Let's determine all cyclic quadrilaterals whose sides are integers, and the numerical value of their area is equal to the numerical value of their perimeter.
Let the sides of a cyclic quadrilateral be denoted by $a, b, c, d$, and its semiperimeter by $s$. According to the Heron's formula for cyclic quadrilaterals, the condition can be written as:

$$
\sqrt{(s-a)(s-b)(s-c)(s-d)}=a+b+c+d
$$

If $s$ were not an integer, then under the square root, each factor would be half of an odd number. However, the product of such numbers cannot be an integer, let alone a perfect square. Let us denote the factors by $x, y, v, z$. These are therefore positive integers, and based on the triangle inequality, they satisfy the following relationship:

$$
x y v z=(x+y+v+z)^{2}
$$

The degree of the left side is higher in the unknowns than the right side, so it is expected that the equation has only a finite number of positive integer solutions. To find these solutions, we adopt the following strategy. We can assume that

$$
x \leq y \leq v \leq z
$$

First, we show that only a finite number of $(x, y)$ pairs can be involved, and for each such pair, we confine the possible values of $v$ to a finite interval. For a fixed $(x, y, v)$, our equation takes the form of a simple quadratic equation in the remaining unknown $z$, which generally does not have a positive integer solution. We sometimes need to verify this by calculation, but in most cases, we can exclude it with the following simple observation:

$$
x+y+v \quad \text { is not prime. }
$$

Indeed, the right side of (1) is greater than 1, so $z$ must be divisible by some prime number $p$. This $p$ divides $(x+y+v+z)$, and thus $(x+y+v)$ as well, but then $x+y+v$ could only be prime if it were equal to $p$, which is excluded by the obvious

$$
z<x+y+v
$$

inequality.

First, let us write down the inequality between the arithmetic and geometric means for the different numbers $x+y+v$ and $z$, then using (1) we get

$$
4(x+y+v) z<(x+y+v+z)^{2}=x y v z
$$

which implies

$$
4(x+y)<(x y-4) v
$$

Since $x+y$ and $v$ are positive, the right side must satisfy $4<x y$, and

$$
4 \frac{x+y}{x y-4}<v
$$

Let us introduce the notations $\lambda=\frac{x+y+v}{z}, \mu=\frac{x+y+v}{v}$. Then $1<\lambda \leq \mu$, and from (1) we have

$$
x y=\frac{(x+y+v+z)^{2}}{z^{2}} \frac{z}{v}=(\lambda+1)^{2} \frac{\mu}{\lambda}
$$

Here, $(\lambda+1)^{2} / \lambda-(\mu+1)^{2} / \mu=(\lambda-\mu)(1-1 / \mu \lambda) \leq 0$, so

$$
x y \leq(\mu+1)^{2}
$$

which implies

$$
x y \leq\left(3+\frac{x}{y}\right)^{2}
$$

and

$$
\sqrt{x y}-2 \leq \mu-1=\frac{x+y}{v}
$$

Combining this with the already known $4<x y$ inequality,

$$
v \leq \frac{x+y}{\sqrt{x y}-2}
$$

Using the floor function, we can combine the upper bound for $v$ with the lower bounds from (2) and (5):

$$
\max \left(y, 1+\left\lfloor 4 \frac{x+y}{x y-4}\right\rfloor\right) \leq v \leq\left\lfloor\frac{x+y}{\sqrt{x y}-2}\right\rfloor
$$

After all these preparations, we can now search for the positive integer solutions of (1) under the assumptions (2) and (4). We have seen that $x y \geq 5$, and from (6) it immediately follows that $x y \leq(3+1)^{2}=$ 16. This condition is not sufficient for (6) to hold, since (6) also states that if $9<x y$, then $x / y$ cannot be too small, i.e., $x$ and $y$ must be somewhat close to each other. In fact, the following positive integer $(x, y)$ pairs satisfy (2), (6), and the $x y \geq 5$ condition:

$$
\begin{array}{llll}
(3,4), & (2,5), & (1,9), & (3,3), \quad(2,4), \quad(1,7), \quad(1,6), \quad(2,3), \quad(1,5)
\end{array}
$$

Each of these pairs defines an interval for $v$ according to (7). Many $v$ values are already excluded from the list based on (3). From the remaining cases, we select those in which (1) has a positive integer solution $z$ that satisfies (4). Considering all cases, (1)-(2)-(4) have 4 solutions in positive integers:

$$
(x, y, v, z)=(4,4,4,4), \quad(2,5,5,8), \quad(1,9,10,10), \quad(3,3,6,6)
$$

which correspond to the following side quadruples:

$$
(a, b, c, d)=(4,4,4,4), \quad(8,5,5,2), \quad(14,6,5,5), \quad(6,6,3,3)
$$

Now we show that these side quadruples determine a total of 7 cyclic quadrilaterals (up to congruence). The $(4,4,4,4)$ side quadruple uniquely determines a square as a cyclic quadrilateral, so it is sufficient to show that each of the remaining 3 quadruples determines 2 cyclic quadrilaterals. After rearrangement, each of the three quadruples is of the form $(a, b, b, c)$, where $a \neq b \neq c$ and $|a-c|<2 b$. From such a quadruple, a cyclic trapezoid can be constructed, and uniquely so. By reflecting the trapezoid over the perpendicular bisector of one of its diagonals, we obtain another cyclic quadrilateral, whose sides are still $(a, b, b, c)$, but now the $b$-length sides share a common vertex. Conversely, if a cyclic quadrilateral has sides of some order $(a, b, b, c)$, then either the $b$-length sides are opposite, in which case we are talking about the already constructed trapezoid; or they share a common vertex, in which case reflecting the sides that share this vertex over the perpendicular bisector of the diagonal that they enclose, we return to the case of the cyclic trapezoid. Therefore, the initial quadrilateral must necessarily be congruent to the already constructed other cyclic quadrilateral.

With this, we have solved the problem, determining all 7 cyclic quadrilaterals whose sides are integers and whose area and perimeter have the same numerical value.

Based on the work of Burcsi Péter (Pápa, Türr István Gymnasium, III. grade)