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

Revision as of 18:21, 2 February 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 ( $(s + \frac{r}{\sqrt{2}}, s + \frac{r}{\sqrt{2}})$ ) since B, D, and E are collinear and contains the diagonal of ABCD. The Pythagorean theorem results in

$AF^2 + EF^2 = AE^2$

$r^2 + (\sqrt{9 + 5\sqrt{2}})^2 = ((s + \frac{r}{\sqrt{2}}) - 0)^2 + ((s + \frac{r}{\sqrt{2}}) - s)^2$

$r^2 + 9 + 5\sqrt{2} = s^2 + rs\sqrt{2} + \frac{r^2}{2} + \frac{r^2}{2}$

$9 + 5\sqrt{2} = s^2 + rs\sqrt{2}$

This implies that $rs = 5$ and $s^2 = 9$ ; dividing gives us $\frac{r}{s} = \frac{5}{9} \Rightarrow B$

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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 * [[2006 AMC 12A Problems]]
-{{AMC box|year=2006|n=12A|num-b=16|num-a=18}}
+{{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}}
 [[Category:Introductory Geometry Problems]]

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

Revision as of 18:21, 2 February 2007

Problem

Solution

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