For a circle of radius 2, the outline equals the area enclosed by that outline.
For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
the outline of the polygon also equals the area enclosed by that outline.

For a sphere of radius 3, the surface area of the sphere equals the volume
enclosed by that surface area. Does that give us any reason to assume
that a platonic solid inscribed by that sphere also has this property
(the surface area being equal to the volume enclosed by that area)?

On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
<dohduhdah@yahoo.com> wrote:

>For a circle of radius 2, the outline equals the area enclosed by that outline.
>For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
>the outline of the polygon also equals the area enclosed by that outline.

Any polygon with each side tangent to the circle can be divided into
triangles by adding lines from its vertices to the centre of the
circle. With the sides of the polygon as base the height of the
triangles equals the radius of the circle.

>For a sphere of radius 3, the surface area of the sphere equals the volume
>enclosed by that surface area. Does that give us any reason to assume
>that a platonic solid inscribed by that sphere also has this property
>(the surface area being equal to the volume enclosed by that area)?

Similar to the 2-dimensional case, the solid polyhedron can be divided
into pyramids with the sphere's centre as top. Then the volume of each
pyramid is 1/3 * height * face area.

On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> <dohd...@yahoo.com> wrote:
>
> >For a circle of radius 2, the outline equals the area enclosed by that outline.
> >For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> >the outline of the polygon also equals the area enclosed by that outline.
> Any polygon with each side tangent to the circle can be divided into
> triangles by adding lines from its vertices to the centre of the
> circle. With the sides of the polygon as base the height of the
> triangles equals the radius of the circle.

https://www.desmos.com/calculator/uwankdkqwp

> >For a sphere of radius 3, the surface area of the sphere equals the volume
> >enclosed by that surface area. Does that give us any reason to assume
> >that a platonic solid inscribed by that sphere also has this property
> >(the surface area being equal to the volume enclosed by that area)?
> Similar to the 2-dimensional case, the solid polyhedron can be divided
> into pyramids with the sphere's centre as top. Then the volume of each
> pyramid is 1/3 * height * face area.

I guess this should be possible to work out with Geogebra instead of desmos in 3D..

On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> <dohd...@yahoo.com> wrote:
>
> >For a circle of radius 2, the outline equals the area enclosed by that outline.
> >For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> >the outline of the polygon also equals the area enclosed by that outline.
> Any polygon with each side tangent to the circle can be divided into
> triangles by adding lines from its vertices to the centre of the
> circle. With the sides of the polygon as base the height of the
> triangles equals the radius of the circle.
> >For a sphere of radius 3, the surface area of the sphere equals the volume
> >enclosed by that surface area. Does that give us any reason to assume
> >that a platonic solid inscribed by that sphere also has this property
> >(the surface area being equal to the volume enclosed by that area)?
> Similar to the 2-dimensional case, the solid polyhedron can be divided
> into pyramids with the sphere's centre as top. Then the volume of each
> pyramid is 1/3 * height * face area.

The argument does make sense though for the 3D case.. so it should hold for
any polygonal shape tangentially centered on a sphere with radius three,
such that these polygonal shapes collectively make up some volume in
3D, like in case of a snub dodacahedron (where there are multiple
polygonal face shapes rather than a single polygonal face shape in the
case of a platonic solid).

https://www.geogebra.org/m/zv2tYKxZ

On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> > On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> > <dohd...@yahoo.com> wrote:
> >
> > >For a circle of radius 2, the outline equals the area enclosed by that outline.
> > >For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> > >the outline of the polygon also equals the area enclosed by that outline.
> > Any polygon with each side tangent to the circle can be divided into
> > triangles by adding lines from its vertices to the centre of the
> > circle. With the sides of the polygon as base the height of the
> > triangles equals the radius of the circle.
> > >For a sphere of radius 3, the surface area of the sphere equals the volume
> > >enclosed by that surface area. Does that give us any reason to assume
> > >that a platonic solid inscribed by that sphere also has this property
> > >(the surface area being equal to the volume enclosed by that area)?
> > Similar to the 2-dimensional case, the solid polyhedron can be divided
> > into pyramids with the sphere's centre as top. Then the volume of each
> > pyramid is 1/3 * height * face area.
> The argument does make sense though for the 3D case.. so it should hold for
> any polygonal shape tangentially centered on a sphere with radius three,
> such that these polygonal shapes collectively make up some volume in
> 3D, like in case of a snub dodacahedron (where there are multiple
> polygonal face shapes rather than a single polygonal face shape in the
> case of a platonic solid).
>
> https://www.geogebra.org/m/zv2tYKxZ

Oh wait, maybe not, I guess the distance from the origin to the center of the faces
will not be equal in case the faces have different (but regular) shapes.
So I guess it would only work for the platonic solids.

On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
<dohduhdah@yahoo.com> wrote:

>On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
>> > On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
>> > <dohd...@yahoo.com> wrote:
>> >
>> > >For a circle of radius 2, the outline equals the area enclosed by that outline.
>> > >For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
>> > >the outline of the polygon also equals the area enclosed by that outline.
>> > Any polygon with each side tangent to the circle can be divided into
>> > triangles by adding lines from its vertices to the centre of the
>> > circle. With the sides of the polygon as base the height of the
>> > triangles equals the radius of the circle.
>> > >For a sphere of radius 3, the surface area of the sphere equals the volume
>> > >enclosed by that surface area. Does that give us any reason to assume
>> > >that a platonic solid inscribed by that sphere also has this property
>> > >(the surface area being equal to the volume enclosed by that area)?
>> > Similar to the 2-dimensional case, the solid polyhedron can be divided
>> > into pyramids with the sphere's centre as top. Then the volume of each
>> > pyramid is 1/3 * height * face area.
>> The argument does make sense though for the 3D case.. so it should hold for
>> any polygonal shape tangentially centered on a sphere with radius three,
>> such that these polygonal shapes collectively make up some volume in
>> 3D, like in case of a snub dodacahedron (where there are multiple
>> polygonal face shapes rather than a single polygonal face shape in the
>> case of a platonic solid).
>>
>> https://www.geogebra.org/m/zv2tYKxZ
>
>Oh wait, maybe not, I guess the distance from the origin to the center of the faces
>will not be equal in case the faces have different (but regular) shapes.
>So I guess it would only work for the platonic solids.

The line from the sphere's centre to the tangent point is
perpendicular to the sphere at that point, so it is also perpendicular
to the face.

On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
> <dohd...@yahoo.com> wrote:
>
> >On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
> >> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> >> > On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> >> > <dohd...@yahoo.com> wrote:
> >> >
> >> > >For a circle of radius 2, the outline equals the area enclosed by that outline.
> >> > >For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> >> > >the outline of the polygon also equals the area enclosed by that outline.
> >> > Any polygon with each side tangent to the circle can be divided into
> >> > triangles by adding lines from its vertices to the centre of the
> >> > circle. With the sides of the polygon as base the height of the
> >> > triangles equals the radius of the circle.
> >> > >For a sphere of radius 3, the surface area of the sphere equals the volume
> >> > >enclosed by that surface area. Does that give us any reason to assume
> >> > >that a platonic solid inscribed by that sphere also has this property
> >> > >(the surface area being equal to the volume enclosed by that area)?
> >> > Similar to the 2-dimensional case, the solid polyhedron can be divided
> >> > into pyramids with the sphere's centre as top. Then the volume of each
> >> > pyramid is 1/3 * height * face area.
> >> The argument does make sense though for the 3D case.. so it should hold for
> >> any polygonal shape tangentially centered on a sphere with radius three,
> >> such that these polygonal shapes collectively make up some volume in
> >> 3D, like in case of a snub dodacahedron (where there are multiple
> >> polygonal face shapes rather than a single polygonal face shape in the
> >> case of a platonic solid).
> >>
> >> https://www.geogebra.org/m/zv2tYKxZ
> >
> >Oh wait, maybe not, I guess the distance from the origin to the center of the faces
> >will not be equal in case the faces have different (but regular) shapes.
> >So I guess it would only work for the platonic solids.
> The line from the sphere's centre to the tangent point is
> perpendicular to the sphere at that point, so it is also perpendicular
> to the face.

Sure, but in 2D it's obviously not working if the sides are not all the same length, so
I don't see how it could be working in 3D.
https://www.desmos.com/calculator/oxrlmp99fn

Can you have two or more different regular polygon shapes that make up a volume such that
the centers of each of these polygons are all equidistant from the origin?

On Wednesday, January 12, 2022 at 1:10:59 AM UTC+1, sobriquet wrote:
> On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
> > On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
> > <dohd...@yahoo.com> wrote:
> >
> > >On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
> > >> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> > >> > On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> > >> > <dohd...@yahoo.com> wrote:
> > >> >
> > >> > >For a circle of radius 2, the outline equals the area enclosed by that outline.
> > >> > >For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> > >> > >the outline of the polygon also equals the area enclosed by that outline.
> > >> > Any polygon with each side tangent to the circle can be divided into
> > >> > triangles by adding lines from its vertices to the centre of the
> > >> > circle. With the sides of the polygon as base the height of the
> > >> > triangles equals the radius of the circle.
> > >> > >For a sphere of radius 3, the surface area of the sphere equals the volume
> > >> > >enclosed by that surface area. Does that give us any reason to assume
> > >> > >that a platonic solid inscribed by that sphere also has this property
> > >> > >(the surface area being equal to the volume enclosed by that area)?
> > >> > Similar to the 2-dimensional case, the solid polyhedron can be divided
> > >> > into pyramids with the sphere's centre as top. Then the volume of each
> > >> > pyramid is 1/3 * height * face area.
> > >> The argument does make sense though for the 3D case.. so it should hold for
> > >> any polygonal shape tangentially centered on a sphere with radius three,
> > >> such that these polygonal shapes collectively make up some volume in
> > >> 3D, like in case of a snub dodacahedron (where there are multiple
> > >> polygonal face shapes rather than a single polygonal face shape in the
> > >> case of a platonic solid).
> > >>
> > >> https://www.geogebra.org/m/zv2tYKxZ
> > >
> > >Oh wait, maybe not, I guess the distance from the origin to the center of the faces
> > >will not be equal in case the faces have different (but regular) shapes.
> > >So I guess it would only work for the platonic solids.
> > The line from the sphere's centre to the tangent point is
> > perpendicular to the sphere at that point, so it is also perpendicular
> > to the face.
> Sure, but in 2D it's obviously not working if the sides are not all the same length, so
> I don't see how it could be working in 3D.
> https://www.desmos.com/calculator/oxrlmp99fn

https://www.desmos.com/calculator/mgeihlxc4s
There is no circle going through the red points (the midpoints of faces of an irregular
polygon).

>
> Can you have two or more different regular polygon shapes that make up a volume such that
> the centers of each of these polygons are all equidistant from the origin?

On 12/01/2022 00:10, sobriquet wrote:
> On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
>> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
>> <dohd...@yahoo.com> wrote:
>>
>>> On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
>>>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
>>>>> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
>>>>> <dohd...@yahoo.com> wrote:
>>>>>
>>>>>> For a circle of radius 2, the outline equals the area enclosed by that outline.
>>>>>> For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
>>>>>> the outline of the polygon also equals the area enclosed by that outline.
>>>>> Any polygon with each side tangent to the circle can be divided into
>>>>> triangles by adding lines from its vertices to the centre of the
>>>>> circle. With the sides of the polygon as base the height of the
>>>>> triangles equals the radius of the circle.
>>>>>> For a sphere of radius 3, the surface area of the sphere equals the volume
>>>>>> enclosed by that surface area. Does that give us any reason to assume
>>>>>> that a platonic solid inscribed by that sphere also has this property
>>>>>> (the surface area being equal to the volume enclosed by that area)?
>>>>> Similar to the 2-dimensional case, the solid polyhedron can be divided
>>>>> into pyramids with the sphere's centre as top. Then the volume of each
>>>>> pyramid is 1/3 * height * face area.
>>>> The argument does make sense though for the 3D case.. so it should hold for
>>>> any polygonal shape tangentially centered on a sphere with radius three,
>>>> such that these polygonal shapes collectively make up some volume in
>>>> 3D, like in case of a snub dodacahedron (where there are multiple
>>>> polygonal face shapes rather than a single polygonal face shape in the
>>>> case of a platonic solid).
>>>>
>>>> https://www.geogebra.org/m/zv2tYKxZ
>>>
>>> Oh wait, maybe not, I guess the distance from the origin to the center of the faces
>>> will not be equal in case the faces have different (but regular) shapes.
>>> So I guess it would only work for the platonic solids.
>> The line from the sphere's centre to the tangent point is
>> perpendicular to the sphere at that point, so it is also perpendicular
>> to the face.
>
> Sure, but in 2D it's obviously not working if the sides are not all the same length, so
> I don't see how it could be working in 3D.
> https://www.desmos.com/calculator/oxrlmp99fn

No, it works for such polygons just the same. I don't get what the desmos link is supposed to show
- the polygon in demo is /inside/ the circle! (It needs to be outside...)

Mike.

>
> Can you have two or more different regular polygon shapes that make up a volume such that
> the centers of each of these polygons are all equidistant from the origin?
>

On 12/01/2022 00:21, sobriquet wrote:
> On Wednesday, January 12, 2022 at 1:10:59 AM UTC+1, sobriquet wrote:
>> On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
>>> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
>>> <dohd...@yahoo.com> wrote:
>>>
>>>> On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
>>>>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
>>>>>> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
>>>>>> <dohd...@yahoo.com> wrote:
>>>>>>
>>>>>>> For a circle of radius 2, the outline equals the area enclosed by that outline.
>>>>>>> For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
>>>>>>> the outline of the polygon also equals the area enclosed by that outline.
>>>>>> Any polygon with each side tangent to the circle can be divided into
>>>>>> triangles by adding lines from its vertices to the centre of the
>>>>>> circle. With the sides of the polygon as base the height of the
>>>>>> triangles equals the radius of the circle.
>>>>>>> For a sphere of radius 3, the surface area of the sphere equals the volume
>>>>>>> enclosed by that surface area. Does that give us any reason to assume
>>>>>>> that a platonic solid inscribed by that sphere also has this property
>>>>>>> (the surface area being equal to the volume enclosed by that area)?
>>>>>> Similar to the 2-dimensional case, the solid polyhedron can be divided
>>>>>> into pyramids with the sphere's centre as top. Then the volume of each
>>>>>> pyramid is 1/3 * height * face area.
>>>>> The argument does make sense though for the 3D case.. so it should hold for
>>>>> any polygonal shape tangentially centered on a sphere with radius three,
>>>>> such that these polygonal shapes collectively make up some volume in
>>>>> 3D, like in case of a snub dodacahedron (where there are multiple
>>>>> polygonal face shapes rather than a single polygonal face shape in the
>>>>> case of a platonic solid).
>>>>>
>>>>> https://www.geogebra.org/m/zv2tYKxZ
>>>>
>>>> Oh wait, maybe not, I guess the distance from the origin to the center of the faces
>>>> will not be equal in case the faces have different (but regular) shapes.
>>>> So I guess it would only work for the platonic solids.
>>> The line from the sphere's centre to the tangent point is
>>> perpendicular to the sphere at that point, so it is also perpendicular
>>> to the face.
>> Sure, but in 2D it's obviously not working if the sides are not all the same length, so
>> I don't see how it could be working in 3D.
>> https://www.desmos.com/calculator/oxrlmp99fn
>
> https://www.desmos.com/calculator/mgeihlxc4s
> There is no circle going through the red points (the midpoints of faces of an irregular
> polygon).
>

There isn't required to be any such circle. You need to start with the circle, and draw a bunch of
tangents that together comprise the polygon. (So the circle "inscribes" the polygon - it touches
each of the lines of the polygon.)

>>
>> Can you have two or more different regular polygon shapes that make up a volume such that
>> the centers of each of these polygons are all equidistant from the origin?

On Wednesday, January 12, 2022 at 1:26:19 AM UTC+1, Mike Terry wrote:
> On 12/01/2022 00:10, sobriquet wrote:
> > On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
> >> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
> >> <dohd...@yahoo.com> wrote:
> >>
> >>> On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
> >>>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> >>>>> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> >>>>> <dohd...@yahoo.com> wrote:
> >>>>>
> >>>>>> For a circle of radius 2, the outline equals the area enclosed by that outline.
> >>>>>> For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> >>>>>> the outline of the polygon also equals the area enclosed by that outline.
> >>>>> Any polygon with each side tangent to the circle can be divided into
> >>>>> triangles by adding lines from its vertices to the centre of the
> >>>>> circle. With the sides of the polygon as base the height of the
> >>>>> triangles equals the radius of the circle.
> >>>>>> For a sphere of radius 3, the surface area of the sphere equals the volume
> >>>>>> enclosed by that surface area. Does that give us any reason to assume
> >>>>>> that a platonic solid inscribed by that sphere also has this property
> >>>>>> (the surface area being equal to the volume enclosed by that area)?
> >>>>> Similar to the 2-dimensional case, the solid polyhedron can be divided
> >>>>> into pyramids with the sphere's centre as top. Then the volume of each
> >>>>> pyramid is 1/3 * height * face area.
> >>>> The argument does make sense though for the 3D case.. so it should hold for
> >>>> any polygonal shape tangentially centered on a sphere with radius three,
> >>>> such that these polygonal shapes collectively make up some volume in
> >>>> 3D, like in case of a snub dodacahedron (where there are multiple
> >>>> polygonal face shapes rather than a single polygonal face shape in the
> >>>> case of a platonic solid).
> >>>>
> >>>> https://www.geogebra.org/m/zv2tYKxZ
> >>>
> >>> Oh wait, maybe not, I guess the distance from the origin to the center of the faces
> >>> will not be equal in case the faces have different (but regular) shapes.
> >>> So I guess it would only work for the platonic solids.
> >> The line from the sphere's centre to the tangent point is
> >> perpendicular to the sphere at that point, so it is also perpendicular
> >> to the face.
> >
> > Sure, but in 2D it's obviously not working if the sides are not all the same length, so
> > I don't see how it could be working in 3D.
> > https://www.desmos.com/calculator/oxrlmp99fn
> No, it works for such polygons just the same. I don't get what the desmos link is supposed to show
> - the polygon in demo is /inside/ the circle! (It needs to be outside...)
>
> Mike.

Yeah, that's why I posted the follow-up with the midpoints highlighted of an irregular polygon.
Ok, not all irregular convex polygon shapes are represented by random points on a unit circle,
But it already shows that for the vast majority of irregular convex polygon shapes that
can be obtained in this fashion, you will not be able to draw a circle through the midpoints
of the edges of such an irregular polygon.

> >
> > Can you have two or more different regular polygon shapes that make up a volume such that
> > the centers of each of these polygons are all equidistant from the origin?
> >

On Wednesday, January 12, 2022 at 1:29:29 AM UTC+1, Mike Terry wrote:
> On 12/01/2022 00:21, sobriquet wrote:
> > On Wednesday, January 12, 2022 at 1:10:59 AM UTC+1, sobriquet wrote:
> >> On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
> >>> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
> >>> <dohd...@yahoo.com> wrote:
> >>>
> >>>> On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
> >>>>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> >>>>>> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> >>>>>> <dohd...@yahoo.com> wrote:
> >>>>>>
> >>>>>>> For a circle of radius 2, the outline equals the area enclosed by that outline.
> >>>>>>> For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> >>>>>>> the outline of the polygon also equals the area enclosed by that outline.
> >>>>>> Any polygon with each side tangent to the circle can be divided into
> >>>>>> triangles by adding lines from its vertices to the centre of the
> >>>>>> circle. With the sides of the polygon as base the height of the
> >>>>>> triangles equals the radius of the circle.
> >>>>>>> For a sphere of radius 3, the surface area of the sphere equals the volume
> >>>>>>> enclosed by that surface area. Does that give us any reason to assume
> >>>>>>> that a platonic solid inscribed by that sphere also has this property
> >>>>>>> (the surface area being equal to the volume enclosed by that area)?
> >>>>>> Similar to the 2-dimensional case, the solid polyhedron can be divided
> >>>>>> into pyramids with the sphere's centre as top. Then the volume of each
> >>>>>> pyramid is 1/3 * height * face area.
> >>>>> The argument does make sense though for the 3D case.. so it should hold for
> >>>>> any polygonal shape tangentially centered on a sphere with radius three,
> >>>>> such that these polygonal shapes collectively make up some volume in
> >>>>> 3D, like in case of a snub dodacahedron (where there are multiple
> >>>>> polygonal face shapes rather than a single polygonal face shape in the
> >>>>> case of a platonic solid).
> >>>>>
> >>>>> https://www.geogebra.org/m/zv2tYKxZ
> >>>>
> >>>> Oh wait, maybe not, I guess the distance from the origin to the center of the faces
> >>>> will not be equal in case the faces have different (but regular) shapes.
> >>>> So I guess it would only work for the platonic solids.
> >>> The line from the sphere's centre to the tangent point is
> >>> perpendicular to the sphere at that point, so it is also perpendicular
> >>> to the face.
> >> Sure, but in 2D it's obviously not working if the sides are not all the same length, so
> >> I don't see how it could be working in 3D.
> >> https://www.desmos.com/calculator/oxrlmp99fn
> >
> > https://www.desmos.com/calculator/mgeihlxc4s
> > There is no circle going through the red points (the midpoints of faces of an irregular
> > polygon).
> >
> There isn't required to be any such circle. You need to start with the circle, and draw a bunch of
> tangents that together comprise the polygon. (So the circle "inscribes" the polygon - it touches
> each of the lines of the polygon.)

Ok, but then the points where the tangents touch the circle will not be the midpoints of the
edges of the irregular polygon.

On 12/01/2022 00:29, Mike Terry wrote:
> On 12/01/2022 00:21, sobriquet wrote:
>> On Wednesday, January 12, 2022 at 1:10:59 AM UTC+1, sobriquet wrote:
>>> On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
>>>> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
>>>> <dohd...@yahoo.com> wrote:
>>>>
>>>>> On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
>>>>>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
>>>>>>> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
>>>>>>> <dohd...@yahoo.com> wrote:
>>>>>>>
>>>>>>>> For a circle of radius 2, the outline equals the area enclosed by that outline.
>>>>>>>> For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
>>>>>>>> the outline of the polygon also equals the area enclosed by that outline.
>>>>>>> Any polygon with each side tangent to the circle can be divided into
>>>>>>> triangles by adding lines from its vertices to the centre of the
>>>>>>> circle. With the sides of the polygon as base the height of the
>>>>>>> triangles equals the radius of the circle.
>>>>>>>> For a sphere of radius 3, the surface area of the sphere equals the volume
>>>>>>>> enclosed by that surface area. Does that give us any reason to assume
>>>>>>>> that a platonic solid inscribed by that sphere also has this property
>>>>>>>> (the surface area being equal to the volume enclosed by that area)?
>>>>>>> Similar to the 2-dimensional case, the solid polyhedron can be divided
>>>>>>> into pyramids with the sphere's centre as top. Then the volume of each
>>>>>>> pyramid is 1/3 * height * face area.
>>>>>> The argument does make sense though for the 3D case.. so it should hold for
>>>>>> any polygonal shape tangentially centered on a sphere with radius three,
>>>>>> such that these polygonal shapes collectively make up some volume in
>>>>>> 3D, like in case of a snub dodacahedron (where there are multiple
>>>>>> polygonal face shapes rather than a single polygonal face shape in the
>>>>>> case of a platonic solid).
>>>>>>
>>>>>> https://www.geogebra.org/m/zv2tYKxZ
>>>>>
>>>>> Oh wait, maybe not, I guess the distance from the origin to the center of the faces
>>>>> will not be equal in case the faces have different (but regular) shapes.
>>>>> So I guess it would only work for the platonic solids.
>>>> The line from the sphere's centre to the tangent point is
>>>> perpendicular to the sphere at that point, so it is also perpendicular
>>>> to the face.
>>> Sure, but in 2D it's obviously not working if the sides are not all the same length, so
>>> I don't see how it could be working in 3D.
>>> https://www.desmos.com/calculator/oxrlmp99fn
>>
>> https://www.desmos.com/calculator/mgeihlxc4s
>> There is no circle going through the red points (the midpoints of faces of an irregular
>> polygon).
>>
>
> There isn't required to be any such circle.  You need to start with the circle, and draw a bunch of
> tangents that together comprise the polygon.  (So the circle "inscribes" the polygon - it touches
> each of the lines of the polygon. >

...should have added, the point at which the circle touches the lines won't in general be the
midpoints of those lines - that only occurs with the case of a regular polygon.

>>>
>>> Can you have two or more different regular polygon shapes that make up a volume such that
>>> the centers of each of these polygons are all equidistant from the origin?
>

On Wednesday, January 12, 2022 at 1:45:28 AM UTC+1, Mike Terry wrote:
> On 12/01/2022 00:29, Mike Terry wrote:
> > On 12/01/2022 00:21, sobriquet wrote:
> >> On Wednesday, January 12, 2022 at 1:10:59 AM UTC+1, sobriquet wrote:
> >>> On Wednesday, January 12, 2022 at 12:24:31 AM UTC+1, Leon Aigret wrote:
> >>>> On Tue, 11 Jan 2022 11:09:42 -0800 (PST), sobriquet
> >>>> <dohd...@yahoo.com> wrote:
> >>>>
> >>>>> On Tuesday, January 11, 2022 at 7:46:30 PM UTC+1, sobriquet wrote:
> >>>>>> On Tuesday, January 11, 2022 at 2:59:28 PM UTC+1, Leon Aigret wrote:
> >>>>>>> On Sat, 8 Jan 2022 15:57:49 -0800 (PST), sobriquet
> >>>>>>> <dohd...@yahoo.com> wrote:
> >>>>>>>
> >>>>>>>> For a circle of radius 2, the outline equals the area enclosed by that outline.
> >>>>>>>> For any regular polygon with sides 3, 4, 5, 6, ... that this circle inscribes,
> >>>>>>>> the outline of the polygon also equals the area enclosed by that outline.
> >>>>>>> Any polygon with each side tangent to the circle can be divided into
> >>>>>>> triangles by adding lines from its vertices to the centre of the
> >>>>>>> circle. With the sides of the polygon as base the height of the
> >>>>>>> triangles equals the radius of the circle.
> >>>>>>>> For a sphere of radius 3, the surface area of the sphere equals the volume
> >>>>>>>> enclosed by that surface area. Does that give us any reason to assume
> >>>>>>>> that a platonic solid inscribed by that sphere also has this property
> >>>>>>>> (the surface area being equal to the volume enclosed by that area)?
> >>>>>>> Similar to the 2-dimensional case, the solid polyhedron can be divided
> >>>>>>> into pyramids with the sphere's centre as top. Then the volume of each
> >>>>>>> pyramid is 1/3 * height * face area.
> >>>>>> The argument does make sense though for the 3D case.. so it should hold for
> >>>>>> any polygonal shape tangentially centered on a sphere with radius three,
> >>>>>> such that these polygonal shapes collectively make up some volume in
> >>>>>> 3D, like in case of a snub dodacahedron (where there are multiple
> >>>>>> polygonal face shapes rather than a single polygonal face shape in the
> >>>>>> case of a platonic solid).
> >>>>>>
> >>>>>> https://www.geogebra.org/m/zv2tYKxZ
> >>>>>
> >>>>> Oh wait, maybe not, I guess the distance from the origin to the center of the faces
> >>>>> will not be equal in case the faces have different (but regular) shapes.
> >>>>> So I guess it would only work for the platonic solids.
> >>>> The line from the sphere's centre to the tangent point is
> >>>> perpendicular to the sphere at that point, so it is also perpendicular
> >>>> to the face.
> >>> Sure, but in 2D it's obviously not working if the sides are not all the same length, so
> >>> I don't see how it could be working in 3D.
> >>> https://www.desmos.com/calculator/oxrlmp99fn
> >>
> >> https://www.desmos.com/calculator/mgeihlxc4s
> >> There is no circle going through the red points (the midpoints of faces of an irregular
> >> polygon).
> >>
> >
> > There isn't required to be any such circle. You need to start with the circle, and draw a bunch of
> > tangents that together comprise the polygon. (So the circle "inscribes" the polygon - it touches
> > each of the lines of the polygon. >
> ..should have added, the point at which the circle touches the lines won't in general be the
> midpoints of those lines - that only occurs with the case of a regular polygon.

Ok, my mistake, in case of an irregular polygon inscribed by a circle of radius 2, the
area equals the outline just the same.

https://www.desmos.com/calculator/fxdzbzehsa

Numerical comparison of things of different units (length, area) is
nonsense. Is a half dozen eggs equal to 6 miles per hour?

--
*********** To reply by e-mail, make w single in address **************

On Wednesday, January 12, 2022 at 10:52:58 AM UTC-8, Ian Gay wrote:
> Numerical comparison of things of different units (length, area) is
> nonsense. Is a half dozen eggs equal to 6 miles per hour?
>
>
>
> --
> *********** To reply by e-mail, make w single in address **************

How do you fit a zero point into the infinitesimal fundamental first real...?

On Wednesday, January 12, 2022 at 7:52:58 PM UTC+1, Ian Gay wrote:
> Numerical comparison of things of different units (length, area) is
> nonsense. Is a half dozen eggs equal to 6 miles per hour?
>
>
>
> --
> *********** To reply by e-mail, make w single in address **************

So basically you're saying that knowing the outline of an area to be finite
doesn't allow us to infer that the area enclosed by that outline must be
finite as well?

I only know of examples of finite area enclosed by an infinite outline
(like the Koch snowflake).

https://en.wikipedia.org/wiki/Koch_snowflake

The thread make me remember a particular example in a lesson in Measure Theory.

....but, yeap, dividing the 'volume' by the 'surface' in 'regular polytopes',
and its relationship with the inradius, nearby the present topic...

https://en.wikipedia.org/wiki/Inscribed_sphere