8.6. Answer: $2 n-4$. First, let's prove that the projection of a convex polyhedron with $n$ faces can have $2 n-4$ sides. Cut off the edge $C D$ of a regular tetrahedron $A B C D$ with a prismatic surface, the lateral edges of which are parallel to $C D$ (Fig. 57). The projection of the resulting polyhedron with $n$ faces onto a plane parallel to the lines $A B$ and $C D$ has $2 n-4$ sides.

Now, let's prove that the projection $M$ of a convex polyhedron with $n$ faces cannot have more than $2 n-4$ sides. The number of sides of the projection onto a plane that is not perpendicular to a face cannot be greater than the number of sides of all other projections. Indeed, in such a projection, each face turns into a side of the polygon; if the projection plane is slightly moved, this side will either remain or split into several sides, while the number of other sides will not change. Therefore, we will consider projections onto planes that are not perpendicular to the faces. In this case, the edges that project into the boundary of the polygon $M$ divide the polyhedron into two parts: "upper" and "lower". Let $p_{1}$ and $p_{2}, q_{1}$ and $q_{2}, r_{1}$ and $r_{2}$ be the number of vertices, edges, and faces in the upper and lower parts (vertices and edges of the section are not counted); $m$ be the number of vertices of the polygon $M$, $m_{1}$ (respectively $m_{2}$) be the number of vertices of $M$ from which at least one edge of the upper (respectively lower) part extends. Since from each vertex of $M$ at least one edge of the upper or lower part extends, $m \leqslant m_{1}+m_{2}$. Now, let's estimate the number $m_{1}$. From each vertex of the upper part, at least three edges extend, so the number of ends of the edges of the upper part is at least $3 p_{1} + m_{1}$. On the other hand, the number of ends of these edges is $2 q_{1}$; therefore, $3 p_{1} + m_{1} \leqslant 2 q_{1}$. Now, let's prove that $p_{1} - q_{1} + r_{1} = 1$. The projections of the edges of the upper part divide the polygon $M$ into several polygons. The sum of the angles of these polygons is $\pi(m-2) + 2 \pi p_{1}$. On the other hand, it is equal to $\sum_{i} \pi(q_{1 i} - 2)$, where $q_{1 i}$ is the number of sides of the $i$-th polygon of the partition; the latter sum is equal to $\pi(m + 2 q_{1}) - 2 r_{1}$. Equating both expressions for the sum of the angles of the polygons, we get the required result. Since $q_{1} = p_{1} + r_{1} - 1$ and $m_{1} + 3 p_{1} \leqslant 2 q_{1}$, then $m_{1} \leqslant 2 r_{1} - 2 - p_{1} \leqslant 2 r_{1} - 2$. Similarly, $m_{2} \leqslant 2 r_{2} - 2$. Therefore, $m \leqslant m_{1} + m_{2} \leqslant 2(r_{1} + r_{2}) - 4 = 2 n - 4$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3f2bc5f5a1af26514cfag-162.jpg?height=471&width=485&top_left_y=297&top_left_x=545)

Fig. 57
8.6. What is the maximum number of sides a coin can have as a projection of a convex polyhedron with $n$ faces?