Geometrical Constructions 2

Home , Centroid

by Pál Ledneczki Ph.D.

Table of contents

1) Pencils of circles, Apollonian problems

2) Approximate rectification of an arc

3) Roulettes

4) Conic sections

5) 3D geometrical constructions

6) Regular and semi-regular polyhedrons

7) Geometrical calculations "When he established the heavens I was there: when he set a compass upon the face of the deep.“ (Proverbs, Chapter 8 par. 27 )

Geometria una et aeterna est in mente Dei refulgens: cuius consortium hominibus tributum inter causas est, cur homo sit imago Dei.

Geometry is one and eternal shining in the mind of God. That share in it accorded to men is one of the reasons that Man is the image of God.

(Kepler, 1571-1630)

God the Geometer, Manuscript illustration.

Geometrical Constructions 2 2 Pencil of Circles

Intersecting (or „elliptical”) pencil Radical center C of coaxial circles

radical axis P

All tangents drawn to the circles For three circles whose centers form of a coaxial pencil from a point a triangle, the three radical axes (of on the radical axis have the the circles taken in pair) concur in a same length. point called the radical center.

Geometrical Constructions 2 Apollonian Problems 3 Conjugate Pencils of Circles

An elliptical pencil of circles is the family of The parabolic pencil of circles is the family of all circles that go through two given points. circles which all have one common point, and thus are all tangent to each other, A hyperbolic pencil of circles is the family of either internally or externally. Also, the all circles that are orthogonal to an elliptical orthogonal set of circles to a parabolic pencil pencil of circles. None of the circles in the is another parabolic pencil. hyperbolic pencil intersect with each other. Geometrical Constructions 2 Apollonian Problems 4 Apollonius of Perga (about 262 - about 190 BC)

Apollonius of Perga was known as 'The Great Geometer'. Little is known of his life but his works have had a very great influence on the development of mathematics, in particular his famous book Conics introduced terms which are familiar to us today such as parabola, ellipse and hyperbola. Perga was a centre of culture at this time and it was the place of worship of Queen Artemis, a nature goddess. When he was a young man Apollonius went to Alexandria where he studied under the followers of Euclid and later he taught there. Apollonius visited Pergamum where a university and library similar to Alexandria had been built. Pergamum, today the town of Bergama in the province of Izmir in Turkey, was an ancient Greek city in Mysia. It was situated 25 km from the Aegean Sea on a hill on the northern side of the wide valley of the Caicus River (called the Bakir river today)

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Apollonius.html

Geometrical Constructions 2 Apollonian Problems 5 Apollonius Circle

DEFITION 1: the set of all points whose distances from two fixed points are in a constant ratio

DEFITION 2: one of the eight circles that is simultaneously tangent to three given circles

http://mathworld.wolfram.com/ApolloniusCircle.html

Geometrical Constructions 2 Apollonian Problems 6 Apollonian Problems on Tangent Circles

Combinatorial approach Options

Circle C 1) (PPP) ; Point (circle of 0 radius) P 2) (PPL) ; Straight line (circle of infinite radius) L 3) (PPC) ;

4) (PLL) ; Apollonian tangent circle problem: choose three from the elements 5) (PLC) : of the set {P, L, C} (an element can be chosen repeatedly) and find the circles tangent to or passing through the given elements. 6) (PCC) :

(In combinatorics: third class combinations with repetitions of three 7) (LLL) ;

elements.) 8) (LLC) :

9) (LCC) : ; subject of our course 10) (CCC) : : not subject of our course

Geometrical Constructions 2 Apollonian Problems 7 1) (PPP) 2) (PPL)

e d P c 1 i d g

P3 f P 2 h l

Geometrical Constructions 2 Apollonian Problems 8 3) (PPC),4)(PLL)

l c 1 d P 1 e c d c P h i h g

f l2 P 2 f e g

Geometrical Constructions 2 Apollonian Problems 9 5) (PLC)

Hint: (PLC) can be reduced to (PPL). An additional point Q can be constructed

as the point of intersection of A1P and the circle through B, A2 and P. In this way two circles, passing through P, tangent to the given circle c and to

the given line l can be constructed. Change the points A1 and A2 to obtain

two more solutions. A1

d c P

c A2 g Q P e B f Q1 Q2 l

Geometrical Constructions 2 Apollonian Problems 10 7) (LLL) 8) (LLC)

l f l3 2 c d e l1 g d c

l f c l 2 1 e

Hint: the solutions are the incircle and Hint: the problem can be reduced to the the excircles of the triangle (PLL) by means of dilatation that means, formed by the lines. The centers draw parallels at the distance of the radius are the points of intersection of of the given circle. The circles passing the bisectors of the interior and through the center of the given circle and exterior angles. tangent to the parallel lines, have the same centers as the circles of solution.

Geometrical Constructions 2 Apollonian Problems 11 Chapter Review

Vocabulary

Geometrical Constructions 2 Apollonian Problems 12 Approximate Rectification of an Arc

B Kochansky’s method: CD = 3r DB ≈ rπ r ≈ rπ O

a r 30°

C A D

Snellius’ method:

P P’ AC = 3r

a O 30° a A C O r AP’ ≈ AP

Geometrical Constructions 2 Approximate rectification of an arc 13 Roulettes

In the most general case roulettes are curves generated by rolling a curve r (rolling courve), without slipping, along another curve b (base curve). The roulette is generated by a point rigidly attached to the rolling curve r. http://www.math.uoc.gr/~pamfilos/eGallery/problems/Roulette2.html

Special roulettes: - Circle rolling along a straight line

- Circle rolling along a circle Straight line rolling along a circle

Geometrical Constructions 2 Roulettes 14 Cycloid

The cycloid is the locus of a point attached to a circle, rolling along a straight line. It was studied and named by Galileo in 1599. Galileo attempted to find the area by weighing pieces of metal cut into the shape of the cycloid. Torricelli, Fermat, and Descartes all found the area. The cycloid was also studied by Roberval in 1634, Wren in 1658, Huygens in 1673, and Johann Bernoulli in 1696. Roberval and Wren found the arc length (MacTutor Archive). Gear teeth were also made out of cycloids, as first proposed by Desargues in the 1630s (Cundy and Rollett 1989). http://mathworld.wolfram.com/Cycloid.html prolate cycloid

curtate cycloid

e System of equations : Tangent:

t x = at − asin t The tangent e is perpendicular to the a segment that connects the point of the y = a − a cost cycloid and the momentary pole M. at M x

Geometrical Constructions 2 Roulettes 15 Hypotrochoid - Hypocycloid

A hypotrochoid is a roulette traced by a point P attached to a circle of radius b rolling around the inside of a fixed circle of radius a, where P is at a distance h from the center of the interior circle. The parametric equations for a hypotrochoid are y ⎛ a − b ⎞ x = ()a − b cost + hcos⎜ t ⎟ ⎝ b ⎠ ⎛ a − b ⎞ y = ()a − b sin t − hsin⎜ t ⎟. a ⎝ b ⎠ b

a : radius of base circle h b : radius of the rolling circle

h : distance of the point and the centre of x rolling circle

Hypocycloid: hypotrochoid with h = b.

In the figure, a : b = 4 : 1 so the curves consist of 4 courses. The curve is closed, if the ratio of radii is a rational number. The name of the hypocycloid in the figure; astroid. http://mathworld.wolfram.com/Hypotrochoid.html

Geometrical Constructions 2 Roulettes 16 Epitrochoid - Epicycloid

The roulette traced by a point P attached to a circle of radius b rolling around the outside of a fixed circle of radius a, where P is at a distance h from the center of the rolling circle.

The parametric equations for an epitrochoid are y ⎛ a + b ⎞ x = ()a + b cost − hcos⎜ t ⎟ ⎝ b ⎠ ⎛ a + b ⎞ b y = ()a + b sin t − hsin⎜ t ⎟. h ⎝ b ⎠ a : radius of base circle a b : radius of the rolling circle

h : distance of the point and the centre of x rolling circle

Epicycloid: epitrochoid with h = b.

In the figure, a : b = 3 : 1 so the curves consist of 3 courses. The curve is closed, if the ratio of radii is a rational number. http://mathworld.wolfram.com/Epitrochoid.html

Geometrical Constructions 2 Roulettes 17 Circle Involute

The roulette traced by a point P attached to a straight line rolling around a fixed circle of radius a, where P is at a distance h from the straight line. y

The parametric equations for an circle involute are h

x = a(cost + t sin t) − hcost y = a(sin t − t cost) − hsin t

a : radius of base circle a h : distance of the point and the rolling x straight line

Archimedean spiral: involute with h = a.

http://mathworld.wolfram.com/CircleInvolute.html

Geometrical Constructions 2 Roulettes 18 Chapter Review

Vocabulary

Geometrical Constructions 2 Roulettes 19 Conic Sections

Ellipse Parabola Hyperbola

Geometrical Constructions 2 Conic Sections 20 Ellipse

t P Definition: r1 r Let two points F1, F2 2 Ellipse is the set of points, major foci and a distance 2a whose sum of distances axis F F be given. 1 2 from F1 and F2 is equal to the given distance. axis minor

r1 + r2 = 2a. Dist(F1, F2 )=2c, a>c.

r1 r2 2a

The ellipse is symmetrical with respect to the straight line F1F2 , to the perpendicular

bisector of F1F2 and for their point of intersection O.

The tangent at a point is the bisector of the external angle of r1 and r2.

Geometrical Constructions 2 Conic Sections 21 Properties of Ellipse

Antipoint E: dist(P,E) = r , dist(F ,E) = 2a. director circle 2 1 Director circle: set of antipoints, circle about a focus with the radius of 2a.

principal circle The director circle is the set of points (antipoints) that are reflections of a focus with respect to all tangents

r1 r2 E of the ellipse. The ellipse is the set of centers of circles P passing through a focus and tangent to a circle, i.e. M the director circle about the other focus. r1 r2 2a O Principal circle: circle about O with the radius of a. r F F 2 1 2 The principal circle is the set of pedal points M of lines

from F2 perpendicular to the tangents of the ellipse.

Under the reflection in the ellipse, a ray emitted from a focus will pass through the other focus.

A point of the ellipse (P), the center of the director

circle (F1) and the antipoint (E) corresponding to the given point, are collinear.

Geometrical Constructions 2 Conic Sections 22 Approximate Construction of Ellipse

Approximate ellipse composed A’ Ellipse with osculating circles of circular arcs C C Osculating circle at C A”

K1 O O L 1 B A Osculating circle at B

Perpendicular bisector of AA” Perpendicular to BC K2

Osculating circle of a curve Let three (different, non collinear) points P1, P2 and P3 tend to the point P0. The three points determine a sequence of circles. If the limiting circle exists, this P1 osculating circle is the best approximating

P2 P3 circle of the curve at the point P0. P0

Geometrical Constructions 2 23 Parabola axis t Definition: Let a point F focus and a straight line d directrix Parabola is the set of be given. The line is not points equidistant from passing through the the focus and the dirctrix. point. P FP = dist(P,d) = PE

F V d E

The parabola is symmetrical with respect to the line, passing through the focus and perpendicular to the directrix. This line is the axis of the parabola.

The tangent at a point is the bisector of the angle Ë FPE.

Geometrical Constructions 2 Conic Sections 24 Properties of Parabola

s i Antipoint E: pedal point of the line passing

a through P perpendicular to d. The set of antipoints is the directrix.

The directrix is the set of points (antipoints) that are reflections of a focus with respect to all tangents of the parabola. The parabola is the set of centers of circles passing through the focus and tangent to a line, i.e. the P directrix.

Tangent at the vertex: set of pedal points M of F lines from F perpendicular to the tangents of M the parabola. d V Under the reflection in the parabola, a ray E emitted from the focus will be parallel to the axis.

Geometrical Constructions 2 Conic Sections 25 Osculating Circle at the Vertex of Parabola

K F

d V

The radius of the osculating circle at the vertex: r = dist(F,d)

Geometrical Constructions 2 Conic Sections 26 Hyperbola

Definition:

axis P 2a

Let two points F1, F2 conjugate Hyperbola is the set of r1 foci and a distance 2a points, whose difference of r1 r2 be given. distances from F1 and F2 is equal to the given distance. r2 traverse F1 F2 axis |r1 - r2|= 2a. Dist(F1, F2 )=2c, a

The hyperbola is symmetrical with respect to the straight line F1F2 , to the

perpendicular bisector of F1F2 and about their point of intersection O.

The tangent at a point is the bisector of the angle of r1 and r2.

Geometrical Constructions 2 Conic Sections 27 Asymptotes of Hyperbola

y Equation of hyperbola: x2 y 2 − = 1 a2 b2

c where b is determined by the Pythagorean b equation: 2 2 2 a x a + b = c . F1 F2 The limit of the ratio y/x can be found from the equation: ⎛ y 2 ⎞ ⎛ b2 b2 ⎞ ⎜ ⎟ = ⎜ − ⎟ lim ⎜ 2 ⎟ lim ⎜ 2 2 ⎟ x →∞ ⎝ x ⎠ x →∞ ⎝ a x ⎠ y b Construction: lim = ± 1) Draw the tangents at the vertices. x →∞ x a

2) Draw the circle with the diameter F1F2. 3) The asymptotes are the lines determined by the center and the points of intersection of the tangents at the vertices and the circle.

Geometrical Constructions 2 Conic Sections 28 Properties of Hyperbola

Antipoint E: dist(P,E) = r2, dist(F1,E) = 2a.

Director circle: set of antipoints, circle about a focus with the radius of 2a.

2a P The director circle is the set of points (antipoints) that director circle are reflections of a focus with respect to all tangents r 2 of the hyperbola. The hyperbola is the set of centers of circles passing through a focus and tangent to a circle, r2 r1 r E i.e. the director circle about the other focus. 1 M r2 O Principal circle: circle about O with the radius of a.

F1 F2 The principal circle is the set of pedal points M of lines

from F2 perpendicular to the tangents of the hyperbola.

Under the reflection in the hyperbola, the line of a ray principal circle emitted from a focus will pass through the other focus.

A point of the ellipse (P), the center of the director

circle (F1) and the antipoint (E) corresponding to the given point, are collinear.

Geometrical Constructions 2 Conic Sections 29 Chapter Review

Vocabulary

Geometrical Constructions 2 Conic Sections 30 Solutions of 3D Geometrical Constructions

Axonometric sketch

Multi-view representation Bold-Italic: will be discussed in Description in pseudo-code this course Italic: will be discussed in other corses offered by our Modeling department Normal: not in our scope Representation by relief

Stereoscopic representation

Perspective

Analytical geometric solution

Computer geometric modeling

Geometrical Constructions 2 3D geometrical constructions 31 Axonometric sketch 1

Front view of P

It is based on 2D representation of 3D z P” Q Cartesian coordinate-system. Points are represented by means of coordinates, such that we measure x, y and z in three independent directions. As a P transformation, it is a 3D ⇒ 2D

degenerated linear mapping that 1 U preserves parallelism. z z R Q’ x The axonometric system is determined O 1 Ux U by the points {O, Ux, Uy, Uz}, the y y 1 image of the origin and the units on the R’ axes x, y and z. x y P’

The axonometric sketch is similar to the Top view of P parallel projection of the figure. Frontal axonometry

Geometrical Constructions 2 3D geometrical constructions 32 Axonometric Sketch 2

Geometrical Constructions 2 3D geometrical constructions 33 Multi View Representation 1

Front view of P 2nd quadrant π 2 1st quadrant π P” 2 P” d2 P x1,2 P

d1 P’ π1 d1 rd Top view of P x1,2 3 quadrant

d2 P’ th π1 4 quadrant

The image planes are also perpendicular. The images are orthogonal projections on image planes.

Geometrical Constructions 2 34 Multi View Representation 2

From the bottom Front view

Side view (from left) Side view (from right)

From back

Top view

Geometrical Constructions 2 3D geometrical constructions 35 Multi View Representation 3

From the bottom Front view

Side view (from the left)

Side view (from the right) From back

Top view

Geometrical Constructions 2 3D geometrical constructions 36 Description in pseudo-code

Pseudo-code (pseudo-language): a form of representation used to provide an outline description of a geometrical algorithm. Pseudo-code contains a mixture of natural language expressions, set algebra notation and symbols of geometrical relations.

An example: let a pair of skew lines and a plane be given. The plane is not parallel to the lines. Let X be an arbitrary point of a. Find the transversal of the lines b* X b*∥ b -parallel to the plane Pa = a 1 a -the shortest one among the transversals that * satisfy the first condition Pb* = b 1 a

Sketch: * a b Pb = b 1 a a B b* l = |Pa, Pb*| t l* P l*∥ l A b Pb

n a n Pa n z l X a l* n a* n 1 l* a*∥a Pb* B = b 1 a* Pa l t B t ∥ n

Geometrical Constructions 2 3D geometrical constructions 37 Modeling

Development Plaster, plastic, wood, etc. Wireframe (string) model

Geometrical Constructions 2 3D geometrical constructions 38 Representation by relief

Relief sculpture - A type of sculpture in which form projects from a background.

Donatello: The feast of Herod Placed in the Siena baptismal font. This work was completed between 1423 and 1427 and may be included in Donatello’s revolutionary innovations in interpreting both the motives of the naturalistic narration and the perspective vision.

In architecture, properties of relief perspective are applied in stage architecture.

http://www2.evansville.edu/studiochalkboard/draw.html http://wwar.world-arts-resources.com/default.html http://www.artlex.com/ArtLex/r/relief.html http://www.metmuseum.org/works_of_art/

Geometrical Constructions 2 3D geometrical constructions 39 Stereoscopic Viewing

The human visual system has a physical configuration that supports two separate images to be gathered (each eye). Because the brain combines these into one, a small, but important mathematical difference exist between these images. This minuscule difference is represented by the figure below.

This effect adds to the depth perception information the brain needs to derive an image with three dimensions. Combining these images makes a three- dimensional image.

http://www.hitl.washington.edu/scivw/EVE/III.A.1.b.StereoscopicViewing.html http://en.wikipedia.org/wiki/Stereoscopy

Geometrical Constructions 2 3D geometrical constructions 40 Stereoscopy

To see stereoscopically, just hide the left image from your left eye and the right image from your right eye by your palms and then try to concentrate your sight on a small object located at the intersection of the line from your left eye to the right figure and the line from your right eye to the left image. The stereoscopically viewed objects are formed in the space in front of the observer and not on the screen.

http://www.pattakon.com/educ/Stereoscopy.htm C:\TEMP\diamonds.exe

Geometrical Constructions 2 3D geometrical constructions 41 Stereogram

Bela Julesz, in front of a picture from his and A. Michael Noll's computer art exhibition, “Computer-Generated Pictures,” held at the Howard Wise Gallery, New York City, in 1965. Photograph courtesy of Rutgers University http://www.eyetricks.com/3dstereo.htm

Geometrical Constructions 2 3D geometrical constructions 42 Perspective

http://employees.oneonta.edu/farberas/arth/arth200/durer_artistdrawingnude.html

http://www2.evansville.edu/studiochalkboard/draw.html http://www.mcescher.com/

Geometrical Constructions 2 3D geometrical constructions 43 Analytical geometric solution

P(x, y, z)

l x = x0 + v1t

y = y0 + v2t

z = x0 + v3t

Point ordered triplet of real number

Straight line linear system of equations

Plane lineal equation of x, y, z variables

Point lying on a plane coordinates of the point satisfy the equation of the plane

Line perpendicular to a plane the normal vector of the plane and the direction vector

of the line are linearly dependent

e.t.c.

Geometrical Constructions 2 3D geometrical constructions 44 Computer geometric modeling

• CAD systems (Computer Aided Design) AutoCAD, Microsystem, ArchiCAD, …

• Computer algebra systems MATHEMATICA, MAPLE, …

Example: construct a regular heptagon. In AutoCAD,

Command: _polygon Enter number of sides <4>: 7

Specify center of polygon or [Edge]: 0,0

Enter an option [Inscribed in circle/Circumscribed about circle] : i

Specify radius of circle: 10

Geometrical Constructions 2 3D geometrical constructions 45 Solid Geometry, Introduction

If we are content to work in two dimensions, we say: all points are in one plane.

If not, we say instead: if [ABC] is a plane, there is a point D not in this plane. (Coxeter: Introduction to Geometry)

A, B, C, D: four non-coplanar points D ABCD: tetrahedron k A, B, C, D: vertices (4) P AB, BC, CA, AD, BD, CD: edges (6) ABC, BCD, CAD, ABD: faces (4) C Euler’s formula: v + e - f = 2 = 4+4-6

AB segment A, B A |AB|= l line A, B (kl) = B = k 1 l intersection of k, l l a = [BCD]plane of B, C, D B P a, P ∈ a lying on, incident A a, A ∉ a not lying on, non-incident

Geometrical Constructions 2 3D geometrical constructions 46 Relations of Spatial Elements

pair of points: A l B determine a line: |AB|= l, determine a distance: dist(A,B) point and line: 1) lying on 2) not lying on determine a plane: [P,l] = P α determine a distance: dist(P,l) = dist(P,P’) P’ (P’ is the orthogonal projection of P on l) l pair of lines: 1) coplanar intersecting a P b determine angles, determine a plane

parallel c determine a distance , determine a plane d

2) non-coplanar, skew determine angles and distance, no plane

n f in common e n z e, n z f

Geometrical Constructions 2 3D geometrical constructions 47 Relations of Spatial Elements

point and plane: 1) lying on 2) not lying on P determine a distance: dist(P,α) = dist(P,P’) α P’ (P’ is the orthogonal projection of P on α) line and plane: 1) lying on 2) parallel l α determine a distance: dist(l, α) = dist(l,l’) l’ (l’ is the orthogonal projection of l on α, l’ 2 l) 3) intersecting (non-perpendicular) α l determine an angle: ∡(l, α) = ∡(l,l’) l’ (l’ is the orthogonal projection of l on α) α* pair of planes: 1) parallel P determine a distance: P α*,dist(P, α) = dist(P,P’) α P’ (P’ is the orthogonal projection of P on α) a 2) intersecting β b determine an angle: l = α 1 β, ∡(α, β) = ∡(a,b) l α (both a α and b β are perpendicular to l)

Geometrical Constructions 2 3D geometrical constructions 48 Selected Theorems on Parallelism

l 1 a =A , l 1 a =A l A A 1 1 2 2 α 2 b 1 2 β l a1 2 a2 l b a1 a2 l, a1, a2: coplanar α l 2 α a 2 b, a 2 c, c a c, b d d 2 2 a β a, b: intersecting b b 2 c c c, d: intersecting b a α =[a,b] b=[c,d] α 2 β α2 β, 2 γ , α b β α g 2 γ G l L β 2 2 g 2 l 2 γ L1 G1 α dist(G1,L1) = dist(G2,L2) α l 1αβ = g α 2 β, γ intersects α β l 2 α, l 2 β γαintersects β g β 1 γ 2 β 1 γ α l g 2 α γ

Geometrical Constructions 2 3D geometrical constructions 49 Selected Theorems on Perpendicularity

Pair of skew lines: Uniqueness of perpendicular line through a point: P b * * through a point P, a* a 2a, b 2b a there is one and n a*, b* intersecting, only one line n a*z b* perpendicular to α a plane α. b* a z b Uniqueness of perpendicular plane through a point: Line and plane: through a point P, n a 1 b =P there is one and n n z a, n z b only one plane α P c α = [a,b] perpendicular to P α b a line n. a c n z c Uniqueness of perpendicular plane through a line: Perpendicular planes: γ through a line l, n α non-perpendicular β bz l β n to a plane α, n there is one and only one plane b β α α z perpendicular to the plane α. α

Geometrical Constructions 2 3D geometrical constructions 50

α Geometrical Loci in 3D

Range of points Pencil of lines in a plane Pencil of planes collinear points concurrent lines

Field of points Boundle of lines Boundle of planes coplanar points concurrent lines concurrent planes

Geometrical Constructions 2 3D geometrical constructions 51 Geometrical Loci in 3D

Set of points at a given distance from a given point Set of points at a given is a sphere. distance from a given line Set of points equidistant is a cylinder. from two given points is the perpendicular bisector E.t.c. plane of the segment.

Geometrical Constructions 2 3D geometrical constructions 52 Geometrical Loci in 3D

Find the set of spatial element that satisfy the conditions as follows: 1) Set of points at a given distance from a given point … 2) Set of planes at a given distance from a given point … 3) Set of planes at a given distance from a given line … 4) Set of planes equidistant from two given points … 5) Set of points at a given distance from two given points … 6) Set of points equidistant from three non-collinear points … 7) Set of points equidistant from a pair of intersecting planes 8) Set of points equidistant from a pair of parallel planes … 9) Set of lines through a point, perpendicular to a given line … 10) Set of lines, parallel to a pair of intersecting planes … 11) Set of lines that form a given acute angle with a given line, through a point of the given line … 12) Set of planes that form a given acute angle with a given line, through a point of the given line … 13) Set of points, whose distance from a point is equal to the distance from a plane … 14) Set of points, whose sum of distances from two given points is equal to a given length … 15) Set of points, whose difference of distances from two given points is equal to a given length …

Geometrical Constructions 2 3D geometrical constructions 53 Loci Problems

1) Find the points of a straight line, whose distance from a given line is equal to a given length.

2) Find the points in the space, whose distance from the point A is r1, from the point B is r2 and from the point C is r3. 3) Let a point A, a straight line b, a plane a and a distance r be given. Find the straight lines intersecting with b, lying in the plane a such that their distance from A is equal to the given length r. 4) Let a straight line a, a plane a and an angle g be given. Find the planes passing through the line a, such that they form an angle equal to g with the plane a. 5) Let a pair of intersecting planes and a straight line be given. Find the planes passing through the line, such that they form equal angles with the given planes. 6) Let a point A, a pair of intersecting planes a, b and an angle g be given. Find the straight lines passing through the point A, parallel to the plane a and form an angle g with b. 7) Let a point A, a pair of skew lines b, c and an angle g be given. Find the straight lines passing through the point A, intersecting with b and form an angle g with c. 8) Let a point A, a pair of skew lines b and c, and two angles g and d be given. Find the lines passing through A such that the angles formed by b and c are the given angles g and d respectively. 9) Let a point A, a straight line b, a plane a and two angles g and d be given. Find the straight lines passing through A, such that the angles formed by the line b and the plane a are equal to the given angles g and d respectively. 10) Let a point A, a straight line b and a plane a and two angles g and d be given. Find the planes passing through A, such that the angles formed by the line b and the plane a are equal to the given angles g and d respectively. 11) Let two pints, a plane and a distance be given. Find the spheres of the given radius, passing through the points and tangent to the given plane.

Geometrical Constructions 2 3D geometrical constructions 54 Solution of 3D Problems

1) Read and reread the problem Show that a tetrahedron has orthocenter if and only if the carefully. opposite edges are perpendicular. B

2) Find the proper type of 1) Let AB and CD perpendicular. O P representation (simple CD perpendicular to [ABP], projection, axonometric CD perpendicular to [ABQ]. sketch, multi-view rep.). (P and Q are the pedal points D of the altitudes from A and B 3) Draw a sketch as if the A respectively.) Q problem was solved. There is only one plane 4) Try to find relations between through AB perpendicular to C the data and the unknown. CD, so AP and BQ are coplanar, so they are intersecting and the orthocenter exists. 5) Write down the solution, use 2) If AP and BQ are intersecting and both AP and BQ are pseudocode. perpendicular to CD, than AB lies in a plane perpendicular to CD. Consequently, AB is perpendicular to CD too. 6) Analyze the solvability and the The condition is necessary and sufficient. number of solution (and write down your results). Exercise: let a pair of skew lines and a plane be given. The plane is not parallel to the lines. Find the shortest transversal 7) Read the question and your of the lines parallel to the plane. answer again.

Geometrical Constructions 2 3D geometrical constructions 55 Geometrical Constructions in 3D

1. Show that the medians of a tetrahedron meet at a point. This centroid of tetrahedron divides the medians in the ratio of 1 : 3. 2. Let four non-coplanar points A, B, C and D be given. Find the plane a such that A and B are on one side of a, C and D are on the other side of a and their distance from a are equal. 3. Slice a cube into three congruent parts such that not only the volumes but the surface areas would be equal. Find the solution if the planes are not parallel to a faces. 4. Mark a pair of opposite vertices of a cube. Find the midpoints of the edges that do not meet at the marked vertices. Show that the midpoints are vertices of a regular hexagon. 5. Find the direction of parallel projection, that projects four non-coplanar points into a parallelogram. 6. Make a hole on a cube such that a cube congruent to the original one can slide through it. 7. Two points and a plane not passing through the points are given. Find the locus of points of tangency of spheres, passing through the points and tangent to the plane 8. The two endpoints of a segment are moving on a pair of perpendicular skew lines. Find the locus of the midpoints of the segment. 9. The three planes of a Cartesian coordinate system reflect a ray of light as mirrors. Prove that the direction of the ray that hits all the three planes is the opposite of the ray. 10. Three non-coplanar circles are pairwise tangential. Prove that the three circles lie on a sphere.

Geometrical Constructions 2 3D geometrical constructions 56 Chapter Review

Vocabulary

Geometrical Constructions 2 3D geometrical constructions 57 Platonic Solids

A regular polyhedron is one whose faces are identical regular polygons.

The solids as drawn in Kepler’s Mysterium Cosmographicum:

tetrahedron octahedron icosahedron cube dodecahedron (Fire) (Air) (Water) (Earth) (Universe)

http://www.math.bme.hu/~prok/RegPoly/index.html

Geometrical Constructions 2 Regular and semi-regular polyhedra 58 Faces around a vertex

Only five regular solids are possible. Schläfli symbol {p, q} means: the faces are regular p-gons, q surrounding each vertex.

{5, 3} P {4, 3}

{3, 5}

P P P

{3, 3} {3, 4}

Geometrical Constructions 2 Regular and semi-regular polyhedra 59 Archimedean Polyhedra

The 13 Archimedean solids are the convex polyhedra that have a similar arrangement of nonintersecting regular convex polygons of two or more different types arranged in the same way about each vertex with all sides the same length (Cromwell 1997, pp. 91-92).

http://mathworld.wolfram.com/ArchimedeanSolid.html

Geometrical Constructions 2 Regular and semi-regular polyhedra 60 Archimedean Polyhedra

Geometrical Constructions 2 Regular and semi-regular polyhedra 61 Fullerains (named after Buckminster Fuller)

The Royal Swedish Academy of Sciences has awarded the 1996 Nobel Prize for Chemistry jointly to:

Professor• Robert F. Curl, Jr., Rice University, Houston, USA

Professor• Sir Harry W. Kroto FRS, University of Sussex, Brighton, UK

Professor• Richard E. Smalley, Rice University, Houston, USA For their Discovery of Fullerenes. In 1985 one of the greatest new discoveries in science was made when chemists Richard Smalley and Harold Kroto discovered the existence of a third form of carbon. Unlike the two other forms of carbon, diamond and graphite, this amazing 60-atom cage molecule was shaped like a soccer ball. A highlight of one of the Both Kroto and Smalley felt it pentagonal rings most appropriate to name it, "buckminsterfullerene" for its striking resemblance to a geodesic dome. A new family of these molecules have since been found called "fullerenes." (Note: Diamond is a molecular network crystal with each carbon bonded to four others in a tetrahedral configuration. Graphite is formed in flat sheets with each carbon Buckminster Fuller's Dome - A highlight of one of the bonded to three others in a Expo '67 Montreal hexagonal rings hexagonal configuration.)

Geometrical Constructions 2 Regular and semi-regular polyhedra 62 Regular Star Polyhedra

Two star polyhedra were discovered by Poinsot in 1809. The others were discovered about 200 years before that by Johannes Kepler (1571-1630), the German astronomer and natural philosopher noted for formulating the three laws of planetary motion, now known as Kepler's laws, including the law that celestial bodies have elliptical, not circular orbits.

Stellation is the process of constructing polyhedron by extending the facial planes past the polyhedron edges of a given polyhedron until they intersect (Wenninger 1989). The set of all possible polyhedron edges of the stellations can be obtained by finding all intersections on the facial planes. The Kepler-Poinsot solids consist of the three dodecahedron stellations and one of the icosahedron stellations, and these are the only stellations of Platonic solids which are uniform polyhedra. http://www.korthalsaltes.com

Geometrical Constructions 2 Regular and semi-regular polyhedra 63 Art and Science

JACOPO DE 'BARBERI: Luca Pacioli, c. 1499 Leonardo's Illustrations for Luca's book. Da This painting shows Fra Luca Pacioli and his Divina Proportione student, Guidobaldo, Duke of Urbino. In the upper Luca Pacioli wrote a book called Da Divina Proportione left is a rhombi-cuboctahedron, and on the table is (1509) which contained a section on the Platonic a dodecahedron on top of a copy of Euclid's Solids and other solids, which has 60 plates of solids Elements. by none other than his student Leonardo da Vinci.

http://www.math.nus.edu.sg/aslaksen/teaching/math-art-arch.shtml#Polyhedra

Geometrical Constructions 2 Regular and semi-regular polyhedra 64 M. C. ESCHER (1902-1972)

Escher made a set of nested Platonic Solids. When he moved to a new studio he have away most of his belongings but took his beloved model.

Stars, 1948 Note the similarity between this polyhedron and Leonardo's illustrations for Pacioli's book

Geometrical Constructions 2 Regular and semi-regular polyhedra 65 Models

Geometrical Constructions 2 Regular and semi-regular polyhedra 66 Chapter Review

Vocabulary

Geometrical Constructions 2 Regular and semi-regular polyhedra 67 Solid Geometry Formulae 1

Solid/Surface Surface Area Volume

c S = 2(ab + bc + ac ) V = abc A b a

3 2 4 r π r S = 4 r π V = 3

= + h S 2 A Ph V = Ah A P

h S = 2 A + Ch = = = π V Ah A r = 2 + π 2 C 2 r 2 rh = r π h

Geometrical Constructions 2 Geometrical Calculations 68 Solid Geometry Formulae 2

Solid/Surface Surface Area Volume

L h Ah S = A + L V = A 3

a Geometrical Calculations L h h S = A + a + L V = ()A + Aa + a A 3

1 l S = A + Cl = h 2 r 2 π h 2 π 2 2 V = r A = r + π r h + r C 3 r

h l 2 2 π h S = π (R + r + (R + r)l) V = ()R2 + Rr + r2 R 3

Geometrical Constructions 2 Geometrical Calculations 69 Solid Geometry Calculations 1

C B 6) The first two steps of a 10-step 1) In the given cube, E is the staircase are shown in the sketch. Find the amount of midpoint of the side CD. E 15 cm Find the measure of the concrete needed to the exposed angle BAE. portion of the staircase. Find the D area of carpet needed to cover 80 cm A the front, top and sides of the 20 cm 2) A standard tennis ball can is a cylinder that holds concrete steps. three tennis balls. Which is greater, the 10 cm 7) A sculpture made of iron has the circumference of the can or its height? If the radius shape of a right square prism topped of a tennis ball is 3.5 cm, what percentage of the by a sphere. The metal in each part of can is occupied by air, out of tennis balls? the sculpture is 2 mm thick. If the 50 cm 3) A right cylinder is inscribed inside a sphere. If outside dimensions are as shown and 3 altitude of the cylinder is 8 cm and volume is 72π the density of iron 7.78 g/cm , cm3, find the radius of the sphere. calculate the mass of the sculpture in kilograms. 4) The sketch shows the pattern 16 cm 16 cm 20 cm (or net) of a pyramid. 10 cm Calculate the surface area and 8) The length, the width and the altitude of a rectangular the volume of the three- prism is directly proportional by 3,4 and 5. If the dimensional figure formed by diagonal of the rectangular prism is 200 cm, find the 2 the pattern. total surface. (cm ) 40 cm

16 cm 9) A tank has the shape shown. 5) A soft drink cup is in the shape of right circular conic Calculate the volume and the frustum. The capacity of the cup is 250 milliliter and surface area of the tank. 30 cm the top and bottom circles are of 6 cm and 4 cm (diameters) respectively. How deep is the cup?

Geometrical Constructions 2 Geometrical Calculations 70 Solid Geometry Calculations 2

10) Determine the ratio of the surface areas 15) A rectangular piece of A/4 size paper (297 mm by of sphere and circumscribed cylinder, 210 mm) can be rolled into cylinder in two and the ratio of volumes of sphere and different directions. If there is no overlapping, circumscribed cylinder. (Diagram on which cylinder has greater volume, the one with Archimedes’ tombstone.) the long side of rectangle as its height, or the one with the short side of the rectangle as its height? 11) A right circular cone has a volume of 210 m3. The 16) A right rectangular prism has a volume of 324 height of the cone is the same length as the diameter cubic units. One edge has measure twice that of a of the base. Find the dimensions of the cone. second edge and nine times that of the third edge. What are the dimensions of the prism? 12) A right cone with radius 6 units and altitude 8 units is 17) The trapezoid given in the figure A given. What is the area of the sphere with the is revolved 360° around the side greatest volume that can be inscribed in the cone? AB. Find the volume of the solid B 13) The pyramid {T,A,B,C} is cut into T body formed. three pieces of equal height by 18) By using the sector given in the planes parallel to the base as figure, a right cone is formed. If shown in the figure. Find the the altitude of the cone is 165 cm, ratios of the three volumes of C find the lateral area of the cone. the pyramid {T,A,B,C}. C A A B 19) Find the ratio between the volume of D 14) Sketched is a pattern for a three 3 cm the pyramid ABCD and the cube. dimensional figure. Use the B dimensions to calculate the 20) The base of a truncated pyramid is an equilateral surface area and the volume of 5 cm triangle of 50 cm side. The height of the solid is the 3D figure formed bz the 20 cm and the angle formed by base and the pattern. lateral face is 60°. Calculate the surface area and the volume.

Geometrical Constructions 2 Geometrical Calculations 71