4. Let $A L, B M, C N$ be the medians of triangle $A B C$, intersecting at point $K$. It is known that a circle can be circumscribed around quadrilateral $C L K M$, and $A B=2$. Find the length of the median $C N$.
4. Answer. $\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b1d77a2f2c14bbc97422g-2.jpg?height=511&width=580&top_left_y=108&top_left_x=184)

Since $M L$ is the midline of triangle $A B C$, we get that $M L \| A B$ and $M L=\frac{1}{2} A B=1$. If $P$ is the intersection point of the median $C N$ and the midline $M L$, then $C P=P N$ and $M P=P L=\frac{1}{2}$.

Let $x$ be the length of the median $C N$, then

$$
C P=\frac{1}{2} x, P K=P N-K N=\frac{1}{2} x-\frac{1}{3} x=\frac{1}{6} x
$$

Since quadrilateral $C L K M$ is cyclic, we get that $\angle M C K=\angle K L M$. Therefore, $\triangle M C P \sim \triangle K L P$ (by two angles). Then $\frac{M P}{K P}=\frac{P C}{P L}$, which means $M P \cdot P L=K P \cdot P C$.

Considering that $M P \cdot P L=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}$ and $K P \cdot P C=\frac{1}{6} x \cdot \frac{1}{2} x=\frac{1}{12} x^{2}$, we get $\frac{1}{12} x^{2}=\frac{1}{4}$, from which $x=\sqrt{3}$ (considering $x>0$).