Find the area of overlap of two triangles Suppose we are given two triangles $ABC$ and $DEF$. We can assume nothing about them other than that they are in the same plane. The triangles may or may not overlap. I want to algorithmically determine the area (possibly $0$) of their overlap; call it $T_{common}$. 
We have a multitude of ways of determining the areas of $ABC$ and $DEF$; among the "nicest" are the Heronian formula, which is in terms of the side lengths alone, and 
$T = \frac{1}{2} \left| \det\begin{pmatrix}x_A & x_B & x_C \\ y_A & y_B & y_C \\ 1 & 1 & 1\end{pmatrix} \right| = \frac{1}{2} \big| x_A y_B - x_A y_C + x_B y_C - x_B y_A + x_C y_A - x_C y_B \big|$
which is in terms of the coordinates alone.
Obviously, there does exist a function from $A,B,C,D,E,F$ to $T_{common}$, but my question is: is there a "nice" (or even not-"nice") expression for $T_{common}$ in terms of the $x$ and $y$ coordinates of $A,B,C,D,E,F$?
I've drawn out on paper what I think are the various cases, but my issues with this approach are: identifying the case is a job in itself, which I can't easily see how to algorithmise ("just look at a picture" doesn't work for a computer); even within each case the algebra is fiddly and error-prone; and I have little confidence that I've enumerated all possible cases and got the computations right!
In my imagination there is a neat approach using ideas from analysis (treating the triangles as functions from $\mathbb{R}^2$ to $\{0,1\}$ and... multiplying them??) but I have no idea whether that's just a flight of fancy or something workable.

Sorry about the comment -- I hit the return key prematurely. 
This isn't really an answer (except in the negative sense).
The common (overlap) area is a function of the coordinates of the 6 points, so it's a mapping from $R^{12}$ into $R$. Think about one of the points moving around, while the other 5 are fixed. When the moving point passes into (or out of) the other triangle, some of the partial derivatives of the area function will be discontinuous (this seems intuitively clear -- the proof is left to someone who has more skill and patience than I do). Anyway, if you believe me, this means that the area function is not smooth. Therefore there can not be some simple formula like Heron's formula that gives the area (because this simple formula would give a smooth function).
