Bend the wire into an arc of a circle.

## Solution

The picture shows that if the wire is bent into a quarter circle, it can be pulled through the corner of the corridor. In this case, it is not difficult to calculate the length of the chord AB, which spans the wire. Let $\mathrm{AX}=\mathrm{OX}=\mathrm{OY}=\mathrm{AY}=\mathrm{x}$. According to the condition, $\mathrm{XT}=1$ - this is the width of the corridor. $\mathrm{OA}=\mathrm{OT}=1+\mathrm{x}$. From triangle OHA, we now get that $(1+x)^{2}=2 x^{2}$, from which we find $x=1+2^{1 / 2}$. From this, we get that $A B=2 x>4$.