A trapezoid's base is $a$, the sum of the other three sides is $d$. What are the lengths of the sides if the area of the trapezoid is maximized?
A cyclic trapezoid is a trapezoid whose sides are chords of a circle. Since the opposite angles of a cyclic quadrilateral sum to $180^{\circ}$, it follows that the trapezoid is isosceles.

Let the larger base of the trapezoid be $AB = a$, the smaller base $DC = c$, and the legs $AD = BC = b$ (Figure 1).

According to the condition:

$$
d = 2b + c
$$

The area of the trapezoid is: $T = \frac{a + c}{2} \cdot m$. By the Pythagorean theorem:

$$
m = \sqrt{b^2 - \left(\frac{a - c}{2}\right)^2}
$$

From (1), $b = \frac{d - c}{2}$; substituting these values, we get:

$$
T = \frac{a + c}{2} \sqrt{\left(\frac{d - c}{2}\right)^2 - \left(\frac{a - c}{2}\right)^2}
$$

Transform the expression, using the known identity $x^2 - y^2 = (x + y)(x - y)$:

$$
T = \frac{a + c}{4} \sqrt{(d + a - 2c)(d - a)}
$$

The value of $(d - a)$ is constant and does not affect the maximum value; $d + a = k$ is the perimeter of the trapezoid. Further transforming the expression, we get:

$$
T = \frac{1}{4} \sqrt{d - a} \sqrt{(a + c)(a + c)(k - 2c)}
$$

Each factor under the second square root is positive, and we can apply the arithmetic mean-geometric mean inequality to the product of three factors:

$$
\begin{gathered}
\sqrt[3]{(a + c)(a + c)(k - 2c)} \leq \\
\leq \frac{(a + c) + (a + c) + (k - 2c)}{3} = \frac{k + 2a}{3}
\end{gathered}
$$

The product is maximized when the factors are equal, i.e., $a + c = k - 2c = d + a - 2c$, from which $3c = d$, and thus $c = \frac{d}{3}$ and $b = \frac{d - c}{2} = \frac{d}{3}$, meaning the legs and the shorter base have the same length.

Such a trapezoid exists if $d > a$ (Figure 2).

The opposite angles are equal because they correspond to equal-length chords, so $AB \parallel CD$, meaning the quadrilateral is indeed a trapezoid.

Katalin Ásványi (Ajka, Bródy I. Gymnasium, 11th grade)

Remarks. 1. The problem can also be solved as an extremum problem using derivatives (see the online solution outline).

2. The unusually high number of zero-point solutions is due to the following incorrect conclusions:

a) "Since the perimeter of the quadrilateral is given, and among quadrilaterals with a given perimeter, the square has the largest area, the quadrilateral must be a square." This reasoning is flawed because not only the perimeter but also one side is given, so (except for the case $d = 3a$) the trapezoid cannot be a square.

b) "Reflecting the cyclic trapezoid over its longer parallel side, we get a hexagon. For a given perimeter, the regular hexagon has the largest area, so the trapezoid has three equal sides, each of length $\frac{d}{3}$." This error is similar to the previous one: the resulting trapezoid has not only a given perimeter but also a predetermined diagonal, so we cannot refer to the general extremum problem.

c) Some students decomposed the area into two independent terms and tried to find the maximum of each term separately, and from this, they attempted to determine the common maximum.

![](https://cdn.mathpix.com/cropped/2024_05_02_fead467f9254c86dacf1g-1.jpg?height=436&width=480&top_left_y=2254&top_left_x=817)

![](https://cdn.mathpix.com/cropped/2024_05_02_fead467f9254c86dacf1g-2.jpg?height=480&width=488&top_left_y=123&top_left_x=816)