Barycentric Coordinates Books∗ 2018 Abstract We've all heard of the term "barycentric coordinates" a couple of times, but this is one of the huge leaps that take incredible amounts of determi- nation to make. This is my shot at bridging this gap. Don't think for a second that barycentric coordinates are elegant. This is a powerful tool, but it is still fundamentally a bash technique. This technique also does have flaws; cyclic quadrilaterals in particular make life miserable.1 ∗Dennis Chen. Email me corrections, problems, etc that could be added to this article at [email protected]. 1See Section 5.7 of http://web.evanchen.cc/handouts/bary/bary-full.pdf. 1 Overview 1 Preliminaries 3 1.1 Notation . 3 1.2 Area Definition . 3 1.3 Mass Points Definition . 4 1.4 Vector Definition . 5 1.5 Normalization . 5 2 Common Points 5 3 Area, Lines, and Circles 7 3.1 Area . 7 3.2 Lines . 8 3.3 Circles . 11 4 Examples 12 5 Exercises 13 6 Hints and Comments 14 7 Extra 15 7.1 Barycentric to Cartesian/Complex . 15 7.2 Conway Notation . 15 7.3 Other Points . 16 7.4 Lines . 17 7.5 Circles . 17 7.6 Parting Shots . 17 2 1 Preliminaries 1.1 Notation Our reference triangle is 4ABC: We denote \A = \CAB; \B = \ABC; \C = \BCA; a = BC; b = CA; c = AB: Also, let P denote the mass of P: 1.2 Area Definition Definition 1. (Signed Areas) We use signed areas for this entire section. What this means that if A; B; C goes counterclockwise, then [ABC] is positive, and if A; B; C goes clockwise, then [ABC] is negative. We first consider the most intuitive definition of barycentric coordinates, which is the area definition. With barycentric coordinates, our reference is a triangle, not a pair of perpendicular axes. Definition 2. (Area Ratios) The barycentric coordinates of P with respect to 4ABC are (x; y; z) such that [BCP ]:[CAP ]:[ABP ] = x : y : z; and x+y+z = 1: It is very important to note that the order of the vertices of 4ABC matter. The coordinates of the same point P are different with respect to 4ABC and 4ACB; for example. (In fact, any permutation of the vertices corresponds to a permutation of the coordinates.) For convenience, we can express this as an x y z ordered ratio; ( x+y+z ; x+y+z ; x+y+z ) can easier be expressed as (x : y : z): (This is because normalizing, or the process of making x + y + z = 1 is fairly easy, and even at some times not needed.) Another important thing to note is that barycentric coordinates always must add to 1: Let's see what happens if we put P outside the triangle. 3 All of a sudden, the sum of the areas of the triangles is not [ABC]: This seems like a problem, until we notice that in the diagram above, [ABC] = x + y − jzj: This is the reason that signed areas are necessary. 1.3 Mass Points Definition Remark 1. We use directed lengths as well for this entire section. Barycentrics are known as an extension of mass points, which may not be apparent at first glance. But the analogy is certainly valid. Let's consider reference triangle 4ABC and a point P: Draw an altitude from A to BC; and draw one from P to BC: Clearly, since the two triangles have the same base, the ratio of [ABC]:[P BC] is the same as the ratio of their altitudes. And PD : AD = a : a + b + c; where P = a + b + c; A = a: We can do this with the other two vertices, which should illustrate the point. We can generalize to zero and negative values if we just use directed lengths. 1 1 Thus, if 4ABC has masses A; B; C; then P has barycentrics ( A : B : 1 C ): (This is due to the reciprocal nature of mass points!) BD z CE Definition 3. P has barycentric coordinates (x; y; z) such that DC = y ; EA = x AF y 2 z ; FB = x : Remark 2. This can be used to prove Ceva's and Menelaus's very easily. 2See http://www.aquatutoring.org/TYMCM%20Mass%20Points.pdf for an introduction to mass points. 4 1.4 Vector Definition Let A;~ B;~ C;~ P~ be the vectors3 with heads A; B; C; P and tail O for some arbi- trary O: (We omit O because our choice of O: is irrelevant. Sometimes, we will define O as P for convenience; other times, we will not specify O:) Definition 4. (Vectors) Then, P~ = xA~ + yB~ + zC;~ x + y + z = 1 where (x; y; z) denotes the barycentric coordinates of P: 1.5 Normalization We say coordinates x; y; z are normalized if x + y + z = 1: If they are not normalized, this means that we have expressed it in the form (kx : ky : kz); where (x; y; z) is normalized, but (kx : ky : kz) are not necessarily. 2 Common Points Theorem 1. (Centroid) The areal4 coordinates of the centroid of 4ABC are 1 1 1 ( 3 ; 3 ; 3 ): Proof. Trivial by the area definition. Theorem 2. (Symmedian Point)5 The barycentric coordinates of the symme- dian point are (a2 : b2 : c2): Proof. It is a well-known property of the symmedian point that its distance to the sides of the triangle is proportional to the lengths of the triangle, i.e. its trilinear coordinates6 are (a : b : c): The conversion is straightforward; we get the barycentric coordinates as (a2 : b2 : c2): Theorem 3. (Incenter) The barycentric coordinates of the incenter of 4ABC a b c are ( a+b+c ; a+b+c ; a+b+c ): Proof. We use the area definition. We note ar br cr [BCP ] = ; [CAP ] = ; [ABP ] = ; 2 2 2 3Numerous resources exist for vectors; you could go to Evan's bary paper and see Appendix A, or you could email me for the chapter on vectors in my book. 4Areal coordinates are another name for barycentric coordinates. The only difference is that areal coordinates must be normalized! 5AKA Lemoine Point. See http://forumgeom.fau.edu/FG2008volume8/FG200812.pdf for basic properties. 6Trilinear coordinates are just ratios of the distances from P to the sides. Like barycentric coordinates, trilinear coordinates are homogenous, i.e. (a : b : c) is the same as (ka : kb : kc); though for some cases you will need to normalize them. 5 (a+b+c)r and [ABC] = 2 : Dividing yields the barycentric coordinates of our in- center as a b c ( ; ; ); a + b + c a + b + c a + b + c as desired. Theorem 4. (Excenter) The A-excenter has barycentric coordinates (−a : b : c); and symmetric expressions exist for the B and C excenters. Proof. Trivially, the perpendiculars from the A − excenter to sides a; b; c all bh have the same absolute distance. Using 2 (4.2), we get −ar br cr [BCE ]:[CAE ]:[ABE ] = : : = −a : b : c; A A A 2 2 2 as desired. Theorem 5. (Orthocenter) The barycentric coordinates of the orthocenter of 4ABC are (tan A : tan B : tan C): Proof. Let H be the orthocenter. We use the area definition. WLOG, the circumcenter has diameter 2; then a = sin A; b = sin B; c = sin C: Then we can let D; E; F be the foots of the altitudes of A; B; C; respectively. Then we use right 4ABD and note that BC = cos B; BD = sin C cos B; HD = sin A cos B cos C cos B cos C; [BCH] = 2 : We use symmetry and note [ACH] = sin B cos A cos C sin C cos A cos B 2 ; [ABH] = 2 : This implies [BCH]:[ACH]:[ABH] = sin A cos B cos C : sinB cos A cos C : sin C cos A cos B = tan A : tan B : tan C; as desired. (The last transition comes from dividing everything by cos A cos B cos C:) Theorem 6. (Circumcenter) The barycentric coordinates of the circumcenter of 4ABC are (sin 2A : sin 2B : sin 2C): Proof. Let the circumcenter be O: We use the area definition. Note that OA = OB = OC: Then we use the Inscribed Angle Theorem and angle chase; this means \BOC = 2A; \COA = 2B; \AOB = 2C: Then we use [ABC] = 1 2 ab sin C (4.4) to get [BCO]:[CAO]:[ABO] = r2 sin 2A : r2 sin 2B : r2 sin 2C = sin 2A : sin 2B : sin 2C; as desired. Remark 3. The orthocenter and circumcenter are not that nice. However, there is some useful stuff you can do with them! 6 3 Area, Lines, and Circles 3.1 Area Theorem 7. (Area, Collinearity, and Concurrency) Given three points P,Q,R with normalized coordinates (x1; y1; z1); (x2; y2; z2); (x3; y3; z3) respectively, x1 y1 z1 [P QR] = x2 y2 z2 : [ABC] x3 y3 z3 Proof. We use Cartesian Coordinates; all points written in Cartesian Coordi- nates will be expressed as [x; y; z] for clarity. Choose O not in the plane determined by A,B,C, such that O = [0; 0; 0];A = [1; 0; 0];B = [0; 1; 0];C = [0; 0; 1]: (We are using a three-dimensional coordinate system!) Then we note the form of the plane containing ABC has the equation x + y + z = 1: (This corresponds to the normalized coordinates of any point in the plane!) Then let the parallelepiped that A,B,C spans (remember their tails are O = [0; 0; 0]; which this time cannot be ignored due to no preservation of generality) be denoted as PABC ; and similarly, let PP QR denote the paral- lelepiped spanned by P;~ Q;~ R:~ Then we use the determinant definition of volume and note that x1 y1 z1 PP QR x2 y2 z2 = : PABC x3 y3 z3 Then we note that by the definition of a parallelepiped, [PP QR] = 2[P QR]h = [PABC ] 2[ABC]h [P QR] [ABC] ; as desired.