Cleverest construction of a dodecahedron / icosahedron?

Question

One can show, as an elementary application of Euler's formula, that there are at most five regular convex polytopes in 3-space. The tetrahedron, cube, and octahedron all admit very intuitive constructions. The cube is a cube, the octahedron is its dual, the tetrahedron has as vertices four pairwise non-adjacent corners of a cube. One can check that that everything you want holds on a single piece of paper.

Does anyone know a correspondingly elementary proof that the dodecahedron or icosahedron exists?

Answer 1

Consider two regular 5gons $ABCDE$ and $AFGHB$ that share an edge. Make yourself clear that there is exactly one way to rotate $AFGHB$ around $AB$ so that $\angle DAH$ becomes rectangular (well, there are two ways: up and down, but we fix the "down" one, so what we see on the "paper" will become the "outsides" of the faces). Once this is done, $D, A, H$ can be viewed as three vertices of a face $DAHI$ of a cube. By symmetry, $B$ is on the midplane of $AH$ and of $DI$. Thus by reflecting $ABCDE$ on that plane we obtain another 5gon that shares edges $BC$ and $HB$ with the previous two (and has $HI$ as a diagonal). In other words: The rotation that was just right to make $\angle DAH=\frac\pi2$ was also just right to produce three contiguous 5gons sharing a vertex! It follows that by applying more symmetries of the cube, we ultimately obtain twelve nicely matching regular 5gons (one for each edge of the cube) that make up a dodekahedron.

(The icosahedron is the dual of the dodekehedron of course).

Answer 2

A very small equilateral triangle on the sphere $S^2$ has angles slightly larger than $60^\circ$, and it's easy to visualize an equilateral spherical triangle with $90^\circ$ angles. By continuity there are equilateral spherical triangles with angles $=72^\circ$, and they all have the same side length $s$. Now start tiling $S^2$ with such triangles, and you will find out that $20$ such triangles will exactly tile the sphere.

(I learned this proof from Milnor who called it an "abstract nonsense proof" of the existence of the icosahedron.)

Update. The validity of the above proof has been questioned in the comments. It has been argued that the tiling might not close up properly and result in a multiple, maybe even infinite, covering of $S^2$. In his talk Milnor had dismissed this possibility on topological grounds. Instead I offer here the following elementary argument, see the acompagning figure:

Begin with an equilateral $72^\circ$-triangle centered at the north pole $N$. Attach such a triangle to each of its sides and insert two such triangles at each of its vertices. The resulting configuration consists of $10$ triangles and is bounded by a polygonal loop $\gamma$. This loop can be characterized as follows: It consists of $6$ arcs of length $s$ zigzaging around the sphere with turning angles $\pm 36^\circ$ at the vertices. Let $M$ be the center of one of these arcs. A rotation $T$ of the sphere by $180^\circ$ around $M$ will interchange in turn the arcs $a$ and $a'$, then $b$ and $b'$, and finally the points $C$ and $C'$. Therefore $T$ will map $\gamma$ onto itself and transport the proper triangulation of the northside of $\gamma$ to its southside.

Answer 3

One of my favorite dodecahedron constructions goes something like this: Begin with two regular pentagons joined along an edge. Cut off the "far" triangles, leaving identical trapezoids joined along the edge. Finally, glue the severed triangles into to fold to create a "pup-tent".

The pup-tent has a perfectly-square base, and placing one such tent on each face of a cube causes trapezoidal faces of the tents to combine with the triangular faces of other tents to (re-)form the pentagonal faces of the dodecahedron.

Answer 4

The icosahedron can be constructed as follows. Consider a right cone $C$ (pyramid) on a regular pentagon, such that the distance from the top vertex $T$ to each of the bottom vertices $B_0, B_1, B_2, B_3, B_4$ equals the distance between each pair nonadjacent vertices $B_i, B_{i+2}$ (here the indices are modulo 5), namely $d(T,B_i)=d(B_j,B_{j+2})$ for all $i,j$.

Clearly, the pyramid $C$ is inscribed in a sphere. Now take the union of the set of the six vertices of $C$ with its antipodal set. The resulting set of 12 points on the sphere is the set of vertices of a regular icosahedron inscribed in the sphere.

Answer 5

Here's a construction of the dodecahedron that satisfies me:

Start with a pentagon surrounded by five other pentagons, like so:

                                                                    

If I stare at this picture for a bit, I become convinced that I can fold up the flaps to form a 3D bowl made of six pentagons, with the relevant 5 edges joined up, and furthermore that this is unique. (For every pair of adjacent pentagons on the outer ring, there's only one pair of angles at which they meet, and since those angles are the same for both of them by symmetry, we can lift each flap by that angle so everything works out.)

Now, I think about one of the divots in the rim of this bowl, say $C$. What angle is that divot between $e$ and $e_1$? It looks similar to the $108^\circ$ angle of one of the pentagons, but is it exactly? Well, I know that the angles on either side of the crease $AC$ are equal by symmetry, and on the $A$ side I can see that a $108^\circ$ angle from the bottom pentagon exactly fills the gap.

So the angle at $C$ is also $108^\circ$, and a regular pentagon would fit there perfectly.

Knowing that the angles of both the zigs and the zags in this zig-zag bowl rim are equal, I can see that the entire border is centrally symmetric - if I inverted a copy of this bowl on the $x$, $y$, and $z$ axes, its rim would line up perfectly with this one:

                                                                    

So if I do that, and place an upside-down bowl on top of my right-side-up bowl, I'll get a polyhedron with 12 pentagons, where each pentagon is surrounded by 5 others. And by my original reasoning, I know that there's only one way to surround a pentagon like that, so my construction must look the same from the perspective of any of its faces.