# Tangent cone of metric graph

I am starting to study some lecture notes about metric geometry and I would appreciate it if someone could some questions regarding the notion of the tangent cone.

Consider 3 half lines joined by their point of origin. You get a network of 3 roads and a junction point. This is an Aleksandrov space of non-positive curvature. It is even geodesically complete. Now, I am having trouble defining the tangent cone at this junction point:

• Is it isometric or BiLipschitz to $$R^k$$ for some $$k>0$$ ?
• What is the Hausdorff dimension of the space at this point ? is it 1 ?
• is the junction point the boundary of the Aleksandrov space ?

• It looks like an exercise. If $X$ is your space and $o$ the joining point (I assume the metric is the geodesic one), it's obvious that $X$ is isometric to its tangent cone at $0$. More generally, if $X$ is a proper metric space and and has a 1-parameter subgroup of non-isometric dilations fixing a point $o$, then every tangent cone of $(X,o)$ is isometric to $(X,o)$ as pointed metric space.
Loosely speaking, if you are standing at the origin then there are only three directions you can travel. So the space of directions $$S_o$$ is only 3 points. Taking the product of the space of directions with $$[0,\infty)$$ and identifying $$S_o \times \{0\}$$ to a point gives you the tangent cone, which in this case is isometric to your starting space.
If you are looking for resources on tangent cones, I would see Nikolaev's paper 'The tangent cone of an Aleksandrov space of curvature $$\leq k$$' or Halbeisen's 'On tangent cones of Alexandrov spaces with curvature bounded below'