If $G$ is a tree with $k$ leaves, then $G$ is the union of $k/2$ pairwise intersecting paths. Suppose $G$ is a tree with $k$ leaves.  Prove that $G$ is the union of paths $P_1, \ldots, P_{\lceil k/2 \rceil}$ s.t. $P_i \cap P_j \ne \emptyset\ \forall i \neq j$ (Ando-Kaneko-Gervacio [1996]).
The proof in the original paper is very dry.  I came up with the following solution but I am not sure if it is complete.  Am I missing something here?
(Prove by construction )
idea : 
Step 1 Get $k$ paths s.t union of those paths will be G.
Step 2 Merge Paths to get the needed result $\lceil k/2 \rceil$ 


*

*Pick any vertex that is not a leaf ( w.l.g. call it $v$) in $G$ run BFS remembering a paths to the leafs.

*Each path has a common starting vertex v. Taking advantage of it. Merge in pair wise manner all paths. if $k$ is even you, after merging we get $k/2$ paths , if $k$ is odd we get $\lceil k/2 \rceil$  paths.

 A: You can run into trouble when you merge paths. Suppose the leaves are $w_1, w_2, \dots, w_k$, so you start with $k$ paths: $(v, \dots, w_i)$ for $i = 1,2, \dots, k$. Merging these paths can be problematic for two reasons:


*

*It may be the case that if we try to merge $(v, \dots, w_1)$ and $(v, \dots, w_2)$ into one path that goes $(w_1, \dots, v, \dots, w_2)$, we don't actually get a simple path as a result. This happens whenever we merge two paths that take the same edge out of $v$, and is sometimes guaranteed to happen no matter how we pair up the paths to merge. (For example, if the same edge out of $v$ leads to more than half the leaves.)

*We can fix this by "simplifying" the paths after we merge them, deleting redundant edges. But then you can't be sure that the paths remain pairwise intersecting: originally, your only guarantee of that was that they all pass through $v$, and after simplifying, the paths might no longer pass through $v$.



It might help you understand the original proof better to realize that if the leaves of a tree $T$ are $w_1, w_2, \dots, w_k$, we can specify a collection of $\left\lceil \frac k2\right\rceil$ paths that cover $T$ by pairing up the leaves and connecting each pair by a path. (If $k$ is odd, one leaf gets used twice.) There is no further freedom in choosing the paths, and each leaf must be the endpoint of a path or else it remains uncovered.
So all that remains is to choose how to pair the leaves. And the claim that drives the proof is this: if the paths $(w_1, \dots, w_2)$ and $(w_3, \dots, w_4)$ are disjoint, then the paths $(w_1, \dots, w_3)$ and $(w_2, \dots, w_4)$ are not disjoint and have a greater total length.
