32. [18] Triangle $A B C$ has $A B=4, B C=6$, and $A C=5$. Let $O$ denote the circumcenter of $A B C$. The circle $\Gamma$ is tangent to and surrounds the circumcircles of triangles $A O B, B O C$, and $A O C$. Determine the diameter of $\Gamma$.
Answer: $\frac{\mathbf{2 5 6} \sqrt{\mathbf{7}}}{\mathbf{1 7}}$. Denote by $\omega, \Gamma_{1}, \Gamma_{2}$, and $\Gamma_{3}$ the circumcenters of triangles $A B C, B O C, C O A$, and $A O B$, respectively. An inversion about $\omega$ interchanges $\Gamma_{1}$ and line $B C, \Gamma_{2}$ and line $C A$, and $\Gamma_{3}$ and line $A B$. This inversion also preserves tangency between generalized circles, so the image of $\Gamma$ is a circle tangent to $A B, B C$, and $C A$. It is the incircle of $A B C$ because it is closer to $O$ than these lines and $A B C$ is acute.

Now we run a few standard calculations. Where $s, r$, and $R$ denote the semiperimeter, inradius, and circumradius of $A B C$, respectively, we have the following:
$$
\begin{array}{l}
{[A B C]=\sqrt{s(s-a)(s-b)(s-c)}=\frac{15 \sqrt{7}}{4} ;} \\
r=[A B C] / s=\sqrt{7} / 2 \\
R=\frac{a b c}{4[A B C]}=\frac{8}{\sqrt{7}} \\
O I^{2}=R(R-2 r)=\frac{8}{7} .
\end{array}
$$

Let $O I$ intersect the incircle of $A B C$ at $P$ and $Q$, with $I$ between $P$ and $O$. Then $O P=r+O I$ and $O Q=r-O I$, and $\overline{P Q}$ is a diameter. Under the inversion, $P$ and $Q$ map to $P^{\prime}$ and $Q^{\prime}$ respectively. Because $P, I, O$, and $Q$ are collinear in that order, $P^{\prime}$ and $Q^{\prime}$ are diametrically opposed on $\Gamma$. It follows that the diameter of $\Gamma$ is
$$
P^{\prime} Q^{\prime}=O P^{\prime}+O Q^{\prime}=\frac{R^{2}}{O P}+\frac{R^{2}}{O Q}=R^{2}\left(\frac{1}{r+O I}+\frac{1}{r-O I}\right)=\frac{2 r R^{2}}{r^{2}-O I^{2}} .
$$

We plug in the values found above to arrive at $\frac{256 \sqrt{7}}{17}$.