10.5. What is the maximum number of 4-element subsets that can be selected from a set of 8 elements such that the intersection of any three of the selected subsets contains no more than one element?
Answer. Eight.

Solution. We will provide two different examples of choosing eight 4-element subsets in a set $\mathrm{X}$ of eight elements, satisfying the condition of the problem. Both examples are constructed as geometric objects.

Example 1. We will consider the elements of $\mathrm{X}$ as the vertices of a unit cube, six of the chosen subsets will be the faces of this cube, and the remaining two will be the vertices of two inscribed regular tetrahedra with edge length $\sqrt{2}$. No two vertices of these tetrahedra lie on the same edge of the cube.

If among any three of the chosen subsets, both tetrahedra are included, or two parallel faces, the intersection is empty. If it is two adjacent faces and a tetrahedron, the intersection of the faces gives an edge, which intersects the tetrahedron at exactly one vertex. If it is three pairwise adjacent faces, their intersection contains the unique vertex of the trihedral angle formed by these faces.

Example 2. We will consider the elements of $\mathrm{X}$ as the vertices of a regular octagon. Number the vertices clockwise from 1 to 8, and mark the quadrilateral $\mathrm{M}$ with vertices numbered $1,2,3,5$. The chosen subsets will be $\mathrm{M}$ and seven other quadrilaterals obtained from $\mathrm{M}$ by rotations of angles $\frac{2 \pi k}{8}, k=1,2, \ldots, 6,7$. If the intersection of any three of them contained at least two vertices, i.e., some side or diagonal of the octagon other than the main ones, then by rotating all three quadrilaterals back to coincide with $\mathrm{M}$, we would obtain three different segments of the same length connecting the vertices of $\mathrm{M}$. The latter is impossible because the lengths of the sides and diagonals of $\mathrm{M}$, measured in sides, are $1,1,2,2,3,4$, and there are no more than two equal lengths among them. If the main diagonal of length 4 is mentioned, it is clear that it is contained in only two of the eight considered quadrilaterals.

We will prove that if in an 8-element set $\mathrm{X}$, nine 4-element subsets are arbitrarily chosen, then the intersection of some three of them contains more than one element. The sum of the cardinalities of the chosen subsets is 36, so one of the elements of $\mathrm{X}$, denoted by $\boldsymbol{x}$, is contained in at least $k \geq 5$ of them. Remove $\boldsymbol{x}$ from these $k$ subsets and consider the $k \geq 5$ resulting 3-element subsets in the 7-element set Y, which is equal to X without $\boldsymbol{x}$. The sum of the cardinalities of the resulting sets is at least 15, so one of the elements of Y, denoted by $\boldsymbol{y}$, is contained in at least three of the $k$ resulting 3-element subsets. Therefore, the pair of elements $\boldsymbol{x}$ and $\boldsymbol{y}$ is contained in at least three of the nine chosen 4-element subsets from $\mathrm{X}$.

Grading Criteria. ($\cdot$) Example for eight subsets: 3 points. ($\cdot$) Proof of the estimate: 4 points. ($\cdot$) Lack of justification for the example: minus 1 point.