# 2006 AMC 12A Problems/Problem 17

## 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 $\overline{BE}$. 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 $r/s$?  $\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qquad \mathrm{(D) \ } \frac{5}{3}\qquad \mathrm{(E) \ } \frac{9}{5}$

## Solutions

### Solution 1

One possibility is to use the coordinate plane, setting $B$ at the origin. Point $A$ will be $(0,s)$ and $E$ will be $\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)$ since $B, D$, and $E$ are collinear and contain a 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 $\frac{r}{s} = \frac{5}{9} \Rightarrow B$.

### Solution 2

First note that angle $\angle AFE$ is right since $\overline{AF}$ is tangent to the circle. Using the Pythagorean Theorem on $\triangle AFE$, then, we see $$AE^2 = 9 + 5\sqrt{2} + r^2.$$

But it can also be seen that $\angle BDA = 45^\circ$. Therefore, since $D$ lies on $\overline{BE}$, $\angle ADE = 135^\circ$. Using the Law of Cosines on $\triangle ADE$, we see $$AE^2 = s^2 + r^2 - 2sr\cos(135^\circ)$$ $$AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)$$ $$AE^2 = s^2 + r^2 + \sqrt{2}sr$$ $$9 + 5\sqrt{2} + r^2 = s^2 + r^2 + \sqrt{2}sr$$ $$9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.$$

Thus, since $r$ and $s$ are rational, $s^2 = 9$ and $sr = 5$. So $s = 3$, $r = \frac{5}{3}$, and $\frac{r}{s} = \frac{5}{9}$.

### Solution 3

(Similar to Solution 1) First, draw line AE and mark a point Z that is equidistant from E and D so that $\angle{DZE} = 90^\circ$ and that line $\overline{AZ}$ includes point D. Since DE is equal to the radius $r$, $DZ=EZ=\frac{r}{\sqrt2}=\frac{r\sqrt2}{2}.$

Note that triangles $\triangle AFE$ and $\triangle AZE$ share the same hypotenuse $(AE)$, meaning that $$AZ^2+EZ^2=AF^2+EF^2$$ Plugging in our values we have: $$\left(s+\frac{r\sqrt{2}}{2}\right)^2+\left(\frac{r\sqrt{2}}{2}\right)^2=\left(\sqrt{9+5\sqrt{2}}\right)^2+r^2$$ $$s^2+sr\sqrt{2}+\frac{r^2}{2}+\frac{r^2}{2}=9+5\sqrt{2}+r^2$$ $$s^2+sr\sqrt{2}=9+5\sqrt{2}$$ By logic $s=3$ and $sr=5 \implies r=5/3.$

Therefore, $\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 