## Task B-2.4.

The lengths of the sides of triangle $A B C$ are $|B C|=27,|A C|=36,|A B|=45$. Let $P$ and $Q$ be points on sides $\overline{A B}$ and $\overline{B C}$, respectively, such that $|A P|=|B Q|$ and $A C \| P Q$. Calculate the area of triangle $P B Q$.
## First Solution.

With the labels as in the figure, $\varangle B P Q = \varangle B A C$ (angles with parallel sides), so $\triangle A B C \sim \triangle P B Q$.

![](https://cdn.mathpix.com/cropped/2024_05_30_0c173970003c85556d5eg-09.jpg?height=451&width=868&top_left_y=2130&top_left_x=523)

Then, $45:(45-y) = 27: y$, which simplifies to $y = \frac{27 \cdot 45}{27 + 45} = \frac{135}{8}$.

The similarity ratio is $k = \frac{y}{27} = \frac{5}{8}$.

1 point

We have $P(\triangle P B Q) = k^2 P(\triangle A B C)$.

The area of triangle $A B C$ is calculated using Heron's formula:

\[
\begin{aligned}
s & = \frac{1}{2}(a + b + c) = \frac{1}{2}(27 + 36 + 45) = 54 \\
P(\triangle A B C) & = \sqrt{s(s - a)(s - b)(s - c)} \\
& = \sqrt{54 \cdot 27 \cdot 18 \cdot 9} = \sqrt{27 \cdot 2 \cdot 27 \cdot 2 \cdot 9 \cdot 9} = 27 \cdot 2 \cdot 9 = 486
\end{aligned}
\]

Thus, $P(\triangle P B Q) = \left(\frac{5}{8}\right)^2 \cdot 486 = \frac{6075}{32}$.

1 point