, 2010

Let us denote the points $A' = AP \cap I_B I_C$, $B' = BP \cap I_A I_C$, $C' = CP \cap I_B I_A$ (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 $\triangle PAC$ and $PBC$ touch $CP$ at $C'$. Therefore, they touch each other at $C'$. Analogously, incircles of $\triangle PAB$, $\triangle PBC$ touch $PB$ at $B'$, and $A'$ - incircles of $\triangle PAC$, $\triangle PAB$.

Let $A''$ be the point where incircle of $\triangle PBC$ touches $BC$. We define $B''$, $C''$ in the same way. Then

Fig. 11

$$AP + BC = PA' + AA' + BA'' + A''C = PB' + AB'' + BB' + B''C = BP + AC.$$

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.