Circles, Isosceles Right Triangles and Squares

Let the midpoint of $AD$ be $O$ , the origin, let $H$ be the center of the square, and let $F$ have coordinates $F(k, k)$ .

Then $EF = 1 - k$ , so $EH = HI = \frac{1}{2}(1 - k)$ , which means $I$ has coordinates $I(k - \frac{1}{2}(1 - k), k + \frac{1}{2}(1 - k)) = I(\frac{3}{2}k - \frac{1}{2}, \frac{1}{2}k + \frac{1}{2})$ .

$I$ is also on the quarter circle $(x + 1)^2 + y^2 = 1$ , so $(\frac{3}{2}k - \frac{1}{2} + 1)^2 + (\frac{1}{2}k + \frac{1}{2})^2 = 1$ , which solves to $k = \frac{1}{5}$ .

The area of the square is then $\frac{1}{2}EF^2 = \frac{1}{2}(1 - k)^2 = \frac{1}{2}(1 - \frac{1}{5})^2 = \frac{8}{25}$ , so $a = 8$ , $b = 25$ , and $a + b = \boxed{33}$ .

Using the above diagram:

$\overline{MC} = \sqrt{2}$ and $\overline{BE} = \overline{BC} - \overline{EC} = 2 - \sqrt{2}x \implies \overline{AH} = \overline{BE} - \overline{JE} = \dfrac{2\sqrt{2} - 3x}{\sqrt{2}}$

$= \cos(\theta)$ and $\overline{MF} = \overline{MC} - \overline{FC} = \sqrt{2} - 2x$

Using the law of cosines on $\triangle{AIM} \implies \overline{IM}^2 = 2(1 - \cos(\theta)) = 3\sqrt{2}x - 2$

Using the Pythagorean theorem on $\triangle{IFM} \implies x^2 + (\sqrt{2} - 2x)^2 = 3\sqrt{2}x - 2$

$\implies 5x^2 - 4\sqrt{2}x + 2 = 3\sqrt{2}x - 2 \implies 5x^2 - 7\sqrt{2}x + 4 = 0 \implies$

$x = \dfrac{7\sqrt{2} \pm 3\sqrt{2}}{10} = \sqrt{2}, \dfrac{2\sqrt{2}}{5}$ and $x \neq \sqrt{2} \implies x = \dfrac{2\sqrt{2}}{5} \implies A_{s} = \dfrac{8}{25}$

$= \dfrac{a}{b} \implies a + b = \boxed{33}$ .

If we draw a rectangle with diagonal IC and sides parallel to the rectangle ABCD, it's easy to see that the ratio of its sides is 3 to 1 (using Rocco's diagram as reference, this rectangle would be IJCK, with K being the intersection point of extrapolated line IG & CD). All we need to do is applying Pythagoras to find the width / height (noted as x) and the area would be double of its square (in fraction form) and finally the answer will appear after summing the nominator and denominator together.

(1 – x)² + (2 – 3x)² = 1²
10x² – 14x + 5 = 1
10x² – 14x = 1 – 5
(x – 0.7)² = 0.7² – 0.4 = 0.09
x = 0.7 ± √0.09
= 0.7 ± 0.3
Since 0 ≤ x < 1, x = 0.4

Area
= 2x²
= 2 × (2/5)²
= 8/25

Answer
= 8 + 25 = 33