A rectangle $K L M N$ is given, where $|K L|=6 \mathrm{~cm}$ and $|M L|=4 \mathrm{~cm}$. Calculate the perimeters of all isosceles triangles $K L X$, if point $X$ lies on side $M N$.

(M. Dillingerová)
In the isosceles triangle $K L X$, $K L$ can be a) a leg or b) the base.

a) It will be triangle $K L X_{1}$, where $|K L|=\left|K X_{1}\right|=6$. Let $\left|N X_{1}\right|=x$. According to the Pythagorean theorem, $4^{2}+x^{2}=6^{2}$, hence $x^{2}=20, x=2 \sqrt{5}$. Then $\left|X_{1} M\right|=6-2 \sqrt{5} \doteq 1.53$.

In triangle $L M X_{1}$, according to the Pythagorean theorem, $(6-2 \sqrt{5})^{2}+4^{2}=\left|L X_{1}\right|^{2}$, hence $\left|L X_{1}\right| \doteq 4.3$ and the perimeter $o \doteq 6+6+4.3=16.3$. The same will apply to triangle $K L X_{2}$, where $|K L|=\left|X_{2} L\right|=6 ;\left|K X_{2}\right|=\left|L X_{1}\right| \doteq 4.3 ; o \doteq 16.3$.

b) If $K L$ is the base of the isosceles triangle, point $X_{3}$ lies at the midpoint of side $M N$. For the length of the leg $r$, we have $r^{2}=3^{2}+4^{2}=25 ; r=5, o=6+5+5=16$.

(3 points for determining the perimeter of triangle $K L X_{1} ; 1$ point for (the same) perimeter of triangle $K L X_{2}$; 2 points for determining the perimeter of triangle $K L X_{3}$.)