The following trivial observation (OK, it’s a version of Axiom of Choice :-)) is still useful:

every family admits a pinning of cardinality not higher than that of the family itself.

It follows easily from my earlier intro how small pinnings lead to cliques. But here let me write down my first steps. Consider 2-coloring:

What can we say about the pinnings when there is no unicolor tetrahedron?

• When then it is necessary and sufficient that there is at least one triangle of each color; and it’s easily possible, hence there may be no unicolor tetrahedron (there is one altogether anyway :-)).
• Now let  .  Any two triangles have at least one vertex in common (because  )).&nbsp Thus if no more than 2 triangles are painted in one of the colors then they admit a 1-element pinning, hence the tetrahedron of the remainng 4 vertices is unicolored (has all its triangular faces colored by the other color). Thus the assumption about the absence of the unicolored tetrahedrons implies that there are at least 3 different triangles painted in each of the colors. And indeed, such coloring, say in  ,  free of unicolored tetrahedrons, is possible:

    blue 

    red 

  And the remaining 4 triangles (for a total of 10) can be painted arbitrarily.

  This is already a result of interest. When    has any finite number of elements then any 5-element subset houses at least 3 triangles of each color which is a significant restriction.

Now let  ,  say  .  Then  ,  contains    different triangles colored blue. If    then there is a vertex, say vertex  ,  which pins at least 2 blue triangles, thus leaving at the most one blue triangle (inside  ).  Thus there are at least 2 more blue triangles within  ,  for a total of at least 5 different blue triangles in the whole  .

Next, let  ,  contain    different triangles colored blue. Then there are 4*3 = 12 incident pairs blue_triangle-vertex, and  ,  which means that there is a vertex, say vertex  ,  which pins at least 3 blue triangles, thus leaving at the most one blue triangle (inside  ).  Thus there are at least 2 more blue triangles within  ,  for a total of at least 6 different blue triangles in the whole  .

In the remaining case of    there are at least 5 different blue triangles in    right away.

Thus there are at least 5 different blue triangles always (under the assumption of no unicolored tetrahedron).

Everything that was said about blue holds also for red. Thus if  ,  and there is no unicolor tetrahedron, then there are at least 5 different triangles of each color.

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