In a canonical polyhedron, all edges are tangent to a unit sphere with the origin at the center of gravity of the points of tangency.
I've been working on exact forms of canonical polyhedra, and the regular pyramid turned out to be an interesting case. For a base with $n$ sides, that polygon has $z$ coordinates of $-tan(\frac{\pi}{2n})$, and the apex is at $(0,0,cot(\frac{\pi}{2n}))$. Here are some examples, each pyramid with its dual. The edges are perpendicular to each other at points tangent to the unit sphere.
The base regular polygon needs a scaling factor that is the root of a particular polynomial. For bases with 3 to 12 sides, here are the roots and their polynomials.
$\begin{array}{ccc} 3 & 1.6329931618554520655 & 2 \sqrt{\frac{2}{3}} \\ 4 & 1.2871885058111652495 & x^4+8 x^2-16 \\ 5 & 1.1690009178779524258 & 5 x^4+40 x^2-64 \\ 6 & 1.1124766546017998270 & 3 x^4+48 x^2-64 \\ 7 & 1.0806190099881630969 & 7 x^6-448 x^2+512 \\ 8 & 1.0607630605786609417 & x^8+32 x^6+64 x^4-1024 x^2+1024 \\ 9 & 1.0475038619623609305 & 3 x^6+72 x^4+384 x^2-512 \\ 10 & 1.0381901644398379801 & 5 x^8+240 x^6+960 x^4-5120 x^2+4096 \\ 11 & 1.0313884014418166375 & 11 x^{10}+88 x^8-2112 x^6-11264 x^4+45056 x^2-32768 \\ 12 & 1.0262650635674814921 & x^8+64 x^6+128 x^4-4096 x^2+4096 \\ \end{array}$
Here's code for calculating the pyramids.
pyramid[basesides_] := Module[{k = Abs[basesides], poly,
sign = Sign[basesides], g},
poly = RootReduce[g /. Last[Solve[myNorm[Mean[Table[{g Cos[2 Pi n/k],
g Sin[2 Pi n/k], -Tan[Pi/(2 k)]}, {n, 0, 1}]]] == 1]]];
{Append[Table[{RootReduce[poly Cos[2 Pi n/k + (sign - 1) Pi/k/2]],
RootReduce[poly Sin[2 Pi n/k + (sign - 1) Pi/k/2]], -sign Tan[
Pi/(2 k)]}, {n, 0, k - 1}], {0, 0, sign Cot[Pi/(2 k)]}],
Polygon[Prepend[Sort[Append[RotateLeft[Append[Reverse[#], k + 1], 1] & /@
Partition[Range[k], 2, 1], {1, k, k + 1}]], Range[k]]]}]
I use Solve in that. Is there a more direct way for generating these polynomials? Is it a known class?
Here are factored forms of the coefficients.
Let $h$ be the height of the pyramid, $r$ the radius of its base and $x$ the distance from the center $O$ of the tangent unit sphere to the base. The apothem of the base is then $rc$, where I introduced the shortcut $c=\cos(\pi/n)$.
The condition that the distance from $O$ to a lateral edge is $1$ gives the equation $$r^2(h-x)^2=r^2+h^2.$$ The condition that the distance from $O$ to a base edge is $1$ gives the equation $$x^2+r^2c^2=1.$$ Finally, the condition that $O$ is the barycenter of the tangency points gives the equation $$h-2x=hc.$$ These three equations can be solved for the unknowns $r$, $h$ and $x$, here's the result: $$ r=\sqrt{2\over c(1+c)},\quad h=2\sqrt{1\over 1-c^2},\quad x=\sqrt{1-c\over 1+c}. $$ Your scaling factor is of course the value of $r$, which can be calculated from the above formula, without solving complicated polynomial equations: $$ r=\sqrt{2\over \cos(\pi/n)(1+\cos(\pi/n))}. $$
I suspect your equations may arise from the search of an exact value for $\cos(\pi/n)$, but have little to do with canonical pyramids.