## Task 4 - 330844

For a triangle, the following conditions are required:

(1) All three side lengths of the triangle have integer measures in centimeters.

(2) The perimeter of the triangle is $50 \mathrm{~cm}$.

Determine the maximum number of triangles that meet these requirements and among which no two triangles are congruent to each other!
If $a, b, c$ with $a \geq b \geq c$ are the integer lengths of three line segments, then these lengths are the side lengths of a triangle if and only if $a < b + c$. Given the condition $a + b + c = 50$ and $a > 16$, $a$ can only be one of the numbers 17, 18, ..., 24.

In the following table, for each of these values of $a$, all pairs $(b, c)$ are found that satisfy the equation $b + c = 50 - a$ and the inequalities $a \geq b \geq c$:

We start with $b = a$ (and thus $c = 50 - 2a$), then we check if decreasing $b$ by 1 and simultaneously increasing $c$ by 1 results in another such pair $(b, c)$; this continues until the condition $b \geq c$ is no longer satisfied.

The number of such pairs found is then recorded in the table.

| $a$ | $b+c=50-a$ | Pairs $(b, c)$ | Number |
| :--- | :--- | :--- | :--- |
| 17 | $b+c=33$ | $(17,16)$ | 1 |
| 18 | $b+c=32$ | $(18,14),(17,15),(16,16)$ | 3 |
| 19 | $b+c=31$ | $(19,12),(18,13),(17,14),(16,15)$ | 4 |
| 20 | $b+c=30$ | $(20,10),(19,11), \ldots, (16,14),(15,15)$ | 6 |
| 21 | $b+c=29$ | $(21,8),(20,9), \ldots, (16,13),(15,14)$ | 7 |
| 22 | $b+c=28$ | $(22,6),(21,7), \ldots, (15,13),(14,14)$ | 9 |
| 23 | $b+c=27$ | $(23,4),(22,5), \ldots, (15,12),(14,13)$ | 10 |
| 24 | $b+c=26$ | $(24,2),(23,3), \ldots, (14,12),(13,13)$ | 12 |

Thus, the table contains the side lengths $a, b, c$ for each triangle of the required type. As evident, there are no two sets of side lengths that would result in congruent triangles (due to the same values for $a, b, c$).

The total number of such triangles is therefore $1 + 3 + 4 + 6 + 7 + 9 + 10 + 12 = 52$.