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I have a pentagonal shape:

enter image description here

I have to find how many different figures can be made, if the star is regarded the same upon rotation and reflection such that each piece can be black or blue.

by each piece i mean the triangles and the central pentagon.

Before I can apply the counting theorem, I need to find the order of the symmetry group of the figure. My guess is that I could superimpose this onto a regular pentagon then apply the elements of the group of symmetries of the regular pentagon to this shape.

The only thing that is preventing me from doing this is that the isosceles triangles have different side lengths, so by rotating the figure by $2 \pi /5$, the space occupied by the figure has changed, hasn't it?

My question is, how should I approach this?

EDIT

Would it be correct to say that this group has 10 symmetries, 5 reflections, 5 rotations and the identity?

I am assuming this is correct

I have deduced the order of each element in the symmetry group.

I have applied the counting theorem and found that the number of orbits is 16.

So the number of possible stars is 16. - Is this correct?

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  • $\begingroup$ If you are not that interested in the Euclidean (metric) structure you could just note that as a graph this is just the inverse of a regular pentagon and as such has (as a graph) the dihedral group of order $10$ as its group of automorphism. Without really checking this I would guess that this is still true in the metric sense. $\endgroup$ – Sebastian Schoennenbeck Mar 13 '15 at 12:33
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    $\begingroup$ What are the "pieces" (vertices, edges, etc.) that you wish to color? $\endgroup$ – Christian Blatter Mar 13 '15 at 13:04
  • $\begingroup$ The triangles/central pentagon. so the order of the fixed set of the identity would be $$|Fix(identity)| = 2^6$$ $\endgroup$ – user214138 Mar 13 '15 at 13:20
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To color just the triangles, I see one coloring with no black triangles, one coloring (up to symmetry) with one black triangle, and two colorings with two black triangles (one with triangles on adjacent sides of the pentagon, one with triangles not on adjacent sides). That's four so far.

The other colorings of the triangles have either on blue triangles, one blue triangle, or two blue triangles. So four of them.

For each of these $8$ colorings I can choose either blue or black for the pentagon. So by (relatively) brute-force counting, there are $16$ possible colorings.

This is just a confirmation of the result. Your approach seems better.

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