Round Rhombus

Let $r_o$ denote the radius of the orange circle and $r_g = kr_o$ the radius of the green circle, where $k > 0$ .

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Relation Between Orange and Green Circles

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The radius of the green circle is twice the radius of the orange circle. That is, $r_g = 2r_o$

To prove that $k = 2$ , consider two circles at the corners of the rhombus:

Since both orange circles have the same radii, then the rectangles and the larger triangles (shared by the center of the large circles) are symmetric. By Pitot's Theorem , the indicated angles are both congruent. Noting that these $30$ - $60$ - $90$ triangles are similar, we can determine the value of $k$ by the following equation $\dfrac{r_g + r_o}{r_o \cdot \csc\left(30^{\circ} + r_g + r_o\right)} = \dfrac{\left(r_g + r_o\right) - r_o}{r_g + 2 \cdot r_o}$ which is $\dfrac{k + 1}{k + 4} = \dfrac{k}{k + 2}$ So we have $r_g = 2r_o$ .

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Final Step

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The final step is to compute the radius of the two dashed circles, which is easy to determine. In terms of $r$ 's, the side length of the rhombus is $r_o \cot\left(30^{\circ}\right) + r_o \tan\left(30^{\circ}\right) + 2\sqrt{r_o \left(r_o + r_g\right)} \cdot 2$ where $r_o = \dfrac{\sqrt{3}}{16}$ and $r_g = \dfrac{\sqrt{3}}{8}$ . Since the radius of the dashed circle is $r_g + r_o$ , its value is $\dfrac{3\sqrt{3}}{16}$ , where $a = 3$ , $b = 3$ and $c = 16$ .

So $abc = \boxed{144}$