PUMaC 2019 · 几何（A 组） · 第 7 题

PUMaC 2019 — Geometry (Division A) — Problem 7

Let ABCD be a trapezoid such that AB ‖ CD and let P = AC ∩ BD , AB = 21, CD = 7, AD = 13, [ ABCD ] = 168. Let the line parallel to AB trough P intersect circumcircle of BCP in X . Circumcircles of BCP and AP D intersect at P, Y . Let XY ∩ BC = Z . If ∠ ADC is a obtuse, then BZ = , where a, b are coprime positive integers. Compute a + b . b

解析

Let ABCD be a trapezoid such that AB ‖ CD and let P = AC ∩ BD , AB = 21, CD = 7, AD = 13, [ ABCD ] = 168. Let the line parallel to AB trough P intersect circumcircle of BCP in X . Circumcircles of BCP and AP D intersect at P, Y . Let XY ∩ BC = Z . If ∠ ADC is a obtuse, then BZ = , where a, b are coprime positive integers. Compute a + b . b Proposed by Aleksa Milojevi´ c. Solution by Igor Medvedev. Answer: 17 168 Solution: The heigh of the trapezoid is = 12. By using Pythagoras theorem we find 14 15 BC = 15. Now we claim BZ = CZ . With this BZ = , so the answer would be 17. 2 Let M be the point of intersection of circumcircle of ∆ AP D and XP other than P . We have ∠ DY M = ∠ DP M = ∠ BP X = ∠ BY X , and similarly ∠ M Y A = ∠ M P A = ∠ XP C = ∠ XY C . Furthermore ∠ ADP = ∠ AP Y = ∠ P CB + ∠ CBP − ∠ Y P B = ∠ Y CB + ∠ CBY − ∠ Y P B = ∠ CBY . Hence triangles ∆ Y DA and ∆ Y BC are similar. The unique spiral similarity between these two triangles has center Y , call it Φ : ∆ Y BC 7 → ∆ Y DA . Let N be the point of intersection of AD and Y M . Then Φ( Z ) = N , because ∠ BY Z = Y Z Y N ∠ BY X = ∠ DY M = ∠ DY N . This means that = . This implies that N Z || M X . Y X Y M CZ DN BZ DN BZ CZ From this we get = . On the other hand, Φ( Z ) = N , so = , hence = , ZB N A ZC N A ZC ZB hence CZ = BZ .