aarjaneiro/planetmath huggingface dataset:
Pappus’s theorem

Pappus’s theorem

Let

A

,

B

,

C

be points on a line (not necessarily in that order) and let

D

,

E

,

F

points on another line (not necessarily in that order). Then the

intersection

points of

A

⁢

D

with

F

⁢

C

,

D

⁢

B

with

C

⁢

E

, and

B

⁢

F

with

E

⁢

A

, are

collinear

.

This is a special case of Pascal’s mystic

hexagram

.

Remark

. Pappus’s

theorem

is a statement about the

incidence relation

between points and lines in any geometric

structure

with points, lines, and an incidence relation between the points and the lines. Generally speaking, an incidence geometry is

Pappian

or

satisfies

the

Pappian property

if the statement of Pappus’s theorem is true. In both

Euclidean

and

affine geometry

, Pappus theorem is true. In plane

projective geometry

, both Pappian and non-Pappian planes exist. Furthermore, it can be shown that every Pappian plane is

Desarguesian

, and the

converse

is true if the plane is finite (the result of Wedderburn’s theorem).

Title

Pappus’s theorem

Canonical name

PappussTheorem

Date of creation

2013-03-22 12:25:01

Last modified on

2013-03-22 12:25:01

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Theorem

Classification

msc 51A05

Synonym

Pappus Theorem

Related topic

PascalsMysticHexagram

Related topic

Collinear

Related topic

Concurrent

Defines

Pappian

Defines

Pappian property

Generated on Fri Feb 9 21:56:17 2018 by

LaTeXML
proof of bisectors theorem

proof of bisectors theorem

Notice that the

triangles

△

⁢

B

⁢

A

⁢

P

and

△

⁢

C

⁢

A

⁢

P

have the same common height

h

, and if

(

B

⁢

A

⁢

P

)

and

(

C

⁢

A

⁢

P

)

denote their respective areas, we have

(

B

⁢

A

⁢

P

)

(

C

⁢

A

⁢

P

)

=

B

⁢

P

⋅

h

/

2

P

⁢

C

⋅

h

/

2

=

B

⁢

P

P

⁢

C

.

On the other hand

(

B

⁢

A

⁢

P

)

=

B

⁢

A

⋅

A

⁢

P

⁢

sin

⁡

B

⁢

A

⁢

P

2

,

(

C

⁢

A

⁢

P

)

=

C

⁢

A

⋅

A

⁢

P

⁢

sin

⁡

C

⁢

A

⁢

P

2

and so

(

B

⁢

A

⁢

P

)

(

C

⁢

A

⁢

P

)

=

B

⁢

A

⋅

A

⁢

P

⁢

sin

⁡

B

⁢

A

⁢

P

/

2

C

⁢

A

⋅

A

⁢

P

⁢

sin

⁡

C

⁢

A

⁢

P

/

2

=

B

⁢

A

⁢

sin

⁡

B

⁢

A

⁢

P

C

⁢

A

⁢

sin

⁡

C

⁢

A

⁢

P

.

We have obtained

B

⁢

P

P

⁢

C

=

B

⁢

A

⁢

sin

⁡

B

⁢

A

⁢

P

C

⁢

A

⁢

sin

⁡

C

⁢

A

⁢

P

,

which is the

generalization

to the

theorem

. In the particular case when

A

⁢

P

is the

bisector

,

∠

⁢

B

⁢

A

⁢

P

=

∠

⁢

C

⁢

A

⁢

P

, and thus

sin

⁡

B

⁢

A

⁢

P

=

sin

⁡

C

⁢

A

⁢

P

. Cancelling out the sines proves the

bisector theorem

.

Title

proof of bisectors theorem

Canonical name

ProofOfBisectorsTheorem

Date of creation

2013-03-22 14:49:25

Last modified on

2013-03-22 14:49:25

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Proof

Classification

msc 51A05

Generated on Fri Feb 9 21:56:21 2018 by

LaTeXML
proof of bisectors theorem

proof of bisectors theorem

Consider

sines law

in

triangles

△

⁢

A

⁢

P

⁢

B

and

△

⁢

A

⁢

P

⁢

C

.

On

△

⁢

A

⁢

P

⁢

B

we have

B

⁢

P

sin

⁡

B

⁢

A

⁢

P

=

A

⁢

B

sin

⁡

A

⁢

P

⁢

B

and on

△

⁢

A

⁢

P

⁢

C

we have

P

⁢

C

sin

⁡

P

⁢

A

⁢

C

=

A

⁢

C

sin

⁡

C

⁢

P

⁢

A

.

Combining the two

relation

gives

B

⁢

P

P

⁢

C

=

A

⁢

B

⁢

sin

⁡

B

⁢

A

⁢

P

/

sin

⁡

A

⁢

P

⁢

B

A

⁢

C

⁢

sin

⁡

P

⁢

A

⁢

C

/

sin

⁡

C

⁢

P

⁢

A

.

However,

∠

⁢

A

⁢

P

⁢

B

+

∠

⁢

C

⁢

P

⁢

A

=

180

∘

, and so

sin

⁡

A

⁢

P

⁢

B

=

sin

⁡

C

⁢

P

⁢

A

. Cancelling gives

B

⁢

P

P

⁢

C

=

A

⁢

B

⁢

sin

⁡

B

⁢

A

⁢

P

A

⁢

C

⁢

sin

⁡

P

⁢

A

⁢

C

,

which is the

generalization

of the

theorem

. When

A

⁢

P

is a

bisector

,

∠

⁢

B

⁢

A

⁢

P

=

∠

⁢

P

⁢

A

⁢

C

and we can cancel further to obtain the

bisector theorem

.

Title

proof of bisectors theorem

Canonical name

ProofOfBisectorsTheorem1

Date of creation

2013-03-22 14:49:28

Last modified on

2013-03-22 14:49:28

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Proof

Classification

msc 51A05

Generated on Fri Feb 9 21:56:23 2018 by

LaTeXML
proof of Ceva’s theorem

proof of Ceva’s theorem

\PMlinkescapephrase

sign of

As in the article on Ceva’s Theorem, we will consider directed line segments.

Let

X

,

Y

and

Z

be points on

B

⁢

C

,

C

⁢

A

and

A

⁢

B

, respectively, such that

A

⁢

X

,

B

⁢

Y

and

C

⁢

Z

are

concurrent

, and let

P

be the point where

A

⁢

X

,

B

⁢

Y

and

C

⁢

Z

meet.
Draw a

parallel

to

A

⁢

B

through the point

C

. Extend

A

⁢

X

until it intersects the parallel at a point

A

′

.

Construct

B

′

in a

similar

way extending

B

⁢

Y

.

The

triangles

△

⁢

A

⁢

B

⁢

X

and

△

⁢

A

′

⁢

C

⁢

X

are similar, and so are

△

⁢

A

⁢

B

⁢

Y

and

△

⁢

C

⁢

B

′

⁢

Y

. Then the following equalities hold:

B

⁢

X

X

⁢

C

=

A

⁢

B

C

⁢

A

′

,

C

⁢

Y

Y

⁢

A

=

C

⁢

B

′

B

⁢

A

and thus

B

⁢

X

X

⁢

C

⋅

C

⁢

Y

Y

⁢

A

=

A

⁢

B

C

⁢

A

′

⋅

C

⁢

B

′

B

⁢

A

=

C

⁢

B

′

A

′

⁢

C

.

(1)

Notice that if

directed segments

are being used then

A

⁢

B

and

B

⁢

A

have opposite signs, and therefore when cancelled change the sign of the expression. That’s why we changed

C

⁢

A

′

to

A

′

⁢

C

.

Now we turn to consider the following

similarities

:

△

⁢

A

⁢

Z

⁢

P

∼

△

⁢

A

′

⁢

C

⁢

P

and

△

⁢

B

⁢

Z

⁢

P

∼

△

⁢

B

′

⁢

C

⁢

P

. From them we get the equalities

C

⁢

P

Z

⁢

P

=

A

′

⁢

C

A

⁢

Z

,

C

⁢

P

Z

⁢

P

=

C

⁢

B

′

Z

⁢

B

which lead to

A

⁢

Z

Z

⁢

B

=

A

′

⁢

C

C

⁢

B

′

.

Multiplying the last expression with (

1

) gives

A

⁢

Z

Z

⁢

B

⋅

B

⁢

X

X

⁢

C

⋅

C

⁢

Y

Y

⁢

A

=

1

and we conclude the proof.

To prove the converse, suppose that

X

,

Y

,

Z

are points on

B

⁢

C

,

C

⁢

A

,

A

⁢

B

respectively and satisfying

A

⁢

Z

Z

⁢

B

⋅

B

⁢

X

X

⁢

C

⋅

C

⁢

Y

Y

⁢

A

=

1

.

Let

Q

be the

intersection

point of

A

⁢

X

with

B

⁢

Y

, and let

Z

′

be the intersection of

C

⁢

Q

with

A

⁢

B

. Since then

A

⁢

X

,

B

⁢

Y

,

C

⁢

Z

′

are concurrent, we have

A

⁢

Z

′

Z

′

⁢

B

⋅

B

⁢

X

X

⁢

C

⋅

C

⁢

Y

Y

⁢

A

=

1

and thus

A

⁢

Z

′

Z

′

⁢

B

=

A

⁢

Z

Z

⁢

B

which implies

Z

=

Z

′

, and therefore

A

⁢

X

,

B

⁢

Y

,

C

⁢

Z

are concurrent.

Title

proof of Ceva’s theorem

Canonical name

ProofOfCevasTheorem

Date of creation

2013-03-22 12:38:54

Last modified on

2013-03-22 12:38:54

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

11

Author

yark (2760)

Entry type

Proof

Classification

msc 51A05

Generated on Fri Feb 9 21:56:25 2018 by

LaTeXML
proof of Menelaus’ theorem

proof of Menelaus’ theorem

First we note that there are two different cases: Either the line connecting

X

,

Y

and

Z

intersects

two sides of the

triangle

or none of them. So in the first case that it intersects two of the triangle’s sides we get the following picture:

From this we follow (

h

1

,

h

2

and

h

3

being undircted):

A

⁢

Z

Z

⁢

B

=

-

h

1

h

2

B

⁢

Y

Y

⁢

C

=

h

2

h

3

C

⁢

X

X

⁢

A

=

h

3

h

1

.

Mulitplying all this we get:

A

⁢

Z

Z

⁢

B

⋅

B

⁢

Y

Y

⁢

C

⋅

C

⁢

X

X

⁢

A

=

-

h

1

⁢

h

2

⁢

h

3

h

2

⁢

h

3

⁢

h

1

=

-

1

.

The second case is that the line connecting

X

,

Y

and

Z

does not intersect any of the triangle’s sides:

In this case we get:

A

⁢

Z

Z

⁢

B

=

-

h

1

h

2

B

⁢

Y

Y

⁢

C

=

-

h

2

h

3

C

⁢

X

X

⁢

A

=

-

h

3

h

1

.

So

multiplication

again yields Menelaus’

theorem

.

Title

proof of Menelaus’ theorem

Canonical

name

ProofOfMenelausTheorem

Date of creation

2013-03-22 12:46:46

Last modified on

2013-03-22 12:46:46

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

4

Author

mathwizard (128)

Entry type

Proof

Classification

msc 51A05

Generated on Fri Feb 9 21:56:27 2018 by

LaTeXML
proof of Pappus’s theorem

proof of Pappus’s theorem

Pappus’s theorem says that if the six vertices of a

hexagon

lie
alternately on two lines, then
the three points of

intersection

of opposite sides are

collinear

.
In the figure, the given lines are

A

11

⁢

A

13

and

A

31

⁢

A

33

,
but we have omitted the letter

A

.

The appearance of the diagram will depend on the order in which the
given points appear on the two lines; two possibilities are shown.

Pappus’s theorem is true in the

affine plane

over any (commutative) field.
A tidy proof is available with the aid of

homogeneous coordinates

.

No three of the four points

A

11

,

A

21

,

A

31

, and

A

13

are
collinear, and therefore we can choose homogeneous coordinates such that

A

11

=

(

1

,

0

,

0

)

A

21

=

(

0

,

1

,

0

)

A

31

=

(

0

,

0

,

1

)

A

13

=

(

1

,

1

,

1

)

That gives us equations for three of the lines in the figure:

A

13

⁢

A

11

:

y

=

z

A

13

⁢

A

21

:

z

=

x

A

13

⁢

A

31

:

x

=

y

.

These lines contain

A

12

,

A

32

, and

A

22

respectively, so

A

12

=

(

p

,

1

,

1

)

A

32

=

(

1

,

q

,

1

)

A

22

=

(

1

,

1

,

r

)

for some scalars

p

,

q

,

r

. So, we get equations for six more lines:

A

31

⁢

A

32

:

y

=

q

⁢

x

A

11

⁢

A

22

:

z

=

r

⁢

y

A

12

⁢

A

21

:

x

=

p

⁢

z

(1)

A

31

⁢

A

12

:

x

=

p

⁢

y

A

11

⁢

A

32

:

y

=

q

⁢

z

A

21

⁢

A

22

:

z

=

r

⁢

x

(2)

By hypothesis, the three lines (

1

) are

concurrent

, and therefore

p

⁢

r

⁢

q

=

1

. But that implies

p

⁢

q

⁢

r

=

1

, and therefore the three lines
(

2

) are concurrent, QED.

Title

proof of Pappus’s theorem

Canonical name

ProofOfPappussTheorem

Date of creation

2013-03-22 13:47:40

Last modified on

2013-03-22 13:47:40

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Proof

Classification

msc 51A05

Generated on Fri Feb 9 21:56:29 2018 by

LaTeXML
proof of trigonometric version of Ceva’s theorem

proof of trigonometric version of Ceva’s theorem

The proof goes by proving the condition imposed on the sines, is

equivalent

to the one imposed on the sides on the normal version of Ceva’s

theorem

.

We want to prove

sin

⁡

A

⁢

C

⁢

Z

sin

⁡

Z

⁢

C

⁢

B

⋅

sin

⁡

B

⁢

A

⁢

X

sin

⁡

X

⁢

A

⁢

C

⋅

sin

⁡

C

⁢

B

⁢

Y

sin

⁡

Y

⁢

B

⁢

A

=

1

if and only if

A

⁢

Z

Z

⁢

B

⋅

B

⁢

X

X

⁢

C

⋅

C

⁢

Y

Y

⁢

A

=

1

.

Now, the

generalization of bisectors theorem

states that

A

⁢

Z

Z

⁢

B

=

C

⁢

A

⁢

sin

⁡

A

⁢

C

⁢

Z

B

⁢

C

⁢

sin

⁡

Z

⁢

C

⁢

B

,

B

⁢

X

X

⁢

C

=

A

⁢

B

⁢

sin

⁡

B

⁢

A

⁢

X

C

⁢

A

⁢

sin

⁡

X

⁢

A

⁢

C

,

C

⁢

Y

Y

⁢

A

=

B

⁢

C

⁢

sin

⁡

C

⁢

B

⁢

Y

A

⁢

B

⁢

sin

⁡

Y

⁢

B

⁢

A

.

Multiplying the three equations, and cancelling all

segments

on the right side gives the desired equivalence.

Title

proof of trigonometric version of Ceva’s theorem

Canonical name

ProofOfTrigonometricVersionOfCevasTheorem

Date of creation

2013-03-22 14:49:20

Last modified on

2013-03-22 14:49:20

Owner

drini (3)

Last modified by

drini (3)

Numerical id

7

Author

drini (3)

Entry type

Proof

Classification

msc 51A05

Generated on Fri Feb 9 21:56:34 2018 by

LaTeXML
trigonometric version of Ceva’s theorem

trigonometric version of Ceva’s theorem

Let

A

⁢

B

⁢

C

be a given

triangle

and

P

any point of the plane. If

X

is the

intersection

point of

A

⁢

P

with

B

⁢

C

,

Y

the intersection point of

B

⁢

P

with

C

⁢

A

and

Z

is the intersection point of

C

⁢

P

with

A

⁢

B

, then

sin

⁡

A

⁢

C

⁢

Z

sin

⁡

Z

⁢

C

⁢

B

⋅

sin

⁡

B

⁢

A

⁢

X

sin

⁡

X

⁢

A

⁢

C

⋅

sin

⁡

C

⁢

B

⁢

Y

sin

⁡

Y

⁢

B

⁢

A

=

1

.

Conversely

, if

X

,

Y

,

Z

are points on

B

⁢

C

,

C

⁢

A

,

A

⁢

B

respectively, and if

sin

⁡

A

⁢

C

⁢

Z

sin

⁡

Z

⁢

C

⁢

B

⋅

sin

⁡

B

⁢

A

⁢

X

sin

⁡

X

⁢

A

⁢

C

⋅

sin

⁡

C

⁢

B

⁢

Y

sin

⁡

Y

⁢

B

⁢

A

=

1

then

A

⁢

D

,

B

⁢

E

,

C

⁢

F

are

concurrent

.

Remarks: All the angles are

directed angles

(counterclockwise is positive), and the intersection points may lie in the prolongation of the segments.

Title

trigonometric version of Ceva’s

theorem

Canonical name

TrigonometricVersionOfCevasTheorem

Date of creation

2013-03-22 14:49:17

Last modified on

2013-03-22 14:49:17

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Theorem

Classification

msc 51A05

Related topic

Triangle

Related topic

Median

Related topic

Centroid

Related topic

Orthocenter

Related topic

OrthicTriangle

Related topic

Cevian

Related topic

Incenter

Related topic

GergonnePoint

Related topic

MenelausTheorem

Related topic

ProofOfVanAubelTheorem

Related topic

VanAubelTheorem

Generated on Fri Feb 9 21:56:36 2018 by

LaTeXML
(partial) linear space

(partial) linear space

A

partial linear space

is a pair

(

P

,

L

)

consisting of a set

P

of points and a set

L

of lines such that

•

every line contains at least two points, and

•

any two distinct points are on at most one line.

If, in

addition

, any two distinct points are on at least one line, then

(

P

,

L

)

is called a

linear space

.

Title

(partial) linear space

Canonical

name

partialLinearSpace

Date of creation

2013-03-22 17:15:32

Last modified on

2013-03-22 17:15:32

Owner

pan (17366)

Last modified by

pan (17366)

Numerical id

8

Author

pan (17366)

Entry type

Definition

Classification

msc 51A05

Generated on Fri Feb 9 21:56:38 2018 by

LaTeXML
polarity

polarity

Definition 1

.

•

Given finite dimensional

vector spaces

V

and

W

, a

duality

of the

projective geometry

P

⁢

G

⁢

(

V

)

to

P

⁢

G

⁢

(

W

)

is an order-reversing

bijection

f

:

P

⁢

G

⁢

(

V

)

→

P

⁢

G

⁢

(

W

)

. If

W

=

V

then we can refer to

f

as a correlation.

•

A correlation of order

2

is called a

polarity

.

•

The set of correlations and

collineations

f

:

P

⁢

G

⁢

(

V

)

→

P

⁢

G

⁢

(

V

)

form a group denoted

P

⁢

Γ

⁢

L

*

⁢

(

V

)

with the operation of

composition

.

Remark 2

.

Dualities are determined by where they map

collinear

triples. Given a map
define on the points of

P

⁢

G

⁢

(

V

)

to the

hyperplanes

of

P

⁢

G

⁢

(

W

)

which maps collinear triples to triples of hyperplanes which intersect in a

codimension

2

subspace

, this specifies a unique duality.

Remark 3

.

A polarity/duality necessarily interchanges points with hyperplanes. In this context points are called “poles” and hyperplanes “polars.”

An alternative

definition

of a duality is a

projectivity

(order-preserving map)

f

:

P

⁢

G

⁢

(

V

)

→

P

⁢

G

⁢

(

V

*

)

.

Through the use of the

fundamental theorem of projective geometry

, dualities and polarities can be identified with

non-degenerate

sesquilinear forms

. (See

Polarities and forms

(

http://planetmath.org/PolaritiesAndForms

).)

Title

polarity

Canonical

name

Polarity

Date of creation

2013-03-22 15:57:58

Last modified on

2013-03-22 15:57:58

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

12

Author

Algeboy (12884)

Entry type

Definition

Classification

msc 51A10

Classification

msc 51A05

Synonym

order reversing

Related topic

SesquilinearFormsOverGeneralFields

Related topic

PolaritiesAndForms

Defines

polarity

Defines

duality

Defines

correlation

Defines

pole

Defines

polar

Generated on Fri Feb 9 21:57:02 2018 by

LaTeXML
Szemerédi-Trotter theorem

Szemerédi-Trotter theorem

The number of incidences of a set of

n

points and a set of

m

lines in the real plane

ℝ

2

is

I

=

O

⁢

(

n

+

m

+

(

n

⁢

m

)

2

3

)

.

Proof

. Let’s consider the points as vertices of a graph, and connect two vertices by an edge if they are

adjacent

on some line. Then the number of edges is

e

=

I

-

m

. If

e

<

4

⁢

n

then we are done. If

e

≥

4

⁢

n

then by

crossing lemma

m

2

≥

cr

⁡

(

G

)

≥

1

64

⁢

(

I

-

m

)

3

n

2

,

and the theorem follows.

Recently, Tóth

[

1

]

extended the theorem to the

complex plane

ℂ

2

. The proof is difficult.

References

1

Csaba D. Tóth.

The Szemerédi-Trotter theorem in the complex plane.

http://www.arxiv.org/abs/math.CO/0305283

arXiv:CO/0305283,
May 2003.

Title

Szemerédi-Trotter theorem

Canonical name

SzemerediTrotterTheorem

Date of creation

2013-03-22 13:21:30

Last modified on

2013-03-22 13:21:30

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

7

Author

bbukh (348)

Entry type

Theorem

Classification

msc 51A20

Classification

msc 05C10

Generated on Fri Feb 9 21:57:32 2018 by

LaTeXML
proof of Desargues’ theorem

proof of Desargues’ theorem

The claim is that if

triangles

A

⁢

B

⁢

C

and

X

⁢

Y

⁢

Z

are

perspective

from
a point

P

, then they are perspective from a line (meaning that
the three points

A

⁢

B

⋅

X

⁢

Y

B

⁢

C

⋅

Y

⁢

Z

C

⁢

A

⋅

Z

⁢

X

are

collinear

) and conversely.

Since no three of

A

,

B

,

C

,

P

are collinear, we can lay down

homogeneous coordinates

such that

A

=

(

1

,

0

,

0

)

B

=

(

0

,

1

,

0

)

C

=

(

0

,

0

,

1

)

P

=

(

1

,

1

,

1

)

By

hypothesis

, there are scalars

p

,

q

,

r

such that

X

=

(

1

,

p

,

p

)

Y

=

(

q

,

1

,

q

)

Z

=

(

r

,

r

,

1

)

The equation for a line through

(

x

1

,

y

1

,

z

1

)

and

(

x

2

,

y

2

,

z

2

)

is

(

y

1

⁢

z

2

-

z

1

⁢

y

2

)

⁢

x

+

(

z

1

⁢

x

2

-

x

1

⁢

z

2

)

⁢

y

+

(

x

1

⁢

y

2

-

y

1

⁢

x

2

)

⁢

z

=

0

,

giving us equations for six lines:

A

⁢

B

:

z

=

0

B

⁢

C

:

x

=

0

C

⁢

A

:

y

=

0

X

⁢

Y

:

(

p

⁢

q

-

p

)

⁢

x

+

(

p

⁢

q

-

q

)

⁢

y

+

(

1

-

p

⁢

q

)

⁢

z

=

0

Y

⁢

Z

:

(

1

-

q

⁢

r

)

⁢

x

+

(

q

⁢

r

-

q

)

⁢

y

+

(

q

⁢

r

-

r

)

⁢

z

=

0

Z

⁢

X

:

(

r

⁢

p

-

p

)

⁢

x

+

(

1

-

r

⁢

p

)

⁢

y

+

(

r

⁢

p

-

r

)

⁢

z

=

0

whence

A

⁢

B

⋅

X

⁢

Y

=

(

p

⁢

q

-

q

,

-

p

⁢

q

+

p

,

0

)

B

⁢

C

⋅

Y

⁢

Z

=

(

0

,

q

⁢

r

-

r

,

-

q

⁢

r

+

q

)

C

⁢

A

⋅

Z

⁢

X

=

(

-

r

⁢

p

+

r

,

0

,

r

⁢

p

-

p

)

.

As claimed, these three points are collinear, since the
determinant

|

p

⁢

q

-

q

-

p

⁢

q

+

p

0

0

q

⁢

r

-

r

-

q

⁢

r

+

q

-

r

⁢

p

+

r

0

r

⁢

p

-

p

|

is zero. (More precisely, all three points are on the line

p

(

q

-

1

)

(

r

-

1

)

x

+

(

p

-

1

)

q

(

r

-

1

)

y

+

(

p

-

1

)

(

q

-

1

)

r

z

=

0

.

)

Since the hypotheses are self-dual, the converse is true also, by
the principle of

duality

.

Title

proof of Desargues’

theorem

Canonical

name

ProofOfDesarguesTheorem

Date of creation

2013-03-22 13:47:51

Last modified on

2013-03-22 13:47:51

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Proof

Classification

msc 51A30

Generated on Fri Feb 9 21:57:37 2018 by

LaTeXML
Veblen-Wedderburn system

Veblen-Wedderburn system

A

Veblen-Wedderburn system

is an

algebraic system

over a set

R

with two

binary operations

+

(called

addition

) and

⋅

(called

multiplication

) on

R

such that

1.

there is a

0

∈

R

, and that

R

is an

abelian group

under

+

, with

0

the

additive identity

2.

R

-

{

0

}

, together with

⋅

, is a loop (we denote

1

as its

identity element

)

3.

⋅

is

right distributive

over

+

; that is,

(

a

+

b

)

⋅

c

=

a

⋅

c

+

b

⋅

c

4.

if

a

≠

b

, then the

equation

x

⋅

a

=

x

⋅

b

+

c

has a unique solution in

x

A Veblen-Wedderburn system is also called a

quasifield

.

Usually, we write

a

⁢

b

instead of

a

⋅

b

.

For any

a

,

b

,

c

∈

R

, by defining a ternary operation

*

on

R

, given by

a

*

b

*

c

:=

a

⁢

b

+

c

,

it is not hard to see that

(

R

,

*

,

0

,

1

)

is a ternary ring. In fact, it is a linear ternary ring because

a

⁢

b

=

a

*

b

*

0

and

a

+

c

=

a

*

1

*

c

.

For example, any field, or more generally, any

division ring

,

associative

or not, is Veblen-Wedderburn. An example of a Veblen-Wedderburn system that is not a division ring is the Hall quasifield.

A well-known fact about Veblen-Wedderburn systems is that, the

projective plane

of a Veblen-Wedderburn system is a

translation plane

, and, conversely, every translation plane can be coordinatized by a Veblen-Wedderburn system. This is the reason why a translation plane is also called a Veblen-Wedderburn plane.

Remark

. Let

R

be a Veblen-Wedderburn system. If the multiplication

⋅

, in addition to be right distributive over

+

, is also left distributive over

+

, then

R

is a

semifield

. If

⋅

, on the other hand, is associative, then

R

is an

abelian

nearfield

(a nearfield such that

+

is

commutative

).

References

1

R. Casse,

Projective Geometry

, An Introduction

, Oxford University Press (2006)

Title

Veblen-Wedderburn system

Canonical name

VeblenWedderburnSystem

Date of creation

2013-03-22 19:15:06

Last modified on

2013-03-22 19:15:06

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Definition

Classification

msc 51A35

Classification

msc 51A40

Classification

msc 51E15

Classification

msc 51A25

Synonym

Veblen-Wedderburn ring

Synonym

quasifield

Synonym

VW system

Synonym

V-W system

Generated on Fri Feb 9 21:58:18 2018 by

LaTeXML