1st Theorem. Let a triangle and its simedian center. ... isosceles trapezoid, so the points ... In the scalene triangle , let K be the simedian.
Ion Patrascu, Florentin Smarandache
Lemoineβs Circles
In Ion Patrascu, Florentin Smarandache: βComplements to Classic Topics of Circles Geometryβ. Brussels (Belgium): Pons Editions, 2016
In this article, we get to Lemoine's circles in a different manner than the known one.
1st Theorem. Let π΄π΅πΆ a triangle and πΎ its simedian center. We take through K the parallel π΄1 π΄2 to π΅πΆ, π΄1 β (π΄π΅), π΄2 β (π΄πΆ); through π΄2 we take the antiparallels π΄2 π΅1 to π΄π΅ in relation to πΆπ΄ and πΆπ΅ , π΅1 β (π΅πΆ) ; through π΅1 we take the parallel π΅1 π΅2 to π΄πΆ, π΅2 β π΄π΅; through π΅2 we take the antiparallels π΅1 πΆ1 to π΅πΆ , πΆ1 β (π΄πΆ) , and through πΆ1 we take the parallel πΆ1 πΆ2 to π΄π΅, πΆ1 β (π΅πΆ). Then: i. πΆ2 π΄1 is an antiparallel of π΄πΆ; ii. π΅1 π΅2 β© πΆ1 πΆ2 = {πΎ }; iii. The points π΄1 , π΄2 , π΅1 , π΅2 , πΆ1 , πΆ2 are concyclical (the first Lemoineβs circle). Proof. i.
The quadrilateral π΅πΆ2 πΎπ΄ is a parallelogram, and its center, i.e. the middle of the segment (πΆ2 π΄1 ), belongs to the simedian π΅πΎ; it follows that πΆ2 π΄2 is an antiparallel to π΄πΆ(see Figure 1).
17
Ion Patrascu, Florentin Smarandache
ii.
iii.
Let {πΎβ²} = π΄1 π΄2 β© π΅1 π΅2 , because the quadrilateral πΎβ²π΅1 πΆπ΄2 is a parallelogram; it follows that πΆπΎβ² is a simedian; also, πΆπΎ is a simedian, and since πΎ, πΎβ² β π΄1 π΄2 , it follows that we have πΎβ² = πΎ. π΅2 πΆ1 being an antiparallel to π΅πΆ and π΄1 π΄2 β₯ π΅πΆ , it means that π΅2 πΆ1 is an antiparallel to π΄1 π΄2 , so the points π΅2 , πΆ1 , π΄2 , π΄1 are concyclical. From π΅1 π΅2 β₯ π΄πΆ , β’π΅2 πΆ1 π΄ β‘ β’π΄π΅πΆ , β’π΅1 π΄2 πΆ β‘ β’π΄π΅πΆ we get that the quadrilateral π΅2 πΆ1 π΄2 π΅1 is an isosceles trapezoid, so the points π΅2 , πΆ1 , π΄2 , π΅1 are concyclical. Analogously, it can be shown that the quadrilateral πΆ2 π΅1 π΄2 π΄1 is an isosceles trapezoid, therefore the points πΆ2 , π΅1 , π΄2 , π΄1 are concyclical.
Figure 1
18
Complements to Classic Topics of Circles Geometry
From the previous three quartets of concyclical points, it results the concyclicity of the points belonging to the first Lemoineβs circle.
2nd Theorem. In the scalene triangle π΄π΅πΆ, let K be the simedian center. We take from πΎ the antiparallel π΄1 π΄2 to π΅πΆ ; π΄1 β π΄π΅, π΄2 β π΄πΆ; through π΄2 we build π΄2 π΅1 β₯ π΄π΅; π΅1 β (π΅πΆ), then through π΅1 we build π΅1 π΅2 the antiparallel to π΄πΆ, π΅2 β (π΄π΅), and through π΅2 we build π΅2 πΆ1 β₯ π΅πΆ, πΆ1 β π΄πΆ , and, finally, through πΆ1 we take the antiparallel πΆ1 πΆ2 to π΄π΅, πΆ2 β (π΅πΆ). Then: i. πΆ2 π΄1 β₯ π΄πΆ; ii. π΅1 π΅2 β© πΆ1 πΆ2 = {πΎ }; iii. The points π΄1 , π΄2 , π΅1 , π΅2 , πΆ1 , πΆ2 are concyclical (the second Lemoineβs circle). Proof. i.
Let {πΎβ²} = π΄1 π΄2 β© π΅1 π΅2 , having β’π΄π΄1 π΄2 = β’π΄πΆπ΅ and β’π΅π΅1 π΅2 β‘ β’π΅π΄πΆ because π΄1 π΄2 Θi π΅1 π΅2 are antiparallels to π΅πΆ , π΄πΆ , respectively, it follows that β’πΎβ²π΄1 π΅2 β‘ β’πΎβ²π΅2 π΄1 , so πΎβ²π΄1 = πΎβ²π΅2 ; having π΄1 π΅2 β₯ π΅1 π΄2 as well, it follows that also πΎβ²π΄2 = πΎβ²π΅1 , so π΄1 π΄2 = π΅1 π΅2 . Because πΆ1 πΆ2 and π΅1 π΅2 are antiparallels to π΄π΅ and π΄πΆ , we
19
Ion Patrascu, Florentin Smarandache
have πΎ"πΆ2 = πΎ"π΅1 ; we noted {πΎ"} = π΅1 π΅2 β© πΆ1 πΆ2 ; since πΆ1 π΅2 β₯ π΅1 πΆ2 , we have that the triangle πΎ"πΆ1 π΅2 is also isosceles, therefore πΎ"πΆ1 = πΆ1 π΅2 , and we get that π΅1 π΅2 = πΆ1 πΆ2 . Let {πΎβ²β²β²} = π΄1 π΄2 β© πΆ1 πΆ2 ; since π΄1 π΄2 and πΆ1 πΆ2 are antiparallels to π΅πΆ and π΄π΅ , we get that the triangle πΎβ²β²β²π΄2 πΆ1 is isosceles, so πΎβ²β²β²π΄2 = πΎβ²β²β²πΆ1 , but π΄1 π΄2 = πΆ1 πΆ2 implies that πΎβ²β²β²πΆ2 = πΎβ²β²β²π΄1 , then β’πΎβ²β²β²π΄1 πΆ2 β‘ β’πΎβ²β²β²π΄2 πΆ1 and, accordingly, πΆ2 π΄1 β₯ π΄πΆ.
Figure 2
ii.
We noted {πΎβ²} = π΄1 π΄2 β© π΅1 π΅2 ; let {π} = π΅2 πΆ1 β© π΅1 π΄2 ; obviously, π΅π΅1 ππ΅2 is a parallelogram; if πΎ0 is the middle of (π΅1 π΅2 ), then π΅πΎ0 is a simedian, since π΅1 π΅2 is an antiparallel to π΄πΆ, and the middle of the antiparallels of π΄πΆ are situated on the
20
Complements to Classic Topics of Circles Geometry
iii.
simedian π΅πΎ . If πΎ0 β K, then πΎ0 πΎ β₯ π΄1 π΅2 (because π΄1 π΄2 = π΅1 π΅2 and π΅1 π΄2 β₯ π΄1 π΅2 ), on the other hand, π΅, πΎ0 , K are collinear (they belong to the simedian π΅πΎ ), therefore πΎ0 πΎ intersects π΄π΅ in π΅, which is absurd, so πΎ0 =K, and, accordingly, π΅1 π΅2 β© π΄1 π΄2 = {πΎ } . Analogously, we prove that πΆ1 πΆ2 β© π΄1 π΄2 = {πΎ }, so π΅1 π΅2 β© πΆ1 πΆ2 = {πΎ }. K is the middle of the congruent antiparalells π΄1 π΄2 , π΅1 π΅2 , πΆ1 πΆ2 , so πΎπ΄1 = πΎπ΄2 = πΎπ΅1 = πΎπ΅2 = πΎπΆ1 = πΎπΆ2 . The simedian center πΎ is the center of the second Lemoineβs circle.
Remark. The center of the first Lemoineβs circle is the middle of the segment [ππΎ ], where π is the center of the circle circumscribed to the triangle π΄π΅πΆ. Indeed, the perpendiculars taken from π΄, π΅, πΆ on the antiparallels π΅2 πΆ1 , π΄1 πΆ2 , π΅1 π΄2 respectively pass through O, the center of the circumscribed circle (the antiparallels have the directions of the tangents taken to the circumscribed circle in π΄, π΅, πΆ). The mediatrix of the segment π΅2 πΆ1 pass though the middle of π΅2 πΆ1 , which coincides with the middle of π΄πΎ , so is the middle line in the triangle π΄πΎπ passing through the middle of (ππΎ) . Analogously, it follows that the mediatrix of π΄1 πΆ2 pass through the middle πΏ1 of [ππΎ ].
21
Ion Patrascu, Florentin Smarandache
References. [1] D. Efremov: Noua geometrie a triunghiului [The New Geometry of the Triangle], translation from Russian into Romanian by Mihai MiculiΘa, Cril Publishing House, Zalau, 2010. [2] Gh. Mihalescu: Geometria elementelor remarcabile [The Geometry of the Outstanding Elements], Tehnica Publishing House, Bucharest, 1957. [3] Ion Patrascu, Gh. Margineanu: Cercul lui Tucker [Tuckerβs Circle], in Alpha journal, year XV, no. 2/2009.
22