First we give the necessary definition: A vertex of a curve $\gamma(r)$ in $\mathbb{R}^2$ is a point where its signed curvature $\boldsymbol{k}_{s}$ has a stationary point, i.e.,where $d\boldsymbol{k}_{s}/dt=0$.

We have Four Vertex Theorem: Every convex simple closed curve in $\mathbb{R}^2$ has at least four vertices.

Now, I have a question: For a convex curve on $\mathbb{R}^2$ plane, we define the distance of two points $(x_1,y_1),(x_2,y_2)$ on the curve is $L=\sqrt{(x_1-x_2)^2 +(y_1-y_2)^2}$. If the distance of these two points is the longest on $\gamma(r)$, would these two points will be vertices?

Obviously, it's true for the elipse.


2 Answers 2


No. In fact a curve in general position does not satisfy this property.

To construct a concrete counterexample, take an ellipse, nearly round one and take its Minkowski sum with a long segment, which does not go in the direction of its axis.

The curve which bounds the obtained convex set does not satisfy your property. Smoothen it if you want a $C^\infty$-curve.

  • 1
    $\begingroup$ Welcome to Math.SE! I see this is your first answer here, hopefully not the last. $\endgroup$
    – user53153
    Feb 25, 2013 at 4:34
  • $\begingroup$ Your counterexample is helpful! @Anton Petrunin $\endgroup$
    – user39843
    Feb 25, 2013 at 5:05

For people who don't visualize Minkowski sums easily, here is a slight modification of the example given by Anton Petrunin. I erased the middle part of an ellipse (original ellipse still shown below in grey), shifted the left part down a bit, and completed the curve with straight lines.

distorted ellipse

The idea is that because of the shift, the diameter of the curve is attained by a pair of points that are close to, but do not coincide with the vertices of the original ellipse. Hence, the curvature is not stationary there.


The corners can be smoothened without affecting anything in the construction, since they are far from attaining the diameter of the curve.

  • $\begingroup$ Thanks,I'm a people who don't visualize Minkowski sums easily. Your counterexample is helpful! $\endgroup$
    – user39843
    Feb 25, 2013 at 5:00
  • $\begingroup$ Anton's counterexample, really. The Minkowski sum with a slanted segment looks about the same as this figure. $\endgroup$
    – user53153
    Feb 25, 2013 at 5:02

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