SOLUTION: Circle O of radius 20 is inscribed in equilateral triangle ABC. Circle P is tangent to circle O and segments AB and BC. Find the radius of circle P. Express your asnwer as a mixed

Make a sketch.


Line segments BA and BC make the angle ABC of 60°.


Circle O is inscribed into this angle and has the radius of 20 units.


    Do not think about the side AC of the equilateral triangle ABC - it does not participate in the solution.


Let D be the tangent point of the circle and the segment BA.


Draw the radius OD to the tangent point D.


The segment OD is the radius of the circle O and its length is 20 units.


In the sketch, draw smaller circle P with the center at the point P and tangent to segments BA and CA, and also
tangent to the circle O.  Let F be the tangent point of the two circles.


It is obvious that the points B, P and O lies are collinear:  they lie on the same straight line
BO which bisect the angle ABC and makes the angles of 30° with its sides BA and BC.


Triangle BDO is the right triangle with the acute angle DBO of 30°.


Hence, its hypotenuse length BO is twice its leg OD, i.e.  BO = 2*20 = 40 units.


Triangle BEP is ALSO the right triangle with the acute angle EBP of 30°.


If r is the radius of the circle P, then the hypotenuse BP is twice the radius r : BP = 2r.


Next, BO consists of three segments  BO = BP + PF + FO.


Based on it, we can write            40 = 2r +  r + 20,


or, simplifying  20 = 3r;  hence   r =   units.


ANSWER.  The radius of the small circle P is   = 6   units.