Problem:
Square ABCD has side length s, a circle centered at E has radius r, and r and s are both rational. The circle passes through D, and D lies on BE. Point F lies on the circle, on the same side of BE as A. Segment AF is tangent to the circle, and AF=9+52​​. What is r/s?
Answer Choices:
A. 21​
B. 95​
C. 53​
D. 35​
E. 59​ Solution:
Let B=(0,0),C=(s,0),A=(0,s),D=(s,s), and E=(s+2​r​,s+2​r​). Apply the Pythagorean Theorem to △AFE to obtain
r2+(9+52​)=(s+2​r​)2+(2​r​)2
from which 9+52​=s2+rs2​. Because r and s are rational, it follows that s2=9 and rs=5, so r/s=5/9.
OR
Extend AD past D to meet the circle at Gî€ =D. Because E is collinear with B and D,â–³EDG is an isosceles right triangle. Thus DG=r2​. By the Power of a Point Theorem,
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