Point Trouble

Let us arbitrarily label the points 1 through 8. Consider any point. If this point were in 4 triangles then there are $(3-1) \times 4 = 8$ vertices which are not this point in these 4 triangles. No 2 of these 8 vertices can be the same point since this would mean that at least 2 of the 4 triangles shared 2 vertices. So we need to choose 8 distinct points from the remaining 7 points, but this is evidently impossible. So the maximum number of triangles that any point can be in is 3. If each point is indeed in 3 triangles then there will be $\dfrac{8 \times 3}{3} = 8$ triangles, so this is our theoretical maximum. We will now show that this can be done:

$123$

$145$

$167$

$258$

$478$

$368$

$246$

$357$

So the answer is $\fbox{8}$

Here's a solution with $8$ triangles: Number the vertices ${n}\bmod{8}$ and draw a triangle with vertices ${n},{n+1},{n+3} \pmod{8}$ for each $n\bmod 8$ . To show that at most $8$ triangles can be drawn, let $T$ be the total number of triangles, let $V$ be the vertex set and let $t(v)$ denote the number of triangles incident on vertex $v$ . Observe that $t(v)\le{3}$ so that

$T=\frac{1}{3} \sum_{v \in V}{t(v)}\le \frac{1}{3}\cdot{8}\cdot{3}=8.$

.

I imagined to take $K_8$ and choosing a triple of vertices at a time. Two triples, or triangles, cannot share a side. Since there are $\binom{8}{2}=28$ edges in $K_8$ and every choice of a triple removes three edges, it is possible to take at most eight triangles. Eight triangles are indeed possible to arrange: consider a cube in which the vertices on the top face are numbered $2,4,6,8$ and the vertices on the bottom face are numbered $1,3,5,7$ , in order that vertex 2 is just above vertex 1 and vertex 4 is just above vertex 3. The triangles 187,765,534,312 lie on the boundary of the cube and share at most one vertex in the bottom face. In a similar way, the triangles 285,247,461,683 share at most one vertex in the top face; every one of them has a side which belongs to the interior of the cube and a side that is the diagonal of a face, with the opposite orientation of the sides of 187,765,534,312 being diagonals of a face, too. This means that the choice 187,765,534,312,285,247,461,683 is just fine.