The Dual Theorem concerning Aubert Line Professor Ion Patrascu, National College "BuzeĹ&#x;ti Brothers" Craiova - Romania Professor Florentin Smarandache, University of New Mexico, Gallup, USA
In this article we introduce the concept of Bobillier transversal of a triangle with respect to a point in its plan; we prove the Aubert Theorem about the collinearity of the orthocenters in the triangles determined by the sides and the diagonals of a complete quadrilateral, and we obtain the Dual Theorem of this Theorem. Theorem 1 (E. Bobillier) Let đ??´đ??ľđ??ś be a triangle and đ?‘€ a point in the plane of the triangle so that the perpendiculars taken in đ?‘€, and đ?‘€đ??´, đ?‘€đ??ľ, đ?‘€đ??ś respectively, intersect the sides đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ at đ??´đ?‘š, đ??ľđ?‘š and đ??śđ?‘š. Then the points đ??´đ?‘š, đ??ľđ?‘š and đ??śđ?‘š are collinear. Proof We note that aria (đ??ľđ?‘€đ??´đ?‘š) aria (đ??śđ?‘€đ??´đ?‘š)
đ??´đ?‘šđ??ľ đ??´đ?‘šđ??ś
=
(see Fig. 1).
Area
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(đ??ľđ?‘€đ??´đ?‘š) = ∙ đ??ľđ?‘€ ∙ đ?‘€đ??´đ?‘š ∙ 2
Ě‚ ). sin(đ??ľđ?‘€đ??´đ?‘š Area
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(đ??śđ?‘€đ??´đ?‘š) = ∙ đ??śđ?‘€ ∙ đ?‘€đ??´đ?‘š ∙ 2
Ě‚ ). sin(đ??śđ?‘€đ??´đ?‘š Since
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Ě‚ ) = 3đ?œ‹ − đ?‘š(đ??´đ?‘€đ??ś Ě‚ ), đ?‘š(đ??śđ?‘€đ??´đ?‘š 2
it explains that Ě‚ ) = − cos(đ??´đ?‘€đ??ś Ě‚ ); sin(đ??śđ?‘€đ??´đ?‘š đ?œ‹ Ě‚ ) = sin (đ??´đ?‘€đ??ľ Ě‚ − ) = − cos(đ??´đ?‘€đ??ľ Ě‚ ). sin(đ??ľđ?‘€đ??´đ?‘š 2 Therefore: Ě‚) đ??´đ?‘šđ??ľ đ?‘€đ??ľ ∙ cos(đ??´đ?‘€đ??ľ (1). = Ě‚) đ??´đ?‘šđ??ś đ?‘€đ??ś ∙ cos(đ??´đ?‘€đ??ś In the same way, we find that: Ě‚) đ??ľđ?‘šđ??ś đ?‘€đ??ś cos(đ??ľđ?‘€đ??ś (2); = ∙ Ě‚) đ??ľđ?‘šđ??´ đ?‘€đ??´ cos(đ??´đ?‘€đ??ľ Ě‚) đ??śđ?‘šđ??´ đ?‘€đ??´ cos(đ??´đ?‘€đ??ś (3). = ∙ Ě‚) đ??śđ?‘šđ??ľ đ?‘€đ??ľ cos(đ??ľđ?‘€đ??ś The relations (1), (2), (3), and the reciprocal Theorem of Menelaus lead to the collinearity of points đ??´đ?‘š, đ??ľđ?‘š, đ??śđ?‘š. Note Bobillier's Theorem can be obtained – by converting the duality with respect to a circle – from the theorem relative to the concurrency of the heights of a triangle. Definition 1 It is called Bobillier transversal of a triangle đ??´đ??ľđ??ś with respect to the point đ?‘€ the line containing the intersections of the perpendiculars taken in đ?‘€ on đ??´đ?‘€, đ??ľđ?‘€, and đ??śđ?‘€ respectively, with sides đ??ľđ??ś, CA and đ??´đ??ľ.
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Note The Bobillier transversal is not defined for any point đ?‘€ in the plane of the triangle đ??´đ??ľđ??ś, for example, where đ?‘€ is one of the vertices or the orthocenter đ??ť of the triangle. Definition 2 If đ??´đ??ľđ??śđ??ˇ is a convex quadrilateral and đ??¸, đ??š are the intersections of the lines đ??´đ??ľ and đ??śđ??ˇ, đ??ľđ??ś and đ??´đ??ˇ respectively, we say that the figure đ??´đ??ľđ??śđ??ˇđ??¸đ??š is a complete quadrilateral. The complete quadrilateral sides are đ??´đ??ľ, đ??ľđ??ś, đ??śđ??ˇ, đ??ˇđ??´, and đ??´đ??ś, đ??ľđ??ˇ and đ??¸đ??š are diagonals. Theorem 2 (Newton-Gauss) The diagonals’ means of a complete quadrilateral are three collinear points. To prove Theorem 2, refer to [1]. Note It is called Newton-Gauss Line of a quadrilateral the line to which the diagonals’ means of a complete quadrilateral belong. Teorema 3 (Aubert) If đ??´đ??ľđ??śđ??ˇđ??¸đ??š is a complete quadrilateral, then the orthocenters đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 of the triangles đ??´đ??ľđ??š, đ??´đ??¸đ??ˇ, đ??ľđ??śđ??¸, and đ??śđ??ˇđ??š respectively, are collinear points. Proof Let đ??´1 , đ??ľ1 , đ??š1 be the feet of the heights of the triangle đ??´đ??ľđ??š and đ??ť1 its orthocenter (see Fig. 2). Considering the power of the point đ??ť1 relative to the circle circumscribed to the triangle ABF, and given the triangle orthocenter’s property according to which its symmetrics to the triangle sides belong to the circumscribed circle, we find that: đ??ť1 đ??´ ∙ đ??ť1 đ??´1 = đ??ť1 đ??ľ ∙ đ??ť1 đ??ľ1 = đ??ť1 đ??š ∙ đ??ť1 đ??š1 .
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This relationship shows that the orthocenter đ??ť1 has equal power with respect to the circles of diameters [đ??´đ??ś], [đ??ľđ??ˇ], [đ??¸đ??š]. As well, we establish that the orthocenters đ??ť2 , đ??ť3 , đ??ť4 have equal powers to these circles. Since the circles of diameters [đ??´đ??ś], [đ??ľđ??ˇ], [đ??¸đ??š] have collinear centers (belonging to the Newton-Gauss line of the đ??´đ??ľđ??śđ??ˇđ??¸đ??š quadrilateral), it follows that the points đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 belong to the radical axis of the circles, and they are, therefore, collinear points.
Notes 1. It is called the Aubert Line or the line of the complete quadrilateral’s orthocenters the line to which the orthocenters đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 belong.
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2. The Aubert Line is perpendicular on the Newton-Gauss line of the quadrilateral (radical axis of two circles is perpendicular to their centers’ line). Theorem 4 (The Dual Theorem of the Theorem 3) If đ??´đ??ľđ??śđ??ˇ is a convex quadrilateral and đ?‘€ is a point in its plane for which there are the Bobillier transversals of triangles đ??´đ??ľđ??ś, đ??ľđ??śđ??ˇ, đ??śđ??ˇđ??´ and đ??ˇđ??´đ??ľ; thereupon these transversals are concurrent. Proof Let us transform the configuration in Fig. 2, by duality with respect to a circle of center đ?‘€. By the considered duality, the lines đ?‘Ž, đ?‘?, đ?‘?, đ?‘‘, đ?‘’ and đ?‘“ correspond to the points đ??´, đ??ľ, đ??ś, đ??ˇ, đ??¸, đ??š (their polars). It is known that polars of collinear points are concurrent lines, therefore we have: đ?‘Ž ∊ đ?‘? ∊ đ?‘’ = {đ??´â€˛ }, đ?‘? ∊ đ?‘? ∊ đ?‘“ = {đ??ľâ€˛ }, đ?‘? ∊ đ?‘‘ ∊ đ?‘’ = {đ??ś ′ }, đ?‘‘ ∊ đ?‘“ ∊ đ?‘Ž = {đ??ˇâ€˛ }, đ?‘Ž ∊ đ?‘? = {đ??¸ ′ }, đ?‘? ∊ đ?‘‘ = {đ??š ′ }. Consequently, by applicable duality, the points đ??´â€˛ , đ??ľâ€˛ , đ??ś ′ , đ??ˇâ€˛ , đ??¸ ′ and đ??š ′ correspond to the straight lines đ??´đ??ľ, đ??ľđ??ś, đ??śđ??ˇ, đ??ˇđ??´, đ??´đ??ś, đ??ľđ??ˇ. To the orthocenter đ??ť1 of the triangle đ??´đ??¸đ??ˇ, it corresponds, by duality, its polar, which we denote đ??´1′ – đ??ľ1′ − đ??ś1′ , and which is the Bobillier transversal of the triangle đ??´â€™đ??śâ€™đ??ˇâ€™ in relation to the point đ?‘€. Indeed, the point đ??śâ€™ corresponds to the line đ??¸đ??ˇ by duality; to the height from đ??´ of the triangle đ??´đ??¸đ??ˇ, also by duality, it correspond its pole, which is the point đ??ś1′ located on đ??´â€™đ??ˇâ€™ such that Ě‚ ′ ) = 900 . đ?‘š(đ??śâ€˛đ?‘€đ??ś 1
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To the height from đ??¸ of the triangle đ??´đ??¸đ??ˇ, it corresponds the point đ??ľ1′ ∈ Ě‚ ′ ) = 900 . đ??´â€˛đ??śâ€˛ such that đ?‘š(đ??ˇâ€˛đ?‘€đ??ľ 1 Also, to the height from đ??ˇ, it corresponds đ??´1′ ∈ đ??śâ€˛đ??ˇâ€™ on C such that đ?‘š(đ??´â€˛ đ?‘€đ??´1′ ) = 900 . To the orthocenter đ??ť2 of the triangle đ??´đ??ľđ??š, it will correspond, by applicable duality, the Bobillier transversal đ??´â€˛2 − đ??ľ2′ − đ??ś2′ in the triangle đ??´â€˛ đ??ľâ€˛ đ??ˇâ€˛ relative to the point đ?‘€. To the orthocenter đ??ť3 of the triangle đ??ľđ??śđ??¸, it will correspond the Bobillier transversal đ??´3 ′ − đ??ľâ€˛ 3 − đ??ś3 ′ in the triangle đ??´â€˛ đ??ľâ€˛ đ??ś ′ relative to the point đ?‘€, and to the orthocenter đ??ť4 of the triangle đ??śđ??ˇđ??š, it will correspond the transversal đ??´â€˛4 − đ??ľ4′ − đ??ś4′ in the triangle đ??ś ′ đ??ˇâ€˛ đ??ľâ€˛ relative to the point đ?‘€. Ě…Ě…Ě…Ě… correspond to the The Bobillier transversals đ??´â€˛đ?‘– − đ??ľđ?‘–′ − đ??śđ?‘–′ , đ?‘– = 1,4 Ě…Ě…Ě…Ě…. collinear points đ??ťđ?‘– , đ?‘– = 1,4 These transversals are concurrent in the pole of the line of the orthocenters towards the considered duality. It results that, given the quadrilateral đ??´â€™đ??ľâ€™đ??śâ€™đ??ˇâ€™, the Bobillier transversals of the triangles đ??´â€™đ??śâ€™đ??ˇâ€™, đ??´â€™đ??ľâ€™đ??ˇâ€™, đ??´â€™đ??ľâ€™đ??śâ€™ and đ??śâ€™đ??ˇâ€™đ??ľâ€™ relative to the point đ?‘€ are concurrent. References [1] Florentin Smarandache, Ion Patrascu: „The Geometry of Homological Trianglesâ€?. The Education Publisher Inc., Columbus, Ohio, USA – 2012. [2] Ion Patrascu, Florentin Smarandache: „Variance on Topics of Plane Geometryâ€?. The Education Publisher Inc., Columbus, Ohio, USA – 2013.
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