Square circle square

Let the side length of the outer square be $1$ . Then the ratio $\dfrac {\overline{IJ}}1 = \overline{IJ} = 1 - \cos 15^\circ$ . Since $2\cos^2 15^\circ - 1 = \cos 30^\circ = \dfrac {\sqrt 3}2$ , $\implies \cos 15^\circ = \dfrac {\sqrt{2+\sqrt 3}}2$ and $\overline{IJ} = \dfrac {2-\sqrt{2+\sqrt 3}}2$ . Therefore $a = \boxed 2$ .

So I kinda cheated on this problem. Sorry :)

$\frac{a-\sqrt{a+\sqrt{a+1}}}{a}$ is always increasing from $0$ to positive infinity, so we know that if we come across an answer that is obviously too big, all numbers above it will also be too big. We also know that $a$ must be an integer (because the answer slot only accepts integers)

If $a=1$ , then $\frac{a-\sqrt{a+\sqrt{a+1}}}{a}$ is negative, and if $a=3$ , then $\frac{a-\sqrt{a+\sqrt{a+1}}}{a}$ is $0.25$ . It is obvious that the ratio cannot be negative, and that $IJ$ is obviously not $\frac{1}{4}$ of the outer square side length. Therefore, $a$ must equal $2$ .

And if we check, $a=2$ gives $\frac{a-\sqrt{a+\sqrt{a+1}}}{a} \approx .03$ , which makes logical sense.