If $ABC$ is acute triangle, prove distance from each vertex to corresponding excentre is less than sum of two greatest side of triangle
Problem
Source: Sharygin final round 2017
Tags: inequalities proposed, geometry
31.07.2017 21:12
WLOG, let $BC$ be the smallest side of $ABC$. Notice that $$AI_A=\frac{AB\cdot AC}{AI}.$$It now suffices to show $$\frac{2bc}{a+b+c}\cos \tfrac{A}{2} >\frac{bc}{b+c}.$$However, $\angle A \le 60^{\circ}$ so $\cos \tfrac{A}{2} \ge \tfrac{\sqrt{3}}{2}$ and $a \le \tfrac{b+c}{2}$. Hence, we may conclude since $AI_A$ is the longest possible join from a vertex to the excenter. $\blacksquare$
31.07.2017 21:24
03.08.2017 04:03
anantmudgal09 wrote: WLOG, let $BC$ be the smallest side of $ABC$. Notice that $$AI_A=\frac{AB\cdot AC}{AI}.$$It now suffices to show $$\frac{2bc}{a+b+c}\cos \tfrac{A}{2} >\frac{bc}{b+c}.$$However, $\angle A \le 60^{\circ}$ so $\cos \tfrac{A}{2} \ge \tfrac{\sqrt{3}}{2}$ and $a \le \tfrac{b+c}{2}$. Hence, we may conclude since $AI_A$ is the longest possible join from a vertex to the excenter. $\blacksquare$ Oops, $AI_A$ is the smallest altitude of the excentral triangle.
03.08.2017 07:04
@wizardmath, thanks a lot for pointing it out. I must be getting old for doing such an error . I hope this is error-free. Notice that $AI_A$ is the longest altitude in the excentral triangle. Note that $\triangle I_ABC \sim \triangle I_AI_CI_B$ so $\measuredangle BI_AC \ge \measuredangle CBI_A$ hence $BC \ge CI_A$. Therefore, $$AI_A< AC+CI_A \le AC+CB,$$as desired.