Dynamic Geometry: P20

Let the angle between the two black lines be $\theta$ . Then $\tan \theta = \frac {55}{48}$ , $\sin \theta = \frac {55}{73}$ , $\cos \theta = \frac {48}{73}$ , and $\tan \frac \theta 2 = \frac {1-\cos \theta}{\sin \theta} = \frac {73-48}{55} = \frac 5{11}$ . We note the center of the orange circle is on the line $y=\frac 5{11}x$ . Let the center of the orange circle be $(u,r)$ . Then $r = \frac 5{11}u$ . Let the center of the green circle (the red point) be $P(x,y)$ . Then $y = r^3 = \frac {125}{1331}u^3$ and

$x = u + \sqrt{(r^3+r)-(r^3-r)} = u + 2r^2 = u + \frac {50}{121}u^2$

Then the area bounded by the blue curve, $y=0$ , and $x=18$ is:

$\begin{aligned} A & = \int_0^{18} y \ dx & \small \blue{\text{Note that }x = u + \frac {50}{121}u^2 \implies dx = \left(1 + \frac {100}{121}u \right)du} \\ & = \int_0^\blue{\frac {11}2} \frac {125}{1331}u^3 \left(1+\frac {100}{121}u\right)du & \small \blue{\text{and }u + \frac {50u^2}{121} = 18 \implies u = \frac {11}2} \\ & = \frac {125}{1331} \left[\frac {u^4}4 + \frac {20u^5}{121} \right]_0^\frac {11}2 \\ & = \frac {6375}{64} \end{aligned}$

Therefore $a+b = 6375+64 = \boxed{6439}$ .