13
$\begingroup$

A monohedron is a convex polyhedron with all faces mutually congruent but with no other symmetry necessarily needed. So obviously, this is a wide class of polyhedra that includes the Platonic solids and isohedra. An earlier post is What are the known convex polyhedra with congruent faces?

Questions: Are there monohedroa with odd numbers of faces (it is known that isohedra necessarily have even numbers of faces — as stated in the MathWorld article Isohedron)? What are the values for the number of edges on a face for which monohedrons are possible? Will relaxing convexity (of the body, not of the faces) have an impact on the answers to these questions?

$\endgroup$
5
  • $\begingroup$ This violates a few of your conditions, but you may find the 7-faced Szilassi polyhedron interesting: en.wikipedia.org/wiki/Szilassi_polyhedron $\endgroup$ Oct 13, 2021 at 12:52
  • 5
    $\begingroup$ Considering the dual of the edge-graph, using the hand-shaking-lemma and that there must be a vertex of degree at most 5, we see that the only option for an odd number of faces is if the monohedron has quadrangle faces. The faces can also not be centrally symmetric as otherwise the whole polyhedron would be a centrally symmetric zonohedron, having even many faces. $\endgroup$
    – M. Winter
    Oct 13, 2021 at 13:04
  • $\begingroup$ Thanks Bill Bradley for the intro to Szilassi polyhedron. I hadn't thought about polyhedra that are not simply connected. But this is a very interesting object - although the faces are not congruent! And thanks M Winter for your analysis; however, I am unable to visualize or draw a polyhedron with odd number of faces and all faces being mutually congruent and non-centrally symmetric quadrangles. $\endgroup$ Oct 14, 2021 at 8:27
  • 4
    $\begingroup$ To add to M. Winter's comment: the four edges of the face cannot include an edge whose length is unique, because then pairs of faces would have to be matched across these edges and the total number of faces would have to be even. Together with the impossibility of centrally symmetric faces that means that the only possible face shapes are kites. $\endgroup$ Oct 17, 2021 at 8:01
  • $\begingroup$ Thanks for pointing this out, Prof Eppstein. $\endgroup$ Oct 17, 2021 at 16:37

3 Answers 3

7
$\begingroup$

The answer to the question in the title is negative in dimension 3: it was shown by Grunbaum that every 3-polytope with congruent facets has an even number of facets. See p. 414 of his book "Convex Polytopes", 2nd edition. The original reference is:

Grunbaum, B. On polyhedra in $\mathbb{E}^3$ having all faces congruent. Bull. Res. Council Israel, 8F (1960), 215-218.

$\endgroup$
1
  • $\begingroup$ Thanks. A statement attributed to Steinhaus is given on page 414: "every congruent faced 3-polytope has an even number of facets". One suspects whether the statement holds for non-convex 3-polytopes as well. I could not yet get to see the paper by Grunbaum you mention. $\endgroup$ Feb 27, 2022 at 14:07
10
$\begingroup$

This is too long for a comment.

The figure shows how to construct an $11$-face convex polyhedron whose faces all quadrilaterals, but not congruent. I think it's only interesting because it shows that there aren't plain combinatorial or topological obstructions to a positive answer to the question. On the other hand I doubt it's possible to deform this specific construction to produce a monohedron (I don't see how to make all the faces have the same number and order of obtuse and acute angles, never mind corresponding edges of equal length).

Start with a hexagonal prism. Cut the top hexagon in $2$ and the bottom one in $3$, to created $5$ degenerate (coplanar) quadrangular faces in place of the two hexagons. Slide G, L and M out a bit to make the faces not coplanar, but since this may destroy the flatness, slide H, I, J, K, C, E, B in a bit, each as necessary, to make all the faces flat again. enter image description here

$\endgroup$
6
  • $\begingroup$ Nice! Guess this is at least a very strong indication that monohedra with odd number of faces exist - maybe the number of faces may be considerably more than 11. And whether simply connectedness and convexity (lack thereof) have anything to do with the question would also hopefully have interesting answers. $\endgroup$ Oct 17, 2021 at 5:35
  • 3
    $\begingroup$ See also en.wikipedia.org/wiki/Herschel_graph for a simpler 9-face example of a polyhedron with an odd number of (non-congruent) quadrilateral faces. $\endgroup$ Oct 17, 2021 at 7:47
  • $\begingroup$ @DavidEppstein Following up on your remark that all faces need to be kites: this means that lengths come in 2 types (S, "short", and L, "long") and opposite edges must have different types. Both Herschel's graph and the 11-face example here have the property that the all edges can be colored S or L in such a way; the coloring is unique (up to permutations) for Herschel's graph, and comes in 2 flavors for the 11-face. $\endgroup$ Nov 17, 2021 at 11:56
  • $\begingroup$ However in both the Herschel and and the two colorings of the 11-face graph, there are vertices containing all four angles of the kite, which adds up to $2\pi$ and is therefore not possible in a convex polyhedron. So an example, if it exists, will have to come from more complex figures. $\endgroup$ Nov 17, 2021 at 13:22
  • $\begingroup$ Nicely done! To recover some congruence characteristics, consider the construction of a quadrangular enneahedron which lends itself to a threefold symmetry; see my answer. $\endgroup$ Jan 5 at 0:29
0
$\begingroup$

As described by User3816, monohedra with an odd number of faces do not exist. But we can construct figures that have congruent edges and all faces with the same number of edges (which would necessarily be four).

The simplest such construction begins with a $D_{3h}$ triangular bipyramid, which has three order-4 equatorial vertices connected by meridians to two order-3 polar vertices. This description allows a degree of freedom for the ratio of the altitude to the circumradius of the triangle formed at the equator; the properties described below are independent of this ratio.

Mark off points A, B, C, one at the midpoint of each equatorial edge. Then cut the solid with three planes patallel the polar axis; one cutting plane passing through A and B, another passing through A and C and the third passing through B and C. This leaves an enneahedron (ennea = Greek for 9) with quadrangular faces.

The cutting planes will intersect the meridians halfway between the equator and the poles; this together with the overall $D_{3h}$ symmetry then guarantees that all faces of the constructed enneahedron are rhombuses and all edges are congruent. The only deviation from a monohedron is that the rhombuses at the poles (left behind from the original bipyramid) have different vertex angles from those ringing the equator (created by the cuts).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.