= Solution
The meetings occur where Bob's constant position $x=\alpha ct_0$ intersects Alice's positive-$x$ branch of the <hyperbolic motion>:
$$
t_\pm=\pm t_0\sqrt{\alpha^2-1}.
$$
Bob is at rest in this <inertial frame>, so \b[his elapsed <proper time>] is
$$
\boxed{\Delta\tau_B=2t_0\sqrt{\alpha^2-1}}.
$$
Parametrize Alice's <world line> by
$$
x=ct_0\cosh\eta,\qquad t=t_0\sinh\eta.
$$
Since $t$ increases with $\eta$, the <proper time> element is
$$
d\tau=\sqrt{dt^2-\frac{dx^2}{c^2}}
=t_0\sqrt{\cosh^2\eta-\sinh^2\eta}\,d\eta=t_0\,d\eta.
$$
At the meetings, $\cosh\eta=\alpha$, giving endpoints $\eta=\pm\operatorname{arcosh}\alpha$. Thus \b[Alice's elapsed proper time] is
$$
\boxed{\Delta\tau_A=2t_0\operatorname{arcosh}\alpha}.
$$
Here $\operatorname{arcosh}$ is the nonnegative real <inverse hyperbolic cosine>. As a check, if $\alpha=\cosh\eta_*$ with $\eta_*>0$, then Bob ages by $2t_0\sinh\eta_*$ and Alice by $2t_0\eta_*$; $\sinh\eta_*>\eta_*$ confirms that Bob again ages more.
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