This is more an extended hint than an answer, but was getting too long for a comment.
Look at the bottom left-hand corner. The angle bisector goes through the centres of the incircle and the smaller circle. If you look at the triangles which are cut off by the bisector, one of the sides of the triangle, and the radii of the two circles you will have similar right-angled triangles.
If you use $d$ for the distance from the vertex to the centre of the small circle, you should be able to express this in terms of the radii of the circles. When you have eliminated $d$ you should get a relationship which allows you to express the smaller radius in terms of the inradius and half the angle at the vertex.