aarjaneiro/planetmath huggingface dataset:
Fermat-Torricelli theorem

Fermat

-

Torricelli theorem

Theorem (Fermat

-

Torricelli).

Let all angles of a

triangle

A

⁢

B

⁢

C

be at most

120

∘

.  Then the inner point

F

of the triangle which makes the sum

A

⁢

F

+

B

⁢

F

+

C

⁢

F

as little as possible, is the point from which the

angle of view

of every side is

120

∘

.

Proof.

Let’s perform the

rotation

of

60

∘

about the point

A

.  When

P

is the image of the point

C

, the triangle

A

⁢

C

⁢

P

is

equilateral

and its angles are

60

∘

.  Let

F

be any inner point of the triangle

A

⁢

B

⁢

C

and

Q

its image in the rotation.  We infer that if the sides of the triangle

A

⁢

B

⁢

C

are all seen from

F

in the angle

120

∘

, then the points

B

,

F

,

Q

,

P

lie on the same line.

Generally, the triangles

A

⁢

P

⁢

Q

and

A

⁢

C

⁢

F

are

congruent

, whence

C

⁢

F

=

Q

⁢

P

.  From the

equilateral triangles

we obtain:

A

⁢

F

+

B

⁢

F

+

C

⁢

F

=

F

⁢

Q

+

B

⁢

F

+

Q

⁢

P

=

B

⁢

F

⁢

Q

⁢

P

Here, the

right hand side

is minimal when the points

B

,

F

,

Q

,

P

are

collinear

, in which case

∠

⁢

C

⁢

F

⁢

A

=

∠

⁢

P

⁢

Q

⁢

A

=

180

∘

-

∠

⁢

A

⁢

Q

⁢

F

=

120

∘

,

∠

⁢

A

⁢

F

⁢

B

=

180

∘

-

∠

⁢

Q

⁢

F

⁢

A

=

120

∘

,

∠

⁢

B

⁢

F

⁢

C

=

360

∘

-

240

∘

=

120

∘

.

Remark.

The point

F

is called the

Fermat point

of the triangle

A

⁢

B

⁢

C

.

References

1

Tero Harju

:

Geometria. Lyhyt kurssi

.  Matematiikan laitos. Turun yliopisto, Turku (2007).

Title

Fermat

-

Torricelli theorem

Canonical name

FermatTorricelliTheorem

Date of creation

2013-03-22 19:36:39

Last modified on

2013-03-22 19:36:39

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51F20

Related topic

CenterOfATriangle

Defines

Fermat point

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

LaTeXML
Heron’s principle

Heron’s principle

Theorem

.

In the

Euclidean plane

, let

l

be a line
and

A

and

B

two points not on

l

.  If

X

is a point of

l

such that the sum

A

⁢

X

+

X

⁢

B

is the least possible,
then the lines

A

⁢

X

and

B

⁢

X

form equal angles with the
line

l

.

This

Heron’s principle

, concerning the

reflection

of
light, is a special case of

Fermat’s principle

in optics.

Proof.

If

A

and

B

are on different sides of

l

, then

X

must be on the line

A

⁢

B

, and the assertion is trivial since the

vertical angles

are equal.  Thus, let the points

A

and

B

be on the same side of

l

.  Denote by

P

and

Q

the points of the line

l

where the normals of

l

set through

A

and

B

intersect

l

, respectively.  Let

C

be the intersection point of the lines

A

⁢

Q

and

B

⁢

P

.  Then,

X

is the point of

l

where the

normal line

of

l

set through

C

intersects

l

.

Justification:  From two pairs of

similar

right triangles

we get the proportion equations

A

⁢

P

:

C

⁢

X

=

P

⁢

Q

:

X

⁢

Q

,

B

⁢

Q

:

C

⁢

X

=

P

⁢

Q

:

P

⁢

X

,

which imply the equation

A

⁢

P

:

P

⁢

X

=

B

⁢

Q

:

X

⁢

Q

.

From this we can infer that also

Δ

⁢

A

⁢

X

⁢

P

∼

Δ

⁢

B

⁢

X

⁢

Q

.

Thus the corresponding angles

A

⁢

X

⁢

P

and

B

⁢

X

⁢

Q

are equal.

We still state that the route

A

⁢

X

⁢

B

is the shortest.  If

X

1

is another point of the line

l

, then

A

⁢

X

1

=

A

′

⁢

X

1

,  and thus we obtain

A

⁢

X

1

⁢

B

=

A

′

⁢

X

1

⁢

B

=

A

′

⁢

X

1

+

X

1

⁢

B

≧

A

′

⁢

B

=

A

′

⁢

X

⁢

B

=

A

⁢

X

⁢

B

.

References

1

Tero Harju

:

Geometria. Lyhyt kurssi

.
Matematiikan laitos. Turun yliopisto (University of Turku), Turku (2007).

Title

Heron’s principle

Canonical

name

HeronsPrinciple

Date of creation

2014-09-15 15:38:36

Last modified on

2014-09-15 15:38:36

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Related topic

Catacaustic

Related topic

PropertiesOfEllipse

Related topic

HeronianMeanIsBetweenGeometricAndArithmeticMean

Generated on Fri Feb 9 22:03:21 2018 by

LaTeXML
lengths of triangle medians

lengths of triangle medians

By the

Apollonius theorem

, in any

triangle

, the

m

a

,

m

b

,

m

c

of the medians (

http://planetmath.org/Median

) of opposing the the sides

a

,

b

,

c

, respectively,
are

m

a

=

1

2

⁢

2

⁢

b

2

+

2

⁢

c

2

-

a

2

,

m

b

=

1

2

⁢

2

⁢

c

2

+

2

⁢

a

2

-

b

2

,

m

c

=

1

2

⁢

2

⁢

a

2

+

2

⁢

b

2

-

c

2

.

Title

lengths of triangle medians

Canonical name

LengthsOfTriangleMedians

Date of creation

2013-03-22 18:26:47

Last modified on

2013-03-22 18:26:47

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Corollary

Classification

msc 51M04

Synonym

lengths of medians

Related topic

ProofOfApolloniusTheorem

Related topic

CommonPointOfTriangleMedians

Related topic

LengthsOfAngleBisectors

Generated on Fri Feb 9 22:03:24 2018 by

LaTeXML
median of trapezoid

median of trapezoid

The

segment

connecting the

midpoints

of the legs (

http://planetmath.org/

Trapezoid

) of a trapezoid, i.e. the median of the trapezoid, is

parallel

to the bases and its length equals the

arithmetic mean

of the legs.

Proof.

Let

A

⁢

B

and

C

⁢

D

be the bases of a trapezoid

A

⁢

B

⁢

C

⁢

D

and

E

the midpoint of the leg

A

⁢

D

and

F

the midpoint of the leg

B

⁢

C

.  Then the median

E

⁢

F

may be determined as vector as follows:

E

⁢

F

→

=

E

⁢

D

→

+

D

⁢

C

→

+

C

⁢

F

→

=

1

2

⁢

A

⁢

D

→

+

D

⁢

C

→

+

1

2

⁢

C

⁢

B

→

=

1

2

⁢

(

A

⁢

D

→

+

D

⁢

C

→

+

C

⁢

B

→

)

+

1

2

⁢

D

⁢

C

→

=

1

2

⁢

A

⁢

B

→

+

1

2

⁢

D

⁢

C

→

=

1

2

⁢

(

A

⁢

B

→

+

D

⁢

C

→

)

The last

expression

tells that

E

⁢

F

→

⁢

∥

A

⁢

B

→

+

D

⁢

C

→

∥

⁢

A

⁢

B

→

and

|

E

⁢

F

→

|

=

|

A

⁢

B

→

+

D

⁢

C

→

|

2

=

|

A

⁢

B

→

|

+

|

D

⁢

C

→

|

2

.  Q.E.D.

Title

median of trapezoid

Canonical

name

MedianOfTrapezoid

Date of creation

2013-03-22 17:46:44

Last modified on

2013-03-22 17:46:44

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51M25

Related topic

MutualPositionsOfVectors

Related topic

MidSegmentTheorem

Related topic

TriangleMidSegmentTheorem

Related topic

HarmonicMeanInTrapezoid

Generated on Fri Feb 9 22:03:26 2018 by

LaTeXML
Napoleon’s theorem

Napoleon’s theorem

Theorem

.

If

equilateral triangles

are erected externally on the three sides of any given

triangle

, then
their centres are the vertices of an equilateral triangle.

If we embed the statement in the complex plane, the proof
is a mere calculation. In the notation
of the figure, we can assume that

A

=

0

,

B

=

1

, and

C

is in the upper half plane. The hypotheses are

1

-

0

Z

-

0

=

C

-

1

X

-

1

=

0

-

C

Y

-

C

=

α

(1)

where

α

=

exp

⁡

π

⁢

i

/

3

, and the

conclusion

we want is

N

-

L

M

-

L

=

α

(2)

where

L

=

1

+

X

+

C

3

M

=

C

+

Y

+

0

3

N

=

0

+

1

+

Z

3

.

From (

1

) and the

relation

α

2

=

α

-

1

, we get

X

,

Y

,

Z

:

X

=

C

-

1

α

+

1

=

(

1

-

α

)

⁢

C

+

α

Y

=

-

C

α

+

C

=

α

⁢

C

Z

=

1

/

α

=

1

-

α

and so

3

⁢

(

M

-

L

)

=

Y

-

1

-

X

=

(

2

⁢

α

-

1

)

⁢

C

-

1

-

α

3

⁢

(

N

-

L

)

=

Z

-

X

-

C

=

(

α

-

2

)

⁢

C

+

1

-

2

⁢

α

=

(

2

⁢

α

-

2

-

α

)

⁢

C

-

α

+

1

-

α

=

(

2

⁢

α

2

-

α

)

⁢

C

-

α

-

α

2

=

3

⁢

(

M

-

L

)

⁢

α

proving (

2

).

Remarks:

The attribution to Napoléon Bonaparte (1769-1821) is
traditional, but dubious. For more on the story, see

http://www.mathpages.com/home/kmath270/kmath270.htm

MathPages.

Title

Napoleon’s

theorem

Canonical name

NapoleonsTheorem

Date of creation

2013-03-22 13:48:50

Last modified on

2013-03-22 13:48:50

Owner

drini (3)

Last modified by

drini (3)

Numerical id

7

Author

drini (3)

Entry type

Theorem

Classification

msc 51M04

Generated on Fri Feb 9 22:03:29 2018 by

LaTeXML
normal of plane

normal of plane

The

perpendicular

or

normal line

of a plane is a special case of the surface normal, but may be defined separately as follows:

A line

l

is a

normal of a plane

π

, if it

intersects

the plane and is perpendicular to all lines

passing through

the

intersection

point in the plane.  Then the plane

π

is a

normal plane

of the line

l

.  The normal plane passing through the

midpoint

(

http://planetmath.org/Midpoint3

) of a

line segment

is the

center normal plane

of the segment.

There is the

Theorem

.

If a line (

l

) a plane (

π

) and is perpendicular to two distinct lines (

m

and

n

) passing through the cutting point (

L

) in the plane, then the line is a normal of the plane.

Proof.

Let

a

be an arbitrary line passing through the point

L

in the plane

π

.  We need to show that

a

⟂

l

.  Set another line of the

plane cutting

the lines

m

,

n

and

a

at the points

M

,

N

and

A

, respectively.  Separate from

l

the equally segments

L

⁢

P

and

L

⁢

Q

.  Then

P

⁢

M

=

Q

⁢

M

and

P

⁢

N

=

Q

⁢

N

,

since any point of the

center normal

of a line segment (

P

⁢

Q

) is equidistant from the

end points

of the segment.  Consequently,

Δ

⁢

M

⁢

N

⁢

P

≅

Δ

⁢

M

⁢

N

⁢

Q

(

SSS

)

.

Thus the segments

P

⁢

A

and

Q

⁢

A

, being corresponding parts of two

congruent

triangles

, are equally long.  I.e., the point

A

is equidistant from the end points of the segment

P

⁢

Q

, and it must be on the

perpendicular bisector

(

http://planetmath.org/CenterNormal

) of

P

⁢

Q

.  Therefore

A

⁢

L

⊥

P

⁢

Q

, i.e.

a

⊥

l

.

Proposition

1.

All of a plane are

parallel

.  If a line is parallel to a normal of a plane, then it is a normal of the plane, too.

Proposition 2.

All normal planes of a line are parallel.  If a plane is parallel to a normal plane of a line, then also it is a normal plane of the line.

Title

normal of plane

Canonical name

NormalOfPlane

Date of creation

2013-04-19 15:03:42

Last modified on

2013-04-19 15:03:42

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Synonym

plane normal

Related topic

AngleBetweenLineAndPlane

Related topic

NormalLine

Related topic

NormalVector

Related topic

Congruence

Related topic

ParallelAndPerpendicularPlanes

Related topic

ParallelismOfTwoPlanes

Defines

normal plane

Defines

center normal plane

Generated on Fri Feb 9 22:03:39 2018 by

LaTeXML
opposing angles in a cyclic quadrilateral are supplementary

opposing angles in a cyclic quadrilateral are supplementary

Theorem 1

.

[Euclid, Book III, Prop. 22]

If a

quadrilateral

is

inscribed

in a circle, then opposite angles of the quadrilateral sum to

180

∘

.

Proof.

Let

A

⁢

B

⁢

C

⁢

D

be a quadrilateral inscribed in a circle

Note that

∠

⁢

B

⁢

A

⁢

D

subtends

arc

B

⁢

C

⁢

D

and

∠

⁢

B

⁢

C

⁢

D

subtends arc

B

⁢

A

⁢

D

. Now, since a

circumferential angle

is half the corresponding

central angle

, we see that

∠

⁢

B

⁢

A

⁢

D

+

∠

⁢

B

⁢

C

⁢

D

is one half of the sum of the two angles

B

⁢

O

⁢

D

at

O

. But the sum of these two angles is

360

∘

, so that

∠

⁢

B

⁢

A

⁢

D

+

∠

⁢

B

⁢

C

⁢

D

=

180

∘

Similarly, the sum of the other two opposing angles is also

180

∘

.
∎

Title

opposing angles in a

cyclic quadrilateral

are

supplementary

Canonical name

OpposingAnglesInACyclicQuadrilateralAreSupplementary

Date of creation

2013-03-22 17:13:31

Last modified on

2013-03-22 17:13:31

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

8

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 51M04

Generated on Fri Feb 9 22:03:42 2018 by

LaTeXML
parallel and perpendicular planes

parallel and perpendicular planes

Theorem 1.

If a plane (

π

) intersects two

parallel planes

(

ϱ

,

σ

), the intersection lines are

parallel

.

Proof.

The intersection lines cannot have common points, because

ϱ

and

σ

have no such ones.  Since the lines are in a same plane

π

, they are parallel.

Theorem 2.

If a plane (

π

) contains the normal (

http://planetmath.org/PlaneNormal

) (

n

) of another plane (

ϱ

), the planes are

perpendicular

(

http://planetmath.org/DihedralAngle

) to each other.

Proof.

Draw in the plane

ϱ

the line

l

cutting the intersection line perpendicularly and cutting also

n

.  Then

l

must be perpendicular to

n

and thus to the whole plane

π

(see the Theorem in the entry

normal of plane

).  Consequently, the

right angle

formed by the lines

n

and

l

is the

normal section

of the

dihedral angle

formed by the planes

π

and

ϱ

.  Therefore,

π

⊥

ϱ

.

Title

parallel and perpendicular planes

Canonical name

ParallelAndPerpendicularPlanes

Date of creation

2013-04-19 15:18:51

Last modified on

2013-04-19 15:18:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Related topic

PlaneNormal

Related topic

NormalOfPlane

Related topic

ParallelismOfTwoPlanes

Generated on Fri Feb 9 22:03:43 2018 by

LaTeXML
parallelism of line and plane

parallelism of line and plane

Parallelity of a line and a plane means that the

angle between line and plane

is 0, i.e. the line and the plane have either no or infinitely many common points.

Theorem

1.

If a line (

l

) is

parallel

to a line (

m

)

contained

in a plane (

π

), then it is parallel to the plane or is contained in the plane.

Proof.

So,

l

|

|

m

⊂

π

.  If

l

⊄

π

,  we can set a set along the

parallel lines

l

and

m

another plane

ϱ

.  The common points of

π

and

ϱ

are on the

intersection

line

m

of the planes.  If

l

would intersect the plane

π

, then it would intersect also the line

m

, contrary to the

assumption

.  Thus

l

|

|

π

.

Theorem 2.

If a plane is set along a line (

l

) which is parallel to another plane (

π

), then the intersection line (

m

) of the planes is parallel to the first-mentioned line.

Proof.

The lines

l

and

m

are in a same plane, and they cannot intersect each other since otherwise

l

would intersect the plane

π

which would contradict the assumption.  Accordingly,

m

|

|

l

.

Title

parallelism of line and plane

Canonical name

ParallelismOfLineAndPlane

Date of creation

2013-03-22 18:47:58

Last modified on

2013-03-22 18:47:58

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Related topic

ParallelismOfTwoPlanes

Generated on Fri Feb 9 22:03:45 2018 by

LaTeXML
parallelogram theorems

parallelogram theorems

Theorem 1.

The opposite sides of a

parallelogram

are

congruent

.

Proof.

In the parallelogram

A

⁢

B

⁢

C

⁢

D

, the line

B

⁢

D

as a

transversal

cuts the

parallel lines

A

⁢

D

and

B

⁢

C

, whence by the theorem of the parent entry (

http://planetmath.org/CorrespondingAnglesInTransversalCutting

) the alternate interior angles

α

and

β

are congruent. And since the line

B

⁢

D

also cuts the parallel lines

A

⁢

B

and

D

⁢

C

, the alternate interior angles

γ

and

δ

are congruent. Moreover, the

triangles

A

⁢

B

⁢

D

and

C

⁢

D

⁢

B

have a common side

B

⁢

D

. Thus, these triangles are congruent (ASA). Accordingly, the corresponding sides are congruent:

A

⁢

B

=

D

⁢

C

and

A

⁢

D

=

B

⁢

C

.
∎

Theorem 2.

If both pairs of opposite sides of a

quadrilateral

are congruent, the quadrilateral is a parallelogram.

Theorem 3.

If one pair of opposite sides of a quadrilateral are both

parallel

and congruent, the quadrilateral is a parallelogram.

Theorem 4.

The

diagonals

of a parallelogram bisect each other.

Theorem 5.

If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.

All of the above theorems hold in

Euclidean geometry

, but not in

hyperbolic geometry

. These theorems do not make sense in

spherical geometry

because there are no parallelograms!

Title

parallelogram theorems

Canonical name

ParallelogramTheorems

Date of creation

2013-03-22 17:15:37

Last modified on

2013-03-22 17:15:37

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Theorem

Classification

msc 51M04

Classification

msc 51-01

Synonym

properties of parallelograms

Related topic

Parallelogram

Related topic

TriangleMidSegmentTheorem

Generated on Fri Feb 9 22:03:47 2018 by

LaTeXML
pivot theorem

pivot theorem

If

A

⁢

B

⁢

C

is a

triangle

,

D

,

E

,

F

points on the sides

B

⁢

C

,

C

⁢

A

,

A

⁢

B

respectively, then the

circumcircles

of the triangles

A

⁢

E

⁢

F

,

B

⁢

F

⁢

D

and

C

⁢

D

⁢

E

have a common point.

Title

pivot theorem

Canonical name

PivotTheorem

Date of creation

2013-03-22 12:10:13

Last modified on

2013-03-22 12:10:13

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Theorem

Classification

msc 51M04

Generated on Fri Feb 9 22:03:49 2018 by

LaTeXML
Plücker’s conoid

Plücker’s conoid

Plücker’s conoid

is a

ruled surface

that results from taking a

straight line

connected

to an axis, rotating it about that axis and moving it

straight

up and down the axis to give the desired number of folds. Being an example of a

right conoid

, Plücker’s conoid is sometimes called a

conical wedge

, or a

conocuneus of Wallis

or even a

cylindroid

.

The Cartesian equation for a conoid with two folds is

z

=

x

2

-

y

2

x

2

+

y

2

. This can be generalized to any desired number

n

of folds as

x

⁢

(

r

,

θ

)

=

r

⁢

cos

⁡

θ

,

y

⁢

(

r

,

θ

)

=

r

⁢

sin

⁡

θ

and

z

⁢

(

r

,

θ

)

=

c

⁢

sin

⁡

(

n

⁢

θ

)

. Plücker’s conoid has applications in mechanical drafting.

References

1

J. Plücker, “On a new

geometry

of space”,

Philosophical Transactions of the Royal Society of London

155

(1965): 725 - 791

2

S. P. Radzevich, “A Possibility of Application of Pliicker’s Conoid for Mathematical Modeling of Contact of Two Smooth Regular Surfaces in the First Order of Tangency”,

Mathematical and Computer Modelling

42

(2005): 999 - 1022

Title

Plücker’s conoid

Canonical name

PluckersConoid

Date of creation

2013-03-22 16:44:02

Last modified on

2013-03-22 16:44:02

Owner

Mravinci (12996)

Last modified by

Mravinci (12996)

Numerical id

4

Author

Mravinci (12996)

Entry type

Definition

Classification

msc 51M04

Classification

msc 51M20

Classification

msc 14J25

Synonym

Plucker’s conoid

Synonym

Plücker conoid

Synonym

Plucker conoid

Synonym

conical wedge

Synonym

conocuneus of Wallis

Synonym

Wallis conocuneus

Generated on Fri Feb 9 22:03:51 2018 by

LaTeXML
pons asinorum

pons asinorum

Pons asinorum

is Latin for “bridge of asses”. During medieval times, this name was given to the fifth

proposition

in the first book of Euclid’s

The

Elements

. In the original Greek, this proposition reads:

T

⁢

ω

~

⁢

ν

ι

⁢

σ

⁢

o

⁢

σ

⁢

κ

⁢

ε

⁢

λ

⁢

ω

~

⁢

ν

τ

⁢

ρ

⁢

ι

⁢

γ

⁢

ω

´

⁢

ν

⁢

ω

⁢

ν

α

⁢

ι

π

⁢

ρ

⁢

o

`

⁢

ς

τ

⁢

η

~

β

⁢

α

´

⁢

σ

⁢

ε

⁢

ι

γ

⁢

ω

⁢

ν

⁢

ι

´

⁢

α

⁢

ι

ι

⁢

σ

⁢

α

⁢

ι

α

⁢

λ

⁢

λ

⁢

η

´

⁢

λ

⁢

α

⁢

ι

⁢

ς

ε

⁢

ι

⁢

σ

⁢

ι

´

⁢

ν

,

κ

⁢

α

⁢

ι

`

π

⁢

ρ

⁢

o

⁢

σ

⁢

ε

⁢

κ

⁢

β

⁢

λ

⁢

η

⁢

θ

⁢

ε

⁢

ι

⁢

σ

⁢

ω

~

⁢

ν

τ

⁢

ω

~

⁢

ν

ι

⁢

σ

⁢

ω

⁢

ν

ε

⁢

υ

⁢

θ

⁢

ε

⁢

ι

⁢

ω

~

⁢

ν

α

⁢

ι

υ

⁢

π

⁢

o

`

τ

⁢

η

`

⁢

ν

β

⁢

α

´

⁢

σ

⁢

ι

⁢

ν

γ

⁢

ω

⁢

ν

⁢

ι

´

⁢

α

⁢

ι

ι

⁢

σ

⁢

α

⁢

ι

α

⁢

λ

⁢

λ

⁢

η

´

⁢

λ

⁢

α

⁢

ι

⁢

ς

ε

⁢

σ

⁢

o

⁢

ν

⁢

τ

⁢

α

⁢

ι

.

A

translation

of this proposition is:

In

isosceles triangles

, the angles at the base equal one another, and, if the equal

straight lines

are produced further, then the angles under the base equal one another.

There are a couple of reasons why this proposition was named pons asinorum:

•

Euclid’s diagram for this proposition looks like a bridge.

•

This is the first nontrivial proposition in

The Elements

and thus tests a student’s ability to understand more advanced

concepts

in

Euclidean geometry

. Therefore, this proposition serves as a bridge from from the

trivial

portion of Euclidean geometry to the nontrivial portion, and the people who cannot cross this bridge are considered to be unintelligent.

For more details, please see

http://planetmath.org/?op=getmsg&id=15847

a post written by rspuzio and

http://planetmath.org/?op=getmsg&id=15849

a post written by Wkbj79.

References

1

Mourmouras, Dimitrios.

The Elements: The original Greek text

. URL:

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

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

2

Wikipedia.

Pons asinorum

. URL:

http://en.wikipedia.org/wiki/Pons_Asinorum

http://en.wikipedia.org/wiki/Pons_Asinorum

Title

pons asinorum

Canonical name

PonsAsinorum

Date of creation

2013-03-22 17:17:31

Last modified on

2013-03-22 17:17:31

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Topic

Classification

msc 51M04

Classification

msc 51-03

Classification

msc 51-00

Classification

msc 01A20

Classification

msc 01A35

Related topic

AnglesOfAnIsoscelesTriangle

Generated on Fri Feb 9 22:03:54 2018 by

LaTeXML
proof of cosines law

proof of cosines law

Let

a

,

b

,

c

be the sides of a

triangle

and

α

,

β

,

γ

its angles, respectively.  By the

projection formula

, one may write the equalities

{

a

=

b

⁢

cos

⁡

γ

+

c

⁢

cos

⁡

β

b

=

c

⁢

cos

⁡

α

+

a

⁢

cos

⁡

γ

c

=

a

⁢

cos

⁡

β

+

b

⁢

cos

⁡

α

.

Multiplying the equalities by

a

,

-

b

and

-

c

, respectively, they read

{

a

2

=

a

⁢

b

⁢

cos

⁡

γ

+

c

⁢

a

⁢

cos

⁡

β

-

b

2

=

-

b

⁢

c

⁢

cos

⁡

α

-

a

⁢

b

⁢

cos

⁡

γ

-

c

2

=

-

c

⁢

a

⁢

cos

⁡

β

-

b

⁢

c

⁢

cos

⁡

α

.

Addition of these yields the sum equation

a

2

-

b

2

-

c

2

=

-

2

⁢

b

⁢

c

⁢

cos

⁡

α

,

i.e.

a

2

=

b

2

+

c

2

-

2

⁢

b

⁢

c

⁢

cos

⁡

α

,

which is the

cosines law

.

Title

proof of cosines law

Canonical name

ProofOfCosinesLaw

Date of creation

2013-03-22 18:27:13

Last modified on

2013-03-22 18:27:13

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Proof

Classification

msc 51M04

Related topic

DerivationOfCosinesLaw

Generated on Fri Feb 9 22:03:56 2018 by

LaTeXML