Final Round

Lemma. Let $C{C}_1$ be the median of a triangle $ABC$. Then the inequality $C{C}_1\le 0.5AB$ is equivalent to the inequality $\angle ACB\ge 90^{\circ }$.

Consider the parallelogram $ADBC$ (see Fig. 1). By the cosine law, from the $△ABC$ and $△CBD$, it follows that $$\cos \angle ACB=\frac{B{C}^2+A{C}^2-2A{B}^2}{2BC\cdot AC}$$ and $$\cos \angle CBD=\frac{B{C}^2+A{C}^2-2C{D}^2}{2BC\cdot AC}.$$ 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 $CD=2C{C}_1$, to the inequality $C{C}_1\le 0.5AB$. 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 $2AC$.