playing with triangles

Let the equal sides of the isosceles triangle be $x$ and the third side be $y.$ First restriction: $x\leq1994$ ,second restriction: $y\leq1994.$ Now,because we have to form a triangle,these conditions need to be fulfilled: $x+x>y$ (triangle inequality), $x+y>x,x+y>x.$ But the second and the third equations are the same,so now we only need to fulfill these two equations: $x+x>y,x+y>x.$ But the second inequality is true as long as $y$ is positive.So,that leaves us us with one equation: $2x>y.$ Now,we take $x=1994\Longrightarrow 3988>y.$ But we know that $y\leq1994.$ Thus,there are 1994 values of $y$ which satisfy the conditions.Thus,1994 triangles can be formed.Now,we take $x=1993\Longrightarrow 3986>y.$ But $y\leq1994.$ Thus,again $y$ has 1994 values.As long as $2x>1994$ $y$ will have 1994 values. $2x>1994$ as long as $x>997.$ Thus, values of $x$ from $1994--998$ $y$ can take up 1994 values.That is a total of $1994*(997)=1988018.$ Now,let us consider values of $x$ for which $2x=1994$ That value is $997\Longrightarrow 1994>y.$ Thus $y$ can have $1993$ values.Now,let us consider cases for which $2x<1994.$ The first value of $x$ that satisfies this condition is $996$ .Putting $x=996$ we get that $1992>y.$ Thus there are 1991 possibilities for $y.$ Calculate the next one and you get that there are $1991$ possibilities for $y.$ This is a series which will continue till $1.$ Thus,total $=1993+1991+......1=\dfrac{997}{2}*1994=994009.$ Thus,total $=1988018+994009=\boxed{2982027}.$