Philippine Mathematical Olympiad

Solution:

Refer to the figure shown below:

Let $V$ and $W$ be the midpoints of $\operatorname{arcs} BD$ and $CD$ respectively. We claim that $VW$ is the desired common tangent. To prove this, let $E$ and $F$ be the orthogonal projections of $V$ and $W$ onto $BC$. Note that $E$ and $F$ are the midpoints of $BD$ and $CD$ respectively. Now we claim that $VE = WF$. To see this, note that
$$\begin{aligned} VE & = BV \sin \angle VBE \\ & = \frac{BX \sin \angle BXV}{\sin \angle BDX} \sin \frac{\angle BXD}{2} \\ & = \frac{BX \sin \frac{\angle BXA}{2}}{\sin \angle BDX} \sin \frac{C}{2} \\ & = \frac{BX}{\sin \angle BDX} \sin^2 \frac{C}{2} \end{aligned}$$
Similarly, we can prove that $WF = \frac{CX}{\sin \angle CDX} \sin^2 \frac{B}{2}$. Thus, to prove the claim, it suffices to prove that $\frac{BX}{CX} = \frac{\sin^2 \frac{B}{2}}{\sin^2 \frac{C}{2}}$. This is because
$$\frac{BX}{CX} = \frac{\sin \angle BAD}{\sin \angle CAD} = \frac{c / BD}{b / CD} = \frac{c(s-c)}{b(s-b)} = \frac{\sin^2 \frac{B}{2}}{\sin^2 \frac{C}{2}}$$
This proves the first claim.

Next, we claim that $VW$ is the desired common tangent. Note that from the first claim, $VWFE$ is a rectangle, since $\angle VEF$ and $\angle WFE$ are both right angles. Let $O_1$ and $O_2$ be the circumcenters of triangles $BDX$ and $CDX$ respectively. Then $O_1V \perp EF$, so since $EF \parallel VW$ we get $O_1V \perp VW$. Likewise, $O_2W \perp VW$ as well, which proves the second claim.

It then follows that $VW$ is the desired common tangent parallel to $BC$, and the required conclusion follows.