For example, lets say you want to find the number of scalene triangles with integer side lengths such that the perimeter is less than or equal to 13.
I could start listing all possible triples in increasing order that add to 1, then those that add to 2, then 3, etc., and then checking if they satisfy the triangle inequality. Or, I could start listing triples that satisfy the triangle inequality and then check if their sum is less than or equal to 13. However, is there any better method that doesn't involve just trying everything, or at the very least is there any method to reduce the number of triples to check?
Best Answer
Let c, a, b be the lengths of the sides AB, BC, and CA of triangle ABC. Use the Ravi substitution: a = x + y, b = y + z , and c = z + x with x, y, z > 0. Then a + b + c = 13 translates to: x + y + y + z + z + x <= 13 ==> x + y + z <= 6.5. Since x,y, and z are positive integers. There are cases to consider:
Case 1: x + y + z = 1. No solution.
Case 2: x + y + z = 2. No solution.
Case 3: x + y + z = 3. This gives: x = y = z = 1. But then a = b = c = 2. We require that the triangle be scalene so the sides have different lengths. So no solution in this case.
Case 4: x + y + z = 4. So either x = y = 1, z = 2 or x = 2, y = z = 1 or y = 2, x = z = 1. So ABC is isosceles. No solution in this case.
Case 5: x + y + z = 5. So either x = 1, y = 1, z = 3 or x = 2, y = 2, z = 1 and permutations of these values. But again ABC is then isosceles. No solution for this case.
Case 6: x + y + z = 6. So x = 1, y = 2, z = 3 and permutations or x = y = z = 2 or x = 1 = y, and z = 4 and permutations. The latter case gives ABC isosceles while the former case renders ABC scalene. So 1 solution for this case.
These are all possible cases. So in summary there is 1 answer: a = x + y = 1 + 2 = 3, b = y + z = 2 + 3 = 5, and c = z + x = 3 + 1 = 4. ABC is a scalene right triangle !