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Claim. The answer to Problem 189 is no. Theorem 3 of Kovač gives an explicit Jordan-measurable coloring of $\mathbb{R}^2$ in $25$ colors such that no color class contains the four vertices of a rectangle of area $1$. Since the question asks for a color class containing the vertices of a rectangle of every area, a single excluded area refutes it. Theorem 4 of the same paper extends the result to $\mathbb{R}^n$ with a different finite coloring, avoiding monochromatic $m$-dimensional boxes of $m$-volume $1$ for every $m\leq n$.

Acceptance. The paper is refereed: Vjekoslav Kovač, Coloring and density theorems for configurations of a given volume, Proc. Lond. Math. Soc. (3) 132 (2026), no. 3, e70143, first posted as arXiv:2309.09973 on 2023-09-18. Thomas Bloom, the site's curator, marks the problem disproved and credits Kovač's coloring on the problem page. Kovač announced a Lean formalization of the coloring, produced by Aristotle (Harmonic) from a LaTeX blueprint Kovač wrote, in the problem's discussion on 2025-12-17. A second development, in Boris Alexeev's repository of formalized Erdős problems from 2025-12-19, names Kovač's blueprint as its source, auto-formalized by Aristotle (Harmonic), and its later Mathlib versions list Kovač as informal author and Aristotle and Kovač as formal authors; it proves the formal-conjectures statement of the problem after Alexeev corrected that statement's definition of a rectangle, which had admitted some trapezoids. The site's label notes this verification. Both developments are linked above at pinned commits: Kovač's file, and the Lean source in Alexeev's repository for the Mathlib version that the formal-conjectures file cites, beside the repository's index of its versions of that proof. Neither has been built or audited by this corpus, so neither is listed as evidence.

What remains. Graham proved the positive statement for right-angled triangles in place of rectangles. The question for parallelograms was open as the parallelogram variant of the formal-conjectures file records; Kovač's Theorem 6 settles it only for parallelograms with a side in one of finitely many fixed directions or with all angles bounded away from zero.