HMMT 二月 2010 · TEAM1 赛 · 第 7 题

英文解析

7. [ 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 .
   In other words, one triangle can be formed from the other by dilating and rotating about one of the 12
   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 .
   °
   10A
   °
   F10
   °P
   ′15
   DQXC
   ′
   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 .