## Task B-3.3.

An isosceles triangle $A B C$ with base $\overline{B C}, |B C|=20$, and legs of length 18 is divided by the line segment $\overline{D E}$ into two parts of equal perimeters and areas. Point $D$ is on the base, and point $E$ is on the leg $\overline{A C}$, and both are different from the vertices of the triangle. Determine the length of $|D C|$.
## Solution.

Let $x=|D C|$. Denote the angle at vertex $C$ as $\gamma$. Since the perimeter of triangle $A B C$ is divided into two equal parts, we have $|D C|+|C E|=28$, so $|C E|=28-x$.

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-16.jpg?height=562&width=896&top_left_y=1952&top_left_x=497)

\[
\begin{aligned}
P_{D C E} & =\frac{1}{2} x \cdot(28-x) \cdot \sin \gamma \\
P_{A B C} & =\frac{1}{2} \cdot 20 \cdot 18 \cdot \sin \gamma
\end{aligned}
\]

Since $P_{A B C}=2 P_{C D E}$, we have

\[
\frac{1}{2} \cdot 20 \cdot 18 \cdot \sin \gamma=2 \cdot \frac{1}{2} x \cdot(28-x) \cdot \sin \gamma
\]

which simplifies to $x(28-x)=180$.

Solving the quadratic equation $x^{2}-28 x+180=0$ gives the solutions $x=10$ and $x=18$.

For $x=10$, we have $28-x=18$, so point $E$ coincides with vertex $A$. Therefore, according to the conditions of the problem, the only solution is $x=18$.