Octahedrons whose faces are congruent quadrangles

Question

There is a octahedron which the faces are all congruent quadrangles. Let $M$ the set of length of edges of faces of the octahedron.

1. Prove that $|M| \le 3$.

2. Prove that all faces have two edges with equal length which meets at one point.

My approach. I think I have to solve this problem with the numbers of faces, edges, and points.

There are $\frac{4 \times 8}{2} = 16$ edges and 8 faces, but we don't know the numbers of points.

By Euler theorem, $v-e+f=2$ and $e=16, f=8$. So $v=10$.

So, the degrees of the points is $(5, 3, 3, ..., 3)$ or $(4, 4, 3, 3, ..., 3)$. (Because the degrees are at least 3.)

Now what do we do?

Answer 1

Thanks @Blue.

I solved this problem in a very simple way, thanks to you.

1) Solved this by contradiction.

If $|M| \ge 4$, the lengths of edges of quadrangles are all different.

Let the four lengths $a, b, c, d$.

Then, it seems to be a contradiction because,

2) It is simmilar from above.

Let the four lengths $a, b, a, c$.

Then, it seems to be a contradiction because,

Answer 2

All right, let's clear up some things about the configuration.

First, any planar graph with all faces quadrilaterals (or even polygons) must be bipartite. How do those parts split? If there are six or more vertices in one of the parts, then, since each of those vertices has degree at least $3$ and each edge has only one of its vertices in the part, that's at least $18$ edges. This contradicts our calculation of $16$ edges, so the two parts must have five vertices each.

Now, count degrees at the vertices in each part. With $16$ total edges over five vertices, that must be four vertices of degree $3$ and one of degree $4$. The same is true of the other part.

Now, this splits into two cases for the configuration. Either the two vertices of degree $4$ are directly connected by an edge, or they're not. Let's see if we can build a planar graph in the former case:

We have several edges left to fill in, but the six faces around the two degree-$4$ vertices give us this much. Number the vertices and label as shown; the vertices of one part are shown in red, and the other part in blue.

Next, what can vertex blue #5 ($U_5$) connect to? It has three edges going to red vertices #2,3,4,5 ($R_2$,$R_3$,$R_4$,$R_5$). In particular, that means it must connect to at least one of $R_2$ or $R_3$. WLOG, that's $R_2$.

Then, there's a quadrilateral with consecutive vertices $U_5$, $R_2$, $U_1$. This must be face $C$, and the fourth vertex is $R_4$. That means it closes with an edge between $U_5$ and $R_4$, our second edge drawn in.

Now, we have some edges in the picture that only have one of their connected faces indicated. First, the triad U2 to R2 to U5 has to be part of a face. It can't be $A$, $B$, or $C$ since those faces have all of their vertices known. It can't be $D$, $E$, or $F$ since those faces have three known vertices each that agree with the list of three we have in at most one place. Therefore, it must be a new face; call it $G$.

Second, the edge from $R_3$ to $U_3$ is part of face $B$. What's its second face? It can't be $A$ or $C$ because those don't have either $R_3$ or $U_3$ as a vertex. It can't be $E$ or $G$ since those have three known vertices and can't include both $R_3$ and $U_3$. It can't be $D$ or $F$ since two faces of the polyhedron can't share adjacent edges. All of the old faces are ruled out, so it has to be a new face $H$.

Third, the edge from $R_4$ to $U_5$ is part of face $C$. What's its second face? Not $A$, $B$, $E$, or $F$ because there aren't enough free vertices. Not $D$ or $G$ because it can't share two adjacent edges with $C$. There are only eight faces, so that leaves us with $H$ as the only option.

How can $R_3$, $U_3$, $R_4$, and $U_5$ be part of the same face $H$? We must have edges from $R_3$ to $U_5$ and $R_4$ to $U_3$.

But that doesn't work. There's no way to connect those edges while remaining planar. Merging vertices $R_1$ and $R_2$ through their connection at $U_2$, that's the equivalent of a $K_{3,3}$ between $(U_1,U_3,U_5)$ and $(R_3,R_4,R_{1,2})$.

This configuration has failed, leaving only one possibility - the two degree-4 vertices aren't connected by an edge, and indeed don't share a face. That configuration works as a polyhedron. A flattened version:

I'm maintaining the color choices; the two caps will meet at corresponding colors when we stack them and open it up.

And that's the configuration you'll need to work with in figuring out how many different edge lengths there can be.