Iranian Mathematical Olympiad

We start by proving a simple lemma.

Lemma 1. Let $X^{\prime }$ be the reflection of the circumcenter of triangle $XYZ$ with respect to side $YZ$. Then $A^{\prime }$ and $X$ are symmetric with respect to the center of the nine-point circle of triangle $XYZ$.

Proof. Let $O$ and $H$ be the circumcenter and orthocenter of triangle $XYZ$, respectively. It is well-known that the center of the nine-point circle is the midpoint of $HO$. On the other hand, it is easy to see that $O{X}^{\prime }=XH$ and $O{X}^{\prime }\parallel XH$. So the quadrilateral $XH{X}^{\prime }O$ is a parallelogram and the proof is complete. $\square$

Using the lemma, it is enough to show that $A^{\prime }$ and $I$ are symmetric with respect to $B_1{C}_1$. To prove it, let $A^{\prime \prime }$ be the reflection of $I$ with respect to $B_1{C}_1$. We claim that $A^{\prime \prime }=A^{\prime }$. We will show that $A_1{A}^{\prime \prime }\perp B_2{C}_2$. By cosine formula in triangles $A_{1}I{B}_2$, $A_{1}I{C}_2$, $A^{\prime \prime }I{B}_1$ and $A^{\prime \prime }I{C}_1$, we get: $$A_1{B}_2^2=r^2+I{B}_2^2-2r\cdot I{B}_2\cos (90^{\circ }+\frac{\angle B}{2}),(1)$$ $$A_1{C}_2^2=r^2+I{C}_2^2-2r\cdot I{C}_2\cos (90^{\circ }+\frac{\angle C}{2}),(2)$$ $$A^{\prime \prime }{B}_2^2=r^2+B_1{B}_2^2-2r\cdot B_1{B}_2\cos (90^{\circ }-\angle A),(3)$$ $$A^{\prime \prime }{C}_2^2=r^2+C_1{C}_2^2-2r\cdot C_1{C}_2\cos (90^{\circ }-\angle A),(4)$$ where $r$ indicates the radius of the incircle of triangle $ABC$. Subtracting equations (1) from (2) and (3) from (4) imply $$\begin{gathered} A_1{B}_2^2-A_1{C}_2^2=I{B}_2^2-I{C}_2^2-2r\left(I{B}_2\cos \left(90^{\circ }+\frac{\angle B}{2}\right)-I{C}_2\cos \left(90^{\circ }+\frac{\angle C}{2}\right)\right), \\ {A}^{\prime \prime }{B}_2^2-A^{\prime \prime }{C}_2^2=B_1{B}_2^2-C_1{C}_2^2-2r\left(B_1{B}_2\cos \left(90^{\circ }-\angle A\right)-C_1{C}_2\cos \left(90^{\circ }-\angle A\right)\right). \end{gathered}$$ Note that $$B_2{C}_1^2-B_1{C}_2^2=r^2+B_1{B}_2^2-r^2-C_1{C}_2^2=B_1{B}_2^2-C_1{C}_2^2.$$ Therefore, we must show $$I{B}_2\cos (90^{\circ }+\frac{\angle B}{2})-I{C}_2\cos (90^{\circ }+\frac{\angle C}{2})=(B_1{B}_2-C_1{C}_2)\cos (90^{\circ }-\angle A).$$ Let $M$ and $N$ be the perpendicular projections of $B_2$ and $C_2$ on $AC$ and $AB$, respectively. It is easy to see that $\angle I{B}_{2}M=90^{\circ }-\frac{\angle B}{2}$ and $\angle I{C}_{2}M=90^{\circ }-\frac{\angle C}{2}$. So $$\begin{array}{rl}I{B}_2\cos (90^{\circ }+\frac{\angle B}{2})-I{C}_2\cos (90^{\circ }+\frac{\angle C}{2})& =I{C}_2\cos (90^{\circ }-\frac{\angle C}{2})-I{B}_2\cos (90^{\circ }-\frac{\angle B}{2})\\ & =(C_{2}N-r)-(B_{2}M-r)\\ & =C_{2}N-B_{2}M.\end{array}$$ On the other hand, $B_1{B}_2-C_1{C}_2=A{C}_2-A{B}_2$ and hence $$(B_1{B}_2-C_1{C}_2)\cos (90^{\circ }-\angle A)=(A{C}_2-A{B}_2)\sin (\angle A)=C_{2}N-B_{2}M$$ This completes the proof.