Saudi Arabian IMO Booklet

Consider the points that arranged as following figure, the other cases will be proved similarly. Denote $R,S$ as intersection of the pairs of lines $PB,QC$ and $PA,QD$. Note that $BMCR$ and $ALDS$ are cyclic quadrilateral. Thus $$\begin{array}{rl}\angle KMN& =\angle BRC=180^{\circ }-(\angle BCR+\angle CBR)\\ & =180^{\circ }-(\angle PBA+\angle QDC)\\ & =180^{\circ }-(\angle PAD+\angle QDA)=\angle ASD=\angle KLN.\end{array}$$ Hence, $K,L,M,N$ belong to the same circle.

Now, consider the following claim: Let be given two circles $(P,R)$, $(Q,R^{\prime })$ and some line cuts them at $B,A,D,C$ such that $AB=CD$ (see the figure). The tangent lines at $A,C$ respectively of $(P),(Q)$ meet at $X$ then $\frac{XP}{XQ}=\frac{R}{R^{\prime }}$.

Indeed, by applying the sine law for triangle $XAC$, we get $$\frac{XA}{XC}=\frac{\sin XCA}{\sin XAC}=\frac{\sin \frac{CQD}{2}}{\sin \frac{APB}{2}}=\frac{CD}{AB}\cdot \frac{R}{R^{\prime }}=\frac{R}{R^{\prime }}=\frac{PA}{QC}.$$ Thus two triangles $XPA$ and $XQC$ are similar, which implies that $\frac{XP}{XQ}=\frac{R}{R^{\prime }}$. The claim is proved.

Back to the problem, denote $X,Y$ as the external and the internal homothety centers of $(P),(Q)$ then $\frac{XP}{XQ}=\frac{YP}{YQ}=k$ with $k$ is the ratio of radius of $(P),(Q)$. It is easy to check that these radiuses are different, otherwise the the tangent lines of $(P),(Q)$ will be parallel and points $K,L,M,N$ will not exist, thus $k\ne 1$. In the other hand, by applying the above claim, we get $$\frac{MP}{MQ}=\frac{NP}{NQ}=\frac{KP}{KQ}=\frac{LP}{LQ}=k.$$ Hence, six points $X,Y,M,N,K,L$ are all belong to the Apollonius circle with ratio $k$ constructing on the segment $PQ$. Thus, the point $X$, which also is the intersection of two common external tangent lines of $(P),(Q)$, is on $(\omega )$. $\square$