Let $G = \langle S \rangle$ be a finitely generated, 1-ended $\delta$-hyperbolic group, let $\Gamma$ be its Cayley graph.
I want to show that for any $n$, there are points $a_n$, $b_n$ such that $d(a_n, b_n) = 2n$, $d(a_n, e) = d(b_n, e) = n$, and there is a path $p$ joining $a_n$, $b_n$ which intersects $B(e,n)$ only at $a_n$ and $b_n$, and is contained in the closed ball $B(e, n+2\delta+2)$.
My ideas so far:
$G$ is 1 ended, so if we remove the compact $B(e,n-1)$, we get an unbounded connected component. There certainly exist points $a \in G$ such that $d(e,a) = n$, since $G$ is an infinite, locally finite graph. But my problem is: what if there is only 1 point at distance $n$ from $e$? Why doesn't this case arise?
Maybe this is something to do with growth functions, since we are looking for the change in number of elements from $|B(n-1)|$ to $|B(n)|$ to be bigger than 1.
Once we have $a_n$ and $b_n$ with the distances desired, I don't see exactly why there is a path that fits within the bounds given either. Maybe if we take a minimal length path from $a_n$ to $b_n$ that doesn't intersect $B(e,n)$ (this certainly exists since you can go away from $e$ from both $a_n$, $b_n$, and you are still connected, right?), call it $p$, this works. It has some length $\geq 2n$. We know the path from $a_n$ to $e$ to $b_n$ is geodesic (by the distances required earlier), call this $q$. I think that the paths $p$ and $q$ must be somewhat close ($p$ seems like a quasigeodesic, since we took minimal length, but this is probably not true), and this probably gives the last result.
I really have little intuition for hyperbolic groups, the only 1 ended finitely generated group I really have a grasp on is $\mathbb{Z}^2$, which is not hyperbolic, so my intuition fails me most of the time. Help appreciated.