After the impact of a pebble, a glass pane cracked in such a way that four smaller triangles with a common vertex at the point of impact were formed. Moreover, it is given that:

- the glass pane had the shape of a rectangle that was $8 \mathrm{dm}$ wide and $6 \mathrm{dm}$ high,
- the triangle on the right had three times the area of the triangle on the left,
- the triangle on the left had half the area of the triangle at the bottom.

Determine the distances of the impact point from the sides of the rectangle.

(E. Novotná)
The triangle on the right and the triangle on the left have one side of the same length, which is a side of the rectangle, and this side is $6 \mathrm{dm}$ long. Since the triangle on the right

![](https://cdn.mathpix.com/cropped/2024_04_17_bf37c56bc0e241535d40g-2.jpg?height=380&width=506&top_left_y=1809&top_left_x=775)

has three times the area of the one on the left, the heights of these triangles to the mentioned side must be in the same ratio. The sum of these heights is equal to the width of the rectangle, which is $8 \mathrm{dm}$. Therefore, the height of the left triangle, i.e., the distance from the impact point to the left side of the rectangle, is $2 \mathrm{dm}$, and the height of the right triangle, i.e., the distance from the impact point to the right side of the rectangle, is $6 \mathrm{dm}$.
From the above, it follows that the left triangle has an area of $6 \mathrm{dm}^{2}\left(\frac{6 \cdot 2}{2}=6\right)$. The triangle at the bottom has an area twice as large, i.e., $12 \mathrm{dm}^{2}$. The side of the bottom triangle, which is also a side of the rectangle, is $8 \mathrm{dm}$ long. This means that the corresponding height of this triangle, i.e., the distance from the impact point to the bottom side of the rectangle, must be $3 \mathrm{dm}\left(\frac{8 \cdot 3}{2}=12\right)$. The height of the rectangle is $6 \mathrm{dm}$, so the distance from the impact point to the top side of the rectangle is also $3 \mathrm{dm}$ $(6-3=3)$.

Another solution. Divide the rectangle into four smaller rectangles by lines passing through the impact point, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_bf37c56bc0e241535d40g-3.jpg?height=380&width=506&top_left_y=667&top_left_x=775)

Each of these smaller rectangles is divided into two congruent triangles by the cracks. It follows that the sum of the areas of the left and right triangles is the same as the sum of the areas of the top and bottom triangles. This sum is half the area of the entire rectangle, i.e., $24 \mathrm{dm}^{2}$. From the second condition in the problem, it follows that the area of the triangle on the left must be $6 \mathrm{dm}^{2}$ and the area of the one on the right must be $18 \mathrm{dm}^{2}$. From the third condition in the problem, it follows that the area of the triangle at the bottom is $12 \mathrm{dm}^{2}$. The area of the triangle at the top is therefore the same.

For all triangles, we know the area and one side. The corresponding heights, i.e., the distances from the impact point to the individual sides of the rectangle, can now be determined in the same way as in the second part of the previous solution.

Evaluation. 3 points for the distances from the impact point to the left and right sides of the rectangle; 3 points for the distances from the impact point to the bottom and top sides of the rectangle. A correct solution without comments is worth a maximum of 3 points.