Trisecting an Angle: Impossible or Possible?

[The legend]

I heard that trisection of any angle is impossible using just the ruler and compass. Is this true?
If yes then what is the proof for that??

Also ... trisecting 180⁰ is surely possible because we can definitely create 60⁰ ... but what about the rest?

 

[Dadface]

This question has been discussed here just recently so try a search. From what I gather lines can be trisected but there is not a general method for all angles.

 

[The legend]

ok tried search and got this link
http://en.wikipedia.org/wiki/Angle_trisection

if you have any other please post.

 

[CRGreathouse]

I heard that trisection of any angle is impossible using just the ruler and compass. Is this true?

It's impossible to trisect an arbitrary angle with a compass and straightedge.

But many angles can be trisected with just those tools. I think that the angles which can be so constructed are precisely those of the form
$$\frac{k\pi}{2^b\cdot1431655765}$$
for integers b and k -- that is,
$$\frac{36k^\circ}{2^b\cdot286331153}$$

This assumes that there are precisely 5 Fermat primes.

If a marked straightedge is allowed instead of a straightedge, then any angle can be trisected. Similarly, a linkage, straightedge, and compass can trisect any angle. Paper folding (oragami under the Huzita-Hatori axioms) can similarly trisect any angle.

 

[The legend]

This origami stuff is cool!
https://www.math.lsu.edu/~verrill/origami/trisect/

Though could you please give me an example of that latex part... since i didnt exactly understand it..

 

[CRGreathouse]

Though could you please give me an example of that latex part... since i didnt exactly understand it..

Pick two whole numbers.

 

[The legend]

ok...
25 and 19

 

[The legend]

please help..

 

[HallsofIvy]

Roughly the idea is this- Imagine that you have a general angle $\theta$ and can trisect it- that is, divide it into three angles each of measure $\theta/3$.
Now let $\phi= \theta/3$ so that the original angle is $\theta= 3\phi$

Pick a point on one side of the original angle, at distance 1 from the vertex, and drop a perpendicular to the opposite side. The distance from the foot of that perpendicular to the vertex is $cos(\theta)= cos(3\phi)$.

From the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) we have $cos(3\phi)= cos(2\phi+ \phi)= cos(2\phi)cos(\phi)- sin(2\phi)sin(\phi)$
$= cos(\phi+ \phi)cos(\phi)- sin(\phi+ \phi)sin(\phi)= (cos^2(\phi)- sin^2(\phi))cos(\phi)- (2sin(\phi)cos(\phi))(sin(\phi)$
$= cos^3(\phi)- sin^2(\phi)cos(\phi)- 2sin^2(\phi)cos(\phi)= cos^3(\phi)- 3sin^2(\phi)cos(\phi)$
Since $sin^2(\phi)= 1- cos^2(\phi)$ that is equal to
$cos^3(\phi)- 3(1- cos^2(\phi))cos(\phi)= 4 cos^3(\phi)- 3 cos(\phi)$

Now, as you take $\theta$ going from 0 to $\pi/2$, the foot of that perpendicular sweeps over all points of the base line. In particular, that length could be, say, 1/3. For that particular $\theta$, we have $4\cos^3(\phi)- 3cos(\phi)= \frac{1}{3}$ or $12 cos^3(\phi)- 9 cos(\phi)- 1= 0$.

Now, drop a perendicular to from the end of the trisector, the line that makes angle $\phi= \theta/3$ with the base. The distance from the vertex to the foot of that perpendicular is $X= cos(\phi)$.

That is, if it were possible to trisect any angle, then it would be possible to construct, with straightedge and compasses, the number X satisfying $12X^3- 9X- 1= 0$.

Well, what kind of number can be "constructed" in this sense? If we take a given line segment to have length "1", we can duplicate that as many times as we like so we can "construct" any integer. We can bisect, trisect, etc. any line segment so we can "construct" numbers of the form 1/n for any n. Duplicating that m times, we can construct line segments of length m/m where m and n are integers. That is, we can "construct" all rational numbers.

And we can do more- given a line segement of length 1, construct a perpendicular at one end and strike of the same lenght, 1, on that. Connect the two endpoints and we have constructed a length of $\sqrt{2}$, an irrational number.

So can we construct all numbers? Or how can we characterize those numbers we can construct? To answer that, we need to define "algebraic" numbers.

A number is said to be "algebraic" if and only if it satifies a polynomial equation with integer coefficients. More than that- a number is "algebraic of order n" if it satisfies a polynomial equation with integer coefficients of degree n but no such equation of lower degree. For example, the numbers that are "algebraic of order 1" are precisely the rational nubmers because x= m/n leads to nx- m= 0 and conversely. $x= \sqrt{2}$ is not rational but does satify $x^2- 2= 0$ and so is "algebraic of degree 2".

In Cartesian geometry, we have linear equations, ax+ by= c, whose graphs are straight lines, and quadratic equations, $x^2+ y^2= r^2$, whose graphs are circles. Since straight edges make lines and compasses make circles, all those numbers that can be "constructed" are constructed form combinations of linear and quadratic equations.

The upshot of all of this is "the numbers that can be 'constructed' with straightedge and compasses are precisely the numbers that are algebraic of order a power of 2"- that is, numbers that are algebraic of order $2^0= 1$, $2^2= 4$, $2^3= 8$, etc.

In the example I gave, $12x^3- 9x- 1= 0$ is of degree 3 and cannot be reduced so x is "algebraic of degree 3", not a power of 2, and so is not "constructible".

All of this is part of "Galois theory" and if you want to know more look that up. But be warned- it is very deep and usually taught as an advanced university course.

Evariste Galois himself, at the age of 18, wrote out "Galois theory" and more on a single night, before being killed in a dual the next day!

(You can also use this argument to show that the other two "impossible constructions" of antiquity are, indeed, impossible. To "square the circle" means to construct a square having the same area as given circle. If we take the radius of the circle to be 1, then its area is $\pi$ and so if we could construct a square of that area, we would have constructed a side length of $\sqrt{\pi}$ and $\pi$ and $\sqrt{\pi}$ are not "algebraic" of any order- they are "transcendental numbers".

The final problem was to "duplicate the cube". That is, given a cube, using three dimensional analogs of the "straight edge" and "compasses", so that we can construct planes and spheres, construct a new cube having exactly twice the volume of first one. If we take one edge of the given cube to be length "1", then the cube has volume 1 and a cube with twice that volume would have volume 2 so edge length $\sqrt[3]{2}$ which is algebraic of order 3, not a power of 2.

By the way, to "duplicate the square" is easy. Take two sides, intersecting at point p and use straightedge and compasses to extend them each on the other side of p a distance equal to the side length. Use those endpoint, and the two vertices of the original square adjacent to p, as vertices of your new square. It has area twice the area of the original square.

Of course, if we take the sides of the original square as length 1, it has area 1 so we are constructing a square of area 2 and so side lengths of $\sqrt{2}$. That number is, as I said before, algebraic of order 2 and so constructible. Roughly the idea is this- Imagine that you have a general angle $\theta$ and can trisect it- that is, divide it into three angles each of measure $\theta/3$.
Now let $\phi= \theta/3$ so that the original angle is $\theta= 3\phi$

Pick a point on one side of the original angle, at distance 1 from the vertex, and drop a perpendicular to the opposite side. The distance from the foot of that perpendicular to the vertex is $cos(\theta)= cos(3\phi)$.

From the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) we have $cos(3\phi)= cos(2\phi+ \phi)= cos(2\phi)cos(\phi)- sin(2\phi)sin(\phi)$
$= cos(\phi+ \phi)cos(\phi)- sin(\phi+ \phi)sin(\phi)= (cos^2(\phi)- sin^2(\phi))cos(\phi)- (2sin(\phi)cos(\phi))(sin(\phi)$
$= cos^3(\phi)- sin^2(\phi)cos(\phi)- 2sin^2(\phi)cos(\phi)= cos^3(\phi)- 3sin^2(\phi)cos(\phi)$
Since $sin^2(\phi)= 1- cos^2(\phi)$ that is equal to
$cos^3(\phi)- 3(1- cos^2(\phi))cos(\phi)= 4 cos^3(\phi)- 3 cos(\phi)$

Now, as you take $\theta$ going from 0 to $\pi/2$, the foot of that perpendicular sweeps over all points of the base line. In particular, that length could be, say, 1/3. For that particular $\theta$, we have $4\cos^3(\phi)- 3cos(\phi)= \frac{1}{3}$ or $12 cos^3(\phi)- 9 cos(\phi)- 1= 0$.

Now, drop a perendicular to from the end of the trisector, the line that makes angle $\phi= \theta/3$ with the base. The distance from the vertex to the foot of that perpendicular is $X= cos(\phi)$.

That is, if it were possible to trisect any angle, then it would be possible to construct, with straightedge and compasses, the number X satisfying $12X^3- 9X- 1= 0$.

Well, what kind of number can be "constructed" in this sense? If we take a given line segment to have length "1", we can duplicate that as many times as we like so we can "construct" any integer. We can bisect, trisect, etc. any line segment so we can "construct" numbers of the form 1/n for any n. Duplicating that m times, we can construct line segments of length m/m where m and n are integers. That is, we can "construct" all rational numbers.

And we can do more- given a line segement of length 1, construct a perpendicular at one end and strike of the same lenght, 1, on that. Connect the two endpoints and we have constructed a length of $\sqrt{2}$, an irrational number.

So can we construct all numbers? Or how can we characterize those numbers we can construct? To answer that, we need to define "algebraic" numbers.

A number is said to be "algebraic" if and only if it satifies a polynomial equation with integer coefficients. More than that- a number is "algebraic of order n" if it satisfies a polynomial equation with integer coefficients of degree n but no such equation of lower degree. For example, the numbers that are "algebraic of order 1" are precisely the rational nubmers because x= m/n leads to nx- m= 0 and conversely. $x= \sqrt{2}$ is not rational but does satify $x^2- 2= 0$ and so is "algebraic of degree 2".

In Cartesian geometry, we have linear equations, ax+ by= c, whose graphs are straight lines, and quadratic equations, $x^2+ y^2= r^2$, whose graphs are circles. Since straight edges make lines and compasses make circles, all those numbers that can be "constructed" are constructed form combinations of linear and quadratic equations.

The upshot of all of this is "the numbers that can be 'constructed' with straightedge and compasses are precisely the numbers that are algebraic of order a power of 2"- that is, numbers that are algebraic of order $2^0= 1$, $2^2= 4$, $2^3= 8$, etc.

In the example I gave, $12x^3- 9x- 1= 0$ is of degree 3 and cannot be reduced so x is "algebraic of degree 3", not a power of 2, and so is not "constructible".

All of this is part of "Galois theory" and if you want to know more look that up. But be warned- it is very deep and usually taught as an advanced university course.

Evariste Galois himself, at the age of 18, wrote out "Galois theory" and more on a single night, before being killed in a dual the next day!

(You can also use this argument to show that the other two "impossible constructions" of antiquity are, indeed, impossible. To "square the circle" means to construct a square having the same area as given circle. If we take the radius of the circle to be 1, then its area is $\pi$ and so if we could construct a square of that area, we would have constructed a side length of $\sqrt{\pi}$ and $\pi$ and $\sqrt{\pi}$ are not "algebraic" of any order- they are "transcendental numbers".

The final problem was to "duplicate the cube". That is, given a cube, using three dimensional analogs of the "straight edge" and "compasses", so that we can construct planes and spheres, construct a new cube having exactly twice the volume of first one. If we take one edge of the given cube to be length "1", then the cube has volume 1 and a cube with twice that volume would have volume 2 so edge length $\sqrt[3]{2}$ which is algebraic of order 3, not a power of 2.

By the way, to "duplicate the square" is easy. Take two sides, intersecting at point p and use straightedge and compasses to extend them each on the other side of p a distance equal to the side length. Use those endpoint, and the two vertices of the original square adjacent to p, as vertices of your new square. It has area twice the area of the original square.

Of course, if we take the sides of the original square as length 1, it has area 1 so we are constructing a square of area 2 and so side lengths of $\sqrt{2}$. That number is, as I said before, algebraic of order 2 and so constructible. Roughly the idea is this- Imagine that you have a general angle $\theta$ and can trisect it- that is, divide it into three angles each of measure $\theta/3$.
Now let $\phi= \theta/3$ so that the original angle is $\theta= 3\phi$

Pick a point on one side of the original angle, at distance 1 from the vertex, and drop a perpendicular to the opposite side. The distance from the foot of that perpendicular to the vertex is $cos(\theta)= cos(3\phi)$.

From the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) we have $cos(3\phi)= cos(2\phi+ \phi)= cos(2\phi)cos(\phi)- sin(2\phi)sin(\phi)$
$= cos(\phi+ \phi)cos(\phi)- sin(\phi+ \phi)sin(\phi)= (cos^2(\phi)- sin^2(\phi))cos(\phi)- (2sin(\phi)cos(\phi))(sin(\phi)$
$= cos^3(\phi)- sin^2(\phi)cos(\phi)- 2sin^2(\phi)cos(\phi)= cos^3(\phi)- 3sin^2(\phi)cos(\phi)$
Since $sin^2(\phi)= 1- cos^2(\phi)$ that is equal to
$cos^3(\phi)- 3(1- cos^2(\phi))cos(\phi)= 4 cos^3(\phi)- 3 cos(\phi)$

Now, as you take $\theta$ going from 0 to $\pi/2$, the foot of that perpendicular sweeps over all points of the base line. In particular, that length could be, say, 1/3. For that particular $\theta$, we have $4\cos^3(\phi)- 3cos(\phi)= \frac{1}{3}$ or $12 cos^3(\phi)- 9 cos(\phi)- 1= 0$.

Now, drop a perendicular to from the end of the trisector, the line that makes angle $\phi= \theta/3$ with the base. The distance from the vertex to the foot of that perpendicular is $X= cos(\phi)$.

That is, if it were possible to trisect any angle, then it would be possible to construct, with straightedge and compasses, the number X satisfying $12X^3- 9X- 1= 0$.

Well, what kind of number can be "constructed" in this sense? If we take a given line segment to have length "1", we can duplicate that as many times as we like so we can "construct" any integer. We can bisect, trisect, etc. any line segment so we can "construct" numbers of the form 1/n for any n. Duplicating that m times, we can construct line segments of length m/m where m and n are integers. That is, we can "construct" all rational numbers.

And we can do more- given a line segement of length 1, construct a perpendicular at one end and strike of the same lenght, 1, on that. Connect the two endpoints and we have constructed a length of $\sqrt{2}$, an irrational number.

So can we construct all numbers? Or how can we characterize those numbers we can construct? To answer that, we need to define "algebraic" numbers.

A number is said to be "algebraic" if and only if it satifies a polynomial equation with integer coefficients. More than that- a number is "algebraic of order n" if it satisfies a polynomial equation with integer coefficients of degree n but no such equation of lower degree. For example, the numbers that are "algebraic of order 1" are precisely the rational nubmers because x= m/n leads to nx- m= 0 and conversely. $x= \sqrt{2}$ is not rational but does satify $x^2- 2= 0$ and so is "algebraic of degree 2".

In Cartesian geometry, we have linear equations, ax+ by= c, whose graphs are straight lines, and quadratic equations, $x^2+ y^2= r^2$, whose graphs are circles. Since straight edges make lines and compasses make circles, all those numbers that can be "constructed" are constructed form combinations of linear and quadratic equations.

The upshot of all of this is "the numbers that can be 'constructed' with straightedge and compasses are precisely the numbers that are algebraic of order a power of 2"- that is, numbers that are algebraic of order $2^0= 1$, $2^2= 4$, $2^3= 8$, etc.

In the example I gave, $12x^3- 9x- 1= 0$ is of degree 3 and cannot be reduced so x is "algebraic of degree 3", not a power of 2, and so is not "constructible".

All of this is part of "Galois theory" and if you want to know more look that up. But be warned- it is very deep and usually taught as an advanced university course.

Evariste Galois himself, at the age of 18, wrote out "Galois theory" and more on a single night, before being killed in a dual the next day!

(You can also use this argument to show that the other two "impossible constructions" of antiquity are, indeed, impossible. To "square the circle" means to construct a square having the same area as given circle. If we take the radius of the circle to be 1, then its area is $\pi$ and so if we could construct a square of that area, we would have constructed a side length of $\sqrt{\pi}$ and $\pi$ and $\sqrt{\pi}$ are not "algebraic" of any order- they are "transcendental numbers".

The final problem was to "duplicate the cube". That is, given a cube, using three dimensional analogs of the "straight edge" and "compasses", so that we can construct planes and spheres, construct a new cube having exactly twice the volume of first one. If we take one edge of the given cube to be length "1", then the cube has volume 1 and a cube with twice that volume would have volume 2 so edge length $\sqrt[3]{2}$ which is algebraic of order 3, not a power of 2.

By the way, to "duplicate the square" is easy. Take two sides, intersecting at point p and use straightedge and compasses to extend them each on the other side of p a distance equal to the side length. Use those endpoint, and the two vertices of the original square adjacent to p, as vertices of your new square. It has area twice the area of the original square.

Of course, if we take the sides of the original square as length 1, it has area 1 so we are constructing a square of area 2 and so side lengths of $\sqrt{2}$. That number is, as I said before, algebraic of order 2 and so constructible.

 

[CRGreathouse]

ok...
25 and 19

So the angle
$$\frac{25\pi}{2^{19}\cdot1431655765}$$
that is,
$$\frac{900^\circ}{2^{19}\cdot286331153}=\frac{225^\circ}{37529996886016}\approx0.000000000005995^\circ$$
can be constructed. Do the same with other choices and you'll get other constructible angles.

 

[filter54321]

You can n-sect any arbitrary angle with a ruler and compass. You just need to get really lucky.

 

[CRGreathouse]

You can n-sect any arbitrary angle with a ruler and compass. You just need to get really lucky.

False.

 

[HallsofIvy]

You can n-sect any arbitrary angle with a ruler and compass. You just need to get really lucky.

Or really ignorant!

 

[The legend]

is the latex code up there not working HallsofIvy?

Anyway ,
Thanks a lot for the reply..
bit of it did fly over my head ... but I'll be reading it again after i learn a bit of new concepts used there.

And thanks to you too CRGreathouse...
though,i never imagined
0.000000000005995 was a constructible angle ..

 

[CRGreathouse]

And thanks to you too CRGreathouse...
though,i never imagined
0.000000000005995 was a constructible angle ..

So, can you tell me if 7 degrees is a constructible angle? 18 degrees? 4 degrees?

 

[robert Ihnot]

It is an interesting fact that the 7 sided figure, the heptagon, can not be constructed by the restrictions of the ancient Greeks. (Construction depends on powers of two and fermat primes: http://www.zefdamen.nl/CropCircles/Constructions/Constructions_en.htm)

However this sort of narrow restriction-as interesting as it is to Algebraists- did not hold back the Greeks since the 7-sided figure is found on many of their vases.

 

[The legend]

4 degrees isn't possible ... (since i know 20 isn't possible)
am trying out for the rest!

 

[The legend]

Another cool thing i found out...

This is perhaps the simplest trisection using a marked straightedge. It was discovered by Archimedes. Given the angle AOX, draw a circle of arbitrary radius centered at O. Extend one side of the angle through the opposite side of the circle at D (top).

Mark off interval BC on the straightedge. BC = OX = radius of the circle.

Slide the straightedge so that B lies on line DOX, C lies on the circle and the straightedge passes through A. Angle CBD is one-third of AOX. Note the element of trial and error inherent in positioning the straightedge.

http://www.uwgb.edu/dutchs/pseudosc/trisect.htm