What's the measure of the radius of the circle below?

Question

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

Answer 1

$\angle AOB = 120^o$

Law of cosines:

$AB^2 = 2R^2 + 2R^2 \cdot \frac 12 = 3R^2 \implies AB = R \sqrt3$

Power point M: $1 \cdot (2R-1) = 2 \cdot (R \sqrt3 - 2) \iff 2R -1 = 2R\sqrt3 - 4 \iff R = \frac3{2(\sqrt3-1)} = \frac{3(\sqrt3+1)}4$

Answer 2

I can't quite follow your work, but you should be able to get a relation with $AM$ and $R$ by using the fact that $OAB$ is a 30-30-120 triangle.

Hint for one possible alternate method: Drop an altitude from $ON$ to $AB$. So $N$ is the midpoint of $AB$. Now focus on the lengths $OM,MN,NB,BO,ON$, writing what you can in terms of the radius $R$.

Let me know if I should elaborate.

Answer 3

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