Chapter 13. Straight line equations Area and barycentric coordinates

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Chapter 13 Straight line equations 13.1 Area and barycentric coordinates Theorem 13.1. If for i = 1,2,3, P i = x i A + y i B + z i C (in absolute barycentric coordinates), then the area of the oriented triangle P 1 P 2 P 3 is x 1 y 1 z 1 P 1 P 2 P 3 = x 2 y 2 z 2 ABC. x 3 y 3 z 3 Example 13.1. (Area of cevian triangle) LetP = (u : v : w) be a point with cevian triangle XYZ. The area of the cevian triangle XYZ is 1 0 v w (v +w)(w+u)(u+v) u 0 w = 2uvw (v +w)(w+u)(u+v). u v 0 If P is an interior point (so that u, v, w are positive), then v + w 2 vw, w + u 2 wu, andu+v 2 uv. It follows that 2uvw (v +w)(w+u)(u+v) 2uvw 2 vw 2 wu 2 uv = 1 4. Equality holds if and only if u = v = w, i.e., P = (1 : 1 : 1) = G, the centroid. The centroid is the interior point with largest cevian triangle.

402 Straight line equations 13.2 Equations of straight lines 13.2.1 Two-point form The area formula has an easy and extremely important consequence: three points P i = (u i,v i,w i ) are collinear if and only if u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 = 0. Consequently, the equation of the line joining two points with coordinates (x 1 : y 1 : z 1 ) and(x 2 : y 2 : z 2 ) is x 1 y 1 z 1 x 2 y 2 z 2 x y z = 0, or (y 1 z 2 y 2 z 1 )x+(z 1 x 2 z 2 x 1 )y +(x 1 y 2 x 2 y 1 )z = 0. Examples (1) The equations of the sidelines BC,CA,AB are respectively x = 0, y = 0, z = 0. (2) Given a point P = (u : v : w), the cevian line AP has equation wy vz = 0; similarly for the other two cevian lines BP and CP. These lines intersect corresponding sidelines at the traces ofp : A P = (0 : v : w), B P = (u : 0 : w), C P = (u : v : 0). (3) The equation of the line joining the centroid and the incenter is 1 1 1 a b c x y z = 0, or (b c)x+(c a)y +(a b)z = 0. (4) The equations of some important lines: Euler line OH cyclic (b2 c 2 )(b 2 +c 2 a 2 )x = 0 OI-line OI Soddy line IG cyclicbc(b c)(b+c a)x = 0 e cyclic (b c)(s a)2 x = 0 Brocard axis OK cyclic b2 c 2 (b 2 c 2 )x = 0 van Aubel line HK cyclic S αα(s β S γ )x = 0

13.2 Equations of straight lines 403 13.2.2 Intersection of two lines The intersection of the two lines p 1 x+q 1 y +r 1 z = 0, p 2 x+q 2 y +r 2 z = 0 is the point (q 1 r 2 q 2 r 1 : r 1 p 2 r 2 p 1 : p 1 q 2 p 2 q 1 ). Proposition 13.2. Three linesp i x+q i y+r i z = 0,i = 1,2,3, are concurrent if and only if p 1 q 1 r 1 p 2 q 2 r 2 p 3 q 3 r 3 = 0. Examples (1) The intersection of the Euler line and the Soddy line is the point (c a)(s b) 2 (a b)(s c) 2 (c 2 a 2 )(c 2 +a 2 b 2 ) (a 2 b 2 )(a 2 +b 2 c 2 ) : : =(c a)(a b) (s b) 2 (s c) 2 (c+a)(c 2 +a 2 b 2 ) (a+b)(a 2 +b 2 c 2 ) : : =(c a)(a b) (s b) 2 a(b c) (c+a)(c 2 +a 2 b 2 ) (b c)(a+b+c) 2 : : =(b c)(c a)(a b) (s b) 2 a (c+a)(c 2 +a 2 b 2 ) (a+b+c) 2 : : = 1 4 (b c)(c a)(a b) (c+a b) 2 4a (c+a)(c 2 +a 2 b 2 ) (a+b+c) 2 : : = 1 4 (b c)(c a)(a b)( 3a4 +2a 2 (b 2 +c 2 )+(b 2 c 2 ) 2 ) : : Writing a 2 = S β +S γ etc., we have 3a 4 +2a 2 (b 2 +c 2 )+(b 2 c 2 ) 2 = 3(S β +S γ ) 2 +2(S β +S γ )(2S α +S β +S γ )+(S β S γ ) 2 =4(S αβ +S γα S βγ ). This intersection has homogeneous barycentric coordinates S βγ +S γα +S αβ : S βγ S γα +S αβ : S βγ +S γα S αβ. This is the reflection of H in O, and is called the delongchamps point L o.

404 Straight line equations 13.3 Perspective triangles Many interesting points and lines in triangle geometry arise from the perspectivity of triangles. We say that two trianglesx 1 Y 1 Z 1 andx 2 Y 2 Z 2 are perspective,x 1 Y 1 Z 1 X 2 Y 2 Z 2, if the linesx 1 X 2,Y 1 Y 2,Z 1 Z 2 are concurrent. The point of concurrency, (X 1 Y 1 Z 1,X 2 Y 2 Z 2 ), is called the perspector. Along with the perspector, there is an axis of perspectivity, or the perspectrix, which is the line joining containing Y 1 Z 2 Z 1 Y 2, Z 1 X 2 X 1 Z 2, X 1 Y 2 Y 1 X 2. We denote this line by L (X 1 Y 1 Z 1,X 2 Y 2 Z 2 ). Homothetic triangles are clearly prespective. If triangles T and T, their perspector is the homothetic center, which we shall denote by 0 (T,T ). Proposition 13.3. A triangle with vertices X = U : v : w, Y = u : V : w, Z = u : v : W, for some U, V, W, is perspective to ABC at (XYZ) = (u : v : w). The perspectrix is the line x u U + y v V + z w W = 0. Proof. The line AX has equation wy vz = 0. It intersects the sideline BC at the point (0 : v : w). Similarly,BY intersectsca at(u : 0 : w) andcz intersectsab at(u : v : 0). These three are the traces of the point (u : v : w). The lineyz has equation (vw VW)x+u(w W)y+u(v V)z = 0. It intersects the sidelinebc at(0 : v V : (w W)). Similarly, the lineszx andxy intersectca and AB respectively at ( (u U) : 0 : w W) and (u U : (v V) : 0). It is easy to see that these three points are collinear on the line x u U + y v V + z w W = 0.

13.3 Perspective triangles 405 The excentral triangle The excentral triangle is perspective with ABC; the perspector is the incenter I: 13.3.1 The Conway configuration I a = a : b : c I b = a : b : c I c = a : b : c I = a : b : c Given triangle ABC, extend (i) CA andba to Y a andz a such that AY a = AZ a = a, (ii) AB and CB toz b andx b such that BZ b = BX b = b, (iii) BC andac tox c andy c such thatcx c = CY c = c. Z a Y a a A a Z Y? X b b B C c X c X c b Y c These points have coordinates Z b Y a = (a+b : 0 : a), Z a = (c+a : a : 0); Z b = ( b : b+c : 0), X b = (0 : a+b : b); X c = (0 : c : c+a), Y c = ( c : 0 : b+c). From the coordinates ofy c andz b, we determine easily the coordinates ofx = BY c CZ b : Y c = c : 0 : b+c = bc : 0 : b(b+c) Z b = b : b+c : 0 = bc : c(b+c) : 0 X = = bc : c(b+c) : b(b+c)

406 Straight line equations Similarly, the coordinates of Y = CZ a AX c, and Z = AX b BY a can be determined. The following table shows that the perspector of trianglesabc andxyz is the point with homogeneous barycentric coordinates ( 1 a : 1 b : 1 c). X = bc : c(b+c) : b(b+c) = 1 1 1 : : b+c b c Y = c(c+a) : ca : a(c+a) = 1 1 1 : : a c+a c Z = b(a+b) : a(a+b) : ab = 1 a : 1 b : 1 a+b? = = 1 1 1 : : a b c

13.4 Perspectivity 407 13.4 Perspectivity Many interesting points and lines in triangle geometry arise from the perspectivity of triangles. We say that two trianglesx 1 Y 1 Z 1 andx 2 Y 2 Z 2 are perspective,x 1 Y 1 Z 1 X 2 Y 2 Z 2, if the linesx 1 X 2,Y 1 Y 2,Z 1 Z 2 are concurrent. The point of concurrency, (X 1 Y 1 Z 1,X 2 Y 2 Z 2 ), is called the perspector. Along with the perspector, there is an axis of perspectivity, or the perspectrix, which is the line joining containing Y 1 Z 2 Z 1 Y 2, Z 1 X 2 X 1 Z 2, X 1 Y 2 Y 1 X 2. We denote this line byl (X 1 Y 1 Z 1,X 2 Y 2 Z 2 ). We justify this in below. If one of the triangles is the triangle of reference, it shall be omitted from the notation. Thus, (XYZ) = (ABC,XYZ) andl (XYZ) = L (ABC,XYZ). Homothetic triangles are clearly prespective. If triangles T and T, their perspector is the homothetic center, which we shall denote by 0 (T,T ). Proposition 13.4. A triangle with vertices X = U : v : w, Y = u : V : w, Z = u : v : W, for some U, V, W, is perspective to ABC at (XYZ) = (u : v : w). The perspectrix is the line x u U + y v V + z w W = 0. Proof. The line AX has equation wy vz = 0. It intersects the sideline BC at the point (0 : v : w). Similarly,BY intersectsca at(u : 0 : w) andcz intersectsab at(u : v : 0). These three are the traces of the point (u : v : w). The lineyz has equation (vw VW)x+u(w W)y+u(v V)z = 0. It intersects the sidelinebc at(0 : v V : (w W)). Similarly, the lineszx andxy intersectca andab respectively at( (u U) : 0 : w W) and(u U : (v V) : 0). These three points are collinear on the trilinear polar of(u U : v V : w W). The triangles XYZ andabc are homothetic if the perspectrix is the line at infinity.

408 Straight line equations 13.4.1 The Schiffler point: intersection of four Euler lines Theorem 13.5. Let I be the incenter of triangle ABC. The Euler lines of the triangles IBC,ICA,IAB are concurrent at a point on the Euler line ofabc, namely, the Schiffler point ( a(b+c a) S c = : b+c b(c+a b) c+a : ) c(a+b c). a+b Proof. LetI be the incenter of triangle ABC. We first compute the equation of the Euler line of the triangle IBC. The centroid of triangleibc is the point(a : a+2b+c : a+b+2c). The circumcenter of triangle is the midpoint of II a. This is the point ( a 2 : b(b+c) : c(b+c)). From these we obtain the equation of the Euler line: x y z 0 = a a+2b+c a+b+2c = (b c)(b+c)x+a(c+a)y a(a+b)z. a 2 b(b+c) c(b+c) The equations of the Euler lines of the triangles ICA and IAB can be obtained by cyclic permutations of a,b,c andx,y,z. Thus the three Euler lines are (b c)(b+c)x + a(c+a)y a(a+b)z = 0, b(b+c)x + (c a)(c+a)y + b(a+b)z = 0, c(b+c)x c(c+a)y + (a b)(a+b)z = 0. Computing the intersection of the latter two lines, we have the point (b+c)x : (c+a)y : (a+b)z = c a b c a b : b b c a b : b c a c c = (c+a)(a b)+bc : b(c+a b) : c(b (c a)) = a(b+c a) : b(c+a b) : c(a+b c). It is easy to verify that this point also lies on the Euler line of IBC given by the first equation: (b c)a(b+c a)+ab(c+a b) ac(a+b c) = a((b c)(b+c a)+b(c+a b) c(a+b c)) = a(b 2 c 2 ab+ca+bc+ab b 2 ca bc+c 2 ) = 0. It is routine to verify that this point also lies on the Euler line of ABC, with equation (b 2 c 2 )(b 2 +c 2 a 2 )x+(c 2 a 2 )(c 2 +a 2 b 2 )y +(a 2 b 2 )(a 2 +b 2 c 2 )z = 0. This shows that the four Euler lines are concurrent.

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