Consider the triangle $E$.

On the coordinate plane, a triangle is placed such that its translations by vectors with integer coordinates do not overlap.
a) Can the area of such a triangle be greater than 1/2?

b) Find the greatest possible area of such a triangle.
a) For example, such a triangle has vertices at points $(0,0), (4/3, 2/3)$, and $(2/3, 4/3)$ (see figure). It is easy to see that the sides of the translated triangles pass through the vertices of the original triangle, and that the area of the triangle is $2/3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1560f9d9b3af2f34b251g-64.jpg?height=688&width=629&top_left_y=382&top_left_x=723)

b) Lemma. If a polygon can be translated by an integer vector without self-intersections, then its area is no more than 1.

Proof. Cut the plane into unit squares using lines of the coordinate grid. Our polygon will be cut into several parts. Translate all these parts (by an integer vector) into one square. By the condition, the resulting figures do not intersect, so their total area is no more than 1.

Let $ABC$ be a triangle satisfying the conditions of the problem. Denote by $A_1, B_1$, and $C_1$ the midpoints of sides $BC, CA$, and $AB$, respectively. Construct the hexagon $AC_1A_1CDE$, where points $D$ and $E$ are obtained by reflecting points $C_1$ and $A_1$ about point $B_1$ (see figure). We will prove that this hexagon can be translated by an integer vector without self-intersections.

![](https://cdn.mathpix.com/cropped/2024_05_06_1560f9d9b3af2f34b251g-64.jpg?height=858&width=972&top_left_y=1566&top_left_x=543)

Assume the contrary: our hexagon overlaps with itself when translated by some integer vector. Consider a point $X$ of the hexagon that, during this movement, lands on a point $Y$ inside the hexagon. If we connect point $B_1$ to all the vertices of the hexagon, we get six small triangles.

Note that points $X$ and $Y$ are in opposite (centrally symmetric relative to $B_1$) triangles. Indeed, any other two triangles can be placed in one large triangle equal to $ABC$ (coinciding with $ABC$, centrally symmetric to $ABC$, or obtained from one of them by translation by some vector). Such a triangle, obviously, also satisfies the conditions of the problem and, therefore, cannot contain "bad" points $X$ and $Y$.

## Suppose, for example, that point $X$ lies in the triangle $C_1A_1B_1$, and $Y$ lies in the triangle $B_1DE$. Translate the triangle $ABC$ so that it overlaps with the trapezoid $AC_1DE$. Now start moving it parallel to $ED$ until it covers the trapezoid $A_1CDE$. Since at each moment it completely contains the triangle $B_1DE$ and fully "sweeps" the triangle $C_1A_1B_1$, there will be a moment when the triangle will cover both point $X$ and point $Y$. But this triangle also satisfies the conditions of the problem. Contradiction. By the lemma, the area of the hexagon (equal to $1.5 S_{ABC}$) is no more than 1. Therefore, $S_{ABC} \leq 2/3$. An example is constructed in a).

## Answer

a) It can; b) $2/3$.