aarjaneiro/planetmath huggingface dataset:
area of a quadrilateral

area of a quadrilateral

Let

a

,

b

,

c

,

d

be the lengths of the sides of a

quadrilateral

and

K

be its area.
Let

s

be the

semiperimeter

.
Then

K

2

=

(

s

-

a

)

⁢

(

s

-

b

)

⁢

(

s

-

c

)

⁢

(

s

-

d

)

-

a

⁢

b

⁢

c

⁢

d

⁢

cos

2

⁡

(

θ

+

ϕ

2

)

where

θ

and

ϕ

are of the quadrilateral.
Letting

d

→

0

we obtain Heron’s formula for the area of a

triangle

.

References

1

C.A. Bretschneider,

Untersuchung der trigonometrischen Relationen des geradlinigen
Viereckes

. Archiv der Math. 2, (1842), 225-261.

2

F. Strehlke,

Zwei neue Sätze vom ebenen und shpärischen Viereck und
Umkehrung des Ptolemaischen Lehrsatzes.

Archiv der Math. 2, (1842) 323-326.

Title

area of a quadrilateral

Canonical name

AreaOfAQuadrilateral

Date of creation

2013-03-22 16:58:22

Last modified on

2013-03-22 16:58:22

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 51N20

Generated on Fri Feb 9 22:11:03 2018 by

LaTeXML
Cartesian coordinates

Cartesian coordinates

The

Cartesian coordinates

of a point in

ℝ

3

for determining its in three-dimensional space are the three real numbers

x

,

y

and

z

, which are called

•

x

-

coordinate

or

abscissa

,

•

y

-coordinate or

ordinate

,

•

z

-coordinate or

applicate

.

The last name “applicate” is rare in English, but its in continental European , as “die Applikate” in German and “aplikaat” in Estonian, are more known.

Similarly, in

ℝ

n

for all

n

=

1

,

2

,

3

,

…

one needs

n

coordinates for specifying the location of a point.

Title

Cartesian coordinates

Canonical

name

CartesianCoordinates

Date of creation

2013-03-22 14:28:59

Last modified on

2013-03-22 14:28:59

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Synonym

rectangular coordinates

Related topic

AnalyticGeometry

Defines

abscissa

Defines

ordinate

Defines

applicate

Generated on Fri Feb 9 22:11:04 2018 by

LaTeXML
confocal

confocal

Two conics are

confocal

if they have coincident foci.

Examples

1.

The family of ellipses

x

2

a

2

+

s

+

y

2

b

2

+

s

=

1

,

where

a

2

>

b

2

and the

parameter

s

is

>

-

b

2

, is confocal.

2.

The family of hyperbolas

x

2

a

2

-

t

-

y

2

t

-

b

2

=

1

,

where

a

2

>

b

2

and the parameter

t

is between

a

2

and

b

2

, is confocal.

Title

confocal

Canonical

name

Confocal

Date of creation

2013-03-22 14:44:58

Last modified on

2013-03-22 14:44:58

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

6

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 51N20

Generated on Fri Feb 9 22:11:33 2018 by

LaTeXML
coordinate way

coordinate way

Let

P

=

(

x

,

y

,

z

)

be a point in

ℝ

3

. If

N

is the projection (

http://planetmath.org/ProjectionOfPoint

) of

P

on the

x

⁢

y

-plane,

A

the projection of

N

on the

x

-axis and

O

the origin, then the

broken line

P

⁢

N

⁢

A

⁢

O

is the

coordinate way

of the point

P

. The lengths of the

line segments

forming the coordinate way are

A

⁢

O

=

|

x

|

,

N

⁢

A

=

|

y

|

,

P

⁢

N

=

|

z

|

.

Title

coordinate way

Canonical

name

CoordinateWay

Date of creation

2013-03-22 17:28:45

Last modified on

2013-03-22 17:28:45

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Classification

msc 51G05

Classification

msc 51-00

Synonym

coordinate broken line

Generated on Fri Feb 9 22:12:03 2018 by

LaTeXML
coordinates of midpoint

coordinates of midpoint

The

coordinates

of the

midpoint

of a

line segment

are the

arithmetic means

of the coordinates of the

endpoints

of the

segment

. Thus, if the endpoints are

(

x

1

,

y

1

)

and

(

x

2

,

y

2

)

,  then the midpoint is

(

x

1

+

x

2

2

,

y

1

+

y

2

2

)

.

For justifying the above coordinates of the midpoint, we know that its

abscissa

x

0

halves on the

x

-axis the line segment between

x

1

and

x

2

. Since the lengths of the half-segments are

x

0

-

x

1

and

x

2

-

x

0

, if

x

1

<

x

2

,  and their opposite numbers, if

x

2

<

x

1

,  in any case we can write

x

0

-

x

1

=

x

2

-

x

0

.

Solving this equation for

x

0

yields:

x

0

=

x

1

+

x

2

2

.  result is gotten for the ordinate of the midpoint.

Title

coordinates of midpoint

Canonical name

CoordinatesOfMidpoint

Date of creation

2013-03-22 17:31:07

Last modified on

2013-03-22 17:31:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Result

Classification

msc 51N20

Classification

msc 51M15

Classification

msc 51-00

Related topic

ConjugateDiametersOfEllipse

Related topic

CentreOfMassOfPolygon

Related topic

Midpoint4

Generated on Fri Feb 9 22:12:05 2018 by

LaTeXML
curve of Agnesi

curve of Agnesi

Given a real constant

c

, the

curve of Agnesi

(often called

witch of Agnesi

in English) is the result of plotting the equation

y

=

8

⁢

c

4

⁢

c

2

+

x

2

in the

Cartesian plane

. If we set

c

=

1

2

, the equation simplifies to

y

=

1

1

+

x

2

. Another way of drawing the curve employs a circle of radius

c

.

In the following diagram, the associated circle is shown in light gray.

(This diagram was made with

Grapher

1.1 for Mac OS X).

This curve was first studied by

Pierre de Fermat

, but

Maria Gaetana Agnesi

later studied it in greater detail and mentioned it in her book

Instituzioni Analitiche

.

Title

curve of Agnesi

Canonical name

CurveOfAgnesi

Date of creation

2013-03-22 16:40:56

Last modified on

2013-03-22 16:40:56

Owner

CompositeFan (12809)

Last modified by

CompositeFan (12809)

Numerical id

5

Author

CompositeFan (12809)

Entry type

Definition

Classification

msc 51N20

Synonym

Agnesi’s curve

Synonym

witch of Agnesi

Synonym

Agnesi’s witch

Synonym

averisera

Synonym

avversiera

Synonym

cubique d’Agnesi

Synonym

agnésienne

Synonym

agnesienne

Related topic

AsymptoteOfLamesCubic

Generated on Fri Feb 9 22:12:08 2018 by

LaTeXML
diametral points

diametral points

Two points

P

1

and

P

2

on the

circumference

of a circle (or on a sphere) are

diametral

, if the

line segment

P

1

⁢

P

2

connecting them

passes through

the centre of the circle (resp. the sphere), i.e. is a diametre (

http://planetmath.org/

Diameter

). Equivalently, the shortest

distance

of the diametral points

P

1

and

P

2

on the circle is maximal on the circle (resp. on the sphere), namely a half of the perimetre (

http://planetmath.org/Perimeter

).

It’s easily justified that a point of a circle (resp. a sphere) has exactly one diametral point.

A circle

c

is a

diametral circle

of a given circle

c

0

, if

c

intersects

c

0

diametrically

, i.e. in two diametral points of

c

0

.

If the equation of

c

0

is

(

x

-

x

0

)

2

+

(

y

-

y

0

)

2

=

r

2

and

(

a

,

b

)

is inside

c

0

, then the equation of the diametral circle

c

with centre

(

a

,

b

)

is given by

(

x

-

a

)

2

+

(

y

-

b

)

2

=

r

2

-

(

x

0

-

a

)

2

-

(

y

0

-

b

)

2

.

Title

diametral points

Canonical name

DiametralPoints

Date of creation

2013-03-22 18:32:14

Last modified on

2013-03-22 18:32:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Classification

msc 51M04

Related topic

Antipodal

Defines

diametral

Defines

diametral circle

Generated on Fri Feb 9 22:12:10 2018 by

LaTeXML
distance from point to a line

distance from point to a line

The

distance

from a point P with

coordinates

(

x

p

,

y

p

)

∈

ℝ

2

to the line with equation

a

⁢

x

+

b

⁢

y

+

c

=

0

is given by

|

a

⁢

x

p

+

b

⁢

y

p

+

c

|

/

a

2

+

b

2

.

Proof

Every point

x

,

y

on the line is at some distance

(

x

-

x

p

)

2

+

(

y

-

y

p

)

2

from P. What we need to find is the minimum such distance. Our problem is

min

(

x

-

x

p

)

2

+

(

y

-

y

p

)

2

subject to

a

⁢

x

+

b

⁢

y

+

c

=

0

This problem is solvable using the

Lagrange multiplier method

. We minimize

(

x

-

x

p

)

2

+

(

y

-

y

p

)

2

+

λ

⁢

(

a

⁢

x

+

b

⁢

y

+

c

)

Calculating the derivatives with respect to

x

,

y

and

λ

and setting them to zero we get three equations:

2

⁢

x

-

2

⁢

x

p

+

λ

⁢

a

=

0

(1)

2

⁢

y

-

2

⁢

y

p

+

λ

⁢

b

=

0

(2)

2

⁢

a

⁢

x

+

2

⁢

b

⁢

y

+

2

⁢

c

=

0

(3)

Solving these leads to

x

p

-

x

=

a

⁢

a

⁢

x

p

+

b

⁢

y

p

+

c

a

2

+

b

2

and

y

p

-

y

=

b

⁢

a

⁢

x

p

+

b

⁢

y

p

+

c

a

2

+

b

2

.
We can now substitute these expressions into

(

x

-

x

p

)

2

+

(

y

-

y

p

)

2

and we get (after some simplification) the desired result.

Title

distance from point to a line

Canonical name

DistanceFromPointToALine

Date of creation

2013-03-22 15:24:30

Last modified on

2013-03-22 15:24:30

Owner

acastaldo (8031)

Last modified by

acastaldo (8031)

Numerical id

7

Author

acastaldo (8031)

Entry type

Result

Classification

msc 51N20

Related topic

DistanceOfNonParallelLines

Related topic

DistanceBetweenTwoLinesInR3

Related topic

Envelope

Related topic

AngleBisectorAsLocus

Generated on Fri Feb 9 22:12:13 2018 by

LaTeXML
envelope

envelope

Two

plane curves

are said to

touch each other

or

have a tangency

at a point if they have a common

tangent line

at that point.

The

envelope

of a family of plane curves is a curve which touches in each of its points one of the curves of the family.

For example, the envelope of the family

y

=

m

⁢

x

-

1

+

m

2

,  with

m

the parameter, may be justified geometrically.  It is the open (

http://planetmath.org/OpenSet

) lower

semicircle

of the

unit circle

.  Indeed, the

distance

of any line

m

⁢

x

-

y

-

1

+

m

2

=

0

of the family from the center of the unit circle is

|

m

⋅

0

-

1

⋅

0

-

1

+

m

2

|

m

2

+

(

-

1

)

2

=

1

,

whence the line is the

tangent

to the circle.

Below, the red curve is the lower semicircle of the unit circle, the black lines belong to the family

y

=

m

⁢

x

-

1

+

m

2

,  and the equation of each line is given.

Title

envelope

Canonical name

Envelope

Date of creation

2013-03-22 17:10:19

Last modified on

2013-03-22 17:10:19

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

23

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Related topic

DistanceFromPointToALine

Defines

envelope

Generated on Fri Feb 9 22:12:27 2018 by

LaTeXML
example needing two Lagrange multipliers

example needing two Lagrange multipliers

Find the semi-axes of the

ellipse

of

intersection

, formed when the plane

z

=

x

+

y

intersects the

ellipsoid

x

2

4

+

y

2

5

+

z

2

25

=

1

.

Let

(

x

,

y

,

z

)

be any point of the ellipsoid.  The square (

http://planetmath.org/SquareOfNumber

)

x

2

+

y

2

+

z

2

of the

distance

of this point from the

midpoint

(

http://planetmath.org/Midpoint3

)

(

0

,

0

,

0

)

has under the constraints

{

g

:=

x

2

4

+

y

2

5

+

z

2

25

-

1

=

0

,

h

:=

x

+

y

-

z

=

0

(1)

the

minimum

and

maximum

values at the

end points

of the semi-axes of the ellipse.  Since we have two constraints, we must take equally many

Lagrange multipliers

,

λ

and

μ

.  A necessary condition of the

extremums

of

f

:=

x

2

+

y

2

+

z

2

is that in to (1), also the equations

{

∂

⁡

f

∂

⁡

x

+

λ

⁢

∂

⁡

g

∂

⁡

x

+

μ

⁢

∂

⁡

h

∂

⁡

x

=

2

⁢

x

+

1

2

⁢

x

⁢

λ

+

μ

=

0

,

∂

⁡

f

∂

⁡

y

+

λ

⁢

∂

⁡

g

∂

⁡

y

+

μ

⁢

∂

⁡

h

∂

⁡

y

=

2

⁢

y

+

2

5

⁢

y

⁢

λ

+

μ

=

0

,

∂

⁡

f

∂

⁡

z

+

λ

⁢

∂

⁡

g

∂

⁡

z

+

μ

⁢

∂

⁡

h

∂

⁡

z

=

2

⁢

z

+

2

25

⁢

z

⁢

λ

-

μ

=

0

,

(2)

are satisfied.  I.e., we have five equations (1), (2) and five unknowns

λ

,

μ

,

x

,

y

,

z

.

The equations (2) give

x

=

-

2

⁢

μ

λ

+

4

,

y

=

-

5

⁢

μ

2

⁢

λ

+

10

,

z

=

25

⁢

μ

2

⁢

λ

+

50

,

which

expressions

may be put into the equation

h

=

0

, and so on.  One obtains the values

λ

1

=

-

10

,

λ

2

=

-

75

17

,

μ

1

=

±

6

⁢

5

19

,

μ

2

=

±

140

17

⁢

646

with which the extremum points

(

x

,

y

,

z

)

can be evaluated.  The corresponding values of

f

are 10 and

75

17

, whence the major semi-axis is

10

and the minor semi-axis

5

⁢

255

17

.

Title

example needing two Lagrange multipliers

Canonical

name

ExampleNeedingTwoLagrangeMultipliers

Date of creation

2013-03-22 18:48:18

Last modified on

2013-03-22 18:48:18

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 51N20

Classification

msc 26B10

Synonym

using Lagrange multipliers to find semi-axes

Related topic

ExampleOfLagrangeMultipliers

Related topic

ExampleOfUsingLagrangeMultipliers

Generated on Fri Feb 9 22:12:33 2018 by

LaTeXML
example of isogonal trajectory

example of isogonal trajectory

Determine the curves which intersect the origin-centered circles at an angle of

45

∘

.

The

differential equation

of the circles

x

2

+

y

2

=

R

2

is

2

⁢

x

⁢

d

⁢

x

+

2

⁢

y

⁢

d

⁢

y

=

0

,  i.e.

x

y

+

d

⁢

y

d

⁢

x

=

0

.

Thus, by the model (2) of the parent entry (

http://planetmath.org/IsogonalTrajectory

), the differential equation of the

isogonal trajectory

reads

x

y

+

y

′

-

tan

⁡

π

4

1

+

y

′

⁢

tan

⁡

π

4

=

0

,

(1)

which can be rewritten as

y

′

=

y

-

x

y

+

x

=

y

x

-

1

y

x

+

1

.

Here, one may take

y

x

:=

t

as a new variable (see ODE types reductible to the variables separable case), when

y

=

x

⁢

t

,

y

′

=

d

⁢

y

d

⁢

x

=

t

+

x

⁢

d

⁢

t

d

⁢

x

,

and in the resulting equation

t

+

x

⁢

d

⁢

t

d

⁢

x

=

t

-

1

t

+

1

one can separate the variables (

http://planetmath.org/SeparationOfVariables

):

1

+

t

1

+

t

2

⁢

d

⁢

t

=

-

d

⁢

x

x

Multiplying here by 2 and integrating then give

2

⁢

arctan

⁡

t

+

ln

⁡

(

1

+

t

2

)

=

-

2

⁢

ln

⁡

x

+

ln

⁡

C

2

≡

-

ln

⁡

x

2

C

2

,

or equivalently

ln

⁡

x

2

+

x

2

⁢

t

2

C

2

=

-

2

⁢

arctan

⁡

t

.

This is

ln

⁡

x

2

+

y

2

C

=

-

arctan

⁡

y

x

,

i.e.

x

2

+

y

2

=

C

⁢

e

-

arctan

⁡

y

x

.

Expressing this in the

polar coordinates

r

,

φ

gives the family of the

integral curves

of the equation (1) in the form

r

=

C

⁢

e

-

φ

.

Consequently, the family of the isogonal trajectories consists of

logarithmic spirals

.

Title

example of isogonal trajectory

Canonical name

ExampleOfIsogonalTrajectory

Date of creation

2013-03-22 18:59:23

Last modified on

2013-03-22 18:59:23

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Example

Classification

msc 51N20

Classification

msc 34A26

Classification

msc 34A09

Synonym

isogonal trajectories of concentric circles

Related topic

IsogonalTrajectory

Generated on Fri Feb 9 22:12:35 2018 by

LaTeXML
example of using Lagrange multipliers

example of using Lagrange multipliers

One to determine the perpendicular distance of the

parallel planes

A

⁢

x

+

B

⁢

y

+

C

⁢

z

+

D

=

0

and

A

⁢

x

+

B

⁢

y

+

C

⁢

z

+

E

=

0

is to use the

Lagrange multiplier method

.  In this case we may to minimise the Euclidean distance of a point

(

x

,

y

,

z

)

of the former plane to a (fixed) point

(

x

0

,

y

0

,

z

0

)

of the latter plane.

Thus we have the equation

A

⁢

x

0

+

B

⁢

y

0

+

C

⁢

z

0

+

E

=

0

which we can subtract from the first plane equation, getting

g

:=

A

⁢

(

x

-

x

0

)

+

B

⁢

(

y

-

y

0

)

+

C

⁢

(

z

-

x

0

)

+

D

-

E

=

0

.

(1)

This is the (only) constraint equation for minimising the square (

http://planetmath.org/SquareOfANumber

)

f

:=

(

x

-

x

0

)

2

+

(

y

-

y

0

)

2

+

(

z

-

x

0

)

2

(2)

of the

distance

of the points.

The polynomial functions

f

and

g

satisfy the differentiability requirements.  Accordingly, we can find the minimising point

(

x

,

y

,

z

)

by considering the system of equations formed by (1) and

{

∂

⁡

f

∂

⁡

x

+

λ

⁢

∂

⁡

g

∂

⁡

x

≡

2

⁢

(

x

-

x

0

)

+

λ

⁢

A

=

0

,

∂

⁡

f

∂

⁡

y

+

λ

⁢

∂

⁡

g

∂

⁡

y

≡

2

⁢

(

y

-

y

0

)

+

λ

⁢

B

=

0

,

∂

⁡

f

∂

⁡

z

+

λ

⁢

∂

⁡

g

∂

⁡

z

≡

2

⁢

(

z

-

z

0

)

+

λ

⁢

C

=

0

.

(3)

We solve from (3) the

differences

x

-

x

0

=

-

A

⁢

λ

2

,

y

-

y

0

=

-

B

⁢

λ

2

,

z

-

z

0

=

-

C

⁢

λ

2

and set them into (1).  It then yields the value

λ

=

2

⁢

(

D

-

E

)

A

2

+

B

2

+

C

2

of the Lagrange multiplier, which we substitute into the preceding three equations obtaining

x

-

x

0

=

A

⁢

(

D

-

E

)

A

2

+

B

2

+

C

2

,

y

-

y

0

=

B

⁢

(

D

-

E

)

A

2

+

B

2

+

C

2

,

z

-

z

0

=

C

⁢

(

D

-

E

)

A

2

+

B

2

+

C

2

.

These values give the minimal distance when put into the

expression

of

f

:

d

=

(

D

-

E

)

2

⁢

(

A

2

+

B

2

+

C

2

)

(

A

2

+

B

2

+

C

2

)

2

.

Hence we have gotten the distance

d

=

|

D

-

E

|

A

2

+

B

2

+

C

2

.

Title

example of using Lagrange multipliers

Canonical

name

ExampleOfUsingLagrangeMultipliers

Date of creation

2013-03-22 18:48:12

Last modified on

2013-03-22 18:48:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 51N20

Classification

msc 26B10

Synonym

example of Lagrange multipliers

Related topic

ParallelismOfTwoPlanes

Related topic

ExampleNeedingTwoLagrangeMultipliers

Generated on Fri Feb 9 22:12:39 2018 by

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generatrices of hyperbolic paraboloid

generatrices of hyperbolic paraboloid

Since the equation

x

2

a

2

-

y

2

b

2

=

2

⁢

z

of

hyperbolic paraboloid

can be gotten by multiplying the equations in the pair

{

x

a

+

y

b

=

2

⁢

z

h

x

a

-

y

b

=

h

,

(1)

of

equations of planes

, the

intersection

line of these planes is

contained

in the

surface

of the hyperbolic paraboloid, for each value of the parameter

h

. So the surface has the family of

generatrices

(= rulings) given by all real values of

h

. The same concerns the other family

{

x

a

-

y

b

=

2

⁢

z

k

x

a

+

y

b

=

k

,

(2)

of lines. It is easily seen that any point of the hyperbolic paraboloid is passed through by exactly two generatrices, one from the family (1) and the other from family (2). Thus the surface is a

doubly ruled surface

.

The latter of the equations (1) tells that all generatrices the first family are

parallel

to the vertical plane

x

a

-

y

b

=

0

,

the so-called

director plane

of the hyperbolic paraboloid; this plane is also a plane of

symmetry

of the surface. According to the latter equation (2), one may say the corresponding things of the alternative director plane

x

a

+

y

b

=

0

.

Note 1.

We can solve from (1) and (2) the

coordinates

of a point

lying on

the surface:

x

=

a

⋅

h

+

k

2

,

y

=

b

⋅

h

-

k

2

,

z

=

h

⁢

k

2

This is a parametric presentation of the hyperbolic paraboloid.

Note 2.

One can check that two distinct lines of one family (1) resp. (2) are never in a same plane, but on the contrary, any line of one family intersects always all lines of the other family (in finity or in

infinity

).

References

1

Lauri Pimiä

:

Analyyttinen geometria

.  Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1958).

Title

generatrices of hyperbolic paraboloid

Canonical name

GeneratricesOfHyperbolicParaboloid

Date of creation

2013-03-22 17:31:56

Last modified on

2013-03-22 17:31:56

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Topic

Classification

msc 51N20

Classification

msc 51M04

Synonym

rulings of hyperbolic paraboloid

Related topic

QuadraticSurfaces

Related topic

GeneratricesOfOneSheetedHyperboloid

Defines

director plane

Generated on Fri Feb 9 22:12:41 2018 by

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harmonic division

harmonic division

•

If the point

X

is on the

line segment

A

⁢

B

and

X

⁢

A

:

X

⁢

B

=

p

:

q

,  then

X

divides

A

⁢

B

internally

in the ratio

p

:

q

.

•

If the point

Y

is on the extension of line segment

A

⁢

B

and

Y

⁢

A

:

Y

⁢

B

=

p

:

q

,  then

Y

divides

A

⁢

B

externally

in the ratio

p

:

q

.

•

If

p

:

q

is the same in both cases, then the points

X

and

Y

divide

A

⁢

B

harmonically

in the ratio

p

:

q

.

Theorem 1.

The

bisectors

of an angle of a

triangle

and its

linear pair

divide the opposite side of the triangle harmonically in the ratio of the

adjacent sides

.

Theorem 2.

If the points

X

and

Y

divide the line segment

A

⁢

B

harmonically in the ratio

p

:

q

, then the circle with

diameter

the segment

X

⁢

Y

(the so-called Apollonius’ circle) is the locus of such points whose

distances

from

A

and

B

have the ratio

p

:

q

.

The latter theorem may be proved by using

analytic geometry

.

Title

harmonic division

Canonical name

HarmonicDivision

Date of creation

2013-03-22 17:34:29

Last modified on

2013-03-22 17:34:29

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Definition

Classification

msc 51N20

Classification

msc 51M04

Related topic

BisectorsTheorem

Related topic

ApolloniusCircle

Defines

harmonically

Defines

divide harmonically

Generated on Fri Feb 9 22:12:54 2018 by

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interactive animation graphics

interactive animation graphics

1

Interactive animation graphics with LiveGraphics3D

For large and/or complex graphics objects

LiveGraphics3D

seems to offer some
useful solutions. A few examples follow: by clicking on the links one can see in real time, interactively
the

rotation

of fairly complex molecules. The conversion of graphics 3D objects

•

http://www.vis.uni-stuttgart.de/ kraus/LiveGraphics3D/examples/molecules.html

Two molecules: created with

M

⁢

a

⁢

t

⁢

h

⁢

e

⁢

m

⁢

a

⁢

t

⁢

i

⁢

c

⁢

a

T

⁢

M

, and then with

LiveGraphics3D

by Martin Kraus

•

http://www.vis.uni-stuttgart.de/ kraus/LiveGraphics3D/examples/molecule2.html

DNA Fragment Interactive Animation : As usual click and hold the mouse button down while moving the mouse around, and the DNA fragment molecule will follow the mouse position! If mouse button is released over the image the molecular fragment will jump up and down in a dance-like motion, or spin around!

•

http://www.wolfram.com/

Producing Graphics with Mathematica

•

http://www.wolfram.com/mathematica/new-in-8/parametric-probability-distributions/

Calculating Parametric Probability Distributions with Mathematica 8

•

Converting Graphics3D Objects: “in order to display any Graphics3D, ContourGraphics, DensityGraphics, or SurfaceGraphics object with LiveGraphics3D, it has to be converted into an appropriate InputForm. This is can be done with the function LiveForm defined in the LiveGraphics3D documentation package. The function also converts lists of Graphics3D objects to animations and replaces all SequenceForms into StringForms to avoid problems with the formatting of SequenceForm.”

Title

interactive animation graphics

Canonical name

InteractiveAnimationGraphics

Date of creation

2013-03-22 19:21:52

Last modified on

2013-03-22 19:21:52

Owner

bci1 (20947)

Last modified by

bci1 (20947)

Numerical id

12

Author

bci1 (20947)

Entry type

Definition

Classification

msc 51N20

Defines

interactive graphics

Defines

molecular geometry

Defines

parametric probability distribution representation

Defines

animation graphics

Defines

LiveGraphics3D

Generated on Fri Feb 9 22:13:05 2018 by

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