Let two squares of side lengths \(a\) and \(b\) with \(b<a\) be positioned so that their centers coincide and that corresponding sides are parallel, as shown. Draw a square \(\mathcal{S}\) so that it is inscribed inside the square of side length \(a\) but circumscribes the square of side length \(b\).

What restrictions must be placed on \(a\) and \(b\) so that such a square exists?

Show that under the conditions laid out in the previous part, square \(\mathcal{S}\) has side length equal to \(\sqrt{ab}\).

NOTE: This discovery was first found by a friend of mine in response to a problem I had created.

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TopNewestLet \( \theta \) be the smaller angle that the diagonal makes with one of the sides. Let \( d \) be the side of the middle square. Then \( a = d (\sin \theta + \cos \theta) \) and \( d = b (\sin \theta + \cos \theta) \). So \( a = b(\sin \theta + \cos \theta)^2 \). It's easy to see that \( 1 \le \sin \theta + \cos \theta \le \sqrt{2}, \) so the condition boils down to \( a \le 2b \).

The area of the square is \( d^2 = b^2 (\sin \theta + \cos \theta)^2 = ab \), so the side is \( \sqrt{ab} \). – Patrick Corn · 2 years, 4 months ago

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2b>a – Сергей Кротов · 2 years, 4 months ago

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