[Combinatorics]
Problem: For each positive integer $n$, let $T(n)$ be the number of triangles with integer side lengths, positive area, and perimeter $n$. For example, $T(6)=1$ since the only such triangle with a perimeter of 6 has side lengths 2,2 and 2 .
Determine the values of $T(10), T(11)$ and $T(12)$.

Solution: Denote the side lengths of a triangle by $a, b$ and $c$, with $0<a \leq b \leq c$.

In order for these lengths to form a triangle, we need $c<a+b$ and $b<a+c$ and $a<b+c$. Since $0<a \leq b \leq c$, then $b<a+c$ and $a<b+c$ follow automatically, so only $c<a+b$ ever needs to be checked.

Instead of directly considering triangles and sets of triangle, we can consider triples $(a, b, c)$ and sets of triples $(a, b, c)$ with the appropriate conditions.

For each positive integer $k \geq 3$, we use the notation $S_{k}$ to denote the set of triples of positive integers $(a, b, c)$ with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=k$.

In this case, $c<a+b$ and $a+b+c=k$, so $c+c<a+b+c=k$, so $2 c<k$ or $c<\frac{1}{2} k$.

Also, if $0<a \leq b \leq c$ and $a+b+c=k$, then $k=a+b+c \leq c+c+c$, so $3 c \geq k$ or $c \geq \frac{1}{3} k$.


Consider $T(10)$, which is the number of elements in $S_{10}$.

We want to find all possible triples $(a, b, c)$ of integers with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=10$.

We need $c<\frac{10}{2}=5$ and $c \geq \frac{10}{3}$. Thus, $c=4$.

Therefore, we need $0<a \leq b \leq 4$ and $a+b=6$.

There are two possibilities: $(a, b, c)=(2,4,4)$ or $(a, b, c)=(3,3,4)$.

Therefore, $T(10)=2$.

Consider $T(11)$. We want to find all possible triples $(a, b, c)$ of integers with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=11$.

We need $c<\frac{11}{2}$ and $c \geq \frac{11}{3}$. Thus, $c=4$ or $c=5$.

If $c=4$, we need $0<a \leq b \leq 4$ and $a+b=7$.

There is only one possibility: $(a, b, c)=(3,4,4)$.

If $c=5$, we need $0<a \leq b \leq 5$ and $a+b=6$.

There are three possibilities: $(a, b, c)=(1,5,5)$ or $(a, b, c)=(2,4,5)$ or $(a, b, c)=(3,3,5)$.

Therefore, $T(11)=4$.

Consider $T(12)$. We want to find all possible triples $(a, b, c)$ of integers with $0<a \leq b \leq c$ and $c<a+b$ and $a+b+c=12$.

We need $c<\frac{12}{2}$ and $c \geq \frac{12}{3}$. Thus, $c=4$ or $c=5$.

If $c=4$, we need $0<a \leq b \leq 4$ and $a+b=8$.

There is only one possibility: $(a, b, c)=(4,4,4)$.



If $c=5$, we need $0<a \leq b \leq 5$ and $a+b=7$.

There are two possibilities: $(a, b, c)=(2,5,5)$ or $(a, b, c)=(3,4,5)$.

Therefore, $T(12)=3$.

Final answer: 2,4,3