Problem (Geometry):
Triangle $A B C$ is inscribed in circle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at point $T$. The tangent to $\omega$ at $A$ intersects the perpendicular bisector of $\overline{A T}$ at point $P$. Given that $A B=14, A C=30$, and $B C=40$, compute $[P B C]$.

Solution:
To begin, denote by $R$ the radius of $\omega$. The semiperimeter of triangle $A B C$ is 42 , and then applying Heron's formula yields

$$
[A B C]=\frac{14 \cdot 30 \cdot 40}{4 R}=\sqrt{42 \cdot 28 \cdot 12 \cdot 2}=168
$$

from which it follows that $R=\frac{14 \cdot 30 \cdot 40}{4 \cdot 168}=25$.

Now consider the point circle with radius zero centered at $T$ in tandem with the circle $\omega$. Because $P A=P T$, it follows that $P$ lies on the radical axis of these circles. Moreover, the midpoints of $\overline{T B}$ and $\overline{T C}$ lie on this radical axis as well. Thus $P$ lies on the midline of $\triangle T B C$ that is parallel to $\overline{B C}$.



To finish, let $O$ denote the center of $\omega$ and $M$ the midpoint of $\overline{B C}$. By considering right triangle $T B O$ with altitude $\overline{B M}$, it follows that $M T \cdot M O=M B^{2}$, but also $M O=\sqrt{O B^{2}-M B^{2}}=\sqrt{25^{2}-20^{2}}=15$, so

$$
M T=\frac{M B^{2}}{M O}=\frac{400}{15}=\frac{80}{3}
$$

Thus the distance from $P$ to $\overline{B C}$ is $\frac{1}{2} M T=\frac{40}{3}$. Finally,

$$
[P B C]=\frac{1}{2} \cdot \frac{40}{3} \cdot B C=\frac{\mathbf{8 0 0}}{\mathbf{3}}
$$

Final answer: $\frac{800}{3}$