Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
James Propp <jamespropp@gmail.com> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this: The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4. Any idea why? What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon. Jim Propp On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral. --- stretchable regular polygons ... a regular polygon with a composite number of edges N = AB can be stretched. Arbitrary lengths can be assigned to the first A edges, while keeping the angles the same as the regular polygon. Then repeat the pattern B times. More generally (?), letting w be the Nth root of 1, e^(2ipi/N), look at sides a,b,c,... as a walk from the Origin, with waypoints 0, a, a+bw, a+bw+cw2, ..., a+bw+cw2+...+zw^(N-1) = 0; and the requirements that the polygon close and not overlap. [Does requiring positive edge lengths guarantee no overlap, for paths that return to 0 in N steps? Maybe a convexity argument?] The minimal polynomial for w is the cyclotomic poly for N, of degree phi(N). Our path-polygon must be a multiple of the cyclo-poly, with positive coefficients (edge-lengths), giving us N-phi(N) parameters to play with. I think the edge-sum of two regular-angle N gons will be another such, even if the periods A and B of the addends are different. Can we make all 120-degree hexagons by edge-adding an ababab and a cdecde? Rich ---- Quoting James Propp <jamespropp@gmail.com>:
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon.
Jim Propp
On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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<< Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. >> Er --- why is that, exactly? Hoped somebody else would ask; but nobody did, so I've had to stick my neck out. Again. WFL On 8/11/17, rcs@xmission.com <rcs@xmission.com> wrote:
I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral.
---
stretchable regular polygons ... a regular polygon with a composite number of edges N = AB can be stretched. Arbitrary lengths can be assigned to the first A edges, while keeping the angles the same as the regular polygon. Then repeat the pattern B times. More generally (?), letting w be the Nth root of 1, e^(2ipi/N), look at sides a,b,c,... as a walk from the Origin, with waypoints 0, a, a+bw, a+bw+cw2, ..., a+bw+cw2+...+zw^(N-1) = 0; and the requirements that the polygon close and not overlap. [Does requiring positive edge lengths guarantee no overlap, for paths that return to 0 in N steps? Maybe a convexity argument?] The minimal polynomial for w is the cyclotomic poly for N, of degree phi(N). Our path-polygon must be a multiple of the cyclo-poly, with positive coefficients (edge-lengths), giving us N-phi(N) parameters to play with. I think the edge-sum of two regular-angle N gons will be another such, even if the periods A and B of the addends are different. Can we make all 120-degree hexagons by edge-adding an ababab and a cdecde?
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon.
Jim Propp
On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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You know the determinant form of the area of a triangle? http://demonstrations.wolfram.com/TheAreaOfATriangleUsingADeterminant/ But if not, just make a the smallest enclosing big rectangle around your triangle and subtract off the right triangles. --ms On 11-Aug-17 20:11, Fred Lunnon wrote:
<< Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. >>
Er --- why is that, exactly?
Hoped somebody else would ask; but nobody did, so I've had to stick my neck out. Again.
WFL
On 8/11/17, rcs@xmission.com <rcs@xmission.com> wrote:
I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral.
---
stretchable regular polygons ... a regular polygon with a composite number of edges N = AB can be stretched. Arbitrary lengths can be assigned to the first A edges, while keeping the angles the same as the regular polygon. Then repeat the pattern B times. More generally (?), letting w be the Nth root of 1, e^(2ipi/N), look at sides a,b,c,... as a walk from the Origin, with waypoints 0, a, a+bw, a+bw+cw2, ..., a+bw+cw2+...+zw^(N-1) = 0; and the requirements that the polygon close and not overlap. [Does requiring positive edge lengths guarantee no overlap, for paths that return to 0 in N steps? Maybe a convexity argument?] The minimal polynomial for w is the cyclotomic poly for N, of degree phi(N). Our path-polygon must be a multiple of the cyclo-poly, with positive coefficients (edge-lengths), giving us N-phi(N) parameters to play with. I think the edge-sum of two regular-angle N gons will be another such, even if the periods A and B of the addends are different. Can we make all 120-degree hexagons by edge-adding an ababab and a cdecde?
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon.
Jim Propp
On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons. http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/ Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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Pict's theorem say that the area of a polygon created by joining lattice points in the plane is given by (# of lattice points inside) + (# of lattice points on boundary)/2 - 1. Andy On Fri, Aug 11, 2017 at 8:11 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. >>
Er --- why is that, exactly?
Hoped somebody else would ask; but nobody did, so I've had to stick my neck out. Again.
WFL
On 8/11/17, rcs@xmission.com <rcs@xmission.com> wrote:
I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral.
---
stretchable regular polygons ... a regular polygon with a composite number of edges N = AB can be stretched. Arbitrary lengths can be assigned to the first A edges, while keeping the angles the same as the regular polygon. Then repeat the pattern B times. More generally (?), letting w be the Nth root of 1, e^(2ipi/N), look at sides a,b,c,... as a walk from the Origin, with waypoints 0, a, a+bw, a+bw+cw2, ..., a+bw+cw2+...+zw^(N-1) = 0; and the requirements that the polygon close and not overlap. [Does requiring positive edge lengths guarantee no overlap, for paths that return to 0 in N steps? Maybe a convexity argument?] The minimal polynomial for w is the cyclotomic poly for N, of degree phi(N). Our path-polygon must be a multiple of the cyclo-poly, with positive coefficients (edge-lengths), giving us N-phi(N) parameters to play with. I think the edge-sum of two regular-angle N gons will be another such, even if the periods A and B of the addends are different. Can we make all 120-degree hexagons by edge-adding an ababab and a cdecde?
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon.
Jim Propp
On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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Pick's theorem, not Pict's (just a typo I'm sure). On Sat, Aug 12, 2017 at 2:38 PM, Andy Latto <andy.latto@pobox.com> wrote:
Pict's theorem say that the area of a polygon created by joining lattice points in the plane is given by (# of lattice points inside) + (# of lattice points on boundary)/2 - 1.
Andy
On Fri, Aug 11, 2017 at 8:11 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. >>
Er --- why is that, exactly?
Hoped somebody else would ask; but nobody did, so I've had to stick my neck out. Again.
WFL
On 8/11/17, rcs@xmission.com <rcs@xmission.com> wrote:
I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral.
---
stretchable regular polygons ... a regular polygon with a composite number of edges N = AB can be stretched. Arbitrary lengths can be assigned to the first A edges, while keeping the angles the same as the regular polygon. Then repeat the pattern B times. More generally (?), letting w be the Nth root of 1, e^(2ipi/N), look at sides a,b,c,... as a walk from the Origin, with waypoints 0, a, a+bw, a+bw+cw2, ..., a+bw+cw2+...+zw^(N-1) = 0; and the requirements that the polygon close and not overlap. [Does requiring positive edge lengths guarantee no overlap, for paths that return to 0 in N steps? Maybe a convexity argument?] The minimal polynomial for w is the cyclotomic poly for N, of degree phi(N). Our path-polygon must be a multiple of the cyclo-poly, with positive coefficients (edge-lengths), giving us N-phi(N) parameters to play with. I think the edge-sum of two regular-angle N gons will be another such, even if the periods A and B of the addends are different. Can we make all 120-degree hexagons by edge-adding an ababab and a cdecde?
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon.
Jim Propp
On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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That's what had been nagging at me. And to think that I once thought to write a paper about Heronian tetrahedra ... Evidently I must be edgy, rather than vertical; perhaps should take more water with my Picks and Scotch? WFL On 8/12/17, Tomas Rokicki <rokicki@gmail.com> wrote:
Pick's theorem, not Pict's (just a typo I'm sure).
On Sat, Aug 12, 2017 at 2:38 PM, Andy Latto <andy.latto@pobox.com> wrote:
Pict's theorem say that the area of a polygon created by joining lattice points in the plane is given by (# of lattice points inside) + (# of lattice points on boundary)/2 - 1.
Andy
On Fri, Aug 11, 2017 at 8:11 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. >>
Er --- why is that, exactly?
Hoped somebody else would ask; but nobody did, so I've had to stick my neck out. Again.
WFL
On 8/11/17, rcs@xmission.com <rcs@xmission.com> wrote:
I always liked the no-equilateral-triangles proof based on the area. Any integer-coordinate triangle in the plane has area either an integer or integer+1/2. Any line segment has length sqrt(N), so its square is an integer. The area of an equilateral triangle of side S is (sqrt3 / 4) S^2, which is irrational when S^2 is integral.
---
stretchable regular polygons ... a regular polygon with a composite number of edges N = AB can be stretched. Arbitrary lengths can be assigned to the first A edges, while keeping the angles the same as the regular polygon. Then repeat the pattern B times. More generally (?), letting w be the Nth root of 1, e^(2ipi/N), look at sides a,b,c,... as a walk from the Origin, with waypoints 0, a, a+bw, a+bw+cw2, ..., a+bw+cw2+...+zw^(N-1) = 0; and the requirements that the polygon close and not overlap. [Does requiring positive edge lengths guarantee no overlap, for paths that return to 0 in N steps? Maybe a convexity argument?] The minimal polynomial for w is the cyclotomic poly for N, of degree phi(N). Our path-polygon must be a multiple of the cyclo-poly, with positive coefficients (edge-lengths), giving us N-phi(N) parameters to play with. I think the edge-sum of two regular-angle N gons will be another such, even if the periods A and B of the addends are different. Can we make all 120-degree hexagons by edge-adding an ababab and a cdecde?
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon.
Jim Propp
On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote: > Here's a beautiful essay by Joel Hamkins that handles 5-gons, > 6-gons, 7-gons, etc. with a uniform argument and then settles the > case of 3-gons by appealing to the fact about 6-gons.
> http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
Suppose we have an equilateral polygon with vertices in the Gaussian integers and common side-length sqrt(L), where L is an integer. Without loss of generality, assume that one of the vertices is at the origin. 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. Hence, by induction, every vertex is divisible by 1 + i, so we can divide throughout and obtain an equilateral polygon with common side-length sqrt(L/2). Hence, we can assume without loss of generality that L is odd. Now colour the vertices of the integer grid red and blue in a checkerboard fashion. Every pair of adjacent vertices in an equilateral polygon of common side length L have opposite colours, so no odd-sided equilateral N-gons can exist. QED. Best wishes, Adam P. Goucher
Very nice! Jim On Sun, Aug 13, 2017 at 9:13 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
Suppose we have an equilateral polygon with vertices in the Gaussian integers and common side-length sqrt(L), where L is an integer. Without loss of generality, assume that one of the vertices is at the origin.
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. Hence, by induction, every vertex is divisible by 1 + i, so we can divide throughout and obtain an equilateral polygon with common side-length sqrt(L/2).
Hence, we can assume without loss of generality that L is odd. Now colour the vertices of the integer grid red and blue in a checkerboard fashion. Every pair of adjacent vertices in an equilateral polygon of common side length L have opposite colours, so no odd-sided equilateral N-gons can exist. QED.
Best wishes,
Adam P. Goucher
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<< 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? WFL On 8/13/17, James Propp <jamespropp@gmail.com> wrote:
Very nice!
Jim
On Sun, Aug 13, 2017 at 9:13 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
Suppose we have an equilateral polygon with vertices in the Gaussian integers and common side-length sqrt(L), where L is an integer. Without loss of generality, assume that one of the vertices is at the origin.
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. Hence, by induction, every vertex is divisible by 1 + i, so we can divide throughout and obtain an equilateral polygon with common side-length sqrt(L/2).
Hence, we can assume without loss of generality that L is odd. Now colour the vertices of the integer grid red and blue in a checkerboard fashion. Every pair of adjacent vertices in an equilateral polygon of common side length L have opposite colours, so no odd-sided equilateral N-gons can exist. QED.
Best wishes,
Adam P. Goucher
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Translate w to the origin. Then L = |v|^2 = x^2 + y^2 is even, and so x and y are either both even or both odd. Then x + iy = (x + iy) ((1 - i)/2) (1 + i) = (((x + y) + i(y - x)) / 2) (1 + i). Since x + y and y - x are both even, the cofactor of 1 + i is an integer. -- Gene On Sunday, August 13, 2017, 10:41:05 AM PDT, Fred Lunnon <fred.lunnon@gmail.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? WFL On 8/13/17, James Propp <jamespropp@gmail.com> wrote:
Very nice!
Jim
On Sun, Aug 13, 2017 at 9:13 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
Suppose we have an equilateral polygon with vertices in the Gaussian integers and common side-length sqrt(L), where L is an integer. Without loss of generality, assume that one of the vertices is at the origin.
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. Hence, by induction, every vertex is divisible by 1 + i, so we can divide throughout and obtain an equilateral polygon with common side-length sqrt(L/2).
Hence, we can assume without loss of generality that L is odd. Now colour the vertices of the integer grid red and blue in a checkerboard fashion. Every pair of adjacent vertices in an equilateral polygon of common side length L have opposite colours, so no odd-sided equilateral N-gons can exist. QED.
Best wishes,
Adam P. Goucher
participants (9)
-
Adam P. Goucher -
Andy Latto -
Eugene Salamin -
Fred Lunnon -
James Propp -
Keith F. Lynch -
Mike Speciner -
rcs＠xmission.com -
Tomas Rokicki