The first time I saw a reference to trilinear coordinates I thought this must be another name for barycentric coordinates. It’s not.

Barycentric coordinates come up often in applications, such as when working with finite element meshes. Trilinear coordinates are less common, at least in my experience, and yet trilinear coordinates simplify a lot of classical geometry problems.

To find the trilinear coordinates of a point relative to a triangle, calculate the distance to each side of the triangle. This triple of numbers is one possible set of trilinear coordinates. So is any multiple of these three points. Trilinear coordinates are homogeneous, like projective coordinates, and are all proportional triples represent the same point.

Here’s how trilinear coordinates relate to recent posts.

My post on the nine point circle mentions the orthocenter. The trilinear coordinates of the orthocenter are

sec *A* : sec *B* : sec *C*.

where *A*, *B*, and *C* are the three angles of the triangle. The center of the nine point circle has trilinear coordinates

cos(*B* − *C*) : cos(*C* − *A*) : cos(*A* − *B*).

The post on incircles and excircles implicitly uses trilinear coodinates. Since the incircle is tangent to all three sides of a triangle, it’s distance to each side is the radius of the circle, and so the incircle’s center has coordinates

1 : 1 : 1.

The three excircles have centers with coordinates

−1 : 1 : 1

1 : −1 : 1

1 : 1 : −1

A point with positive coordinates is inside the triangle and a point with one or two negative coordinates lies outside the triangle. More specifically, if the first coordinate is negative, the point lies outside the triangle and inside the sector determined by angle *A*, and similarly for the other coordinates. A point cannot have all three trilinear coordinates negative: if a point lies inside the sector of each angle of the triangle, then it’s inside the triangle.

The vertices of the equilateral triangle described by Morley’s theorem have coordinates

1 : 2 cos(*C*/3) : 2 cos(*B*/3)

2 cos(*C*/3) : 1 : 2 cos(*A*/3)

2 cos(*B*/3) : 2 cos(*A*/3) : 1