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: Obviously, each of the hexagon have area of $\frac{3\sqrt{3}}{2}$ , which gives a total of $6\sqrt{3}$ . Now we calculate the area of the intersection of two hexagons. There are two cases :
Case 1: The hexagons are on the opposite sides
In this case, we see that the two hexagons' intersection is the middle square plus two triangles on its side. The square have area $1$ . Each of the triangles are 120-30-30, and hence its short side is $\frac{\sqrt{3}}{3}$ because its longer side is $1$ . Therefore, each of the small triangles have area of $\frac{\sqrt{3}}{12}$ , so each of the intersections have $1+\frac{\sqrt{3}}{6}$ area. Since there are $2$ pairs of hexagons on opposite side of each other, that gives a total area of $2+\frac{\sqrt{3}}{3}$
Case 2: The hexagons are adjacent [asy] pair[] A,B; path h1, h2,h3,h4; h1 = shift((sqrt(3)-1)/2*dir(90))*polygon(6); h2 = shift((sqrt(3)-1)/2*dir(0))*rotate(90)*polygon(6); h3 = shift((sqrt(3)-1)/2*dir(-90))*polygon(6); h4 = shift((sqrt(3)-1)/2*dir(180))*rotate(90)*polygon(6); B.cyclic=true; B.push((0,0)); int indexSquare[] = sequence(4); int indexPolygon[] = sequence(12); for(int i : indexSquare){ A.push(1/sqrt(2)*dir(45+90*i)); } draw(A[0]--A[1]--A[2]--A[3]--cycle); fill(h1,green); fill(h2,red); fill(buildcycle(h2,h1),yellow); draw(h1); draw(h2); draw(h3); draw(h4); for(int i : sequence(4)) { for(int j : sequence(3)) { B.push(B[3*i+j]+((sqrt(3)-1.5)*(j-1)*(j)+4-2*sqrt(3))*dir(90*i+60+30*j*(1.5*(j-1)-1))); } } for(int i=0; i <12; ++i) { draw(shift(0.5,0.5-sqrt(3))*(B[i])--shift(0.5,0.5-sqrt(3))*(B[i + 1]),linewidth(2)); } [/asy]
In this case, their intersection is a convex hexagon. We cut it in half and we're left with a quadrilateral with two sides with length one (because it shares these sides with one of the hexagons, and for this reason the angle between the two sides is $120$ degrees) and one short side and one long side. Here, we divide the area even further. We draw a line that divide the quadrilateral into a triangle with side lengths $1$ $1$ , and $\sqrt{3}$ (because it is 30-30-120) and another triangle with unknown side lengths. Simple angle chasing yields the triangle with unknown side lengths is a 15-75-90 triangle. If you remembered your sines and cosines, you know that $\sin{15}=\frac{\sqrt{6}-\sqrt{2}}{4}$ or you can calculate it by using the fact that $\sin{15}=\sin{45-30}$ . Also, $\sin{75}=\frac{\sqrt{6}+\sqrt{2}}{4}$ . Now, because the 15-75-90 triangle have one side with length $\sqrt{3}$ (it shares the side with the other triangle) and it corresponds to the 75 degrees angle, we have $\frac{\sqrt{3}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{x}{\frac{\sqrt{6}-\sqrt{2}}{4}}$ $\implies x=\frac{\sqrt{3}{(\sqrt{6}-\sqrt{2})}}{\sqrt{6}+\sqrt{2}}=2\sqrt{3}-3$
where $x$ denote the shorter leg of the 15-75-90 triangle. Hence, that triangle have area $\frac{\sqrt{3}}{2} x=\frac{\sqrt{3}}{2}(2\sqrt{3}-3)=\frac{6-3\sqrt{3}}{2}$ . The isosceles 120-30-30 triangle obviously have area $\frac{\sqrt{3}}{4}$ , and that gives a total of $\frac{12-5\sqrt{3}}{4}$ . We multiply that by $2$ because $\frac{12-5\sqrt{3}}{4}$ is only half of the intersection of the hexagons. We multiply the result by $4$ because there are $4$ pairs of adjacent hexagons. That gives $24-10\sqrt{3}$ , which is it for Case 2.
Now that we got the total area of the intersection of two hexagons, we find the area of the intersection of three hexagons. Note that the intersection of any three hexagons is just the middle square plus one small triangle on the side. Again, the middle square have area $1$ and the triangle have area $\frac{\sqrt{3}}{12}$ , so the total is $1+\frac{\sqrt{3}}{12}$ . We multiply that by $4\choose3$ to get $4+\frac{\sqrt{3}}{3}$
Then, we calculate the area of the intersection of all four hexagons, which is just the middle square with area $1$
Finally, we apply PIE. The total area is $6\sqrt{3}-[(2+\frac{\sqrt{3}}{3})+(24-10\sqrt{3})]+(4+\frac{\sqrt{3}}{3})-1=16\sqrt{3}-23$ ,
so our final answer is $16+3-23=\boxed{4}$ . ~ Ddk001

Answer: B 4