I was thinking of interesting and difficult geometry problems to attempt to solve when I came up with this:
Given and isosceles triangle with given base b and height h, find at what height, a, the area under a is equal to one third the area of the original triangle, in terms of b and h. Here is a diagram of the problem. Hope that you find this interesting and come up with a solution in terms of b and h.
Have fun!
The upper triangle is similar to the larger triangle and their area ratio is $\frac{2}{3}$. Their length ratio is therefore $\frac{\sqrt2}{\sqrt3}$. $$\frac{h-a}{h} = \frac{\sqrt2}{\sqrt3}$$ $$h - a = \frac{\sqrt{2}\cdot h}{\sqrt3}$$ $$a = h - \frac{\sqrt{2}\cdot h}{\sqrt3}$$
$$a = h\left (1 - \frac{\sqrt2}{\sqrt3} \right)$$