1. $A B C$ is an isosceles triangle with $A B=2$ and $\measuredangle A B C=90^{\circ} . D$ is the midpoint of $B C$ and $E$ is on $A C$ such that the area of $A E D B$ is twice the area of $E C D$. Find the length of $D E$.
Answer: $\sqrt[\sqrt{17}]{3}$ Let $F$ be the foot of the perpendicular from $E$ to $B C$. We have $[A E D B]+[E D C]=$ $[A B C]=2 \Rightarrow[E D C]=\frac{2}{3}$. Since we also have $[E D C]=\frac{1}{2}(E F)(D C)$, we get $E F=F C=\frac{4}{3}$. So $F D=\frac{1}{3}$, and $E D=\frac{\sqrt{17}}{3}$ by the Pythagorean Theorem.