A pentagon cannot be embedded on an integer lattice of any number of dimensions. Suppose it could. Let one of the vertices be at the origin, and the other two vectors represented by x, y. Then cos^2(3pi/5) = (x.y)^2/((x.x)(y.y)) = a rational number. But cos^2(3pi/5) is irrational. I believe that the only regular polygons embeddable in Z^n are the triangle, square and hexagon. (I made an equivalent assumption to reduce the latest Project Euler problem to a number-theoretic bash. Unfortunately, it still takes 400 CPU-hours... :/) Let's see if we can prove it. cos^2(pi - 2pi/n) being rational is equivalent to the real and imaginary parts of e^(2 pi i/n) being roots of quadratic equations. And that is equivalent to e^(2 pi i/n) having minimal (cyclotomic) polynomial of degree <= 2, i.e. phi(n) <= 2. But the only n for which phi(n) <= 2 are 1, 2, 3, 4 and 6. QED. Corollary: The only regular polytopes embeddable in Z^n (for any n) are: Line segment {(0), (1)} Hexagon {(0,1,2),(0,2,1),{1,2,0},(2,1,0),(2,0,1),(1,0,2)} Simplex {(1,0,0,0,...,0), (0,1,0,0,...,0), ... (0,0,0,0,...,1)} Hypercube {(±1, ±1, ..., ±1)} Orthoplex {(±1,0,0,...,0), (0,±1,0,...,0), ... (0,0,0,...,±1)} 24-cell {(±2,0,0,0), (0,±2,0,0), (0,0,±2,0), (0,0,0,±2), (±1,±1,±1,±1)} Sincerely, Adam P. Goucher http://cp4space.wordpress.com
----- Original Message ----- From: Bill Gosper Sent: 05/26/13 09:33 AM To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Platonic regular polyhedra with integer vertices ?
Cris Moore>What is the smallest number of dimensions in which we can embed a pentagon with integer vertices?
More generally, for an n-gon, the argument below suggests (I think) that we need at least p-1 dimensions where p is the largest prime factor of n. We can do a hexagon in three dimensions, but not, I think, in two.
Cris
On May 25, 2013, at 11:09 PM, Bill Gosper wrote:
Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center.
Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix.
No, That was DanA.
http://www.tweedledum.com/rwg/rectarith12.pdf (end) illustrates hexagons in R^3, followed immediately by
"How many dimensions before we see (regular) octagons, pentagons, dodecagons, ...? We won't! Even in an infinite dimensional grid, the only regular polygons of finite size are triangles, hexagons, and squares. We could probably prescribe the nth coordinate of the kth vertex of a regular pentagon, say, but infinitely many coordinates would be nonzero, resulting in infinite size. And it would only approach regularity as we consider higher and higher dimensions." --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun