Continuing my line of posts explaining complex proofs by skipping the hard part, this post gives a proof that certain angles cannot be trisected. Terry Tau recently had a post on the same subject, which contained a very long proof using only basic geometry and a very short proof using Galois theory. The proof I use here is between the two he presents in both difficulty and length. It's a remarkable proof in that there's very little geometry per se, most of the proof would usually be classified as algebra.

As a crash course on what we mean to say an angle can be trisected: We start by assuming you have a line of length one, a ruler and a compass. A number $n$ is "constructible" if, starting with these things, you can construct a line of length $n$. It's simple to see that if $a$ and $b$ are constructible, then $a+b$ is as well (just stick the two lines together). This means we can repeatedly add 1 to itself to generate all the natural numbers.

It's also possible to construct $ab$ and $a/b$, although this piece is harder to show. To start things off on the trisection proof, let's place a limit on what numbers can be constructed:

**Lemma 1:** All constructible numbers are the solution to either a linear or quadratic equation with constructible coefficients

To construct some points we can either:

- Draw two lines and find their intersection
- Draw a line and a circle, and find their intersection
- Draw two circles, and find their intersection

The intersection will always be a solution to an equation of degree at most two. For example, if we have the lines $y=m_1 x + b_1$ and $y = m_2 x + b_2$ their intersection has x coordinate $m_1 x + b_1= m_2 x + b_2$ or $(m_1 - m_2) x + (b_1 - b_2)=0$, so $x$ is the solution to a linear equation with both coefficients constructible.

In cases 2 and 3, the x and y coordinates will be solutions to quadratics.

**Lemma 2**: The only constructible points are solutions to equations of degree $2^n$ for some integer $n$

For example, we could find the point which satisfies $x^2-2=0$ which is $x=\sqrt{2}$. Now that we can construct $\sqrt{2}$, so we can use it in our equation; for example we can find $x=\sqrt[4]{2}$ by solving $x^2-\sqrt{2}=0$.

We can repeat this process to find the eigth root, sixteenth root etc. but we can only find roots that are powers of two. More generally, we can only construct equations of degree two; their coefficients might themselves be solutions to quadratics so we have an equation of degree 4 over our original set, but we can never create an equation of degree 3 or 5 or 379.

**Lemma 3**: If an angle $\theta$ is constructible, then $\cos\theta$ is constructible.

$\cos\theta$ represents the adjacent side of a right triangle with hypotenuse 1 and adjacent angle $\theta$; since we can draw a line of length 1 we can use that as a hypotenuse, and if we could draw the angle $\theta$ we can construct the entire triangle.

**Proposition**: $\cos 20^\circ$ is not constructible

Once we've proven this, we'll have proven that $20^\circ$ is not constructible, and so an angle of $60^\circ$ cannot be trisected.

I'm sure you all remember the triple angle formula $\cos 3\theta = 4 \cos ^3\theta-3\cos\theta$, plugging in $\cos 60=1/2$ and rearranging we get that $4\cos ^3\theta - 3\cos\theta - 1/2 = 0$; to make it more obvious I'll substitute $x$ for $\cos\theta$: $4x^3 - 3x - 1/2 = 0$. So $x$ is the solution to this cubic equation, but Lemma 1 told us that constructible points are only solutions to equations of degree $2^n$. So $x$ is not constructible. QED.

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