HYACINTHOS 26734

[Antreas P. Hatzipolakis]:

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

Denote:

MaMbMc = the midheight triangle
(ie Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.)

Maa, Mab, Mac = the reflections of Ma in BC, CA, AB, resp.
Mba, Mbb, Mbc = the reflections of Mb in BC, CA, AB, resp.
Mca, Mcb, Mcc =  the reflections of Mc in BC, CA, AB, resp.

(O), (Oa), (Ob), (Oc) = the circumcircles of ABC, MaaMabMac, MbaMbbMbc, McaMcbMcc, resp.

R1 = the radical axis of (O), (Oa)
R2 = the radical axis of (O), (Ob)
R3 = the radical axis of (O), (Oc)

 

A*B*C* = the triangle bounded by MabMac, MbaMbc, McaMcb

A**B**C** = the triangle bounded by R1, R2, R3

1. A'B'C', A*B*C* are homothetic. The homothetic center lies on the Euler line of ABC
2. A'B'C', A**B**C** are homothetic
3. A*B*C*, A**B**C** are homothetic

4. ABC, A**B**C** are perspective

5. ABC, A*B*C* are orthologic.
The orthologic center (ABC, A*B*C*) is the O
The orthologic center (A*B*C*, ABC) ?
6. ABC, A**B**C** are orthologic.

The orthologic center (ABC, A**B**C**) is the O
The orthologic center ( A**B**C**, ABC) is the N

 

 

 

[César Lozada]:

 

1.       A'B'C', A*B*C* are homothetic.. The homothetic center lies on the Euler line of ABC ✓

Z1 = X(25)

 

2.        A'B'C', A**B**C** are homothetic ✓

Z2 = X(6)X(25) ∩ X(389)X(6623)

= SB*SC*(SB+SC)*(3*SA-16*R^2+3* SW) : : (barycentrics)

= On lines: {6, 25}, {185, 5893}, {389, 6623}, {403, 14831}, {1112, 3917}, {1597, 5462}, {5020, 11470}, {5943, 8889}

= [ -0.113177512764019, -0.59721933358836, 4.106359795667404 ]

 

3.        A*B*C*, A**B**C** are homothetic ✓

Z3 = X(5)X(13382) ∩ X(6)X(25)

= (SB+SC)*((6*R^2-3*SW)*S^2+(16* R^2-3*SW)*SB*SC) : : (barycentrics)

= On lines: {5, 13382}, {6, 25}, {5446, 5944}, {5893, 13474}

= [ -0.856900021303265, -2.00021877937257, 5.420923646689970 ]

 

4.       A*B*C*, A**B**C** are homothetic ✓

Z4= X(6)

 

5.       ABC, A*B*C* are orthologic. The orthologic center (ABC, A*B*C*) is the O✓. The orthologic center (A*B*C*, ABC) ?

Z5 = X(5)

 

6.       ABC, A**B**C** are orthologic.. The orthologic center (ABC, A*B*C*) is the O✓. The orthologic center (A**B**C**, ABC) is the N ✗

Z6 = X(3)X(6) ∩ X(5)X(13382)

= (SB+SC)*(3*SA^2-16*R^2*SA-3* SB*SC) : : (barycentrics)

= 5*X(3)+3*X(52) = X(3)+3*X(389) = 7*X(3)+9*X(568) = 13*X(3)+3*X(6243) = X(3)-3*X(9729) = X(3)-9*X(9730) = 11*X(3)-3*X(10625) = 25*X(3)-9*X(13340) = 5*X(3)-3*X(13348) = X(52)-5*X(389) = 7*X(52)-15*X(568) = 13*X(52)-5*X(6243) = X(52)+5*X(9729) = X(52)+15*X(9730) = 11*X(52)+5*X(10625) = 5*X(52)+3*X(13340) = 7*X(389)-3*X(568) = 13*X(389)-X(6243) = X(389)+3*X(9730) = 11*X(389)+X(10625) = 5*X(389)+X(13348) = 3*X(568)+7*X(9729) = X(568)+7*X(9730)

= On lines: {3, 6}, {5, 13382}, {30, 12002}, {51, 3146}, {143, 12103}, {185, 3091}, {373, 12111}, {542, 9825}, {546, 5462}, {631, 14831}, {632, 5892}, {1154, 12108}, {1173, 7464}, {1204, 5422}, {1216, 14869}, {1493, 11802}, {3066, 12174}, {3088, 5476}, {3090, 5890}, {3523, 14531}, {3525, 5562}, {3529, 3567}, {3544, 15024}, {3627, 5946}, {3628, 10219}, {3818, 9815}, {3819, 5889}, {3857, 15026}, {5072, 12162}, {5076, 10575}, {5609, 11806}, {5640, 11381}, {5663, 12811}, {6756, 11645}, {10095, 12102}, {10257, 12242}, {11440, 15018}, {11444, 15082}, {11459, 12045}, {11562, 15027}, {11818, 14864}, {12086, 15019}, {12812, 13363}, {13367, 15053}, {13417, 15021}, {13474, 13570}

= midpoint of X(i) and X(j) for these {i,j}: {5, 13382}, {52, 13348}, {389, 9729}, {5462, 13630}, {5890, 6688}, {6102, 11793}

= reflection of X(11695) in X(12006)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (185, 15043, 5943), (389, 9730, 9729), (5892, 6102, 11793), (11432, 13346, 5097)

= [ 3.960026783192471, 3.74724411965174, -0.781286116247964 ]

 

César Lozada