Circles centred on vertices of hex, tangential to each other

[wazzawayne]

Hi,
For the below problem, I just guessed that the circles are identical based on the diagram!
But, how would I normally arrive at the conclusion that all the circles are identical within the few seconds that would be allowed while solving this problem???

Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

[Atekihcan]

...how would I normally arrive at the conclusion that all the circles are identical within the few seconds that would be allowed while solving this problem?

Because if they are not identical, then all the circles cannot touch the middle circle.
Try drawing some figure such that all conditions are satisfied but the circles are not identical, you'll see that it is not possible. This can be proved mathematically but that will be time waste.

[GMATGuruNY]

Shaded region = hexagon - circle areas.

Hexagon:
Since every side of the 6 triangles shown above is composed of two radii, the 6 triangles are both congruent and equilateral.
Since the perimeter of the regular hexagon is 36, each side of the hexagon -- and thus each triangle side -- has a length of 6.
In an equilateral triangle, A = (s²√3)/4:
(6²)(√3)/4 = 9√3.
Since the hexagon is composed of 6 of these triangles:
Hexagon area = 6(9√3) = 54√3.

Circle areas:
Since r=3, the area of each circle = π(3²) = 9π.
The central angle of each red sector = (angle of equilateral triangle) + (angle of equilateral triangle) = 60+60 = 120 degrees.
Since 120/360 = 1/3, each red sector constitutes 1/3 of a circle.
Thus, the 6 red sectors -- along with the circle with center O -- constitute 3 entire circles:
3(9Ï€) = 27Ï€.

Shaded region = 54√3 - 27π.

The correct answer is E.