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What is a proper solution for this problem? It has been bugging me for a good while.

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Hint: The key word is homothety.

First just draw any circle with center at $C$ which touch bot $l$ and $m$. Then draw a line through peak and $P$. Mark where this line cuts circle with $A$ and $B$. Now draw a parallel through $P$ to $CA$ (or $CB$). Mark a point where this parallel cuts angle bisector. This is the center of circle(s) you want.

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  • $\begingroup$ Very clever, thanks. $\endgroup$ – Siegmeyer Oct 31 '17 at 10:43

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