Another solution

[Bella]

Before I learned Professor Loh's solution using combinatorics, I solved the problem by counting equilateral triangles (5), all the isosceles triangles (14), and then the triangles with unequal sides (3).

[audrey]

@Bella Nice! That's definitely a valid way of doing it-- splitting into cases based on whether the triangles are equilateral, isosceles, or scalene (no two sides with the same lengths) is a good choice, because then it makes it harder to accidentally miss a triangle. Just goes to show that there are often multiple ways to solve a problem!