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

By the condition in the statement, the points $A$, $B$, $P$, and $Q$ lie on some circle ${\omega }_1$, and the points $C$, $D$, $P$, and $Q$ lie on some circle ${\omega }_2$. The line $PQ$ is the radical axis of these two circles, and we proceed to prove that $M$ lies on this radical axis. Let $MA$ meet ${\omega }_1$ again at $A^{\prime }$, and let $MD$ meet ${\omega }_2$ again at $D^{\prime }$; if, say, $MA$ is tangent to ${\omega }_1$, then $A^{\prime }=A$. Then $\angle (M{A}^{\prime },A^{\prime }B)=\angle (A{A}^{\prime },A^{\prime }B)=\angle (AP,PB)=\angle (CP,PD)=\angle (C{D}^{\prime },D^{\prime }D)=\angle (C{D}^{\prime },D^{\prime }M)$, and $$\begin{array}{rl}\angle (M{A}^{\prime },MB)& =\angle (MA,MB)=\angle (MA,MC)+\angle (MC,MB)\\ & =\angle (MB,MD)+\angle (MC,MB)=\angle (MC,MD)\\ & =\angle (MC,M{D}^{\prime }),\end{array}$$ showing that the triangles $M{A}^{\prime }B$ and $M{D}^{\prime }C$ are similar and have opposite orientations. Hence $M{A}^{\prime }/M{D}^{\prime }=MB/MC=MD/MA$, so $MA\cdot M{A}^{\prime }=MD\cdot M{D}^{\prime }$.

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We still have to decide whether such a point $M$ exists, and deal with the exceptional cases. This can be done in several different ways, e.g., by investigating the orientation-changing similarity transformation mapping $A$ to $D$ and $B$ to $C$. We present a more geometrical proof. Let the diagonals $AC$ and $BD$ cross at $K$. Choose points $F$, on the diagonal $AC$, and $G$, on the diagonal $BD$, so that the configurations $(A,F,K,C)$ and $(D,K,G,B)$ are similar. If $AD$ and $BC$ are not parallel, then $F\ne K\ne G$. By similarity, it is sufficient to find the point $M$ such that the triangles $MKF$ and $MGK$ are similar and have opposite orientations. If $KF\ne KG$, one may choose $M$ to be the point where $FG$ crosses the tangent at $K$ to the circle $FGK$. The exceptional cases are $AD\parallel BC$ and $KF=KG$ (in which case $AC=BD$ by similarity). These cases may be considered as limit cases: in the former case, $M$ tends to $K$ and this can be dealt with along the above lines; in the latter, $M$ becomes an ideal point, so the lines $PQ$ are all parallel. In fact, both cases can be dealt with independently and are relatively easy.