Some pairs {P,Q}={X_i,X_j}={2.112}, {3.1301}, {4,110}, {5,?}, {6,110}
A3B3C3 is also cyclologic to ABC. Cyclology center lies on circumcircle of A3B3C3.
If P=X(32) then Q=?
P=x(32) then Q=(a^2 (a - b) (a + b) (a - c) (a + c) (a^2 b^2 - a^2 c^2 - b^2 c^2) (a^2 b^2 - a^2 c^2 + b^2 c^2) (b^6 + a^4 c^2 + a^2 c^4) (a^4 b^2 + a^2 b^4 + c^6):.... :....)
P=(a^3:b^3:c^3 ) then Q=(a^2 (a - b) (a + b) (a - c) (a + c) (a^2 b^2 - a^2 c^2 - b^2 c^2) (a^2 b^2 - a^2 c^2 + b^2 c^2) (b^6 + a^4 c^2 + a^2 c^4) (a^4 b^2 + a^2 b^4 + c^6):....:....)
Very nice sir but I have a another way mapping, but urs is quite simple and elegant, thank u sir
Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...
Some pairs {P,Q}={X_i,X_j}={2.112}, {3.1301}, {4,110}, {5,?}, {6,110}
ReplyDeleteA3B3C3 is also cyclologic to ABC. Cyclology center lies on circumcircle of A3B3C3.
ReplyDeleteIf P=X(32) then Q=?
ReplyDeleteP=x(32) then Q=(a^2 (a - b) (a + b) (a - c) (a + c) (a^2 b^2 - a^2 c^2 -
ReplyDeleteb^2 c^2) (a^2 b^2 - a^2 c^2 + b^2 c^2) (b^6 + a^4 c^2 +
a^2 c^4) (a^4 b^2 + a^2 b^4 + c^6):.... :....)
P=(a^3:b^3:c^3 ) then Q=(a^2 (a - b) (a + b) (a - c) (a + c) (a^2 b^2 - a^2 c^2 -
ReplyDeleteb^2 c^2) (a^2 b^2 - a^2 c^2 + b^2 c^2) (b^6 + a^4 c^2 +
a^2 c^4) (a^4 b^2 + a^2 b^4 + c^6):....:....)
Very nice sir but I have a another way mapping, but urs is quite simple and elegant, thank u sir
ReplyDelete