Question 535964
<pre>
Let the two equal sides have length a=b and the base have length c

The perimeter = a+b+c = a+a+c = 2a+c = 99

By the triangle inequality we always have a + c > a true so the only
triangular inequality we have to be concerned about is

a + a > c 

or

2a > c

and

c > 0

So putting everything we have together:

(1)     2a+c = 99
(2)       2a > c
(3)        c > 0

Solve (1) for c

(1)     2a+c = 99
(4)        c = 99 - 2a  

Substitute (4) into (2)

(2)       2a > c
          2a > 99 - 2a
          4a > 99
(5)        a > 24.75

Substitute (4) into (3)

(3)        c > 0
     99 - 2a > 0
         -2a > -99
(6)        a < 49.5

Putting (5) and (6) together

        24.75 < a < 49.75

So a can be any integer from 25 to 49, inclusive.

There are 49 integers from 1 thru 49.
We subtract the number of integers 34 or less, and there are 24
of them, so the number of triangles satisfying the given
conditions is 49 - 24 or 25.

That's the answer, 25.

Here they all are, where "a" can be any integer from 25 
through 49, inclusive.

     a  b  c
 1.  25 25 49
 2.  26 26 47
 3.  27 27 45
 4.  28 28 43
 5.  29 29 41
 6.  30 30 39
 7.  31 31 37
 8.  32 32 35
 9.  33 33 33
10.  34 34 31
11.  35 35 29
12.  36 36 27
13.  37 37 25
14.  38 38 23
15.  39 39 21
16.  40 40 19
17.  41 41 17
18.  42 42 15
19.  43 43 13
20.  44 44 11
21.  45 45 9
22.  46 46 7
23.  47 47 5
24.  48 48 3
25.  49 49 1

Edwin</pre>