 Solution

At that simultaneous entrance-clock event, $x_G=0$ and $t_G=-vL/c^2$. Its runner-frame position isThe runner is at $x^{\prime }=0$, so the entrance is still ahead by distance $\gamma {\beta }^{2}L$ at the runner-frame emission time. The simultaneously measured distance between the two tunnel ends is $x_E^{\prime }-x_G^{\prime }=\gamma L(1-{\beta }^2)=L/\gamma$, consistent with length contraction.