Question 1189710
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The tutor greenestamps has a great answer. I'll provide a (slightly) different viewpoint.


I'll use the letter notation he has set up.
Here's what the drawing looks like with those points mentioned.
<img width = "35%" src = "https://i.imgur.com/Gyt7sPs.png">


Let circle N have a radius of 1. This means EN = ND = 1.
<font color=blue>Goal: Find the length of FM</font>
This will effectively give the ratio FM/EN, aka the ratio of the two radii
Note how FM/EN = FM/1 = FM


As mentioned by the other tutor, the 120 degree angle ABC is bisected into two smaller 60 degree angles
angle ABM = angle MBC = 60
This leads to triangles BEN and BFM being 30-60-90 triangles
The points of tangency E and F have the 90 degree angles at those locations.


The useful formulas for any 30-60-90 triangle are that
hypotenuse = 2*(short leg)
long leg = sqrt(3)*(short leg)
That second formula rearranges to
short leg = (long leg)/sqrt(3) = (long leg)*sqrt(3)/3


For triangle BEN we have EB as the short leg and EN as the long leg
short leg = (long leg)*sqrt(3)/3
EB = (EN)*sqrt(3)/3
EB = (1)*sqrt(3)/3
EB = sqrt(3)/3


And,
hypotenuse = 2*(short leg)
BN = 2*EB
BN = 2*sqrt(3)/3 is the hypotenuse of triangle BEN 


Let x be the radius of circle M
FM = x
DM = x as well since they're both radii of the same circle M
Now focus on triangle BFM 


Since it is also a 30-60-90 triangle we can say
short leg = (long leg)*sqrt(3)/3
BF = (FM)*sqrt(3)/3
BF = x*sqrt(3)/3
and also
hypotenuse = 2*(short leg)
BM = 2*(BF)
BM = 2x*sqrt(3)/3


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The key things to recap is that we found
BN = 2*sqrt(3)/3
ND = 1
DM = x



Now break segment BM into smaller pieces
BM = BN + ND + DM
BM = 2*sqrt(3)/3 + 1 + x


We have
BM = 2x*sqrt(3)/3
and
BM = 2*sqrt(3)/3 + 1 + x


Set those two right hand sides equal to one another. Solve for x.


2*sqrt(3)/3 + 1 + x = 2x*sqrt(3)/3
2*sqrt(3) + 3 + 3x = 2x*sqrt(3) ............. multiply both sides by 3 to clear out the fractions
2*sqrt(3) + 3 = 2x*sqrt(3)-3x
2*sqrt(3) + 3 = x(2*sqrt(3)-3)
x = (2*sqrt(3) + 3)/(2*sqrt(3)-3)



Let's multiply top and bottom of that expression by 2*sqrt(3)+3 to rationalize the denominator


{{{x = (2*sqrt(3) + 3)/(2*sqrt(3)-3)}}}


{{{x = ((2*sqrt(3) + 3)(2*sqrt(3)+3))/((2*sqrt(3)-3)(2*sqrt(3)+3))}}}


{{{x = (21+12sqrt(3))/(12-9)}}}


{{{x = (3(7+4sqrt(3)))/(3)}}}


{{{x = 7+4sqrt(3)}}}


I recommend using a tool like WolframAlpha to check your work.


If EN = 1, then FM = x = 7+4*sqrt(3)


Therefore, the ratio of the radius of circle M to circle N is exactly 7+4*sqrt(3)
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