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

m (→‎Solution: correct asy)
m (→‎Solution: more typos)
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label("C",C,NE);
 
label("C",C,NE);
 
label("D",D,SE);
 
label("D",D,SE);
−
label("E",E,W);
+
label("E",E,N);
 
label("F",F,S);
 
label("F",F,S);
 
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);
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<cmath>
 
<cmath>
 
EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</cmath>
 
EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</cmath>
−
Hence the answer to this problem is <math>25+7</math>, or $\boxed{032}.
+
Hence the answer to this problem is <math>25+7</math>, or <math>\boxed{032}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:59, 23 March 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

[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 \[AD \ge 20\sqrt {7} .\] On the other hand, 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. Hence \[AD = 20\sqrt {7}.\] It follows that $A,D,E$ are collinear, and also that $ADC$ and $ACF$ are $30-60-90$ triangles. Hence $AF = 15\sqrt {7}$, and \[EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}\] Hence the answer to this problem is $25+7$, or $\boxed{032}$.

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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