HMMT November

Since $\angle J=\angle A=90^{\circ}$ and $A M=J S, J A M S$ must be a rectangle. In addition, $\angle M+\angle S=270^{\circ}$, so $\angle M=\angle S=135^{\circ}$. Therefore, $\angle E S M=\angle E M S=45^{\circ}$, which means $M E S$ is an isosceles right triangle. Note that $A M E$ and $J S E$ are congruent, which means that $[J A E S]=[J A E]+[J S E]=[J A E]+[A M E]>[A M E]$, so $A E$ cannot be our diagonal. Similarly, $J E$ cannot be our diagonal. Diagonals $S A$ and $J M$ bisect rectangle $J A M S$, so they also cannot bisect the pentagon. Thus, the only diagonal that can bisect $[J A M E S]$ is $M S$, which implies $[J A M S]=[M E S]$. We know $[J A M S]=J A \cdot A M$ and $[M E S]=\frac{M E \cdot E S}{2}$, and $M E=E S=\frac{J A}{\sqrt{2}}$, which implies $$J A \cdot A M=\frac{J A^{2}}{4} \Longrightarrow \frac{A M}{J A}=\frac{1}{4}$$ Finally, $E M$ and $M S$ are both $\frac{1}{\sqrt{2}}$ the length of $S M=J A$. This means that $A M$ is our shortest side and $J A$ is our longest side, so $\frac{1}{4}$ is our answer.