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Ok, I completely revised my question. For those interested about my purpose with this question, see the older versions.

So, I would like to calculate the area of a cross-section of a tetrahedron. The only things I know are the $4$ points, with each their own $(x,y,z)$ coordinates.

Under this tetrahedron is a line, the $x$-axis. The cross-sections should be perpendicular to the $x$-axis. A cross-section of a tetrahedron can either be a triangle or a quadrilateral. Between the projection of the two left-most points it will be a triangle, and also between the projections of the third and fourth point. Between the second and third point it will be a quadrilateral. For illustrative purposes:

enter image description here

My question is, how can I find the area of a cross-section perpendicular to the $x$-axis?

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    I once worked out something similar for the easier case of slicing a $d$-cube perpendicular to $(1,1,\cdots,1)$ (which is symmetric and divides into $d$ intervals, with ${d\choose k}$ vertices on one side and ${d\choose k+1}$ on the other) and seem to recall I needed to use $\min()$ functions. Anyway if I get a chance I'll look at it more tomorrow.2012-03-17

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Ok, I more or less solved it -- but I'm not very happy with the solution. I would like to end up with an expression for the area between the first and second points, dependent on $y$ of course. This should have the form $a + by + cy^2$ (in this case I chose to project it on the $y$-axis instead of the $x$-axis). I explain why.

Lets create a tetrahedron (having volume 1) with the points $P_1 = (1,0,2/3)$, $P_2 = (4,1,0)$, $P_3 = (1,2,4)$ and $P_4 = (0,3,1)$. Note that the $y$-position is respectively $0,1,2,3$. The tetrahedron looks like this:

enter image description here

For the cross-sections between $P_1$ en $P_2$, we need points on the lines $P_1 P_2$, $P_1 P_3$ and $P_1 P_4$ (so three points in total). The equations for the relevant parts of these lines are

$\begin{align} P_1 P_2 & = (1-y)P_1 + y P_2 & y \in [0,1] \\ P_1 P_3 & = (1-\frac{y}{2})P_1 + \frac{y}{2} P_3 & y \in [0,\frac{1}{2}] \\ P_1 P_4 & = (1-\frac{y}{3})P_1 + \frac{y}{3} P_3 & y \in [0,\frac{1}{3}] \\ \end{align}$

The yellow cross-section above is for $y=0.8$. The area of the triangle can now be calculated as follows. Define two sides of the yellow triangle (using the blue dots):

$\begin{align} Side1 = P_1 P_2 (0.8) - P_1 P_3 (0.8)\\ Side2 = P_1 P_4 (0.8) - P_1 P_2 (0.8) \end{align}$

The area now follows from the cross product between these two, i.e. $A = \frac{1}{2} | Side1 \times Side2 |$.

If you apply this procedure for $y = 0$ with small steps until $y = 1$ and plot the area as a function of $y$, you end up with (blue solid line):

enter image description here

Which is actually the same as the first part of the uniform quadratic B-Spline, i.e. $.5 t^2$ (yellow dashed line). Perhaps you know that these B-Splines can be defined as projections of simplices, so they are also called "Simplex Splines".

New Question: I should be able to derive this $.5 t^2$ somehow, using the steps described above. How should I do this?

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    Ah, got it! Just writing out the equations (i.e. substituting the coordinates) does the trick, of course.2012-03-18