2006 AMC 12A Problem 17

Problem 17 of 25IntermediateGeometry

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}

A.

Segment

AF

is tangent to the circle, and

AF = \sqrt{9 + 5\sqrt{2}}.

What is

\frac{r}{s}?

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Solution

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.

Tagged: power of a point · tangent line · coordinate geometry