Let $ABCD$ be an [isosceles trapezoid](https://artofproblemsolving.com/wiki/index.php/Isosceles_trapezoid) with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The [diagonals](https://artofproblemsolving.com/wiki/index.php/Diagonal) 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](https://artofproblemsolving.com/wiki/index.php/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 $m + n$.
Key observation. $AD = 20\sqrt{7}$.
Proof 1. By the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/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 $25+7=\boxed{032}$.