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 $m + n$ .

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

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.