Jura has sticks of lengths $2 \mathrm{~cm}, 3 \mathrm{~cm}, 3 \mathrm{~cm}, 3 \mathrm{~cm}, 4 \mathrm{~cm}, 5 \mathrm{~cm}, 5 \mathrm{~cm}, 5 \mathrm{~cm}, 6 \mathrm{~cm}, 6 \mathrm{~cm}$, and $9 \mathrm{~cm}$. She assembles them into the sides of triangles such that no stick is part of the side of two or more triangles. She can use as many sticks as she wants, but she must not break them, and the used sticks must lie entirely on the boundary of the triangle. Jura claims that it is possible to use the sticks to form the sides of three triangles with the same perimeter. Is she right? What is the largest possible perimeter that these triangles could have?

(M. Dillingerová)
To have triangles with the maximum possible perimeter, we need to use as many of the longest rods as possible.

All the rods together measure $51 \mathrm{~cm}$. If we used all of them, the perimeter of one triangle would be $51: 3=17(\mathrm{~cm})$. In one of these triangles, there would need to be a rod $9 \mathrm{~cm}$ long, and the remaining two sides would total $8 \mathrm{~cm}$. This would mean that the sum of the lengths of two sides in the triangle would be less than the length of the third side, so we cannot use the $9 \mathrm{~cm}$ rod for a triangle with a perimeter of $17 \mathrm{~cm}$. Of course, it cannot be used for a triangle with an even smaller perimeter either.

The remaining rods have a total length of $42 \mathrm{~cm}$, so the perimeter of one triangle would be $14 \mathrm{~cm}$. This is already possible to achieve, and can be done in any of the following ways (all values are in $\mathrm{cm}$):

$\bullet 6,6,2 ; 5,5,4 ; 5,3+3,3$.

- $6,5,3 ; \quad 5,5,4 ; \quad 6,3+3,2$.
- $6,5,3 ; \quad 5,5,4 ; \quad 6,3,3+2$.
- $6,5,3 ; \quad 6,5,3 ; \quad 5,4+2,3$.
- $6,5,3 ; \quad 6,5,3 ; \quad 5,4,2+3$.

In all five cases, the triangle inequality is satisfied for all the triangles thus assembled, so Jura is right and the maximum possible perimeters are $14 \mathrm{~cm}$.

Evaluation. 1 point for calculating the maximum perimeter of $17 \mathrm{~cm}$; 2 points for excluding the $9 \mathrm{~cm}$ rod; 1 point for finding the perimeter of $14 \mathrm{~cm}$; 1 point for distributing the rods to individual perimeters; 1 point for some verification of the triangle inequality in this distribution. For experimentally finding one solution and determining the perimeters of $14 \mathrm{~cm}$ without explaining why the solver does not use the $9 \mathrm{~cm}$ rod, award a total of 3 points.