B3. Let $a$ and $b$ be real numbers such that $\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}}=1$. Determine all possible values of the expression

$$
(a+b)\left(\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}\right)
$$

## Problems for 2nd Year

Time for solving: 120 minutes. In section A, we will award two points for the correct answer and deduct one point for an incorrect answer. Write your answers for section A in the left table, leave the right table blank.
![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-06.jpg?height=168&width=802&top_left_y=647&top_left_x=640)
B3. We multiply the equality by $\left(1+a^{2}\right)\left(1+b^{2}\right) \neq 0$ and simplify to obtain the equivalent equality $a^{2} b^{2}=1$. Therefore, $a b= \pm 1$. We also simplify the given expression:

$$
\begin{aligned}
(a+b)\left(\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}\right) & =(a+b) \frac{a+a b^{2}+b+a^{2} b}{\left(1+a^{2}\right)\left(1+b^{2}\right)}=(a+b) \frac{(a+b)(1+a b)}{\left(1+a^{2}\right)\left(1+b^{2}\right)}= \\
& =\frac{\left(a^{2}+b^{2}+2 a b\right)(1+a b)}{1+a^{2}+b^{2}+a^{2} b^{2}}
\end{aligned}
$$

If $a b=-1$, the value of this expression is 0. If $a b=1$, the value of this expression is

$$
\frac{\left(a^{2}+b^{2}+2\right) \cdot 2}{a^{2}+b^{2}+2}=2
$$

We can verify that these two values are indeed achievable. The value 0 is achieved, for example, when $a=1$ and $b=-1$, and the value 2 when $a=b=1$.
Writing the equivalent equality $a^{2} b^{2}=1$ ..... 1 point
Writing the equivalent equalities $a b= \pm 1$ ..... 1 point
Simplified form of the expression ( $\frac{\left(a^{2}+b^{2}+2 a b\right)(1+a b)}{1+a^{2}+b^{2}+a^{2} b^{2}}$ ) ..... 2 points
Calculated value of the expression when $a b=-1$ ..... 1 point
Calculated value of the expression when $a b=1$ ..... 1 point

## Solutions for 2nd Year

In section A, a correct answer will be awarded 2 points, while an incorrect answer will result in the deduction of 1 point. To avoid a potential negative final score, each contestant is given 3 initial points.

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-20.jpg?height=142&width=234&top_left_y=520&top_left_x=914)

Justifications: