Four given segments have a common starting point. In what arrangement (in a plane) of these segments will the area of the quadrilateral defined by their endpoints as vertices be the largest?
I. solution. Let the common starting point of the given segments be $O$, and their endpoints in a certain position be $A, B, C, D$, labeling the segments so that $A B C D$ does not form a crossed quadrilateral.

![](https://cdn.mathpix.com/cropped/2024_05_02_a4ab403ca9a6042cc7fdg-1.jpg?height=458&width=590&top_left_y=243&top_left_x=754)

The area of this, as is well known (Figure 2),

$$
t=\frac{1}{2} \cdot A C \cdot B D \cdot \sin \varphi
$$

where $\varphi$ is the angle between the diagonals $A C$ and $A D$.

By changing the relative positions of the segments, it always holds that

$$
A C \leq A O+O C, \quad B D \leq B O+O D, \quad(0 \leq) \sin \varphi \leq 1
$$

thus

$$
t \leq \frac{1}{2}(A O+O C)(B O+O D)
$$

Here, equality holds when the factors in (2) independently reach their maximum values, i.e., if the segments $O A$ and $O C$, as well as $O B$ and $O D$, lie on each other's extensions—meaning that $O$ becomes the intersection point of the diagonals $A C$ and $B D$—and if $\varphi=90^{\circ}$, i.e., the diagonals are perpendicular to each other. In this case, the maximum value of $t$ depends only on the lengths of the segments and their pairing. 

The 4 segments can be paired in three different ways, because the first chosen segment can be extended by any of the remaining 3, and then the remaining 2 segments form the other pair. Let the lengths of the 4 segments be $p, q, r, s$ such that $p \geq q \geq r \geq s(>0)$, so the maximum value of the double area is one of the following 3 products:

$$
2 t_{1}=(p+q)(r+s), \quad 2 t_{2}=(p+r)(q+s), \quad 2 t_{3}=(p+s)(q+r)
$$

Subtracting the first from the third, and then the second, the difference can be transformed as follows:

$$
2\left(t_{3}-t_{1}\right)=(p-r)(q-s), \quad 2\left(t_{3}-t_{2}\right)=(p-q)(r-s)
$$

Neither of these is negative, so $t_{3} \geq t_{1}, t_{3} \geq t_{2}$, thus the largest area quadrilateral is obtained when the longest segment is paired with the shortest segment (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_a4ab403ca9a6042cc7fdg-1.jpg?height=401&width=461&top_left_y=1797&top_left_x=775)

Figure 3

II. solution. With the above notation, the areas of the triangles $O A B, O B C, O C D$, and $O D A$ are each maximized when the two sides starting from $O$ are perpendicular to each other. These 4 conditions can be satisfied simultaneously by filling the $360^{\circ}$ angle around point $O$ with the right angles at $O$ of the four triangles, and in this case, the area of the quadrilateral $A B C D$ is also the largest, equal to the sum of the areas of the four right triangles. In this arrangement, 2-2 segments lie on each other's extensions, and the diagonals of the quadrilateral $A B C D$ are formed by them.

From the above 3 pairings of segments, the largest area is obtained when one diagonal is the sum of the smallest and the largest segment. It is known that among two rectangles of equal perimeter, the one with the smaller difference (in absolute value) between the sides has the larger area. In other words, if the sum of the factors in two (two-factor) products is the same, then the product is larger where the factors differ less. In our case, the two factors are the sums of the two pairs of the 4 given segments, and this difference is the smallest when the smallest segment is paired with the largest.