Barycentric Coordinates We Generalize the Homogeneous

Home , Centroid, Circumscribed circle, Euler line, Incenter, Trilinear coordinates

Barycentric Coordinates

We generalize the homogeneous coordinates to be applicable more easily to ﬁnite triangles.

Consider homogeneous coordinates piece by piece: [a : b : c] = a[1 : 0 : 0] + b[0 : 1 : 0] + c[0 : 0 : 1]. If b and c were zero, we’d be at the ideal point for the x-axis, while the ideal point for the y-axis has a and c zero. If a and b are zero we’re at the origin. If only one or none of the coordinates are zero, then the a, b, and c tell you how “close” you are to the corresponding point. For instance, holding a and b fixed and letting c get larger, you get closer and closer to the origin, while holding a and c fixed and letting b get large moves you vertically toward the ideal point on the y-axis, while keeping x constant. We can actually do this with any three points. Given three non-collinear points A, B, and C in the plane (let’s stick with the Euclidean plane for now) and a point P in the plane, we can give barycentric coordinates for the point P as [a : b : c]. These will be defined so that if we put a mass of a at A, b and B, and c at C, then the center of mass of these masses will be at P . But the center of mass is easy to find! If the points are at (xA, yA), (xB, yB), and (xC , yC ) respectively, the x-coordinate of the center of mass is the weighted sum of the x-coordinates divided by the total mass: (axA + bxB + cxC )/(a + b + c). Similarly for the y-coordinates. Points in the actual Euclidean plane cannot have a+b+c = 0, though those seem like pretty reasonable projective points. Sometimes, we take advantage of this fact and “normalize” the coordinates by dividing through by a + b + c so that the sum of the barycentric coordinates is 1. Barycentric coordinates are a generalization of homogeneous coordiantes. Homogeneous coordinates are barycentric coordinates where the original three points A, B, and C are the origin and the ideal points of the axes. Barycentric coordinates are sometimes called areal coordinates, because they relate to area. In 4ABC with point P that has barycentric coordinates [a : b : c], the ratio of the areas of the triangles 4PBC : 4P AC : 4P AB is the same as the ratio a : b : c. In particular, if a = 0 then the point P is on BC and divides it in the internal ratio b : c. We can also allow barycentric coordinates to be negative. In that case, the point ends up outside the triangle. Here are some neat points in barycentric coordinates: • The centroid of a triangle has barycentric coordinates [1 : 1 : 1]. The centroid is where the medians cross. • The incenter of a triangle, the center of the inscribed circle, has barycentric coordinates [sin(A) : sin(B) : sin(C)]. • The circumcenter of a triangle, the center of the circumscribed circle, has barycentric coordinates [sin(2A) : sin(2B) : sin(2C)] = [(1 − cos(B) cos(C)) : (1 − cos(A) cos(C)) : (1 − cos(A) cos(B))]. • The orthocenter, where the three altitudes meet, has barycentric coordinates [tan(A): tan(B) : tan(C)].

1 Barycentric coordinates have a lot of stuﬀ going on with them, and if you want to study triangles in any serious way you will deﬁnitely learn a lot about them (and their complementary trilinear coordinates which measure the relative distances to the sides of a triangle instead of to the vertices). What we’re going to need most is to be able to tell when three points are collinear, given their barycentric coordinates.

Theorem: Let P , Q, and R be points, with normalized (i.e. total = 1) barycentric coordinates [ap : bp : cp], [aq : bq : cq], and [ar : br : cr] with respect to triangle 4ABC. Then   ap bp cp area(4P QR) = det  aq bq cq  area(4ABC) ar br cr .

Corollary: The points are collinear if and only if this determinant is zero.

The theorem is not too hard to prove if you know a little about determinants. The ﬁrst thing to know is that the area of a triangle with vertices (xA, yA), (xB, yB), and (xC , yC ) is  x x x  1 A B C det y y y . This can easily be veriﬁed using the shoelace method. The second 2  A B C  1 1 1 thing is that for matrices M and N, det(MN) = det(M) det(N). This is not trivial to prove, so we’ll skip it. But given these two facts the theorem is easy to prove because we just write down the determinant formula for the area of 4P QR and it factors into the determinant area for 4ABC and the matrix given in the theorem.

Example: The orthocenter, circumcenter, and centroid are collinear. Proof: put their coordinates in a matrix, determinant is easily seen to be zero. This line is called the Euler line of a triangle.