ADGEOM 4653 * ADGEOM 4654 * ADGEOM 4660

#4653

 

Dear geometers,

 

Let ABC be a triangle with centroid G and orthocenter H.

 

P=X(381) is midpoint of GH which is the center of orthocentroidal circle.

 

Let Pa,Pb,Pc be the X(381) of the triangles PBC, PCA, PAB respectively.

 

Then APa,BPb,CPc are concurrent at.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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#4654

 

[Tran Quang Hung]:

Let ABC be a triangle with centroid G and orthocenter H.

P=X(381) is midpoint of GH which is the center of orthocentroidal
circle.

Let Pa,Pb,Pc be the X(381) of the triangles PBC, PCA, PAB respectively.

Then APa,BPb,CPc are concurrent.

Which is this point?

*** APa,BPb,CPc are concurrent at

W = 1/((4 a^4-5 a^2 b^2+b^4-5 a^2 c^2-2 b^2 c^2+c^4)(2 a^4-4 a^2 b^2+2
b^4-4 a^2 c^2-b^2 c^2+2 c^4)) : .... : ....

on the line X(381)X(14483), with (6 - 9 - 13) - search numbers
(0.125506022546977, 4.78594924329040, 0.269389149223193)

Angel Montesdeoca

 

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#4660

 

 

Dear geometers,

I have seen general problem for this,

G and H be the centroid and orthocenter of ABC respectively.

Let ABC be a triangle and P is a point on Euler line of ABC which has Shinagawa coefficients P=kG+H.

Let Ga,Ha be the centroid and orthocenter of PBC respectively.

Pa is a point on Euler line of PBC which has Shinagawa coefficients Pa=kGa+Ha.
 
Define similarly the points Pb and Pc.

Then APa,BPb and CPc are concurrent. Which is this concurrent point in term of k?

Best regards,
Tran Quang Hung.