2006 AMC 12A Problems/Problem 17

Problem

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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 $\overline{BE}$ . Point $F$ lies on the circle, on the same side of $\overline{BE}$ as $A$ . Segment $AF$ is tangent to the circle, and $AF=\sqrt{9+5\sqrt{2}}$ . What is $r/s$ ?

$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qquad \mathrm{(D) \ } \frac{5}{3}\qquad \mathrm{(E) \ } \frac{9}{5}$

Solution

One possibility is to use the coordinate plane, setting B at the origin. Point A will be ( $s$ , 0) and E will be ( $(s + \frac{r}{\sqrt{2}}, s + \frac{r}{\sqrt{2}})$ ) since B, D, and E are collinear and contains the diagonal of ABCD. The Pythagorean theorem results in