Solution
= Solution
The cups have equal temperatures up to the intervention at $t=1$. For the later recrossing, subtract the formula for $T_2$ from that for $T_1$ and set the result to zero:
$$
-e^{-a(t-1)}+\frac1a\left(1-e^{-a(t-2)}\right)=0.
$$
Writing $z=e^{-a(t-2)}$ gives $z=(1+ae^{-a})^{-1}$. Hence the nontrivial meeting time is
$$
\boxed{t^*=2+\frac1a\log(1+ae^{-a})}.
$$