10. The incircle of a triangle $A B C$ is tangent to $B C$ at $D$. Let $H$ and $\Gamma$ denote the orthocenter and circumcircle of $\triangle A B C$. The $B$-mixtilinear incircle, centered at $O_{B}$, is tangent to lines $B A$ and $B C$ and internally tangent to $\Gamma$. The $C$-mixtilinear incircle, centered at $O_{C}$, is defined similarly. Suppose that $\overline{D H} \perp \overline{O_{B} O_{C}}, A B=\sqrt{3}$ and $A C=2$. Find $B C$.
Answer:
$$
\sqrt{\frac{1}{3}(7+2 \sqrt{13})}
$$

Let the $B$-mixtilinear incircle $\omega_{B}$ touch $\Gamma$ at $T_{B}, B A$ at $B_{1}$ and $B C$ at $B_{2}$. Define $T_{C} \in \Gamma, C_{1} \in C B$, $C_{2} \in C A$, and $\omega_{C}$ similarly. Call $I$ the incenter of triangle $A B C$, and $\gamma$ the incircle.
We first identify two points on the radical axis of the $B$ and $C$ mixtilinear incircles:
- The midpoint $M$ of arc $B C$ of the circumcircle of $A B C$. This follows from the fact that $M, B_{1}$, $T_{B}$ are collinear with
$$
M B^{2}=M C^{2}=M B_{1} \cdot M T_{B}
$$
and similarly for $C$.
- The midpoint $N$ of $I D$. To see this, first recall that $I$ is the midpoint of segments $B_{1} B_{2}$ and $C_{1} C_{2}$. From this, we can see that the radical axis of $\omega_{B}$ and $\gamma$ contains $N$ (since it is the line through the midpoints of the common external tangents of $\left.\omega_{B}, \gamma\right)$. A similar argument for $C$ shows that the midpoint of $I D$ is actually the radical center of the $\omega_{B}, \omega_{C}, \gamma$.

Now consider a homothety with ratio 2 at $I$. It sends line $M N$ to the line through $D$ and the $A$-excenter $I_{A}$ (since $M$ is the midpoint of $I I_{A}$, by "Fact 5 "). Since $D H$ was supposed to be parallel to line $M N$, it follows that line $D H$ passes through $I_{A}$; however a homothety at $D$ implies that this occurs only if $H$ is the midpoint of the $A$-altitude.
Let $a=B C, b=C A=2$ and $c=A B=\sqrt{3}$. So, we have to just find the value of $a$ such that the orthocenter of $A B C$ lies on the midpoint of the $A$-altitude. This is a direct computation with the Law of Cosines, but a more elegant solution is possible using the fact that $H$ has barycentric coordinates $\left(S_{B} S_{C}: S_{C} S_{A}: S_{A} S_{B}\right)$, where $S_{A}=\frac{1}{2}\left(b^{2}+c^{2}-a^{2}\right)$ and so on. Indeed, as $H$ is on the $A$-midline we deduce directly that
$$
S_{B} S_{C}=S_{A}\left(S_{B}+S_{C}\right)=a^{2} S_{A} \Longrightarrow \frac{1}{4}\left(a^{2}-1\right)\left(a^{2}+1\right)=\frac{1}{2} a^{2}\left(7-a^{2}\right) .
$$

Solving as a quadratic in $a^{2}$ and taking the square roots gives
$$
3 a^{4}-14 a^{2}-1=0 \Longrightarrow a=\sqrt{\frac{1}{3}(7+2 \sqrt{13})}
$$
as desired. 1