In triangle $ABC$, a line parallel to the base is drawn through an arbitrary point $G$ on the altitude $AD$, intersecting $AB$ at $E$ and $AC$ at $F$. On the extension of $EF$, we measure $EA$ from $E$ and $AF$ from $F$, such that $EH = EA$ and $FI = AF$. Finally, we connect $H$ to $B$ and $I$ to $C$. Given the sides of the triangle $a, b, c$, and $AD = h$, and $AG = x$, the questions are:

$1^{\circ}$. What is the area of trapezoid $BHI C$?

$2^{\circ}$. Determine $x$ such that the area of the trapezoid is maximized.
![](https://cdn.mathpix.com/cropped/2024_05_02_966e1e76262838051677g-1.jpg?height=393&width=542&top_left_y=129&top_left_x=781)

$1^{\circ}$. The area of the trapezoid:

$$
\begin{aligned}
& t=\frac{1}{2}(B C+H I) \cdot D G \\
& B C=a, D G=h-x \\
& H I=A E+E F+F A
\end{aligned}
$$

Since

$$
A E F \triangle \sim A B C \triangle
$$

it follows that

$$
A E+E F+F A: a+b+c=x: h
$$

from which

$$
A E+E F+F A=H I=\frac{(a+b+c) x}{h}
$$

and thus

$$
t=\frac{1}{2}\left(a+\frac{(a+b+c) x}{h}\right)(h-x)
$$

$2^{\circ}$. (1) can also be written as:

$$
t=-\frac{a+b+c}{2 h} x^{2}+\frac{b+c}{2} x+\frac{a h}{2}
$$

$t$ is maximized when

$$
x=\frac{(b+c) h}{2(a+b+c)}
$$

(János Prohászka, Esztergom.)

The problem was also solved by: Devecis M., Döme B., Juvancz I., Kohn B., Krisztián Gy., Rehberger Z., Sasvári G., Spitzer Ö.