Trees in a forest stand on the points of a regular triangular grid with side length $d$ (that is, if the trees $A, B, C$ form a triangle with no other trees inside, then $A B=B C=C A=d$). They want to demarcate a rectangular part of the forest such that the corners of the rectangle are trees, the distance between the trees on two opposite sides is $d$, and there are as many trees inside the rectangle as on its perimeter. Is this possible?
11 I. solution. Two adjacent equilateral triangles in the triangular lattice form a rhombus with $60^{\circ}$ acute angles and diagonals of lengths $d$ and $d \sqrt{3}$, which are perpendicular to each other. Therefore, every lattice line in the lattice has a perpendicular lattice line at a distance of $d$, and this lattice constant is $d \sqrt{3}$. Every rectangle in the lattice with sides $d$ and $d \sqrt{3}$ has one lattice point in its interior, which is the center.

![](https://cdn.mathpix.com/cropped/2024_05_02_a8e2415cc102009cf38ag-1.jpg?height=356&width=585&top_left_y=337&top_left_x=759)

Let the sides of the enclosed rectangle be $b d$ and $c d \sqrt{3}$, where $b$ and $c$ are natural numbers. Thus, there are $b, c, b, c$ trees on the sides (counting the starting point for each side but not the endpoint), totaling $2(b+c)$ trees. The lines connecting the pairs of opposite boundary lattice points divide the forest section into $b c$ rectangles with sides $d$ and $d \sqrt{3}$. Each of these rectangles has a tree at its center, and the lines intersect at $(b-1)(c-1)$ internal points, each of which has a tree. This accounts for all the internal trees, whose number is $b c + (b-1)(c-1)$. The question is whether there exists a pair of natural numbers $b, c$ such that

$$
2(b+c) = b c + (b-1)(c-1)
$$

From this,

$$
b = \frac{3c-1}{2c-3} = 1 + \frac{c+2}{2c-3}
$$

Since $c+2$ is positive, $2c-3$ must also be positive, so $c > 3/2$, and since $c$ is an integer, $c \geq 2$. Furthermore, for the second term on the right side,

$$
\frac{c+2}{2c-3} = \frac{1}{2} \left( \frac{2c+4}{2c-3} \right) = \frac{1}{2} \left( 1 + \frac{7}{2c-3} \right)
$$

The second term in the last parentheses must be an odd integer. This is only possible for integers $c > 1$ if the denominator is 1 or 7. Thus, $c_1 = 2$ and $c_2 = 5$, and $b_1 = 5$ and $b_2 = 2$, so the condition is satisfied.

The two solutions are different in shape but have the same area of the forest section. The longer side is $5: 2 \sqrt{3} \approx 1.44$ times, or $5 \sqrt{3}: 2 \approx 4.33$ times, the shorter side, but the area in both cases is $b c d^2 \sqrt{3} = 10 \sqrt{3} d^2$. (From the diagram, we see that they can be transformed into each other by stretching along one side.)

Seprödi László (Budapest, Fáy A. g. III. o. t.)

II. solution. If we sacrifice the symmetry of (1) and write $c = b + e$ (where $e$ is an integer), we get a quadratic equation in $b$:

$$
2b^2 + 2(e-3)b + (1-3e) = 0
$$

From this, $b$ is rational only if the discriminant is a perfect square:

$$
4(e-3)^2 - 8(1-3e) = 4(e^2 + 7)
$$

Thus, if $e^2 + 7 = f^2$, where $f$ is an integer. Then $f^2 - e^2 = (f-e)(f+e) = 7$, so $f-e = 7/(f+e)$, and similarly to the first solution, $e = \pm 3$ and $|f| = 4$. Now, with these two values, (2) gives the positive solution $b, c = 2, 5$ again.

Benczúr András (Budapest, Fazekas M. gyak. g. III. o. t.)

Remarks. 1. By breaking down the equation (1), reducing it to zero, and multiplying the terms containing the unknowns by 2, we can factorize them:

$$
\begin{gathered}
2bc - 3b - 3c + 1 = 0 \\
4bc - 6b - 6c + 2 = 2b(2c-3) - 3(2c-3) - 7 = 0 \\
(2b-3)(2c-3) = 7
\end{gathered}
$$

From this, we can similarly conclude as in the first solution.

Kóta József (Tatabánya, Árpád g. III. o. t.)[^0]

2. Several students started from the observation that in a "small" lattice rectangle, there are fewer lattice points inside than on the perimeter, while in a "large" rectangle, there is a surplus inside. They then observed the increase in the number of internal lattice points when one side is increased, and thus arrived at one or both solutions. However, it is not clear from their work whether there are further solutions. These solutions also answered the question affirmatively, so they are correct. We evaluated these solutions with 2 points.
3. In the original problem statement, the triangle $ABC$ had to be considered equilateral. With fixed $A$ and $B$ such that $AB = d$, there are infinitely many $C$ such that the triangle $ABC$ has no additional trees. In all these cases, the height from $C$ is $d \sqrt{3} / 2$.

[^0]:    ${ }^{1} \mathrm{The}$ parenthetical remark in the original problem statement has been removed. - Ed.

    ${ }^{2}$ A lattice line is any line that passes through two (and thus infinitely many) lattice points.