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