Denote $H'$ intersection of $CH$ and $A'B'$, $O'$ intersection of $CO$ and $A_1B_1$ It is well known that $ CH'\perp A'B'$ and $CO'\perp A_1B_1$. So $CC'O'H'$ is cyclic. Consider inversion with th pole $C$ and power $k = CA'*CA_1=CB'*CB_1$. Line $A'H'B'$ transforms in $(A_1HB_1C)$ Line $A_1O'B_1$ transforms in $(A'OB'C)$ It can easily to see $I(O')= O, I(H')= H$ So circle $(CC'O'H')$ transforms in line $OH$ which is perpendicular to diameter $(CC'O'H')$ - $CC'$. So $OH\perp CC'$.

Here's another solution: Since $A_1A'B'B_1$ is cyclic, $C'A'\cdot C'B'=C'A_1\cdot C'B_1$, so $C'$ lies on the radical axis of the circles $(CA_1B_1),(CA'B')$. This means that $CC'$ is perpendicular to the line of centers of these two circles (call this $(*)$). However, the center of $(CA_1B_1)$ is the midpoint of $CH$, while the center of $(CA'B')$ is the midpoint of $OH$, so $OH$ is parallel to the line of centers, and thus perpendicular to $CC'$, according to $(*)$.

The official solution is the same as grobber's one. It is quiate unusual, since they needed to present a notion of radical axe which is out of school education.

Myth wrote: The official solution is the same as grobber's one. It is quiate unusual, since they needed to present a notion of radical axe which is out of school education. Maybe L. Emel'yanov was expecting the contestants to deduce the radical axe idea by themselves

Very funny!

Myth wrote: 11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$. More general (but equally trivial): Let P and Q be two isogonal conjugate points with respect to triangle ABC. Let X and Y be the orthogonal projections of the point P on the sides BC and CA of triangle ABC, and let X' and Y' be the orthogonal projections of the point Q on these sides. The lines XY and X'Y' intersect each other at the point C'. Prove that $CC^{\prime}\perp PQ$. The above problem 11.4 is a particular case of this if P = H and Q = O. Both Prowler's and Grobber's solutions of problem 11.4 work similarly for the general version (one just has to apply the well-known result that the points X, Y, X', Y' are concyclic). darij

darij grinberg wrote: Myth wrote: 11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$. More general (but equally trivial)... Say it only for yourself. If this problem appeared as 11.4 it means that the jury had some reasons for that.

grobber wrote: The center of $(CA'B')$ is the midpoint of $OH$. Small typo here: it should be the midpoint of $OC$. A very nice solution, grobber!