Radical Axis of Lemoine Circles Ion Patrascu, professor, “FraČ›ii BuzeČ™tiâ€? National College, Craiova, Romania Florentin Smarandache, professor, New Mexico University, USA
In this article, we emphasize the radical axis of the Lemoine Circles. For the start, let us remind:
Theorem 1. The parallels taken through the simmedian center đ??ž of a triangle to the sides of the triangle determine on them six concyclic points (The First Lemoine Circle).
Theorem 2. The antiparallels taken through the simmedian center of a triangle to the sides of a triangle determine on them six concyclic points (The Second Lemoine Circle).
Remark 1. If đ??´đ??ľđ??ś is a scalene triangle and đ??ž is its simmedian center, then đ??ż, the center of the First Lemoine Circle, is the middle of the segment [đ?‘‚đ??ž], where đ?‘‚ is the center of the circumscribed circle, and the center of the Second Lemoine Circle is đ??ž. It follows that the radical axis of Lemoine circles is perpendicular on the line of the centers đ??żđ??ž, therefore on the line đ?‘‚đ??ž.
Proposition 1. The radical axis of Lemoine Circles is perpendicular on the line đ?‘‚đ??ž raised in the simmedian center đ??ž. 1
Proof. Let đ??´1 đ??´2 be the antiparallel to đ??ľđ??ś taken through đ??ž, then đ??žđ??´1 is the radius đ?‘…đ??ż2 of the Second Lemoine Circle; we have: đ?‘…đ??ż2 =
đ?‘Žđ?‘?đ?‘? đ?‘Ž2 +đ?‘?2 +đ?‘? 2
.
Figura 1 Let đ??´1′ đ??´â€˛2 be the Lemoine parallel taken to đ??ľđ??ś; we evaluate the power of đ??ž towards the First Lemoine Circle. We have: ⃗⃗⃗⃗⃗⃗⃗⃗ đ??žđ??´1′ ∙ ⃗⃗⃗⃗⃗⃗⃗⃗ đ??žđ??´â€˛2 = đ??żđ??ž 2 − đ?‘…đ??ż21 .
(1)
Let đ?‘† be the simmedian leg from đ??´; it follows that: đ??žđ??´â€˛1 đ??ľđ?‘†
=
đ??´đ??ž đ??´đ?‘†
−
đ??žđ??´â€˛2 đ?‘†đ??ś
.
We obtain: đ??žđ??´1′ = đ??ľđ?‘† ∙ but
đ??ľđ?‘† đ?‘†đ??ś
=
đ?‘?2 đ?‘?2
and
đ??´đ??ž đ??´đ?‘†
đ??´đ??ž đ??´đ?‘†
=
and đ??žđ??´â€˛2 = đ?‘†đ??ś ∙ đ?‘?2 +đ?‘? 2 đ?‘Ž2 +đ?‘?2 +đ?‘? 2
đ??´đ??ž đ??´đ?‘†
. 2
,
Therefore: 2
2 đ?‘?2 đ?‘? 2
đ??´đ??ž −đ?‘Ž ⃗⃗⃗⃗⃗⃗⃗⃗ đ??žđ??´1′ ∙ ⃗⃗⃗⃗⃗⃗⃗⃗ đ??žđ??´â€˛2 = −đ??ľđ?‘† ∙ đ?‘†đ??ś ∙ ( ) = (đ?‘?2 đ??´đ?‘†
+đ?‘?
(đ?‘?2 +đ?‘? 2 )2
∙ 2 )2 (đ?‘Ž 2
+đ?‘?2 +đ?‘? 2 )2
= −đ?‘…đ??ż22 .
(2)
We draw the perpendicular in đ??ž on the line đ??żđ??ž and denote by đ?‘ƒ and đ?‘„ its intersection to the First Lemoine Circle; we have ⃗⃗⃗⃗⃗ đ??žđ?‘ƒ ∙ ⃗⃗⃗⃗⃗⃗ đ??žđ?‘„ = −đ?‘…đ??ż22 ; by the other hand, đ??žđ?‘ƒ = đ??žđ?‘„ (đ?‘ƒđ?‘„ is a chord which is perpendicular to the diameter passing through đ??ž). It follows that đ??žđ?‘ƒ = đ??žđ?‘„ = đ?‘…đ??ż2 , so đ?‘ƒ and đ?‘„ are situated on the Second Lemoine Circle. Because đ?‘ƒđ?‘„ is a chord which is common to the Lemoine Circles, it follows that đ?‘ƒđ?‘„ is the radical axis.
Comment 1. After equalizing relations (1) and (2) or by the Pythagorean theorem in the triangle đ?‘ƒđ??žđ??ż, we can calculate đ?‘…đ??ż1 . It is known that: 3đ?‘Ž2 đ?‘?2 đ?‘? 2
đ?‘‚đ??ž 2 = đ?‘…2 − (đ?‘Ž2
+đ?‘?2 +đ?‘? 2 )2
,
1
and since đ??żđ??ž = đ?‘‚đ??ž, we find that: 2
1
đ?‘Ž2 đ?‘?2 đ?‘? 2
4
+đ?‘?2 +đ?‘? 2 )2
đ?‘…đ??ż21 = ∙ [đ?‘…2 + (đ?‘Ž2
].
Remark 2. The proven Proposition regarding the radical axis of the Lemoine Circles is a particular case of the following Proposition, which we leave it to the reader to prove.
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Proposition 2. If đ?’ž(đ?‘‚1 , đ?‘…1 ) Ĺ&#x;i đ?’ž(đ?‘‚2 , đ?‘…2 ) are two circles such as the power of center đ?‘‚1 towards đ?’ž(đ?‘‚2 , đ?‘…2 ) is −đ?‘…12 , then the radical axis of the circles is the perpendicular in đ?‘‚1 on the line of centers đ?‘‚1 đ?‘‚2 .
Bibliography. [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.
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