Bulgaria

Since $/\asymp PB{C}_1=/\asymp O{C}_1{A}_1$ and $/\asymp O{C}_1{A}_1=/\asymp OB{A}_1$ (why?), then $/\asymp PBA=/\asymp OBC$. Analogously $/\asymp PAB=/\asymp OAC$. Denote by $M$ and $N$ the projections of $P$ on $BC$ and $AC$. Since $$B{C}_1\cdot BH=BO\cdot BP\cos \angle OBA\cos \angle PBA=B{A}_1\cdot BM,$$ then the points $H$, $C_1$, $A_1$ and $M$ lie on a circle with center at the midpoint $Q$ of $OP$ (since the bisectors of $H{C}_1$ and $M{A}_1$ pass through $Q$). We get in the same way that the points $H$, $C_1$, $N$ and $B_1$ lie on a circle with center $Q$. Hence the points $A_1$, $B_1$, $C_1$, $H$, $M$ and $N$ are con-cyclic.