[ Properties and characteristics of a parallelogram ] [ Vectors help solve the task. ]

Given points $A(-6; -1), B(1; 2)$, and $C(-3; -2)$. Find the coordinates of vertex $M$ of the parallelogram $A B M C$.
The first method.

The coordinates of the midpoint $K\left(x_{0} ; y_{0}\right)$ of the diagonal $B C$ of the parallelogram $A B M C$ are the arithmetic means of the corresponding coordinates of the endpoints of the segment $B C$, i.e.,

$$
x_{0}=\frac{1-3}{2}=-1, y_{0}=\frac{2-2}{2}=0
$$

Since the diagonals of a parallelogram bisect each other, $K\left(x_{0} ; y_{0}\right)$ is the midpoint of the segment with endpoints at $A(-6 ;-1)$ and $M\left(x_{1} ; y_{1}\right)$. Therefore,

$$
x_{0}=\frac{-6+x_{1}}{2}=-1, y_{0}=\frac{-1+y_{1}}{2}=0
$$

From this, we find that $x_{1}=4, y_{1}=1$.

The second method.

Let $x_{1}, y_{1}$ be the coordinates of point $M$. If $A B M C$ is a parallelogram, then $\overrightarrow{B M}=\overrightarrow{A C}$, and since

$$
\overrightarrow{B M}=\overrightarrow{\left(x_{1}-1 ; y_{1}-2\right)}, \overrightarrow{A C}=\overrightarrow{(-3-(-6) ;-2-(-1)}=\overrightarrow{(3 ; 1)}
$$

then

$$
x_{1}-1=3, y_{1}-2=-1
$$

From this, we find that $x_{1}=4, y_{1}=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-39.jpg?height=510&width=800&top_left_y=2418&top_left_x=640)

## [ Dirichlet's Principle (finite number of points, lines, etc.). ]  [ Tables and tournaments (other). ] Colorings  [ Cardinality of sets. Bijective mappings ] Problem 21979 Topics: Estimation + example  a) What is the maximum number of fields on an $8 \times 8$ board that can be painted black so that in each corner of three fields, there is at least one unpainted field?  b) What is the minimum number of fields on an $8 \times 8$ board that can be painted black so that in each corner of three fields, there is at least one black field?

## Solution

a) Example. 32 fields can be painted (for example, in stripes).

Estimate. Divide the board into 16 squares $2 \times 2$. In each of them, no more than two fields can be painted (otherwise, a black corner will appear). Therefore, a total of no more than 32 fields can be painted.

b) The problem reduces to a) by replacing black fields with unpainted and vice versa.

## Answer

a)-b) 32 fields.