Let $P$ be a three-dimensional convex polytope with $N$ faces; $O$ a point outside $P$. What is the maximal number $f(N)$ of vertices of $P$ which may be seen from $O$?
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$\begingroup$ Maybe the pyramid over a regular polygon would be a good example. $\endgroup$– F. C.Commented Nov 12, 2020 at 7:42
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1$\begingroup$ @F.C. This shows that $f(N)\ge N$. $\endgroup$– YCorCommented Nov 12, 2020 at 8:47
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1$\begingroup$ Oh, I see that I had somehow missed the point. But then one can truncate the top vertex and iterates this kind of truncation, adding each time 2 vertices and one face. $\endgroup$– F. C.Commented Nov 12, 2020 at 9:13
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$\begingroup$ F.C. is right; this is clearly optimal, as a convex polytope with $N$ facets cannot have more than $2N-4$ vertices, by Euler's formula (and relation $e\geq 3v/2$). $\endgroup$– Ilya BogdanovCommented Nov 12, 2020 at 10:14
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1$\begingroup$ @FedorPetrov As F.C. mentions, you may start with a tetrahedron all of whose vertices are visible, and on each stage truncate a top vertex creating a new face and increasing the number of (visible) vertices by 2. (A top vertex here is a vertex such that all its three faces are visible.) So the answer is $2N-4$. $\endgroup$– Ilya BogdanovCommented Nov 13, 2020 at 8:35
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1 Answer
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Every polyhedron can be projectively transformed so that all its vertices are "beyond a single face":
Then every vertex of of the transformed $P$ can be seen from a single point outside. So, assuming that your question is asking for the maximum over all polyhedra with $N$ faces, your question is equivalent to "what is the maximal number of vertices of a polyhedron with $N$ faces".
And if I am not missing something, then the answer here is $2N-4$ (see also the comments below your question, or see this paper).
