# Calculating the area of a cross-section of a tetrahedron

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:

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

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Let $X_i=(x_i,y_i,z_i)$ for $1\le i\le4$ & $x_1\le x_2\le x_3\le <x_4$ not all equal. Take a cross section $x=x_k+t(x_{k+1}-x_k)=(1-t)x_k+tx_{k+1}$ between $x_k$ and $x_{k+1}$ for $t\in(0,1)$. Tetrahedron $T$ in homegeneous coordinates is $$T=\left\{\sum_{i=1}^{4}t_iX_i~\Big\lvert~ t_i\ge0,~\sum t_i\le1\right\}$$ and our cross section has the additional constraint $$\sum t_ix_i=x=x_k+t(x_{k+1}-x_k)=(1-t)x_k+tx_{k+1}.$$ Might be easier if we could map $T$ onto the standard simplex & the $x$ axis to a line. How'd you make the nice img? – bgins Mar 17 '12 at 22:32
Thanks @bgins, I'll have a look at that. I thought of something similar myself, at least for the first part (triangle). Represent the three edges going out from Point $1$ (= $P_1$)as $(1-t)P_1 + tP2$, and the same for the other points. Then take $t \in [0,1]$ for the first edge, and scale for the other two. This way you can find the three points of a triangle, and calculate the area with e.g. the cross product of two edges of this triangle. Problem is, I'd like to get a result like $a+bx+cx^2$ for the first part (and a separate one for the second and third). I made the image in Inkscape :) – Ailurus Mar 17 '12 at 22:42
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. – bgins Mar 17 '12 at 23:10

## 1 Answer

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:

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):

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. – Ailurus Mar 18 '12 at 1:32