Inequality for 5 points in the plane

I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE

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rrusczyk

Mar 24, 2025

SmartGroot

Mar 26, 2025

k a March Highlights and 2025 AoPS Online Class Information

jlacosta 0

Mar 2, 2025

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jlacosta

Mar 2, 2025

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sum(ab/4a^2+b^2) <= 3/5

truongphatt2668 3

N 5 minutes ago by Nguyenhuyen_AG

Source: I remember I read it somewhere

Let $a,b,c>0$ . Prove that:
$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$

3 replies

1 viewing

truongphatt2668

Monday at 1:23 PM

Nguyenhuyen_AG

5 minutes ago

Inspired by old results

sqing 2

N 15 minutes ago by sqing

Source: Own

Let $a,b$ be reals such that $a^2+b^2 +ab =1.$ Prove that $9\geq (a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq \frac{19-8\sqrt 3}{27}$ Let $a,b$ be reals such that $a^3+b^3 -2ab =1.$ Prove that $(a^2-a+1)(b^2-b+1) (a^2-ab+b^2) \geq 1$

2 replies

1 viewing

sqing

23 minutes ago

sqing

15 minutes ago

Painting Beads on Necklace

amuthup 45

N an hour ago by maromex

Source: 2021 ISL C2

Let $n\ge 3$ be a fixed integer. There are $m\ge n+1$ beads on a circular necklace. You wish to paint the beads using $n$ colors, such that among any $n+1$ consecutive beads every color appears at least once. Find the largest value of $m$ for which this task is $\emph{not}$ possible.

Carl Schildkraut, USA

45 replies

amuthup

Jul 12, 2022

maromex

an hour ago

Inclusion Exclusion Principle

chandru1 1

N an hour ago by onofre.campos

How does one prove the identity $1=\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}2^{n-k}$ This easy via the binomial theorem for the quantity is just $(2-1)^{k}$ , but how do we arrive at this using the I-E-P?

1 reply

chandru1

Dec 4, 2020

onofre.campos

an hour ago

No more topics!

Inequality for 5 points in the plane

WakeUp 3

N Oct 26, 2019 by Plops

Source: Japanese MO Finals 2000

Given five points $A,B,C,D,E$ in a plane, no three of which are collinear, prove the inequality
$AB+BC+CA+DE\le AD+AE+BD+BE+CD+CE$

3 replies

WakeUp

Feb 10, 2011

Plops

Oct 26, 2019

Inequality for 5 points in the plane

G H J

Reveal topic tags

G H BBookmark kLocked kLocked NReply

Source: Japanese MO Finals 2000

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WakeUp

1347 posts

WakeUp

#1 h Feb 10, 2011, 10:55 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Given five points $A,B,C,D,E$ in a plane, no three of which are collinear, prove the inequality
$AB+BC+CA+DE\le AD+AE+BD+BE+CD+CE$

Z K Y

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SomePig

499 posts

SomePig

#2 h Feb 12, 2011, 11:20 AM • 2 Y

Y by Adventure10, Mango247

Let F,G,H be the intersections of AD,BE; AD,CE; BD,CE respectively.
AB ≤ AF+BF
BC ≤ BH+CH
CA ≤ CD+DH+GH+FG+EF+AE
DE ≤ DG+EG
AB+BC+CA+DE ≤ (AF+FG+DG)+AE+(BH+DH)+(BF+EF)+CD+(CH+GH+EG)
= AD+AE+BD+BE+CD+CE

Z K Y

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DVDthe1st

341 posts

DVDthe1st

#3 h Jun 25, 2014, 3:17 PM • 2 Y

Y by mathematiculperson, Adventure10

The above proof is wrong. This is because $D$ might be between $A$ and $G$ , and thus $AF+FG+DG$ will be more than $AD$ .

Here is my solution:

Call $BC$ and $DE$ bad if segments $BD,CE$ don't intersect, and segments $BE,CD$ don't intersect. This happens iff line $BC$ passes through segment $DE$ , or vice versa.

Suppose segments $BD$ and $CE$ intersect. By triangle inequality, we get $BD+CE\ge DE+BC$ . Then, $AD+AE+BD+BE+CD+CE=(AD+CD)+(AE+BE)+(BD+CE)\ge AC+AB+DE+BC$ , where equality cannot hold since no three points are collinear.

Also, if $A$ is inside triangle $DBC$ , then by triangle inequality (applied a few times) we get $DB+DC\ge AB+AC$ . Then $AD+AE+BD+BE+CD+CE=(AD+AE)+(BE+CE)+(DB+DC)\ge DE+BC+AB+AC$ .

Now we draw lines $AB,AC,BC$ . This divides the plane into seven regions: triangle $ABC$ , regions where the ex-circles are (call them $A_1,B_1,C_1$ , and the remaining regions (let $A_2$ be right next to $A$ etc.)

Suppose line $DE$ doesn't cut segment $BC$ (since it cannot cut all of $AB,AC,BC$ ). Then either $D,E$ are on the same side of line $BC$ (then $BC$ and $DE$ are not bad, and we finish by the first case), or $D,E$ are on different sides. Suppose (WLOG) that $D$ is in $A_1$ (since if it were in $B_2$ or $C_2$ we are done by the second case). Then $E$ must be in either $B_1$ (which means $DE,AB$ are not bad) or $C_1$ (which means $DE,AC$ are not bad). Thus we are done.

Z K Y

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Plops

946 posts

Plops

#4 h Oct 26, 2019, 5:13 AM • 1 Y

Y by Adventure10

I am having trouble applying the triangle inequality to get

DVDthe1st wrote:

$DB+DC\ge AB+AC$

Z K Y

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