The cross-section of a building's floor is a symmetric pentagon on a vertical axis, which can be divided into a rectangle and an isosceles triangle standing on it. The common base of the two parts is $2a$, their heights are $b$ and $m$ respectively. For the purpose of designing a residential room to be built, determine the base, height, and area of the largest rectangle that can be inscribed in the pentagon. - Construct the vertices of the rectangle.
1 It is obvious that the inscribed rectangle $T$ will also be symmetric with respect to the axis of symmetry of the pentagon $K$ cross-section, with two vertices of $T$ lying on the slanted sides of $K$, allowing for their lower endpoints as well.

![](https://cdn.mathpix.com/cropped/2024_05_02_a1050abd3da3e772c893g-1.jpg?height=488&width=585&top_left_y=271&top_left_x=759)

Let the base of $T$ be $x$, and its height be $y$. The triangle above $T$ is similar to the triangular upper part of $K$, hence

$$
x:(b+m-y)=2 a: m
$$

From this, the base of $T$, and then the area $t$ of $T$ are

$$
\begin{gathered}
x=\frac{2 a}{m}(b+m-y) \\
t=x y=\frac{2 a}{m}\left[(b+m) y-y^{2}\right]=\frac{2 a}{m}\left[\left(\frac{b+m}{2}\right)^{2}-\left(y-\frac{b+m}{2}\right)^{2}\right]= \\
=\frac{a(b+m)^{2}}{2 m}-\frac{2 a}{m}\left(y-\frac{b+m}{2}\right)^{2}
\end{gathered}
$$

This takes its maximum value when the subtrahend is 0, i.e., if

$$
y=\frac{b+m}{2}, \quad \text { and then } \quad x=\frac{a}{m}(b+m), \quad t_{\max }=\frac{a(b+m)^{2}}{2 m}
$$

Our result is valid only if the found height is greater than the height $b$ of the lower part of $K$:

$$
y \geq b, \quad \frac{b+m}{2} \geq b, \quad \text { i.e., } \quad m \geq b
$$

If $m < b$, the problem needs to be addressed differently. For the rectangle $T$ with area $t_{\max }$, one of its upper vertices, $C$, can be determined by the direction from the midpoint $F$ of the lower side $AB$. Denoting the angle of inclination by $\alpha$,

$$
\operatorname{tg} \alpha=\frac{y}{x / 2}=\frac{2 y}{x}=(b+m): \frac{a}{m}(b+m)=\frac{m}{a}
$$

which is equal to the angle of inclination of the slanted sides of $K$. Thus, $FC$ is parallel to the other slanted side of $K$.

Tóth Tibor (Szolnok, Verseghy F. Gymn., III. grade) Schlosser Attila (Esztergom, Temesvári Pelbárt Gymn., IV. grade)

Note. Several people completed $K$ to a triangle by extending its slanted sides to the base of the rectangle, then sought the rectangle with the largest area inscribed in this triangle, with the condition that it must also be inside the original $K$.[^0]


[^0]:    ${ }^{1}$ We were saddened to learn of the passing of the problem setter, Theodor Kasper, who was a lecturer at the Meisseni (NDK) Technical College and a dear friend of our journal. He was keenly interested in the work of our readers and in competitions for students, providing ideas for new problems. We remember him with fondness. (The Editorial Board)