2008 AIME I Problems/Problem 10

Revision as of 20:08, 19 June 2014 by Mathgeek2006 (talk | contribs) (Solution 3)


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]

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, $AF=\frac{CF\sqrt{3}}2=15\sqrt{7}$.


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 $G$. $ADG$ 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 $25 + 7 = \boxed{032}$

See also

2008 AIME I (ProblemsAnswer KeyResources)
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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