Valentines Day Problem 1.

Using the diagram above and the law of cosines on $\triangle{CEF}$ with included $m\angle{ECF} = 45^{\circ}$ we have:

$(r - \dfrac{a}{2})^2 = (\sqrt{2}a - r)^2 + \dfrac{a^2}{4} - 2(\dfrac{a}{2})(\dfrac{\sqrt{2}a - r}{\sqrt{2}} ) \implies$

$4r^2 - 4ar + a^2 = 8a^2 - 8\sqrt{2}ar - 4r^2 + a^2 - 2\sqrt{2}a(\sqrt{2}a - r) \implies$

$-4ar = -6\sqrt{2}ar + 4a^2 \implies 2(3\sqrt{2} - 2)ar = 4a^2 \implies r = \dfrac{3\sqrt{2} + 2}{7}a$

$\implies \dfrac{A_{c}}{A_{s}} = \dfrac{(3\sqrt{2} + 2)^2}{49}\pi = \dfrac{2(11 + 6\sqrt{2})}{49}\pi =$

$\dfrac{\alpha(\beta + \gamma\sqrt{\alpha})}{\omega}\pi \implies \alpha + \beta + \gamma + \omega = \boxed{68}$ .

Label the diagram as follows, and let the sides of the square be $s$ and the radius of the blue circle be $r$ .

By the properties of a square, $EG = EB = EC = EH = \cfrac{s}{2}$ , $GA = \cfrac{\sqrt{2}s}{2}$ , and $\angle EGF = 45°$ .

Then $EF = HF - HE = r - \cfrac{s}{2}$ and $FG = FA - GA = r - \cfrac{\sqrt{2}s}{2}$ .

By the law of cosines on $\triangle EFG$ , $EF^2 = EG^2 + FG^2 - 2 \cdot EG \cdot FG \cdot \cos \angle EGF$ , or $\bigg(r - \cfrac{s}{2}\bigg)^2 = \bigg(\cfrac{s}{2} \bigg)^2 + \bigg(r - \cfrac{\sqrt{2}s}{2} \bigg)^2 - 2 \cdot \bigg(\cfrac{s}{2} \bigg) \cdot \bigg(r - \cfrac{\sqrt{2}s}{2} \bigg) \cdot \cos 45°$ , which solves to $r = \cfrac{2 + 3\sqrt{2}}{7}s$ .

Therefore, $\cfrac{A_c}{A_s} = \cfrac{\pi r^2}{s^2} = \cfrac{\pi (\frac{2 + 3\sqrt{2}}{7}s)^2}{s^2} = \cfrac{2(11 + 6\sqrt{2})}{49}\pi$ , so that $\alpha = 2$ , $\beta = 11$ , $\gamma = 6$ , $\omega = 49$ , and $\alpha + \beta + \gamma + \omega = \boxed{68}$ .