HMMT 二月 2024 · 几何 · 第 10 题

HMMT February 2024 — Geometry — Problem 10

Suppose point P is inside quadrilateral ABCD such that ∠ P AB = ∠ P DA, ∠ P AD = ∠ P DC, ∠ P BA = ∠ P CB, and ∠ P BC = ∠ P CD. If P A = 4 , P B = 5 , and P C = 10 , compute the perimeter of ABCD .

解析

Suppose point P is inside quadrilateral ABCD such that ∠ P AB = ∠ P DA, ∠ P AD = ∠ P DC, ∠ P BA = ∠ P CB, and ∠ P BC = ∠ P CD. If P A = 4 , P B = 5 , and P C = 10 , compute the perimeter of ABCD . Proposed by: Rishabh Das √ 9 410 Answer: 5 Solution: D A 5 4 8 ′ N P M P 4 5 10 B C ◦ First of all, note that the angle conditions imply that ∠ BAD + ∠ ABC = 180 , so the quadrilateral is a trapezoid with AD ∥ BC . Moreover, they imply AB and CD are both tangent to ( P AD ) and ( P BC ) ; in particular AB = CD or ABCD is isosceles trapezoid. Since the midpoints of AD and BC clearly lie on the radical axis of the two circles, P is on the midline of the trapezoid. ′ ′ ′ Reflect △ P AB over the midline and translate it so that D = B and C = A . Note that P is still ′ ′ on the midline. The angle conditions now imply P DP C is cyclic, and P P bisects CD . This means ′ ′ 10 · 4 = P C · CP = P D · DP = 5 · P D , so P D = 8 . ′ Now P DP C is a cyclic quadrilateral with side lengths 10 , 8 , 5 , 4 in that order. Using standard cyclic quadrilateral facts (either law of cosines or three applications on Ptolemy on the three possible quadri- √ √ 2 410 410 ′ ′ laterals formed with these side lengths) we get CD = and P P = . Finally, note that P P 5 2 is equal to the midline of the trapezoid, so the final answer is √ 9 410 ′ 2 · CD + 2 · P P = . 5