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<p>We give an alternative proof of the statement, by using elimination from algebraic geometry, that the only set <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(S\subset \mathbb {R}^2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>S</mi> <mo>⊂</mo> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mn>2</mn> </msup> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(\left| S\right| =6\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfenced close="|" open="|"> <mi>S</mi> </mfenced> <mo>=</mo> <mn>6</mn> </mrow> </math></EquationSource> </InlineEquation> such that all subsets that form a triangle are isosceles triangles, is the regular pentagon with its center. Our proof can be extended to answer some related questions raised by Erdős.</p>

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A note on Erdős’s mysterious remark

  • 期刊论文

  • 发表日期：

    2026年1月26日

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  • Zoltán Kovács

摘要

We give an alternative proof of the statement, by using elimination from algebraic geometry, that the only set \(S\subset \mathbb {R}^2\) S ⊂ R 2 , \(\left| S\right| =6\) S = 6 such that all subsets that form a triangle are isosceles triangles, is the regular pentagon with its center. Our proof can be extended to answer some related questions raised by Erdős.