8
$\begingroup$

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 $f(n)$?

$\endgroup$
0

1 Answer 1

Reset to default
7
$\begingroup$

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.

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.

$\endgroup$
2
  • $\begingroup$ Nice observation on the pentagonal prism. Is 10 vertices the minimal vertex count? $\endgroup$ Commented 20 hours ago
  • $\begingroup$ Come to think of it, we might want to test the troligonal prism, which would beat even the bipyramid. $\endgroup$ Commented 18 hours ago

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.