Geometrical Constructions 1

by Pál Ledneczki Ph.D. Geometrical Constructions 1 „Let no one destitute of geometry enter my doors” (Plato, 427-347 BC)

RAFFAELLO: School of Athens, Plato & Aristotle

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 2 © Pál Ledneczki Geometrical Constructions 1

Drawing Instruments

For note taking and constructions For drawings (assignments)

- loose-leaf white papers of the size A4 - three A2 sheets of drawing paper (594 mm × 420 mm) - pencils of 0.3 (H) and 0.5 (2B) - set of drawing pens (0.2, 0.4, 0.7) - eraser, white, soft - set square (two triangles, 30 cm) - set square (two triangles, 20 cm) - A1 sheet of drawing paper for folder

- pair of compasses with joint for pen frame 10 folder - French-curves (title box on drawings) - set of color pencils 15 I -1 NAME (A1 size of 15 Date Signature GR # - drawing board paper folded into two) 30 30 60 30 lettering

Name Architectural Representation

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Contents

1) Basic constructions

2) Loci problems

3) Geometrical transformations, symmetries

4) Pencils of circles, Apollonian problems

5) Regular polygons, golden ratio

6) Affine mapping, axial affinity

7) Conic sections

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 4 © Pál Ledneczki Geometrical Constructions 1 1 Basic Constructions

Use of Set Squares and Pair of Compasses

copying angle bisecting an angle

perpendicular bisector

A B of a segment

parallel lines by perpendicular sliding ruler line by rotating ruler

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Triangles, Some Properties

C Triangle inequalities ? a a+b > c, b+c > a, c+a > b b Sum of interior angles ß a+? a B a + ß + ?= 180° c A

Any exterior angle of a triangle is equal to the sum of the two interior angles non adjacent to the given exterior angle.

About the sides and opposite angles, if and only if b > a than ß > a for any pair of sides and angles opposite the sides.

Classification of triangles; acute, obtuse, right, isosceles, equilateral

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Theorems on Right Triangle

Theorem of Pythagoras In a right-angled triangle, the sum of the squares b a of the legs is equal to the square of the hypotenuse.

c Converse: If the sum of the squares of two sides in a triangle a2 + b2 = c2 is equal to the square of the third side, then the triangle is a right triangle.

C

Theorem of Thales In a circle, if we connect two endpoints of a diameter with any point of the circle, we get a right A B angle. Converse: If a segment AB subtends a right angle at the point C, then C is a point on the circle with diameter AB.

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Triangles, Special Points, Lines and Circles

O The three altitudes of a triangle meet at a point O. This point is the orthocenter of the triangle. The orthocenter of a right triangle is the vertex at the right angle.

The three medians of a triangle meet at a point G (point of gravity). This G point is the centroid of the triangle. The centroid trisect the median such that the segment connecting the vertex and the centroid is the 2/3-rd of the corresponding median.

C The three perpendicular bisectors of the sides of a triangle meet at a point C. This point is the center of the circumscribed circle of the triangle.

e

G O The three points O, G, and C are collinear. The line passing through them is C the Euler’s line. The ratio of the segments OG and GC is equal to 2 to 1.

The three bisectors of angles of a triangle meet at a point P. This point is the center of the inscribed circle. (Two exterior bisectors of angles and the P interior bisector of the third angle also meet at a point. About this point a circle tangent to the lines of the triangle can be drawn.)

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Theorems on Circles

Circles are similar Two circles are always similar. Except the case of congruent or concentric circles, two circles have two C2 centers of similitude C1 and C2. C1

C 2 Theorem on the angles at the circumference and at the C1 center The angle at the center is the double the angle at the " circumference on the same arc. " " Corollaries:

2" A chord subtends equal angles at the points on the A " same of the two arcs determined by the chord. B In a cyclic quadrilateral, the sum of the opposite angles is 180°. B1

A P 1 Circle power

A2 On secants of a circle passing through a point P, the B2 product of segments is equal to the square of the 2 2 2 2 2 T tangential segment: PA1 +PB1 =PA2 +PB2 =PT

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Parallel Transversals, Division of a Segment

Theorem on parallel transversals: the ratios of the y corresponding segments of arms of an angle cut by parallel x lines are equal. x’ y’ x:y = x’:y’, x:x’ = y:y’

Divide the segment AB by a point C such that AC:CB = p:q. A u Find the point C. p=3 l Solution: 1) draw an arbitrary ray r from A C q=5 2) measure an arbitrary unit u from A onto the ray p + q times, get the point B’ B’ 3) connect B and B’ r B 4) draw parallel l to BB’ through the endpoint of segment p

5) C is the point of intersection of l and AB

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Ratios of Segments in Triangle

In a right triangle,

h 2 pq, b2 cp, a2 cq. a c = = = b hc The altitude assigned to the hypotenuse is the geometrical mean of the two segments of the hypotenuse. p q c A leg is the geometrical mean of the hypotenuse and the orthogonal projection of the leg on the hypotenuse.

In an arbitrary triangle, B q a p b = . q c p c

C The bisector of an angle in a triangle divides the A b c opposite side at the ratio of the adjacent sides.

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Quadrilaterals

Cyclic quadrilateral Circumscribed quadrilateral Diagonals and bimedians a b "

d M $ c

" +$ =180° a +c = b + d coincidence of midpoints

Kite (Deltoid) Trapezium Rhombus

a a

a a

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Euclid of Alexandria (about 325BC-265BC)

More than one thousand editions of The Elements have been published since it was first printed in 1482. BL van der Waerden assesses the importance of the Elements: ‘Almost from the time of its writing and lasting almost to the present, the Elements has exerted a continuous and major influence on human affairs. It was the primary source of geometric reasoning, theorems, and methods at least until the advent of non-Euclidean geometry in the 19th century. It is sometimes said that, next to the Bible, the "Elements" may be the most translated, published, and studied of all the books produced in the Western world.’

Quotations by Euclid: This is a detail from the fresco The School of Athens by Raphael There is no royal road to geometry. A youth who had begun to read geometry with Euclid, when he had learnt the first proposition, inquired, "What do I get by learning these things?" So Euclid called a slave and said "Give him threepence, since he must make a gain out of what he learns." That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles. [the 5th postulate] http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Euclid.html

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Geometrical Constructions

Euclidean construction (instruments: straight edge and a pair of compasses) : We may fit a ruler to two given points and draw a straight line passing through them We may measure the distance of two points by compass and draw a circle about a given point We may determine the point of intersection of two straight lines We may determine the points of intersection of a straight line and a circle We may determine the points of intersection of two circles Formal solution of constructions problems: Sketch Draw a sketch diagram as if the problem was solved Plan Try to find relations between the given data and the unknown elements, make a plan Algorithm Write down the algorithm of the solution Discussion Analyze the problem in point of view of conditions of solvability and the number of solutions

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 14 © Pál Ledneczki Geometrical Constructions 1 How to Solve It? (according to George Polya, 1887 - 1985)

First. You have to understand the problem.

What is the unknown? What are the data? What is the condition? Is it possible to satisfy the condition? Is the condition sufficient to determine the unknown? Or is it insufficient? Or redundant? Or contradictory? Draw a figure. Introduce suitable notation. Separate the various parts of the condition. Can you write them down?

Second. Find the connection between the data and the unknown. You may be obliged to consider auxiliary problems if an immediate connection cannot be found. You should obtain eventually a plan of the solution.

Have you seen it before? Or have you seen the same problem in a slightly different form? Do you know a related problem? Do you know a theorem that could be useful? Look at the unknown! And try to think of a familiar problem having the same or a similar unknown. Here is a problem related to yours and solved before. Could you use it? Could you use its result? Could you use its method? Should you introduce some auxiliary element in order to make its use possible? Could you restate the problem? Could you restate it still differently? Go back to definitions.

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Polya.html

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How to Solve It?

If you cannot solve the proposed problem do not let this failure afflict you too much but try to find consolation with some easier success, try to solve first some related problems; then you may find courage to attack your original problem again. Do not forget that human superiority consists in going around an obstacle that cannot be overcome directly, in devising some suitable auxiliary problem when the origin alone appears insoluble. Try to solve first some related problem. Could you imagine a more accessible related problem? A more general problem? A more special problem? An analogous problem? Could you solve a part of the problem? Keep only a part of the condition, drop the other part; how far is the unknown then determined, how can it vary? Could you derive something useful from the data? Could you think of other data appropriate to determine the unknown? Could you change the unknown or data, or both if necessary, so that the new unknown and the new data are nearer to each other? Did you use all the data? Did you use the whole condition? Have you taken into account all essential notions involved in the problem?

Third. Carry out your plan. Carrying out your plan of the solution, check each step. Can you see clearly that the step is correct? Can you prove that it is correct?

Looking Back

Fourth. Examine the solution obtained. Can you check the result? Can you check the argument? Can you derive the solution differently? Can you see it at a glance? Can you use the result, or the method, for some other problem? Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 16 © Pál Ledneczki Geometrical Constructions 1 1 Basic Constructions

Constructions, Example

Construct parallelogram if two sides and one Construct parallelogram if one side and two of the diagonals are given. diagonals are given. (Your solution.)

Sketch: d1 b

a constructible ? Algorithm: 1. construct triangle with the sides of

{a, b, d1} 2. reflect the triangle with respect to the midpoint of d1 Construction:

b a d d1 a b 1

Discussion: if the given segments satisfy the triangle-inequality, the problem is solvable. There are two isometric solutions of opposite sense. Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 17 © Pál Ledneczki Geometrical Constructions 1 2 Loci Problems

Determining locus

Locus in Latin means location. The plural is loci. A locus of points is the set of points, and only those points, that satisfy the given conditions.

The locus of points at a given distance from a given point is a circle whose center is the given point and whose radius is the given distance.

The locus of points equidistant from two given points is the perpendicular bisector of the line segment joining the two points.

The locus of points equidistant from the sides of a given angle is the bisector of the angle.

The locus of points equidistant from two given intersecting lines is the bisectors of the angles formed by the lines. Etc.

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 18 © Pál Ledneczki Geometrical Constructions 1 2 Loci Problems

Theorems on Loci

If a segment AB subtends a given angle at the point C, C then the locus of C consists of two arcs of circles of the same radius, symmetrical with respect to the segment.

A B This theorem is the converse of the theorem on angles at circumference, and generalization of the converse of Thales theorem.

A and B are fixed points. P is a moving point such that l = PA : PB is constant, than the locus of P is a circle.

P Hint to the proof: PH and PK are angle bisectors ËHPK = 90°

A H B K An angle bisector in a triangle divides the opposite side at the ratio of the adjacent sides AH AP = = ? constant, H and K are fixed HB PB (This circle is called Apollonian circle.) Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 19 © Pál Ledneczki Geometrical Constructions 1 2 Loci Problems

Exercises on Loci

Problem 1 : find the locus of points at which two given circles subtend equal angles. The centers of homothecy Sketch: are obviously points of the locus. R O1 O2 Because of the similarity r of circles, C1 C 2 PO r 1 = = constant PO R P 2 Algorithm: 1. find the centres of similarity Discussion: 1) circles with equal radii; perpendicular 2. draw the circle with the diameter C1C2 bisector of C1C2 2) circles with different radii, disjoint or Construction: touching; see the construction 3) circles with different radii, partially overlapping; the arc of circle outside the union of the given circles C2 C1 4) circles with different radii, one contains the other, tangents; the point of contact P 5) circles with different radii, no point in common, overlapping; no solution Problem 2 : given the base, the vertical angle and the height to the base. Find the triangle.

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 20 © Pál Ledneczki Geometrical Constructions 1 3 Transformational Geometry

Transformational Geometry

By a transformation of the plane, we mean a one-to-one correspondence P1 P’ among all the points in the plane, that is, a rule for associating pairs of points with the understanding that each pair has a first member P and a second member P’ and that every point occurs as the first member of just one pair and also as the second member of just one pair. Points that are assigned to themselves are called invariant or fixed points of the transformation.

Transformations we shall discuss in this course: Isometries direct (displacement, sense-preserving) rotation translation opposite (reversal, sense-reversing) reflection Similarities homothecy (dilatation) Affinities axial Collineations central-axial (Non-linear transformations)

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 21 © Pál Ledneczki Geometrical Constructions 1 3 Transformational Geometry

Isometries

Direct isometry Q” (translation+rotation) P Q

R” R Q’

P” P’ Preserve:

R’ distance þ Reverse isometry angle measure þ midpoint þ (translation+rotation+reflection) parallelism þ collinearity þ

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Similarities

C

Homothecy (dilatation)

P

Similarity (dilatation+reflection+ rotation+translation) P’

Preserve: distance ý P* angle measure þ midpoint þ parallelism þ collinearity þ

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Affinities

P’ P

Preserve: distance ý P” angle measure ý axial midpoint þ parallelism þ collinearity þ

P*

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Collineations

center Preserve: distance ý P Central-axial angle measure ý collineation midpoint ý

P” parallelism ý collinearity þ

collineation P’ P*

axis

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(Non-linear transformation)

P

P’

Topological

Any transformation preserving neighborhood is called topological.

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Isometries, Invariant Points

Reflection Translation B B B’ B’

A A’ A A’ C C C’ C’ Invariant points: Invariant points: points of the mirror line i

Reflection in a point, half-turn Rotation about a point C’ A

B B C’ O A’ A’

B’ a A C O C B’ Invariant point: center of symmetry Invariant point: center of rotation If an isometry has more than one invariant point, it must be either the identity or a reflection.

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 27 © Pál Ledneczki Geometrical Constructions 1 3 Transformational Geometry

An Exercise on Isometry

In an equilateral triangle, show that the sum of the distances of an internal point from the sides is a constant. C Proof: 1) draw l through P parallel to the Sketch: x+y base

2) reflect the upper triangle and the segment y for l

y 3) x+y = x+y’, x and y’ are collinear x B’ A’ P l 4) the altitudes in the small triangle through A’ and C are x+y y’ 5) the sum of the three segments x+y+z is equal to the length of the altitude of the triangle ABC z C’ Conclusion: the sum of the distances is the constant length of the altitude of the triangle.

Remark: the statement is true for the points of A B the sides too. Problem: In an acute triangle ABC, locate a point P whose distances from A, B, C have the smallest possible sum. (Do research on FERMAT point.)

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An Exercise on Similarity

A’,B’,C’ are the mid-points of the sides BC, CA, AB of a triangle ABC. G, O, C are the centroid, orthocenter and circumcenter of the triangle ABC. Prove that G is the center of similitude of the triangles A’B’C’ and ABC, C is the ortocenter of the triangle A’B’C’, and hence G, O and C are collinear. C C

B’ B’ A’ A’ G = G’ G = G’ O 1 2 C=O’

A A B B C’ C’ The two triangles ABC and A’B’C’ are similar with respect to the center G=G’ and the ratio of - 2:1, consequently O’ = C, G and O are collinear. (The altitudes of the smaller triangle coincide with the perpendicular bisectors of the larger triangle.) A

Problem: Given a triangle ABC, construct a square such hat two vertices lie on BA produced, CA produced and the opposite side along BC. B C

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 29 © Pál Ledneczki Geometrical Constructions 1 4 Apollonian Problems

Pencil of Circles

Intersecting (or „elliptic”) pencil Radical center C of coaxal circles

radical axis P

C

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

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 30 © Pál Ledneczki Geometrical Constructions 1 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

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 31 © Pál Ledneczki Geometrical Constructions 1 4 Apollonian Problems

Apollonius Circle

DEFITION 1: the set of all points whose distances from two fixed points are in a constant ratio (see problem on page 19)

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

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

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 32 © Pál Ledneczki Geometrical Constructions 1 4 Apollonian Problems

Apollonian Problems on Tangent Circles

Combinatorial approach Options

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

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

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

elements.) 8) (LLC) þ

9) (LCC) ý þ subject of our course 10) (CCC) ý ý not subject of our course

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 33 © Pál Ledneczki Geometrical Constructions 1 4 Apollonian Problems

1) (PPP) 2) (PPL)

•

P1

ƒ ‚ • P1 ‡ ‚ …

P2

P3 „ P2 † l

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 34 © Pál Ledneczki Geometrical Constructions 1 4 Apollonian Problems

3) (PPC), 4) (PLL)

• l1 ‚ P1 ƒ c ‚ • P † ‡ † …

„ l2 P2 „ ƒ …

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 35 © Pál Ledneczki Geometrical Constructions 1 4 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

‚ • P

c A2 … Q P ƒ B „ Q1 Q2 l

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 36 © Pál Ledneczki Geometrical Constructions 1 4 Apollonian Problems

7) (LLL) 8) (LLC)

l „ l3 2 • ƒ ‚ l1 … ‚ c

„ • l2 l1 ƒ

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

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 37 © Pál Ledneczki Geometrical Constructions 1 5 Regular Polygons, Golden Ratio

Golden Ratio

Divide a segment in such a way that the ratio of the larger part to the smaller is equal to the ratio of the whole to the larger part.

Golden rectangle Golden triangle 1 t = t 1-t t 2 + t - 1 = 0 5 - 1 t = 1-t 2 36° Construction of t t 1 1 ½

t ½ t t 1-t 36° 36° 72° t 1 t

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 38 © Pál Ledneczki Geometrical Constructions 1 5 Regular Polygons, Golden Ratio

Regular Polygons

A polygon is a many-sided shape. A regular polygon is one in which all of the sides and angles are equal. Some examples are shown below.

Only certain regular polygons are "constructible" using the classical Greek tools of the compass and straightedge(see Topic 1). According to Gauss´ theorem, a regular n-gon can be constructed, if and only if the odd prime factors of n are distinct “Fermat primes”

2k Fk = 2 + 1.

20 21 22 23 24 F0 = 2 + 1 = 3, F1 = 2 + 1 = 5, F2 = 2 + 1 = 17, F3 = 2 + 1 = 257, F4 = 2 + 1 = 65537,

and it is known, that Fk is composit for 5 £ k £16. http://en.wikipedia.org/wiki/Constructible_polygon http://en.wikipedia.org/wiki/Tilings_of_regular_polygons http://mathforum.org/dr.math/faq/formulas/faq.regpoly.html

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 39 © Pál Ledneczki Geometrical Constructions 1 5 Regular Polygons, Golden Ratio

Approximate Construction of Regular Polygons

Approximate construction of 1) Draw a diameter AB of the circumscribed circle regular heptagon: 2) Construct an equilateral triangle with the base of the diameter C 3) Divide the diameter into n=7 equal parts 4) Project the second point of division from the vertex C of the triangle onto the circle 5) segment a is the approximate length of the inscribed regular polygon (heptagon)

About the accuracy of the approximate construction, if the radius of the circle is 10 cm,

an is the length of a side of the inscribed polygon, calculated in analytical geometry, A 1 2 3 4 5 6 7 B tn is the length of a side calculated by trigonometry (the “exact” value). a n 3 4 5 6 7 8 9 10 11 12 13

an 10 3 10 2 11.75 10.00 8.69 7.68 6.89 6.23 5.70 5.26 4.87

tn 10 3 10 2 11.76 10.00 8.67 7.65 6.84 6.18 5.64 5.18 4.79

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Golden Kite an Dart

“Sharp” triangle “Flat” triangle

36° 36° 36°

72° kite

dart

http://goldennumber.net/penrose.htm

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Golden Spiral

O: center of “dilative rotation” or “spiral similarity”

o Nautilus shell

The true spiral is closely approximated by the artificial spiral formed by circular quadrants inscribed in the successive squares.

http://www.geocities.com/jyce3/spiral.htm http://www.sciencenews.org/articles/20050402/mathtrek.asp Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 42 © Pál Ledneczki Geometrical Constructions 1 5 Regular Polygons, Golden Ratio

Constructions on Regular Pentagon

Given: O, A Given: O, a Given: A,B

A ƒ ‚ … „ ƒ O O R/2 R/2 • „ R • „ ƒ r r • ‚ ‚ a M r r A AB/2 B

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 43 © Pál Ledneczki Geometrical Constructions 1 6 Affine Mapping, Axial Affinity

Affine Collineation

In affine geometry, parallelism plays an important role. An affine transformation preserves neither distances nor angles but the parallelism of lines.

The two chessboards are in affine relation.

P’ P

The circle and ellipse are equivalent in affinity.

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 44 © Pál Ledneczki Geometrical Constructions 1 6 Affine Mapping, Axial Affinity

Affine Triangles

Any two triangles are related by a unique affine collineation.

C AB A' B' C* = AQ A' Q' BC B'C' = BR B' R' C’ P R R* R’

A P’ Q B B’ Q’

A’ Q* B* * * (AQB) @ çæ A' Q B ÷ö The ratios can be transferred by è ø means of parallel transversals: (BRC ) @ çæB' R*C*÷ö è ø

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Axial Affinity

Oblique Orthogonal

l

P h P

T=T’ a T=T’ a

A=A’

P’ h’ P’

An axial affinity is uniquely determined by the axis (invariant line) and l’ a pair of points. The axis, direction and ratio PT:P’T’ also determines the axial affinity. The axial affinity is called orthogonal if the direction PP’ is perpendicular to the axis.

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Circle and Ellipse in Axial Affinity

Orthogonal 2 Oblique axis

(P) (d2)

P b d2 P* d a 1 axis axis1

P

a = radius of greater circle = (P) half of the major axis b = radius of smaller circle = half of the minor axis {d1,(d2)} pair of perpendicular (P) ] P orthogonal affinity diameters of the circle axis: axis1 Y ratio: b:a {d1,d2} pair of conjugate diameters of the P*]P orthogonal affinity ellipse

axis: axis2 ratio: a:b

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 47 © Pál Ledneczki Geometrical Constructions 1 6 Affine Mapping, Axial Affinity

Constructions on Ellipse: Ritz’ Costruction

Let the ellipse be given by a pair of conjugate axes. Construct the major and minor axes.

L a L Q* Q* P -90° Q M b Q -90° P O N N O

The construction can be derived from the concentric circles method. Rotate Q by -90°. Draw and extend the line segment Q*P. Find the midpoint M of the segment Q*P. Draw semicircle about M with the radius MO, find the points of intersection with |Q*P|, L and N. The lines of axes are LO and NO, the half length of axes are a=LP and b=PN.

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 48 © Pál Ledneczki Geometrical Constructions 1 6 Affine Mapping, Axial Affinity Constructions on Ellipse: Invariant Pair of Right Angles in Oblique Axial Affinity

P P b O axis A2 A1 O

A1 A2

P’

P’

The perpendicular bisector b of PP’ intersects the axis at O. The circle about O through P and P’ intersects the axis at The axes of an ellipse can be found by means of invariant pair of right angles A1 and A2. The affinity transformation preserves the right angle at P. construction.

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Approximate Construction of Ellipse

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

K1 O O L 1 B A Osculating circle at B

Perpendicular bisector of AA” Perpendicular to BC K2

L2

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

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Conic Sections

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Ellipse

t P Definition: r1 Let two points F1, F2 r2 Ellipse is the set of points, major foci and a distance 2a whose sum of distances axis F F be given. 1 2 from F1 and F2 is equal to the given distance. axis minor

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

r1 r2 2a

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

bisector of F1F2 and for their point of intersection O.

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

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Properties of Ellipse

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

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

r1 r2 E of the ellipse. The ellipse is the set of centers of circles P passing through a focus and tangent to a circle, i.e. M the director circle about the other focus. r1 r2 2a O Principal circle: circle about O with the radius of a. r F F 2 1 2 The principal circle is the set of pedal points M of lines from F2 perpendicular to the tangents of the ellipse.

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

A point of the ellipse (P), the center of the director circle (F1) and the antipoint (E) corresponding to the given point, are collinear.

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

F V d E

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

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

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Properties of Parabola

Antipoint E: pedal point of the line passing

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

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

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

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Osculating Circle at the Vertex of Parabola

axis

K F

d V

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

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Hyperbola

t

Definition: axis P 2a

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

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

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

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

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Asymptotes of Hyperbola

y Equation of hyperbola: x2 y2 - = 1 a2 b2

c where b is determined by the Pythagorean b equation: 2 2 2 a x a + b = c . F1 F2 The limit of the ratio y/x can be found from the equation: æ y2 ö æ b2 b2 ö ç ÷ = ç - ÷ lim ç 2 ÷ lim ç 2 2 ÷ x ®¥ è x ø x ®¥ è a x ø y b Construction: lim = ± 1) Draw the tangents at the vertices. x ®¥ x a

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

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 58 © Pál Ledneczki Geometrical Constructions 1 7 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 P The director circle is the set of points (antipoints) that director circle r are reflections of a focus with respect to all tangents 2 of the hyperbola. The hyperbola is the set of centers of circles passing through a focus and tangent to a circle, r2 r1 r E i.e. the director circle about the other focus. 1 M r2 O Principal circle: circle about O with the radius of a.

F1 F2 The principal circle is the set of pedal points M of lines from F2 perpendicular to the tangents of the hyperbola.

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

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

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

Budapest University of Technology and Economics ? Faculty of Architecture ? Department of Architectural Representation 2005 Fall Semester 59 © Pál Ledneczki