Maximum number of non-congruent triangles with side lengths integers and less than $9$ 
Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.

I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ of them but after that I gave up. Is there any systematic approach to question?
 A: We do some systematic counting.
First the isosceles (including equilateral) ones. If the base is $1$, other two sides can be any from $1$ to $8$. If base is $2$, other two can be any from $2$ to $8$. If base is $3$ - $2$ to $8$. Base $4$ - $3$ to $8$. Base $5$ - $3$ to $8$. Base $6$ - $4$ to $8$. Base $7$ - $4$ to $8$. Base $8$ - $5$ to $8$. Total $$8+7+7+6+6+5+5+4=48$$
For scalene, we write the triplet of side lengths in decreasing order respecting triangle inequality.

*

*$[8,7,(6 \,\text{to}\, 2)]. [8,6,(5\,\text{to}\, 3)]. [8, 5, 4]. [8,4,✗].$

*$[7,6,(5 \,\text{to}\, 2)]. [7,5,(4 \,\text{to}\, 3)]. [7, 4, ✗].$

*$[6,5,(4 \,\text{to}\, 2)]. [6,4,3]. [6, 3, ✗].$

*$[5,4,(3 \,\text{to}\, 2)]. [5,3,✗].$

*$[4,3,2]. [4,2,✗].$

*$[3,2,✗].$
Sum : $5+3+1+4+2+3+1+2+1=22$
Grand total : $\boxed{48+22=70}$
A: Yes, a systematic approach is to count the number of triples of three integers which satisfy the triangle inequality. All that this requires is that the two shorter lengths must add up to more than the greatest length.
Longest side 8
The other sides can then be 8 and 8,8 and 7,...,8 and 1 OR 7 and 7, 7 and 6,...,7+2 OR etc.
The total number is $8+6+4+2=20$.
Longest side 7
The total number is $7+5+3+1=16$
Longest side 6,5, ...
The totals for these are $12,9,6,4,2,1$.
Grand total
We must now add these  numbers, obtaining $70$.
A generalisation
Note the numbers we added above are the squares plus twice the triangular numbers:-
$$1+4+9+16+2(1+3+6+10).$$
We can generalises this result to other bounds on the side lengths.
A: The number of non congruent non degenerate triangles
with integer sides with largest side equal to $\,n+1\,$
is the OEIS sequence A002623.
For $\,n=7\,$ the answer is $70.$ The OEIS entry has a
comment which justifies the interpretation as count of
triangles

Also number of triples (x,y,z) with 0 < x <= y <= z <= n + 1, x + y > z. - Ralf Steiner, Feb 06 2020

The formula

A002620(n+3) = a(n+1) - a(n). - Michael Somos, Sep 04 1999

means that the first differences of the triangle counts
is the quarter-squares sequence A002620.
Alternatively, the
OEIS sequence A173196
has two extra zeros at the beginning of the sequence.
Thus it is the number of non congruent non degenerate
triangles with integer sides with largest side $\,<n.\,$
The value for $\,n=9\,$ is again $70$. The
triangle counts arranged as a number triangle
OEIS sequence A003983
with $\,n\,$ the largest side and $\,m\,$ the
smallest side makes the pattern clear.
 n\m 1  2  3  4  5  6  7  8  9 10   row sum accum sum
 -+------------------------------   ------- ---------
 1|  1                                  1        1
 2|  1  1                               2        3          
 3|  1  2  1                            4        7
 4|  1  2  2  1                         6       13
 5|  1  2  3  2  1                      9       22
 6|  1  2  3  3  2  1                  12       34
 7|  1  2  3  4  3  2  1               16       50
 8|  1  2  3  4  4  3  2  1            20       70
 9|  1  2  3  4  5  4  3  2  1         25       95
10|  1  2  3  4  5  5  4  3  2  1      30      125

A: Put $L$ to be the longest side, an $x,y$ the other two.

We can limit ourselves in considering $x \le y$.

Then we shall have
$$
\left\{ \matrix{
  0 \le x \le y \le L \hfill \cr 
  L \le x + y \hfill \cr}  \right.
$$
which is the area within the triangle $VCB$
in this sketch

and which is equivalent to triangle $OVA$, which
in turn is half of the triangle $OAC$.
Now, the number of integer points inside the right triangle $OAC$ with sides $[0,L]$, including the borders i.e. the equalities
is
$$
N_{\,OAC}  = 1 + 2 +  \cdots  + \left( {L + 1} \right)
 = \left( \matrix{  L + 2 \cr   2 \cr}  \right)
$$
while those on the segment $OV$ are
$$
N_{\,OV}  = \left\lfloor {L/2} \right\rfloor  + 1
$$
So the number we are looking for is
$$ \bbox[lightyellow] {  
\eqalign{
  & N = \left( {{{N_{\,OAC}  - N_{\,OV} } \over 2}} \right) + N_{\,OV}
  = {1 \over 2}\left( {N_{\,OAC}  + N_{\,OV} } \right) =   \cr 
  &  = {1 \over 2}\left( {\left( \matrix{  L + 2 \cr   2 \cr}  \right)
 + \left\lfloor {L/2} \right\rfloor  + 1} \right) \cr} 
}$$
which for $L= 1, \ldots,10$ gives
$$
2, 4, 6, 9, 12, 16, 20, 25, 30, 36
$$
If instead you are not going to consider the degenerated cases $x=0, \; x+y=L$ then it is easy
to deduct from the above the corresponding side lengths.
