OM = 28 , and AM = 75, compute the perimeter of 4 BP C . Proposed by: Jeffrey Lu Answer: 192 Solution: The point P has many well-known properties, including the property that ∠ BAP = ∠ ACP and ∠ CAP = ∠ BAP . We prove this for completeness. √ ′ Invert at A with radius AB · AC and reflect about the A -angle bisector. Let P be the image of P . The angle conditions translate to ′ • P lies on line AM ′ • P lies on the line parallel to BC that passes through the reflection of A about BC (since P lies on the circle with diameter AO ) ′ ′ ′ In other words, P is the reflection of A about M . Then BP ‖ AC and CP ‖ AB , so the circumcircles of 4 ABP and 4 ACP are tangent to AC and AB , respectively. This gives the desired result. ′ ′ Extend BP and CP to meet the circumcircle of 4 ABC again at B and C , respectively. Then ′ ′ ′ ′ ′ ∠ C BA = ∠ ACP = ∠ BAP , so BC ‖ AP . Similarly, CB ‖ AP , so BCB C is an isosceles trapezoid. ′ ′ In particular, this means B P = CP , so BP + P C = BB . Now observe that ∠ ABP = ∠ CAP = ′ ′ ′ ∠ BAM , so if AM meets the circumcircle of 4 ABC again at A , then AA = BB . Thus the perimeter ′ ′ of 4 BP C is BP + P C + BC = BB + BC = AA + BC . Now we compute. We have √ √ 2 2 BC = 2 AO − OM = 2 81 · 25 = 90 and Power of a Point gives 2 2 BM 45 ′ M A = = = 27 . AM 75 ′ Thus AA + BC = 75 + 27 + 90 = 192.