14th Turkish Mathematical Olympiad

Let $r$ be the inradius and $I$ the incenter of the triangle $ABC$. The triangle $ABC$ is similar to the triangle $A{B}_1{C}_1$ with scale factor $\cos \angle A$. Let $X$ and $Y$ be the points of tangency of the incircle of the triangle $A{B}_1{C}_1$ with the line segments $A{C}_1$ and $A{B}_1$, respectively. As $AX/A{T}_B=\cos \angle A$ by similarity, we conclude that the line segment $T_{B}X$ is perpendicular to the line segment $AX$, and therefore, passes through the point $O_A$. In other words, the line $T_B{O}_A$ is perpendicular to the line $AB$. Similarly, the line $T_C{O}_A$ is perpendicular to the line $AC$. On the other hand, the lines $I{T}_C$ and $I{T}_B$ are perpendicular to the lines $AB$ and $AC$, respectively. It follows that quadrilateral $I{T}_B{O}_A{T}_C$ is a parallelogram.

Since $I{T}_C=I{T}_B=r$, this parallelogram is equilateral. This implies that $T_B{O}_A=O_A{T}_C=r$, and finishes the proof.