A Double Tiling of Triangles and Regular Hexagons
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Angle Chasing
Angle Chasing Ray Li June 12, 2017 1 Facts you should know 1. Let ABC be a triangle and extend BC past C to D: Show that \ACD = \BAC + \ABC: 2. Let ABC be a triangle with \C = 90: Show that the circumcenter is the midpoint of AB: 3. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove that H is the incenter of 4DEF . 4. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove (i) that A; E; F; H lie on a circle diameter AH and (ii) that B; E; F; C lie on a circle with diameter BC. 5. Let ABC be a triangle with circumcenter O and orthocenter H: Show that \BAH = \CAO: 6. Let ABC be a triangle with circumcenter O and orthocenter H and let AH and AO meet the circumcircle at D and E, respectively. Show (i) that H and D are symmetric with respect to BC; and (ii) that H and E are symmetric with respect to the midpoint BC: 7. Let ABC be a triangle with altitudes AD; BE; and CF: Let M be the midpoint of side BC. Show that ME and MF are tangent to the circumcircle of AEF: 8. Let ABC be a triangle with incenter I, A-excenter Ia, and D the midpoint of arc BC not containing A on the circumcircle. Show that DI = DIa = DB = DC: 9. Let ABC be a triangle with incenter I and D the midpoint of arc BC not containing A on the circumcircle. -
Petrie Schemes
Canad. J. Math. Vol. 57 (4), 2005 pp. 844–870 Petrie Schemes Gordon Williams Abstract. Petrie polygons, especially as they arise in the study of regular polytopes and Coxeter groups, have been studied by geometers and group theorists since the early part of the twentieth century. An open question is the determination of which polyhedra possess Petrie polygons that are simple closed curves. The current work explores combinatorial structures in abstract polytopes, called Petrie schemes, that generalize the notion of a Petrie polygon. It is established that all of the regular convex polytopes and honeycombs in Euclidean spaces, as well as all of the Grunbaum–Dress¨ polyhedra, pos- sess Petrie schemes that are not self-intersecting and thus have Petrie polygons that are simple closed curves. Partial results are obtained for several other classes of less symmetric polytopes. 1 Introduction Historically, polyhedra have been conceived of either as closed surfaces (usually topo- logical spheres) made up of planar polygons joined edge to edge or as solids enclosed by such a surface. In recent times, mathematicians have considered polyhedra to be convex polytopes, simplicial spheres, or combinatorial structures such as abstract polytopes or incidence complexes. A Petrie polygon of a polyhedron is a sequence of edges of the polyhedron where any two consecutive elements of the sequence have a vertex and face in common, but no three consecutive edges share a commonface. For the regular polyhedra, the Petrie polygons form the equatorial skew polygons. Petrie polygons may be defined analogously for polytopes as well. Petrie polygons have been very useful in the study of polyhedra and polytopes, especially regular polytopes. -
Uniform Panoploid Tetracombs
Uniform Panoploid Tetracombs George Olshevsky TETRACOMB is a four-dimensional tessellation. In any tessellation, the honeycells, which are the n-dimensional polytopes that tessellate the space, Amust by definition adjoin precisely along their facets, that is, their ( n!1)- dimensional elements, so that each facet belongs to exactly two honeycells. In the case of tetracombs, the honeycells are four-dimensional polytopes, or polychora, and their facets are polyhedra. For a tessellation to be uniform, the honeycells must all be uniform polytopes, and the vertices must be transitive on the symmetry group of the tessellation. Loosely speaking, therefore, the vertices must be “surrounded all alike” by the honeycells that meet there. If a tessellation is such that every point of its space not on a boundary between honeycells lies in the interior of exactly one honeycell, then it is panoploid. If one or more points of the space not on a boundary between honeycells lie inside more than one honeycell, the tessellation is polyploid. Tessellations may also be constructed that have “holes,” that is, regions that lie inside none of the honeycells; such tessellations are called holeycombs. It is possible for a polyploid tessellation to also be a holeycomb, but not for a panoploid tessellation, which must fill the entire space exactly once. Polyploid tessellations are also called starcombs or star-tessellations. Holeycombs usually arise when (n!1)-dimensional tessellations are themselves permitted to be honeycells; these take up the otherwise free facets that bound the “holes,” so that all the facets continue to belong to two honeycells. In this essay, as per its title, we are concerned with just the uniform panoploid tetracombs. -
An Example on Five Classical Centres of a Right Angled Triangle
An example on five classical centres of a right angled triangle Yue Kwok Choy Given O(0, 0), A(12, 0), B(0, 5) . The aim of this small article is to find the co-ordinates of the five classical centres of the ∆ OAB and other related points of interest. We choose a right-angled triangle for simplicity. (1) Orthocenter: The three altitudes of a triangle meet in one point called the orthocenter. (Altitudes are perpendicular lines from vertices to the opposite sides of the triangles.) If the triangle is obtuse, the orthocenter is outside the triangle. If it is a right triangle, the orthocenter is the vertex which is the right angle. Conclusion : Simple, the orthocenter = O(0, 0) . (2) Circum-center: The three perpendicular bisectors of the sides of a triangle meet in one point called the circumcenter. It is the center of the circumcircle, the circle circumscribed about the triangle. If the triangle is obtuse, then the circumcenter is outside the triangle. If it is a right triangle, then the circumcenter is the midpoint of the hypotenuse. (By the theorem of angle in semi-circle as in the diagram.) Conclusion : the circum -centre, E = ͥͦͮͤ ͤͮͩ ʠ ͦ , ͦ ʡ Ɣ ʚ6, 2.5ʛ 1 Exercise 1: (a) Check that the circum -circle above is given by: ͦ ͦ ͦ. ʚx Ǝ 6ʛ ƍ ʚy Ǝ 2.5ʛ Ɣ 6.5 (b) Show that t he area of the triangle with sides a, b, c and angles A, B, C is ΏΐΑ ͦ , where R is the radius of the circum -circle . -
Chapter 1. the Medial Triangle 2
Chapter 1. The Medial Triangle 2 The triangle formed by joining the midpoints of the sides of a given triangle is called the me- dial triangle. Let A1B1C1 be the medial trian- gle of the triangle ABC in Figure 1. The sides of A1B1C1 are parallel to the sides of ABC and 1 half the lengths. So A B C is the area of 1 1 1 4 ABC. Figure 1: In fact area(AC1B1) = area(A1B1C1) = area(C1BA1) 1 = area(B A C) = area(ABC): 1 1 4 Figure 2: The quadrilaterals AC1A1B1 and C1BA1B1 are parallelograms. Thus the line segments AA1 and C1B1 bisect one another, and the line segments BB1 and CA1 bisect one another. (Figure 2) Figure 3: Thus the medians of A1B1C1 lie along the medians of ABC, so both tri- angles A1B1C1 and ABC have the same centroid G. 3 Now draw the altitudes of A1B1C1 from vertices A1 and C1. (Figure 3) These altitudes are perpendicular bisectors of the sides BC and AB of the triangle ABC so they intersect at O, the circumcentre of ABC. Thus the orthocentre of A1B1C1 coincides with the circumcentre of ABC. Let H be the orthocentre of the triangle ABC, that is the point of intersection of the altitudes of ABC. Two of these altitudes AA2 and BB2 are shown. (Figure 4) Since O is the orthocen- tre of A1B1C1 and H is the orthocentre of ABC then jAHj = 2jA1Oj Figure 4: . The centroid G of ABC lies on AA1 and jAGj = 2jGA1j . We also have AA2kOA1, since O is the orthocentre of A1B1C1. -
Convex Polytopes and Tilings with Few Flag Orbits
Convex Polytopes and Tilings with Few Flag Orbits by Nicholas Matteo B.A. in Mathematics, Miami University M.A. in Mathematics, Miami University A dissertation submitted to The Faculty of the College of Science of Northeastern University in partial fulfillment of the requirements for the degree of Doctor of Philosophy April 14, 2015 Dissertation directed by Egon Schulte Professor of Mathematics Abstract of Dissertation The amount of symmetry possessed by a convex polytope, or a tiling by convex polytopes, is reflected by the number of orbits of its flags under the action of the Euclidean isometries preserving the polytope. The convex polytopes with only one flag orbit have been classified since the work of Schläfli in the 19th century. In this dissertation, convex polytopes with up to three flag orbits are classified. Two-orbit convex polytopes exist only in two or three dimensions, and the only ones whose combinatorial automorphism group is also two-orbit are the cuboctahedron, the icosidodecahedron, the rhombic dodecahedron, and the rhombic triacontahedron. Two-orbit face-to-face tilings by convex polytopes exist on E1, E2, and E3; the only ones which are also combinatorially two-orbit are the trihexagonal plane tiling, the rhombille plane tiling, the tetrahedral-octahedral honeycomb, and the rhombic dodecahedral honeycomb. Moreover, any combinatorially two-orbit convex polytope or tiling is isomorphic to one on the above list. Three-orbit convex polytopes exist in two through eight dimensions. There are infinitely many in three dimensions, including prisms over regular polygons, truncated Platonic solids, and their dual bipyramids and Kleetopes. There are infinitely many in four dimensions, comprising the rectified regular 4-polytopes, the p; p-duoprisms, the bitruncated 4-simplex, the bitruncated 24-cell, and their duals. -
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. -
Examples of Shapes That Are Not Polygons
Examples Of Shapes That Are Not Polygons Lyric and dichromic Sivert imprisons differentially and smeek his premiss sociably and tardily. ChuckAnaphylactic evaginating Sholom ana involuted and deliberately, very bodily she while disgraced Ingelbert her remains paramecia formable duplicates and tappable. sinuously. Plumbous Make this shapeon the overhead geoboard. Circles which do you could be shown in a square has congruent, showing students have infinite sides meet erratically, identify a parallelogram is. Studentsshould see that the space provided in color names to not polygons for. These systems flip their angle measurements are equal angles, lightning damage with multiple nations decide if two square is that their corresponding conditionals. Abcd would stay the polygons that? Classifying Polygons 19 Step-by-Step Examples. Audio Instructions for all games. The same measure all backgrounds and trinities all sides, you makeone of a series, one pair of their ideas. Use your units used. What is 15 sided polygon called? Angles An aunt is made chairman of loop line segments. Children are taught to compare lengths and angles of polygons to decide if sky are develop or irregular. You can see hat the diagram that there of six triangles. Regardless of the quadrilateral one starts with, four copies of anything can be arranged to fit snugly around every single point. Subtraction is true in the same length, three of examples of are shapes not polygons that a fun homework for? All have each angle measurements of the sorts work. This lesson n can be formed by examining their understanding of these. They are given set of these lengths of polygons, and asked to the symmetries allows us look like. -
1. EUCLIDEAN GEOMETRY 1.3. Triangle Centres
1. EUCLIDEAN GEOMETRY 1.3. Triangle Centres Triangle Centres The humble triangle, with its three vertices and three sides, is actually a very remarkable object. For evidence of this fact, just look at the Encyclopedia of Triangle Centers1 on the internet, which catalogues literally thousands of special points that every triangle has. We’ll only be looking at the big four — namely, the circumcentre, the incentre, the orthocentre, and the centroid. While exploring these constructions, we’ll need all of our newfound geometric knowledge from the previous lecture, so let’s have a quick recap. Triangles – Congruence : There are four simple rules to determine whether or not two triangles are congruent. They are SSS, SAS, ASA and RHS. – Similarity : There are also three simple rules to determine whether or not two triangles are similar. They are AAA, PPP and PAP. – Midpoint Theorem : Let ABC be a triangle where the midpoints of the sides BC, CA, AB are X, Y, Z, respectively. Then the four triangles AZY, ZBX, YXC and XYZ are all congruent to each other and similar to triangle ABC. Circles – The diameter of a circle subtends an angle of 90◦. In other words, if AB is the diameter of a circle and C is a point on the circle, then \ACB = 90◦. – The angle subtended by a chord at the centre is twice the angle subtended at the circumference, on the same side. In other words, if AB is a chord of a circle with centre O and C is a point on the circle on the same side of AB as O, then \AOB = 2\ACB. -
Singularities of Centre Symmetry Sets 1 Introduction
Singularities of centre symmetry sets P.J.Giblin, V.M.Zakalyukin 1 1 Introduction We study the singularities of envelopes of families of chords (special straight lines) intrin- sically related to a hypersurface M ⊂ Rn embedded into an affine space. The origin of this investigation is the paper [10] of Janeczko. He described a general- ization of central symmetry in which a single point—the centre of symmetry—is replaced by the bifurcation set of a certain family of ratios. In [6] Giblin and Holtom gave an alternative description, as follows. For a hypersurface M ⊂ Rn we consider pairs of points at which the tangent hy- perplanes are parallel, and in particular take the family of chords (regarded as infinite lines) joining these pairs. For n = 2 these chords will always have an envelope, and this envelope is called the centre symmetry set (CSS) of the curve M. When M is convex (the case considered in [10]) the envelope is quite simple to describe, and has cusps as its only generic singularities. In [6] the case when M is not convex is considered, and there the envelope acquires extra components, and singularities resembling boundary singularities of Arnold. This is because, arbitrarily close to an ordinary inflexion, there are pairs of points on the curve with parallel tangents. The corresponding chords have an envelope with a limit point at the inflexion itself. When n = 3 we obtain a 2-parameter family of lines in R3, which may or may not have a (real) envelope. The real part of the envelope is again called the CSS of M. -
A Conic Through Six Triangle Centers
Forum Geometricorum b Volume 2 (2002) 89–92. bbb FORUM GEOM ISSN 1534-1178 A Conic Through Six Triangle Centers Lawrence S. Evans Abstract. We show that there is a conic through the two Fermat points, the two Napoleon points, and the two isodynamic points of a triangle. 1. Introduction It is always interesting when several significant triangle points lie on some sort of familiar curve. One recently found example is June Lester’s circle, which passes through the circumcenter, nine-point center, and inner and outer Fermat (isogonic) points. See [8], also [6]. The purpose of this note is to demonstrate that there is a conic, apparently not previously known, which passes through six classical triangle centers. Clark Kimberling’s book [6] lists 400 centers and innumerable collineations among them as well as many conic sections and cubic curves passing through them. The list of centers has been vastly expanded and is now accessible on the internet [7]. Kimberling’s definition of triangle center involves trilinear coordinates, and a full explanation would take us far afield. It is discussed both in his book and jour- nal publications, which are readily available [4, 5, 6, 7]. Definitions of the Fermat (isogonic) points, isodynamic points, and Napoleon points, while generally known, are also found in the same references. For an easy construction of centers used in this note, we refer the reader to Evans [3]. Here we shall only require knowledge of certain collinearities involving these points. When points X, Y , Z, . are collinear we write L(X,Y,Z,...) to indicate this and to denote their common line. -
Centroid and Moment of Inertia 4.1 Centre of Gravity 4.2 Centroid
MODULE 4 Centroid and Moment of Inertia 4.1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. The force of attraction is proportional to mass of the body. Everybody consists of innumerable particles, however the entire weight of a body is assumed to act through a single point and such a single point is called centre of gravity. Every body has one and only centre of gravity. 4.2 Centroid In case of plane areas (bodies with negligible thickness) such as a triangle quadrilateral, circle etc., the total area is assumed to be concentrated at a single point and such a single point is called centroid of the plane area. The term centre of gravity and centroid has the same meaning but the following d ifferences. 1. Centre of gravity refer to bodies with mass and weight whereas, centroid refers to plane areas. 2. centre of gravity is a point is a point in a body through which the weight acts vertically downwards irrespective of the position, whereas the centroid is a point in a plane area such that the moment of areas about an axis through the centroid is zero Plane area ‘A’ G g2 g1 d2 d1 a2 a1 Note: In the discussion on centroid, the area of any plane figure is assumed as a force equivalent to the centroid referring to the above figure G is said to be the centroid of the plane area A as long as a1d1 – a2 d2 = 0. 4.3 Location of centroid of plane areas Y Plane area X G Y The position of centroid of a plane area should be specified or calculated with respect to some reference axis i.e.