7. Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. Let $H$ be the orthocenter of $A B C$. Find the radius of the circle with nonzero radius tangent to the circumcircles of $A H B, B H C, C H A$.
Answer:
$\frac{65}{4}$
Solution 1. We claim that the circle in question is the circumcircle of the anticomplementary triangle of $A B C$, the triangle for which $A B C$ is the medial triangle.
Let $A^{\prime} B^{\prime} C^{\prime}$ be the anticomplementary triangle of $A B C$, such that $A$ is the midpoint of $B^{\prime} C^{\prime}, B$ is the midpoint of $A^{\prime} C^{\prime}$, and $C$ is the midpoint of $A^{\prime} B^{\prime}$. Denote by $\omega$ the circumcircle of $A^{\prime} B^{\prime} C^{\prime}$. Denote by $\omega_{A}$ the circumcircle of $B H C$, and similarly define $\omega_{B}, \omega_{C}$.
Since $\angle B A^{\prime} C=\angle B A C=180^{\circ}-\angle B H C$, we have that $\omega_{A}$ passes through $A^{\prime}$. Thus, $\omega_{A}$ can be redefined as the circumcircle of $A^{\prime} B C$. Since triangle $A^{\prime} B^{\prime} C^{\prime}$ is triangle $A^{\prime} B C$ dilated by a factor of 2 from point $A^{\prime}, \omega$ is $\omega_{A}$ dilated by a factor of 2 from point $A^{\prime}$. Thus, circles $\omega$ and $\omega_{A}$ are tangent at $A^{\prime}$.
By a similar logic, $\omega$ is also tangent to $\omega_{B}$ and $\omega_{C}$. Therefore, the circumcircle of the anticomplementary triangle of $A B C$ is indeed the circle that the question is asking for.
Using the formula $R=\frac{a b c}{4 A}$, we can find that the circumradius of triangle $A B C$ is $\frac{65}{8}$. The circumradius of the anticomplementary triangle is double of that, so the answer is $\frac{65}{4}$.
Solution 2. It is well-known that the circumcircle of $A H B$ is the reflection of the circumcircle of $A B C$ over $A B$. In particular, the circumcircle of $A H B$ has radius equal to the circumradius $R=\frac{65}{8}$. Similarly, the circumcircles of $B H C$ and $C H A$ have radii $R$. Since $H$ lies on all three circles (in the question), the circle centered at $H$ with radius $2 R=\frac{65}{4}$ is tangent to each circle at the antipode of $H$ in that circle.