Sublinear clique transversals under a large-clique hypothesis: is $\tau(G)=o_c(n)$? (Erdős #611)

For a graph $G$ on $n$ vertices, the clique transversal number $\tau(G)$ is the minimum number of vertices meeting every maximal clique of $G$ (an inclusion-maximal complete subgraph). Fix $c>0$. Is it true that if every maximal clique of $G$ has at least $cn$ vertices, then $\tau(G)=o_c(n)$ — that is, $\tau(G)/n\to0$ as $n\to\infty$ for each fixed $c$? More precisely, estimate the least $k_c(n)$ such that whenever every maximal clique of $G$ has at least $k_c(n)$ vertices, $\tau(G)<(1-c)n$.

Acceptance. FULLY RESOLVES: a complete proof (machine-checkable preferred, else full written) that for every fixed $c>0$, having all maximal cliques of size $\geq cn$ forces $\tau(G)=o_c(n)$; OR a disproof — an explicit family of graphs with all maximal cliques of size $\geq cn$ yet $\tau(G)\gg n$, with proof. A determination of the threshold growth $k_c(n)$ up to constant factors also fully resolves the quantitative form. ADVANCES: improve either known bound on $k_c(n)$ beyond those in the background — the lower bound $k_c(n)\geq n^{c'/\log\log n}$, or the transversal bound $\tau(G)\leq n-(kn)^{1/2}$ — with proof; OR prove the sublinear-transversal conclusion for a natural subclass of graphs with proof; OR compute $\tau(G)$ exactly for explicit large-clique families up to a new size via reproducible ILP search with a certificate, charting $k_c(n)$ empirically. Deliver the proof file, the improved-bound proof, or the search code plus certified results.