nguoidep_68

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

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.