2013 All-Russian Olympiad Final Round Grade 10 Day 2 P7

[tanmoy]

The incircle of triangle $ ABC $ has centre $I$ and touches the sides $ BC $, $ CA $, $ AB $ at points $ A_1 $, $ B_1 $, $ C_1 $, respectively. Let $ I_a $, $ I_b $, $ I_c $ be excentres of triangle $ ABC $, touching the sides $ BC $, $ CA $, $ AB $ respectively. The segments $ I_aB_1 $ and $ I_bA_1 $ intersect at $ C_2 $. Similarly, segments $ I_bC_1 $ and $ I_cB_1 $ intersect at $ A_2 $, and the segments $ I_cA_1 $ and $ I_aC_1 $ at $ B_2 $. Prove that $ I $ is the center of the circumcircle of the triangle $ A_2B_2C_2 $.

[rah4927]

I haven't really tried out this problem, but it seems that $AA_2,BB_2,CC_2$ are concurrent. Furthermore, it seems(this one I am not so sure about) that $\Delta ABC$ is similar to $\Delta A_2B_2C_2$ , but not directly.

[tanmoy]

I haven't really tried out this problem, but it seems that $AA_2,BB_2,CC_2$ are concurrent. Furthermore, it seems(this one I am not so sure about) that $\Delta ABC$ is similar to $\Delta A_2B_2C_2$ , but not directly.

Nope

[tanmoy]

$\text{My solution}:
$

Notice that $\triangle A_{1}B_{1}C_{1}$ and $\triangle I_{a}I_{b}I_{c}$ are homothetic,so,we get $\displaystyle{\frac{B_{2}I_{a}}{B_{2}C_{1}}=\frac{I_{a}I_{c}}{A_{1}C_{1}}=\frac{I_{a}I_{b}}{A_{1}B_{1}}=\frac{C_{2}I_{a}}{C_{2}B_{1}}}$.
$\therefore$ $B_{2}$ and $C_{2}$ are symmetric WRT $AI$.
Similarly, we can prove that $C_{2}$ and $A_{2}$ are symmetric WRT $BI$,
so $IA_{2}=IB_{2}=IC_{2}$ $\Rightarrow$ $I$ is the circumcenter of $\triangle A_{2}B_{2}C_{2}$.