Lemoine Circles Radius Calculus

Transcription

1 Lemoine Circles Radius Calculus Ion Pătrașcu, professor, Frații Buzești National College,Craiova, Romania Florentin Smarandache, professor, New Mexico University, U.S.A. For the calculus of the first Lemoine Circle we will first prove: 1 st Theorem (E. Lemoine 1873). The first Lemoine circle divides the sides of a triangle in segments proportional to the squares of the triangle s sides. Each extreme segment is proportional to the corresponding adjacent side, and the chord-segment in the Lemoine circle is proportional to the square of the side that contains it. We will prove that BC c = C B 1 a = B 1C b. In figure 1, K represents the symmedian center of the triangle ABC, and A 1 A ; B 1 B ; C 1 C represent Lemoine parallels. Figure 1 1

2 The triangles BC A 1 ; CB 1 A and KC A 1 have heights relative to the sides BC ; B 1 C and C B 1 equal (A 1 A BC). Hence: Area BA 1 C = Area KC A 1 = Area CB 1 A. (1) BC C B 1 B 1 C On the other hand: A 1 C and B 1 A being antiparallels with respect to AC and AB, it follows that ΔBC A 1 ~ΔBAC and ΔCB 1 A ~ΔCAB, likewise KC AC implies: ΔKC B 1 ~ΔABC. We obtain:. Area BC A 1 = BC ; Area KC B 1 = C B 1 ; Area CB 1 A = CB 1 Area ABC c Area ABC a Area ABC b. () If we note Area ABC = S, we obtain from the relations (1) and (): BC c = C B 1 a = B 1C b. Consequences. C1. According to the 1 st theorem we find that: BC = ac a +b +c ; B 1C = C. We find that: B 1 C = A C 1 = A 1B a 3 b 3 c ab a +b +c ; B 1C = 3, meaning that: a 3 a +b +c. The chords determined by the first Lemoine circle on the triangle s sides are proportional to the cubes of the sides. Due to this property, the first Lemoine circle is known in England by the name of triplicate ratio circle, Tucker. 1 st Proposition. The radius of the first Lemoine circle, R L1 is given by the formula: R L1 = 1 4 R (a +b +c )+ a b c (a +b +c ), (3)

3 where R represents the radius of the circle inscribed in the triangle ABC. Let L be the center of the first Lemoine circle that is known to represent the middle of the segment (OK) O being the center of the circle inscribed in the triangle ABC. Considering C1, we obtain BB 1 = a (c +a ) a +b +c. Taking into account the power of point B in relation to the first Lemoine circle, we have: BC BB 1 = BT LT, (BT is the tangent traced from B to the first Lemoine circle see Figure 1) Hence: R L1 = BL BC BB 1. (4) The median theorem in triangle BOK implies that: BL = (BK +BO ) OK. 4 It is known that K = (a +c ) S b a +b +c ; S b = ac m b a +c, where S b and m b are the lengths of the symmedian and the median from B, and OK = R see (3). Consequently: BK = a c (a +c ) a b c (a +b +c ), and 4BL = R + 4a c (a +c )+a b c (a +b +c ). 3a b c (a +b +c ), As: BC BB 1 = a c (a +c ), by replacing in (4) we obtain formula (3). (a +b +c ) nd Proposition. The radius of the second Lemoine circle, R L, is given by the formula: R L = abc a +b +c (5) 3

4 Figure In Figure, A 1 A ; B 1 B ; C 1 C are Lemoine antiparallels traced through symmedian center K that is the center of the second Lemoine circle, thence: R L = KA 1 = KA. If we note with S and M the feet of the symmedian and the median from A, it is known that: AK = b +c. KS a From the similarity of triangles AA A 1 and ABC, we have:. A 1A BC = AK AM. But: AK = b +c and AS = bc m AS a +b +c b +c a. A 1 A = R L, BC = a, consequently: R L = AK a, and as AK = bc m a formula (5) is the consequence. m a a +b +c Observations. 1. If we use tgω = obtain: R L = R tgω. 4S a +b +c, ω being the Brocard angle (see []), we 4

5 . If, in Figure 1, we denote with M 1 the middle of the antiparallel B C 1, which is equal to R L (due to their similarity), we thus find from the rectangular triangle LM 1 C 1 that: LC 1 = LM 1 + M 1 C 1, but LM 1 = 1 4 a and M 1 C = 1 R L ; it follows that: R L1 = 1 4 (R + R L ) = R (1 + 4 tg ω). We obtain: R L1 = R 1 + tg ω. 3 rd Proposition. The chords determined by the sides of the triangle in the second Lemoine circle are respectively proportional to the opposing angles cosines. KC B 1 is an isosceles triangle, KC B 1 = KB 1 C = A; as KC = R L we have that cos A = C B 1 R L, deci C B 1 cos A = R L, similary:. A C 1 cos B = B A 1 cos C = R L. Remark. Due to this property of the Lemoine s second circle, in England this circle is known as the cosine circle. References. [1] D. Efremov, The new geometry of the triangle, translated in Romanian from Russian by Mihai Micutite, GIL Publiching, Zalau, 010. [] F. Smarandache and I. Patrascu, The Geometry of Homological Triangles, The Education Publisher, Ohio, USA, 01. [3] I. Patrascu and F. Smarandache, Variance on Topics of Plane Geometry, Educational Publisher, Ohio, USA,