CEMC Cayley

Suppose that the crease intersects $PS$ at $X$, $QR$ at $Y$, and the line $PR$ at $Z$. We want to determine the length of $XY$. Since $P$ folds on top of $R$, then line segment $PZ$ folds on top of line segment $RZ$, since after the fold $Z$ corresponds with itself and $P$ corresponds with $R$. This means that $PZ=RZ$ and $PR$ must be perpendicular to $XY$ at point $Z$. Since $PS=RQ$ and $SR=QP$, then right-angled triangles $\triangle PSR$ and $\triangle RQP$ are congruent (side-angle-side). Therefore, $\angle XPZ=\angle YRZ$. Since $PZ=RZ$, then right-angled triangles $\triangle PZX$ and $\triangle RZY$ are congruent too (angle-side-angle). Thus, $XZ=ZY$ and so $XY=2XZ$. Since $\triangle PSR$ is right-angled at $S$, then by the Pythagorean Theorem, $PR=\sqrt{PS^{2}+SR^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{100}=10$ since $PR>0$. Since $PZ=RZ$, then $PZ=\frac{1}{2}PR=5$. Now $\triangle PZX$ is similar to $\triangle PSR$ (common angle at $P$ and right angle), so $\frac{XZ}{PZ}=\frac{RS}{PS}$ or $XZ=\frac{5 \cdot 6}{8}=\frac{30}{8}=\frac{15}{4}$. Therefore, $XY=2XZ=\frac{15}{2}$, so the length of the fold is $\frac{15}{2}$ or 7.5.