Harvard-MIT Math Tournament

Solution:

It is easy to see that we will want one vertex of the triangle to be where the diameter meets the semicircle, so the diameter is divided into segments of length $x$ and $1 - x$, where $x$ is the length of the leg on the diameter. The other leg of the triangle will be the geometric mean of these two numbers, $\sqrt{x(1 - x)}$. Therefore the area of the triangle is $\frac{x \sqrt{x(1 - x)}}{2}$, so it will be maximized when $\frac{d}{dx}\left(x^{3} - x^{4}\right) = 3x^{2} - 4x^{3} = 0$, or $x = 3/4$. Therefore the maximum area is $\frac{3 \sqrt{3}}{32}$.