The Congruent Dilemma

Each of these triangles can be unique described by the lengths of sides from shortest to longest, $10 \leq a \leq b \leq c \leq 22.$ Note that a number may be repeated! The number of possible choices for 3 combinations-with-repetition out of a set of 13 values can be found with a "stars-and-bars" approach: $N = \binom{15}{3} = \frac{15\cdot 14\cdot 13}{3\cdot 2\cdot 1} = 5 \cdot 7 \cdot 13 = 455.$

However, the triangle inequality requires that $c < a + b$ . We must therefore exclude the few cases where $c \geq a + b$ . They are easily enumerated:

• $a = b = 10$ ; $c = 20, 21, 22$ ( 3 cases).

• $a = 10; b = 11$ ; $c = 21, 22$ ( 2 cases).

• $a = 10; b = 12$ ; $c = 22$ ( 1 case).

• $a = b = 11$ ; $c = 22$ ( 1 case).

Excluding these seven cases leaves $N - 6 = \boxed{448}$ triples of side lengths that form legitimate, non-degenerate triangles.