Petr, Martin, and Jirka hit a special target that had only three fields with mutually different values. Each of the boys threw a total of ten times and always hit the target. The point gain from the first eight throws was the same for all three boys. In the last two shots, Jirka hit twice the field with the smallest possible value, Martin hit twice the field with the middle value, and Petr hit twice the field with the highest value. The arithmetic mean of all of Martin's throws was 1 greater than Jirka's mean, and Petr's mean was 1 greater than Martin's mean.

Determine all possible values of the fields on the target, knowing that one of them was 12.

(E. Novotná)
The arithmetic mean of all Martin's throws was 1 greater than the mean of Jirka's throws, and each of the boys threw a total of ten times. Therefore, Martin's total sum was 10 greater than Jirka's total sum. These total sums differed only by the sums of the last two throws - Jirka hit the field with the smallest possible value twice, Martin hit the field with the median value twice. Therefore, the field with the median value was 5 greater than the field with the smallest value.

Similarly, it can be reasoned that the field with the largest value was 5 greater than the field with the median value. One of the values of these three fields was 12, but we do not know whether it was the smallest, median, or largest. The following three possibilities for the values of the fields on the target are considered:

- $12,17,22$,
- $7,12,17$,
- $2,7,12$.

Note. The sum of the first eight throws of any of the boys is denoted by $S$, and the unknown values on the target are denoted by $j, m$, and $p$, where $j<m<p$. With this notation, the beginning of the previous solution can be written as follows:

$$
\begin{aligned}
\frac{S+2 m}{10} & =\frac{S+2 j}{10}+1, \\
2 m & =2 j+10, \\
m & =j+5 .
\end{aligned}
$$

Similarly, it can be reasoned that $p=m+5$.

Scoring suggestion. 4 points for discovering and justifying the dependencies between the values of the fields on the target; 2 points for listing all three possibilities for the values on the target.