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We know in $\mathbb{R}^2$ a polygon with the least number of vertices whose diagonals enclose an interior region is a pentagon.

What about in $\mathbb{R}^3$? The least number of vertices that a polyhedron can have, such that its diagonal faces enclose an interior solid region?

Note: "interior" means the solid does not intersect the polyhedron surface. Well, we may consider another problem that the interior solid can have points or edges on the polyhedron surfaces, but no faces.

If $f(n)$ denotes the least number in $\mathbb{R}^n$, is there a formula for $f(n)$?

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  • $\begingroup$ @DavidK Yes, diagonal faces. Have fixed. $\endgroup$ Commented 11 hours ago

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In the Archimedean-solid truncated tetrahedron, with twelve vertices, four of the body diagonals enclose the edges of an internal regular tetrahedron. Each of the diagonal faces is parallel to a triangular faces of the outer polyhedron.

However, this is not the minimal vertex count. A pentagonal prism, with ten vertices, yields a fully enclosed pentagonal bipyramid when we draw the diagonal face from each basal edge to the opposing vertex on the opposite base. And a triangular prism gives a similar result with only six vertices, enclosing a triangular bipyramid.

If we allow the smaller polygon to intersect the larger one at vertices, the pentagonal bipyramid works with seven vertices. In this case a pentagram is made from the diagonals of the equatorial pentagon, and the bipyramd built from this and sharing the same ap8ces as the original byprramid encloses a smaller pentagonal bipyramid.

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  • $\begingroup$ Nice observation on the pentagonal prism. Is 10 vertices the minimal vertex count? $\endgroup$ Commented yesterday
  • $\begingroup$ Come to think of it, we might want to test the troligonal prism, which would beat even the bipyramid. $\endgroup$ Commented yesterday
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I come up with an idea that a regular tetrahedron can be enclosed by diagonal faces of a polyhedron of $7$ vertices. As shown in the figure, we want $ABCD$ to be the enclosed tetrahedron. The vertices of an <span class=$7$-vertices polyhedron that contains a tetrahedron" />

Define the 7 vertices as follows. Point 1 is in the plane region bounded by rays $CB$ and $DB$; 2 is on ray $BC$; 3 is on ray $BD$; 4 is on ray $CA$; 5 is on ray $AB$; 6 is on ray $DA$; 7 is in the plane region bounded by rays $AC$ and $AD$.

(1) $\triangle BCD$ is contained in the triangle 1,2,3. Points 4,6 on one side of this plane; 5,7 on the other side.

(2) $\triangle ABC$ is contained in the triangle 2,4,5. Points 1,6 on one side and 3,7 on the other side.

(3) $\triangle ABD$ is contained in the triangle 3,5,6. Points 1,4 on one side and 2,7 on the other side.

(4) $\triangle ACD$ is contained in the triangle 4,6,7, provided that 4,6 are sufficiently far from point $A$. Points 2,3 on one side and 1,5 on the other side.

With at least one point on each side of the plane, the triangle is on a diagonal face, not surface face. Thus, the tetrahedron is inside the polyhedron.

So far I don't know if $7$ is the least number. If not, the desired polyhedron must have at least $4$ diagonal faces to enclose the interior solid, each face containing $3$ vertices. If no three diagonal planes intersect at a line, then a vertex lies on at most two planes (faces). But one can check that $6$ vertices are not enough.

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  • $\begingroup$ Six vertices is enough to enclose a triangular bipyramid instead, to which my answer alludes. $\endgroup$ Commented 10 hours ago

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