August 2017 - mathfun@mailman.xmission.com

[math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Bill Gosper 14 Aug '17

14 Aug '17

(png detached) ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Mon, Aug 14, 2017 at 12:05 PM To: mathfuneavesdroppers(a)googlegroups.com Rohan insists that 3 is too small a horizontal magnifier for these triangular trinskys, and that someone should buckle down and add TrinskyCircularize to Corey's old package of Minsky tools. (MinskyCircularize is what mapped the green figures to the black ones in https://youtu.be/lXsVWwPa7bc .) Indeed, compare 3 with 2√3: gosper.org/trispiralrt12.png <http://gosper.org/trispiralrt12.png> So much for "only gridpoints". --rwg On Mon, Aug 7, 2017 at 4:03 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > More generally, there is no threefold rotational symmetry. > Julian's function can produce only gridpoints: > ListPlot[{3, 1} # & /@ > NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ > {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ > {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, > {2, 1}, # != -{1, 4} &], > AspectRatio -> 1, Axes -> False, Frame -> True] > gosper.org/trinsky3spiral.png > And all the arguments to the Floors are merely rational. --rwg > >

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[math-fun] Complete graph K19 with all distances at powers of d
by Ed Pegg Jr 14 Aug '17

14 Aug '17

The root d=1.15096... from d^6−d^2−1=0 seems to have some unique properties. At https://math.stackexchange.com/questions/2390420/ I build K_19 with it so that all distances are integral powers of d. Are there any other values that can build a power-clique this large? Zak's triangle, http://community.wolfram.com/groups/-/m/t/851275 , is a triangle that can be divided into 6 similar but non-congruent copies of itself. The value d can also build odd objects in 2D space, as seen at https://math.stackexchange.com/questions/1828794/ --Ed Pegg Jr

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Re: [math-fun] Odd-sided equilateral integer polygons
by Dan Asimov 14 Aug '17

14 Aug '17

Also regular hexagons, e.g. (1 0 -1) (0 1 -1) (-1 1 0) (-1 0 1) (0 -1 1) (1 -1 0) —Dan ----- 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(a)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) -----

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Re: [math-fun] Odd-sided equilateral integer polygons
by Keith F. Lynch 14 Aug '17

14 Aug '17

Fred Lunnon <fred.lunnon(a)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). My apologies. It was from me, as I think was clear from context. My mailer doesn't play well with this list. I always have to be careful when setting up the headers, and this once, as I was in a hurry, I screwed up. I accidentally left in the From: line from the mail I was responding to.

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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Keith F. Lynch 13 Aug '17

13 Aug '17

James Propp <jamespropp(a)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.

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[math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Bill Gosper 13 Aug '17

13 Aug '17

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: >> << 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(a)xmission.com <rcs(a)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(a)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(a)keithlynch.net> wrote: >>>> >>>>> James Propp <jamespropp(a)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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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 12 Aug '17

12 Aug '17

Maybe this mistake will lead to some insight(s). The symmetric group S_n does act by permutations on the standard basis vectors in R^n, and each of these has expression as an integer matrix. S_n certainly has elements of order n, for instance (12...n). Thus this element of S_n corresponds to a rotation of R^n that is of order n. So, (12...n) generates the cyclic group, isomorphic to Z/n, of integer-matrix rotations of R^n. Let (K,L,...,M) be an integer lattice point such that the iterates of this point under the group {(12...n)^k | 0 <= k <= n-1} are all distinct. Then: What is the orbit by this group of an integer lattice point (K,L,...,M) ??? Especially in the case n = p prime, if we assume the orbit {(12...n)^k (KL...M) | 0 <= k <= n-1} has cardinality = n but does not lie in any 2-plane of R^n, then . . . . . . what is it??? Simple case: n = p = 5. Let g denote the rotation of R^5 corresponding to (12...n). Since 5 is odd there is always one eigenvalue = 1, so we consider the rotation on the R^4 perpendicular to an eigenvector of this value. But a rotation g of R^4 of order 5 can always be resolved into two rotations, one on each of two perpendicular 2-planes, say E_1 and E_2. if we let tau = exp(2pi*i/5) then the rotations on E_1 and E_2 of order 5, thought of as complex numbers, are given by (z, w) |—> (tau^j z, tau^k w) where 1 <= i, j <= 4. Suppose we take as generator g the one with j = 1, k = 2: (z, w) |—> (tau z, tau^2 w) . Then the five rotations generated by g, as viewed on the torus {(exp(2pi*si), exp(2pi*ti)) | (s,t) in [0,1] x [0,1]} form the vertices of the (self-dual) tiling of the square torus by 5 squares. —Dan Jim Propp wrote: ----- I remember making the same mistake (confusing an n-simplex with a regular n-gon), and drawing the same sort of incorrect conclusions (a regular heptagon can be constructed with straightedge and compass in seven dimensions) when I was in college. If there's a compendium of commonly made mistakes in higher mathematics, this one should probably be added to the list! On Sat, Aug 12, 2017 at 1:51 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > Gene is right that my argument failed. I was unduly confident that the > action of the symmetric group S_n [as rotations of the regular n-simplex > centered at the origin in R^n] could be parlayed into a regular lattice > n-gon, but this is not to be, thanks to a web page I found this morning: > > Theorem: > -------- > (i) The rectangle and octagon are the only two equiangular lattice > polygons in Z^2.( Since the equiangular octagon is not equilateral, the > square is the only regular lattice polygon in Z^2.) > > (ii) The square, triangle and hexagon are the only regular lattice > polygons in Z^d for any dimension d. The square can be embedded in Z^2 > while the triangle and hexagon require Z^3. > ----- > > Proofs and more at <http://dynamicsofpolygons.org/PDFs/LatticePolygons.pdf > >From: Eugene Salamin > >Sent: Aug 11, 2017 5:16 PM > > >On Friday, August 11, 2017, 5:04:38 PM PDT, Dan Asimov < > dasimov(a)earthlink.net> wrote: > > > >Here is a one way to attack the n-dimensional case of regular polygons > >whose vertices lie in Z^n. > > > >Suppose there's a regular p-gon P, p in {3,4,5,...} with vertices in Z^n. > > > >If n >= p then R^p \sub R^n, so the standard basis vectors of R^p show > >that such a p-gon always exists. > > > >----------------------------------------------------------------- > > > >But these standard basis vectors form a (p-1)-simplex, and for p>3, do > not lie in a common 2-plane. For example, in Z^4, the basis vectors > (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) form a tetrahedron. -----

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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 12 Aug '17

12 Aug '17

Gene is right that my argument failed. I was unduly confident that the action of the symmetric group S_n [as rotations of the regular n-simplex centered at the origin in R^n] could be parlayed into a regular lattice n-gon, but this is not to be, thanks to a web page I found this morning: Theorem: -------- (i) The rectangle and octagon are the only two equiangular lattice polygons in Z^2.( Since the equiangular octagon is not equilateral, the square is the only regular lattice polygon in Z^2.) (ii) The square, triangle and hexagon are the only regular lattice polygons in Z^d for any dimension d. The square can be embedded in Z^2 while the triangle and hexagon require Z^3. ----- Proofs and more at <http://dynamicsofpolygons.org/PDFs/LatticePolygons.pdf>. —Dan >From: Eugene Salamin >Sent: Aug 11, 2017 5:16 PM >On Friday, August 11, 2017, 5:04:38 PM PDT, Dan Asimov <dasimov(a)earthlink.net> wrote: > >Here is a one way to attack the n-dimensional case of regular polygons >whose vertices lie in Z^n. > >Suppose there's a regular p-gon P, p in {3,4,5,...} with vertices in Z^n. > >If n >= p then R^p \sub R^n, so the standard basis vectors of R^p show >that such a p-gon always exists. > >----------------------------------------------------------------- > >But these standard basis vectors form a (p-1)-simplex, and for p>3, do not lie in a common 2-plane. For example, in Z^4, the basis vectors (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) form a tetrahedron. -----

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[math-fun] Square-free subsets of the integer grid
by Keith F. Lynch 12 Aug '17

12 Aug '17

We've been discussing the fact that squares are the only regular polygons on the integer grid. I'd like to poke some holes in that. The question is how many holes are needed? In other words, I want to remove just enough points from the integer grid (aka square lattice on the XY plane, aka gaussian integers) such that no square polygons exist on the remaining points. I'm mostly interested in a grid that extends forever in all four quadrants. (It's not interesting that if you remove points from everwhere but a line you can't have any squares.) Of course no repeating pattern of removals will work. I'm pretty sure the proportion that are removed must increase without limit as the grid gets larger, as happens with gaussian primes. (Gaussian primes themselves obviously allow squares, due to their fourfold symmetry. Whether they allow squares in just a single octant (e.g. A+Bi where A is a non-negative integer and B is a non-negative integer no greater than A) is an interesting question, but doesn't involve a grid that extends forever in all eight octants.) I'd like to maximize the asymptotic density, i.e. have as few holes as possible out to each distance, roughly speaking. Has anyone ever worked on this before? Keep in mind that the sides of squares on the integer grid don't need to be parallel to, or at a 45 degree angle to, the axes. Squares can sneak in at lots of different angles.

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Re: [math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
by Dan Asimov 12 Aug '17

12 Aug '17

Here is a one way to attack the n-dimensional case of regular polygons whose vertices lie in Z^n. Suppose there's a regular p-gon P, p in {3,4,5,...} with vertices in Z^n. If n >= p then R^p \sub R^n, so the standard basis vectors of R^p show that such a p-gon always exists. So we shall assume n < p. Let the vertices of P be V(j) in R^n, j = 1,...,p. Now consider the n x n matrix M whose rows are V(j+1)-V(j), j = 1,...,n. As Rich points out, these row vectors V(j+1)-V(j) all have length sqrt(N) for some N in Z+. Thus we may write: M = sqrt(N) M_1 for an n x n matrix M_1 whose rows have unit length. Hence M and M_1 have the same eigenvalues. Now to find the eigenvalues of M_1, we may use any basis we choose, so let the first row be (1, 0, ..., 0); let the second row be (cos(2pi/p), sin(2pi/p), 0, ..., 0); ...; let the jth row be (cos(2(j-1)pi/p), sin(2(j-1)pi/p), 0, ..., 0). Then it's easy to see that this matrix has sin(2pi/p) as an eigenvalue, and hence so does M. But M is an integer matrix, so the roots of its characteristic polynomial are algebraic of degree <= n. So in cases where sin(2pi/p) is algebraic of degree > n, we have reached a contradiction. E.g., for a pentagon in Z^3, sin(2pi/5) is algebraic of degree 4. Since 4 > 3, the 3 x 3 integer matrix M in this case cannot have sin(2pi/5) as an eigenvalue. E.g., for the case of p prime, the pth cyclotomic polynomial is of degree p-1, which shows that a regular p-gon cannot have its vertices on Z^(p-2). The general case that's easy to state is the following. (Note that p is not necessarily prime.) ----- Let phi denote the Euler phi function. If phi(p) > n then there does not exist any p-gon in R^n with vertices in Z^n. ----- For, phi(p) is the degree of the algebraic number sin(2pi/p). —Dan ----- 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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