66th Belarusian Mathematical Olympiad

Without loss of generality we can assume that $Q$ belongs to the segment $PC$. First, note that if the incircle of the triangle $ABC$ touches the sides $AB$, $BC$, $CA$ at the points $C_1$, $A_1$, $B_1$, respectively, then $$\begin{array}{rl}A{B}_1& =A{C}_1=\frac{AB+AC-BC}{2},\\ B{A}_1& =B{C}_1=\frac{BA+BC-AC}{2},\\ C{A}_1& =C{B}_1=\frac{CA+CB-AB}{2}.\end{array}$$ Let ${\omega }_1$ and ${\omega }_2$ denote the incircles of the triangles $ABP$ and $APC$, respectively. Let internal tangent $l$ (different from $AP$) to the circles ${\omega }_1$ and ${\omega }_2$ meet the side $BC$ at $Q^{\prime }$. We show that $I_1$, $P$, $Q^{\prime }$, and $I_2$ are concyclic, which gives $Q^{\prime }\equiv Q$. Let $R$ be the intersection point of the lines $l$ and $AP$. Since $I_1$ and $I_2$ lie on the internal and external bisectors of the angle $\angle P{Q}^{\prime }R$, respectively, we have $\angle I_1{Q}^{\prime }{I}_2=90^{\circ }$.



Similarly, we show that $\angle I_{1}P{I}_2=90^{\circ }$. Therefore, the quadrilateral $I_{1}P{Q}^{\prime }{I}_2$ is cyclic, so $Q^{\prime }\equiv Q$. Using the formulae mentioned above, we have $DP=(PB+PA-AB)/2$ and $PE=(PC+PA-AC)/2$, from lemma of Problem A.7 it follows that $DP=QE=PE-PQ$. Thus, $$PQ=PE-DP=(AB+PC-AC-PB)/2.$$