## Task B-3.6.

A rectangle is given with side lengths $a$ and $b$, where $a = \frac{4}{7} b$. Determine the probability that a randomly selected point within the given rectangle is closer to the shorter side and is less than $\frac{a}{8}$ away from it.
Solution.

The lines $A I, B J, C G$ and $D F$ are the angle bisectors of the angles at vertices $A, B, C$ and $D$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-19.jpg?height=654&width=1108&top_left_y=427&top_left_x=385)

Since every point on the angle bisector is equidistant from its sides, the points closer to side $a$ are located in the eastern part as shown in the sketch of the rectangle.

The points that are less than $\frac{a}{8}$ away from side $a$ are the points within the congruent trapezoids $A B J I$ and $C D F G$.

Triangle $A B E$ is an isosceles right triangle, so its height is $\frac{a}{2}$.

Since triangles $A B E$ and $I J E$ are similar with a similarity coefficient $k=\frac{\frac{3 a}{8}}{\frac{a}{2}}=\frac{3}{4}$, we conclude that the length of segment $\overline{I J}$ is $\frac{3}{4} a$.

Let's calculate the sum of the areas of these trapezoids $A B J I$ and $C D F G$.

$$
P_{A B J I}+P_{C D F G}=2 P_{A B J I}=2 \cdot \frac{a+\frac{3}{4} a}{2} \cdot \frac{a}{8}=\frac{7}{32} a^{2}
$$

The probability of the desired event is:

$$
p=\frac{\frac{7}{32} a^{2}}{a b}=\frac{7 a}{32 b}=\frac{7}{32} \cdot \frac{4}{7}=\frac{1}{8}
$$

Note: For the first 6 points, the student can clearly mark the angle bisectors and the region favorable for the desired event on the diagram instead of describing it.

If the student has not drawn the angle bisectors and incorrectly marked the region, but has accurately calculated the probability that a point belongs to their marked region, the student can receive a maximum of 2 points.