A nine-member selection committee chooses from three candidates. Each of them ranks them, giving 3 points to their first choice, 2 points to their second choice, and 1 point to their third choice. Summing up the candidates' points, it turned out that the order was clear, with all three having different point totals. One member of the committee noticed that if the election had been conducted such that each of the nine only chose one candidate and gave that candidate 1 point, then the order of the candidates would have been reversed. How many points did the candidates originally receive?
Solution. Let the candidates be $A, B$, and $C$ in the order of the first voting results. The total number of points distributed is $9 \cdot (3+2+1) = 54$.

If the order is reversed in the second vote, this means that $C$, who lost in the first vote, will now be first. This is only possible if $C$ received at least 3 points from at least 4 voters in the first instance, so $C$'s points are at least $4 \cdot 3 + 5 \cdot 1 = 17$. $C$ could not have received first place from 5 voters, because in that case, with the guaranteed 15 points, there would be at least $4 \cdot 1$ more points, totaling at least 19 points; but then $C$ could not have been last in the first vote, as 19 is more than the average score of the candidates, which is 18. Since the first vote resulted in a clear order, 18 cannot be $C$'s score either. Thus, $C$ has exactly 17 points, in which case the winner $A$ has 19 points, and $B$ has 18 points.

The table below shows that this is indeed possible. $C$ gets four first places, while $B$ gets three, and $A$ gets two. The other values in the table can be determined uniquely from the conditions.

|  | 1. | 2. | 3. | 4. | 5. | 6. | 7. | 8. | 9. | points |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $C$ | 3 | 3 | 3 | 3 | 1 | 1 | 1 | 1 | 1 | **17** |
| $B$ | 1 | 2 | 1 | 1 | 3 | 3 | 3 | 2 | 2 | **18** |
| $A$ | 2 | 1 | 2 | 2 | 2 | 2 | 2 | 3 | 3 | **19** |