, 2011

Consider circles $\omega_1$ and $\omega_2$ centred at $A$ and $B$ respectively such that both $\omega_1$ and $\omega_2$ are orthogonal to $\omega$. Let $I$ be one of the intersection points of $\omega_1$ and $\omega_2$. Note that $I$ must exist because $AB$ is outside $\omega$. Then invert the diagram through a circle centred at $I$ with any radius. $\omega'_1$ and $\omega'_2$ (the images of $\omega_1$ and $\omega_2$) are straight lines passing through the centre of $\omega'$ (since they are orthogonal and angles between circles are preserved by inversion). $\omega'_1$ passes through the centre of circle $AX'_iQ'_i$ and $\omega'_2$ passes through the centre of circle $BX'_iQ'_i$ for the same reason. Hence $Q'_i$ is $X'_i$ reflected in line $\omega'_1$ and $X'_{i+1}$ is the reflection of $Q'_i$ in the line $\omega'_2$. Thus $\angle X'_iOX'_{i+1} = 2\theta$ where $\theta$ is the angle between the lines $\omega'_1$ and $\omega'_2$. Thus $X'_0 = X'_k$ is equivalent to $k\theta = 2n\pi$, for some integer $n$. This means that consecutive terms in the sequence of points $X'_i$ are determined by a rotation through a constant angle. Hence this result will be true irrespective of the choice of $X'_0$. There is a one-to-one correspondence between $X_i$ and $X'_i$ and hence the same is true if $X_0 = X_k$ for some positive integer $k$.