# Find number of triangles with integral sides and side lengths ≤ 2n

Find number of triangles with integral sides and side lengths less than or equal to $$2n$$. I approached this method by recursion.

Say $$A_{2n}\$$is the number of triangles with integral sides and side lengths less than or equal to $$2n$$. $$A_{2n-1}\$$ is the number of triangles with integral sides and side lengths less than or equal to $$2n-1$$ .

So $$A_{2n}=A_{2n-1}+\left( \text{number of triangles having at least one side equal to} 2n \right) .$$

How to count the number of triangles having at least one side equal to $$2n$$?

Also, is there any other better method to this other than generalization as well?

• Perhaps it would be easier to count up instead of counting down... – abiessu Dec 17 '18 at 12:54
• Welcome to Math.SE! Please use MathJax. – GNUSupporter 8964民主女神 地下教會 Dec 17 '18 at 12:55

Number of triangles having at least one side equal to $$2n$$ is the number of couple $$(a,b)$$ with $$a\ge b$$ and $$a+b > 2n$$. That is, \begin{aligned} \sum^{2n}_{b=1}\sum^{2n}_{a = \max(b,2n + 1-b)} 1 &= \sum^{2n}_{b=1} (2n+1 - \max(b,2n+1-b))\\ &=\sum^{2n}_{b=n+1} ((2n+1) - b) + \sum^{n}_{b=1} (2n+1 - (2n+1-b))\\ &=\sum^{1}_{c=n}c + \sum^{n}_{b=1} b\hspace{5em}(\text{where } c = 2n+1-b)\\ &= n(n+1) \end{aligned} In order to argue recursively, you would need to compute $$A_{2n-1}$$ in similar fashion as well.