Two Remarkable Ortho-Homological Triangles Prof. Ion Pătraşcu - The National ollege Fraţii Buzeşti, raiova, Romania Prof. Florentin Smarandache University of New Mexico, U.S.. In a previous paper [5] we have introduced the ortho-homological triangles, which are triangles that are orthological and homological simultaneously. In this article we call attention to two remarkable ortho-homological triangles (the given triangle B and its first Brocard s triangle), and using the Sondat s theorem relative to orthological triangles, we emphasize on four important collinear points in the geometry of the triangle. Definition The first Brocard triangle of a given triangle B is the triangle formed by the projections of the symmedian center of the triangle B on its perpendicular bisectors. Observation In figure we note with K the symmedian center, O', OB ', O ' the perpendicular bisectors of the triangle B and B the first Brocard s triangle. B B O K B Fig. Theorem If B is a given triangle and B is its first triangle Brocard, then the triangles B and B are ortho-homological.
We ll perform the proof of this theorem in two stages. I. We prove that the triangles B and B are orthological. The perpendiculars from, B, on B, respective B are perpendicular bisectors in the triangle B, therefore are concurrent in O, the center of the circumscribed circle of triangle B which is the orthological center for triangles B and B. II. We prove that the triangles B and B are homological, that is the lines, BB, are concurrent. To continue with these proves we need to refresh some knowledge and some helpful results. Definition In any triangle B there exist the points Ω and Ω ' and the angle ω such that: m( Ω B) Ω B Ω ω m Ω ' B Ω ' Ω ' B ω ( ) ω Ω Ω ω ω B Fig. The points Ω and Ω ' are called the first, respectively the second point of Brocard and ω is called the Brocard s angle. Lemma In the triangle B let Ω the first point of Brocard and { "} { "} B" c " a Ω B, then:
ria B" B "sinω () ria " "sin( ω) () From () and () we find: ria B" B sinω ria " sin( ω) (3) On the other side, the mentioned triangles have the same height built from, therefore: ria B" B" ria " " (4) From (3) and (4) we have: B" B sinω " sin( ω) (5) pplying the sinus theorem in the triangle Ω and in the triangle BΩ, it results: Ω sin ( ω) sin Ω (6) Ω B sinω sin BΩ (7) Because m Ω o 80 ( ) ( ) o m BΩ 80 From the relations (6) and (7) we find: sinω sin sin ( ω ) B (8) sin pplying the sinus theorem in the triangle B leads to: sin B (9) sin B The relations (5), (8), (9) provide us the relation: B" c " a Remark By making the notations: { B" } BΩ and { " } Ω B we obtain also the relations: B" a " b and B" b B" c Lemma In a triangle B, the Brocard s evian BΩ, symmedian from and the median from are concurrent. 3
K 3 K K It is known that the symmedian K of triangle B intersects B in the point such b B" a that. We had that the evian BΩ intersects in B " such that. B c B" b The median from intersects B in ' and B' '. B ' B" Because, the reciprocal of eva s theorem ensures the concurrency ' B" B of the lines BΩ, K and '. Lemma 3 Give a triangle B and ω the Brocard s angle, then ctgω ctg + ctgb + ctg (9) From the relation (8) we find: a sin sin ( ω) sinω (0) b sin From the sinus theorem in the triangle B we have that a sin b sin B sin sinω Substituting it in (0) it results: sin ( ω) sin B sin Furthermore we have: sin( ω) sin cosω sinω cos sin sinω sin cosω sinω cos () sin B sin Dividing relation () by sin sinω and taking into account that sin sin( B+ ), and sin( B + ) sin B cos+ sin cos B we obtain relation (5) Lemma 4 If in the triangle B, K is the symmedian center and K, K, K 3 are its projections on the sides B,, B, then: F E B Fig. 3 4
KK KK KK3 tgω a b c : Let the symmedian in the triangle B, we have: B ria B, ria where E and F are the projection of on respectively B. F c It results that E b KK3 F From the fact that KK3 F and KK E we find that KK E KK 3 lso: KK KK KK, and similarly:, consequently: b c a b KK KK KK 3 () a b c The relation () is equivalent to: akk bkk ckk3 akk+ bkk + ckk3 a b c a + b + c Because akk+ bkk + ckk3 ria B S, we have: KK KK KK3 S a b c a + b + c If we note H, H, H 3 the projections of, B, on B,, B, we have ctg H bccos BH S H B Fig. 4 From the cosine s theorem it results that : b + c a bc cos, and therefore 5
b + c a ctg 4S Taking into account the relation (9), we find: a + b + c ctgω, 4S then 4S tgω a + b + c and then KK S tgω. a a + b + c Lemma 5 The evians, BB, are the isotomics of the symmedians, BB, in the triangle B. : J Ω K J K B Fig. 5 In figure 5 we note J the intersection point of the evians from the Lemma. Because K B, we have that ' KK atgω. On the other side from the right triangle ' B ' we have: tg B' tgω, consequently the point, the vertex of the first B' triangle of Brocard belongs to the evians BΩ. We note { J' } K ', and evidently from K B it results that JJ ' is the median in the triangle J K, therefore J ' J' K. 6
We note with ' the intersection of the evians with B, because K ' and J ' is a median in the triangle K it results that ' is a median in triangle ' therefore the points ' and are isometric. Similarly it can be shown that BB and ' are the isometrics of the symmedians ' BB and. The second part of this proof: Indeed it is known that the isometric evians of certain concurrent evians are concurrent and from Lemma 5 along with the fact that the symmedians of a triangle are concurrent, it results the concurrency of the evians, BB, and therefore the triangle B and the first triangle of Brocard are homological. The homology s center (the concurrency point) of these evians is marked in some works with Ω" with and it is called the third point of Brocard. From the previous proof, it results that Ω" is the isotomic conjugate of the symmedian center K. Remark The triangles B and B (first Brocard triangle) are triple-homological, since first time the evians B, B, are concurrent (in a Brocard point), second time the evians, B, B are also concurrent (in the second Brocard point), and third time the evians, BB, are concurrent as well (in the third point of Brocard). Definition 3 It is called the Tarry point of a triangle B, the concurrency point of the perpendiculars from, B, on the sides B,, B of the Brocard s first triangle. Remark 3 The fact that the perpendiculars from the above definition are concurrent results from the theorem and from the theorem that states that the relation of triangles orthology is symmetric. We continue to prove the concurrency using another approach that will introduce supplementary information about the Tarry s point. We ll use the following: Lemma 6: The first triangle Brocard of a triangle and the triangle itself are similar. From K (see Fig. ), similarly B and O' B it results that ( ) m KO 90 ( ) ( ) m KBO m KO 90 and therefore the first triangle of Brocard is inscribed in the circle with OK as diameter (this circle is called the Brocard circle). Because 7
( ) m O 80 B and, B,, O are concyclic, it results that B B, similarly m( B' O' ) 80, it results that m( BO ) m( ) but BO B, therefore B and the triangle B is similar wit the triangle B. Theorem The orthology center of the triangle B and of the first triangle of Brocard is the Tarry s point T of the triangle B, and T belongs to the circumscribed circle of the triangle B. We mark with T the intersection of the perpendicular raised from B on with the perpendicular raised from on B and let We have { B '} BT, { '} B T. ( ' ') 80 ( ) m B T m B B K S B O T B B Fig 6 8
But because of Lemma 6 B. It results that m( B' T' ) 80, therefore m( BT' ) + m( B) 80 Therefore T belongs to the circumscribed circle of triangle B If { ' } B T and if we note with T ' the intersection of the perpendicular raised from on B with the perpendicular raised from B on, we observe that m( B' T' ' ) m( B) therefore m( BT' ) + m( B) and it results that T ' belongs to the circumscribed triangle B. Therefore T T ' and the theorem is proved. Theorem 3 If through the vertexes, B, of a triangle are constructed the parallels to the sides B, respectively B of the first triangle of Brocard of this triangle, then these lines are concurrent in a point S (the Steiner point of the triangle) We note with S the polar intersection constructed through to B with the polar constructed through B to (see Fig. 6). We have m( SB) 80 m( B) (angles with parallel sides) because m( B) m, we have m SB m, ( ) 80 therefore SB are concyclic. Similarly, if we note with S ' the intersection of the polar constructed through to B B we find that the points S ' B are concyclic. with the parallel constructed through to Because the parallels from to B contain the points,, ' are on the circumscribed circle of the triangle, it results that S S' Remark 4 Because S B and B T, it results that SS and the points SS, ', and the theorem is proved. ( ) 90 m ST, but S and T belong to the circumscribed circle to the triangle B, consequently the Steiner s point and the Tarry point are diametric opposed. 9
Theorem 4 In a triangle B the Tarry point T, the center of the circumscribed circle O, the third point of Brocard Ω " and Steiner s point S are collinear points The P. Sondat s theorem relative to the orthological triangles (see [4]) says that the points T, O, Ω " are collinear, therefore the points: T, O, Ω ", S are collinear. References [] Roger. Johnson dvanced Euclidean Geometry, Dover Publications, Inc. Mineola, New York, 007. [] F. Smarandache Multispace & Multistructure, Neutrospheric Trandisciplinarity (00 ollected Papers of Sciences), Vol IV. North-European Scientific Publishers, Hanko, Findland, 00. [3]. Barbu Teoreme fundamentale din geometria triunghiului. Editura Unique, Bacău, 008. [4] I. Pătraşcu, F. Smarandache http://www.scribd.com/doc/3373308 - n application of Sondat Theorem Regarding the Orthohomological Triangles. [5] Ion Pătrașcu, Florentin Smarandache, Theorem about Simulotaneously Orthological and Homological Triangles, http://arxiv.org/abs/004.0347. 0