## Task 22/75

How many different triangles are there where the measure of the perimeter is 50 and the measures of the sides are natural numbers?
Let $a, b, c$ be the sides of the triangle, where (w.l.o.g.) $a \geq b \geq c$. From $a+b+c=50$ and $a<b+c$ it follows that $3a \geq 50$, thus

$$
\frac{50}{3} \leq a < 25 \quad \text{and} \quad c \geq 50 - 2a
$$

Due to the required integer values of $a, b, c$, only the 8 values $17, 18, 19, \ldots, 24$ are possible for $a$. For $c$, the following variations are possible (where $n(a)$ denotes their number):

| $a$ | $c$ | $n(a)$ |
| :---: | :---: | :---: |
| 17 | 16 | 1 |
| 18 | $14 \ldots 16$ | 3 |
| 19 | $12 \ldots 15$ | 4 |
| 20 | $10 \ldots 15$ | 6 |
| 21 | $8 \ldots 14$ | 7 |
| 22 | $6 \ldots 14$ | 9 |
| 23 | $4 \ldots 13$ | 10 |
| 24 | $2 \ldots 13$ | 12 |

For each variation of $c$, $b$ is uniquely determined by $b=50-a-c$. Therefore, there are a total of

$$
\sum_{a=17}^{24} n(a) = 52
$$

different triangles. Among them, no two can be the same, since in each case $a$ is the largest and $c$ is the smallest side. Thus, $a \geq b \geq c$ always holds.

Additional: The formation rule for $n(a)$ is:

$$
n(a) = \frac{1}{4} \left[6a - 97 + (-1)^a \right]
$$