Problem: Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution: To make things simpler, let's take only the original sheet and the 30 degree rotated sheet. Then the diagram is this;
[asy]  size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/  [/asy]
The area of this diagram is the original square plus the area of the four triangles that 'jut' out of the square. Because the square is rotated $30^{\circ}$ , each triangle is a 30-60-90 triangle. Similarly, the triangles that are bounded on the inside of the original square outside of the rotated square are also congruent 30-60-90 triangles. Noting this, we can do some labelling:
[asy]  size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ label("$y$",(0.1,-0.05)); label("$x$",(0.4,0.05)); label("$y\sqrt{3}$",(0.8,-0.05)); label("$\frac{x}{2}$",(0.22,-0.12)); label("$\frac{x\sqrt{3}}{2}$",(0.5,-0.15)); label("$2y$",(0.8,0.15)); label("$y$",(1.05,0.1)); label("$\frac{x}{2}$",(1.12,0.22));  [/asy]
Since the side lengths of the squares must be the same, and they are both 6, we have a system of equations; \[y+x+y\sqrt{3} = 6\] \[\frac{x\sqrt{3}}{2} + 2y + \frac{x}{2} = 6\]
We solve this to get $x = 6-2\sqrt{3}$ and $y = 3-\sqrt{3}$
The area of each triangle is $\frac{x}{2} \cdot \frac{x\sqrt{3}}{2} \cdot \frac{1}{2}  = 6\sqrt{3} - 9$ by plugging in $x$
The rotated 60 degree square is the same thing as rotating it 30 degrees counterclockwise, so it's triangles that jut out of the square will be congruent to the triangles we have found, and therefore they will have the same area.
Unfortunately, when drawing all three squares, we see the two triangles overlap; take the very top for example.
[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.3657,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw(A--E--C--cycle); draw(B--D--G--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",E,N); label("$F$",F,N); label("$G$",G,N); [/asy]
The area of this shape is twice the area of each of the triangles that we have already found minus the area of the small triangle that is overlapped by the two by PIE. Now we only need to find the area of $\bigtriangleup BCF$
$\angle GBD \cong \angle ECA \cong 30^{\circ}$ and by symmetry $\bigtriangleup BCF$ is isosceles, so it is a 30-30-120 triangle. If we draw a perpendicular, we split it into two 30-60-90 triangles;
[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.3657,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw(A--E--C--cycle); draw(B--D--G--cycle); draw(F--(0.5,0)); label("$A$",A,S); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",E,N); label("$F$",F,N); label("$G$",G,N); label("$H$",(0.5,0),S); [/asy]
By symmetry, the distance from A to the edge of the square is equal to the distance from D to the edge of the square is equal to $y$ . AC = BD = $x$ , and the side length of the square is 6, so we use PIE to obtain \[x+x-BC = 6-y-y \implies BC = 12 - 6\sqrt{3}\] To find the height of $\bigtriangleup BFC$ , we see that $HC = \frac{BC}{2} = 6-3\sqrt{3}$ . Then by 30-60-90 triangles, $HF = \frac{HC}{\sqrt{3}} = 2\sqrt{3} - 3$ . Finally, the area of $\bigtriangleup BFC = \frac{BC \cdot HF}{2} = 21\sqrt{3}-36$
Putting it all together, the area of the entire diagram is the area of the square plus four of these triangle-triangle intersections. The area of these intersections by PIE is $2 \cdot [ACE] - [BFC] = 12\sqrt{3}-18-(21\sqrt{3}-36) = 18-9\sqrt{3}$ . 
Therefore the total area is $36 + 4 \cdot(18-9\sqrt{3}) = 36 + 72 - 36 \sqrt{3} = 108 - 36\sqrt{3}$
Thus $a + b + c = 108+36+3 = 147 = \boxed{147}$

Answer: E 147