9.3. Find five different numbers if all possible sums of triples of these numbers are equal to $3,4,6$, $7,9,10,11,14,15$ and 17. The numbers do not have to be integers.
Answer: $-3,2,4,5,8$.

Solution. Let the required numbers be $a<b<c<d<e$.

Notice that each number appears in exactly six triplets of numbers, so by adding all ten sums of triplets and dividing by 6, we get the sum of all five numbers, which is 16. Clearly, the smallest sum is $a+b+c=3$, from which $d+e=16-a-b-c=13$, and the largest sum is $c+d+e=17$, from which $a+b=16-c-d-e=-1$. Therefore, $c=a+b+c-a-b=c+d+e-d-e=4$.

The next largest after $a+b+c$ is $a+b+d$, the terms of which in ascending order are no greater than the terms of any other triplet different from $a+b+c$, hence $d=4-a-b=5$. The second to last largest before $c+d+e$ is $b+d+e$, the terms of which in ascending order are no less than the terms of any other triplet different from $c+d+e$, hence $b=15-d-e=2$.

Now we find $a=a+b-b=-1-2=-3, \quad e=d+e-d=13-5=8$. In total:
$a=-3, b=2, c=4, d=5, e=8$.

Grading criteria. The sum of all numbers is found: 2 points. $c=a+b+c-a-b=c+d+e-d-e=4$ is found: 1 point. It is justified that the next largest after $a+b+c$ is $a+b+d$, and the second to last largest before $c+d+e$ is $b+d+e$, and $b$ and $d$ are found: 2 points. Finding $a$ and $e$: 1 point.