This problem is inspired from #5175 by Tran Quang Hung.
Let P,Q be two points on Kiepert hyperbola of ABC. Circles APQ, BPQ, CPQ intersects the Kiepert hyperbola of ABC at A', B',C' resp.
R=X(2)-Centroid of of A'B'C' lies on Kiepert hyperbola of ABC.
So we have a transformation T, T(P,Q)->R, Let's show it with * operation. Then we have:
P*Q=R, P*R=Q, Q*R=P
A set of {P,Q,R} is {X(2), X(4), X(9180)}
ReplyDeleteX(2)*X(4)= X(9180)
X(2)*X(9180)=X(4)
X(4)*X(9180)=X(2)
But ı'm not sure how to define the fourth element S (to be identiy).