February 20th, 2018, 09:50 AM  #1 
Newbie Joined: Dec 2017 From: Spain Posts: 16 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(n3)/2)+n (diagonals formula) with n=6 (6(63)/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 
February 27th, 2018, 08:58 AM  #2 
Newbie Joined: Dec 2017 From: Spain Posts: 16 Thanks: 1 
Hi? Someone??

March 25th, 2018, 08:49 PM  #3  
Newbie Joined: Mar 2018 From: Jackson MS Posts: 1 Thanks: 0  Quote:
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  

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