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.

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). − − − − − −

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 midpoint 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 isotomic 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.

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 .

P b P c

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

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

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. We call this point K the symmedian point of triangle ABC for the reason below.

C′

B′

B D C

A′ Z

Since A′ is equidistant from B and C, we construct the circle A′(B) = A′(C) and extend the sides AB and AC to meet this circle again at Z and Y respectively. Note that

∠(A′Y, A′B′)= π 2(π α γ)= π 2β, − − − − and similarly, ∠(A′C′, A′Z′)= π 2γ. Since ∠(A′B′, A′C′)= π 2α, we have − − ∠(A′Y, A′Z)=∠(A′Y, A′B′)+ ∠(A′B′, A′C′)+ ∠(A′C′, A′Z) =(π 2β)+(π 2α)+(π 2γ) − − − =π 0 mod π. ≡

This shows that Y , A′ and Z′ are collinear, so that (i) AA′ is a median of triangle AY Z, (ii) AY Z and ABC are similar. It follows that AA′ is the isogonal line of the A-median. We say that it is a symmedian of triangle ABC. Similarly, the BB′ and CC′ are the symmedians isogonal to B- and C-medians. The lines AA′, BB′, CC′ therefore intersect at the isogonal conjugate of the centroid G. This we call the symmedian point of triangle ABC.

7If triangle ABC is acute, this is the Gergonne point of A′B′C′. 10.3 Examples of isogonal conjugates 329

10.3.3 The Gergonne point and the insimilicenter T+ Theorem 10.1. The isogonal conjugate of the Gergonne point is the insimilicenter of the circumcircle and the incircle:

T+ = Ge∗.

E F ′ E′

T+ F I O Ge

D′ B D C

Proof. Consider the intouch triangle DEF of triangle ABC. (1) If D′ is the reﬂection of D in the bisector AI, then (i) D′ is a point on the incircle, and (ii) the lines AD and AD′ are isogonal with respect to A.

A A

E E F ′ E′ F F G I e I P

D′ D′ B D C B D C

(2) Likewise, E′ and F ′ are the reﬂections of E and F in the bisectors BI and CI respectively, then (i) these are points on the incircle, (ii) the lines BE′ and CF ′ are isogonals of BE and CF with respect to angles B and C. Therefore, the lines AD′, BE′, and CF ′ concur at the isogonal conjugate of the Ger- gonne point. (3) In fact, E′F ′ is parallel to BC. This follows from 330 Isotomic and isogonal conjugates

E F ′ E′

F I

D′ B D C

(ID, IE′) =(ID, IE) + (IE, IE′) =(ID, IE) + 2(IE, IB) =(ID, IE) + 2((IE, AC) + (AC,IB)) π =(ID, IE) + 2(AC, IB) since (IE,AC)= 2 B =(π C) + 2 C + − 2 =B + C = A (mod π) −

Similarly,

(ID, IF ′) =(ID, IF ) + (IF, IF ′) =(ID, IF ) + 2(IF, IC) =(ID, IF ) + 2((IF, AB) + (AB,IC)) π =(ID, IF ) + 2(AB, IC) since (IF,AB)= 2 C = (π B) 2 B + − − − 2 = (B + C)= A (mod π) −

(4) Similarly, F ′D′ and D′E′ are parallel to CA and AB respectively. It follows that D′E′F ′ is homothetic to ABC. The ratio of homothety is r : R. Therefore, the center of homothety is the point which divides OI in the ratio R : r. This is the insimilicenter of (O) and (I). 10.3 Examples of isogonal conjugates 331

10.3.4 The Nagel point and the exsimilicenter T − Theorem 10.2. The isogonal conjugate of the Nagel point is the exsimilicenter of the circumcircle and the incircle:

T = Na∗. −

D1′ Cc

O Bb I Q Na

E1′ F1′

B Aa C

Proof. (1) Consider the A-cevian of the Nagel point, which joins the vertex A to the point of tangency Aa of the excircle with BC. This contains the antipode D1 on the incircle of the point of tangency D with BC. Therefore, the reflection of AAa in the bisector AI contains the reflection D1′ of D1. D1′ is clearly on the incircle. Indeed, it is the antipode of D′, the reflection of D in AI. (2) Likewise, if we consider the isogonal lines of the BBb and CCc in the respective angle bisectors, these contains E1′ and F1′ which are the antipodes of E′ and F ′ on the incircle.

The lines AD1′ , BE1′ , and CF1′ concur at the isogonal conjugate of the Nagel point. 332 Isotomic and isogonal conjugates

D1 D1′

D′ C B D Aa

Clearly, D1′ E1′ F1′ and D′E′F ′ are oppositely congruent at the O. Since D′E′F ′ is homothetic to ABC, so are D1′ E1′ F1′ and ABC, with ratio of homothety r : R. The center of homothety is the point which divides OI in the ratio R : r. This is− the exsimilicenter − of (O) and (I).

Exercise

1. The triangles D1′ E1′ F1′ is perspective with the intouch triangle DEF . The perspector is the orthocenter of DEF . (Hint: Show that DD1′ is parallel to the bisector IA). A

D1′ E

F I

E1′ F1′

B D C 10.4 Isogonal conjugate of an inﬁnite point 333

10.4 Isogonal conjugate of an inﬁnite point

Theorem 10.3. Given a triangle ABC and a line ℓ, let ℓa, ℓb, ℓc be the parallels to ℓ through A, B, C respectively, and ℓa′ , ℓb′ , ℓc′ their reﬂections in the angle bisectors AI, BI, CI respectively. The lines ℓa′ , ℓb′ , ℓc′ intersect at a point on the circumcircle of triangle ABC.

A A

ℓb′ I P ℓb′ I O P

ℓa′ B C B C

ℓ ℓa ℓb ℓc ℓc′ ℓ ℓb ℓc ℓc′

Proof. (1) Let P be the intersection of ℓb′ and ℓc′ .

(BP, PC) =(ℓb′ , ℓc′ )

=(ℓb′ , IB)+(IB, IC)+(IC, ℓc′ )

=(IB, ℓb)+(IB, IC)+(ℓc, IC) =(IB, ℓ)+(IB, IC)+(ℓ, IC) =2(IB, IC) π A =2 + 2 2 =(BA, AC) (mod π).

Therefore, ℓb′ and ℓc′ intersect at a point on the circumcircle of triangle ABC. (2) Similarly, ℓa′ and ℓb′ intersect at a point P ′ on the circumcircle. Clearly, P and P ′ are the same point since they are both on the reflection of ℓb in the bisector IB. (3) Therefore, the three reflections ℓa′ , ℓb′ , and ℓc′ intersect at the same point on the circumcircle. This point is the isogonal conjugate of the infinite point of ℓ.

Theorem 10.4. A point with homogeneous barycentric coordinates (x : y : z) lies on the circumcircle of triangle ABC if and only if

a2yz + b2zx + c2xy =0. 334 Isotomic and isogonal conjugates

Proposition 10.1. The isogonal conjugates of the inﬁnite points of two perpendicular lines are antipodal points on the circumcircle.

ℓ′ Q

ℓ I O P

B C

Proof. If P and Q are the isogonal conjugates of the inﬁnite points of two perpendicular lines ℓ and ℓ′ through A, then AP and AQ are the reﬂections of ℓ and ℓ′ in the bisector AI.

π (AP, AQ)=(AP, IA)+(IA, AQ)= (ℓ, IA) (IA, ℓ′)= (ℓ, ℓ′)= . − − − 2 Therefore, P and Q are antipodal points.

Exercise 1. Find the locus of the isotomic conjugates of points on the circumcircle. 8 L x y z 2. Let be the line u + v + w =0, intersecting the side lines BC, CA, AB of triangle ABC at U, V , W respectively.

(a) Find the equation of the perpendiculars to BC, CA, AB at U, V , W respectively. 9 (b) Find the coordinates of the vertices of the triangle bounded by these three perpendiculars. 10 (c) Show that this triangle is perspective with ABC at a point P on the circumcircle. 11 (d) Show that the Simson line of the point P is parallel to L.

8The line a2x + b2y + c2z =0 perpendicular to the Euler line. 9 2 2 (SB v + SC w)x + a wy + a vz =0, etc. 10 2 2 2 2 2 2 ( S u + SABuv + SBC vw + SCAwu : b (c uv SAuw S + Bvw) : c (b uw SAuv SC vw). 2 11 − a − − − − P = 2 : : . −a vw+SBuv+SC uw · · · · · · 10.4 Isogonal conjugate of an inﬁnite point 335

3. Let P and Q be antipodal points on the circumcircle. The lines P Q• and QP • joining each of these points to the isotomic conjugate of the other intersect orthogonally on the circumcircle.

P •

P Q O

B C

Q•