A GENERALIZATION OF THE ISOGONAL POINT

Size: px

Start display at page:

Download "A GENERALIZATION OF THE ISOGONAL POINT"

Diana Floyd
6 years ago
Views:

1 INTERNATIONAL JOURNAL OF GEOMETRY Vol. 1 (2012), No. 1, A GENERALIZATION OF THE ISOGONAL POINT PETRU I. BRAICA and ANDREI BUD Abstract. In this paper we give a generalization of the isogonal point in terms of concurrency, starting with the ideea of "breaking" the equilateral triangles constructed on the sides of one triangle, rotating them with the same acute angle and compressing with the same ratio the broken sides. 1. Introduction We will denote by E 2 the euclidean space. De nition 1.1. For X 2 E 2 xed and 2 ( ; ), we call rotation of center X and angle of Y 2 E 2, the point z R x; (Y ), having the following properties: (i) m(]y XZ) ;! (ii) XY XZ.! De nition 1.2. Let O 2 E 2 a xed point and k 2 Rnf0g. We call homothety with center O and ratio k an application: with the following properties: H 0;k : E 2! E 2 : H 0;k (M) M 0 (1) H 0;k (O) O; (2) If M 6 0, then the points O; M; M 0 are collinear; (3) If k > 0, then M 0 2 [OM, and if k < 0, then O 2 [MM 0 ]; (4) OM 0 jkj OM. Keywords and phrases: Triangle, isogonal point (2010)Mathematics Subject Classi cation: 51M04, 51M25, 51M30

2 42 Petru I. Braica and Andrei Bud 2. Main Results If we consider an arbitrary triangle ABC and the points C a (H A;k R A; )(B), C b (H B;k R B; )(A), A b (H B;k R B; )(C), A c (H C;k R C; )(B), B c (H C;k R C; )(A), B a (H A;k R A; )(C) with k 2 R and 2 ( ; ) xed. Lemma 2.1. If C 0 ; B 0 are points in the interior of the triangle ABC with 4ABC 0 4ACB 0 and CC 0 \ AB fc 1 g, also BB 0 \ AC fb 1 g, then the following identity is true: BA 0 A 0 C with y not m(]abc 0 ) m(]acb 0 ). Proof. We have B 1 A B 1 C S 4BB 0 C S 4BAB 0 sin C sin(b + y) sin B sin(c + y) ; BC B0 C sin(c + y) AB AB 0 sin(a + x) sin C sin y sin(a + x) sin A sin x sin(c + y) ; where x not m(]bac 0 ) m(]cab 0 ) (see Figure 1): Similarly, we have C 1 A C 1 B S 4ACC 0 S 4CC 0 B AC0 AC sin(a + x) sin y sin B sin(a + x) BC 0 BC sin(b + y) sin x sin A sin(b + y) : Now, using the Ceva s Theorem, in the triangle ABC, one obtains A 0 C BA 0 B 1C B 1 A C 1A C 1 B sin B sin(c + y) sin C sin(b + y) : Theorem 2.1. Let us consider now an arbitrary triangle ABC and the points C a (H A;k R A; )(B), C b (H B;k R B;k )(A), A b (H B;k R B; )(C), A c (H C;k R C; )(B), B c (H C;k R C; )(A), B a (H A;k R A; )(C), with k 2 R, 2 ( ; ) xed. Using this points, de ned above, we consider the following intersections: BB c \ CC b fp a g, AP a \ BC fp A g, BB a \ AA b fp c g, CP c \ AB fp C g, CC a \ AA c fp b g, BP b \

3 A generalization of the isogonal point 43 AC fp B g. The concurrency of the cevians AP A, BP B, CP C takes place in the point T ;k. Proof. Applying Lema 2.1 one obtains: BP A P A C CP B P B A AP C P C B sin C sin(b + ) sin B sin(c + ) sin A sin(c + ) sin C sin(a + ) sin B sin(a + ) sin A sin(b + ) : Now, evaluating the product BP A P A C CP B P B A AP C P C B 1 and from the reciprocal of Ceva s Theorem, we have the concurrency in T ;k (see Figure 2). Corrolary 2.1. For 60 one obtains the rst Torricelli point, and for 60 one obtains the second Torricelli point, replacing k 1. Corrolary 2.2. For k 1(2 cos ), one obtains: C a C b ; B a B c ; A c A b and the ABC a, BA b C, CB a A are similar isosceles triangles and the Theorem of Kiepert takes place.[1]: Corrolary 2.3. From Lemma 2:2 one obtaines that the geometrical locus of the intersection point CC 0 \ BB 0 is the line AA 0, where BA 0 A 0 C sin C sin(b + y) sin B sin(c + y) :

4 44 Petru I. Braica and Andrei Bud Theorem 2.2. Let us consider now an arbitrary triangle ABC and the following points C a (H A;k R A; )(B), C b (H B;k R B; )(A), A b (H B;k R B; )(C),A c (H C;k R C; )(B), B c (H C;k R C; )(A), B a (H A;k R A; )(C) with k 2 R, 2 ( ; ) xed. Using this points, we consider the following intersections: fq a g CC a \BB a, fq b g AA b \CC b, fq c g BB c \ AA c, AQ a \ BC fq A g, BQ b \ AC fq B g, CQ c \ AB fq C g. The concurrency of lines AQ A, BQ B, CQ C takes place in the point T ;k. Proof. We make the following notations: m(]c a BA) m(]c b AB) m(]b a CA) m(]b c AC) m(]a c BC) m(]a b CB). Applying Lemma 2.1 one obtains: BQ A Q A C CQ B Q B A AQ C Q C B sin C sin(b + ) sin B sin(c + ) sin A sin(c + ) sin C sin(a + ) sin B sin(a + ) sin A sin(b + ) : Now, applying again the reciprocal of the Ceva s Theorem, one obtains the required concurrency in the point T;k (See Figure 3). Remark 2.1. The corrolary 2.1, 2.2 and 2.3 takes place also for Theorem 2.2. Remark 2.2. The barycentric coordinates for the points T k; and T k; can remain an open problem. Remark 2.3. For k 1, Theorem 2.1 become Theorem 1.1 from [4].

5 A generalization of the isogonal point 45 References [1] Barbu, C., Fundamental Theorems of Triangle Geometry (Romanian), Ed. Unique, Bac¼au, [2] Nicolescu, L. and Bosko, V., Practical Problems of Geometry (Romanian), Ed. Tehnic¼a, Bucureşti, [3] Mathematical Gazette (Romanian), Bucharest. [4] Braica, P. and Pop, O. T., An Extension of Torricelli s Theorem (Romanian). Received: February 2, ELEMENTARY SCHOOL "GRIGORE MOISIL" MILENIULUI 1, SATU MARE, ROMANIA address: petrubr@yahoo.com ICHB BUCUREŞTI, ROMANIA address: and_rei_95@yahoo.com

Two applications of the theorem of Carnot

Two applications of the theorem of Carnot Zoltán Szilasi Abstract Using the theorem of Carnot we give elementary proofs of two statements of C Bradley We prove his conjecture concerning the tangents to