THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

Let $M$, $P$ and $Q$ be the midpoints of $[BC]$, $[CA]$ and $[AB]$, respectively.

The circumcircle of triangle $A_1{B}_1{C}_1$ is the Euler circle. Point $M$ lies on this circle.

It is enough to prove now that $[A_{1}M]$ is a common chord of the three circles, $(A_1{B}_1{C}_1)$, $(AKL)$ and $(DE{A}_1)$.

The segments $[MK]$ and $[ML]$ are midlines of the triangles $BC{C}_1$ and $BC{B}_1$ respectively, hence $MK\parallel C{C}_1\perp AB$ and $ML\parallel B{B}_1\perp AC$. So, the circle $(AKL)$ has diameter $[AM]$ and therefore passes through $M$.

Finally, we prove that the quadrilateral $D{A}_{1}ME$ is cyclic. From the cyclic quadrilaterals $ADB{A}_1$ and $AEC{A}_1$, $\overline{A{A}_{1}D}\equiv \overline{ABD}$ and $\overline{A{A}_{1}E}\equiv \overline{ACE}\equiv \overline{ABD}$, so $\overline{D{A}_{1}E}=2\overline{ABD}=180^{\circ }-2\overline{DAB}$. We notice now that $DQ=AB/2=MP$, $QM=AC/2=PE$ and $$\overline{DQM}=\overline{DQB}+\overline{BQM}=2\overline{DAB}+\overline{BAC},$$ $$\overline{EPM}=\overline{EPC}+\overline{CPM}=2\overline{EAC}+\overline{CAB},$$ so $\Delta MPE\equiv \Delta DQM$ (S.A.S.). This leads to $\overline{DME}=\overline{DMQ}+\overline{QMP}+\overline{PME}=\overline{DMQ}+\overline{BQM}+\overline{QDM}=180^{\circ }-\overline{DQB}=180^{\circ }-2\overline{DAB}$. Since $m(\overline{D{A}_{1}E})=m(\overline{DME})$, the quadrilateral $D{A}_{1}ME$ is cyclic.