A Challenging Problem of Uniform Polyhedron With Right Kite Faces by H.C.R

In general, the ratio of unequal edges say $a$ & $b$ ( $a<b$ ) of a uniform polyhedron having $2n+2$ vertices lying on a spherical surface, $4n$ edges & $2n$ congruent faces each as a right kite, is given by following formula (see the detailed analysis of uniform trapezohedron by HC Rajpoot )

$\boxed{\frac ab=\sqrt{\tan\frac{\pi}{n}\tan\frac{\pi}{2n}}}$

As per given problem, no. of edges $2n+2=1002$ i.e. $n=500$ , setting the value of $n$ in above general formula

$\frac ba=\frac{1}{\sqrt{\tan\frac{\pi}{500}\tan\frac{\pi}{2\cdot 500}}}\approx 225.0772278$