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Area of Square Inside Two Concentric Squares

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 $$b$$.

1. What restrictions must be placed on $$a$$ and $$b$$ so that such a square exists?

2. 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.

Note by David Altizio
2 years, 11 months ago

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Let $$\theta$$ 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 $$\sqrt{ab}$$. · 2 years, 11 months ago