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Trilinear Coordinates

\begin{figure}\begin{center}\BoxedEPSF{trilinear_area.epsf scaled 800}\end{center}\end{figure}

Given a Triangle $\Delta ABC$, the trilinear coordinates of a point $P$ with respect to $\Delta ABC$ are an ordered Triple of numbers, each of which is Proportional to the directed distance from $P$ to one of the side lines. Trilinear coordinates are denoted $\alpha:\beta:\gamma$ or $(\alpha,\beta,\gamma)$ and also are known as Barycentric Coordinates, Homogeneous Coordinates, or ``trilinears.''

In trilinear coordinates, the three Vertices $A$, $B$, and $C$ are given by $1:0:0$, $0:1:0$, and $0:0:1$. Let the point $P$ in the above diagram have trilinear coordinates $\alpha:\beta:\gamma$ and lie at distances $a'$, $b'$, and $c'$ from the sides $BC$, $AC$, and $AB$, respectively. Then the distances $a'=k\alpha$, $b'=k\beta$, and $c'=k\gamma$ can be found by writing $\Delta_a$ for the Area of $\Delta BPC$, and similarly for $\Delta_b$ and $\Delta_c$. We then have

$\displaystyle \Delta$ $\textstyle =$ $\displaystyle \Delta_a+\Delta_b+\Delta_c={\textstyle{1\over 2}}aa'+{\textstyle{1\over 2}}bb'+{\textstyle{1\over 2}}cc'$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(ak\alpha+bk\beta+ck\gamma) = {\textstyle{1\over 2}}k(a\alpha+b\beta+c\gamma).$ (1)

k\equiv {2\Delta\over a\alpha +b\beta +c\gamma},
\end{displaymath} (2)

where $\Delta$ is the Area of $\Delta ABC$ and $a$, $b$, and $c$ are the lengths of its sides. When the values of the coordinates are taken as the actual lengths (i.e., the trilinears are chosen so that $k=1$), the coordinates are known as Exact Trilinear Coordinates.

Trilinear coordinates are unchanged when each is multiplied by any constant $\mu$, so

t_1:t_2:t_3=\mu t_1:\mu t_2:\mu t_3.
\end{displaymath} (3)

When normalized so that
\end{displaymath} (4)

trilinear coordinates are called Areal Coordinates. The trilinear coordinates of the line
\end{displaymath} (5)

\end{displaymath} (6)

where $d_i$ is the Point-Line Distance from Vertex $i$ to the Line.

Trilinear coordinates for some common Points are summarized in the following table, where $A$, $B$, and $C$ are the angles at the corresponding vertices and $a$, $b$, and $c$ are the opposite side lengths.

Point Trilinear Center Function
Centroid $M$ $\csc A$, $1/a$
Circumcenter $O$ $\cos A$
de Longchamps Point $\cos A-\cos B\cos C$
Equal Detour Point $\sec({\textstyle{1\over 2}}A)\cos({\textstyle{1\over 2}}B)\cos({\textstyle{1\over 2}}C)+1$
Feuerbach Point $F$ $1-\cos(B-C)$
Incenter $I$ 1
Isoperimetric Point $\sec({\textstyle{1\over 2}}A)\cos({\textstyle{1\over 2}}B)\cos({\textstyle{1\over 2}}C)-1$
Lemoine Point $a$
Nine-Point Center $N$ $\cos(B-C)$
Orthocenter $H$ $\cos B\cos C$
Vertex $A$ $1:0:0$
Vertex $B$ $0:1:0$
Vertex $C$ $0:0:1$

To convert trilinear coordinates to a vector position for a given triangle specified by the $x$- and $y$-coordinates of its axes, pick two Unit Vectors along the sides. For instance, pick

$\displaystyle \hat{\bf a}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}a_1\\  a_2\end{array}\right]$ (7)
$\displaystyle \hat{\bf c}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}c_1\\  c_2\end{array}\right],$ (8)

where these are the Unit Vectors $BC$ and $AB$. Assume the Triangle has been labeled such that $A={\bf x}_1$ is the lower rightmost Vertex and $C={\bf x}_2$. Then the Vectors obtained by traveling $l_a$ and $l_c$ along the sides and then inward Perpendicular to them must meet

\left[{\matrix{x_1\cr y_1\cr}}\right]+l_c\left[{\matrix{c_1\... a_2}}\right]-k\alpha\left[{\matrix{a_2\cr -a_1\cr}}\right].
\end{displaymath} (9)

Solving the two equations
$\displaystyle x_1+l_cc_1-k\gamma c_2$ $\textstyle =$ $\displaystyle x_2l_aa_1-k\alpha a_2$ (10)
$\displaystyle y_1+l_cc_2+k\gamma c_1$ $\textstyle =$ $\displaystyle y_2l_aa_2+k\alpha a_1,$ (11)


$\displaystyle l_c$ $\textstyle =$ $\displaystyle {k\alpha(a_1c_1+a_2c_2)-\gamma k({c_1}^2+{c_2}^2)+c_2(x_1-x_2)+c_1(y_2-y_1)\over a_1c_2-a_2c_1}$ (12)
$\displaystyle l_a$ $\textstyle =$ $\displaystyle {k\alpha({a_1}^2+{a_2}^2)-\gamma k(a_1c_1+a_2c_2)+a_2(x_1-x_2)+a_1(y_2-y_1)\over a_1c_2-a_2c_1}.$ (13)

But $\hat{\bf a}$ and $\hat{\bf c}$ are Unit Vectors, so

$\displaystyle l_c$ $\textstyle =$ $\displaystyle {k\alpha(a_1c_1+a_2c_2)-\gamma k+c_2(x_1-x_2)+c_1(y_2-y_1)\over a_1c_2-a_2c_1}$ (14)
$\displaystyle l_a$ $\textstyle =$ $\displaystyle {k\alpha-\gamma k(a_1c_1+a_2c_2)+a_2(x_1-x_2)+a_1(y_2-y_1)\over a_1c_2-a_2c_1}.$ (15)

And the Vector coordinates of the point $\alpha:\beta:\gamma$ are then
{\bf x}={\bf x}_1+l_c\left[{\matrix{c_1\cr c_2\cr}}\right]-k\gamma\left[{\matrix{c_2\cr -c_1\cr}}\right].
\end{displaymath} (16)

See also Areal Coordinates, Exact Trilinear Coordinates, Orthocentric Coordinates, Power Curve, Quadriplanar Coordinates, Triangle, Trilinear Polar


Boyer, C. B. History of Analytic Geometry. New York: Yeshiva University, 1956.

Casey, J. ``The General Equation--Trilinear Co-Ordinates.'' Ch. 10 in A Treatise on the Analytical Geometry of the Point, Line, Circle, and Conic Sections, Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd ed., rev. enl. Dublin: Hodges, Figgis, & Co., pp. 333-348, 1893.

Coolidge, J. L. A Treatise on Algebraic Plane Curves. New York: Dover, pp. 67-71, 1959.

Coxeter, H. S. M. Introduction to Geometry, 2nd ed. New York: Wiley, 1969.

Coxeter, H. S. M. ``Some Applications of Trilinear Coordinates.'' Linear Algebra Appl. 226-228, 375-388, 1995.

Kimberling, C. ``Triangle Centers and Central Triangles.'' Congr. Numer. 129, 1-295, 1998.

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© 1996-9 Eric W. Weisstein