
Math Events Math Events, Competitions, Meetups  Local, Regional, State, National, International 
 LinkBack  Thread Tools  Display Modes 
October 30th, 2007, 04:07 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, 08:57 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. 

Tags 
problem, triangle 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Triangle problem  jiasyuen  Algebra  1  December 19th, 2013 03:12 AM 
Triangle Problem  Jakarta  Algebra  6  April 3rd, 2013 08:26 AM 
Right Triangle problem  ro_ch  Algebra  3  March 9th, 2013 06:11 PM 
Help with a triangle problem  JohnA  Algebra  12  March 16th, 2012 06:50 AM 
triangle problem  Kath  Algebra  4  February 28th, 2009 03:25 PM 