$ P$ is a point on the exterior of a circle centered at $ O$. The tangents to the circle from $ P$ touch the circle at $ A$ and $ B$. Let $ Q$ be the point of intersection of $ PO$ and $ AB$. Let $ CD$ be any chord of the circle passing through $ Q$. Prove that $ \triangle PAB$ and $ \triangle PCD$ have the same incentre.
Problem
Source: CWMO 2001, Problem 6
Tags: geometry, incenter, geometry unsolved
Mr Green
27.12.2008 05:56
We have $ QP.QO = QA.QB = QC.QD$ so $ PCOD$ is inscribed so $ \angle CPO = \angle OPD$ Let $ PO$ intersect $ (O)$ at $ I,K$ $ (P,Q,I,K) = - 1$ and $ \angle IDK = 90^o$ so $ DO$ bisector $ \angle PDC$ so $ I$ is incenter of $ PCD$ Easy to prove $ I$ is incenter of $ PAB$ So $ PAB$ and $ PCD$ have the same incenter
vsathiam
25.03.2017 05:50
Let PO intersect circle O at X and E and let D be an arbitrary point on circle O. Let line DQ be extended past Q to intersect the circle again at a point C. By construction, (E, X; Q, P) is a harmonic bundle. Also, angle EDX is a right angle. Hence, DX bisects angle QDP and circle O is the apollonian circle of segment EX with respect to Q. Hence, BX bisects angle ABP, AX bisects QAP, and CX bisects angle QCP. Hence the incenter of triangle PAB is point X, which also happens to be the incenter of triangle PCD.
Tumon2001
25.03.2017 20:33
Let segment $OP $ intersect the circle at $M $. Observe that $OCPD $ is cyclic. Since, $OC = OD$, so, $OP $ bisects $\angle CPD $. Also, $\angle PDC = \angle MOC = 2\angle MDC$ (since $O $ is the center). Thus, $MD $ bisects $\angle PDC $. Hence, $M $ is the incenter of both $\Delta PAB $ and $\Delta PCD $. This proves the result.