## Task 5 - 321245

Determine the maximum number of triangles with integer side lengths and a perimeter of 1993, among which no two triangles are congruent to each other.
Let the three side lengths be $a, b$, and $c$ with $0 < a \leq b \leq c$. Given that $a + b + c = 1993$, we have $c_{\max }=\left\lfloor\frac{1993}{2}\right\rfloor=996$. Due to the sorting by size, we also have $c_{\text {min }}=\left\lceil\frac{1993}{3}\right\rceil=665$, because then $a=b=664$. Therefore, we have:

$$
665 \leq c \leq 996
$$

Furthermore, we have:

$$
a+b=1993-c
$$

To determine the number of triangles for a given $c$, we need to find the minimum and maximum $a$. $a$ is minimal when $b$ is maximal, and the maximal $b$ is equal to $c$. Therefore,

$$
a_{\min }=1993-2 c
$$

$a$ is maximal when it is equal to $b$, or 1 less than $b$ if $a+b$ is odd. Thus,

$$
a_{\max }=\left\lfloor\frac{1993-c}{2}\right\rfloor
$$

We distinguish between the two cases, whether $c$ is even or odd. Let the number of respective triangles be $A_{c}$.

1. $c=2 m$ with $m=333 \ldots 498$. Then,

$$
\begin{gathered}
a_{\min }=1993-4 m \\
a_{\max }=996-m \\
A_{2 m}=996-m-(1993-4 m)+1=3 m-996
\end{gathered}
$$

2. $c=2 m+1$ with $m=332 \ldots 497$. Then,

$$
\begin{gathered}
a_{\min }=1991-4 m \\
a_{\max }=996-m \\
A_{2 m+1}=996-m-(1991-4 m)+1=3 m-994
\end{gathered}
$$

The total number of non-congruent triangles is then

$$
\begin{gathered}
A=\sum_{m=333}^{498}(3 m-996)+\sum_{m=332}^{497}(3 m-994) \\
A=\sum_{m=1}^{166}(3 m)+\sum_{m=1}^{166}(3 m-1)
\end{gathered}
$$

$$
\begin{gathered}
A=2 \cdot 3 \sum_{m=1}^{166} m-166 \\
A=3 \cdot 166 \cdot 167-166=166 \cdot(3 \cdot 167-1)=166 \cdot 500 \\
A=83000
\end{gathered}
$$

Therefore, there are exactly 83000 non-congruent triangles.