Convexity limit angles for isoptics

Given an oval C in the plane, the α\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha $$\end{document}-isoptic Cα\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C_\alpha $$\end{document} of C is the plane curve composed of the points from which C can be seen under the angle π-α\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\pi -\alpha $$\end{document}. We consider isoptics of ovals parametrized with the support function p(t)=a+cosnt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p(t)=a+\cos n t$$\end{document}, n∈N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\in \mathbb {N}$$\end{document}, and present an example of an oval such that when α\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha $$\end{document} increases, the α\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha $$\end{document}-isoptics begin to be convex, then lose their convexity and finally are convex again along a curve intersecting the isoptics orthogonally. Next we give an example of a curve from the same family, for which the curvature of the isoptics changes its sign three times. These changes occur on the symmetry axes of the oval C and coincide with the orthogonal trajectories which start at the points with extremal curvature. Finally, we formulate the hypothesis concerning the general case where we expect n-1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n-1$$\end{document} convexity limit angles for the isoptics of an oval parametrized by p(t)=a+cosnt\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p(t)=a+\cos n t$$\end{document}.


Introduction
Isoptic curves were defined in 1704 by Philipe de La Hire, as mentioned in Lawrence (1972) p. 58. These curves remain interesting, as can be seen for example in Kunkli et al. (2013) and Dana-Picard (2020). Isoptics also have applications in mechanics and optics (see Weiss and Martini 2000;Wunderlich 1971).
Isoptics of convex curves are not necessarily convex. For isoptics of an ellipse, in Miernowski and Mozgawa (1997) was introduced the notion of the limit angle, described below. The inspiration to study this problem came about when we noticed certain irregularities. These were on graphs of the dual curves to isoptics of the oval, with support function p(t) = 10 + cos 3t, see Skrzypiec (2018).
The results presented below have application to optics and architecture (i.e., lighting and shades).
Remark 1 An angle α 0 for which C α is convex for α < α 0 and concave for α 0 < α (α near α 0 ) is called a limit angle (Miernowski and Mozgawa 1997). For the sake of clarity and generality, we call it a convexity limit angle, and extend its use also in the case where the sign of the curvature is considered only along a curve intersecting the isoptic orthogonally.
Proof Using the parametrization of isoptics from Cieślak et al. (1991), we prove that the curve C given by with the support function p(t) = a + cos 3t, where a > 8 for some values of a, is the example needed in our theorem. Let us notice that this curve has three axes of symmetry. One of them is the Xaxis. We obtain the others by rotations by the angles 2 3 π and 4 3 π . Each of these axes of symmetry passes through points of extremal curvature of C. Isoptics C α are symmetric with respect to the same axes of symmetry as C. That is why it is enough to study them in the neighbourhood of one of these symmetry axes. Without loss of generality, we choose the X -axis.
We notice that the denominator of this curvature is always positive since the tangent vector to the isoptic is always non-zero, and for β ∈ 0, π 2 we have cos β > 0. Along the positive X -axis the numerator of the curvature of the isoptic C α is given by Hence our problem simplifies to finding roots of functions f 1 (a, α) = −a + 6 cos α 2 + 2 cos 3α 2 (4) and f 2 (a, α) = a − 12 cos α 2 + 4 cos 3α 2 .
The denominator of the curvature of C α at t = − α 2 can be written as so the function f 1 has no roots. Moreover, f 1 (a, α) < 0 for a > 8 and α ∈ (0, π) since it has no roots and f 1 (a, 0) < 0. For the function f 2 let us substitute x = cos α 2 . Since α ∈ (0, π), we consider the resulting polynomial for x ∈ (0, 1] with the parameter a > 8. To find the number of distinct roots of polynomial v 2 we apply the Sturm theorem (see Serret 1866; Sturm 1829) on the interval (0, 1]. Constructing the Sturm sequence the signs of their values for a ∈ (8, 8 √ 2) are presented in Table 1.
Therefore, based on the Sturm theorem we conclude that the polynomial v 2 has two zeroes in the interval (0, 1] if the parameter a ∈ (8, 8 √ 2).
Similar computations as those above yield that along the negative X -axis the curvature of isoptics of the given oval is always positive.
Proof We prove that the curve C with support function p(t) = a+cos 4t, a ∈ (15, 17), is the example needed in our theorem. We note that this curve has four axes of symmetry. Two of them, the X and Y axes, pass through the points of maximal curvature of C. The other two are the lines y = x and y = −x, passing through the points of minimal curvature of C. Isoptics C α are symmetric with respect to the same axes of symmetry as C. Moreover, the first derivative of the curvature of isoptics is zero along these symmetry axes. Similar to the development above for p(t) = a + cos 3t, we can show that orthogonal trajectories which start at z k π 4 are straight lines and lie on the axes of symmetry of the oval C. That is why we will study them along two orthogonal trajectories (see Fig. 1). The first of them starts at z(0) and lies on the positive X -axis. It consists of the points z π 4 The second trajectory starts at z( π 4 ) and lies on the line y = x. It is the set of the points z α π 4 − α 2 , α ∈ (0, π) . Along the positive X -axis the curvature of the isoptic C α = {z α (t), t ∈ (0, 2π)} is given by k α − α 2 = − cos α 2 (4 − a + 8 cos α + 3 cos 2α)(−8 + a − 16 cos α + 9 cos 2α) (4 − a + 8 cos α + 3 cos 2α) 3 .
Similarly to the development above, we can consider the functions g 1 (a, α) = 4 + a + 8 cos α + 3 cos 2α (10) and g 2 (a, α) = −8 − a − 16 cos α + 9 cos 2α (11) and substitute x = cos α. Then we can consider the polynomials and with the parameter a > 15 for x ∈ (−1, 1]. The function g 1 has no roots. Let us construct the Sturm sequence of w 2 The signs of its values are presented in Table 3. Hence for a ∈ (15, 17) we have Z (−1)− Z (1) = 1 and based on the Sturm theorem we conclude that the polynomial w 2 has one root in the interval (−1, 1]. If a > 17 then Z (−1) − Z (1) = 0 which implies that w 2 has no roots between −1 and 1.

General case for p(t) = a + cos nt
For a > n 2 − 1 the curve C parametrized by the support function p(t) = a + cos nt is an oval and has n axes of symmetry. Isoptics of such ovals also have n symmetry axes, passing through the points of extremal curvature of C and coinciding with the orthogonal trajectories of the evolutions of the ovals. Those isoptics are invariant with respect to rotations about the origin by the angles 2kπ n , where k = 1, . . . , n. That is why we consider the curvature of the isoptics of the oval with the support function p(t) = a + cos nt, where a > n 2 − 1 along two half-lines.
-The first half-line starts at z(0) and lies on the positive X -axis. It starts on the given oval, at the point in which it has maximal curvature. It is a set of the points z α − α 2 , α ∈ (0, π) and we denote it I. -The second half-line starts at z π n and passes though the points z α π n − α 2 for α ∈ (0, π). It starts on the given oval, at the point in which it has minimal curvature. We denote it II.
For the curvature Along the trajectory I we obtain and We need to calculate the number of roots of the functions f 1 (α, a, n) and f 2 (α, a, n). The functions vary with α, and have parameters a and n. From the form of the denominator we observe that f 1 has no roots. We can also check it applying the same procedure for f 1 as we will use for f 2 , below. We perform a change of variables in f 2 , in order to obtain a polynomial of a real variable. We will do this for two cases: n odd and n even.
Using the trigonometric identity for sin nx we get Since sin u > 0 for u ∈ (0, π 2 ), using the Pythagorean trigonometric identity and the substitution x = cos u, with x ∈ (0, 1), for the function f 2 sin u we obtain the polynomial We now need to calculate the number of zeroes for x ∈ (0, 1).
We can repeat the above algorithm for even n. However, if we do it in a slightly different way we will obtain polynomials of lower degrees. Noticing that n = 2k for k ∈ Z, let us consider the function Substituting x = cos α where x ∈ (−1, 1) and using the trigonometric identity for cos nx, the function f 2 2 sin α 2 becomes the polynomial v 2 (x, a, n, k) = a + (4k 2 + 1) Our approach would be similar to the previous cases, applying the Sturm theorem for finding the number of roots. However, since the polynomial v 2 (x, a, n, k) depends on the degree n, we need to generalize this theorem. Along the trajectory II, after similar computations we get the following formulae n α π n − α 2 = −g 1 (α, a, n) · g 2 (α, a, n) and d α π and g 2 (α, a, n) = −2a sin α 2 − (n + 1) 2 sin (n − 1)α 2 + (n − 1) 2 sin (n + 1)α 2 .
The function g 1 (α, a, n) has no roots. We want to transform the function g 2 (α, a, n) into a real variable polynomial. For n odd using the substitution x = cos α 2 we obtain the following polynomial for the interval (0, 1). For n even substituting x = cos u we obtain where n = 2k and for which we are looking for the number of roots in the interval (−1, 1). This number of zeroes depends on the parameter a and we need to find such a 1 (n), that for a ∈ n 2 − 1, a 1 (n) the polynomials v 2 have the largest possible number of zeros in the appropriate intervals.
Whereas results for arbitrary n are as yet unknown, those for some fixed values of n are. We also put forward a hypothesis concerning the general case.
The maximal number of convexity limit angles along the orthogonal trajectories starting at the points of C of the largest (I) and the smallest (II) curvature, are obtained from computer experimentation, and are presented in Table 4.
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