Waste less time on Facebook — follow Brilliant.

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
3 years, 1 month ago

No vote yet
1 vote


Sort by:

Top Newest

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 \( \sqrt{ab} \). Patrick Corn · 3 years, 1 month ago

Log in to reply

2b>a Сергей Кротов · 3 years, 1 month ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...