
3. Find all real $a, b, c$, such that

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

(Pavel Novotný)


Solution. We show that the only solution is $a=1, b=4$ a $c=3$.

Let $s=b+c \geqslant 7$. Substituting $a=5-b$ and $c=s-b$ the first condition gives

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

thus

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

The equation has a real solution iff the discriminant $4(s+5)^{2}-12\left(s^{2}-1\right) \geqslant 0$. This yields $s^{2}-5 s-14 \leqslant 0$, or $(s+2)(s-7) \leqslant 0$. Since $s \geqslant 7$, there must be $s=7$. If we substitute to (5) we get

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

with the only solution $b=4$. Then $a=1$ and $c=3$.

## Second Round of the 62nd Czech and Slovak Mathematical Olympiad (January 15th, 2013)

![](https://cdn.mathpix.com/cropped/2024_04_17_5816354d2b4b35f9c465g-15.jpg?height=234&width=554&top_left_y=471&top_left_x=728)
