5. For a convex polyhedron, the internal dihedral angle at each edge is acute. How many faces can the polyhedron have?
5. For each of the faces, consider the vector of the external normal, i.e., a vector perpendicular to this face and directed outward from the polyhedron.

$1^{\circ}$. Let's prove that the angle between any two external normals is obtuse or straight. Suppose this is not the case, and there exist two faces $\Gamma_{1}$ and $\Gamma_{2}$, the external normals of which form an angle not greater than $\pi / 2$. Then the faces $\Gamma_{1}$ and $\Gamma_{2}$ belong to half-planes $\Pi_{1}$ and $\Pi_{2}$, which form a dihedral angle of at least $\pi / 2$.

Take a point $P$ on the face $\Gamma_{2}$. Let $P^{\prime}$ be the projection of the point $P$ onto the plane of the face $\Gamma_{1}$. Since the angle between

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-335.jpg?height=412&width=504&top_left_y=1217&top_left_x=741)

Fig. 167 the external normals to the faces $\Gamma_{1}$ and $\Gamma_{2}$ is not greater than $\pi / 2$, and our polyhedron is convex, the point $P^{\prime}$ lies outside the polygon $\Gamma_{1}$. Therefore, there exists a line containing some edge $r$ of the face $\Gamma_{1}$ and separating $P^{\prime}$ from $\Gamma_{1}$. The polyhedron lies inside the acute dihedral angle corresponding to the edge $r$, but $P$ lies outside this dihedral angle (Fig. 167). Contradiction.

$2^{\circ}$. It remains to show that in space, there do not exist more than four vectors, the pairwise angles between which are obtuse or straight.

Suppose this is not the case, and $\vec{u}_{0}, \vec{u}_{1}, \vec{u}_{2}, \vec{u}_{3}, \vec{u}_{4}$ are five vectors, the pairwise angles between which are obtuse or straight. Introduce a rectangular coordinate system such that the $O z$ axis is directed along $\vec{u}_{0}$.

Denote by $\vec{v}_{i}$ the projection of the vector $\vec{u}_{i}$ onto the plane $O x y$. Let $\vec{u}_{i}=\left(x_{i}, y_{i}, z_{i}\right)$, then $x_{0}=y_{0}=0, z_{0}>0$ and $v_{0}=0$.

The condition that all angles are obtuse or straight is equivalent to all scalar products being negative. In particular,

$$
z_{0} z_{i}=\left(\vec{u}_{0}, \vec{u}_{i}\right) < 0 \quad \text{for} \quad 1 \leqslant i \leqslant 4.
$$

Since $z_{0} > 0$, it follows that $z_{i} < 0$ for $1 \leqslant i \leqslant 4$. Now consider the scalar products between the projections:

$$
\left(\vec{v}_{i}, \vec{v}_{j}\right)=\left(\vec{u}_{i}, \vec{u}_{j}\right)-z_{i} z_{j} < 0 \quad \text{for} \quad 1 \leqslant i, j \leqslant 4.
$$

Thus, $\vec{v}_{1}, \vec{v}_{2}, \vec{v}_{3}, \vec{v}_{4}$ are four vectors on the plane, the angles between which are obtuse or straight. But this is impossible.

Therefore, a polyhedron has four faces. Note that a polyhedron with four faces can only be a tetrahedron.

Idea of another solution. First, prove that at each vertex of the polyhedron, three faces meet. For this, it is sufficient to prove that the sum of the dihedral angles of any $n$-hedral angle is greater than $\pi(n-2)$. Since an $n$-hedral angle can be cut into $n-2$ trihedral angles, it is sufficient to prove this statement for a trihedral angle. Now, it can be shown that all planar angles at each vertex are acute. From this, it is not difficult to deduce that all faces are triangles. Now, it is not difficult to verify that the polyhedron is a tetrahedron.