Dear friends,

Given a triangle ABC and a point P, we denote by Ba and Ca the intersections of the line parallel to BC with AB and BC respectively. The other intersections Cb, Ab and Ac, Bc are defined cyclically. The lines through the midpoints of AcAb and BcCb, BaBc and CaAc, CbCa and AbBa concur in a point Q.

If P={p, q, r}, then Q={(2p+q+r)(p^2+2p(q+r)+qr, , }.

Example: If P=X(2), Q=X(2). If P=X(1), Q={(2a+b+c)(a^2+2a(b+c)+bc), , }. The latter for a=6, b=9 and c=13 gives kx=1.93330058.. (non ETC).

I confirmed that this image was not a quadrangle point.

Best regards,

Seiichi.