TITLE: Prove that the circumcenter of $\triangle PIQ$ is on the hypotenuse $AC$.
QUESTION [6 upvotes]: In right angled $\triangle ABC$ with $\angle B=90 ^{\circ}$, $BD$ is an altitude on $AC$. $P,Q,I$ are the incenters of $\triangle ABD,\triangle CBD$ and $\triangle ABC$ respectively. Prove that the circumcenter of $\triangle PIQ$ lies on the hypotenuse $AC$.
This question was asked today in the Indian National Mathematics Olympiad (INMO) 2015 in which I appeared.I totally had no idea and infact have no idea on how to proceed. Any help would be appreciated.

REPLY [3 votes]: We note that $\triangle ABC \sim \triangle ADB \sim \triangle BDC$, with corresponding segments within these triangles having lengths in the proportion $1 : \sin C : \sin A$. In particular, if $r$ is the radius of incircle $\bigcirc I$ of $\triangle ABC$, then $r\sin C$ and $r \sin A$ are radii of incircles $\bigcirc P$ and $\bigcirc Q$, respectively.

With $J$ and $K$ (and $L$) the points of tangency of $\bigcirc I$ with the sides of $\triangle ABC$ as shown, we see that $\square BJIK$ is a square of side-length $r$. The perpendiculars from $J$ and $K$ to $\overline{BD}$ have length $r\sin C$ and $r\sin A$, respectively; these match the lengths of perpendiculars from $P$ and $Q$ to $\overline{BD}$, so that $\overline{JP}$ and $\overline{KQ}$ are parallel to $\overline{BD}$. Moreover, $\overline{JP}$ is the hypotenuse of a right triangle with a leg of length $r \sin C$ opposite an angle congruent to $\angle C$; that hypotenuse must have length $r$. Likewise for $\overline{KQ}$.
The above shows that points $P$ and $Q$ are translates of $J$ and $K$ by distance $r$ in direction $\overrightarrow{BD}$. But point $L$ is a translate of $I$ in the same way. Consequently,
$$|\overline{LP}| = |\overline{IJ}| = r = |\overline{IK}| = |\overline{LQ}| \qquad\text{and}\qquad |\overline{LI}| = r$$
so that the circumcenter of $\triangle IPQ$ is $L$. $\square$