# Prove that $O$ is the center of circle inscribed in triangle $O_1,O_2,O_3$ iff circles $k_1,k_2,k_3$ have same radius

Circles $k_1,k_2,k_3$ touch one another from outside and all of them touch circle $k$ from inside. Let $O_1,O_2,O_3,O$ be the centers of $k_1,k_2,k_3,k$ respectively. Prove that $O$ is the center of circle inscribed in triangle $O_1,O_2,O_3$ iff circles $k_1,k_2,k_3$ have same radius.

I easy proved that it is true if $k_1,k_2,k_3$ have same radius. Problem is to prove that if $k_1,k_2,k_3$ do not have same radius then $O$ cannot be the center of circle inscribed in triangle $O_1,O_2,O_3$. I tried in several ways, but I cannot prove this.

You have that $O$ is the incenter of $O_1 O_2 O_3$, so assuming that $\Gamma_1\cap\Gamma_2=P_3$ and so on, we have $OP_3\perp O_1 O_2$. Assuming $R_i > R_j$, we have $OO_i > OO_j$, so $OO_i+R_i > OO_j+R_j$, and $O$ cannot be the center of the circle externally tangent to $\Gamma_1,\Gamma_2,\Gamma_3$. To let that happen, we must have $R_1=R_2=R_3$.