Triple Circles, but actually, there is only one

Geometry Level 5

Circle $\omega_1$ is tangent to $AB$ and $AC$ , $\omega_2$ is tangent to $BA$ and $BC$ , and $\omega_3$ is tangent to $CB$ and $CA$ . Those three circles have equal radii and are inside $\triangle ABC$ . If $\omega_1, \omega_2$ and $\omega_3$ have a common point $P$ , which of the answers is true?

$\textbf{Note:}$ $I$ is the incenter. $O$ is the circumcenter, $H$ is the orthocenter, $G$ is the centroid, $I_a$ is the excenter for the A-excircle, and $A' , B', C'$ are the centers of circles $\omega_1, \omega_2 , \omega_3$ respectively.

\triangle ABC \sim \triangle C'B'A'

Points

P, I, O

are collinear Points

P, I, H

form an equilateral triangle Points

P, I, H

are collinear

P, G, I_a

are collinear

1 solution

Alan Yan
Sep 13, 2015

EDIT: In the diagram, $C'$ should be $A'$ and $B'$ should be $C'$ and $A'$ should be $'B$ .

Notice how $\triangle A'B'C' \sim \triangle ABC$ . We can prove this by using the fact that $A'B' || AB , A'C' || AC , B'C' || BC$ and start angle chasing.

Thus, there must be a point inside $\triangle A'B'C'$ that is the center of homothety of the two triangles that maps $A$ to $A'$ , $B$ to $B'$ , and $C$ to $C'$ .

Notice how by connecting $A$ to $A'$ and likewise for $B$ and $C$ we get the three angle bisectors. These intersect at $I$ .

This means that $I$ must be the center of homothety of $\triangle ABC$ and $\triangle A'B'C'$ .

Now notice how $P$ is equidistant from $A', B', C'$ which implies it is the circumcenter of $\triangle A'B'C'$ . This implies that $I$ maps $O$ to $P$ . Because homothety preserves the collinearity of mapped points, we know that $\boxed{P, I, O \text{ are collinear} }$

Triple Circles, but actually, there is only one

1 solution

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