Triangle Problem

nguoidep_68 · October 30th, 2007, 03:07 AM

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

johnny · November 4th, 2007, 07:57 AM

Since AM+BN+CP=MB+NC+PA, we can say that the half-length 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.

October 30th, 2007, 03:07 AM	#1
nguoidep_68 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
$nguoidep_68 is offline$

November 4th, 2007, 07:57 AM	#2
johnny Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0	Since AM+BN+CP=MB+NC+PA, we can say that the half-length 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.
$johnny is offline$

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