1. Affine transfer
Let F(x)=Mx+b be an invertible affine map from triangle T to triangle T′. Every triangle formed by three sites has its area multiplied by |det M|. Taking the minimum over triples and then the maximum over configurations preserves this same factor. Since F is bijective, both feasible constructions and upper bounds transfer.
Aₙ(T′)/area(T′) = Aₙ(T)/area(T).
For the right triangle (0,0),(1,0),(0,1), use F(x,y)=(x+y/2,√3 y/2). Its determinant is √3/2, and its image is our equilateral triangle with area √3/4.
Attaining constructions
In the right triangle, put s=√2. For n=5 take the following five points; for n=6 use the six rational points below. Apply F to obtain exact equilateral-triangle configurations. Checking all 10 or 20 determinants gives minimum twice-area 3−2√2 or 1/8 respectively before transformation.
n=5: (0,0), (2−s,0), (0,1), (2−s,s−1), (3−2s,s−1).
n=6: (1/4,0), (3/4,0), (0,3/8), (3/4,1/4), (0,1), (1/4,1/2).
2. Classical optima and attribution
For a unit-area triangle the optima are 3−2√2 at n=5 and 1/8 at n=6. The five-point result is credited to Peng (1989), with proofs for n=5,6 by Yang, Zhang and Zeng (1994). Multiplying by √3/4 gives the displayed results. These are literature results, not new breakthroughs by this site.
De Comité–Delahaye: historical correction and references ↗
Sudermann-Merx (2026), §2.1 and Table 1: affine normalization and coordinates ↗
Only the classical n=5,6 results are used here; this page does not rely on the newer computational claims for n=7,8.
3. Precision and the discussion correction
√3/(2+2√2)² = √3/(12+8√2) = (√3/4)(3−2√2) ≈ 0.0742932342850689.
The first formula proposed by 大小糖 #181 is correct. Its subsequent denominator 6+4√2 is missing a factor of two. We thank the author for drawing attention to the value; numerical formula recognition alone is not an optimality proof. Original discussion ↗
We archive these cases as continuously solved and retain all historical records. The verifier stores twice the area; the page shows area. Both exact optima are irrational, whereas a polygon of finite-decimal points has rational triangle areas, so no nine-decimal certificate attains these optima exactly. This does not prove which certificate is best on the fixed grid, nor justify rounding a record into an exact optimum.