What is the maximum length of a space diagonal in an icosahedron with side length 4? For those who don't know, an icosahedron is a 3d polyhedron with 12 vertices and 20 faces. Here's a pic:
I am really asking the length from vertex to the opposite vertex. Thanks in advance. I might be able to solve it and will put my solution in the comments.
On
Here's a guide to finding the solution:
This icosahedron can be inscribed in a sphere of some radius $r$; your answer is $2r$. So the question is "What radius?"
In such a sphere, we can take the north and south poles as two of the vertices; the others, in two groups of five, lie on two lines of latitude, at latitudes $\pm c$, for some value $c$. For one, the points are at longitudes that are multiples of $2\pi/5$; for the other, they are those same longitudes, but with $2\pi/10$ added. In particular, we can say that at latitude $c$, there's a point $A$ at longitude $0$; at latitude $-c,$ there are points $B$, $C$ at longitude $\pm 2\pi/10$.
The distance from the pole to the point $A$ is $4$; the distance from $B$ to $C$ is also $4.$
The coordinates of $A$ are $$ A = (r \cos c, r \sin c, 0) $$
(I'm putting the equator on the $xz$ plane, so the north pole is $(0, 1, 0)$, OK?)
The squared distance from $A$ to the pole is therefore $$ 16 = r^2 \cos^2 c + (1-r\sin c)^2 = r^2 -2r \sin c + 1 $$ which is one equation in $r$ and $c$.
The coordinates of $B$ and $C$ are $$ B = (r \cos c \cos 2\pi/10, - r \sin c , r\cos c \sin 2\pi/10)\\ C = (r \cos c \cos 2\pi/10, - r \sin c , -r\cos c \sin 2\pi/10) $$ The squared distance between them is $$ 16 = (2r \cos c \sin \pi/5)^2 $$ so that $$ 2 = r \cos c \sin \pi/5 $$
From this, and $$ 16 = r^2 -2r \sin c + 1 $$
you can solve for $r$ and $c$ (but particularly $r$), and then you have the answer.
I found this in a Math League book.
The icosahedron can be thought of as 3 polyhedra glued together. One has 2 regular pentagonal bases of side length 4. The other 2, glued to the bases, are pyramids with a pentagonal base and all edges with length 4. The sum of the heights of all these is the distance between opposite vertices. For the pyramids, the distance from a base vertex to the center is $\frac{2}{sin 36} = \sqrt{8+\frac{8}{\sqrt{5}}}$. This forms one leg of a right triangle with 4 as the hypotenuse and the height as the other leg, so the height of the pyramid is $\sqrt{8-\frac{8}{\sqrt{5}}}$. For the other part, let ABC be on of the triangular faces, with A on the top base and BC on the bottom base. Here's a pic to visualize the triangle:
Let M be the midpoint of BC. Since A "overhangs" segment BC slightly, AM = $2\sqrt{3}$ is not equal to the height of the polyhedra. Rather AM is the is the hypotenuse of a right triangle whose legs are the height and the difference d between the distance from A to the center of the top base and M to the center of the bottom base. The first distance, as we found above is $\sqrt{8+\frac{8}{\sqrt{5}}}$. The second distance is $\frac{2}{tan 36} = \sqrt{4+\frac{8}{\sqrt{5}}}$. Their difference is $d = \sqrt{4-\frac{8}{\sqrt{5}}}$. Thus, we get $h = \sqrt{8+\frac{8}{\sqrt{5}}}$.
Adding the 2 heights of the pyramids, we get the distance between opposite vertices to be $\sqrt{40+8\sqrt{5}}$, or $7.6085$ to the ten thousandth place.