Romania — 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 $AIA_1$, $BIB_1$ and $CIC_1$, 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 $I_bI_c$, $I_cI_a$ and $I_aI_b$, respectively. To show them collinear, we evaluate the ratio $A_2I_b/A_2I_c$ and the like, and refer to the converse of Menelaus' theorem.

To evaluate the ratio $A_2I_b/A_2I_c$, we apply the Menelaus theorem to triangle $I_aI_bI_c$ and line $A_1A_2$. Let the latter meet the lines $I_aI_b$ and $I_aI_c$ at $P$ and $Q$, respectively, to write
$$\frac{A_2I_b}{A_2I_c} = \frac{PI_b}{PI_a} \cdot \frac{QI_a}{QI_c}.$$
We evaluate $PI_a$, $PI_b$, $QI_a$ and $QI_c$ as follows. Consider the cyclic quadrangles $IA_1CP$ and $BICI_a$ to infer that the triangles $IPI_a$ and $IA_1B$ are similar and get thereby
$$PI_a = A_1B \cdot \frac{II_a}{IB} = \frac{s-c}{\sin \frac{C}{2}}.$$
Since $I_aI_b = c/\sin \frac{C}{2}$, we obtain $PI_b = |2c-s|/\sin \frac{C}{2}$. Next, consider the cyclic quadrangles $IA_1QB$ and $BICI_a$ to deduce that the triangles $IQI_a$ and $IA_1C$ are similar and get thereby
$$QI_a = A_1C \cdot \frac{II_a}{IC} = \frac{s-b}{\sin \frac{B}{2}}.$$
Since $I_aI_c = b/\sin \frac{B}{2}$, we obtain $QI_c = |2b-s|/\sin \frac{B}{2}$. Consequently,
$$\frac{A_2I_b}{A_2I_c} = \frac{s-b}{s-c} \cdot \frac{|2c-s|}{|2b-s|}$$
and the conclusion follows.