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Difference between revisions of "2008 AIME I Problems/Problem 10"

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Let <math>ABCD</math> be an [[isosceles trapezoid]] with <math>\overline{AD}||\overline{BC}</math> whose angle at the longer base <math>\overline{AD}</math> is <math>\dfrac{\pi}{3}</math>. The [[diagonal]]s have length <math>10\sqrt {21}</math>, and point <math>E</math> is at distances <math>10\sqrt {7}</math> and <math>30\sqrt {7}</math> from vertices <math>A</math> and <math>D</math>, respectively. Let <math>F</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AD}</math>. The distance <math>EF</math> can be expressed in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>.
 
Let <math>ABCD</math> be an [[isosceles trapezoid]] with <math>\overline{AD}||\overline{BC}</math> whose angle at the longer base <math>\overline{AD}</math> is <math>\dfrac{\pi}{3}</math>. The [[diagonal]]s have length <math>10\sqrt {21}</math>, and point <math>E</math> is at distances <math>10\sqrt {7}</math> and <math>30\sqrt {7}</math> from vertices <math>A</math> and <math>D</math>, respectively. Let <math>F</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AD}</math>. The distance <math>EF</math> can be expressed in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>.
  
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__TOC__
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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=== Solution 2 ===
 
=== Solution 2 ===
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No restrictions are set on the lengths of the bases, so let <math>\angle CAF = 30^{\circ}</math>. Since <math>CAF</math> is a <math>30-60-90</math> triangle, <math>AF=\frac{CF\sqrt{3}}2=15\sqrt{7}</math>.
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No restrictions are set on the lengths of the bases, so for calculational simplicity let <math>\angle CAF = 30^{\circ}</math>. Since <math>CAF</math> is a <math>30-60-90</math> triangle, <math>AF=\frac{CF\sqrt{3}}2=15\sqrt{7}</math>.
 
<center><math>EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}</math></center>
 
<center><math>EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}</math></center>
−
The answer is <math>25+7=\boxed{032}</math>.
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The answer is <math>25+7=\boxed{032}</math>. Note that while this is not rigorous, the above solution shows that <math>\angle CAF = 30^{\circ}</math> is indeed the only possibility.  
  
 
== See also ==
 
== See also ==

Revision as of 16:36, 19 April 2008

Problem

Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The diagonals have length $10\sqrt {21}$, and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$, respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$. The distance $EF$ can be expressed in the form $m\sqrt {n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$.

Solution

Solution 1

[asy] size(300); defaultpen(1); pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0); draw(F--C--B--A); draw(E--A--D--C); draw(A--C,dashed); draw(circle(A,abs(C-A)),dotted); label("\(A\)",A,S); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,SE); label("\(E\)",E,N); label("\(F\)",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); [/asy]

Applying the triangle inequality to $ADE$, we see that $AD \ge 20\sqrt {7}$. However, if $AD$ is strictly greater than $20\sqrt {7}$, then the circle with radius $10\sqrt {21}$ and center $A$ does not touch $DC$, which implies that $AC > 10\sqrt {21}$, a contradiction. Therefore, $AD = 20\sqrt {7}$.

It follows that $A$, $D$, and $E$ are collinear, and also that $ADC$ and $ACF$ are $30-60-90$ triangles. Hence $AF = 15\sqrt {7}$, and

$EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}$

Finally, the answer is $25+7=\boxed{032}$.

Solution 2

No restrictions are set on the lengths of the bases, so for calculational simplicity let $\angle CAF = 30^{\circ}$. Since $CAF$ is a $30-60-90$ triangle, $AF=\frac{CF\sqrt{3}}2=15\sqrt{7}$.

$EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}$

The answer is $25+7=\boxed{032}$. Note that while this is not rigorous, the above solution shows that $\angle CAF = 30^{\circ}$ is indeed the only possibility.

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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