## Task 21/63

A convex polyhedron with 53 vertices and 19 faces is modified by cutting off all its vertices with plane cuts, such that each cut exactly captures one vertex and no cut intersects or touches another.

How many edges, vertices, and faces does the resulting polyhedron have?
Let $e$ be the number of vertices, $f$ the number of faces, and $k$ the number of edges of the given polyhedron. $E, F$, and $K$ are the corresponding numbers for the new polyhedron.

If $k_{i}$ edges meet at the $i$-th vertex of the given polyhedron, then by cutting off this vertex, a polygon with $k_{i}$ sides is created, which is a new face, $k_{i}$ new edges, and $k_{i}$ new vertices. None of the original faces and edges are lost, but the original vertex is.

Since each edge connects two vertices, it follows that the number of new edges is $2k$ and the number of new vertices is $2k$, while the number of new faces is $e$. Therefore, we have

$$
K=3k \quad ; \quad E=2k \quad ; \quad F=f+e
$$

From Euler's polyhedral formula $e+f=k+2$, it follows that $k=e+f-2=53+19-2=70$. Thus,

$$
K=210 \quad ; \quad E=140 \quad ; \quad F=72
$$

Indeed, the Euler's polyhedral formula also holds for these numbers: $E+F=212=K+2$.

Note: A convex polyhedron with 53 vertices and 19 faces as described in the problem does not exist, because according to Euler's polyhedral formula, it would have to have 70 edges. In every vertex, at least 3 edges meet, each connecting 2 vertices. Therefore, $k \geq \frac{3}{2} e$ must hold, which the "polyhedron" does not satisfy.