#5185
Dear friends,
I introduce four new points on Euler line:
I found these points trying to solve a problem proposed by Kadir Altintas.
Given a triangle ABC and a point P, call DEF its circuncevian triangle. The circumcenters of six triangles PBD, PDC, PCE, PEA, PAF and PFB lie on the same conic. When this conic is a circle?
Reference:
http://garciacapitan.blogspot.com/2019/02/four-points-on-euler-line.html
Best regards,
Francisco Javier Garcia Capitan
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#5188
Hello Francisco Javier,
If P has trilinear coordinates tri_P = [x,y,z]
then trilinear coordinates of D,E,F are :
tri_D = [-a*y*z, (c*y + b*z)*y, (c*y + b*z)*z]
tri_E = [(c*x + a*z)*x, -b*x*z, (c*x + a*z)*z]
tri_F = [(b*x + a*y)*x, (b*x + a*y)*y, -c*x*y]
The six points of the conic have trilinear coordinates :
tri_M1 = [(a^3*x - a*b^2*x - a*c^2*x + a^2*c*z + b^2*c*z - c^3*z)*a,
-(a^3*b*x - a*b^3*x + a*b*c^2*x - a^4*y + 2*a^2*b^2*y - b^4*y + 2*a^2*c^2*y + 2*b^2*c^2*y - c^4*y + 2*a^2*b*c*z),
-(a^3*c*x + a*b^2*c*x - a*c^3*x + a^2*b^2*z - b^4*z + a^2*c^2*z + 2*b^2*c^2*z - c^4*z)]
tri_M2 = [(a^3*x - a*b^2*x - a*c^2*x + a^2*b*y - b^3*y + b*c^2*y)*a,
-(a^3*b*x - a*b^3*x + a*b*c^2*x + a^2*b^2*y - b^4*y + a^2*c^2*y + 2*b^2*c^2*y - c^4*y),
-(a^3*c*x + a*b^2*c*x - a*c^3*x + 2*a^2*b*c*y - a^4*z + 2*a^2*b^2*z - b^4*z + 2*a^2*c^2*z + 2*b^2*c^2*z - c^4*z)]
tri_M3 = [(a^4*x - a^2*b^2*x - 2*a^2*c^2*x - b^2*c^2*x + c^4*x + a^3*b*y - a*b^3*y - a*b*c^2*y),
-(a^3*x - a*b^2*x - a*c^2*x + a^2*b*y - b^3*y + b*c^2*y)*b,
-(2*a*b^2*c*x + a^2*b*c*y + b^3*c*y - b*c^3*y - a^4*z + 2*a^2*b^2*z - b^4*z + 2*a^2*c^2*z + 2*b^2*c^2*z - c^4*z)]
tri_M4 = [(a^4*x - 2*a^2*b^2*x + b^4*x - 2*a^2*c^2*x - 2*b^2*c^2*x + c^4*x + a^3*b*y - a*b^3*y - a*b*c^2*y - 2*a*b^2*c*z),
-(a^2*b*y - b^3*y + b*c^2*y - a^2*c*z - b^2*c*z + c^3*z)*b,
-(a^2*b*c*y + b^3*c*y - b*c^3*y - a^4*z + a^2*b^2*z + 2*a^2*c^2*z + b^2*c^2*z - c^4*z)]
tri_M5 = [(a^4*x - 2*a^2*b^2*x + b^4*x - 2*a^2*c^2*x - 2*b^2*c^2*x + c^4*x - 2*a*b*c^2*y + a^3*c*z - a*b^2*c*z - a*c^3*z),
(a^4*y - 2*a^2*b^2*y + b^4*y - a^2*c^2*y - b^2*c^2*y - a^2*b*c*z + b^3*c*z - b*c^3*z),
(a^2*b*y - b^3*y + b*c^2*y - a^2*c*z - b^2*c*z + c^3*z)*c]
tri_M6 = [(a^4*x - 2*a^2*b^2*x + b^4*x - a^2*c^2*x - b^2*c^2*x + a^3*c*z - a*b^2*c*z - a*c^3*z),
-(2*a*b*c^2*x - a^4*y + 2*a^2*b^2*y - b^4*y + 2*a^2*c^2*y + 2*b^2*c^2*y - c^4*y + a^2*b*c*z - b^3*c*z + b*c^3*z),
-(a^3*x - a*b^2*x - a*c^2*x + a^2*c*z + b^2*c*z - c^3*z)*c]
Midpoints of segments M1M4, M2M5 and M3M6 are the same point :
tri_M14 = [(2*a^4*x - 3*a^2*b^2*x + b^4*x - 3*a^2*c^2*x - 2*b^2*c^2*x + c^4*x + a^3*b*y - a*b^3*y - a*b*c^2*y + a^3*c*z - a*b^2*c*z - a*c^3*z),
-(a^3*b*x - a*b^3*x + a*b*c^2*x - a^4*y + 3*a^2*b^2*y - 2*b^4*y + 2*a^2*c^2*y + 3*b^2*c^2*y - c^4*y + a^2*b*c*z - b^3*c*z + b*c^3*z),
-(a^3*c*x + a*b^2*c*x - a*c^3*x + a^2*b*c*y + b^3*c*y - b*c^3*y - a^4*z + 2*a^2*b^2*z - b^4*z + 3*a^2*c^2*z + 3*b^2*c^2*z - 2*c^4*z)]
The circumcenter of the triangle M1M2M3 has trilinear coordinates :
tri_M = [(a^5*x^2*y - a^3*b^2*x^2*y - a^3*c^2*x^2*y - a*b^2*c^2*x^2*y + 2*a^4*b*x*y^2 - 2*a^2*b^3*x*y^2 - 2*a^2*b*c^2*x*y^2 - b^3*c^2*x*y^2 + b*c^4*x*y^2 + a^3*b^2*y^3 - a*b^4*y^3 - a*b*c^3*x^2*z + 2*a^4*c*x*y*z - 2*a^2*b^2*c*x*y*z - 3*a^2*c^3*x*y*z - b^2*c^3*x*y*z + c^5*x*y*z + a^3*b*c*y^2*z - a*b^3*c*y^2*z - a^2*b*c^2*x*z^2),
-y*(a^4*b*x^2 - a^2*b^3*x^2 + 2*a^3*b^2*x*y - 2*a*b^4*x*y + 2*a*b^2*c^2*x*y + a^2*b^3*y^2 - b^5*y^2 + a^2*b*c^2*y^2 + 2*b^3*c^2*y^2 - b*c^4*y^2 + a^3*b*c*x*z - a*b^3*c*x*z + a*b*c^3*x*z - a^4*c*y*z + 2*a^2*b^2*c*y*z - b^4*c*y*z + 2*a^2*c^3*y*z + 2*b^2*c^3*y*z - c^5*y*z + a^2*b*c^2*z^2),
-(a^4*c*x^2*y + a^2*b^2*c*x^2*y - a^2*c^3*x^2*y + 2*a^3*b*c*x*y^2 + a*b^3*c*x*y^2 - a*b*c^3*x*y^2 + a^2*b^2*c*y^3 - a^2*b*c^2*x^2*z + a^3*b^2*x*y*z - a*b^4*x*y*z + a^3*c^2*x*y*z + 2*a*b^2*c^2*x*y*z - a*c^4*x*y*z + a^2*b^3*y^2*z - b^5*y^2*z + 2*a^2*b*c^2*y^2*z + 2*b^3*c^2*y^2*z - b*c^4*y^2*z - a^3*b*c*x*z^2 - a^4*c*y*z^2 + a^2*b^2*c*y*z^2 - b^4*c*y*z^2 + 2*a^2*c^3*y*z^2 + 2*b^2*c^3*y*z^2 - c^5*y*z^2)]
We have a circle when M14 = M ...or (by computing quadrance between these two points) (x,y,z) satisfy the sextic form equation :
a^9*b*c*x^4*y^2 - 2*a^7*b^3*c*x^4*y^2 + a^5*b^5*c*x^4*y^2 - 4*a^7*b*c^3*x^4*y^2 + 2*a^5*b^3*c^3*x^4*y^2 - 2*a^3*b^5*c^3*x^4*y^2 + 5*a^5*b*c^5*x^4*y^2 - 2*a^3*b*c^7*x^4*y^2 + a^10*c*x^3*y^3 - 4*a^6*b^4*c*x^3*y^3 + 4*a^4*b^6*c*x^3*y^3 - a^2*b^8*c*x^3*y^3 - 4*a^8*c^3*x^3*y^3 - 2*a^6*b^2*c^3*x^3*y^3 - 2*a^2*b^6*c^3*x^3*y^3 + 6*a^6*c^5*x^3*y^3 - 4*a^4*c^7*x^3*y^3 + 2*a^2*b^2*c^7*x^3*y^3 + a^2*c^9*x^3*y^3 + 2*a^9*b*c*x^2*y^4 - 4*a^7*b^3*c*x^2*y^4 + 4*a^3*b^7*c*x^2*y^4 - 2*a*b^9*c*x^2*y^4 - 4*a^7*b*c^3*x^2*y^4 + 2*a^5*b^3*c^3*x^2*y^4 - 4*a^3*b^5*c^3*x^2*y^4 + 2*a*b^7*c^3*x^2*y^4 - 4*a^3*b^3*c^5*x^2*y^4 + 4*a^3*b*c^7*x^2*y^4 + 2*a*b^3*c^7*x^2*y^4 - 2*a*b*c^9*x^2*y^4 + a^8*b^2*c*x*y^5 - 4*a^6*b^4*c*x*y^5 + 4*a^4*b^6*c*x*y^5 - b^10*c*x*y^5 - 2*a^4*b^4*c^3*x*y^5 + 2*b^8*c^3*x*y^5 - 2*a^4*b^2*c^5*x*y^5 - 2*b^4*c^7*x*y^5 + b^2*c^9*x*y^5 - a^5*b^5*c*y^6 + 2*a^3*b^7*c*y^6 - a*b^9*c*y^6 - 2*a^3*b^5*c^3*y^6 + 2*a*b^7*c^3*y^6 - a*b^5*c^5*y^6 - 2*a^7*b^2*c^2*x^4*y*z + 4*a^5*b^4*c^2*x^4*y*z - 2*a^3*b^6*c^2*x^4*y*z + 8*a^5*b^2*c^4*x^4*y*z - 6*a^3*b^2*c^6*x^4*y*z + a^10*b*x^3*y^2*z - 2*a^8*b^3*x^3*y^2*z + 2*a^4*b^7*x^3*y^2*z - a^2*b^9*x^3*y^2*z - 8*a^8*b*c^2*x^3*y^2*z + 6*a^6*b^3*c^2*x^3*y^2*z + 4*a^4*b^5*c^2*x^3*y^2*z - 2*a^2*b^7*c^2*x^3*y^2*z + 18*a^6*b*c^4*x^3*y^2*z + 2*a^4*b^3*c^4*x^3*y^2*z - 4*a^2*b^5*c^4*x^3*y^2*z - 16*a^4*b*c^6*x^3*y^2*z + 2*a^2*b^3*c^6*x^3*y^2*z + 5*a^2*b*c^8*x^3*y^2*z + a^11*x^2*y^3*z - 4*a^7*b^4*x^2*y^3*z + 2*a^5*b^6*x^2*y^3*z + 3*a^3*b^8*x^2*y^3*z - 2*a*b^10*x^2*y^3*z - 6*a^9*c^2*x^2*y^3*z - 4*a^7*b^2*c^2*x^2*y^3*z + 10*a^5*b^4*c^2*x^2*y^3*z + 14*a^7*c^4*x^2*y^3*z + 2*a^5*b^2*c^4*x^2*y^3*z - 8*a^3*b^4*c^4*x^2*y^3*z - 16*a^5*c^6*x^2*y^3*z + 4*a^3*b^2*c^6*x^2*y^3*z + 6*a*b^4*c^6*x^2*y^3*z + 9*a^3*c^8*x^2*y^3*z - 2*a*b^2*c^8*x^2*y^3*z - 2*a*c^10*x^2*y^3*z + 2*a^10*b*x*y^4*z - 3*a^8*b^3*x*y^4*z - 2*a^6*b^5*x*y^4*z + 4*a^4*b^7*x*y^4*z - b^11*x*y^4*z - 6*a^8*b*c^2*x*y^4*z + 4*a^6*b^3*c^2*x*y^4*z + 2*a^4*b^5*c^2*x*y^4*z + 4*a^6*b*c^4*x*y^4*z - 10*a^4*b^3*c^4*x*y^4*z + 2*a^2*b^5*c^4*x*y^4*z + 4*b^7*c^4*x*y^4*z + 4*a^4*b*c^6*x*y^4*z + 4*a^2*b^3*c^6*x*y^4*z - 2*b^5*c^6*x*y^4*z - 6*a^2*b*c^8*x*y^4*z - 3*b^3*c^8*x*y^4*z + 2*b*c^10*x*y^4*z + a^9*b^2*y^5*z - 2*a^7*b^4*y^5*z + 2*a^3*b^8*y^5*z - a*b^10*y^5*z - 2*a^5*b^2*c^4*y^5*z - 2*a^3*b^4*c^4*y^5*z + 4*a*b^6*c^4*y^5*z - 4*a*b^4*c^6*y^5*z + a*b^2*c^8*y^5*z - 4*a^3*b^3*c^5*x^4*z^2 - 2*a^8*b^2*c*x^3*y*z^2 + 4*a^6*b^4*c*x^3*y*z^2 - 2*a^4*b^6*c*x^3*y*z^2 + 12*a^6*b^2*c^3*x^3*y*z^2 - 4*a^2*b^6*c^3*x^3*y*z^2 - 14*a^4*b^2*c^5*x^3*y*z^2 + 4*a^2*b^2*c^7*x^3*y*z^2 - 4*a^9*b*c*x^2*y^2*z^2 + 4*a^7*b^3*c*x^2*y^2*z^2 + 2*a^5*b^5*c*x^2*y^2*z^2 - 2*a*b^9*c*x^2*y^2*z^2 + 18*a^7*b*c^3*x^2*y^2*z^2 + 4*a^5*b^3*c^3*x^2*y^2*z^2 - 6*a^3*b^5*c^3*x^2*y^2*z^2 - 28*a^5*b*c^5*x^2*y^2*z^2 + 4*a^3*b^3*c^5*x^2*y^2*z^2 + 2*a*b^5*c^5*x^2*y^2*z^2 + 18*a^3*b*c^7*x^2*y^2*z^2 + 4*a*b^3*c^7*x^2*y^2*z^2 - 4*a*b*c^9*x^2*y^2*z^2 - 2*a^10*c*x*y^3*z^2 - 2*a^8*b^2*c*x*y^3*z^2 + 6*a^6*b^4*c*x*y^3*z^2 - 2*b^10*c*x*y^3*z^2 + 9*a^8*c^3*x*y^3*z^2 + 4*a^6*b^2*c^3*x*y^3*z^2 - 8*a^4*b^4*c^3*x*y^3*z^2 + 3*b^8*c^3*x*y^3*z^2 - 16*a^6*c^5*x*y^3*z^2 + 2*a^4*b^2*c^5*x*y^3*z^2 + 10*a^2*b^4*c^5*x*y^3*z^2 + 2*b^6*c^5*x*y^3*z^2 + 14*a^4*c^7*x*y^3*z^2 - 4*a^2*b^2*c^7*x*y^3*z^2 - 4*b^4*c^7*x*y^3*z^2 - 6*a^2*c^9*x*y^3*z^2 + c^11*x*y^3*z^2 - 2*a^9*b*c*y^4*z^2 + 2*a^7*b^3*c*y^4*z^2 + 2*a^3*b^7*c*y^4*z^2 - 2*a*b^9*c*y^4*z^2 + 4*a^7*b*c^3*y^4*z^2 - 4*a^5*b^3*c^3*y^4*z^2 - 4*a^3*b^5*c^3*y^4*z^2 + 4*a*b^7*c^3*y^4*z^2 + 2*a^3*b^3*c^5*y^4*z^2 - 4*a^3*b*c^7*y^4*z^2 - 4*a*b^3*c^7*y^4*z^2 + 2*a*b*c^9*y^4*z^2 - 8*a^4*b^3*c^4*x^3*z^3 + 4*a^7*b^2*c^2*x^2*y*z^3 - 4*a^3*b^6*c^2*x^2*y*z^3 - 14*a^5*b^2*c^4*x^2*y*z^3 - 2*a*b^6*c^4*x^2*y*z^3 + 12*a^3*b^2*c^6*x^2*y*z^3 + 4*a*b^4*c^6*x^2*y*z^3 - 2*a*b^2*c^8*x^2*y*z^3 + 5*a^8*b*c^2*x*y^2*z^3 + 2*a^6*b^3*c^2*x*y^2*z^3 - 4*a^4*b^5*c^2*x*y^2*z^3 - 2*a^2*b^7*c^2*x*y^2*z^3 - b^9*c^2*x*y^2*z^3 - 16*a^6*b*c^4*x*y^2*z^3 + 2*a^4*b^3*c^4*x*y^2*z^3 + 4*a^2*b^5*c^4*x*y^2*z^3 + 2*b^7*c^4*x*y^2*z^3 + 18*a^4*b*c^6*x*y^2*z^3 + 6*a^2*b^3*c^6*x*y^2*z^3 - 8*a^2*b*c^8*x*y^2*z^3 - 2*b^3*c^8*x*y^2*z^3 + b*c^10*x*y^2*z^3 + a^9*c^2*y^3*z^3 + 2*a^7*b^2*c^2*y^3*z^3 - 2*a^3*b^6*c^2*y^3*z^3 - a*b^8*c^2*y^3*z^3 - 4*a^7*c^4*y^3*z^3 + 4*a*b^6*c^4*y^3*z^3 + 6*a^5*c^6*y^3*z^3 - 2*a^3*b^2*c^6*y^3*z^3 - 4*a*b^4*c^6*y^3*z^3 - 4*a^3*c^8*y^3*z^3 + a*c^10*y^3*z^3 - 4*a^5*b^3*c^3*x^2*z^4 - 6*a^6*b^2*c^3*x*y*z^4 - 2*a^2*b^6*c^3*x*y*z^4 + 8*a^4*b^2*c^5*x*y*z^4 + 4*a^2*b^4*c^5*x*y*z^4 - 2*a^2*b^2*c^7*x*y*z^4 - 2*a^7*b*c^3*y^2*z^4 - 2*a^3*b^5*c^3*y^2*z^4 + 5*a^5*b*c^5*y^2*z^4 + 2*a^3*b^3*c^5*y^2*z^4 + a*b^5*c^5*y^2*z^4 - 4*a^3*b*c^7*y^2*z^4 - 2*a*b^3*c^7*y^2*z^4 + a*b*c^9*y^2*z^4 = 0
Dominique
dominique laurain
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#5210
Hello,
I recall the original subject : given a reference triangle ABC and a point P inside it, the pedal triangle of P is DEF (vertices are intersections of AP,BP,CP lines with circumcircle).
Then six points M1,M2,M3,M4,M5,M6 are circumcenters of triangles PBD, PDC,PCE,PEA,PAF,PFB.
The original post is : the six points are lying on an ellipse.
I have found more about that ellipse : be x:y:z trilinear coordinates of P then center of the ellipse is the point with barycentric coordinates c*(a*y + b*x) , a*(b*z + c*y) and b*(a*z + c*x) with respect to the points M1,M3,M5
The ellipse is the inellipse of a triangle vertices H1,H2,H3 with barycentric coordinates
z*(a*y + b*x),x*(b*z + c*y),-y*(a*z + c*x),
-z*(a*y + b*x),x*(b*z + c*y),y*(a*z + c*x)
and z*(a*y + b*x),-x*(b*z + c*y),y*(a*z + c*x)
with respect to M1,M3,M5.
The foci of the ellipse are the points which can be deduced from Linfield theorem (see https://en.wikipedia.org/wiki/Marden%27s_theorem) using the cutting ratios (a/b)*(y/x) , (a/c)*(z/x) and 1 given complex coordinates z1,z2,z3 of the M1,M3,M5 points.
Cuting ratios for edges of H1H2H3 triangle were used to deduce Hi points from Mi points :
H1M3 / H2M3 = (c*y)/(b*z) => (c*y + b*z) M3 = (b*z) H1 + (c*y) H2
H2M5 / H3M5 = (a*z)/(c*x) => (a*z + c*x) M5 = (c*x) H2 + (a*z) H3
H3M1 / H1M1 = (b*x)/(a*y) => (b*x + a*y) M1 = (a*y) H3 + (b*x) H1
Dominique
dominique laurain
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#5211
Dear all,
I would like to add a comment:
Four circles (M1M2M3), (M1M5M6), (M2M6M4), (M3M4M5) are concurrent.
Four circles (M4M5M6), (M4M2M3), (M5M3M1), (M6M1M2) are also concurrent.
This holds whenever M1M4, M2M5 and M3M6 share the same midpoint. I discovered this fact when solving another problem by Kadir Altintas.
Best regards,
Vu Thanh Tung
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#5212
Dear Vu Thanh Tung and Francisco,
The opposite sides of the hexagon are parallel and hence the
circumcenters M1,M2,M3,M4,M5,M6 are lying on a conic
and we can prove with center Q that is the mid point of the diagonals
M1M4, M2M5, M3M6 and also mid point of PO (O=circumenter of ABC).
If we try to find when the conic is a circle by finding the locus of P
when QM1 = QM2 = . . .
we get three coaxial circles and the points are X(5000), X(5001).
The contruction of these circles is as follows:
Let A1B1C1 be the orthic triangle
Let A2B2C2 be the cevian triangle of K the lemoine point.
Let A3, B3, C3 be the intersections of the line sides of A2B2C2
with the sides of ABC. A3B3C3 is a line the trilinear polar of K.
The circumcircles of the triangles AA1A3, BB1B3, CC1C3
are the three circles that give the two points iff ABC is acute angled
for which the conic of M1,M2,M3,M4,M5,M6 is a circle.
Nikolaos Dergiades
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#5213
Dear all,
About the concurrency, we can have generalization.
Let A , B , C , D ,E , F be 6 points on a circle (O).
Let P be another point on the plane.
Let M1 , M2 , M3 , M4 , M5 , M6 be circumcenter of PAB, PBC, PCD , PDE , PEF, PFA.
Then:
(1) M1M4 , M2M5, M3M6 are concurrent at a point X
(2) if additionally AD , BE , CF are concurrent at point K then:
(2.1) O , K , X are collinear.
(2.2) if <OPK = 90° (or P on the circle of diameter OK) then X is midpoint of OK
(2.3)(original problem) if M=K then X is also the midpoint of M1M4 , M2M5, M3M6.
Best regards,
Vu Thanh Tung
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#5214
Dear all,
The concurrency problem was already posted in AoPS:
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#5219
The ellipse is degenerated to a circle when foci of the ellipse collapse into the same point.
P point has trilinear coordinates x:y:z
My equation in complex numbers from the Linfield theorem is :
f'(z) = alpha*z^2 + beta*z + gamma
with f(z) = (z-z1)^i (z-z2)^j (z-z3)^k
whose roots are
z1 = ( z*(a*y + b*x)*m1 + x*(b*z + c*y)*m3 - y*(a*z + c*x)*m5 ) /(2*b*x*z)
z2 = (-z*(a*y + b*x)*m1 + x*(b*z + c*y)*m3 + y*(a*z + c*x)*m5 ) /(2*c*x*y)
z3 = ( z*(a*y + b*x)*m1 - x*(b*z + c*y)*m3 + y*(a*z + c*x)*m5 ) /(2*a*y*z)
equivalent complex numbers in the complex plane for H1,H2,H3
m1,m3,m5 are complex numbers for M1,M3,M5
i = (a/b)*(y/x) ; j = (a/c)*(z/x) ; k = 1
alpha = i + j + k ; beta = -((z2 + z3)*i + (z3 + z1)*j + (z1 + z2)*k) ; gamma = z2*z3*i + z3*z1*j + z1*z2*k
Square discriminant of the equation f'(z) = 0 (whose roots are foci) is :
disc2 = (c*x*y*(m3 - m5)^2 + b*x*z*(m1 - m3)^2 + a*y*z*(m1 - m5)^2)*(b*x + a*y)*(c*x + a*z)*(c*y + b*z)/(b^2*c^2*x^3*y*z)
So unique root when disc2 = 0 or
c*x*y*(m3 - m5)^2 + b*x*z*(m1 - m3)^2 + a*y*z*(m1 - m5)^2 = 0
Dominique
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