TITLE: How to prove collinearity of circumcenters
QUESTION [5 upvotes]: 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

REPLY [2 votes]: 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.

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 $X^*$).



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.



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.