50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

Let us denote the points $A^{\prime }=AP\cap I_B{I}_C$, $B^{\prime }=BP\cap I_A{I}_C$, $C^{\prime }=CP\cap I_B{I}_A$ (see Fig. 11).

$I_P=P$ implies that $I_{A}P$ is a bisector of $\angle BPC$ and $\angle I_B{I}_A{I}_C$, therefore two couples of lines $I_A{I}_B$, $I_A{I}_C$, and also $BP$, $CP$ are symmetric with respect to $I_{A}P$. Thus, $\angle (I_A{I}_B,CP)=\angle (BP,I_A{I}_C)$ (oriented angles).

By analogy, $\angle (I_B{I}_C,AP)=\angle (CP,I_B{I}_A)$, $\angle (I_C{I}_A,BP)=\angle (AP,I_C{I}_B)$, hence, $\angle (I_A{I}_B,CP)=\angle (BP,I_A{I}_C)=-\angle (AP,I_C{I}_B)=\angle (CP,I_B{I}_A)\Rightarrow CP\perp I_B{I}_A$.

Feet of perpendicular from incenter to the side of triangle coincide with the point where incircle touches this side, hence, incircles of $△PAC$ and $PBC$ touch $CP$ at $C^{\prime }$. Therefore, they touch each other at $C^{\prime }$. Analogously, incircles of $△PAB$, $△PBC$ touch $PB$ at $B^{\prime }$, and $A^{\prime }$ - incircles of $△PAC$, $△PAB$.

Let $A^{\prime \prime }$ be the point where incircle of $△PBC$ touches $BC$. We define $B^{\prime \prime }$, $C^{\prime \prime }$ in the same way. Then

Fig. 11

$$AP+BC=P{A}^{\prime }+A{A}^{\prime }+B{A}^{\prime \prime }+A^{\prime \prime }C=P{B}^{\prime }+A{B}^{\prime \prime }+B{B}^{\prime }+B^{\prime \prime }C=BP+AC.$$

Following the same lines, we get $AP+BC=CP+AB$. Hence, $P$ is such a point that $AP-BP=AC-BC$, $BP-CP=BA-CA$, $CP-AP=CB-AB$, and we are done.