# Number of integer triangles with longest side $2m+1$

Prove that there are $(m+1)^2$ distinct triangles with integer side lengths and greatest side equal to $2m+1$ units. Also show that $3m+1$ of these triangles are isosceles or equilateral.

I think we would have to start with the property of triangles that the sum of any two sides is greater than the third side and then make cases satisfying this condition, and then add them all up, but I am not sure about this method. Can anyone help me out?

• Its true....the following four triangles are possible for the condition m=1.....(3,2,3).....(3,3,3).....(3,2,2).....(3,1,3)....where the nos. in bracket represent the side lengths of the triangle. – SirXYZ Oct 27 '16 at 15:48
• You're very right. For some reason I thought I read right triangles. – Nitin Oct 27 '16 at 15:49

You're on the right track. If the maximum side length is $2m+1$, the sum of the other two side lengths $j$ and $k$ must satisfy $j + k > 2m+1$. So if you fix a value of one side $1 \leq j \leq 2m+1$, how many possible values are there for $k$?

If $j = 1$, then $k = 2m+1$ is the only valid value, since if it was any greater then $2m+1$ wouldn't be the biggest value. If it was any smaller than $j+k < 1 + 2m+1 = 2m+2$, so we wouldn't have a triangle.

In general, you can see if you think through this for the other $j$ in the same way, that there are $j$ possibilities for each $j$. We can represent this sum as

$$\sum_{j=1}^{2m+1} j = (m+1)(2m+1)$$

But wait, you say. This is bigger than $(m+1)^2$. That can't be right. And it isn't, because we double counted some elements. In fact, we for every pair $(j,k)$ where $j$ isn't equal to $k$, we counted this twice. Check for yourself by drawing out the table of the valid $j,k$ values for some small $m$.

One way to get rid of this is to figure out how many things we double counted and subtract that number from the total we obtained above. That's possible but requires a bit of thinking.

What we'll do is to count the number that were NOT double counted, add that to the total, and then divide by two. This way, we'll double count everything and then divide that number to get the (single) count we want.

So now we ask - when can we have $j = k$? If you look at the $m=2$ case again you see the max side length must be $5$. Thus we can have $(5,3,3)$ but not $(5,2,2)$, since $4 = 2 + 2 < 5 < 3 + 3 = 6$. In general, we can do this every time $j > \frac{1}{2}(2m+1)$, or whenever $j \geq m+1$.

Thus, we have $m+1$ total instances of $(2m+1, j, j)$, so we add that now to $(m+1)(2m+1)$. This gives $1(m+1) + (2m+1)(m+1) = [(2m+1)+1](m+1) = 2(m+1)^2$.

Since this is now double the number of possible triangles, we have $\frac{2(m+1)^2}{2} = (m+1)^2$ possible triangles with integer sides and maximum length $2m+1$. Done!

Can you try a similar approach to count the number of isosceles or equilateral triangles?

For a given $m$ let $k$ denote the shortest side of the triangle, which may range from 1 to $2m+1$.

• If $1\le k\le m+1$ there are $k$ possible triangles that can be formed with the given $m$ and $k$: the middle side may range from $2(m+1)-k$ (because of the triangle inequality) to $2m+1$.
• If $m+1<k\le2m+1$ there are $2(m+1)-k$ possible triangles. Here the middle side may range from $k$ to $2m+1$.

Hence the number of distinct integer triangles with longest side $2m+1$ is $$\sum_{k=1}^{m+1}k+\sum_{k=m+2}^{2m+1}(2(m+1)-k)$$ $$=\frac{(m+1)(m+2)}2+\sum_{k=1}^mk$$ $$=\frac{(m+1)(m+2)}2+\frac{m(m+1)}2=(m+1)^2$$

Integer isosceles and equilateral triangles with longest side $2m+1$ fall into two classes:

• Isosceles triangles with base $2m+1$; the short equal legs may range from $m+1$ to $2m+1$, for a total of $m+1$ triangles
• Isosceles triangles with two legs of length $2m+1$; the short base may range from 1 to $2m$, for a total of $2m$ triangles (we stop at $2m$ because the equilateral triangle has already been counted in the preceding case)

Hence the number of integer triangles with the given longest side that are isosceles or equilateral is $3m+1$.