aarjaneiro/planetmath huggingface dataset:
Poincaré disc model

Poincaré disc model

The

Poincaré disc model

for

ℍ

2

is the disc

{

(

x

,

y

)

∈

ℝ

2

:

x

2

+

y

2

<

1

}

in which a point is

similar

to the Euclidean point and a line must be one of the following:

•

a

diameter

(excluding its

endpoints

) of the

unit circle

;

•

an arc (excluding its endpoints) of a circle such that it intersects the unit circle at two distinct points and the two circles are

perpendicular

at both intersection points.

The Poincaré disc model has the drawback that lines in the model do not resemble Euclidean lines; however, it has the advantage that it is angle preserving. That is, the Euclidean of an angle within the model is the

angle measure

in

hyperbolic geometry

. For this reason, this model is also referred to as the

conformal disc model

. (See the entry conformal for more details.)

Some points outside of the Poincaré disc model are important for constructions within the model. The following is an example of such:

Let

ℓ

be a line in the Poincaré disc model that is not a diameter of the circle. The

pole

of

ℓ

is the intersection of the Euclidean lines that are

tangent

(

http://planetmath.org/TangentLine

) to the circle at the endpoints of

ℓ

.

Note that this matches the definition of pole for the

Beltrami-Klein model

. Also, poles are important for the same reason that they are important in the Beltrami-Klein model: Given a line

ℓ

that is not a diameter of the Poincaré disc model, one constructs a line perpendicular to

ℓ

by considering Euclidean lines

passing through

P

⁢

(

ℓ

)

. Thus, two

disjointly parallel lines

ℓ

and

m

that are not diameters of the Poincaré disc model, one constructs their common perpendicular by connecting their poles. It is actually much easier to do this construction by finding the poles of the two lines, finding the common perpendicular with respect to the Beltrami-Klein model, then converting the common perpendicular to the Poincaré disc model. See the entry on converting between the Beltrami-Klein model and the Poincaré disc model for more details.

In all pictures in this entry from this point on, blue segments are lines in the Beltrami-Klein model, and red arcs are lines in the Poincaré disc model.

Below is a picture of two disjointly parallel lines

ℓ

and

m

in the Poincaré disc model, neither of which is a diamter of the unit circle:

Their poles can be found:

The common perpendicular with respect to the Beltrami-Klein model can be found:

From this, the common perpendicular

n

with respect to the Poincaré disc model can be found:

Title

Poincaré disc model

Canonical name

PoincareDiscModel

Date of creation

2013-03-22 17:06:56

Last modified on

2013-03-22 17:06:56

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

25

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 51M10

Classification

msc 51-00

Synonym

conformal disc model

Related topic

ConvertingBetweenTheBeltramiKleinModelAndThePoincareDiscModel

Related topic

ConvertingBetweenThePoincareDiscModelAndTheUpperHalfPlaneModel

Generated on Fri Feb 9 22:06:20 2018 by

LaTeXML
proof of AAA (hyperbolic)

proof of AAA (hyperbolic)

Following is a proof that AAA holds in

hyperbolic geometry

.

Proof.

Suppose that we have two

triangles

△

⁢

A

⁢

B

⁢

C

and

△

⁢

D

⁢

E

⁢

F

such that all three pairs of corresponding angles are

congruent

, but that the two triangles are not congruent. Without loss of generality, let us further assume that

ℓ

⁢

(

A

⁢

B

)

<

ℓ

⁢

(

D

⁢

E

)

, where

ℓ

is used to denote length. (Note that, if

ℓ

⁢

(

A

⁢

B

)

=

ℓ

⁢

(

D

⁢

E

)

, then the two triangles would be congruent by ASA.) Then there are three cases:

1.

ℓ

⁢

(

A

⁢

C

)

>

ℓ

⁢

(

D

⁢

F

)

2.

ℓ

⁢

(

A

⁢

C

)

=

ℓ

⁢

(

D

⁢

F

)

3.

ℓ

⁢

(

A

⁢

C

)

<

ℓ

⁢

(

D

⁢

F

)

Before investigating the cases,

△

⁢

D

⁢

E

⁢

F

will be placed on

△

⁢

A

⁢

B

⁢

C

so that the following are true:

•

A

and

D

correspond

•

A

,

B

, and

E

are

collinear

•

A

,

C

, and

F

are collinear

Now let us investigate each case.

Case 1: Let

G

denote the

intersection

of

B

⁢

C

¯

and

E

⁢

F

¯

Note that

∠

⁢

A

⁢

B

⁢

C

and

∠

⁢

C

⁢

B

⁢

E

are

supplementary

. By

hypothesis

,

∠

⁢

A

⁢

B

⁢

C

and

∠

⁢

D

⁢

E

⁢

F

are congruent. Thus,

∠

⁢

C

⁢

B

⁢

E

and

∠

⁢

D

⁢

E

⁢

F

are supplementary. Therefore,

△

⁢

B

⁢

E

⁢

G

contains two angles which are supplementary, a

contradiction

.

Case 2:

Note that

∠

⁢

A

⁢

B

⁢

C

and

∠

⁢

C

⁢

B

⁢

E

are supplementary. By hypothesis,

∠

⁢

A

⁢

B

⁢

C

and

∠

⁢

D

⁢

E

⁢

F

are congruent. Thus,

∠

⁢

C

⁢

B

⁢

E

and

∠

⁢

D

⁢

E

⁢

F

are supplementary. Therefore,

△

⁢

B

⁢

C

⁢

E

contains two angles which are supplementary, a contradiction.

Case 3: This is the most interesting case, as it is the one that holds in

Euclidean geometry

.

Note that

∠

⁢

A

⁢

B

⁢

C

and

∠

⁢

C

⁢

B

⁢

E

are supplementary. By hypothesis,

∠

⁢

A

⁢

B

⁢

C

and

∠

⁢

D

⁢

E

⁢

F

are congruent. Thus,

∠

⁢

C

⁢

B

⁢

E

and

∠

⁢

D

⁢

E

⁢

F

are supplementary. Similarly,

∠

⁢

B

⁢

C

⁢

F

and

∠

⁢

D

⁢

F

⁢

E

are supplementary. Thus,

B

⁢

C

⁢

F

⁢

E

is a

quadrilateral

whose

angle sum

is exactly

2

⁢

π

radians, a contradiction.

Since none of the three cases is possible, it follows that

△

⁢

A

⁢

B

⁢

C

and

△

⁢

D

⁢

E

⁢

F

are congruent.

∎

Title

proof of AAA (hyperbolic)

Canonical name

ProofOfAAAhyperbolic

Date of creation

2013-03-22 17:08:46

Last modified on

2013-03-22 17:08:46

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 51M10

Generated on Fri Feb 9 22:06:32 2018 by

LaTeXML
Saccheri quadrilateral

Saccheri quadrilateral

In

hyperbolic geometry

, a

Saccheri quadrilateral

is a

quadrilateral

such that one set of opposite sides (called the legs (

http://planetmath.org/Leg

))

congruent

, the other set of opposite sides (called the bases (

http://planetmath.org/Base9

))

disjointly parallel

, and, at one of the bases, both angles are

right angles

. Since the

angle sum

of a

triangle

in hyperbolic geometry is

strictly

less than

π

radians, the angle sum of a quadrilateral in hyperbolic geometry is strictly less than

2

⁢

π

radians. Thus, in any Saccheri quadrilateral, the angles that are not right angles must be acute.

The discovery of Saccheri quadrilaterals is attributed to Giovanni Saccheri.

The common

perpendicular

to the bases of a Saccheri quadrilateral always the quadrilateral into two congruent

Lambert quadrilaterals

. In other , every Saccheri quadrilateral is

symmetric about

the common perpendicular to its bases. Thus, the two

acute angles

of a Saccheri quadrilateral are also congruent.

The legs of a Saccheri quadrilateral are disjointly parallel since one of the bases is a common perpendicular. Therefore, Saccheri quadrilaterals are

parallelograms

. Note also that Saccheri quadrilaterals are

right trapezoids

as well as

isosceles trapezoids

.

Below are some examples of Saccheri quadrilaterals in various models. In each example, the Saccheri quadrilateral is labelled as

A

⁢

B

⁢

C

⁢

D

, and the common perpendicular to the bases is drawn in cyan.

•

The

Beltrami-Klein model

:

In the following example, green lines indicate verification of acute angles by using the poles. (Recall that most other models of hyperbolic geometry are angle preserving. Thus, verification of angle measures is unnecessary in those models.)

•

The Poincaré disc model:

•

The

upper half plane model

:

Title

Saccheri quadrilateral

Canonical

name

SaccheriQuadrilateral

Date of creation

2013-03-22 17:08:20

Last modified on

2013-03-22 17:08:20

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

16

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 51M10

Classification

msc 51-00

Synonym

Saccheri’s quadrilateral

Related topic

IsoscelesTrapezoid

Related topic

RightTrapezoid

Generated on Fri Feb 9 22:06:48 2018 by

LaTeXML
center normal and center normal plane as loci

center normal and center normal plane as loci

Theorem

1.

In the

Euclidean plane

, the

center normal

of a

line segment

is the locus of the points which are equidistant from the both

end points

of the

segment

.

Proof.

Let

A

and

B

be arbitrary given distinct points.

1

∘

.

Let

P

be a point equidistant from

A

and

B

.  If

P

∈

A

⁢

B

,  then

P

is trivially on the center normal of

A

⁢

B

.  Thus suppose that

P

∉

A

⁢

B

.  In the

triangle

P

⁢

A

⁢

B

, let the

angle bisector

of

∠

⁢

P

intersect

the side (

http://planetmath.org/Triangle

)

A

⁢

B

in the point

D

.  Then we have

Δ

⁢

P

⁢

D

⁢

A

≅

Δ

⁢

P

⁢

D

⁢

B

(SAS)

,

whence

∠

⁢

P

⁢

D

⁢

A

=

∠

⁢

P

⁢

D

⁢

B

=

90

∘

,

D

⁢

A

=

D

⁢

B

.

Consequently, the point

P

is always on the center normal of

A

⁢

B

.

2

∘

.

Let

Q

be any point on the center normal and

D

the

midpoint

of the line segment

A

⁢

B

.  We can assume that

Q

≠

D

.  Then we have

Δ

⁢

Q

⁢

D

⁢

A

≅

Δ

⁢

Q

⁢

D

⁢

B

(SAS)

,

implying that

Q

⁢

A

=

Q

⁢

B

.

Thus

Q

is equidistant from

A

and

B

.

Theorem 2.

In the

Euclidean space

, the

center normal plane

of a line segment is the locus of the points which are equidistant from the both end points of the segment.

Proof.

Change “center normal” in the preceding proof to “center normal plane”.

Title

center normal and center normal plane as loci

Canonical name

CenterNormalAndCenterNormalPlaneAsLoci

Date of creation

2013-03-22 18:48:51

Last modified on

2013-03-22 18:48:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M15

Classification

msc 51N05

Classification

msc 51N20

Synonym

center normal as locus

Synonym

center normal plane as locus

Related topic

SAS

Related topic

CircumCircle

Related topic

AngleBisectorAsLocus

Generated on Fri Feb 9 22:06:57 2018 by

LaTeXML
circle with given center and given radius

circle with given center and given radius

Task.

Draw the circle having a given point

O

as its center and a given

line segment

of length

A

⁢

B

as its radius. This construction must be performed with constraints in the spirit of Euclid: One must not take the length of

A

⁢

B

¯

between the tips of the

compass

(i.e. (

http://planetmath.org/Ie

), one must pretend that the compass is collapsible (

http://planetmath.org/CollapsibleCompass

)). This means than one may only draw arcs that are of circles with the center and one point of the

circumference

known.

Solution.

1.

Draw an arc of the circle

a

through

A

with center

O

and an arc of the circle

o

through

O

with center

A

. These arcs must

intersect

each other. Let one of the

intersection

points be

C

.

2.

Draw the lines

C

⁢

A

↔

and

C

⁢

O

↔

.

3.

Draw an arc of the circle

b

through

B

and with center

A

. Let

D

be the intersection point of

b

and the line

C

⁢

A

↔

.

4.

Draw an arc of the circle

c

through

C

and with center

D

. Let

E

be the intersection point of

d

and the line

C

⁢

O

↔

with

E

≠

C

.

5.

Draw the circle

e

through

E

and with center

O

. This is the required circle.

A justification for this construction is that

O

⁢

E

=

C

⁢

E

-

C

⁢

O

=

C

⁢

D

-

C

⁢

A

=

A

⁢

D

=

A

⁢

B

.

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

References

1

E. J. Nyström

:

Korkeamman geometrian alkeet sovellutuksineen

.  Kustannusosakeyhtiö Otava, Helsinki (1948).

Title

circle with given center and given radius

Canonical name

CircleWithGivenCenterAndGivenRadius

Date of creation

2013-03-22 17:14:03

Last modified on

2013-03-22 17:14:03

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

30

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Generated on Fri Feb 9 22:07:00 2018 by

LaTeXML
compass and straightedge construction of center of given circle

compass and straightedge construction of center of given circle

Given a circle in the

Euclidean plane

, one can construct its center (

http://planetmath.org/Center8

) using

compass

and straightedge as follows:

1.

Draw a chord. Label its

endpoints

as

A

and

B

.

2.

Construct the

perpendicular bisector

of

A

⁢

B

¯

in order to find the two points

C

and

D

where it intersects the circle.

3.

Construct the perpendicular bisector of

C

⁢

D

¯

to determine the

midpoint

O

of

C

⁢

D

¯

.

O

is the center of the circle.

A justification for these constructions is supplied in the entry construct the center of a given circle.

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

Title

compass and straightedge construction of center of given circle

Canonical name

CompassAndStraightedgeConstructionOfCenterOfGivenCircle

Date of creation

2013-03-22 17:13:44

Last modified on

2013-03-22 17:13:44

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Related topic

ConstructTheCenterOfAGivenCircle

Generated on Fri Feb 9 22:07:03 2018 by

LaTeXML
compass and straightedge construction of duplicating an angle

compass and straightedge construction of duplicating an angle

In all of the pictures supplied, the angle in red is the one that is to be duplicated.

One can duplicate an angle using compass and straightedge as follows:

1.

Draw one of the rays of the new angle.

2.

With one point of the compass at the vertex (

http://planetmath.org/Vertex5

) of the given angle, draw an arc that

intersects

both of its rays.

3.

With one point of the compass at the point where the vertex of the new angle is supposed to be, draw an arc that intersects the ray of the new angle and extends far from the ray in one direction. The compass needs to be open the same amount as it was in the previous step.

4.

Use the compass to the

distance

between the two points of

intersection

of the given angle and its arc. This length needs to be used in the next step.

5.

With one point on the intersection of the ray of the new angle and its arc, draw an arc so that it intersects the other arc that was drawn for the new angle.

6.

Draw the ray from the vertex of the angle to the intersection of the two arcs.

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

Title

compass and straightedge construction of duplicating an angle

Canonical name

CompassAndStraightedgeConstructionOfDuplicatingAnAngle

Date of creation

2013-03-22 17:13:20

Last modified on

2013-03-22 17:13:20

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Synonym

compass and straightedge construction of copying an angle

Defines

duplicating an angle

Defines

copying an angle

Defines

duplicate an angle

Defines

copy an angle

Defines

duplicate the angle

Defines

copy the angle

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

LaTeXML
compass and straightedge construction of geometric mean

compass and straightedge construction of geometric mean

Given

line segments

of lengths

a

and

b

, one can construct a line segment of length

a

⁢

b

using

compass

and straightedge as follows:

1.

Draw a line segment of length

a

. Label its

endpoints

A

and

C

.

2.

Extend the line segment past

C

.

3.

Mark off a line segment of length

b

such that one of its endpoints is

C

. Label its other endpoint as

B

.

4.

Construct the

perpendicular bisector

of

A

⁢

B

¯

in order to find its

midpoint

M

.

5.

Construct a

semicircle

with center

M

and radii

A

⁢

M

¯

and

B

⁢

M

¯

.

6.

Erect the

perpendicular

to

A

⁢

B

¯

at

C

to find the point

D

where it

intersects

the semicircle. The line segment

D

⁢

C

¯

is of the desired length.

This construction is justified because, if

A

⁢

D

¯

and

B

⁢

D

¯

were drawn, then the two smaller

triangles

would be

similar

, yielding

A

⁢

C

D

⁢

C

=

D

⁢

C

B

⁢

C

.

Plugging in

A

⁢

C

=

a

and

B

⁢

C

=

b

gives that

D

⁢

C

=

a

⁢

b

as desired.

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

Title

compass and straightedge construction of geometric mean

Canonical name

CompassAndStraightedgeConstructionOfGeometricMean

Date of creation

2013-03-22 17:14:55

Last modified on

2013-03-22 17:14:55

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Related topic

ConstructionOfCentralProportion

Generated on Fri Feb 9 22:07:10 2018 by

LaTeXML
compass and straightedge construction of perpendicular

compass and straightedge construction of perpendicular

Let

P

be a point and

ℓ

be a line in the

Euclidean plane

. One can construct a line

m

perpendicular

to

ℓ

and

passing through

P

. The construction given here yields

m

in any circumstance: Whether

P

∈

ℓ

or

P

∉

ℓ

does not matter. On the other hand, the construction looks quite different in these two cases. Thus, the sequence of pictures on the left (in which

ℓ

is in red) is for the case that

P

∉

ℓ

, and the sequence of pictures on the right (in which

ℓ

is in green) is for the case that

P

∈

ℓ

.

1.

With one point of the

compass

on

P

, draw an arc that intersects

ℓ

at two points. Label these as

Q

and

R

.

2.

Construct the

perpendicular bisector

of

Q

⁢

R

¯

. This line is

m

.

This construction is justified because

Q

and

R

are constructed so that

P

is equidistant from them and thus lies on the perpendicular bisector of

Q

⁢

R

¯

.

In the case that

P

∉

ℓ

, this construction is referred to as

dropping the perpendicular

from

P

to

ℓ

. In the case that

P

∈

ℓ

, this construction is referred to as

erecting the perpendicular

to

ℓ

at

P

.

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

Title

compass and straightedge construction of perpendicular

Canonical name

CompassAndStraightedgeConstructionOfPerpendicular

Date of creation

2013-03-22 17:14:01

Last modified on

2013-03-22 17:14:01

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

5

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Related topic

ProjectionOfPoint

Defines

drop the perpendicular

Defines

dropping the perpendicular

Defines

erect the perpendicular

Defines

erecting the perpendicular

Generated on Fri Feb 9 22:07:13 2018 by

LaTeXML
compass and straightedge construction of regular triangle

compass and straightedge construction of regular triangle

One can construct a

regular triangle

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.

Draw an arc of the circle with center

P

and radius

P

⁢

Q

¯

.

3.

Draw an arc of the circle with center

Q

and radius

P

⁢

Q

¯

to find a point

R

where it

intersects

the arc from the previous step.

4.

Draw the regular triangle

△

⁢

P

⁢

Q

⁢

R

.

This construction is justified by the following:

•

P

⁢

Q

¯

≅

P

⁢

R

¯

since they are both radii of the circle from step 2;

•

P

⁢

Q

¯

≅

Q

⁢

R

¯

since they are both radii of the circle from step 3;

•

Thus,

△

⁢

P

⁢

Q

⁢

R

is an

equilateral triangle

;

•

In

Euclidean geometry

, any equilateral triangle is automatically a regular triangle. Therefore,

△

⁢

P

⁢

Q

⁢

R

is a regular triangle.

This construction is based off of the one that Euclid provides in

The

Elements

as the first

proposition

of the first book. Please see

http://planetmath.org/?op=getmsg;id=15600

this post for more details.

This construction also yields a method for constructing a

60

∘

angle using compass and straightedge.

Note that, with the exception of actually drawing the sides of the

triangle

, only the compass was used in this construction. Since regular triangles tessellate, repeated use of this construction provides a way to find infinitely many points on a line given two points on a line using just a compass.

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

Title

compass and straightedge construction of regular triangle

Canonical

name

CompassAndStraightedgeConstructionOfRegularTriangle

Date of creation

2013-03-22 17:19:10

Last modified on

2013-03-22 17:19:10

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

10

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51-00

Synonym

compass and straightedge construction of equilateral triangle

Synonym

compass and straightedge construction of equiangular triangle

Defines

compass and straightedge construction of

60

∘

angle

Generated on Fri Feb 9 22:07:16 2018 by

LaTeXML
compass and straightedge construction of similar triangles

compass and straightedge construction of similar triangles

Let

a

>

0

and

b

>

0

. If

line segments

of lengths

a

and

b

are

constructible

, one can construct a line segment of length

a

⁢

b

using compass and straightedge as follows:

1.

Draw a line segment of length

a

. Label its

endpoints

as

C

and

D

.

2.

Extend the line segment past both

C

and

D

3.

Erect the

perpendicular

to

C

⁢

D

↔

at

C

.

4.

Use the compass to determine a point

E

on the erected perpendicular such that

C

⁢

E

=

1

.

5.

Use the compass to determine a point

F

on

C

⁢

E

→

such that

C

⁢

F

=

b

.

Note that the pictures indicate that

b

>

1

, but the exact same procedure works if

0

<

b

≤

1

.

6.

Connect the points

D

and

E

.

7.

Copy the angle

∠

⁢

C

⁢

D

⁢

E

at

F

to form

similar triangles

. Label the

intersection

of the constructed ray and

C

⁢

D

↔

as

G

.

Note that, if

0

<

b

<

1

, then

F

will be between

C

and

E

, and

G

will be between

C

and

D

. Also, if

b

=

1

, then

F

=

E

and

G

=

D

.

This construction is justified by the following:

•

Since the angle

∠

⁢

C

⁢

F

⁢

G

was copied from the angle

∠

⁢

C

⁢

E

⁢

D

and the two

triangles

share the angle

∠

⁢

D

⁢

C

⁢

E

, then the two triangles

△

⁢

C

⁢

E

⁢

D

and

△

⁢

C

⁢

F

⁢

G

are

similar

;

•

Since

△

⁢

C

⁢

E

⁢

D

∼

△

⁢

C

⁢

F

⁢

G

, we have that

C

⁢

E

C

⁢

F

=

C

⁢

D

C

⁢

G

;

•

Plugging in

C

⁢

D

=

a

,

C

⁢

E

=

1

, and

C

⁢

F

=

b

yields that

C

⁢

G

=

a

⁢

b

.

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

Title

compass and straightedge construction of similar triangles

Canonical name

CompassAndStraightedgeConstructionOfSimilarTriangles

Date of creation

2013-03-22 17:16:02

Last modified on

2013-03-22 17:16:02

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 51M15

Classification

msc 51F99

Generated on Fri Feb 9 22:07:20 2018 by

LaTeXML