Classify the regular polygons that fit about a common vetex
$begingroup$
I know this can be solved using a (quasi-)symmetric groups approach from crystallography, but I wish to solve it with a more simple approach, number theory motivated.
I wish to classify the regular polygons for which several copies can fit around a common vertex.
The inner verices angle of an $n$ polygon is $pi-frac {2pi}{n}$, and we wish it to be equal to a fraction of a circle $=frac {2pi}{m}$. A quick algebra brings us to $mn=2m+2n $ or $m(n-2)=2n$. So this may hold for any $n$ s.t. $n-2|2n$. This means that $n-2=(n-2,2n)=(n-2,n-2+n+2)|n+2$. Now $(n+2,n-2)=1$ for odd $n$'s and $(n+2,n-2)=2$ for even ones. This means that only $n=3,4$ are possible, indeed 4 squares and 6 triangles fit around a common vertex.
But three hexagons can also fit, why did we miss this option?
Thanks a lot.
number-theory symmetry
asked Jan 6 at 16:32
galragalra
400212
$endgroup$
add a comment |
$begingroup$
I know this can be solved using a (quasi-)symmetric groups approach from crystallography, but I wish to solve it with a more simple approach, number theory motivated.
I wish to classify the regular polygons for which several copies can fit around a common vertex.
The inner verices angle of an $n$ polygon is $pi-frac {2pi}{n}$, and we wish it to be equal to a fraction of a circle $=frac {2pi}{m}$. A quick algebra brings us to $mn=2m+2n $ or $m(n-2)=2n$. So this may hold for any $n$ s.t. $n-2|2n$. This means that $n-2=(n-2,2n)=(n-2,n-2+n+2)|n+2$. Now $(n+2,n-2)=1$ for odd $n$'s and $(n+2,n-2)=2$ for even ones. This means that only $n=3,4$ are possible, indeed 4 squares and 6 triangles fit around a common vertex.
But three hexagons can also fit, why did we miss this option?
Thanks a lot.
number-theory symmetry
asked Jan 6 at 16:32
galragalra
400212
$endgroup$
add a comment |
$begingroup$
I know this can be solved using a (quasi-)symmetric groups approach from crystallography, but I wish to solve it with a more simple approach, number theory motivated.
I wish to classify the regular polygons for which several copies can fit around a common vertex.
The inner verices angle of an $n$ polygon is $pi-frac {2pi}{n}$, and we wish it to be equal to a fraction of a circle $=frac {2pi}{m}$. A quick algebra brings us to $mn=2m+2n $ or $m(n-2)=2n$. So this may hold for any $n$ s.t. $n-2|2n$. This means that $n-2=(n-2,2n)=(n-2,n-2+n+2)|n+2$. Now $(n+2,n-2)=1$ for odd $n$'s and $(n+2,n-2)=2$ for even ones. This means that only $n=3,4$ are possible, indeed 4 squares and 6 triangles fit around a common vertex.
But three hexagons can also fit, why did we miss this option?
Thanks a lot.
number-theory symmetry
asked Jan 6 at 16:32
galragalra
400212
$endgroup$
I know this can be solved using a (quasi-)symmetric groups approach from crystallography, but I wish to solve it with a more simple approach, number theory motivated.
I wish to classify the regular polygons for which several copies can fit around a common vertex.
The inner verices angle of an $n$ polygon is $pi-frac {2pi}{n}$, and we wish it to be equal to a fraction of a circle $=frac {2pi}{m}$. A quick algebra brings us to $mn=2m+2n $ or $m(n-2)=2n$. So this may hold for any $n$ s.t. $n-2|2n$. This means that $n-2=(n-2,2n)=(n-2,n-2+n+2)|n+2$. Now $(n+2,n-2)=1$ for odd $n$'s and $(n+2,n-2)=2$ for even ones. This means that only $n=3,4$ are possible, indeed 4 squares and 6 triangles fit around a common vertex.
But three hexagons can also fit, why did we miss this option?
Thanks a lot.
number-theory symmetry
number-theory symmetry
asked Jan 6 at 16:32
galragalra
400212
asked Jan 6 at 16:32
galragalra
400212
asked Jan 6 at 16:32
galragalra
400212
asked Jan 6 at 16:32
galragalra
400212
asked Jan 6 at 16:32
galragalra
400212
400212
add a comment |
add a comment |
1 Answer
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$begingroup$
It is not the case that $(n+2,n-2)=2$ for even $n$. If $nequiv 2pmod{4}$, then $4$ divides both $n+2$ and $n-2$.
Note that you can simplify the reasoning a bit by just using the fact that if $a|b$ and $a|c$, then $a|b+c$ rather than the more general fact about GCDs. In particular, you know
$$n-2|2n.$$
It is also clear that
$$n-2|-2n+4.$$
Therefore,
$$n-2|4.$$
This takes you immediately to your answer without needing to break into cases.
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
$endgroup$
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
add a comment |
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1 Answer
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$begingroup$
It is not the case that $(n+2,n-2)=2$ for even $n$. If $nequiv 2pmod{4}$, then $4$ divides both $n+2$ and $n-2$.
Note that you can simplify the reasoning a bit by just using the fact that if $a|b$ and $a|c$, then $a|b+c$ rather than the more general fact about GCDs. In particular, you know
$$n-2|2n.$$
It is also clear that
$$n-2|-2n+4.$$
Therefore,
$$n-2|4.$$
This takes you immediately to your answer without needing to break into cases.
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
$endgroup$
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
add a comment |
$begingroup$
It is not the case that $(n+2,n-2)=2$ for even $n$. If $nequiv 2pmod{4}$, then $4$ divides both $n+2$ and $n-2$.
Note that you can simplify the reasoning a bit by just using the fact that if $a|b$ and $a|c$, then $a|b+c$ rather than the more general fact about GCDs. In particular, you know
$$n-2|2n.$$
It is also clear that
$$n-2|-2n+4.$$
Therefore,
$$n-2|4.$$
This takes you immediately to your answer without needing to break into cases.
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
$endgroup$
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
add a comment |
$begingroup$
It is not the case that $(n+2,n-2)=2$ for even $n$. If $nequiv 2pmod{4}$, then $4$ divides both $n+2$ and $n-2$.
Note that you can simplify the reasoning a bit by just using the fact that if $a|b$ and $a|c$, then $a|b+c$ rather than the more general fact about GCDs. In particular, you know
$$n-2|2n.$$
It is also clear that
$$n-2|-2n+4.$$
Therefore,
$$n-2|4.$$
This takes you immediately to your answer without needing to break into cases.
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
$endgroup$
It is not the case that $(n+2,n-2)=2$ for even $n$. If $nequiv 2pmod{4}$, then $4$ divides both $n+2$ and $n-2$.
Note that you can simplify the reasoning a bit by just using the fact that if $a|b$ and $a|c$, then $a|b+c$ rather than the more general fact about GCDs. In particular, you know
$$n-2|2n.$$
It is also clear that
$$n-2|-2n+4.$$
Therefore,
$$n-2|4.$$
This takes you immediately to your answer without needing to break into cases.
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
answered Jan 6 at 16:38
Milo BrandtMilo Brandt
40k476140
40k476140
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
add a comment |
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
$begingroup$
Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks!
$endgroup$
– galra
Jan 6 at 16:53
add a comment |
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