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Isodynamic Points of the Triangle Isodynamic points of the triangle A file of the Geometrikon gallery by Paris Pamfilos It is in the nature of an hypothesis, when once a man has conceived it, that it assimilates every thing to itself, as proper nourishment; and, from the first moment of your begetting it, it generally grows the stronger by every thing you see, hear, read, or understand. L. Sterne, Tristram Shandy, II, 19 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 Trilinear coordinates 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 triangles. 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).
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