ADGEOM 1560 * ADGEOM 1572 * ADGEOM 1574

#1560

 

Dear geometers,

Let ABC be a triangle and P is the point such that pedal triangle DEF of P with respect to ABC is cevian triangle also. Let (Ia),(Ib),(Ic) be excircles of ABC. The tangent orther than BC to (Ia) from D touches (Ia) at X. Similarly we have Y,Z. Then AX, BY,CZ are concurrent at a point Q.

Is the point Q on a cubic ?

Best regards,

Tran Quang Hung.

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#1572

Dear Tran Quang Hung,

Actually, AX, BY, CZ are concurrent for any cevian triangle DEF.  This is a nice variation of the Lozada perspector, using excircles instead of the incircle.  For example, if DEF is the cevian triangle of X(1), the lines concur in X(56).  For X(2), the lines concur in X(7).  For X(7), the lines concur in X(479).

This definitely deserves further exploration!

Best regards,
Randy Hutson

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#1574

In the preamble to X(6056) in ETC (Lozada Perspectors), we can obtain this new variation by:

If P = p : q : r (trilinears), then the perspector is P* = [a/(b + c - a)]p^² : [b/(a + c - b)]q^² : [c/(a + b - c)]r^².

Best regards,
Randy Hutson