# An example

Yes, here is a list of rational coordinates lying on the unit sphere, the convex hull of which is combinatorially equivalent to a regular dodecahedron. This polyhedron is invariant under reflections in three orthogonal hyperplanes (having a symmetry group $C_2 \times C_2 \times C_2$ of order 8, much smaller than the order-120 symmetry group of a regular dodecahedron):

```
verts = {
{304/425, 297/425, 0},
{1, 0, 0},
{52/173, 132/173, -99/173},
{52/173, 132/173, 99/173},
{54/175, 22/175, -33/35},
{54/175, 22/175, 33/35},
{44/125, -108/125, -9/25},
{44/125, -108/125, 9/25},
{3236/4325, -1452/4325, -99/173},
{3236/4325, -1452/4325, 99/173},
{-304/425, -297/425, 0},
{-1, 0, 0},
{-52/173, -132/173, 99/173},
{-52/173, -132/173, -99/173},
{-54/175, -22/175, 33/35},
{-54/175, -22/175, -33/35},
{-44/125, 108/125, 9/25},
{-44/125, 108/125, -9/25},
{-3236/4325, 1452/4325, 99/173},
{-3236/4325, 1452/4325, -99/173}
}
```

Mathematica confirms that the convex hull has twelve pentagonal faces:

```
In[4]:= InputForm@MeshCells[Region`Mesh`MergeCells[ConvexHullMesh[verts]], 2]
Out[4]//InputForm=
{Polygon[{4, 17, 19, 15, 6}], Polygon[{1, 4, 6, 10, 2}],
Polygon[{1, 2, 9, 5, 3}], Polygon[{1, 3, 18, 17, 4}],
Polygon[{2, 10, 8, 7, 9}], Polygon[{6, 15, 13, 8, 10}],
Polygon[{3, 5, 16, 20, 18}], Polygon[{11, 12, 20, 16, 14}],
Polygon[{5, 9, 7, 14, 16}], Polygon[{7, 8, 13, 11, 14}],
Polygon[{12, 19, 17, 18, 20}], Polygon[{11, 13, 15, 19, 12}]}
```

and that the vertices lie on the unit sphere:

```
In[3]:= Map[Norm, verts]
Out[3]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
```

# How I found the solution

The main idea was to use *stereographic projection* to transform the unit sphere onto the plane. This reduces the question to the following equivalent problem:

Does there exist an embedding of the dodecahedral graph into $\mathbb{Q}^2$ such that the five vertices of each face are all concyclic?

Now, twenty points are a lot to contend with, so I decided to instead search for a symmetrical solution which has the following three symmetries:

- reflectional symmetry about the x-axis;
- reflectional symmetry about the y-axis;
- inversive symmetry in an appropriate-sized circle centred on the origin.

This reduced the problem to finding six positive rationals $a,b,c,d,x,y > 0$ such that:

- $y < b < d$ and $ad > bc$;
- $a < x < \sqrt{c^2 + d^2}$;
- $(0, y)$ lies on the circle through $(c, d), (-c, d), (a, b), (-a, b)$;
- $(x, 0)$ lies on the circle through $(a, b)$ and $(c, d)$ which intersects the origin-centred circle of radius $\sqrt{c^2 + d^2}$ at right-angles;
- $(x, 0)$ lies on the circle through $(a, b), (0, y), (a, -b), (0, -y)$;

I wrote an integer-arithmetic C program to search through all possibilities of positive integers $a,b,c,d < 1000$ satisfying the inequality constraints. For each of these 4-tuples, it computes $x$ and $y$ (from the penultimate and antepenultimate constraints, respectively), checks whether they're rational (using an analytic formula), and then verifies the last condition.

All of the solutions found by my program were integer scaled-up versions of the same primitive solution:

```
$ ./dodeca
22, 21, 22, 54, 40/1, 10/1
44, 42, 44, 108, 80/1, 20/1
66, 63, 66, 162, 120/1, 30/1
88, 84, 88, 216, 160/1, 40/1
110, 105, 110, 270, 200/1, 50/1
132, 126, 132, 324, 240/1, 60/1
154, 147, 154, 378, 280/1, 70/1
176, 168, 176, 432, 320/1, 80/1
198, 189, 198, 486, 360/1, 90/1
220, 210, 220, 540, 400/1, 100/1
242, 231, 242, 594, 440/1, 110/1
264, 252, 264, 648, 480/1, 120/1
286, 273, 286, 702, 520/1, 130/1
308, 294, 308, 756, 560/1, 140/1
330, 315, 330, 810, 600/1, 150/1
352, 336, 352, 864, 640/1, 160/1
374, 357, 374, 918, 680/1, 170/1
396, 378, 396, 972, 720/1, 180/1
```

(The numbers in order are the values of $a, b, c, d, x, y$ in order; the latter two are rendered as rationals because they're not *a priori* integers.)

I took this primitive solution and computed the vertices as points in $\mathbb{Q}[i]$ (identifying the plane with the complex numbers). I divided all points by the Gaussian integer $c + di$ to eliminate the arbitrary scale factor, resulting in a set of rational points invariant under inversion in the **unit** circle. After stereographically projecting back to the unit sphere, this inversion invariance corresponds to a reflectional symmetry through the equator.

**EDIT:** After porting the code to the GPU and expanding the search range to 10000, the number of known primitive solutions has increased to 3.

**Much later EDIT:** There exists infinitely many primitive solutions.