On a regular triangular lattice with unit side length, we have marked four lattice points that are the vertices of a parallelogram $\mathcal{P}$; the area of $\mathcal{P}$ is $\sqrt{3}$ units. What can the sum of the lengths of the lattice segments lying inside the interior of $\mathcal{P}$ be?
For the cases, adjust the task according to how many sides of the parallelogram lie on grid lines.

1. Case: All four sides lie on grid lines.

Since the area of the parallelogram is $\sqrt{3}$, it contains 4 grid triangles. Thus, the total length of the grid lines inside the parallelogram is 3 (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_ad23a2c4da9a134f978eg-1.jpg?height=526&width=374&top_left_y=325&top_left_x=862)

2. Case: Two sides (let's call them bases) lie on grid lines. Since the height of the parallelogram is at least $\frac{\sqrt{3}}{2}$ (since the distance between two parallel grid lines is at least this much), the length of the bases can only be 1 or 2. 2.a Case: The bases are 2 units long.

Notice that if we "translate" a parallelogram by one unit (for example, from the dotted to the solid one in Figure 2), the sum of the internal grid lines does not change (since the "falling out" and "entering" triangles are congruent, translatable into each other, and contain the same length of grid lines). This is only not true if the resulting parallelogram has all its sides on grid lines, because in this case, a side "disappears" from the interior of the parallelogram, or the total length decreases by 1 (for example, when translating from the solid to the starred parallelogram in Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_ad23a2c4da9a134f978eg-1.jpg?height=517&width=1195&top_left_y=1281&top_left_x=454)

Thus, we can translate all parallelograms that fit this case into a parallelogram with all sides on grid lines, and during this process, the sum of the internal grid lines decreases by 1. Therefore, in this case, the total length of the internal grid lines is $3+1=4$.

2.b Case: The bases are 1 unit long (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_ad23a2c4da9a134f978eg-1.jpg?height=518&width=341&top_left_y=2023&top_left_x=881)

In this case, it is still true that we can freely translate the parallelogram, but at the end (when we get a parallelogram with all sides on grid lines), the sum of the grid lines decreases by 2. Thus, in this case, the total length of the internal grid lines is $3+2=5$.

3. Case: The parallelogram has no sides on grid lines. Here, it is also true that we can translate the parallelogram (not necessarily by one unit) until we get one with one side on a grid line (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_ad23a2c4da9a134f978eg-2.jpg?height=374&width=643&top_left_y=285&top_left_x=730)

Then we get a parallelogram from Case 2 (since the translation does not change the area of the parallelogram), which can be either:

- If the base is 2, the sum of the internal grid lines decreases by 2 during the translation, so initially it was $4+2=6$.
- If the base is 1, the sum of the internal grid lines decreases by 1 during the translation, so initially it was $5+1=6$.

Thus, in this case, the total length of the internal grid lines is always 6.

Therefore, in a parallelogram with an area of $\sqrt{3}$, the total length of the grid lines can be 3, 4, 5, or 6.

Viktor Tóth (Táncsics Mihály Gymnasium, Kaposvár, 12th grade)