A Set of Points

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A Set of Points

Postby measure » Tue Sep 26, 2017 8:45 pm UTC

Consider the set of points in R2 defined by the following construction rules:
    Rule 1: The points (0,0) and (1,0) are in the set.
    Rule 2: For any pair of points A and B in the set, the point created by rotating B 1/5 of a turn (2π/5) about A is also in the set.
This is sufficient to create a minimal, non-trivial (countably infinite) set of points such that the set has 5-way symmetry about each point in the set.

Let the notation B^A mean point B rotated around point A, B^A2 means B rotated twice (2/5) around A, etc.
    A = (0,0)
    B = (1,0)
    C = B^A2
    D = B^A3
    E = B^A4
    F = A^B
    G = E^C3
    H = C^B2
    I = D^B3
    J = F^H4
    K = G^I4
    L = K^J = (-4,0)
There might be a faster way, but this shows that the point (-4,0) is in the set using 27 applications of Rule 2. I'm not particularly interested in the process of proving that point L above is the point (-4,0), which could be done by considering the points as complex numbers and doing a bunch of algebra.
    Rule 1': The complex numbers 0 + 0i and 1 + 0i are in the set.
    Rule 2': A & B -> A + (B - A)*X where X is the unit vector cos(2π/5) + i*sin(2π/5)
My question is whether the point (-1,0) is in the set. I've been unable to reach it though the trial-and-error construction I used to reach (-4,0) above, and I wrote a program that searched exhaustively up through ~6 steps, but the solution space quickly becomes too large for my patience. I would like either a construction like the one above to show that (-1,0) is in the set, or a proof (formal or otherwise) that it is not in the set.

I believe the set is dense in the sense that for any point P not necessarily in the set and for any positive real ε, there exists a point Q in the set such that |P-Q| < ε. (Dis)proof of this would be welcome as well.

EDIT: I found a faster way of reaching (-4,0)
    A = (0,0)
    B = (1,0)
    C = B^A
    D = B^C
    E = B^C2
    F = D^A3
    G = E^F = (-4,0)
    <8 moves>
Last edited by measure on Mon Oct 02, 2017 2:23 pm UTC, edited 1 time in total.

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Re: A Set of Points

Postby notzeb » Tue Sep 26, 2017 11:38 pm UTC

(-1,0) is not in your set. If we let ζ be a fifth root of unity, then every point in your set is an element of the ring Z[ζ] (that is, it can be written as a sum a + bζ + cζ2 + dζ3 + eζ4 for some integers a, b, c, d, e). Algebraically, your rule 2 produces the point A + ζ(B-A) = (1-ζ)A + ζB, and modulo (1-ζ) this is congruent to B. So modulo (1-ζ), every point you can produce ends up being congruent to one of the two starting points. Since -1 is not congruent to either 0 or 1 modulo (1-ζ) (since the norm of (1-ζ) is 5), you can't reach it. (On the other hand, -4 is congruent to 1 modulo (1-ζ), since 1 - (-4) = 5 = (1-ζ)(1-ζ2)(1-ζ3)(1-ζ4).)

Is this the only obstruction? Can we find a way to reach any point in Z[ζ] which is congruent to either 0 or 1 modulo (1-ζ)? (If it helps, the congruence class of a + bζ + cζ2 + dζ3 + eζ4 modulo (1-ζ) is the congruence class of a+b+c+d+e modulo 5.)

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Re: A Set of Points

Postby measure » Wed Sep 27, 2017 3:00 pm UTC

Thanks notzeb! That's just what I was looking for. I figured some sort of complex analysis was the way to go here, but I'm not familiar enough with it to notice that congruence.

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