Romanian Mathematical Olympiad

First solution for a). Since $ME \perp OC$, it suffices to show that the quadrilateral $MEOF$ is cyclic.

(1) Since $ME \parallel AO$, we have $\angle MEF = \angle OAE$, therefore it remains to prove that $\angle MOD = \angle OAE$. This follows from similarity: $\triangle MOD \sim \triangle OAE$, because $\angle DMO = \angle EOA = 90^\circ$ and $\frac{DM}{MO} = \frac{EO}{OA} = \frac{1}{2}$.

(2)

Second solution for a). By Menelaus' theorem in $\triangle CDO$, with the transversal $AFE$, we get: $\frac{DF}{FO} = \frac{AD}{AC} = \frac{1}{4}$. Let $AB = 4a$. In triangle $MDO$, we have $MD = a$, $MO = 2a$, and $DO = a\sqrt{5}$, so $DF = \frac{a\sqrt{5}}{5}$. Therefore, $DO \cdot DF = MD^2$, hence, by the converse of the hypotenuse leg theorem, we obtain $MF \perp DO$.

Third solution for a). Since $\frac{CA}{CE} = 2\sqrt{2} = \frac{OA}{AD}$, and $\angle ACE = \angle OAD = 45^\circ$, the triangles $CAE$ and $AOD$ are similar. Then $\angle AOD = \angle CAE = \angle DAF$, so $\triangle DAF \sim \triangle DOA$ (AA), hence $MD^2 = DA^2 = DF \cdot DO$, and thus $MF \perp DO$.

First solution for b). Let $G, R$ be the midpoints of segments $AB$ and $MO$, respectively. Since $CMGO$ is a parallelogram, $R$ is the midpoint of $CG$. $EMFO$ is cyclic, therefore $\angle EFO = \angle EMO = 45^\circ = \angle OBA$, so $ABOF$ is cyclic, with $G$ the center of its circumscribed circle. Thus $GF = GO$, $RF = RO$, so $RG$ is the perpendicular bisector of segment $OF$. Since $C \in RG$, we get $CF = CO$.

Second solution for b). Let $G$ be the midpoint of $AB$. Analogous to (2), the triangles $MOD$ and $ACG$ are similar, so $\angle ACG + \angle CDO = \angle MOD + \angle CDO = 90^\circ$, thus $CG \perp OD$. Let $H$ be the intersection of the lines $CG$ and $OD$. Since $OG \parallel CD$, we get $\frac{OH}{HD} = \frac{OG}{CD} = \frac{2}{3}$, consequently $\frac{OH}{OD} = \frac{2}{5}$. In the right triangle $MOD$, we have $OM = 2MD$, $OD = MD\sqrt{5}$ and $OF = \frac{OM^2}{OD} = \frac{4\sqrt{5}}{5}MD$, therefore $\frac{OF}{OD} = \frac{4}{5}$. It follows that $H$ is the midpoint of $OF$, so $CH$ is both an altitude and median in the triangle $COF$, hence $CF = CO$.