HYACINTHOS 28626

[Tran Quang Hung]:

 

 

Let ABC be a triangle with NPC center N.

 

The circle (NBC) meets AB, AC again at Ba,Ca, resp.

 

Define similarly points Ac, Bc, Cb, Ab.

 

Let A' be the circumcenter of the triangle AAbAc. 

Define similarly points B' and C'.

 

Then the circumcenter of the triangle A'B'C' lies on the Euler line of ABC. 

 

Which is this point?

 

 

--------------------------------------------------------------------------------------------

 

 

[Ercole Suppa]

 

Dear Tran Quang Hung,

 

the circumcenter of triangle A'B'C' is

 

W = EULER LINE INTERCEPT OF X(1154)X(14143)

 

= 2 a^16-9 a^14 (b^2+c^2)-(b^2-c^2)^6 (b^2+c^2)^2+a^12 (19 b^4+30 b^2 c^2+19 c^4)+a^2 (b^2-c^2)^4 (b^6+5 b^4 c^2+5 b^2 c^4+c^6)-a^10 (29 b^6+41 b^4 c^2+41 b^2 c^4+29 c^6)+a^4 (b^2-c^2)^2 (9 b^8-4 b^6 c^2-11 b^4 c^4-4 b^2 c^6+9 c^8)+a^8 (35 b^8+20 b^6 c^2+28 b^4 c^4+20 b^2 c^6+35 c^8)+a^6 (-27 b^10+17 b^8 c^2+b^6 c^4+b^4 c^6+17 b^2 c^8-27 c^10) : : (barys)

 

= 4 S^4 + (7 R^4-4 SB SC-4 R^2 SW) S^2 + 43 R^4 SB SC-36 R^2 SB SC SW+8 SB SC SW^2 : : (barys)

=  As a point on the Euler line, X() has Shinagawa coefficients  {7 R^4+4 S^2-4 R^2 SW,43 R^4-4 S^2-36 R^2 SW+8 SW^2}

= lies on these lines: {2,3}, {1154,14143}, {1157,20424}, {3574,6150}, {6288,14072}, {18016,18400}, {21975,23237}, {22051,25044}

 

= reflection of X(i) in X(j) for these {i,j}: {3,10126}, {10285,5}, {14142,140}, {20030,15335}, {20120,546}

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: {15335,20030,381}

 

 

= (6-8-13) search numbers [8.03400265791470230, 7.14328883314957956, -5.01269055238750012]

 

 

Best regards

Ercole Suppa