Sides of rectangle with maximal area inscribed a given triangle?
The problem can be tackled through the AM-GM inequality. For simplicity, let us regard the $ABC$ triangle as the union of two right triangles having legs $c_1,h$ and $c_2,h$, with $c_1+c_2=c$.
The height $y$ of the inscribed rectangle fixes its area
$$ A = y\left(c-\frac{y}{h}c_1-\frac{y}{h}c_2\right)=\frac{y}{h}\left(1-\frac{y}{h}\right)hc $$
and $\frac{y}{h}\left(1-\frac{y}{h}\right)$ attains its maximum value when $y=\frac{h}{2}$. In such a case $A=\frac{hc}{4}=\frac{[ABC]}{2}$.
The fact that the area of the rectangle cannot exceed half the area of the triangle also follows from a nice origami argument: $ABC\setminus\text{rectangle}$ is made by three triangles that can be folded along three sides of the rectangle, and when we actually fold them...
