BarycentricsFormulas

1. Formulas in Barycentrics

Consider the triangle of reference ABC and a point P, with absolute barycentric coordinates (see
BarycentricCoordinates.html ) w.r. to ABC: P(p_x, p_y, p_z). Denote by (P_x, P_y, P_z) etc. the
corresponding trilinear coordinates. The basic relations between barycentric and cartesian
coordinates are expressed through the vector-relations:

P = p_x*A+p_y*B+p_z*C .

Writing these equations in matrix form we get:

If N denotes the above matrix then its inverse M = N^-1 (use Maxima) is:

The denominator of the factor equals to twice the area of the triangle. Notice that this matrix relation
does not express a linear relation between the two sets of coordinates, because of
the third coordinate of the column vector on the right side, which is always 1, even for the sum of
two points. There is though a particular instance of interest, when we consider the line between
two points P, Q:
S(t)=(1-t)P+tQ.
For the corresponding barycentric coordinates we get, using the previous matrix M.

Since r = t/(t-1) is the signed ratio r = SP/SQ (with inverse given by the same function: t = r/(r-1)), we see that the point
(s(t)_x, s(t)_y, s(t)_z), on the line of PQ, divides the segment PQ in this ratio r = SP/SQ = t/(t-1).
Also the difference of coordinates are involved in interesting formulas. In fact, from the very first equation we have:

Where we put d_x = p_x-q_x, etc.. These d's satisfying d_x+d_y+d_z = 0. Take in account last equation
and consider now the inner product of P-Q with itself giving:

Taking the origin to be the center O of the circumcircle of triangle ABC, we have, R being the radius
and denoting with {A,B,C} also the measure of the angles of the triangle of reference:

Taking into account the equation d_x+d_y+d_z=0, only the second term contributes to the product, thus:

This, by the sinus-equation, giving in turn (a,b,c being the side-lengths of the triangle):

From the equation d_x+d_y+d_z=0, we have that -d_yd_z = (1/2)(d_y²+d_z²-d_x²) and similar equations for
the cyclic permutations of the symbols (x->y->z->x). Introducing this into the above equation and
refactoring we get the well known formula ([YiuGN], p.87):

2. Relations involving the trilinears

Taking into account the relation of (absolute x) barycentrics to (absolute X) trilinears, D denoting the area of the triangle:

we get the corresponding formula expressing the distance in terms of the differences (D_x, D_y, D_z)
of the trilinear coordinates of the two points:

And since aD_x+bD_y+cD_z=0, with D_x = P_x-Q_x etc., we have equivalently:

Last equation, taking into account the well known R = abc/(4D) and the sine equation a = 2Rsin(A) etc., leads to [Carnoy p.84]:

From this, assuming Q fixed and P variable on the circle with center at Q and radius r, we obtain the equation of the circle
in trilinears:

This obtains its usual for trilinears homogeneous form by multiplying where appropriate with 1 written
as 1 = (aP_x+bP_y+cP_z)/(2D). In fact, expanding D_x² = (P_x-Q_x)² , etc. and multiplying the linear in
Q_x,Q_y,Q_z terms with the last expression of the unit and also multiplying the quadratic terms with the
square of the expression we obtain the equation of the circle in the form [Carnoy p. 131]:

The coefficients (k,l,m) are functions of the coordinates of the center and the radius of the circle.

In particular, for the circumcircle of the triangle of reference, we obtain easily its equation from the fact that the vertices,
with (non absolute) trilinears (1/a,0,0), (0,1/b,0), (0,0,1/c) satisfy the equation => k = sin(2A)/a, etc..
After simplifying we get the equation (see also Trilinears.html ) :
                                                                       ayz + bzx  + cxy = 0.
Remark-1 It can be proved ([Carnoy p. 133], [Koehler p. 173]) that equation:
                                                                      sin(2A)x² + sin(2B)y² + sin(2C)z² = 0,
represents the Conjugate circle of the triangle of reference (see Autopolar.html ). Thus, since the difference of the
equations of two circles represents their radical axis the linear part of the general equation:
                                                                       kx + ly + mz =0,
represents the radical axis of the general circle and the conjugate circle of the triangle of reference ABC ([LoneyII p. 69],
see also Remark-2 of Section-5 of ProjectiveCoordinates.html ).

Remark-2 From the above form of the general equation follows that all circles pass through the intersection points of the curves:

The first equation representing the line at infinity, this implies that the two points representing the solutions of the equations
are the complex imaginary circular points at infinity.
See also the file AreaInBarycentrics.html computing the area of a triangle, whose vertices are given in barycentric
coordinates.

Bibliography

[Carnoy] Carnoy, Joseph. Geometrie Analytique Paris, Gauthier-Villars
[Koehler] Koehler, J. Exercices de Geoemetrie Analytique, Premiere partie Gauthier-Villars, Paris 1886
[LoneyII] Loney, S. L. Coordinate Geometry, vol. II Trilinear Coordinates Macmillan, London 1934
[YiuGN] Yiu, P. GeometryNotes020402 (pdf). http://www.math.fau.edu/yiu/GeometryNotes020402.pdf

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1. Formulas in Barycentrics

2. Relations involving the trilinears

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Bibliography