Point $P$ is in the middle of the interval that surrounded in curve

Question

Let $c$ be a close curve such that $c$ does not intersect itself, $c\in \mathbb{R}^2$ (in the plane), show that for all point $P$ that surrounded in $c$ there are two points $A,B$ on $c$ such that $P$ is in the middle of the interval of $A$ & $B$

As you can see English isn't my first language.

Answer 1

1. I'm going to assume that the curve is actually smooth, i.e., that I can write $c : [0, 1] \to R^2$ and that $c'(t)$ exists and is nonzero for every $t$.

2. I'm then going to scale up or down the whole problem so that the length of the curve is exactly 2\pi.

3. Now I'm going to reparameterize and change the domain to $[0, 2\pi]$ so that $\| c'(t) \| = 1$ for every $t$.

4. And I'm going to shift the parameterization so that $c(0)$ is a point of the curve with minimal $y$ coordinate, i.e., "it's at the bottom of the curve."

N.B.: my old grad-school colleague Glenn Davis has pointed out that the smoothness assumption is probably not necessary here; I made it mostly so that the remainder of the argument would be easy to express. My best guess is that "finite length" for the curve is enough to make everything here work, and it's possible that there's even a very similar proof for infinite-length curves, but I don't see all the details after 5 minutes of thought.

That's it for the setup. Now (shamelessly imitating Whitney's proof of the Whitney-Graustein theorem), I note that if $P$ is "inside" the curve described by $c$, then the curve $$ h: [0, 2\pi] \to S^1 : t \mapsto \frac{c(t) - P}{\| c(t) - P \|} $$ (which is essentially a map from the circle to the circle) must have degree $d = \pm 1$. (This "degree" is just the winding number of the curve $c$ around the point $P$; for a simple closed curve, this is either 0, for points "outside" the curve, or $1$, for points inside, or $-1$ (if the curve $c$ is parameterized clockwise instead of counterclockwise)).

I'm going to do a proof by contradiction, by assuming that $P$ is not the midpoint of any chord of the curve $c$.

First, I define, for $s, t \in [0, 2\pi]$, \begin{align} H(s, t) &= \frac{c(s) + c(t)}{2} \\ K(s, t) &= \frac{P - H(s, t)}{\| P - H(s, t) \|} \end{align}

$H$ represents the midpoint of the segment between two points of the curve, and $K$ is the unit vector pointing from $P$ to this midpoint. $K$ is well defined (i.e., the denominator is never $0$) by the contradiction hypotheses.

Now I want to look at $K(s, t)$ on the triangle where $s \le t$. That triangle has three sides: on the first, $s = 0$, and $t$ ranges from $0$ to $2 \pi$. On the second, $t = 2\pi$ and $s$ ranges from $0$ to $2\pi$. On the third, $s$ and $t$ together range from $2\pi$ down to $0$.

If we traverse the first side, the point $H(s, t)$ traces out a curve in the plane (basically, a scaled-down-by-half-around-the-point $c(0)$ version of $c$), and by hypothesis, this misses $P$, so $K$ traces out some path on $S^1$, with some winding number $s$.

When we traverse the second side, the point $H(s, t)$ traces out the SAME curve, with the same winding number for $K$.

And when we traverse the third side, the point $H(s, t)$ traverses the original curve $c$ ... but in reverse .. , so that $K$ has winding number $-d$. So the total winding number gotten by following around the boundary of the triangle is $-d + 2s$ for some $s$; since $d$ is $\pm 1$, this is an odd number.

Now consider shrinking that triangle in the domain of $H$, shrinking towards the point $(0, 0)$. Because everything in sight is continuous, the winding number for $K$ around these concentric smaller triangles must also be odd. But when the triangle becomes nearly a point, the curve traced out by $H(s, t)$ becomes a very small curve in the plane right near $c(0)$, and thus the winding number for $K$ on this tiny triangle becomes $0$.

Since the winding number is a continuous function which starts out odd and ends up at 0, we have a contradiction.

Thanks for asking this question -- it's fun to find another place where that Whitney proof works out well!

One final note from Glenn: the essential idea I've cribbed from Whitney's paper seems to appear in a paper by Hopf that Whitney cites, so maybe I'm over-crediting Whitney.

Answer 2

Let $P$ be inside the closed curve $c$ and take an arbitrary line through $P$. Let $A$ and $B$ be the intersections of that line with that curve.

If $PA=PB$ we are done. Without loss of generality let's assume $PA<PB$. Now rotate the line through $P$. As the line rotates smoothly, the points $A$ and $B$ also move smoothly around the curve. When the line has been rotated $180°$ the points $A$ and $B$ will have changed positions, so now $PA>PB$, so $PA$ went from being smaller than $PB$ to being greater than $PB$. At some time during the rotation, those two values must have become equal, and that is where $P$ was the midpoint of $\overline{AB}$.

Note that proof is not rigorous, but I do not see how to make this rigorous without using calculus. I do not know what level of math you are using, so I cannot tidy this up for you. Do be aware that I left out one possibility, namely that the line through $P$ may intersect curve $c$ more than twice. Again, I'll leave it up to you to close this gap, according to the mathematical resources now allowed to you.

Answer 3

This is really a reply to @Rory Daulton's answer, because I need a schematic picture to show what I'm saying.

Consider the blue point $P$ with the two green lines through it. We'd like to show that this is a midpoint of some chord of the curve. @Rory's suggestion is that we consider a path in angle space, $\theta(t)$ with the property that the endpoints of some chord through $P$ with angle $\theta(t)$ can be made to vary continuously as a function of $t$, Now in this picture, the red segment is a part of the $x$-axis, and the curve in the shaded blue area is the graph of $y = f(x)$, where $$ f(x) = \begin{cases} x^2 \sin(\frac{1}{x^2}) & x \ne 0 \\ 0 & x = 0 \end{cases}. $$ Note that $f$ is everywhere differentiable, and hence the overall blue curve can be made smooth.

Further note that for any near-horizontal chord through $P$ -- between the green lines, say -- the left end of the chord must lie in the red portion of the curve, or the two nearby blue pieces. If the right end is at a point where $f > 0$, then the left end must be in the left-most blue bit adjacent to the red; if the right end is at a point where $f < 0$, then the left end must be at the right-hand blue bit adjacent to the red. These two "bits" are separated by at least the half length of the red piece (as drawn). Now let's suppose that the blue shaded region is much farther from $P$ than is the red piece, so that the chord we're looking for is not horizontal. (I drew them close together to make it fit nicely). As we rotate the chord from the SW-NE green line to the NW-SE green line, the "right hand chord end" must at some time $t_0$ be at the "center" of that wiggly sine bit. But since in every interval containing that point, there are places where $f > 0$ and places where $f < 0$, we know that for some $t_1$ arbitrarily close to $t_0$, the left end is in the right "bit", and for some $t_2$ arbitrarily close to $t_0$, the left end is in the left "bit". Since those bits are distant from one another, the "left end" function cannot be a continuous function of $t$.

Hence the "rotating a line" argument really cannot be made to work.