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

Revision as of 18:09, 8 December 2013

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 $r/s$ ?

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

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 draw the line $\overline{AE}$ . Since $\overline{AF}$ is tangent to the circle, $\angle AFE$ is a right angle. 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 $\angle ADE = 45^\circ$ since $E$ lies on $BD$ . Therefore, $\angle ADE = 135^\circ$ . Using the Law of Cosines on $\triangle ADE$ , we see

\[AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)\] (Error compiling LaTeX. Unknown error_msg)

$AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)$ $AE^2 = s^2 + r^2 + \sqrt{2}sr$ $9 + 5\sqrt{2} + r^2 = s^2 + r^2 + \sqrt{2}sr$ $9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.$

Thus, since $r$ and $s$ are rational, $s^2 = 9$ and $sr = 5$ . So $s = 3$ and $r = \frac{5}{3}$ and $\frac{r}{s} = \frac{5}{9}$ .

@@ Line 7: / Line 7: @@
 == Solution ==
+===Solution 1===
 One possibility is to use the [[coordinate plane]], setting <math>B</math> at the origin. Point <math>A</math> will be <math>(0,s)</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 contain a diagonal of <math>ABCD</math>. The [[Pythagorean theorem]] results in
@@ Line 18: / Line 19: @@
 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>.
+===Solution 2===
+First draw the line <math>\overline{AE}</math>. Since <math>\overline{AF}</math> is tangent to the circle, <math>\angle AFE</math> is a right angle. Using the Pythagorean Theorem on <math>\triangle AFE</math>, then, we see
+<cmath>AE^2 = 9 + 5\sqrt{2} + r^2.</cmath>
+But it can also be seen that <math>\angle ADE = 45^\circ</math> since <math>E</math> lies on <math>BD</math>. Therefore, <math>\angle ADE = 135^\circ</math>. Using the Law of Cosines on <math>\triangle ADE</math>, we see
+<cmath>AE^2 &= s^2 + r^2 - 2sr\cos(135^\circ)</cmath>
+<cmath>AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)</cmath>
+<cmath>AE^2 = s^2 + r^2 + \sqrt{2}sr</cmath>
+<cmath>9 + 5\sqrt{2} + r^2 = s^2 + r^2 + \sqrt{2}sr</cmath>
+<cmath>9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.</cmath>
+Thus, since <math>r</math> and <math>s</math> are rational, <math>s^2 = 9</math> and <math>sr = 5</math>. So <math>s = 3</math> and <math>r = \frac{5}{3}</math> and <math>\frac{r}{s} = \frac{5}{9}</math>.
 == 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
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