The distance $AD = d$. On this, we mark the distance $AB = x$ (variable) and construct the equilateral triangle $ABC$. At point $C$, we erect a perpendicular to $BC$, and at point $D$, we erect a perpendicular to $AD$, which intersect at point $E$. Examine the variation of the area of the quadrilateral $ACED$ as $0 \leq x \leq d$.
I. Solution.

The quadrilateral $A C E D$ is divided into three triangles: $\triangle A B C, \triangle B D E, \triangle B C E$.

The area of $\triangle A B C$ is $t_{1}=\frac{x^{2} \sqrt{3}}{4}$.

The altitudes of $\triangle A B C$ intersect at point $O$. The altitude to side $B C$ intersects $D E$ at point $F$. Since $C O \| D E$ and $A F \| C E$, the quadrilateral $C O F E$ is a parallelogram, thus

$$
E F=C O=\frac{2}{3} C H=\frac{2}{3} \cdot \frac{x \sqrt{3}}{2}=\frac{x}{\sqrt{3}}
$$

and

$$
C E=O F=A F-A O
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_3020345954be729e1175g-1.jpg?height=461&width=372&top_left_y=642&top_left_x=863)

Since $\angle D A F = 30^{\circ}$, $A F = A D : \frac{\sqrt{3}}{3} = \frac{2 d}{\sqrt{3}}$ and $D F = \frac{d}{\sqrt{3}}$.

Therefore,

$$
C E=\frac{2 d}{\sqrt{3}}-\frac{x}{\sqrt{3}}=\frac{2 d-x}{\sqrt{3}}
$$

The area of the right triangle $\triangle B D E$ is

$$
t_{2}=\frac{B D \cdot D E}{2}=\frac{d-x}{2} \cdot \frac{d+x}{\sqrt{3}}=\frac{d^{2}-x^{2}}{2 \sqrt{3}}
$$

The area of the right triangle $\triangle B C E$ is

$$
t_{3}=\frac{B C \cdot C E}{2}=\frac{x(2 d-x)}{2 \sqrt{3}}
$$

The area of the quadrilateral $A C E D$ is

$$
\begin{aligned}
t=t_{1}+t_{2}+t_{3}= & \frac{x^{2} \sqrt{3}}{4}+\frac{d^{2}-x^{2}}{2 \sqrt{3}}+\frac{2 d x-x^{2}}{2 \sqrt{3}}=-\frac{x^{2}}{4 \sqrt{3}}+\frac{d x}{\sqrt{3}}+\frac{d^{2}}{2 \sqrt{3}}= \\
& =\frac{2 d^{2}+4 d x-x^{2}}{4 \sqrt{3}}=\frac{6 d^{2}-(2 d-x)^{2}}{4 \sqrt{3}}
\end{aligned}
$$

As $x$ increases from 0 to $d$, $(2 d-x)^{2}$ decreases continuously from $4 d^{2}$ to $d^{2}$, so the value of $t$ increases continuously from $\frac{d^{2}}{2 \sqrt{3}}$ to $\frac{5 d^{2}}{4 \sqrt{3}}$. The change is represented by an ascending arc of a parabola, whose upper vertex is at $x=2 d$. Kálmán Mogyoróssy (Br. Kemény Zsigmond g. VI. o. Bp. VI.).

II. Solution. As $x$ varies from 0 to $d$, the vertex $C$ moves along the fixed line $A L$ from $A$ to $C'$. The line $A L$ forms a $60^{\circ}$ angle with $A D$ and $A C' = A D = d$. The line $C E$ remains parallel to the altitude $A F$ and its boundary positions are $A F$ and $C' E'$.

The initial area of the quadrilateral $A C E D$ is equal to the area of $\triangle A D F$, which is $\frac{d}{2} \cdot \frac{d}{\sqrt{3}}=\frac{d^{2}}{2 \sqrt{3}}$.

The other boundary value of the quadrilateral's area is the area of $A C' E' D$. This consists of two parts: the equilateral triangle $\triangle A C' D$ and the right triangle $\triangle D C' E'$.

The area of the former is $\frac{d^{2} \sqrt{3}}{4}$, and the area of the latter is equal to that of $\triangle A D F$, which is $\frac{d^{2}}{2 \sqrt{3}}$.

The maximum area of the quadrilateral is:

$$
\frac{a^{2} \sqrt{3}}{4}+\frac{d^{2}}{2 \sqrt{3}}=\frac{3 d^{2}}{4 \sqrt{3}}+\frac{2 d^{2}}{4 \sqrt{3}}=\frac{5 d^{2}}{4 \sqrt{3}}
$$

Dezső Szittyai (Wágner g. V. o. Rákospalota).