2008 AIME I Problems/Problem 10

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$",E,W); label("$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}.

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2008 AIME I Problems/Problem 10

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