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

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__TOC__
 
== Problem ==
 
== Problem ==
 
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__
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== Solution 1 ==
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== Solution ==
 
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=== Solution 1 ===
 
 
<center><asy>
 
<center><asy>
 
size(300);
 
size(300);
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clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle);
 
clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle);
 
</asy></center>
 
</asy></center>
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Applying the [[triangle inequality]] to <math>ADE</math>, we see that <math>AD \ge 20\sqrt {7}</math>. However, if <math>AD</math> is strictly greater than <math>20\sqrt {7}</math>, then the circle with radius <math>10\sqrt {21}</math> and center <math>A</math> does not touch <math>DC</math>, which implies that <math>AC > 10\sqrt {21}</math>, a contradiction. Therefore, <math>AD = 20\sqrt {7}</math>.
 
  
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It follows that <math>A</math>, <math>D</math>, and <math>E</math> are collinear, and also that <math>ADC</math> and <math>ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and
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'''Key observation.''' <math>AD = 20\sqrt{7}</math>.
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<center><math>EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</math></center>
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 +
'''Proof 1.''' By the [[triangle inequality]], we can immediately see that <math>AD \geq 20\sqrt{7}</math>. However, notice that <math>10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}</math>, so by the law of sines, when <math>AD = 20\sqrt{7}</math>, <math>\angle ACD</math> is right and the circle centered at <math>A</math> with radius <math>10\sqrt{21}</math>, which we will call <math>\omega</math>, is tangent to <math>\overline{CD}</math>. Thus, if <math>AD</math> were increased, <math>\overline{CD}</math> would have to be moved even farther outwards from <math>A</math> to maintain the angle of <math>\frac{\pi}{3}</math> and <math>\omega</math> could not touch it, a contradiction.
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'''Proof 2.''' Again, use the triangle inequality to obtain <math>AD \geq 20\sqrt{7}</math>. Let <math>x = AD</math> and <math>y = CD</math>. By the law of cosines on <math>\triangle ADC</math>, <math>2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0</math>. Viewing this as a quadratic in <math>y</math>, the discriminant <math>\Delta</math> must satisfy <math>\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}</math>. Combining these two inequalities yields the desired conclusion.
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This observation tells us that <math>E</math>, <math>A</math>, and <math>D</math> are collinear, in that order.
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Then, <math>\triangle ADC</math> and <math>\triangle ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and
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<center><math>EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</math>.</center>
 
Finally, the answer is <math>25+7=\boxed{032}</math>.
 
Finally, the answer is <math>25+7=\boxed{032}</math>.
  
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=== Solution 2 ===
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== Solution 2 ==
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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>.
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<center><math>EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}</math></center>
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Extend <math>\overline {AB}</math> through <math>B</math>, to meet <math>\overline {DC}</math> (extended through <math>C</math>) at <math>G</math>. <math>ADG</math> is an equilateral triangle because of the angle conditions on the base.
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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.  
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If <math>\overline {GC} = x</math> then <math>\overline {CD} = 40\sqrt{7}-x</math>, because <math>\overline{AD}</math> and therefore <math>\overline{GD}</math> <math>= 40\sqrt{7}</math>.
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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>,  
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and <math>\overline{CF} = \frac{40\sqrt{21} - \sqrt{3}x}{2}</math>
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 +
Similarly <math>CAF</math> is a 30-60-90 triangle and thus <math>\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}</math>.
 +
 
 +
Equating and solving for <math>x</math>, <math>x = 30\sqrt{7}</math> and thus <math>\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5\sqrt{7}</math>.
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<math>\overline{ED}-\overline{FD} = \overline{EF}</math>
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<math>30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7}</math> and <math> 25 + 7 = \boxed{032}</math>
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How do you assume <math>\overline{AD}</math> <math>= 40\sqrt{7}</math>.
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 +
~polya_mouse.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 15:11, 16 November 2024

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 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]

Key observation. $AD = 20\sqrt{7}$.

Proof 1. By the triangle inequality, we can immediately see that $AD \geq 20\sqrt{7}$. However, notice that $10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}$, so by the law of sines, when $AD = 20\sqrt{7}$, $\angle ACD$ is right and the circle centered at $A$ with radius $10\sqrt{21}$, which we will call $\omega$, is tangent to $\overline{CD}$. Thus, if $AD$ were increased, $\overline{CD}$ would have to be moved even farther outwards from $A$ to maintain the angle of $\frac{\pi}{3}$ and $\omega$ could not touch it, a contradiction.

Proof 2. Again, use the triangle inequality to obtain $AD \geq 20\sqrt{7}$. Let $x = AD$ and $y = CD$. By the law of cosines on $\triangle ADC$, $2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0$. Viewing this as a quadratic in $y$, the discriminant $\Delta$ must satisfy $\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}$. Combining these two inequalities yields the desired conclusion.

This observation tells us that $E$, $A$, and $D$ are collinear, in that order.

Then, $\triangle ADC$ and $\triangle ACF$ 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

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

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

By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2}$, 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 $\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5\sqrt{7}$.

$\overline{ED}-\overline{FD} = \overline{EF}$

$30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7}$ and $25 + 7 = \boxed{032}$

How do you assume $\overline{AD}$ $= 40\sqrt{7}$.

~polya_mouse.

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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