3. Determine all triples of real numbers $a, b, c$ that satisfy the conditions

$$
a^{2}+b^{2}+c^{2}=26, \quad a+b=5 \quad \text { and } \quad b+c \geq 7
$$

The written part of the school round in category A takes place

on Tuesday, December 11, 2012

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies and school MF tables. Calculators, laptops, and any other electronic aids are not allowed. These details will be communicated to the students before the start of the competition.

## Solutions to the problems of the written part of the school round in category $A$
3. We will show that the task is satisfied by a unique triplet of numbers: $a=1, b=4$, and $c=3$.

Let $s=b+c \geq 7$. By substituting $a=5-b$ and $c=s-b$ into the first condition, we get

$$
(5-b)^{2}+b^{2}+(s-b)^{2}=26
$$

and thus

$$
3 b^{2}-2(s+5) b+s^{2}-1=0
$$

This quadratic equation with parameter $s$ has solutions in the set of real numbers if and only if its discriminant satisfies the inequality $4(s+5)^{2}-12\left(s^{2}-1\right) \geq 0$. After simplification, we get $s^{2}-5 s-14 \leq 0$ or $(s+2)(s-7) \leq 0$. Since $s \geq 7$, it must be that $s=7$. Substituting into (5) yields

$$
3 b^{2}-24 b+48=0
$$

This equation has the unique solution $b=4$. It is then easy to compute $a=1$ and $c=3$.

Another solution. If we guess the satisfying triplet $a=1, b=4$, and $c=3$, we can easily justify its uniqueness by showing that the value of the non-negative sum

$$
S=(a-1)^{2}+(b-4)^{2}+(c-3)^{2}
$$

must be zero for any satisfying triplet. Under the conditions of the problem, we have

$$
\begin{aligned}
S & =\left(a^{2}+b^{2}+c^{2}\right)-2(a+b)-6(b+c)+\left(1^{2}+4^{2}+3^{2}\right)= \\
& =26-2 \cdot 5-6(b+c)+26=42-6(b+c) \leq 42-6 \cdot 7=0
\end{aligned}
$$

so indeed $S=0$, and thus $a=1, b=4$, and $c=3$.

For a complete solution, award 6 points. For the step-by-step derivation as in the first solution, award one point for expressing $a=5-b, c=s-b$, where $s \geq 7$, a second point for the equation $3 b^{2}-2(s+5) b+s^{2}-1=0$, two points for the inequality $4(s+5)^{2}-12\left(s^{2}-1\right) \geq 0$, and one point for solving it. The sixth point is then for computing $b$, $a$, and $c$. For simply guessing the satisfying triplet, award 1 point.