HMMT 二月 2011 · TEAM2 赛 · 第 9 题
HMMT February 2011 — TEAM2 Round — Problem 9
题目详情
- [ 25 ] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its circum- center. Let M be the midpoint of segment BC . Let the tangents to ω at B and C intersect at X . Prove that ∠ OAM = ∠ OXA . (Hint: use SAS similarity).
解析
- [ 25 ] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its circum- center. Let M be the midpoint of segment BC . Let the tangents to ω at B and C intersect at X . Prove that ∠ OAM = ∠ OXA . (Hint: use SAS similarity). Team Round B A O H M B C D X Solution: π OM OC Note that 4 OM C ∼ 4 OCX since ∠ OM C = ∠ OCX = . Hence = , or, equivalently, 2 OC OX OM OA = . By SAS similarity, it follows that 4 OAM ∼ 4 OXA . Therefore, ∠ OAM = ∠ OXA . OA OX