Smallest cylinder into which a regular tetrahedron can fit?

Question

Given a regular tetrahedron (as shown) of edge length $b$, determine the diameter $d$ of the smallest right circular cylinder (pipe) of infinite length along which the tetrahedron can slide.

Answer 1

Let

• $v_1, v_2, v_3, v_4$ be the vertices of a regular tetrahedron $T$ of side length $b$.
• $C$ be an infinite cylinder of diameter $d$ containing $T$.

WOLOG, choose the coordinate system such that

• the centroid of $T$ is the origin.
• $v_1$ is located on the +ve $z$-axis.
• $v_2$ falls on the $xz$-plane.

Up to labeling of vertices, the location of the $4$ vertices are:

$$\begin{cases} \vec{v}_1 &= \frac{b}{\sqrt{24}}(0,0,3)\\ \vec{v}_2 &= \frac{b}{\sqrt{24}}(\sqrt{8},0,-1)\\ \vec{v}_3 &= \frac{b}{\sqrt{24}}(-\sqrt{2},\sqrt{6},-1)\\ \vec{v}_4 &= \frac{b}{\sqrt{24}}(-\sqrt{2},-\sqrt{6},-1)\\ \end{cases}$$ Identify vectors in $\mathbb{R}^3$ as $3\times 1$ column vectors and consider following $3 \times 3$ matrix constructed using outer product of $\vec{v}_i$ and their transposes $\vec{v}_i^T$. By brute force, one can show that

$$\sum_{i=1}^4 \vec{v}_i \otimes \vec{v}_i^T = \frac{b^2}{2} I_3$$

As a consequence of this, for any unit vector $\hat{n}$, we have

$$\sum_{i=1}^4 |\vec{v}_i\cdot \hat{n}|^2 = \sum_{i=1}^4 \left|\vec{v}_i^T \hat{n}\right|^2 = \sum_{i=1}^4 {\rm Tr}\left((\vec{v}_i \otimes \vec{v}_i^T)( \hat{n}\otimes \hat{n}^T) \right) = \frac{b^2}{2} {\rm Tr}\left(\hat{n}\otimes \hat{n}^T\right) = \frac{b^2}{2}$$

Together with the identity $\sum_{i=1}^4 \vec{v}_i = \vec{0}$, this leads to

$$ \sum_{1 \le i < j \le 4} |\hat{n}\cdot(\vec{v}_i - \vec{v}_j)|^2 = \frac12 \sum_{i=1}^4\sum_{j=1}^4 |\hat{n}\cdot(\vec{v}_i - \vec{v}_j)|^2 = \left(\sum_{i=1}^4 |\hat{n}\cdot\vec{v}_i|^2\right)\left(\sum_{j=1}^4 1\right) = 2b^2$$

Let $P$ be a plane passing through the origin whose normal is pointing along the axis of $C$. Orthogonal project $T$ onto $P$. Let $\vec{u}_i \in P$ be the projected image of $\vec{v}_i$ and $\ell_{ij} = |\vec{u}_i - \vec{u}_j|$

Take any two unit vectors $\hat{n}_1$, $\hat{n}_2$ from $P$ orthogonal to each other. It is easy to see

$$\ell_{ij}^2 = |\vec{u}_i - \vec{u}_j|^2 = |\hat{n}_1\cdot (\vec{v}_i - \vec{v}_j)|^2 + |\hat{n}_2\cdot (\vec{v}_i - \vec{v}_j)|^2$$

Apply result from above, we find

$$\sum_{1\le i < j \le 4} \ell_{ij}^2 = 4b^2$$

Translate the origin to where the axis of $C$ intersect $P$. In the new coordinate system, we have $$|\vec{u}_i| \le \frac{d}{2}$$

This implies

$$\sum_{1\le i < j \le 4} \ell_{ij}^2 = \frac12 \sum_{i=1}^4\sum_{j=1}^4 |\vec{u}_i - \vec{u}_j|^2 = \left(\sum_{i=1}^4 |\vec{u}_i |^2\right)\left(\sum_{j=1}^4 1\right) - \left| \sum_{k=1}^4\vec{u}_k\right|^2\\ \le 4 \sum_{i=1}^4 |\vec{u}_i |^2 \le 16 \left(\frac{d}{2}\right)^2 = 4d^2$$

Since $\ell_{ij}$ is invariant under this sort of change of coordinate system, we get

$$4b^2 \le 4d^2 \quad\implies\quad b \le d$$

So the diameter $d$ of the cylinder $C$ is bounded from below by $b$.

On the other hand, if we construct a line passing through $\frac12\left(\vec{v}_1+\vec{v}_2\right)$ and $\frac12\left(\vec{v}_3+\vec{v}_4\right)$ and fatten it to a cylinder of diameter $b$. One can verify this cylinder contains $T$.

Combine these two observations, $b$ is the smallest diameter we seek.