On 2017-08-11 17:25, Mike Speciner wrote:

You know the determinant form of the area of a triangle? http://demonstrations.wolfram.com/TheAreaOfATriangleUsingADeterminant/

Here's my vector valued areatri[p1_, p2_, p3_] := Cross[p1 - p3, p2 - p1]/2 E.g., the general formula: In[531]:= areatri[{x1, y1, z}, {x2, y2, z}, {x3, y3, z}] Out[531]= {0, 0, 1/2 (-x2 y1 + x3 y1 + x1 y2 - x3 y2 - x1 y3 + x2 y3)} For an arbitrary polygon in the complex plane, areaPolygon[L_List] := Total[MapThread[Im[#1\[Conjugate]*#2] &, {L, RotateLeft[L]}]]/2 (Likewise = integers/2). Showing off for a regular pentagon, In[532]:= E^(2 I π Range@5/5) Out[532]= {E^(2 I π/5), E^(4 I π/5), E^(-4 I π/5), E^(-2 I π/5), 1} In[533]:= areaPolygon@% Out[533]= 5/2 Im[E^(2 I π/5)] In[534]:= ComplexExpand@% Out[534]= 5/2 √(5/8 + √5/8) Vs the builtin In[541]:= Area@RegularPolygon@5 Out[541]= 5/4 √(1/2 (5 + √5)) --rwg

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:

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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:

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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.

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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>:

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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:

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James Propp <jamespropp@gmail.com <javascript:;>> wrote:

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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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