, 2016

Solution:

Let the $B$-mixtilinear incircle $\omega_{B}$ touch $\Gamma$ at $T_{B}$, $BA$ at $B_{1}$ and $BC$ at $B_{2}$. Define $T_{C} \in \Gamma$, $C_{1} \in CB$, $C_{2} \in CA$, and $\omega_{C}$ similarly. Call $I$ the incenter of triangle $ABC$, 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 $BC$ of the circumcircle of $ABC$. This follows from the fact that $M, B_{1}$, $T_{B}$ are collinear with
$$MB^{2} = MC^{2} = MB_{1} \cdot MT_{B}$$
and similarly for $C$.
- The midpoint $N$ of $ID$. 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 $\omega_{B}, \gamma$). A similar argument for $C$ shows that the midpoint of $ID$ is actually the radical center of the $\omega_{B}, \omega_{C}, \gamma$.

Now consider a homothety with ratio $2$ at $I$. It sends line $MN$ to the line through $D$ and the $A$-excenter $I_{A}$ (since $M$ is the midpoint of $II_{A}$, by "Fact 5"). Since $DH$ was supposed to be parallel to line $MN$, it follows that line $DH$ passes through $I_{A}$; however a homothety at $D$ implies that this occurs only if $H$ is the midpoint of the $A$-altitude.

Let $a = BC$, $b = CA = 2$ and $c = AB = \sqrt{3}$. So, we have to just find the value of $a$ such that the orthocenter of $ABC$ 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 $S_{A} = \frac{1}{2}(b^{2} + c^{2} - a^{2})$ and so on. Indeed, as $H$ is on the $A$-midline we deduce directly that
$$S_{B}S_{C} = S_{A}(S_{B} + S_{C}) = a^{2}S_{A} \Longrightarrow \frac{1}{4}(a^{2} - 1)(a^{2} + 1) = \frac{1}{2}a^{2}(7 - a^{2})$$
Solving as a quadratic in $a^{2}$ and taking the square roots gives
$$3a^{4} - 14a^{2} - 1 = 0 \Longrightarrow a = \sqrt{\frac{1}{3}(7 + 2\sqrt{13})}$$