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?

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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.