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, 2012

Geometry Difficulty 5.2 AIME, harder Prove it Hong Kong

Let $ABC$ be a scalene triangle (with no two sides equal). A circle touching sides $AB$ and $BC$ intersects the median $AM$ at $P$ and $Q$ . Another circle touching sides $AC$ and $BC$ also intersects $AM$ at $P$ and $Q$ . Prove that the two circles are the same circle.

Solution

Let $\Gamma_1$ be the circle touching $AB$ and $BC$ , and let $\Gamma_2$ be the circle touching $AC$ and $BC$ . Let $\Gamma_1$ touch $AB$ and $BC$ at $X$ and $D$ respectively. Let $\Gamma_2$ touch $AC$ and $BC$ at $Y$ and $E$ respectively.

Suppose on the contrary that $\Gamma_1 \neq \Gamma_2$ . Then $D \neq E$ since otherwise both circles pass through three same points. By using powers, we have
$MD^2 = MP \times MQ = ME^2.$
This implies $M$ is the midpoint of $D$ and $E$ . As $M$ is also the midpoint of $B$ and $C$ , we have $BD = CE$ .
Similarly, we have $AX^2 = AP \times AQ = AY^2$ , and so $AX = AY$ . It follows that
$AB = AX + BX = AX + BD = AY + CE = AY + CY = AC,$
contradicting the assumption that $\triangle ABC$ is scalene. Therefore, $\Gamma_1 = \Gamma_2$ , which is the incircle of $\triangle ABC$ .

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