Two rings

Let's say that the two rings are given by $x^2+y^2=1$ and $x^2+z^2=1$ ; at the points of intersection we have $y^2=z^2$ . If the width of the rings is $a$ , then the coordinates of the points of contact are $y=\pm \frac{a}{2}$ and $z=a-1$ . It is required that $\frac{a^2}{4}=(a-1)^2$ , with the smaller solution $a=\frac{2}{3}$ . The ratio we seek is $\frac{2}{a}=\boxed{3}$ .

A nice problem, charming and playful!

The picture on the left shows a side view of the two rings, and the picture on the right shows the top view.

Let $r$ be the radius of one ring and $w$ be the width of one ring. Since $r = AB = A'B'$ , $BC = B'C'$ , and $\angle C = \angle C'$ , $\triangle ABC \cong \triangle A'B'C'$ by hypotenuse-leg theorem, which means $AC = r - w = A'C'$ . Therefore, $2(r - w) = w$ , or $2r = 3w$ .

The ratio of the diameter to the width is then $\frac{2r}{w} = \frac{3w}{w} = \boxed{3}$ .