Consider for $n\in\mathbb{N}, n\geq 6$ the discrete triangle $\nabla=\{(i,j)\in \{1,\ldots,n-1\}^2 \mid j\leq n-i\}$. This is basically the lower "half" of a chess board if you cut it along the diagonal which runs from the upper left corner to the lower right corner and throw away half squares. Now, I consider sets of tuples of the form $\Delta(k,l,m):=\{(k,m),(l,m),(k,n-(l-1))\}$ where $(k,l,m)\in\Gamma:=\{(x,y,z)\in\{1,\ldots,n-1\}^3 \mid l> k; m\leq n-(l-1) \}$ and I am wondering about the maximal number of squares in $\nabla$ which can be covered by a set $\{ \Delta(k,l,m)\}_{(k,l,m)\in\Gamma'}$ with $\Gamma'\subset \Gamma$ and for all distinct $(k,l,m),(k',l',m')\in \Gamma'$ we have $\Delta(k,l,m)\cap \Delta(k',l',m')=\emptyset $. The condition implies simply that a square in $\nabla$ is not covered twice by the $\Delta(k,l,m)\}_{(k,l,m)\in\mathcal{S}'}$.
The problem is equivalent to the maximal clique problem of a vertex induced subgraph $K'$ of the Kneser graph $K(\frac{n(n-1)}{2},3)$, where the underlying set aka the vertex set of $K'$ is $$\{\Delta\in\mathrm{Choice}(\nabla,3) | \Delta_1,\Delta_2,\Delta_3\in\Gamma\}$$
where $\Delta_1,\Delta_2,\Delta_3\in\Gamma$ losely means that the implied parameter triples are in $\Gamma$ and $\mathrm{Choice}(\nabla,3)$ that $3$ elements are chosen from $\nabla$ without replacement. The neighborhood relationships are given by $K(\frac{n(n-1)}{2},3)$. This part in the Wikipedia article seems to resolve the question for Kneser graphs https://en.wikipedia.org/wiki/Kneser_graph#Cliques but I am wondering whether anyone sees a direct application of results on Kneser graphs to this Problem.
The graph $K'=(V',E')$, as defined hereinabove, has $\binom{n}{3}$ vertices, since it can be seen as the symmetric half of the projection to (wlog) the $x_1$ plane of $6$ symmetric points in a 6-cut cube, where the symmetries are given by the cut planes and each tetrahedron contains exaclty one point. Additionally, $\Delta(k,l,m)\cap \Delta(k',l',m')\neq \emptyset$ iff two values of $(k,l,m)$ and $(k',l',m')$ are equal. Therefore, two vertices $v,v'$ in $K'$ are adjacent iff the corresponding 3-sets $(k,l,m)$ and $(k',l',m')$ share at most one element. Consquently, by this identity, $K'$ is regular with valency $\frac{(n-4)(n-3)(n+4)}{6}$.
Any suggestions which property $K'$ might inherit from $K$, which could give me a quantitative expression for the maximal clique size in $K'$(aside from cliques in $K'$ being cliques in $K$)? The graph $K'$ is a lot smaller than $K$, i.e., $|K'|=o(|K|)$ in Landau notation, so I do not see any way to proceed directly.
EDIT: Found an error in my designs. Deleted corresponding text passage here and added results on path regarding related clique problem.I will probably develop it here as it moves forward.
