See http://www.mathlinks.ro/Forum/viewtopic.php?t=25876 post #3. darij

I think I have it. Since $\angle BA_1A_2 = const$ line $A_1A_2$ goes through constant point $M$ on circle. Same holds for $A_1A_3$ for some point $N$. So we have two projective map $A_2\mapsto A_1$ and $A_1\mapsto A_3$ wich give us another projective map $A_2\mapsto A_3$ fro line $AB$ to line $AC$. Since $A\mapsto A$ this map is perspective, so line $A_2A_3$ goes throught some constant point. Now it is not difficult to see that is orthocenter of triangle $ABC$.

I suppose that $AA_1\cap (BC)\ne \emptyset$. I note $u=m(\widehat {BAA_1}),\ v=m(\widehat {CAA_1})\ (u+v=A),\ D\in AH\cap BC$. $1.\ \dfrac {A_2B}{A_2A}=\dfrac {\sin u}{\sin (v+B)}\cdot \dfrac {\cos B}{\cos A},\ \dfrac {A_3C}{A_3A}=\dfrac {\sin v}{\sin (u+C)}\cdot \dfrac {\cos C}{\cos A}$ $2.\ \sin u\cdot \sin B +\sin v\cdot \sin C=\sin A\cdot \sin (u+C)$ $3.\ H\in A_2A_3\Longleftrightarrow \dfrac {A_2B}{A_2A}\cdot \tan B +\dfrac {A_3C}{A_3A}\cdot \tan C =\tan A$ \[ \overline {\underline {\left| \ \blacktriangle\ A.S.O.\ \blacktriangle\ \right| }} \]\[(1^{\circ}\wedge \ 2^{\circ}\Longrightarrow \ 3^{\circ})\]

Let the altitudes from $B$ respectively $C$ intersect the circumcircle at B' respectively C'. Then points $(C',A_2,A_1)$ and $(A_1,A_3,B')$ are colliniar. Hence by Pascal's theorem apllied to $ABB'A_1C'C$ we conclude that points $A_1,H,A_3$ are collinar.

My solution started like Saylors one but ended with an angle chaise, Anyway Saylors solution was done by like 5 of the contyestants.