HMMT 二月 2005 · 几何 · 第 10 题

HMMT February 2005 — Geometry — Problem 10

Let AB be the diameter of a semicircle Γ. Two circles, ω and ω , externally tangent 1 2 to each other and internally tangent to Γ, are tangent to the line AB at P and Q , respectively, and to semicircular arc AB at C and D , respectively, with AP < AQ . ◦ 6 6 6 6 Suppose F lies on Γ such that F QB = CQA and that ABF = 80 . Find P DQ in degrees. 1

解析

Let AB be the diameter of a semicircle Γ. Two circles, ω and ω , externally tangent 1 2 to each other and internally tangent to Γ, are tangent to the line AB at P and Q , respectively, and to semicircular arc AB at C and D , respectively, with AP < AQ . ◦ 6 6 6 6 Suppose F lies on Γ such that F QB = CQA and that ABF = 80 . Find P DQ in degrees. Solution: 35 Extend the semicircle centered at O to an entire circle ω , and let the reflection of F ′ ′ over AB be F . Then CQF is a straight line. Also, the homothety centered at C taking ω into ω takes P to a point X on ω and AB to the parallel line tangent to ω 1 at X . Therefore, X is the midpoint of semicircle AXB , and C , P , and X lie on a line. Similarly, D , Q , and X lie on a line. So, ◦ ◦ 6 6 6 6 6 45 = XCB = P CB = P CQ + QCB = P CQ + 10 , ′ ′ ◦ ◦ ◦ 6 6 6 6 6 6 since QCB = F CB = F AB = F AB = 90 − ABF = 10 . Thus P CQ = 35 . ◦ 6 6 6 We will show that P CQ = P DQ to get that P DQ = 35 . 6 Note that XP Q subtends the sum of arcs AC and BX , which is equal to arc XC . 6 6 6 6 Therefore XP Q = CDX , so CDQP is cyclic and P CQ = P DQ . The conclusion follows. C D F A B P O Q F’ X 4