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Chapter 10 Isotomic and Isogonal Conjugates

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Chapter 10 Isotomic and Isogonal Conjugates Chapter 10 Isotomic and isogonal conjugates 10.1 Isotomic conjugates The Gergonne and Nagel points are examples of isotomic conjugates. Two points P and Q (not on any of the side lines of the reference triangle) are said to be isotomic conjugates if their respective traces are symmetric with respect to the midpoints of the corresponding sides. Thus, BX = X′C, CY = Y ′A, AZ = Z′B. B X′ Z X Z′ P P • C A Y Y ′ We shall denote the isotomic conjugate of P by P •. If P =(x : y : z), then 1 1 1 P • = : : =(yz : zx : xy). x y z 320 Isotomic and isogonal conjugates 10.1.1 The Gergonne and Nagel points 1 1 1 Ge = s a : s b : s c , Na =(s a : s : s c). − − − − − − Ib A Ic Z′ Y Y Z I ′ Na Ge B C X X′ Ia 10.1 Isotomic conjugates 321 10.1.2 The isotomic conjugate of the orthocenter The isotomic conjugate of the orthocenter is the point 2 2 2 2 2 2 2 2 2 H• =(b + c a : c + a b : a + b c ). − − − Its traces are the pedals of the deLongchamps point Lo, the reflection of H in O. A Z′ L Y o O Y ′ H• Z H B C X X′ Exercise 1. Let XYZ be the cevian triangle of H•. Show that the lines joining X, Y , Z to the midpoints of the corresponding altitudes are concurrent. What is the common point? 1 2. Show that H• is the perspector of the triangle of reflections of the centroid G in the sidelines of the medial triangle. A Z′ H • Y ′ G B C X′ 1 The centroid. Apply Menelaus theorem to triangle AAH D, and transversal MX, where M is the mid- point of the altitude AAH . 322 Isotomic and isogonal conjugates 10.1.3 Congruent-parallelians point Given triangle ABC, we want to construct a point P the three lines through which parallel to the sides cut out equal intercepts. Let P = xA + yB + zC in absolute barycentric coordinates. The parallel to BC cuts out an intercept of length (1 x)a. It follows that the three intercepts parallel to the sides are equal if and only if − 1 1 1 1 x :1 y :1 z = : : . − − − a b c The right hand side clearly gives the homogeneous barycentric coordinates of I•, the iso- tomic conjugate of the incenter I. 2 This is a point we can easily construct. Now, translating into absolute barycentric coordinates: 1 1 I• = [(1 x)A + (1 y)B + (1 z)C]= (3G P ). 2 − − − 2 − we obtain P =3G 2I•, and can be easily constructed as the point dividing the segment − I•G externally in the ratio I•P : PG = 3 : 2. The point P is called the congruent- − parallelians point of triangle ABC. 3 B P G I• C A Exercise 1. Calculate the homogeneous barycentric coordinates of the congruent-parallelian point and the length of the equal parallelians. 4 2. Let A′B′C′ be the midway triangle of a point P . The line B′C′ intersects CA at B = B′C′ CA, C = B′C′ AB, a ∩ a ∩ C = C′A′ AB, A = C′A′ BC, b ∩ b ∩ A = A′B′ BC, B = A′B′ CA. c ∩ c ∩ Determine P for which the three segments BaCa, CbAb and AcBc have equal lengths. 2 The isotomic conjugate of the incenter appears in ETC as the point X75. 3 This point appears in ETC as the point X192. 4(ca + ab bc : ab + bc ca : bc + ca ab). The common length of the equal parallelians is 2abc . − − − ab+bc+ca 10.1 Isotomic conjugates 323 10.1.4 Yff-Brocard points Consider a point P = (u : v : w) with traces X, Y , Z satisfying BX = CY = AZ = µ. This means that w u v a = b = c = µ. v + w w + u u + v Elimination of u, v, w leads to 0 µ a µ − − 0= b µ 0 µ =(a µ)(b µ)(c µ) µ3. − − − − − − µ c µ 0 − − Indeed, µ is the unique positive root of the cubic polynomial (a t)(b t)(c t) t3. − − − − This gives the point 1 1 1 c µ 3 a µ 3 b µ 3 P = − : − : − . b µ c µ a µ − − − ! A A Z′ Y ′ Z Y P • P B C B C X X′ The isotomic conjugate 1 1 1 b µ 3 c µ 3 a µ 3 P • = − : − : − c µ a µ b µ − − − ! has traces X′, Y ′, Z′ that satisfy CX′ = AY ′ = BZ′ = µ. These points are called the Yff-Brocard points. 5 They were briefly considered by A. L. Crelle. 6 Exercise 1. Justify that the polynomial f(t) := (a t)(b t)(c t) t3 has a unique positive root. − − − − 2. Is it possible to find collinear points X on BC, Y on CA, and Z on AB with BX CX = CY AY = AZ BZ? − − − 5P. Yff, An analogue of the Brocard points, Amer. Math. Monthly, 70 (1963) 495 – 501. 6A. L. Crelle, 1815. 324 Isotomic and isogonal conjugates 10.2 Isogonal conjugates 10.2.1 Reflections and isogonal conjugates Consider a point P with reflections P a, P b, P c in the sidelines BC, CA, AB. Let Q be a point on the line isogonal to AP with respect to angle A, i.e., the lines AQ and AP are symmetric with respect to the bisector of angle BAC. A P b P c P Q B C Clearly, the triangles AQP b and AQP c are congruent, so that Q is equidistant from P b and P c. For the same reason, any point on a line isogonal to BP is equidistant from P c and a P . It follows that the intersection P ∗ of two lines isogonal to AP and BP is equidistant a b c from the three reflections P , P , P . Furthermore, P ∗ is on a line isogonal to CP . For this reason, we call P ∗ the isogonal conjugate of P . It is the center of the circle of reflections of P . A P b P c P P ∗ B C P a 10.2.2 The pedal circle Clearly, (P ∗)∗ = P . Moreover, the circles of reflections of P and P ∗ are congruent, since, a a in Figure ??, the trapezoid PP ∗Pa∗P being isosceles, PPa∗ = P ∗P . It follows that the 10.2 Isogonal conjugates 325 pedals of P and P ∗ on the sidelines all lie on the same circle with center the midpoint of PP ∗. We call this the common pedal circle of P and P ∗. A Qb P b P c Qc P M Q = P ∗ B C Qa P a Exercise 1. The perpendiculars from the vertices of ABC to the corresponding sides of the pedal triangle of a point P concur at the isogonal conjugate of P . 2. Given a point P with isogonal conjugate P ∗, let X, Y , Z be the pedals of P on the sidelines BC, CA, AB of triangle ABC. If the circle X(P ∗) intersects BC at X1, X2, Y (P ∗) intersects CA at Y1, Y2, and Z(P ∗) intersects AB at Z1, Z2, then the six points X1, X2, Y1, Y2, Z1, Z2 are on a circle center P . 10.2.3 Coordinates of isogonal conjugate Let P = (x : y : z) be a point other than the vertices of triangle ABC. Its isogonal conjugate is the point 2 2 2 P ∗ =(a yz : b zx : c xy). (10.1) Proof. We show that the reflection of the cevian AP in the bisector of angle A intersects b2 c2 the line BC at the point X′ =(0: y : z ). Let X be the A-trace of P , with ∠BAP = θ. This is the point X = (0 : y : z) = (0 : S S : c2) in Conway’s notation. It follows that S S : c2 = y : z. If α − θ − α − θ − 326 Isotomic and isogonal conjugates A θ θ B C X X′ the reflection of AX (in the bisector of angle A) intersects BC at X′, we have X′ = (0 : 2 2 2 2 2 2 b2 c2 b : Sα Sθ)=(0: b c : c (Sα Sθ)) = (0 : b z : c y)=(0: y : z ). Similarly, the− reflections− of the cevians− BP and CP− in the respective angle bisectors intersect CA at a2 c2 a2 b2 Y ′ = ( x : 0 : z ) and AB at Z′ = ( x : y : 0). These points X′, Y ′, Z′ are the traces of the point P ∗ whose coordinates are given in (10.1) above. Example. Isogonal conjugate pairs: P O G Ge Na P ∗ H K T+ T − 10.3 Examples of isogonal conjugates 327 10.3 Examples of isogonal conjugates 10.3.1 The circumcenter and orthocenter The circumcenter O and the orthocenter H are isogonal conjugates, since π ∠(AO, AC)= β = ∠(AH, AB), 2 − − and similarly ∠(BO,BA)= ∠(BH,BC), ∠(CO,CB)= ∠(CH,CA). − − Since OOa = AH, AOOaH is a parallelogram, and HOa = AO. This means that the circle of reflections of O is congruent to the circumcircle. Therefore, the circle of reflections of H is the circumcircle, and the reflections of H lie on the circumcircle. A Hb Hc O H B C Ha 328 Isotomic and isogonal conjugates 10.3.2 The symmedian point and the centroid Consider triangle ABC together with its tangential triangle A′B′C′, the triangle bounded by the tangents of the circumcircle at the vertices. A simple application of the Ceva theorem 7 shows that the lines AA′, BB′, CC′ are concurrent at a point K.
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