Uniform Polyhedron with 500 congruent right kite faces!

Question

The diagram above shows a uniform polyhedron having 502 vertices exactly lying on a spherical surface, 1000 edges & 500 congruent right kite faces each having two unequal edges $a$ & $b$ given $b>a$. How to find out the ratio $\frac{b}{a}$ of unequal edges of this uniform polyhedron?

Note: The conditions, that 500 right kite faces are congruent & 502 vertices exactly lies on a spherical surface, are governing conditions which fully describe this uniform polyhedron (trapezohedron) & will produce the expression of ratio $\frac{b}{a}$. This is also feasible for 2n no. of congruent right kite faces.

Edit/JL: Below there are Mathematica images of the resulting polyhedron (not yet confirmed by the OP) with $n=5$ and $n=17$. The faces are rendered using a non-trivial opacity setting so that we can see through them to some extent. Because projecting the object to a plane distorts the angles a bit, it may not be entirely clear that the 4-gonal faces all have two 90 degree angles - you have to take my word for that :-)

Answer 1

Here is the simplest solution of generalized polyhedron with right kite faces covering up countless problems like the above problem based on simple geometry.

In general, the ratio of unequal edges $\color{blue}{a}$ & $\color{blue}{b}$ $\space \color{blue}{\forall \space (a\leq b)}$ of any $\color{blue}{\text{uniform polyhedron}}$ having $\color{blue}{2n \space (\forall \space n\geq 3)}$ no. of $\color{blue}{\text{congruent right kite faces}}$ is given as $$\bbox[4pt, border: 1px solid blue;]{\color{red}{\frac{a}{b}=\sqrt{\tan\left(\frac{\pi}{n}\right)\tan\left(\frac{\pi}{2n}\right)}}}$$

I had analysed it for a generalized case here Uniform polyhedrons with right kite faces by HCR

Answer 2

The vertices are $$N=(0,0,1), S=(0,0,-1)$$ for the poles, and $$A_k=\left(\sqrt{1-h^2}\cos \frac{k\pi}{250},\sqrt{1-h^2}\sin\frac{k\pi}{250},(-1)^kh\right)$$ with $0\le k<500$ for the points near the equator (for some $0<h<1$). Then from several applications of Pythagoras$$b^2={(1-h^2)+(1-h)^2}=2-2h $$ $$a^2+b^2=(1-h^2)+(1+h)^2=2+2h $$ so that $a^2=4 h$. But $a$ is also the distance between $A_0$ and $A_1$, hence $$\begin{align}4h=a^2&=(1-h^2)\left(\cos\frac\pi{250}-1\right)^2 +(1-h^2)\sin^2\frac \pi{250}+(2h)^2\\ &=(1-h^2)(2-2\cos\frac\pi{250})+4h^2\\ &=2-2\cos\frac\pi{250}+h^2(2+2\cos\frac\pi{250})\end{align} $$ This gives you a quadratic equation for $h$, but one solution $h=1$ is trivial (and clearly useless). After dividing it away, we get $$ h=\frac{1-\cos\frac\pi{250}}{1+\cos\frac\pi{250}}$$ and from this also $b$ and $a$ via the above equations. Especially, $$\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}-1=\frac{2+2h}{2-2h}-1=\frac{2h}{1-h}=\frac1{\cos\frac\pi{250}}-1, $$ $$\frac ab=\sqrt{\frac1{\cos\frac\pi{250}}-1}. $$