How to prove collinearity of circumcenters Let $A_{1}A_{2}A_{3}$ be a non-isosceles triangle with incenter I. Let $C_{i}$
, $i = 1, 2, 3$, be the
smaller circle through $I$ tangent to $A_{i}A_{i+1}$ and $A_{i}A_{i+2}$ (the addition of indices being mod 3). Let
$B_{i}$
, $i = 1, 2, 3$, be the second point of intersection of $C_{i+1}$ and $C_{i+2}$. Prove that the circumcenters
of the triangles $A_{1}B_{1}I, A_{2}B_{2}I, A_{3}B_{3}I$ are collinear.
This is an IMO shortlisted question , I have the solution which uses inversion to proof this. However I am not able to visulaize it properly as the diagram, which I have drawn as per my understanding clearly shows that the points are not collinear. Please use a diagram and explain the answer in more elegant manner.
Here is the solution


 A: This answer will be a clarification and simplification of the given solution.  As such it will not be a complete proof in itself, on the assumption that a student of the IMO can fill in some of the steps.

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*Notation: $(X)$ and $(XYZ)$ will denote the circle with center $X$ and circumcircle of $X,Y,Z$ respectively.  $X'$ will denote the inversion of point $X$ with respect to the incircle $(I)$ (the OP proof uses the notation $X^*$).



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*The sketch above is the requested diagram.  The aim of the proof is to show that the three circles $(A_iIB_i)$ (red) are coaxal, in which case their centers are collinear.  The circles are coincident at $I$, so they are coaxal if they concur at a second point.  The circles invert in the incircle to become three lines and, if we can prove these lines are concurrent, we are done.



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*Since the constructions involved can result in a cluttered diagram, the above sketch shows the constructions associated with vertex $A_1$.  It is based on a common construction that shows the relationship of incircle, excircle, angle bisector at $A_1$, and the circle $(A_2IA_3)$ (blue). (see e.g. this note by Even Chen )  Also shown is $C_1$ (red) and the inversions of $C_1$ and $(A_2IA_3)$.  Convince yourself that these lines $KL$ and $A'_2A'_3$ are parallel (hint: they are both perpendicular to $A_1I$), and that the inversion of $C_i$ is the line $B'_{i+1}B'_{i+2}$


*Now consider the triangles $\triangle A'_1A'_2A'_3$ and $\triangle B'_1B'_2B'_3$.  As shown in the previous step, these triangles have pairwise parallel sides.  These sides pairwise intersect on the projective line at infinity, so by Desargues theorem there is a point $P'$ such that the triangles are perspective from $P'$.  But this means that the lines $A_iB_i$ are concurrent at $P'$, which means that the circles $C_i$ are concurrent at $P=P''$.  They are concurrent at the two points $I,P$, and thus are coaxal, which is what we wanted to show.
