Nice problem. * You can consider intersection points both in the positive part and negative part and you get two perspectors P1 and P2. * The line P1P2 goes through X(n) for n=1, 57, 1360, 1699, 1709, 1768, 1836, 3358, 4312, 5805, 7956, 7965, 8727, 11219, 11246, 11372, 13226, 15299, 17728, 24465 * The circle through the circumcenter O and also through P1, P2 goes through X(7330).
Nice problem.
ReplyDelete* You can consider intersection points both in the positive part and negative part and you get two perspectors P1 and P2.
* The line P1P2 goes through X(n) for n=1, 57, 1360, 1699, 1709, 1768, 1836, 3358, 4312, 5805, 7956, 7965, 8727, 11219, 11246, 11372, 13226, 15299, 17728, 24465
* The circle through the circumcenter O and also through P1, P2 goes through X(7330).
ABC is orthologic to A1B1C1 and A2B2C2. In both case orthology center is X(1)-incenter of ABC.
ReplyDeleteX(1)X(3) line of ABC is Euler Line of A1B1C! and A2B2C2.
ReplyDelete