#5137
Dear geometers,
Let ABC be a triangle.
d is a line passing through circumcenter of ABC.
A' is reflection of A in line d.
A" is reflection of A' in BC.
Define similarly the points B', B", C', C".
Then triangles ABC and A"B"C" are symmetric in point P on NPC of ABC.
Circle (A"B"C") passes through orthocenter H of ABC.
Then Simson line d1 of H wrt A"B"C" is perpendicular to line d.
The line passing through H and is parallel to d which meets circle (A"B"C") at H'.
Steiner line d2 of H' wrt A"B"C" passes through circumcenter O of ABC.
Orthopole Q of d2 wrt ABC lies on d1.
If d=Euler line, d=OI line, which is the point P, H'and Q?
Best regards,
Tran Quang Hung.
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#5144
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#5144
[Tran Quang Hung]:
Let ABC be a triangle.
d is a line passing through circumcenter of ABC.
A' is reflection of A in line d.
A" is reflection of A' in BC.
Define similarly the points B', B", C', C".
Then triangles ABC and A"B"C" are symmetric in point P on NPC of ABC.
Circle (A"B"C") passes through orthocenter H of ABC.
Then Simson line d1 of H wrt A"B"C" is perpendicular to line d.
The line passing through H and is parallel to d which meets circle (A"B"C") at H'.
Steiner line d2 of H' wrt A"B"C" passes through circumcenter O of ABC.
Orthopole Q of d2 wrt ABC lies on d1.
If d=Euler line, d=OI line, which is the point P, H'and Q?
*** If d=Euler line:
P=X(125).
H' =X(17511).
Q=X(3258) = reflection of X(125) in Euler line.
*** If d=OI line:
P=X(11).
H'= a^6+6 a^4 b c-2 a^5 (b+c)-2 a^3 b c (b+c)-(b-c)^4 (b+c)^2+a^2 b c (3 b^2-5 b c+3 c^2)+a (b-c)^2 (2 b^3-3 b^2 c-3 b c^2+2 c^3) : ... : ...
lies on lines X(i)X(j) for these {i, j}: {4,8}, {11,901}, {100,3259}, {149,513}, {497,14115}, {528,14513}, {953,5840}, {2829,14511}, {3025,13274}, {3583,10774}, {4380,17036}, {5854,17101}, {6075,10707}, {6789,11813}, {8047,23813}, {13273,13756},
and is the reflection of X(i) in X(j), for these {i, j}: {100,3259}, {901,11} .
Q=X(3259) = complement of X(901).
*** If d=Brocard axis:
P=X(115).
H' = a^2 (a^8 b^2 c^2-2 a^2 b^4 c^4 (b^2+c^2)-a^6 (b^6+b^4 c^2+b^2 c^4+c^6)+a^4 (b^8+4 b^4 c^4+c^8)-b^2 c^2 (b^8-3 b^6 c^2+3 b^4 c^4-3 b^2 c^6+c^8)) : ... : ...,
lies on lines X(i)X(j) for these {i, j}: {4,69}, {99,2679}, {115,805}, {148,512}, {543,14509}, {671,6071}, {2549,14113}, {2698,23698}, {2794,14510}, {14061,22103},
and is the reflection of X(i) in X(j), for these {i, j}: {99,2679}, {805,115}.
Q=X(2679).
Angel Montesdeoca
Let ABC be a triangle.
d is a line passing through circumcenter of ABC.
A' is reflection of A in line d.
A" is reflection of A' in BC.
Define similarly the points B', B", C', C".
Then triangles ABC and A"B"C" are symmetric in point P on NPC of ABC.
Circle (A"B"C") passes through orthocenter H of ABC.
Then Simson line d1 of H wrt A"B"C" is perpendicular to line d.
The line passing through H and is parallel to d which meets circle (A"B"C") at H'.
Steiner line d2 of H' wrt A"B"C" passes through circumcenter O of ABC.
Orthopole Q of d2 wrt ABC lies on d1.
If d=Euler line, d=OI line, which is the point P, H'and Q?
*** If d=Euler line:
P=X(125).
H' =X(17511).
Q=X(3258) = reflection of X(125) in Euler line.
*** If d=OI line:
P=X(11).
H'= a^6+6 a^4 b c-2 a^5 (b+c)-2 a^3 b c (b+c)-(b-c)^4 (b+c)^2+a^2 b c (3 b^2-5 b c+3 c^2)+a (b-c)^2 (2 b^3-3 b^2 c-3 b c^2+2 c^3) : ... : ...
lies on lines X(i)X(j) for these {i, j}: {4,8}, {11,901}, {100,3259}, {149,513}, {497,14115}, {528,14513}, {953,5840}, {2829,14511}, {3025,13274}, {3583,10774}, {4380,17036}, {5854,17101}, {6075,10707}, {6789,11813}, {8047,23813}, {13273,13756},
and is the reflection of X(i) in X(j), for these {i, j}: {100,3259}, {901,11} .
Q=X(3259) = complement of X(901).
*** If d=Brocard axis:
P=X(115).
H' = a^2 (a^8 b^2 c^2-2 a^2 b^4 c^4 (b^2+c^2)-a^6 (b^6+b^4 c^2+b^2 c^4+c^6)+a^4 (b^8+4 b^4 c^4+c^8)-b^2 c^2 (b^8-3 b^6 c^2+3 b^4 c^4-3 b^2 c^6+c^8)) : ... : ...,
lies on lines X(i)X(j) for these {i, j}: {4,69}, {99,2679}, {115,805}, {148,512}, {543,14509}, {671,6071}, {2549,14113}, {2698,23698}, {2794,14510}, {14061,22103},
and is the reflection of X(i) in X(j), for these {i, j}: {99,2679}, {805,115}.
Q=X(2679).
Angel Montesdeoca
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