On an $n \times n$ chessboard, we write the number 1 in the first field of the first row, and in the remaining fields of the first row and the first column, we write the numbers $2,3, \ldots, n$ respectively. In every other field, we write the number that is one less than the sum of the numbers at the beginning of the row and the column. Then, we place a point mass of the written number grams at the center of each field. Where is the center of mass of this point system? (We consider the chessboard to be weightless.)

Where is the center of mass if we fill the inner fields of the table with the product of the two numbers instead of the reduced sum?
The center of mass (center of gravity) of a system of points - as known from physics - is the point to which, if we imagine the total mass of the system (denoted by the usual symbol $\sum m$), the moment about any (rotational) axis is equal to the sum of the moments of the individual masses of the system. The moment is given by the product of the mass and the distance (arm) from the axis (with sign). The position of the center of mass can be determined by the distances from two different direction axes (the axes, of course, being in the plane). It is convenient to choose the coordinate axes such that the distance from the center of each field is a multiple of the distance to the nearest point.

I. Place the chessboard in the usual Cartesian coordinate system and choose its unit length such that the center of the field common to the 1st row and 1st column has coordinates $(1,1)$, and the center of the field common to the 2nd row and 2nd column has coordinates $(2,2)$. Then, for the center of the field common to the $i$-th row and the $j$-th column (numbered from left to right in the usual way in chess), the coordinates are clearly $(j, i)$, and the mass of the material point placed there in the first mass point system is $j+i-1$. Thus, for each material point, the arm with respect to the $x$ and $y$ axes is given by its ordinate and abscissa, respectively, and due to the obvious symmetry of the point system, the two coordinates of the center of mass $T$ are equal, so it is sufficient to determine the abscissa $x_{T}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_770dc4a79e7b4d3f6d36g-1.jpg?height=1022&width=1008&top_left_y=790&top_left_x=553)

To find the sum of the masses, consider the sum of the masses in each column. Their values form an arithmetic sequence with a difference of 1, increasing upwards in each column. In the first column,

$$
s_{1}=1+2+\ldots+n=\frac{n}{2}(1+n)
$$

The terms in the 2nd column are each 1 greater, since the number of the field with center $(j, i)$ can be written as $j+(i-1)$, so $s_{2}=s_{1}+n$, similarly $s_{3}=s_{2}+n=s_{1}+2 n, \ldots$, and the column sums also form an arithmetic sequence. The sum of the last column is

$$
s_{n}=n+(n+1)+\ldots+(2 n-1)=\frac{n}{2}(3 n-1)
$$

or, after rearrangement:

$$
s_{n}=\frac{n}{2}\{(1+n)+2(n-1)\}=s_{1}+(n-1) n
$$

Thus, as the sum of the $n$ sums,

$$
\sum m=\frac{n}{2}\left\{\frac{n}{2}(1+n)+\frac{n}{2}(3 n-1)\right\}=\frac{n^{2}}{4} \cdot 4 n=n^{3}
$$

In the moment of the point system about the $y$ axis, the arm (abscissa) of each column's masses is common and equal to the column number, so based on the above,

$$
\sum m_{k} x_{k}=s_{1} \cdot 1+s_{2} \cdot 2+s_{3} \cdot 3+\ldots+s_{n} \cdot n=
$$

$$
\begin{gathered}
=\left(s_{1}+0 \cdot n\right) \cdot 1+\left(s_{1}+1 \cdot n\right) \cdot 2+\left(s_{1}+2 \cdot n\right) \cdot 3+\ldots+\left\{s_{1}+(n-1) n\right\} \cdot n= \\
=s_{1}(1+2+3+\ldots+n)+n\{(0 \cdot 1+1 \cdot 2+2 \cdot 3+\ldots+(n-1) n)\}
\end{gathered}
$$

(The $k$ index runs through the fields of the chessboard in some order.)

The second expression in parentheses can be transformed as follows:

$$
\begin{gathered}
\left(1^{2}-1\right)+\left(2^{2}-2\right)+\left(3^{2}-3\right)+\ldots+\left(n^{2}-n\right)= \\
=\left(1^{2}+2^{2}+\ldots+n^{2}\right)-(1+2+\ldots+n)= \\
=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}=\frac{n(n+1)}{6}(2 n-2)=\frac{(n+1) n(n-1)}{3}
\end{gathered}
$$

Thus,

$$
\sum m_{k} x_{k}=\frac{n^{2}(n+1)^{2}}{4}+\frac{n^{2}(n+1)(n-1)}{3}=\frac{n^{2}(n+1)}{12}(7 n-1)
$$

With this, the equation and solution for $x_{T}$ are

$$
x_{T} \cdot \sum m=\sum m_{k} x_{k}, \quad x_{T}=y_{T}=\frac{\sum m_{k} x_{k}}{\sum m}=\frac{(n+1)(7 n-1)}{12 n}
$$

(The center of mass does not fall into the center of any field of the board for any $n(>1)$, because $x_{T}$ cannot be an integer, since $n$ is relatively prime to both $n+1$ and $7 n-1$.)

II. For the second point system, we proceed similarly. The sum of the masses in the first column remains $s_{1}$, the sum of the masses in the 2nd column is $2 s_{1}$, since we wrote twice the number of the 1st column for each term, similarly the sum of the masses in the 3rd, ..., $n$-th column is $3 s_{1}, \ldots, n s_{1}$, so

$$
\begin{gathered}
\sum m=s_{1}+2 s_{1}+3 s_{1}+\ldots+n s_{1}= \\
=s_{1}(1+2+\ldots+n)=s_{1}^{2}=\frac{n^{2}(n+1)^{2}}{4}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_770dc4a79e7b4d3f6d36g-2.jpg?height=775&width=764&top_left_y=1412&top_left_x=664)

Figure 2

The distance from the $y$-axis is again common for each column, so

$$
\begin{gathered}
\sum m_{k} x_{k}=s_{1} \cdot 1+\left(2 s_{1}\right) \cdot 2+\left(3 s_{1}\right) \cdot 3+\ldots+\left(n s_{1}\right) \cdot n= \\
=s_{1}\left(1^{2}+2^{2}+\ldots+n^{2}\right)=\frac{n^{2}(n+1)^{2}(2 n+1)}{12}
\end{gathered}
$$

and thus the coordinates of the center of mass $U$ are:

$$
x_{U}=y_{U}=\frac{\sum m_{k} x_{k}}{\sum m}=\frac{2 n+1}{3}
$$

Zoltán Papp (Debrecen, Fazekas M. Gymnasium, 4th grade)

Remarks. 1. At first glance, it would be more convenient to choose the lower left corner of the chessboard as the origin of the coordinate system. As a result of the more complicated calculation, we would have found the coordinates to be 1/2 smaller:

$$
x_{T}^{\prime}=y_{T}^{\prime}=x_{T}-\frac{1}{2}=\frac{7 n^{2}-1}{12 n}, \quad x_{U}^{\prime}=y_{U}^{\prime}=\frac{4 n-1}{6}
$$

Zoltán Papp

2. The sum $\sum m$ can also be simply formed by grouping parallel to any of the diagonals.