(solved by Anca Arnautu and Adrien Laroche). Let $\mathscr{P}$ be the parabola in the plane with equation $y=x^{2}$. Let $\Gamma_{1}$ be the circle of diameter 1 tangent internally to $\mathscr{P}$ at the origin. By recurrence, we define $\Gamma_{n+1}$ as the circle tangent to $\Gamma_{n}$ and twice to $\mathscr{P}$. Calculate the diameter of $\Gamma_{2007}$.
. Let $r_{n}$ be the radius of $\Gamma_{n}$ and $d_{n}$ its diameter. Set $s_{n}=d_{1}+d_{2}+\cdots+d_{n}$, and let $x_{n}$ be the positive abscissa of the point of tangency of $\Gamma_{n}$ and $\mathscr{P}$ (where $\Gamma_{n}$ is tangent to $\mathscr{P}$ at $\left(x_{n}, x_{n}^{2}\right)$ and $\left(-x_{n}, x_{n}^{2}\right)$). Then, the center of $\Gamma_{n}$ is $\left(0, s_{n-1}+r_{n}\right)$. The slope of the tangent line to $\mathscr{P}$ at $\left(x_{n}, x_{n}^{2}\right)$ is $2 x_{n}$, and the slope of the normal line is $-\frac{1}{2 x_{n}}$. Since the center of $\Gamma_{n}$ lies on this normal, we have $\frac{x_{n}^{2}-\left(s_{n-1}+r_{n}\right)}{x_{n}-0}=-\frac{1}{2 x_{n}}$, and $x_{n}^{2}-s_{n-1}-r_{n}=-\frac{1}{2}$. Furthermore, $r_{n}^{2}=\left(x_{n}-0\right)^{2}+\left(x_{n}^{2}=s_{n-1}-r_{n}\right)^{2}$, so $x_{n}^{2}=r_{n}^{2}-\frac{1}{4}$ and $r_{n}^{2}-r_{n}-s_{n-1}+\frac{1}{4}=0$. Solving this equation gives $r_{n}=\frac{1+2 \sqrt{s_{n-1}}}{2}$ and $d_{n}=1+2 \sqrt{s_{n-1}}$. Then we prove by strong induction that $d_{n}=2 n-1$. The answer to the question is therefore 4013.