PUMaC 2011 · 几何（A 组） · 第 4 题

PUMaC 2011 — Geometry (Division A) — Problem 4

[ 4 ] Let ABC be a triangle with AB = 15 , BC = 17, CA = 21, and incenter I . If the circumcircle of triangle IBC intersects side AC again at P , find CP .

解析

First solution : We claim that BP is perpendicular to AI . Let M be the intersection of lines BP and AI . We have that ∠ IBM = ∠ IBP = ∠ ICP . Also, ∠ BIM = ∠ ABI + ∠ IAB , so 1 ◦ ∠ IBM + ∠ BIM = ∠ ICP + ∠ ABI + ∠ IAB = ( ∠ A + ∠ B + ∠ C ) = 90 . 2 2 Thus, BP is indeed perpendicular to AI . Thus, triangles ABM and AP M are congruent, so AP = AB = 15, so P C = AC − AP = 21 − 15 = 6 . Figure 5: Problem 4 diagram. Second solution : Let ω be the circumcircle of triangle IBC . We claim that ω is the circle with diameter IO , where O is the excenter of ABC corresponding to A . Draw the external angle bisectors at vertices B and C . These two lines intersect at O . Moreover, since IC is perpendicular to CO and IB is perpendicular to BO , so quadrilateral IBOC is cyclic, and its circumcircle is precisely ω . Thus, since I and O lie on the angle bisector of ∠ BAC , the circle ω lies symmetric to ∠ BAC . Thus, if AB = 15, then AP = 15 as well, from which is follows that CP = AC − AP = 21 − 15 = 6 . ′