CHARMING PROOFS a Journey Into Elegant Mathematics
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Investigating Centers of Triangles: the Fermat Point
Investigating Centers of Triangles: The Fermat Point A thesis submitted to the Miami University Honors Program in partial fulfillment of the requirements for University Honors with Distinction by Katherine Elizabeth Strauss May 2011 Oxford, Ohio ABSTRACT INVESTIGATING CENTERS OF TRIANGLES: THE FERMAT POINT By Katherine Elizabeth Strauss Somewhere along their journey through their math classes, many students develop a fear of mathematics. They begin to view their math courses as the study of tricks and often seemingly unsolvable puzzles. There is a demand for teachers to make mathematics more useful and believable by providing their students with problems applicable to life outside of the classroom with the intention of building upon the mathematics content taught in the classroom. This paper discusses how to integrate one specific problem, involving the Fermat Point, into a high school geometry curriculum. It also calls educators to integrate interesting and challenging problems into the mathematics classes they teach. In doing so, a teacher may show their students how to apply the mathematics skills taught in the classroom to solve problems that, at first, may not seem directly applicable to mathematics. The purpose of this paper is to inspire other educators to pursue similar problems and investigations in the classroom in order to help students view mathematics through a more useful lens. After a discussion of the Fermat Point, this paper takes the reader on a brief tour of other useful centers of a triangle to provide future researchers and educators a starting point in order to create relevant problems for their students. iii iv Acknowledgements First of all, thank you to my advisor, Dr. -
Properties of Equidiagonal Quadrilaterals (2014)
Forum Geometricorum Volume 14 (2014) 129–144. FORUM GEOM ISSN 1534-1178 Properties of Equidiagonal Quadrilaterals Martin Josefsson Abstract. We prove eight necessary and sufficient conditions for a convex quadri- lateral to have congruent diagonals, and one dual connection between equidiag- onal and orthodiagonal quadrilaterals. Quadrilaterals with both congruent and perpendicular diagonals are also discussed, including a proposal for what they may be called and how to calculate their area in several ways. Finally we derive a cubic equation for calculating the lengths of the congruent diagonals. 1. Introduction One class of quadrilaterals that have received little interest in the geometrical literature are the equidiagonal quadrilaterals. They are defined to be quadrilat- erals with congruent diagonals. Three well known special cases of them are the isosceles trapezoid, the rectangle and the square, but there are other as well. Fur- thermore, there exists many equidiagonal quadrilaterals that besides congruent di- agonals have no special properties. Take any convex quadrilateral ABCD and move the vertex D along the line BD into a position D such that AC = BD. Then ABCD is an equidiagonal quadrilateral (see Figure 1). C D D A B Figure 1. An equidiagonal quadrilateral ABCD Before we begin to study equidiagonal quadrilaterals, let us define our notations. In a convex quadrilateral ABCD, the sides are labeled a = AB, b = BC, c = CD and d = DA, and the diagonals are p = AC and q = BD. We use θ for the angle between the diagonals. The line segments connecting the midpoints of opposite sides of a quadrilateral are called the bimedians and are denoted m and n, where m connects the midpoints of the sides a and c. -
On the Fermat Point of a Triangle Jakob Krarup, Kees Roos
280 NAW 5/18 nr. 4 december 2017 On the Fermat point of a triangle Jakob Krarup, Kees Roos Jakob Krarup Kees Roos Department of Computer Science Faculty of EEMCS University of Copenhagen, Denmark Technical University Delft [email protected] [email protected] Research On the Fermat point of a triangle For a given triangle 9ABC, Pierre de Fermat posed around 1640 the problem of finding a lem has been solved but it is not known by 2 point P minimizing the sum sP of the Euclidean distances from P to the vertices A, B, C. whom. As will be made clear in the sec- Based on geometrical arguments this problem was first solved by Torricelli shortly after, tion ‘Duality of (F) and (D)’, Torricelli’s ap- by Simpson in 1750, and by several others. Steeped in modern optimization techniques, proach already reveals that the problems notably duality, however, Jakob Krarup and Kees Roos show that the problem admits a (F) and (D) are dual to each other. straightforward solution. Using Simpson’s construction they furthermore derive a formula Our first aim is to show that application expressing sP in terms of the given triangle. This formula appears to reveal a simple re- of duality in conic optimization yields the lationship between the area of 9ABC and the areas of the two equilateral triangles that solution of (F), straightforwardly. Devel- occur in the so-called Napoleon’s Theorem. oped in the 1990s, the field of conic optimi- zation (CO) is a generalization of the more We deal with two problems in planar geo- Problem (D) was posed in 1755 by well-known field of linear optimization (LO). -
Shape Skeletons Creating Polyhedra with Straws
Shape Skeletons Creating Polyhedra with Straws Topics: 3-Dimensional Shapes, Regular Solids, Geometry Materials List Drinking straws or stir straws, cut in Use simple materials to investigate regular or advanced 3-dimensional shapes. half Fun to create, these shapes make wonderful showpieces and learning tools! Paperclips to use with the drinking Assembly straws or chenille 1. Choose which shape to construct. Note: the 4-sided tetrahedron, 8-sided stems to use with octahedron, and 20-sided icosahedron have triangular faces and will form sturdier the stir straws skeletal shapes. The 6-sided cube with square faces and the 12-sided Scissors dodecahedron with pentagonal faces will be less sturdy. See the Taking it Appropriate tool for Further section. cutting the wire in the chenille stems, Platonic Solids if used This activity can be used to teach: Common Core Math Tetrahedron Cube Octahedron Dodecahedron Icosahedron Standards: Angles and volume Polyhedron Faces Shape of Face Edges Vertices and measurement Tetrahedron 4 Triangles 6 4 (Measurement & Cube 6 Squares 12 8 Data, Grade 4, 5, 6, & Octahedron 8 Triangles 12 6 7; Grade 5, 3, 4, & 5) Dodecahedron 12 Pentagons 30 20 2-Dimensional and 3- Dimensional Shapes Icosahedron 20 Triangles 30 12 (Geometry, Grades 2- 12) 2. Use the table and images above to construct the selected shape by creating one or Problem Solving and more face shapes and then add straws or join shapes at each of the vertices: Reasoning a. For drinking straws and paperclips: Bend the (Mathematical paperclips so that the 2 loops form a “V” or “L” Practices Grades 2- shape as needed, widen the narrower loop and insert 12) one loop into the end of one straw half, and the other loop into another straw half. -
Downloaded from Bookstore.Ams.Org 30-60-90 Triangle, 190, 233 36-72
Index 30-60-90 triangle, 190, 233 intersects interior of a side, 144 36-72-72 triangle, 226 to the base of an isosceles triangle, 145 360 theorem, 96, 97 to the hypotenuse, 144 45-45-90 triangle, 190, 233 to the longest side, 144 60-60-60 triangle, 189 Amtrak model, 29 and (logical conjunction), 385 AA congruence theorem for asymptotic angle, 83 triangles, 353 acute, 88 AA similarity theorem, 216 included between two sides, 104 AAA congruence theorem in hyperbolic inscribed in a semicircle, 257 geometry, 338 inscribed in an arc, 257 AAA construction theorem, 191 obtuse, 88 AAASA congruence, 197, 354 of a polygon, 156 AAS congruence theorem, 119 of a triangle, 103 AASAS congruence, 179 of an asymptotic triangle, 351 ABCD property of rigid motions, 441 on a side of a line, 149 absolute value, 434 opposite a side, 104 acute angle, 88 proper, 84 acute triangle, 105 right, 88 adapted coordinate function, 72 straight, 84 adjacency lemma, 98 zero, 84 adjacent angles, 90, 91 angle addition theorem, 90 adjacent edges of a polygon, 156 angle bisector, 100, 147 adjacent interior angle, 113 angle bisector concurrence theorem, 268 admissible decomposition, 201 angle bisector proportion theorem, 219 algebraic number, 317 angle bisector theorem, 147 all-or-nothing theorem, 333 converse, 149 alternate interior angles, 150 angle construction theorem, 88 alternate interior angles postulate, 323 angle criterion for convexity, 160 alternate interior angles theorem, 150 angle measure, 54, 85 converse, 185, 323 between two lines, 357 altitude concurrence theorem, -
On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem
Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, pp.15-27 ON THE STANDARD LENGTHS OF ANGLE BISECTORS AND THE ANGLE BISECTOR THEOREM G.W INDIKA SHAMEERA AMARASINGHE ABSTRACT. In this paper the author unveils several alternative proofs for the standard lengths of Angle Bisectors and Angle Bisector Theorem in any triangle, along with some new useful derivatives of them. 2010 Mathematical Subject Classification: 97G40 Keywords and phrases: Angle Bisector theorem, Parallel lines, Pythagoras Theorem, Similar triangles. 1. INTRODUCTION In this paper the author introduces alternative proofs for the standard length of An- gle Bisectors and the Angle Bisector Theorem in classical Euclidean Plane Geometry, on a concise elementary format while promoting the significance of them by acquainting some prominent generalized side length ratios within any two distinct triangles existed with some certain correlations of their corresponding angles, as new lemmas. Within this paper 8 new alternative proofs are exposed by the author on the angle bisection, 3 new proofs each for the lengths of the Angle Bisectors by various perspectives with also 5 new proofs for the Angle Bisector Theorem. 1.1. The Standard Length of the Angle Bisector Date: 1 February 2012 . 15 G.W Indika Shameera Amarasinghe The length of the angle bisector of a standard triangle such as AD in figure 1.1 is AD2 = AB · AC − BD · DC, or AD2 = bc 1 − (a2/(b + c)2) according to the standard notation of a triangle as it was initially proved by an extension of the angle bisector up to the circumcircle of the triangle. -
Polygrams and Polygons
Poligrams and Polygons THE TRIANGLE The Triangle is the only Lineal Figure into which all surfaces can be reduced, for every Polygon can be divided into Triangles by drawing lines from its angles to its centre. Thus the Triangle is the first and simplest of all Lineal Figures. We refer to the Triad operating in all things, to the 3 Supernal Sephiroth, and to Binah the 3rd Sephirah. Among the Planets it is especially referred to Saturn; and among the Elements to Fire. As the colour of Saturn is black and the Triangle that of Fire, the Black Triangle will represent Saturn, and the Red Fire. The 3 Angles also symbolize the 3 Alchemical Principles of Nature, Mercury, Sulphur, and Salt. As there are 3600 in every great circle, the number of degrees cut off between its angles when inscribed within a Circle will be 120°, the number forming the astrological Trine inscribing the Trine within a circle, that is, reflected from every second point. THE SQUARE The Square is an important lineal figure which naturally represents stability and equilibrium. It includes the idea of surface and superficial measurement. It refers to the Quaternary in all things and to the Tetrad of the Letter of the Holy Name Tetragrammaton operating through the four Elements of Fire, Water, Air, and Earth. It is allotted to Chesed, the 4th Sephirah, and among the Planets it is referred to Jupiter. As representing the 4 Elements it represents their ultimation with the material form. The 4 angles also include the ideas of the 2 extremities of the Horizon, and the 2 extremities of the Median, which latter are usually called the Zenith and the Nadir: also the 4 Cardinal Points. -
The Arbelos in Wasan Geometry: Chiba's Problem
Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.8, (2019), Issue 2, pp.98-104 THE ARBELOS IN WASAN GEOMETRY: CHIBA’S PROBLEM HIROSHI OKUMURA Abstract. We consider a sangaku problem involving an arbelos with two congruent circles, and give several conditions where the two congruent circles appear. Also we show several pairs of congruent circles and Archimedean circles. 2010 Mathematical Subject Classification: 01A27, 51M04, 51N20. Keywords and phrases: arbelos, Archimedean circle, non-Archimedean congruent circles. 1. INTRODUCTION We consider the arbelos appeared in Wasan geometry, and consider an arbelos formed by three semicircles α, β and γ with diameters AO , BO and AB , respectively for a point O on the segment AB . The radii of α and β are denoted by a and b, respectively. We consider the following sangaku .(in 1880 [5] (see Figure 1 (نproblem proposed by Chiba ( ઍ༿ӫ࣏҆ δ ε γ α h β B O H A Figure 1. Problem 1. For a point H on the segment AO, let h be the perpendicular to AB at H. Let δ be the circle touching α and β externally and h, and let ε be the incircle of the curvilinear triangle made by α, γ and h. If δ and ε touch and δ is congruent to the circle of diameter AH, find the radius of ε in terms of a and b. Circles of radius rA = ab /(a + b) are said to be Archimedean. In this paper we generalize the problem and consider two circles touching at a point on a perpendicular to the line AB . -
Sink Or Float? Thought Problems in Math and Physics
AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 33 33 Sink or Float? Sink or Float: Thought Problems in Math and Physics is a collection of prob- lems drawn from mathematics and the real world. Its multiple-choice format Sink or Float? forces the reader to become actively involved in deciding upon the answer. Thought Problems in Math and Physics The book’s aim is to show just how much can be learned by using everyday Thought Problems in Math and Physics common sense. The problems are all concrete and understandable by nearly anyone, meaning that not only will students become caught up in some of Keith Kendig the questions, but professional mathematicians, too, will easily get hooked. The more than 250 questions cover a wide swath of classical math and physics. Each problem s solution, with explanation, appears in the answer section at the end of the book. A notable feature is the generous sprinkling of boxes appearing throughout the text. These contain historical asides or A tub is filled Will a solid little-known facts. The problems themselves can easily turn into serious with lubricated aluminum or tin debate-starters, and the book will find a natural home in the classroom. ball bearings. cube sink... ... or float? Keith Kendig AMSMAA / PRESS 4-Color Process 395 page • 3/4” • 50lb stock • finish size: 7” x 10” i i \AAARoot" – 2009/1/21 – 13:22 – page i – #1 i i Sink or Float? Thought Problems in Math and Physics i i i i i i \AAARoot" – 2011/5/26 – 11:22 – page ii – #2 i i c 2008 by The -
Archimedes and the Arbelos1 Bobby Hanson October 17, 2007
Archimedes and the Arbelos1 Bobby Hanson October 17, 2007 The mathematician’s patterns, like the painter’s or the poet’s must be beautiful; the ideas like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics. — G.H. Hardy, A Mathematician’s Apology ACBr 1 − r Figure 1. The Arbelos Problem 1. We will warm up on an easy problem: Show that traveling from A to B along the big semicircle is the same distance as traveling from A to B by way of C along the two smaller semicircles. Proof. The arc from A to C has length πr/2. The arc from C to B has length π(1 − r)/2. The arc from A to B has length π/2. ˜ 1My notes are shamelessly stolen from notes by Tom Rike, of the Berkeley Math Circle available at http://mathcircle.berkeley.edu/BMC6/ps0506/ArbelosBMC.pdf . 1 2 If we draw the line tangent to the two smaller semicircles, it must be perpendicular to AB. (Why?) We will let D be the point where this line intersects the largest of the semicircles; X and Y will indicate the points of intersection with the line segments AD and BD with the two smaller semicircles respectively (see Figure 2). Finally, let P be the point where XY and CD intersect. D X P Y ACBr 1 − r Figure 2 Problem 2. Now show that XY and CD are the same length, and that they bisect each other. -
Advanced Euclidean Geometry
Advanced Euclidean Geometry Paul Yiu Summer 2016 Department of Mathematics Florida Atlantic University July 18, 2016 Summer 2016 Contents 1 Some Basic Theorems 101 1.1 The Pythagorean Theorem . ............................ 101 1.2 Constructions of geometric mean . ........................ 104 1.3 The golden ratio . .......................... 106 1.3.1 The regular pentagon . ............................ 106 1.4 Basic construction principles ............................ 108 1.4.1 Perpendicular bisector locus . ....................... 108 1.4.2 Angle bisector locus . ............................ 109 1.4.3 Tangency of circles . ......................... 110 1.4.4 Construction of tangents of a circle . ............... 110 1.5 The intersecting chords theorem ........................... 112 1.6 Ptolemy’s theorem . ................................. 114 2 The laws of sines and cosines 115 2.1 The law of sines . ................................ 115 2.2 The orthocenter ................................... 116 2.3 The law of cosines .................................. 117 2.4 The centroid ..................................... 120 2.5 The angle bisector theorem . ............................ 121 2.5.1 The lengths of the bisectors . ........................ 121 2.6 The circle of Apollonius . ............................ 123 3 The tritangent circles 125 3.1 The incircle ..................................... 125 3.2 Euler’s formula . ................................ 128 3.3 Steiner’s porism ................................... 129 3.4 The excircles .................................... -
4 3 E B a C F 2
See-Saw Geometry and the Method of Mass-Points1 Bobby Hanson February 27, 2008 Give me a place to stand on, and I can move the Earth. — Archimedes. 1. Motivating Problems Today we are going to explore a kind of geometry similar to regular Euclidean Geometry, involving points and lines and triangles, etc. The main difference that we will see is that we are going to give mass to the points in our geometry. We will call such a geometry, See-Saw Geometry (we’ll understand why, shortly). Before going into the details of See-Saw Geometry, let’s look at some problems that might be solved using this different geometry. Note that these problems are perfectly solvable using regular Euclidean Geometry, but we will find See-Saw Geometry to be very fast and effective. Problem 1. Below is the triangle △ABC. Side BC is divided by D in a ratio of 5 : 2, and AB is divided by E in a ration of 3 : 4. Find the ratios in which F divides the line segments AD and CE; i.e., find AF : F D and CF : F E. (Note: in Figure 1, below, only the ratios are shown; the actual lengths are unknown). B 3 E 5 4 F D 2 A C Figure 1 1My notes are shamelessly stolen from notes by Tom Rike, of the Berkeley Math Circle available at http://mathcircle.berkeley.edu/archivedocs/2007 2008/lectures/0708lecturespdf/MassPointsBMC07.pdf . 1 2 Problem 2. In Figure 2, below, D and E divide sides BC and AB, respectively, as before.