The plane does not intersect at least one edge in each face.

## Solution

The answer is 66. We need to prove that this number is possible, and that a larger number is impossible. First, let's prove the latter. Each face of the polyhedron is intersected by the plane in no more than two edges, meaning that at least one edge belonging to the face remains unintersected. Therefore, in each face, the unintersected edges constitute no less than a third of the total number of edges (in that face). Summing up all these inequalities for all faces, and considering that each edge belongs to two faces, we obtain the desired statement. We will provide an example of a convex polyhedron, 66 edges of which are intersected by a certain plane, not passing through its vertices. Consider a regular 100-gon in the plane; consider a segment of its contour - a broken line \(L\) consisting of 32 sides; denote its vertices as \( \mathrm{C}_{1}, \mathrm{C}_{1}, \ldots, \mathrm{C}_{33} \). Through the center \(O\) of the 100-gon, draw a segment \( \mathrm{AB} \) perpendicular to the plane such that \(A\) and \(B\) lie on opposite sides of the plane at equal distances from it. Construct a polyhedron: two of its faces are triangles \( \mathrm{AC}_{1} \mathrm{~B} \) and \( \mathrm{AC}_{33} \mathrm{~B} \), and the other faces are triangles of the form \( \mathrm{AC}_{\mathrm{i}} \mathrm{C}_{\mathrm{i}+1} \) and \( \mathrm{BC}_{\mathrm{i}} \mathrm{C}_{\mathrm{i}+1} \). This polyhedron has exactly 99 edges. It is not difficult to understand that all edges of this polyhedron of the form \( \mathrm{AC}_{\mathrm{i}} \) and \( \mathrm{BC}_{\mathrm{i}} \) (there are 66 such edges) can be intersected by one plane.
The number of edges of a convex polyhedron is 99. What is the maximum number of edges that a plane, not passing through its vertices, can intersect?

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