Harvard-MIT November Tournament

Solution:

$72^{\circ}$

Because this polyhedron is regular, all vertices must look the same. Let's consider just one vertex. Each triangle has a vertex angle of $60^{\circ}$, so we must have fewer than $6$ triangles; if we had $6$, there would be $360^{\circ}$ at each vertex and you wouldn't be able to "fold" the polyhedron up (that is, it would be a flat plane). It's easy to see that we need at least $3$ triangles at each vertex, and this gives a triangular pyramid with only $4$ faces. Having $4$ triangles meeting at each vertex gives an octahedron (two square pyramids with the squares glued together) with $8$ faces. Therefore, an icosahedron has $5$ triangles meeting at each vertex, so rotating by $\frac{360^{\circ}}{5} = 72^{\circ}$ gives another identical icosahedron.

Alternate solution:

Euler's formula tells us that $V - E + F = 2$, where an icosahedron has $V$ vertices, $E$ edges, and $F$ faces. We're told that $F = 20$. Each triangle has $3$ edges, and every edge is common to $2$ triangles, so $E = \frac{3 \times 20}{2} = 30$. Additionally, each triangle has $3$ vertices, so if every vertex is common to $n$ triangles, then $V = \frac{3 \times 20}{n} = \frac{60}{n}$. Plugging this into the formula, we have $\frac{60}{n} - 30 + 20 = 2$, so $\frac{60}{n} = 12$ and $n = 5$. Again this shows that the rotation is $\frac{360^{\circ}}{5} = 72^{\circ}$.