SOLUTION: In the diagram attached below, Circles M and N are tangent to each other, and to Line AB and Line BC. If Angle ABC=120°, what is the ratio of the radius of Circle M to the radius

Question 1189710: In the diagram attached below, Circles M and N are tangent to each other, and to Line AB and Line BC. If Angle ABC=120°, what is the ratio of the radius of Circle M to the radius of circle N

Diagram: https://imgur.com/a/4dfUycL
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let r and R be the radii of circles N and M, respectively; the problem asks us to find R/r.

Draw segment BM, intersecting circle M at D.

Segment BM bisects angle ABC, so angles ABM and CBM are each 60 degrees.

Draw segment NE with E the point of tangency of circle N to line BC. Draw segment MF with F the point of tangency of circle M to line BC.

Triangles BEN and BFM are similar 30-60-90 right triangles.

The ratio of the hypotenuse to the long leg in a 30-60-90 right triangle is .

In triangle BFM, FM is the long leg; the hypotenuse is

So in triangle BFM,

Solve for R/r:

The current equation involves r/R on the left; take reciprocals to get an equation involving R/r:

ANSWER: The ratio of the radius of circle M to the radius of circle N is

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

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.

Let circle N have a radius of 1. This means EN = ND = 1.
Goal: Find the length of FM
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

----------------------------------------------------

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

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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