## Task B-1.6.

Point $P$ is located inside an equilateral triangle such that it is at distances of $3 \mathrm{~cm}, 4 \mathrm{~cm}$, and $5 \mathrm{~cm}$ from its sides, respectively. A line parallel to the side of the triangle closest to point $P$ is drawn through point $P$. Calculate the length of the side of the given triangle and the ratio in which the drawn parallel divides the area of the triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-04.jpg?height=631&width=762&top_left_y=1312&top_left_x=550)

(The distances from point $P$ to the sides of the triangle are scored)

From point $P$, we draw lines to the vertices of the triangle. If we denote the side length of the triangle by $a$, the area of the desired triangle can be calculated as

$$
P=\frac{1}{2} a \cdot 3+\frac{1}{2} a \cdot 4+\frac{1}{2} a \cdot 5=6 a \text {. }
$$

Since the given triangle is equilateral, the area can also be calculated using the formula $P=\frac{a^{2} \sqrt{3}}{4}$.

Therefore, $\frac{a^{2} \sqrt{3}}{4}=6 a$.

From this, the side length of the triangle is $a=\frac{24}{\sqrt{3}} \text{ cm} = 8 \sqrt{3} \text{ cm}$,

and the height of the triangle is $\frac{a \sqrt{3}}{2}=12 \text{ cm}$.

1 point

A line parallel to side $\overline{A C}$ is drawn through point $P$. It divides the triangle $A B C$ into an equilateral triangle $D B E$ with a height of $12 \text{ cm} - 3 \text{ cm} = 9 \text{ cm}$ and an isosceles trapezoid $A D E C$ with a height of $3 \text{ cm}$. The side length of triangle $D B E$ can be calculated from the height:

$$
\frac{x \sqrt{3}}{2}=9, \quad x=\frac{18}{\sqrt{3}}=6 \sqrt{3} \text{ cm}
$$

The area of triangle $D B E$ is $P_{1}=(6 \sqrt{3})^{2} \cdot \frac{\sqrt{3}}{4}=27 \sqrt{3} \text{ cm}^{2}$.

The area of trapezoid $A D E C$ is $P_{2}=\frac{8 \sqrt{3}+6 \sqrt{3}}{2} \cdot 3=21 \sqrt{3} \text{ cm}^{2}$.

The desired ratio is $\frac{P_{1}}{P_{2}}=\frac{27 \sqrt{3}}{21 \sqrt{3}}=\frac{9}{7}$ or $\frac{P_{2}}{P_{1}}=\frac{7}{9}$.