Lymnmeatlypamgfm

2022-05-03

A stained glass window consists of nine squares of glass in a 3x3 array. Of the nine squares, k are red, the rest blue. A set of windows is produced such that any possible window can be formed in just one way by rotating and/or turning over one of the windows in the set. Altogether there are more than 100 red squares in the set. Find k.

first, there are 8 Isometries of a square.

Identity, three rotations (90,-90,180) four reflections (vertical, horizontal, two diagonal axis). let G be the permutation group, then |G|=8, and I can find fix(g) for every g.

can someone give me a hint of how to proceed from there.

first, there are 8 Isometries of a square.

Identity, three rotations (90,-90,180) four reflections (vertical, horizontal, two diagonal axis). let G be the permutation group, then |G|=8, and I can find fix(g) for every g.

can someone give me a hint of how to proceed from there.

Aliana Sexton

Beginner2022-05-04Added 20 answers

I upvoted the first answer but I would like to show how to compute the cycle index $Z(G)$ of the group $G$ of symmetries of the square and apply the Polya Enumeration Theorem to this problem.

We need to enumerate and factor into cycles the eight permutations that contribute to $Z(G).$

There is the identity, which contributes

${a}_{1}^{9}.$

The two 90 degree rotations contribute

$2{a}_{1}{a}_{4}^{2}.$

The 180 degree rotation contributes

${a}_{1}{a}_{2}^{4}.$

The vertical and horizontal reflections contribute

$2{a}_{1}^{3}{a}_{2}^{3}.$

The reflections in a diagonal contribute

$2{a}_{1}^{3}{a}_{2}^{3}.$

This yields the cycle index

$Z(G)=\frac{1}{8}({a}_{1}^{9}+2{a}_{1}{a}_{4}^{2}+{a}_{1}{a}_{2}^{4}+4{a}_{1}^{3}{a}_{2}^{3}).$

As we are interested in the red squares we evaluate

$Z(G)(1+R)$

to get

$1/8\phantom{\rule{thinmathspace}{0ex}}{(1+R)}^{9}+1/2\phantom{\rule{thinmathspace}{0ex}}{(1+R)}^{3}{({R}^{2}+1)}^{3}+1/8\phantom{\rule{thinmathspace}{0ex}}(1+R){({R}^{2}+1)}^{4}\phantom{\rule{0ex}{0ex}}+1/4\phantom{\rule{thinmathspace}{0ex}}(1+R){({R}^{4}+1)}^{2}$

which is

${R}^{9}+3\phantom{\rule{thinmathspace}{0ex}}{R}^{8}+8\phantom{\rule{thinmathspace}{0ex}}{R}^{7}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{6}+23\phantom{\rule{thinmathspace}{0ex}}{R}^{5}+23\phantom{\rule{thinmathspace}{0ex}}{R}^{4}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{3}+8\phantom{\rule{thinmathspace}{0ex}}{R}^{2}+3\phantom{\rule{thinmathspace}{0ex}}R+1.$

This is the classification of the orbits according to the number of red squares. Differentiate and multiply by $R$ to obtain the total count of the squares, which yields

$9\phantom{\rule{thinmathspace}{0ex}}{R}^{9}+24\phantom{\rule{thinmathspace}{0ex}}{R}^{8}+56\phantom{\rule{thinmathspace}{0ex}}{R}^{7}+96\phantom{\rule{thinmathspace}{0ex}}{R}^{6}+115\phantom{\rule{thinmathspace}{0ex}}{R}^{5}+92\phantom{\rule{thinmathspace}{0ex}}{R}^{4}+48\phantom{\rule{thinmathspace}{0ex}}{R}^{3}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{2}+3\phantom{\rule{thinmathspace}{0ex}}R$

which matches the accepted answer.

We need to enumerate and factor into cycles the eight permutations that contribute to $Z(G).$

There is the identity, which contributes

${a}_{1}^{9}.$

The two 90 degree rotations contribute

$2{a}_{1}{a}_{4}^{2}.$

The 180 degree rotation contributes

${a}_{1}{a}_{2}^{4}.$

The vertical and horizontal reflections contribute

$2{a}_{1}^{3}{a}_{2}^{3}.$

The reflections in a diagonal contribute

$2{a}_{1}^{3}{a}_{2}^{3}.$

This yields the cycle index

$Z(G)=\frac{1}{8}({a}_{1}^{9}+2{a}_{1}{a}_{4}^{2}+{a}_{1}{a}_{2}^{4}+4{a}_{1}^{3}{a}_{2}^{3}).$

As we are interested in the red squares we evaluate

$Z(G)(1+R)$

to get

$1/8\phantom{\rule{thinmathspace}{0ex}}{(1+R)}^{9}+1/2\phantom{\rule{thinmathspace}{0ex}}{(1+R)}^{3}{({R}^{2}+1)}^{3}+1/8\phantom{\rule{thinmathspace}{0ex}}(1+R){({R}^{2}+1)}^{4}\phantom{\rule{0ex}{0ex}}+1/4\phantom{\rule{thinmathspace}{0ex}}(1+R){({R}^{4}+1)}^{2}$

which is

${R}^{9}+3\phantom{\rule{thinmathspace}{0ex}}{R}^{8}+8\phantom{\rule{thinmathspace}{0ex}}{R}^{7}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{6}+23\phantom{\rule{thinmathspace}{0ex}}{R}^{5}+23\phantom{\rule{thinmathspace}{0ex}}{R}^{4}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{3}+8\phantom{\rule{thinmathspace}{0ex}}{R}^{2}+3\phantom{\rule{thinmathspace}{0ex}}R+1.$

This is the classification of the orbits according to the number of red squares. Differentiate and multiply by $R$ to obtain the total count of the squares, which yields

$9\phantom{\rule{thinmathspace}{0ex}}{R}^{9}+24\phantom{\rule{thinmathspace}{0ex}}{R}^{8}+56\phantom{\rule{thinmathspace}{0ex}}{R}^{7}+96\phantom{\rule{thinmathspace}{0ex}}{R}^{6}+115\phantom{\rule{thinmathspace}{0ex}}{R}^{5}+92\phantom{\rule{thinmathspace}{0ex}}{R}^{4}+48\phantom{\rule{thinmathspace}{0ex}}{R}^{3}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{2}+3\phantom{\rule{thinmathspace}{0ex}}R$

which matches the accepted answer.

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