11. Let $\Omega(I, r)$ be the incircle of $\triangle A B C$. Let $D, E$, and $F$ denote the points where $\Omega$ touches $B C, A C$, and $A B$, respectively. Let $P, Q$, and $R$ denote the midpoints of $E F, D F$, and $D E$ respectively. We prove that $\Omega_{a}$ passes through $Q$ and $R$. Since $\triangle I Q D \sim \triangle I D B$ and $\triangle I R D \sim \triangle I D C$, we obtain $I Q \cdot I B=I R \cdot I C=r^{2}$. We conclude that $B, C, Q$, and $R$ lie on a single circle $\Gamma_{a}$. Moreover, since the power of $I$ with respect to $\Gamma_{a}$ is $r^{2}$, it follows for a tangent $I X$ from $I$ to $\Gamma_{a}$ that $X$ lies on $\Omega$ and hence $\Omega$ is perpendicular to $\Gamma_{a}$. From the uniqueness of $\Omega_{a}$ it follows that $\Omega_{a}=\Gamma_{a}$. Thus $\Omega_{a}$ contains $Q$ and $R$. Similarly $\Omega_{b}$ contains $P$ and $R$ and $\Omega_{c}$ contains $P$ and $Q$. Hence, $A^{\prime}=P, B^{\prime}=Q$ and $C^{\prime}=R$. Therefore the radius of the circumcircle of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is half the radius of $\Omega$.