HMMT 二月 2015 · 团队赛 · 第 4 题

HMMT February 2015 — Team Round — Problem 4

[ 15 ] (Convex) quadrilateral ABCD with BC = CD is inscribed in circle Ω; the diagonals of ABCD meet at X . Suppose AD < AB , the circumcircle of triangle BCX intersects segment AB at a point − − → − − → Y 6 = B , and ray CY meets Ω again at a point Z 6 = C . Prove that ray DY bisects angle ZDB . − − → (We have only included the conditions AD < AB and that Z lies on ray CY for everyone’s convenience. With slight modifications, the problem holds in general. But we will only grade your team’s solution in this special case.)

解析

[ 15 ] (Convex) quadrilateral ABCD with BC = CD is inscribed in circle Ω; the diagonals of ABCD meet at X . Suppose AD < AB , the circumcircle of triangle BCX intersects segment AB at a point − − → − − → Y 6 = B , and ray CY meets Ω again at a point Z 6 = C . Prove that ray DY bisects angle ZDB . − − → (We have only included the conditions AD < AB and that Z lies on ray CY for everyone’s convenience. With slight modifications, the problem holds in general. But we will only grade your team’s solution in this special case.) Answer: N/A This is mostly just angle chasing. The conditions AD < AB (or ∠ ABD < ∠ ADB ) − − → and the assumption Z ∈ CY are not crucial, as long as we’re careful with configurations (for example, 1 DY may only be an external angle bisector of ∠ ZDB in some cases), but it’s the easiest to visualize. In this case Y and Z lie between A and B , on the respective segment/arc. We’ll prove Y is the − − → incenter of △ ZDB ; it will follow that ray DY indeed (internally) bisects ∠ ZDB . It suffices to prove the following two facts: 2 • BY is the internal angle bisector of ∠ DBZ . • ZY is the internal angle bisector of ∠ BZD , since CB = CD . Indeed (more explicitly), arcs BC and CD are equal, so ∠ BZC = ∠ CZD , i.e. Y Z bisects ∠ BZD .