# Problem 3 – Hint 3

Theorem: if a parabola and a circle intersect at four points, the sum of polar angles $\alpha_1+\alpha_2+\alpha_3+\alpha_4=2\pi n,$ where $n$ is an integer (see the figure below). Proof: without restricting the generality of the foregoing, let the circle be of unit radius and centered at the origin, given by $x^2+y^2=1$. Let us consider the complex plane $z=x+\mathrm i y$; then, the formula of the circle takes form $z\bar z=1$, where $\bar z=x-\mathrm i y$ is the complex conjugate of $z$ so that $x=\frac 12 (z+\bar z)$ and $y=\frac {\mathrm i}2 (\bar z-z)$. Then the formula of the parabola $ax=b+cy-y^2$ is rewritten as $2a(z+\bar z)=2b+2c\mathrm i(\bar z-z)+(\bar z-z)^2$. The intersection points satisfy both the equation of this parabola, and the equation of the circle from where we substitute $\bar z=1/z$; we end up in a fourth order algebraic equation for $z$, with the zeroth order term being equal to 1. This means that according to the Vieta’s formula, the product of the four roots $z_1z_2z_3z_4=1$. All these roots are on the unit circle and therefore $z_i=\mathrm e^{\mathrm i\alpha_i}$; consequently, $\mathrm e^{\mathrm i(\alpha_1+\alpha_ 2+\alpha_ 3+\alpha_ 4)}=1$ which proves the theorem.

Using this theorem, there is no need for long mathematical calculations!