Generating Lemoine Circles Professor Ion Patrascu, Fratii Buzesti National College, Craiova, Romania Professor Florentin Smarandache, New Mexico University, USA
In this paper, we generalize the theorem relative to the first circle of Lemoine and thereby highlight a method to build Lemoine circles. Firstly, we review some notions and results.
Definition 1. 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.
Proposition 1. In the triangle đ??´đ??ľđ??ś , the cevian đ??´đ?‘† , đ?‘† ∈ (đ??ľđ??ś), is a simedian if and only if
đ?‘†đ??ľ đ?‘†đ??ś
đ??´đ??ľ 2
=( ) . đ??´đ??ś
For Proof, see [2].
Definition 2. It is called a simedian center of a triangle (or Lemoine point) the intersection of triangle’s simedians.
Theorem 1. 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 circle - 1873). A Proof of this theorem can be found in [2].
Definition 3. We assert that in a scalene triangle đ??´đ??ľđ??ś the line đ?‘€đ?‘ , where đ?‘€ ∈ đ??´đ??ľ and đ?‘ ∈ đ??´đ??ś, is an anti-parallel to đ??ľđ??ś if âˆ˘đ?‘€đ?‘ đ??´ ≥ âˆ˘đ??´đ??ľđ??ś.
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Lemma 1. In the triangle đ??´đ??ľđ??ś, let đ??´đ?‘† be a simedian, đ?‘† ∈ (đ??ľđ??ś). If đ?‘ƒ is the middle of the segment (đ?‘€đ?‘ ), having đ?‘€ ∈ (đ??´đ??ľ) and đ?‘ ∈ (đ??´đ??ś), belonging to the simedian đ??´đ?‘†, then đ?‘€đ?‘ and đ??ľđ??ś are anti-parallels.
Proof. We draw through đ?‘€ and đ?‘ , đ?‘€đ?‘‡ âˆĽ đ??´đ??ś and đ?‘ đ?‘… âˆĽ đ??´đ??ľ, đ?‘…, đ?‘‡ ∈ (đ??ľđ??ś), see Figure 1. Let {đ?‘„} = đ?‘€đ?‘‡ ∊ đ?‘ đ?‘…; since đ?‘€đ?‘ƒ = đ?‘ƒđ?‘ and đ??´đ?‘€đ?‘„đ?‘ is a parallelogram, it follows that đ?‘„ ∈ đ??´đ?‘†.
Figure 1. Thales' Theorem provides the relations: đ??´đ?‘ đ??ľđ?‘… đ??´đ??ľ đ??ľđ??ś (1); (2). = = đ??´đ??ś đ??ľđ??ś đ??´đ?‘€ đ??śđ?‘‡ From (1) and (2), by multiplication, we obtain: đ??´đ?‘ đ??´đ??ľ đ??ľđ?‘… (3). ∙ = đ??´đ?‘€ đ??´đ??ś đ?‘‡đ??ś Using again Thales' Theorem, we obtain: đ??ľđ?‘… đ??ľđ?‘†
=
đ??´đ?‘„ đ??´đ?‘†
(4),
2
đ?‘‡đ??ś đ?‘†đ??ś
=
đ??´đ?‘„ đ??´đ?‘†
(5).
From these relations, we get đ??ľđ?‘… đ??ľđ?‘†
=
đ?‘‡đ??ś đ?‘†đ??ś
(6) or
đ??ľđ?‘† đ?‘†đ??ś
=
đ??ľđ?‘… đ?‘‡đ??ś
(7).
In view of Proposition 1, the relations (7) and (3) drive to
đ??´đ?‘ đ??´đ??ľ
=
đ??´đ??ľ đ??´đ??ś
, which
shows that ∆đ??´đ?‘€đ?‘ ~∆đ??´đ??śđ??ľ, so âˆ˘đ??´đ?‘€đ?‘ ≥ âˆ˘đ??´đ??ľđ??ś, therefore đ?‘€đ?‘ and đ??ľđ??ś are antiparallels in relation to đ??´đ??ľ and đ??´đ??ś.
Remark. 1. The reciprocal of Lemma 1 is also valid, meaning that if đ?‘ƒ is the middle of the anti-parallel đ?‘€đ?‘ to đ??ľđ??ś, then đ?‘ƒ belongs to the simedian from đ??´.
Theorem 2. (Generalization of Theorem 1) Let đ??´đ??ľđ??ś be a scalene triangle and đ??ž its simedian center. We take đ?‘€ ∈ đ??´đ??ž and draw đ?‘€đ?‘ âˆĽ đ??´đ??ľ, đ?‘€đ?‘ƒ âˆĽ đ??´đ??ś, where đ?‘ ∈ đ??ľđ??ž, đ?‘ƒ ∈ đ??śđ??ž. Then: i.
đ?‘ đ?‘ƒ âˆĽ đ??ľđ??ś;
ii.
đ?‘€đ?‘ , đ?‘ đ?‘ƒ and đ?‘€đ?‘ƒ intersect the sides of triangle đ??´đ??ľđ??ś in six concyclic points.
Proof. In triangle đ??´đ??ľđ??ś, let đ??´đ??´1 , đ??ľđ??ľ1 , đ??śđ??ś1 the simedians concurrent in đ??ž (see Figure 2). We have from Thales' Theorem that: From relations (1) and (2), it follows that
đ??ľđ?‘ đ?‘ đ??ž
=
đ??śđ?‘ƒ đ?‘ƒđ??ž
đ??´đ?‘€ đ?‘€đ??ž
=
đ??ľđ?‘ đ?‘ đ??ž
(1);
đ??´đ?‘€ đ?‘€đ??ž
=
đ??śđ?‘ƒ đ?‘ƒđ??ž
(2).
(3), which shows that đ?‘ đ?‘ƒ âˆĽ
đ??ľđ??ś. Let đ?‘…, đ?‘†, đ?‘‰, đ?‘Š, đ?‘ˆ, đ?‘‡ be the intersection points of the parallels đ?‘€đ?‘ , đ?‘€đ?‘ƒ, đ?‘ đ?‘ƒ of the sides of the triangles to the other sides. Obviously, by construction, the quadrilaterals đ??´đ?‘†đ?‘€đ?‘Š; đ??śđ?‘ˆđ?‘ƒđ?‘‰; đ??ľđ?‘…đ?‘ đ?‘‡ are parallelograms. The middle of the diagonal đ?‘Šđ?‘† falls on đ??´đ?‘€, so on the simedian đ??´đ??ž, and from Lemma 1 we get that đ?‘Šđ?‘† is an anti-parallel to đ??ľđ??ś. Since đ?‘‡đ?‘ˆ âˆĽ đ??ľđ??ś, it follows that đ?‘Šđ?‘† and đ?‘‡đ?‘ˆ are antiparallels, therefore the points đ?‘Š, đ?‘†, đ?‘ˆ, đ?‘‡ are concyclic (4). 3
Figure 2. Analogously, we show that the points đ?‘ˆ, đ?‘‰, đ?‘…, đ?‘† are concyclic (5). From đ?‘Šđ?‘† and đ??ľđ??ś anti-parallels, đ?‘ˆđ?‘‰ and đ??´đ??ľ anti-parallels, we have that âˆ˘đ?‘Šđ?‘†đ??´ ≥ âˆ˘đ??´đ??ľđ??ś and âˆ˘đ?‘‰đ?‘ˆđ??ś ≥ âˆ˘đ??´đ??ľđ??ś , therefore: âˆ˘đ?‘Šđ?‘†đ??´ ≥ âˆ˘đ?‘‰đ?‘ˆđ??ś , and since đ?‘‰đ?‘Š âˆĽ đ??´đ??ś , it follows that the trapeze đ?‘Šđ?‘†đ?‘ˆđ?‘‰ is isosceles, therefore the points đ?‘Š, đ?‘†, đ?‘ˆ, đ?‘‰ are concyclic (6). The relations (4), (5), (6) drive to the concyclicality of the points đ?‘…, đ?‘ˆ, đ?‘‰, đ?‘†, đ?‘Š, đ?‘‡, and the theorem is proved.
Remarks. 2. For any point đ?‘€ found on the simedian đ??´đ??´1 , 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 Theorem 2 generalizes the Theorem 1 because we get the second in the case the parallels are taken to the sides through the simedian center đ?‘˜.
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4. We get a circle built as in Theorem 2 from the first Lemoine circle by homothety of pole đ?‘˜ and of ratio đ?œ† ∈ â„?. 5. The centers of Lemoine circles built as above belong to the line đ?‘‚đ??ž, where đ?‘‚ is the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś.
Bibliography. [1] Exercices de GÊomÊtrie, par F.G.M., Huitième Êdition, Paris VIe, Librairie GÊnÊrale, 77, Rue Le Vaugirard. [2] Ion Paatrascu, Florentin Smarandache: Variance on topics of Plane Geometry, Educational Publishing, Columbus, Ohio, 2013.
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