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Exact area.

edited Sep 28, 2017 at 0:45

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I don't know if this is the optimal, but an isosceles triangle with base and height $\sqrt{2}$ overlaps $2 \left(\sqrt{2}-1\right) \approx 0.828427$ when placed as below, and so improves over $\frac{3}{4}$:

[![SquareTri][1]][1]

answered Sep 28, 2017 at 0:32

Joseph O'Rourke

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