It is a fact (that one can verify, for example, by plugging in n=5 into the formulae on the Wolfram Mathworld page on antiprisms) that the height of a (regular) pentagonal antiprism (i.e., the pentagonal antiprism with all edges the same length and top and bottom faces regular pentagons) is equal to the circumradius of its base. Does anyone know a (relatively) short proof of this fact that lends some nice insight into why this equality occurs? That reveals it as something more essential than a coincidence about the trigonometric functions of angles like $\pi/5$? I have the intuition that the fact that if you augment the pentagonal antiprism with two pentagonal bipyramids you get a regular icosahedron ought to provide a quick proof of the antiprism height = base circumradius equality, but I have been unable to crystallize this intuition into a short, elegant proof. Thanks for any insight you can provide.
To elaborate somewhat, the nicest/most direct proof I know of this equality is as follows. We first derive the exact value of the circumradius $r$ of the regular pentagon with edge-length 1 (details below); it turns out to be $r=\sqrt{\frac{5+\sqrt{5}}{10}}$. From this, the height $p$ of the equilateral pentagonal pyramid with edge-length 1 is $p = \sqrt{1-r^2} = \sqrt{\frac{5-\sqrt{5}}{10}}$. Now consider the regular icosahedron with side length 1, with one vertex at the "north pole" of its circumscribed sphere of radius $R$. The plane $P$ through the five vertices nearest to (but not at) the north pole cuts the icosahedron in a regular pentagon which is the top base of a pentagonal antiprism of height $h$, as well as the base of a pentagonal pyramid with apex at the north pole. Considering the radius of the sphere from its center to the north pole, it is cut into two pieces by $P$, of lengths $h/2$ and $p$, respectively. That is, $R = h/2 + p$. On the other hand, considering the radius from the center of the sphere to one of the vertices in plane $P$, we get a right triangle with hypotenuse $R$ and legs $h/2$ and $r$. Hence, $R^2 = (h/2)^2 + r^2$. Squaring the first equation, equating the right hand sides, and solving for $h$ yields $$h = \frac{r^2-p^2}{p} = \frac{\frac{2\sqrt{5}}{10}}{\sqrt{\frac{5-\sqrt{5}}{10}}} = \frac{\sqrt{2}}{\sqrt{5-\sqrt{5}}} = \sqrt{\frac{5+\sqrt{5}}{10}} = r.$$
Does anyone know or can anyone find a more insightful/less computational proof of this equality?
And here's the most elementary way I know of deriving $r$. I presume the usual well-known occurrences of the golden ratio as in this diagram:
Evidently from the diagram, $r^2 = a^2+\frac{1}{4}$ and $\frac{1}{\phi^2} = (a-\frac{r}{\phi^2})^2 + \frac{1}{4}$. Expanding the latter equation, substituting the first, and clearing denominators yields
$r^2(\phi^4 + 1) - 2\phi^2ar = \phi^2$. Moving the remaining term with $a$ to the right, squaring, and and substituting from the first equation again yields $r^4(\phi^4+1)^2 - 2r^2\phi^2(\phi^4+1) + \phi^4 = 4\phi^4(r^2-\frac{1}{4})r^2$. Collecting powers of $r$ gives us $r^4(\phi^8 - 2\phi^4+1) + r^2(-2\phi^6 + \phi^4 - 2\phi^2) + \phi^4 = 0$. Substituting the value of $\phi$ yields, with some calculation, $r^4\left(\frac{35+15\sqrt{5}}{2}\right) + r^2\left(\frac{-35-15\sqrt{5}}{2}\right) + \frac{7+3\sqrt{5}}{2} = 0,$ and dividing by the constant term produces the seemingly miraculous simplification $5r^4 - 5r^2 +1 = 0$, whence $r^2 = \frac{5\pm\sqrt{5}}{10}$. Examination of the magnitudes of the two roots then makes it clear that $r = \sqrt{\frac{5+\sqrt{5}}{10}}$. A less computational demonstration of the value of $r$ would be welcome also.