When working on an exercise, I stumbled upon the following "intuitive fact" which I couldn't prove:
Let $\alpha, \beta:[a, b] \to \mathbb{R}^{2}$ be two smooth curves with $\alpha(s) = \lambda(s)\beta(s)$ where $\lambda(s) \geq 1$ and $\lambda(a) = \lambda(b) = 1$. The trace of $\alpha$ is an arc of a circle of radius $r$. The tangent vectors of both cuves coincide at the ends of the domain.
Conjecture: There exists a point $s\in[a, b]$ with curvature $\kappa_{\beta}(s) \geq \frac{1}{r}$
The intuition is that $\beta$ has to "bend" more in order to get to the same point.
My attempt
I was trying to use an angular function for the curve $\beta$, $\theta'(s) = \kappa_{\beta}(s)$ since it traslates to $\theta(a) - \theta(b) = $"measure of the arc described by $\alpha$", and seems to relate both givens and the curvature which I'm trying to bound. However, the inequalities come out the other way, giving me an upper bound instead of a lower bound, which is a clear sign that there's some property that I'm not exploiting.
Edit:
- In my exercise, the curvature of $\beta$ does not vanish. Use that hypothesis to show the result if necessary.
- As pointed out in the comments, degenerate cases are avoided with a little stronger but still reasonable condition that I've added in the statement. For example, in my exercise the curve $\beta$ is also simple.