How many isosceles triangles with integer sides are possible such that sum of two of the side is 12 ?

Let the sides of the isosceles triangle be (a, a, b), where a and b are integers.

Case 1: The sum of the two equal sides is 12

a + a = 12 ⇒ a = 6

a + a = 12 ⇒ a = 6

Triangle inequality: 6 + 6 > b ⇒ b < 12

Also b > 0 and integer.

So, b = 1, 2, 3, …, 11

So, Total = 11 triangles

Case 2: One equal side + base = 12 

a + b = 12 ⇒ b = 12 − a

Triangle inequality: a + a > b ⇒ 2a > 12 − a ⇒ 3a > 12 ⇒ a > 4

Also: b > 0 ⇒ 12 − a > 0 ⇒ a < 12

So, a = 5,6,7,8,9,10,11

Hence, Total = 7 triangles

But when a=6, the triangle is (6,6,6), which was already counted in Case 1. So subtract 1 duplicate.

So final count: 11 + 7 - 1 = 17