HMMT 二月 2010 · TEAM1 赛 · 第 7 题
HMMT February 2010 — TEAM1 Round — Problem 7
题目详情
- [ 25 ] Point P lies inside a convex pentagon AF QDC such that F P DQ is a parallelogram. Given that ◦ ◦ ∠ F AQ = ∠ P AC = 10 , and ∠ P F A = ∠ P DC = 15 . What is ∠ AQC ?
解析
- [ 25 ] Point P lies inside a convex pentagon AF QDC such that F P DQ is a parallelogram. Given that ◦ ◦ ∠ F AQ = ∠ P AC = 10 , and ∠ P F A = ∠ P DC = 15 . What is ∠ AQC ? π ′ ′ Answer: Let C be the point such that there is a spiral similarity between 4 AF P and 4 AQC . 12 In other words, one triangle can be formed from the other by dilating and rotating about one of the Team Round A ′ triangle’s vertices (in this case, A ). We will show that C is C , so our answer will be ∠ AQC = ′ ◦ ′ ∠ AQC = ∠ AF P = 15 . By the spiral similarity theorem, 4 AF Q ∼ 4 AP C (this is intuitive by ◦ ′ looking at a diagram), so ∠ P AC = ∠ F AQ = 10 , so to show that C is C , it is sufficient to show that ′ ◦ ∠ P DC = 15 . ◦ A 10 ◦ 10 F P ◦ 15 ′ C X Q D ′ Let X be the fourth point of the parallelogram F P C X (see the above diagram). The angle between ◦ ◦ ′ ′ lines F P and F A is 15 . Since XC ‖ F P , the angle between F A and XC is 15 as well. In addition, ′ ◦ ′ ′ ′ the angle between QA and QC is ∠ AQC = 15 , so ∠ XC Q = ∠ F AQ . Further, because F P C X is a ′ ′ ′ QC QC QC QA ′ parallelogram, = . By similar triangles 4 AF P and 4 AQC , = . By SAS similarity, ′ XC F P F P F A ′ ′ ◦ there is a spiral similarity between 4 XC Q and 4 F AQ , so ∠ F QX = ∠ AQC = 15 . ′ ∼ Note that the segments F P , XC , and QD are all parallel and equal in length. Therefore, 4 F QX = ′ ′ ◦ ′ ◦ 4 P DC are congruent, and ∠ P DC = 15 as well. So C is C , and ∠ AQC = 15 .