An Inequality with Altitudes and Angle Bisectors

Problem 1

An Inequality with Altitudes and Angle Bisectors

Solution

WLOG, $a\ge b\ge c,\,$ implying $h_c=\max (h_a,h_b,h_c)\,$ and $\ell_a=\min (\ell_a,\ell_b\ell_c).\,$ We have

$\displaystyle h_cc=2[\Delta ABC] = b\ell_a\sin\frac{A}{2}+c\ell_a\sin\frac{A}{2},$

that is $\displaystyle\ell_a(b+c)\sin\frac{A}{2}=h_cc.\,$ Suffice it to show that $\displaystyle\ell_a(b+c)\sin\frac{A}{2}\ge c,\,$ which is equivalent to $\displaystyle\frac{b+c}{c}\ge\frac{1}{\displaystyle\sin\frac{A}{2}}.\,$ But $A\ge B\ge C\,$ (Euclid I.18), so that $\displaystyle A\ge\frac{\pi}{3},\,$ $\displaystyle \frac{A}{2}\ge\frac{\pi}{6}\,$ and, subsequently, $\displaystyle \sin\frac{A}{2}\ge\frac{1}{2},\,$ i.e., $\displaystyle 2\ge\frac{1}{\displaystyle\sin\frac{A}{2}}.\,$ Also $\displaystyle\frac{b+c}{c}\ge 2.\,$ By transitivity, $\displaystyle\ell_a(b+c)\sin\frac{A}{2}\ge c\,$ and, hence, $h_c\ge\ell_a.$

Acknowledgment

The problem and the solution above have been kindly posted by Leo Giugiuc at the CutTheKnotMath face book page. The problem is by L. Panaitopol; it has been published at the Romanian Mathematical Magazine.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

 62813355

Search by google: