Simple geometry problem

[bonanova]

Welcome back, Denizens.

Sharpen your pencils and brush up on your sines and cosines.

Here's a simple geometry / trig puzzle to start the new BD site off.

1. Draw two circles with radii R > r that are centered on the positive x-axis and tangent at the origin.
   
2. Draw an isosceles triangle, in the upper half-plane, whose base is the portion of the x-axis between the circles -- that is, outside the smaller circle and inside the larger circle -- and whose apex touches the larger circle.
   
3. Draw a third circle, also in the upper half-plane, that does not intersect any of the three existing shapes but is tangent to all of them.
   

For the third circle, determine

1. Its radius r_o
   
2. The x-coordinate of its center
   

Have fun!

[crazy]

The figure implies R = 2r and Base of the triangle = R.

Is it intentional?

[bonanova]

Not at all. I agree that's how it looks, but that was chance at work.

I intended for r to be any value between 0 and R.

It might be interesting to get a solution first with r=R/2 and then generalize.

Or even simpler, the extreme cases of r equalling 0 or R.

[bushindo2]

Welcome back, Denizens.

Sharpen your pencils and brush up on your sines and cosines.

Here's a simple geometry / trig puzzle to start the new BD site off.

1. Draw two circles with radii R > r that are centered on the positive x-axis and tangent at the origin.
   
2. Draw an isosceles triangle, in the upper half-plane, whose base is the portion of the x-axis between the circles -- that is, outside the smaller circle and inside the larger circle -- and whose apex touches the larger circle.
   
3. Draw a third circle, also in the upper half-plane, that does not intersect any of the three existing shapes but is tangent to all of them.
   

For the third circle, determine

1. Its radius r_o
   
2. The x-coordinate of its center
   

Have fun!

This is a well constructed (and well phrased) problem. Thanks bonanova for sharing. It's great to have Brainden back.

[bushindo2]

This is a well constructed (and well phrased) problem. Thanks bonanova for sharing. It's great to have Brainden back.

Simpler solution

[bonanova]

Nicely done ... and the interesting property of the third circle is that it's in all cases centered above the left end of the triangle's base. I was looking around for a geometrical proof of that property.

[bushindo2]

Nicely done ... and the interesting property of the third circle is that it's in all cases centered above the left end of the triangle's base. I was looking around for a geometrical proof of that property.

The geometrical proof for the property you mentioned can be derived from two equations for a circle and one equation for the shortest distance to a line. The solution for x₀ can be simplified to the quadratic equation in post #4. I suppose it is relatively straightforward to plug x₀ = 2r into that quadratic equation and show that 2r is always a solution.

So, do I get the coveted bonanova gold star?

[bonanova]

Of course! Here it is [thanks to Plainglazed for a tip].