PUMaC 2021 · 团队赛 · 第 15 题
PUMaC 2021 — Team Round — Problem 15
题目详情
- Let △ ABC be an acute triangle with angles ∠ BAC = 70 , ∠ ABC = 60 , let D, E be the feet of perpendiculars from B, C to AC, AB , respectively, and let H be the orthocenter of ABC . ◦ Let F be a point on the shorter arc AB of circumcircle of ABC satisfying ∠ F AB = 10 and let G be the foot of perpendicular from H to AF . If I = BF ∩ EG and J = CF ∩ DG , compute the angle ∠ GIJ . 2
解析
- Let △ ABC be an acute triangle with angles ∠ BAC = 70 , ∠ ABC = 60 , let D, E be the feet of perpendiculars from B, C to AC, AB , respectively, and let H be the orthocenter of ABC . ◦ Let F be a point on the shorter arc AB of circumcircle of ABC satisfying ∠ F AB = 10 and let G be the foot of perpendicular from H to AF . If I = BF ∩ EG and J = CF ∩ DG , compute the angle ∠ GIJ . Proposed by: Aleksa Milojevic Answer: 60 This problem is an instance of a more general statement which states the following (using somewhat different notation). Let A, B be the two intersections of circles Γ , Γ and let 1 2 C, D, E be points on Γ , F, G, H on Γ such that triplets ( A, C, F ), ( A, D, G ) and ( A, E, H ) 1 2 are collinear. Moreover, let I = F G ∩ CD, J = F H ∩ CE . Show that ∠ CJI = ∠ AHG . One can relate this version to the one in the problem statement by setting A, G, H as the starting triangle ABC , making Γ its circumcircle and Γ pass through the orthocenter. 2 1 In the solution, we use Miquel’s theorem extensively. This theorem states that for a quadri- lateral XY ZT with U = XT ∩ Y Z and V = XY ∩ ZT , the circumcircles of the triangles U XY, U T Z, V Y Z, V XT pass through a single point. Let us now begin the solution: the goal is to show that △ GIJ is similar to △ ACB . Once we ◦ show this, it will be clear that ∠ GIJ = ∠ ABC = 60 . ◦ As a preliminary step, note that ADHEG is cyclic, as ∠ AGH = ∠ AEH = ∠ ADH = 90 . Let M be the intersection of this circle and of the circumcircle of ABC . Lemma. F GIJM is cyclic. .....solution to be completed..... To show this, it is key that BCIJ is cyclic. After that, it is simple to show using spiral similarity that AGH is similar to CIJ with the center in B . To show BCIJ is cyclic, first note that BEHJ is cyclic by Miquel theorem on the complete quadrilateral ACEHF J . Let K be the intersection of DE and GH . K lies on IJ by Desargues. The quadrilateral EHBK is cyclic because of Miquel on ADEGHK and thus the pentagon BEHJK is cyclic. Applying Miquel on CDEIJK one gets CIJB is cyclic, which completes the proof. 8