Harvard-MIT Mathematics Tournament

Solution:

Suppose first that $D$ lies between $B$ and $C$. Let $ABC$ be inscribed in circle $\omega$, and extend $AD$ to intersect $\omega$ again at $E$. Note that $A$ subtends a quarter of the circle, so in particular, the chord through $C$ perpendicular to $BC$ and parallel to $AD$ has length $BC = 5$. Therefore, $AD = 5 + DE$. By power of a point, $6 = BD \cdot DC = AD \cdot DE = AD^2 - 5AD$, implying $AD = 6$, so the area of $ABC$ is $\frac{1}{2} BC \cdot AD = 15$.

If $D$ does not lie between $B$ and $C$, then $BC = 1$, so $A$ lies on a circle of radius $\sqrt{2}/2$ through $B$ and $C$. But then it is easy to check that the perpendicular to $BC$ through $D$ cannot intersect the circle, a contradiction.