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

Solution 4 - Alcumus

Let $B=(0,0)$ , $C=(s,0)$ , $A=(0,s)$ , $D=(s,s)$ , and $E=\left(s+\frac{r}{\sqrt{2}},s+\frac{r}{\sqrt{2}} \right)$ . Apply the Pythagorean Theorem to $\triangle AFE$ to obtain\[ r^2+\left(9+5\sqrt{2}\right)=\left(s+\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}\right)^2, \]from which $9+5\sqrt{2}=s^2+rs\sqrt{2}$ . Because $r$ and $s$ are rational, it follows that $s^2=9$ and $rs=5$ , so $r/s = \boxed{5/9}$ .

OR

Extend $\overline{AD}$ past $D$ to meet the circle at $G \ne D$ . Because $E$ is collinear with $B$ and $D$ , $\angle EDG = 45^\circ.$ Also, $ED = EG,$ which implies $\angle EGD = 45^\circ$ , so $\triangle EDG$ is an isosceles right triangle. Thus $DG = r\sqrt{2}$ . By the Power of a Point Theorem,\begin{align*} 9+5\sqrt{2} &= AF^2 \\ &= AD\cdot AG\\ & = AD\cdot \left(AD+DG\right) \\ &= s\left(s+r\sqrt{2}\right) \\ &= s^2+rs\sqrt{2}.\end{align*}As in the first solution, we conclude that $r/s=\boxed{5/9}$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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