Harvard-MIT Mathematics Tournament

Solution:

Given a polygon $P_{1} P_{2} \cdots P_{k}$, let $\left[P_{1} P_{2} \cdots P_{k}\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\overline{A C}$, and let $E$ be the intersection of $\overline{A D}$ and $\overline{B^{\prime} C}$. Then we end up with the pentagon $A C D E B^{\prime}$, depicted above. Let's suppose, without loss of generality, that $A B C D$ has area 1. Then $\triangle A E C$ must have area $\frac{3}{10}$, since
$$\begin{aligned} \{[A B C D] \} & =[A B C]+[A C D] \\ & =\left[A B^{\prime} C\right]+[A C D] \\ & =\left[A B^{\prime} E\right]+2[A E C]+[E D C] \\ & =\left[A C D E B^{\prime}\right]+[A E C] \\ & =\frac{7}{10}[A B C D]+[A E C], \end{aligned}$$
That is, $[A E C]=\frac{3}{10}[A B C D]=\frac{3}{10}$.

Since $\triangle E C D$ is congruent to $\triangle E A B^{\prime}$, both triangles have area $\frac{1}{5}$. Note that $\triangle A B^{\prime} C$, $\triangle A B C$, and $\triangle C D A$ are all congruent, and all have area $\frac{1}{2}$. Since $\triangle A E C$ and $\triangle E D C$ share altitude $\overline{D C}$, $\frac{D E}{E A}=\frac{[D E C]}{[A E C]}=\frac{2}{3}$. Because $\triangle C A E$ is isosceles, $C E=E A$. Let $A E=3 x$. Then $C E=3 x$, $D E=2 x$, and $C D=x \sqrt{9-4}=x \sqrt{5}$. Then $\frac{A D}{D C}=\frac{A E+E D}{D C}=\frac{3+2}{\sqrt{5}}=\sqrt{5}$.