Star Stumper

My solution is similar to Ruslan's. Consider the isosceles triangle formed by the left point A of the star, the upper point B of the star, and the point C where the $120^o$ angle is given. Then $AB = \sqrt{2}$ , the height of this triangle over AB is $h= \tan(30^o)\times\frac{AB}{2}$ $=\frac{\sqrt{6}}{6}$ , and the area is $F=\frac{(AB)h}{2}=\frac{\sqrt{3}}{6}$ . Considering the square formed by the points of the star, we see that the shaded area is $2-4F=\frac{6-2\sqrt{3}}{3}$ .

Let the side of the star is x.The midpoints make a triangle with the star .By cos rule’ 2 = 2x^2 - 2x^2 cos 120, so x=√(2/3), and the area of the triangle formed by the midpoints and sides of the star is (2/3)(1/2√3)/2 =1/6 √3 shaded area = 4 – 4(area of the kite) = 4 – 4(1/2 + 1/6 √3) = 2 – (2/3) √3 = (6 - 2√3)/3. So a+b+c = 11