| ethics. Mental arithmetic, etc. |
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On a straight line, ten points were placed at equal intervals, and they occupied a segment of length $a$. On another straight line, one hundred points were placed at the same intervals, and they occupied a segment of length $b$. How many times is $b$ greater than $a$?
On a segment of length $a$, 9 intervals fit, and on a segment of length $b-99$. Therefore, $b=11a$.

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-18.jpg?height=108&width=1584&top_left_y=229&top_left_x=0)

Author: Folkpor

In a hexagon, five angles are $90^{\circ}$, and one angle is $-270^{\circ}$ (see the figure). Using a ruler without divisions, divide it into two equal-area polygons.

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-18.jpg?height=357&width=600&top_left_y=502&top_left_x=732)

## Solution

First method. Extend the hexagon $A B C D E F$ to a rectangle $A B M F$ (see Fig. a). We use the fact that any line passing through the center of symmetry of a figure divides it into two equal parts. Therefore, if we draw a line through the points of intersection of the diagonals of rectangles $A B M F$ and $C M E D$, it will divide each of these rectangles into two equal-area parts. Consequently, this line will divide the given hexagon into two equal-area polygons.

Comment. Another possible "solution" (see Fig. b). But for the figure given in the problem, this division results in 3 parts.

Since the problem does not specify that the cut must be a straight line, this "solution" can be easily corrected as follows.

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-18.jpg?height=411&width=626&top_left_y=1438&top_left_x=41)

Fig. a

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-18.jpg?height=409&width=600&top_left_y=1436&top_left_x=729)

Fig. b

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-18.jpg?height=403&width=640&top_left_y=1445&top_left_x=1372)

Fig. c

Second method. Extend $C D$ to intersect $A F$ at point $G$ (see Fig. c), and extend $E D$ to intersect $A B$ at point $H$. Find the center $O$ of rectangle $D E F G$ by drawing its diagonals. Similarly, find the center $O_{1}$ of rectangle $A G D H$. The line $O O_{1}$ intersects $D G$ at some point $P$. It remains to find the center $O_{2}$ of rectangle $A B C G$ and draw the line $P O_{2}$. By the reasoning outlined in the first solution, segment $X P$ divides rectangle $A B C G$ into two equal parts, and therefore, the broken line $X P Y$ will divide the given hexagon into two equal-area parts.