5 Let $D$ be a dodecahedron which can be inscribed in a sphere with radius $R$. Let $I$ be an icosahedron which can also be inscribed in a sphere of radius $R$. Which has the greater volume, and why?
Note: A regular polyhedron is a geometric solid, all of whose faces are congruent regular polygons, in which the same number of polygons meet at each vertex. A regular dodecahedron is a polyhedron with 12 faces which are regular pentagons and a regular icosahedron is a polyhedron with 20 faces which are equilateral triangles. A polyhedron is inscribed in a sphere if all of its vertices lie on the surface of the sphere.
The illustration below shows a dodecahdron and an icosahedron, not necessarily to scale.
Solution 1: $D$ has the greater volume.
The key idea that we will use is a beautiful relationship between $I$ and $D$; they are $d u a l$ polyhedra. To understand this, note that $D$ has 12 faces, 30 edges, and 20 vertices. (The 12 pentagonal faces produce 60 edges and 60 vertices, but these are, respectively, double- and triple-counted. Hence there are $60 / 2=30$ edges and $60 / 3=20$ vertices). Likewise, $I$ has 20 faces, 30 edges, and 12 vertices. Notice that $I$ has as many vertices as $D$ has faces, and vice-versa.

Consequently, if we join each center of the faces of an icosahedron to the centers of adjacent faces, we will get a dodecahedron, and vice-versa. (For the other platonic solids, the cube and octahedron are duals of each other, and the tetrahedron is its own dual).
We will use duality to prove an important lemma.
Lemma: If a dodecahedron and an icosahedron can be inscribed in the same sphere, then they circumscribe the same sphere as well.

To see why this is true, imagine $I$ and $D$ both sitting inside a sphere of radius $R$ with center $O$, which circumscribes both polyhedra. Thus all the vertices of both polyhedra lie on the surface of this sphere. Because the polyhedra are dual, we can place the polyhedra so that for every vertex $A$ of $I$, the line $A O$ is perpendicular to and passes through the center of a face of $D$. Likewise, for every vertex $B$ of $D$, the line $B O$ is perpendicular to and passes through the center of a face of $I$.

The illustration below is an imperfect depiction of a portion of this situation. The left picture is a "top view," as seen by an observer looking directly down at the center of a face of $D$. Point $A$ is a vertex of $I$, and $B$ is a vertex of $D$.

The picture on the right is a "side view" (not-to-scale), showing the projections of $A$ and $B$ down to the center $O$ of the sphere. Let $C O$ be the perpendicular projection of $A O$ onto $B O$. Note that point $C$ is the center of a face of the icosahedron $I$ (in the Top View, you cannot see $C$ because it is directly "underneath" $B$ ), and $h=C O$ is the radius of the inscribed sphere of $I$. Also $r=A C$ is the radius of the circumscribed circle about a face of $I$.

Let $E O$ be the perpendicular projection of $B O$ on to $A O$. Clearly $E O=C O=h$ (since $A O=$ $B O=R$ ), but by duality, $E$ must be a center of a face of the dodecahedron $D$. Consequently, the radius of the inscribed sphere of $D$ must also equal $h$.

Now it is a simple matter to compute and compare the two volumes. Notice that $D$ consists of 12 pyramids with regular pentagonal bases (each with circumradius $r$ ) and height $h$. Likewise, $I$ consists of 20 pyramids with equilateral triangular bases (each with circumradius $r$ ) and height $h$. Recall that the volume of a pyramid with base area $B$ and height $h$ is $B h / 3$. Now we can easily compare the two volumes. ${ }^{1}$
$$
\frac{\operatorname{vol}(D)}{\operatorname{vol}(I)}=\frac{\frac{h}{3} \times 12 \times \text { area of pentagon }}{\frac{h}{3} \times 20 \times \text { area of triangle }}
$$
${ }^{1}$ We are using the fact that a regular $n$-gon that can be inscribed in a circle of radius $r$ can be dissected into $n$ "pie slices" that are isosceles triangles with sides of length $r$ and vertex angle $(360 / n)^{\circ}$. Thus the area of the regular $n$-gon is $n \times\left(r^{2} / 2\right) \times \sin (360 / n)^{\circ}$.
$$
\begin{array}{l}
=\frac{3 \times \text { area of pentagon }}{5 \times \text { area of triangle }} \\
=\frac{3 \times 5 \times \frac{r^{2}}{2} \sin 72^{\circ}}{5 \times 3 \times \frac{r^{2}}{2} \sin 120^{\circ}} \\
=\frac{\sin 72^{\circ}}{\sin 60^{\circ}} \\
>1
\end{array}
$$

Solution 2: (sketch) Instead of showing that the inscribed spheres have equal radii, we can use duality and the "top view" used in the previous solution to show that the radius of the circumscribed circle of each face of $D$ is equal to the radius of the circumscribed circle of each face of $I$. In other words, the two circles depicted below have the same radii. This fact is attributed to the ancient Greek mathematician Apollonius.

Once this is known, it is easy to deduce that the two polyhedra have identical inscribed sphere radii, and the rest of the solution proceeds as before.