p/2 < AC + BD < p

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

p/2 < AC + BD < p

G H J

Reveal topic tags

triangle inequality

L

G H BBookmark kLocked kLocked NReply

Source: South Africa 2001

The post below has been deleted. Click to close.

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

3660 posts

Arne

#1 h Oct 1, 2005, 12:26 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

$ABCD$ is a convex quadrilateral with perimeter $p$ . Prove that $\dfrac{1}{2}p < AC + BD < p.$ (A polygon is convex if all of its interior angles are less than $180^\circ$ .)

Z K Y

The post below has been deleted. Click to close.

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

606 posts

Jan

#2 h Oct 1, 2005, 6:36 PM • 1 Y

Y by Adventure10

We consider two possibilities: $AD > BC$ and $BC > AD$

If $AD > BC$ :
Triangle inequality:
$AC < AB + BC$ and $BD < BC + CD$
So we see that $AC + BD < AB + 2BC + CD < AB + BC + CD + AD = p$

If $BC > AD$
Triangle inequality:
$AC < AD + CD$ and $BD < AD + AB$
So again, $AC + BD < AB + 2AD + CD < AB + BC + CD + AD = p$

Z K Y

The post below has been deleted. Click to close.

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

3660 posts

Arne

#3 h Oct 1, 2005, 6:41 PM • 1 Y

Y by Adventure10

What about the other side of the inequality?

Z K Y

The post below has been deleted. Click to close.

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

606 posts

Jan

#4 h Oct 1, 2005, 7:08 PM • 2 Y

Y by Adventure10, Mango247

I'll try later...

Z K Y

The post below has been deleted. Click to close.

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

606 posts

Jan

#5 h Oct 1, 2005, 8:30 PM • 2 Y

Y by Adventure10, Mango247

We call $S$ the intersection of $AC$ and $BD$

Again triangle inequality:
$AB < AS + BS$ , and $CD < DS + CS \Rightarrow AB+CD < AC + BD$ (1)

Triangle inequality:
$AD < AS + DS$ , and $BC < BS + CS \Rightarrow AD + BC < AC + BD$ (2)

We make the sum of (1) and (2):
$p < 2AC + 2 BD \iff \dfrac{p}{2} < AC + BD$

So combine my two posts and get:

$\dfrac{p}{2} < AC + BD < p$

QED !

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a