Problem:
In triangle ABC,AB=AC=100, and BC=56. Circle P has radius 16 and is tangent to AC and BC. Circle Q is externally tangent to circle P and is tangent to AB and BC. No point of circle Q lies outside of △ABC. The radius of circle Q can be expressed in the form m−nk, where m,n, and k are positive integers and k is the product of distinct primes. Find m+nk.
Solution:
Let q be the radius of circle Q. The perpendicular from A to BC has length 4252−72=4⋅24, and so sinB=sinC=2524. Thus
tan2C=tan2B=1+cosBsinB=43.
Place the figure in a Cartesian coordinate system with B=(0,0),C=(56,0), and A=(28,96). Note that circles P and Q are both tangent to BC, and their centers P and Q lie on the angle bisectors of angles C and B, respectively. Thus P=(56−4⋅16/3,16) and Q=(4q/3,q). Also, PQ=q+16, and so
PQ2=(34(q+16)−56)2+(q−16)2=(q+16)2,
which yields (4q−104)2=576q, and thus q2−88q+676=0. The roots of this equation are 44±635, but 44+635 is impossible because it exceeds the base of △ABC. Hence q=44−635, and the requested number is 44+6⋅35=254.
