Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.
Problem
Source: Philippine MO 2022/4
Tags: geometry, incenter, exterior angle, angle bisector
P2nisic
18.03.2022 16:32
Let $I_A,I_B,I_C$ be the excenter of $ABC$ and$X,Y,Z$ be the midpoint of $BC,AC,AB$. Taylor circle of $I_AI_BI_C$ gives $P_CQ_CP_AQ_AP_BQ_B$ are on a circle. $P_CQ_C//I_AI_B$ $XQ_B//AB\Rightarrow X,Y,Q_B \text{ collinear}$ $XP_C=XQ_B$ If $J$ is the center of $P_CQ_CP_AQ_AP_BQ_B$ then it is the incenter of $XYZ$ $J,I,G$ are collinear. I can not find a proof form the last one. Edit:yes actually is easy.
cadaeibf
18.03.2022 20:06
P2nisic wrote: $J,I,G$ are collinear. I can not find a proof form the last one. The proof is simply by a homothety with center $G$ and factor $-\frac{1}{2}$, which brings $A\mapsto X,B\mapsto Y,C\mapsto Z$, and the inventer of $ABC$ gets mapped to the incenter $J$ of $XYZ$. This is also known as the Spieker point of $ABC$.
sarjinius
28.03.2022 10:49
Let $M_A, M_B, M_C$ be the midpoints of $BC, CA, AB$ respectively and let $I_A, I_B, I_C$ be the $A, B, C$-excenters of $\triangle ABC$ respectively. Note that $\angle BP_AC = 90^{\circ}$, so $M_AB = M_AC = M_AP_A$. So $\angle BM_AP_A = 180^{\circ} - 2\angle CBI_A = 180^{\circ} - 2\left(90^{\circ} - \frac{\angle ABC}{2}\right) = \angle ABC$. Hence, $M_AP_A \parallel AB$, and since $M_AM_B \parallel AB$, we get $P_A \in M_AM_B$. It can be proved analogously that $Q_B \in M_AM_B, Q_A, Q_C \in M_CM_A$, and $P_B, P_C \in M_BM_C$. Then $M_AQ_B = M_AM_B + M_BQ_B = M_CA + M_BA = M_CQ_C + M_AM_C = M_AQ_C$. Similarly, we can show that $M_BP_A = M_BP_C$ and $M_CQ_A = M_CP_B$. Thus, the angle bisector of $\angle M_BM_AM_C$ is the perpendicular bisector of both $P_AQ_A$ and $Q_BQ_C$. Similarly, we can show that the angle bisector of $\angle M_CM_BM_A$ is the perpendicular bisector of both $P_CP_A$ and $P_BQ_B$, and the angle bisector of $\angle M_AM_CM_B$ is the perpendicular bisector of both $Q_AP_B$ and $P_CQ_C$. Let $I_M$ be the incenter of $\triangle M_AM_BM_C$. Then $I_MP_A = I_MQ_A = I_MP_B = I_MQ_B = I_MQ_C = I_MP_C$, making $I_M$ the desired center of the six points. A homothety centered at $G$ takes $\triangle ABC$ to $\triangle M_AM_BM_C$, and this same homothety must take $I$ to $I_M$. Thus, $I_M$ lies on line $IG$, and we are done.