12.1. The equation of a common tangent line to the graphs of $f(x)$ and $g(x)$ at points $\left(x_{1}, f\left(x_{1}\right)\right)$ and $\left(x_{2}, g\left(x_{2}\right)\right)$ has the form

$$
y=f\left(x_{1}\right)+f^{\prime}\left(x_{1}\right)\left(x-x_{1}\right)=g\left(x_{2}\right)+g^{\prime}\left(x_{2}\right)\left(x-x_{2}\right)
$$

Hence $f^{\prime}\left(x_{1}\right)=g^{\prime}\left(x_{2}\right)$ and $f\left(x_{1}\right)-f^{\prime}\left(x_{1}\right) x_{1}=g\left(x_{2}\right)-g^{\prime}\left(x_{2}\right) x_{2}$. Since $f^{\prime}(x)=2 x-2 a$ and $g^{\prime}(x)=-2 x$, we get that $x_{1}+x_{2}=a$ and $x_{1}^{2}+x_{2}^{2}=1$. Then $x_{1} x_{2}=\frac{a^{2}-1}{2}$ and hence $x_{1}$ and $x_{2}$ are the roots of the quadratic equation $x^{2}-a x+\frac{a^{2}-1}{2}=0$. Since the graphs of $f(x)=x^{2}-2 a x$ and $g(x)=-x^{2}-1$ have two common tangent lines, it follows that they are disjoint, $x_{1} \neq x_{2}$,
$a^{2}<2$, and the tangent points are $M\left(x_{1}, f\left(x_{1}\right)\right), N\left(x_{2}, f\left(x_{2}\right)\right), P\left(x_{2}, g\left(x_{2}\right)\right)$ and $Q\left(x_{1}, g\left(x_{1}\right)\right)$. Then

$$
\begin{aligned}
P Q^{2} & =\left(x_{1}-x_{2}\right)^{2}+\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)^{2} \\
& =\left(x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}\right)^{2}\left(1+x_{1}+x_{2}\right)=\left(2-a^{2}\right)\left(1+a^{2}\right) \\
M Q & =\left|f\left(x_{1}\right)-g\left(x_{1}\right)\right|=\left|x_{1}^{2}-2 a x_{1}+x_{1}^{2}+1\right|=2-a^{2}
\end{aligned}
$$

and similarly

$$
M N^{2}=\left(2-a^{2}\right)\left(1+a^{2}\right), N P=2-a^{2}
$$

Hence $M N P Q$ is a parallelogram with perimeter

$$
2\left(\sqrt{\left(2-a^{2}\right)\left(1+a^{2}\right)}+2-a^{2}\right)=6
$$

Then $\sqrt{\left(2-a^{2}\right)\left(1+a^{2}\right)}=1+a^{2}$ and we get that $a= \pm \frac{\sqrt{2}}{2}$.
Problem 12.1. Find all real numbers $a$ such that the graphs of the functions $x^{2}-2 a x$ and $-x^{2}-1$ have two common tangent lines and the perimeter of the quadrilateral with vertices at the tangent points is equal to 6 .

Oleg Mushkarov, Nikolai Nikolov