The sum of the two legs and one of the parallel sides of an isosceles trapezoid is $d$, and the angle at the other parallel side is $\alpha$. Construct the isosceles trapezoid with the largest possible area from these data.
I. solution: We can solve our task by calculation. Let's denote the length of the leg of an arbitrary isosceles trapezoid that meets the conditions by $x$. The area $t(x)$ of the trapezoids, which depends on $x$, can be decomposed into a rectangle and two right-angled triangles (Figure 1):

![](https://cdn.mathpix.com/cropped/2024_05_02_7473a435e91d9bcf75feg-1.jpg?height=330&width=509&top_left_y=323&top_left_x=816)

Figure 1

$$
t(x)=(d-2 x) x \sin \alpha + x^{2} \sin \alpha \cos \alpha = (\sin \alpha \cos \alpha - 2 \sin \alpha) x^{2} + (d \sin \alpha) x
$$

(The formula correctly gives the area even for $\alpha \geq 90^{\circ}$, because in this case the second term is 0 or negative.)

This is a quadratic function in $x$, where the coefficient of the quadratic term is $\sin \alpha(\cos \alpha - 2) < 0$, because the first factor is positive in the interval $0 < \alpha < 180^{\circ}$, and the second factor is negative.

Therefore, the function $t(x)$ has a maximum at

$$
x_{0}=\frac{-d \sin \alpha}{2 \sin \alpha(\cos \alpha - 2)}=\frac{d}{2(2-\cos \alpha)}
$$

Accordingly, the leg $x_{0}$ can be constructed as the fourth proportional from the ratio pair

$$
(2-\cos \alpha): \frac{d}{2}=1: x_{0}
$$

(Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_7473a435e91d9bcf75feg-1.jpg?height=708&width=632&top_left_y=1389&top_left_x=752)

Figure 2

$$
\alpha=90^{\circ} \quad \text { in which case } \quad x=\frac{d}{4}
$$

thus the trapezoid becomes a rectangle with sides $\frac{d}{4}$ and $\frac{d}{2}$.

Zoltán Daróczy (Debrecen, Ref. g. Ill. o. t.)

II. solution: The problem can also be solved purely geometrically. If we cut a trapezoid that meets the conditions (other than having the maximum area) with a line through the midpoints of the parallel sides, we get a right-angled trapezoid where the sum of the slanted side and one of the parallel sides is $\frac{d}{2}$, and the other parallel side has an angle $\alpha$. We need to determine the one with the largest area from these.

![](https://cdn.mathpix.com/cropped/2024_05_02_7473a435e91d9bcf75feg-2.jpg?height=537&width=602&top_left_y=127&top_left_x=751)

Figure 3

Let $ABCD$ be a trapezoid with the desired properties (Figure 3). Extend the slanted side $AD$ by the distance $DC$. Thus, $AE=\frac{d}{2}$. Let $EC$ intersect the line $AB$ at $F$. Then, the opposite vertex of all trapezoids with the desired properties lies on the segment $EF$. Draw the midpoint $G$ of the perpendicular side $BC$. All such midpoints lie on a line starting from $F$. Let this line intersect $AE$ at $H$ and the side $CD$ of the trapezoid $ABCD$ or its extension at $K$. By reflecting the triangle $GCK$ over $G$, we see that the area of the trapezoid $ABCD$ is smaller than the area of the triangle $AFH$ by the area of the triangle $DHK$. Thus, the trapezoid with the largest area is obtained by drawing a line parallel to $AF$ from $H$ and then drawing a perpendicular from the intersection of this line with $EF$ to $AF$.

This procedure is valid regardless of whether $\alpha$ is an acute, right, or obtuse angle.

Remark: Some solvers thought the construction was feasible by constructing the isosceles trapezoid with the maximum area and angle $\alpha$ in a circle of arbitrary radius and then constructing a smaller or larger similar trapezoid corresponding to the given $d$. Of course, this is incorrect, as it solves the problem for the case where the radius of the circle circumscribed around the trapezoid is constant along with $\alpha$. In this case, however, changing the leg also changes $d$, whereas in the given problem, where $d$ is constant, changing the leg changes the radius of the circle circumscribed around the trapezoid.