Romanian Mathematical Olympiad

Triangles $\Delta MEP$ and $\Delta MEN$ are congruent, therefore $MN=MP$. Denote by $O_1$ the midpoint of $HF$. It follows that $M{O}_1=\frac{AC}{2}=\frac{a\sqrt{3}}{2}$ and, since $NP=a$, the triangle $MNP$ must be equilateral. Let $O$ be the midpoint of $AC$, $M_1$ the midpoint of $AO$ and $\{L_1\}=LK\cap M{O}_1$. Since $M{M}_1\parallel EO\parallel LK\parallel C{O}_1$, we have $M{L}_1/M{O}_1=M_{1}K/M_{1}C=2/3$ so $L_1$ is the centroid of the triangle $MNP$.

Notice that $AL=AG=\frac{3\sqrt{2a}}{4}$ thus $ALGK$ is a rhombus, which implies $AG\perp KL$. We also have $C{O}_{1}LK$ parallelogram, therefore $LK\parallel C{O}_1$ and $KL\perp M{O}_1$. $HF\perp (ACGE)$ so $KL\perp HF$ which implies $KL\perp (MNP)$. It follows that the pyramid $MNPQ$ is regular. From the similarity of the triangles $\Delta M{L}_{1}Q$ and $\Delta M{O}_{1}C$ we have $Q{L}_1=\frac{a\sqrt{6}}{3}$ forcing the tetrahedron $MNPQ$ to be regular.