Amaze, amaze, amaze!!!! (Classic)

An alien—more specifically, an Eridian—named Rocky, and needs to pass a solid xenonite crystal to his human friend named Ryland Grace, in a neighboring spaceship. The crystal is shaped like a regular tetrahedron, as shown below, and all its edges have length 1.

To safely transport the crystal, Rocky needs to create a long cylindrical tunnel between the spaceships, and then orient the tetrahedron so that it fits through the tunnel. It’s okay if the crystal fits snugly inside the tunnel—in this case, it can slide along without any friction. What is the minimum possible radius for Rocky's tunnel so that the crystal will fit through it?

Consider the regular unit tetrahedron, with one of its equilateral unit triangle faces lying in the $xy$-plane, in particular, let's say its vertices are at $(0,0,0),$ $(1,0,0),$ $(1/2, \sqrt{3}/2, 0)$ and $(1/2, \sqrt{3}/6, \sqrt{6}/3).$ If we take any plane that is parallel to the $xy$-plane, say the plane $z = h$, and intersect it with the tetrahedron, we will get an equilateral triangle with side length $s=1-h.$ This would be equivalent to orienting the tetrahedron so that one of its faces was always in the cross sectional plane of the cylindrical tunnel. In this case, the largest possible sidelength of the equilateral triangles was a side length of $s=1,$ which is circumscribed by a circle of radius $r_3 = \frac{\sqrt{3}}{3}.$ This is a pretty good start, but let's see whether any other orientations will give us a better answer.

If we take any plane that is perpendicular to the $xy$-plane, say the plane $x=h,$ and intersect it with the tetrahedron, then by diligent application of ratios, we will get an triangle with vertices \begin{align*} v_1&=(h,0,0), \\ v_2&=(h, \max\{h,1-h\} \sqrt{3},0), \text{ and }\\ v_3&=(h, \max\{h, 1-h\}\frac{\sqrt{3}}{3}, 2\max\{h, 1-h\}\frac{\sqrt{6}}{3}).\end{align*} Therefore we see that this triangle is isoceles, since $$d_{12} = \max\{h,1-h\} \sqrt{3}$$ and $$d_{13} = \sqrt{ h^2/3 + 24 h^2/9 } = \max\{h, 1-h\}\sqrt{3},$$ while the final length is given by $$d_{23} = \sqrt{4h^2/3 + 24h^2/9} = 2\max\{h, 1-h\}.$$ In this, case the largest possible sidelengths are given by the isoceles triangle with sidelengths $\sqrt{3}/2 - \sqrt{3}/2 - 1.$ Let's pretend like two of the side with length 1 are located at the points $(-1/2,0)$ and $(1/2,0)$, from which with some symmetry and Pythagorean allusions we get that the third vertex is at the point $(0, \sqrt{2}/2).$ The centroid of this triangle is at the point $(0, \sqrt{2}/6).$ The circumscribing circle has radius $$r_2 = \sqrt{ \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{6}\right)^2 } = \sqrt{ \frac{1}{4} + \frac{2}{36} } = \frac{\sqrt{11}}{6} \lt \frac{\sqrt{3}}{3}.$$

To recap, we have covered the cases where three vertices of the tetrahedron fit snugly against the edge of the cylindrical tunnel, in which case we need $r_3 = \frac{\sqrt{3}}{3}.$ We have also covered when two vertices of the tetrahedron fit snugly against the edge of the cylindrical tunnel. We can rule out a case where none of the vertices fit snugly against the edge of the cylindrical tunnel, since we could definitely reduce the radius until at least one of the vertices fit snugly against the cylindrical tunnel. So the only case that we have left to solve for is when only a single vertex fits snugly against the cylindrical tunnel.

Let's assume that the vertex at $(1/2, \sqrt{3}/6, \sqrt{6}/3)$ is the one that is riding snugly a long the cylindrical tunnel and that the at any moment, without loss of generality a cross section of the cylindrical tunnel through this vertex cuts out a triangle whose other vertices are at $(h,0,0)$ and $(k, \max\{k, 1-k\}\sqrt{3},0)$ for some $0\leq h, k \leq 1.$ In particular, this would mean that the centroid of the triangle is at $$C(h,k) = \left( \frac{2h+2k+1}{6}, \frac{\sqrt{3}(6\max\{k,1-k\}+1)}{18}, \frac{\sqrt{6}}{9} \right).$$ The distance to the centroid to the point $(1/2, \sqrt{3}/6, \sqrt{6}/3)$ is \begin{align*}d(h,k) &= \sqrt{ \left(\frac{1 - h -k}{3}\right)^2 + \left(\frac{\sqrt{3}(1-3\max\{k, 1-k\})}{9}\right)^2 + \left( \frac{2\sqrt{6}}{9} \right)^2 } \\ &= \sqrt{ \frac{(h+k)^2}{9} + \frac{1}{27} \left( 9 \max \{k, 1-k\}^2 - 6 \max \{k, 1-k\} + 1 \right) + \frac{8}{27} } \\ &= \sqrt{ \frac{(h+k)^2}{9} + \frac{1}{3} \max \{k, 1-k\}^2 - \frac{2}{9} \max \{k, 1-k\} + \frac{1}{3} }\end{align*} Let's assume that $k \leq \frac{1}{2}$ for the time being, then we get \begin{align*}d_1(h,k) &= \sqrt{ \frac{h^2 + 2hk + k^2}{9} + \frac{k^2}{3} - \frac{2k}{9} + \frac{1}{3} } \\ &= \sqrt{ \frac{4}{9} \left( k - \frac{1-h}{4} \right)^2 + \frac{h^2}{12} + \frac{h}{12} + \frac{11}{36} }\\ & \geq \frac{\sqrt{11}}{6},\end{align*} since $0 \leq h \leq 1$ and $(k - (1-h)/4)^2 \geq 0$ as well. On the other hand, if we assume that $k \geq \frac{1}{2},$ then we get \begin{align*}d_2(h,k) &= \sqrt{ \frac{h^2 + 2hk +k^2}{9} + \frac{(1-k)^2}{3} - \frac{2 (1-k)}{9} + \frac{1}{3} } \\ &= \sqrt{ \frac{4}{9} \left( k - \frac{2 - h}{4} \right)^2 + \frac{h^2}{12} + \frac{h}{9} + \frac{1}{3}} \\ &\geq \frac{\sqrt{3}}{3} \geq \frac{\sqrt{11}}{6},\end{align*} since $0 \leq h \leq 1$ and $(k - (2-h)/4)^2 \geq 0.$ So we see that even if we can get an orientation where only one vertex is flush with the cylindrical wall, we cannot get a circumscribed radius that is any smaller than $r_2 = \frac{\sqrt{11}}{6}.$ Therefore, we see that the smallest possible radius for Rocky's tunnel is $$\frac{\sqrt{11}}{6} \approx 0.552770798393\dots$$ units.