Polyhedron Puzzles

As summer approaches, JRMF's plan is to develop a slew of activities to test in the fall. To that end, I'll be trying to post activities more frequently but with less of a mathematical dive. Today we'll look at some puzzles involving making polyhedra from a set of faces.

Setup

For this activity, you'll want a set of polygons that can be used to build polyhedra. The most flexible medium I know of is paper or cardstock so that you're not constrained to whatever pieces come in commercial sets, but the medium is difficult to work in. At the rigid end of the spectrum, we have the usual suspects:

Geometiles ($85 for a set of 96)
Polydrons ($30 for a set of 50)

I happened to meet Cristina and Mircea Draghicescu this weekend, who gifted me some of their ITSPHUN polygons to play with. I think these happen to be a Goldilocks manipulative for this activity, so they're what I'll be using.

They are hard to beat for ease of assembly and disassembly and have enough flex that we can fudge our way to some shapes that are topologically correct but geometrically can't be realized with regular polygons. Below are three examples of polyhedra, with the third being an example of a fudge:

The premise of the activity is simple. For a set of polygons, students need to decide whether a (possibly fudged) polyhedron can be built from them. To allow for the aforementioned fudging, instead of defining a polyhedron, we'll instead define a couple rules that a finished structure needs to satisfy:

A hook on each polygon must link with a hook from exactly one other polygon (i.e., no hooks go unused, link a shape to itself, or form a juncture of 3+ polygons)
Two polygons can be linked at most once to one another

While not everything that can be constructed in this way need be a polyhedron, students will almost certainly construct polyhedra since non-polyhedra are hard to achieve with small numbers of polygons and require awkward bends.

Assuming the rest of the hooks will eventually get used, here are three examples of good linkages, where the last is an example of a fudge:

Here are three examples of bad linkages:

The challenges below are mostly possible with some impossible arrangements peppered in. They feature mostly triangles and squares with a few pentagons and hexagons.

Challenges

For which of the following polygon sets can you find a polyhedron-like structure that follows the rules above?

$4$ triangles
$1$ square, $4$ triangles
$3$ squares, $2$ triangles
$5$ triangles
$6$ triangles
$6$ squares
$4$ squares, $2$ triangles
$1$ pentagon, $5$ triangles
$1$ hexagon, $5$ squares
$1$ pentagon, $2$ squares, $3$ triangles,
$7$ triangles
$3$ squares, $4$ triangles
$5$ squares, $2$ triangles
$1$ pentagons, $5$ squares, $1$ triangle
$2$ pentagons, $5$ squares
$3$ pentagons, $4$ squares
$2$ hexagons, $3$ squares, $2$ triangles
$8$ triangles
$4$ hexagons, $4$ triangles
$2$ squares, $6$ triangles
$4$ squares, $4$ triangles
$6$ squares, $2$ triangles
$1$ pentagon, $7$ triangles
$1$ pentagon, $3$ squares, $4$ triangles
$1$ hexagon, $3$ squares, $4$ triangles
$1$ hexagon, $4$ squares, $3$ triangles
$1$ hexagon, $5$ squares, $2$ triangles
$1$ hexagon, $7$ squares
$1$ hexagon, $2$ pentagons, $2$ squares, $3$ triangles
$1$ hexagon, $2$ pentagons, $3$ squares, $2$ triangles

Notes

If two structures are "convex enough" and feature the same polygon, that polygon can be removed from each and they can be fused at their holes. This sort of fusing can be useful for using one structure to build another.
If a polygon has $s$ edges, there must be at least $s$ other polygons to link to those edges. This rules out things like $1$ hexagon, $5$ squares.
Since edges of polygons pair up to make the edges of polyhedron, the total number of edges among the polygons must be even. This rules out things like $7$ triangles or $1$ pentagon, $3$ squares, $4$ triangles.

Taking $f$ to be the number of faces, $e$ the number of edges, and $v$ the number of vertices on a hypothetical polyhedron:

If the number of polygons is kept small, you shouldn't leave Euler's formula territory, where $v-e+f = 2$. (Since our rules don't allow the construction of nonorientable surfaces, we can always at least say $v-e+f \geq 2$.) If our $f$ polygons have $s_1, s_2, ..., s_f$ edges, then $e = (s_1 + ... + s_f)/2$, from which we can find $v$.
Since face $i$ is involved in $s_i$ vertices, then we also have
\[ s_1 + ... + s_f = d_1 + ... + d_v \]
where $d_j$ is the degree of vertex $j$ (the number of faces adjacent to vertex $j$). Since at least $3$ faces must come together to form a vertex, we have $d_j \geq 3$ and thus
\[ 2e = s_1 + ... + s_f \geq 3v \]
so $e \geq 3v/2$. If in a situation where Euler's formula applies,
\[ 2e \geq 3v = 3(e-f+2) = 3e-3f+6, \]
and hence
\[ f \geq \frac{e+6}{3} \]
This gives a restriction on the number of edges the polygons are allowed to have relative to the total number of polygons. For example, if we tried to build a structure with $5$ squares, we would have $f = 5$ but $e = 4\cdot6/2 = 12$, giving $(e+6)/3 = 6.$

Extra Challenges

Of the structures that can be built, which can be built without fudging? (i.e., which can be made so that every polygon remains flat)
If a construction requires fudging, can you see a way to make it by replacing some (or all) regular polygons with irregular polygons of the same type? (e.g., an isosceles triangle instead of an equilateral triangle or a parallelogram instead of a square.)

This post was sponsored by the Julia Robinson Mathematics Festival

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