36*. A polyhedron is called regular if all its faces are regular and congruent polygons to each other and all its polyhedral angles are also regular and congruent to each other ${ }^{2}$. Investigate the possibility of constructing simple regular polyhedra with $n$-sided faces and $m$-sided angles and determine the possible number of their types, using the following considerations.

Divide the entire space into $f$ congruent regular $n$-sided angles with a common vertex $S$, with $m$ angles at each edge, and express the magnitudes of their dihedral angles in terms of $n$ and $f$. If a regular $n$-sided pyramid is constructed from the found magnitude of the dihedral angle, with the base being a regular $n$-sided polygon - a face of the desired polyhedron, then from such congruent pyramids, a regular polyhedron can be assembled (for example, a cube is assembled from six regular quadrilateral pyramids with a common vertex at the center of the cube). It can be proven that any regular polyhedron can be obtained by such a construction. (The last remark can be used without proof in solving this and the following problems.)



## 44
36. Let the faces of a regular polyhedron be regular $n$-gons, and the polyhedral angles at the vertices be regular $m$-gonal angles; obviously, $m \geqslant 3, n \geqslant 3$. Divide the space into $f$ congruent regular $n$-gonal angles with a common vertex $S$ such that around each edge $[S A)$ there are $m$ angles. Then the magnitude of each of these $f$ polyhedral angles is $\frac{4 \pi}{f}$. Since this magnitude is the difference between the sum of the magnitudes of its dihedral angles and the number $\pi(n-2)$, the sum of all dihedral angles of the $n$-gonal angle is $\frac{4 \pi}{f} + \pi(n-2) = \frac{4 + nf - 2f}{f} \pi$. One dihedral angle has a magnitude that is $n$ times smaller, i.e., $\frac{4 + nf - 2f}{nf} \pi$.

On the other hand, since at each of the $p$ edges of our $n$-gonal angles there are $m$ congruent dihedral angles, the magnitude of each of them is $\frac{2 \pi}{m}$. Thus,

$$
\begin{gathered}
\frac{2 \pi}{m} = \frac{4 + nf - 2f}{nf} \pi \\
2nf + 2mf - mnf = 4m \\
f = \frac{4m}{2m + 2n - mn}
\end{gathered}
$$

## 168

Since $f > 0, m > 0$, from (1) we have $2m + 2n - mn > 0$,

$$
\frac{2}{m} + \frac{2}{n} > 1
$$

Thus, the division of space into congruent polyhedral angles is possible only if the numbers $m \geqslant 3$ and $n \geqslant 3$ satisfy the inequality (2). Clearly, only five cases are possible: 1) $m=3$ and $n=3: \frac{2}{3} + \frac{2}{3} > 1$. From (1) we get $f=4$, the polyhedron is bounded by 4 triangles - this is a tetrahedron. The corresponding values of the numbers $a$ and $p$ are obtained from the formulas in problems 27 and 28: for the tetrahedron $a=6, p=4$.

2) For $m=3, n=4$ we get: $\frac{2}{3} + \frac{2}{4} > 1, f=6$. The polyhedron is bounded by six squares - this is a hexahedron or cube. Here $a=12, p=8$.
3) For $m=4, n=3$ we also get $\frac{2}{4} + \frac{2}{3} > 1 ; f=8$. The polyhedron is bounded by eight triangles - this is an octahedron. In it, $a=12, p=6$.
4) For $m=3, n=5$ we have: $\frac{2}{3} + \frac{2}{5} > 1 ; f=12$. The polyhedron is bounded by twelve pentagons - this is a dodecahedron. In it, $a=30, p=20$.
5) For $m=5, n=3 ; \frac{2}{5} + \frac{2}{3} > 1, f=20$. The polyhedron is bounded by 20 triangles - this is an icosahedron. In it, $a=30, p=12$.

For $m=4$ and $n=4$ we have: $\frac{2}{4} + \frac{2}{4} = 1$, and the inequality (2) is not satisfied. For further increases in the values of $m$ and $n$, the expression on the left side of the inequality (2) becomes even smaller, and it cannot be valid. It can be shown that any regular polyhedron can be obtained by the described construction, and our considerations show that there are only five types of regular polyhedra in space.