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Prove that a regular curve parametrized by arc length such that all its normal lines pass through a single point is contained in a circumference.

Suppose $\alpha:I\rightarrow \mathbb{R}^3$ is a curve such that all its normal lines pass through $P$. Then we can write that for every $s\in I$ there exist a $r_a$ such that:

$$\alpha(s)+r_s n(s)=P$$

or

$$r_s n(s)=P-\alpha(s)$$

then

$$\alpha'(s) \cdot (P-\alpha(s))=\alpha'(s)\cdot r_s n(s)=0$$

since $(P-\alpha(s))'=-\alpha'(s)$ which implies that the distance $|P-\alpha |$ is constant.

My question is, how do I prove that $\alpha$ is in a circumference not in a sphere.

Thanks!

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To show that the curve lies on a plane, it is enough to show that it is torsion-free (I'll let you figure out why) or that its binormal vector is constant (same). In your situation, the quickest way to show this is probably to use the Frenet-Serret formulas.

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  • $\begingroup$ I know that I should show that the curve is torsion-free and I think I can derive this from the equation $ \alpha(s)+r_s n(s)=P$ but I don't know why $r_s$ is a derivable function to apply the rule for multiplication. $\endgroup$ – YTS Sep 22 '14 at 21:29
  • $\begingroup$ Just differentiate again. Assume $\alpha$ is arclength-parametrized, for starters. $\endgroup$ – Ted Shifrin Sep 23 '14 at 0:11
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    $\begingroup$ If I'm not mistaken, $r_s = (P - \alpha_s)\cdot n(s)$,right? So $r_s$ is a differentiable function of $s$. I think you know how to proceed. $\endgroup$ – Seub Sep 23 '14 at 1:21

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