Honzík had four identical triangles. He was arranging them into various shapes by placing the triangles $\mathrm{k}$ side by side with sides of the same length. First, he arranged a shape from three triangles as shown in the picture, which had a perimeter of $43 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f09179dc58e2e20ca09dg-2.jpg?height=354&width=623&top_left_y=2270&top_left_x=722)

Then he disassembled the shape and arranged another shape from three triangles, which had a perimeter of $35 \mathrm{~cm}$. Finally, he arranged another shape from all four triangles as shown in the picture, and this shape had a perimeter of $46 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f09179dc58e2e20ca09dg-3.jpg?height=380&width=734&top_left_y=378&top_left_x=664)

Determine the lengths of the sides of the triangles.

(E. Semerádová)
In the used triangles, we can distinguish the mutually equal sides by color - for example, we can use red, green, and blue, and the corresponding lengths in centimeters we denote as $\check{c}, z$, and $m$:
![](https://cdn.mathpix.com/cropped/2024_04_17_f09179dc58e2e20ca09dg-3.jpg?height=416&width=1444&top_left_y=1066&top_left_x=310)

In the perimeter of the last figure, each color is represented exactly twice. This perimeter is equal to

$$
2 \cdot(\check{c}+z+m)=46,
$$

so the perimeter of one triangle is $23 \mathrm{~cm}$. In the perimeter of the first figure, there are three red sides, one green, and one blue, which is the same as two red sides and the sides of the perimeter of one triangle. This perimeter is equal to

$$
3 \check{c}+z+m=2 \check{c}+(\check{c}+z+m)=43 .
$$

The perimeter of one triangle is $23 \mathrm{~cm}$, so $2 \check{c}=43-23=20(\mathrm{~cm})$ and the length of one side of the triangle is $\check{c}=10(\mathrm{~cm})$.

The second figure (which is not shown in the problem statement) is composed of three triangles according to the same rules as the first, for example, like this:

![](https://cdn.mathpix.com/cropped/2024_04_17_f09179dc58e2e20ca09dg-3.jpg?height=414&width=619&top_left_y=2214&top_left_x=724)

The perimeter of this figure consists of one red, one green, and three blue sides, which is the same as the sides of the perimeter of one triangle plus two additional blue sides. This perimeter is given as

$$
\check{c}+z+3 m=(\check{c}+z+m)+2 m=35 .
$$

Similarly to above, we determine the length of the second side of the triangle: $2 m=35-23=12(\mathrm{~cm})$, so $m=6(\mathrm{~cm})$. Knowing the perimeter and two sides of the triangle, we can express the length of the third side: $z=23-10-6=7(\mathrm{~cm})$.

The lengths of the sides of Honza's triangles are $10 \mathrm{~cm}, 6 \mathrm{~cm}$, and $7 \mathrm{~cm}$. (The result is consistent with the triangle inequality.)

Scoring suggestion. 2 points for determining the perimeter of the triangle; 2 points for determining the length of one side; 1 point each for determining the lengths of the remaining two sides.

Note. For the second figure, other compositions can also be considered - in each case, one side of the triangle appears multiple times in the perimeter of the figure. With the given notation, at most the lengths of sides $z$ and $m$ would be swapped, which is equivalent to a different notation at the beginning. It is not necessary to analyze all possibilities in solving the problem.