AoPS - Problem

Source: Mathematics Regional Olympiad of Mexico Southeast 2020 P5

Tags: geometry, circumcircle, Inequality, Tangents, inequalities

AlanLG

23.10.2021 04:53

Let $ABC$ an acute triangle with $\angle BAC\geq 60^\circ$ and $\Gamma$ it´s circumcircule. Let $P$ the intersection of the tangents to $\Gamma$ from $B$ and $C$. Let $\Omega$ the circumcircle of the triangle $BPC$. The bisector of $\angle BAC$ intersect $\Gamma$ again in $E$ and $\Omega$ in $D$, in the way that $E$ is between $A$ and $D$. Prove that $\frac{AE}{ED}\leq 2$ and determine when equality holds.

Let radius of $\Gamma$ be $R$. For any point $X$, let $I_{\Gamma}: X\to X'$. Since $B\equiv B', C\equiv C'$, $I_{\Gamma}: \Omega\to BC$. So $D', P'\in BC$. And $\frac{AE}{ED}=\frac{AE}{\frac{R^2\cdot D'E}{OD'\cdot OE}}=\frac{AE\cdot OD'}{R\cdot D'E}\le 2\iff \frac{AE}{2R}\le\frac{D'E}{OD'}$ Since $\angle A=60^{\circ}, P'E\ge OP'\implies D'E^2=D'P'^2+P'E^2\ge D'P'^2+OP'^2=OD'^2\implies D'E\ge OD'$. So $\frac{AE}{2R}=\cos\angle OEA\le 1\le\frac{D'E}{OD'}$ Equality occurs iff $\angle OEA=0^{\circ}, \angle A=60^{\circ}\iff \triangle ABC$ is equilateral.