Euclid Contest April 2015 Problem 8B
I cannot type it as it has a diagram along with it, which might mess up my interpretation.
This was a tough problem, I cannot solve it completely. HINTS ONLY
I believe it has something to do with Homethecy of the circle. And the scaling factor.
But I am not sure how to use homethecy with circles, what can I do? Hints please.
Hint. Draw mid-perpendicular of YZ, also perpendicular lines of two circles from this line to XZ. Then consider 3 similar triangles, and we have
$$\frac{\sqrt{a^2-\left(b/2\right)^2}-R}{R}=\frac{a}{b/2}=\frac{\sqrt{a^2-\left(b/2\right)^2}-(2R+r)}{r}.$$
Simplify the equality, then you could get the answer.
BTW, the condition $b<2a$ is not necessary, since for any triangle, summation of any two edges is larger than the other, i.e. $|YZ|<|XZ|+|XY|$.