Posted on Apr 3, 2026
An alien, more specifically an Eridian, named Rocky needs to pass a solid xenonite crystal to his human friend 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 is 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 the tunnel so that the crystal will fit through it?
To minimize the cylinder radius, we want the smallest possible circle that contains an orthogonal projection of the tetrahedron. So pick an axis direction, project the tetrahedron onto the perpendicular plane, and take the circumradius of that projection.
Place the regular tetrahedron with vertices at \((\pm1,\pm1,\pm1)\) and scale by \(1/(2\sqrt{2})\) so every edge has length 1.
Choose the cylinder axis to be the line through the midpoints of a pair of opposite edges; in these coordinates, that can be the \(x\)-axis.
Projecting onto the \(yz\)-plane gives the four points
\[ \left(\pm\frac{1}{2\sqrt{2}}, \pm\frac{1}{2\sqrt{2}}\right), \]
which form a square of side length \(1/\sqrt{2}\). The smallest circle containing that square has radius
\[ r=\frac{\text{diagonal}}{2}=\frac{\left(1/\sqrt{2}\right)\sqrt{2}}{2}=\frac{1}{2}. \]
This orientation puts all four projected vertices on the circle, so it is snug. Any tilt breaks the symmetry and increases the required covering circle, so this is minimal.
Minimum possible radius: \( \boxed{\tfrac{1}{2}} \).
Rotate the tetrahedron and watch the shadow radius. The display uses an orthographic projection.
Solid view
Shadow view
Next, Rocky wants to transport a solid crystal shaped like a regular dodecahedron to his human friend. As before, each edge has length 1.
This time, the long tunnel between the space ships can be any right prismatic shape, not necessarily a cylinder. Once again, Rocky needs the crystal to fit through the tunnel, and it is okay if that fit is snug.
What is the minimum possible cross-sectional area for the tunnel so that the crystal will fit through it?
For a right prism, the crystal fits iff one orthogonal projection of the dodecahedron fits inside the cross-section. So the minimum possible cross-sectional area equals the minimum area of an orthogonal projection of the dodecahedron.
Use the standard coordinates for a regular dodecahedron:
where \(\varphi=(1+\sqrt{5})/2\). In this model the edge length is \(2/\varphi\), so scale by \(\varphi/2\) to make the edges length 1.
By symmetry, it suffices to check the three main symmetry axes: through opposite faces, through opposite vertices, and through opposite edges. The smallest shadow comes from the 2-fold symmetry axis through opposite edges. Take that axis as the \(x\)-axis and project onto the \(yz\)-plane.
The convex hull of the projected vertices is an octagon with vertices
\[ \left(\pm\frac{1+\varphi}{2}, 0\right),\quad \left(\pm\frac{1}{2}, \pm\frac{1+\varphi}{2}\right),\quad \left(\pm\frac{\varphi}{2}, \pm\frac{\varphi}{2}\right), \]
listed in cyclic order. Its area (shoelace formula) is
\[ A=\frac{5}{2}+\sqrt{5}=\frac{5+2\sqrt{5}}{2}. \]
Projections along the other symmetry axes are larger, so this is minimal.
Minimum possible cross-sectional area: \( \boxed{\frac{5+2\sqrt{5}}{2}} \).
Rotate the dodecahedron and watch the cross-sectional area of its shadow.
Solid view
Shadow view