# A question on Newton's “theorem about ovals”

This is a question about a result from Newton's Principia. It says, roughly, that the if you intersect lines $ax + by + c$ with a smooth, closed, convex curve, then the area of the curve that the line cuts off cannot be expressed as an algebraic function of $a, b$ and $c$.

My question is regarding Newton's proof. There are a couple versions out there, but I'll summarize the one on Wikipedia.

Fix a point $P$ inside the curve, and fix a line $L$ through $P$. Construct a spiral, $f$ as follows: draw another line $M$ through $P$, and let $\theta$ be the angle this makes with $L$. Thus $L$ and $M$ close off a sector of the curve, with some area $A(\theta)$. If we plot $(A(\theta), \theta)$ in polar coordinates, we get a spiral. Newton observes that this spiral intersects any line through $P$ infinitely many times. Thus the spiral cannot be algebraic, for if it were then there would be infinitely many solutions to a polynomial.

Fine. I agree with Newton's argument that the spiral cannot be algebraic, but how does this imply the original statement? More specifically, the proof is only successful because we allow the spiral to continue infinitely. But why can't we make a periodic function $B(\theta)$ such that $B$ expresses the area of the sector with angle $\theta$ for $\theta \in [0, 2\pi]$?

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Here is how I read the text, without too much background knowledge.

The area $A(\theta)$ has to be monotonically increasing, as it integrates area. After $n$ complete turns it will have covered $n$ times the complete area. The derivative of that function will be periodic in the angle, as the derivative corresponds to a function of the radius, and the radius is $2\pi$-periodic.

The same is true for other shapes like e.g. the square. But in those cases, the original curve will not be smooth, so the area will only be a piecewise algebraic function of the angle $\theta$. For sufficiently smooth curves, if the area cut off is algebraic at all, it has to be algebraic over the whole domain of the angle.

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How can a function be both periodic and monotone (and not be constant)? –  Kopper Oct 23 '12 at 15:18
@JayKopper, you are right, I was thinking about the derivative there. Fixed my formulation to actually express what I had in mind. Thanks for catching this. –  MvG Oct 23 '12 at 17:12
Ok, then I agree. But I still have my original question. The theorem states, as I understand it, that there is no polynomial $P$ with $P(f,r,s,t)=0$, where $f(r,s,t)$ is the area of the curve cut off by the line $rx+sy=t.$ The proof seems to provide a function $g(r,s,t)$ which has that property, but $g\neq f$ –  Kopper Oct 23 '12 at 22:38
@JayKopper, your notation is confusing: why is $f$ a parameter to $P$? As I understand things, Newton talks about two lines and the angle between them, and the translation to a single line is later interpretation, probably using $\theta=\pi$. If $f$ measures the same thing as $g$, wouldn't you agree that $f=g$? –  MvG Oct 24 '12 at 6:20
Isn't that precisely the definition of "$f$ is an algebraic function"? Further, I wouldn't claim $f$ and $g$ measure the same thing, merely that they agree for $\theta\in[2\pi ]$. For example, if we take the unit circle and a ray from the origin making angle $\theta$ with the poisitive $x-$axis, I would say that the area of the sector is $f(\theta) = \theta$ for $\theta \in [0, 2\pi)$ and is continued to the real line via $f(\theta + 2\pi) = f(\theta).$ –  Kopper Oct 24 '12 at 12:02

I have a partial answer, but am still stuck. I've seen the material presented several different ways. For example, Arnol'd says (pg. 132), as a remark,

If an oval is algebraically squarable, then the area of a sector cut from the oval by an angle whose vertex is in the oval is also an algebraic function of the two lines that constitute the angle...

He alludes to a proof which I give (ish) below

Proposition 1. Let $C$ be a smooth convex curve and $L$ be the line $ax + by = c.$ Let $f(a,b,c)$ be the function whose value is the area of $C$ cut off by the line $L$. Fix a point $P$ and a ray $M$ in $C$. Denote by $g(\theta)$ the area of the sector made by $M$ and the ray at angle $\theta$ from $M.$ Then if $f$ is an algebraic function, so is $g.$

This is proved quickly by drawing the line $L$ between the two points on $C$ which are intersected by $M$ and the ray at angle $\theta.$ Then $g(\theta)$ is the area of a triangle plus the area cut off by $L.$ The area of the triangle is obviously algebraic in $a,b,c,$ and by assumption $f$ is, so it follows that $g$ must be too.

This is, as I understand it, a different proposition from this next one:

Proposition 2. Same setup, but let $h(\theta)$ be the function that gives the area swept out by rotating the ray in Prop (1) through angle $\theta,$ i.e. $h:\mathbb{R}\rightarrow\mathbb{R}$ with $h(\theta) = g(\theta)$ for $\theta \in [0, 2 \pi),$ and $h(2\pi n + \theta ) = nk + h(\theta),$ where $k$ is the total area of the curve. Then $f$ algebraic implies $h$ algebraic in $a,b,c$ implies $h$ is algebraic in $\theta.$

Arnol'd's proof, similar to Newton's, shows that $h$ is not algebraic, and so $f$ cannot be. But I can't figure out the proof of Prop (2)

I think the idea is that it follows from Prop (1) since we can write $h(\theta) = nk + g(\theta),$ with $n=\lfloor \theta/2 \pi \rfloor$. It then suffices to show that $\lfloor \theta/2 \pi \rfloor$ is algebraic in $\theta.$ But this is clearly not true, so I am stuck.

Perhaps I need to write $\theta$ in terms of the lines it is the angle between?

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