Two Lengths

1. A, B and C are in a straight line. AB=BD=CD=1. AD=AC. Find BC.
2. Perpendicular lines CA and CB are both tangent to circle O with radius 4. AB passes through O, the center of the circle. AO intersects the circle at D. BO intersects the circle at E. If BE=1, find AD.

1. From Eva Zerz: The possible distances can easily be computed as 0, (-1+sqrt(5))/2 and (1+sqrt(5))/2 (if "A, B, C in a straight line" does not infer that B has to be between A and C). For if A(-1/0) and B(0/0), then C(c/0), D(d1/d2) with d1^2+d2^2=1, (d1-c)^2+d2^2=1 and (d1+1)^2+d2^2=(c+1)^2. The first two equations give c(c-2d1)=0, hence either c=0 or c=2d1. The first and third equation yield 2d1+1=c^2+2c, which results in the golden ratio equation c^2+c-1=0 when 2d1 is replaced by c.
2. From Dennis Yelle:

   Call the tangents to the circle F and G, so we have:
   
                           C
   
                       F       G
   
                    A  D   O   E   B
   
   CFGO is a square, all angles are 90 degrees and all lengths
   are equal to 4.
   Triangles AFO and OGB are similar right triangles.
   OG = 4, OB = 5, therefore GB = 3.  (3*3 + 4*4 = 5*5)
   AO / FO = OB / GB
   AO = OB / GB * FO = 5 / 3 * 4 = 6.6667
   AD = AO - DO = 6.6667 - 4 = 2.6667