Number of maximum degree vertices in a tree Let $T$ be a tree of order $n$, where $T$ has $\Delta(T) \geq 3$ leaves. Show that $T$ has exactly one vertex of maximum degree $\Delta(T)$.
I am stuck on how to prove this statement. 
 A: let $T$ be the tree with vertex set $V$ and edge set $E$.
a basic property of trees is that $|E| = |V|- 1$. also, the sum over the degrees of all vertices is equal to $2|E|$, which implies: $$\frac{\sum_{u\in V}\deg(u)}{2}= |V|-1$$.
let $l$ be the amount of leaves which is also the maximum degree of the graph, let $c$ be a vertex of maximum degree $l$ we know exists. since the degree for leaves is $1$, and the degree of $c$ is $l$, we can extract them from the sum of degrees as $2l$:
$$ \frac{2l + \sum_{u\in V \setminus (L+c)}\deg(u)}{2}=|V|-1 $$
a bit of arrangement leads to:
$$ \frac{\sum_{u\in V \setminus (L+c)}\deg(u)}{2}=|V|-l-1 $$
since each $u\in V \setminus (L+c)$ is not a leaf, they all have degree at least 2, also note that $|\{u\in V \setminus (L+c)\}| = |V|-l-1$, so the only way for the equation to hold is if the degree of each $u\in V \setminus (L+c)$ is exactly $2$, implying there can only be a single vertex of maximum degree greater or equal to $3$.
A: I don't think I understand your question.  What would it say about this tree graph?

A: Suppose that there is 2 or more vertices of maximum degree $\Delta (T)$, then the number of leaves will be at least $\Delta (T) -2$, since (I'll consider it a directed graph now just to use the terms indegree and out degree, but it also works in ordinary graphs) every vertex in a tree has its indegree $1$ or $0$, so its out degree is at least $\Delta (T)-1$, and each edge coming OUT of this vertex is connected to at least $1$ leaf. (2 vertices can't be connected to the same leaf also).
