HMMT 二月 2011 · TEAM1 赛 · 第 11 题
HMMT February 2011 — TEAM1 Round — Problem 11
题目详情
- [ 20 ] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its cir- cumcenter. Let M be the midpoint of segment BC . Let the circumcircle of triangle AOM intersect ω again at D . If H is the orthocenter of triangle ABC , prove that ∠ DAH = ∠ M AO .
解析
- [ 20 ] Let ABC be a non-isosceles, non-right triangle, let ω be its circumcircle, and let O be its cir- cumcenter. Let M be the midpoint of segment BC . Let the circumcircle of triangle AOM intersect ω again at D . If H is the orthocenter of triangle ABC , prove that ∠ DAH = ∠ M AO . Solution: First solution. A O H M B C D X Let X be the intersection of the line tangent to ω at B with the line tangent to ω at C . Note that π OM OC OM OA 4 OM C ∼ 4 OCX since ∠ OM C = ∠ OCX = . Hence = , or, equivalently, = . By 2 OC OX OA OX SAS similarity, it follows that 4 OAM ∼ 4 OXA . Therefore, ∠ OAM = ∠ OXA . We claim now that ∠ OAD = ∠ OAX . By the similarity 4 OAM ∼ OXA , we have that ∠ OAX = ∠ OM A . Since AOM D is a cyclic quadrilateral, we have that ∠ OM A = ∠ ODA . Since OA = OD , we have that ∠ ODA = ∠ OAD . Combining these equations tells us that ∠ OAX = ∠ OAD , so A , D , and X are collinear. Finally, since both AH and OX are perpendicular to BC , it follows that AH ‖ OX , so ∠ DAH = ∠ DXO = ∠ AXO = ∠ M AO , as desired. Second solution. Let Y be the intersection of BC with the line tangent to ω at A . Then the circumcircle of triangle AOM has diameter OY , so AD is perpendicular to OY because the radical axis of two circles is perpendicular to the line between their centers. Since Y is on the polar of A , it follows that A is on the polar of Y , so AD ⊥ OX implies that AD is the polar of Y , i.e. AD is the symmedian from A in triangle ABC . Hence AD and AM are isogonal. Since AH and AO are also isogonal, the desired conclusion follows immediately. Remark: with sufficient perserverance, angle-chasing solutions involving only the points given in the diagram may also be devised.