Problem: Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$ , where $m$ $n$ , and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$ [asy]         import geometry;         unitsize(3cm);         draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle);         draw(shift((1/2,1-sqrt(3)/2))*polygon(6));         draw(shift((1/2,sqrt(3)/2))*polygon(6));         draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6));         draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); 		draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); [/asy] $\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad  \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$

Solution: [asy] import geometry;         unitsize(3cm);         draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle);         draw(shift((1/2,1-sqrt(3)/2))*polygon(6));         draw(shift((1/2,sqrt(3)/2))*polygon(6));         draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6));         draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); 		draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2));         draw((1,1)--(3-sqrt(3),3-sqrt(3)),linewidth(2));         draw((0,1)--(sqrt(3)-2,3-sqrt(3)),linewidth(2));         draw((0,0)--(sqrt(3)-2,sqrt(3)-2),linewidth(2));         draw((1,0)--(3-sqrt(3),sqrt(3)-2),linewidth(2));         draw((0, sqrt(3))--(0, -sqrt(3)+1),linewidth(2));         draw((1, sqrt(3))--(1, -sqrt(3)+1),linewidth(2));         draw((1-sqrt(3),0)--(sqrt(3),0),linewidth(2));         draw((1-sqrt(3),1)--(sqrt(3),1),linewidth(2));         label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);         label("$B$", (1, sqrt(3)), NE);         label("$C$", (1,1), SW);         label("$D$", (1,0), NW);         label("$E$", (1, -sqrt(3)+1), SE);         label("$F$", (0, -sqrt(3)+1), SW);         label("$G$", (0, sqrt(3)), NW); [/asy]
Divide this polygon into a 'cross' and $8$ triangles like $\triangle ABC$ $BD = 2 \cdot \frac{ \sqrt{3} }{2} = \sqrt{3}$ $BE = 2BD - CD = 2\sqrt{3} - 1$ $EF = 1$ $[BEFG] = 1 \cdot (2\sqrt{3} - 1) = 2\sqrt{3} - 1$ $[\text{cross}] = 2\cdot [BEFG] - 1 = 2(2\sqrt{3} - 1)- 1 = 4\sqrt{3} - 3$
$\angle ABC = 120^{\circ} - 90^{\circ} = 30^{\circ}$ $\angle ACB = \frac{90^{\circ}}{2} = 45^{\circ}$ $\angle BAC = 180^{\circ} - 30^{\circ} - 45^{\circ} = 105^{\circ}$ $BC = BD - CD = \sqrt{3} - 1$ . Draw a perpendicular line $AG$ to side $BC$
$BG + CG = BC$ $AG \cdot \sqrt{3} + AG = \sqrt{3} - 1$ $AG = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$ $[ABC] = \frac12 \cdot AG \cdot BC = \frac12 (2 - \sqrt{3})(\sqrt{3} - 1)$
$[\text{12-sided outer polygon}] = [\text{cross}] + 8 \cdot [ABC] = 4\sqrt{3} - 3 + 8 \cdot \frac12 (2 - \sqrt{3})(\sqrt{3} - 1) = 4\sqrt{3} - 3 + 4 (3\sqrt{3} - 5) = 16\sqrt{3}-23$
\[16+3-23=\boxed{4}\]

Answer: B 4