It Is All Circles

For the first equilateral triangle with inscribed circle whose radius $r_{1} = 1$ we have $\dfrac{1}{2\sqrt{3}}x_{1} = 1 \implies x _{1}= 2\sqrt{3}$ .

Using the tangents to the second circle the midpoint of the second equilateral triangle is $3\sqrt{3} \implies$ the side of the second equilateral triangle is $x_{2} = 6\sqrt{3} \implies r_{2} = 3$ . The two triangles are in proportion as is the two circles with common ratio $3$ .

Using the tangents to the third circle the midpoint of the second equilateral triangle is $9\sqrt{3} \implies$ the side of the second equilateral triangle is $x_{3} = 18\sqrt{3} \implies r_{3} = 3$ .

In general the midpoint of the $n$ th equilateral triangle is $3^{n - 1}\sqrt{3} \implies$ the side of the $n$ th equilateral triangle is $x_{n} = 2\sqrt{3} * 3^{n - 1} \implies r_{n} = 3^{n - 1} \implies$

The area of the $n$ th triangle with height $h_{n} = \dfrac{\sqrt{3}}{2}(2\sqrt{3} * 3^{n - 1})$ is $A^{*}_{n} =3^{2n - 1}\sqrt{3}$ and the sum of the areas of the $n$ circles is $A^{**}_{n} = \pi\sum_{j = 1}^{n} r_{j}^2 = \pi\sum_{j = 1}^{n} 9^{j - 1} = \dfrac{(3^{2n} - 1)\pi}{8}$

$\implies$ The desired area $A_{n} = 3^{2n - 1}(\sqrt{3} - \dfrac{3\pi}{8}) + \dfrac{\pi}{8} = \dfrac{3^{2n - 1}}{8}(8\sqrt{3} - 3\pi) + \dfrac{\pi}{8}$ .

For $n = 1954 \implies A = \dfrac{3^{3907}}{2^3}(2^3\sqrt{3} - 3\pi) + \dfrac{\pi}{2^3} = \dfrac{a^b}{c^a}(c^a\sqrt{a} - a) + \dfrac{\pi}{c^a} \implies a + b + c = \boxed{3912}$ .