Κυριακή 14 Ιουλίου 2024

ETC APH-FJGC-EULER

X(44234) = 1ST HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

X(44898) = 2ND HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

X(45306) = 3RD HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

X(45307) = 4TH HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

X(45308) = 5TH HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

X(64480) = 6TH HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

Barycentrics    -a^16 b^2 + 5 a^14 b^4 - 9 a^12 b^6 + 5 a^10 b^8 + 5 a^8 b^10 - 9 a^6 b^12 + 5 a^4 b^14 - a^2 b^16 - a^16 c^2 - 2 a^14 b^2 c^2 + 3 a^12 b^4 c^2 + 13 a^10 b^6 c^2 - 23 a^8 b^8 c^2 + 13 a^6 b^10 c^2 - 6 a^4 b^12 c^2 + 4 a^2 b^14 c^2 - b^16 c^2 + 5 a^14 c^4 + 3 a^12 b^2 c^4 - 28 a^10 b^4 c^4 + 17 a^8 b^6 c^4 + 17 a^6 b^8 c^4 - 7 a^4 b^10 c^4 - 12 a^2 b^12 c^4 + 5 b^14 c^4 - 9 a^12 c^6 + 13 a^10 b^2 c^6 + 17 a^8 b^4 c^6 - 42 a^6 b^6 c^6 + 8 a^4 b^8 c^6 + 28 a^2 b^10 c^6 - 9 b^12 c^6 + 5 a^10 c^8 - 23 a^8 b^2 c^8 + 17 a^6 b^4 c^8 + 8 a^4 b^6 c^8 - 38 a^2 b^8 c^8 + 5 b^10 c^8 + 5 a^8 c^10 + 13 a^6 b^2 c^10 - 7 a^4 b^4 c^10 + 28 a^2 b^6 c^10 + 5 b^8 c^10 - 9 a^6 c^12 - 6 a^4 b^2 c^12 - 12 a^2 b^4 c^12 - 9 b^6 c^12 + 5 a^4 c^14 + 4 a^2 b^2 c^14 + 5 b^4 c^14 - a^2 c^16 - b^2 c^16 + 4 a^13 b c OH S - 6 a^11 b^3 c OH S - 6 a^9 b^5 c OH S + 10 a^7 b^7 c OH S + 6 a^5 b^9 c OH S - 12 a^3 b^11 c OH S + 4 a b^13 c OH S - 6 a^11 b c^3 OH S + 24 a^9 b^3 c^3 OH S - 12 a^7 b^5 c^3 OH S - 24 a^5 b^7 c^3 OH S + 30 a^3 b^9 c^3 OH S - 12 a b^11 c^3 OH S - 6 a^9 b c^5 OH S - 12 a^7 b^3 c^5 OH S + 36 a^5 b^5 c^5 OH S - 18 a^3 b^7 c^5 OH S + 12 a b^9 c^5 OH S + 10 a^7 b c^7 OH S - 24 a^5 b^3 c^7 OH S - 18 a^3 b^5 c^7 OH S - 8 a b^7 c^7 OH S + 6 a^5 b c^9 OH S + 30 a^3 b^3 c^9 OH S + 12 a b^5 c^9 OH S - 12 a^3 b c^11 OH S - 12 a b^3 c^11 OH S + 4 a b c^13 OH S : :

See Antreas Hatzipolakis and Francisco Javier García Capitán, euclid 6388.

X(64480) lies on these lines: {2, 3), {542, 44123}, {1989, 8106}, {8115, 45016}, {13415, 18374}, {15360, 24650}, {32225, 44125}


X(64481) = 7TH HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

Barycentrics    -a^16 b^2 + 5 a^14 b^4 - 9 a^12 b^6 + 5 a^10 b^8 + 5 a^8 b^10 - 9 a^6 b^12 + 5 a^4 b^14 - a^2 b^16 - a^16 c^2 - 2 a^14 b^2 c^2 + 3 a^12 b^4 c^2 + 13 a^10 b^6 c^2 - 23 a^8 b^8 c^2 + 13 a^6 b^10 c^2 - 6 a^4 b^12 c^2 + 4 a^2 b^14 c^2 - b^16 c^2 + 5 a^14 c^4 + 3 a^12 b^2 c^4 - 28 a^10 b^4 c^4 + 17 a^8 b^6 c^4 + 17 a^6 b^8 c^4 - 7 a^4 b^10 c^4 - 12 a^2 b^12 c^4 + 5 b^14 c^4 - 9 a^12 c^6 + 13 a^10 b^2 c^6 + 17 a^8 b^4 c^6 - 42 a^6 b^6 c^6 + 8 a^4 b^8 c^6 + 28 a^2 b^10 c^6 - 9 b^12 c^6 + 5 a^10 c^8 - 23 a^8 b^2 c^8 + 17 a^6 b^4 c^8 + 8 a^4 b^6 c^8 - 38 a^2 b^8 c^8 + 5 b^10 c^8 + 5 a^8 c^10 + 13 a^6 b^2 c^10 - 7 a^4 b^4 c^10 + 28 a^2 b^6 c^10 + 5 b^8 c^10 - 9 a^6 c^12 - 6 a^4 b^2 c^12 - 12 a^2 b^4 c^12 - 9 b^6 c^12 + 5 a^4 c^14 + 4 a^2 b^2 c^14 + 5 b^4 c^14 - a^2 c^16 - b^2 c^16 - 4 a^13 b c OH S + 6 a^11 b^3 c OH S + 6 a^9 b^5 c OH S - 10 a^7 b^7 c OH S - 6 a^5 b^9 c OH S + 12 a^3 b^11 c OH S - 4 a b^13 c OH S + 6 a^11 b c^3 OH S - 24 a^9 b^3 c^3 OH S + 12 a^7 b^5 c^3 OH S + 24 a^5 b^7 c^3 OH S - 30 a^3 b^9 c^3 OH S + 12 a b^11 c^3 OH S + 6 a^9 b c^5 OH S + 12 a^7 b^3 c^5 OH S - 36 a^5 b^5 c^5 OH S + 18 a^3 b^7 c^5 OH S - 12 a b^9 c^5 OH S - 10 a^7 b c^7 OH S + 24 a^5 b^3 c^7 OH S + 18 a^3 b^5 c^7 OH S + 8 a b^7 c^7 OH S - 6 a^5 b c^9 OH S - 30 a^3 b^3 c^9 OH S - 12 a b^5 c^9 OH S + 12 a^3 b c^11 OH S + 12 a b^3 c^11 OH S - 4 a b c^13 OH S : :

See Antreas Hatzipolakis and Francisco Javier García Capitán, euclid 6388.

X(64481) lies on these lines: {2, 3}, {542, 44124}, {1989, 8105}, {8116, 45016}, {13414, 18374}, {15360, 24651}, {32225, 44126}


X(64482) = 8TH HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

Barycentrics    -2 a^12 + 6 a^10 b^2 - 13 a^8 b^4 + 5 a^6 b^6 + 7 a^4 b^8 - 5 a^2 b^10 + 2 b^12 + 6 a^10 c^2 - 2 a^8 b^2 c^2 + 11 a^6 b^4 c^2 - 19 a^4 b^6 c^2 + 7 a^2 b^8 c^2 - 9 b^10 c^2 - 13 a^8 c^4 + 11 a^6 b^2 c^4 + 6 a^4 b^4 c^4 + 22 b^8 c^4 + 5 a^6 c^6 - 19 a^4 b^2 c^6 - 30 b^6 c^6 + 7 a^4 c^8 + 7 a^2 b^2 c^8 + 22 b^4 c^8 - 5 a^2 c^10 - 9 b^2 c^10 + 2 c^12 - 2 a^10 W + 5 a^8 b^2 W + 6 a^6 b^4 W - 7 a^4 b^6 W - 4 a^2 b^8 W + 2 b^10 W + 5 a^8 c^2 W - 30 a^6 b^2 c^2 W + 14 a^4 b^4 c^2 W + 25 a^2 b^6 c^2 W - 8 b^8 c^2 W + 6 a^6 c^4 W + 14 a^4 b^2 c^4 W - 46 a^2 b^4 c^4 W + 6 b^6 c^4 W - 7 a^4 c^6 W + 25 a^2 b^2 c^6 W + 6 b^4 c^6 W - 4 a^2 c^8 W - 8 b^2 c^8 W + 2 c^10 W : : where W^2 = a^4 - a^2 b^2 + b^4 - a^2 c^2 - b^2 c^2 + c^4

See Antreas Hatzipolakis and Francisco Javier García Capitán, euclid 6388.

X(64482) lies on these lines: {2, 3}, {2028, 31862}, {3413, 6321}


X(64483) = 9TH HATZIPOLAKIS-GARCÍA CAPITÁN-EULER POINT

Barycentrics    -2 a^12 + 6 a^10 b^2 - 13 a^8 b^4 + 5 a^6 b^6 + 7 a^4 b^8 - 5 a^2 b^10 + 2 b^12 + 6 a^10 c^2 - 2 a^8 b^2 c^2 + 11 a^6 b^4 c^2 - 19 a^4 b^6 c^2 + 7 a^2 b^8 c^2 - 9 b^10 c^2 - 13 a^8 c^4 + 11 a^6 b^2 c^4 + 6 a^4 b^4 c^4 + 22 b^8 c^4 + 5 a^6 c^6 - 19 a^4 b^2 c^6 - 30 b^6 c^6 + 7 a^4 c^8 + 7 a^2 b^2 c^8 + 22 b^4 c^8 - 5 a^2 c^10 - 9 b^2 c^10 + 2 c^12 + 2 a^10 W - 5 a^8 b^2 W - 6 a^6 b^4 W + 7 a^4 b^6 W + 4 a^2 b^8 W - 2 b^10 W - 5 a^8 c^2 W + 30 a^6 b^2 c^2 W - 14 a^4 b^4 c^2 W - 25 a^2 b^6 c^2 W + 8 b^8 c^2 W - 6 a^6 c^4 W - 14 a^4 b^2 c^4 W + 46 a^2 b^4 c^4 W - 6 b^6 c^4 W + 7 a^4 c^6 W - 25 a^2 b^2 c^6 W - 6 b^4 c^6 W + 4 a^2 c^8 W + 8 b^2 c^8 W - 2 c^10 W : : where W^2 = a^4 - a^2 b^2 + b^4 - a^2 c^2 - b^2 c^2 + c^4

See Antreas Hatzipolakis and Francisco Javier García Capitán, euclid 6388.

X(64483) lies on these lines: {2, 3}, {2029, 31863}, {3414, 6321}


Σάββατο 13 Ιουλίου 2024

SAME CENTROID

Let ABC be a triangle and A'B'C' the cevian triangle of O

Denote

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

Ha, Hb, Hc = the orthocenters of OMbMc, OMcMa, OMaMb, resp,

The triangles ABC and HaHbHc share the same centroid G

APH

Francisco Javier García Capitán A triangle of orthocenters with centroid G
FJGC

Παρασκευή 12 Ιουλίου 2024

AN ORTHOCENTER

Let ABC be a triangle and A'B'C' the orthic triangle

Denote

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

Ha, Hb, Hc = the orthocenters of HMbMc, HMcMa, HMaMb, resp.

Then
Orthocenter of HaHbHc = Nine Point Circle Center N

Euclid 6368
Romantics of Geometry

PROOF


The perpendicular from Mb to HMc intersects B'C' at A". In the triangle BB'C' we have MbA" // BC' and Mb = midpoint of BB', therefore A" is the midpoint of B'C'.
Similarly in the triangle CC'B' the perpendicular from Mc to HMb passes through the midpoint of B'C'. Therefore A" is the orthocenter Ha of HMbMc.
Similarly the orthocenters Hb, Hc of HMcMa, HMaMb are the midpoints of C'A', A'B', resp., that is HaHbHc is the medial triangle of A'B'C'

The orthocenter of HaHbHc is the circumcenter of A'B'C'

The circumcenter of A'B'C' is the Nine Point Circle center N of ABC.
H of HaHbHc = O of A'B'C' = N of ABC
QED

Douglas Hofstadter, FOREWORD

Douglas Hofstadter, FOREWORD In: Clark Kimberling, Triangle Centers and Central Triangles. Congressus Numerantum, vol. 129, August, 1998. W...