Cheresanov E.

On a 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, this is a triangle with 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 shifted 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_82cd6cdb4fc29ee56532g-28.jpg?height=709&width=637&top_left_y=0&top_left_x=716)

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 with 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 \(A C_1 A_1 C D E\), 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_82cd6cdb4fc29ee56532g-28.jpg?height=849&width=964&top_left_y=1205&top_left_x=547)

Assume the contrary: our hexagon overlaps 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.

Notice that points \(X\) and \(Y\) are in opposite (centrally symmetric with respect 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, consequently, cannot contain "bad" points \(X\) and \(Y\).

Suppose, for example, that point \(X\) lies in triangle \(C_1 A_1 B_1\), and \(Y\) lies in triangle \(B_1 D E\). Translate triangle \(ABC\) so that it overlaps trapezoid \(A C_1 D E\). Now start moving it parallel to \(ED\) until it covers trapezoid \(A_1 C D E\). Since at each moment it completely contains triangle \(B_1 D E\) and fully "sweeps" triangle \(C_1 A_1 B_1\), 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 part (a).

| Problem 107764 Topics: | Arbitrary Polygons | ] | Difficulty: 5  Classes: 9,10,1 |
| :---: | :---: | :---: | :---: |
|  | [ Lines and curves dividing figures into equal areas ] |  |  |
|  | Continuity considerations | ] |  |
|  | Convex hull and supporting lines (planes). | ] |  |
|  | Reconfigurations | ] |  |
| Author: Proizvooov.V. |  |  |  |
| {Consider an arbitrary polygon (not necessarily convex).  a) Is there always a chord of the polygon that divides it into two equal areas?  b) Prove that any polygon can be divided by some chord into parts, each of which has an area of at least 1/3 of the area of the polygon. (A chord of a polygon is a segment whose endpoints lie on the contour of the polygon, and which itself lies entirely within the polygon, including the contour.)} |  |  |  |
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## Solution

a) An example is a polygon consisting of three identical squares (halls) connected by thin curved corridors (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_82cd6cdb4fc29ee56532g-29.jpg?height=700&width=688&top_left_y=1039&top_left_x=685)

Let \(S\) be the area of the polygon, 0.3S be the area of one hall (0.1S be the total area of all corridors). If a chord intersects only a corridor, then on one side of it are two halls, and their area is more than half the area of the polygon. If a chord intersects one of the halls, then it does not intersect the "intersection" of the three corridors, and then again in one part are two halls.

b) Choose a direction of the chord that is not parallel to the sides and diagonals of the polygon, and consider this direction as vertical. Then the chord can pass through no more than one vertex of the polygon. If no internal point of the chord is a vertex of the polygon, the chord divides the polygon exactly into 2 parts; otherwise, it divides it exactly into 3 parts.

If the area of the smaller part is greater than or equal to \(S / 3\), the problem is solved. Otherwise, move the chord perpendicular to the vertical in the direction in which the area of the smaller part of the polygon increases. During the movement, the chord may "hit" an internal vertex (but not two vertices at once).

If the chord does not encounter any obstacles on its path, the area of the smallest part changes continuously. If at some point it reaches \(S / 3\), the problem is solved. Otherwise, the chord will hit an internal vertex. At this point, two corridors may converge or diverge (left figure), then the chord, exiting the corridor, will jump up or, entering the corridor, will jump down (or "jump" to another part of the polygon, as in the right figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_82cd6cdb4fc29ee56532g-30.jpg?height=345&width=832&top_left_y=-1&top_left_x=613)

Of the three resulting parts of the polygon, the largest must have an area greater than \(S / 3\). Direct the vertical chord into the largest part, then the two smaller parts will merge. If their area becomes greater than or equal to \(S / 3\), the problem is solved. Otherwise, continue moving the chord until it hits the next vertex, and so on.

The area of the smaller part is always increasing, so we cannot pass through the same internal vertex twice. But there are a finite number of internal vertices, so the process will end.

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