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I'm trying to show that a closed, smooth plane curve with curvature 1 is a circle.

The Frenet equations are:

\begin{align} t' &= kn \\ n' &= -kt - \tau b \\ b' &= \tau n. \end{align}

Now, I've shown that if $\alpha(t)$ is a plane curve, then $\tau = 0$.

Thus, our equations give:

$$t'' = -t$$ and $$b'=0.$$

From the first equation we have that,

\begin{align} t'' &= t'\\ \implies \alpha'''(s)&=-\alpha'(s)\\ \implies \alpha(s) &= c_1sinx+c_2cosx+c_3. \end{align}

It seems like I may be on the right track... though I have not yet used the fact that $\alpha$ is a closed curve. Or is there a different approach I should take?

*Also, I have moved the other direction: I have previously shown that a circle possesses curvature $\frac{1}{r}$ where $r$ is its radius. A circle of radius one, then has curvature 1. This proof doesn't work the other direction.

*For reference, I'm almost exclusively using Do Carmo's Differential Geometry of Curves and Surfaces

Thx!

Update:

No one has answered with anything yet, but here's another idea:

Do Carmo asks that we prove the existence of an Osculating Circle. Here's an idea for that:

Let there be two points $s_0$ and $s_1$ on $\alpha(s)$. Let their corresponding normals intersect at point which we will define as the center of a circle. Let ${s_0}$ approach ${s_1}$. Now show the limiting position of the intersection of normals becomes the center of an osculating circle (i.e., a circle with a tangent that coincides with the tangent of our curve at $s_1$) and the limiting position of the circle become the osculating circle. We may then prove that the radius of this circle is $\frac{1}{k(s)}$ Is this correct and how may I approach this?

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    $\begingroup$ Once you derived the ansatz for $\alpha$ you just have to compute the curvature in terms of the constants. This will give you a relation between the $c_i$. Then you should be able to directly show it is a circle by constructing a middle point and computing the distance. $\endgroup$ – Nikolas Kuhn Mar 4 '16 at 10:23
  • $\begingroup$ Showing that the curvature of the curve at a point is the reciprocal of the radius of the osculating circle can be easily done with elementary co-ordinate geometry and elementary calculus. Compute the co-ordinates of the intersection of the normals at $s_0$ and $s_1$ and let $s_0$ approach $s_1$ $\endgroup$ – DanielWainfleet Mar 4 '16 at 19:15
  • $\begingroup$ Are you sure that the curve is assumed plane to begin with? It seems so obvious then: The conditions $t'=n$, $n'=-t$ and $t(0)=(1,0)$, $n(0)=(0,1)$ enforce $\alpha(s)=\alpha(0)+(\sin s,\cos s)$. $\endgroup$ – Christian Blatter Mar 6 '16 at 13:15
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$\alpha$ has unit speed so that $$ n'=-kT=-T$$ where $T=\alpha'$ Hence $$ (\alpha +n)'=T-T =0 $$ So $\alpha +n=C\Rightarrow |\alpha -C|=1 $

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