Bariz - Problem

Source: 2020 Instagram Math Olympiad IGMO - Shortlist g3

Tags: conics, ellipse, geometry, equal segments

parmenides51

23.03.2021 08:54

For any given ellipse $\omega$ , let P be a point external to it. $A$ and $B$ are points on $\omega$ such that $P A$ and $P B$ are tangent to $\omega$ . C, D $\omega$ and E are points on $\omega$ such that $AC = CD = DE = EB$. Lines $P C, P D, P E$ meets $\omega$ again at points $F, G, H$, respectively. $M_1, M_2, M_3$ are midpoints of $CF, DG$ and $EH$, respectively. Prove that $P, A, B, M_1, M_2$ and $M_3$ lie on an ellipse. by @pepemaths

In fact this is true that by varying a point $X$ on $\omega$, taking $Y=PX\cap \omega$ and $M$ the midpoint of $XY$, the locus of $M$ is an ellipse which passes through $P$. To do this consider an affinity bringing the ellipse to a circle, of center $O$. Since $M'$ is still the midpoint of $X'Y'$, $OM'$ is perpendicular to $X'Y'$. Therefore, since $\angle P'M'O'=\frac{\pi}{2}$ for all these points $M'$, they lie on the circle of diameter $P'O'$. Reversing the affinity, the points all lie on an ellipse, which comprises $P$ and $A$ and $B$ (note that $A$ is the midpoint of chord $AA$)