Problem 1205

Solution:

By the Law of Sines we have $\sin \angle A = \frac{X Y}{A P} = \frac{4}{5}$. Let $I$, $T$, and $Q$ denote the center of $\omega$, the point of tangency between $\omega$ and $\Gamma$, and the center of $\Gamma$ respectively. Since we are told $A B C$ is acute, we can compute $\tan \frac{\angle A}{2} = \frac{1}{2}$. Since $\angle E A I = \frac{\angle A}{2}$ and $\overline{A E}$ is tangent to $\omega$, we find $r = \frac{A E}{2} = \frac{15}{2}$.

Let $H$ be the foot of the altitude from $A$ to $\overline{B C}$. Define $h_{T}$ to be the homothety about $T$ which sends $\Gamma$ to $\omega$. We have $h_{T}(\overline{A Q}) = \overline{D I}$, and conclude that $A$, $T$, and $D$ are collinear. Now since $\overline{A P}$ is a diameter of $\Gamma$, $\angle P A T$ is right, implying that $D T H P$ is cyclic. Invoking Power of a Point twice, we have $225 = A E^{2} = A T \cdot A D = A P \cdot A H$. Because we are given radius of $\Gamma$ we can find $A P = 10$ and $A H = \frac{45}{2} = h_{a}$.

If we write $a$, $b$, $c$, $s$ in the usual manner with respect to triangle $A B C$, we seek $B D \cdot D C = (s-b)(s-c)$. But recall that Heron's formula gives us
$$\sqrt{s(s-a)(s-b)(s-c)} = K$$
where $K$ is the area of triangle $A B C$. Writing $K = r s$, we have $(s-b)(s-c) = \frac{r^{2} s}{s-a}$. Knowing $r = \frac{15}{2}$, we need only compute the ratio $\frac{s}{a}$. By writing $K = \frac{1}{2} a h_{a} = r s$, we find $\frac{s}{a} = \frac{h_{a}}{2 r} = \frac{3}{2}$.

Now we compute our answer,
$$\frac{r^{2} s}{s-a} = \left(\frac{15}{2}\right)^{2} \cdot \frac{\frac{s}{a}}{\frac{s}{a} - 1} = \frac{675}{4}.$$