This post is about a 3D geometry Riddler puzzle involving spheres and tetrahedra! Here is the problem:

We want to create a new gift for fall, and we have a lot of spheres, of radius 1, left over from last year’s fidget sphere craze, and we’d like to sell them in sets of four. We also have a lot of extra tetrahedral packaging from last month’s Pyramid Fest. What’s the smallest tetrahedron into which we can pack four spheres?

Here is my solution:

[Show Solution]

We’ll solve this problem using old school (read: high school) geometry. Rather than starting with four spheres and looking for the smallest tetrahedron that contains them, we’ll solve the equivalent problem of starting with a fixed tetrahedron and looking for the largest spheres we can pack inside. We’ll use a tetrahedron of unit sidelength. Place one vertex at the origin $\vec O(0,0,0)$, one vertex on the $x$-axis $\vec P(1,0,0)$, and one face in the $xy$-plane. This leads to a vertex at $\vec Q(\frac{1}{2},\frac{\sqrt{3}}{2},0)$ because $OPQ$ is an equilateral triangle. To find the final vertex $R$, we know by symmetry that its $x$ and $y$ coordinates are the same as those of the centroid of $OPQ$, so $x=\frac{1}{2}$ and $y=\frac{\sqrt{3}}{6}$ (it’s at one third the altitude!). To find the final coordinate, we look for $z$ such that $x^2+y^2+z^2=1$, since all sides have length $1$. This leads us to the final vertex of the tetrahedron: $\vec R(\frac{1}{2},\frac{\sqrt{3}}{6},\frac{\sqrt{6}}{3})$.

The center of the tetrahedron can be found by looking for a point of the form $\alpha(\vec P+ \vec Q+\vec R)$ (again by symmetry) whose $x$-coordinate is $\frac{1}{2}$. Doing this leads us to the center: $\vec C(\frac{1}{2},\frac{\sqrt{3}}{6},\frac{\sqrt{6}}{12})$. Next, let’s look for the center of the sphere closest to the origin. By symmetry, it must be of the form $\vec C_1 = \beta \vec C$. We want to find $\beta$ such that it’s $x$-distance from the point $x=\frac{1}{2}$ is $r$ (so it just touches the sphere closest to $\vec P$), and we also want its $z$ coordinate to be $r$ (so it’s a distance $r$ from the $xy$ plane, because it’s a tangent point of the sphere). For a reference, see the figure below. I’m saying that $C_1$ (center of the sphere) should be on the line $OC$ and we should have $|C_1M| = |C_1A_1| = r$.

The equations just described are:

\begin{align}

\frac{1}{2} \beta &= \frac{1}{2}-r\\

\frac{\sqrt{6}}{12} \beta &= r

\end{align}

Solving this system of equations yields $\beta=\frac{6-\sqrt{6}}{5}$ and $r=\frac{\sqrt{6}-1}{10}\approx 0.145$. So this is the radius of the largest sphere we can use when the tetrahedron has sidelength $1$. If the radius is instead $1$, then we must scale the tetrahedron accordingly. The resulting sidelength will be:

\[

s = \frac{10}{\sqrt{6}-1} = 2\sqrt{6}+2 \approx 6.89898

\]Here is a scale diagram of the entire pyramid with the four spheres inside:

Do the tetrahedrons have to be perfectly symmetrical in faces, edges, and vertices? If so, we don’t have to account for the top sphere sinking into the bottom three’s “divot”.