China Girls' Mathematical Olympiad

If $ABCD$ is a square, then $\frac{y}{x}=\frac{1+\sqrt{3}}{2}$.

Now we prove that $\frac{y}{x}\le \frac{1+\sqrt{3}}{2}$.

Denote by $P_1,Q_1,R_1,S_1$ the midpoints of $DA$, $AB$, $BC$, $CD$, and by $E$, $F$, $G$, $H$ the midpoints of $SP$, $PQ$, $QR$, $RS$. Then $P_1{Q}_1{R}_1{S}_1$ is a parallelogram.

Now draw lines $P_{1}E$, $S_{1}E$, and denote by $M$, $N$ the midpoints of $DP$, $DS$. Then $$\begin{array}{rl}D{S}_1& =S_{1}N=DN=EM,\\ D{P}_1& =P_{1}M=MD=EN,\end{array}$$ and $$\begin{array}{rl}\angle P_{1}D{S}_1& =360^{\circ }-60^{\circ }-60^{\circ }-\angle PDS\\ & =240^{\circ }-(180^{\circ }-\angle END)=60^{\circ }+\angle END\\ & =\angle EN{S}_1=\angle EM{P}_1.\end{array}$$ So we have $△D{P}_1{S}_1\cong △M{P}_{1}E\cong △NE{S}_1$. Hence, $△E{P}_1{S}_1$ is equilateral.

By the same argument, $△G{Q}_1{R}_1$ is also equilateral. Now let $U$, $V$ be the midpoints of $P_1{S}_1$, $Q_1{R}_1$, respectively. We then obtain $$\begin{array}{rl}EG& \le EU+UV+VG=\frac{\sqrt{3}}{2}{P}_1{S}_1+P_1{Q}_1+\frac{\sqrt{3}}{2}{Q}_1{R}_1\\ & =P_1{Q}_1+\sqrt{3}{P}_1{S}_1=\frac{1}{2}BD+\frac{\sqrt{3}}{2}AC,\end{array}$$ and also $$FH\le \frac{1}{2}AC+\frac{\sqrt{3}}{2}BD.$$

Taking the sum of these two inequalities, we have $y\le \frac{1+\sqrt{3}}{2}x$, i.e. $$\frac{y}{x}\le \frac{1+\sqrt{3}}{2}.$$