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shitij agarwal

Let two equal sides of an isosceles

be p units each and let remaining sides be q units.
Case1: p>q
q can take values 1,2,3,.....,p-1(if p-1>0), condition for p,p,q be sides of a

is automatically satisfied here. For each positive integer p>1,
we have p-1 isosceles

s i.e.

_{[p=2 ]}

^{[1994 ]} (p-1)

=1+2+3....+1993
=1998721

Case2:p<q
In order that p,p,q may be sides of

we must have 2p>q i.e.
p<q<2p
If p is even say 2m ,then q can take value 1,2,...m-1
If p is odd say 2m-1,then q can take values 1,2,...,m-1=(p-1)/2
No. of possible isosceles triangles is
(1-1)/2+(3-1)/2+....+(1993-1)/2+1+2+3+....
for q=1994,p+q>q is true.
Also ,we must have q/2<p<q
There are in all isosceles

_{[q even ]}

(q-2)/2 + _{[q odd ]}

(q-1)/2

s
1

q

1994 1

q

1994
i.e.(1+2+3.....+996)+(1+2+3+...+996) triangles
=993012 triangles
Total no of isosceles triangles=1998721+993012
=2991733
Hope it will help u ,I gave my best to solve this
TC.......
SHITIJ