trisection of angle

trisection of angle

Given an angle of measure (

http://

planetmath

.org/AngleMeasure

)

α

such that

0

<

α

≤

π

2

, one can

construct

an angle of measure

α

3

using a

compass

and a ruler (

http://planetmath.org/MarkedRuler

) with one mark on it as follows:

1.

Construct a circle

c

with the vertex (

http://planetmath.org/Vertex5

)

O

of the angle as its center. Label the

intersections

of this circle with the rays of the angle as

A

and

B

. Mark the length

O

⁢

B

on the ruler.

2.

Draw the ray

A

⁢

O

→

.

3.

Use the

marked ruler

to determine

C

∈

c

and

D

∈

A

⁢

O

→

such that

C

⁢

D

=

O

⁢

B

and

B

,

C

, and

D

are

collinear

. Draw the

line segment

B

⁢

D

¯

. Then the

angle measure

of

∠

⁢

C

⁢

D

⁢

O

is

α

3

. (The line segment

O

⁢

C

¯

is drawn in red. Having this line segment drawn is useful for reference purposes for the justification of the construction.)

Let

m

denote the measure of an angle. Then this construction is justified by the following:

•

Since

∠

⁢

A

⁢

O

⁢

B

is an

exterior angle

of

△

⁢

B

⁢

O

⁢

D

, we have that

m

⁢

(

∠

⁢

A

⁢

O

⁢

B

)

=

m

⁢

(

∠

⁢

O

⁢

B

⁢

D

)

+

m

⁢

(

∠

⁢

O

⁢

D

⁢

B

)

;

•

Since

O

⁢

C

=

O

⁢

B

=

C

⁢

D

, we have that

△

⁢

B

⁢

O

⁢

C

and

△

⁢

O

⁢

C

⁢

D

are

isosceles triangles

;

•

Since the angles of an isosceles triangle are

congruent

,

m

⁢

(

∠

⁢

O

⁢

B

⁢

C

)

=

m

⁢

(

∠

⁢

O

⁢

C

⁢

B

)

and

m

⁢

(

∠

⁢

C

⁢

O

⁢

D

)

=

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

;

•

Since

∠

⁢

O

⁢

C

⁢

B

is an exterior angle of

△

⁢

O

⁢

C

⁢

D

, we have that

m

⁢

(

∠

⁢

O

⁢

C

⁢

B

)

=

m

⁢

(

∠

⁢

C

⁢

O

⁢

D

)

+

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

;

•

Note that

∠

⁢

O

⁢

B

⁢

C

=

∠

⁢

O

⁢

B

⁢

D

and

∠

⁢

O

⁢

D

⁢

B

=

∠

⁢

C

⁢

D

⁢

O

;

•

Thus,

α

=

m

⁢

(

∠

⁢

A

⁢

O

⁢

B

)

=

m

⁢

(

∠

⁢

O

⁢

B

⁢

D

)

+

m

⁢

(

∠

⁢

O

⁢

D

⁢

B

)

=

m

⁢

(

∠

⁢

O

⁢

B

⁢

C

)

+

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

=

m

⁢

(

∠

⁢

O

⁢

C

⁢

B

)

+

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

=

m

⁢

(

∠

⁢

C

⁢

O

⁢

D

)

+

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

+

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

=

3

⁢

m

⁢

(

∠

⁢

C

⁢

D

⁢

O

)

.

Note that, since angles of measure

π

6

,

π

3

, and

π

2

are constructible using compass and straightedge, this procedure can be extended to trisect any angle of measure

β

such that

0

<

β

≤

2

⁢

π

:

•

If

0

<

β

≤

π

2

, then use the construction given above.

•

If

π

2

<

β

≤

π

, then trisect an angle of measure

β

-

π

2

and add on an angle of measure

π

6

to the result.

•

If

π

<

β

≤

3

⁢

π

2

, then trisect an angle of measure

β

-

π

and add on an angle of measure

π

3

to the result.

•

If

3

⁢

π

2

<

β

≤

2

⁢

π

, then trisect an angle of measure

β

-

3

⁢

π

2

and add on an angle of measure

π

2

to the result.

This construction is attributed to Archimedes.

References

1

Rotman, Joseph J.

A First Course in Abstract Algebra

. Upper Saddle River, NJ: Prentice-Hall, 1996.

Title

trisection of angle

Canonical name

TrisectionOfAngle

Date of creation

2013-03-22 17:16:35

Last modified on

2013-03-22 17:16:35

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Algorithm

Classification

msc 01A20

Classification

msc 51M15

Related topic

VariantsOnCompassAndStraightedgeConstructions

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