A is the set of all triangles whose perimeter is 2013. B is the set of all triangles whose perimeter is 2016. Which set has more triangles.

Question

Let A be the set of all triangles whose lengths of sides are integers and whose perimeter is $2013$. Let B be the set of all triangles whose lengths of sides are integers and perimeter is $2016$. Which of the two sets has more elements (triangles). Explain it in details.

What I tried:

I couldn't count the number of triangles for both sets. Supposing that triangles exist they must comply with the triangle inequality. According to the triangle inequality for set A the sides of the triangles are from 1 to 1006, for set B the sides of the triangles are from 2 to 1007. In both cases the possible length side values are 1006.

Can somebody give me an idea?

Answer 1

Hint: categorize the triangles by the longest side. For a perimeter of $2016$ the longest side cannot be longer than $1007$ because the triangle inequality will fail. It can't be shorter than $672$ because another side will be longer. If I tell you the longest side is $1234$ how many triangles can you form? Find the rule for the number given the longest side and add up the results for longest sides from $672$ to $1007$

Answer 2

Take a triangle from $ A $ and add $1 $ to each side. The triangle inequality still holds (because $1 <1+1 $) and you have an element from $ B$. This map is injective but not surjective because $ B $ has flat triangles (since $2016 $ is even).

Edit: As noted by @Daniel Schepler in the comments, there are no non-flat triangles with a side of length $1 $ in $ B $, so the map above is actually a bijection if flat triangles are forbidden.