HYACINTHOS 24884

 

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the pedal triangle of N.

Denote:

N* = the NPC center of A'B'C'.

Na, Nb, Nc = the reflections of N* in AH, BH, CH, resp.

N1, N2, N3 = the reflections of Na, Nb, Nc in BC, CA, AB, resp.

The Euler line of N1N2N3 passes through N and N*.

Which points are the N,N* wrt triangle N1N2N3 ?

 

[César Lozada]:

 

Let N w/r to  N1N2N3 = Q w/r to ABC. Then:

Q = Midpoint of X(442) and  X(548)

= (5*sin(A/2)-6*sin(3*A/2))*cos( (B-C)/2)+(-cos(A)+1/2)*cos(B-C )-sin(A/2)*cos(3*(B-C)/2)-5*co s(A)-3*cos(2*A)-3 : : (trilinears)

= (23*R+14*r)*X(3)+(R+2*r)*X(4)

= [23*R+14*r, -21*R-10*r] (Shinagawa coefficients)

= midpoint of X(i) and X(j) for these {i,j}: {442,548}, {3651,10021}

= on line {2,3}

= [ 6.126289114526334, 5.24060788625781, -2.814966722975488 ]

 

Let N* w/r to  N1N2N3 = Q* w/r to ABC. Then:

Q* = Midpoint of X(5) and X(3651)

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

=  (11*R+6*r)*X(3)+(R+2*r)*X(4)

= [11*R+6*r, -9*R-2*r]  (Shinagawa coefficients)

= midpoint of X(i) and X(j) for these {i,j}: {3,5499}, {5,3651}, {6175,8703}

= reflection of X(i) in X(j) for these (i,j): (5428,3530), (6841,3628), (10021,140)

= On lines: {2,3}, {79,5432}, {484,3649}, {2771,6684}, {3035,3647}, {5433,5441}, {6690,6701}

= [ 5.470215334856322, 4.58626484578421, -2.059156719722993 ]

 

Regards,

César Lozada

[APH]:

 

Thanks, César !

The problem can be phrased equivalently:

Let ABC be a triangle, A'B'C' the pedal triangle of N and A"B"C" the pedal triangle of H (orthic triangle).

Denote:

N* = the NPC center of A'B'C'.

N1, N2, N3 = the reflections of N* in A", B", C", resp.

The Euler line of N1N2N3 passes through N and N*.

Which points are the N,N* wrt triangle N1N2N3 ?

Now, if A*, B*, C* are points on the lines N*A", N*B", N*C" dividing the segments N*A", N*B", N*C" in same ratio, then the Euler line of A*B*C* passes through N*,N.

Special case:

A*, B*, C* = A", B", C", resp. :

Let ABC be a triangle, A'B'C' the pedal triangle of N and A"B"C" the pedal triangle of H (orthic triangle).

Denote:

N* = the NPC center of A'B'C'.

The Euler line of A"B"C" passes through N,N*.

N is the circumcenter of A"B"C". Which point is the N* wrt triangle A"B"C"?

APH