Number of ways of making a tangential quadrilateral such that its perimeter is $24$ and side lengths are integers from $1$ to $12$ My attempt is as follows:
Label the sides of the quadrilateral as $a$, $b$, $c$ and $d$ in a counterclockwise order. Since the quadrilateral is tangential and its perimeter is $24$, $a+c=b+d=\frac{24}{2}=12$.
The number of noncommutative ways of choosing $a+c=12$ is $11$ ($1+11$, $2+10$, ... $5+7$, $6+6$, $7+5$, ... , $10+2$, $11+1$) and the same applies for $b+d$. The number of choosing two pairs to form a quadrilateral is therefore $11 \cdot 11=121$.
The number $121$ is overcounted since we counted every rotation of the quadrilateral as a new one. For example, $a = 5$,$b = 1$ ,$c = 7$,$d = 11$ describes the same quadrilateral as $a = 11$,$b = 5$,$c = 1$,$d = 7$. There are $4$ rotations, so we counted every quadrilateral $4$ times, exept the case where $a=b=c=d=6$, which we correctly counted only once. The true number is therefore $\frac{121-1}{4}+1=31$
I somehow have a feeling that this number is too low, and I suspect that there is a mistake in my reasoning for the overcounting process. Is there a better way of approaching this problem?

Indeed you over-divided the numbers. For example $(4,8; 4,8)$ is only counted twice so it should be divided by $2$ while you divided it by $4$.
The systematic way to count this is to let $a$ to be the smallest number.
$a=1: b$ has $11$ choices.
$a=2: b$ has $9$ choices (because $d\geq 2$ as well).
$a=3: b$ has $7$ choices.
And so on until $a=6$ where $b$ has $1$ choices. In this way rotation won't matter and we always get a unique combination.
Therefore totally $11+9+7+...+1=36$.
