darij grinberg wrote: Extension: Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC. Darij Actually this is a big hint . For the moment I will give only a very concise sketch. a) Is enough to prove that B',O and I are collinear. b) One can prove (is not very difficult, indeed) that the bisectors (of both angles and sides) meet on c(ABC). c) Then using Pascal theorem one can prove the claim.

This is highly interesting. Could you tell me some details of steps b) and c)? I cannot complete your solution. On the other hand, my own solution is very long. Thanks! Darij

What sprmnt is saying is, I think, that $OX$ and $AB'$ meet on $(ABC)$, and this is a simple angle chase (the same holds, of course, for $OZ$ and $B'C$). All we need to do is compute the angle between $AB'$ and the perp. bisector of $AB$. On the other hand, we know that the bisectors of $\angle A,\angle C$ meet the perpendicular bisectors of $BC,AB$ (respectively) on $(ABC)$.

mmm i don't know if this is ok or not, because it seems to be quite easy... what i had thought is exactly what grobber said: $OX$ and $AB'$ meet on $(ABC)$ (let $M$ be that intersection), and the same holds for $OZ$ and $CB'$ (they meet in a point $N$, in $(ABC)$... let $P$ and $Q$ be the second intersections of $OX$ and $OZ$ with $(ABC)$ (i.e., the intersection of the perpendicular bisectors of $AB$ and $BC$ with the circumcircle), respectively. Now you just have to notice that in the cyclic hexagon $PCNQAM$ the intersections of the opposite sides lie on a line...and those intersections are $PC\cap AQ = I$, $CN\cap AM = B'$ and $NQ\cap MP = O$, so B' lies on the line OI and we are done

darij grinberg wrote: This is highly interesting. Could you tell me some details of steps b) and c)? Darij Tanks to Grobber and RamLaf, I sould have only to attach a picture. BTW, here follow the arguments which prove the claims. If K (opposite to A) and H are the common points to c(ABC) and the axis of BC, then AI pass through K. furtermore, as BCZ is isosceles, the angle bisector of <ZBA pass through H. Similarly for all the other bisectors. PS Darij, what contest is 239MO?

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Wow, this was a cook. Probably I should have thought about the problem for a longer time, and not at midnight. Anyway, all of your solutions are correct and nice. My own solution used nine-point centers, Kosnita points and isogonal conjugates and was so monstrous that I don't want even to mention it. A note to your post, Sprmnt21: sprmnt21 wrote: If K (opposite to A) and H are the common points to c(ABC) and the axis of BC, then AI pass through K. furtermore, as BCZ is isosceles, the angle bisector of <ZBA pass through H. I guess <ZBA should be < ZCA. Again, your proofs are very nice. Now I have another synthetic proof, based on your ideas: Let the lines AX and AB' meet the circumcircle of triangle ABC at the points U and M (apart from A). Since the line AB' bisects the angle XAC, we have < MAC = < UAM, and thus, replacing chordal angles by arcs on the circumcircle of triangle ABC, we have arc MC = arc UM. On the other hand, the point X lies on the perpendicular bisector of the segment AB, and thus < XBA = < BAX, or, in other words, < CBA = < BAU. In terms of arcs, this yields arc CA = arc BU. Together with arc MC = arc UM, this implies arc MA = arc MC + arc CA = arc UM + arc BU = arc BM, what yields that the point M is the midpoint of the arc ACB on the circumcircle of triangle ABC. In other words, the point M is the midpoint of the segment $I_aI_b$, where $I_a$, $I_b$, $I_c$ are the excenters of triangle ABC. Similarly, the point N where the line CB' meets the circumcircle of triangle ABC (apart from C) is the midpoint of the segment $I_bI_c$. Now, applying the Pappos theorem to the collinear point triplets (C, M, $I_a$) and (A, N, $I_c$), we see that the points $CN \cap AM$, $MI_c \cap NI_a$ and $I_aA \cap I_cC$ are collinear. But $CN \cap AM = B^{\prime}$. Further, $MI_c \cap NI_a$ is the point of intersection of the lines joining the vertices $I_c$ and $I_a$ of triangle $I_aI_bI_c$ with the midpoints M and N of the opposite sides $I_aI_b$ and $I_bI_c$; in other words, the point $MI_c \cap NI_a$ is the centroid of triangle $I_aI_bI_c$. Finally, the point $I_aA \cap I_cC$ is the point I, the orthocenter of triangle $I_aI_bI_c$. Hence, we have obtained that the point B' is collinear with the centroid and the orthocenter of triangle $I_aI_bI_c$. In other words, the point B' lies on the Euler line of triangle $I_aI_bI_c$. But this Euler line is the line OI. Hence, the point B' lies on the line OI. Similarly, we can show that the points C' and A' lie on the line OI, and the problem is solved. sprmnt21 wrote: Darij, what contest is 239MO? The Open Olympiad of the 239th Mathematical Lyceum of St. Petersburg, Russia. This is a very special olympiad, organized by enthusiasts, and containing very hard problems. Darij

The perpendicular bisectors of the sides $AB$ and $BC$ of a triangle $ABC$ meet the lines $BC$ and $AB$ at the points $X$ and $Z$, respectively. The angle bisectors of the angles $XAC$ and $ZCA$ intersect at a point $B'$. Similarly, define two points $C'$ and $A'$. Prove that the points $A'$, $B'$, $C'$ lie on one line through the incenter $I$ of triangle $ABC$.