3
$\begingroup$

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: regular pentagon with pentagram 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.

$\endgroup$
  • 1
    $\begingroup$ Observation: A regular pentagonal antiprism is actually a regular icosahedron (RI) from which five faces surrounding each of two opposite veerices have been truncated. If you can prove that an edge of a RI inscribed in a sphere intercepts a great circular arc measuring $\arctan 2$ (correct), you're good. $\endgroup$ – Oscar Lanzi Jan 18 '17 at 1:06
0
$\begingroup$

OK, so I will post an answer based on Oscar Lanzi's comment, but I am not going to accept it, as I don't quite think it provides that "aha! I see why it must be true" feeling for this question. However, it is at least much less computationally intensive. Consider this diagram of an icosahedron with unit edge length and center O: diagram of icosahedron Point M is the center of pentagon FGJKI and N is the center of ADHEC. So MN is the height of a pentagonal antiprism and MI is the circumradius of its base. As O is the midpoint of MN, the desired condition from the original question, namely MN = MI, is equivalent to $\tan IOM = 2$. As Oscar Lanzi points out, since OM extends to vertex L of the icosahedron, this is further equivalent to showing that the angle $\theta$ subtended by an edge of an icosahedron satisfies $\tan\theta = 2$.

Now draw OP to the midpoint of edge IC of the icosahedron. Bisecting all of the "equatorial" edges of the icosahedron, evidently OP is the circumradius of a regular decagon PQR... with edge length PQ = 1/2, since PQ joins the midpoints of two edges of an equilateral triangle of edge length 1. Looking up well known formulas and special angles, the circumradius of this regular decagon is $OP = \frac{1/2}{2\sin18^\circ} = 1/(\sqrt5-1)$. Hence $\tan IOP = (\sqrt5-1)/2$, but evidently angle $\theta$ is twice angle IOP. So using the double-angle formula for tangent, $$\tan\theta = \frac{\sqrt5-1}{1-\left(\frac{\sqrt5-1}2\right)^2} = \frac{\sqrt5-1}{1-\frac{6-2\sqrt5}4} = \frac{\sqrt5-1}{\frac{\sqrt5-1}2}=2$$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.