Problem

Source: 2024 Mexican Mathematical Olympiad, Problem 3

Tags: geometry, hexagon, dodecagon, Mexico



Let $ABCDEF$ a convex hexagon, and let $A_1,B_1,C_1,D_1,E_1$ y $F_1$ be the midpoints of $AB,BC,CD,$ $DE,EF$ and $FA$, respectively. Construct points $A_2,B_2,C_2,D_2,E_2$ and $F_2$ in the interior of $A_1B_1C_1D_1E_1F_1$ such that both 1. The sides of the dodecagon $A_2A_1B_2B_1C_2C_1D_2D_1E_2E_1F_2F_1$ are all equal and 2. $\angle A_1B_2B_1+\angle C_1D_2D_1+\angle E_1F_2F_1=\angle B_1C_2C_1+\angle D_1E_2E_1+\angle F_1A_2A_1=360^\circ$, where all these angles are less than $180 ^\circ$, Prove that $A_2B_2C_2D_2E_2F_2$ is cyclic. Note: Dodecagon $A_2A_1B_2B_1C_2C_1D_2D_1E_2E_1F_2F_1$ is shaped like a 6-pointed star, where the points are $A_1,B_1,C_1,D_1,E_1$ y $F_1$.