3. Given are 12 planks, each with a length of $13 \mathrm{dm}$. Each plank is divided into several pieces, from which 13 triangles are made. Each of them is made from pieces with lengths of $3 \mathrm{dm}, 4 \mathrm{dm}$, and $5 \mathrm{dm}$.

How should the 12 planks be cut?
Solution. The perimeter of each of the triangles is

## $3 \mathrm{dm}+4 \mathrm{dm}+5 \mathrm{dm}=12 \mathrm{dm}$.

The sum of all perimeters is $13 \cdot 12 \mathrm{dm}=156 \mathrm{dm}$, which is equal to the sum of the lengths of all the sticks. This means that all the sticks are cut in such a way that each piece obtained is used.

The number 13 can be written as a sum of 3's, 4's, and 5's in the following three ways

| I way: | $13=3+3+3+4$ | $(x$-sticks) |
| :--- | :--- | :--- |
| II way: | $13=4+4+5$ | $(y$-sticks) |
| III way: | $13=3+5+5$ | $(z$-sticks) |

Let $x$-sticks be cut in the first way, $y$-sticks be cut in the second way, and $z$-sticks be cut in the third way.

Now we get the equations

$$
\begin{array}{ll}
x+y+z=12, & \text { because the number of sticks is } 12 \\
3 x+z=13, & \text { there are } 13 \text { pieces with a length of } 3 \mathrm{~cm} \\
x+2 y=13, & \text { there are } 13 \text { pieces with a length of } 4 \mathrm{~cm} \\
y+2 z=13, & \text { there are } 13 \text { pieces with a length of } 5 \mathrm{~cm} .
\end{array}
$$

It is not difficult to determine the values of $x, y$, and $z$. Indeed,

$$
\begin{aligned}
& x=13-2 y \\
& z=13-3 x=13-3(13-2 y)=6 y-26
\end{aligned}
$$

from which we get

$$
\begin{aligned}
& 13-2 y+y+6 y-26=12 \\
& 5 y-13=12
\end{aligned}
$$

Thus, $y=5$ sticks need to be cut into pieces of $4 \mathrm{dm}, 4 \mathrm{dm}, 5 \mathrm{dm}$, $x=3$ sticks need to be cut into four pieces with lengths of $3 \mathrm{dm}$, $3 \mathrm{dm}, 3 \mathrm{dm}, 4 \mathrm{dm}$, and $z=4$ sticks need to be cut into three pieces with lengths of $3 \mathrm{dm}, 5 \mathrm{dm}, 5 \mathrm{dm}$.