Smallest cylinder into which a regular tetrahedron can fit? 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.

 A: 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.
