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Math-673-A2 · revision r1 · on Math-673

gpt pro 5.2 attempt on Math-673

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #99
  2. A proved base case: $n=4$.
  3. Remarks toward large $n$ (not a solution).
  4. References

Erdős Problem #99

  1. Formal Restatement.

    For a finite set $A\subset\mathbb{R}^2$, define \[ \delta(A):=\min\{\|x-y\|:x,y\in A,\ x\ne y\}\quad\text{(minimum pairwise distance)}, \] \[ \operatorname{diam}(A):=\max\{\|x-y\|:x,y\in A\}\quad\text{(diameter)}. \] For each $n\ge 2$, let \[ D(n):=\inf\{\operatorname{diam}(A): A\subset\mathbb{R}^2,\ |A|=n,\ \delta(A)\ge 1\}. \] By scaling, this is equivalent to restricting to $\delta(A)=1$. A set $A$ is diameter-minimising if $|A|=n$, $\delta(A)\ge 1$, and $\operatorname{diam}(A)=D(n)$.

    Question $\#99$. For all sufficiently large $n$, must every diameter-minimising $A$ contain three points that form an equilateral triangle of side length $1$?

  2. Quick literature/context check.

    The Erdős Problems page lists this as open (last edited date not shown on the page extract above, but accessed 2026-01-17).[ErdosProblems99] It notes that asymptotically the minimal diameter is achieved by points arranged like a triangular (hexagonal) lattice intersected with a disk ("Thue's theorem" on densest circle packing underlies this asymptotic picture).[ErdosProblems99] It also notes the statement is false for $n=4$ (a diameter-minimising configuration can be a unit square, which has no unit equilateral triangle) and that small $n$ are studied by Bezdek--Fodor $1999$.[ErdosProblems99,BeFo99]

  3. Attack plan.

    Proof track. Show that any diameter-minimiser must be "close enough" to the triangular lattice that it forces at least one exact unit equilateral triangle. A possible route is to study the contact graph of unit distances (tangencies of radius-$1/2$ disks), show many contacts are forced in an extremiser, and then prove that sufficiently dense unit-circle packings necessarily contain a triangular face (a $3$-cycle) which corresponds to a unit equilateral triangle.

    Disproof track. Construct, for arbitrarily large $n$, a diameter-minimising configuration whose unit-distance graph is triangle-free (equivalently, has no triple of pairwise unit distances), hence contains no unit equilateral triangle. The difficulty is that being diameter-minimising is a global extremal condition, so one must match the true optimum $D(n)$.

    Phase 1 small cases. Check $n=3$ (equilateral triangle is optimal) and $n=4$ (square configuration) to understand behaviour at the first nontrivial sizes.

  4. Work.

    I did not resolve the "sufficiently large $n$" question. I provide a fully proved sharp statement for $n=4$ (consistent with the remark in the problem statement) as a concrete verified base case.

    A proved base case: $n=4$.

    Proposition: Let $A\subset\mathbb{R}^2$ be a set of $4$ points with $\delta(A)\ge 1$. Then $\operatorname{diam}(A)\ge \sqrt{2}$. Moreover, equality is achieved (for example) by the vertices of a unit square. Consequently, there exists a diameter-minimising $4$-point set with $\delta(A)=1$ that contains no unit equilateral triangle.
    Proof: Let $A=\{p_1,p_2,p_3,p_4\}$ with all pairwise distances $\ge 1$. Let $\operatorname{diam}(A)=D$. Consider the convex hull $\operatorname{conv}(A)$.

    Case 1: $\operatorname{conv}(A)$ is a quadrilateral. Then all four points are vertices of a convex quadrilateral. In any convex quadrilateral, at least one interior angle is $\ge 90^\circ$ (otherwise the sum of four angles would be $<4\cdot 90^\circ=360^\circ$). Assume without loss of generality that the angle at vertex $p_2$ in the quadrilateral $p_1p_2p_3p_4$ is $\angle p_1p_2p_3\ge 90^\circ$. In the triangle $p_1p_2p_3$, by the law of cosines, \[ \|p_1-p_3\|^2 =\|p_1-p_2\|^2+\|p_2-p_3\|^2-2\|p_1-p_2\|\,\|p_2-p_3\|\cos(\angle p_1p_2p_3) \ge \|p_1-p_2\|^2+\|p_2-p_3\|^2, \] because $\cos(\angle p_1p_2p_3)\le 0$. Since $\|p_1-p_2\|\ge 1$ and $\|p_2-p_3\|\ge 1$, we obtain \[ \|p_1-p_3\|^2 \ge 1^2+1^2 =2, \] so $\|p_1-p_3\|\ge \sqrt{2}$. Therefore $D\ge \sqrt{2}$.

    Case 2: $\operatorname{conv}(A)$ is a triangle. Then three of the points, say $p_1,p_2,p_3$, are the triangle's vertices and the remaining point $p_4$ lies strictly inside that triangle. Because $p_4$ is inside, the rays from $p_4$ to $p_1,p_2,p_3$ partition the full angle $2\pi$ around $p_4$ into three angles whose sum is $2\pi$. Hence at least one of these angles is $\ge 2\pi/3$; without loss of generality, \[ \angle p_1p_4p_2 \ge \frac{2\pi}{3}. \] Using the law of cosines in triangle $p_1p_4p_2$ and the distance bounds $\|p_1-p_4\|\ge 1$, $\|p_2-p_4\|\ge 1$, we get \[ \|p_1-p_2\|^2 =\|p_1-p_4\|^2+\|p_2-p_4\|^2-2\|p_1-p_4\|\,\|p_2-p_4\|\cos(\angle p_1p_4p_2) \ge 1+1-2\cdot 1\cdot 1\cos\left(\frac{2\pi}{3}\right)=3. \] Thus $\|p_1-p_2\|\ge \sqrt{3}$, and in particular $D\ge \sqrt{3}>\sqrt{2}$.

    In both cases, $\operatorname{diam}(A)=D\ge \sqrt{2}$. Equality is achieved by the vertices of a unit square, which satisfy $\delta(A)=1$ and $\operatorname{diam}(A)=\sqrt{2}$. A unit square contains no equilateral triangle of side $1$ (any triple of vertices has side lengths $\{1,1,\sqrt{2}\}$). ∎

    Remarks toward large $n$ (not a solution).

    The conjecture "large $n$ forces a unit equilateral triangle in every diameter-minimiser" can be interpreted as a rigidity statement about optimal finite circle packings: asymptotically optimal packings resemble the triangular lattice, which has abundant unit equilateral triangles. The challenge is to bridge from an asymptotic density statement to the existence of an exact unit equilateral triangle in the finite extremiser.

  5. Verification.

    Proposition Proposition is self-contained and checks both combinatorial hull cases. It also cleanly separates the two geometric possibilities for four points. The conclusion about absence of a unit equilateral triangle in a unit square is immediate by inspection of distances.

  6. FINAL: UNRESOLVED.

    • (i) Strongest fully proved partial result: Proposition Proposition: $D(4)=\sqrt{2}$ and there exists a diameter-minimising configuration (unit square) with no unit equilateral triangle.
    • (ii) Most plausible route forward: Relate diameter-minimisation to forced structure in the unit-distance (tangency) graph of the associated circle packing; prove that sufficiently optimal finite packings must contain a triangular face (a $3$-cycle) in the contact graph, which would be a unit equilateral triangle.
    • (iii) First place the proof attempt breaks: Current arguments from asymptotic optimality (triangular lattice density) do not directly imply the existence of an exact unit equilateral triangle in the finite extremiser, because one may imagine perturbations that preserve diameter while breaking exact unit-triangle equalities.
    • (iv) What a counterexample would have to look like: An infinite family of diameter-minimising sets $A_n$ with $\delta(A_n)=1$ and no triple of points forming an equilateral triangle of side $1$.

  7. Completion estimate.

    COMPLETION: 20%.

References

  • [Er86] P. Erdős, On some metric and combinatorial geometric problems, Discrete Mathematics 60 $1986$, 147--153.
  • [FiRe92] P. C. Fishburn and J. A. Reeds, Unit distances between vertices of a convex polygon, Computational Geometry 2 $1992$, no. 2, 81--91.
  • [EFPR93] P. Erdős, Z. Füredi, J. Pach, and I. Z. Ruzsa, The grid revisited, Discrete Mathematics 111 $1993$, 189--196.
  • [GK15] L. Guth and N. H. Katz, On the Erdős distinct distances problem in the plane, Annals of Mathematics 181 $2015$, 155--190.
  • [BeFo99] A. Bezdek and F. Fodor, Minimal diameter of certain sets in the plane, Journal of Combinatorial Theory, Series A 85 $1999$, no. 1, 105--111.