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Difference between revisions of "2004 AIME II Problems/Problem 7"

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(solutions)
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<math> ABCD </math> is a rectangular sheet of paper that has been folded so that corner <math> B </math> is matched with point <math> B' </math> on edge <math> AD. </math> The crease is <math> EF, </math> where <math> E </math> is on <math> AB </math> and <math> F </math> is on <math> CD. </math> The dimensions <math> AE=8, BE=17, </math> and <math> CF=3 </math> are given. The perimeter of rectangle <math> ABCD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math>
 
<math> ABCD </math> is a rectangular sheet of paper that has been folded so that corner <math> B </math> is matched with point <math> B' </math> on edge <math> AD. </math> The crease is <math> EF, </math> where <math> E </math> is on <math> AB </math> and <math> F </math> is on <math> CD. </math> The dimensions <math> AE=8, BE=17, </math> and <math> CF=3 </math> are given. The perimeter of rectangle <math> ABCD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math>
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
=== Solution 1 (synthetic) ===
 +
<center><asy>
 +
pointpen = black; pathpen = black +linewidth(0.7);
 +
pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0);
 +
D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle);
 +
D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G);
 +
D(E--MP("B'",G)--F--B,dashed);
 +
MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0));
 +
</asy></center>
 +
Since <math>EF</math> is the perpendicular bisector of <math>\overline{BB'}</math>, it follows that <math>BE = B'E</math> (by SAS). By the [[Pythagorean Theorem]], we have <math>AB' = 15</math>. Similarly, from <math>BF = B'F</math>, we have
 +
<center><math>
 +
\begin{align*}
 +
BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \
 +
BC  &= \frac{70}{3}
 +
\end{align*}
 +
</math></center>
 +
Thus the perimeter of <math>ABCD</math> is <math>2\left(25 + \frac{70}{3}\right) = \frac{290}{3}</math>, and the answer is <math>m+n=\boxed{293}</math>.
 +
 
 +
=== Solution 2 (analytic) ===
 +
Let <math>A = (0,0), B=(0,25)</math>, so <math>E = (0,8)</math> and <math>F = (l,22)</math>, and let <math>l = AD</math> be the length of the rectangle. The [[slope]] of <math>EF</math> is <math>\frac{14}{l}</math> and so the equation of <math>EF</math> is <math>y -8 = \frac{14}{l}x</math>. We know that <math>EF</math> is perpendicular to and bisects <math>BB'</math>. The slope of <math>BB'</math> is thus <math>\frac{-l}{14}</math>, and so the equation of <math>BB'</math> is <math>y -25 = \frac{-l}{14}x</math>. Let the point of intersection of <math>EF, BB'</math> be <math>G</math>. Then the y-coordinate of <math>G</math> is <math>\frac{25}{2}</math>, so
 +
<center> <math>
 +
\begin{align*}
 +
\frac{14}{l}x &= y-8 = \frac{9}{2}\
 +
\frac{-l}{14}x &= y-25 = -\frac{25}{2}\
 +
\end{align*}
 +
</math> </center>
 +
Dividing the two equations yields
 +
<center><math>l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}</math></center>
 +
The answer is <math>\boxed{293}</math> as above.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2004|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2004|n=II|num-b=6|num-a=8}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 17:00, 8 June 2008

Problem

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Solution 1 (synthetic)

[asy] pointpen = black; pathpen = black +linewidth(0.7); pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G); D(E--MP("B'",G)--F--B,dashed); MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0)); [/asy]

Since $EF$ is the perpendicular bisector of $\overline{BB'}$, it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem, we have $AB' = 15$. Similarly, from $BF = B'F$, we have

$\begin{align*}

BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \ BC &= \frac{70}{3}

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Thus the perimeter of $ABCD$ is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$, and the answer is $m+n=\boxed{293}$.

Solution 2 (analytic)

Let $A = (0,0), B=(0,25)$, so $E = (0,8)$ and $F = (l,22)$, and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$. We know that $EF$ is perpendicular to and bisects $BB'$. The slope of $BB'$ is thus $\frac{-l}{14}$, and so the equation of $BB'$ is $y -25 = \frac{-l}{14}x$. Let the point of intersection of $EF, BB'$ be $G$. Then the y-coordinate of $G$ is $\frac{25}{2}$, so

$\begin{align*}

\frac{14}{l}x &= y-8 = \frac{9}{2}\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Dividing the two equations yields

$l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}$

The answer is $\boxed{293}$ as above.

See also

2004 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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