Pyramid Volumes Sum

Question

Let ABCD be a tetrahedron whose edges have length 5. Consider the set of pyramids of the form AMNP where M lies on AB, N lies on AC, and P lies on AD such that AM, AN, and AP $\in \{1,2,3,4\}$. Compute the sum of the volumes of all such pyramids.

I attempted this problem by assigning coordinates and using shoelace theorem. I ended up getting a form that I found the expected value of, but it was completely incorrect. Also, I was stumbling around with my solution, assuming things.

Essentially, I am stuck. It would be great if anyone could provide a thread to get me started.

Answer 1

Let

• $|AP| = u, |AQ| = v, |AR| = w$.
• Given any 4 points $X,Y,Z,T$, let $\Delta_{XYZT}$ be the volume of tetrahedron $XYZT$.

We will use the fact under affine transform, the ratio of volumes of different shapes remain the same. If one apply an affine transform to map the vertices of original tetrahedron to new positions $$(A, B, C, D) \mapsto (A'B'C'D') = ((0,0,0),(5,0,0),(0,5,0),(0,0,5))$$ Vertices $P,Q,R$ will get mapped to following positions

$$(P,Q,R) \mapsto (P',Q',R') = ((u,0,0),(0,v,0),(0,0,w))$$ This tells us

$$\Delta_{APQR} : \Delta_{ABCD} = \Delta_{A'P'Q'R'} : \Delta_{A'B'C'D'} = \frac16 uvw : \frac16 5^3\\ \implies \Delta_{APQD} = \frac{\Delta_{ABCD}}{125} uvw $$ Summing over all possible pyramids is equivalent to summing $u,v,w$ over $\{1,2,3,4\}$. This means

$$\verb/SUM/ = \frac{\Delta_{ABCD}}{125}\sum_{u=1}^4\sum_{v=1}^4\sum_{w=1}^4 uvw = \frac{\Delta_{ABCD}}{125} (1+2+3+4)^3 = 8\Delta_{ABCD}$$

If you look up the volume of a regular tetrahedron on wiki (or just compute it by hand), you will find

$$\Delta_{ABCD} = \frac{5^3}{6\sqrt{2}}\quad\implies\quad\verb/SUM/ = 8\Delta_{ABCD} = \frac{250\sqrt{2}}{3}$$

Answer 2

You may know that one of the easiest way to describe a regular tetrahedron with coordinates is to select 4 vertices among the 8 vertices of a cube in this way:

$$A(0,0,0), \ B(0,a,a), \ C(a,0,a), \ D(a,a,0)$$

One can check easily that all points are at the same distance, equal to $\sqrt{a^2+a^2}$. As this distance must be equal to 5, we must take:

$$a=5/\sqrt{2}$$

The volume of tetrahedron defined by

$$\|\vec{AM}\|=m/\sqrt{2}, \ \ \|\vec{AN}\|=n/\sqrt{2}, \ \ \|\vec{AP}\|=p/\sqrt{2}$$

with $1 \le m,n,p \le 4$ is classicaly given by:

$$V_{m,n,p}=\dfrac{1}{3!}\det \begin{pmatrix}|&|&|\\\vec{AM}&\vec{AN}&\vec{AP}\\|&|&|\end{pmatrix}=\dfrac16\det \begin{pmatrix} 0&n/\sqrt{2}&p/\sqrt{2}\\ m/\sqrt{2}&0&p/\sqrt{2}\\ m/\sqrt{2}&n/\sqrt{2}&0\end{pmatrix}=\frac{mnp}{6 \sqrt{2}}$$

Can you take it from here ?