If an isosceles triangle has positive integers as its lengths then find the number of such triangles none of whose lengths exceed
When you say "lengths" you mean the total length of all three sides, right? An isosceles triangle has two sides of equal length and another side. Calling the two equal lengths "x" and the third side "y", the total length is 2x+ y so you are looking for all possible integers x and y such that 2x+ y= 1994. So y= 1994- 2x= 2(997- x) which means that y must be an even number.
In order to have 3 sides, x and y must be positive (not 0) so x can be any integer from 1 to 996 which then gives a unique value for y.
There are a lot more possibilities than these. Assuming that the total length of all sides may not exceed 1994, this does not mean that the total length has to equal 1994. For example a triangle with legs of 1, 1, and 2 satisfies the requirement.
Let y = length of the base and x = the lengths of the two equal sides. The requirement is that 2x + y <= 1994. Let's look at what happens for different values of y:
If y = 1, the x can be any value from 1 to (1994-1)/2 rounded down to an integer, or 996
if y= 2 then x can be any value from 1 to (1994-2)/2 = 996
if y = 3 then x can be any integer from 1 to 995
if y = 4 then x can be any integer from 1 to 995
if y = 1989 the x can be any integer from 1 to 2
if y = 1990 then x can be any value from 1 to 2
if y = 1991 then x = 1
if y = 1992 then x = 1.
So the total of all possibilities is .
However, given the strange wording of the question perhaps OP means that no single leg may exceed length of 1994. If that's what he meant then there are 1994 choices for length of leg x and 1994 choices for leg y, so the total number of possibilities is 1994 x 1994 = 3,976,036.