nine points on a sphere 
Show with proof that there exist $9$ points on the unit sphere (centred at the origin) so that each of the $9$ points has exactly $4$ equidistinct nearest neighbours.

I found a solution to this problem online, which is shown below, but I would like to know how to come up with this solution.
Place $3$ of the points around the circle at positions $(1,0,0),(-1/2, \pm \sqrt{3}/2, 0)$. Choose $c$ with $0 < c < 1$ and set $ r = \sqrt{1-c^2}$ and place three points along each of the two circles at positions $(-r, 0, \pm c)$ and $(1/2 r, \pm \sqrt{3}/2 r, \pm c)$. Choose $c=\sqrt{5}/3$ to yield the required points.
Then one can check that the distance between nearest neighbours is $2/\sqrt{3}$ (isn't there a way better than a tedious brute-force check)?

How did the answerer know to choose those points and how did they know to choose $c=\sqrt{5}/3$?

 A: Here is a way to come up with an answer: consider the 4-regular planar graph of 9 vertices, and we look at the ones with a lot of symmetry. Make it planar is to make it easier to calculate the distance. Since there are 9 vertices, we consider the one that is "symmetry" with "threefold" (that is, rotate 120 degrees and be identical, in some sense.)
So let us draw 9 points, with threefold symmetry

Since we need 4-regular, it is better to have the triangle $ABC$ drawn, otherwise they are "floating outside" and "unstable". Furthermore, we assign three more vertices to $A,B,C$ (remember if we have a lot of triangles, it is easier to calculate the distance!), so we get this

After this, we can construct the regular graphs with symmetry easily. $E$ must connect to four points, $ED$ will cause a hexagon, $EI$ will cause a triangle, and $EH$ will cause a matching, but we want fore degree, so we choose a hexagon and a triangle, like this
Now the problem is, how to put them on a sphere such that all the points connected are equal? If you think it in 3D, you may think it something as this: $ABC-EGI$ is a prism, and $F,H,D$ is points outside and they form the pyramids to the surface, like this. (there is an error in the picture, the label of $G$ and $I$ are reversed)
All the edges in black are equal. Now we need to adjust the height to make the nine points on a sphere. The $D,F,H$ are $(1,0,0),(-1/2, \pm \sqrt{3}/2, 0)$. So that is how the answer choose the nine points. To choose $c=\sqrt 5/3$, we only need to solve the equations of $ED=EG$.
