Subfield: Combinatorics


Question: 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 .
If $m$ is a positive integer with $m \geq 3$, prove that $T(2 m)=T(2 m-3)$.

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 $c \geq \frac{1}{3} k$.

We show that $T(2 m)=T(2 m-3)$ by creating a one-to-one correspondence between the triples in $S_{2 m}$ and the triples $S_{2 m-3}$.

Note that $S_{2 m}$ is the set of triples $(a, b, c)$ of positive integers with $0<a \leq b \leq c$, with $c<a+b$, and with $a+b+c=2 m$.

Also, $S_{2 m-3}$ is the set of triples $(A, B, C)$ of positive integers with $0<A \leq B \leq C$, with $C<A+B$, and with $A+B+C=2 m-3$.

Consider a triple $(a, b, c)$ in $S_{2 m}$ and a corresponding triple $(a-1, b-1, c-1)$.

We show that $(a-1, b-1, c-1)$ is in $S_{2 m-3}$ :

* Since $(a, b, c)$ is in $S_{2 m}$, then $c<\frac{1}{2}(2 m)=m$. This means that $b \leq c \leq m-1$, so $a=2 m-b-c \geq 2$. Therefore, $a-1, b-1$ and $c-1$ are positive integers since $a, b$ and $c$ are positive integers with $2 \leq a \leq b \leq c$.
* Since $2 \leq a \leq b \leq c$, then $1 \leq a-1 \leq b-1 \leq c-1$, so $0<a-1 \leq b-1 \leq c-1$.
* Since $a+b+c=2 m$, then $c=2 m-(a+b)$ so $a+b$ and $c$ have the same parity.

Since $c<a+b$, then $c \leq a+b-2$. (In other words, it cannot be the case that $c=a+b-1$.) Therefore, $c-1 \leq(a-1)+(b-1)-1$; that is, $c-1<(a-1)+(b-1)$.

$*$ Since $a+b+c=2 m$, then $(a-1)+(b-1)+(c-1)=2 m-3$.

Therefore, $(a-1, b-1, c-1)$ is in $S_{2 m-3}$, since it satisfies all of the conditions of $S_{2 m-3}$. Note as well that two different triples in $S_{2 m}$ correspond to two different triples in $S_{2 m-3}$. Thus, every triple in $S_{2 m}$ corresponds to a different triple in $S_{2 m-3}$.

Thus, $T(2 m) \leq T(2 m-3)$.

Consider a triple $(A, B, C)$ in $S_{2 m-3}$ and a corresponding triple $(A+1, B+1, C+1)$. We show that $(A+1, B+1, C+1)$ is in $S_{2 m}$ :

* Since $(A, B, C)$ is in $S_{2 m-3}$, then $A, B$ and $C$ are positive integers, so $A+1, B+1$ and $C+1$ are positive integers.
* Since $0<A \leq B \leq C$, then $1<A+1 \leq B+1 \leq C+1$, so $0<A+1 \leq B+1 \leq C+1$.
* Since $C<A+B$, then $C+1<(A+1)+(B+1)-1$ so $C+1<(A+1)+(B+1)$.
* Since $A+B+C=2 m-3$, then $(A+1)+(B+1)+(C+1)=2 m$.

Therefore, $(A+1, B+1, C+1)$ is in $S_{2 m}$.

Note again that two different triples in $S_{2 m-3}$ correspond to two different triples in $S_{2 m}$. Thus, every triple in $S_{2 m-3}$ corresponds to a different triple in $S_{2 m}$.

Therefore, $T(2 m-3) \leq T(2 m)$.

Since $T(2 m) \leq T(2 m-3)$ and $T(2 m-3) \leq T(2 m)$, then $T(2 m)=T(2 m-3)$.