ADGEOM 5178 * ADGEOM 5181

#5178

Let ABC be a triangle P=X(35).

A'B'C' is pedal triangle of P.

Then reflections of Euler lines of triangles PBC, PCA, PAB in PA', PB', PC' respectively are concurrent.

Which is this concurrent point?

Best regards,
Tran Quang Hung.

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#5181

 

[Tran Quang Hung]

 

Let ABC be a triangle P=X(35).

 

A'B'C' is pedal triangle of P.

 

Then reflections of Euler lines of triangles PBC, PCA, PAB in PA', PB', PC' respectively are concurrent.

 

Which is this concurrent point?

 

Best regards,

Tran Quang Hung.

 

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[Ercole Suppa]

 

 

Then reflections of Euler lines of triangles PBC, PCA, PAB in PA', PB', PC' respectively concur at point:

 

 

X = midpoint of X(21) and X(3871)

 

= a^2 (a^5-b^5-b^4 c+b^3 c^2+b^2 c^3-b c^4-c^5-a^4 (b+c)-2 a^3 (b^2+b c+c^2)+a (b^2+b c+c^2)^2+2 a^2 (b^3+b^2 c+b c^2+c^3)) :: (barys)

 

= 2 R S^2 + (5 a R^2-b R^2-c R^2-a SB+b SB+c SB-a SC+b SC+c SC)S -2 R SB SC :: (barys)

 

= on lines X(i)X(j) for these {i,j}: {1,27086},{8,21},{30,11491},{35,758},{100,442},{191,3811},{404,11281},{1392,6767},{1621,6675},{2475,3085},{2646,4996},{2802,3746},{3241,10267},{3647,3935},{3649,14882},{3651,5758},{3874,14799},{3957,14798},{4421,6175},{4428,15671},{5010,16126},{5428,5844},{5703,11507},{7676,17768},{9780,11517},{10528,15680},{11510,20057},{12867,13615},{13607,17009},{15674,19843},{25440,26725}

 

= midpoint of X(i) and X(j) for these {i,j}: {21,3871}

 

= reflection of X(i) in X(j) for these {i,j}: {24390,6675}

 

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

 

= (6-9-13) search numbers [2.26338369616380577, 2.58854507743396244, 0.803956183916033209]

 

 

Best regards

Ercole Suppa