HMMT 十一月 2019 · 团队赛 · 第 6 题

HMMT November 2019 — Team Round — Problem 6

[45] Let ABCD be an isosceles trapezoid with AB = 1, BC = DA = 5, CD = 7. Let P be the intersection of diagonals AC and BD , and let Q be the foot of the altitude from D to BC . Let P Q intersect AB at R . Compute sin ∠ RP D .

解析

[45] Let ABCD be an isosceles trapezoid with AB = 1, BC = DA = 5, CD = 7. Let P be the intersection of diagonals AC and BD , and let Q be the foot of the altitude from D to BC . Let P Q intersect AB at R . Compute sin ∠ RP D . Proposed by: Milan Haiman 4 Answer: 5 Let M be the foot of the altitude from B to CD . Then 2 CM + AB = CD = ⇒ CM = 3. Then DM = 4 and by the Pythagorean theorem, BM = 4. Thus BM D is a right isosceles triangle i.e. π π π ∠ BDM = ∠ P DC = . Similarly, ∠ P CD = . Thus ∠ DP C = , which means quadrilateral P QDC is 4 4 2 4 cyclic. Now, sin ∠ RP D = sin ∠ DCQ = sin ∠ M CB = . 5 An alternate solution is also possible: 2 2 2 2 2 2 2 2 Note that AC ⊥ BD since AB + CD = 1 + 7 = 5 + 5 = BC + DA . Thus P is the foot of the altitude from D to AC . Since D is on the circumcircle of 4 ABC , line P QR is the Simson line of D . Thus R is the foot from D to AB . Then from quadrilateral RAP D being cyclic we have ∠ RP D = ∠ RAD . So 4 sin ∠ RP D = . 5