2006 AMC 12A Problem 17

Question

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

\frac{r}{s}?

\dfrac{1}{2}

\dfrac{5}{9}

\dfrac{3}{5}

\dfrac{5}{3}

\dfrac{9}{5}

The tangent length satisfies

AF^2 = AE^2 - r^2

Place

B

at the origin; then

9 + 5\sqrt{2} = s^2 + rs\sqrt{2},

and match rational and irrational parts

Set

B = (0, 0),

C = (s, 0),

A = (0, s),

D = (s, s),

so that

E = \left(s + \tfrac{r}{\sqrt{2}},\ s + \tfrac{r}{\sqrt{2}}\right)

lies on ray

BD.

Since

AF

is tangent to the circle,

AF^2 = AE^2 - r^2.

Computing

AE^2

and simplifying gives

9 + 5\sqrt{2} = s^2 + rs\sqrt{2}.

Because

r

and

s

are rational, the rational and irrational parts match:

s^2 = 9

and

rs = 5.

Thus

s = 3,\ r = \tfrac{5}{3},

and

\frac{r}{s} = \tfrac{5}{9}.

Thus, the correct answer is B.

Problem 17 in other years