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Problem 1207

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claims/: The 1 claim page of Problem 1207, one per claimant's result; the problem's standing derives from them.


Statement. Let Pd(n)P_d(n) be such that in any set of nn points in Rd\mathbb{R}^d there exist at least Pd(n)P_d(n) many points which do not contain an isosceles triangle. Estimate Pd(n)P_d(n) - in particular, is it true that

P2(n)<n1−cP_2(n)<n^{1-c}

for some constant c>0c>0?

Status. Open. The site's label is OPEN (page last edited 7 September 2026). Its remark of that date credits [LPZ26] with answering the particular question and keeps the problem open for the general estimate of Pd(n)P_d(n).

Source. erdosproblems.com/1207, accessed 2026-09-04 and 2026-10-07. Cite as: T. F. Bloom, Erdős Problem #1207, https://www.erdosproblems.com/1207.

References.

  • [BMP05] Brass, Peter and Moser, William and Pach, János, Research problems in discrete geometry. (2005), xii+499.
  • [Er80] Erdős, Paul, A survey of problems in combinatorial number theory. Ann. Discrete Math. (1980), 89-115.
  • [LPZ26] S. Lee, C. Pohoata, and D. G. Zhu, The Minkowski grid has robustly many repeated distances. arXiv:2607.05374 (2026).
  • [PaTa02] Pach, János and Tardos, Gábor, Isosceles triangles determined by a planar point set. Graphs Combin. 18 (2002), 769-779.

Formalization. Statement in formal-conjectures.

Current assessment

The site's formulation asks for estimates of Pd(n)P_d(n), the size of an isosceles-free subset that every set of nn points in Rd\mathbb{R}^d is guaranteed to contain, and in particular whether P2(n)<n1−cP_2(n)<n^{1-c} for some c>0c>0. The site's label is OPEN.

Known bounds. By the site's remarks, Erdős [Er80] attributes the problem to Riddell and notes that Pd(n)>nϵdP_d(n)>n^{\epsilon_d} with ϵd→0\epsilon_d\to 0 as d→∞d\to\infty: a subset with all distances distinct contains no isosceles triangle, so Pd(n)≥fd(n)≥n1/(3d−3)−o(1)P_d(n)\ge f_d(n)\ge n^{1/(3d-3)-o(1)}, with fd(n)f_d(n) as in Problem 1208. In the plane, the bound of [PaTa02], Theorem 1 on the number of isosceles triangles, combined with random deletion, gives P2(n)≫εn2e/(5e−1)−εP_2(n)\gg_\varepsilon n^{2e/(5e-1)-\varepsilon} for every ε>0\varepsilon>0, where 2e/(5e−1)≈0.43182e/(5e-1)\approx0.4318. The case d=1d=1 asks for sets without three-term arithmetic progressions, the case k=3k=3 of Problem 201.

Claims. One result is claimed from outside the project. Lee, Pohoata and Zhu's Minkowski grid construction (arXiv, 6 July 2026, found with the help of ChatGPT) gives a planar set of nn points in which every subset of at least n1−δn^{1-\delta} points contains an isosceles triangle, so P2(n)<n1−cP_2(n)<n^{1-c} for some c>0c>0. It answers the particular question and leaves the general estimate open; it is a preprint without journal publication or Lean proof, so it stays claimed.

Not recorded as a claim. Section 5.3 of [BMP05] asserts that the regular polygon gives P2(n)≪n1/2P_2(n)\ll n^{1/2} but gives no details, and the site disputes the assertion: the isosceles triangles on the vertices of a regular polygon correspond to three-term progressions among the vertex indices, so the polygon bounds P2(n)P_2(n) only by about r3(n)r_3(n), the largest size of a progression-free subset of {1,…,n}\{1,\ldots,n\}. The assertion has no proof to record.

Search scope: the site's problem page and the arXiv record and text of [LPZ26]. Not searched: zbMATH and MathSciNet.

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