Inside the rectangle $A B C D$, points $E$ and $F$ are located such that the segments $E A, E D, E F, F B, F C$ are all congruent. The side $A B$ is $22 \, \text{cm}$ long, and the circumcircle of triangle $A F D$ has a radius of $10 \, \text{cm}$.

Determine the length of side $B C$.

(L. Růžičková)

Hint. Where does the center of the circumcircle of triangle $A F D$ lie?
Point $E$ is equally distant from points $A$ and $D$, and point $F$ is equally distant from points $B$ and $C$. Segments $A D$ and $B C$ are opposite sides of a rectangle. Therefore, points $E$ and $F$ lie on the common axis of segments $A D$ and $B C$. Point $E$ is equally distant from all vertices of triangle $A F D$, so it is the center of the circumscribed circle of this triangle. The representation of the points from the problem looks as follows (points $P$ and $R$ are the intersections of the axis $E F$ with sides $A D$ and $B C$, i.e., the midpoints of these sides):

![](https://cdn.mathpix.com/cropped/2024_04_17_174124d98c91f026d57bg-4.jpg?height=488&width=682&top_left_y=1629&top_left_x=687)

Triangles $A P E, D P E, B R F$, and $C R F$ are congruent right triangles, one of whose legs is half the length of the sought side of the rectangle, and the sizes of the remaining two sides can be derived from the problem statement. For example, in triangle $A P E$, the hypotenuse $A E$ has a length of $10 \, \text{cm}$ and the leg $P E$ has a length of $(22-10): 2=6 \, \text{cm}$. According to the Pythagorean theorem, we have

$$
|P A|^{2}+6^{2}=10^{2}
$$

thus $|P A|^{2}=64$ and $|P A|=8 \, \text{cm}$. The length of side $B C$ is $16 \, \text{cm}$.

Note. In the given image, we implicitly suggest that side $A B$ is longer than $B C$. This is confirmed by the following calculation, but it can also be noticed directly: The hypotenuse $A E$ in the right triangle $A P E$ is longer than the leg $P A$, which is half of side $B C$. If side $B C$ were longer than side $A B$, segment $A E$ would also be longer than half of side $A B$. However, we know that $|A E|=10 \, \text{cm}$ and $\frac{1}{2}|A B|=11 \, \text{cm}$, so this situation cannot occur.