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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)
Geometrical Constructions 23
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
Geometrical Constructions 24
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
Geometrical Constructions 25
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
Geometrical Constructions 26
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
Geometrical Constructions 27
1) (PPP) 2) (PPL)
P1
P3
P2
P1
P2
l
Apollonian Problems
Geometrical Constructions 28
3) (PPC), 4) (PLL)
P1
P2
c
l1
l2
P
Apollonian Problems
Geometrical Constructions 29
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
Geometrical Constructions 210
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
Geometrical Constructions 211
Chapter Review
Vocabulary
Apollonian Problems
Geometrical Constructions 212
Conic Sections
Conic Sections
Ellipse Parabola Hyperbola
Geometrical Constructions 213
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
Geometrical Constructions 214
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
Geometrical Constructions 215
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
Geometrical Constructions 216
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
Geometrical Constructions 217
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
Geometrical Constructions 218
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
Geometrical Constructions 219
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
Geometrical Constructions 220
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
Geometrical Constructions 221
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
Geometrical Constructions 222
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
Geometrical Constructions 223
Chapter Review
Vocabulary
Conic Sections
Geometrical Constructions 224
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
Geometrical Constructions 225
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
Geometrical Constructions 226
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
Geometrical Constructions 227
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
Geometrical Constructions 228
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
a : radius of base circle
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
Geometrical Constructions 229
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
−−=−+=
a : radius of base circle
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
Geometrical Constructions 230
Chapter Review
Vocabulary
Roulettes
Geometrical Constructions 231
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
Geometrical Constructions 232
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
3D geometrical constructions
Geometrical Constructions 233
Axonometric Sketch 2
3D geometrical constructions
Geometrical Constructions 234
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
Geometrical Constructions 235
Multi View Representation 2
Front view
Top view
Side view (from left)
From the bottom
Side view (from right)
From back
3D geometrical constructions
Geometrical Constructions 236
Multi View Representation 3
Front view
Top view
Side view (from the left)
From the bottom
Side view (from the right)
From back
3D geometrical constructions
Geometrical Constructions 237
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*
3D geometrical constructions
Geometrical Constructions 238
Modeling
Development Plaster, plastic, wood, etc. Wireframe (string) model
3D geometrical constructions
Geometrical Constructions 239
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.
3D geometrical constructions
Geometrical Constructions 240
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
3D geometrical constructions
Geometrical Constructions 241
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
3D geometrical constructions
Geometrical Constructions 242
Stereogram
http://www.eyetricks.com/3dstereo.htm
3D geometrical constructions
Geometrical Constructions 243
Perspective
http://employees.oneonta.edu/farberas/arth/arth200/durer_artistdrawingnude.html
http://www.mcescher.com/http://www2.evansville.edu/studiochalkboard/draw.html
3D geometrical constructions
Geometrical Constructions 244
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.
3D geometrical constructions
Geometrical Constructions 245
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
3D geometrical constructions
Geometrical Constructions 246
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
3D geometrical constructions
Solid Geometry, IntroductionSolid Geometry, Introduction
Geometrical Constructions 247
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
3D geometrical constructions
Geometrical Constructions 248
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
3D geometrical constructions
Geometrical Constructions 249
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
3D geometrical constructions
Geometrical Constructions 250
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
3D geometrical constructions
Geometrical Constructions 251
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
3D geometrical constructions
Geometrical Constructions 252
Geometrical Loci in 3D
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.
3D geometrical constructions
Geometrical Constructions 253
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 …
3D geometrical constructions
Geometrical Constructions 254
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
3D geometrical constructions
Geometrical Constructions 255
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.
3D geometrical constructions
Geometrical Constructions 256
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.
3D geometrical constructions
Geometrical Constructions 257
Chapter Review
Vocabulary
3D geometrical constructions
Geometrical Constructions 258
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
Geometrical Constructions 259
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}
Regular and semi-regular polyhedra
Geometrical Constructions 260
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
Regular and semi-regular polyhedra
Geometrical Constructions 261
Archimedean Polyhedra
Regular and semi-regular polyhedra
Geometrical Constructions 262
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 and semi-regular polyhedra
Geometrical Constructions 263
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
Regular and semi-regular polyhedra
Geometrical Constructions 264
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
Regular and semi-regular polyhedra
Geometrical Constructions 265
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.
Regular and semi-regular polyhedra
Geometrical Constructions 266
Models
Regular and semi-regular polyhedra
Geometrical Constructions 267
Chapter Review
Vocabulary
Regular and semi-regular polyhedra
Geometrical Constructions 268
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
Geometrical Constructions 269
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
Geometrical Calculations
Geometrical Constructions 270
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
Geometrical Calculations
Geometrical Constructions 271
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
Geometrical Calculations
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