(9) - Outcoming

Let $M$ be the origin $(0,0)$ of the $xy$ -plane, the coordinates of $N$ be $(x,y)$ , the radius of the light orange be $r$ , $AM=5$ , $AP = 6$ , hence $A(-5,0)$ , $P(1,0)$ , the center of the blue semicircle $Q(-2,0)$ , $BP=4$ , hence $B(5,0)$ , the center of orange semicircle $R(3,0)$ . Using Pythagorean theorem we have:

$\begin{cases} MN^2 = (x-0)^2 + (y-0)^2 = (5-r)^2 & \implies x^2 + y^2 = r^2 - 10r + 25 & ...(1) \\ QN^2 = (x+2)^2 + (y-0)^2 = (3+r)^2 & \implies x^2 + 4x+ y^2 = r^2 +6r + 5 & ...(2) \\ RN^2 = (x-3)^2 + (y-0)^2 = (2+r)^2 & \implies x^2 -6x+ y^2 = r^2 +4r -5 & ...(3) \end{cases}$

$\begin{cases} (2)-(1): & 4x = 16r - 20 & \implies x = 4r - 5 & ...(4) \\ (1)-(3): & 6x = -14r + 30 & \implies 3x = -7r + 15 & ...(5) \end{cases}$

From $3(4)-(5): \quad 0 = 19 r -30 \implies r = \dfrac {30}{19}$ . From $(4): \quad x = 4\times \dfrac {30}{19} -5 = \dfrac {25}{19}$ .

Now we have:

$\begin{aligned} \frac {NP}{MN} & = \frac {\sqrt{(x-1)^2+y^2}}{\sqrt{x^2+y^2}} \\ & = \frac {\sqrt{{\color{#3D99F6}x^2+y^2}-2{\color{#D61F06}x}+1}}{\sqrt{\color{#3D99F6}x^2+y^2}} & \small \color{#3D99F6} \text{Recall }x^2 + y^2 = (5-r)^2 = \left(\frac {65}{19}\right)^2 \\ & = \frac {\sqrt{{\color{#3D99F6}\left(\frac {65}{19}\right)^2}-2{\color{#D61F06}\left(\frac {25}{19}\right)}+1}}{\color{#3D99F6}\frac {65}{19}} & \small \color{#D61F06} \text{and }x = \frac {25}{19} \\ & = \frac {6\sqrt{101}}{19} \end{aligned}$

Then $p^2(m+n) = 101^2(6+65) = \boxed{724271}$ .

Since $\frac{BP}{PA} = \frac{2}{3}$ , let $BP = 4$ and $PA = 6$ , so that the radius of the red semicircle is $r_r = 2$ , the radius of the blue semicircle is $r_b = 3$ , and the radius of the green semicircle is $r_g = 5$ .

Since the circles are all tangent to each other, the radius of the orange circle can be found by using Destartes' Theorem , $\frac{1}{r_o} = \frac{1}{2} + \frac{1}{3} - \frac{1}{5} + 2\sqrt{\frac{1}{2 \cdot 3} - \frac{1}{2 \cdot 5} - \frac{1}{3 \cdot 5}}$ , which solves to $r_o = \frac{30}{19}$ .

Let $O$ be the center of the red semicircle. Then $ON = r_r + r_o = 2 + \frac{30}{19} = \frac{65}{19}$ and $OM = r_b - r_r = 5 - 2 = 3$ . Also, $MN = r_b - r_o = 5 - \frac{30}{19} = \frac{65}{19}$ , $MP = r_b - 2r_r = 5 - 2 \cdot 2 = 1$ , and $OP = r_r = 2$ .

By the law of cosines on $\triangle MNO$ , $\cos O = \frac{MO^2 + NO^2 - MN^2}{2 \cdot MO \cdot NO} = \frac{3^2 + (\frac{68}{19})^2 - (\frac{65}{19})^2}{2 \cdot 3 \cdot \frac{68}{19}} = \frac{8}{17}$ .

By the law of cosines on $\triangle PNO$ , $NP = \sqrt{PO^2 + NO^2 - 2 \cdot PO \cdot NO \cdot \cos O} = \sqrt{2^2 + (\frac{68}{19})^2 - 2 \cdot 2 \cdot \frac{68}{19} \cdot \frac{8}{17}} = \frac{6 \sqrt{101}}{19}$ .

Therefore, $\frac{NP}{MN} = \frac{\frac{6 \sqrt{101}}{19}}{\frac{65}{19}} = \frac{6 \sqrt{101}}{65}$ , so $m = 6$ , $p = 101$ , and $n = 65$ , which means $p^2(m + n) = 101^2(6 + 65) = \boxed{724271}$ .