HYACINTHOS 28564

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle.

We construct the external Vecten squares based on the sides of the triangle:

BCCaBa, CAAbCb, ABBcAc with centers A', B', C', resp.

CaBa, AbCb, BcAc bound a triangle A"B"C"
AbAc, BcBa, CaCb bound a triangle A0B0C0

The Euler lines of A'B'C', A"B"C", A0B0C0 are concurrent (*)

(*) Tran Quang Hung

Which is the point of concurrence wrt triangle ABC ?

I think the same is true if the Vecten squares BCCaBa, CAAbCb, ABBcAc are erected internally the triangle.

Point of concurrence?  

 

 

[Peter Moses]:

Hi Antreas

Outer:

 

= REFLECTION OF X(4) IN X(591)

= 5 a^4-4 a^2 b^2-b^4-4 a^2 c^2+2 b^2 c^2-c^4+4 (2 a^2-b^2-c^2) S:: 

 

= 2 X[1991] - 3 X[3524], 7 X[488] - 4 X[6311], X[5871] - 4 X[9733].

 

= lies on these lines: {2,372}, {3,5861}, {4,591}, {30,1160}, {148,22601}, {193,9541}, {194,13678}, {371,13712}, {376,524}, {490,1588}, {492,23249}, {754,8982}, {1270,6560}, {1271,6396}, {1991,3524}, {3593,6564}, {3594,7375}, {5591,6398}, {5871,9733}, {6231,9880}, {9770,13674}, {10783,12305}, {23269,23311}

.
= reflection of X(i) in X(j) for these {i,j}: {4, 591}, {5861, 3}.

**********

Inner, presumably replace S with -S:

 

= REFLECTION OF X(4) IN X(1991) 

 
= 5 a^4-4 a^2 b^2-b^4-4 a^2 c^2+2 b^2 c^2-c^4-4 (2 a^2-b^2-c^2) S:: 

 

= 2 X[591] - 3 X[3524], 7 X[487] - 4 X[6315], X[5870] - 4 X[9732].

 

lies on these lines: {2,371}, {3,5860}, {4,1991}, {30,1161}, {69,9541}, {148,22630}, {194,13798}, {372,13835}, {376,524}, {489,1587}, {591,3524}, {1270,6200}, {1271,6561}, {3592,7376}, {3595,6565}, {5590,6221}, {5870,9732}, {6230,9880}, {9770,13794}, {10784,12306}, {23275,23312}.

= reflection of X(i) in X(j) for these {i,j}: {4, 1991}, {5860, 3}.

Best regards,
Peter Moses.