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

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== Problem ==
 
== Problem ==
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{{image}}
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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>?
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 +
 
 +
<asy>
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real s = 90;
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real r = 50;
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pair e = (r/sqrt(2),r/sqrt(2));
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pair f = (4.34, 74.58);
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draw((-s, 0) -- (-s,-s) -- (0, -s) -- (0,0) -- (-s, 0));
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draw(circle(e,r));
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draw((-s,0) -- f);
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dot(e);
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dot(f);
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label("A", (-s,0), W);
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label("B", (-s,-s), W);
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label("C", (0,-s), E);
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label("D", (0,0), SW);
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label("E", e, E);
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label("F", f, N);
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 +
</asy>
  
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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>?
 
  
 
<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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== Solution ==
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== Solutions ==
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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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=== 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
 +
 
 +
<cmath>AF^2 + EF^2 = AE^2</cmath>
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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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<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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<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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=== Solution 2 ===
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First note that angle <math>\angle AFE</math> is right since <math>\overline{AF}</math> is tangent to the circle. Using the Pythagorean Theorem on <math>\triangle AFE</math>, then, we see
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<cmath>AE^2 = 9 + 5\sqrt{2} + r^2.</cmath>
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But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</math>, <math>\angle ADE = 135^\circ</math>. Using the Law of Cosines on <math>\triangle ADE</math>, we see
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<cmath>AE^2 = s^2 + r^2 - 2sr\cos(135^\circ)</cmath>
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<cmath>AE^2 = s^2 + r^2 - 2sr\left(-\frac{1}{\sqrt{2}}\right)</cmath>
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<cmath>AE^2 = s^2 + r^2 + \sqrt{2}sr</cmath>
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<cmath>9 + 5\sqrt{2} + r^2 = s^2 + r^2 + \sqrt{2}sr</cmath>
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<cmath>9 + 5\sqrt{2} = s^2 + \sqrt{2}sr.</cmath>
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 +
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>, <math>r = \frac{5}{3}</math>, and <math>\frac{r}{s} = \frac{5}{9}</math>.
 +
 
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=== Solution 3 ===
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(Similar to Solution 1)
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First, draw line AE and mark a point Z that is equidistant from E and D so that <math>\angle{DZE} = 90^\circ</math> and that line <math>\overline{AZ}</math> includes point D. Since DE is equal to the radius <math>r</math>, <math>DZ=EZ=\frac{r}{\sqrt2}=\frac{r\sqrt2}{2}.</math>
 +
 
 +
Note that triangles <math>\triangle AFE</math> and <math>\triangle AZE</math> share the same hypotenuse <math>(AE)</math>, meaning that
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<cmath>AZ^2+EZ^2=AF^2+EF^2</cmath>
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Plugging in our values we have:
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<cmath>\left(s+\frac{r\sqrt{2}}{2}\right)^2+\left(\frac{r\sqrt{2}}{2}\right)^2=\left(\sqrt{9+5\sqrt{2}}\right)^2+r^2</cmath>
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<cmath>s^2+sr\sqrt{2}+\frac{r^2}{2}+\frac{r^2}{2}=9+5\sqrt{2}+r^2</cmath>
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<cmath>s^2+sr\sqrt{2}=9+5\sqrt{2}</cmath>
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By logic <math>s=3</math> and <math>sr=5 \implies r=5/3.</math>
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Therefore, <math>\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}</math>
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===Solution 4 - Alcumus===
  
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<math>AF^2 + EF^2 = AE^2</math>
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Let <math>B=(0,0)</math>, <math>C=(s,0)</math>, <math>A=(0,s)</math>, <math>D=(s,s)</math>, and <math>E=\left(s+\frac{r}{\sqrt{2}},s+\frac{r}{\sqrt{2}} \right)</math>. Apply the Pythagorean Theorem to <math>\triangle AFE</math> to obtain <cmath>
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r^2+\left(9+5\sqrt{2}\right)=\left(s+\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}\right)^2,
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</cmath> from which <math>9+5\sqrt{2}=s^2+rs\sqrt{2}</math>. Because <math>r</math> and <math>s</math> are rational, it follows that <math>s^2=9</math> and <math>rs=5</math>, so <math>r/s = \boxed{5/9}</math>.
  
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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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OR
  
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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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Extend <math>\overline{AD}</math> past <math>D</math> to meet the circle at <math>G \ne D</math>. Because <math>E</math> is collinear with <math>B</math> and <math>D</math>, <math>\angle EDG = 45^\circ.</math> Also, <math>ED = EG,</math> which implies <math>\angle EGD = 45^\circ</math>, so <math>\triangle EDG</math> is an isosceles right triangle. Thus <math>DG = r\sqrt{2}</math>. By the Power of a Point Theorem,
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<cmath>\begin{align*}
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9+5\sqrt{2} &= AF^2 \\
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&= AD\cdot AG\\
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& = AD\cdot \left(AD+DG\right) \\
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&=
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s\left(s+r\sqrt{2}\right) \\
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&= s^2+rs\sqrt{2}.\end{align*}</cmath> As in the first solution, we conclude that <math>r/s=\boxed{5/9}</math>.
  
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<math>9 + 5\sqrt{2} = s^2 + rs\sqrt{2}</math>
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===Solution 5 - Answer Choices===
  
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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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(Last minute solution) We roughly measure the distance of <math>r</math> and the distance of <math>y</math>. Since <math>r</math> is clearly less than <math>s</math>, we can eliminate answer choices (D) and (E). Next, if we compare the distances, <math>r</math> seems to be just a little more than half of <math>s</math>, thus eliminating answer choice (A). <math>r</math> is only a little bit bigger than half of <math>s</math>, so we can reasonably assume that their ratio is less than <math>\frac{3}{5}</math>. That leaves us with answer choice <math>\boxed{C}</math> , or <math>\frac{5}{9}</math>.
−
<!--Todo: geometric proof without coordinate plane-->
 
  
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== See also ==
+
==Video Solution by SpredTheMathLove==
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* [[2006 AMC 12A Problems]]
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https://www.youtube.com/watch?v=N0DCok2ha4M
  
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{{AMC box|year=2006|n=12A|num-b=16|num-a=18}}
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== See Also ==
 +
{{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}}
 +
{{MAA Notice}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Latest revision as of 22:29, 28 September 2023

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 $r/s$?


[asy] real s = 90; real r = 50; pair e = (r/sqrt(2),r/sqrt(2)); pair f = (4.34, 74.58); draw((-s, 0) -- (-s,-s) -- (0, -s) -- (0,0) -- (-s, 0)); draw(circle(e,r)); draw((-s,0) -- f); dot(e); dot(f); label("A", (-s,0), W); label("B", (-s,-s), W); label("C", (0,-s), E); label("D", (0,0), SW); label("E", e, E); label("F", f, N); [/asy]


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

Solutions

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

Solution 3

(Similar to Solution 1) First, draw line AE and mark a point Z that is equidistant from E and D so that $\angle{DZE} = 90^\circ$ and that line $\overline{AZ}$ includes point D. Since DE is equal to the radius $r$, $DZ=EZ=\frac{r}{\sqrt2}=\frac{r\sqrt2}{2}.$

Note that triangles $\triangle AFE$ and $\triangle AZE$ share the same hypotenuse $(AE)$, meaning that \[AZ^2+EZ^2=AF^2+EF^2\] Plugging in our values we have: \[\left(s+\frac{r\sqrt{2}}{2}\right)^2+\left(\frac{r\sqrt{2}}{2}\right)^2=\left(\sqrt{9+5\sqrt{2}}\right)^2+r^2\] \[s^2+sr\sqrt{2}+\frac{r^2}{2}+\frac{r^2}{2}=9+5\sqrt{2}+r^2\] \[s^2+sr\sqrt{2}=9+5\sqrt{2}\] By logic $s=3$ and $sr=5 \implies r=5/3.$

Therefore, $\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}$

Solution 4 - Alcumus

Let $B=(0,0)$, $C=(s,0)$, $A=(0,s)$, $D=(s,s)$, and $E=\left(s+\frac{r}{\sqrt{2}},s+\frac{r}{\sqrt{2}} \right)$. Apply the Pythagorean Theorem to $\triangle AFE$ to obtain \[r^2+\left(9+5\sqrt{2}\right)=\left(s+\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}\right)^2,\] from which $9+5\sqrt{2}=s^2+rs\sqrt{2}$. Because $r$ and $s$ are rational, it follows that $s^2=9$ and $rs=5$, so $r/s = \boxed{5/9}$.

OR

Extend $\overline{AD}$ past $D$ to meet the circle at $G \ne D$. Because $E$ is collinear with $B$ and $D$, $\angle EDG = 45^\circ.$ Also, $ED = EG,$ which implies $\angle EGD = 45^\circ$, so $\triangle EDG$ is an isosceles right triangle. Thus $DG = r\sqrt{2}$. By the Power of a Point Theorem, \begin{align*} 9+5\sqrt{2} &= AF^2 \\ &= AD\cdot AG\\ & = AD\cdot \left(AD+DG\right) \\ &= s\left(s+r\sqrt{2}\right) \\ &= s^2+rs\sqrt{2}.\end{align*} As in the first solution, we conclude that $r/s=\boxed{5/9}$.

Solution 5 - Answer Choices

(Last minute solution) We roughly measure the distance of $r$ and the distance of $y$. Since $r$ is clearly less than $s$, we can eliminate answer choices (D) and (E). Next, if we compare the distances, $r$ seems to be just a little more than half of $s$, thus eliminating answer choice (A). $r$ is only a little bit bigger than half of $s$, so we can reasonably assume that their ratio is less than $\frac{3}{5}$. That leaves us with answer choice $\boxed{C}$ , or $\frac{5}{9}$.

Video Solution by SpredTheMathLove

https://www.youtube.com/watch?v=N0DCok2ha4M

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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