We have a circle. Construct the largest possible area rectangle, where side $AB$ is tangent to the circle, and side $CD$ is a chord of the circle.
Let the distance from the center of the circle to the chord $CD$ be denoted by $x$, $CE$ by $y$, the area of the quadrilateral by $t$, and the radius of the circle by $r$, then

$$
x^{2}+y^{2}=r^{2}
$$

$$
(x+r) y=\frac{t}{2}
$$

Squaring both sides of (2):

$$
(x+r)^{2} y^{2}=\frac{t^{2}}{4}
$$

Substituting the value of $y^{2}$ from (1) into (3):

$$
(x+r)^{2}\left(r^{2}-x^{2}\right)=\frac{t^{2}}{4}
$$

or

$$
(r+x)(r+x)(r+x)(r-x)=\frac{t^{2}}{4}
$$

or multiplying both sides of the equation by 3:

$$
(r+x)(r+x)(r+x)(3 r-3 x)=\frac{3 t^{2}}{4}
$$

But the sum of the factors on the left side of the equation $(6r)$ is a constant number, so why is the area of the quadrilateral the largest possible when the factors are equal (see note); $t$ is therefore maximum when

$$
3 r-3 x=r+x
$$

or

$$
x=\frac{r}{2} \text {. }
$$

According to this, the side $CD$ of the largest area quadrilateral is obtained by erecting a perpendicular chord at the midpoint of any radius $OP$; the side $AB$ of the quadrilateral is obtained by extending the radius $OP$ beyond the center of the circle to the circumference and drawing a tangent at the point thus obtained.

Note. If the constant number $k$ is decomposed into positive addends in different ways, then the product of these addends is the largest when they are equal. We only prove this theorem for the case where the number $k$ is decomposed into four parts. Let the parts, whose sum is $k$, be $x, y, u, v$. We show that if

$$
x_{1}+y_{1}+u_{1}+v_{1}=k
$$

and, for example, $x_{1}$ is not equal to $y_{1}$, then the product $x_{1} y_{1} u_{1} v_{1}$ cannot be the largest value of the product $x y u v$. Let the parts of the number $k$ be:

$$
\frac{x_{1}+y_{1}}{2}, \frac{x_{1}+y_{1}}{2}, u_{1}, v_{1}
$$

then under the condition (1):

$$
\frac{x_{1}+y_{1}}{2}+\frac{x_{1}+y_{1}}{2}+u_{1}+v_{1}=k
$$

but it is known that:

$$
\frac{x_{1}+y_{1}}{2} \cdot \frac{x_{1}+y_{1}}{2}>x_{1} y_{1}
$$

and thus:

$$
\frac{x_{1}+y_{1}}{2} \cdot \frac{x_{1}+y_{1}}{2} u_{1} v_{1}>x_{1} y_{1} u_{1} v_{1}
$$

We see that the product cannot be maximum if the factors are not equal.

The problem was solved by: Beck Ferencz, Friedmann Bernát, Szabó István, Szabó Károly.