Complements to Classic Topics of Circles Geometry”

Ion Patrascu, Florentin Smarandache

Radical Axis of Lemoine’s Circles

In Ion Patrascu, Florentin Smarandache: “Complements to Classic Topics of Circles Geometry”. Brussels (Belgium): Pons Editions, 2016 Complements to Classic Topics of Circles Geometry

In this article, we emphasize the radical axis of the Lemoine’s circles. For the start, let us remind:

1st Theorem.

The parallels taken through the simmedian center 퐾 of a triangle to the sides of the triangle determine on them six concyclic points (the first Lemoine’s circle).

2nd Theorem.

The antiparallels taken through the triangle’s simmedian center to the sides of a triangle determine six concyclic points (the second Lemoine’s circle).

1st Remark.

If 퐴퐵퐶 is a scalene triangle and 퐾 is its simmedian center, then 퐿 , the center of the first Lemoine’s circle, is the middle of the segment [푂퐾], where 푂 is the center of the circumscribed circle, and

31 Ion Patrascu, Florentin Smarandache the center of the second Lemoine’s circle is 퐾 . It follows that the radical axis of Lemoine’s circles is perpendicular on the line of the centers 퐿퐾, therefore on the line 푂퐾.

1st Proposition.

The radical axis of Lemoine’s circles is perpendicular on the line 푂퐾 raised in the simmedian center 퐾.

Proof.

Let 퐴1퐴2 be the antiparallel to 퐵퐶 taken through

퐾, then 퐾퐴1 is the radius 푅퐿2 of the second Lemoine’s circle; we have: 푎푏푐 푅 = . 퐿2 푎2+푏2+푐2

Figure 1

Complements to Classic Topics of Circles Geometry

′ ′ Let 퐴1퐴2 be the Lemoine’s parallel taken to 퐵퐶; we evaluate the power of 퐾 towards the first Lemoine’s circle. We have: ⃗⃗⃗⃗⃗⃗⃗⃗′ ⃗⃗⃗⃗⃗⃗⃗⃗′ 2 2 퐾퐴1 ∙ 퐾퐴2 = 퐿퐾 − 푅퐿1. (1) Let 푆 be the simmedian leg from 퐴; it follows that: 퐾퐴′ 퐴퐾 퐾퐴′ 1 = − 2 . 퐵푆 퐴푆 푆퐶 We obtain: 퐴퐾 퐴퐾 퐾퐴′ = 퐵푆 ∙ and 퐾퐴′ = 푆퐶 ∙ , 1 퐴푆 2 퐴푆 퐵푆 푐2 퐴퐾 푏2+푐2 but = and = . 푆퐶 푏2 퐴푆 푎2+푏2+푐2 Therefore: 퐴퐾 2 −푎2푏2푐2 퐾퐴⃗⃗⃗⃗⃗⃗⃗⃗′ ∙ 퐾퐴⃗⃗⃗⃗⃗⃗⃗⃗′ = −퐵푆 ∙ 푆퐶 ∙ ( ) = ; 1 2 퐴푆 (푏2 + 푐2)2 2 (푏2+푐2) = −푅2 . (2) (푎2+푏2+푐2)2 퐿2 We draw the perpendicular in 퐾 on the line 퐿퐾 and denote by 푃 and 푄 its intersection to the first Lemoine’s circle. ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 2 We have 퐾푃 ∙ 퐾푄 = −푅퐿2 ; by the other hand, 퐾푃 = 퐾푄 (푃푄 is a chord which is perpendicular to the diameter passing through 퐾).

It follows that 퐾푃 = 퐾푄 = 푅퐿2 , so 푃 and 푄 are situated on the second Lemoine’s circle. Because 푃푄 is a chord which is common to the Lemoine’s circles, it obviously follows that 푃푄 is the radical axis.

Ion Patrascu, Florentin Smarandache

Comment.

Equalizing (1) and (2), or by the Pythagorean theorem in the triangle 푃퐾퐿, we can calculate 푅퐿1. 3푎2푏2푐2 It is known that: 푂퐾2 = 푅2 − , and (푎2+푏2+푐2)2 1 since 퐿퐾 = 푂퐾, we find that: 2 1 푎2푏2푐2 푅2 = ∙ [푅2 + ]. 퐿1 4 (푎2+푏2+푐2)2

2nd Remark.

The 1st Proposition, ref. the radical axis of the Lemoine’s circles, is a particular case of the following Proposition, which we leave to the reader to prove.

2nd Proposition.

If 풞(푂1, 푅1) şi 풞(푂2, 푅2) are two circles such as 2 the power of center 푂1 towards 풞(푂2, 푅2) is −푅1 , then the radical axis of the circles is the perpendicular in

푂1 on the line of centers 푂1푂2.

References.

[1] F. Smarandache, Ion Patrascu: The Geometry of Homological Triangles, Education Publisher, Ohio, USA, 2012. [2] Ion Patrascu, F. Smarandache: Variance on Topics of Plane Geometry, Education Publisher, Ohio, USA, 2013.

Complements to Classic Topics of Circles Geometry

Generating Lemoine’s circles

In this paper, we generalize the theorem relative to the first Lemoine’s circle and thereby highlight a method to build Lemoine’s circles. Firstly, we review some notions and results.

1st Definition.

It is called a simedian of a triangle the symmetric of a median of the triangle with respect to the internal bisector of the triangle that has in common with the median the peak of the triangle.

1st Proposition.

In the triangle 퐴퐵퐶, the cevian 퐴푆, 푆 ∈ (퐵퐶), is a 푆퐵 퐴퐵 2 simedian if and only if = ( ) . For Proof, see [2]. 푆퐶 퐴퐶

2nd Definition.

It is called a simedian center of a triangle (or Lemoine’s point) the intersection of triangle’s simedians.

Ion Patrascu, Florentin Smarandache

1st Theorem.

The parallels to the sides of a triangle taken through the simedian center intersect the triangle’s sides in six concyclic points (the first Lemoine’s circle - 1873). A Proof of this theorem can be found in [2].

3rd Definition.

We assert that in a scalene triangle 퐴퐵퐶 the line 푀푁, where 푀 ∈ 퐴퐵 and 푁 ∈ 퐴퐶, is an antiparallel to 퐵퐶 if ∢푀푁퐴 ≡ ∢퐴퐵퐶.

1st Lemma.

In the triangle 퐴퐵퐶 , let 퐴푆 be a simedian, 푆 ∈ (퐵퐶). If 푃 is the middle of the segment (푀푁), having 푀 ∈ (퐴퐵) and 푁 ∈ (퐴퐶), belonging to the simedian 퐴푆, then 푀푁 and 퐵퐶 are antiparallels.

Proof.

We draw through 푀 and 푁, 푀푇 ∥ 퐴퐶 and 푁푅 ∥ 퐴퐵, 푅, 푇 ∈ (퐵퐶) , see Figure 1. Let {푄} = 푀푇 ∩ 푁푅 ; since 푀푃 = 푃푁 and 퐴푀푄푁 is a parallelogram, it follows that 푄 ∈ 퐴푆.

Complements to Classic Topics of Circles Geometry

Figure 1.

Thales's Theorem provides the relations: 퐴푁 퐵푅 = ; (1) 퐴퐶 퐵퐶 퐴퐵 퐵퐶 = . (2) 퐴푀 퐶푇 From (1) and (2), by multiplication, we obtain: 퐴푁 퐴퐵 퐵푅 ∙ = . (3) 퐴푀 퐴퐶 푇퐶 Using again Thales's Theorem, we obtain: 퐵푅 퐴푄 = , (4) 퐵푆 퐴푆 푇퐶 퐴푄 = . (5) 푆퐶 퐴푆 From these relations, we get 퐵푅 푇퐶 = , (6) 퐵푆 푆퐶 or 퐵푆 퐵푅 = . (7) 푆퐶 푇퐶 In view of Proposition 1, the relations (7) and (3) 퐴푁 퐴퐵 lead to = , which shows that ∆퐴푀푁~∆퐴퐶퐵 , so 퐴퐵 퐴퐶 ∢퐴푀푁 ≡ ∢퐴퐵퐶.

Ion Patrascu, Florentin Smarandache

Therefore, 푀푁 and 퐵퐶 are antiparallels in relation to 퐴퐵 and 퐴퐶.

Remark.

1. The reciprocal of Lemma 1 is also valid, meaning that if 푃 is the middle of the antiparallel 푀푁 to 퐵퐶, then 푃 belongs to the simedian from 퐴.

2nd Theorem. (Generalization of the 1st Theorem) Let 퐴퐵퐶 be a scalene triangle and 퐾 its simedian center. We take 푀 ∈ 퐴퐾 and draw 푀푁 ∥ 퐴퐵, 푀푃 ∥ 퐴퐶, where 푁 ∈ 퐵퐾, 푃 ∈ 퐶퐾. Then: i. 푁푃 ∥ 퐵퐶; ii. 푀푁, 푁푃 and 푀푃 intersect the sides of triangle 퐴퐵퐶 in six concyclic points.

Proof.

In triangle 퐴퐵퐶, let 퐴퐴1, 퐵퐵1, 퐶퐶1 the simedians concurrent in 퐾 (see Figure 2). We have from Thales' Theorem that: 퐴푀 퐵푁 = ; (1) 푀퐾 푁퐾 퐴푀 퐶푃 = . (2) 푀퐾 푃퐾 From relations (1) and (2), it follows that 퐵푁 퐶푃 = , (3) 푁퐾 푃퐾 which shows that 푁푃 ∥ 퐵퐶.

Complements to Classic Topics of Circles Geometry

Let 푅, 푆, 푉, 푊, 푈, 푇 be the intersection points of the parallels 푀푁, 푀푃, 푁푃 of the sides of the triangles to the other sides. Obviously, by construction, the quadrilaterals 퐴푆푀푊; 퐶푈푃푉; 퐵푅푁푇 are parallelograms. The middle of the diagonal 푊푆 falls on 퐴푀, so on the simedian 퐴퐾, and from 1st Lemma we get that 푊푆 is an antiparallel to 퐵퐶. Since 푇푈 ∥ 퐵퐶 , it follows that 푊푆 and 푇푈 are antiparallels, therefore the points 푊, 푆, 푈, 푇 are concyclic (4).

Figure 2.

Analogously, we show that the points 푈, 푉, 푅, 푆 are concyclic (5). From 푊푆 and 퐵퐶 antiparallels, 푈푉 and 퐴퐵 antiparallels, we have that ∢푊푆퐴 ≡ ∢퐴퐵퐶 and ∢푉푈퐶 ≡ ∢퐴퐵퐶 , therefore: ∢푊푆퐴 ≡ ∢푉푈퐶 , and since 39

Ion Patrascu, Florentin Smarandache

푉푊 ∥ 퐴퐶, it follows that the trapeze 푊푆푈푉 is isosceles, therefore the points 푊, 푆, 푈, 푉 are concyclic (6). The relations (4), (5), (6) drive to the concyclicality of the points 푅, 푈, 푉, 푆, 푊, 푇, and the theorem is proved.

Further Remarks.

2. For any point 푀 found on the simedian 퐴퐴1, by performing the constructions from hypothesis, we get a circumscribed circle of the 6 points of intersection of the parallels taken to the sides of triangle. 3. The 2nd Theorem generalizes the 1st Theorem because we get the second in the case the parallels are taken to the sides through the simedian center 푘. 4. We get a circle built as in 2nd Theorem from the first Lemoine’s circle by homothety of pole 푘 and of ratio 휆 ∈ ℝ. 5. The centers of Lemoine’s circles built as above belong to the line 푂퐾, where 푂 is the center of the circle circumscribed to the triangle 퐴퐵퐶.

References.

[1] Exercices de Géométrie, par F.G.M., Huitième édition, Paris VIe, Librairie Générale, 77, Rue Le Vaugirard. [2] Ion Patrascu, Florentin Smarandache: Variance on topics of Plane Geometry, Educational Publishing, Columbus, Ohio, 2013. 40