Squaring the Circle with Charles Barkley

N.B. This post title is likely to be an even wider stretch in terms of obscure references than usual, but it is meant to evoke the recent CapitalOne ad with Charles Barkley driving Samuel L. Jackson and Spike Lee to "In-the-Annapolis" for the NCAA tournament.

The two larger squares are congruent, and the smaller square makes a 45 degree angle with one of the larger squares. Both larger squares touch the circle at one corner, while the smaller square touches the circle at two corners.

How many times greater is the area of one of the larger squares than the area of the smaller square?

Without loss of generality, let's assume that the circle has radius $1$ and is centered at the origin of the $xy$-plane. Let $x$ be the side length of the larger square and let $y$ be the side length of the smaller square. Since the two larger squares are each incident on the circle at a single point, the shared edge of the two larger squares must sit along the $y$-axis and we know that for instance the upper right corner of the uppermost larger square is at $(\sqrt{1-x^2}, x).$

Assuming the dashed line implies that the uppermost corner of the smaller square lies on the extended line between the two upper corners of upper larger square. Thus, it must lie at the point $(-\sqrt{1-x^2}, x)$. Since the isoceles right triangle bounded by the dashed line, the upper larger square and the smaller square has hypotenuse of length $y$, the length of the dashed line must be $y / \sqrt{2},$ so we have the following relation: $$x + \frac{y}{\sqrt{2}} = 2\sqrt{1-x^2}.$$

Given that the smaller square has side length $y,$ the other point of the smaller square that is incident on the circle must be at $(-\sqrt{1-x^2} - \frac{y}{\sqrt{2}}, x - \frac{y}{\sqrt{2}}).$ Since this point is incident on the circle, we must additionally have $$\left(-\sqrt{1-x^2} - \frac{y}{\sqrt{2}}\right)^2 + \left(x - \frac{y}{\sqrt{2}}\right)^2 = 1.$$ Solving for $y / \sqrt{2}$ in the first equation and plugging into the second we see \begin{align*} 1 &= (x-3\sqrt{1-x^2})^2 + 4(x - \sqrt{1-x^2})^2\\ &= x^2 - 6x\sqrt{1-x^2} + 9(1-x^2) + 4(1 - 2x\sqrt{1-x^2} + (1-x^2))\\ &= 13 - 14x\sqrt{1-x^2} -8x^2, \end{align*} or equivalently $$49x^2(1-x^2) = 36 - 48x^2 + 16x^4 \,\,\Rightarrow\,\, 0 = 36 - 97x^2 + 65x^4.$$ Solving the quadratic equation we get $$x^2 = \frac{97 \pm \sqrt{97^2 - 4 \cdot 65 \cdot 36}}{2 \cdot 65} = \frac{97 \pm 7}{130}.$$ Ignoring nonsensical negative values, we get that $x_\pm = \sqrt{ \frac{97 \pm 7}{130} },$ which implies that $$y_\pm = \sqrt{2} \left( 2 \sqrt{1 - \frac{97 \pm 7}{130}} - \sqrt{ \frac{97 \pm 7}{130}} \right) = \frac{1}{\sqrt{65}} \left( \sqrt{132 \mp 28} - \sqrt{97 \pm 7} \right).$$ So in particular, when $x = x_+ = \sqrt{\frac{104}{130}} = \frac{2}{\sqrt{5}}$ implies a value of $$y = y_+ = \frac{\sqrt{104} - \sqrt{104}}{\sqrt{65}} = 0,$$ which doesn't seem to be the answer we were looking for. However, when $x = x_- = \sqrt{\frac{90}{130}} = \frac{3}{\sqrt{13}},$ we have $$y = y_- = \frac{\sqrt{160} - \sqrt{90}}{\sqrt{65}} = \frac{4\sqrt{10} - 3\sqrt{10}}{\sqrt{65}} = \frac{\sqrt{2}}{\sqrt{13}},$$ which makes one of the larger squares $(x / y)^2 = 4.5$ times larger than the smaller sqaure.