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
$$\begin{aligned} \angle KMN &= \angle BRC = 180^\circ - (\angle BCR + \angle CBR) \\ &= 180^\circ - (\angle PBA + \angle QDC) \\ &= 180^\circ - (\angle PAD + \angle QDA) = \angle ASD = \angle KLN. \end{aligned}$$
Hence, $K, L, M, N$ belong to the same circle.

Now, consider the following claim: Let be given two circles $(P, R)$, $(Q, R')$ 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 $\frac{XP}{XQ} = \frac{R}{R'}$.

Indeed, by applying the sine law for triangle $XAC$, we get
$$\frac{XA}{XC} = \frac{\sin XCA}{\sin XAC} = \frac{\sin \frac{CQD}{2}}{\sin \frac{APB}{2}} = \frac{CD}{AB} \cdot \frac{R}{R'} = \frac{R}{R'} = \frac{PA}{QC}.$$
Thus two triangles $XPA$ and $XQC$ are similar, which implies that $\frac{XP}{XQ} = \frac{R}{R'}$. 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 $k \neq 1$. 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 $(\omega)$. $\square$