Center on line IG

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Center on line IG

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Philippine MO 2022/4

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

InternetPerson10

450 posts

InternetPerson10

#1 h Mar 18, 2022, 4:53 AM • 2 Y

Y by HWenslawski, son7

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 $IG$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

406 posts

#2 h Mar 18, 2022, 1:32 PM • 2 Y

Y by HWenslawski, son7

Let $I_A,I_B,I_C$ be the excenter of $ABC$ and $X,Y,Z$ be the midpoint of $BC,AC,AB$ .

:arrow:

Taylor circle of $I_AI_BI_C$ gives $P_CQ_CP_AQ_AP_BQ_B$ are on a circle.

:arrow:

$P_CQ_C//I_AI_B$

:arrow:

$XQ_B//AB\Rightarrow X,Y,Q_B \text{ collinear}$

:arrow:

$XP_C=XQ_B$

:arrow:

If $J$ is the center of $P_CQ_CP_AQ_AP_BQ_B$ then it is the incenter of $XYZ$

:arrow:

$J,I,G$ are collinear.

I can not find a proof form the last one.

Edit:yes actually is easy.

This post has been edited 1 time. Last edited by P2nisic, Mar 18, 2022, 6:08 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

700 posts

#3 h Mar 18, 2022, 5:06 PM

Y by

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 $ABC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

239 posts

#4 h Mar 28, 2022, 7:49 AM • 2 Y

Y by InternetPerson10, teomihai

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.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a