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: Note that each of the green sections is a rectangle, so its interior angles are all $90^{\circ}.$ Since $\angle{ABC}=120^{\circ}$ , every one of the orange sections is a $30-60-90$ right triangle.
Define $x$ to be the distance from the corner of the square with side length $1$ to the corner of the larger blue square. Due to the sides of the two squares being parallel to each other, the large blue triangle is a $45-45-90$ right triangle. By $AAA$ similarity, the smaller blue triangles are also $45-45-90,$ and have side lengths of $\frac{x\sqrt{2}}{2}, \frac{x\sqrt{2}}{2},$ and $x$ . By $30-60-90$ triangle relations, the largest altitude of the orange triangle is $\frac{x\sqrt{6}}{2}.$
Now, we can find the height of the hexagon to obtain an equation in terms of $x$ . Consider a hexagon with side length $1$ , where point $P$ is the foot of the perpendicular dropped from $B$ , bisecting $\angle{ABC}$ [asy]         import geometry;         import fontsize;         defaultpen(fontsize(9pt));         unitsize(3cm);         draw(shift((1/2,1-sqrt(3)/2))*polygon(6));         draw((1,1-sqrt(3))--(1, 1)--cycle,linewidth(1));         draw((3/2,1-sqrt(3)/2)--(1, 1-sqrt(3)/2)--cycle,linewidth(1));         draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 +0.1)--(1, 1-sqrt(3)/2+0.1)--cycle,linewidth(1));         draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 -0.1)--(1, 1-sqrt(3)/2-0.1)--cycle,linewidth(1));         label("$A$", (1, 1), N);         label("$P$", (1, 1-sqrt(3)/2), W);         label("$C$", (1, 1-sqrt(3)), S);         label("$B$", (3/2, 1-sqrt(3)/2), E);         label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), NW);         label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), SW); [/asy] Note that triangles $\triangle{APB}$ and $\triangle{CPB}$ are congruent $30-60-90$ triangles, by SAS congruence. Since the side length of this hexagon is $1$ , the length of $AP$ is $\frac{1}{2}\cdot \sqrt{3} =\frac{\sqrt{3}}{2}$ , by $30-60-90$ triangle relations. The height of the hexagon is twice this value, or $\sqrt{3}.$
The height is also equal to the sum of the values along the long blue line, in the first diagram. Therefore, $1+\frac{x\sqrt{2}}{2}+\frac{x\sqrt{6}}{2}=\sqrt{3}.$ Solving and rationalizing, $x=2\sqrt{2}-\sqrt{6}.$
The area of the dodecagon is equal to the sum of the areas of the four rectangles, eight orange triangles, and purple square. In terms of $x$ , this is \[8\cdot \frac{1}{2}\cdot \frac{x\sqrt{2}}{2}\cdot \frac{x\sqrt{6}}{{2}}+4\cdot 1\cdot\frac{x\sqrt{6}}{2}+(x\sqrt{2}+1)^2.\]
Plugging in $x=2\sqrt{2}-\sqrt{6}$ , the area of the dodecagon is $16\sqrt{3}-23$ . Therefore, the answer is $16+3-23=$ $\boxed{4}.$

Answer: B 4