Ratios!!

$\overline{OC} = \dfrac{\sqrt{5}}{2}s \implies \overline{PC} = s$

and

$a^2 + (b - \dfrac{s}{2})^2 = \dfrac{s^2}{4}$

$(a - s)^2 + (b - s)^2 = s^2$

$\implies$

$a^2 + b^2 - bs = 0$

$-1| (a^2 + b^2 - 2bs - 2as = -s^2)$

$\implies b + 2a = s \implies b = s - 2a \implies a(5a - 2s) = 0$ and $a \neq 0 \implies$ $a = \dfrac{2s}{5} \implies b = \dfrac{s}{5} \implies$

$P(\dfrac{2s}{5}, \dfrac{s}{5})$ and $C(s,s) \implies m_{PC} = \dfrac{4}{3} \implies y - s = \dfrac{4}{3}(x - s)$ and $y = 0 \implies$

$x = \dfrac{s}{4} \implies \overline{QD} = \dfrac{3s}{4} \implies A_{\triangle{QCD}} = \dfrac{3}{8}s^2 \implies$

$\dfrac{A_{\triangle{QCD}}}{A_{ABCD}} = \dfrac{3}{8} = \dfrac{\alpha}{\beta} \implies \alpha + \beta = \boxed{11}$ .

Solution 1:

Without loss of generality, let the side of the square be $1$ . Also let $O$ be the center of the semicircle and draw in $OC$ and $OP$ :

Since $\angle OBC = \angle OPC = 90°$ , $OB = OP = \frac{1}{2}$ , and $OC = OC$ , $\triangle OBC \cong \triangle OPC$ by HL congruence, so $\angle BCO = \angle PCO$ by corresponding parts.

Let $\alpha = \angle BCO = \angle PCO$ , $\beta = \angle QCD$ , and $b = QD$ . From $\triangle OBC$ , $\tan \alpha = \frac{\frac{1}{2}}{1} = \frac{1}{2}$ , and from $\triangle QCD$ , $\tan \beta = \frac{b}{1} = b$ .

By the angle sum of $\angle BCD$ , $2\alpha + \beta = 90°$ . Substituting $\tan \alpha = \frac{1}{2}$ and $\tan \beta = b$ and then dividing by $2$ gives $\tan^{-1} \frac{1}{2} + \frac{1}{2} \tan^{-1} b = 45°$ .

Letting $T = \tan(\frac{1}{2} \tan^{-1} b)$ , then $\tan(\tan^{-1} \frac{1}{2} + \tan^{-1} T) = \tan 45°$ , or applying the tangent sum equation, $\cfrac{\frac{1}{2} + T}{1 - \frac{1}{2}T} = 1$ , which solves to $T = \frac{1}{3}$ .

Therefore, $b = \tan(2 \tan^{-1} T) = \cfrac{2T}{1 - T^2} = \cfrac{2 \cdot \frac{1}{3}}{1 - (\frac{1}{3})^2} = \cfrac{3}{4}$ , so that $\cfrac{A_{\triangle QCD}}{A_{ABCD}} = \cfrac{\frac{1}{2} \cdot \frac{3}{4} \cdot 1}{1^2} = \cfrac{3}{8}$ , which means $\alpha = 3$ , $\beta = 8$ , and $\alpha + \beta = \boxed{11}$ .

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Solution 2:

Without loss of generality, let $AB = BC = CD = AD = 1$ , and place the whole diagram on a coordinate graph so that $A$ is at $(0, 0)$ and $C$ is at $(1, 1)$ .

The equation of the semicircle is then $x^2 + (y - \frac{1}{2})^2 = \frac{1}{4}$ , and the equation of the line $CQ$ with slope $m$ is $y = m(x - 1) + 1$ .

Substituting $y = m(x - 1) + 1$ into $x^2 + (y - \frac{1}{2})^2 = \frac{1}{4}$ and solving for $x$ gives $(m^2 + 1)x^2 + (m - 2m^2)x + (m^2 - m) = 0$ .

For a tangent line, the discriminant of $x$ will be $0$ , so $(m - 2m^2)^2 - 4(m^2 + 1)(m^2 - m) = -m(3m - 4) = 0$ , which solves to $m = \frac{4}{3}$ for $m > 0$ .

The equation of the line $CQ$ is then $y = \frac{4}{3}(x - 1) + 1$ , which solving $0 = \frac{4}{3}(x - 1) + 1$ gives an $x$ -intercept of $x = \frac{1}{4} = AQ$ .

Therefore, $QD = 1 - AQ = 1 - \frac{1}{4} = \frac{3}{4}$ , so that $\cfrac{A_{\triangle QCD}}{A_{ABCD}} = \cfrac{\frac{1}{2} \cdot \frac{3}{4} \cdot 1}{1^2} = \cfrac{3}{8}$ , which means $\alpha = 3$ , $\beta = 8$ , and $\alpha + \beta = \boxed{11}$ .