THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

First solution. Let the line $EF$ meet the circle $C_{1}BE$ again at $T$. Then $$\angle (T{C}_1,C_{1}B)=\angle (TE,EB)=\angle (FE,EA)=\angle (F{A}_1,A_{1}A),$$ so $T{C}_1$ and $F{A}_1$ are parallel. Similarly, $$\angle (TB,B{C}_1)=\angle (TE,E{C}_1)=\angle (FE,E{C}_1)=\angle (FC,C{C}_1)=\angle (FD,D{A}_1),$$ so $TB$ and $FD$ are parallel. Thus, the corresponding sides of the triangles $A_{1}DF$ and $C_{1}BT$ are parallel. Since the vectors $\overrightarrow{C_{1}B}$ and $\overrightarrow{A_{1}D}$ are counter-directed, these triangles are homothetical under a homothety of negative ratio. The centre of this homothety $O$ lies on the lines $BD$, $A_1{C}_1$, and $TF=EF$, so the three are concurrent.



Second solution. Let the lines $AB$ and $CD$ meet at $S$, and let the lines $A_{1}F$ and $C_{1}E$ meet at $K$. Since $$\begin{array}{rl}\angle (EK,FK)& =\angle (E{C}_1,EF)+\angle (EF,A_{1}F)=\angle (C{C}_1,CF)+\angle (EA,A{A}_1)\\ & =\angle (BC,FS)+\angle (ES,BC)=\angle (ES,FS),\end{array}$$ the points $E$, $F$, $S$, and $K$ are concyclic. Thus $\angle (KS,KF)=\angle (ES,EF)=\angle (EA,EF)=\angle (A_{1}A,A_{1}F)$, showing that $KS$ is parallel to both $AD$ and $BC$. This shows the triangles $A_{1}DF$ and $C_{1}BE$ in perspective from the ideal point of the three parallel lines, so the lines $A_1{C}_1$, $BD$ and $EF$ are concurrent, by Desargues' theorem; since the segments $A_1{C}_1$ and $BD$ have a non-empty intersection, the point of concurrency is not ideal.