Problem 10-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)

Answer: 1.
Solution. We will provide two solutions to the problem.

First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:

$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow \quad(t-1)\left(t^{2}-(\lambda-1) t+(10-\lambda)\right)=0
$$

From this, it is clear that for any $\lambda$ there is a root $t=1$, so the value $x=1$ works. It remains to prove that there are no other values that are roots for any $\lambda$ (although this is obvious). Indeed, $t^{2}-(\lambda-1) t+(10-\lambda)=0$ implies $t=\frac{\lambda-1 \pm \sqrt{\lambda^{2}+2 \lambda-39}}{2}$. Take any pair of values of $\lambda$ for which the discriminant takes the same positive value, for example, for $\lambda=10$ and $\lambda=-12$ we have $t \in\{0,9\}$ and $t \in\{-11,-2\}$ - there are no intersections. Therefore, the answer is $x=1$.

Second Solution. Subtract the second equation from the first, transforming it, we get $(a-1)(b-1)(c-1)=0$. From this, it follows that one of $a, b, c$ is equal to one. Other $x$ do not work, as the triples $(a, b, c)=(4,1,1)$ and $(a, b, c)=(0,9,1)$ satisfy the condition.

| Grade | Score | Content Criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Not proven why there are no others besides 1, or there is a minor error in the proof. |
| $+/-$ | 16 | Significant errors in the proof (several transitions with division by possibly zero, misunderstanding of the condition (with necessary calculations for the proof)). |
| $-/+$ | 10 | Two specific cases are considered, which show that it can only be equal to 1. But there is no proof that $= 1$. |
| .- | 6 | Only one case is found (it is claimed that $\in\{1,4\}$ or $\in\{0,1,9\}$. |
| - | 0 | The solution is completely incorrect/only the answer. |