2006 AMC 12A Problems/Problem 17

Problem

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)$ since $B, D$ , and $E$ are collinear and contains the diagonal of $ABCD$ . The Pythagorean theorem results in

$AF^2 + EF^2 = AE^2$

$r^2 + \left(\sqrt{9 + 5\sqrt{2}}\right)^2 = \left(\left(s + \frac{r}{\sqrt{2}}\right) - 0\right)^2 + \left(\left(s + \frac{r}{\sqrt{2}}\right) - s\right)^2$

$r^2 + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}$

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

This implies that $rs = 5$ and $s^2 = 9$ ; dividing gives us $\frac{r}{s} = \frac{5}{9} \Rightarrow B$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2006 AMC 12A Problems/Problem 17

Problem

Solution

See also