HYACINTHOS 28021

[Tran Quang Hung]:

 

 

Let ABC be a triangle.

 

Let A'B'C' is the orthic triangle.

 

Let A*,B*,C* be the 1st isodynamic point of AB'C',BC'A',CA'B', resp...

 

Then Euler line of the triangles AB*C*, BC*A*, CA*B* are concurrent.

 

Similar problem is true with 2nd isodynamic point.

 

Which are the concurrence points?

[Randy Hutson]:


Dear Tran Quang Hung,

The Euler lines are concurrent in X(13), which is also the center of equilateral triangle A*B*C*, and the perspector of ABC and A*B*C*.  If we call A**B**C** the triangle obtained using 2nd isodynamic points, the Euler lines are concurrent in X(14), which is also the center of equilateral triangle A**B**C**, and the perspector of ABC and A**B**C**.  Also A*B*C* and A**B**C** are perspective at X(115) and inversely similar with similitude center X(4).

The mid-triangle of A*B*C* and A**B**C** is perspective to ABC at X(671).

The cross-triangle of A*B*C* and A**B**C** is perspective to ABC at a point X, not in ETC.   

 

X = (3 a^8 - 7 a^6 (b^2 + c^2) + a^4 (7 b^4 + 5 b^2 c^2 + 7 c^4) - a^2 (5 b^6 - b^4 c^2 - b^2 c^4 + 5 c^6) + (b^2 - c^2)^2 (2 b^4 + b^2 c^2 + 2 c^4)) / (b^2 + c^2 - a^2) : :   (Barycentrics)

= lies on the lines: {4,542}, {25,6054}, {98,275}, {99,317}, {107,11005}, {114,6353}, {115,3087}, {147,6995}, {250,403}, {297,5182}, {393,5477}, {458,7879} et al.  

 

(6,9,13) values: (6.433598230792300, 12.133135468118008, -7.728551564078381).

The cross-triangle of A*B*C* and A**B**C** is perspective to A'B'C' at X(542).
The side-triangle of A*B*C* and A**B**C** is degenerate, lying on the orthic axis, with centroid X(1637).

Also, A*B*C* is the reflection of the pedal triangle of X(15) in X(11542), and A**B**C** is the reflection of the pedal triangle of X(16) in X(11543).

Best regards,
Randy Hutson