Geometrical Constructions 2 - epab.bme.huepab.bme.hu/geoc2/GC2_Lecture_notes_11_Spring.pdfGeometrical Constructions 2 2 "When he established the heavens I was there: when he set a

Geometrical Constructions 2by Pál Ledneczki Ph.D.

Table of contents

1) Pencils of circles, Apollonian problems

2) Conic sections

3) Approximate rectification of an arc

4) Roulettes

5) Methods of representation in 3D

6) 3D geometrical constructions

7) Regular and semi-regular polyhedrons

8) Geometrical calculations

Geometrical Constructions 22

"When he established the heavens I was there: when he set a compass upon the face of the deep.“

(Proverbs, Chapter 8 par. 27 )

God the Geometer, Manuscript illustration.

Geometria una et aeterna est in mente Dei refulgens: cuiusconsortium 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)


Pencil of Circles

Apollonian Problems

radicalaxis

Intersecting (or „elliptic”) pencilof coaxal circles

P

For three circles whose centers form a triangle, the three radical axes (of the circles taken in pair) concur in a point called the radical center.

Radical center C

All tangents drawn to the circles of a coaxal pencil from a point on the radical axis have the same length.

C


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

Apollonian Problems

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


Apollonius Circle

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

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

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

Apollonian Problems

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


Apollonian Problems on Tangent Circles

Combinatorial approach

Circle C

Point (circle of 0 radius) P

Straight line (circle of infinite radius) L

Options

1) (PPP)

2) (PPL)

3) (PPC)

4) (PLL)

5) (PLC)

6) (PCC)

7) (LLL)

8) (LLC)

9) (LCC)

10) (CCC)

Apollonian tangent circle problem: choose three from the elements

of the set {P, L, C} (an element can be chosen repeatedly) and fid the

circles tangent to or passing through the given elements.

(In combinatorics: third class combinations with repetitions of three

elements.)

subject of our course

not subject of our course

Apollonian Problems


1) (PPP) 2) (PPL)

P1

P3

P2

P1

P2

l

Apollonian Problems


3) (PPC), 4) (PLL)

P1

P2

c

l1

l2

P

Apollonian Problems


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

A2

B

Q

P

P

c

l

Q1 Q2

Apollonian Problems


7) (LLL) 8) (LLC)

l1

l2l3

Hint: the solutions are the incircle andthe excircles of the triangle formed by the lines. The centers are the points of intersection of the bisectors of the interior and exterior angles.

Hint: the problem can be reduced to the(PLL) by means of dilatation that means, draw parallels at the distance of the radius of the given circle. The circles passing through the center of the given circle and tangent to the parallel lines, have the same centers as the circles of solution.

l1

l2

c

Apollonian Problems


Chapter Review

Vocabulary

Apollonian Problems


Conic Sections

Conic Sections

Ellipse Parabola Hyperbola


Ellipse

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.

2a

r1 r2

r1r2

F2F1

P

Let two points F1, F2

foci and a distance 2abe given.

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

Ellipse is the set of points, whose sum of distances from F1 and F2 is equal to the given distance.

r1 + r2 = 2a.

Definition:

t

major axis

min

or

axis

Conic Sections


Properties of Ellipse

2a

r1

r2

r1 r2

r2

F1 F2

P

O

E

director circle

principal circle

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

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

The director circle is the set of points (antipoints) that are reflections of a focus with respect to all tangents of the ellipse. The ellipse is the set of centers of circles passing through a focus and tangent to a circle, i.e. the director circle about the other focus.

Principal circle: circle about O with the radius of a.

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.

M

Conic Sections


Osculating circle at B

Osculating circle of an Ellipse

BL1

L2

Osculating circle at CC

O

Perpendicular to BC

P1

P2 P3P0

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 osculating circle is the best approximating circle of the curve at the point P0.

Osculating circle of a curve

Conic Sections

A


Perpendicular bisector of AA”

Approximate Construction of Ellipse

A

A’

A”

O

C

K1

K2

Approximate ellipse composed

of two circular arcs

Conic Sections

Approximate ellipse composed

of three circular arcs (Five-Center Method)

L1

L2

K

B

C


Parabola

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.

Let a point F focus and a straight line d directrixbe given. The line is not passing through the point.

Parabola is the set of points equidistant from the focus and the dirctrix.

axis

F

d

P

V

E

Definition:t

FP = dist(P,d) = PE

Conic Sections


Properties of Parabola

Antipoint E: pedal point of the line passing 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 directrix.

Tangent at the vertex: set of pedal points M of lines from F perpendicular to the tangents of the parabola.

Under the reflection in the parabola, a ray emitted from the focus will be parallel to the axis.

d

F

P

E

axi

s

VM

Conic Sections


Osculating Circle at the Vertex of Parabola

d

Faxi

s

V

K

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

Conic Sections


Hyperbola

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.

Let two points F1, F2

foci and a distance 2abe given.

Dist(F1, F2 ) =2c, a<c.

Hyperbola is the set of points, whose difference of distances from F1 and F2 is equal to the given distance.

|r1 - r2| = 2a.

Definition:

2a

r1

r2

r1

r2

F2F1

P

t

traverseaxis

conju

gat

eax

is

Conic Sections


Asymptotes of Hyperbola

12

2

2

2

=−by

ax

.222 cba =+

ab

xy

xb

ab

xy

x

xx

±=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∞→

∞→∞→

lim

limlim 2

2

2

2

2

2

F2F1

y

a

cb

x

Equation of hyperbola:

where b is determined by the Pythagorean

equation:

The limit of the ratio y/x can be found from the

equation:

Construction:1) Draw the tangents at the vertices.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 thevertices and the circle.

Conic Sections


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.

The director circle is the set of points (antipoints) that are reflections of a focus with respect to all tangents of the hyperbola. The hyperbola is the set of centers of circles passing through a focus and tangent to a circle, i.e. the director circle about the other focus.

Principal circle: circle about O with the radius of a.

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 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.

2a

r2

r1

r2

F1 F2

P

O

E

director circle

principal circle

r2

Mr1

Conic Sections


Chapter Review

Vocabulary

Conic Sections


Approximate Rectification of an Arc

A

O

C

30°

≈ rπ

r

r

B

D

Kochansky’s method:

CD = 3r

DB ≈ rπ

Snellius’ method:

AC = 3r

α ≤ 30°

AP’ ≈ APO

α

P P’

C rA

Approximate rectification of an arc


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

Roulettes

http://www.math.uoc.gr/~pamfilos/eGallery/problems/Roulette2.html


curtate 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

Cycloid

x

t

y

System of equations :

taaytaatx

cossin

−=−=

Tangent:

The tangent e is perpendicular to the segment that connects the point of the cycloid and the momentary pole M.

a

at M

e

Roulettes

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


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

( )

( ) .sinsin

coscos

⎟⎠⎞

⎜⎝⎛ −

−−=

⎟⎠⎞

⎜⎝⎛ −

+−=

tb

bahtbay

tb

bahtbax

a : radius of base circle

b : radius of the rolling circle

h : distance of the point and the centre ofrolling circle

Hypocycloid: hypotrochoid with h = b.

x

a b

h

In the figure, a : b = 1 : 4 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

y

Roulettes

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


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

( )

( ) .sinsin

coscos

⎟⎠⎞

⎜⎝⎛ +

−+=

⎟⎠⎞

⎜⎝⎛ +

−+=

tb

bahtbay

tb

bahtbax


b : radius of the rolling circle

h : distance of the point and the centre ofrolling circle

In the figure, a : b = 1 : 3 so the curves consist of 3 courses. The curve is closed, if the ratio of radii is a rational number.

Epicycloid: epitrochoid with h = b.

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

x

y

a

bh

Roulettes

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


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.

The parametric equations for an circle involute are

thtttaythtttax

sin)cos(sincos)sin(cos

−−=−+=


h : distance of the point and the rolling straight line

Archimedean spiral: involute with h = a.

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

x

y

a

h

Roulettes

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


Chapter Review

Vocabulary

Roulettes


Methods of Representation in 3D

Axonometric sketch

Multi-view representation

Description in pseudo-code

Modeling

Representation by relief

Stereoscopic representation

Perspective

Analytical geometric solution

Computer geometric modeling

Bold-Italic: will be discussed in this course

Italic: will be discussed in other course offered by our department

Normal: not in our scope

3D geometrical constructions


It is based on 2D representation of 3D

Cartesian coordinate-system. Points are

represented by means of coordinates,

such that we measure x, y and z in

three independent directions. As a

transformation, it is a 3D ⇒ 2D

degenerated linear mapping that

preserves parallelism.

The axonometric system is determined

by the points {O, Ux, Uy, Uz}, the

image of the origin and the units on the

axes x, y and z.

The axonometric sketch is similar to the

parallel projection of the figure.

Top view of P

1

Front view of P

Axonometric sketch 1

y

x

z

P

Q

R’

P’

R

Uz

Ux

Uy

O

1

x

z

Frontal axonometry

Q’

1

P”

y



Axonometric Sketch 2



Multi View Representation 1

P”

P’

P

π2

π1

1st quadrant2nd quadrant

3rd quadrant

4th quadrant

x1,2

π2

π1

P”

P’

P

x1,2

d1

d1

d2

d2

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

Front view of P

Top view of P



Front view

Top view

Side view (from left)

From the bottom

Side view (from right)

From back




Front view

Top view

Side view (from the left)

From the bottom

Side view (from the right)

From back



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. Find the transversal of the lines

-parallel to the plane

-the shortest one among the transversals that satisfy the first condition

ab

αX

A

B b*

Pa

Pb*

Pb

n

t

l

l*

Sketch:

Let X be an arbitrary point of a.

b* X b*∥ b

Pa = a 1 α

Pb* = b* 1 α

Pb = b 1 α

l = |Pa, Pb*|

l* Pb l*∥ l

n α n Pa n ⊥ l

a* n 1 l* a* ∥ a

B = b 1 a*

t B t ∥ n

a*



Modeling

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



Representation by relief

Donatello: The feast of HerodPlaced 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.

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

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

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/

http://www.thais.it/scultura/image/sch00094.htm


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.htmlhttp://en.wikipedia.org/wiki/Stereoscopy


http://www.hitl.washington.edu/scivw/EVE/III.A.1.b.StereoscopicViewing.html

http://en.wikipedia.org/wiki/Stereoscopy


Stereoscopy

To see stereoscopically, just hide the left image from your left eye and the right imagefrom 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.

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


http://www.pattakon.com/educ/Stereoscopy.htm


Stereogram

http://www.eyetricks.com/3dstereo.htm


http://www.eyetricks.com/3dstereo.htm


Perspective

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

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


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

http://www.mcescher.com/

http://www2.evansville.edu/studiochalkboard/draw.html

http://employees.oneonta.edu/farberas/arth/Images/ARTH_214images/Durer/Durer_perspNude.jpg


Analytical geometric solution

P

P(x, y, z)

x = x0 + v1t

y = y0 + v2t

z = x0 + v3t

l

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.



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>: i

Specify radius of circle: 10



AB segment A, B

|AB| = l line A, B

(kl) = B = k 1 l intersection of k, l

a = [BCD] plane of B, C, D

P a, P ∈ a lying on, incident

A a, A ∉ a not lying on, non-incident

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)

l

A

B

C

D

k

A, B, C, D: four non-coplanar points

ABCD: tetrahedron

A, B, C, D: vertices (4)

AB, BC, CA, AD, BD, CD: edges (6)

ABC, BCD, CAD, ABD: faces (4)

Euler’s formula: v + e - f = 2 = 4+4-6

P


Solid Geometry, IntroductionSolid Geometry, Introduction


Relations of Spatial Elements

pair of points: 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] = α

determine a distance: dist(P,l) = dist(P,P’)

(P’ is the orthogonal projection of P on l)

pair of lines: 1) coplanar

intersecting determine angles, determine a plane

parallel determine a distance , determine a plane

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

in common

n z e, n z f

A Bl

l

P

P’

a bP

cd

efn



point and plane: 1) lying on2) not lying on

determine a distance: dist(P,α) = dist(P,P’)

(P’ is the orthogonal projection of P on α)line and plane: 1) lying on

2) paralleldetermine a distance: dist(l, α) = dist(l,l’)

(l’ is the orthogonal projection of l on α, l’ 2 l)

3) intersecting (non-perpendicular)determine an angle: ∡(l, α) = ∡(l,l’)

(l’ is the orthogonal projection of l on α)pair of planes: 1) parallel

determine a distance: P α*, dist(P, α) = dist(P,P’)

(P’ is the orthogonal projection of P on α)2) intersecting

determine an angle: l = α 1 β, ∡(α, β) = ∡(a,b)

(both a α and b β are perpendicular to l)

Relations of Spatial Elements

P

P’

α

l

l’

P

P’

α

l

l’

α*

β

l

α

α

α

a

b



α 2 b

l b

l 2 α

a 2 c, b 2 da, b: intersecting

c, d: intersecting

α =[a,b] 2 b=[c,d]

α 2 β

g 2 l

dist(G1,L1) = dist(G2,L2)

α 2 β, γ intersects α

γ intersects β

α 1 γ 2 β 1 γ

Selected Theorems on Parallelism

l 1 a1 =A1, l 1 a2 =A2

a1 2 a2

l, a1, a2: coplanar

a 2 b, a 2 c,

b 2 c

α2 β, α2 γ ,

β 2 γ

α 1 β = g

l 2 α, l 2 β

l 2 g

A1 A2l

a1 a2

ab

c

γ

α

β

α

β

g

l

β

α

l

β

α ab

cd

β

α

gG2

L2

G1L1

l

β

αγ



n z αβ n

β z α

Selected Theorems on Perpendicularity

Pair of skew lines:

a

ba*

b*

a*2a, b*2ba*, b* intersecting,

a*z b*

a z b

Uniqueness of perpendicular line through a point:

through a point P,there is one and only one line nperpendicular toa plane α.

Uniqueness of perpendicular plane through a point:

through a point P,there is one and only one plane αperpendicular toa line n.

α

n

P

n

P

Uniqueness of perpendicular plane through a line:

through a line l,non-perpendicularto a plane α,there is one and only one plane βperpendicular tothe plane α.

l β

P

n

n

Line and plane:

Perpendicular planes:

abc

a 1 b =P

n z a, n z bc γ = [a,b]

n z c

γ β

γ

α

α

α



Geometrical Loci in 3D

Range of points

collinear points

Pencil of lines in a plane

concurrent lines

Pencil of planes

Field of points

coplanar points

Boundle of lines

concurrent lines

Boundle of planes

concurrent planes




Set of points at a given distance from a given point is a sphere.

E.t.c.

Set of points at a given distance from a given line is a cylinder.

Set of points equidistant from two given points is the perpendicular bisector plane of the segment.




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 …



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 α and a distance r be given. Find the straight lines intersecting with b, lying in the plane α such that their distance from A is equal to the given length r.

4) Let a straight line a, a plane α and an angle γ be given. Find the planes passing through the line a, such that they form an angle equal to γ with the plane α.

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 α, β and an angle γ be given. Find the straight lines passing through the point A, parallel to the plane α and form an angle γ with β.

7) Let a point A, a pair of skew lines b, c and an angle γ be given. Find the straight lines passing through the point A, intersecting with b and form an angle γ with c.

8) Let a point A, a pair of skew lines b and c, and two angles γ and δ be given. Find the lines passing through A such that the angles formed by b and c are the given angles γ and δ respectively.

9) Let a point A, a straight line b, a plane α and two angles γ and δ be given. Find the straight lines passing through A, such that the angles formed by the line b and the plane α are equal to the given angles γ and δrespectively.

10) Let a point A, a straight line b and a plane α and two angles γ and δ be given. Find the planes passing through A, such that the angles formed by the line b and the plane α are equal to the given angles γ and δrespectively.

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.

Loci Problems



1) Read and reread the problem carefully.

2) Find the proper type of representation (simple projection, axonometric sketch, multi-view rep.).

3) Draw a sketch as if the problem was solved.

4) Try to find relations between the data and the unknown.

5) Write down the solution, use pseudocode.

6) Analyze the solvability and the number of solution (and write down your results).

7) Read the question and your answer again.

Show that a tetrahedron has orthocenter if and only if the opposite edges are perpendicular.

Solution of 3D Problems

A

B

C

D

P

Q

O1) Let AB and CD perpendicular.

CD perpendicular to [ABP], CD perpendicular to [ABQ]. (P and Q are the pedal pointsof the altitudes from A and Brespectively.)There is only one plane through AB perpendicular to CD, so AP and BQ are coplanar,so they are intersecting and the orthocenter exists.

2) If AP and BQ are intersecting and both AP and BQ areperpendicular to CD, than AB lies in a plane perpendicular toCD. Consequently, AB is perpendicular to CD too.

The condition is necessary and sufficient.

Exercise: let a pair of skew lines and a plane be given. The plane is not parallel to the lines. Find the shortest transversal of the lines parallel to the plane.



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 α such that A and Bare on one side of α, C and D are on the other side of α and their distance from α 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.



Chapter Review

Vocabulary



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

Regular and semi-regular polyhedra

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


Faces around a vertex

P

P

P

PP

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

{4, 3}

{5, 3}

{3, 3}{3, 4}

{3, 5}



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


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


Archimedean Polyhedra



Fullerains (named after Buckminster Fuller)

A highlight of one of the pentagonal rings

A highlight of one of the hexagonal rings

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.

Buckminster Fuller's Dome -Expo '67 Montreal

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.Both Kroto and Smalley felt it 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 bonded to three others in a hexagonal configuration.)



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


http://www.korthalsaltes.com/


Art and Science

JACOPO DE 'BARBERI: Luca Pacioli, c. 1499

This painting shows Fra Luca Pacioli and his student, Guidobaldo, Duke of Urbino. In the upper left is a rhombi-cuboctahedron, and on the table is a dodecahedron on top of a copy of Euclid's Elements.

Leonardo's Illustrations for Luca's book. DaDivina Proportione

Luca Pacioli wrote a book called Da Divina Proportione(1509) which contained a section on the Platonic Solids and other solids, which has 60 plates of solids by none other than his student Leonardo da Vinci.

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


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


M. C. ESCHER (1902-1972)

Stars, 1948 Note the similarity betweenthis polyhedron and Leonardo'sillustrations for Pacioli's book

Escher made a set of nested Platonic Solids. When he moved to a new studio he have awaymost of his belongings but took his belovedmodel.



Models



Chapter Review

Vocabulary



Solid Geometry Formulae 1

Solid/Surface Surface Area Volume

Ah

r

h

r

a b

c

A)(2 acbcabS ++=

π24 rS =

P

C rhrChASππ 22

22 +=

=+=A

PhAS += 2

abcV =

34 3πrV =

AhV =

hrAhVπ2=

==

Geometrical Calculations


Solid Geometry Formulae 2

Solid/Surface Surface Area Volume

A

h

A

ah

A

h l

r

R

r

h

L LAS +=

LaAS ++=L

222

21

rhrr

ClAS

++=

=+=

ππC

l))(( 22 lrRrRS +++= π

3AhV =

( )aAaAhV ++=3

3

2 hrVπ

=

( )22

3rRrRhV ++=

π



Solid Geometry Calculations 1

A

BC

D

E1) In the given cube, E is the

midpoint of the side CD. Find the measure of the angle BAE.

2) A standard tennis ball can is a cylinder that holds three tennis balls. Which is greater, the circumference of the can or its height? If the radius of a tennis ball is 3.5 cm, what percentage of the can is occupied by air, out of tennis balls?

3) A right cylinder is inscribed inside a sphere. If altitude of the cylinder is 8 cm and volume is 72πcm3, find the radius of the sphere.

4) The sketch shows the pattern (or net) of a pyramid. Calculate the surface area and the volume of the three-dimensional figure formed by the pattern.

10 cm

16 cm 16 cm

5) A soft drink cup is in the shape of right circular conic frustum. The capacity of the cup is 250 milliliter and the top and bottom circles are of 6 cm and 4 cm (diameters) respectively. How deep is the cup?

6) The first two steps of a 10-step staircase are shown in the sketch. Find the amount of concrete needed to the exposed portion of the staircase. Find the area of carpet needed to cover the front, top and sides of the concrete steps.

80 cm 20 cm

15 cm

7) A sculpture made of iron has the shape of a right square prism topped by a sphere. The metal in each part of the sculpture is 2 mm thick. If the outside dimensions are as shown and the density of iron 7.78 g/cm3, calculate the mass of the sculpture in kilograms.

10 cm

20 cm

50 cm

8) The length, the width and the altitude of a rectangular prism is directly proportional by 3,4 and 5. If the diagonal of the rectangular prism is cm, find the total surface. (cm2)

9) A tank has the shape shown. Calculate the volume and the surface area of the tank.

40 cm

16 cm

30 cm

200



Solid Geometry Calculations 2

10) Determine the ratio of the surface areas of sphere and circumscribed cylinder, and the ratio of volumes of sphere and circumscribed cylinder. (Diagram on Archimedes’ tombstone.)

11) A right circular cone has a volume of 210 m3. The height of the cone is the same length as the diameter of the base. Find the dimensions of the cone.

12) A right cone with radius 6 units and altitude 8 units is given. What is the area of the sphere with the greatest volume that can be inscribed in the cone?

13) The pyramid {T,A,B,C} is cut into three pieces of equal height by planes parallel to the base as shown in the figure. Find the ratios of the three volumes of the pyramid {T,A,B,C}.

T

A B

C

14) Sketched is a pattern for a three dimensional figure. Use the dimensions to calculate the surface area and the volume of the 3D figure formed bz the pattern.

3 cm

5 cm

15) A rectangular piece of A/4 size paper (297 mm by 210 mm) can be rolled into cylinder in two different directions. If there is no overlapping, which cylinder has greater volume, the one with the long side of rectangle as its height, or the one with the short side of the rectangle as its height?

16) A right rectangular prism has a volume of 324 cubic units. One edge has measure twice that of a second edge and nine times that of the third edge. What are the dimensions of the prism?

17) The trapezoid given in the figure is revolved 360° around the side AB. Find the volume of the solid body formed.

A

B

18) By using the sector given in the figure, a right cone is formed. If the altitude of the cone is 165 cm, find the lateral area of the cone.

19) Find the ratio between the volume of the pyramid ABCD and the cube.

20) The base of a truncated pyramid is an equilateral triangle of 50 cm side. The height of the solid is 20 cm and the angle formed by base and the lateral face is 60°. Calculate the surface area and the volume.

B

AC

D


Documents

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