Let's highlight one of the triples $\{A, B, C\}$. Each of the other triples contains exactly one of the points $A, B, C$. Let's consider two cases.

1) All other triples contain point $A$. Then there are no more than the number of non-intersecting pairs formed from 1954 points, which is no more than 977. This number of triples is achieved if we divide 1954 points into 977 pairs and add point $A$ to each of them.
2) There is a triple $\{A, D, E\}$ containing $A$, and a triple containing $B$. The latter triple must intersect with the previous one - we can assume it has the form $\{B, D, F\}$. Consider any triple different from the three mentioned. If it contains $A$, then it also contains $F$ (since it intersects with $\{B, D, F\}$), so there is no more than one such triple. Similarly, there is no more than one "additional" triple containing $B$ (it must contain $E$), and no more than one containing $C$. Thus, in this case, there are no more than six triples.

## Answer

977 triples.
There are 1955 points. What is the maximum number of triples that can be chosen from them so that any two triples have exactly one point in common?

#