Re: [math-fun] Odd-sided equilateral integer polygons
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote:
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining. My apologies if I over-explain, since I'm not sure quite what your question is. Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.) Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity. If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage. Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't). I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble! Fred Lunnon On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote:
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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"(envelope-from <kfl@panix.com>)" in source text reveals probable author to be Keith F. Lynch. WFL On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't).
I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble!
Fred Lunnon
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote:
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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This is a lovely proof, but I don't think you need to bring Gaussian integers into the picture to express it. The key step is: Checkerboard-color the integer lattice. The square of the distance between two points is of the form i^2+j^2, which has the same parity as i+j. That means any given distance can either be the distance between pairs of same-color points or pairs of opposite-color points, but certainly not both. Now if the shared distance between successive points has i+j odd, then the points must alternate colors, so there are an even number of them; while if i+j is even, then the points are all the same color, and the points of a given color form a sub-copy of the integer lattice (rotated by 45 degrees and expanded by a factor of sqrt(2)). So rotate and shrink to get a smaller but similar equilateral polygon, etc. Nice proof, Adam. --Michael On Sun, Aug 13, 2017 at 3:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
"(envelope-from <kfl@panix.com>)" in source text reveals probable author to be Keith F. Lynch.
WFL
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't).
I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble!
Fred Lunnon
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote:
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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-- Forewarned is worth an octopus in the bush.
What happens in Z^3? We can still checkerboard-color the lattice, and i^2+j^2+k^2 has the same parity as i+j+k. When i+j+k is odd, the proof goes through; but when i+j+k is even, there's no obvious analogue of rotating by 45 degrees and shrinking by sqrt(2). Jim Propp On Sunday, August 13, 2017, Michael Kleber <michael.kleber@gmail.com> wrote:
This is a lovely proof, but I don't think you need to bring Gaussian integers into the picture to express it.
The key step is: Checkerboard-color the integer lattice. The square of the distance between two points is of the form i^2+j^2, which has the same parity as i+j. That means any given distance can either be the distance between pairs of same-color points or pairs of opposite-color points, but certainly not both.
Now if the shared distance between successive points has i+j odd, then the points must alternate colors, so there are an even number of them; while if i+j is even, then the points are all the same color, and the points of a given color form a sub-copy of the integer lattice (rotated by 45 degrees and expanded by a factor of sqrt(2)). So rotate and shrink to get a smaller but similar equilateral polygon, etc.
Nice proof, Adam.
--Michael
On Sun, Aug 13, 2017 at 3:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
"(envelope-from <kfl@panix.com>)" in source text reveals probable author to be Keith F. Lynch.
WFL
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't).
I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble!
Fred Lunnon
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote:
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Duh: there's a triangle joining (1,0,0), (0,1,0), and (0,0,1). Jim Propp On Sunday, August 13, 2017, James Propp <jamespropp@gmail.com> wrote:
What happens in Z^3?
We can still checkerboard-color the lattice, and i^2+j^2+k^2 has the same parity as i+j+k. When i+j+k is odd, the proof goes through; but when i+j+k is even, there's no obvious analogue of rotating by 45 degrees and shrinking by sqrt(2).
Jim Propp
On Sunday, August 13, 2017, Michael Kleber <michael.kleber@gmail.com> wrote:
This is a lovely proof, but I don't think you need to bring Gaussian integers into the picture to express it.
The key step is: Checkerboard-color the integer lattice. The square of the distance between two points is of the form i^2+j^2, which has the same parity as i+j. That means any given distance can either be the distance between pairs of same-color points or pairs of opposite-color points, but certainly not both.
Now if the shared distance between successive points has i+j odd, then the points must alternate colors, so there are an even number of them; while if i+j is even, then the points are all the same color, and the points of a given color form a sub-copy of the integer lattice (rotated by 45 degrees and expanded by a factor of sqrt(2)). So rotate and shrink to get a smaller but similar equilateral polygon, etc.
Nice proof, Adam.
--Michael
On Sun, Aug 13, 2017 at 3:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
"(envelope-from <kfl@panix.com>)" in source text reveals probable author to be Keith F. Lynch.
WFL
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't).
I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble!
Fred Lunnon
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote: > Now, if L is even, then for any pair of adjacent vertices v, w in > Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
suppose P is a convex polygon with vertices at the usual integer (2d) lattice points (Z2) Is P necessarily similar to a polygon T obtained by (1) deleting finitely many points from Z2, and then (2) computing the Voronoi cells of the remaining points? if the answer is yes, then up to similarity, every convex polygon with Z2 vertices has (up to similarity) a Voronoi deletion birthday. for example, the square has birthday zero. a "picket" shaped polygon of area 1.25 is the unique polygon at birthday one. three more are born at day two, including a pentagon that is obviously the "correct" resolution of the Major League Baseball impossible home plate problem, described here http://mathworld.wolfram.com/HomePlate.html another example: what is the birthday of the 45-45-90 triangle? i worked with a high school student, Michael Nizsenson, on related problems recently. On Sun, Aug 13, 2017 at 5:00 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
This is a lovely proof, but I don't think you need to bring Gaussian integers into the picture to express it.
The key step is: Checkerboard-color the integer lattice. The square of the distance between two points is of the form i^2+j^2, which has the same parity as i+j. That means any given distance can either be the distance between pairs of same-color points or pairs of opposite-color points, but certainly not both.
Now if the shared distance between successive points has i+j odd, then the points must alternate colors, so there are an even number of them; while if i+j is even, then the points are all the same color, and the points of a given color form a sub-copy of the integer lattice (rotated by 45 degrees and expanded by a factor of sqrt(2)). So rotate and shrink to get a smaller but similar equilateral polygon, etc.
Nice proof, Adam.
--Michael
On Sun, Aug 13, 2017 at 3:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
"(envelope-from <kfl@panix.com>)" in source text reveals probable author to be Keith F. Lynch.
WFL
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't).
I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble!
Fred Lunnon
On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.com> wrote:
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i.
C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Thane Plambeck tplambeck@gmail.com http://counterwave.com/
participants (4)
-
Fred Lunnon -
James Propp -
Michael Kleber -
Thane Plambeck