NMO Selection Tests for the Balkan and International Mathematical Olympiads

The problem amounts to showing collinearity of the antipodes $A_2$, $B_2$ and $C_2$ of $I$ in the circles $AI{A}_1$, $BI{B}_1$ and $CI{C}_1$, respectively. In the sequel, we use the following standard notations: $I_a$, $I_b$ and $I_c$ are the centers of the excircles tangent to the sides $BC$, $CA$ and $AB$, respectively; $a$, $b$ and $c$ are the lengths of the sides $BC$, $CA$ and $AB$, respectively; and $s=(a+b+c)/2$ is the semiperimeter of the triangle $ABC$. Clearly, the points $A_2$, $B_2$ and $C_2$ lie on the lines $I_b{I}_c$, $I_c{I}_a$ and $I_a{I}_b$, respectively. To show them collinear, we evaluate the ratio $A_2{I}_b/A_2{I}_c$ and the like, and refer to the converse of Menelaus' theorem.

To evaluate the ratio $A_2{I}_b/A_2{I}_c$, we apply the Menelaus theorem to triangle $I_a{I}_b{I}_c$ and line $A_1{A}_2$. Let the latter meet the lines $I_a{I}_b$ and $I_a{I}_c$ at $P$ and $Q$, respectively, to write $$\frac{A_2{I}_b}{A_2{I}_c}=\frac{P{I}_b}{P{I}_a}\cdot \frac{Q{I}_a}{Q{I}_c}.$$ We evaluate $P{I}_a$, $P{I}_b$, $Q{I}_a$ and $Q{I}_c$ as follows. Consider the cyclic quadrangles $I{A}_{1}CP$ and $BIC{I}_a$ to infer that the triangles $IP{I}_a$ and $I{A}_{1}B$ are similar and get thereby $$P{I}_a=A_{1}B\cdot \frac{I{I}_a}{IB}=\frac{s-c}{\sin \frac{C}{2}}.$$ Since $I_a{I}_b=c/\sin \frac{C}{2}$, we obtain $P{I}_b=|2c-s|/\sin \frac{C}{2}$. Next, consider the cyclic quadrangles $I{A}_{1}QB$ and $BIC{I}_a$ to deduce that the triangles $IQ{I}_a$ and $I{A}_{1}C$ are similar and get thereby $$Q{I}_a=A_{1}C\cdot \frac{I{I}_a}{IC}=\frac{s-b}{\sin \frac{B}{2}}.$$ Since $I_a{I}_c=b/\sin \frac{B}{2}$, we obtain $Q{I}_c=|2b-s|/\sin \frac{B}{2}$. Consequently, $$\frac{A_2{I}_b}{A_2{I}_c}=\frac{s-b}{s-c}\cdot \frac{|2c-s|}{|2b-s|}$$ and the conclusion follows.