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?

Use this diagram.

The equilateral triangle $ABC$ has sides of length $4$, so $CM=2$ and $AM=2\sqrt 3$.
The area you are trying to find is shaded and is clearly twice the area of triangle $AEM$. We know the base, $AM$, so we want the height $EF$. If we let $x=EF$ then $MF=x$ due to $45-45-90$ triangle $EFM$. We also have $AF=\sqrt 3x$ due to $30-60-90$ triangle $AEF$. Therefore we get
$$AF+MF=AM$$
$$\sqrt 3x+x=2\sqrt 3$$
$$x=\frac{2\sqrt 3}{\sqrt 3+1}$$
$$=\frac{2\sqrt 3(\sqrt 3-1)}{(\sqrt 3+1)(\sqrt 3-1)}$$
$$=\frac{6-2\sqrt 3}{2}$$
$$=3-\sqrt 3$$
Therefore the area we want is twice the area of triangle $AEM$:
$$Area=2\cdot \frac 12bh$$
$$=2\cdot \frac 12 \cdot 2\sqrt 3 \cdot (3-\sqrt 3)$$
$$=6\sqrt 3-6$$
