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Difference between revisions of "2006 AMC 12A Problems/Problem 17"

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== Problem ==
 
== Problem ==
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Square <math>ABCD</math> has side length <math>s</math>, a circle centered at <math>E</math> has radius <math>r</math>, and <math>r</math> and <math>s</math> are both rational. The circle passes through <math>D</math>, and <math>D</math> lies on <math>\overline{BE}</math>. Point <math>F</math> lies on the circle, on the same side of <math>\overline{BE}</math> as <math>A</math>. Segment <math>AF</math> is tangent to the circle, and <math>AF=\sqrt{9+5\sqrt{2}}</math>. What is <math>r/s</math>?
 
Square <math>ABCD</math> has side length <math>s</math>, a circle centered at <math>E</math> has radius <math>r</math>, and <math>r</math> and <math>s</math> are both rational. The circle passes through <math>D</math>, and <math>D</math> lies on <math>\overline{BE}</math>. Point <math>F</math> lies on the circle, on the same side of <math>\overline{BE}</math> as <math>A</math>. Segment <math>AF</math> is tangent to the circle, and <math>AF=\sqrt{9+5\sqrt{2}}</math>. What is <math>r/s</math>?
  
 
<math> \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}</math>
 
<math> \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}</math>
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<center>[[Image:AMC12_2006A_17.png]]</center>
  
 
== Solution ==
 
== Solution ==

Revision as of 19:12, 8 November 2007

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

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

See also

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
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