**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 $.

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**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.

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

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**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.

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

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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.

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?)

ReplyDeleteI 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.

ReplyDeleteNot 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.