Proving Inequalities for Cyclic Quadrilaterals

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Homework Statement


Let ABCD be a convex cyclic quadrilateral. Prove that

|AB-CD|+|AD-BC| \geq 2|AC-BD|

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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ehrenfest said:
First, isn't a cyclic quadrilateral always convex?

Hi ehrenfest! :smile:

Yes … "convex" seems unnecessary!
 
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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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