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Let $\alpha$ be a plane oval of constant width. Show that the sum of the radii of curvature $(\frac{1}{\kappa})$ at opposite points is a constant independent of the choice of points.

Any hints on starting the proof?

Attempt:

Let P be some point on the oval, and let $\mathbf{t}$ be the tangent to the oval at that point. Then let P' be the point on the opposite side of the curve with the same tangent line as P.

Then the curvature is the same at P and P'. Since the length between the two points is always constant (width) no matter what P and P' are picked the curvature will be the same.

Ash
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  • I'm not sure how an ellipse can be a figure of constant width if it's not a circle. There are figures of constant width that are not circles, surely, but general ellipses? ¶ The radius of curvature of a curve at a point is the radius of a circle that fits the curve at that point to second order. For instance, a parabola, at its apex, has a radius of curvature equal to twice the distance between the apex and the focus, because a circle of that radius fits the parabola at the apex to second order (has identical second derivative, if the parabola is vertical). – Brian Tung Mar 03 '16 at 00:55
  • Note that an ellipse does not satisfy the proposition: Its radius of curvature is minimum at the ends of the major axis (where its width is maximum), and maximum at the ends of the minor axis (where its width is minimum). – Brian Tung Mar 03 '16 at 01:01
  • @BrianTung Thank you for that clarification. I edited my post and added my attempt to the problem, but do you have any hints that could help me solve this? I am still new to proofs, and need a little help with the direction. Any help is greatly appreciated. – Ash Mar 03 '16 at 02:05
  • This really isn't going to lead to a proper proof. When you say "the same tangent line as $P$", I assume you mean that the tangent lines at $P$ and $P'$ are parallel, not identical. These lines are then separated by the constant width of the figure. However, it does not follow that their curvatures are identical. See this Wikipedia article, for instance: Sharper "corners" are opposite flatter "sides", and their curvatures are different. The proposition claims that though those curvatures vary, their sum is constant. – Brian Tung Mar 03 '16 at 03:04
  • Is this for a class? Where did you see this exercise? Without knowing more about the context in which you encountered this problem, it's difficult to suggest a viable path for you to proceed toward a proof. – Brian Tung Mar 03 '16 at 03:04
  • BTW, the right tag is definitely differential geometry. This is far from a multivariable calculus course. – Ted Shifrin Mar 03 '16 at 03:09
  • @TedShifrin I have been self studying from a book named Elements of Differential Geometry that was recommended to me by a professor, and he said it wasn't very different from multivariable calculus but I have been struggling with a lot of the exercises in the book. Thank you for all your help! – Ash Mar 03 '16 at 06:31

1 Answers1

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Here's a hint: Let $\alpha(s)$ be an arclength parametrization of the curve, and let $\beta(s)$ parametrize the curve by the opposite point. Note that $\beta$ will not (in general) be an arclength parametrization. Get a formula for $\beta$ in terms of $\alpha(s), T(s), N(s)$, and then calculate the curvature of $\beta$ (remembering that it is not arclength parametrized).

Ted Shifrin
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