Regular curve which normal lines pass through a fixed point

Question

I have the following question:

Assume that $α$ is a regular curve in $R^2$ and all the normal lines of the curve pass though the origin. Prove that $α$ is contained in a circle around the origin. (Recall the normal line at $α(t)$ is the line through $α(t)$ pointing in the direction of the normal vector $N(t)$.)

My attempt:

I know that if $α(t)$ is contained in a circle, then the curvature is constant. So I have to prove that the curvature at $α(t)$ is constant. I also know that the equations of the normal lines through the origin that passes through arbitrary points on the curve $α(t)$ denoted as $(t, α(t))$ have the equations:

$$α(t) = \frac{t}{α'(t)}$$

I'm confused on how to proceed from here.

Answer 1

More than a year has passed. I hope my solution still helps.

I suppose you know the Frenet frame and the Frenet-Serret formulas. Let me denote the unit tangent vector and unit normal vector at $\alpha(t)$ by $\mathbf{T}(t)$ and $\mathbf{N}(t)$. Finally, denote $\mathbf{B}(t)=\mathbf{T}(t)\times\mathbf{N}(t)$.

My strategy is to prove that

1. $||\alpha||$ is constant, so that (the trace of) $\alpha$ is contained in some sphere centered around $\mathbf{0}=(0,0,0)$.
2. $\mathbf{B}(t)$ is constant, so that the curve is a plane curve, i.e., it lies on some plane.

For 1, we differentiate $||\alpha|| ^2$ with respect to $t$: \begin{align*} \frac{d||\alpha(t)||^2}{dt}=\frac{d}{dt}\alpha(t)\cdot\alpha(t)=2\alpha(t)\cdot\alpha'(t) \end{align*} Since the normal line at $\alpha(t)$ passes through $\alpha(t)$ and $\mathbf{0}$, thus the vector $\alpha(t)-\mathbf{0}=\alpha(t)$ is parallel to the normal vector $\mathbf{N}(t)$, and so it is orthogonal to the tangent vector $\mathbf{T}(t)=\alpha'(t)$. Therefore $\alpha(t)\cdot\alpha'(t)=0$. Hence, $||\alpha||^2$ is constant, and so is $||\alpha||$.

For 2, we differentiate $\mathbf{B}(t)$: \begin{align*} \frac{d\mathbf{B}(t)}{dt}&=\mathbf{T}'(t)\times\mathbf{N}(t)+\mathbf{T}(t)\times\mathbf{N}'(t) \\ &=\alpha''(t)\times\frac{\alpha''(t)}{|\alpha''(t)|}+\alpha'(t)\times(\lambda\alpha'(t)) \\ &=0+\lambda(\alpha'(t)\times\alpha'(t)) \\ &=0 \end{align*} Here I write $\mathbf{N}(t)=\lambda\alpha(t)$ because $\alpha(t)$ is parallel to $\mathbf{N}(t)$ as aforementioned. This proves 2.

Since the intersection of a sphere and a plane is precisely a circle, we conclude that (the trace of) $\alpha$ is contained in a circle. Moreover, since the normal lines are contained in the plane, thus the centre of the circle is the origin. $\qquad\square$

Answer 2

We have $\alpha'(t) \perp N(t) \parallel \alpha(t)$ for all $t$, so $\alpha' \cdot \alpha = 0$,

therefore

$$\frac{d\|\alpha\|}{dt} = \frac{\alpha\cdot\alpha'}{\|\alpha\|} = 0,$$

so $\|\alpha\|$ is constant. $~~~~\square$

Answer 3

If all normal lines pass through a fixed point, then for all $s \in I$, there is a $\lambda = \lambda(s)$ such that

$$\alpha(s) + \lambda(s)n(s) = p,$$ where $p$ is our supposed fixed point. So, deriving we are left with:

$$\alpha'(s) + \lambda'(s)n(s) + \lambda(s)(-k(s)t(s) - \tau(s)b(s)) = 0$$ $$t + \lambda'n - \lambda k t - \lambda \tau b = 0 \implies (1-\lambda k)t + \ldots$$

But $\{t,n,b\}$ form an orthogonal basis, and so it is true that $(1 - \lambda k) = 0, \lambda' = 0$ and $\lambda \tau = 0$.

So $\lambda' = 0 \implies \lambda = \text{constant}$. Also, $k = \frac{1}{\lambda}$, $\tau = 0$.

See the curvature is constant and equal to $\frac{1}{\lambda}$, which is the curvature for a circle when $\lambda = R$, $R$ being the circle's radius. And so we can contain $\alpha(s)$ in a circle centered on the origin - it need only have radius $R > \lambda$.

Answer 4

Hint:

Assume WLOG that $\alpha$ is parametrized by arc-length. The normal line at the point $\alpha(s)$ is given by:

$$\alpha(s)+\lambda n(s),$$

where $n$ is the unit normal.

By your hypothesis, for each $s$ there exists a $\lambda(s)$ such that

$$\alpha(s)+\lambda(s)n(s)=0.$$

Hence

$$\|\alpha(s)\|=|\lambda(s)|,$$

so it suffices to see that $\lambda$ is constant.

Answer 5

Let $\alpha=\beta \mathbb{n}$ where $\beta$ is a function of $s$. Now by definition $$\mathbb{t}=\frac{d\alpha}{ds}=\frac{d\beta}{ds}\mathbb{n}-\beta\kappa\mathbb{t}$$

This implies that $\frac{d\beta}{ds}=0$ and that $\kappa=-\frac{1}{\beta}$