Harvard-MIT Mathematics Tournament

Solution:
Let $h$ be the height of the cylinder. Then the volume of each piece is half the volume of the cylinder, so it is $\frac{1}{2} \pi h$. The base of the piece has area $\pi$, and the ellipse formed by the cut has area $\pi \cdot 1 \cdot \sqrt{1+\frac{h^{2}}{4}}$ because its area is the product of the semiaxes times $\pi$. The rest of the area of the piece is half the lateral area of the cylinder, so it is $\pi h$. Thus, the value of $m$ is
$$\begin{aligned} \frac{\pi+\pi \sqrt{1+h^{2} / 4}+\pi h}{\pi h / 2} & =\frac{2+2 h+\sqrt{4+h^{2}}}{h} \\ & =\frac{2}{h}+2+\sqrt{\frac{4}{h^{2}}+1} \end{aligned}$$
a decreasing function of $h$ whose limit as $h \rightarrow \infty$ is $3$. Therefore the greatest lower bound of $m$ is $3$.