Trisecting an angle

[Thunderbird4!]

I searched for a thread about this and find it hard to believe there isn't one. Oh well... I shall continue.

I've been told time and time again that it is impossible to trisect an angle using a compass and a straightedge, but I just can't accept this.
Let us take a circle (a 360 degree angle). We then bisect the circle (as most of us learned in high school, bisecting an angle can be done with a straightedge and a compass relatively easily). Using our compass/straightedge combination we can find the midpoint of the diameter we drew earlier and draw a line perpendicular to the diameter (bisecting a 180 degree angle). We then have four 90 degree arcs of the circle, or four equal sections of our original angle. If we continue the process we increase the number of equal sections by doubling. We have 1 (the circle), 2, 4, 8, 16, 32, 64...

This sequence can be equated to 2^x and if we want one third of the circle (trisected) this number must be a multiple of 3.
Therefore: 2^x=3y
Since I know someone out there will be pedantic about this, x is restricted to positive integers.

Is it safe of me to assume that at some point 2^x will be divisible by 3? Have I made any mistakes mathematically in this process?
If yes to the former and no to the later have I not just shown proof of concept that an angle can be trisected?

[imatrendytotebag]

Is it safe of me to assume that at some point 2x will be divisible by 3? Have I made any mistakes mathematically in this process?
If yes to the former and no to the later have I not just shown proof of concept that an angle can be trisected?

First of all, it has been conclusively proven that, while certain angles can be trisected (like a 90 degree angle), there is no algorithm that will trisect an arbitrary angle. Therefore, no matter how clever you hope to be, you will NEVER be able to come up with an angle trisection construction. It simply can't be done.

Secondly, you ask about 2^x being divisible by three. By the unique factorization theorem, any number that is a power of two is not equal to any number times 3. In particular, the only factors of 2^x are smaller powers of 2. So 2^x = 3y will NEVER be solvable in the integers.

Lastly, as for the actual proof that such a construction does not exist, I am at a loss. I do know it has something to do with looking at the coordinate plane and seeing which points can be constructed using only a straightedge and compass, and seeing that the points necessary to determine 1/3 of an arbitrary angle are simply not constructible.

For example, suppose we have the points (0,0) and (1,0). Now we can draw a line through those two points, and a circle centered at (0,0) and through (1,0), and we get an intersection at (-1,0). Using this method repeatedly it is straightforward to see how we can get every lattice point on the x-axis, and since we can construct perpendiculars it is quickly obvious that every lattice point on the plane is constructible (as in it is the intersection of 2 circles, 2 lines, or a circle and a line, that are constructible).

Then using the equations for lines and circles we see what types of real numbers can be constructed. At this point I am not familiar enough with galois theory to continue the proof.

[Thunderbird4!]

The whole unique factorization thing sent me off on a tangent and I think I may have that formula incorrect. I was assuming the number of sections was equal to 2 to the power of the number of lines drawn (consistent for 0 lines, 1 line and 2 lines). However, with 3 lines we don't have equal sections. Therefore, I think I have to amend the formula to 2x=3y. Where x is equal to a power of 2. Meaning it is 2^x+1 which changes nothing...

0 lines, 1 section (2⁰)
1 line, 2 sections (2¹)
2 lines, 4 sections (2²)
4 lines, 8 sections (2³)
8 lines, 16 sections (2⁴)
Continuing where the number of sections is double the number of lines which must be double the previous number of lines. As you say the number of sections will never be divisible by 3 into a whole number (since you can't have a fraction of a section).

All of the above reasoning correct (this has totally blown my mind... I would've sworn it could be done)?

[3.14159265...]

Say you want to trisect the 60 degree angle.

You already have that cos(pi/3)=1/2. (This is why you can trisect the 180 degree angle, because cos(pi/3) has an answer in radicals.)

Now consider cos(pi/9) (this is a third of the 60 degree angle).

Using cosine identities, expand Cos(3x).

You will get a third degree polynomial. Manipulate that to be able to use Eisenstein criterion to show thats its irreducible, and thus the 60 degree angle not trisectible.

THANKS Baldadonis

[antonfire]

The proof isn't too hard, actually, assuming a fact that's intuitively clear, but a bit hard to prove. I'll give a sketch here.

Denote by S the set of all constructable lengths. Then, if you start with two perpendicular lines for the axes and some length of 1, a point (x,y) is constructable iff |x| and |y| are both constructable lengths. An angle is constructable, then, if and only if its sin and cos are constructable. Now, how can you construct points? Only by constructing lines or circles and intersecting them. But the equation for a circle at (x₀,y₀) of radius r₀ is (x-x₀)²+(y-y₀)²=r₀². It's a quadratic in x and y, with coefficients in terms of x₀, y₀, r₀. The equation for another circle/line is also quadratic, so we the system of equations for one of these intersections is at most quadratic in x and y. So, the solution can be expressed in terms of the original parameters and square roots.

In short, the constructable lengths are exactly the things that are expressible using only the integers, +,*,/,- and square roots. (The above is only a sketch of a proof of this - you should work out the details.) However, consider c=cos(pi/9). We know that cos(3x)=4cos(x)³-3cos(x). Thus 1/2=cos(pi/3)=4c³-3c. This polynomial is irreducible.

Now, if you believe that the solution to this cubic equation can't be expressed in terms of square roots, then we are done. The angle pi/3 is constructable, but the angle pi/9 isn't. So you can't always trisect an angle. The way to prove that fact involves field extensions. If you care, the degree of the Q[a] over Q is a power of 2 when a is constructable (which follows easily from the above), and the degree of Q[c] over Q is 3. So c is not constructable.


Edit: I was beaten to it.

[Cosmologicon]

Now consider cos(pi/9) (this is a third of the 60 degree angle).

Using cosine identities, expand Cos(3x).

You will get a third degree polynomial. Manipulate that to be able to use Eisenstein criterion to show thats its irreducible, and thus the 60 degree angle not trisectible.

What? How does that show that it's not constructible? x² - 2 meets the Eisenstein criterion, but sqrt(2) is constructible.

[antonfire]

It's an order 3 polynomial. Minimal polynomials for constructable numbers have orders that are powers of 2.

[Cosmologicon]

Oh okay. In that case the only argument I have is with your assertion that that fact is intuitively clear.

[Indon]

All of the above reasoning correct (this has totally blown my mind... I would've sworn it could be done)?

I think I came to the same conclusion as you - that it is impossible to trisect an angle using an arbitrary number of clever bisecting operations; you can just never get 1/3'rd of an angle that way.

I'm inclined to agree that this is because no power of 2 is divisible by 3, because of the way powers of two factorize.

[MartianInvader]

0 lines, 1 section (2⁰)
1 line, 2 sections (2¹)
2 lines, 4 sections (2²)
4 lines, 8 sections (2³)
8 lines, 16 sections (2⁴)

I think what you need to consider is the portion to take by looking at some number of these sub-angles. That is, each little section is of size 1/2^x, and you can use any number of them - so you'd want something of the form
a/2^x = 1/3

Where both x and a are integers. This'll never happen (for example, no matter how high x is, a/2^x will have a finite decimal expansion, while 1/3 does not). What you CAN do is use this method to get arbitrarily close to 1/3 of an angle (that is, for any number I tell you, like "one billionth of a degree", you can use this method to find an angle that is within that distance from the true trisection). This is not, however, the same thing as actually constructing a trisected angle.

[orangeperson]

Instead of using those useless things with a spike on one end and a pencil on the other that you call a compass, use a directional compass. Most of these have degree measures on them. Using one you can find the value of the angle, divide it by three, put two dots down at 1/3 and 2/3 of the angle, and draw straight lines from the vertex to the marks you made.

[3.14159265...]

Gee Willikers, WHY DIDN"T I THINK OF THAT.

[adlaiff6]

Instead of using those useless things with a spike on one end and a pencil on the other that you call a compass, use a directional compass. Most of these have degree measures on them. Using one you can find the value of the angle, divide it by three, put two dots down at 1/3 and 2/3 of the angle, and draw straight lines from the vertex to the marks you made.

Yeah, because when we were all greeks, we could totally build a directional compass as large as we wanted and we could have used that to build all of our architecture.

antonfire pretty much nailed it.

[Macbi]

So what numbers can we divide angles by, clearly every power of 2, and noting that divides by three, but what about dividing into 5? Pentsecting?

[antonfire]

Good question. The answer has to do with Chebyshev polynomials, T_n(x). An angle a is constructable iff cos(a) is. cos(n*a)=T_n(cos(a)). So, you can always n-sect an angle only if T_n(x)-c=0 has constructable roots for every constructable c between -1 and 1. By handwaving, you can always find such a c with T_n(x)-c irreducible, and so if n is not a power of 2, if r is a root of T_n(b)-c, the degree of r over Q[c] is not a power of 2, and so the degree of r over Q is not a power of two, and so r is not constructable.

So, assuming you can always find a constructable c between -1 and 1 such that T_n(x)-c is irreducible, you can't do much beyond bisecting angles. That fact seems true (hell, I'm pretty sure it's true for rational c, even), but I'm too lazy to try to prove it right now.

For specific cases, it's easy to come up with examples using Eisenstein's criterion: T₅(x)=16x⁵-10x³+5x, so T₅(x)-5/7 is irreducible. So, in particular, the angle arccos(5/7) is constructable, but the angle arccos(5/7)/5 is not.

[Cosmologicon]

Okay, so you can't construct 1/3 or 1/5 of an arbitrary angle.... What's the smallest prime p greater than 2 such that you can p-sect an arbitrary angle? Actually what are the first couple? After that I'm sure it's in OEIS.

[antonfire]

Like I said, I suspect there are none.

Fine, fine, I'll prove it, sheesh. Let p be an odd prime.

cos(px)+i sin(px)=exp(i px)=(cos(x)+i sin(x))^p=cos(x)^p+p*(...)+(i sin(x))^p.

Since p is odd, the last term doesn't contribute to the real part, so cos(px)=cos(x)^p+p*(...), i.e. T_p(x)=x^p+p*(...). So, the leading coefficient is 1 (mod p), and the rest of the coefficients are 0 (mod p). Choose a prime q>p, pick a k such that -q*c+k is divisible by p but not by p^2, where c is the constant term (yeah, yeah, c is actually 0, but I don't need to show that). We can choose a k between 0 and p, inclusive. Then by Eisenstein's criterion with p as the prime q*T_p(x)-c is irreducible. Thus T_p(x)-c/q is irreducible, so any root of it has degree p. 0<c<=p<q, so 0<c/q<1.

So, arccos(c/q) is constructable, but arccos(c/q)/p is not, since it's a root of T_p(x)-c/q, and so has degree p, and p is not a power of 2.


So, you can't p-sect an angle for any odd prime p, so you can't n-sect an angle for any n that is not a power of two. Ugh.

[Cosmologicon]

Well done, and thanks! Good to know!

In case anyone wants to know, apparently the primes p for which you can p-sect a right angle are the Fermat primes. This means that the numbers n for which you can n-sect a right angle are numbers for which phi(n) is a power of 2.

[Unakau]

I remember sitting down during a one-hour study hall. I took the entire time trying to trisect an angle on sketchpad.

The next day, I came up with the idea of finding a value of 2^x where x is a positive integer and 2^x is divisible by three. No such luck, as I found this to be impossible.

My next idea is what enlightened me. I realize that the reason an angle could be bisected was because you could find the middle point of an arc. I then pondered as to whether or not you could divide an arc into three equal arcs. I tried this, but then I went on the internet and found that the problem itself was not only unsolved, but unsolvable altogether.

I was pretty naive to think that I made a breakthrough in the eight grade, but I learned a few cool things along the way. The whole 2^x being divisible by three thing were cool, and I somehow tried a bunch of stuff with hexagons, which helped me learn more about them. I learned how to construct a regular pentagon, which was cool. What I went after turned out to be a foolish goal, but I'm glad I put myself through that.

[themandotcom]

I think that the reason you can't trisect an angle is because you can't exactly solve a 5th degree polynomial by hand.

[antonfire]

You're in the right ballpark (it has to do with roots of polynomials), but no. See the above posts for the actual reason. (Or, rather, for the usual proof.) If you'd like anything specific explained, feel free to ask.

[ptveite]

In fact, I think that you can n-sect a 90 degree angle (or at least you can construct a regular n-gon) exactly when n=2ⁱp₁p₂...p_j where the p_k's are distinct Fermat primes 2^(2^n)+1

[btilly]

Just one addendum. You can trisect an arbitrary angle with a compass and a straightedge marked in 2 places. See, for instance, http://www.geom.uiuc.edu/docs/forum/angtri/ or http://www.math.niu.edu/~rusin/known-ma ... ck/trisect for constructions.

Of course the trick is that the marked straightedge allows constructions that can't be done with one that is unmarked.

If this subject interests you, you should read Geometric Constructions by George Edward Martin.

[J Spade]

So if it can't be done for an odd prime, then what about just any old odd number? 9 doesn't work...

[btilly]

So if it can't be done for an odd prime, then what about just any old odd number? 9 doesn't work...

If what can't be done?

I'm sure it is clear in your head, but without knowing which post you're responding to nobody else can make heads or tails of what you're asking. (I suspect that ptveite already answered your question, but I'm not sure since I don't know what your question is.)

[J Spade]

n-secting an arbitrary angle.

It can't be done with an odd prime, as proven, but what else won't work? For example, 9, 12, and any multiple of 3.

[btilly]

n-secting an arbitrary angle.

It can't be done with an odd prime, as proven, but what else won't work? For example, 9, 12, and any multiple of 3.

Since it is easy to add angles together with ruler and compass, if n divides m, then being able to m-sect an arbitrary angle implies that you can n-sect it. (Just m-sect it, then add m/n copies together.)

Therefore you can't n-sect an arbitrary angle for any n that is a multiple of an odd prime. That just leaves the powers of 2. And those you can do by repeated bisections.

[MathGeek]

Through all of my travels around the Internet, I have discovered that the smallest possible angle that can be constructed when angle measurement is restricted to the whole numbers and degrees is an angle of 3 degrees.

For example...

1) Construct a regular pentagon ABCDE.
2) Construct segment BD and DE.
3) Construct triangle DEF such DE = EF = FD.
4) Copy <BDF and bisect <BDF and let G lie such <BDG ~= <GDF
5) Copy <GDF and bisect <GDF and let H lie such <GDH ~= <HDF

I have no proof to assert that 3 degrees is the smallest angle that can be constructed w/ compass & straightedge & maintain a whole number degrees of measurement, yet I am sure somebody has done it.

There are trigonometric constants for it to verify the possibility.

[antonfire]

The above techniques + your favorite computer algebra system make it easy to check that neither 1 degree nor 2 degrees are constructable angles. If T_n(x) denotes the Chebyshev polynomial of degree n (that is, cos(nx)=T_n(cos(x))), then cos(1) is a root of T₆₀(x)-1/2, which has no irreducible factors of degree a power of two. cos(2) is a root of T₃₀(x)-1/2, which also has no irreducible factors of degree a power of two. So, neither cos(1) nor cos(2) are constructable.

In particular, this means that the constructable whole-degree angles are precisely multiples of three degrees.


Yes, this is inelegant, but I'm too lazy to try to come up with a better solution.

[evilbeanfiend]

n-secting an arbitrary angle.

It can't be done with an odd prime, as proven, but what else won't work? For example, 9, 12, and any multiple of 3.

Since it is easy to add angles together with ruler and compass, if n divides m, then being able to m-sect an arbitrary angle implies that you can n-sect it. (Just m-sect it, then add m/n copies together.)

Therefore you can't n-sect an arbitrary angle for any n that is a multiple of an odd prime. That just leaves the powers of 2. And those you can do by repeated bisections.

its pretty obvious that any odd number must have at least 1 prime factor that is odd. of course you can get arbitrarily close to an n-sect in the same way that a binary string can get arbitrarily close to representing an real number

[glup.up]

i found this from mathematics magazine vol.81 no.1 feb 2008

the title is "a brief history of impossibility"

i quote this from there..

suppose we wish to trisect <BOE, which we may assume the central angle of arc BE in a circle (with center O). ...Draw BC parallel to OE and then draw CA with the property that DA=OB

then <DOA = 1/3<BOE

[antonfire]

Ugh, more obvious demonstration, not involving degree 60 polynomials: If you can construct 1 degree or 2 degrees, you can construct 20 degrees, which we've previously shown is impossible.

*hangs head in shame*

Honestly, factoring T₆₀-1/2? What was I thinking?


Anyway, glup.up, that's not a complete construction. What is D? What is A?

[glup.up]

hope it help

[Micron]

hope it help

Not really, since you don't know >OBC or point D. Given point D you can prove that it defines the trisection of >BOE but that doesn't help you find it.

You cannot "draw CA" until you have point A. Without knowing D, A could be anywhere on the line OE between E and some point to the right such that OE = EA. Basically you're asking us to take every point on the arc BE, find a point on the line OE which is 1 radius distant from the current point, and then pick the one which allows us to draw a line through some unknown point D. It can't be done.

Try it; remove points A and D and the lines that depend on them, then reconstruct the diagram.

[antonfire]

Well, it can, it just uses a different set of laws for what you're allowed to do in a construction.

if you allow two ends of a compass to slide along the circle and line (at fixed distance OB) until they point at C, you're set. Yes, this involves picking something out of an infinite set, but so does taking the intersection between two lines.