Q..Q..

let the sides of the triangle be a,a,b:
we know for it to be a triangle 2a>b

acc to problem a can have max value of 1994
when a is 1, b can be 1(1 triangle possible)
when a is 2,b can be 1,2,3(3 triangles possible)
when a is 997 b can have 1993 values
but when 997<a≤1994 b will have 1994 values for each...

so total no of triangles is\sum_{n=1}^{997}{2n-1} + 997 x 1994

the answer turns out to be 2982027

@Ketan: the answer is 2991733. I hav not seen ur proof. will see later n tell u.

Let the 2 equal sides of an isosceles triangle be p units each and let the remaining sides be q units.

Taking Case1: p>q then q can take values 1,2,3,...p-1(iff p-1>0) and hence condition for(p,p,q be theing the sides of a triangle) is satisfied for each +ve integer p>1, we can have p-1 isosceles triangle is =
1+2+3+.......1993=1987021

Case2: p<q. In order that p,p,q be the sides of the triangle we must have 2p>q i.e.p,q,2p.
If P is even, say 2m then q can take values 1,2,..m-1. if p is odd say 2m-1 then q can take values 1,2,,...m-1=(p-1)/2. Number of possible triangles is (1-1)/2+(3-1)/2...(1993-1)/2+1+2+3...for q=1994.p+q>p is true.
Since we have q/2<p<q
If q is even there are q-(q-1)/2-(q-1)/2 possible values of p.
If q is odd, there are (q-1)-(q-1)/2 valus for p.

So there are in total (1+2+...996)+(1+2+...996) triangles.
So the toal no.of isosceles triangles is 1987021+993012=2980033.

@Ketan: i m getting this . i think the answer given in the book is wrong. Can anyone tell the correct answr. Ketan , you may be right.