Relationships Between Six Excircles
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Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 249:36 AM Jul 249:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 249:36 AM Jul 249:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 249:36 AM Jul 249:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 249:36 AM Jul 249:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point. -
Figures Circumscribing Circles Tom M
Figures Circumscribing Circles Tom M. Apostol and Mamikon A. Mnatsakanian 1. INTRODUCTION. The centroid of the boundary of an arbitrarytriangle need not be at the same point as the centroid of its interior. But we have discovered that the two centroids are always collinear with the center of the inscribed circle, at distances in the ratio 3 : 2 from the center. We thought this charming fact must surely be known, but could find no mention of it in the literature. This paper generalizes this elegant and surprising result to any polygon that circumscribes a circle (Theorem 6). A key ingredient of the proof is a link to Archimedes' striking discovery concerning the area of a circular disk [4, p. 91], which for our purposes we prefer to state as follows: Theorem 1 (Archimedes). The area of a circular disk is equal to the product of its semiperimeter and its radius. Expressed as a formula, this becomes A = Pr, (1) where A is the area, P is the perimeter, and r is the radius of the disk. First we extend (1) to a large class of plane figures circumscribing a circle that we call circumgons, defined in section 2. They include arbitrarytriangles, all regular polygons, some irregularpolygons, and other figures composed of line segments and circular arcs. Examples are shown in Figures 1 through 4. Section 3 treats circum- gonal rings, plane regions lying between two similar circumgons. These rings have a constant width that replaces the radius in the corresponding extension of (1). We also show that all rings of constant width are necessarily circumgonal rings. -
Special Isocubics in the Triangle Plane
Special Isocubics in the Triangle Plane Jean-Pierre Ehrmann and Bernard Gibert June 19, 2015 Special Isocubics in the Triangle Plane This paper is organized into five main parts : a reminder of poles and polars with respect to a cubic. • a study on central, oblique, axial isocubics i.e. invariant under a central, oblique, • axial (orthogonal) symmetry followed by a generalization with harmonic homolo- gies. a study on circular isocubics i.e. cubics passing through the circular points at • infinity. a study on equilateral isocubics i.e. cubics denoted 60 with three real distinct • K asymptotes making 60◦ angles with one another. a study on conico-pivotal isocubics i.e. such that the line through two isoconjugate • points envelopes a conic. A number of practical constructions is provided and many examples of “unusual” cubics appear. Most of these cubics (and many other) can be seen on the web-site : http://bernard.gibert.pagesperso-orange.fr where they are detailed and referenced under a catalogue number of the form Knnn. We sincerely thank Edward Brisse, Fred Lang, Wilson Stothers and Paul Yiu for their friendly support and help. Chapter 1 Preliminaries and definitions 1.1 Notations We will denote by the cubic curve with barycentric equation • K F (x,y,z) = 0 where F is a third degree homogeneous polynomial in x,y,z. Its partial derivatives ∂F ∂2F will be noted F ′ for and F ′′ for when no confusion is possible. x ∂x xy ∂x∂y Any cubic with three real distinct asymptotes making 60◦ angles with one another • will be called an equilateral cubic or a 60. -
Generalized Gergonne and Nagel Points
Generalized Gergonne and Nagel Points B. Odehnal June 16, 2009 Abstract In this paper we show that the Gergonne point G of a triangle ∆ in the Euclidean plane can in fact be seen from the more general point of view, i.e., from the viewpoint of projective geometry. So it turns out that there are up to four Gergonne points associated with ∆. The Gergonne and Nagel point are isotomic conjugates of each other and thus we find up to four Nagel points associated with a generic triangle. We reformulate the problems in a more general setting and illustrate the different appearances of Gergonne points in different affine geometries. Darboux’s cubic can also be found in the more general setting and finally a projective version of Feuerbach’s circle appears. Mathematics Subject Classification (2000): 51M04, 51M05, 51B20 Key Words: triangle, incenter, excenters, Gergonne point, Brianchon’s theorem, Nagel point, isotomic conjugate, Darboux’s cubic, Feuerbach’s nine point circle. 1 Introduction Assume ∆ = {A,B,C} is a triangle with vertices A, B, and C and the incircle i. Gergonne’s point is the locus of concurrency of the three cevians connecting the contact points of i and ∆ with the opposite vertices. The Gergonne point, which can also be labelled with X7 according to [10, 11] has frequently attracted mathematician’s interest. Some generalizations have been given. So, for example, one can replace the incircle with a concentric 1 circle and replace the contact points of the incircle and the sides of ∆ with their reflections with regard to the incenter as done in [2]. -
Cevians, Symmedians, and Excircles Cevian Cevian Triangle & Circle
10/5/2011 Cevians, Symmedians, and Excircles MA 341 – Topics in Geometry Lecture 16 Cevian A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension). B cevian C A D 05-Oct-2011 MA 341 001 2 Cevian Triangle & Circle • Pick P in the interior of ∆ABC • Draw cevians from each vertex through P to the opposite side • Gives set of three intersecting cevians AA’, BB’, and CC’ with respect to that point. • The triangle ∆A’B’C’ is known as the cevian triangle of ∆ABC with respect to P • Circumcircle of ∆A’B’C’ is known as the evian circle with respect to P. 05-Oct-2011 MA 341 001 3 1 10/5/2011 Cevian circle Cevian triangle 05-Oct-2011 MA 341 001 4 Cevians In ∆ABC examples of cevians are: medians – cevian point = G perpendicular bisectors – cevian point = O angle bisectors – cevian point = I (incenter) altitudes – cevian point = H Ceva’s Theorem deals with concurrence of any set of cevians. 05-Oct-2011 MA 341 001 5 Gergonne Point In ∆ABC find the incircle and points of tangency of incircle with sides of ∆ABC. Known as contact triangle 05-Oct-2011 MA 341 001 6 2 10/5/2011 Gergonne Point These cevians are concurrent! Why? Recall that AE=AF, BD=BF, and CD=CE Ge 05-Oct-2011 MA 341 001 7 Gergonne Point The point is called the Gergonne point, Ge. Ge 05-Oct-2011 MA 341 001 8 Gergonne Point Draw lines parallel to sides of contact triangle through Ge. -
Searching for the Center
Searching For The Center Brief Overview: This is a three-lesson unit that discovers and applies points of concurrency of a triangle. The lessons are labs used to introduce the topics of incenter, circumcenter, centroid, circumscribed circles, and inscribed circles. The lesson is intended to provide practice and verification that the incenter must be constructed in order to find a point equidistant from the sides of any triangle, a circumcenter must be constructed in order to find a point equidistant from the vertices of a triangle, and a centroid must be constructed in order to distribute mass evenly. The labs provide a way to link this knowledge so that the students will be able to recall this information a month from now, 3 months from now, and so on. An application is included in each of the three labs in order to demonstrate why, in a real life situation, a person would want to create an incenter, a circumcenter and a centroid. NCTM Content Standard/National Science Education Standard: • Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships. • Use visualization, spatial reasoning, and geometric modeling to solve problems. Grade/Level: These lessons were created as a linking/remembering device, especially for a co-taught classroom, but can be adapted or used for a regular ed, or even honors level in 9th through 12th Grade. With more modification, these lessons might be appropriate for middle school use as well. Duration/Length: Lesson #1 45 minutes Lesson #2 30 minutes Lesson #3 30 minutes Student Outcomes: Students will: • Define and differentiate between perpendicular bisector, angle bisector, segment, triangle, circle, radius, point, inscribed circle, circumscribed circle, incenter, circumcenter, and centroid. -
G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter
Regents Exam Questions Name: ________________________ G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter 1 Which geometric principle is used in the 3 In the diagram below, point B is the incenter of construction shown below? FEC, and EBR, CBD, and FB are drawn. If m∠FEC = 84 and m∠ECF = 28, determine and state m∠BRC . 1) The intersection of the angle bisectors of a 4 In the diagram below of isosceles triangle ABC, triangle is the center of the inscribed circle. AB ≅ CB and angle bisectors AD, BF, and CE are 2) The intersection of the angle bisectors of a drawn and intersect at X. triangle is the center of the circumscribed circle. 3) The intersection of the perpendicular bisectors of the sides of a triangle is the center of the inscribed circle. 4) The intersection of the perpendicular bisectors of the sides of a triangle is the center of the circumscribed circle. 2 In the diagram below of ABC, CD is the bisector of ∠BCA, AE is the bisector of ∠CAB, and BG is drawn. If m∠BAC = 50°, find m∠AXC . Which statement must be true? 1) DG = EG 2) AG = BG 3) ∠AEB ≅ ∠AEC 4) ∠DBG ≅ ∠EBG 1 Regents Exam Questions Name: ________________________ G.CO.C.10: Centroid, Orthocenter, Incenter and Circumcenter www.jmap.org 5 The diagram below shows the construction of the 9 Triangle ABC is graphed on the set of axes below. center of the circle circumscribed about ABC. What are the coordinates of the point of intersection of the medians of ABC? 1) (−1,2) 2) (−3,2) 3) (0,2) This construction represents how to find the intersection of 4) (1,2) 1) the angle bisectors of ABC 10 The vertices of the triangle in the diagram below 2) the medians to the sides of ABC are A(7,9), B(3,3), and C(11,3). -
Half-Anticevian Triangle of the Incenter
International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 c IJCDM Volume 2, 2017, pp.55-71. Received 20 February 2017. Published on-line 28 February 2017 web: http://www.journal-1.eu/ c The Author(s) This article is published with open access1. Computer Discovered Mathematics: Half-Anticevian Triangle of the Incenter Sava Grozdeva, Hiroshi Okumurab and Deko Dekovc 2 a VUZF University of Finance, Business and Entrepreneurship, Gusla Street 1, 1618 Sofia, Bulgaria e-mail: [email protected] b Department of Mathematics, Yamato University, Osaka, Japan e-mail: [email protected] cZahari Knjazheski 81, 6000 Stara Zagora, Bulgaria e-mail: [email protected] web: http://www.ddekov.eu/ Abstract. By using the computer program ”Discoverer” we study the Half- Anticevian triangle of the Incenter. Keywords. Half-Anticevian triangle, triangle geometry, notable point, computer discovered mathematics, Euclidean geometry, “Discoverer”. Mathematics Subject Classification (2010). 51-04, 68T01, 68T99. 1. Introduction Let P be an arbitrary point in the plane of triangle ABC. Denote by P aP bP c the Anticevian triangle of point P . Let Ma be the midpoint of segment AP a and define Mb and Mc similarly. We call triangle MaMbMc the Half-Anticevian Triangle of Point P . Figure 1 illustrates the definition. In figure 1, P is an arbitrary point, P aP bP c is the Anticevian triangle, and MaMbMc is the Half-Anticevian Triangle of P . Note that the Half-Anticevian Triangle of the Centroid is the Medial Triangle. We encourage the students, teachers and researchers to investigate the Half- Anticevian triangles of other notable points, e.g. -
Nagel , Speiker, Napoleon, Torricelli
Nagel , Speiker, Napoleon, Torricelli MA 341 – Topics in Geometry Lecture 17 Centroid The point of concurrency of the three medians. 07-Oct-2011 MA 341 2 Circumcenter Point of concurrency of the three perpendicular bisectors. 07-Oct-2011 MA 341 3 Orthocenter Point of concurrency of the three altitudes. 07-Oct-2011 MA 341 4 Incenter Point of concurrency of the three angle bisectors. 07-Oct-2011 MA 341 5 Symmedian Point Point of concurrency of the three symmedians. 07-Oct-2011 MA 341 6 Gergonne Point Point of concurrency of the three segments from vertices to intangency points. 07-Oct-2011 MA 341 7 Spieker Point The Spieker point is the incenter of the medial triangle. 07-Oct-2011 MA 341 8 Nine Point Circle Center The 9 point circle center is midpoint of the Euler segment. 07-Oct-2011 MA 341 9 Mittenpunkt Point The mittenpunkt of ΔABC is the symmedian point of the excentral triangle 05-Oct-2011 MA 341 001 10 Nagel Point Nagel point = point of concurrency of cevians to points of tangency of excircles 05-Oct-2011 MA 341 001 11 Nagel Point 05-Oct-2011 MA 341 001 12 The Nagel Point Ea has the unique property of being the point on the perimeter that is exactly half way around the triangle from A. = AB+BEaa AC+CE =s Then =- =- =- =- BEaa s AB s c and CE s AC s b BE sc- a = - CEa s b 07-Oct-2011 MA 341 13 The Nagel Point Likewise CE sa- b = - AEb s c AE sb- c = - BEc s a Apply Ceva’s Theorem AE BE CE sbscsa--- cab=´´=1 --- BEcab CE AE s a s b s c 07-Oct-2011 MA 341 14 The Nagel Segment 1. -
Barycentric Coordinates in Olympiad Geometry
Barycentric Coordinates in Olympiad Geometry Max Schindler∗ Evan Cheny July 13, 2012 I suppose it is tempting, if the only tool you have is a hammer, to treat everything as if it were a nail. Abstract In this paper we present a powerful computational approach to large class of olympiad geometry problems{ barycentric coordinates. We then extend this method using some of the techniques from vector computations to greatly extend the scope of this technique. Special thanks to Amir Hossein and the other olympiad moderators for helping to get this article featured: I certainly did not have such ambitious goals in mind when I first wrote this! ∗Mewto55555, Missouri. I can be contacted at igoroogenfl[email protected]. yv Enhance, SFBA. I can be reached at [email protected]. 1 Contents Title Page 1 1 Preliminaries 4 1.1 Advantages of barycentric coordinates . .4 1.2 Notations and Conventions . .5 1.3 How to Use this Article . .5 2 The Basics 6 2.1 The Coordinates . .6 2.2 Lines . .6 2.2.1 The Equation of a Line . .6 2.2.2 Ceva and Menelaus . .7 2.3 Special points in barycentric coordinates . .7 3 Standard Strategies 9 3.1 EFFT: Perpendicular Lines . .9 3.2 Distance Formula . 11 3.3 Circles . 11 3.3.1 Equation of the Circle . 11 4 Trickier Tactics 12 4.1 Areas and Lines . 12 4.2 Non-normalized Coordinates . 13 4.3 O, H, and Strong EFFT . 13 4.4 Conway's Formula . 14 4.5 A Few Final Lemmas . 15 5 Example Problems 16 5.1 USAMO 2001/2 . -
Conic Construction of a Triangle from Its Incenter, Nine-Point Center, and a Vertex
Conic Construction of a Triangle From Its Incenter, Nine-point Center, and a Vertex Paul Yiu Abstract. We construct a triangle given its incenter, nine-point center and a vertex by locating the circumcenter as an intersection of two rectangular hyper- bolas. Some special configurations leading to solutions constructible with ruler and compass are studied. The related problem of construction of a triangle given its circumcenter, incenter, and one vertex is revisited, and it is established that such a triangle exists if and only if the incenter lies inside the cardioid relative to the circumcircle. 1. Introduction In this note we solve the construction problem of a triangle ABC given its in- center I, its nine-point center N, and one vertex A. This is Problem 35 in Harold Connelly’s list of construction problems [1]. We shall show that the triangle is in general not constructible with ruler and compass, but can be easily obtained by in- tersecting conics. Some special configurations of I, N, A from which the triangle can be constructible with ruler and compass are studied in details in 2, 6 – 8. §§ 2. The case of collinear I, N, A In this case, the triangle ABC is isosceles, with AB = AC. The incircle and the nine-point circle are tangent to each other internally at the midpoint of the base BC. We set up a rectangular coordinate system with N at the origin, I at ( d, 0), and A at (a, 0) for a > 0, and seek the inradius r. The circumradius is − A r R = 2(d + r). -
Lemmas in Euclidean Geometry1 Yufei Zhao [email protected]
IMO Training 2007 Lemmas in Euclidean Geometry Yufei Zhao Lemmas in Euclidean Geometry1 Yufei Zhao [email protected] 1. Construction of the symmedian. Let ABC be a triangle and Γ its circumcircle. Let the tangent to Γ at B and C meet at D. Then AD coincides with a symmedian of △ABC. (The symmedian is the reflection of the median across the angle bisector, all through the same vertex.) A A A O B C M B B F C E M' C P D D Q D We give three proofs. The first proof is a straightforward computation using Sine Law. The second proof uses similar triangles. The third proof uses projective geometry. First proof. Let the reflection of AD across the angle bisector of ∠BAC meet BC at M ′. Then ′ ′ sin ∠BAM ′ ′ BM AM sin ∠ABC sin ∠BAM sin ∠ABD sin ∠CAD sin ∠ABD CD AD ′ = ′ sin ∠CAM ′ = ∠ ∠ ′ = ∠ ∠ = = 1 M C AM sin ∠ACB sin ACD sin CAM sin ACD sin BAD AD BD Therefore, AM ′ is the median, and thus AD is the symmedian. Second proof. Let O be the circumcenter of ABC and let ω be the circle centered at D with radius DB. Let lines AB and AC meet ω at P and Q, respectively. Since ∠PBQ = ∠BQC + ∠BAC = 1 ∠ ∠ ◦ 2 ( BDC + DOC) = 90 , we see that PQ is a diameter of ω and hence passes through D. Since ∠ABC = ∠AQP and ∠ACB = ∠AP Q, we see that triangles ABC and AQP are similar. If M is the midpoint of BC, noting that D is the midpoint of QP , the similarity implies that ∠BAM = ∠QAD, from which the result follows.