## Task A-1.1.

On the side $\overline{A B}$ of triangle $A B C$, there are points $P_{1}, P_{2}$ and $P_{3}$ such that

$$
\left|A P_{1}\right|=\left|P_{1} P_{2}\right|=\left|P_{2} P_{3}\right|=\left|P_{3} B\right|=\frac{1}{4}|A B|
$$

Parallel lines to side $\overline{B C}$ are drawn through these points, dividing the triangle into four parts. The area of the part located between the parallels through $P_{2}$ and $P_{3}$ is 5.

What is the area of triangle $A B C$?
## First Solution.

Let $Q_{1}, Q_{2}$, and $Q_{3}$ be the points where the parallels through $P_{1}, P_{2}$, and $P_{3}$ intersect the side $\overline{C A}$, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-01.jpg?height=653&width=1099&top_left_y=1421&top_left_x=410)

Since $B C \| P_{i} Q_{i}$ for $i=1,2,3$, we conclude that

$$
\varangle A B C=\varangle A P_{1} Q_{1}=\varangle A P_{2} Q_{2}=\varangle A P_{3} Q_{3}
$$

Furthermore, triangles $A B C, A P_{1} Q_{1}, A P_{2} Q_{2}$, and $A P_{3} Q_{3}$ share the same angle at vertex $A$, so by the $\mathrm{K}-\mathrm{K}$ similarity theorem, we conclude that these triangles are similar.

Let $X$ and $Y$ be the areas of triangle $A P_{1} Q_{1}$ and quadrilateral $P_{1} P_{2} Q_{2} Q_{1}$, respectively.

Since $|A B|=4\left|A P_{1}\right|$, the area of triangle $A B C$ is $16 X$.

Since $\left|A P_{2}\right|=2\left|A P_{1}\right|$, it follows that $X+Y=4 X$.

1 point

Furthermore, since $\left|A P_{3}\right|=3\left|A P_{1}\right|$, we have $X+Y+5=9 X$.

1 point

From the last two equations, we get $4 X+5=9 X$, i.e., $X=1$. 3 points

Therefore, the area of triangle $A B C$ is 16.