Iranian Mathematical Olympiad

Without loss of generality, we can suppose that ${\omega }_1$ is shorter than ${\omega }_2$. Let $l^{\prime }$ be the common external tangent of ${\omega }_1^{\prime }$ and ${\omega }_2$ and let it intersect $l$ at $O$. Let $OA$ be tangent to ${\omega }_1$ at $A$, $B:={\omega }_1\cap {\omega }_1^{\prime }$, $C:={\omega }_1^{\prime }\cap l^{\prime }$, $D:=l^{\prime }\cap {\omega }_2$, $E:={\omega }_2\cap {\omega }_2^{\prime }$ and let $OF$ be tangent to ${\omega }_2^{\prime }$ at $F$. We must prove that $A$, $O$ and $F$ are collinear. Let the common internal tangent of ${\omega }_1$ and ${\omega }_1^{\prime }$ meet $l$ at $O^{\prime }$, so $$O^{\prime }X\cdot O^{\prime }Y=O^{\prime }{B}^2=O^{\prime }{X}^{\prime }\cdot O^{\prime }{Y}^{\prime }$$ Thus the power of $O^{\prime }$ with respect to ${\omega }_2$ and ${\omega }_2^{\prime }$ is equal to this value. Hence the circle $C_1$ with center $O^{\prime }$ and radius $O^{\prime }B$ passes through $E$ and is orthogonal to all arcs ${\omega }_1$, ${\omega }_2$, ${\omega }_1^{\prime }$ and ${\omega }_2^{\prime }$. We also have $O{A}^2=OX\cdot OY=O{D}^2$ and $O{F}^2=O{X}^{\prime }\cdot O{Y}^{\prime }=O{C}^2$, so the circle $C_2$ with center $O$ and radius $OA$ is orthogonal to both ${\omega }_1$ and ${\omega }_2$ and also the circle $C_3$ with center $O$ and radius $OF$ is orthogonal to both ${\omega }_1^{\prime }$ and ${\omega }_2^{\prime }$. We use the following classic lemma without proving it. Lemma 1. Let circle $\omega$ be orthogonal to circles $C_1$ and $C_2$ and intersect them in $\{A_1,B_1\}$ and $\{A_2,B_2\}$, respectively. Let $O$, $O_1$ and $O_2$ be the centers of $\omega$, $C_1$ and $C_2$, respectively. Suppose that the right angles $\angle O{A}_{1}O$ and $\angle O{A}_{2}O$ have different orientations; i.e. one of them is clockwise and the other one is not. Then the pairs $\{A,A^{\prime }\}$ and $\{B,B^{\prime }\}$ are direct anti-homologous points and pairs $\{A,B^{\prime }\}$ and $\{A^{\prime },B\}$ are inverse anti-homologous points for $C_1$ and $C_2$. $\square$ Using the lemma for circle $C_2$, we deduce that $A$ and $D$ are inverse anti-homologous points for ${\omega }_1$ and ${\omega }_2$ (we do not check the conditions here for brevity). Using the lemma again for circle $C_1$, we get that $B$ and $E$ are also inverse anti-homologous points for ${\omega }_1$ and ${\omega }_2$. So, the quadrilateral $ADEB$ is cyclic. Similarly, by using the lemma twice for circles $C_1$ and $C_3$, we get that pairs $\{B,E\}$ and $\{C,F\}$ are inverse anti-homologous points for ${\omega }_1^{\prime }$ and ${\omega }_2^{\prime }$. So the quadrilateral $BEFC$ is also cyclic. According to the lemma for circle $C_1$, points $B$ and $E$ are direct anti-homologous points for ${\omega }_1^{\prime }$ and ${\omega }_2$, and so are $C$ and $D$. Hence, the quadrilateral $BCDE$ is also cyclic. By the last three paragraphs, we deduce that all points $A$, $B$, $\dots$, $F$ lie on the circumcircle of triangle $BCD$, which we call $C_4$ (checking distinctness of the points is not mentioned here). Now, circles $C_2$, $C_3$ and $C_4$ have a line of symmetry, so their intersections and centers also have a line of symmetry. It follows that $O$, $A$ and $F$ are collinear and the problem is solved.

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Alternative solution.

Second Solution. To prove that $A$, $B$, $\dots$, $F$ lie on a circle, we can use the following lemma. Lemma 2. If a circle $\omega$ is tangent to circles $C_1$ and $C_2$, then the points of tangency are anti-homologous points for $C_1$ and $C_2$. If none of $C_1$ and $C_2$ contains the other and $\omega$ is tangent to them both internally or both externally, then the points of tangency are direct anti-homologous points. $\square$ Note that these lemmas remain true if one of the circles is a line (considered as a circle with infinite radius). If we suppose that interior of $l^{\prime }$ (considered as a circle) is the half-plane not containing $X$, then according to lemma 2 for circle ${\omega }_1^{\prime }$, points $B$ and $C$ are direct anti-homologous for circles ${\omega }_1$ and $l^{\prime }$. By applying lemma 1 for circle $C_2$, points $A$ and $D$ are also anti-homologous points for ${\omega }_1$ and $l^{\prime }$. Using the fact that the arcs are all in one side of $l$, it is easily seen that $A$ and $D$ are also direct anti-homologous points, so $ABCD$ is a cyclic quadrilateral. Similarly, by considering anti-homologous points for pairs of circles $\{{\omega }_2,{\omega }_1^{\prime }\}$, $\{l^{\prime },{\omega }_2^{\prime }\}$ we conclude that quadrilaterals $BCDE$ and $CDEF$ are also cyclic, so points $A$, $B$, $C$, $D$, $E$ and $F$ lie on a circle which we call $C_4$. The rest of the proof is the same as the first solution. $\square$

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Alternative solution.

Third Solution. Define points $A$, $B$, $\dots$, $F$ and circles $C_1$, $C_2$ and $C_3$ similar to the first solution. We can suppose that $X$ is between $X^{\prime }$ and $O$, without loss of generality. Thus, $C_1$ and $C_2$ are disjoint and $C_3$ contains both of them. According to lemma 1 and the fact that ${\omega }_1$ lies entirely on one side of $l$, points $A$ and $B$ are direct anti-homologous points for $C_2$ and $C_1$. Similarly, $B$ and $C$ are inverse anti-homologous points for $C_1$ and $C_3$ and points $C$ and $D$ are direct anti-homologous points for $C_3$ and $C_2$. Pairs of points $\{D,E\}$ and $\{E,F\}$ are also anti-homologous in a similar manner as are pairs $\{A,B\}$ and $\{B,C\}$, respectively. Our motivation is that two anti-homologous points are related to each other by an inversion depending only on the two circles. So the problem says that the composition of six particular inversions is the identity. With this intuition in mind, we move all points to $C_1$. Let $A^{\prime }$, $C^{\prime }$, $D^{\prime }$ and $F^{\prime }$ be points on $C_1$ such that pairs $\{A,A^{\prime }\}$ and $\{D,D^{\prime }\}$ are direct homologous points for circles $C_1$ and $C_2$, and pairs $\{C,C^{\prime }\}$ and $\{F,F^{\prime }\}$ are inverse homologous points for circles $C_1$ and $C_3$, respectively. We have $$O^{\prime }{D}^{\prime }\parallel OD\parallel OC\parallel O^{\prime }{C}^{\prime }$$ so, $A^{\prime }{D}^{\prime }$ passes through $O^{\prime }$. The lines $A^{\prime }B$ and $D^{\prime }E$ pass through the direct homothetic center of $C_1$ and $C_2$, which lies on $l$. The lines $B{C}^{\prime }$ and $E{F}^{\prime }$ pass through the inverse homothetic center of $C_1$ and $C_3$, which also lies on $l$. So, by Pascal's theorem for the cyclic hexagon $A^{\prime }B{C}^{\prime }{D}^{\prime }E{F}^{\prime }$, we conclude that $A^{\prime }{F}^{\prime }$ also passes through $O^{\prime }$. Hence, we similarly have $$OA\parallel O^{\prime }{A}^{\prime }\parallel O^{\prime }{F}^{\prime }\parallel OF,$$ so $O$, $A$ and $F$ are collinear and we are done. $\square$