I. solution. Using the notations of Figure 1, let the midpoint of the diagonal $BD$ be $F$, the midpoint of the diagonal $AC$ be $G$, the intersection of the diagonals be $M$, and finally $\angle DMC = \gamma$. The radius of the circle is $r = 13 \, \text{cm}$, and the distance of the diagonals from the center of the circle is $OF = OG = d = 5 \, \text{cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_95bcbc8e1bac0a451388g-1.jpg?height=588&width=591&top_left_y=297&top_left_x=756)

Figure 1

The diagonals of the trapezoid are chords in the circle. The perpendicular bisector of a chord passes through the center, so $\angle OFB = 90^\circ$. Applying the Pythagorean theorem in the right triangle $OFB$:

$$
\begin{aligned}
BF & = \sqrt{r^2 - d^2} = \sqrt{13^2 - 5^2} = \\
& = \sqrt{144} = 12 \, \text{cm}
\end{aligned}
$$

Thus,

$$
AC = BD = 2BF = 24 \, \text{cm}
$$

The area of the trapezoid is therefore:

$$
T = \frac{1}{2} \cdot AC \cdot BD \cdot \sin \gamma = \frac{1}{2} \cdot 24 \cdot 24 \cdot \sin \gamma = 288 \cdot \sin \gamma
$$

Since the value of $\sin \gamma$ is at most 1, the area is maximized when $\sin \gamma = 1$, i.e., $\gamma = 90^\circ$. In this case, the area is: $T_{\max} = 288 \, \text{cm}^2$.

II. solution. The radius of the circle is $13 \, \text{cm}$, and the diagonals of the cyclic quadrilateral are $5 \, \text{cm}$ away from the center of the circle. Draw one of the diagonals and connect its endpoints to the center of the circle (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_95bcbc8e1bac0a451388g-1.jpg?height=415&width=420&top_left_y=1698&top_left_x=839)

Figure 2

The two lines thus formed are of equal length, as they are both radii of the circle, forming an isosceles triangle with the diagonal. Bisect the triangle by drawing a perpendicular from the vertex opposite the base to the base. The resulting triangles are right triangles, as two of their sides are perpendicular to each other, and they are congruent, as their corresponding sides are of equal length. One leg of such a right triangle is the distance from the center of the circle to one of the diagonals of the cyclic quadrilateral, which is $5 \, \text{cm}$, and the hypotenuse is the radius of the circle, which is $13 \, \text{cm}$. The other leg, which is half the diagonal, can be calculated using the Pythagorean theorem: $12 \, \text{cm}$.

The diagonals of a cyclic quadrilateral are of equal length, so both diagonals of the cyclic quadrilateral are $24 \, \text{cm}$ long. The cyclic quadrilateral can be divided into a rectangle whose one diagonal coincides with one of the diagonals of the cyclic quadrilateral (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_95bcbc8e1bac0a451388g-2.jpg?height=406&width=414&top_left_y=133&top_left_x=842)

Figure 3

The Pythagorean theorem can also be applied to the two adjacent sides and one diagonal of a rectangle, stating that the sum of the squares of the two adjacent sides is equal to the square of the diagonal. The product of these two sides is equal to the area of the rectangle, and thus to the area of the cyclic quadrilateral. Denoting the two sides by $a$ and $b$, we write the inequality between the geometric and quadratic means:

$$
\begin{aligned}
\sqrt{ab} & \leq \sqrt{\frac{a^2 + b^2}{2}} = \frac{24}{\sqrt{2}} \\
ab & \leq \frac{24^2}{2}
\end{aligned}
$$

According to the inequality, the product of two numbers with a given sum of squares is maximized when the two numbers are equal. Therefore, in the rectangle with the maximum area, the two adjacent sides are of equal length, making the rectangle a square. The area of a square is half the product of its diagonals, which in our case is $\frac{24^2}{2} = 288 \, \text{cm}^2$.

Thus, the maximum area of the cyclic quadrilateral is $288 \, \text{cm}^2$.

Czett Mátyás (Zalaegerszegi Zrínyi Miklós Gimn., 10th grade)
A $13 \mathrm{~cm}$ radius circle is inscribed in a trapezoid, and we know that the diagonals are $5 \mathrm{~cm}$ away from the center of the circle. What is the maximum possible area of the trapezoid?