[math-fun] Find a polyhedron with all vertices on its circumsphere, but without all faces touching its insphere
There is one with 6 faces, anyhow. (Far below if you want to try to find one without spoilage.) —Dan Let T be a skinny isosceles triangle with apex A. Let the polyhedron P be the result of truncating [the product T x [0, 1] of T with the unit interval] by cutting off a neighborhood N of {A} x [0, 1], thereby creating a new face — a skinny rectangle R. Then P has all 8 of its vertices on its circumsphere C. But clearly C would not touch the rectangle R if N is small enough. —Dan
Or simpler, just take any rectangular prism that is not a cube. Works in two dimensions too. Andy On Mon, May 27, 2019, 03:52 Dan Asimov <dasimov@earthlink.net> wrote:
There is one with 6 faces, anyhow.
(Far below if you want to try to find one without spoilage.)
—Dan
Let T be a skinny isosceles triangle with apex A.
Let the polyhedron P be the result of truncating [the product T x [0, 1] of T with the unit interval] by cutting off a neighborhood N of {A} x [0, 1], thereby creating a new face — a skinny rectangle R.
Then P has all 8 of its vertices on its circumsphere C. But clearly C would not touch the rectangle R if N is small enough.
—Dan
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There are lots of solutions to this. Any elongated prism will work. The insphere is not unique, but it can be made unique by tapering one end, forcing the insphere to the other end. The smallest is a triangular prism, with 5 faces. You can taper one end if you want to force a unique insphere. The face at the smaller end doesn't touch it. A specific instance of this is a tetrahedron with one truncated vertex. Just snip a tiny piece off of one vertex, leaving a little triangle that's parallel to the opposite face. The result is a 5-sided tapered triangular prism that satisfies the condition. More symmetrical, albeit larger, solutions can also be found. For instance, almost any Archimedean solid will work. All vertices are the same, so they all touch the circumsphere. But the faces differ, and in most cases some faces have their center closer to the center of the polyhedron than others, so only those faces touch the insphere. Example: a truncated tetrahedron (3.6.6). This has 8 faces (4 hexagons and 4 triangles). Only the hexagons touch the insphere: http://www.karzes.com/polyhedra/polyhedron.html?ph=3.6.6 As I said, I believe most of the other Archimedean solids work as well. Tom Andy Latto writes:
Or simpler, just take any rectangular prism that is not a cube. Works in two dimensions too.
Andy
On Mon, May 27, 2019, 03:52 Dan Asimov <dasimov@earthlink.net> wrote:
There is one with 6 faces, anyhow.
(Far below if you want to try to find one without spoilage.)
—Dan
...
Let T be a skinny isosceles triangle with apex A.
Let the polyhedron P be the result of truncating [the product T x [0, 1] of T with the unit interval] by cutting off a neighborhood N of {A} x [0, 1], thereby creating a new face — a skinny rectangle R.
Then P has all 8 of its vertices on its circumsphere C. But clearly C would not touch the rectangle R if N is small enough.
—Dan
Actually, I believe *all* Archimedean solids satisty the condition (note that they exclude the Platonic solids). All vertices in an Archimedean solid are the same distance from the center of the polyhedron. There are at least two different polygons among the faces, and only the largest of them will touch the insphere. Tom Tom Karzes writes:
There are lots of solutions to this. Any elongated prism will work. The insphere is not unique, but it can be made unique by tapering one end, forcing the insphere to the other end. The smallest is a triangular prism, with 5 faces. You can taper one end if you want to force a unique insphere. The face at the smaller end doesn't touch it.
A specific instance of this is a tetrahedron with one truncated vertex. Just snip a tiny piece off of one vertex, leaving a little triangle that's parallel to the opposite face. The result is a 5-sided tapered triangular prism that satisfies the condition.
More symmetrical, albeit larger, solutions can also be found. For instance, almost any Archimedean solid will work. All vertices are the same, so they all touch the circumsphere. But the faces differ, and in most cases some faces have their center closer to the center of the polyhedron than others, so only those faces touch the insphere.
Example: a truncated tetrahedron (3.6.6). This has 8 faces (4 hexagons and 4 triangles). Only the hexagons touch the insphere:
http://www.karzes.com/polyhedra/polyhedron.html?ph=3.6.6
As I said, I believe most of the other Archimedean solids work as well.
Tom
Andy Latto writes:
Or simpler, just take any rectangular prism that is not a cube. Works in two dimensions too.
Andy
On Mon, May 27, 2019, 03:52 Dan Asimov <dasimov@earthlink.net> wrote:
There is one with 6 faces, anyhow.
(Far below if you want to try to find one without spoilage.)
—Dan
...
Let T be a skinny isosceles triangle with apex A.
Let the polyhedron P be the result of truncating [the product T x [0, 1] of T with the unit interval] by cutting off a neighborhood N of {A} x [0, 1], thereby creating a new face — a skinny rectangle R.
Then P has all 8 of its vertices on its circumsphere C. But clearly C would not touch the rectangle R if N is small enough.
—Dan
doesn’t any square prism other than a cube also work? - Cris
On May 27, 2019, at 1:51 AM, Dan Asimov <dasimov@earthlink.net> wrote:
There is one with 6 faces, anyhow.
(Far below if you want to try to find one without spoilage.)
—Dan
Let T be a skinny isosceles triangle with apex A.
Let the polyhedron P be the result of truncating [the product T x [0, 1] of T with the unit interval] by cutting off a neighborhood N of {A} x [0, 1], thereby creating a new face — a skinny rectangle R.
Then P has all 8 of its vertices on its circumsphere C. But clearly C would not touch the rectangle R if N is small enough.
—Dan
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participants (4)
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Andy Latto -
Cris Moore -
Dan Asimov -
Tom Karzes