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October 30th, 2007, 03:07 AM  #1 
Newbie Joined: Oct 2007 From: China Posts: 3 Thanks: 0  Triangle Problem
Suppose that M ;N;P are three points lying respectively on the edges AB;BC;CA of a triangle ABC such that AM+BN+CP=MB+NC+PA Prove that area(MNP)<=area(ABC)*1/4 
November 4th, 2007, 07:57 AM  #2 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
Since AM+BN+CP=MB+NC+PA, we can say that the halflength of the perimeter is (1/2)(P), thus (1/2)(P)=AM+BN+CP=MB+NC+PA, where P is the perimeter of the triangle. Also, P=2(AM+BN+CP)=2(MB+NC+PA). Using Heron's theorem (where A is the area of the triangle), A=sqrt((1/2)(P)((1/2)(P)AB)((1/2)(P)BC)((1/2)(P)CA)) Well, this is the part of the solution that I've got. I'll put more thinking on it later. 

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