Task B-2.8. In an isosceles triangle $ABC$ with a base length of $18 \, \text{cm}$ and a leg length of $41 \, \text{cm}$, inscribe an isosceles triangle $DEF$ of maximum area, such that the bases of the two triangles are parallel, and the vertex of the inscribed triangle is at the midpoint of the base of the given triangle. Determine the lengths of the sides of triangle $DEF$.

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The task is to inscribe an isosceles triangle $DEF$ with maximum area in an isosceles triangle $ABC$ where the base $BC$ is $18 \, \text{cm}$ and the legs $AB$ and $AC$ are $41 \, \text{cm}$. The bases of the two triangles are parallel, and the vertex $D$ of the inscribed triangle $DEF$ is at the midpoint of the base $BC$ of the given triangle $ABC$. Determine the lengths of the sides of triangle $DEF$.
Solution. Sketch

![](https://cdn.mathpix.com/cropped/2024_05_30_cf2c4989ba380bc8684dg-11.jpg?height=726&width=358&top_left_y=1071&top_left_x=843)

The height $|CF|$ of the isosceles triangle $ABC$ is $\sqrt{41^2 - 9^2} = 40 \text{ cm}$. Let $z$ be the height of triangle $DEF$. From the similarity of triangles $ABC$ and $EDC$, we have:

\[
\begin{aligned}
18: 40 & = x: (40 - z) \\
z & = \frac{360 - 20x}{9}
\end{aligned}
\]

The area of triangle $DEF$ is

\[
\begin{aligned}
P & = \frac{xz}{2} \\
P & = \frac{x(360 - 20x)}{18} \\
P & = -\frac{10}{9} x^2 + 20x
\end{aligned}
\]

The maximum area is achieved for

\[
x = -\frac{b}{2a} = \frac{20}{\frac{20}{9}} = 9 \text{ cm}
\]

The length of the leg of this triangle is

\[
y = \sqrt{\left(\frac{x}{2}\right)^2 + z^2} = \sqrt{\left(\frac{9}{2}\right)^2 + 20^2} = \frac{41}{2} \text{ cm}
\]

Note: Award 1 point for the correct solution.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

14th February 2012.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.