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Uniform Polyhedron

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 :-)

enter image description here

enter image description here

Jyrki Lahtonen
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Harish Chandra Rajpoot
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    Is there a reason to think that a diagram with 5 quadrangles would fully describe the polyhedron (or the asked ratio)? It may, but don't you think it would be a miracle if it did? We are not exactly discussing a platonic solid here :-) – Jyrki Lahtonen Apr 22 '15 at 20:16
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    I suppose the very acute angles are all at the north and south pole and the length $a$ wiggle about the equator? – Hagen von Eitzen Apr 22 '15 at 20:58
  • I think Coxeter has something to say about polyhedra like this. – David K Apr 22 '15 at 21:19
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    @JyrkiLahtonen: Kindly, change your way of thinking. The conditions, that all 500 right kite faces are congruent & all 502 vertices lie on a spherical surface, are fully governing conditions to produce the expression of the ratio $\frac{b}{a}$ of unequal edges. It is also possible for 2n no. of right kite faces, can you imagine this? – Harish Chandra Rajpoot Apr 23 '15 at 10:33
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    @Hagen von Eitzen: Your supposition is absolutely right! – Harish Chandra Rajpoot Apr 23 '15 at 10:37
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    Harish Chandra Rajpoot: Apologies. I think I now understood what the question is about. I solved it numerically for $n=5$ and $n=17$, and took the liberty of adding the resulting images to your question. Please confirm that this is the kind of picture the answerers should have in mind when thinking about it. If so, then I will certainly recommend the question for reopening. – Jyrki Lahtonen Apr 25 '15 at 06:37

2 Answers2

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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, $$\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}-1=\frac{2+2h}{2-2h}-1=\frac{2h}{1-h}=\frac1{\cos\frac\pi{250}}-1, $$ $$\frac ab=\sqrt{\frac1{\cos\frac\pi{250}}-1}. $$

Hagen von Eitzen
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    It is not clear to me the faces generated this way are right angled kites. e.g. it is not obvious why the 4 points $A_0, A_1, A_2$ and $N$ are coplanar... – achille hui Apr 25 '15 at 23:37
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    @achillehui: A good comment! $A_0A_1\perp A_1S$ by symmetry/design. By Thales' theorem $A_1N\perp A_1S$. Therefore $A_1S$ must be normal to the plane determined by $N,A_0$ and $A_1$. Another symmetry (reflect the whole picture w.r.t. the plane of the meridian $NA_1S$) implies that $A_1S$ is also normal to the plane determined by $N,A_2$ and $A_1$. This implies that $N,A_0,A_1,A_2$ are all coplanar. – Jyrki Lahtonen Apr 30 '15 at 11:40
  • @JyrkiLahtonen cool, this settle the missing piece of this answer. – achille hui Apr 30 '15 at 16:10
  • @achillehui Honestly, I didn't even intend to show that there exists such a polyhedron at all (one might make oneself clear that there does by continuity arguments, for example). Instead I only wanted to determine $h$ (and then $a$ and $b$) from the given assumptions that all points are on the sphere, the obvious rotational symmetry holds for the non-poles, and how the longer side and the long diagonal of the presumed kite depend on $h$ (which is the one paramter determining the vertical position of all non-poles). – Hagen von Eitzen Apr 30 '15 at 18:18
  • Hagen, the problem that concerned both Achille and me was the following: Continuity does make it clear that there are points $A_0,A_1,\ldots$ with longitudes $i\pi/250$ alternating between the hemispheres such that $A_{2i}A_{2i+1}\perp A_{2i}N$ and $A_{2i}A_{2i+1}\perp A_{2i+1}S$ - just solve for the variable $h$ in your equations. But the coplanarity of the kites was not immediately obvious. Consider the following small change. Assume that we prescribed, instead of kites, faces with two 89 degree angles. – Jyrki Lahtonen May 10 '15 at 11:24
  • (cont'd) Continuity would still prove the existence of a value of $h$ such that the angles would have those fixed values. But this time instead of 250 4-gons on both hemispheres you would get 500 triangles. The kites would have a small bend along the diagonals. – Jyrki Lahtonen May 10 '15 at 11:26
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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

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Harish Chandra Rajpoot
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    I object to your continued reference of this polyhedron as a "uniform" polyhedron. It is not uniform as the term is commonly used in the mathematical literature: it is not vertex-transitive nor are the faces regular polygons. Instead, this polyhedron has a polyhedral graph equivalent to one of an infinite family of dual antiprisms, but is not strictly such due to its different geometry. – heropup Jun 15 '15 at 00:08
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    Your objection is unreasonable & not applicable at all. Perhaps you have not read properly the conditions of the questions. Here uniformity stands only for the fact that all the vertices are exactly lying on a spherical surface. There are absolutely infinite no. of such polyhedrons/trapezohedrons having congruent right kite faces. Kindly go through the link I provided hereby & notice two conditions I applied to find all the parameters. Moreover, I have not applied any terminology, other facts. My derivations are straight forward based on simple geometry. There is no complexity as such. – Harish Chandra Rajpoot Jun 16 '15 at 08:53