Silk Road Mathematics Competition

Question

In a triangle ABC with incenter I, let P be the intersection point of the bisector of the angle A with the circumcircle other than A, D the point of tangency of the incircle to the side BC, and Q the intersection point of PD with the circumcircle other than P. Show that PI = QI if PD is equal to the inradius. FIGURE_0

Accepted Answer

FIGURE_0 In the triangle BIA, BIP = IBA + IAB = B/2 + A/2. On the other hand, PBI = PBC + CBI = A/2 + B/2, as PBC = PAC = A/2. Hence, in the triangle BPI, BP = IP. Since QPB = BPD and BQP = BAP = A/2 = PAC = PBC = PBD, the triangles QPB and BPD are similar. In particular, PDPB = BPQP. Combining this with the fact BP = IP, gives PDPI = PIPQ. Therefore, the triangles DPI and IPQ are similar, and QIPI = IDPD = 1.