Article: CJB/2010/121 Some Lines through the Centroid G Christopher Bradley Abstract: Using areal co-ordinates we catalogue some lines through the centroid G of a triangle ABC. Co-ordinates of points and equations of lines are given.

1. The Symmedian line

A C' B'

N M

Ka KmG K

B L C

G(1,1,1) Ka K (b^2+c^2 (a^2,b^2,c^2) Km -a^2,..) (b^2+c^2,..)

A' Fig. 1 The equation of the line through the three symmedian points is (b2 – c2)x + (c2 – a2)y + (a2 – b2)z = 0. (1) Km is the symmedian of the Medial triangle and Ka is the symmedian of the Anticomplementary triangle. Ka has co-ordinates (b2 + c2 – a2, c2 + a2 – b2, a2 + b2 – c2) and Km has co-ordinates (b2 + c2, c2 + a2, a2 + b2). K, of course, has co-ordinates (a2, b2, c2).

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2. The Euler line

A

deL

O Sch G N H

B C

Fig. 2 The Centroid G(1, 1, 1), the Circumcentre O(a2(b2 + c2 – a2), .., ..), the Orthocentre H(1/(b2 + c2 – a2), .., ..), the Nine-Point Centre N(a2(b2 + c2) – (b2 – c2)2, .., ..), deLongchamps point deL(– 3a4 + 2a2(b2 + c2) + (b2 – c2)2, .., ..), Schiffler’s point Sch(a/(cos B + cos C), .., ..). The equation of the Euler line is

(b2 + c2 – a2)(b2 – c2)x + (c2 + a2 – b2)(c2 – a2)y + (a2 + b2 – c2)(a2 – b2)z = 0. (2)

Schiffler’s point is the intersection of the Euler’s line of ABC, IBC, ICA, IAB, where I is the incentre.

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3. The Incentre Centroid line

A

G I Na Sp

B C

Fig. 3 This line passes through G(1, 1, 1), I(a, b, c), the Spieker Centre Sp(b + c, c + a, a + b) and Nagel’s point (b + c – a, c + a – b, a + b – c). It is the analogue of the Euler line when O is replaced by I. It is also the analogue of the Symmedian line when the symmedian point is replaced with the incentre. The Spieker Centre is the centroid of the three sides of the triangle with no internal material present. It is also the incentre of the medial triangle, as its co-ordinates imply. These last two properties are, in fact, the same property. The equation of this line is (b – c)x + (c – a)y + (a – b)z = 0 (3)

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4. The Mittenpunkt line

A

F M N

Mi G Ni Mi* E Ge

B L D C

Fig. 4 This line contains the Mittenpunkt Mi(a(b + c – a), b(c + a – b), c(a + b – c)), the isogonal conjugate of the Mittenpunkt Mi*(a/(b + c – a), b/(c + a – b), c/(a + b – c)), Gergonne’s point Ge(1/(b + c – a), 1/(c + a – b), 1/(a + b – c)), the centroid G(1, 1, 1), the centre of the Nine-Point Conic (with Cevian points G and Ge) Ni(a(b + c) – (b – c)2, b(c + a) – (c – a)2, c(a + b) – (a – b)2)

As it is an analogue of the Euler line (when Mi replaces O) one has GGe = 2MiG and MiNi = NiGe. It also has many other less important points lying on it.

Its equation is (b + c – a)(b – c)x + (c + a – b)(c – a)y + (a + b – c)(a – b)z = 0. (4)

It should be noted that we can obtain countless lines through G by using a point P other than G (not on the sides of ABC) and drawing the Cevians APD, BPE, CPF, constructing the conic DEFLMN and finding its centre Q. Then PGQ is a straight line and QP = 3GQ.

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5. The Centroid Feuerbach line

A

Feu

O IS I ES G

B C

Fig. 5a

In Fig. 5 IS and ES are the centres of similitude of the circumcircle and the incircle and so lie on the line IO and are harmonic conjugates with respect to O and I. IS has co-ordinates (a2(b + c – a), b2(c + a – b), c2(a + b – c)) and ES has co-ordinates (a2/(b + c – a), b2/(c + a – b), c2(a + b – c)). The line OI IS ES has equation bc(b – c)(b + c – a)x + ca(c – a)(c + a – b)y + ab(a – b)(a + b – c)z = 0. (5a)

IS also lies on the line joinin G to Feurbach’s point Fe, the point where the nine-point circle touches the incircle. Feurbach’s co-ordinates are ((b + c – a)(b – c)2, (c + a – b)(c – a)2, (a + b – c)(a – b)2). The line G Fe has equation (b – c)(a(b + c) – b2 – c2)x + (c – a)(b(c + a) – c2 – a2)y + (a – b)(c(a + b) – a2 – b2)z = 0. (5b)

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A

Feu

O IS I ES G N FeH

B C

Fig. 5b

It is also the case that G ES passes through the harmonic conjugate FeH of Feu with respect to I (the incentre) and N(the nine-point centre). The co-ordinates of FeH are (bc(b + c)2/(b + c – a), ca(c + a)2/ (c + a – b), ab(a + b)2/(a + b – c)) and the line G FeH ES has equation (b + c – a)(b – c)(a(b + c) + b2+ c2)x + (c + a – b)(c – a)(b(c + a) + c2 + a2)y + (a + b – c)(a – b)(c(a + b) + a2 + b2)z = 0. (5c)

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