Homogeneous Barycentric Coordinates

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1 hapter 9 Homogeneous arycentric oordinates 9. bsolute and homogeneous barycentric coordinates The notion of barycentric coordinates dates back to. F. Möbius ( ). Given a reference triangle, we put at the vertices,, masses u, v, w respectively, and determine the balance point. The masses at and can be replaced by a single mass v + w at the point = v +w. Together with the mass at, this can be replaced by a mass u + v + w v+w at the point P which divides in the ratio P : P = v + w : u. This is the point with absolute barycentric coordinate u +v +w, provided u + v + w 0. We also say that u+v+w the balance point P has homogeneous barycentric coordinates (u : v : w) with reference to. 9.. The centroid The midpoints of the sides are D = + 2, E = + 2 The centroid G divides each median in the ratio 2:. Thus,, F = +. 2 G = +2D 3 = This is the absolute barycentric coordinate of G (with reference to ). Its homogeneous barycentric coordinates are simply G =(::). triple (u : v : w) with u + v + w =0does not represent any finite point on the plane. We shall say that it represents an infinite point. See?.

2 40 Homogeneous arycentric oordinates F G E Z I D 9..2 The incenter The bisector divides the side in the ratio : = c : b. This gives = b+c. Note that has length ca. Now, in triangle, the bisector I divides b+c b+c ca in the ratio I : I = c : = b + c : a. It follows that b+c I = a +(b + c) a + b + c = a + b + c. a + b + c The homogeneous barycentric coordinates of the incenter are I =(a : b : c) The barycenter of the perimeter onsider the barycenter (center of mass) of the perimeter of triangle. The edges,, can be replaced respectively by masses a, b, c at their midpoint D = +, 2 E = +, and F = +. With reference to the medial triangle DEF, this has coordinates 2 2 a : b : c. Since the sidelengths of triangle DEF are in the same proportions, this barycenter is the incenter of the medial triangle, also called the Spieker center S p of. F E S p D

3 9. bsolute and homogeneous barycentric coordinates 4 The center of mass of the perimeter is therefore the point S p = a D + b E + c F a + b + c = a + + b + + c a + b + c (b + c) +(c + a) +(a + b) =. 2(a + b + c) In homogeneous barycentric coordinates, 9..4 The Gergonne point S p =(b + c : c + a : a + b). We follow the same method to compute the coordinates of the Gergonne point G e. Here, = s b and = s c, so that = (s b)b +(s c). a Z G e I The ratio G e : G e, however, is not immediate obvious. It can nevertheless be found by applying the Menelaus theorem to triangle with transversal Z. Thus, G e G e Z Z =. From this, Therefore, G e G e = Z Z = a s c s a s b = a(s a) (s b)(s c). (s b)(s c) + a(s a) G e = (s b)(s c)+a(s a) (s b)(s c) +(s a)(s c) +(s a)(s b) =. (s b)(s c)+a(s a)

4 42 Homogeneous arycentric oordinates The homogeneous barycentric coordinates of the Gergonne point are 9.2 evian triangle G e = (s b)(s c) :(s c)(s a) :(s a)(s b) = : :. s a s b s c It is clear that the calculations in the preceding section applies in the general case. We summarize the results in the following useful alternative of the eva theorem. Theorem 9. (eva). Let,, Z be points on the lines,, respectively. The lines,, Z are collinear if and only if the given points have coordinates of the form = (0 : y : z), = (x : 0 : z), Z = (x : y : 0), for some x, y, z. If this condition is satisfied, the common point of the lines,, Z is P =(x : y : z). P P P P P Remarks. () The points,, Z are called the traces of P. We also say that Z is the cevian triangle of P (with reference to triangle ). Sometimes, we shall adopt the more functional notation for the cevian triangle and its vertices: cev(p ): P =(0:y : z), P =(x :0:z), P =(x : y :0). (2) The point P divides the segment in the ratio P : = x : x + y + z. (3) It follows that the areas of the oriented triangles P and are in the ratio Δ(P) : Δ() = x : x + y + z. This leads to the following interpretation of homogeneous barycentric coordinates: the homogeneous barycentric coordinates of a point P can be taken as the proportions of (signed) areas of oriented triangles: P =Δ(P):Δ(P):Δ(P).

5 9.2 evian triangle 43 O 9.2. The circumcenter onsider the circumcenter O of triangle. Since O =2α, the area of triangle O is 2 O O sin O = 2 R2 sin 2α. Similarly, the areas of triangles O and O are respectively 2 R2 sin 2β and 2 R2 sin 2γ. It follows that the circumcenter O has homogeneous barycentric coordinates ΔO :ΔO :ΔO = 2 R2 sin 2α : 2 R2 sin 2β : 2 R2 sin 2γ =sin2α :sin2β :sin2γ = a cos α : b cos β : c cos γ = a b2 + c 2 a 2 : b c2 + a 2 b 2 : c a2 + b 2 c 2 2bc 2ca 2ab = a 2 (b 2 + c 2 a 2 ):b 2 (c 2 + a 2 b 2 ):c 2 (a 2 + b 2 c 2 ) The Nagel point and the extouch triangle The -excircle touches the side at a point such that = s c and = s b. From this, the homogeneous barycentric coordinates of are 0:s b : s c; similarly for the points of tangency and Z of the - and -excircles: =(0 : s b : s c), =(s a :0:s c), Z =(s a : s b :0), From these we conclude that,, and Z concur. Their common point is called the Nagel point and has coordinates N a =(s a : s b : s c). The triangle Z is called the extouch triangle.

6 44 Homogeneous arycentric oordinates I b I c Z N a s c s b s b s c The orthocenter and the orthic triangle For the orthocenter H with traces,, Z on,, respectively, we have = c cos β, = b cos γ. This gives : = c cos β : b cos γ = cos β b similarly for the other two traces. I a : cos γ ; c Z H = 0 : b cos β : c cos γ a c = : 0 : cos α cos γ a b Z = : : 0 cos α cos β H = a cos α : b cos β : c cos γ = b 2 c 2 a 2 : c 2 +a 2 b 2 : a 2 +b 2 c 2

7 9.3 Homotheties 45 The triangle Z is called the orthic triangle. 9.3 Homotheties Let P be a given point, and k a real number. The homothety with center P and ratio k is the transformation h(p, k) which maps a point to the point such that P = k P. Equivalently, divides P in the ratio P : = k : k, and h(p, k)() =( k)p + k. P k k 9.3. Superiors and inferiors The homotheties h(g, 2) and h ( G, 2) are called the superior and inferior operations respectively. Thus, sup(p ) and inf(p ) are the points dividing P and the centroid G according to the ratios PG : Gsup(P )= :2, PG : Ginf(P )= 2:. P G sup(p ) inf(p ) Proposition 9.. IfP =(u : v : w) in homogeneous barycentric coordinates, then sup(p )= (v + w u : w + u v : u + v w), inf(p )= (v + w : w + u : u + v). Proof. In absolute barycentric coordinates, sup(p )= 3G 2P 2(u + v + w) = ( + + ) u + v + w (u + v + w)( + + ) 2(u + v + w) = u + v + w u + v + w (v + w u) +(w + u v) +(u + v w) =. u + v + w Therefore, sup(p )=(v + w u : w + u v : u + v w) in homogeneous barycentric coordinates. The case for inferior is similar.

8 46 Homogeneous arycentric oordinates Example. () The superior of the incenter is the Nagel point. The inferior of the incenter is the Spieker center, the barycenter of the perimeter of the triangle. (2) The nine-point center, being the midpoint of O and H, is the inferior of O. H From the homogeneous barycentric of O, we obtain N G O N = b 2 (c 2 + a 2 b 2 )+c 2 (a 2 + b 2 c 2 ): : = a 2 (b 2 + c 2 ) (b 2 c 2 ) 2 : :. L o 9.4 rea and barycentric coordinates Theorem 9.2. If for i =, 2, 3, P i = x i + y i + z i (in absolute barycentric coordinates), then the area of the oriented triangle P P 2 P 3 is x y z ΔP P 2 P 3 = x 2 y 2 z 2 x 3 y 3 z 3 Δ. Theorem 9.3 (Routh theorem). If,, Z are points on the lines,, respectively such that : = λ :, : = μ :, Z : Z = ν :, then the cevian lines,, Z bound a triangle with area (λμν ) 2 (μν + μ +)(νλ + ν +)(λμ + λ +) Δ. ν R Z μ P Q λ Proof. In homogeneous barycentric coordinates with reference to triangle, =(0::λ), =(μ :0:), Z =(:ν :0).

9 9.4 rea and barycentric coordinates 47 Those of P, Q, R can be worked out easily: P = Z Q = Z R = =(μ :0:) Z =(:ν :0) =(0::λ) Z =(:ν :0) =(0::λ) =(μ :0:) P =(μ : μν :) Q =(:ν : νλ) R =(λμ ::λ) This means that the absolute barycentric coordinates of P, Q, R are P = (μ + μν + ), μν + μ + Q = ( + ν + νλ), νλ + ν + R = (λμ + + λ). λμ + λ + From these, μ μν ν νλ λμ λ rea(pqr)= (μν + μ +)(νλ + ν +)(λμ + λ +) Δ (λμν ) 2 = (μν + μ +)(νλ + ν +)(λμ + λ +) Δ.

Menelaus and Ceva theorems

Menelaus and Ceva theorems hapter 21 Menelaus and eva theorems 21.1 Menelaus theorem Theorem 21.1 (Menelaus). Given a triangle with points,, on the side lines,, respectively, the points,, are collinear if and only if = 1. W Proof.