How many isosceles triangles with integer sides are possible such that sum of…

Key idea: use the triangle inequality and consider two cases for which two sides sum to 12.

Case 1 — the two equal sides sum to 12:

If the equal sides add to 12, each equal side is 6, so the triangle is (6, 6, x) with integer x. The triangle inequality 6 + 6 > x implies x < 12, and x must be a positive integer, so x = 1, 2, ..., 11. That gives 11 possible triangles.

Case 2 — two unequal sides sum to 12:

Possible unordered pairs that sum to 12 are: 1 and 11, 2 and 10, 3 and 9, 4 and 8, 5 and 7.

• For each pair (a, b) with a + b = 12, check (a, a, b) and (b, b, a).

• 1 and 11: (1,1,11) is invalid (2 ≤ 11), (11,11,1) is valid → 1 triangle.

• 2 and 10: (2,2,10) is invalid (4 < 10), (10,10,2) is valid → 1 triangle.

• 3 and 9: (3,3,9) is invalid (6 < 9), (9,9,3) is valid → 1 triangle.

• 4 and 8: (4,4,8) is invalid (8 = 8 is degenerate), (8,8,4) is valid → 1 triangle.

• 5 and 7: (5,5,7) is valid (10 > 7), (7,7,5) is valid → 2 triangles.

From the unequal-side pairs we get 1 + 1 + 1 + 1 + 2 = 6 valid isosceles triangles.

Total count = 11 (from the equal-sides case) + 6 (from unequal-side pairs) = 17.

Answer: 17