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I'm going through Solution 1 for problem G2 from IMO 2006 (https://www.imo-official.org/problems/IMO2006SL.pdf).

In the penultimate paragraph they conclude that homothety $h$ takes circle $(ABP)$ to circle $(CDQ)$. Why is that? Why is point $S$ the homothetic center of these two circles? How does tangency of just one line $SB$ to these two circles lead to this?

Thanks.

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Well, I had a problem with that conclusion also. But in the next step they say $AB$ goes to $DC$ (specificly $A$ gose to $D$).

Say $O$ is a center of big and $O'$ a center of small circle. Then $$\angle AOB = {1\over 2}\angle AXB = {1\over 2}\angle DQC = \angle DO'C$$ so $$\angle OBA = \angle O'CD$$ which means that $O$ goes to $O'$ and we are done.

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