The Prisoner's Dilemma

Solution.

We received a solution from William Keehn and a partial solution from Jesse Waite of Fort Leavenworth. Both submitters realized that one straightforward way to control the numbers of sides of the cross-sections is to tip back the top-right vertex of the tetrahedron in the figure associated with the problem, straight toward its leftmost horizontal edge. This tipping is equivalent to rotating the tetrahedron around that leftmost edge (before starting to slide it through the cutting plane). When you do this, the cross sections have three sides until that top-right vertex reaches the cutting plane, and four sides thereafter.

Finishing the problem off just amounts to computing the angle at which to tilt the tetrahedron so the top-right vertex reaches the cutting plane exactly halfway along the passage of the tetrahedron through the cutting plane. Then the average number of sides of the cross-section will be 3.5, as desired. To do this, we reduce the three-dimensional problem to one in the plane, by looking directly along the rotation axis (the left horizontal edge). Note the projection of the tetrahedron along this direction is not an equilateral triangle, but rather an isosceles triangle: all four edges proceeding from left to right in the figure are foreshortened in this view. The top two end up superimposed on the altitude of the top-left face, which maintains its length because it is perpendicular to the axis we are looking along. Thus, assuming the edge length of the tetrahedron is one, the foreshortened edges in the projection end up with a length of $\sqrt{3}/2$, the height of an equilateral triangle with unit sides.

In other words, from the side, the tipped tetrahedron ends up looking like Figure 2, where $T$ is the top-right vertex, $R$ corresponds to the left-hand edge of the tetrahedron that has projected onto a single point (since we are looking directly along that axis), and $B$ is the bottom-right vertex. From this point of view, the cutting plane becomes a straight vertical line. Three different positions of
the cutting plane relative to the tetrahedron are indicated in red in Figure 2: the initial contact at point $B$, the final exit at the line through $R$ and $R^{\prime }$, and exactly half-way at the line through $T$ and $T^{\prime }$. The blue dashed horizontal line is the line through $B$ perpendicular to the cutting plane; you can think of the point $B$ as traveling along this line or equivalently of the cutting plane as sliding along this line, staying perpendicular to it.

We want to solve for the tipping angle, represented by $\theta$ in Figure 2. The condition on the average number of faces corresponds to the requirement that $B{T}^{\prime }=\frac{1}{2}B{R}^{\prime }$. From the right triangles formed in the diagram, we have that $B{T}^{\prime }=\sin \theta$ and $B{R}^{\prime }=(\sqrt{3}/2)\sin (\beta +\theta )$. Substituting to eliminate $B{R}^{\prime }$ and $B{T}^{\prime }$, we have $\sin \theta =\sqrt{3}\sin (\beta +\theta )/4$.

To finish, we need the sum formula for sine, namely $\sin (\beta +\theta )=\sin \beta \cos \theta +\cos \beta \sin \theta$. To take advantage of this formula, we need the values of $\sin \beta$ and $\cos \beta$. Fortunately, they are not too difficult to calculate because we know all of the sides of isosceles triangle $BTR$. Dropping a perpendicular from $R$ to the midpoint of $BT$, we find that $\sin \beta =\sqrt{2/3}$ and $\cos \beta =\sqrt{1/3}$. Putting all these ingredients together, $4\sin \theta =\sqrt{2}\cos \theta +\sin \theta$. Solving, we determine that $\theta ={\tan }^{-1}(\sqrt{2}/3)\approx {25.24}^{\circ }$.

As for the dodecahedron, Jesse noticed that if you pass it through a cutting plane perpendicular to the line joining two opposite vertices, all of the cross sections are either triangles or hexagons, so that average is less than six. On the other hand, if you pass it through a plane perpendicular to the line joining the centers of opposite faces, then the cross sections are pentagons except for a band around the middle where they are decagons. You can look up (or derive) a coordinate system for a regular dodecahedron where there are only four different $z$-coordinates for the vertices: $\pm \varphi \pm 1$, which means that the bottom face must be perpendicular to the $z$-axis. Using these coordinates, the height of the entire dodecahedron is $2\varphi +2$ of which $2\varphi -2$ is the height of the band where the cross sections are decagons (and so there are pentagon cross sections for the remaining four units of height). Thus, the average number of sides of a cross section in this direction is
Therefore, the average number of cross-section sides depends on the orientation, like it does with the tetrahedron and octahedron.

That’s also the case with the icosahedron. Here we observe that the cross sections perpendicular to a line joining the midpoint of two opposite edges are all octagons (except for finitely many individual cross sections, which don’t contribute to the average, since there are infinitely many octagonal cross sections). However, cutting perpendicular to the line joining two opposite vertices again results in pentagons and decagons. In this direction, the icosahedron conveniently decomposes into two pentagonal pyramids and a pentagonal antiprism. Again, one can look up the heights of these solids to find that the average number of sides in this direction is$$S_v=\frac{10\sqrt{1+1/\sqrt{5}}+5\sqrt{1-1/\sqrt{5}}}{\sqrt{1+1/\sqrt{5}}+\sqrt{1-1/\sqrt{5}}}.$$

The first step in simplifying this formidable-looking expression is to clear the denominator by multiplying top and bottom by $\sqrt{1+1/\sqrt{5}}-\sqrt{1-1/\sqrt{5}}$. Following that trick with a bit of algebra will show that $S_v=5\varphi \approx 8.09\ne 8$, so again the average depends on orientation.