Prove that if MST T1 has k edges with weight 1 then T2 has also k edges with weight 1

Question

We have two different minimum spanning trees ($T_1$ and $T_2$) of $G$. The graph $G$ has edges with weight $1$ or $2$. How can I prove that if $T_1$ has $k$ edges with weight equal to $1$ then $T_2$ has also $k$ edges with weight equal to $1$?

Answer 1

Let $n$ denote the number of vertices in $G$. Let $k_1$ be the number of edges with weight one in $T_1$. Then the total weight of $T_1$ is $k_1 + 2(n - k_1)$. Similarly, the total weight of $T_2$ is $k_2 + 2(n - k_2)$ where $k_2$ denotes the number of edges with weight one in $T_2$. Since both $T_1$ and $T_2$ are minimum spanning trees we must have $k_1 + 2(n - k_1) = k_2 + 2(n - k_2)$. After cancelling out some terms we get $-k_1 = -k_2$ and hence $k_1 = k_2$.