fact that the three main geodesic of the triangle $\alpha\beta\gamma$ do not share any curves. However, now all three main geodesics are closer to $w$ than $c$, which contradicts our choice of $c$. Therefore, the length of $[w^{-1}, w^{+1}]_{S \backslash w}$ is at most $5M$.

Using a similar argument we can show the geodesic in ${\mathcal{C}}(S \backslash v_{\beta\gamma})$ connecting $v_{\beta\gamma}^{-1}$ to the appropriate vertex in $[c, v_{\beta\gamma}]_S$ is $\leq 5M$. Now consider the geodesic in ${\mathcal{C}}(S \backslash c)$ connecting $c_0$ to the second vertex, $x$, of $[c, v_{\beta\gamma}]_S$. Consider the path $[x, v_{\beta\gamma}]_S \cup [v_{\beta\gamma}, \beta_0]_S \cup [\beta_0, v_{\alpha\beta}]_S \cup [v_{\alpha\beta}, c_0]_S$. $c$ cannot be in anywhere in this path, otherwise it would contradict how we chose $c$ or $v_*$. So we can apply the Bounded Geodesic Image Theorem and get that $d_{S \backslash c}(c', x) \leq 4M$. Therefore the path from $\{v_{\beta\gamma}, v_{\beta\gamma}^{-1}\}$ to $\{c, c_0\}$ in the pants graph is less than or equal to $16(5M) + 5M + 4M$. A similar argument can be made for the other two sides of the triangle $\alpha\beta\gamma$, so $\{c, c_0\}$ can be taken to be a center of the triangle. Since $M \leq 100$ the triangle $\alpha\beta\gamma$ is $8,900$-centered at 