# How many triangles with integral side lengths are possible, provided their perimeter is $36$ units?

How many triangles with integral side lengths are possible, provided their perimeter is $36$ units?

My approach:

Let the side lengths be $a, b, c$; now,

$$a + b + c = 36$$

Now, $1 \leq a, b, c \leq 18$.

Applying multinomial theorem, I'm getting $187$ which is wrong.

Please help.

• Did you take the triangle inequality into account? – J. M. isn't a mathematician Jul 13 '12 at 12:55
• Yes, I did. When one side is of $1$ unit, then the max possible values for other two will be $18$ i.e. $(1, 18, 18)$ – Bazinga Jul 13 '12 at 12:56
• $$18+18+1\neq36$$ – J. M. isn't a mathematician Jul 13 '12 at 13:00
• You didn't say how you did it. Possibly permutations of $a,b,c$ are considered to give the same triangle; did you account for that (after all drawing the the triangle elsewhere or in a different orientation does not count as giving a different triangle either, I suppose)? – Marc van Leeuwen Jul 13 '12 at 13:10
• I counted the number of positive integer solutions $\{a,b,c\}$ to $a+b+c=36$; it is 108. But of course they are not all triangles. – i. m. soloveichik Jul 13 '12 at 13:18

## 1 Answer

The number of triangles with perimeter $n$ and integer side lengths is given by Alcuin's sequence $T(n)$. The generating function for $T(n)$ is $\dfrac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$. Alcuin's sequence can be expressed as

$$T(n)=\begin{cases}\left[\frac{n^2}{48}\right]&n\text{ even}\\\left[\frac{(n+3)^2}{48}\right]&n\text{ odd}\end{cases}$$

where $[x]$ is the nearest integer function, and thus $T(36)=27$. See this article by Krier and Manvel for more details. See also Andrews, Jordan/Walch/Wisner, these two by Hirschhorn, and Bindner/Erickson.

• What if the perimeter is 18? Then T(N) isn't an integer. – MCCCS Mar 30 '17 at 6:53
• Did you notice the use of the nearest integer function? – J. M. isn't a mathematician Mar 30 '17 at 9:04