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((0.5,sqrt(3))--(0.5,1-sqrt(3)),linewidth(2));         draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2));         draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2));         draw((1, sqrt(3))--(1,1),linewidth(2));         label("$O$",(0.5,0.5),SE);         dot((0.5,0.5));         label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);         label("$B$", (1, sqrt(3)), NE);         label("$C$", (1,1), E);         label("$D$", (1/2, sqrt(3)), N); [/asy]
Begin by dividing the figure as shown above. Clearly, the entire figure has 8-fold symmetry. Therefore, we can calculate the area of $ODBA$ and multiply it by 8. We split $[ODBA]$ into $[ODBC]+[ABC]$
Knowing the side length of the hexagon is $1$ , we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is $2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.$ Thus, $OD=\sqrt{3}-\frac{1}{2}$ and $BC=\sqrt{3}-1.$ Recognizing $DB=\frac{1}{2}$ and $ODBC$ is a trapezoid, \[8\cdot[ODBC]=2\left(\sqrt{3}-\frac{1}{2}+\sqrt{3}-1\right)=4\sqrt{3}-3.\]
Next, we aim to find $[ABC]$ . By angle chasing, we find $\angle A=105^\circ,$ $\angle B=30^\circ,$ and $\angle C = 45^\circ.$ We can use the law of sines to find $AB$
\[\frac{\sin(105^\circ)}{\sqrt{3}-1}=\frac{\sin(45^\circ)}{AB}\implies AB=\frac{\sqrt{6}-\sqrt{2}}{2\sin(105^\circ)}.\]
We may not know what $\sin(105^\circ)$ is by memory, but we can cleverly calculate it using a common trig identity:
\begin{align*}     \sin(105^\circ)&=\sin(60^\circ+45^\circ),\\     &=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ),\\     &=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2},\\     &=\frac{\sqrt{6}+\sqrt{2}}{4}. \end{align*}
With some simplification, we'll find $AB=4-2\sqrt{3}$ . Now, we can easily calculate $8\cdot [ABC]$ as \[8\cdot\frac{1}{2}\cdot(\sqrt{3}-1)(4-2\sqrt{3})\sin(30^\circ)=12\sqrt{3}-20.\]
Thus, the area of the dodecagon is $8\cdot [ODBA] = 8\cdot [ODBC] + 8 \cdot [ABC] =4\sqrt{3}-3+12\sqrt{3}-20=16\sqrt{3}-23.$
Finally, we find \[16+3-23=\boxed{4}.\]

Answer: B 4