75th 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: $△MOD\sim △OAE$, 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 $△CDO$, 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, $DO\cdot DF=M{D}^2$, 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 $△DAF\sim △DOA$ (AA), hence $M{D}^2=D{A}^2=DF\cdot DO$, 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 $OF=\frac{O{M}^2}{OD}=\frac{4\sqrt{5}}{5}MD$, 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 $CF=CO$.