SMOOTH BODIES of CONSTANT WIDTH in E 2 Sasˇo Bogoev

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SMOOTH BODIES of CONSTANT WIDTH in E 2 Sasˇo Bogoev Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, Vol.8, (2019), Issue 2, pp.74-90 SMOOTH BODIES OF CONSTANT WIDTH IN E2 Sa soˇ Bogoev ABSTRACT . This paper is concerned with another possible construction of two-dimensional bodies of constant width in Euclidean plane. We derive a new parametrization of their boundary curves and give a connection to the standard support function. Considering this parametrization, we also obtain an area formula and show some examples in the sequel. At the end, we take a step back to the Reuleaux polygons in an attempt to represent them holding on to our approach, although they are not smooth and are outside the scope of the title of the paper. 1. I NTRODUCTION Let K E2 be a compact convex body (set), and D its diameter. We say the body K is of constant⊂ width D if it has exactly one pair of points (refer to them as opposite or antipodal points) at distance D in all directions. This definition implies that K is to be strictly convex too. The bodies of constant width have been studied by many authors, but their properties are still subject to analysis. In brief, we only mention a few of their basic features. For instance, the Barbier’s theorem states that any curve of constant width D has length of Dπ. Certainly, the boundary of our body K is a curve of constant width and we denote it by ∂K, as usual. A simple example of a curve of constant width is the circle, but there exist infinitely many others which are commonly named as orbiforms. Another widely known result refers to the area and claims that among all bodies of a given constant width, the disk and the Reuleaux triangle have the largest and (Blaschke–Lebesgue theorem) the smallest area respectively. Further, the only centrally symmetric body of constant width is the disk, and so on. In the sequel, we denote the area of K by A(K) and its perimeter by L(K). The main focus herein is on the bodies of constant width which boundaries are smooth curves of at least class C1. In spite of the fact that our construction is based on a union of two (or more) curves of class C2, the resulting curves might not have C2 continuity at the two (or more) join points. However, regardless of this classification, since C1 continuity is preserved, we simply say K is a smooth body of constant width. 2010 Mathematics Subject Classification: 52A10, 53A04. Keywords : constant width, isoperimetric, curve, body, set, construction, parametrization, area, Reuleaux polygons. 74 SMOOTH BODIES OF CONSTANT WIDTH IN E2 2. P ARAMETRIZATION Now, let’s start speculating. Suppose we have some body K of constant width which diameter is D, and a line segment, let say AB , of the same length. By the definition above, and due to the strict convexity, we are able to fit the whole AB into the interior of K, except the two endpoints A and B which will belong to the boundary of K. What is more important is that we are restricted to do this placement in a unique way for every single direction and hence, we conclude ∂K could be well–defined by the position of AB , taking the direction as a parameter. Moreover, for any plane body of constant width which boundary is a curve of class C2, it is proven that the sum of the radii of curvature at any two opposite points (in our case A and B) equals to the diameter D. This fact leads us to an idea to start a rotation of AB without leaving the interior of K. Although the center of this rotation F will vary, it must lie somewhere on AB in every moment (follows from the mentioned sum of the radii of curvature and the properties of the distance as a metric function). After a rotation of AB by π, we arrive at the same direction which we started from and therefore, its position must match the starting one. The trajectories of the points A and B bring us back the boundary of K. Since we were in the interior of K all the time, and the points A and B were on its boundary, we covered the whole area A(K) as well with this rotation. The conclusion is that we can perform exactly the same movement of AB even if K is not present. All we need to know is how the center of the rotation F changes when we change the direction of AB . Recall that we are dealing with a smooth ∂K, which means its derivative exists everywhere. Finally, let’s fix a coordinate system and place K such that one of its diameters, say AB , lies on the x-axis, and the point A is at the origin, i.e. A = (0, 0) and B = (D, 0) , as shown in Figure 1. FIGURE 1. Rotation of the diameter AB . After a rotation by α, AB arrives at the position A′B′. Denote the points A′ and B′ by their coordinates (xA, yA) and (xB, yB), where all xA, yA, xB, yB are some functions of the parameter α. 75 S. BOGOEV At this position, we are to continue with a rotation which center must be exactly F, as previously explained. Following the point F, we define the function f : [0, π] [0, D] by f (α) = d(A′, F) , → where d is the standard metric in E2 (distance). The smoothness of ∂K (it is of class C1) implies continuity of f almost everywhere on [0, π]. By definition, the function f is also non-negative, bounded by D and therefore, it is measurable. While having this in mind, in order to simplify the analysis, for now we assume f is continuous everywhere on [0, π]. Considering Figure 1, let dl A and dl B be positive lengths, and dα small enough. Then, we may write dl A = f (α) dα , dl = (D f (α)) dα . B − Under this assumptions, the Barbier’s theorem follows immediately, π π L(K) = (dl A + dl B) = D dα = Dπ . Z0 Z0 We track the points A and B separately, and Figure 1 shows that yA(α) is negative, yB(α) is positive on the interval (0, π), while the differential functions dx A and dx B are positive and negative on the same interval respectively. Thus, the differentials of the coordinates of A, B are: π dx = dl cos ( α) = dl sin α = f (α) sin α dα ; A A 2 − A π dy = dl sin ( α) = dl cos α = f (α) cos α dα ; A − A 2 − − A − π (2.1) dx = dl cos ( α) = dl sin α = (D f (α)) sin α dα ; B − B 2 − − B − − π dy = dl sin ( α) = dl cos α = (D f (α)) cos α dα . B B 2 − B − Taking in account the starting values xA(0) = yA(0) = yB(0) = xB(π) = yA(π) = yB(π) = 0, and xA(π) = xB(0) = D, the parametric equations of the coordinates of the points A and B are: α xA(α) = f (t) sin t dt ; Z0 α yA(α) = f (t) cos t dt ; − Z0 α (2.2) xB(α) = D (D f (t)) sin t dt ; − Z0 − α yB(α) = (D f (t)) cos t dt . Z0 − The graph constructed by the union of the points (xA, yA) and (xB, yB) coincide with ∂K or, if we are more formal, ∂K = xA, yA : α [0, π] xB, yB : α [0, π] . Notice that the x-axis splits this union into! two curves such∈ that each S ! of them shows∈ the trajectory of A and B separately. At α = π, the two points interchange their starting coordinates and probably this is the moment when we give up on achieving C2 continuity. Namely, it is easy to observe 76 SMOOTH BODIES OF CONSTANT WIDTH IN E2 = = D that only in case when f (0) f (π) 2 , the starting and the ending curvature match each other at the join points. Having this in mind, since in the start we picked out the diameter AB arbitrarily, if we reparametrize the same curve, we infer that the function f does not even have to be continuous necessarily. Anyway, what is important is that when the points A and B finish their journey, after integrating over [0, π], by (2.1) we conclude π π dx A = xA(π) xA(0) = D f (α) sin α dα = D , Z0 − ⇐⇒ Z0 π π (2.3) dy A = yA(π) yA(0) = 0 f (α) cos α dα = 0 . Z0 − ⇐⇒ Z0 In fact, the last two relations are constraints on f which must hold for any smooth curve of constant width. These constraints obtain the curve to be closed. From this point, if we try to get back step by step aiming to reconstruct ∂K, we will find we have no other obstacle except one which hinder us to guarantee the smoothness of the curve. Let’s suppose we are equipped with a non-negative continuous function f on [0, π], bounded by the length of the segment we rotate, say D, and assume the equations (2.3) hold too. In order to eliminate prospective singular points on the curve (2.2), we are looking for the candidates. By use of (2.1), we set the next conditions: = = x′A(α) f (α) sin α 0 ; y′ (α) = f (α) cos α = 0 ; A − x′ (α) = (D f (α)) sin α = 0 ; B − − y′ (α) = (D f (α)) cos α = 0 .
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