Isodynamic points of the

A file of the Geometrikon gallery by Paris Pamfilos

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Contents (Last update: 09‑03‑2021)

1 Apollonian circles of a triangle 1

2 Isodynamic points, Brocard and Lemoine axes 3

3 Given Apollonian circles 6

4 Given isodynamic points 8

5 of the isodynamic points 11

6 Same isodynamic points and same circumcircle 12

7 A matter of uniqueness 14

1 Apollonian circles of a triangle

Denote by {푎 = |퐵퐶|, 푏 = |퐶퐴|, 푐 = |퐴퐵|} the lengths of the sides of triangle 퐴퐵퐶. The “Apollonian circle of the triangle relative to side 퐵퐶” is the locus of points 푋 which satisfy

A

X

D' BCD

Figure 1: The Apollonian circle relative to the side 퐵퐶 the condition |푋퐵|/|푋퐶| = 푐/푏. By its definition, this circle passes through the vertex 퐴 and the traces {퐷, 퐷′} of the bisectors of 퐴̂ on 퐵퐶 (see figure 1). Since the bisectors are orthogonal, 퐷퐷′ is a diameter of this circle. Analogous is the definition of the Apollo‑ nian circles on sides 퐶퐴 and 퐴퐵. The following theorem ([Joh60, p.295]) gives another characterization of these circles in terms of Pedal . 1 Apollonian circles of a triangle 2

Theorem 1. The Apollonian circle 훼, passing through 퐴, is the locus of points 푋, such that their pedal triangles are isosceli relative to the vertex lying on 퐵퐶 (see figure 2).

A α F X D

BCE

Figure 2: The pedals of points 푋 on the Apollonian circles of the triangle are isosceli

Proof. This follows from the sine formula |퐸퐷| |푋퐵| sin(퐵)̂ 푐 ⋅ sin(퐵)̂ = = = 1. |퐸퐹| |푋퐶| sin(퐶)̂ 푏 ⋅ sin(퐶)̂

C Ο κ γ

F EA B G η μ χ I

Figure 3: Points {퐴, 퐺, 퐼} are collinear

Exercise 1. If 훾 is the Apollonian circle through 퐶 of the triangle 퐴퐵퐶 intersecting again the circumcircle 휅(푂) of 퐴퐵퐶 in 퐺 and 퐼 is the diametral to 푂 on the circle 휇 = (퐴푂퐵), show that {퐴, 퐺, 퐼} are collinear (see figure 3). For 퐹 = 퐶퐺 ∩ 퐴퐵 show that (퐶, 퐺) ∼ (퐹, 퐼) are harmonic pairs and 푂퐹 is the polar of 퐼 relative to 훾. Hint: Consider the circle 휒(퐼, |퐼퐵|) and see first that the three circles {훾, 휅, 휒} are pairwise orthogonal, which implies that 퐼 is on the radical axis of {훾, 휅}. Exercise 2. With the definitions of the preceding exercise, show that triangles {퐴퐵퐶, 퐴퐺퐽, 퐺퐵퐽} are similar, where 퐽 = 퐶퐼 ∩ 퐸푂 (see figure 4). Hint: Angle chasing: 퐴퐽퐺̂ = 퐺퐽퐵̂ = 퐴푂퐵/2̂ = 퐴퐶퐵,̂ 퐴퐺퐽̂ = 퐴퐵퐶,̂ 퐽퐺퐵̂ = 퐶퐴퐵.̂

Exercise 3. With the definitions of the two previous exercises, define the similarities {푓퐴, 푓퐵, 푓퐶}, where 푓퐴 has center at 퐴, rotation angle 퐴̂ and ratio 푘퐴 = 퐴퐵/퐴퐶, the others being defined by cyclic permutation of the letters {퐴, 퐵, 퐶}. Show that the composition 푓퐶 ∘ 푓퐵 ∘ 푓퐴 coincides with the symmetry at 퐵.

Hint: By the similarity of the triangles of the previous exercise, 푔 = 푓퐵 ∘ 푓퐴 has 푔(퐺) = 퐺. Also 푔(퐵) = 퐴. Hence 푔 is the similarity with center at 퐺, rotation angle 퐵퐺퐴̂ and ra‑ tio 퐺퐵/퐺퐴 = (퐺퐵/퐺퐽) ⋅ (퐺퐽/퐺퐴) = (퐴퐵/퐴퐶) ⋅ (퐵퐶/퐵퐴). Its composition with 푓퐶 defines ℎ = 푓퐶 ∘ 푔 = 푓퐶 ∘ 푓퐵 ∘ 푓퐴 which leaves 퐵 fixed. Since the similarities build a group, the composition is a similarity with fixed point 퐵. The angle is the sum of the angles of the similarities {푔, 푓퐶}, which is 휋. Finally, the ratio is the product of ratios, which turns out to be 1, thereby proving the claim. 2 Isodynamic points, Brocard and Lemoine axes 3

C γ O κ J E A B μ G

I

Figure 4: Three similar triangles

2 Isodynamic points, Brocard and Lemoine axes

Taking the ratios of the sides of a triangle 퐴퐵퐶, in a given orientation, we can define three constants

γ C'

β I2 α

B' A

I A' 1 B O C

δ Lemoine Brocard

Figure 5: Isodynamic points {퐼1, 퐼2} of the triangle 퐴퐵퐶

푘푎 = |퐴퐵|/|퐴퐶|, 푘푏 = |퐵퐶|/|퐵퐴|, 푘푐 = |퐶퐴|/|퐶퐵|, And three corresponding Apollonian circles {훼, 훽, 훾} on the sides of the triangle

푎 = |퐵퐶|, 푏 = |퐶퐴|, 푐 = |퐴퐵|, w.r. to the constants 푘푎, 푘푏, 푘푐. Theorem 2. The three Apollonian circles {훼, 훽, 훾} of the triangle form an intersecting pencil of circles passing through two points {퐼1, 퐼2} (see figure 5). Proof. In fact, assume for the moment that 퐼 is one common point of {훼, 훾}. Then, we have |퐼퐵|/|퐼퐶| = |퐴퐵|/|퐴퐶| and |퐼퐴|/|퐼퐵| = |퐶퐴|/|퐶퐵|. Multiplying the equations side by side we 2 Isodynamic points, Brocard and Lemoine axes 4 obtain the relation∶ |퐼퐴|/|퐼퐶| = |퐴퐵|/|퐶퐵|. Thus 퐼 belongs also to circle 훽. Since the circles {훼, 훽, 훾} are orthogonal to the circumcircle 훿 of the triangle 퐴퐵퐶, the inverted of 퐼 w.r. to the circumcircle 훿 of 퐴퐵퐶 is also a common point of the three circles. Assuming that |퐴퐵| < |퐶퐴| < |퐶퐵| , implies that the two circles 훼, 훾 intersect. In fact, in that case 퐴 is inside 훾 and 퐵 is outside, but inside 훼 . Thus, the two circles 훼, 훾 must intersect. In the case of isosceles triangle one of the three circles is the medial line of its base and the other two are symmetric w.r. to this line and the circles intersect again.

The points {퐼1, 퐼2} are called “isodynamic” points of the triangle 퐴퐵퐶. The line 퐼1퐼2 is called “Brocard axis” of the triangle. The orthogonal to it, line of centers of the circles {훼, 훽, 훾} is called “Lemoine axis” of the triangle. The Isodynamic points play an important role in the “geometry of the triangle” ([Gal13], [Yiu13]). Here are some related properties in which participate concepts studied in this branch of geometry. In these the concept of Inversion transformation plays a central role. Theorem 3. The following properties hold for every triangle 퐴퐵퐶 ∶ 1. The centers {퐴′, 퐵′, 퐶′} of the three Apollonian circles lie respectively on the side‑lines {퐵퐶, 퐶퐴, 퐴퐵} of the triangle and {퐴′퐴, 퐵′퐵, 퐶′퐶} are respectively tangent to the circum‑ circle 훿 of 퐴퐵퐶. 2. The Apollonian circles are orthogonal to the circumcircle of 퐴퐵퐶. 3. The center 푂 of the circumcircle 훿 of 퐴퐵퐶 is on the radical axis of the pencil of Apollonian circles and the isodynamic points are inverse with respect to 훿.

4. Each one of the inversions {푓푎, 푓푏, 푓푐} w.r.t. {훼, 훽, 훾} permutes these circles, hence the three circles intersect pairwise at an angle of measure 휋/3. 5. Each center of the three circles is a “similarity center” of the two others. 6. The tangency of 퐴′퐴 to 훿 at 퐴 implies that 퐴′퐴 is the harmonic conjugate of the “sym‑ median” through 퐴 of w.r. to the sides {퐴퐵, 퐴퐶}. 7. The previous property implies that the “Lemoine line”, carrying {퐴′, 퐵′, 퐶′} is the “trilinear polar” of the “symmedian point” 퐾 of the triangle 퐴퐵퐶. 8. The pedal triangles of the isodynamic points are equilateral triangles (See Figure 6). Proof. Nrs 1‑2 follow from the orthogonality of 훿 to the Apollonian circles. Nr‑3 is a general property of intersecting pencils of circles and their orthogonal to it circles. Nr‑4 e.g. for circle 훼. The inversion relative to it maps the circle 훾 passing through {퐼1, 퐼2, 퐶} to circle 훽 passing through 퐼1, 퐼2, 퐵 and vice versa. Nr‑5 e.g. for circle 훼 follows from nr‑4, since the center of a circle whose inversion interchanges two circles is a similarity center of the two circles. Nr‑6 this is a well known property of the symmedians. Nr‑7 follows from the definition of the “trilinear polar” of a point w.r. to a triangle. Nr‑8 follows from theorem 2.

Exercise 4. The only points whose pedal triangles w.r.t △퐴퐵퐶 are equilateral are the isodynamic points {퐼1, 퐼2} of △퐴퐵퐶. Exercise 5. In figure 6 show that the line pairs {(퐴′퐵′, 퐴″퐵″), (퐵′퐶′, 퐴″퐶″), (퐴′퐶′, 퐵″퐶″)} form the same angle.

Exercise 6. The only inversions permuting the vertices of the triangle 퐴퐵퐶 are the ones relative to the circumcircle 휅 of the triangle and the Apollonian circles. 2 Isodynamic points, Brocard and Lemoine axes 5

B''

Ι2 A α β B' C''

C' Ι1

A'' B A' C

Figure 6: The two equilateral pedals of the isodynamic points of 퐴퐵퐶

Hint: Such an inversion maps necessarily the circumcircle 휅 of △퐴퐵퐶 to itself. If it fixes all three vertices, then it fixes 휅 and is the inversion relative to 휅 . If it interchanges two vertices, {퐵, 퐶} say, then the inversion must fix the third vertex 퐴 and coincides with the inversion relative to the Apollonian circle through 퐴.

A κ P P'

O BC

Figure 7: Common Apollonian circle of two triangles {퐴퐵퐶, 푃퐴푃′}

Exercise 7. Show that the Apollonian circle 휅 of △퐴퐵퐶 through 퐴 is simultaneously the Apol‑ lonian circle through 퐴 of the triangle 푃퐴푃′, for every point 푃 and its inverse relative to 휅 (see figure 7). Remark 1. Since the isodynamic points are inverse relative to the circumcircle 휅 of △퐴퐵퐶 one is inside, conventionally denoted by 퐼1 and the other 퐼2 is outside 휅. Next proposi‑ tions give formulate some characteristic properties of these points.

Theorem 4. The inversion relative to the circle 휆(퐼2) which is orthogonal to the circumcircle 휅 of △퐴퐵퐶 and has its center at the outer isodynamic point 퐼2 , maps the vertices of △퐴퐵퐶 to the vertices of an equilateral 퐴′퐵′퐶′ inscribed in 휅 (see figure 8) and the Apollonian circles to the symmetry axes of the equilateral. Proof. Next equalities of ratios, resulting from similar triangles in figure 8, from properties ′ ′ ′ ′ of inversions and from properties of the isodynamic point 퐼2, show that 퐵 퐶 = 퐶 퐴 . Analogous equalities of ratios can be used to show also the equality of the other sides of △퐴′퐵′퐶′. ′ ′ ′ ′ 퐵퐶 퐶 퐵 퐴 퐶 퐴퐶 ′ ′ 퐵퐶 ′ ′ ′ 퐴퐶 ′ = ′ , ′ = ⇒ 퐶 퐵 = 퐼2퐵 , 퐴 퐶 = 퐼2퐶 ⇒ 퐼2퐶 퐼2퐵 퐼2퐶 퐼2퐴 퐼2퐶 퐼2퐴 ′ ′ ′ 퐶 퐵 퐵퐶 퐼2퐴 퐼2퐵 ′ ′ = ⋅ ⋅ ′ 퐴 퐶 퐴퐶 퐼2퐶 퐼2퐶 ′ 퐵퐶 퐵퐴 퐼2퐵 = ⋅ ⋅ ′ 퐴퐶 퐵퐶 퐼2퐶 ′ 퐵퐴 퐼2퐵 퐵퐴 퐼2퐶 퐵퐴 퐴퐶 = ⋅ ′ = ⋅ = ⋅ = 1. 퐶퐴 퐼2퐶 퐶퐴 퐼2퐵 퐶퐴 퐴퐵

The Apollonian circles being orthogonal to line 퐼1퐼2 and passing through 퐼2 map to lines orthogonal to 퐼1퐼2 etc. 3 Given Apollonian circles 6

A' λ κ A O

I 1 B' B I 2 C' C

′ ′ ′ Figure 8: Inversion w.r.t. 휆(퐼2) mapping △퐴퐵퐶 to △퐴 퐵 퐶

Exercise 8. The anti‑inversion relative to the circle 휇(퐼1) with center at the inner isodynamic ′ point 퐼1 and diameter the minimal chord 퐷퐷 of the circumcircle 휅 through 퐼1, maps the vertices of △퐴퐵퐶 to the vertices of an equilateral 퐴′퐵′퐶′ inscribed in 휅 (see figure 9) and the Apollonian circles to the symmetry axes of the equilateral.

λ D' B' μ κ A C' O I1

C I2 B

D A'

′ ′ ′ Figure 9: Antiinversion w.r.t. 휇(퐼1) mapping △퐴퐵퐶 to the equilateral △퐴 퐵 퐶

Hint: Same reasoning as in the preceding theorem.

3 Given Apollonian circles

Next propositions lead to a sort of inverse to theorem 3, giving the description of triangles which have the same Apollonian circles, and consequently the same isodynamic points.

α β I2 γ Ο Ο β O γ α X4

X3 I1

X1

X2 δ

Figure 10: Inversions w.r. to the circles {훼, 훽, 훾} 3 Given Apollonian circles 7

Theorem 5. Given three circles {훼(푂훼), 훽(푂훽), 훾(푂훾)} passing through the points {퐼1, 퐼2} and cutting each other pairwise under an angle of 휋/3, consider the three inversions {푓훼, 푓훽, 푓훾} relative to the corresponding circles (see figure 10). Then 1. Each inversion interchanges the two other circles.

2. The compositions 푓훽 ∘ 푓훼 = 푓훾 ∘ 푓훽 are equal and the successive transforms of a point 푋1, 푋2 = 푓훼(푋1), 푋3 = 푓훽(푋2), 푋4 = 푓훾(푋3) form a cyclic quadrangle whose circumcircle is orthogonal to the three given circles.

Proof. Nr‑1 is equivalent with the intersection condition of the three circles by the angle of 휋/3. Nr‑2 follows by considering the circle 훿 passing through {푋1, 푋2, 푋3}. Since {푋1, 푋2} are inverse w.r. to 훼 the circle 훿 is orthogonal to 훼. Analogous argument shows that it is orthogonal to 훽 and consequently to all circles of the pencil generated by {훼, 훽} hence also to 훾. This implies 푋4 ∈ 훿 and that 푋1 is the inverse of 푋4 w.r. to 훽 thereby proving the claim.

α I β 2 γ

Οβ

Ογ

Oα X4

X1 I1 X2=X3

Figure 11: Constructing a triangle 퐴퐵퐶 with given isodynamic points {퐼1, 퐼2}

Corollary 1. With the definitions and notation of the preceding theorem, when point 푋1 is on the circle 훾, then 푋2 = 푋3 ∈ 훽 and 푋4 ∈ 훼 and the sides of the triangle 푋1푋2푋4 pass through the centers {푂훼, 푂훽, 푂훾} of the three circles (see figure 11). Corollary 2. With the definitions and conventions of the preceding corollary, the Apollonian circles of the triangle 푋1푋2푋4 are the circles {훼, 훽, 훾}. Conversely, if the triangle 푋1푋2푋4 has {훼, 훽, 훾} as Apollonian circles, then it is constructible as in the preceding corollary.

Proof. Consider one of these circles, 훼 say, passing through 푋4 and having its center on 푋1푋2 latter being points inverse w.r. to 훼. Then 훼 intersects the line 푋1푋2 at diametral points {푈, 푉}, which, by a well known property of inversion, are “harmonic conjugate” w.r. to {푋1, 푋2}. From the orthogonality of {푋4푈, 푋4푉} follows that these are the bisectors of the angle 푋1̂푋4푋2, hence the validity of the claim for the circle 훼. Analogous is the proof for the other two circles. The converse is trivial.

Figure 12 displays triangles sharing the same Apollonian circles and resulting by the procedure of corollary 3, to be proved in the next section. According to this, we take equilaterals homothetic to 퐴1퐵1퐶1 w.r.t. point 퐼2 and applying to these (to their vertices) the inversion w.r.t. circle 휀. The Apollonian circles are the inverses {훼, 훽, 훾} relative to 휀 of the lines {훼′, 훽′, 훾′}. Next section discusses the details of this procedure. 4 Given isodynamic points 8

α γ' β'

A

C1 β α' C

I2 A1

B B 1 I 1 γ ε

Figure 12: Triangles sharing the same Apollonian circles

4 Given isodynamic points

If we are asked to construct triangles with the given isodynamic points {퐼1, 퐼2} but pos‑ sibly different triples of Apollonian circles, we first must notice that these triples ofcir‑ cles are selected among the members of the intersecting pencil 풫 of all circles passing through {퐼1, 퐼2}. From the intersection angle condition of 휋/3 for these triples it fol‑ lows that a single circle of the triple determines completely the other two. In the fol‑ lowing 휀(푋, 푌) will denote a circle with center 푋, passing through 푌. Applying the inversion 푓휀 relative to 휀(퐼1, 퐼2), the pencil 풫 transforms to the pencil of lines through 퐼2 and the admissible triples of Apollonian circles {훼, 훽, 훾} transform to triples of lines

γ' γ

ε

Ι2 Ι1

α' β

α β'

Figure 13: Admissible triples of Apollonian circles and their inverses relative to 휀

′ ′ ′ {훼 = 푓휀(훼), 훽 = 푓휀(훽), 훾 = 푓휀(훾)} through 퐼2 intersecting at angles of measure 휋/3 (see figure 13). Next propositions show that the triangles having the triple {훼, 훽, 훾} for Apollonian circles correspond under the inversion 푓휀 to equilateral triangles with sym‑ metry axes the lines 훼′, 훽′, 훾′ . To see this we use the well known property of inversions 4 Given isodynamic points 9

α' Y κ α

Y' P' X X' O P

Figure 14: Inverseness is invariant under inversion

([Joh60, p.55], [Ped90, p.96]) according to which “The property of inverseness is invariant un‑ der inversion”. This property guarantees that, if points {푋, 푌} are inverse w.r. to a circle 훼 and the whole figure is inverted w.r. to a circle 휅, then the 휅−inverses {푋′, 푌′} are inverse w.r. to the 휅−inverse circle 훼′ (see figure 13). This has an interesting corollary which I formulate as a lemma. The following arguments use elementary properties of “inversions” and pencils of circles (coaxal circles), an account of which can be found in Johnson [Joh60, p.28], Pedoe [Ped90, p.106] and also in the file Inversion of this gallery.

Y' α' X' γ' γ ε χ' χ

X I1 Y

Ο α α

I2 Figure 15: Inversion w.r. to a circle is conjugate to reflection w.r. to a line

Lemma 1. For an inversion 푓훼 w.r. to a circle 훼 there is an inversion 푓휀 w.r. to a circle 휀 such ′ that the conjugate transformation 푓휀 ∘ 푓훼 ∘ 푓휀 is a reflection 푟훼′ on the line 훼 = 푓휀(훼).

Proof. For the proof select two points {퐼1, 퐼2} on the circle 훼(푂훼) and consider the non‑ intersecting pencil 풫 of (coaxal) circles {훾} with limits these two points, all of them be‑ ing then orthogonal to 훼 (see figure 15). Consider also the inversion 푓휀 w.r. to the circle ′ 휀(퐼1, 퐼2) and define the line 훼 = 푓휀(훼) passing through 퐼2. Since inversions leave invari‑ ant the circles orthogonal to the circle of inversion, a point 푋 of the plane 푋 ≠ 푂훼 de‑ ′ fines a unique member 훾 ∋ 푋 of the pencil 풫 and 푋 = 푓훼(푋) ∈ 훾 too. Then, the circle ′ ′ ′ ′ 휒 = (푋푋 퐼1) maps under 푓휀 to a line 휒 = 푌푌 which is orthogonal to 훼 , since 휒 pass‑ ′ ′ ing through two 훼−inverse points is orthogonal to 훼. The points {푌 = 푓휀(푋), 푌 = 푓휀(푋 )} then, are reflections of each other on 훼′, as claimed.

Theorem 6. Let {퐼1, 퐼2} be the isodynamic points of the triangle 퐴퐵퐶 and 푓휀 the inversion relative to the circle 휀(퐼1, 퐼2). The images {퐴1 = 푓휀(퐴), 퐵1 = 푓휀(퐵), 퐶1 = 푓휀(퐶)} of the vertices of 퐴퐵퐶 ′ ′ ′ form an 퐴1퐵1퐶1 and the images {훼 = 푓휀(훼), 훽 = 푓휀(훽), 훾 = 푓휀(훾)} of the 4 Given isodynamic points 10

Apollonian circles {훼, 훽, 훾} of the triangle 퐴퐵퐶 are the symmetry axes of 퐴1퐵1퐶1 meeting in the center 퐼2 of the triangle (see figure 16).

C2 β

B2

β' I α' 1 A C A1 ε α 1

γ B I2 γ' C A 2 B1

Figure 16: Inversion of vertices of △퐴퐵퐶 to vertices of equilateral 퐴1퐵1퐶1

Proof. The proof results from the general properties of inversions and the previous dis‑ cussion (see figure 16). The Apollonian circles {훼, 훽, 훾} transform respectively to lines ′ ′ ′ {훼 , 훽 , 훾 } through 퐼2, intersecting pairwise at 휋/3. Since {퐵, 퐶} are inverse w.r. to 훼, ′ their images {퐵1 = 푓휀(퐵), 퐶1 = 푓휀(퐶)}, by lemma 1, are inverse w.r. to 훼 = 푓휀(훼), hence ′ ′ are points reflected in 훼 . It follows that 퐵1퐶1 is orthogonal to 훼 and analogously the ′ ′ ′ other sides of 퐴1퐵1퐶1 have the lines {훼 , 훽 , 훾 } as perpendicular bisectors.

β'

Β2 γ' Α 2 C C1 A1 I A 1 I2

ε B α'

B1

C2

Figure 17: Generation of all triangles with the same isodynamic points

The theorem implies immediately next corollary giving a way to generate all possible triangles with given isodynamic points {퐼1, 퐼2} (see figure 17).

Corollary 3. All triangles 퐴퐵퐶, with given isodynamic points {퐼1, 퐼2}, result by applying the inversion 푓휀 relative to the circle 휀(퐼1, 퐼2) to the vertices {퐴1, 퐵1, 퐶1} of equilateral triangles 퐴1퐵1퐶1 centered in 퐼2 ∶ {퐴 = 푓휀(퐴1), 퐵 = 푓휀(퐵1), 퐶 = 푓휀(퐶1)}.

Corollary 4. Given three points in general position {퐼1, 퐼2} and 퐴, there is precisely one triangle 퐴퐵퐶 having 퐴 as vertex and {퐼1, 퐼2} as isodynamic points (see figure 18). 5 Trilinear coordinates of the isodynamic points 11

B B 1

κ' κ

I C I 1 2 C Α 1 1

ε A

Figure 18: Triangle from the isodynamics {퐼1, 퐼2} and vertex 퐴

Proof. To see this notice first that the circumcircle 휅 of the requested triangle is uniquely defined. This, because it is the unique member of the pencil 풫 of circles of non‑intersecting type with limit points the given {퐼1, 퐼2}, passing through 퐴. By theorem 6 the inverse ′ 휅 = 푓휀(휅) relative to the circle 휀(퐼1, 퐼2) carries 퐴1 = 푓휀(퐴), which is a vertex of a unique ′ ′ equilateral 퐴1퐵1퐶1 inscribed in 휅 , whose centroid 퐼2 coincides with the center of 휅 . The requested triangle 퐴퐵퐶 has vertices the inverses via 푓휀 of the vertices of this equilat‑ eral. Remark 2. The pencil 풫 consisting of the circles orthogonal to the Apollonian circles is called “Schoute pencil” of the triangle 퐴퐵퐶. As noticed above, the isodynamic points are its “limit points”1

Exercise 9. Show that the equilateral of theorem 6 and that of theorem 4 are homothetic. Show that, more general, any inversion w.r.t. a circle centered at 퐼2 maps the vertices of △퐴퐵퐶 to the vertices of an equilateral and all these equilaterals are homothetic.

5 Trilinear coordinates of the isodynamic points

The appearance of the equilateral here offers the means to calculate easily these kindof coordinates for {퐼1, 퐼2}. The result is 휋 휋 휋 (푥 ∶ 푥 ∶ 푥 ) = (sin (퐴̂ + ), sin (퐵)̂ + ), sin (퐶̂ + )) 1 2 3 3 3 3 휋 휋 휋 (푥′ ∶ 푥′ ∶ 푥′ ) = (sin (퐴̂ − ), sin (퐵̂ − ), sin (퐶̂ − )) . 1 2 3 3 3 3

The formulas result from a calculation of the ratios (푥1 ∶ 푥2 ∶ 푥3) of the distances of 퐼1 from the sides of 퐴퐵퐶. For this we notice 휋 퐵퐼̂퐶 = 퐴̂ + since 1 3 퐵퐼̂1퐶 = 2휋 − 퐵퐼̂1퐴 − 퐴퐼̂1퐶

= 2휋 − (휋 − 퐵퐴퐼̂1 − 퐴퐵퐼̂1) − (휋 − 퐶퐴퐼̂1 − 퐼1̂퐶퐴) 휋 = 퐴̂ + (퐷퐸퐼̂ + 퐹퐸퐼̂ ) = 퐴̂ + . 1 1 3 1Thanks to Dan Reznik for the suggestion to refer here to the Schoute pencil and its limit points. 6 Same isodynamic points and same circumcircle 12

A

t 1 F D x x 2 3 I 1

t x t 2 1 3

B C E

Figure 19: Trilienear coordinates 푥1 ∶ 푥2 ∶ 푥3 of 퐼1

Last equality resulting from the cyclic quadrangles 퐷퐵퐸퐼1 and 퐹퐼1퐸퐶. Calculating the double of the area of triangles {퐼1퐵퐶, 퐼1퐶퐴} and dividing gives:

2(퐼1퐵퐶) = 푥1 ⋅ 퐵퐶 = 푡2 ⋅ 푡3 ⋅ sin(퐴̂ + 휋/3),

2(퐼1퐶퐴) = 푥2 ⋅ 퐶퐴 = 푡3 ⋅ 푡1 ⋅ sin(퐵̂ + 휋/3) ⇒ 푥 ⋅ 퐵퐶 푡 ⋅ sin(퐴̂ + 휋/3) 1 = 2 ⇒ 푥2 ⋅ 퐶퐴 푡1 ⋅ sin(퐵̂ + 휋/3) 푥 sin(퐴̂ + 휋/3) 1 = , 푥2 sin(퐵̂ + 휋/3) where the last simplification is due to the property of the isodynamic point 푡2/푡3 = 퐴퐵/퐴퐶, etc. The formulas for 퐼2 result by analogous reasoning.

6 Same isodynamic points and same circumcircle

Figure 20 shows some members of the family 풯 of triangles {퐴퐵퐶} sharing the same iso‑ dynamic points {퐼1, 퐼2} and the same circumcircle 휅 but not the same Apollonian circles.

A κ' 1 A κ Ι 1 B Ι 1 2 C ε B

C 1

Figure 20: The same isodynamic points {퐼1, 퐼2} and same circumcircle 6 Same isodynamic points and same circumcircle 13

It is created by rotating the equilateral 퐴1퐵1퐶1 in its circumcircle about 퐼2 and taking from the resulting triangles {퐴′퐵′퐶′} the inverses (of their vertices) relative to the circle 휀 (corollary 4). On the occasion of this figure I would like to mention some facts pertaining tothe “Geometry of the triangle” and whose proofs can be found in [Pam04]. 1. As noticed in the figure the triangles of the family 풯 have their sides enveloping an ellipse. This is the “Brocard ellipse” of the triangle 퐴퐵퐶, inscribed in 퐴퐵퐶 and all other members of the family 풯 . 2. The reason of this enveloping property is simple to explain, provided one has some knowledge of elementary projective geometry. There is namely a “projectivity” 푓 푓 푓 푓 ′ mapping the circle 휅 onto 휅 and also mapping {퐴1 ↦ 퐴, 퐵1 ↦ 퐵, 퐶1 ↦ 퐶}. This 푓 maps all the equilaterals inscribed in 휅′ to the triangles of the family 풯 . The common inscribed circle of the equilaterals maps then to this “Brocard ellipse” also inscribed to all members of 풯 .

f(X)

S

X B1 B

κ' κ

I2 λ I1 C

C1 Α1

ε A η

′ Figure 21: Projectivity 푓 coinciding with the inversion 푓휀 on the points of 휅

3. The map 푓 can be simply described as the “homology” with center 퐼1 , fixed axis coinciding with the radical axis 휂 of the circles {휅, 휅′} and a constant “cross ratio” 푘 = (푋푓 (푋)푆퐼1), where 푆 = 휂 ∩ 푋푓 (푋), proved also to coincide with the inversion ′ 푓휀 on the points of the circle 휅 but not on the points of 휅. Figure 21 shows the property of 푓 to have the line 푋푓 (푋) pass through 퐼1 and shows also the “Brocard” ellipse as image 푓 (휆) of the incircle of the equilateral.

4. The equilaterals are “orbital” triples of points resulting by repeated application of the ∘ rotation 휌 by 120 about 퐼2 ∶ {퐴1, 퐵1 = 휌(퐴1), 퐶1 = 휌(퐵1), 퐴1 = 휌(퐶1)}. This behavior is transferred by conjugation by 푓 to all the triangles of the family 풯 . Latter are then orbital w.r. to the projectivity 휎 = 푓 ∘ 휌 ∘ 푓 −1, mapping 휅 onto itself and recycling the vertices of the triangles: {퐴, 퐵 = 휎(퐴), 퐶 = 휎(퐵), 퐴 = 휎(퐶)}.

5. All these triangles 퐴퐵퐶 ∈ 풯 have the same “symmedian point” 퐾 = 푓 (퐼2), which is a fixed point of the periodic projectivity 휎. The “homography axis” of 휎 coincides with the “Lemoine axis” of the triangle, which is also the same for all triangles 퐴퐵퐶 ∈ 풯 . Figure 22 shows two triangles {퐴퐵퐶, 퐴′퐵′퐶′} of the family 풯 , their common sym‑ median point 퐾 and also illustrates the characteristic property of the “homography axis” of 휎 which guarantees that, for two arbitrary points {퐴, 퐵} of 휅 and their im‑ ages {퐴′ = 휎(퐴), 퐵′ = 휎(퐵)}, the intersections 푃 = 퐴퐵′ ∩ 퐴′퐵 lie on the homogra‑ phy axis of 휎. 7 A matter of uniqueness 14

ε

B κ A' Lemoine

C' I C K I 1 P 2 B'

A

Figure 22: Intersections {푃 = 퐴퐵′ ∩ 퐴′퐵} are on the homography axis

6. By the way, and because was mentioned in nr‑3, every pair of circles {휅′, 휅} defines ′ a circle 휀 interchanging them through the corresponding inversion 푓휀 ∶ 푓휀(휅 ) = 휅, ′ satisfying, of course also 푓휀(휅) = 휅 . The two circles define also a projectivity 푓 leav‑ ′ ing fixed the center 퐼1 of 휀 and the radical axis 휂 of the circles {휅 , 휅} and coinciding with 푓 on the points of 휅′. Figure 23 shows the image 푓 (휅) of 휅 under 푓 , which is a conic and never coincides with 휅′ , except in the case the two circles are equal 2 and 푓휀 is a reflection. Thus, while 푓휀 is always an involution satisfying 푓휀 = 푒, the projectivity 푓 for non equal circles is never an involution 푓 2 ≠ 푒. Maps such as 푓 , having an axis 휂 of fixed points, an isolated fixed point 퐼1 and having images

ε

I1 XYS κ' f(κ) κ=f(κ') η

′ Figure 23: Projectivity 푓 coinciding with the inversion 푓휀 on the points of 휅

푌 = 푓 (푋) such that the lines 푋푌 ∋ 퐼1 and the cross ratio (푋푌푆퐼1) = 푘 is constant, where 푆 = 푋푌 ∩ 휂, are called “homologies” and play a fundamental role in projec‑ tive geometry ([VY10, I,p.72]).

7 A matter of uniqueness

A matter of uniqueness, concerning the reduction to equilaterals, arises from thepossi‑ bility to interchange the roles of {퐼1, 퐼2}. In fact, the same procedure can be applied using ′ ″ ″ ″ the circle 휀 (퐼2, 퐼1) instead of 휀(퐼1, 퐼2) and defining the equilateral 퐴 퐵 퐶 inscribed in 휅″, which this time is the inverse of 휅 relative to 휀′ (see figure 24). It is though easy to see that the two equilaterals produce the same triangle 퐴퐵퐶 ∶

′ ″ ′ ″ ′ ″ 퐴 = 푓휀(퐴 ) = 푓휀′ (퐴 ), 퐵 = 푓휀(퐵 ) = 푓휀′ (퐵 ), 퐶 = 푓휀(퐶 ) = 푓휀′ (퐶 ).

This is proved by the following lemma, which uses the definitions and conventions adopted so far. 7 A matter of uniqueness 15

′ ′ ″ ″ ′ Lemma 2. The points {퐴 ∈ 휅 , 퐴 ∈ 휅 } are respectively the inverses {퐴 = 푓휀(퐴), 퐴 = 푓휀′ (퐴)} ″ ′ of the same point 퐴 ∈ 휅, if and only if, the lines {퐼1퐴 , 퐼2퐴 } intersect at a point 퐷 on the perpendicular bisector of the segment 퐼1퐼2.

A''

κ'' κ D A Α' κ' λ ε' B'' I B I B' 1 2

ε C' C

C''

′ ″ Figure 24: 퐴 = 푓휀(퐴 ) = 푓휀′ (퐴 ) ...

If we trust this lemma, the claim follows at once, since starting from the isosceles 퐼1퐷퐼2 ″ ′ resulting from the lines {퐼1퐴 , 퐼2퐴 } we obtain the other vertices of the equilaterals by rotating these lines in opposite orientation and by the same angle of 120∘, respectively 240∘, implying that they intersect again in a point of the perpendicular bisector.

μ A

A'' A' ε A''' ε' D

I 1 I2 η

′ ″ Figure 25: 푓휀(퐴 ) = 푓휀′ (퐴 ) = 퐴

The lemma in turn, is a corollary of the more general following one (see figure 25). ′ Lemma 3. Given the equal circles {휀, 휀 } centered respectively in {퐼1, 퐼2} with radius |퐼1퐼2|, the ′ ″ inversions {푓휀, 푓휀′ } relative to these circles, map the points {퐴 , 퐴 } respectively to the same point ″ ′ 퐴, if and only if, the lines {퐼1퐴 , 퐼2퐴 } intersect in a point 퐷 on the perpendicular bisector of the segment 퐼1퐼2. ″ Proof. The lemma follows by applying lemma 1. By it inverting {퐴, 퐴 = 푓휀′ (퐴)} relative to 휀 we obtain two points {퐴′, 퐴‴}, which are inverse w.r. to the inverse of 휀′ relative ′ ′ ‴ to 휀. But this inverse 푓휀(휀 ) is the perpendicular bisector of 퐼1퐼2 and {퐴 , 퐴 } are then reflected w.r. to this line. The argument can be reversed to show also the converse. BIBLIOGRAPHY 16

Bibliography

[Gal13] William Gallatly. The modern geometry of the triangle. Francis Hodgsonn, London, 1913.

[Joh60] Roger Johnson. Advanced . Dover Publications, New York, 1960.

[Pam04] P. Pamfilos. On Some Actions of 퐷3 on a Triangle. Forum Geometricorum, 4:157– 186, 2004.

[Ped90] D Pedoe. A course of Geometry. Dover, New York, 1990.

[VY10] Oswald Veblen and John Young. Projective Geometry vol. I, II. Ginn and Company, New York, 1910.

[Yiu13] Paul Yiu. Introduction to the Geometry of the Triangle. http://math.fau.edu/Yiu/ Geometry.html, 2013.

Related material 1. Barycentric coordinates 2. Cross Ratio 3. Inversion transformation 4. Pedal triangles 5. Projectivities

Any correction, suggestion or proposal from the reader, to improve/extend the exposition, is welcome and could be send by e‑mail to: [email protected]