aarjaneiro/planetmath huggingface dataset:
mountain pass theorem

mountain pass theorem

Let

X

a real

Banach space

and

F

∈

C

1

⁢

(

X

,

ℝ

)

. Consider

K

a

compact

metric space

, and

K

*

⊂

K

a closed nonempty subset of

K

. If

p

*

:

K

*

→

X

is a

continuous mapping

, set

𝒫

=

{

p

∈

C

(

K

,

X

)

;

p

=

p

*

on

K

*

}

.

Define

c

=

inf

p

∈

𝒫

⁡

max

t

∈

K

⁡

F

⁢

(

p

⁢

(

t

)

)

.

Assume that

c

>

max

t

∈

K

*

⁡

F

⁢

(

p

*

⁢

(

t

)

)

.

(1)

Then there exists a

sequence

(

x

n

)

in

X

such that

(i)

lim

n

→

∞

⁡

F

⁢

(

x

n

)

=

c

;

(ii)

lim

n

→

∞

⁡

∥

F

′

⁢

(

x

n

)

∥

=

0

.

The name of this

theorem

is a

consequence

of a simplified visualization for the objects from theorem. If we consider the set

K

*

=

{

A

,

B

}

, where

A

and

B

are two villages,

𝒫

is the set of all the routes from

A

to

B

, and

F

⁢

(

x

)

represents the altitude of point

x

; then the

assumption

(

1

) is

equivalent

to say that the villages

A

and

B

are

separated

with a mountains chain. So, the

conclusion

of the theorem tell us that exists a route between the villages with a

minimal

altitude. With other words exists a “mountain pass” .

Title

mountain pass theorem

Canonical name

MountainPassTheorem

Date of creation

2013-03-22 15:19:19

Last modified on

2013-03-22 15:19:19

Owner

ncrom (8997)

Last modified by

ncrom (8997)

Numerical id

8

Author

ncrom (8997)

Entry type

Theorem

Classification

msc 49J40

Generated on Fri Feb 9 21:48:30 2018 by

LaTeXML
equation of catenary via calculus of variations

equation of catenary via calculus of variations

Using the mechanical principle that the

centre of mass

itself as low as possible, determine the equation of the curve formed by a

l

when supported at its ends in the points

P

1

=

(

x

1

,

y

1

)

and

P

2

=

(

x

2

,

y

2

)

.

We have an

isoperimetric problem

to minimise

∫

P

1

P

2

y

⁢

𝑑

s

(1)

under the constraint

∫

P

1

P

2

𝑑

s

=

l

,

(2)

where both the path integrals are taken along some curve

c

.  Using a

Lagrange multiplier

λ

, the task changes to a free problem

∫

P

1

P

2

(

y

-

λ

)

⁢

𝑑

s

=

∫

x

1

x

2

(

y

-

λ

)

⁢

1

+

y

′

⁣

2

⁢

|

d

⁢

x

|

=

min

!

(3)

(cf. example of

calculus of variations

).

The Euler–Lagrange

differential equation

(

http://planetmath.org/EulerLagrangeDifferentialEquation

), the necessary condition for (3) to give an

extremal

c

, reduces to the

Beltrami identity

(

y

-

λ

)

⁢

1

+

y

′

⁣

2

-

y

′

⋅

(

y

-

λ

)

⋅

y

′

1

+

y

′

⁣

2

≡

y

-

λ

1

+

y

′

⁣

2

=

a

,

where

a

is a

constant of integration

.  After solving this equation for the

derivative

y

′

and

separation of variables

, we get

±

d

⁢

y

(

y

-

λ

)

2

-

a

2

=

d

⁢

x

a

which may become clearer by notating

y

-

λ

:=

u

;  then by integrating

±

d

⁢

u

u

2

-

a

2

=

d

⁢

x

a

we choose the new constant of integration

b

such that

x

=

b

when

u

=

a

:

±

∫

a

u

d

⁢

u

u

2

-

a

2

=

∫

b

x

d

⁢

x

a

We can write two

equivalent

(

http://planetmath.org/Equivalent3

) results

ln

⁡

u

+

u

2

-

a

2

a

=

+

x

-

b

a

,

ln

⁡

u

-

u

2

-

a

2

a

=

-

x

-

b

a

,

i.e.

u

+

u

2

-

a

2

a

=

e

+

x

-

b

a

,

u

-

u

2

-

a

2

a

=

e

-

x

-

b

a

.

Adding these allows to eliminate the square roots and to obtain

u

=

a

2

⁢

(

e

x

-

b

a

+

e

-

x

-

b

a

)

,

or

y

-

λ

=

a

⁢

cosh

⁡

x

-

b

a

.

(4)

This is the sought form of the equation of the chain curve.  The

constants

λ

,

a

,

b

can then be determined for putting the curve to pass through the given points

P

1

and

P

2

.

References

1

E. Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset IV. Johdatus variatiolaskuun

.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1946).

Title

equation of catenary via calculus of variations

Canonical name

EquationOfCatenaryViaCalculusOfVariations

Date of creation

2013-03-22 19:12:07

Last modified on

2013-03-22 19:12:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Derivation

Classification

msc 49K05

Classification

msc 49K22

Classification

msc 47A60

Related topic

Catenary

Related topic

CalculusOfVariations

Related topic

LeastSurfaceOfRevolution

Generated on Fri Feb 9 21:48:39 2018 by

LaTeXML
least surface of revolution

least surface of revolution

The points

P

1

=

(

x

1

,

y

1

)

and

P

2

=

(

x

2

,

y

2

)

have to be by an arc

c

such that when it rotates around the

x

-axis, the area of the surface of revolution (

http://planetmath.org/SurfaceOfRevolution

) formed by it is as small as possible.

The area in question, expressed by the

path integral

A

=

2

⁢

π

⁢

∫

P

1

P

2

y

⁢

𝑑

s

,

(1)

along

c

, is to be minimised; i.e. we must minimise

∫

P

1

P

2

y

⁢

𝑑

s

=

∫

x

1

x

2

1

+

y

′

⁣

2

⁢

|

d

⁢

x

|

.

(2)

Since the integrand in (2) does not explicitly depend on

x

, the Euler–Lagrange

differential equation

(

http://planetmath.org/EulerLagrangeDifferentialEquation

) of the problem, the necessary condition for (2) to give an

extremal

c

, reduces to the

Beltrami identity

y

⁢

1

+

y

′

⁣

2

-

y

′

⋅

y

⁢

y

′

1

+

y

′

⁣

2

≡

y

1

+

y

′

⁣

2

=

a

,

where

a

is a

constant of integration

.  After solving this equation for the

derivative

y

′

and

separation of variables

, we get

±

d

⁢

y

y

2

-

a

2

=

d

⁢

x

a

,

by integration of which
we choose the new constant of integration

b

such that

x

=

b

when

y

=

a

:

±

∫

a

y

d

⁢

y

y

2

-

a

2

=

∫

b

x

d

⁢

x

a

We can write two

equivalent

(

http://planetmath.org/Equivalent3

) results

ln

⁡

y

+

y

2

-

a

2

a

=

+

x

-

b

a

,

ln

⁡

y

-

-

2

a

2

a

=

-

x

-

b

a

,

i.e.

y

+

y

2

-

a

2

a

=

e

+

x

-

b

a

,

y

-

y

2

-

a

2

a

=

e

-

x

-

b

a

.

Adding these yields

y

=

a

2

⁢

(

e

x

-

b

a

+

e

-

x

-

b

a

)

=

a

⁢

cosh

⁡

x

-

b

a

.

(3)

From this we see that the extremals

c

of the problem are catenaries.  It means that the least surface of revolution in the question is a catenoid.

Title

least surface of revolution

Canonical name

LeastSurfaceOfRevolution

Date of creation

2013-03-22 19:12:11

Last modified on

2013-03-22 19:12:11

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Example

Classification

msc 49K05

Classification

msc 53A05

Classification

msc 26B15

Related topic

MinimalSurface

Related topic

EquationOfCatenaryViaCalculusOfVariations

Related topic

Catenary

Related topic

MinimalSurface2

Related topic

CalculusOfVariations

Related topic

SurfaceOfRevolution2

Generated on Fri Feb 9 21:48:42 2018 by

LaTeXML
Lagrange multiplier method

Lagrange multiplier method

The Lagrange multiplier method is used when one needs to find the extreme or

stationary points

of a

function

on a set which is a subset of the domain.

Method

Suppose that

f

⁢

(

𝐱

)

and

g

i

⁢

(

𝐱

)

,

i

=

1

,

…

,

m

(

𝐱

∈

ℝ

n

) are

differentiable functions

that map

ℝ

n

↦

ℝ

, and we want to solve

min

⁡

f

⁢

(

𝐱

)

,

max

⁡

f

⁢

(

𝐱

)

such that

g

i

⁢

(

𝐱

)

=

0

,

i

=

1

,

…

,

m

By a

calculus

theorem

, if the constaints are independent, the

gradient

of

f

,

∇

⁡

f

, must

satisfy

the following equation at the stationary points:

∇

⁡

f

=

∑

i

=

1

m

λ

i

⁢

∇

⁡

g

i

The constraints are said to be independent iff all the gradients of each constraint are

linearly independent

, that is:

{

∇

⁡

g

1

⁢

(

𝐱

)

,

…

,

∇

⁡

g

m

⁢

(

𝐱

)

}

is a set of linearly independent vectors on all points where the constraints are verified.

Note that this is

equivalent

to finding the stationary points of:

f

⁢

(

𝐱

)

-

∑

i

=

1

m

λ

i

⁢

(

g

i

⁢

(

𝐱

)

)

for

𝐱

in the domain and the

Lagrange multipliers

λ

i

without

restrictions

.

After finding those points, one applies the

g

i

constraints to get the actual stationary points subject to the constraints.

Title

Lagrange multiplier method

Canonical

name

LagrangeMultiplierMethod

Date of creation

2013-03-22 12:25:10

Last modified on

2013-03-22 12:25:10

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

10

Author

cvalente (11260)

Entry type

Definition

Classification

msc 49K30

Related topic

ExampleOfCalculusOfVariations

Related topic

IsoperimetricProblem

Defines

Lagrange multiplier

Generated on Fri Feb 9 21:48:46 2018 by

LaTeXML
interval halving

interval halving

Interval halving is an efficient method for solving equations. The requirements for using this method are that we have an equation

f

⁢

(

x

)

=

0

where

f

⁢

(

x

)

is a

continuous function

, and two values

x

1

and

x

2

such that

f

⁢

(

x

1

)

⁢

f

⁢

(

x

2

)

<

0

.
Since

f

⁢

(

x

1

)

and

f

⁢

(

x

2

)

have opposite signs, we know by the

intermediate value theorem

that there exists a solution

x

^

and that

x

1

≤

x

^

≤

x

2

, and with only

n

+

1

function evaluations we can find a shorter

interval

of length

ϵ

=

|

x

1

-

x

2

|

⁢

2

-

n

that contains

x

^

. If we take the solution

x

s

=

(

x

1

+

x

2

)

/

2

we get an error

|

x

s

-

x

^

|

<

ϵ

.

Algorithm

IntervalHalving

(

x

1

,

x

2

,

f

⁢

(

x

)

,

ϵ

)

Input

:

x

1

,

x

2

,

f

⁢

(

x

)

as above.

ϵ

is the length of the desired interval

Output

: A solution to the equation

f

⁢

(

x

)

=

0

that lies in an interval of length

<

ϵ

. Requires

f

⁢

(

x

1

)

⁢

f

⁢

(

x

2

)

<

0

s

⁢

t

⁢

a

⁢

r

⁢

t

←

m

⁢

i

⁢

n

⁢

(

x

1

,

x

2

)

e

⁢

n

⁢

d

←

m

⁢

a

⁢

x

⁢

(

x

1

,

x

2

)

m

⁢

i

⁢

d

⁢

d

⁢

l

⁢

e

←

(

s

⁢

t

⁢

a

⁢

r

⁢

t

+

e

⁢

n

⁢

d

)

/

2

while

e

⁢

n

⁢

d

-

s

⁢

t

⁢

a

⁢

r

⁢

t

<

ϵ

begin

if

f

⁢

(

m

⁢

i

⁢

d

⁢

d

⁢

l

⁢

e

)

⁢

f

⁢

(

s

⁢

t

⁢

a

⁢

r

⁢

t

)

<

0

then

e

⁢

n

⁢

d

←

m

⁢

i

⁢

d

⁢

d

⁢

l

⁢

e

else

s

⁢

t

⁢

a

⁢

r

⁢

t

←

m

⁢

i

⁢

d

⁢

d

⁢

l

⁢

e

m

⁢

i

⁢

d

⁢

d

⁢

l

⁢

e

←

(

s

⁢

t

⁢

a

⁢

r

⁢

t

+

e

⁢

n

⁢

d

)

/

2

end

x

s

←

(

s

⁢

t

⁢

a

⁢

r

⁢

t

+

e

⁢

n

⁢

d

)

/

2

// the solution

The algorithm works by taking an interval

[

x

1

,

x

2

]

and dividing it into two intervals of equal size. Look at a point in the middle,

x

m

. If the function changes sign in the interval

[

x

1

,

x

m

]

, that is

f

⁢

(

x

1

)

⁢

f

⁢

(

x

m

)

<

0

, then by the intermediate value theorem we have a solution in this interval. If not, then the solution is in the interval

[

x

m

,

x

2

]

. Repeat this until you get a sufficiently smal interval.

Interval halving is very useful in that it only requires the function to be

continuous

. More efficient methods like Newton’s method require that the function is

differentiable

and that we have to have a good starting point.
Interval halving has

linear convergence

while Newton’s method has

quadratic convergence

.

Title

interval halving

Canonical name

IntervalHalving

Date of creation

2013-03-22 14:20:02

Last modified on

2013-03-22 14:20:02

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

12

Author

mathcam (2727)

Entry type

Algorithm

Classification

msc 49M15

Synonym

bisection algorithm

Generated on Fri Feb 9 21:48:48 2018 by

LaTeXML
interval halving converges linearly

interval halving converges linearly

Theorem 1

.

The

interval halving

algorithm converges linearly.

Proof.

To see that interval halving (or bisection) converges linearly we use the alternative definition of

linear convergence

that says that

|

x

i

+

1

-

x

i

|

<

c

⁢

|

x

i

-

x

i

-

1

|

for some constant

1

>

c

>

0

.

In the case of interval halving,

|

x

i

+

1

-

x

i

|

is the length of the

interval

we should search for the solution in and has

x

i

+

2

as its

midpoint

. We have then that this interval has half the length of the previous interval which means,

𝑙𝑒𝑛𝑔𝑡ℎ

i

+

1

=

1

2

⁢

𝑙𝑒𝑛𝑔𝑡ℎ

i

. Thus

c

=

1

/

2

and we have exact linear convergence.
∎

Title

interval halving converges linearly

Canonical name

IntervalHalvingConvergesLinearly

Date of creation

2013-03-22 14:21:02

Last modified on

2013-03-22 14:21:02

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 49M15

Generated on Fri Feb 9 21:48:50 2018 by

LaTeXML
nth root by Newton’s method

nth root by Newton’s method

The Newton’s method is very suitable for computing approximate values of higher

n

th

roots (

http://planetmath.org/NthRoot

) of positive numbers (and odd roots of

negative numbers

!).

The general recurrence formula

x

k

+

1

=

x

k

-

f

⁢

(

x

k

)

f

′

⁢

(

x

k

)

of the method for determining the zero of a function

f

, applied to

f

⁢

(

x

)

:=

x

n

-

α

whose zero is

α

n

, reads

x

k

+

1

=

1

n

⁢

[

(

n

-

1

)

⁢

x

k

+

α

x

k

n

-

1

]

.

(1)

For a

radicand

α

, beginning from some initial value

x

0

and using (1) repeatedly with successive values of

k

, one obtains after a few steps a sufficiently accurate value of

α

n

if

x

0

was not very far from the searched root.

Especially for cube root

α

3

, the formula (1) is

x

k

+

1

=

1

3

⁢

[

2

⁢

x

k

+

α

x

k

2

]

.

(2)

For example, if one wants to compute

2

3

and uses

x

0

=

1

, already the fifth step gives

x

5

=

1.259921049894873

which decimals.

Title

nth root by Newton’s method

Canonical

name

NthRootByNewtonsMethod

Date of creation

2013-03-22 19:09:38

Last modified on

2013-03-22 19:09:38

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Example

Classification

msc 49M15

Classification

msc 65H05

Classification

msc 26A06

Synonym

cube root of 2

Related topic

NthRoot

Generated on Fri Feb 9 21:49:01 2018 by

LaTeXML
secant method

secant method

The

secant method

which is

similar

to the

Newton-Raphson method

is used to find the

extremum

value for a

function

over an

interval

for which the defined function has only one extremum. If there is more then one

minimum

or

maximum

, then convergence is not guaranteed.

The advantage over the Newton-Raphson method is that the secant method does not require the

second derivative

so only one function is used, the

derivative

. However, two initial guesses are needed. The

algorithm

to find the extremum is to

iterate

using the following expression

x

k

=

x

k

-

f

′

⁢

(

x

k

)

⁢

x

k

-

x

k

-

1

f

′

⁢

(

x

k

)

-

f

′

⁢

(

x

k

-

1

)

An analytical example is given below for the

simple function

f

⁢

(

x

)

=

x

2

+

1

taking the derivative yields

f

′

⁢

(

x

)

=

2

⁢

x

So for the initial guesses of

x

0

=

3

and

x

1

=

5

,

x

2

, the first

iteration

evaluates to

x

2

=

5

-

f

′

⁢

(

x

1

)

⁢

[

5

-

3

f

′

⁢

(

x

1

)

-

f

′

⁢

(

x

0

)

]

The derivatives at these two points are

f

′

⁢

(

x

1

)

=

10

f

′

⁢

(

x

0

)

=

6

giving us the value for the first iteration of

x

2

=

0

.

To check when we are done iterating, we need one more iteration for comparison, so

x

3

=

x

2

-

f

′

⁢

(

x

2

)

⁢

[

0

-

5

f

′

⁢

(

x

2

)

-

f

′

⁢

(

x

1

)

]

which is

x

3

=

0

Since

x

3

-

x

2

=

0

, we are done and our extremum is

0

.

Not every function is so easy to iterate analytically and we must resort to numerical means. The attached file,

\PMlinktofile

secantMethod1.msecantMethod1.m, shows how to iterate using matlab on the more complicated function

f

⁢

(

x

)

=

x

⁢

100

-

x

2

One still must be careful when using the secant method since the above function has a maximum and a minimum on the interval of [-10,10] and you will not get convergence if your initial guesses are -2 and 2. However, on the interval of [0,10], there is only one extremum, so choose guesses of 5 and 6. Then the matlab function

converges

to the extremum of

5

*

2

=

7.0711

which is a maximum as the figure below shows

Figure 1:

Example Function

0.1

References

[1] Tragesser, S. ”

Optimization

”, lecture notes, University of Colorado at Colorado Springs, Spring 2006.

Title

secant method

Canonical name

SecantMethod

Date of creation

2013-03-22 15:39:22

Last modified on

2013-03-22 15:39:22

Owner

bloftin (6104)

Last modified by

bloftin (6104)

Numerical id

10

Author

bloftin (6104)

Entry type

Algorithm

Classification

msc 49M15

Related topic

MethodsToFindExtremum

Generated on Fri Feb 9 21:49:03 2018 by

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Plateau’s Problem

Plateau’s Problem

The “’s Problem” is the problem of finding the surface with minimal area among all surfaces which have the same prescribed boundary.

This problem is named after the Belgian physicist Joseph (1801-1883) who experimented with soap films. As a matter of fact if you take a wire (which represents a

closed curve

in three-dimensional space) and dip it in a solution of soapy water, you obtain a soapy surface which has the wire as boundary.
It turns out that this surface has the minimal area among all surfaces with the same boundary, so the soap film is a solution to the ’s Problem.

Jesse Douglas (1897-1965) solved the problem by proving the existence of such

minimal surfaces

. The solution to the problem is achieved by finding an

harmonic

and

conformal

parameterization of the surface.

The extension of the problem to higher dimensions (i.e. for

k

-dimensional surfaces in

n

-dimensional space) turns out to be much more difficult to study.
Moreover while the solutions to the original problem are always

regular

it turns out that the solutions to the extended problem may have singularities if

n

≥

8

.
To solve the extended problem the theory of

perimeters

(De Giorgi) for
boundaries and the theory of

rectifiable
currents

(Federer and Fleming) has been developed.

Title

Plateau

’s Problem

Canonical name

PlateausProblem

Date of creation

2013-03-22 13:38:13

Last modified on

2013-03-22 13:38:13

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

12

Author

paolini (1187)

Entry type

Topic

Classification

msc 49Q05

Related topic

MinimalSurface2

Related topic

LeastSurfaceOfRevolution

Defines

minimal surface

Generated on Fri Feb 9 21:49:04 2018 by

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adjacent

adjacent

Let

θ

be an angle of a

triangle

. A side of the triangle is

adjacent

to

θ

if it is one of the sides (

http://planetmath.org/Angle

) of

θ

.

When a phrase such as “adjacent of an angle” is used, one must determine from context whether it refers to this

definition

of adjacent or the other definition of adjacent (

http://planetmath.org/Adjacent2

). Note that the latter is specifically for

right triangles

.

Title

adjacent

Canonical

name

Adjacent1

Date of creation

2013-03-22 17:10:44

Last modified on

2013-03-22 17:10:44

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 51-00

Synonym

adjacent side

Generated on Fri Feb 9 21:49:24 2018 by

LaTeXML
alternate proof of parallelogram law

alternate proof of parallelogram law

Proof of this is simple, given the

cosine law

:

c

2

=

a

2

+

b

2

-

2

⁢

a

⁢

b

⁢

cos

⁡

ϕ

where

a

,

b

, and

c

are the lengths of the sides of the

triangle

, and angle

ϕ

is the corner angle opposite the side of length

c

.

Let us define the largest

interior angles

as angle

θ

.
Applying this to the

parallelogram

, we find that

d

1

2

=

u

2

+

v

2

-

2

⁢

u

⁢

v

⁢

cos

⁡

θ

d

2

2

=

u

2

+

v

2

-

2

⁢

u

⁢

v

⁢

cos

⁡

(

π

-

θ

)

Knowing that

cos

⁡

(

π

-

θ

)

=

-

cos

⁡

θ

we can add the two expressions together, and find ourselves with

d

1

2

+

d

2

2

=

2

⁢

u

2

+

2

⁢

v

2

-

2

⁢

u

⁢

v

⁢

cos

⁡

θ

+

2

⁢

u

⁢

v

⁢

cos

⁡

θ

d

1

2

+

d

2

2

=

2

⁢

u

2

+

2

⁢

v

2

which is the theorem we set out to prove.

Title

alternate proof of parallelogram law

Canonical name

AlternateProofOfParallelogramLaw

Date of creation

2013-03-22 12:43:52

Last modified on

2013-03-22 12:43:52

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Related topic

ProofOfParallelogramLaw2

Generated on Fri Feb 9 21:49:25 2018 by

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alternative proof of necessity direction of equivalent conditions for triangles (hyperbolic and spherical)

alternative proof of necessity direction of equivalent conditions for triangles (hyperbolic and spherical)

The following is a proof that, in

hyperbolic geometry

and

spherical geometry

, an

equiangular triangle

△

⁢

A

⁢

B

⁢

C

is automatically

equilateral

(

http://planetmath.org/EquilateralTriangle

) (and therefore

regular

(

http://planetmath.org/RegularTriangle

)). It better the proof of sufficiency supplied in the entry

equivalent conditions for triangles

and is slightly shorter than the proof of necessity supplied in the same entry.

Proof.

Assume that

△

⁢

A

⁢

B

⁢

C

is

equiangular

.

Since

∠

⁢

A

≅

∠

⁢

B

≅

∠

⁢

C

, AAA yields that

△

⁢

A

⁢

B

⁢

C

≅

△

⁢

B

⁢

C

⁢

A

. By CPCTC,

A

⁢

B

¯

≅

A

⁢

C

¯

≅

B

⁢

C

¯

. Hence,

△

⁢

A

⁢

B

⁢

C

is equilateral.

∎

Title

alternative proof of necessity direction of equivalent conditions for triangles (hyperbolic and spherical)

Canonical name

AlternativeProofOfNecessityDirectionOfEquivalentConditionsForTriangleshyperbolicAndSpherical

Date of creation

2013-03-22 17:12:55

Last modified on

2013-03-22 17:12:55

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

5

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 51-00

Generated on Fri Feb 9 21:49:28 2018 by

LaTeXML
angle bisector

angle bisector

For every angle, there exists a line that divides the angle into two equal parts.
This line is called the

angle bisector

.

The interior

bisector

of an angle is the line or

line segment

that divides it into two equal angles on the same side as the angle.

The exterior bisector of an angle is the line or line segment that divides it into two equal angles on the opposite side as the angle.

For a

triangle

, the point where the angle bisectors of the three angles meet is called the

incenter

.

Title

angle bisector

Canonical

name

AngleBisector

Date of creation

2013-03-22 13:10:38

Last modified on

2013-03-22 13:10:38

Owner

giri (919)

Last modified by

giri (919)

Numerical id

4

Author

giri (919)

Entry type

Definition

Classification

msc 51-00

Synonym

interior angle bisector

Synonym

exterior angle bisector

Generated on Fri Feb 9 21:49:29 2018 by

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