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.My question is whether the point (-1,0) is in the set.
Rule 2': A & B -> A + (B - A)*X where X is the unit vector cos(2π/5) + i*sin(2π/5)
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)