229. The length of the edge of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ is 1. On the extension of the edge $A D$ beyond point $D$, a point $M$ is chosen such that $|A M| = 2 \sqrt{2 / 5}$. Point $E$ is the midpoint of the edge $A_{1} B_{1}$, and point $F$ is the midpoint of the edge $D D_{1}$. What is the maximum value that the ratio $|M P| / |P Q|$ can take, where point $P$ lies on the segment $A E$, and point $Q$ lies on the segment $C F$?
229. The lines $A E$ and $C F$ are perpendicular. Let $Q_{1}$ be the projection of $Q$ onto the plane $A B B_{1} A_{1} \cdot Q_{1}$ lies on the segment $B L$, where $L$ is the midpoint of $A A_{1}$. Let $N$ be the intersection point of $A E$ and $L B$. It is not difficult to find that $|A N|=\frac{1}{\sqrt{5}}$. Denote: $|A P|=\frac{1}{\sqrt{5}}+x,\left|N Q_{1}\right|=y$. Then $|P M|^{2}=\frac{8}{5}+\left(\frac{1}{\sqrt{5}}+x\right)^{2},|P Q|^{2}=x^{2}+y^{2}+1, \frac{|P M|^{2}}{|Q P|^{2}}$ reaches its maximum value when $y=0$. It remains to find the maximum value of the fraction $\frac{\frac{9}{5}+\frac{2}{\sqrt{5}} x+x^{2}}{x^{2}+1}$. This value is achieved when $x=\frac{1}{\sqrt{5}}$.

Answer: $\sqrt{2}$.