[Geometry]
Problem: Let $T=13$. If $r$ is the radius of a right circular cone and the cone's height is $T-r^{2}$, let $V$ be the maximum possible volume of the cone. Compute $\pi / V$.

Solution: The cone's volume is $\frac{1}{3} \pi r^{2}\left(T-r^{2}\right)$. Maximizing this is equivalent to maximizing $x(T-x)$, where $x=r^{2}$. Using the formula for the vertex of a parabola (or the AM-GM inequality), the maximum value occurs when $x=\frac{T}{2}$. Hence $V=\frac{1}{3} \pi \cdot \frac{T}{2} \cdot \frac{T}{2}=\frac{\pi T^{2}}{12}$, and $\pi / V=12 / T^{2}$. With $T=13, V=\frac{\mathbf{1 2}}{\mathbf{1 6 9}}$.

Final answer: $\frac{12}{169}$