PUMaC 2011 · 几何（A 组） · 第 3 题

PUMaC 2011 — Geometry (Division A) — Problem 3

[ 4 ] Let P Q and P R be tangents to a circle ω with diameter AB so that A, Q, R, B lie on ω in that order. Let H be the projection of P onto AB and let AR and P H intersect at S . If ◦ ◦ ∠ QP H = 30 and ∠ HP R = 20 , find ∠ ASQ in degrees.

解析

We claim that P S = P R . To see this, let x = ∠ RAB . Then, calculating arc measures gives ◦ ◦ that ∠ P RS = 90 − x . Also, from right triangle ASH , we have that ∠ P SR = ∠ ASH = 90 − x . ◦ Thus, P R = P S . It also follows from the angles in triangle P SR that x = 10 . Now, since P R and P Q are tangents to the circle, we have P Q = P S = P R . Thus, there is a circle centered at P passing through Q, S , and R . Then we can obtain that 1 1 ◦ ◦ ◦ ∠ QSA = ∠ SQR + ∠ QRS = ∠ SP R + ∠ SP Q = 15 + 10 = 25 . 2 2 ◦ It is interesting to note that the points Q, S, B are collinear because ∠ RQB = 10 = ∠ RAB . ◦ Hence ∠ AQB = 90 , from which another solution can be found.