My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
February 20th, 2018, 10:50 AM   #1
Newbie
 
Joined: Dec 2017
From: Spain

Posts: 18
Thanks: 1

Friends and strangers

Six points A, B, C, D, E, F with different distances pairwise are given so that no three of them lie on the same line. Consider all possible triangles with vertices on the points A, B, C, D, E, F.
Prove that there is at least one triangle between these, where it's longest side is at the same time the smallest side of another triangle.

What I've found out so far:
Considering ABCDEF as a hexagon:
The first thing is to notice that there are 15 lines (12 diagonals+6 sides):
(n(n-3)/2)+n (diagonals formula) with n=6
(6(6-3)/2)+n=15

With these 15 lines, you can form 20 triangles:
Since every triangle has three lines and each of the 15 lines takes part on 4 triangles:
15*(4/3)=20

I don't know now how to move forward, but I'm sure that the solution has something to do with the theorem of friends and strangers:

https://en.wikipedia.org/wiki/Theore..._and_strangers

Does anybody have an idea of how to continue, every advice or idea is welcomed.
Thanks
MIKI14 is offline  
 
February 27th, 2018, 09:58 AM   #2
Newbie
 
Joined: Dec 2017
From: Spain

Posts: 18
Thanks: 1

Hi? Someone??
MIKI14 is offline  
March 25th, 2018, 09:49 PM   #3
Newbie
 
Joined: Mar 2018
From: Jackson MS

Posts: 1
Thanks: 0

Quote:
Originally Posted by MIKI14 View Post
Six points A, B, C, D, E, F with different distances pairwise are given so that no three of them lie on the same line. Consider all possible triangles with vertices on the points A, B, C, D, E, F.

Prove that there is at least one triangle between these, where it's longest side is at the same time the smallest side of another triangle.



What I've found out so far:

Considering ABCDEF as a hexagon:

The first thing is to notice that there are 15 lines (12 diagonals+6 sides):

(n(n-3)/2)+n (diagonals formula) with n=6

(6(6-3)/2)+n=15



With these 15 lines, you can form 20 triangles:

Since every triangle has three lines and each of the 15 lines takes part on 4 triangles:

15*(4/3)=20



I don't know now how to move forward, but I'm sure that the solution has something to do with the theorem of friends and strangers:



https://en.wikipedia.org/wiki/Theore..._and_strangers



Does anybody have an idea of how to continue, every advice or idea is welcomed.

Thanks


Assuming it is a regular hexagon- you need to also consider the case of an irregular polygon as well. Hummm - I love a good proof


Sent from my iPhone using Tapatalk
lathampatterson is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
friends, strangers



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Hello Friends ^_^ Safofoh Math 2 October 24th, 2016 11:11 PM
Hello, friends Copernicus New Users 0 June 23rd, 2016 11:56 AM
Hi friends Safofoh Number Theory 2 November 26th, 2015 12:31 AM
Hi friends... Arun kumar New Users 1 July 23rd, 2014 07:57 AM
Hi friends... Arun kumar New Users 1 July 23rd, 2014 03:46 AM





Copyright © 2018 My Math Forum. All rights reserved.