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Does there exist for every $\\varepsilon>0$ a contraction $f\\colon E\\to \\mathbb{R}^{2}$ such that $\\lambda^{2}(f(E))\\geq \\lambda^{2}(E)-\\varepsilon$ and $f(E)$ is a polygon? We answer this question in the negative by constructing a bounded, simply connected open counterexample. 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