  Consider branching Brownian motion in which we begin with one particle at the
origin, particles independently move according to Brownian motion, and
particles split into two at rate one. It is well-known that the right-most
particle at time $t$ will be near $\sqrt{2} t$. Roberts considered the
so-called consistent maximal displacement and showed that with high
probability, there will be a particle at time $t$ whose ancestors stayed within
a distance $ct^{1/3}$ of the curve $s \mapsto \sqrt{2} s$ for all $s \in [0,
t]$, where $c = (3 \pi^2)^{1/3}/\sqrt{2}$. We consider the question of how
close the trajectory of a particle can stay to the curve $s \mapsto (\sqrt{2} +
\varepsilon) s$ for all $s \in [0, t]$, where $\varepsilon> 0$ is small. We
find that there is a phase transition, with the behavior changing when $t$ is
of the order $\varepsilon^{-3/2}$. This result allows us to determine, for
branching Brownian motion in which particles have a drift to the left of
$\sqrt{2} + \varepsilon$ and are killed at the origin, the position at which a
particle needs to begin at time zero for there to be a high probability that
the process avoids extinction until time $t$.
