DOCSLIB.ORG

Mixed Intersection of Cevians and Perspective Triangles

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Mixed Intersection of Cevians and Perspective Triangles 15TH INTERNATIONAL CONFERENCE ON GEOMETRY AND GRAPHICS ©2012 ISGG 1–5 AUGUST, 2012, MONTREAL, CANADA MIXED INTERSECTION OF CEVIANS AND PERSPECTIVE TRIANGLES Boris ODEHNAL Dresden University of Technology, Germany ABSTRACT: Any point P in the plane of a triangle ∆ = (A,B,C) can be considered as the intersection of certain Cevians, whether P is a triangle center or not. Taking two centers, say Y and Z, we find six Cevians with a total of 11 points of intersection, among them ∆’s vertices and Y and Z. The remaining six points form two triangles ∆1 and ∆2 which are both perspective to the base triangle ∆ and to each other. The three centers of perspectivity are triangle centers of ∆ and collinear, independent of the choice of Y and Z. Furthermore any pair out of ∆, ∆1, and ∆2 has the same perspectrix which yields a closed chain of Desargues’ (103,103) configurations. Then special affine appearances of Desargues’ configurations can be obtained by a suitable choice of Y and Z. Any choice of a fixed center Y leads to exactly one center Z as the fourth point of intersection of two conic sections circumscribed to ∆ such that the perspectors P1, P2, and consequently P12 are points at infinity. For fixed center Y we obtain a quadratic transformation in ∆’s plane which transforms central lines to central conic sections. Keywords: triangle, Cevian, perspective triangles, Desargues configuration, triangle center, cross- point 1. INTRODUCTION C Many triangle centers appear as the intersection of certain Cevians. For example the incenter X1 X4 is the intersection of the interior angle bisectors h of a triangle ∆ with vertices A, B, C. The cen- B m troid X2 can be found as the common point of hA X2 B the medians, whereas the circumcenter X3 is the mA C C meet of the bisectors of ∆’s edges. The ortho- m h center X4 comes along as the intersection of the A B altitudes of ∆. Here and in the following, centers Figure 1: Is it a mistake? of ∆ are labelled according to C. Kimberling’s list, cf. [4, 6]. What happens if one makes the erroneous centers, say Y and Z, and intersect the respective construction by mixing Cevians of different cen- Cevians we obtain, besides the two centers Y and ters when intersecting them? Figure 1 shows Z, the three vertices of ∆. Further we find six such a mistake: The mixed intersection of me- points which can be arranged in many ways in ∆ dians and altitudes in produces a triangle. two perspective triangles ∆1 and ∆2. Usually three arbitrarily chosen Cevians will In Section 2 we shall show that these triangles not meet in a common point, except maybe in are perspective to each other with perspector P12. cases where the side lengths of ∆ fulfil certain They are also perspective to ∆ with perspectors relations. However, if we start with two triangle P1 and P2, cf. Figure 6. All the three perspec- Paper #89 tors are triangle centers of ∆, provided Y and Z b := CA, and c := AB denote the side lengths are centers of ∆. Moreover, the perspectors P1, of ∆. We denote the line joining two points P P2, and P12 are collinear and consequently the and Q by [P,Q]. A point Z = (z0 : z1 : z2) is perspectrix is the same for any pair of triangles a center exactly if z0 = f (a,b,c) is a homoge- out of the three. So we obtain a closed chain neous function of the side lengths a, b, and c of three Desarguesian (103,103) configurations. of the base triangle ∆ with z1 = f (b,c,a) and This will be discussed in Section 4. An Exam- z2 = f (c,a,b). Later we use the symbol ζ in or- ple is displayed in Figure 6. Special affine ver- der to indicate that a homogeneous function f is sions of (103,103) configurations are described transformed via the cyclic shift: If f = f (a,b,c), in Section 5. In Section 3 we show that the per- then f ζ = f (b,c,a). Thus a triangle center X = spector P is the crosspoint of Y and Z. This ξ ξ ξ ξ ξ ζ 12 ( 0 : 1 : 2) is characterized by i+1 = i for gives a new access to the crosspoints of triangle i ∈{0,1,2} and i is counted modulo 3. Simi- centers. larly a line in ∆’s plane is called a central line if its homogeneous trilinear coordinates follow C the same rules as those of centers. Note that ζ applies to any cyclically ordered triplet. 2.PERSPECTIVE TRIANGLES AND A1 THEIR PERSPECTORS B1 Assume Y =(ξ0 : ξ1 : ξ2) and Z =(η0 : η1 : η2) ∆ C2 ∆ ∆1 are triangle centers of . Thenwelookatthefol- lowing points of intersection of Cevians through Y Y and Z: Z C1 ∆2 A1:=[C,Y] ∩ [B,Z],A2:=[B,Y] ∩ [C,Z], B A2 2 B1:=[A,Y] ∩ [C,Z],B2:=[C,Y] ∩ [A,Z], (1) A B C1:=[B,Y] ∩ [A,Z],C2:=[A,Y] ∩ [B,Z]. ∆ ∆ Figure 2: The triangles 1 and 2 whose vertices We define two triangles collecting the intersec- are the mixed intersections of Cevians of Y and tion points of wrong pairs of Cevians by letting Z. ∆1 =(A1,B1,C1) and ∆2 =(A2,B2,C2) as illus- trated in Figure 2. If we fix Y then the mapping q : Z → P12 is No we can show: quadratic and sends centers to centers and cen- tral lines are mapped to central conics. The map- Theorem 2.1. The triangles ∆1 and ∆2 are per- ping q is birational, i.e., its inverse is also ra- spective to the base triangle ∆. The perspectors tional. It turns out that q is a composition of P1 and P2 are triangle centers of ∆. the isogonal conjugation with a collineation. We pay our attention to q in Section 6. Proof. With the above prerequisites we find the In the following we use homogeneous trilin- vertices of ∆1 and ∆2 as ear coordinates (p0 : p1 : p2) in order to repre- sent points in the plane of ∆. The vertices of A1 =(ξ0η0 : ξ1η0 : ξ0η2), ∆ are the base points and thus their coordinate ξ η ξ η ξ η vectors are A =(1:0:0), B =(0:1:0), and B1 =( 1 0 : 1 1 : 2 1), (2) C =(0:0:1), cf. [4, 6]. Further we let a := BC, C1 =(ξ0η2 : ξ2η1 : ξ2η2), 2 C can be found as intersection of [A,Ai] and [B,Bi] and we arrive at P1 =(ξ0ξ1η0η2 : ξ1ξ2η1η0 : ξ2ξ0η2η1) P12 = X6 and (4) P2 P2 =(ξ0ξ2η0η1 : ξ1ξ0η1η2 : ξ2ξ1η2η0). X4 P1 and P2 are centers for ξi and ηi (with i ∈ X ∆ ∆ 2 2 {0,1,2}) are center functions, i.e., they are ho- mogeneous and cyclic symmetric in a, b, and c. ∆1 ζ For example (ξ0ξ1η0η2) = ξ1ξ2η1η0 and like- P1 wise for all the other coordinate functions of P1 and P2, respectively. A B Obviously P1 and P2 are triangle centers for Figure 3: The three perspectors P1, P2, and P12 any choice of centers Y and Z. It can be seen at constructed out of the centroid Y = X2 and the once that Y = Z results in P1 = P2. Furthermore ∆ orthocenter Z = X4. if Y = X1, i.e., the incenter of and Z is an arbi- η η C trary center not equal to X1, then P1 =( 0 2 : η1η0 : η2η0) and P2 = (η0η1 : η1η2 : η2η0). The mappings Z 7→ P and Z 7→ P are birational P 1 2 2 for they are compositions of the isogonal conju- P12 gation (x0 : x1 : x2) 7→ (x1x2 : x2x0 : x0x1) with collineations in ∆’s plane with X1 for a fixed ∆2 point and shifting ∆’s vertices in clockwise or X 28 counter clockwise direction, respectively. X 3 It is easy to show that the following holds: ∆ ∆ ∆1 Theorem 2.2. The triangles 1 and 2 are per- ∆ P 1 spective. The perspector P12 is a triangle center that is collinear with P1 and P2 from Theorem A B 2.1, except Y = Z, where P1 = P2 = P12 = Y = Z. Proof. The lines [A ,A ], [B ,B ], and [C ,C ] Figure 4: The three perspectors P , P , and P 1 2 1 2 1 2 1 2 12 are concurrent. In order to prove this, we com- constructed out of the circumcenter Y = X and 3 pute the trilinear coordinates of the three lines the center Z = X . 28 and show that they are linearly dependent. The point of concurrency can be computed as and the intersection of any pair of the above given ξ η ξ η ξ η A2 =( 0 0 : 0 1 : 2 0), lines which gives B2 =(ξ0η1 : ξ1η1 : ξ1η2), (3) P12 =(ξ0η0(ξ2η1 + ξ1η2) : C2 =(ξ2η0 : ξ1η2 : ξ2η2). ξ1η1(ξ0η2 + ξ2η0) : (5) Then we show that the lines [A1,A2], [B1,B2], ξ2η2(ξ1η0 + ξ0η1)), and [C1,C2] are concurrent by computing their trilinears and showing the linear dependency. which is obviously a triangle center of ∆ for the The perspectors Pi of ∆ and ∆i (with i ∈{1,2}) coordinate function i + 1 is the ζ -image of the 3 coordinate function i, for i ∈{0,1,2} and i is of P and Q, respectively. Then define ∆3 = counted modulo 3. (A′′′,B′′′,C′′′) by letting The three perspectors P1, P2, and P12 are A′′′ :=[A,A′′] ∩ [B′,C′], collinear, for their trilinear coordinate vectors ′′′ ′′ ′ ′ are linearly dependent. B :=[B,B ] ∩ [C ,A ], C′′′ :=[C,C′′] ∩ [A′,B′]. If Y = Z, then P1 = P2 = Z as outlined above.
Load more

Recommended publications
  • Self-Inverse Gemini Triangles
    INTERNATIONAL JOURNAL OF GEOMETRY Vol. 9 (2020), No. 1, 25 - 39 SELF-INVERSE GEMINI TRIANGLES CLARK KIMBERLING and PETER MOSES Abstract. In the plane of a triangle ABC, every triangle center U = u : v : w (barycentric coordinates) is associated with a central triangle having A-vertex AU = −u : v + w : v + w. The triangle AU BU CU is a self-inverse Gemini triangle. Let mU (X) be the image of a point X under the collineation that maps A; B; C; G respectively onto AU ;BU ;CU ;G, where G is the centroid of ABC. Then mU (mU (X)) = X: Properties of the self-inverse mapping mU are presented, with attention to associated conics (e.g., Jerabek, Kiepert, Feuerbach, Nagel, Steiner), as well as cubics of the types pK(Y; Y ) and pK(U ∗ Y; Y ). 1. Introduction The term Gemini triangle, introduced in the Encyclopedia of Triangle Centers (ETC [7], just before X(24537)), applies to certain triangles defined by barycentric coordinates. In keeping with the meaning of gemini, these triangles tend to occur in pairs. Specifically, for a given reference triangle ABC with sidelengths a; b; c, a Germini triangle A0B0C0 is a cen- tral triangle ([6], pp. 53-56) with A-vertex given by (1.1) A0 = f(a; b; c): g(b; c; a): g(b; a; c); where f and g are center functions ([6], p. 46). Because A0B0C0 is a central triangle, the vertices B0 and C0 can be read from (1.1) as B0 = g(c; b; a): f(b; c; a): g(c; a; b) C0 = g(a; b; c): g(a; c; b): f(c; a; b): Now suppose that U = u : v : w is a triangle center, and let 0 −u v + w v + w 1 MU = @ w + u −v w + u A : u + v u + v −w Keywords and phrases Triangle, Gemini triangle, barycentric coordinates, collineation, conic, cubic, circumcircle, Jerabek, Kiepert, Feuerbach, Nagel, Steiner (2010)Mathematics Subject Classification: 51N15, 51N20 Received: 13.10.2019.
    [Show full text]
  • A Geometric Proof of the Siebeck–Marden Theorem Beniamin Bogosel
    A Geometric Proof of the Siebeck–Marden Theorem Beniamin Bogosel Abstract. The Siebeck–Marden theorem relates the roots of a third degree polynomial and the roots of its derivative in a geometrical way. A few geometric arguments imply that every inellipse for a triangle is uniquely related to a certain logarithmic potential via its focal points. This fact provides a new direct proof of a general form of the result of Siebeck and Marden. Given three noncollinear points a, b, c ∈ C, we can consider the cubic polynomial P(z) = (z − a)(z − b)(z − c), whose derivative P (z) has two roots f1, f2. The Gauss–Lucas theorem is a well-known result which states that given a polynomial Q with roots z1,...,zn, the roots of its derivative Q are in the convex hull of z1,...,zn. In the simple case where we have only three roots, there is a more precise result. The roots f1, f2 of the derivative polynomial are situated in the interior of the triangle abc and they have an interesting geometric property: f1 and f2 are the focal points of the unique ellipse that is tangent to the sides of the triangle abc at its midpoints. This ellipse is called the Steiner inellipse associated to the triangle abc. In the rest of this note, we use the term inellipse to denote an ellipse situated in a triangle that is tangent to all three of its sides. This geometric connection between the roots of P and the roots of P was first observed by Siebeck (1864) [12] and was reproved by Marden (1945) [8].
    [Show full text]
  • An Area Inequality for Ellipses Inscribed in Quadrilaterals
    1 Introduction It is well known(see [CH], [K], or [MP]) that there is a unique ellipse in- scribed in a given triangle, T , tangent to the sides of T at their respective midpoints. This is often called the midpoint or Steiner inellipse, and it can be characterized as the inscribed ellipse having maximum area. In addition, one has the following inequality. If E denotes any ellipse inscribed in T , then Area (E) π , (trineq) Area (T ) ≤ 3√3 with equality if and only if E is the midpoint ellipse. In [MP] the authors also discuss a connection between the Steiner ellipse and the orthogonal least squares line for the vertices of T . Definition: The line, £, is called a line of best fit for n given points z1, ..., zn n 2 in C, if £ minimizes d (zj ,l) among all lines l in the plane. Here d (zj,l) j=1 denotes the distance(Euclidean)P from zj to l. In [MP] the authors prove the following results, some of which are also proven in 1 n Theorem A: Suppose zj are points in C,g = zj is the centroid, and n j=1 n n 2 2 2 P Z = (zj g) = zj ng . j=1 − j=1 − (a)P If Z = 0, thenP every line through g is a line of best fit for the points z1, ..., zn. (b) If Z = 0, then the line, £, thru g that is parallel to the vector from 6 (0, 0) to √Z is the unique line of best fit for z1, ..., zn.
    [Show full text]
  • Coincidence of Centers for Scalene Triangles
    Forum Geometricorum b Volume 7 (2007) 137–146. bbb FORUM GEOM ISSN 1534-1178 Coincidence of Centers for Scalene Triangles Sadi Abu-Saymeh and Mowaffaq Hajja Abstract.Acenter function is a function Z that assigns to every triangle T in a Euclidean plane E a point Z(T ) in E in a manner that is symmetric and that respects isometries and dilations. A family F of center functions is said to be complete if for every scalene triangle ABC and every point P in its plane, there is Z∈F such that Z(ABC)=P . It is said to be separating if no two center functions in F coincide for any scalene triangle. In this note, we give simple examples of complete separating families of continuous triangle center functions. Regarding the impression that no two different center functions can coincide on a scalene triangle, we show that for every center function Z and every scalene triangle T , there is another center function Z, of a simple type, such that Z(T )=Z(T ). 1. Introduction Exercise 1 of [33, p. 37] states that if any two of the four classical centers coin- cide for a triangle, then it is equilateral. This can be seen by proving each of the 6 substatements involved, as is done for example in [26, pp. 78–79], and it also follows from more interesting considerations as described in Remark 5 below. The statement is still true if one adds the Gergonne, the Nagel, and the Fermat-Torricelli centers to the list. Here again, one proves each of the relevant 21 substatements; see [15], where variants of these 21 substatements are proved.
    [Show full text]
  • Some Properties of Rectangle and a Random Point
    SOME PROPERTIES OF RECTANGLE AND A RANDOM POINT QUANG HUNG TRAN Abstract. We establish a relationship between the two important central lines of the triangle, the Euler line and the Brocard axis, in a configuration with an arbitrary rectangle and a random point. The classical Cartesian coordinate system method shows its strength in these theorems. Along with that, some related problems on rectangles and a random point are proposed with similar solutions using Cartesian coordinate system. 1. Introduction In the geometry of triangles, the Euler line [8] plays an important role and is almost the most classic concept in this field. The Euler line passes through the centroid, orthocenter, and circumcenter of the triangle. The symmedian point [9] of a triangle is the concurrent point of its symmedian lines, the Brocard axis [10] is the line connecting the circumcircle and the symmedian point of that triangle. The Brocard axis is also a central line that plays an important role in the geometry of triangle. The coordinate method was invented by Ren´eDescartes from the 17th century [1,2], up to now, the classical coordinate method of Descartes is still one of the most important method of mathematics and geometry. In this paper, we apply Cartesian coordinate system to solve an interesting the- orem for an arbitrary rectangle and a random point in which two important central lines in the triangle are mentioned as follows. The idea of using the Cartesian coor- dinate system is also used in a similar way to solve the new problems we introduced in Section 3.
    [Show full text]
  • Bicentric Pairs of Points and Related Triangle Centers
    Forum Geometricorum b Volume 3 (2003) 35–47. bbb FORUM GEOM ISSN 1534-1178 Bicentric Pairs of Points and Related Triangle Centers Clark Kimberling Abstract. Bicentric pairs of points in the plane of triangle ABC occur in con- nection with three configurations: (1) cevian traces of a triangle center; (2) points of intersection of a central line and central circumconic; and (3) vertex-products of bicentric triangles. These bicentric pairs are formulated using trilinear coordi- nates. Various binary operations, when applied to bicentric pairs, yield triangle centers. 1. Introduction Much of modern triangle geometry is carried out in in one or the other of two homogeneous coordinate systems: barycentric and trilinear. Definitions of triangle center, central line, and bicentric pair, given in [2] in terms of trilinears, carry over readily to barycentric definitions and representations. In this paper, we choose to work in trilinears, except as otherwise noted. Definitions of triangle center (or simply center) and bicentric pair will now be briefly summarized. A triangle center is a point (as defined in [2] as a function of variables a, b, c that are sidelengths of a triangle) of the form f(a, b, c):f(b, c, a):f(c, a, b), where f is homogeneous in a, b, c, and |f(a, c, b)| = |f(a, b, c)|. (1) If a point satisfies the other defining conditions but (1) fails, then the points Fab := f(a, b, c):f(b, c, a):f(c, a, b), Fac := f(a, c, b):f(b, a, c):f(c, b, a) (2) are a bicentric pair.
    [Show full text]
  • Permutation Ellipses
    Journal for Geometry and Graphics Volume 24 (2020), No. 2, 233–247 Permutation Ellipses Clark Kimberling1, Peter J. C. Moses2 1Department of Mathematics, University of Evansville, Evansville, Indiana 47722, USA [email protected] 2Engineering Division, Moparmatic Co., Astwood Bank, Nr. Redditch, Worcestershire, B96 6DT, UK [email protected] Abstract. We use homogeneous coordinates in the plane of a triangle to define a family of ellipses having the centroid of the triangle as center. The family, which includes the Steiner circumscribed and inscribed ellipses, is closed under many operations, including permutation of coordinates, complements and anticomple- ments, duality, and inversion. Key Words: barycentric coordinates, Steiner circumellipse, Steiner inellipse, com- plement and anticomplement, dual conic, inversion in ellipse, Frégier ellipse MSC 2020: 51N20, 51N15 1 Introduction In 1827, something revolutionary happened in triangle geometry. A descendant of Martin Luther named Möbius introduced barycentric coordinates. Suddenly, points and lines in the plane of a triangle ABC with sidelengths a, b, c had a new kind of representation, and traditional geometric relationships found new algebraic kinds of expression. The lengths a, b, c came to be regarded as variables or indeterminates, and a criterion for collinearity of three points became the zeroness of a determinant, as did the concurrence of three lines. More recently, algebraic concepts have been imposed on the objects of triangle geometry. You can make up a suitable function f(a, b, c) and say “Let P be the point with barycentric coordinates f(a, b, c): f(b, c, a): f(c, a, b)” and then use algebra to find geometrically notable properties of P .
    [Show full text]
  • Steiner's Hat: a Constant-Area Deltoid Associated with the Ellipse
    KoG•24–2020 R. Garcia, D. Reznik, H. Stachel, M. Helman: Steiner’s Hat: a Constant-Area Deltoid ... https://doi.org/10.31896/k.24.2 RONALDO GARCIA, DAN REZNIK Original scientific paper HELLMUTH STACHEL, MARK HELMAN Accepted 5. 10. 2020. Steiner's Hat: a Constant-Area Deltoid Associated with the Ellipse Steiner's Hat: a Constant-Area Deltoid Associ- Steinerova krivulja: deltoide konstantne povrˇsine ated with the Ellipse pridruˇzene elipsi ABSTRACT SAZETAKˇ The Negative Pedal Curve (NPC) of the Ellipse with re- Negativno noˇziˇsnakrivulja elipse s obzirom na neku nje- spect to a boundary point M is a 3-cusp closed-curve which zinu toˇcku M je zatvorena krivulja s tri ˇsiljka koja je afina is the affine image of the Steiner Deltoid. Over all M the slika Steinerove deltoide. Za sve toˇcke M na elipsi krivulje family has invariant area and displays an array of interesting dobivene familije imaju istu povrˇsinui niz zanimljivih svoj- properties. stava. Key words: curve, envelope, ellipse, pedal, evolute, deltoid, Poncelet, osculating, orthologic Kljuˇcnerijeˇci: krivulja, envelopa, elipsa, noˇziˇsnakrivulja, evoluta, deltoida, Poncelet, oskulacija, ortologija MSC2010: 51M04 51N20 65D18 1 Introduction Main Results: 0 0 Given an ellipse E with non-zero semi-axes a;b centered • The triangle T defined by the 3 cusps Pi has invariant at O, let M be a point in the plane. The Negative Pedal area over M, Figure 7. Curve (NPC) of E with respect to M is the envelope of • The triangle T defined by the pre-images Pi of the 3 lines passing through points P(t) on the boundary of E cusps has invariant area over M, Figure 7.
    [Show full text]
  • An Elementary Proof of Marden's Theorem
    An Elementary Proof of Marden’s Theorem Dan Kalman What I call Marden’s Theorem is one of my favorite results in mathematics. It es- tablishes a fantastic relation between the geometry of plane figures and the relative positions of roots of a polynomial and its derivative. Although it has a much more general statement, here is the version that I like best: Marden’s Theorem. Let p(z) be a third-degree polynomial with complex coefficients, and whose roots z1,z2, and z3 are noncollinear points in the complex plane. Let T be the triangle with vertices at z1,z2, and z3. There is a unique ellipse inscribed in T and tangent to the sides at their midpoints. The foci of this ellipse are the roots of p(z). I call this Marden’s Theorem because I first read it in M. Marden’s wonderful book [6]. But this material appeared previously in Marden’s earlier paper [5]. In both sources Marden attributes the theorem to Siebeck, citing a paper from 1864 [8]. In- deed, Marden reports appearances of various versions of the theorem in nine papers spanning the period from 1864 to 1928. Of particular interest in what follows below is an 1892 paper by Maxime Bocherˆ [1]. In his presentation Marden states the theorem in a more general form than given m1 above, corresponding to the logarithmic derivative of a product (z − z1) (z − m2 m3 z2) (z − z3) where the only restriction on the exponents m j is that they be nonzero, and with a general conic section taking the place of the ellipse.
    [Show full text]
  • 104.08 Finding Triangles with Given Circum-Medial Triangle
    The Mathematical Gazette March 2020 104.08 Finding triangles with given circum-medial triangle Let △A1B1C1 be an arbitrary triangle and P an arbitrary point not on its circumcircle Σ. The ‘other’ intersection points of the ‘cevians’ A1PB,,1P C1P with Σ, the points AB,,,C make the so-called circumcevian triangle of △A1B1C1 with respect to PP. If is the centroid of △A1B1C1 then △ABC is called * the circum-medial triangle of △A1B1C1 (see Figure 1: P = centroid G1). C1 A B1 B G1 A1 C FIGURE 1: How to reverse the process of constructing the circum-medial triangle? A recent paper [1] was concerned with iterating the construction of the circum-medial triangle (it turned out that the shape of the triangles converges to equilaterality). Here, we reverse the process. Specifically, given △ABC we look for △A1B1C1 such that △ABC is the circum-medial triangle of △A1B1C1. It turns out that there are in general two solutions, namely, the circumcevian triangles of △ABC with respect to the foci of the Steiner inellipse of △ABC. Theorem 1: Given △ABCGand a point 1 not on its circumcircle Σ, let △A1B1C1 be its circumcevian triangle with respect to G1. Then G1 is the centroid of △A1B1C1 if, and only if, it is a focus of the Steiner inellipse of △ABC. Remark: Since an ellipse has two foci, this yields the two mentioned solutions (see Figure 2), except in the case where △ABC is equilateral, when * Unfortunately, not every language has such short and precise terms for the corresponding triangles. For instance in German there is no such term, one would have to describe the whole process with many words.
    [Show full text]
  • Dao Thanh Oai, Generalizations of Some Famous Classical Euclidean
    International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 c IJCDM September 2016, Volume 1, No.3, pp.13-20. Received 15 July 2016. Published on-line 22 July 2016 web: http://www.journal-1.eu/ c The Author(s) This article is published with open access1. Generalizations of some famous classical Euclidean geometry theorems Dao Thanh Oai Cao Mai Doai, Quang Trung, Kien Xuong, Thai Binh, Vietnam e-mail: [email protected] Abstract. In this note, we introduce generalizations of some famous classical Euclidean geometry theorems. We use problems in order to state the theorems. Keywords. Euclidean geometry, Triangle geometry. Mathematics Subject Classification (2010). 51-04, 68T01, 68T99. Problem 1 ([1],[2], A generalization the Lester circle theorem associated with the Neuberg cubic). Let P be a point on the Neuberg cubic. Let PA be the reflection of P in line BC, and define PB and PC cyclically. It is known that the lines APA, BPB, CPC concur. Let Q(P ) be the point of concurrence. Then two Fermat points, P , Q(P ) lie on a circle. Remark: Let P is the circumcenter, it is well-know that Q(P ) is the Nine point center, the conjucture becomes Lester circle theorem, you can see the Lester circle theorem in([3],[4]). Problem 2 ([5],[6], A generalization of the Parry circle theorem associated with two isogonal conjugate points). Let a rectangular circumhyperbola of ABC, let L be the isogonal conjugate line of the hyperbola. The tangent line to the hyperbola at X(4) meets L at point K.
    [Show full text]
  • An Area Inequality for Ellipses Inscribed in Quadrilaterals
    Journal of Mathematical Inequalities Volume 4, Number 3 (2010), 431–443 AN AREA INEQUALITY FOR ELLIPSES INSCRIBED IN QUADRILATERALS ALAN HORWITZ (Communicated by S. S. Gomis) Area(E) Abstract. If E is any ellipse inscribed in a convex quadrilateral, –D, then we prove that Area(–D) π , and equality holds if and only if –D is a parallelogram and E is tangent to the sides of –D 4 at the midpoints. We also prove that the foci of the unique ellipse of maximal area inscribed in a parallelogram, –D, lie on the orthogonal least squares line for the vertices of –D. This does not hold in general for convex quadrilaterals. 1. Introduction It is well known (see [1], [4], or [5]) that there is a unique ellipse inscribed in a given triangle, T , tangent to the sides of T at their respective midpoints. This is often called the midpoint or Steiner inellipse, and it can be characterized as the inscribed ellipse having maximum area. In addition, one has the following inequality. If E denotes any ellipse inscribed in T ,then Area(E) π √ , (1) Area(T) 3 3 with equality if and only if E is the midpoint ellipse. In [5] the authors also discuss a connection between the Steiner ellipse and the orthogonal least squares line for the vertices of T . DEFINITION 1. The line, £, is called a line of best fitforn given points z1,...,zn n 2 in C,if£ minimizes ∑ d (z j,l) among all lines l in the plane. Here d (z j,l) denotes j=1 the distance(Euclidean) from z j to l .
    [Show full text]