QG-P10 2^nd Quasi Orthocenter

Let P1.P2.P3.P4 be a Quadrigon and let S be its Diagonal Crosspoint QG-P1.

Let Hi i+1 = Orthocenter of Triangle Pi.Pi+1.S (i = cyclic sequence 1,2,3,4) .

Now is H12.H23.H34.H41 a parallelogram. Remarkable is that each vertex of the Reference Quadrigon lies on a sideline of the parallelogram and that the sides of this parallelogram are perpendicular to the diagonals of the Reference Quadrigon.

QG-P10 is the Center of this parallelogram.

CT-Coordinates QG-P10 in 3 QA-Quadrigons (only coordinates of 1^st Quadrigon point are given)

(-a² (p² + p q + q r + r²) S_A + (c² p q + a² p r + c² p r + a² q r) S_C + 4 (2 p² + p q + p r + q r + r²) Δ² :

-b² (a² + c²) p r – (a² – c²) q r S_C + S_A ((a² – c²) p q + (p² + r²) S_C) + 8 p r Δ² :

S_A (c² p q + a² p r + c² p r + a² q r + (p² + p q + q r) S_B) – c² r² S_C + 4 r (p + 2 r) Δ² )

CT-Coordinates QG-P10 in 3 QL-Quadrigons (only coordinates of 1^st Quadrigon point are given)

(a² m (b² l² + c² l² – 2 b² l m + b² m²) n – a² m² n² S_A – 2 c² l m² n S_C + S_B (-l m (b² m² + a² n² + b² n²) + l² (m² + n²) S_C) – 4 (l – 2 m) m n (2 l – m + n) Δ² :

b² (a² + b² + c²) l m² n – b² l² n² S_B – m (b² l² + c² l² + b² m²) n S_C + S_A (-l m (b² m² + a² n² + b² n²) + m² (l² + n²) S_C) – 4 l n (l m – 2 l n + m n) Δ² :

c² l m (b² m² – 2 b² m n + a² n² + b² n²) – m (b² l² + c² l² + b² m²) n S_B + S_A (-2 a² l m² n + (l² + m²) n² S_B) – c² l² m² S_C + 4 l m (2 m – n) (l – m + 2 n) Δ²)

CT-Area of QG-P10-Triangle in the QA-environment (equals 4 * area QA-Morley Triangle)

((a² q r + b² p r+ c² p q)²) / (8 Δ (p + q) (p + r) (q + r) (p + q + r))

CT-Area of QG-P10-Triangle in the QL-environment

0 (because vertices are collinear)

–

DT-Coordinates QG-P10 in 3 QA-Quadrigons (only coordinates of 1^st Quadrigon point are given)

(-(Sc (p²-q²-r²)-2 Sb q²) (-c² p²+a² r²) :

-Sa² p² (p²+q²-r²)+Sc² r² (p²-q²-r²)-S² q² (p²-q²+r²)-2 Sb q² (Sa p²+Sc r²) :

(Sa (p²+q²-r²)+2 Sb q²) (c² p²-a² r²)}

DT-Coordinates QG-P10 in 3 QL-Quadrigons (only coordinates of 1^st Quadrigon point are given)

(Sb m² (Sb (l²-n²)+Sc (l²-m²)) : Sa a² l² m²+Sc c² m² n²-Sa Sc m⁴-S² l² n² : -Sb m² (Sb (l²-n²)+Sa (m²-n²)))

DT-Area of QG-P10-Triangle in the QA-environment (equals 4 * area QA-Morley Triangle)

-(Sa p²+Sb q²+Sc r²)² / (2S (-p+q+r) (p+q-r) (p-q+r) (p+q+r))

DT-Area of QG-P10-Triangle in the QL-environment

0 (because vertices are collinear)

Properties

• QG-P8, QG-P9, QG-P10, QG-P11 are collinear on QG-L5, 2^nd QG-Quasi Euler Line.
• QG-P10 is the Orthocenter of the 2^nd QG-Quasi Diagonal Triangle: QG-Tr2.
• QG-P10 is the Reflection of QG-P5 in QA-P1 (QA-Centroid).
• QG-P10 is the Reflection of QG-P15 in QG-P7.
• QG-P10 is the Midpoint of QG-P1 and QG-P6.
• The three QA-versions of QG-P10 define a circle with center QA-P15 (note Eckart Schmidt).
• The three QA-versions of QG-P10 form a Triangle with centroid QA-P14 (note Eckart Schmidt).
• The three QL-versions of QG-P10 are collinear on the line QL-L6 (Quasi Ortholine).
• QA-P2 (Euler-Poncelet Point) lies on the circumcircle of the triangle formed by the three QA-versions of QG-P10. QA-P15 is the circumcenter.
• The line through the 1^st and 2^nd Quasi Orthocenters (QG-P6 and QG-P10) is perpendicular to the Newton Line (QL-L1).
• When the Quadrigon is cyclic the Maltitudes concur in QG-P10 (a Maltitude is a line drawn from the midpoint of a side perpendicular to the opposite side of a Quadrigon). This point in a cyclic Quadrigon is also known as the Anticenter.
• Let Hi be the Orthocenters of the Component Triangles Pj.Pk.Pl of the Reference Quadrigon. Now QG-P10 is the Diagonal Crosspoint of the Quadrigon formed by the Midpoints(Pi, Hi) (Eckart Schmidt, September 18, 2012).
• The QG-P10 Triple Triangle in a Quadrangle is Orthologic wrt all QA-Component Triangles. See QA-Tr-1.

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