Proving Inequalities for Cyclic Quadrilaterals

  • Thread starter ehrenfest
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In summary, the problem asks to prove that for a convex cyclic quadrilateral ABCD, the expression |AB-CD|+|AD-BC| \geq 2|AC-BD| holds true. The attempt at a solution involves using triangle inequalities for the various triangles formed in the quadrilateral, but no satisfactory solution was found.
  • #1
ehrenfest
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Homework Statement


Let ABCD be a convex cyclic quadrilateral. Prove that

[tex]|AB-CD|+|AD-BC| \geq 2|AC-BD|[/tex]

Homework Equations


The Attempt at a Solution


First, isn't a cyclic quadrilateral always convex?

http://en.wikipedia.org/wiki/Cyclic_quadrilateral
 
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  • #2
ehrenfest said:
First, isn't a cyclic quadrilateral always convex?

Hi ehrenfest! :smile:

Yes … "convex" seems unnecessary!
 
  • #3
Putnam is supposed to be for fun Ehrenfest. If you ask for help on every problem that you can't immediately solve... how are you having fun? The pleasure is all in finding the aha! moment yourself.
 
  • #4
tiny-tim said:
Hi ehrenfest! :smile:

Yes … "convex" seems unnecessary!

So, I can get triangle inequalities for the triangles ABC, BCD, ACD, ABD. Put there are 12 of them and I tried to play around with so they would look similar to the inequality in the problem statement but I did not get very far.
 

FAQ: Proving Inequalities for Cyclic Quadrilaterals

What is "Putnam and Beyond prob 117"?

"Putnam and Beyond prob 117" refers to a specific problem from the book "Putnam and Beyond" by Razvan Gelca and Titu Andreescu. This book is a collection of challenging mathematical problems and their solutions, primarily aimed at students preparing for the William Lowell Putnam Mathematical Competition.

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"Putnam and Beyond prob 117" is considered to be a difficult problem, as it is taken from a book that is intended for students preparing for a highly competitive mathematical competition. It may require advanced mathematical knowledge and problem-solving skills to solve.

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