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

(Problem)
 
(7 intermediate revisions by 4 users not shown)
Line 2: Line 2:
 
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>?
  
[asy]
 
pair A,B,C,D,I,F;
 
A=(0,10); B=(0,0); C=(10,0); D=(10,10);
 
  
I=(14,13); F=(11,17);
+
<asy>
draw(A--B--C--D--cycle,linewidth(0.7));
+
real s = 90;
draw(Circle(I,5),linewidth(0.7));
+
real r = 50;
draw(A--F,linewidth(0.7));
+
pair e = (r/sqrt(2),r/sqrt(2));
label("<math>A</math>",A,NW);
+
pair f = (4.34, 74.58);
label("<math>B</math>",B,SW);
+
draw((-s, 0) -- (-s,-s) -- (0, -s) -- (0,0) -- (-s, 0));
label("<math>C</math>",C,SE);
+
draw(circle(e,r));
label("<math>D</math>",D,SW);
+
draw((-s,0) -- f);
label("<math>F</math>",F,N);
+
dot(e);
label("<math>E</math>",I,E);
+
dot(f);
dot(I);
 
[/asy]
 
  
 +
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>
  
<center>[[Image:AMC12_2006A_17.png]]</center>
 
  
 
<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>
Line 65: Line 67:
  
 
Therefore, <math>\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}</math>
 
Therefore, <math>\frac{\frac{5}{3}}{3}=\frac{5}{9}=\boxed{B}</math>
 +
 +
===Solution 4 - Alcumus===
 +
 +
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>
 +
r^2+\left(9+5\sqrt{2}\right)=\left(s+\frac{r}{\sqrt{2}}\right)^2+\left(\frac{r}{\sqrt{2}}\right)^2,
 +
</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>.
 +
 +
OR
 +
 +
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,
 +
<cmath>\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*}</cmath> As in the first solution, we conclude that <math>r/s=\boxed{5/9}</math>.
 +
 +
===Solution 5 - Answer Choices===
 +
 +
(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>.
 +
 +
==Video Solution by SpredTheMathLove==
 +
https://www.youtube.com/watch?v=N0DCok2ha4M
  
 
== See Also ==
 
== See Also ==

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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png