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Source. The question on pp. 122--123 of P. Erdős, Set-theoretic, measure-theoretic, combinatorial, and number-theoretic problems concerning point sets in Euclidean space, Real Anal. Exchange 4 (1978/79), no. 2, 113--138, doi:10.2307/44151159, as identified on the source card. Pages are those of the journal print; the question is unnumbered.

Read depth. Claims checked: the passage was read clause by clause on pp. 122--123.

Statement

The question (p. 122, quoted). "Is it true that there is an absolute constant $C$ so that if $S$ has planar measure greater than $C$ then $S$ contains the vertices of a triangle area [sic] $1$?"

Here $S$ is a plane set and the triangle is one of area $1$. The example (pp. 122--123): the open disc $S=\{|z|<2\cdot3^{-3/4}\}$ contains no triangle of area $1$, because the triangle of largest area inscribed in a circle is equilateral, and an equilateral triangle inscribed in the circle of radius $2\cdot3^{-3/4}$ has area exactly $1$. Its area is $4\pi3^{-3/2}$, and the paper suggests that this may be the correct value of $C$, adding that it has no real evidence for this. It notes that such problems can also be posed in higher dimensions (p. 123).

Proof pointer

None for the question. The example rests on the inscribed-triangle fact the paper cites as well known: an equilateral triangle inscribed in a circle of radius $R$ has area $3\sqrt3R^2/4$, which is $1$ at $R=2\cdot3^{-3/4}$, and no triangle with vertices in the open disc reaches that area.

Dependencies

None.

Bears on

• Problem 352: the question is the problem's, which the site cites to this paper among others. The site asks for some $c>0$ with every measurable set of measure at least $c$ containing such a triangle; a constant works for one wording exactly when some constant works for the other. The disc example shows that a constant $C$ in the paper's wording is at least $4\pi3^{-3/2}$, and a constant $c$ in the site's wording exceeds it, since the open disc itself has that measure; the paper gives no upper bound and does not answer the question.