Puzzle:

A square has its corner points on an outer circle with its lines as tangents of an inner circe. All three shapes have the same middle point.

What is the radius of the outter circle r₂, given that the inner circle has the radius r₁

Show solution

The answer is \(\sqrt{2*r_1^2}\).

First, recognize that the point at which the square touches the inner circle is the middle point of the square line and is r₁ distanced from the middle point.

Since it is a square that means that the length of each side can then be described as \(2*r_1\)

Since we know the length of the sides of the square we can now use the pythagorean theorem to determine the length of the squares corner points which are on the outer circle to the inner point, which is the radius of the outer circle.

The radius of the outer circle is therefore \(r_1^2 + r_1^2 = r_2^2 \Rightarrow sqrt(2*r_1^2) = r_2\)