Sunday, March 19, 2006

Barycentric Coordinates. Topic: Analytic Geometry. Level: Olympiad.

Definition: We define barycentric coordinates (mass points) as follows. Given a triangle $ ABC $, we have $ A(1,0,0) $, $ B(0,1,0) $, and $ C(0,0,1) $.

Let $ d(P,l) $ denote the signed distance from a point $ P $ to a line $ l $.

A point $ Q(x,y,z) $ in the plane is represented by

$ \left(\frac{d(Q,BC)}{d(A,BC)}, \frac{d(Q,CA)}{d(B,CA)}, \frac{d(Q,AB)}{d(C,AB)}\right) $.

It's easy to show that all points $ Q(x,y,z) $ satisfy $ x+y+z = 1 $.


Theorem: Three points $ P_i(x_i,y_i,z_i) $ for $ i = 1,2,3 $ are collinear iff

$ \left|\begin{array}{ccc} x_1 \quad y_1 \quad z_1 \\ x_2 \quad y_2 \quad z_2 \\ x_3 \quad y_3 \quad z_3 \end{array} \right| $ = 0.


Comment: The proof of this is pretty standard. It works the same way as with Cartesian coordinates (you can use vectors and cross products).


Problem: (WOOT Message Board) Let $D, E, F$ lie on sides $BC, CA, AB$, respectively, of triangle $ABC$. Suppose that $D, E, F$ are collinear. Prove that the midpoints of $AD, BE, CF$ are collinear.

Solution: We use barycentric coordinates with triangle $ ABC $.

Since $ D $ lies on $ BC $, we know $ d(D, BC) = 0 $ so the $ x $-coordinate will be zero. Hence the coordinates of $ D $ have the form $ (0,t_1,1-t_1) $ for some real $ t_1 $.

Similarly, $ E $ is of the form $ (1-t_2, 0, t_2) $ and $ F $ is $ (t_3, 1-t_3, 0) $ for real $ t_2, t_3 $.

Since $ D, E, F $ are collinear, we know the determinant of the matrix with the three points is zero. This comes out to be

$ t_1t_2t_3-(1-t_1)(1-t_2)(1-t_3) = 0 $. (1)

We can find the midpoints by simply averaging the points (work this out rigorously if you want). These are

$ \left(\frac{1}{2},\frac{t_1}{2},\frac{1-t_1}{2}\right) $, $ \left(\frac{1-t_2}{2},\frac{1}{2},\frac{t_2}{2}\right) $, and $ \left(\frac{t_3}{2},\frac{1-t_3}{2},\frac{1}{2}\right) $.

If we take the determinant of the matrix with these three points, we find that it is equivalent to

$ \frac{t_1t_2t_3-(1-t_1)(1-t_2)(1-t_3)}{4} $

after simplification. But by (1) we know that it zero as well, which means the midpoints are indeed collinear. QED.


Comment: Barycentric coordinates are very powerful, and give easy proofs to some results which would be very difficult without them.


Practice Problem: (Menelaus' Theorem) Let $ D, E, F $ be points on sides $ BC , CA, AB $ (possibly extended), respectively, of triangle $ ABC $. Prove that $ D,E,F $ are collinear iff $ BD \cdot CE \cdot AF = -DC \cdot AE \cdot BF $ with signed distances.


  1. It's strange trying to find an intuitive justification for barycentric coordinates, though... quite odd. How does the technique extend to area ratios? (Cross product?)

  2. I think if you make an affine transformation from ABC to R^3 that preserves the coordinates of A,B, and C it's just three arbitrary points in the plane x+y+z = 1, so you can use the same method as with space coordinates.

    Not completely certain about this method yet, but it seems like it works.

    Another Idea: (from MathWorld) If you write the equation of a line with two of the points, you can see that the third point has to make the determinant zero in order to be on the line as well.