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

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== Solution ==
 
== Solution ==
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One possibility is to use the [[coordinate plane]], setting B at the origin. Point A will be (<math>s</math>, 0) and E will be (<math>(s + \frac{r}{\sqrt{2}}, s + \frac{r}{\sqrt{2}})</math>) since B, D, and E are [[collinear]] and contains the diagonal of ABCD. The [[Pythagorean theorem]] results in
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One possibility is to use the [[coordinate plane]], setting <math>B</math> at the origin. Point <math>A</math> will be <math>(s, 0)</math> and <math>E</math> will be <math>\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)</math> since <math>B, D</math>, and <math>E</math> are [[collinear]] and contains the diagonal of <math>ABCD</math>. The [[Pythagorean theorem]] results in
  
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<math>AF^2 + EF^2 = AE^2</math>
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<cmath>AF^2 + EF^2 = AE^2</cmath>
  
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<math>r^2  + (\sqrt{9 + 5\sqrt{2}})^2  = ((s + \frac{r}{\sqrt{2}}) - 0)^2 + ((s + \frac{r}{\sqrt{2}}) - s)^2</math>
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<cmath>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</cmath>
  
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<math>r^2  + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}</math>
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<cmath>r^2  + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}</cmath>
  
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<math>9 + 5\sqrt{2} = s^2 + rs\sqrt{2}</math>
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<cmath>9 + 5\sqrt{2} = s^2 + rs\sqrt{2}</cmath>
  
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This implies that <math>rs = 5</math> and <math>s^2 = 9</math>; dividing gives us <math>\frac{r}{s} = \frac{5}{9} \Rightarrow B</math>
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This implies that <math>rs = 5</math> and <math>s^2 = 9</math>; dividing gives us <math>\frac{r}{s} = \frac{5}{9} \Rightarrow B</math>.
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<!--Todo: geometric proof without coordinate plane-->
 
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
 
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{{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 16:57, 28 October 2007

Problem

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

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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