aarjaneiro/planetmath huggingface dataset:
compass and straightedge construction of square

compass and straightedge construction of square

One can construct a square with sides of a given length

s

using

compass

and straightedge as follows:

1.

Draw a

line segment

of length s. Label its

endpoints

P

and

Q

.

2.

Extend the line segment past

Q

.

3.

Erect the

perpendicular

to

P

⁢

Q

→

at

Q

.

4.

Using the line drawn in the previous step, mark off a line segment of length

s

such that one of its endpoints is

Q

. Label the other endpoint as

R

.

5.

Draw an arc of the circle with center

P

and radius

P

⁢

Q

¯

.

6.

Draw an arc of the circle with center

R

and radius

Q

⁢

R

¯

to find the point

S

where it intersects the arc from the previous step such that

S

≠

Q

.

7.

Draw the square

P

⁢

Q

⁢

R

⁢

S

.

This construction is justified because

P

⁢

S

=

P

⁢

Q

=

Q

⁢

R

=

Q

⁢

S

, yielding that

P

⁢

Q

⁢

R

⁢

S

is a

rhombus

. Since

∠

⁢

P

⁢

Q

⁢

R

is a

right angle

, it follows that

P

⁢

Q

⁢

R

⁢

S

is a square.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title

compass and straightedge construction of square

Canonical name

CompassAndStraightedgeConstructionOfSquare

Date of creation

2013-03-22 17:19:13

Last modified on

2013-03-22 17:19:13

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

5

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Generated on Fri Feb 9 22:07:23 2018 by

LaTeXML
construct the center of a given circle

construct the center of a given circle

[Euclid, Book III, Prop. 1]

Find the center (

http://planetmath.org/Center8

) of a given circle.

Since, in

Euclidean geometry

, a

circle has one center

only, it suffices to construct a point that is a center of the given circle.

Draw any chord

A

⁢

B

¯

in the circle, and construct the

perpendicular bisector

of

A

⁢

B

¯

, intersecting

A

⁢

B

¯

in

C

, and the circle in

D

,

E

.

Let

O

be the center of the circle; we will show that

O

is the

midpoint

of

D

⁢

E

¯

. Note that in the diagram below,

O

is purposely drawn not to lie on

D

⁢

E

¯

; the proof shows that this position is impossible and that in fact

O

lies on

D

⁢

E

¯

. It then follows easily that in fact

O

is the midpoint of

D

⁢

E

¯

.

Since

O

is the center of the circle, it follows that

O

⁢

A

=

O

⁢

B

. Since

D

⁢

E

¯

bisects

A

⁢

B

¯

, we see in addition that

A

⁢

C

=

B

⁢

C

.

△

⁢

A

⁢

C

⁢

O

and

△

⁢

B

⁢

C

⁢

O

share their third side,

O

⁢

C

¯

. So by SSS,

△

⁢

A

⁢

C

⁢

O

≅

△

⁢

B

⁢

C

⁢

O

, and thus, using CPCTC,

∠

⁢

A

⁢

C

⁢

O

≅

∠

⁢

B

⁢

C

⁢

O

. But

∠

⁢

A

⁢

C

⁢

O

+

∠

⁢

B

⁢

C

⁢

O

=

180

∘

, so

∠

⁢

A

⁢

C

⁢

O

and

∠

⁢

B

⁢

C

⁢

O

are each

right angles

. Thus

O

in fact lies on

D

⁢

E

¯

.

However, since

O

is the center of the circle, it must be equidistant from

D

and

E

, and thus

O

is the midpoint of

D

⁢

E

¯

.

Title

construct the center of a given circle

Canonical name

ConstructTheCenterOfAGivenCircle

Date of creation

2013-03-22 17:13:41

Last modified on

2013-03-22 17:13:41

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

9

Author

rm50 (10146)

Entry type

Derivation

Classification

msc 51M15

Classification

msc 51-00

Related topic

CompassAndStraightedgeConstructionOfCenterOfGivenCircle

Generated on Fri Feb 9 22:07:25 2018 by

LaTeXML
construction of central proportional

construction of central proportional

Task.

Given two

line segments

p

and

q

. Using

compass

and straightedge, construct the central proportional (the geometric mean) of the line segments.

Solution.

Set the line segments

A

⁢

D

=

p

and

D

⁢

B

=

q

on a line so that

D

is between

A

and

B

. Draw a half-circle with

diameter

A

⁢

B

(for finding the centre, see the entry

midpoint

). Let

C

be the point where the

normal line

of

A

⁢

B

passing through

D

intersects

the arc of the half-circle. The line segment

C

⁢

D

is the required central proportional. Below is a picture that illustrates this solution:

(For more details on the procedure to create this picture, see

compass and straightedge construction of geometric mean

.)

Proof.

By Thales’

theorem

, the

triangle

A

⁢

B

⁢

C

is a

right triangle

. Its height

C

⁢

D

this triangle into two smaller right triangles which have equal angles with the triangle

A

⁢

B

⁢

C

and thus are

similar

(

http://planetmath.org/SimilarityInGeometry

). Accordingly, we can write the proportion equation concerning the

catheti

of the smaller triangles

p

:

C

⁢

D

=

C

⁢

D

:

q

.

The equation shows that

C

⁢

D

is the central proportional of

p

and

q

.

Note.

The word

catheti

(in sing.

cathetus

) the two shorter sides of a right triangle.

Title

construction of central proportional

Canonical name

ConstructionOfCentralProportional

Date of creation

2013-03-22 17:34:14

Last modified on

2013-03-22 17:34:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Algorithm

Classification

msc 51M15

Related topic

GoldenRatio

Related topic

CompassAndStraightedgeConstructionOfGeometricMean

Related topic

ConstructionOfFourthProportional

Defines

cathetus

Defines

catheti

Generated on Fri Feb 9 22:07:28 2018 by

LaTeXML
construction of contraharmonic mean of two segments

construction of contraharmonic mean of two segments

Let

a

and

b

two

line segments

(and their ).  The

contraharmonic mean

x

:=

a

2

+

b

2

a

+

b

=

(

a

2

+

b

2

)

2

a

+

b

,

satisfying the

proportion equation

a

+

b

a

2

+

b

2

=

a

2

+

b

2

x

,

can be constructed geometrically (

http://

planetmath

.org/GeometricConstruction

) as the third proportional of the segments

a

+

b

and

a

2

+

b

2

, the latter of which is gotten as the

hypotenuse

of the

right triangle

with

catheti

a

and

b

.  See the

construction of fourth proportional

.

(The dotted lines are

parallel

, with declivity

45

∘

.)

Title

construction of contraharmonic mean of two segments

Canonical

name

ConstructionOfContraharmonicMeanOfTwoSegments

Date of creation

2013-03-22 19:12:34

Last modified on

2013-03-22 19:12:34

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Example

Classification

msc 51M15

Classification

msc 51-00

Related topic

ContraharmonicProportion

Related topic

HarmonicMeanInTrapezoid

Generated on Fri Feb 9 22:07:30 2018 by

LaTeXML
construction of fourth proportional

construction of fourth proportional

Task.

Given three

line segments

a

,

b

and

c

.  Using

compass

and straightedge, construct the

fourth proportional

of the line segments.

Solution.

Draw an angle (

α

) and denote its vertex (

http://planetmath.org/Angle

) by

P

.  Separate from one side (

http://planetmath.org/Angle

) of the angle the line segments

P

⁢

A

=

a

and

A

⁢

B

=

b

, and from the other side of the angle the line segment

P

⁢

C

=

c

.  Draw the line

A

⁢

C

and another line

parallel

to it

passing through

B

.  If the last line intersects the other side of the angle in the point

D

, then the line segment

C

⁢

D

=

x

is the required fourth proportional:

a

:

b

=

c

:

x

Justification: the

intercept theorem

.

The below picture illustrates this solution:

Note.

The special case

c

=

b

gives the third proportional

x

of

a

and

b

:

a

:

b

=

b

:

x

Title

construction of fourth proportional

Canonical name

ConstructionOfFourthProportional

Date of creation

2013-03-22 18:49:59

Last modified on

2013-03-22 18:49:59

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Application

Classification

msc 51M15

Classification

msc 51M04

Related topic

ConstructionOfCentralProportion

Related topic

ProportionEquation

Generated on Fri Feb 9 22:07:43 2018 by

LaTeXML
construction of regular 2⁢n-gon from regular n-gon

construction of regular

2

⁢

n

-gon from regular

n

-gon

Given a

regular

n

-gon (

http://planetmath.org/RegularPolygon

), one can construct a regular

2

⁢

n

-gon using

compass

and straightedge. This procedure will be demonstrated by starting with a

regular pentagon

; the procedure will thus produce a

regular decagon

.

The procedure is as follows:

1.

Bisect two of the

interior angles

of the

regular polygon

. These

angle bisectors

will intersect at the center (

http://planetmath.org/Center9

) of the regular polygon.

2.

Connect each vertex of the regular polygon to the center.

3.

Construct the circumscribed circle of the regular polygon.

4.

Bisect each of the

central angles

of the circle to obtain the points where the angle bisectors intersect the circle.

5.

Connect the dots to form the regular

2

⁢

n

-gon. In the picture below, all drawn figures except for the original

polygon

, the circle, and the formed polygon are drawn in cyan to emphasize the three figures that are not dashed.

This construction is justified because the

triangles

formed by the drawn radii of the circle and the drawn (blue) polygon are

congruent

by SAS (note that all of the central angles have measure (

http://planetmath.org/AngleMeasure

)

360

∘

2

⁢

n

), giving that all of the sides and all of the interior angles of the drawn polygon are congruent.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title

construction of regular

2

⁢

n

-gon from regular

n

-gon

Canonical name

ConstructionOfRegular2ngonFromRegularNgon

Date of creation

2013-03-22 17:19:32

Last modified on

2013-03-22 17:19:32

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

19

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Generated on Fri Feb 9 22:07:46 2018 by

LaTeXML
construction of tangent

construction of tangent

Task.

Using the

compass

and straightedge, construct to a given circle the

tangent lines

through a given point outside the circle.

Let

O

be the centre of the given circle and

P

the given point. With

O

⁢

P

as

diameter

, draw the circle (see

midpoint

). If

A

and

B

are the points where this circle

intersects

the given circle, then by Thales’

theorem

, the angles

O

⁢

A

⁢

P

and

O

⁢

B

⁢

P

are

right angles

. According to the definition of the

tangent of circle

, the lines

A

⁢

P

and

B

⁢

P

are required

tangents

.

The

line segment

A

⁢

B

is

The convex angle

A

⁢

P

⁢

B

is called a

tangent angle

(or

tangent-tangent angle

) of the given circle and the convex angle

A

⁢

O

⁢

B

the

corresponding

central angle

. It is apparent that a tangent angle and the corresponding central angle are

supplementary

.  The chord

A

⁢

B

is the

tangent chord

corresponding the tangent angle and the point

P

(see equation of tangent chord (

http://planetmath.org/EquationOfTangentOfCircle

)!).

The tangent angle is the

angle of view

of the line segment

A

⁢

B

from the point

P

.

Note that if a circle is

inscribed

in a

polygon

, then the angles of the polygon are tangent angles of the circle and the centre of the circle is the common

intersection

point of the

angle bisectors

.

Title

construction of tangent

Canonical name

ConstructionOfTangent

Date of creation

2013-03-22 17:36:04

Last modified on

2013-03-22 17:36:04

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Related topic

Incircle

Related topic

AngleBisectorAsLocus

Defines

tangent angle

Defines

tangent-tangent angle

Defines

tangent chord

Generated on Fri Feb 9 22:07:59 2018 by

LaTeXML
criterion for constructibility of regular polygon

criterion for constructibility of regular polygon

Theorem 1

.

Let

n

be an integer with

n

≥

3

. Then a

regular

n

-gon (

http://planetmath.org/RegularPolygon

) is

constructible

(

http://planetmath.org/Constructible2

) if and only if a

primitive

n

th root of unity (

http://planetmath.org/PrimitiveRootOfUnity

) is a constructible number.

Proof.

First of all, note that a is a constructible number if and only if

cos

⁡

(

2

⁢

π

n

)

+

i

⁢

sin

⁡

(

2

⁢

π

n

)

is a constructible number. See the entry on roots of unity for more details. Therefore, without loss of generality, only the constructibility of the number

cos

⁡

(

2

⁢

π

n

)

+

i

⁢

sin

⁡

(

2

⁢

π

n

)

will be considered.

Sufficiency: If a regular

n

-gon is constructible, then so is the angle whose vertex (

http://planetmath.org/Vertex5

) is the center (

http://planetmath.org/Center9

) of the

polygon

and whose rays

pass through

adjacent vertices

of the polygon. The measure (

http://planetmath.org/AngleMeasure

) of this angle is

2

⁢

π

n

.

By the

theorem on constructible angles

,

sin

⁡

(

2

⁢

π

n

)

and

cos

⁡

(

2

⁢

π

n

)

are constructible numbers. Note that

i

is also a constructible number. Thus,

cos

⁡

(

2

⁢

π

n

)

+

i

⁢

sin

⁡

(

2

⁢

π

n

)

is a constructible number.

Necessity: If

ω

=

cos

⁡

(

2

⁢

π

n

)

+

i

⁢

sin

⁡

(

2

⁢

π

n

)

is a constructible number, then so is

ω

m

for any integer

m

.

On the

complex plane

, for every integer

m

with

0

≤

m

<

n

, construct the point corresponding to

ω

m

. Use

line segments

to connect the points corresponding to

ω

m

and

ω

m

+

1

for every integer

m

with

0

≤

m

<

n

. (Note that

ω

0

=

1

=

ω

n

.) This forms a regular

n

-gon.
∎

Title

criterion for constructibility of regular polygon

Canonical name

CriterionForConstructibilityOfRegularPolygon

Date of creation

2013-03-22 17:18:40

Last modified on

2013-03-22 17:18:40

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 51M15

Classification

msc 12D15

Related topic

RegularPolygon

Related topic

RootOfUnity

Related topic

TheoremOnConstructibleAngles

Generated on Fri Feb 9 22:08:02 2018 by

LaTeXML
geometric constructions by Euclid

geometric constructions by Euclid

The

geometric constructions

using

compass

and straightedge consist of three fundamental tasks as given in Euclid’s

The

Elements

(in ancient Greek

Σ

⁢

τ

⁢

o

⁢

ι

⁢

χ

⁢

ε

⁢

ι

´

⁢

α

, transliterated

Stoikheia

). These fundamental tasks are as follows:

1.

Drawing a line through two given points.

2.

Drawing a circle having a given point as its center and

passing through

another given point.

3.

Setting a plane passing through three given non-collinear points, where one performs tasks based on the two preceding tasks.

Example.

The usual task of drawing a circle with a given point as its center and with a given

line segment

as its radius (a fundamental task in many textbooks) can be to Euclid’s fundamental tasks (one needs five circles!).

Remark.

It can be proven that all geometric constructions with compass and straightedge are possible using

only

the compass. (See e.g. (

http://planetmath.org/Eg

)

compass and straightedge construction of parallel line

.)

In the text of Euclid, the constructions are not listed separately, but are combined with the

theorems

as

propositions

. A way to tell whether a proposition is a theorem or a construction is to go to the end of
the proof and see if it ends with QED, in which case it is a theorem, or with QEF, in which case it
is a construction. Note that QEF is an abbreviation for the Latin phrase

quod erat faciendum

, meaning ‘which was to be done’.

Here is a list of the geometric constructions to be found in

The Elements

:

•

I 1 Given a line segment, construct an

equilateral triangle

having that segment as a side.

•

I 2 Given a point and a line segment, construct a line segment having the given point as
an

endpoint

and equal in length to the given line segment.

•

I 3 Given two line segments, produce a line segment whose length is the

difference

of
the lengths of the two given line segments.

•

I 9 Bisect a given angle.

•

I 10 Bisect a given line segment.

•

I 11 Given a line and a point on this line, construct a line

orthogonal

to the given
line passing through the given point.

•

I 12 Given a line and a point not on this line, construct a line orthogonal to the given
line passing through the given point. (i.e. Find the

projection

of a point on a line.)

•

III 1 Construct the center of a given circle.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

References

http://www.physics.ntua.gr/Faculty/mourmouras/euclid/

Online edition of Euclid’s

The Elements

in Greek prepared by D. E. Mourmouras.

Title

geometric constructions by Euclid

Canonical

name

GeometricConstructionsByEuclid

Date of creation

2013-03-22 17:12:32

Last modified on

2013-03-22 17:12:32

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

20

Author

pahio (2872)

Entry type

Topic

Classification

msc 51M15

Classification

msc 51-00

Synonym

geometric construction

Defines

QEF

Defines

Q.E.F.

Generated on Fri Feb 9 22:08:04 2018 by

LaTeXML
index of entries on compass and straightedge constructions

index of entries on compass and straightedge constructions

The following entries discuss

compass and straightedge constructions

:

•

geometric constructions by Euclid

•

perpendicular bisector

•

midpoint

•

compass and straightedge construction of perpendicular

•

compass and straightedge construction of

angle bisector

•

compass and straightedge construction of regular triangle

•

compass and straightedge construction of duplicating an angle

•

compass and straightedge construction of center of given circle

•

construct the center of a given circle

•

construction of tangent

•

circumcenter

()

•

compass and straightedge construction of parallel line

•

compass and straightedge construction of square

•

n

-

section

of

line segment

with compass and straightedge (

http://

planetmath

.org/NSectionOfLineSegmentWithCompassAndStraightedge

)

•

circle with given center and given radius

•

compass and straightedge construction of similar triangles

•

compass and straightedge construction of geometric mean

•

construction of fourth proportional

•

construction of central proportional

•

compass and straightedge construction of inverse point

•

compass and straightedge construction of regular pentagon

•

construction of

regular

2

⁢

n

-gon from regular

n

-gon (

http://planetmath.org/ConstructionOfRegular2nGonFromRegularNGon

)

•

compass and straightedge construction (

http://planetmath.org/CompassAndStraightedgeConstruction

)

•

constructible numbers

•

theorem on constructible numbers

•

theorem on constructible angles

•

criterion for constructibility of regular polygon

•

classical problems of constructibility

Title

index of entries on compass and straightedge constructions

Canonical name

IndexOfEntriesOnCompassAndStraightedgeConstructions

Date of creation

2013-03-22 17:11:15

Last modified on

2013-03-22 17:11:15

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

30

Author

Wkbj79 (1863)

Entry type

Topic

Classification

msc 51M15

Classification

msc 51-00

Related topic

HighschoolMathematics

Generated on Fri Feb 9 22:08:07 2018 by

LaTeXML
midpoint

midpoint

The

concept

of

midpoint

of

line segment

(

http://planetmath.org/Midpoint

) is a special case of the midpoint of a curve or arbitrary figure in

ℝ

2

or

ℝ

3

.

A point

T

is a

midpoint

of the figure

f

, if for each point

A

of

f

there is a point

B

of

f

such that

T

is the midpoint of the line segment

A

⁢

B

.  One says also that

f

is

symmetric about

the point

T

.

Given the equation of a curve in

ℝ

2

or of a

surface

f

in

ℝ

3

, one can, if , take a new point

T

for the origin by using the linear substitutions of the form

x

:=

x

′

+

a

,

y

:=

y

′

+

b

etc.

Thus one may test whether the origin is the midpoint of

f

by checking whether

f

always contains along with any point

(

x

,

y

,

z

)

also the point

(

-

x

,

-

y

,

-

z

)

.

It is easily verified the

Theorem.

If the origin is the midpoint of a

quadratic curve

or a

quadratic surface

, then its equation has no terms of degree (

http://planetmath.org/BasicPolynomial

) 1.

Similarly one can verify the generalisation, that if the origin is the midpoint of an

algebraic curve

or surface of degree

n

, the equation has no terms of degree

n

-

1

,

n

-

3

and so on.

Note.

Some curves and surfaces have infinitely many midpoints (see quadratic surfaces (

http://planetmath.org/QuadraticSurfaces

)).

References

1

Felix Iversen

:

Analyyttisen geometrian oppikirja

. Tiedekirjasto Nr. 19.  Second edition.  Kustannusosakeyhtiö Otava, Helsinki (1963).

Title

midpoint

Canonical name

Midpoint1

Date of creation

2015-04-25 17:39:00

Last modified on

2015-04-25 17:39:00

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Definition

Classification

msc 51M15

Classification

msc 51-00

Synonym

centre

Synonym

center

Generated on Fri Feb 9 22:08:09 2018 by

LaTeXML
perpendicular bisector

perpendicular bisector

Let

A

⁢

B

¯

be a

line segment

in a plane (we are assuming the

Euclidean plane

). A

bisector

of

A

⁢

B

¯

is any line that

passes through

the

midpoint

of

A

⁢

B

¯

. A

perpendicular bisector

of

A

⁢

B

¯

is a bisector that is

perpendicular

to

A

⁢

B

¯

.

It is an easy exercise to show that a line

ℓ

is a perpendicular bisector of

A

⁢

B

¯

iff every point lying on

ℓ

is equidistant from

A

and

B

. From this, one concludes that

the

perpendicular bisector of a line segment is always unique.

A basic way to

construct

the perpendicular bisector

ℓ

given a line segment

A

⁢

B

¯

using the standard

ruler and compass construction

is as follows:

1.

use a compass to draw the circle

C

1

centered at point

A

with radius the length of

A

⁢

B

¯

, by fixing one end of the compass at

A

and the movable end at

B

,

2.

similarly, draw the circle

C

2

centered at

B

with the same radius as above, with the compass fixed at

B

and movable at

A

,

3.

C

1

and

C

2

intersect at two points, say

P

,

Q

(why?)

4.

with a ruler, draw the line

P

⁢

Q

↔

=

ℓ

,

5.

then

ℓ

is the perpendicular bisector of

A

⁢

B

¯

.

Figure 1:

The construction of a perpendicular bisector

(Note: we assume that there is always an ample supply of compasses and rulers of varying sizes, so that given any positive real number

r

, we can find a compass that opens wider than

r

and a ruler that is longer than

r

).

One of the most common use of perpendicular bisectors is to find the center of a circle constructed from three points in a Euclidean plane:

Given three non

collinear points

X

,

Y

,

Z

in a Euclidean plane, let

C

be the unique circle determined by

X

,

Y

,

Z

. Then the center of

C

is located at the

intersection

of the perpendicular bisectors of

X

⁢

Y

¯

and

Y

⁢

Z

¯

.

Title

perpendicular bisector

Canonical

name

PerpendicularBisector

Date of creation

2013-03-22 16:29:03

Last modified on

2013-03-22 16:29:03

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

18

Author

CWoo (3771)

Entry type

Definition

Classification

msc 51M15

Classification

msc 51N20

Classification

msc 51N05

Synonym

center normal

Related topic

Circumcircle

Defines

bisector

Generated on Fri Feb 9 22:08:11 2018 by

LaTeXML