B2. Let $A B C D$ be a square with side length 1 . Points $X$ and $Y$ are on sides $B C$ and $C D$ respectively such that the areas of triangles $A B X, X C Y$, and $Y D A$ are equal. Find the ratio of the area of $\triangle A X Y$ to the area of $\triangle X C Y$.
Solution :
Let $a=B X$ and $b=C Y$. The equality of the triangle areas gives us:
$$
\frac{a}{2}=\frac{(1-a) b}{2}=\frac{1-b}{2} .
$$

The first and third equations give $a+b=1$, and the second and third thus imply $b^{2}+b-1=0$. Solving gives $b=\frac{-1 \pm \sqrt{5}}{2}$; since $b>0$, it follows that $b=\frac{-1+\sqrt{5}}{2}$, whence $a=\frac{3-\sqrt{5}}{2}$. Thus
Area of $\triangle A X Y=1-3($ Area of $\triangle A B X)=1-\frac{3 a}{2}=\frac{3 \sqrt{5}-5}{4}$,
and so
$$
\frac{\text { Area of } \triangle A X Y}{\text { Area of } \triangle C X Y}=\left(\frac{3 \sqrt{5}-5}{4}\right) /\left(\frac{3-\sqrt{5}}{4}\right)=\sqrt{5} \text {. }
$$

The answer is $\sqrt{5}$.