Problem (Geometry):
Let $T=20$. In $\triangle L E O, \sin \angle L E O=\frac{1}{T}$. If $L E=\frac{1}{n}$ for some positive real number $n$, then $E O=$ $n^{3}-4 n^{2}+5 n$. As $n$ ranges over the positive reals, compute the least possible value of $[L E O]$.

Solution:
Note that $[L E O]=\frac{1}{2}(\sin \angle L E O) \cdot L E \cdot E O=\frac{1}{2} \cdot \frac{1}{T} \cdot \frac{1}{n} \cdot\left(n^{3}-4 n^{2}+5 n\right)=\frac{n^{2}-4 n+5}{2 T}$. Because $T$ is a constant, the least possible value of $[L E O]$ is achieved when the function $f(n)=n^{2}-4 n+5$ is minimized.



This occurs when $n=-(-4) /(2 \cdot 1)=2$, and the minimum value is $f(2)=1$. Hence the desired least possible value of $[L E O]$ is $\frac{1}{2 T}$, and with $T=20$, this is $\frac{1}{40}$.

Final answer: $\frac{1}{40}$