Problem: Circles $\omega_1$ $\omega_2$ , and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two.  Points $P_1$ $P_2$ , and $P_3$ lie on $\omega_1$ $\omega_2$ , and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$ , where $P_4 = P_1$ .  See the figure below.  The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$ .  What is $a+b$
[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); [/asy]
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$

Solution: Let $O_1$ $O_2$ , and $O_3$ be the centers of $\omega_1$ $\omega_2$ , and $\omega_3$
Let $P_1P_2 = a$ $\angle O_2O_1P_1= \alpha$
\[[O_1P_1O_2P_2O_3P_3] = [P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = [O_1O_2O_3] + 3 \cdot [O_1P_1O_2]\]
\[[P_1P_2P_3] + 3 \cdot [O_1P_1P_3] = \frac12 \cdot a \cdot \frac{ \sqrt{3} }{2} a + 3 \cdot \frac12 \cdot 4 \cdot a \cdot \sin 30^{\circ} = \frac{ \sqrt{3} }{4} a^2 + 3a\]
\[[O_1O_2O_3] + 3 \cdot [O_1P_1O_2] = \frac12 \cdot 8 \cdot \frac{ \sqrt{3} }{2} \cdot 8 + 3 \cdot \frac12 \cdot 4 \cdot 8 \cdot \sin \alpha = 16 \sqrt{3} + 48 \sin \alpha\]
\[\frac{ \sqrt{3} }{4} a^2 + 3a = 16 \sqrt{3} + 48 \sin \alpha\]
\[\sin \alpha = \frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3}\]
By the Law of Cosine from $\triangle O_2P_1P_2$ $O_2P_1^2 = a^2 + 4^2 - 2 \cdot a \cdot 4 \cdot \cos 30^{\circ} = a^2 + 16 - 4a \sqrt{3}$
By the Law of Cosine from $\triangle O_1O_2P_1$ $O_2P_1^2 = 4^2 + 8 ^ 2 - 2 \cdot 4 \cdot 8 \cdot \cos \alpha = 80 - 64 \cos \alpha$
\[a^2 + 16 - 4a \sqrt{3} = 80 - 64 \cos \alpha\]
\[\cos \alpha = -(\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)\]
\[\because \sin^2 \alpha + \cos^2 \alpha = 1\]
\[\therefore (\frac{ \sqrt{3} }{192} a^2 + \frac{a}{16} - \frac{ \sqrt{3} }{3})^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1\]
\[[ \frac{1}{\sqrt{3}} (\frac{ a^2 }{64}  + \frac{ \sqrt{3} }{16} a - 1)]^2 + (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 1\]
\[(\frac{ a^2 }{64} + \frac{ \sqrt{3} }{16} a - 1)^2 + 3 \cdot (\frac{a^2}{64} - \frac{ \sqrt{3} }{16} a - 1)^2 = 3\]
\[(a^2 + 4 \sqrt{3} a - 64)^2 + 3 \cdot (a^2 - 4 \sqrt{3} a - 64)^2 = 3 \cdot 64^2\]
\[(a^2 - 64)^2 + 2 (a^2 - 64) \cdot 4 \sqrt{3} a + (4 \sqrt{3} a)^2 + 3(a^2 - 64)^2 - 6 (a^2 - 64) \cdot 4 \sqrt{3} a + 3(4 \sqrt{3} a)^2 = 3 \cdot 64^2\]
\[4 (a^2 - 64)^2 - 4 (a^2 - 64) \cdot 4 \sqrt{3} a + 192 a^2 - 3 \cdot 64^2 = 0\]
\[4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a) + 192(a^2 - 64) = 0\]
\[4 (a^2 - 64)(a^2 - 64 - 4 \sqrt{3} a + 48) = 0\]
\[(a^2 - 64)(a^2 - 4 \sqrt{3} a - 16) = 0\]
\[a^2 -64 = 0 \quad or \quad a^2 - 4 \sqrt{3} a - 16 = 0\]
If $a^2 =64$ $a= 8$ $P_1O_2^2 = 8^2 + 4^2 - 2 \cdot 8 \cdot 4 \cdot \cos 30^{\circ} = 80 - 32 \sqrt{3} < 16$ $P_1O_2 < 4$ , meaning that $P_1$ is inside the circle.
However, $P_1$ is not inside the circle.
\[So, \quad a^2 - 4 \sqrt{3} a - 16 = 0\]
\[a = \frac{ 4 \sqrt{3} + \sqrt{ (4 \sqrt{3})^2 - 4(-16) } }{2} = 2 \sqrt{3} + 2 \sqrt{7}\]
\[[P_1P_2P_3] = a^2 \cdot \frac{ \sqrt{3} }{4} = (2 \sqrt{3} + 2 \sqrt{7})^2 \cdot \frac{ \sqrt{3} }{4} = \sqrt{300} + \sqrt{252}\]
Therefore, the answer is $300 + 252 = \boxed{552}$

Answer: D 552