Angles, Arcs and Areas!

By the exterior angle theorem on $\triangle ADC$ , $\angle DAC = \angle BDA - \angle DCA = 60° - 45° = 15°$ .

By the law of cosines on $\triangle ADC$ , $\cfrac{AD}{\sin \angle DCA} = \cfrac{DC}{\sin \angle DAC}$ , or $\cfrac{AD}{\sin 45°} = \cfrac{1}{\sin 15°}$ , so $AD = \cfrac{\sin 45°}{\sin 15°} = \sqrt{3} + 1$ .

Now construct equilateral $\triangle EBD$ where $E$ is on $AD$ :

Then $\angle BEA = 180° - \angle BED = 180° - 60° = 120°$ and $AE = AD - ED = \sqrt{3} + 1 - 2 = \sqrt{3} - 1$ .

Since $\angle AEB = \angle ADC = 120°$ , and $\cfrac{BE}{EA} = \cfrac{AD}{DC} = \sqrt{3} + 1$ , $\triangle BEA \sim \triangle ADC$ by SAS similarity, so $\angle EBA = \angle DAC = 15°$ and $\angle EAB = \angle DCA = 45°$ .

That means $\angle ABC = \angle EBA + \angle EBD = 15° + 60° = 75°$ and $\angle BAC = \angle EAB + \angle DAC = 45° + 15° = 60°$ .

Since $\angle ABC = 75°$ , $m(\widehat{ARC}) = 2 \cdot 75° = 150° = \cfrac{150°\pi}{180°} = \cfrac{5\pi}{6}$ .

Since $\triangle ABC$ is circumscribed in the circle, the circle's radius is $R = \cfrac{BC}{2 \sin \angle BAC} = \cfrac{3}{2 \sin 60°} = \sqrt{3}$ , and its area is $A_c = \pi R^2 = 3\pi$ .

Therefore, $A_c + m(\widehat{ARC}) = 3\pi + \cfrac{5\pi}{6} = \cfrac{23\pi}{6}$ , which means $a = 23$ , $b = 6$ , and $a + b = \boxed{29}$ .

$\sin(15^{\circ}) = \sin(45^{\circ} - 30^{\circ}) = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}$

Using the law of sines on $\triangle{ADC} \implies$

$\dfrac{\overline{AD}}{\sin(45^{\circ})} = \dfrac{1}{\sin(15^{\circ})} \implies \overline{AD} = \dfrac{2}{\sqrt{3} - 1} = \sqrt{3} + 1$

and

$\dfrac{\overline{AC}}{\sin(120^{\circ})} = \dfrac{\sqrt{3} + 1}{\sin(45^{\circ})} \implies \overline{AC} = \dfrac{3 + \sqrt{3}}{\sqrt{2}}$

Using the law of cosines on $\triangle{ABC}$ with included $\angle{C} = 45^{\circ} \implies$
$\overline{AB} = \sqrt{6}$

Using the law of sines on $\triangle{ABD} \implies$

$\dfrac{\sqrt{6}}{\sin(60^{\circ})} = \dfrac{\sqrt{3} + 1}{\sin(\theta)} \implies \sin(\theta) = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \implies \theta = 75^{\circ}$

$\implies m(\widehat{ARC}) = 150^{\circ}$ and converting to radians $\implies \boxed{m(\widehat{ARC}) = \dfrac{5\pi}{6}}$

To find the area $A_{c}$ of the above circle we use coordinates below with $m\angle{BAD} = 45^{\circ} \implies m\angle{BAC} = 60^{\circ}$

$A(0,0), B(\sqrt{6}\cos(60^{\circ}), \sqrt{6}\sin(60^{\circ})) = (\sqrt{\dfrac{3}{2}}, \dfrac{3}{\sqrt{2}})$ and $C(\dfrac{3 + \sqrt{3}}{\sqrt{2}}, 0)$

$(1) x_{0}^2 + y_{0}^2 = r^2$

$(2) x_{0}^2 - (3 + \sqrt{3})\sqrt{2}x_{0} + 6 + 3\sqrt{3} + y_{0}^2 = r^2$

$(3) x_{0}^2 - \sqrt{6}x_{0} + y_{0}^2 - 3\sqrt{2}y_{0} + 6 = r^2$

Subtracting $(1)$ from $(2)$ we obtain $x_{0} = \dfrac{(3 + \sqrt{3})\sqrt{2}}{4}$

Subtracting $(1)$ from $(3)$ we obtain $y_{0} = \dfrac{(3 -\sqrt{3})\sqrt{2}}{4}$

$\implies r^2 = 3 \implies r = \sqrt{3} \implies \boxed{A_{c} = 3\pi} \implies$ $A_{c} + m(\widehat{ARC}) = \dfrac{23\pi}{6} = \dfrac{a\pi}{b}$

$\implies a + b = \boxed{29}$ .