# Problem 4. (3 points)

In triangle $A B C$, the median $B M$ is drawn, in triangle $A B M$ - the median $B N$, in triangle $B N C$ - the median $N K$. It turns out that $N K \perp B M$. Find $A B: A C$.

Answer: $\frac{1}{2}$
Solution:

Let $\vec{b}=\overrightarrow{A B}$ and $\vec{a}=\overrightarrow{A C}$.

$\overrightarrow{N K}=\overrightarrow{A K}-\overrightarrow{A N}=\frac{\vec{a}+\vec{b}}{2}-\frac{\vec{a}}{4}=\frac{\vec{a}+2 \vec{b}}{4}$.

$\overrightarrow{B M}=\overrightarrow{A M}-\overrightarrow{A B}=\frac{\vec{a}}{2}-\vec{b}$

$0=\overrightarrow{B M} \cdot \overrightarrow{N K}=\frac{\vec{a}^{2}-4 \vec{b}^{2}}{8}$, from which $|\vec{a}|=2|\vec{b}|$, that is, $A B: A C=\frac{1}{2}$.