Saudi Arabia Mathematical Competitions

We shall prove that $A_1$ is the orthocenter of triangle $I_{a}BC$. Indeed, we have $$\begin{array}{rl}\widehat{PE{I}_a}& =180^{\circ }-\widehat{PQA}-\widehat{ECQ}\\ & =180^{\circ }-\left(90^{\circ }-\frac{1}{2}\hat{A}\right)-\left(90^{\circ }-\frac{1}{2}\hat{C}\right)\\ & =\frac{1}{2}(\hat{A}+\hat{C})=90^{\circ }-\frac{1}{2}\hat{B}=\widehat{PB{I}_a}\end{array}$$ hence quadrilateral $BE{I}_{a}P$ is cyclic. Since $\widehat{BP{I}_a}=90^{\circ }$, it follows $BE\perp C{I}_a$. In an analogous way we get $CD\perp BI$, hence $A_1$ is the orthocenter of triangle $I_{a}BC$.



It follows that $B{A}_1\parallel CI$, where $I$ is the incenter of triangle $ABC$. Also, $C{A}_1\parallel BI$, hence $BIC{A}_1$ is a parallelogram. Similarly, $AIB{C}_1$ is a parallelogram, hence $A{C}_1{A}_{1}C$ is a parallelogram. We obtain that the segments $A{A}_1$ and $C{C}_1$ have the same midpoint. In an analogous way, the segments $B{B}_1$ and $A{A}_1$ have the same midpoint, and the conclusion follows.



Remark. It is clear that $I_b,A,I_c$ and $I_c,B,I_a$, and $I_a,C,I_b$ are collinear. As in the previous solution, $A_1$ is the orthocenter of triangle $I_{a}BC$, hence $I_a{A}_1\perp BC$. Similarly, $I_b{B}_1\perp CA$ and $I_c{C}_1\perp AB$. The triangles $ABC$ and $I_a{I}_b{I}_c$ are orthological, that is the perpendicular lines through $A,B,C$ on $I_b{I}_c,I_c{I}_a$, and $I_a{I}_b$, respectively, are concurrent (as internal bisectors of triangle $ABC$ ). It follows that also, $I_a{A}_1,I_b{B}_1,I_c{C}_1$ are concurrent.