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 aa and bb so that such a square exists?

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

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

#Geometry #Square #Proofs

Note by David Altizio
7 years ago

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Comments

2b>a

Сергей Кротов - 7 years ago

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

The area of the square is d2=b2(sin⁡θ+cos⁡θ)2=ab d^2 = b^2 (\sin \theta + \cos \theta)^2 = ab , so the side is ab \sqrt{ab} .

Patrick Corn - 7 years ago
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