Complements to Classic Topics - UNM Gallup

1 downloads 0 Views 844KB Size Report
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

Lemoine’s Circles

In Ion Patrascu, Florentin Smarandache: β€œComplements to Classic Topics of Circles Geometry”. 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 𝐴𝐡𝐢 a triangle and 𝐾 its simedian center. We take through K the parallel 𝐴1 𝐴2 to 𝐡𝐢, 𝐴1 ∈ (𝐴𝐡), 𝐴2 ∈ (𝐴𝐢); through 𝐴2 we take the antiparallels 𝐴2 𝐡1 to 𝐴𝐡 in relation to 𝐢𝐴 and 𝐢𝐡 , 𝐡1 ∈ (𝐡𝐢) ; through 𝐡1 we take the parallel 𝐡1 𝐡2 to 𝐴𝐢, 𝐡2 ∈ 𝐴𝐡; through 𝐡2 we take the antiparallels 𝐡1 𝐢1 to 𝐡𝐢 , 𝐢1 ∈ (𝐴𝐢) , and through 𝐢1 we take the parallel 𝐢1 𝐢2 to 𝐴𝐡, 𝐢1 ∈ (𝐡𝐢). Then: i. 𝐢2 𝐴1 is an antiparallel of 𝐴𝐢; ii. 𝐡1 𝐡2 ∩ 𝐢1 𝐢2 = {𝐾 }; iii. The points 𝐴1 , 𝐴2 , 𝐡1 , 𝐡2 , 𝐢1 , 𝐢2 are concyclical (the first Lemoine’s circle). Proof. i.

The quadrilateral 𝐡𝐢2 𝐾𝐴 is a parallelogram, and its center, i.e. the middle of the segment (𝐢2 𝐴1 ), belongs to the simedian 𝐡𝐾; it follows that 𝐢2 𝐴2 is an antiparallel to 𝐴𝐢(see Figure 1).

17

Ion Patrascu, Florentin Smarandache

ii.

iii.

Let {𝐾′} = 𝐴1 𝐴2 ∩ 𝐡1 𝐡2 , because the quadrilateral 𝐾′𝐡1 𝐢𝐴2 is a parallelogram; it follows that 𝐢𝐾′ is a simedian; also, 𝐢𝐾 is a simedian, and since 𝐾, 𝐾′ ∈ 𝐴1 𝐴2 , it follows that we have 𝐾′ = 𝐾. 𝐡2 𝐢1 being an antiparallel to 𝐡𝐢 and 𝐴1 𝐴2 βˆ₯ 𝐡𝐢 , it means that 𝐡2 𝐢1 is an antiparallel to 𝐴1 𝐴2 , so the points 𝐡2 , 𝐢1 , 𝐴2 , 𝐴1 are concyclical. From 𝐡1 𝐡2 βˆ₯ 𝐴𝐢 , −𝐡2 𝐢1 𝐴 ≑ −𝐴𝐡𝐢 , −𝐡1 𝐴2 𝐢 ≑ −𝐴𝐡𝐢 we get that the quadrilateral 𝐡2 𝐢1 𝐴2 𝐡1 is an isosceles trapezoid, so the points 𝐡2 , 𝐢1 , 𝐴2 , 𝐡1 are concyclical. Analogously, it can be shown that the quadrilateral 𝐢2 𝐡1 𝐴2 𝐴1 is an isosceles trapezoid, therefore the points 𝐢2 , 𝐡1 , 𝐴2 , 𝐴1 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 Lemoine’s circle.

2nd Theorem. In the scalene triangle 𝐴𝐡𝐢, let K be the simedian center. We take from 𝐾 the antiparallel 𝐴1 𝐴2 to 𝐡𝐢 ; 𝐴1 ∈ 𝐴𝐡, 𝐴2 ∈ 𝐴𝐢; through 𝐴2 we build 𝐴2 𝐡1 βˆ₯ 𝐴𝐡; 𝐡1 ∈ (𝐡𝐢), then through 𝐡1 we build 𝐡1 𝐡2 the antiparallel to 𝐴𝐢, 𝐡2 ∈ (𝐴𝐡), and through 𝐡2 we build 𝐡2 𝐢1 βˆ₯ 𝐡𝐢, 𝐢1 ∈ 𝐴𝐢 , and, finally, through 𝐢1 we take the antiparallel 𝐢1 𝐢2 to 𝐴𝐡, 𝐢2 ∈ (𝐡𝐢). Then: i. 𝐢2 𝐴1 βˆ₯ 𝐴𝐢; ii. 𝐡1 𝐡2 ∩ 𝐢1 𝐢2 = {𝐾 }; iii. The points 𝐴1 , 𝐴2 , 𝐡1 , 𝐡2 , 𝐢1 , 𝐢2 are concyclical (the second Lemoine’s circle). Proof. i.

Let {𝐾′} = 𝐴1 𝐴2 ∩ 𝐡1 𝐡2 , having −𝐴𝐴1 𝐴2 = −𝐴𝐢𝐡 and −𝐡𝐡1 𝐡2 ≑ −𝐡𝐴𝐢 because 𝐴1 𝐴2 Θ™i 𝐡1 𝐡2 are antiparallels to 𝐡𝐢 , 𝐴𝐢 , respectively, it follows that βˆ’πΎβ€²π΄1 𝐡2 ≑ βˆ’πΎβ€²π΅2 𝐴1 , so 𝐾′𝐴1 = 𝐾′𝐡2 ; having 𝐴1 𝐡2 βˆ₯ 𝐡1 𝐴2 as well, it follows that also 𝐾′𝐴2 = 𝐾′𝐡1 , so 𝐴1 𝐴2 = 𝐡1 𝐡2 . Because 𝐢1 𝐢2 and 𝐡1 𝐡2 are antiparallels to 𝐴𝐡 and 𝐴𝐢 , we

19

Ion Patrascu, Florentin Smarandache

have 𝐾"𝐢2 = 𝐾"𝐡1 ; we noted {𝐾"} = 𝐡1 𝐡2 ∩ 𝐢1 𝐢2 ; since 𝐢1 𝐡2 βˆ₯ 𝐡1 𝐢2 , we have that the triangle 𝐾"𝐢1 𝐡2 is also isosceles, therefore 𝐾"𝐢1 = 𝐢1 𝐡2 , and we get that 𝐡1 𝐡2 = 𝐢1 𝐢2 . Let {𝐾′′′} = 𝐴1 𝐴2 ∩ 𝐢1 𝐢2 ; since 𝐴1 𝐴2 and 𝐢1 𝐢2 are antiparallels to 𝐡𝐢 and 𝐴𝐡 , we get that the triangle 𝐾′′′𝐴2 𝐢1 is isosceles, so 𝐾′′′𝐴2 = 𝐾′′′𝐢1 , but 𝐴1 𝐴2 = 𝐢1 𝐢2 implies that 𝐾′′′𝐢2 = 𝐾′′′𝐴1 , then βˆ’πΎβ€²β€²β€²π΄1 𝐢2 ≑ βˆ’πΎβ€²β€²β€²π΄2 𝐢1 and, accordingly, 𝐢2 𝐴1 βˆ₯ 𝐴𝐢.

Figure 2

ii.

We noted {𝐾′} = 𝐴1 𝐴2 ∩ 𝐡1 𝐡2 ; let {𝑋} = 𝐡2 𝐢1 ∩ 𝐡1 𝐴2 ; obviously, 𝐡𝐡1 𝑋𝐡2 is a parallelogram; if 𝐾0 is the middle of (𝐡1 𝐡2 ), then 𝐡𝐾0 is a simedian, since 𝐡1 𝐡2 is an antiparallel to 𝐴𝐢, and the middle of the antiparallels of 𝐴𝐢 are situated on the

20

Complements to Classic Topics of Circles Geometry

iii.

simedian 𝐡𝐾 . If 𝐾0 β‰  K, then 𝐾0 𝐾 βˆ₯ 𝐴1 𝐡2 (because 𝐴1 𝐴2 = 𝐡1 𝐡2 and 𝐡1 𝐴2 βˆ₯ 𝐴1 𝐡2 ), on the other hand, 𝐡, 𝐾0 , K are collinear (they belong to the simedian 𝐡𝐾 ), therefore 𝐾0 𝐾 intersects 𝐴𝐡 in 𝐡, which is absurd, so 𝐾0 =K, and, accordingly, 𝐡1 𝐡2 ∩ 𝐴1 𝐴2 = {𝐾 } . Analogously, we prove that 𝐢1 𝐢2 ∩ 𝐴1 𝐴2 = {𝐾 }, so 𝐡1 𝐡2 ∩ 𝐢1 𝐢2 = {𝐾 }. K is the middle of the congruent antiparalells 𝐴1 𝐴2 , 𝐡1 𝐡2 , 𝐢1 𝐢2 , so 𝐾𝐴1 = 𝐾𝐴2 = 𝐾𝐡1 = 𝐾𝐡2 = 𝐾𝐢1 = 𝐾𝐢2 . The simedian center 𝐾 is the center of the second Lemoine’s circle.

Remark. The center of the first Lemoine’s circle is the middle of the segment [𝑂𝐾 ], where 𝑂 is the center of the circle circumscribed to the triangle 𝐴𝐡𝐢. Indeed, the perpendiculars taken from 𝐴, 𝐡, 𝐢 on the antiparallels 𝐡2 𝐢1 , 𝐴1 𝐢2 , 𝐡1 𝐴2 respectively pass through O, the center of the circumscribed circle (the antiparallels have the directions of the tangents taken to the circumscribed circle in 𝐴, 𝐡, 𝐢). The mediatrix of the segment 𝐡2 𝐢1 pass though the middle of 𝐡2 𝐢1 , which coincides with the middle of 𝐴𝐾 , so is the middle line in the triangle 𝐴𝐾𝑂 passing through the middle of (𝑂𝐾) . Analogously, it follows that the mediatrix of 𝐴1 𝐢2 pass through the middle 𝐿1 of [𝑂𝐾 ].

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 MiculiΘ›a, 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 [Tucker’s Circle], in Alpha journal, year XV, no. 2/2009.

22