Hint

Physicsknight wrote: A hexagon $A_1A_2A_3A_4A_5A_6$ has no four concyclic vertices, and its diagonals $A_1A_4$, $A_2A_5$ and $A_3A_6$ concur. Let $l_i $ be the radical axis of circles $A_iA_{i+1}A_{i-2} $ and $A_iA_{i-1}A_{i+2} $ (the points $A_i $ and $A_{i+6} $ coincide). Prove that $l_i, i=1,\cdots,6$, concur. We show that $l_1,l_2,l_6$ are concurrent. We invert about $A_1$ with arbitrary radius. We have the rephrased problem: Inverted Problem wrote: $A_1,A_2,A_3,A_5,A_6$ are points and $A_4$ lies on the radical axis of $\odot(A_1A_2A_5),\odot(A_1A_3A_6)$ and none of the four points are concyclic. Let $A_2A_5 \cap A_3A_6 = Z$. Let $X$ and $Y$ be the second intersections of $A_2A_4$ and $A_6A_4$ with $\odot(A_2A_3A_6)$ and $\odot(A_2A_5A_6)$. Prove that $A_1Z$ is the radical axis of $\odot(A_1A_2X)$ and $\odot(A_1A_6Y)$. Fix $A_1,A_2,A_3,A_5,A_6$ and animate $A_4$ on the radical axis of $\odot(A_1A_2A_5),\odot(A_1A_3A_6)$. We have $A_4\mapsto X\mapsto A_1A_2X\cap A_1Z$ and $A_4\mapsto Y \mapsto A_1A_6Y\cap A_1Z$ are projective maps. Therefore we need to show for three points of $A_4$. If $A_4 = A_1$ the circles became lines through $A_1$ so we are done in this case. Now we choose $A_4$ on $A_2A_5$. Here we get $\odot(A_1A_6Y)\equiv \odot(A_1A_3A_6)$. Let $A_1Z$ meet $\odot(A_1A_3A_6)$ at $V$. We have $A_1Z\cdot ZV = A_3Z\cdot ZA_5 = A_2Z\cdot ZX$ so $V,A_1,A_2,X$ are concyclic and we are done. Similarly we can choose $A_4$ on $A_3A_6$ and we are done similarly. This finishes the three cases for the problem and we are done.

Stats: Only 1 student solved this on the contest- AoPS User: DragonEye (India) who successfully bary bashed this problem on the contest

Is there any other solution without animating $A_6$ ? The official solution is also animating $A_6$.