give a closed plane curve C with k>0 that is not convex. (k=curvature) Solution
ID: 3077328 • Letter: G
Question
give a closed plane curve C with k>0 that is not convex. (k=curvature)Explanation / Answer
A simple closed curve is convex if and only if k > 0 or k < 0 everywhere along the curve. Proof. Assume first that g is a naturally parameterized convex curve. Leta(t) be a continuous direction angle for the tangent t(t). As we know, a = k, thus, it suffices to show that a is a weakly monotonous function. This follows if we show that if a takes the same value at two different parameters t ,t , then a is constant on the interval [t ,t ]. The rotation number of a simple curve is +1, hence the image of t covers the whole unit circle. As a consequence, we can find a point at which t(t ) = -t(t ) = -t(t ). If the tangent lines at t ,t ,t were different, then one of them would be between the others and this tangent would have points of the curve on both sides. This contradicts convexity, hence two of these tangents say the tangents at P = g(t ) and Q = g(t ) coincide. We claim that the segment PQ is an arc of g. It is enough to prove that this segment is in the image of g. Assume to the contrary that a point XePQ is not covered by g. Drawing a line e $ PQ through X, we can find at least two intersection points R and S of e and the curve, since e separates P and Q and g has two essentially disjoint arcs connecting P to Q. Since PQ is a tangent of g, the point R and S must lie on the same side of it. As a consequence, we get that one of the triangles PQR and PQS, say the first one is inside the other. However, this leads to a contradiction, since for such a configuration the tangent through S necessarily separates two vertices of the triangle PQR, which lie lie on the curve. If g is defined on the interval [a,b], then g(a) = g(b) is either on the segment PQ or not. The first case is not possible, because then a would be constant on the intervals [a,t ] and [t ,b], yielding a(a) = a(t ) = a(t ) = a(b) and rotation number = (a(b) - a(b))/2p = 0. In the second case a is constant on the interval [t ,t ], as we wanted to show. [red] Now to prove the converse, assume that g is a simple closed curve with k > 0 everywhere and assume to the contrary that g is not convex (the case k < 0 can be treated analogously). Then we can find a point P = g(t ), such that the tangent at P has curve points on both of its sides. Let us find on each side a curve point, say Q = g(t ) and R = g(t ) respectively, lying at maximal distance from the tangent at P. Then the tangents at P,Q and R are different and parallel. Since the unit tangent vectors t(t ), i=1,2,3 hav parallel directions, two of them, say t(t ) and t(t ) must be equal. A = g(t ) and B = g(t ) divide the curve into two arcs. Denoting by K and K 10 the total curvatures of these arcs, we deduce that these total curvatures have the form K = 2k p, K = 2k p, where k ,k e Z, since the unit tangent at the ends of the arcs are equal. On the other hand, we have k +k = 1 by the Umlaufsatz and k > 0, k > 0 by the assumption k > 0. This is possible only if one of the total curvatures K or K is equal to zero. Since k > 0, this means that k = 0 along one of the arcs between A and B. But then this arc would be a straight line segment, implying that the tangents at A and B coincide. The contradiction proves the theorem.Related Questions
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