Geometrical Constructions 2
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.
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2
Pencil of Circles
Intersecting (or „elliptical”) pencil
of coaxial circles
radical
axis
Radical center C
P
C
All tangents drawn to the circles
of a coaxial pencil from a point
on the radical axis have the
same length.
Geometrical Constructions 2
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.
3
Apollonian Problems
Conjugate Pencils of Circles
The parabolic pencil of circles is the family of
circles which all have one common point,
and thus are all tangent to each other,
either internally or externally. Also, the
orthogonal set of circles to a parabolic pencil
is another parabolic pencil.
An elliptical pencil of circles is the family of
all circles that go through two given points.
A hyperbolic pencil of circles is the family of
all circles that are orthogonal to an elliptical
pencil of circles. None of the circles in the
hyperbolic pencil intersect with each other.
Geometrical Constructions 2
4
Apollonian Problems
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
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5
Apollonian Problems
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
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Apollonian Problems
Apollonian Problems on Tangent Circles
Combinatorial approach
Options
Circle
C
Point (circle of 0 radius)
P
Straight line (circle of infinite radius)
L
1)
(PPP)
;
2)
(PPL)
;
3)
(PPC)
;
4)
(PLL)
;
5)
(PLC) :
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)
:
10) (CCC)
:
Apollonian tangent circle problem: choose three from the elements
of the set {P, L, C} (an element can be chosen repeatedly) and find
;
subject of our course
:
not subject of our course
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7
Apollonian Problems
1) (PPP)
2) (PPL)
c
P1
c
e
i
d
P1
d
g
P2
f
P3
P2
h
l
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8
Apollonian Problems
3)
(PPC),
4)
c
(PLL)
l1
P1
d
e
d
c
P
g
h
f
P2
g
Geometrical Constructions 2
i
h
c
l2
e
f
9
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
A1
two more solutions.
P
d c
c
g
P
A2
Q
e
Q1
Geometrical Constructions 2
B
f
Q2
10
l
Apollonian Problems
7)
(LLL)
8)
(LLC)
f
l2
l3
dl
e
d
f c
1
g
c
l2
l1
Hint: the solutions are the incircle and
the excircles of the triangle
formed by the lines. The centers
are the points of intersection of
the bisectors of the interior and
exterior angles.
Geometrical Constructions 2
c
e
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.
11
Apollonian Problems
Chapter Review
Vocabulary
Geometrical Constructions 2
12
Apollonian Problems
Approximate Rectification of an Arc
B
Kochansky’s method:
CD = 3r
DB ≈ rπ
r
≈ rπ
O
a
30°
C
r
A
D
Snellius’ method:
AC = 3r
P P’
a
C
Geometrical Constructions 2
O
r
a O 30°
A
AP’ ≈ AP
13
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
Geometrical Constructions 2
Straight line rolling along a circle
14
Roulettes
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).
prolate cycloid
http://mathworld.wolfram.com/Cycloid.html
curtate cycloid
y
e
System of equations :
t
at
a
M
Geometrical Constructions 2
x
x = at − a sin t
y = a − a cos t
15
Tangent:
The tangent e is perpendicular to the
segment that connects the point of the
cycloid and the momentary pole M.
Roulettes
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 ) cos t + h cos⎜
t⎟
b
⎠
⎝
⎛a−b ⎞
y = (a − b )sin t − h sin ⎜
t ⎟.
b
⎠
⎝
a
a : radius of base circle
b
h
b : radius of the rolling circle
x
h : distance of the point and the centre of
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
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Roulettes
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 ) cos t − h cos⎜
t⎟
b
⎝
⎠
⎛a+b ⎞
y = (a + b )sin t − h sin ⎜
t ⎟.
⎝ b ⎠
b
h
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
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Roulettes
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 (cos t + t sin t ) − h cos t
y = a (sin t − t cos t ) − h sin 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
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Roulettes
Chapter Review
Vocabulary
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Roulettes
Conic Sections
Ellipse
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Parabola
20
Hyperbola
Conic Sections
Ellipse
t
Definition:
r1
F1
minor
axis
Let two points F1, F2
foci and a distance 2a
be given.
P
r2
major
axis
F2
Ellipse is the set of points,
whose sum of distances
from F1 and F2 is equal to
the given distance.
r1 + r2 = 2a.
Dist(F1, F2 ) =2c, a>c.
r2
r1
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.
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Conic Sections
Properties of Ellipse
director 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.
principal circle
r1
P
r1
2a
r2
O
F1
r2 E
M
r2
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.
F2
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.
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Conic Sections
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
A
L1
B
Osculating circle at B
Perpendicular bisector of AA”
Perpendicular to BC
K2
L2
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
P1
P2
Geometrical Constructions 2
P0
P3
23
axis
Parabola
t
Let a point F focus and
a straight line d directrix
be given. The line is not
passing through the
point.
Definition:
Parabola is the set of
points equidistant from
the focus and the dirctrix.
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.
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Conic Sections
Properties of Parabola
axis
Antipoint E: pedal point of the line passing
through P perpendicular to d. The set of
antipoints is the directrix.
P
Tangent at the vertex: set of pedal points M of
lines from F perpendicular to the tangents of
the parabola.
F
M
d
V
E
Geometrical Constructions 2
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.
Under the reflection in the parabola, a ray
emitted from the focus will be parallel to the
axis.
25
Conic Sections
axis
Osculating Circle at the Vertex of Parabola
K
F
d
V
The radius of the osculating circle at the vertex: r = dist(F,d)
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26
Conic Sections
Hyperbola
Let two points F1, F2
foci and a distance 2a
be given.
conjugate
axis
t
2a
P
r1
r2
r1
r2
traverse
F2 axis
F1
Dist(F1, F2 ) =2c, a<c.
Definition:
Hyperbola is the set of
points, whose difference of
distances from F1 and F2 is
equal to the given distance.
|r1 - r2| = 2a.
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.
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Conic Sections
Asymptotes of Hyperbola
y
Equation of hyperbola:
x2
y2
− 2 =1
2
a
b
where b is determined by the Pythagorean
c
F1
b
a
F2
equation:
x
a2 + b2 = c2 .
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 the
vertices and the circle.
Geometrical Constructions 2
28
⎛ y2 ⎞
⎛ b2 b2 ⎞
lim ⎜⎜ 2 ⎟⎟ = lim ⎜⎜ 2 − 2 ⎟⎟
x →∞ x
x ⎠
⎝ ⎠ x →∞ ⎝ a
y
b
lim = ±
x →∞ x
a
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.
2a
director circle
r1
r1 E
r2
P
r2
r2
M
Principal circle: circle about O with the radius of a.
O
F1
F2
principal circle
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.
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.
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29
Conic Sections
Chapter Review
Vocabulary
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30
Conic Sections
Solutions of 3D Geometrical Constructions
ƒAxonometric sketch
ƒMulti-view representation
ƒDescription in pseudo-code
ƒModeling
Bold-Italic: will be discussed in
this course
Italic: will be discussed in other
corses offered by our
department
Normal: not in our scope
ƒRepresentation by relief
ƒStereoscopic representation
ƒPerspective
ƒAnalytical geometric solution
ƒComputer geometric modeling
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31
3D geometrical constructions
Axonometric sketch 1
Front view of P
z
It is based on 2D representation of 3D
P”
Cartesian coordinate-system. Points are
Q
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
z
preserves parallelism.
O
The axonometric system is determined
Uy
by the points {O, Ux, Uy, Uz}, the
image of the origin and the units on the
axes x, y and z.
y
1
y
1 Ux
Q’
R
x
1
R’
x
P’
The axonometric sketch is similar to the
Top view of P
parallel projection of the figure.
Geometrical Constructions 2
Uz
Frontal axonometry
32
3D geometrical constructions
Axonometric Sketch 2
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33
3D geometrical constructions
Multi View Representation 1
Front view of P
2nd quadrant
π2
1st quadrant
π2
P”
P”
d2
P
x1,2
P
d1
P’
d1
3rd quadrant
x1,2
d2
P’
π1
Top view of P
4th quadrant
The image planes are also perpendicular. The images are orthogonal projections on image
planes.
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34
π1
Multi View Representation 2
Front view
From the bottom
Side view (from left)
Side view (from right)
From back
Top view
Geometrical Constructions 2
35
3D geometrical constructions
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
36
3D geometrical constructions
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
Let X be an arbitrary point of a.
b*
-parallel to the plane
a
Pb
a*
A
Pb* = b* 1 a
Pb = b 1 a
b
B
b*
l = |Pa, Pb*|
t
X
l*
n
Pb*
Pa
a
l*
Pb
n
a
a*
l *∥ l
n
n 1 l*
Pa
n
z
l
a*∥a
B = b 1 a*
l
t
Geometrical Constructions 2
b* ∥ b
Pa = a 1 a
-the shortest one among the transversals that
satisfy the first condition
Sketch:
X
37
B
t∥n
3D geometrical constructions
Modeling
Development
Geometrical Constructions 2
Plaster, plastic, wood, etc.
38
Wireframe (string) model
3D geometrical constructions
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/
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39
3D geometrical constructions
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 threedimensional image.
http://www.hitl.washington.edu/scivw/EVE/III.A.1.b.StereoscopicViewing.html
http://en.wikipedia.org/wiki/Stereoscopy
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40
3D geometrical constructions
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
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41
C:\TEMP\diamonds.exe
3D geometrical constructions
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
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42
3D geometrical constructions
Perspective
http://employees.oneonta.edu/farberas/arth/arth200/durer_artistdrawingnude.html
http://www.mcescher.com/
http://www2.evansville.edu/studiochalkboard/draw.html
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43
3D geometrical constructions
Analytical geometric solution
P(x, y, z)
P
x = x0 + v1t
l
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.
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44
3D geometrical constructions
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
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3D geometrical constructions
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)
Euler’s formula: v + e - f = 2 = 4+4-6
C
A
l
B
Geometrical Constructions 2
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
46
3D geometrical constructions
Relations of Spatial Elements
pair of points:
A
point and line:
1) lying on
B
l
determine a line: |AB| = l,
determine a distance: dist(A,B)
2) not lying on
determine a plane: [P,l] = α
P
determine a distance:
P’
dist(P,l) = dist(P,P’)
(P’ is the orthogonal projection of P on l)
l
pair of lines:
1) coplanar
intersecting
parallel
P
a
b
determine angles, determine a plane
determine a distance , determine a plane
c
d
2) non-coplanar, skew
determine angles and distance, no plane
n
f
n
e
Geometrical Constructions 2
in common
47
z
e, n
zf
3D geometrical constructions
Relations of Spatial Elements
point and plane: 1) lying on
2) not lying on
P
determine a distance:
α
P’
line and plane:
1) lying on
2) parallel
(P’ is the orthogonal projection of P on α)
l
α
determine a distance:
l’
l’
1) parallel
α*
P
α*, dist(P, α) = dist(P,P’)
(P’ is the orthogonal projection of P on α)
β
l
α
Geometrical Constructions 2
∡(l, α) = ∡(l,l’)
determine a distance: P
P’
a
2 l)
(l’ is the orthogonal projection of l on α)
α
2) intersecting
b
dist(l, α) = dist(l,l’)
(l’ is the orthogonal projection of l on α, l’
3) intersecting (non-perpendicular)
l determine an angle:
α
pair of planes:
dist(P,α) = dist(P,P’)
determine an angle:
(both a
48
α and b
l = α 1 β, ∡(α, β) = ∡(a,b)
β are perpendicular to l)
3D geometrical constructions
Selected Theorems on Parallelism
l A1
l 1 a1 =A1, l 1 a2 =A2
A2
a1
a1
a2
β
a2
α
l
l
l, a1, a2: coplanar
b
b
α2 β,
β
γ
β
2
c
a
α
α2 γ ,
g
β
γ
G2 l
G1
α
α
β
α
l
g
l
2 α, l 2 β
2
Geometrical Constructions 2
β
b
c, d: intersecting
α =[a,b]
L2
L1
α
49
2
b=[c,d]
2 b
g2l
α
dist(G1,L1) = dist(G2,L2)
2 β, γ intersects α
γ intersects β
γ
g
b
a, b: intersecting
α
α 1β=g
l
b
2α
a 2 c, b 2 d
d
β
2c
2
l
α
2 b, a 2 c,
a
c
a
2
α 1γ
2β1γ
3D geometrical constructions
Selected Theorems on Perpendicularity
Pair of skew lines:
b
a*
a
a*2a, b*2b
a*, b* intersecting,
zb
az b
a
b*
*
*
n
a 1 b =P
P
n
c
z a, n z b
n
zc
b
c
a
α = [a,b]
β
n
α
α
n
zα
n
b
β
zα
Geometrical Constructions 2
through a point P,
there is one and
only one plane α
perpendicular to
a line n.
n
α
P
Uniqueness of perpendicular plane through a line:
Perpendicular planes:
γ
only one line n
perpendicular to
a plane α.
Uniqueness of perpendicular plane through a point:
Line and plane:
α
Uniqueness of perpendicular line through a point:
P
through a point P,
n
there is one and
through a line l,
non-perpendicular
to a plane α,
there is one and
only one plane b
perpendicular to
the plane α.
50
l β
α
3D geometrical constructions
Geometrical Loci in 3D
Range of points
Pencil of lines in a plane
collinear points
concurrent lines
Pencil of planes
Field of points
Boundle of lines
Boundle of planes
coplanar points
concurrent lines
concurrent planes
Geometrical Constructions 2
51
3D geometrical constructions
Geometrical Loci in 3D
Set of points at a given
distance from a given point
is a sphere.
Set of points at a given
distance from a given line
is a cylinder.
E.t.c.
Geometrical Constructions 2
52
Set of points equidistant
from two given points is
the perpendicular bisector
plane of the segment.
3D geometrical constructions
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
53
3D geometrical constructions
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
54
3D geometrical constructions
Solution of 3D Problems
1) Read and reread the problem
carefully.
Show that a tetrahedron has orthocenter if and only if the
opposite edges are perpendicular.
B
2) Find the proper type of
representation (simple
projection, axonometric
sketch, multi-view rep.).
1) Let AB and CD perpendicular.
O
P
CD perpendicular to [ABP],
CD perpendicular to [ABQ].
(P and Q are the pedal points
D
of the altitudes from A and B
A
Q
respectively.)
There is only one plane
through AB perpendicular to
C
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 are
perpendicular to CD, than AB lies in a plane perpendicular to
CD. Consequently, AB is perpendicular to CD too.
The condition is necessary and sufficient.
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.
Geometrical Constructions 2
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.
55
3D geometrical constructions
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
56
3D geometrical constructions
Chapter Review
Vocabulary
Geometrical Constructions 2
57
3D geometrical constructions
Platonic Solids
A regular polyhedron is one whose faces are identical regular polygons.
The solids as drawn in Kepler’s Mysterium Cosmographicum:
tetrahedron
(Fire)
octahedron
icosahedron
cube
dodecahedron
(Air)
(Water)
(Earth)
(Universe)
http://www.math.bme.hu/~prok/RegPoly/index.html
Geometrical Constructions 2
58
Regular and semi-regular polyhedra
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}
P
{3, 5}
P
P
P
{3, 3}
{3, 4}
Geometrical Constructions 2
59
Regular and semi-regular polyhedra
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
60
Regular and semi-regular polyhedra
Archimedean Polyhedra
Geometrical Constructions 2
61
Regular and semi-regular polyhedra
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.
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
Buckminster Fuller's Dome bonded to three others in a
Expo '67 Montreal
hexagonal configuration.)
•
•
•
A highlight of one of the
pentagonal rings
A highlight of one of the
hexagonal rings
Geometrical Constructions 2
62
Regular and semi-regular polyhedra
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
http://www.korthalsaltes.com
polyhedra.
Geometrical Constructions 2
63
Regular and semi-regular polyhedra
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. Da
Divina 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
Geometrical Constructions 2
64
Regular and semi-regular polyhedra
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
65
Regular and semi-regular polyhedra
Models
Geometrical Constructions 2
66
Regular and semi-regular polyhedra
Chapter Review
Vocabulary
Geometrical Constructions 2
67
Regular and semi-regular polyhedra
Solid Geometry Formulae 1
Solid/Surface
Surface Area
c
A
a
S = 2 ( ab + bc + ac )
V
= abc
b
S = 4r π
2
r
S = 2 A + Ph
h
A
3
π
4
r
V =
3
V = Ah
P
h
A
Volume
r
Geometrical Constructions 2
C
S = 2 A + Ch =
= 2 r 2 π + 2 π rh
68
V = Ah =
= r 2π h
Geometrical Calculations
Solid Geometry Formulae 2
Solid/Surface
Volume
S = A + L
h
L
Surface Area
Ah
3
V
=
V =
h
A+
3
A
a
L
Geometrical Calculations
S = A+a+L
h
A
S = A +
l
h
A
r
C
1
Cl =
2
= r 2π + π r
h2 + r
2
l
2
2
S = π ( R + r + ( R + r )l )
R
Geometrical Constructions 2
69
Aa + a
)
2
r πh
V =
3
r
h
(
V=
πh
3
(R
2
+ Rr + r 2 )
Geometrical Calculations
Solid Geometry Calculations 1
B
C
1) In the given cube, E is the
midpoint of the side CD.
Find the measure of the
angle BAE.
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.
E
D
A
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?
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.
3) A right cylinder is inscribed inside a sphere. If
altitude of the cylinder is 8 cm and volume is 72π
cm3, 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 threedimensional figure formed by
the pattern.
16 cm
16 cm
10 cm
80 cm
20 cm
10 cm
50 cm
20 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 200 cm, find the
total surface. (cm2)
40 cm
9) A tank has the shape shown.
Calculate the volume and the
surface area of the tank.
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?
Geometrical Constructions 2
15 cm
70
16 cm
30 cm
Geometrical Calculations
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.)
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?
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.
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?
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?
17) The trapezoid given in the figure
is revolved 360° around the side
AB. Find the volume of the solid
body formed.
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}.
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.
Geometrical Constructions 2
T
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.
C
A
A
19) Find the ratio between the volume of A
the pyramid ABCD and the cube.
B
3 cm
C
D
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.
5 cm
71
B
Geometrical Calculations