The question
We are given a set of size $m$, which we can assume to be the set $[m] = \{1,2,\dots,m\}$.
Say that a sequence $(A_i,B_i,C_i)_{i=1}^k$ triple-splits the set $[m]$ if:
- For each $i$, $(A_i, B_i, C_i)$ is a partition of $[m]$: $A_i,B_i,C_i$ are pairwise disjoint subsets of $[m]$, and $A_i \cup B_i \cup C_i = [m]$.
- For every $3$-element subset $\{x,y,z\} \subset [m]$, there is an $i$ such that $A_i \times B_i \times C_i$ contains a permutation of $\{x,y,z\}$. Iin other words, there is an $i$ such that the three elements $x,y,z$ are in different parts of the partition $(A_i, B_i,C_i)$.
What is the smallest $k$, in terms of $m$, for which we can find such a sequence? I am fine with asymptotic answers, and I am interested in both upper bounds (ideally constructive ones) and lower bounds better than $\Omega(\log m)$.
This can be stated in terms of hypergraphs. What is the smallest $k$ for which we can cover the complete $3$-uniform hypergraph on $m$ vertices with $k$ tripartite $3$-uniform hypergraphs?
Motivation
This is based on my solution to "I don't want the smallest one, I want the second-smallest one" on Puzzling.SE. A sequence that triple-splits $[m]$ would give us a solution to the problem of finding the third-smallest element of $\{x_1,x_2,\dots,x_m\}$. Specifically, the formula would be $$ \min\Big\{\max\{\min\{x_i : i \in A_1\}, \min\{x_i : i \in B_1\}, \min\{x_i : i \in C_1\}\}, \\ \qquad\max\{\min\{x_i : i \in A_2\}, \min\{x_i : i \in B_1\}, \min\{x_i : i \in C_2\}\}, \\ \dots \\ \qquad\max\{\min\{x_i : i \in A_k\}, \min\{x_i : i \in B_k\}, \min\{x_i : i \in C_k\}\}\Big\}. $$ Each max is a max of three distinct elements from $\{x_1, \dots, x_m\}$, so each max is at least the third-smallest element. On the other hand, if the sequence triple-splits $[m]$, and $x_a,x_b,x_c$ are the three smallest elements, then there has to be some $i$ for which $A_i, B_i, C_i$ each contain one of $a,b,c$. For that $i$, the max we'll get will be $\max\{x_a,x_b,x_c\}$: exactly the third-smallest element.