A square is cut into two parts by a line passing through its center. What is the geometric locus of the centroids of the two parts as the cutting line takes on all possible positions?[^0]


[^0]:    ${ }^{1}$ The mechanical center of gravity of the quadrilateral plate $A B C D$ lies on the line connecting the centers of gravity of the triangular plates $A C B$ and $A C D$. See also the solution to problem 1246, K. M. L. 29 (1964/11) p. 117.
I. solution. Place a coordinate system on the diagram, taking the midlines of the square as the axes. Let the equation of the bisector line be $y=m x$. Due to symmetry, it suffices to consider the values $-1 \leq m \leq 1$ and to look at the centroid of the lower part of the bisected square, since the upper part can be obtained from this by a $180^{\circ}$ rotation, and if the bisector segment forms a smaller angle with the $Y$-axis (and thus $|m|>1$), we are faced with the mirror image of an already examined case with respect to one of the diagonals.

Let the lower two vertices of the square be $A(-3 ;-3)$ and $B(3 ;-3)$, then the other two vertices of the lower trapezoid are $C(3 ; 3 m)$ and $D(-3 ;-3 m)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0b9e10b3659dc3e78b1eg-1.jpg?height=710&width=680&top_left_y=458&top_left_x=709)

By bisecting the trapezoid with the diagonal $A C$, the centroid $S_{1}(1 ; m-2)$ of the triangle $A C B$ and the centroid $S_{2}(-1 ;-1)$ of the triangle $A C D$ lie on the line segment that is divided into thirds by the constant $O A$ weight line. The coordinates of the sought centroid $S(x, y)$ of the trapezoid lie on the line connecting these two centroids, thus they satisfy the equation of the line $S_{1} S_{2}$:

$$
y=\frac{m-1}{2}(x+1)-1
$$

Similarly, after bisecting with the diagonal $B D$, the coordinates of the centroid $S$ also satisfy the equation of the line connecting the centroid $S_{3}(-1 ;-m-2)$ of the triangle $B D A$ and the centroid $S_{4}(1 ;-1)$ of the triangle $B D C$:

$$
y=\frac{m+1}{2}(x-1)-1
$$

From these two relationships,

$$
x=m, \quad y=\frac{m^{2}-3}{2}
$$

Thus, the geometric locus of the centroid of the trapezoid $A B C D$ (by eliminating $m$) is the

$$
y=\frac{x^{2}}{2}-\frac{3}{2}, \quad-1 \leq x \leq 1
$$

parabolic arc.

![](https://cdn.mathpix.com/cropped/2024_05_02_0b9e10b3659dc3e78b1eg-1.jpg?height=702&width=616&top_left_y=1994&top_left_x=684)

The vertex of the parabola is the lower quarter point of the perpendicular bisector of $A B$, which is $(0 ;-3 / 2)$. The axis of the parabola is this perpendicular bisector, the focus $F_{0}$ is the point $(0 ;-1)$, which is the lower third point of the perpendicular bisector, and the directrix intersects the perpendicular bisector at the lower sixth point. The endpoints of the parabolic arc are $S_{2}$ and $S_{4}$, since when $m=-1$, the triangle $A B C$ degenerates into the side $A B$ of the square, and when $m=1$, the triangle $A B D$ degenerates into the side $A B$ of the square, and the mass of the trapezoid decreases to 0.

The tangent to the parabolic arc at $S_{2}$ and $S_{4}$ is at a $45^{\circ}$ angle to the axis, or parallel to the diagonal from $B$ and $A$, respectively. Therefore, the complete geometric locus described in the first paragraph, which is a closed curve consisting of 4 parabolic arcs with 4 axes, has a tangent at the connection points.

Domokos Zsuzsanna (Makó, József A. g. III. o. t.)

II. solution. Using the notation of the previous solution, we again only examine the centroid $S$ of the trapezoid $A B C D$ cut by lines perpendicular to the side $A B$ of the square. Let the other two vertices of the square be $B_{1}$ and $A_{1}$, and the midpoint of the segment $A B$ be $F$. Then $S_{2}$, as the point that divides the segment $A O$ in the ratio 2:1 from $O$, is also the centroid of the triangle $A B A_{1}$, so $S_{2}$ is the point that divides the segment $A_{1} F$ in the ratio 2:1 from $F$, and similarly, $S_{4}$ is the point that divides the segment $B_{1} F$ in the ratio 2:1 from $F$. The trapezoid $S_{2} S_{4} S_{1} S_{3}$ is thus a scaled-down version of the trapezoid $A_{1} B_{1} C D$ by a factor of 1/3 from $F$. It suffices, therefore, to examine the geometric locus of the intersection point $E$ of the diagonals of the latter trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_02_0b9e10b3659dc3e78b1eg-2.jpg?height=626&width=616&top_left_y=804&top_left_x=749)

Let the perpendicular from $E$ to $A_{1} B_{1}$ intersect $A_{1} B_{1}$ at $E_{1}$. From the similar triangles $A_{1} E D$ and $C E B_{1}$, and from the angle formed by the parallel lines $A_{1} B_{1}$ and $D$, we get:

$$
\frac{A_{1} D}{C B_{1}}=\frac{D E}{E B_{1}}=\frac{A_{1} E_{1}}{E_{1} B_{1}}
$$

However, $A_{1} E_{1} + E_{1} B_{1} = A_{1} D + D A = A_{1} D + C B_{1}$, so it must be that $E_{1} B_{1} = C B_{1}$ (and $A_{1} E_{1} = A_{1} D$). Let the perpendicular from $O$ to $B_{1} C$ intersect $B_{1} C$ at $G$, and the perpendicular from $E$ to $O G$ intersect $O G$ at $E_{2}$. Then, by rotating the square $B_{1} F_{1} O G$ by $90^{\circ}$ around its center, $F_{1} E_{2}$ goes to $O C$, so the two lines are perpendicular to each other.

It is known that the line $E_{1} E_{2}$, which is parallel to the parallel sides of the trapezoid, is bisected by the intersection point $E$ of the diagonals. Draw a line parallel to $C D$ from $E_{1}$, then a parallelogram is formed between $O F_{1}$ and $E_{1} E_{2}$, and the median from $E$ bisects the segment of the parallelogram that lies within it and is perpendicular to it, since it is parallel to $O C$; this median also bisects the segment $F_{1} E_{2}$, because equal segments lie outside the parallelogram at both ends, so this is the perpendicular bisector of $E_{1} E_{2}$, thus $E E_{2} = E F_{1}$.

This shows that $E$ lies on the arc of the parabola that lies within the square $A A_{1} B_{1} B$, which has focus $F_{1}$ and directrix the line $O G$. If $E$ is any point on this arc, $D$ is the intersection of the lines $A A_{1}$ and $B_{1} E$, and $C$ is the intersection of $D O$ and $B B_{1}$, then, according to the previous reasoning, the intersection of the diagonals of the trapezoid $A_{1} B_{1} C D$ lies on the parabolic arc and on the line $B_{1} D$, so it is the point $E$. By scaling this arc by a factor of 1/3 from $F$, we arrive at the geometric locus described in the previous solution.

Balogh Kálmán (Budapest, Fazekas M. gyak. g. II. o. t.) with additions.