Antreas Hatzipolakis[APH]:
Let A'B'C' be the Gergonne triangle of triangle ABC (ie the cevian triangle of
the Gergonne point).Draw tangent lines from the vertices A,B,C to the incircle of A''B'C' and
denote A1,A2.B1,B2.C1,C2 the intersections of these tangent lines with
the sides of the triangle ABC.
a. Prove that the six points A1,A2.B1,B2.C1,C2 are concyclic.
b. Let T be that circle. Prove that the incenters of ABC, A'B'C' and the center
of T are collinear.
Reference:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=624821Which point is the center of the circle?
PS: If we replace the Gergonne point with a variable point P, which is
the locus of P such that the six points lie on a conic?
And for which points other than Gergonne point the conic is a circle?
[CL]
Dear Antreas:
The problem is simpler to prove (algebraically) if we start from the inner triangle, i.e., making A’B’C’ as the reference triangle and ABC as the tangential triangle of A’B’C’.
They are concyclic on a circle with center Z having trilinear coordinates:
Z = a*(a^8-2*(b+c)*a^7-2*(b^2-b*c+c^2)*a^6+2*(b+c)*(3*b^2-2*b*c+3*c^2)*a^5
-2*b*c*(b-c)^2*a^4-2*(b+c)*(3*b^4-4*b^3*c+4*b^2*c^2-4*b*c^3+3*c^4)*a^3
+2*(b^2+c^2)*(b^4-b*c*(b+c)^2+c^4)*a^2+2*(b^2-c^2)*(b-c)*a*(b^4+c^4)
-(b^2-c^2)^2*(b-c)^2*(b^2+c^2)) : :
= 1+2*sin(A/2)*(2*cos(A)*cos(B)*cos(C)*sin(A/2)+cos((B-C)/2)*cos(2*A)) : :
= on line (1,3) //this proves (b)
= R^2*X(1)-(R^2+2*R*r+r^2)*X(3)
= ( 16.810275174475530, 14.17526023700343, -13.931565762699010 )
The circle has radius rr satisfying:
rr^2=R^2*(-2*cos(A)*cos(B-C)+4-8*sin(A/2)*cos((B-C)/2)+4*sin(3*A/2)*cos((B-C)/2)-4*cos(A)+2*cos(B-C)+cos(2*A))*(cot(B/2)+cot(C/2))^2/(sin(B)+sin(C)+2*sin(A)+sin(2*A)-2*cos(3*A/2)*cos((B-C)/2))^2
No ETC-centers lie on this circle.
Regards
César Lozada
[APH]:
Dear César
Thanks !!Original Reference:
P. Dolgirev, Interesting circle. Journal of Classical Geometry, vol 3, p. 55
http://jcgeometry.org/Articles/Volume3/JCG2014V2pp53-55.pdfAPH
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