XXIX Rioplatense Mathematical Olympiad

First we recall that since $ABCDEFGHI$ is a regular 9-gon all of its sides are equal and each of its angles is equal to $\frac{7\cdot 180^{\circ }}{9}=140^{\circ }$; likewise, all sides of $ABJKLM$ are equal and each of its angles is equal to $120^{\circ }$.

By symmetry, $\angle KEL=\angle LGK$, so it suffices to show that $\angle HMG=\angle LGK$.

Triangle $AIH$ is isosceles with $\angle AIH=140^{\circ }$, thus $\angle IHA=\angle IAH=\frac{180^{\circ }-140^{\circ }}{2}=20^{\circ }$. But we also have $\angle IAM=\angle IAB-\angle MAB=140^{\circ }-120^{\circ }=20^{\circ }$. This implies that $A$, $M$, $H$ are collinear.

Now let us consider quadrilateral $AHGB$. We have $\angle AHG=140^{\circ }-20^{\circ }=120^{\circ }=\angle BAH$; furthermore $AB=HG$. Hence $BAGH$ is an isosceles trapezoid and in particular $\angle ABG=180^{\circ }-\angle HAB=60^{\circ }$. But we also have $\angle ABL=\frac{1}{2}\angle ABJ=60^{\circ }$, so this time we obtain that $B$, $L$, $G$ are collinear.

Notice that line $BL$ is the perpendicular bisector of $MK$ (because it is an axis of symmetry). Since $G$ lies on that line, we find that $GM=GK$, and thus triangles $△GML$ and $△GKL$ are congruent, because the corresponding sides are equal. Hence $\angle LGM=\angle LGK$. But we also have $\angle HMG=\angle LGM$.



because $HM$ and $GL$ are parallel. Therefore $\angle HMG=\angle LGK=\angle KEL$, and we are done.