Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the possible number of n sided polygons(every face is the same regular polygon) that touching their corners to sphere surface and also touching each other ? I would like to know the relation between n sides polygon and possible placing number.

And also I would like to find a (one side of polygon ) depends on r, n (number of sides), m (total number of polygons that placed on the sphere).

Examples: Can I place 17 regular triangles on a sphere surface?

Can I place 7 squares on a sphere surface?

Can I place 10 regular hexagons on a sphere surface?

I do not know the related tag about my question. Please feel free to retag it. Thanks a lot for answers.

enter image description here

share|improve this question
    
nice question ! How did it came up ? –  Selim Ghazouani Jun 12 '12 at 11:26
    
Thanks. It is just my idea. I would like to learn the related subjects to solve the question.. It is very interesting subject for me. –  Mathlover Jun 12 '12 at 11:34
1  
Would you like your regular polygons to form a polyhedron? If so, the answer is that there are only five regular polyhedra, the so-called, Platonic solids: dodecahedron, cube, tetrahedron, octahedron, and icosahedron. So it would have to be a subset of these. If you don't require them to be a polyhedron, then I'm not quite sure I understand the question. –  J. Loreaux Jun 12 '12 at 13:02
    
J.Loreaux:Thanks for references. Yes I mentioned the polyhedrons that were created with same regular polygons. Is it possible to create other platonic solids that are different from dodecahedron, cube, tetrahedron, octahedron, and icosahedron? If not, How can it be proved that is impossible? –  Mathlover Jun 12 '12 at 13:29
    
From en.wikipedia.org/wiki/Platonic_solid "a Platonic solid is a regular, convex polyhedron." To prove there are only five, one argues that all vertices must have some number of the same regular polygon meeting, and the sum of the angles must add to less than $360^\circ$. The only choices are then $3,4$ or $5$ triangles, $3$ squares, or $3$ pentagons. Then construction of the known ones shows they all exist. A search for inscribed sphere looking for information on other polyhedra that have one didn't turn up anything for me. –  Ross Millikan Jun 12 '12 at 13:54

1 Answer 1

up vote 2 down vote accepted

There is a special case where any $r$, $n$ and $m$ will work for some $a$: if you place the polygons along a single strip, $a$ just needs to be small enough so that the strip doesn't intersect itself. More generally, for any planar arrangement of polygons, if no vertex lies in the interior of the figure, you can embed it on the sphere for any small enough $a$.

Given your figure I don't think this is what you're looking for, so we'll require that (hypothesis $H_1$) some vertex lies in the interior, or equivalently there should be a loop of $k\ge 3$ faces around some vertex. Then you must have $2\pi/k>\pi-2\pi/n$ because of the sphere's curvature. There are exactly 5 $(n,k)$ pairs satisfying those constraints (in particular $3\le n\le 5$, so hexagons won't work).

For each of these pairs there is a Platonic solid: $$\begin{matrix} (n,k)&\text{Platonic solid $S_{nk}$}&a/r&m_0\\ \hline (3,3)&\text{tetrahedron}&\sqrt{8/3}&4\\ (3,4)&\text{octahedron}&\sqrt 2&8\\ (3,5)&\text{icosahedron}&\sqrt{2-2/\sqrt 5}&20\\ (4,3)&\text{cube}&2/\sqrt 3&6\\ (5,3)&\text{dodecahedron}&4/(\sqrt 3\cdot(1+\sqrt 5))&12 \end{matrix}$$

$(n,k)$ uniquely determines the angle between faces and thus the ratio $a/r$, in an injective way as can be seen above. So $k$ must be constant for all vertices lying in the interior.

Platonic solids are solutions, but is every solution a subset of a Platonic solid?

This is always true when $(n,k)=(3,3)$. In the other cases, we must require that the figure be connected, but this still leaves a degree of freedom in rotation around a vertex: so we must require that any two faces are path-connected through edges. Note that because the vertices lie on a sphere and edge length is constant, two edges from the figure are either equal or disjoint except for at most one point.

But given the sphere and a face $F$, the only faces sharing an edge with $F$ are the neighbors of $F$ in the unique Platonic solid containing $F$. So if we define the equivalence relation $F\sim G$ iff $F$ and $G$ are in the same Platonic solid, the connectivity hypothesis implies that all faces are in the same equivalence class, hence in the same Platonic solid. QED.

If we don't require the connectivity constraint, we must at least require that ($H_2$) the interior of the cones around the center of the sphere containing each face is disjoint (otherwise you can "stack" faces above each other, which is probably not what you want). The sum of the area of their intersections with the sphere is proportional to $m$ and maximal for the Platonic solid, proving that the Platonic solid maximizes $m$.

So, conditional to ($H_1$) and ($H_2$), the feasible $(a,r,n,m)$ are those where $a/r$ matches that of $S_{nk}$ for some $k$, and $k\le m\le m_0(S_{nk})$.

Thus:

  • you can place 17 triangles, by removing 3 triangles from an icosahedron
  • you can't place 7 squares
  • you can't place 10 hexagons
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.