Find the radius of the middle circle

[xcrunner448]

Homework Statement

There are 3 circles, each tangent to 2 lines and to each other (as in the picture). The radius of the right (largest) circle is 8, and the radius of the left (smallest) circle is 4. What is the radius of the middle circle?

The Attempt at a Solution

I tried using similar triangles to solve the problem, using the right triangles formed by the bottom line, the radii of the circles, and the angle bisector (which goes through the centers of all of the circles). However, I could not find any relation that would give me an equation to find x. I also tried to somehow calculate the angle between the two lines, but that got really complicated. I am not sure what other method I could use to solve this, and any help would be appreciated.

 

[chaoseverlasting]

Can you show us? From the graph, we can at least see three equal ratios. Let the distance from the vertex to the first circle be y. Working on from there get the three equal ratios from which you should be able to get two equations. You can always eliminate y and solve for x.

 

[Filip Larsen]

Using similar triangles should get you the answer (hint: you have to solve two equations with two unknowns).

 

[xcrunner448]

Ok I think I got it now. Using y as the distance from the vertex to the center of the first circle, I got y/4 = (4+x)/x and y-8-x = 4+x. From the second equation I got y=2x+12, and plugging that into the first equation I found that x=4*sqrt(2). I guess I just didn't see that when I first tried to solve it. Thank you!

In the meantime I think I solved it another way, although it is much more complex. After quite a bit of work I found that the angle between the two tangent lines for tangential circles of radii a and b (with b>a) is 2*arcsin((b-a)/(a+b)). I did this for each pair of circles (b=8, a=x and b=x, a=4) and set them equal to each other, giving 2*arcsin((8-x)/(8+x))=2*arcsin((x-4)/(x+4)), which led to the proportion (8-x)/(8+x)=(x-4)/(x+4). When I solved that I got x=4*sqrt(2), the same answer as above. It was a lot more work than the similar triangles method, but it worked. Also, I thought it was interesting that the radius of the middle circle is always the geometric mean of the radii of the two outer circles. (x=4*sqrt(2)=sqrt(32)=sqrt(4*8))

 

[SammyS]

You can do it directly with the right triangles indicated on the figure I modified using your figure.
The hypotenuse of the red triangle is x+4, its vertical leg is x‒4 .
The hypotenuse of the blue triangle is x+8, its vertical leg is 8‒x .