Let $H$ be the orthocenter of $\triangle ABC$. Then $\angle CHA=180-\angle CB_1A$, so $B_1CHA$ is cyclic. Hence $\angle HB_1A=\angle HCA=90-\angle B_1A_1C_1$. Thus $H$ is the circumcenter of $\triangle A_1B_1C_1$. Let $K,N$ be the orthocenter and nine-point center of $\triangle A_1B_1C_1$. Then $N$ is the midpoint of $HK$. Let $A_2B_2C_2$ be the medial triangle of $\triangle A_1B_1C_1$. Let $O$ be the circumcenter of $\triangle ABC$. It is easy to see that triangles $ABC$ and $A_2B_2C_2$ are homothetic about $H$. Thus $\triangle HON\sim\triangle HAA_2$. So $\angle HNO=90$ and finally $HO=OK$.

oneplusone wrote: It is easy to see that triangles $ABC$ and $A_2B_2C_2$ are homothetic about $H$. My solution is same as yours. But I think u made a tying mistake in your proof. $ ABC $ and $ A_2B_2C_2 $ are not homothetic about $ H $. There exists a spiral similarity centered at $ H $ takes $ ABC $ to $ A_2B_2C_2 $.

The same solution as yours and oneplusone's ^_^ , maybe it's just another saying of spiral similar because he mentioned $\triangle HON \sim \triangle HAA_2$

My solution starts from the construction of such triangle $\triangle ABC$: take $\triangle A_1B_1C_1$ with $O_1$ circumcenter and rotate the figure of angle $\alpha$ around $O_1$, getting the new triangle $\triangle A'B'C'$; next, let $A_1B_1\cap A'B'=C$ and, similarly $B$ and $C$ and we claim that $\triangle ABC$ is the subject triangle. For this see that $A_1CO_1BA'$ is cyclic ($\widehat{A_1CA'}=\widehat{A_1O_1A'}=\widehat{A_1BA'}=\alpha$), in a similar way $B'B_1AO_1C$ and $AC'C_1BO_1$ are cyclic. Since $\triangle A_1O_1A'=\triangle B_1O_1B'=\triangle C_1O_1C'$, the three circles are equal and we find ourselves within the so called 'the 5 lei coin problem', see post #8 :here. If $O_a, O_b, O_c$ are the circumcenters of the three circles, see that $\triangle O_aO_bO_c=\triangle ABC$ and, as per the above problem, the orthocenter of each of them is the circumcenter of the other. Since there is a spiral similarity sending $\triangle A_1B_1C_1$ to $\triangle O_aO_bO_c$, we see, that the same sends $H_{A_1}$ to $O$, inferring that $\triangle H_{A_1}O_1O \sim \triangle A_1O_1O_a$, done. N.B. there has been a wrong statement, now amended. Best regards, sunken rock

Attachments:

This problem is almost the same as problem 81 IMO Longlist 1992. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005928&sid=9281bd38146e737a227c0109de6e1633#p2005928