3D GEOMETRY 3D GEOMETRY

Σχετικά έγγραφα
2 Composition. Invertible Mappings

On a four-dimensional hyperbolic manifold with finite volume

Volume of a Cuboid. Volume = length x breadth x height. V = l x b x h. The formula for the volume of a cuboid is

Section 8.3 Trigonometric Equations

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

derivation of the Laplacian from rectangular to spherical coordinates

Homework 8 Model Solution Section

Every set of first-order formulas is equivalent to an independent set

Fractional Colorings and Zykov Products of graphs

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

the total number of electrons passing through the lamp.

EE512: Error Control Coding

C.S. 430 Assignment 6, Sample Solutions

CHAPTER 12: PERIMETER, AREA, CIRCUMFERENCE, AND 12.1 INTRODUCTION TO GEOMETRIC 12.2 PERIMETER: SQUARES, RECTANGLES,

Finite Field Problems: Solutions

Homework 3 Solutions

[1] P Q. Fig. 3.1

CRASH COURSE IN PRECALCULUS

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

Lecture 2. Soundness and completeness of propositional logic

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

Section 7.6 Double and Half Angle Formulas

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Concrete Mathematics Exercises from 30 September 2016

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

ΚΥΠΡΙΑΚΟΣ ΣΥΝΔΕΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY 21 ος ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ Δεύτερος Γύρος - 30 Μαρτίου 2011

Potential Dividers. 46 minutes. 46 marks. Page 1 of 11

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

ΚΥΠΡΙΑΚΗ ΜΑΘΗΜΑΤΙΚΗ ΕΤΑΙΡΕΙΑ IΔ ΚΥΠΡΙΑΚΗ ΜΑΘΗΜΑΤΙΚΗ ΟΛΥΜΠΙΑΔΑ ΑΠΡΙΛΙΟΥ 2013 Β & Γ ΛΥΚΕΙΟΥ.

Partial Differential Equations in Biology The boundary element method. March 26, 2013

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

Areas and Lengths in Polar Coordinates

Matrices and Determinants

Section 9.2 Polar Equations and Graphs

Εγχειρίδια Μαθηµατικών και Χταποδάκι στα Κάρβουνα

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Example Sheet 3 Solutions

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

Math221: HW# 1 solutions

Areas and Lengths in Polar Coordinates

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

TMA4115 Matematikk 3

Other Test Constructions: Likelihood Ratio & Bayes Tests

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Reminders: linear functions

Math 6 SL Probability Distributions Practice Test Mark Scheme

New bounds for spherical two-distance sets and equiangular lines

5.4 The Poisson Distribution.

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

A Note on Intuitionistic Fuzzy. Equivalence Relation

Strain gauge and rosettes

HISTOGRAMS AND PERCENTILES What is the 25 th percentile of a histogram? What is the 50 th percentile for the cigarette histogram?

Second Order Partial Differential Equations

10/3/ revolution = 360 = 2 π radians = = x. 2π = x = 360 = : Measures of Angles and Rotations

Αλγόριθμοι και πολυπλοκότητα NP-Completeness (2)

14 Lesson 2: The Omega Verb - Present Tense

Right Rear Door. Let's now finish the door hinge saga with the right rear door

Advanced Subsidiary Unit 1: Understanding and Written Response

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is

ω ω ω ω ω ω+2 ω ω+2 + ω ω ω ω+2 + ω ω+1 ω ω+2 2 ω ω ω ω ω ω ω ω+1 ω ω2 ω ω2 + ω ω ω2 + ω ω ω ω2 + ω ω+1 ω ω2 + ω ω+1 + ω ω ω ω2 + ω

Partial Trace and Partial Transpose

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Solutions to Exercise Sheet 5

4.6 Autoregressive Moving Average Model ARMA(1,1)

PARTIAL NOTES for 6.1 Trigonometric Identities

The Simply Typed Lambda Calculus

ΠΑΝΕΠΙΣΤΗΜΙΟ ΠΕΙΡΑΙΑ ΤΜΗΜΑ ΝΑΥΤΙΛΙΑΚΩΝ ΣΠΟΥΔΩΝ ΠΡΟΓΡΑΜΜΑ ΜΕΤΑΠΤΥΧΙΑΚΩΝ ΣΠΟΥΔΩΝ ΣΤΗΝ ΝΑΥΤΙΛΙΑ

Συστήματα Διαχείρισης Βάσεων Δεδομένων

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

Inverse trigonometric functions & General Solution of Trigonometric Equations

ΑΚΑ ΗΜΙΑ ΕΜΠΟΡΙΚΟΥ ΝΑΥΤΙΚΟΥ ΜΑΚΕ ΟΝΙΑΣ ΣΧΟΛΗ ΜΗΧΑΝΙΚΩΝ ΠΤΥΧΙΑΚΗ ΕΡΓΑΣΙΑ

Parametrized Surfaces

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Review Test 3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

( ) 2 and compare to M.

Trigonometric Formula Sheet

KANGOUROU MATHEMATICS

Paper Reference. Paper Reference(s) 1776/04 Edexcel GCSE Modern Greek Paper 4 Writing. Thursday 21 May 2009 Afternoon Time: 1 hour 15 minutes

Mean bond enthalpy Standard enthalpy of formation Bond N H N N N N H O O O

The challenges of non-stable predicates

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Ordinary Level

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

Chapter 3: Ordinal Numbers

1 String with massive end-points

Integrals in cylindrical, spherical coordinates (Sect. 15.7)

ΦΥΛΛΟ ΕΡΓΑΣΙΑΣ Α. Διαβάστε τις ειδήσεις και εν συνεχεία σημειώστε. Οπτική γωνία είδησης 1:.

Lecture 34 Bootstrap confidence intervals

Solution Series 9. i=1 x i and i=1 x i.

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 24/3/2007

Instruction Execution Times

Spherical Coordinates

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

Κάθε γνήσιο αντίγραφο φέρει υπογραφή του συγγραφέα. / Each genuine copy is signed by the author.

Γ ΓΥΜΝΑΣΙΟΥ & Α ΛΥΚΕΙΟΥ

Transcript:

1 Studying a document and learning new vocabulary 1. Read the following document We live in a three-dimensional world. Every object you can see or touch has three dimensions that can be measured: length, width, and height. The room you are sitting in can be described by these three dimensions. Even you can be described by these three dimensions. There are many types of three-dimensional shapes. You've surely seen spheres, cylinders and cubes before. In this lesson, you will learn about particular solids called polyhedra and you will also learn about two special types of polyhedra: prisms and pyramids. Polyhedra Dice, match boxes, soccer balls are all examples of polyhedra. Generally, a three-dimensional shape whose faces are polygons is known as a polyhedron. This term comes from the Greek words poly, which means "many," and hedron, which means "face". So, quite literally, a polyhedron is a three-dimensional object with many faces. The faces of a cube are squares. The faces of a rectangular prism are rectangles. And the faces of a truncated icosahedron are pentagons and hexagons. The other parts of a polyhedron are its edges, the line segments along which two faces intersect, and its vertices, the points at which three or more faces meet. Prisms A prism is a polyhedron for which the top and bottom faces (known as the bases) are congruent polygons, and all other faces (known as the lateral faces) are parallelograms or rectangles. Technically, when the sides are rectangles, the shape is known as a right prism, indicating that the lateral faces meet the sides of the base at right angles. In this lesson, when we use the term prism, we mean a right prism. But there are other types of prisms, too. A prism is described by the shape of its base. For instance, a rectangular prism (cuboid) has bases that are rectangles, and a pentagonal prism has bases that are pentagons. Pyramids A pyramid is a polyhedron for which the base is a polygon and all lateral faces are triangles. In this lesson, we will only concern ourselves with pyramids whose lateral faces are congruent that is, they are the same size and shape. Technically, when the lateral faces are congruent triangles, the shape is known as a right pyramid, indicating that the apex the vertex at which the lateral faces meet is directly above the center of the base. In this lesson, when we use the term pyramid, we mean a right pyramid. But there are other types of pyramids, too. A pyramid is typically described by the shape of its base. For instance, a triangular pyramid has a base that is a triangle, and a hexagonal pyramid has a base that is a hexagon. For any solid, the surface area is exactly the area of all of the outside surfaces of a three-dimensional object. And volume is all of the space inside a three-dimensional object. Adapted from http://www.learner.org

2. Complete the following grid Find out the English word corresponding to its definition, or conversely, give a defintion of the given English keyword, and add its French version: English keyword or expression English definition French keyword or expression 1 The distance value between two points 2 The distance value that is lateral (side to side) 3 The height 4 A sphere 5 A three-dimensional figure with two circular bases that are parallel and congruent 6 A polyhedron (plur. polyhedra) 7 A prism 8 A polyhedron whose faces are rectangles 9 A right pyramid 10 Said for two figures that have the same size and shape, i.e., one can be transformed into the other by an isometry 11 One of the surfaces of a polyhedron 12 The segment along which two faces meet 13 The point at which two or more edges meet 3. In pairs, compare your answers and correct your mistakes, if needed.

2 Building polyhedra with polydrons You are given polydrons composed of equilateral triangles, squares, regular pentagons and regular hexagons, all polygons with sides of the same length. These polydrons will enable you to build polyhedra. Step 1 a) Build any polyhedron you want. b) Describe them as precisely as you can. Step 2 a) Inside the collection of polyhedra you have, keep only the ones which are convex and have the same regular polygon for each of their faces. b) Can you build other polyhedra under these conditions? Step 3 a) Inside the collection of polyhedra you have, keep only the ones which are convex, have the same regular polygon for each of their faces and the same number of faces meeting together at each vertex. b) How many polyhedra can you build under these conditions? c) Define them as precisely as you can.

3 Polyhedra, polyhedra!!! 1. Any polygon can be defined through a relation involving the number of its vertices (denoted V) and the number of its sides (denoted S). What is this relation? 2. a) Conjecture an equivalent relation from the following Platonic polyhedra: b) Sketch other polyhedra and check if the formula is still valid.

4 Studying an authentical document Timaeus is a theoretical treatise of Plato in the form of a Socratic dialogue, written circa 360 BC. The work puts forward speculation on the nature of the physical world. The triangles 53c-55d Τὰ τρίγωνα (...) every plane rectilinear figure is composed of triangles; (...) Then let us choose two triangles (...) one isosceles, the other having (...) its hypotenuse twice the lesser side. When two such triangles are joined at the diagonal, and this is repeated three times, and the triangles rest their diagonals and shorter sides on the same point as a centre, a single equilateral triangle is formed out of six triangles; and four equilateral triangles, if put together, make out (...) the first solid form (...). The second species of solid is formed out of the same triangles, which unite as eight equilateral triangles and form (...) the second body (...). And the third body is made up of 120 triangular elements, forming twelve solid angles, each of them (...) having altogether twenty bases, each of which is an equilateral triangle. The one element [that is, the triangle which has its hypotenuse twice the lesser side] having generated these figures, generated no more; but the isosceles triangle produced the fourth elementary figure, which is compounded of four such triangles, joining their right angles in a centre, and forming one equilateral quadrangle. Six of these united form eight solid angles (...); the figure of the body thus composed is a cube, having six plane quadrangular equilateral bases. There was yet a fifth combination which God used in the delineation of the universe. http://www.ellopos.net/elpenor/physis/plato-timaeus/triangles.asp Questions 1. Give a definition of an isosceles triangle. 2. Sketch the second kind of triangles described by Plato: the other having (...) its hypotenuse twice the lesser side. 3. Make the construction of an equilateral triangle as defined by Plato. 4. What is the first solid form described by Timaeus? 5. Describe the second body and the third body. 6. Explain: The one element [that is, the triangle which has its hypotenuse twice the lesser side] having generated these figures, generated no more. 7. Explain what Plato calls an equilateral quadrangle. 8. How is generated a cube for Plato? 9. What is the fifth combination which God used in the delineation of the universe?

5 To go further with the Timaeus 1. Read the excerpt of Plato s Timaeus 55d 57d. 2. Write a short summary of that text. 3. By which logical argument did Plato assign one solid respectively to earth and fire? Timaeus 55d - 57d BUT, leaving this enquiry, let us proceed to distribute the elementary forms, which have now been created in idea, among the four elements. To earth, then, let us assign the cubical form; for earth is the most immoveable of the four and the most plastic of all bodies, and that which has the most stable bases must of necessity be of such a nature. Now, of the triangles which we assumed at first, that which has two equal sides is by nature more firmly based than that which has unequal sides; and of the compound figures which are formed out of either, the plane equilateral quadrangle has necessarily, a more stable basis than the equilateral triangle, both in the whole and in the parts. Wherefore, in assigning this figure to earth, we adhere to probability; and to water we assign that one of the remaining forms which is the least moveable; and the most moveable of them to fire; and to air that which is intermediate. Also we assign the smallest body to fire, and the greatest to water, and the intermediate in size to air; and, again, the acutest body to fire, and the next in acuteness to, air, and the third to water. Of all these elements, that which has the fewest bases must necessarily be the most moveable, for it must be the acutest and most penetrating in every way, and also the lightest as being composed of the smallest number of similar particles: and the second body has similar properties in a second degree, and the third body in the third degree. Let it be agreed, then, both according to strict reason and according to probability, that the pyramid is the solid which is the original element and seed of fire; and let us assign the element which was next in the order of generation to air, and the third to water. τὰ τέσσαρα γένη καὶ τοῦτον μὲν μεθετέον τὰ δὲ γεγονότα νῦν τῷ λόγῳ γένη διανείμωμεν εἰς πῦρ καὶ γῆν καὶ ὕδωρ καὶ ἀέρα. γῇ μὲν δὴ τὸ κυβικὸν εἶδος δῶμεν [55e] ἀκινητοτάτη γὰρ τῶν τεττάρων γενῶν γῆ καὶ τῶν σωμάτων πλαστικωτάτη μάλιστα δὲ ἀνάγκη γεγονέναι τοιοῦτον τὸ τὰς βάσεις ἀσφαλεστάτας ἔχον βάσις δὲ ἥ τε τῶν κατ ἀρχὰς τριγώνων ὑποτεθέντων ἀσφαλεστέρα κατὰ φύσιν ἡ τῶν ἴσων πλευρῶν τῆς τῶν ἀνίσων τό τε ἐξ ἑκατέρου συντεθὲν ἐπίπεδον ἰσόπλευρον ἰσοπλεύρου τετράγωνον τριγώνου κατά τε μέρη καὶ καθ ὅλον στασιμωτέρως ἐξ ἀνάγκης βέβηκεν. [56a] διὸ γῇ μὲν τοῦτο ἀπονέμοντες τὸν εἰκότα λόγον διασῴζομεν ὕδατι δ αὖτῶν λοιπῶν τὸ δυσκινητότατον εἶδος τὸ δ εὐκινητότατον πυρί τὸ δὲ μέσον ἀέρι καὶ τὸ μὲν σμικρότατον σῶμα πυρί τὸ δ αὖ μέγιστον ὕδατι τὸ δὲ μέσον ἀέρι καὶ τὸ μὲν ὀξύτατον αὖ πυρί τὸ δὲ δεύτερον ἀέρι τὸ δὲ τρίτον ὕδατι. ταῦτ οὖν δὴ πάντα τὸ μὲν ἔχον ὀλιγίστας βάσεις εὐκινητότατον ἀνάγκη πεφυκέναι [56b] τμητικώτατόν τε καὶ ὀξύτατον ὂν πάντῃ πάντων ἔτι τε ἐλαφρότατον ἐξ ὀλιγίστων συνεστὸς τῶν αὐτῶν μερῶν τὸ δὲ δεύτερον δευτέρως τὰ αὐτὰ ταῦτ ἔχειν τρίτως δὲ τὸ τρίτον. ἔστω δὴ κατὰ τὸν ὀρθὸν λόγον καὶ κατὰ τὸν εἰκότα τὸ μὲν τῆς πυραμίδος στερεὸν γεγονὸς εἶδος πυρὸς στοιχεῖον καὶ σπέρμα τὸ δὲ δεύτερον κατὰ γένεσιν εἴπωμεν ἀέρος τὸ δὲ τρίτον ὕδατος. http://www.ellopos.net/elpenor/physis/plato-timaeus/triangles.asp

6 Working in groups Group 1: slicing a cube 1. Sketch a cube with edge a. 2. Compute the radius of its circumscribing sphere. 3. Sketch the 4 diameters of the circumscribed sphere. They slice the cube into solids. Describe them precisely. 4. Evaluate the volume of each of these solids. 5. Conclusion: build the general formula giving the volume of such a solid. 6. Prepare an oral presentation. Group 2: volume of a tetrahedron 1. Remind the formula giving the area of a triangle. The aim of this activity is now to find out the formula giving the volume a tetrahedron. Consider a tetraedron with area of one face B, height h and volume V. 2. Sketch a picture. 3. Cut it with three planes each of them parallel to a face and passing through the midpoints of the edges. 4. Define the different solids obtained from the initial tetrahedron. 5. a) Explain why three of them are congruent. b) Evaluate their volume denoted t, in terms of V. 6. Consider the fourth solid. a) What solid is obtained by completing it with one solid of volume t? b) Evaluate the volume of this fourth solid denoted p. 7. Deduce the expression of V, in terms of B and h. 8. Prepare an oral presentation. Group 3: properties of regular tetrahedra (1) 1. Sketch a regular tetrahedron called ABCD with edge a. 2. Property of the edges a) Prove that line (CD) is perpendicular to plane (ABI) with I the midpoint of [CD]. b) Deduce that line (CD) is perpendicular to line (AB). c) Explain the general property in a regular tetrahedron you ve proved. 3. Properties of the medians Definition: a median of a tetrahedron is a line segment connecting a vertex with the centroid of the opposite face.

a) What is the centroid in a triangle? b) Naming S the centroid of BCD, prove that line (CD) is perpendicular to line (AS). Similarly, we would prove that line (BC) is perpendicular to line (AS). c) What can you deduce for the median (AS)? d) Explain the general property in a regular tetrahedron you ve proved. e) Prove that the four medians in a tetrahedraon are concurrent and that their intersection point is the centroid of the tetrahedron. 4. Prepare an oral presentation. Group 4: properties of regular tetrahedra (2) 1. Sketch a regular tetrahedron called ABCD with edge a and name I, J, K, L M and N the midpoints of its sides [AB], [AC], [AD], [BC], [BD] and [CD]. Definition 1: a bimedian of a tetrahedron is a line segment connecting the midpoints of opposite edges. 2. Define the bimedians of ABCD. 3. Property n 1 Prove that the bimedians bisect each other. 4. Property n 2 Definition 2: the perpendicular bisector plane (or bisecting plane) of a segment is the plane that cut that segment orthogonally at its midpoint. Consequence: the bisecting plane of segment [AB] is the set of all the points that are equidistant from A and B. a) Prove that (CDI) is the bisecting plane of [AB]. b) Deduce that line (IN) is perpendicular to line (AB). c) Prove that line (IN) is perpendicular to line (CD). d) Explain the general property you ve proved. 5. Property n 3 Prove that the bimedians intersect at the centroid of the tetrahedron. 6. Prepare an oral presentation. 7 Oral presentations