2006 AMC 12A Problems/Problem 17

Revision as of 23:24, 17 March 2015 by NormanWho (talk | contribs) (Solution 2)

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 $r/s$?

$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{5}{9}\qquad \mathrm{(C) \ } \frac{3}{5}\qquad \mathrm{(D) \ } \frac{5}{3}\qquad \mathrm{(E) \ }  \frac{9}{5}$

AMC12 2006A 17.png

Solution

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

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions

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