 # A stained glass window consists of nine squares of glass in a 3x3 array. Of the nine squares, k are Lymnmeatlypamgfm 2022-05-03 Answered
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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Aliana Sexton
I upvoted the first answer but I would like to show how to compute the cycle index $Z\left(G\right)$ 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 $Z\left(G\right).$

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
$Z\left(G\right)=\frac{1}{8}\left({a}_{1}^{9}+2{a}_{1}{a}_{4}^{2}+{a}_{1}{a}_{2}^{4}+4{a}_{1}^{3}{a}_{2}^{3}\right).$
As we are interested in the red squares we evaluate
$Z\left(G\right)\left(1+R\right)$
to get
$1/8\phantom{\rule{thinmathspace}{0ex}}{\left(1+R\right)}^{9}+1/2\phantom{\rule{thinmathspace}{0ex}}{\left(1+R\right)}^{3}{\left({R}^{2}+1\right)}^{3}+1/8\phantom{\rule{thinmathspace}{0ex}}\left(1+R\right){\left({R}^{2}+1\right)}^{4}\phantom{\rule{0ex}{0ex}}+1/4\phantom{\rule{thinmathspace}{0ex}}\left(1+R\right){\left({R}^{4}+1\right)}^{2}$
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
$9\phantom{\rule{thinmathspace}{0ex}}{R}^{9}+24\phantom{\rule{thinmathspace}{0ex}}{R}^{8}+56\phantom{\rule{thinmathspace}{0ex}}{R}^{7}+96\phantom{\rule{thinmathspace}{0ex}}{R}^{6}+115\phantom{\rule{thinmathspace}{0ex}}{R}^{5}+92\phantom{\rule{thinmathspace}{0ex}}{R}^{4}+48\phantom{\rule{thinmathspace}{0ex}}{R}^{3}+16\phantom{\rule{thinmathspace}{0ex}}{R}^{2}+3\phantom{\rule{thinmathspace}{0ex}}R$