I am trying to decipher the following "proof" of $\mathbb{R}^2$ is not quasi-isometric to the $3-$regular tree $T_3$ which I scribbled down in a sleep deprived state during a lecture at some point. There quite possibly might be some typos, indeed in my handwritten notes, I somehow switched between $f$ and $\phi$ for the function halfway through.
Suppose we have a q.i.
$\phi :\mathbb{R}^2\to T_3$
Consider the triangle $ABC$ with edges $\alpha \beta \gamma$. $\alpha$ opposite to $A$ and so on. There exists $r$ depending on the q.i. constants such that $\phi(\alpha)$ is $r$-dense in the geodesic between $\phi(B)$ and $\phi(C)$.
I am completely baffled about the following line. How do we know this $m$ exists?
So $m\in [\phi(A),\phi(B)]\cap [\phi(B),\phi(C)]\cap [\phi(C),\phi(A)]$ ([x,y] being the geodesics between 2 points.) Therefore, there exists $P\in \alpha$, $Q\in \beta$, $R\in \gamma$ such that $d_{T_3}(\phi(P),m)\leq r$.
After this I have just written "triangle sufficiently large" and similar inequalities $d_{T_3}(\phi(Q),m)\leq r$ and $d_{T_3}(\phi(R),m)\leq r$.
Any help on what's going on in the proof is appreciated. Thanks!
The idea is that you are mapping a euclidean triangle to a tree, so you have to collapse that triangle to a (coarse)tripod, and with large enough triangles that will contradict the existence of a quasi-isometry.
Draw a picture of a triangle in $T_3$ to find $m$, and draw a picture of what is happening to the triangles whith sufficiently large triangles. That should explain what is happening and give you enough to fill in the details.
(Really you should pretty much always have a picture of what is going on when working with stuff like this.)