[alogo] Centroid Characterization and related observations

Consider a point D and its cevians AE, BF, CG w.r to triangle ABC. Show that the only point for which the three ratios AD/AE = BD/BF = CD/CG = k (constant) is the centroid of the triangle.


Use the trilinear coordinates (x:y:z) of point D to calculate all relevant segments:
z'/y' = EEB/EEC = z/y, z'/sinB + y'/sinC = a => ( D = -(z*sinB + y*sinC)), z' = -z*a/D =>
z' = (a*z*sinB*sinC)/(z*sinC+y*sinB),
y' = (a*y*sinB*sinC)/(z*sinC+y*sinB).
BE/EC = DBD/DDC = z*sinC/(y*sinB) = z*c/(y*b),
AD/AE = (z/sinB+y/sinC)/a ,
AE2 = [bc/(zc+yb)2][(b2+c2-a2)yz + bc(y2+z2)].
Setting AD/AE = BD/BF = CD/CG = k, gives a linear system for the trilinears (denoting u=sinA, v=sinB, w=sinC):

[0_0] [0_1] [0_2]

R being the circumradius, and using the sinus-theorem a=2RsinA, ...etc. Thus, the trilinears of D are (1/sinA : 1/sinB : 1/sinC) = (1/a : 1/b : 1/c) i.e. those of the centroid.
One can handle analogously the question of determining points D such that the ratios AD/AE , BD/BF , CD/CG , satisfy a generalization of the previous relation. Namely, assuming f(a,b,c) to be a function of the side-lengths of the triangle, one can ask for those D that the corresponding ratios satisfy:
AD/AE = kf(a,b,c), BD/BF = kf(b,c,a), CD/CG = kf(c,a,b) (cyclically permute a,b,c in f). The previous reasoning gives:

[0_0] [0_1] [0_2] [0_3]

Since trilinears are defined modulo a multiplicative constant, one can consider the last column as representing the trilinears (g(a,b,c)) of the point D with the given property. For example, taking f(a,b,c) = 1/bc, we find as solution the trilinears ((b+c-a)/a : (c+a-b)/b : (a+b-c)/c ), which characterize the Nagel point of the triangle (see Nagel.html ). Below is a small list of examples with f's and the corresponding g's:
f = 1 ---> g = 1/a (centroid)
f = a ---> g = (b+c-a)/a (Nagel)
f = 1/a ---> g = (a(b+c)-bc)/a (X(192))
f = b+c ---> g = 1 (incenter)
f = 1/(b+c) ---> g = ((a+c)(b+c)+(b-a)(b+a))/a (???)
For some simple functions f the corresponding g might not be contained in the known list of triangle centers.
In order to find the constant k, for which the relations AD/AE = kf(a,b,c), BD/BF = kf(b,c,a), CD/CG = kf(c,a,b) hold, one can use absolute trilinears, satisfying ax+by+cz = 2S (S being the area of the triangle). Thus, in the last example, with f(a,b,c) = b+c, giving the incenter, the absolute trilinears are (r : r : r) , r being the inradius. Substitute this in one of the three linear equations above to get the corresponding k = 2Rr/(abc).
Having the trilinears (x : y : z) of a point, it is easy to find the corresponding function f(a,b,c). It suffices to multiply with the first matrix above and factor out the terms f(a,b,c), .. etc. Thus, denoting by {f(a,b,c)} the tripples resulting by cyclic permutation of a,b,c, the invertible map:
{f(a,b,c)} -> {g(a,b,c) = (-f(a,b,c)+f(b,c,a)+f(c,a,b))/a}
establishes a relationship between the various triangle centers and the tripples of ratios (AD/AE, BD/BF, CD/CG).

See Also


Return to Gallery

Produced with EucliDraw©