Cleverest construction of a dodecahedron / icosahedron? 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?
 A: Martin Gardner once pointed out (from another source, which I don't remember) that you can construct a pop-up dodecahedron in the following way.  Draw a pentagon, surrounded by five other pentagons, each of which shares a side with the central pentagon.  Cut out two such figures, scoring each of the shared edges.
Now, stack the figures one on top of the other, such that the centers are on top of each other, but the surrounding pentagons are "half a pentagon out of phase."  Carefully holding the figures down, wind a rubber band around the perimeter of the stacked figures, alternating above and below the two figures.  When you release the stack, the assembly should pop up to form a dodecahedron.
I should mention that I've never tried it, but it seems plausible. :-)
A: 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.
A: 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.
A: 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.
A: $\renewcommand{\phi}{\varphi}$Another construction of a regular icosahedron: It's easy to check that if $\phi = \frac{1}{2}(1 + \sqrt{5})$ is the golden ratio, the twelve vertices of three golden rectangles in the coordinate planes
$$
(0, \pm1, \pm\phi),\qquad
(\pm\phi, 0, \pm1),\qquad
(\pm1, \pm\phi, 0),
$$
determine twenty equilateral triangles of side length two, with five triangles meeting at each vertex. The blue triangles in the stereogram meet at $(\phi, 0, 1)$.
(Amusing side fact: The rectangles' edges are linked like Borromean rings.)

A: Euclid's Method
I posted an answer above which is a method I thought of. The answer here is based on Euclid's XIII book of its classical "Elements" collection.
Euclid's method to build a dodecahedron is
to start with a cube and collocate  at each side of the cube a ``roof'' shaped structure. The figure below
 
illustrates Euclid's idea. 
 We observe that each plane face is a pentagon with two of
its sides in the triangular face of the roof-like shape and the other tree
sides in the trapezoidal of an adjacent roof-like shape. The common edge
between the two polygons is a side of the cube.   For example see
that, after joining the blue triangle with the blue trapezoidal we should
get a regular pentagon.
We need to find the vertices of the dodecahedron, and impose the following two conditions: 


*

*The pentagons are regular. That is, their sides and angles are  all
congruent.

*Pentagonal faces are flat. That is we need to verify that the
triangle of a 
    roof-like structure with a trapezium of an adjacent roof-like structure (
    see the blue shapes in the figure)  are in the same plane.
Let us assume that the cube has the collection of eight vertices $(\pm 1, \pm 1, \pm 1)$.
Here, the notation $(\pm a, \pm b, \pm c)$ indicates all the 8 combinations
of the form $(a,b,c)$ with either a ``$+$'', or a $"-"$ sign in front of each coordinate.
These vertices also belong to the dodecahedron but we need to find a few (12) more.
We see first that all the points are sitting in sphere of radius $\sqrt{3}$. 
This adds an extra constraint to the rest of the points. Their size should be
$\sqrt{3}$.
It is a well known fact that the diagonals $d$ on a pentagon
are related to their side length $a$ by the equation:
\begin{eqnarray*}
  d = \phi a.
\end{eqnarray*}
where $\phi$ is the Golden's Ratio  ($\phi=(1+\sqrt{5})/2$.)
Initially we  want to find the top two vertices of the blue plane in Figure~\ref{dodecacube}. 
Let us focus on the top roof-like structure which we show in this figure

Consider a bisector plane through the roof-like structure going through the top side
$BC$ and the middle line on the base $NM$ as shown in the figure.
Also, trace perpendiculars from $BO$ and $CP$ from the top to the base of the structure.
We have the following measures as functions of the Golden ration $\phi$:
\begin{eqnarray*}
  BC &=& OP = a = \frac{2}{\phi} \\
MN &=& 2 \\
ON &=&PM = \frac{2 - a}{2} = \frac{2-2/\phi}{2 } =  \frac{\phi-1}{\phi} \\
CM &=& \sqrt{(CD)^2 - (DM)^2}=\sqrt{a^2 - 1} = 
\sqrt{\frac{4}{\phi^2} - 1} = \frac{\sqrt{4 - \phi^2}}{\phi} \\
PC &=& \sqrt{(CM)^2 - (PM)^2}= \sqrt{ \frac{4-\phi^2}{\phi^2} - 
\frac{(\phi-1)^2}{\phi^2}} = \frac{\sqrt{3 -2 \phi^2 + 2 \phi}}{\phi} \\
\end{eqnarray*}
Now, since $3 - 2 \phi^2 + 2 \phi = 1$  (this could be verified by direct
  evaluation or by using the equation $\phi^2 - \phi - 1 = 0$, which is used sometimes
to define the Golden ratio. That is, $3-2 \phi^2 + 2 \phi 
= 1 - 2(\phi^2 - \phi - 1) = 1$)  we find that the altitude of the
roof-like structure is $PC = h = \frac{1}{\phi}$, we name $b=ON=PM=(\phi-1)/\phi=1-1/\phi$ 
which is another important element to compute the other 12 vertices of the
dodecahedron. Both $h$ and $b$ are offsets from the cube in orthogonal
directions from each other. We rewrite them here for clarity of exposition since
they will be used repeatedly in the computations below.
\begin{eqnarray*}
  h &=& \frac{1}{\phi} \\
  b &=& \frac{\phi-1}{\phi} = 1 - \frac{1}{\phi} =  \frac{1}{\phi^2},
\end{eqnarray*}
since $\phi=1 + 1/\phi$.
We have all the elements to complete the other 12 vertices of the dodecahedron.
We do this in 6 couples as follows:


*

*Top and bottom vertices:  \rm We find the vertices $B$ and $C$
and their   reflections with respect to the $XY$ plane at the
bottom.   The vertex at $B$ has an $x$ coordinate of
$-1+ON=-1+b=-1+1-1/\phi=-1/\phi$.   The $y$ coordinate is $y=0$, and
the $z$ coordinate is $1+h=1+1/\phi$.   The vertex $C$ is  the
vertex $B$ plus the coordinate    $(a,0,0)=(2/\phi, 0, 0)$. That is,
the vertex at $C$ is   $(1/\phi, 0,1+1/\phi)$.   The bottom vertices
are a reflection where the sign of $z$ is changed. That is   the top
and bottom vertices are:
\begin{eqnarray*}
      (\pm 1/\phi, \, 0 \,  , \pm (1 + 1/\phi ) )   \end{eqnarray*}   It can be verified that all these points sit in a sphere of radius
$\sqrt{3}$.

*Left and right vertices : \rm All these vertices have the same
$z$ coordiante $z=0$. The left front vertices share the   same $x$
coordinate $x=-1-h=-1-1/\phi$. The $y$ coordinate of the left front 
vertex is $1-b=1-1+1/\phi=1/\phi$. The back coordinate has
$y=-1+b=-1/\phi$.   The right vertices are a reflection with respect
to the $YZ$ plane (change sign of $x$) of these two. That is, in
summary the left and right vertices are given by the combinations:
\begin{eqnarray*}
      (\pm (-1 - 1/\phi), \,  \pm 1/\phi \, ,   0  )   \end{eqnarray*}   Again, all these points are collocated in a sphere
of radius $\sqrt{3}$.

*The front and back vertices : \rm The front top and bottom   $x$
coordinates are $x=0$, the front top and bottom $y$ coordiantes are 
located at the same $y$ coordinate $y=1+h=1+1/\phi$. The front top
vertex is located at $z = 1-b=1-(1-1/\phi)=1/\phi$, while the front
bottom vertex is $-1+b=-1+(1-1/\phi)=-1/\phi$. The back vertices are
a reflection of these by switching the sign of $y$. That is the
front and back vertices are given by   \begin{eqnarray*}
      (0, \pm (1+1/\phi) , \pm 1/\phi  )   \end{eqnarray*}   These four points are located in a sphere of radius $\sqrt{3}$.


This completes the 20 vertices of the dedecahedron.
Finally we need to verify that the faces of the triangles and the trapezoids
are sitting in the same plane for each of the pentagons.  To verify
this we find the normal vectors to a triangle and a contiguous trapezoid.
For example in the first figure we find the normal to the two blue polygons.
Normals are easy constructed by taking cross products. For example the blue
triangle can be seen as the triangle $\triangle CED$ in the roof-like (second figure) structure. 
Consider the points (vectors) top-right  $C$, cube top-front-right $E$ and
cube point $D$ top-back-right. That is,
\begin{eqnarray*}
C &=& (1/\phi, 0, 1+1/\phi) \\
E &=& (1,1,1) \\
D &=& (1,-1,1)
\end{eqnarray*}
Two vectors involved are
\begin{eqnarray*}
  \overrightarrow{CE} &=& E - C =  \left ( 1 - \frac{1}{\phi}, 1 , -\frac{1}{\phi} \right ) \\
  \overrightarrow{CD} &=& D - C =  (1-1/\phi, -1, -1/\phi)
\end{eqnarray*}
Then
\begin{eqnarray*}
  n_{CED} = \overrightarrow{CE} \times \overrightarrow{CD} = 
  -2 \left ( \frac{1}{\phi} , 0 , \frac{ \phi -1}{\phi} \right )
  = \frac{-2}{\phi} (1, 0, \phi-1).
\end{eqnarray*}
Likewise for the blue trapezoid we consider the three points:
$E$ front-top-right, $F$ back,top-right, and the center-right-front
on the roof-like structure $G$.
\begin{eqnarray*}
E &=& (1,1,1) \\
F &=& (1,-1,1) \\
G &=& (1+1/\phi \, ,  \, 1/\phi \, , \,  0)
\end{eqnarray*}
We then build the cross product
Two vectors involved are
\begin{eqnarray*}
  \overrightarrow{EF} &=& F - E =  (0,-2,0) \\
  \overrightarrow{EG} &=& G - E =  \left ( \frac{1}{\phi},\frac{1}{\phi}-1 , -1 \right )
\end{eqnarray*}
\begin{eqnarray*}
  n_{EFG} = \overrightarrow {EF} \times \overrightarrow{EG} =  
  \left ( 2, 0 , \frac{2}{\phi} \right ) 
  = 2 \left ( 1, 0 , \frac{1}{\phi} \right ).
\end{eqnarray*}
and since $1/\phi = \phi-1$ we see that the vectors $n_{CED}$ and $n_{EFG}$ are
colinear. That is the blue polygons are sitting at the same plane. Due to symmetries
all the other polygons satisfy the same requirement and then the dodecahedron
is built.
Next figure

 shows a dodecahedron inscribed in a sphere and
a dodecahedron as a spherical polyhedron. They were plotted with TiKz using the coordinates found above. 
I advertised in my first question on the dodecahedron that I would post a similar answer for the icosahedron. I have failed to do so, but if there is some interest I would post that one using the same idea in my first answer about finding the first point as the intersection of two circles and building the points in a domino-effect from there. I would also post an Euclide's approach for the icosahedron.  
Thanks.
A: 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).
A: I could restate lots of the former answers. But to add further things only: did you think about antiprisms?
The icosahedron is nothing but a pentagonal antiprism with attached pentagonal pyramids on either side.
The dodecahedron can be obtained - at least combinatorically - from a decagonal antiprism with attached tip-truncated decagonal pyramids, provided the latteral faces of antiprism and pyramid here become coplanar.
In fact, those zig-zags of these antiprisms each are the so-called Petrie polygons of these specific polyhedra.
--- rk
A: 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.
A: It seems to me that there is a simple construction of a regular icosahedron based on a truncated regular octahedron. Truncate symmetrically the vertices of the octahedron producing square sections in place of the former vertices. Then note that on the eight fomer faces there are hexagones, and that by joining alternatively the vertices of a hexagone we get an equilateral triangle. Now use the freedom in choosing the depth of the cuts that make the symmetric truncation to ensure that the diagonal of each square has the same length as the side of any equilateral triangle. Then it is easy to see that by joining vertices alternatively all around an initial vertex we get five identical equilateral triangles. Symmetry of the construction ensueres that proceding systematically we get in all 12 vertices, each surrounded by 5 equilateral triangles, everything symmetric, so we have a regular icosahedron. An easy way to get the 12 is to note that each edge of the octahedron contributes one vertex to the final figure (remember the alternation in joining vertices). Duality gives of course a regular dodecahedron.  
