Inscribing Squares Since 2K

Construction: Let $O$ be the centre of the circle. Join $OA$ , $OE$ and $OF$ . Let $J$ be a point on side $EF$ such that $OJ \perp EF$ . $OJ$ intersects side $AB$ at $I$ .

Let the side lengths of the larger and smaller square be $A$ and $a$ respectively. Then $(2R)^2 = A^2 + A^2 \implies 2R^2 = A^2$ .

We have $OE = OF = R$ , so $\Delta EOF$ is isosceles. Hence, $OJ$ is also the median to side $EF$ . $\implies EJ = JF = \frac{EF}{2} = \frac{a}{2}$ .

Also, $\angle AIO = 90^{\circ}$ . $\implies AI = IB = \frac{A}{2}$ and $\angle IAO = 45^{\circ} \implies AI = IO \implies IO = \frac{A}{2}$ .

Now, since $\Delta EJO$ is right angled at $J$ , applying the Pythagorean Theorem,

$EJ^2 + JO^2 = OE^2 \implies \left(\frac{a}{2}\right)^2+\left(IJ\ +\ IO\right)^2=R^2 \implies \left(\frac{a}{2}\right)^2+\left(a\ +\ \frac{A}{2}\right)^2=R^2 \implies a^2\left(\frac{1}{4}\right)+a^2\left(1+\ \frac{A}{a}\cdot \frac{1}{2}\right)^2=\frac{A^2}{2}$

Dividing throughout by $a^2$ , we have:

$\left(\frac{1}{4}\right)+\left(1+\ \frac{A}{a}\cdot \frac{1}{2}\right)^2=\left(\frac{A}{a}\right)^2\cdot \frac{1}{2}$

Let $\frac{A}{a} = k$ , then the above equation becomes:

$\frac{1}{4}+\left(1+\ \frac{k}{2}\right)^2=\frac{k^2}{2}$

Solving the above quadratic equation, we have $k=5$ or $-1$ . We reject $-1$ as the ratio of the sides cannot be negative. So, $k = 5 \implies \boxed{\frac{A}{a} = 5}$ . Hence, ratio of their areas $= \frac{A^2}{a^2} = \left(\frac{A}{a}\right)^2 = 5^2 = \boxed{25}$