When shooting with a bow, the performance of the archer is also tracked. It is calculated by removing the best and the worst attempt from all attempts, and the arithmetic mean of the remaining scores is calculated.

Friends Petr, Jirka, Michal, and Zdeněk shot one arrow each in four rounds. Each shot was rated with a whole number from 0 to 10. In each round, the sum of the ratings of all boys was 32 points, but no two boys had the same rating in any round.

In the following table, only some of the data from the described competition are filled in. Complete the missing data.

(M. Dillingerová)

|  | 1st round | 2nd round | 3rd round | 4th round | performance |
| :---: | :---: | :---: | :---: | :---: | :---: |
| Petr |  |  |  | 5 | 10 |
| Jirka |  |  | 9 | 10 | 7.5 |
| Michal |  |  | 5 |  | 8 |
| Zdeněk |  |  |  |  | 8.5 |
| total | 32 | 32 | 32 | 32 | - |

Hint. Start with Petr.
Because Peter's performance was 10, he had to score 10 points in each of the first three rounds. Because the sum of the ratings in the 3rd round was 32 points, Zdeněk must have hit 8 points in this round. Because the sum of the ratings in the 4th round was 32 points, the sum of the ratings of Michal and Zdeněk in this round must have been 17 points. Because no two boys had the same rating in any round, they could have had in this round

a) either Michal 9 and Zdeněk 8 points,

b) or Michal 8 and Zdeněk 9 points.

Let's assume possibility a) and try to complete the table

|  | 1st round | 2nd round | 3rd round | 4th round | performance |
| :---: | :---: | :---: | :---: | :---: | :---: |
| Peter | 10 | 10 | 10 | 5 | 10 |
| Jirka |  |  | 9 | 10 | 7.5 |
| Michal |  |  | 5 | 9 | 8 |
| Zdeněk |  |  | 8 | 8 | 8.5 |
| total | 32 | 32 | 32 | 32 | - |

For Zdeněk's performance to be $8.5=17: 2$, he must have hit 9 points in each of the first two rounds. For Michal's performance to be $8=16: 2$ and to not have the same rating as Zdeněk in any round, he must have hit 8 points in each of the first two rounds. For the sum of the ratings in the 1st and 2nd rounds to be 32 points, Jirka must have hit 5 points in these two rounds. In this case, however, his performance would not be 7.5 (but only 7). Therefore, possibility a) could not have occurred.

Let's assume possibility b) and try to complete the table

|  | 1st round | 2nd round | 3rd round | 4th round | performance |
| :---: | :---: | :---: | :---: | :---: | :---: |
| Peter | 10 | 10 | 10 | 5 | 10 |
| Jirka |  |  | 9 | 10 | 7.5 |
| Michal |  |  | 5 | 8 | 8 |
| Zdeněk |  |  | 8 | 9 | 8.5 |
| total | 32 | 32 | 32 | 32 | - |

For Jirka's performance to be $7.5=15: 2$, he must have hit 6 points in one of the first two rounds and 6 or fewer points in the other. For Michal's performance to be $8=16: 2$ and to not have the same rating as Peter, he must have hit 8 points in one of the first two rounds and 8 or 9 points in the other.

If Jirka hit 6 points in the same round as Michal hit 8, then Zdeněk must have hit 8 points in the same round (to make the sum of the ratings in that round equal to 32 points). This would mean that Michal and Zdeněk had the same rating, so this possibility could not have occurred.

Therefore, Jirka must have hit 6 points in a different round than Michal hit 8. Let's assume he did so in the 1st round and try to complete the table (from the previous, it follows that Michal must have hit 9 points in the same round)

|  | 1st round | 2nd round | 3rd round | 4th round | performance |
| :---: | :---: | :---: | :---: | :---: | :---: |
| Peter | 10 | 10 | 10 | 5 | 10 |
| Jirka | 6 |  | 9 | 10 | 7.5 |
| Michal | 9 | 8 | 5 | 8 | 8 |
| Zdeněk |  |  | 8 | 9 | 8.5 |
| total | 32 | 32 | 32 | 32 | - |

For the sum of the ratings in the 1st round to be 32 points, Zdeněk must have hit 7 points in this round. For Zdeněk's performance to be $8.5=17: 2$, he must have hit 9 points in the second round. For the sum of the ratings in the 3rd round to be 32 points, Jirka must have hit 5 points in this round. Since 5 is less than 6, Jirka's performance is consistent with the problem statement. We have found one valid solution to the problem:

|  | 1st round | 2nd round | 3rd round | 4th round | performance |
| :---: | :---: | :---: | :---: | :---: | :---: |
| Peter | 10 | 10 | 10 | 5 | 10 |
| Jirka | 6 | 5 | 9 | 10 | 7.5 |
| Michal | 9 | 8 | 5 | 8 | 8 |
| Zdeněk | 7 | 9 | 8 | 9 | 8.5 |
| total | 32 | 32 | 32 | 32 | - |

However, Jirka could have hit 6 points in the 2nd round. In this case, the final table would have the ratings for the 1st and 2nd rounds swapped.