What's the measure of the radius of the circle below? For reference: In the figure, calculate $R$. If : $\overset{\LARGE{\frown}}{AB}= 120°$, $CM = 1$ and $MB = 2$
(Answer: $\frac{3}{4}(\sqrt{3}+1$))
My progress:

Draw OP
Th.Chords:
$CM.MP = BM.AM \implies 1.[(R-1)+(R)] = 2.AM\\
\boxed{2AM = 2R-1}: \boxed{R=AM+ \frac{1}{2}}\\
\triangle AOB (isósceles):$\
Draw $AE$
$\implies$
$\triangle EAB(right): 
AE^2+(2+AM)^2 = 4R^2\\
AE^2 +4+4AM + AM^2 = 4R^2\\
AE^2 + 4+8R-4 + R^2 - R+\frac{1}{4} = 4R^2\\
4AE^2+16+32R - 16+4R^2-4R+1 = 16R^2\\
4AE^2+28R-12R^2+1 = 0 \implies\\
AE^2 = 12R^2-28R-1\\$
...?
I have not found another relationship with AM


Extend $\overline{CMO}$ to meet the circle at E and Extend $\overline{AO}$ to meet the circle at $D$. As $\triangle ABD$ is a 30-60-90 triangle $AB=R\sqrt{3}$. Since $BM=2$ and $AM=R\sqrt{3}-2$.
Now, look at $\triangle ACM$ and $\triangle BEM$ they are similar.
$\frac{1}{2}=\frac{R\sqrt{3}-2}{2R-1}$
After solving for R you get,
$R=\frac{3}{2(\sqrt{3}-1)}$
by getting the decimal value for R,
$R=2.04$
The decimal value of the answer you have posted is also 2.04
Then the proof will be true
