Dividing of an angle

**grottvald** · August 22nd 2010, 12:34 PM

How can I show that a given angle can not be divided into six equal parts using a compass and ruler?
I can't see why this couln't be done? I have som trouble understaning the problem with the problem so to say. Could someone spread som light over this matter?

**Soroban** · August 22nd 2010, 12:47 PM

Hello, grottvald!

How can I show that a given angle cannot be divided into six equal parts
using a compass and ruler?

If your "ruler" is an unmarked straightedge, it is not possible.

If it were possible, you could pair the six angles into three equal angles
. . and you have trisected an angle . . . an impossible contruction.

**Plato** · August 22nd 2010, 12:49 PM

Trisecting an angle is impossible.
Therefore, dividing an angle into six equivalent parts is impossible.

**HallsofIvy** · August 23rd 2010, 03:17 AM

The proof that a triangle cannot be trisected, that Plato links to, is very deep.

First, a number is said to be "algebraic of order n" if and only if it satisifies a polynomial equation with integer coefficients, of order n, but no such equation of lower order. For example, any rational number can be written $x= \frac{m}{n}$ where m and n are integers so we can write nx-m= 0, a polynomial of order 1 with integer coefficients. Conversely, if ax+ b= 0, with a and b integers, then $x= -\frac{b}{a}$ so x is irrational: the numbers that are "algebraic of order 1" are precisely the rational numbers.

$x= \sqrt{2}$ is NOT a rational number, so is not algebraic of order 1, but does satisfy $x^2= 2$ and so is algebraic of order 2.

There are also numbers, such as e or [itex]\pi[/itex] that are not algebraic of any order- they are called "transcendental numbers".

Now, think in terms of "constructing" numbers. That is, given a line segment taken to have length "1", what other lengths can we construct using only compasses and straightedge? Since we can duplicate a length as many times as we wish, we can construct any integer. Since we can divide a length into many equal lengths, we can construct any number of the form $\frac{1}{n}$ . And since we can take as many copies of those as we wish, we can construct any number of the form $\frac{m}{n}$ - we can construct any rational number, any number that is algebraic of order 1.

But, given a line segment of length 1, we can construct a perpendicular at one end, strike off an equal length on the perpendicular and connect the two endpoints to make a right triangle- we will have constructed a length $\sqrt{2}$ . More generally, it is possible to construct any length that is algebraic of order 2.

Using a lot of very deep mathematics, it is possible to show, roughly speaking, using the fact that the straight edge gives lines corresponding to linear equations and the compasses circles corresponding to quadratic equations, that a number is "constructible" in this sense if and only if it is "algebraic of order a power of 2".

Now, suppose it were possible to tri-sect any angle. Call one of the constructed angle " $\theta$ " so that the original angle is $3\theta$ . We can take one side of the smaller angle to be 1, drop a perpendicular to the opposite side and so construct a length $cos(\theta)$ .

Doing the same for the large triangle we would have a length $cos(3\theta)$ . But $cos(3\theta)= cos(2\theta+ \theta)= cos(2\theta)cos(\theta)- sin(2\theta)sin(\theta)= (cos^2(\theta)- sin^2(\theta))(cos(\theta)- (2sin(\theta)cos(\theta))sin(\theta)$ $= cos^3(\theta)- sin^2(\theta)cos(\theta)- 2sin^2(\theta)cos(\theta)$ $= cos^3(\theta)- 3sin^2(\theta)cos(\theta)= cos^3(\theta)- 3(1- cos^2(\theta)cos(\theta)$ $= 4cos^3(\theta)- 3cos(\theta)= 0$ .

That is, if it were possible, given any angle [itex]3\theta[/itex], to trisect the angle, constructing the angle [itex]\theta[/itex], then it would be possible to construct $cos(\theta)$ from $cos(3\theta)$ - and that satisfies a cubic equation. Since 3 is not a power of two, it is not possible to construct a length that is algebraic of order 3.

(For some $\theta$ it might be that the cubic equation is reducible to a lower degree equation so that $\theta$ is algebaic of order 2 or 1- for example it certainly is possible to trisect a 90 degree angle simply because it is possible to construct a 30 degree angle,- but there will exist some $/theta$ for which this is not true- it is not possible to trisect every angle.

As Soroban and Plato say, because it is not possible to trisect an angle, it is not possible to divide one into six equal parts- If it were, just doubling the original angle and then dividing that into six equal parts would give a trisection of the original angle.

By the way, this same analysis can be applied to the other "classic constructions" of antiquity:

Duplicating a cube: Given a cube, using three-dimensonal analogues of the compass and straight edge (able to construct a plane through any three points and a sphere given the center and radius), construct a cube having exactly twice the volume of the original sphere. Take the radius of the original sphere to be 1. It has volume $\frac{4}{3}\pi$ . If it were possible to construct a sphere of twice that volume, $\frac{8}{3}\pi$ , then we would have constructed a radius of length $\sqrt[3]{2}$ and that number is algebraic of power 3, not a multiple of 2.

Squaring the circle: Given a circle, construct, with straightedge and compasses, a square having the same area.
If we take the radius of the circle to be 1, then its area is $\pi$ . A square having the same area would have to have side length $\sqrt{\pi}$ . $\pi$ itself is transcendental- it is not algebraic of any order so neither is $\sqrt{\pi}$ .

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