Τρίτη 30 Ιουλίου 2024
Πέμπτη 18 Ιουλίου 2024
Τρίτη 16 Ιουλίου 2024
FJGC, An unique orthocentroidal system
Francisco Javier García Capitán, An unique orthocentroidal system
FJGC
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^4See 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^4See 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
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
Francisco Javier García Capitán A triangle of orthocenters with centroid G
FJGC
Παρασκευή 12 Ιουλίου 2024
AN ORTHOCENTER
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...
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