HYACINTHOS 26410

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.

Ma, Mb, Mc = the midpoints of AN, BN, CN, resp.

P = the Poncelet Point of (ABCN)
(ie the point the NPCs of ABC, NBC, NCA, NAB are concurrent at).

The circumcircles of PNaMa, PNbMb, PNcMc are coaxial.

Second intersection?

[César Lozada]:

Q2 = X(5)X(930) ∩ X(546)X(1154)

= (cos(2*A)-cos(4*A)+7/2)*cos(B- C)+2*(cos(A)-cos(3*A))*cos(2*( B-C))-3*(cos(2*A)-1)*cos(3*(B- C))+cos(5*A)+3*cos(A)-3*cos(3* A) : : (trilinears)

= 24*S^4+2*(11*R^4-(5*SA+3*SW)* R^2-2*(5*SA+SW)*(SB+SC))*S^2-( 5*R^4+2*R^2*SW-4*SW^2)*(SB+SC) *SA : : (barycentrics)

= On lines: {5, 930}, {137, 5501}, {546, 1154}, {10285, 12026}

= {X(137), X(5501)}-Harmonic conjugate of X(8254)

= [ -3.153893667415306, -3.20282776337338, 7.313650010742246 ]

 

César Lozada