Area of the overlap between a triangle and a square $ABC$ is an equilateral triangle, each side has length 4.  
$M$ is the midpoint of $\overline{BC}$, and $\overline{AM}$ is a diagonal of square $ALMN$. Find the area of the region common to both $ABC$ and $ALMN$.
I'm not sure how to solve this problem, or even where to start. Any hints?

I'm assuming that this is the figure you mean.
                         
First off, notice that the overlapping area is a kite shape, and since kites are orthodiagonal quadrilaterals, we can find the area using the formula
$$
A=\frac{p\cdot q}{2}
$$
Where $p$ and $q$ are the kite's diagonals. The long diagonal is clearly just the height of the triangle, which is just
$$r\cos\:(30^\circ)=\frac{4\sqrt {3}}{2}=p$$
The shorter diagonal is a bit more challenging to find. The easiest to understand, in my opinion, is to find the positions where the sides of the square intersect the sides of the triangle (you only need to find one). This can be solved as a system of equations, since we know the angles from the center diagonal of the triangle and square to be $30^\circ$ and $45^\circ$ respectively:
$$
a\sin (45^\circ) = b\sin (30^\circ)
$$$$
a\cos (45^\circ) = -b\cos (30^\circ)
$$
Once you've solved for $a$ and $b$, the second diagonal of the kite is just
$$
2a\sin (45^\circ)=q
$$ or $$
2b\sin (30^\circ)=q
$$
Hope this helped :-)
