What are the bounds between which $x^{2}+2 x y$ varies, if $x^{2}+y^{2}=1$?
Given the condition $x^{2}+y^{2}=1$, there exists a real number $\alpha$ such that $x=\sin \alpha$ and $y=\cos \alpha$. The expression under consideration thus becomes

$$
x^{2}+2 x y=\frac{1-\cos 2 \alpha}{2}+\sin 2 \alpha=\frac{1+(2 \sin 2 \alpha-\cos 2 \alpha)}{2}
$$

We will show that the set of values of the function in the parentheses is the closed interval $[-\sqrt{5}, \sqrt{5}]$.

Since $\left(\frac{2}{\sqrt{5}}\right)^{2}+\left(\frac{1}{\sqrt{5}}\right)^{2}=1$, there exists a $\beta$ such that $\cos \beta=\frac{2}{\sqrt{5}}$ and $\sin \beta=\frac{1}{\sqrt{5}}$. With this,

$$
f(\alpha)=\sin (2 \alpha-\beta)=\frac{2}{\sqrt{5}} \sin 2 \alpha-\frac{1}{\sqrt{5}} \cos 2 \alpha
$$

the set of values of this function is the interval $[-1,1]$, which is equivalent to our statement. Therefore, the values of $x^{2}+2 x y$ span the interval $\left[\frac{1}{2}-\frac{\sqrt{5}}{2}, \frac{1}{2}+\frac{\sqrt{5}}{2}\right]$.

David Papp (Budapest, Szent István Gymnasium, 11th grade)