Is this triangulation consistent with Sperner's Lemma?

Question

Since any two triangles which intersect have an edge or a vertex in common, the triangulation is simplicial.

However, I am concerned about triangle $A$. Is every sub-triangle supposed to have a vertex in common with the main outer triangle?

Answer 1

The triangle immediately to the left of triangle $A$ (lets call it $X$) and the large triangle to its right (the one that includes side $RB$) intersect, but have no common vertex or side. So this is not a simplical subdivision. The same could be said for triangle $X$ and the triangle directly below it (the one without the label $A$ in its interior) as well.

Answer 2

No, this is not a simplicial triangulation.

But yes, triangles are allowed to have nothing on the boundary of the original triangle (though they are of course inside that triangle). Note by the way that only five triangles in the above triangulation have a vertex in common with the original triangle, which is not the same as having a vertex on the boundary of that triangle. But in any case there is no requirement of this kind; in a typical simplicial triangulation, most triangles will actually be entirely in the interior of the original triangle.