International Mathematical Olympiad

We first prove that triangles $ADQ$ and $CDB$ are similar. Since $ABCD$ is cyclic, we have $\angle DAQ=\angle DCB$. By the tangency of $AC$ to the circle $AQD$ we also have $\angle CBD=\angle CAD=\angle AQD$. The claimed similarity is proven.

Let $R$ be the midpoint of $CD$. Points $N$ and $R$ correspond in the proven similarity, and so $\angle QNA=\angle BRC$.



Let $K$ be the second common point of line $CD$ with circle $ABR$ (i.e., if $CD$ intersects circle $ABR$, then $K\ne R$ is the other intersection; otherwise, if $CD$ is tangent to $CD$, then $K=R$). In both cases, we have $\angle BAK=\angle BRC=\angle QNA$; that indicates that $AK$ is tangent to circle $ANQ$. It can be showed analogously that $BK$ is tangent to circle $BMP$.

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Alternative solution.

We present a second solution, without using the condition that $ABCD$ is cyclic. Again, $M$ and $N$ can be any points on lines $BC$ and $AD$ such that $BM:MC=DN:NA$.

Let $AB$ and $CD$ meet at $T$ (if $AB\parallel CD$ then $T$ is their common ideal point). Let $CD$ meet the tangent to the circle $ANQ$ at $A$, and the tangent to the circle $BMP$ at $B$ at points $K_1$ and $K_2$, respectively.



Let $I$ and $J$ be the ideal points of $AD$ and $BC$, respectively. Notice that the pencils ($AD$, $AC$, $AT$, $A{K}_1$) and ($QA$, $QD$, $QI$, $QN$) of lines are congruent, because $\angle K_{1}AD=\angle AQN$, $\angle CAD=\angle AQD$ and $\angle IAT=\angle IQT$. Hence,

$$(D,C;T,K_1)=(AD,AC;AT,A{K}_1)=(QA,QD;QI,QN)=(A,D;I,N)=\frac{DN}{NA}.$$

It can be obtained analogously that

$$(D,C;T,K_2)=(BD,BC;BT,B{K}_2)=(PC,PB;PJ,PM)=(C,B;I,N)=\frac{BM}{MC}.$$

From $BM:MC=DN:DA$ we get $(D,C;T,K_1)=(D,C;T,K_2)$ and hence $K_1=K_2$.