Lengths of the angle bisectors of a triangle with perimeter

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Lengths of the angle bisectors of a triangle with perimeter

G H J

Reveal topic tags

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: VAIMO 2, German Pre-TST 2002

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3647 posts

orl

#1 h Jul 17, 2011, 12:43 AM • 2 Y

Y by Adventure10, Mango247

Prove: If $x, y, z$ are the lengths of the angle bisectors of a triangle with perimeter 6, than we have:

$\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \geq 1.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4148 posts

#2 h Jul 17, 2011, 10:22 PM • 1 Y

Y by Adventure10

$x^2=\frac{4bcs(s-a)}{(b+c)^2} \ , \ y^2= \frac{4cas(s-b)}{(c+a)^2} \ , \ z^2= \frac{4abs(s-c)}{(a+b)^2}$

By AM-GM $(b+c)^2 \ge 4bc \ , \ (c+a)^2 \ge 4ca \ , \ (a+b)^2 \ge ab$

$\Longrightarrow \ s \left ( \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \right) \ge \frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}$

By AM-HM $\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c} \ge \frac{9}{ (s-a)+(s-b)+(s-c)}=\frac{9}{s}$

$\Longrightarrow \ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \ge \frac{9}{s^2}=1.$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a