Bakayev E.V.

In triangle $\mathrm{ABC}$, medians $\mathrm{BK}$ and $\mathrm{CN}$ are drawn, intersecting at point M. What is the maximum number of sides of quadrilateral ANMK that can have a length of 1?
Method 1. Suppose that at least three sides of the quadrilateral ANMK are equal to 1. There are only two fundamentally different cases. 1) $\mathrm{AN}=\mathrm{NM}=\mathrm{MK}=1$. Then $\mathrm{NB}=1, \mathrm{MB}=2$, so $\mathrm{MN}+\mathrm{NB}=\mathrm{MB}$. 2) $\mathrm{KA}=\mathrm{AN}=\mathrm{NM}=1$. Then $\mathrm{AC}=2, \mathrm{NC}=3$, so $\mathrm{NA}+\mathrm{AC}=\mathrm{NC}$. In both cases, a contradiction with the triangle inequality is obtained.

Method 2. If more than two sides of the quadrilateral are equal to 1, then either $\mathrm{AK}=\mathrm{NA}$ or $\mathrm{KM}=\mathrm{MN}$. In the first case, triangle ABC is isosceles. In the second case, $\mathrm{BK}=\mathrm{CN}=3$, so in this case, triangle ABC is also isosceles. From this, it follows that $\mathrm{AK}=\mathrm{KM}=\mathrm{MN}=\mathrm{NA}$, meaning that $AKMN$ is a rhombus. This is a contradiction, as lines AK and NM are not parallel.

Method 3. Let $\mathrm{L}$ be the midpoint of side $\mathrm{BC}$. If $\mathrm{NM}=\mathrm{NA}=\mathrm{NB}$, then triangle AMB is a right triangle.

Triangle LMK is similar to triangle AMB with a similarity ratio of 0.5, so MK $=\mathrm{LM}$. Therefore, in triangle AMK, the hypotenuse is larger than in triangle LMK, meaning $\mathrm{AK}>\mathrm{LK}=\mathrm{NA}$. Thus, in the quadrilateral ANMK, there are no three sides equal to each other.

The case when $\mathrm{KM}=\mathrm{KA}$ is analogous, and if $\mathrm{KM} \neq \mathrm{KA}$ and $\mathrm{NM} \neq \mathrm{NA}$, then among the segments NM, NA, KM, and KA, there are also no three equal to each other.

## Answer

2 sides.