# Compute #different ways of drawing a diagonal line on every face of a regular cube.

Let $X=\{$ways of drawing a diagonal line on every face of a regular cube$\}$. Group $G$ is the rotation group of regular cube. $|G|=24$.
We want to compute #different orbits of $X$ under group action $G$.

I tried to equate one specific way of drawing a diagonal line in a particular face with coloring that face in red and the other way with coloring in blue. Thus we can consider coloring the regular cube with 2 colors. (I don't know if this way of thinking is accurate.) By invoking Pólya enumeration theorem, I find #orbits=10.
However, I can only find 8 ways by drawing. Is there something wrong with my method?

• I think $8$ is the correct answer. I didn't try drawing them, but that's what I get using Burnside's lemma. Drawing a diagonal on each face is not the same as coloring each face red or blue. Consider a 90 degree rotation about the axis through the centers of the top and bottom face: the color of the top face does not change, but the direction of a diagonal line drawn on the top face does change. – bof Oct 19 '17 at 5:28
• My calculation: $$(1\cdot2^6+6\cdot0+3\cdot2^4+6\cdot2^3+8\cdot2^2)/24=8.$$ For coloring faces red or blue, it's the same except that the $6\cdot0$ term becomes $6\cdot2^3$ so the final result is $10$ instead of $8.$ – bof Oct 19 '17 at 5:34

That method failed because diagonals are reversed by 90 degree rotations but colours are not.

The correct answer is 8. They are:

• Four triangles forming a tetrahedron on the surface of the cube.
• Two triangles sharing an edge, plus a separate line.
• Two triangles not sharing an edge.
• One triangle, plus a separate V.
• Two separate tripods centred at opposite corners.
• Two separate S shapes.
• Two separate Z shapes.
• A quadrilateral around the cube, plus two separate edges on opposing faces.

Consider the graph whose vertices are the vertices of the cube and whose edges are the diagonals of the facets of the cube. This graph is the disjoint union of a red $K_4$ and a blue $K_4$. Choosing a diagonal on each facet means choosing on each facet either the red or the blue diagonal, among them $r\in[0,3]$ red diagonals.

If $r=0$ only blue diagonals are chosen, forming the blue $K_4$.

If $r=1$ one edge of the blue $K_4$ is replaced by the red diagonal on the same facet.

If $r=2$ two edges of the blue $K_4$ are replaced by the red diagonals on the respective facets. These two red diagonals either form a path of length $2$ on the red $K_4$, or they are opposite edges of the red $K_4$.

If $r=3$ the three chosen red diagonals form a star, a triangle, or a path of length $3$ on the red $K_4$. The three remaining edges of the blue $K_4$ then automatically form a figure of the same kind.

All figures produced in these ways are mirror symmetric, apart from the pairs of paths of length $3$ . Both of them either form an $S$, or both of them form a $Z$.

The total number of different figures therefore is $8$.