## 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.
Source: rec.puzzles, Eva Zerz in July 1995, Dennis Yelle in February, 1996.
Solutions were received from Madhuram Rajkumar, Le Nguyen, Joseph DeVincentis, Al Zimmermann, Philippe Fondanaiche, Robert McQuaid, and Mark Morse.
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```

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