2018 AMC 12B Problem #25

Let $A$ be the center of $\omega_3$ , $C$ be the center of $\omega_2$ , and $B$ be the intersection of segments $P_2P_3$ and $AC$ . We will then examine $\triangle AP_3B$ and $\triangle CP_2B$ .

Let $r$ be the radius of either circle. Then $AP_3 = CP_2 = r$ and $AC = 2r$ . Since $P_2P_3$ is tangent to circle $\omega_2$ , $\angle CP_2B = 90°$ . Since $P_1P_3$ is tangent to circle $\omega_3$ , $\angle AP_3P_1 = 90°$ , and since $\triangle P_1P_2P_3$ is an equilateral triangle, $\angle P_1P_3P_2 = 60°$ , which means $\angle AP_3B$ $=$ $\angle AP_3P_1 - \angle P_1P_3P_2$ $=$ $90° - 60°$ $=$ $30°$ .

By law of sines on $\triangle AP_3B$ , $\frac{\sin \angle ABP_3}{AP_3} = \frac{\sin \angle AP_3B}{AB}$ or $\frac{\sin \angle ABP_3}{r} = \frac{\sin 30°}{AB}$ , and by law of sines on $\triangle CP_2B$ , $\frac{\sin \angle CBP_2}{CP_2} = \frac{\sin \angle CP_2B}{BC}$ or $\frac{\sin \angle CBP_2}{r} = \frac{\sin 90°}{BC}$ . Since $\angle ABP_3 = \angle CBP_2$ as vertical angles, we can combine the two ratios and arrive at $\frac{\sin 30°}{AB} = \frac{\sin 90°}{BC}$ , and simplifying this gives $BC = 2AB$ . Since $AB + BC = AC = 2r$ , $AB + 2AB = 2r$ , so $AB = \frac{2}{3}r$ , and $BC = 2AB = 2\frac{2}{3}r = \frac{4}{3}r$ .

Using $\frac{\sin \angle ABP_3}{r} = \frac{\sin 30°}{AB}$ once again along with $AB = 2r$ , we find that $\angle ABP_3 = \sin^{-1}\frac{3}{4}$ . Since the sum of the angles of $\triangle AP_3B$ is $180°$ , $\angle BAP_3$ $=$ $180° - \angle AP_3B - \angle ABP_3$ $=$ $180° - 30° - \sin^{-1}\frac{3}{4}$ $=$ $150° - \sin^{-1}\frac{3}{4}$ . Therefore, $\sin \angle BAP_3$ $=$ $\sin (150° - \sin^{-1}\frac{3}{4})$ $=$ $\sin 150° \cos(\sin^{-1}\frac{3}{4}) - \cos 150° \sin(\sin^{-1}\frac{3}{4})$ $=$ $\frac{\sqrt{7} + 3\sqrt{3}}{8}$ .

By law of sines on $\triangle AP_3B$ once again, $\frac{\sin \angle AP_3B}{AB} = \frac{\sin \angle BAP_3}{BP_3}$ or $\frac{\sin 30°}{\frac{2}{3}r} = \frac{\frac{\sqrt{7} + 3\sqrt{3}}{8}}{BP_3}$ , which means $BP_3 = (\frac{\sqrt{7} + 3\sqrt{3}}{6})r$ .

By Pythagorean's Theorem, $BP_2 = \sqrt{BC^2 - CP_2^2} = \sqrt{(\frac{4}{3}r)^2 - r^2} = \frac{\sqrt{7}}{3}r$ .

Since $P_2P_3 = BP_2 + BP_3$ , $P_2P_3$ $=$ $\frac{\sqrt{7}}{3}r + (\frac{\sqrt{7} + 3\sqrt{3}}{6})r$ $=$ $\frac{r}{2}(\sqrt{7} + \sqrt{3})$ .

The area of an equilateral triangle formed by side lengths $P_2P_3$ would then be $A$ $=$ $\frac{\sqrt{3}}{4}(\frac{r}{2}(\sqrt{7} + \sqrt{3}))^2$ $=$ $\frac{r^2}{16}(10\sqrt{3} + 6\sqrt{7})$ .

In the problem, $r = 4$ , so $A = \frac{4^2}{16}(10\sqrt{3} + 6\sqrt{7}) = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252}$ , so $a = 300$ and $b = 252$ , and $a + b = 300 + 252 = \boxed{552}$ .