The triangle $ABC$ is divided into sub-simplices in a non-trivial way (there exists at least one point on each of the segments $AB$, $AC$, $BC$ other than $A$, $B$ and $C$).
We label the vertices by numbers in $\{1, 2, 3\}$ according to the following rules:
• $A$ gets $2$ or $3$, $B$ gets $1$ or $3$ and $C$ gets $1$ or $2$.
• Any vertex on $AB$ gets $3$.
• Any vertex on $AC$ gets $2$.
• Any vertex on $BC$ gets $1$.
• Any other vertex gets 1, 2 or 3.
Is it true that there must exists a triangle labeled by 1, 2, 3? Prove or find a counterexample.
I tend to believe that this proposition is true, and should be proved by a reduction to Sperner's Lemma. However - I cannot put my mind into finding a solution.
I would love to see a solution if this is actually true!
It is true.
Pick one on vertex from each of $AB$, $AC$, and $BC$, none of which is equal to $A$, $B$, or $C$. By the given conditions, these have distinct labels. Now apply Sperner's lemma with those as the three outer vertices.
For example, here is a triangulation meeting the given conditions, with one vertex selected on each of the edges (the enlarged ones). Note that a triangulation is a purely combinatorial object; you could just as well view it as a triangulation of the circle. Then to apply Sperner's lemma, all you need is some 3 vertices on the circle, of 3 distinct colors, and so that the vertices along the arcs between them only contain the 2 colors of the endpoints.