<s>[INST] Find the side of an equilateral triangle inscribed in a circle. 
Excuse the poor drawing. 

$\triangle CDE$ is an equilateral triangle inscribed inside a circle, with side length $16$. Let $F$ be the midpoint of $DE$. Points $G$ and $H$ are on the circle so that $\triangle FGH$ is an equilateral. Find the side length of $\triangle FGH$ and express it as $a\sqrt{b}-c$ where $a, b, c$ are positive integers. 

My attempt
$$\theta \text{ in the Large } \triangle = \theta \text{ in the Small } \triangle = 60^{\circ}$$
$$(\text{side of the Large } \triangle) \cdot \sin 60^{\circ} + (\text{side of the Small } \triangle )\cdot \sin 60^{\circ} = 2\cdot(\text{radius of circle})$$
Solving for the small side, I got $\dfrac{64}{3}-16$, which doesn't match the required form. 
Thanks for any help.
 [/INST] Let $R$ be the radius of circle.  Draw a perpendicular to any of sides of large equilateral $\triangle CDE$ to obtain a right triangle by which we can calculate the side length $L$ of large equilateral triangle as follows $$L=2R\cos 30^\circ=16$$  $$\implies 2R\frac{\sqrt{3}}{2}=16 \implies \color{blue}{R=\frac{16}{\sqrt{3}}}$$ Let $l$ be side length of small equilateral $\triangle FGH$ & draw a perpendicular say $AM$, from center $A$ passing through $F$, to the opposite side $GH$ of small $\triangle FGH$ Then we have $$AM=AF+FM=R\sin 30^\circ+l\sin 60^\circ=\frac{16}{\sqrt{3}}\frac{1}{2}+l\frac{\sqrt{3}}{2}=\color{red}{\frac{16+3l}{2\sqrt{3}}}$$ in right $\triangle AMG$ applying pythagorean theorem as follows $$AG^2=AM^2+GM^2$$ $$\left(\frac{16}{\sqrt{3}}\right)^2=\left(\frac{16+3l}{2\sqrt{3}}\right)^2+\left(\frac{l}{2}\right)^2$$  $$ 12l^2+96l-768=0$$ Solving for $l$ as follows $$l=\frac{-96\pm\sqrt{(96)^2-4(12)(-768)}}{2(12)}$$ $$l=\frac{-96\pm96\sqrt{5}}{24}=-4\pm 4\sqrt{5}$$ but $l>0$ hence the side of the small equilateral $\triangle FGH$ $$l=\color{red}{4\sqrt{5}-4}$$ Now, comparing the side $l$ with $\color{red}{a\sqrt{b}-c}$ we get $$\color{blue}{a=4, b=5, c=4}$$
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