Correct solutions came from Philippe Fondanaiche, France; Farid Lian, Colombia; Jerome Cherry, USA; Ron Welch, USA; Juan Carlos Marivela, Spain; Jerome Cherry, USA; John Snyder, USA.
Let x denote the point of tangency of circles T and L. Triangle abt is a right triangle, so (ax + xt)2 = bt2 + ab2 and (1/4 + xt)2 = bt2 + (1/4)2 . Use bt + xt = 1/2 so solve for xt = 1/6. Let a = angle abs and r be the radius of S. The Law of Cosines gives
(r + 1/4)2 = (1/2 -
r)2 + (1/4)2 - 2(1/4)(1/2
- r) cos a,
(r + 1/6)2 = (1/2 - r)2
+ (1/3)2 - 2(1/3)(1/2 -
r) cos(p/2 - a),
so cos a = (6r - 1)/(2r - 1), cos(p/2 - a) = sin(a) = (6r - 1)/(2r - 1). Now use cos2(a) + sin2(a) = 1 to solve for r = 1/12. The result now follows easily.