Let P be a point in triangle ABC such that angles APB, BPC, and CPA are all 120 degrees. Can lines AB, AC, BC, PA, PB, and PC all have integral lengths?

Lets name the sides opposite to A,B,C as a,b,c,& let PA be x,PB be y,PC be z,we assume x,y,z are integers

a^2=y^2+z^2+yz

b^2=x^2+z^2+xz

c^2=x^2+y^2+xy

If x=y=t then

c=(sqrt 3)*t

SO nither can be integer at the same time

x,y,z must be diffrent integers

c will be integer only if R.H.S is a perfect square

R.H.S can be taken as quadratic eqn either in x or in y

lets take in x

SO D=y^2 - 4*y^2= -3*y^2

D must be 0 for R.H.S to be perfect square,this indicates that y=0

Hence pt P will coincide with one of the vertex of the triangle

So we can extend the same reasoning to all pairs of x,y,z

and  hence  we deduce that for  a,b,or c to be integers all  among

x,y,z can't be integers

So all a,b,c,x,y,z can't be integers at same time

Posted by Nishant on 2007-04-13 02:58:50