HYACINTHOS 25424

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

A', B', C' = the reflections of I in BC, CA, AB, resp.

Ma, Mb, Mc = the midpoints of AA', BB', CC, resp.

M1, M2, M3 = the midpoints of AI, BI, CI, resp.

The circumcircles of the trianges IMaM1, IMbM2, IMcM3 are coaxial.

2nd intersection?


[Tran Quang Hung]:

Dear Mr Antreas, this problem is true for NPC center N

Let ABC be a triangle.

Denote:

A', B', C' = the reflections of N in BC, CA, AB, resp.

Ma, Mb, Mc = the midpoints of AA', BB', CC, resp.

M1, M2, M3 = the midpoints of AN, BN, CN, resp.

The circumcircles of the trianges NMaM1, NMbM2, NMcM3 are coaxial.

 

and I think Napoleon-Feuerbach cubic is locus of P for this problem is true ?

[César Lozada]:

 

 

>>I think Napoleon-Feuerbach cubic is locus of P for this problem is true ?

 

Proved. Moreover, N, P and Q2 (the 2nd point of intersection) are collinear.

 

Conjecture: Q2(P)=N for P = ({Feuerbach cubic} /\ {circumcircle of ABC} ) - {A, B, C}

 

For P=I. See #25422

 

For P=0, Q2(O) = N

 

For P=X(54), Q2(P) = X(11804)

 

For P=N:

 

Q2(N) = midpoint of X(3) and X(1263)

= 8*S^4-2*(3*R^4+(SB+SC)*(2*SA+2 *SW-5*R^2))*S^2-SA*(27*R^4-22* SW*R^2+4*SW^2)*(SB+SC) (barycentrics)

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

= On lines: {3,1263}, {5,49}, {30,137}, {128,3628}, {140,6592}, {549,930}

= midpoint of X(i) and X(j) for these {i,j}: {3,1263}, {5,1141}

= reflection of X(i) in X(j) for these (i,j): (128,3628), (6592,140)

= [ -0.011003000623983, 0.60814053458003, 3.224722419793887 ]

 

César Lozada