Harvard-MIT Mathematics Tournament

Solution:

Answer: $\sqrt[3]{2}-1$. Let $AD = x$, angle $ABC = t$. We also have $\angle BCA = 90 - t$ and $\angle DCA = 90 - 2t$ so that $\angle ADC = 2t$. Considering triangles $ABE$ and $ADC$, we obtain, respectively,
$\cos(t) = 1/(1 + x)$ and $\cos(2t) = x$. By the double angle formula we get, $(1 + x)^3 = 2$.

Alternatively, construct $M$, the midpoint of segment $BC$, and note that triangles $ABC$, $EBA$, and $MBD$ are similar. Thus, $AB^2 = BC \cdot BE = BC$. In particular,
$$AB = \frac{BC}{AB} = \frac{AB}{BE} = \frac{BD}{BM} = \frac{2BD}{BC} = \frac{2}{AB^2}$$
from which $AB = \sqrt[3]{2}$ and $AD = \sqrt[3]{2} - 1$.