ADGEOM 2795 * ADGEOM 2799 * ADGEOM 2800 * ADGEOM 2801 * ADGEOM 2803

 

#2795

 

Dear Geometers

 

I found a strange property of Euler lines as follow.

Given triangle ABC and a point P.

PA,PB,PC intersect the circle (PBC),(PCA),(PAB) at D,E,F

Euler lines of DBC,ECA,FAB cut BC,CA,AB at X,Y,Z, respectively

Then X,Y,Z, are collinear. The line XYZ also passes through the centroid of triangle ABC.

Is this known?

 

Best regard

Sincerely

Ngo Quang Duong

 

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#2799

Dear Ngo Quang Duong,

 

Good work!  One more property:

ABC and the triangle bounded by the three Euler lines

are perspective at a point on the circumcircle of ABC.

The collinear line is the axis of perspectivity. 

Please refer to the attachment file.(with different notations)

I think they are theorems on a general projective plane.

Maybe I post later!

 

Best regards,

Tsihong Lau

 

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#2800

 

Dear Ngo Quang Duong and Tsihong Lau.

The three Euler lines are concurrent (at a point on the circumcircle of ABC)  if only if P  lies on Euler-Morley quintic (Q003 in related curves of Bernard Gibert).

Writing Q as Q(P)=Intersection of the three lines of Euler  , the occurrence of {i,j} in the following list means that Q(X(i)) = X(j): {1, 100}, {2, 1296}, {4, 110}, {1113, 1114}, {1114, 1113}.

Q(X(1156))=(a/((b - c) (a^2 + b^2 + 4 b c + c^2 - 2 a (b + c))):...:...)

Best regards,
Angel Montesdeoca

 

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#2801

 

Dear  Ngo Quang Duong,

 

Perhaps we have a more strange property

that is very very nice to be true.

CONJECTURE

Given a triangle ABC and a point P,

PA, PB, PC intersect the circles (PBC), (PCA), (PAB) at D, E, F respectively.

La is the line that connects the triangle centers X(m), X(n)  of triangle DBC

and similarly define the lines Lb, Lc of triangles ECA, FAB.

If the lines La, Lb, Lc meet the lines BC, CA, AB at X, Y, Z

respectively then X,Y, Z are collinear.

 

I have checked it graphically for 3 combinations

m = 1, n = 2

m = 1, n = 3

m = 2, n = 3  that is yours.

 

Best regards

Nikos Dergiades

 

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#2803

 

Dear Angel, Ngo Quang Duong and Tsihong Lau,

 

Let ABC be a triangle and I its incenter.

Denote:

D= the antigonal conjugate of A wrt IBC.

E= the antigonal conjugate of B wrt ICA.

F= the antigonal conjugate of C wrt IAB.

Euler lines of DBC, ECA and FAB concur in X(104), the antipode of X(100).

 

Best regards, Seiichi Kirikami.