Dynamic Geometry: P36

Let the black equilateral triangle be inscribed in the white space, and draw $\triangle ABC$ , as follows:

By the properties of an equilateral triangle, if the sides of the large pink equilateral triangle are $2$ , then $AC = \cfrac{2\sqrt{3}}{3}$ .

That makes $DC = AC - AD = \cfrac{2\sqrt{3}}{3} - 1$ , and $CB = 2 \cdot DC = 2\bigg(\cfrac{2\sqrt{3}}{3} - 1\bigg)$

If the radius of a cyan circle is $r$ , then $AB = r + 1$ .

By the law of cosines on $\triangle ABC$ , $AB^2 = AC^2 + CB^2 - 2 \cdot AC \cdot CB \cdot \cos \angle ACB$ , or

$(r + 1)^2 = \bigg(\cfrac{2\sqrt{3}}{3}\bigg)^2 + \bigg(2\bigg(\cfrac{2\sqrt{3}}{3} - 1\bigg) \bigg)^2 - 2 \cdot \cfrac{2\sqrt{3}}{3} \cdot 2\bigg(\cfrac{2\sqrt{3}}{3} - 1\bigg) \cdot \cfrac{1}{2}$

which solves to $r = 2\sqrt{2 - \sqrt{3}} - 1$ .

Therefore, $a = 2$ , $b = 3$ , $c = 1$ , and $a + b + c = \boxed{6}$ .