Belarus — Final Round

Lemma. Let $CC_1$ be the median of a triangle $ABC$. Then the inequality $CC_1 \le 0.5 AB$ is equivalent to the inequality $\angle ACB \ge 90^\circ$.

Consider the parallelogram $ADBC$ (see Fig. 1). By the cosine law, from the $\triangle ABC$ and $\triangle CBD$, it follows that
$$\cos \angle ACB = \frac{BC^2 + AC^2 - 2AB^2}{2BC \cdot AC}$$
and
$$\cos \angle CBD = \frac{BC^2 + AC^2 - 2CD^2}{2BC \cdot AC}.$$
Since the sum of these angles is equal to $180^\circ$, we see that $\angle ACB \ge 90^\circ$ iff $\cos \angle CBD \ge \cos \angle ACB$. But this inequality is equivalent to the inequality $CD \le AB$ or, since $CD = 2CC_1$, to the inequality $CC_1 \le 0.5AB$. The proof is complete.

Fig. 1

Fig. 2

Let $P, Q, T$ be the midpoints of the segments $KN, KM, LM$, respectively (see Fig. 2). We have $AC \le AP + PQ + QT + TC$. By condition, $\angle NAK \ge 90^\circ$ and $\angle MCL \ge 90^\circ$, then from the lemma it follows that $AP \le 0.5KN$ and $CT \le 0.5LM$. Moreover, $PQ = 0.5NM$ and $QT = 0.5KL$ as the midlines of the respective triangles. Therefore $AC \le 0.5KN + 0.5NM + 0.5KL + 0.5LM$, i.e. the perimeter of $KLMN$ is greater than or equal to $2AC$.