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

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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>.
 
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>. Note that while this is not rigorous, the above solution shows that <math>\angle CAF = 30^{\circ}</math> is indeed the only possibility.  
+
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.
 +
 
 +
=== Solution 3 ===
 +
 
 +
Extend <math>\overline {AB}</math> through <math>B</math>, to meet <math>\overline {DC}</math> extended through <math>C</math> at <math>E</math>. <math>ADE</math> is an equilateral triangle because of the angle conditions on the base.
 +
 
 +
If <math>\overline {EC} = x</math> then <math>\overline {CD} = 40\sqrt{7}-x</math>, because <math>\overline{AD}</math> and therefore <math>\overline{ED}</math> <math>= 40\sqrt{7}</math>.
 +
 
 +
By simple angle chasing, <math>CFD</math> is a 30-60-90 triangle and thus <math>\overline{FD} = \frac{40\sqrt{7)-x}{2}</math> and <math>\overline{CF} = /frac {40\sqrt{21} - \sqrt{3}x}{2}</math>
 +
 
 +
Similarly <math>CAF</math> is a 30-60-90 triangle and thus <math>\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}</math>.</center>
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 +
Equating and solving for <math>x</math>, <math>x = 30/sqrt{7}</math> and thus <math>\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5sqrt{7}</math>.</center>
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 +
<math>\overline{ED}-\overline{FD} = \overline{EF}</math></center>
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<math>30\sqrt{7} - 5\sqrt{7} = 25sqrt\{7}</math> and <math> 25 + 7 = \boxed{032}</math>
  
 
== See also ==
 
== See also ==

Revision as of 00:54, 30 June 2011

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]

Assuming that $ADE$ is a triangle and applying the triangle inequality, we see that $AD > 20\sqrt {7}$. 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. As a result, A, D, and E are collinear. Therefore, $AD = 20\sqrt {7}$.

Thus, $ADC$ 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.

Solution 3

Extend $\overline {AB}$ through $B$, to meet $\overline {DC}$ extended through $C$ at $E$. $ADE$ is an equilateral triangle because of the angle conditions on the base.

If $\overline {EC} = x$ then $\overline {CD} = 40\sqrt{7}-x$, because $\overline{AD}$ and therefore $\overline{ED}$ $= 40\sqrt{7}$.

By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\overline{FD} = \frac{40\sqrt{7)-x}{2}$ (Error compiling LaTeX. Unknown error_msg) and $\overline{CF} = /frac {40\sqrt{21} - \sqrt{3}x}{2}$

Similarly $CAF$ is a 30-60-90 triangle and thus $\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}$. Equating and solving for $x$, $x = 30/sqrt{7}$ and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5sqrt{7}$. $\overline{ED}-\overline{FD} = \overline{EF}$

$30\sqrt{7} - 5\sqrt{7} = 25sqrt\{7}$ (Error compiling LaTeX. Unknown error_msg) and $25 + 7 = \boxed{032}$

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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