A convex cover for closed unit curves has area at least 0.0975

TL;DR

This paper establishes a new lower bound of 0.0975 for the minimal area of convex sets covering all closed plane curves of unit length, improving previous bounds through geometric and numerical methods.

Contribution

It introduces a combined geometric and numerical approach to improve the lower bound on the area of convex covers for unit curves.

Findings

Lower bound of 0.0975 for convex covers of unit curves

Improved previous lower bound of 0.096694

Bounds for convex hulls of specific shapes like circle, triangle, and rectangle

Abstract

We combine geometric methods with numerical box search algorithm to show that the minimal area of a convex set on the plane which can cover every closed plane curve of unit length is at least 0.0975. This improves the best previous lower bound of 0.096694. In fact, we show that the minimal area of convex hull of circle, equilateral triangle, and rectangle of perimeter $1$ is between 0.0975 and 0.09763.

Figures

Equations

$$0.096694\le \beta \le \mathrm{0.11023.}$$

$$0.0975\le \beta ,$$

$$0.0975\le {\beta }^{\prime }\le \beta .$$

$${\beta }^{\prime }\approx \mathrm{0.09762742.}$$

$$f(x_1,y_1,x_2,y_2,\theta )>0.0975,\forall x_1,y_1,x_2,y_2,\theta .$$

$$0\le x_1\le 0.05,0\le y_1\le 0.04,-0.17\le x_2\le 0.17,-0.13\le y_2\le 0.13,0\le \theta \le \frac{2\pi }{3}.$$

$$\psi (x_1,y_1)\ge \psi (x_1,0),\forall x_1,y_1.$$

$$\psi (x_1,y_1)\ge \psi (x_1,0)\ge \psi (0.05,0)>0.0975,$$

$$\psi (x_1,y_1)\ge \psi (0,y_1)\ge \psi (0,0.04)>0.0975,$$

$$\mathcal{A}(X)>\frac{1}{2}\left(2r\cos \left(\frac{\pi }{500}\right)+\frac{\sqrt{3}}{9}\right)\left(\sqrt{x_2^2+y_2^2}\right)+\frac{S(F)}{2}+\frac{\pi {\left(\frac{\sqrt{3}}{18}\right)}^2}{2}>0.0975$$

$$|f(x_1+{ϵ}_1,y_1+{ϵ}_2,x_2+{ϵ}_3,y_2+{ϵ}_4,\theta +{ϵ}_5)-f(x_1,y_1,x_2,y_2,\theta )|\le \sum _{i=1}^5{ϵ}_i{C}_i,$$

$$C=\max \left[\underset{t\to -\infty }{\lim }\frac{g(t)}{t},\underset{t\to +\infty }{\lim }\frac{g(t)}{t}\right]<\infty ,$$

$$|g(t+ϵ)-g(t)|\le Cϵ,\forall t,\forall ϵ>0.$$

$$\underset{x_1\to -\infty }{\lim }\frac{g(x_1)}{x_1}=\underset{x_1\to +\infty }{\lim }\frac{g(x_1)}{x_1}\le \frac{0.386+0.0375}{2}<C_1,$$

$$|f(x_1+{ϵ}_1,y_1,x_2,y_2,\theta )-f(x_1,y_1,x_2,y_2,\theta )|\le C_1{ϵ}_1.$$

$$\underset{x_2\to -\infty }{\lim }\frac{g(x_2)}{x_2}=\underset{x_2\to +\infty }{\lim }\frac{g(x_2)}{x_2}\le \frac{2r+1/3}{2}<C_3,$$

$$\underset{y_2\to -\infty }{\lim }\frac{g(y_2)}{y_2}=\underset{y_2\to +\infty }{\lim }\frac{g(y_2)}{y_2}\le \frac{1/3+0.4625}{2}<C_4.$$

$$|f(x_1,y_1,x_2+{ϵ}_3,y_2,\theta )-f(x_1,y_1,x_2,y_2,\theta )|\le C_3{ϵ}_3,$$

$$|f(x_1,y_1,x_2,y_2+{ϵ}_4,\theta )-f(x_1,y_1,x_2,y_2,\theta )|\le C_4{ϵ}_4.$$

$$|f(x_1,y_1+{ϵ}_2,x_2,y_2,\theta )-f(x_1,y_1,x_2,y_2,\theta )|\le C_2{ϵ}_2.$$

$$|f(x_1,y_1+{ϵ}^{\ast },x_2,y_2,\theta )-f(x_1,y_1,x_2,y_2,\theta )|\le C_2{ϵ}^{\ast }.$$

$$|f(x_1,y_1+{ϵ}^{\ast }+{\delta }_2,x_2,y_2,\theta )-f(x_1,y_1+{ϵ}^{\ast },x_2,y_2,\theta )|\le C_2{\delta }_2$$

$$|f(x_1,y_1,x_2,y_2,\theta +{ϵ}_5)-f(x_1,y_1,x_2,y_2,\theta )|\le C_5{ϵ}_5.$$

$$|(S(T_1{X}_1{X}_2)+S(T_2{X}_3{X}_4)+S(T_3{X}_5{X}_6))-(S(T_1^{\prime }{X}_1{X}_2)+S(T_2^{\prime }{X}_3{X}_4)+S(T_3^{\prime }{X}_5{X}_6))|.$$

$$|\mathcal{A}(R,F,T^{\prime })-\mathcal{A}(R,F,T)|<3\cdot 0.46402\cdot \frac{\sqrt{3}}{18}{ϵ}_5<0.134{ϵ}_5=C_5{ϵ}_5.$$

$$f(z^{\ast })-d_1{C}_1-d_2{C}_2-d_3{C}_3-d_4{C}_4-d_5{C}_5\ge 0.0975,$$

$$f(z^{\ast })-\frac{0.05}{2}{C}_1-\frac{0.04}{2}{C}_2-\frac{0.17-(-0.17)}{2}{C}_3-\frac{0.13-(-0.13)}{2}{C}_4-\frac{2\pi /3}{2}{C}_5>0.0975,$$

$$S_{min}\approx \mathrm{0.09762742.}$$

$$\frac{1}{2}{h}_1ϵ+\frac{1}{2}{h}_2ϵ+Q_1{Q}_2ϵ=\frac{ϵ}{2}(h_1+h_2+2Q_1{Q}_2)$$

$$f(w)-f(w+ϵ)\le \mathcal{A}(R^{\prime },F^{\prime },T)-\mathcal{A}(R^{\prime \prime },F^{\prime },T)\le \frac{ϵ}{2}(0.386+0.25)=0.318ϵ,$$

Full text

1. Introduction

In 1966, Leo Moser [9] posed a question to determine the smallest area of plane region $S$ which can cover every curve of length $1$, assuming that the curve may be translated and rotated to fit inside the region. This problem, known as “Moser’s worm problem”, remains open, and it is not even known that the smallest-area region with this property exists - an alternative possibility is that the infimum of possible areas of $S$ can not be actually attained. The current record belonging to Norwood and Poole [10], who constructed a (nonconvex) set $S$ with this property and area at most $0.260437$. From the lower bounds perspective, it is known only that area of any such set $S$ must be strictly positive [8].

Several variants of Moser’s worm problem has been studied in the literature. In particular, one can modify:

1. (1)

the allowed covering regions (e.g. triangle, rectangle, convex, non-convex) 2. (2)

the sets of curves to be covered (e.g. closed curves, convex curves)

If one insists for the covering set $S$ to be convex, Laidecker and Poole [7] used Blachke Selection Theorem to show that the solution to Moser’s worm problem exists. If $\alpha$ is the area of this smallest convex covering set, it is known that $0.232239\leq\alpha\leq 0.27091$, where the upper bound was found by Wang [13] in 2006, while the lower bound was proved by Khandhawit, Sriswasdi and Pagonakis [6] in 2013.

This work studies the version of Moser’s worm problem, asking for convex region $S$ of minimal area $\beta$, which can cover every closed curve of length $1$. H.G. Eggleton [3] proved in 1957 that a triangle covers all closed unit curves if and only if it can cover the circle of perimeter $1$. Therefore, the smallest triangle which covers every closed unit curve is the equilateral triangle of side ${\displaystyle\frac{\sqrt{3}}{\pi}}$ and area $\frac{3\sqrt{3}}{4\pi^{2}}\approx 0.13162$. In 1972, Shaer and Wetzel [11] proved that the smallest rectangle with this property has side lengths $\frac{1}{\pi}$ and $\sqrt{\frac{1}{4}-\frac{1}{\pi^{2}}}$ and area about $0.12274$. In 2006, Furedi and Wetzel [5] found a cover in the form of pentagon with area less than $0.11222$. Recently, Wichiramala [14] discovered a hexagonal cover with area less than $0.11023$.

Minimum-area convex cover for a set of curves is, equivalently, the minimum-area convex hull of these curves. In 1973, Chakerian and Klamkin [2] proved that the convex hull of the circle with perimeter $1$ and the line segment of length $\frac{1}{2}$ is at least $0.0963275$, thus providing the first lower bound for $\beta$. In 2006, Furedi and Wetzel [5] proved that the convex hull of the circle with perimeter 1 and the $0.0261682\times 0.4738318$ rectangle is at least $0.0966675$. In 2011, Furedi and Wetzel [5] improved the lower bound by replacing the $0.0261682\times 0.4738318$ rectangle by curvilinear rectangle. This gives the area of convex hull about $0.096694$. Hence, the best published bounds for $\beta$ were, before this work,

[TABLE]

In an unpublished work, Som-am [12] used the Brass grid search method [1] to show that the minimal-area convex hull of the line segment of length $\frac{1}{2}$, circle with perimeter $1$, and the equilateral triangle with perimeter $1$, is about $0.096905$.

The main result of this paper is the following one.

Theorem 1**.**

Any convex set $S$ on the plane which can cover circle of perimeter $1$, equilateral triangle of perimeter $1$, and rectangle of size $0.0375\times 0.4625$ (and perimeter $1$) has area at least $0.0975$.

Theorem 1 immediately implies that

[TABLE]

which is an improvement comparing the best published lower bound $0.096694$, as well as comparing an unpublished lower bound $0.096905$.

If $\beta^{\prime}$ is the minimal area of a set which can cover circle, equilateral triangle, and any rectangle of perimeter $1$, then Theorem 1 states that

[TABLE]

Our computation shows that the actual value of $\beta^{\prime}$ is about

[TABLE]

The bound in Theorem 1 is slightly weaker, because we need some margin to allow rigorous analysis of our numerical algorithms. We also show rigorously that $\beta^{\prime}\leq 0.09763$. This implies that, to improve lower bound for $\beta$ to $0.09763$ and beyond, different approach is required.

This paper is organized as follows. In Section 2, we give some basic definitions, and prove preliminary lemmas by geometric methods. These lemmas, together with the box-search numerical algorithm, are applied for proving Theorem 1 in Section 3 and 4. Some concluding discussion is presented in Section 5.

2. Geometric analysis

Assume that $C$ is a circle with radius $r={\displaystyle\frac{1}{2\pi}}$, $R$ is a rectangle with side $u\times v$ such that $u+v={\displaystyle\frac{1}{2}}$ and $u=0.0375$, and $T$ is an equilateral triangle of side ${\displaystyle\frac{1}{3}}$. Remark that $C$, $R$, and $T$ are convex polygons in $\mathbb{R}^{2}$. Our aim is to prove that, no matter how $C$, $R$, and $T$ are placed in $\mathbb{R}^{2}$, the area of their convex hull is at least $0.0975$.

Let $F$ be a regular $500$-gon inscribed in the circle, such that the sides of $R$ are parallel to some longest diagonals of $F$. We will call the union $X=F\cup R\cup T$ a configuration. For any configuration $X$, let $\mathcal{H}(X)$ denote the convex hull of $X$, and $\mathcal{A}(X)$ the area of $\mathcal{H}(X)$.

Let us put a coordinate center $(0,0)$ at the center of $F$, and let X axis and Y axis be parallel to the longer and shorter sides of the rectangle, respectively. Let $R_{0}(x_{1},y_{1})$ be the center of $R$. We can orient the axes in such a way that $x_{1}\geq 0$ and $y_{1}\geq 0$. The vertices of $R$ are defined by $R_{1}\left(x_{1}-{\left(\frac{\frac{1}{2}-u}{2}\right)},y_{1}+{\frac{u}{2}}\right)$, $R_{2}\left(x_{1}+{\left(\frac{\frac{1}{2}-u}{2}\right)},y_{1}+{\frac{u}{2}}\right)$, $R_{3}\left(x_{1}+{\left(\frac{\frac{1}{2}-u}{2}\right)},y_{1}-{\frac{u}{2}}\right)$, and $R_{4}\left(x_{1}-{\left(\frac{\frac{1}{2}-u}{2}\right)},y_{1}-{\frac{u}{2}}\right)$. Let $T_{0}(x_{2},y_{2})$ be the center of $T$ and $\theta$ be the angle between X axis and $T_{0}T_{1}$,

Then $T_{1}(x_{2}+\frac{\sqrt{3}}{9}\cos\theta,y_{2}+\frac{\sqrt{3}}{9}\sin\theta),$ $T_{2}(x_{2}+\frac{\sqrt{3}}{9}\cos(\theta+\frac{2\pi}{3}),y_{2}+\frac{\sqrt{3}}{9}\sin(\theta+\frac{2\pi}{3}))$ and $T_{3}(x_{2}+\frac{\sqrt{3}}{9}\cos(\theta+\frac{4\pi}{3}),y_{2}+\frac{\sqrt{3}}{9}\sin(\theta+\frac{4\pi}{3}))$ are the vertices of triangle $T$.

In summary, the location of $F$, $R$, and $T$ is fully described by $5$ parameters: $x_{1},y_{1},x_{2},y_{2}$, and $\theta$.

Let $f:\mathbb{R}^{5}\rightarrow\mathbb{R}$ be a function which maps vector $(x_{1},y_{1},x_{2},y_{2},\theta)$ to the area $\mathcal{A}(X)$ of the convex hull of the corresponding configuration $X$. Clearly, $f$ is a continuous function. Because $F$ is a subset of $C$, Theorem 1 would follow from the inequality

[TABLE]

The following result of Fary and Redei [4] plays an important role in our analysis

Lemma 2**.**

[4]** Let $S_{1}$ and $S_{2}$ be two bounded convex sets in ${\mathbb{R}}^{2}$. If $S_{1}$ is translated along a line with constant velocity, then the volume of the convex hull of $S_{1}$ and $S_{2}$ is a convex function of time.

Corollary 3**.**

Function $f$ is a convex function in each of the coordinates $x_{1},y_{1},x_{2},y_{2}$.

Proof.

Convexity of $f$ with respect to $x_{1}$ follows from Lemma 2 with $S_{2}$ being the convex hull of $F$ and $T$, while $S_{1}=R$ moving along the X axis. Convexity of $f$ with respect to $y_{1}$, $x_{2}$, and $y_{2}$ follows from Lemma 2 in a similar way. ∎

Lemma 4**.**

Let $Z$ be a region of points $z=(x_{1},y_{1},x_{2},y_{2},\theta)$ in ${\mathbb{R}}^{5}$ satisfying the inequalities

[TABLE]

If $f(z)>0.0975$ for all $z\in Z$, then in fact $f(z)>0.0975$ for all $z\in{\mathbb{R}}^{5}$.

Proof.

Let $\psi(x_{1},y_{1})$ be the area of convex hull of $F$ and $R$ only. Lemma 2 implies that $\psi(x_{1},y_{1})$ is a convex function in both coordinates. Assume that $x_{1}\geq 0.05$. By symmetry, $\psi(x_{1},y_{1})=\psi(x_{1},-y_{1})$, hence

[TABLE]

Also, by symmetry, $\psi(x_{1},0)=\psi(-x_{1},0)$, hence convexity of $\psi(x_{1},0)$ implies that $\psi(x_{1},0)\geq\psi(0,0)$, and that $\psi(x_{1},0)$ is non-decreasing in $x_{1}$ for $x_{1}\geq 0$. Hence, $x_{1}\geq 0.05$ implies that

[TABLE]

where the last equality is verified directly. For similar reasons,

[TABLE]

whenever $y_{1}\geq 0.04$.

From symmetry, we may assume that $x_{1}\geq 0$ and $y_{1}\geq 0$. Hence, either $f(z)>0.0975$, or we may assume that $0\leq x_{1}\leq 0.05$, and $0\leq y_{1}\leq 0.04$.

Next, assume that $|x_{2}|\geq 0.17$. Then $\sqrt{x_{2}^{2}+y_{2}^{2}}\geq|x_{2}|\geq 0.17$. Let $C_{1}$ be the incircle of $T$ with radius $\frac{\sqrt{3}}{18}$ and center $(x_{2},y_{2})$, see Figure 2.

Let $l$ be the line segment between $(0,0)$ and $(x_{2},y_{2})$. Next, let points $H,G\in C_{1}$ and $E,K\in F$ be such that line segments $HG$ and $EK$ are perpendicular to $l$, and pass through $(x_{2},y_{2})$ and $(0,0)$, respectively, see Figure 2. Then $EKGH$ is trapezoid with base lengths $|HG|=\frac{\sqrt{3}}{9}$ and $|EK|\geq 2r\cos\left(\frac{\pi}{500}\right)$, where $r=\frac{1}{2\pi}$. The area of $F$ is $S(F)=500\frac{r^{2}}{2}\sin\left(\frac{2\pi}{500}\right)$. Thus,

[TABLE]

To prove the bound for $y_{2}$, we need the following claim.

Claim 1. If there is a point $P^{\prime}\in T$ with $y$-coordinate $y^{*}\geq 0.13+\frac{\sqrt{3}}{18}$, then $f(z)>0.0975$.

Indeed, let $R^{\prime}=(0,-r)$, and $R_{1},R_{2},R_{3},R_{4}$ be the vertices of the rectangle, see Figure 3.

Because $P^{\prime},R_{1},R_{2},R_{3},R_{4},R^{\prime}\in\mathcal{H}(X)$, we have $\mathcal{A}(X)>\mathcal{A}(\{P^{\prime},R_{1},R_{2},R_{3},R_{4},R^{\prime}\})$ $=u(\frac{1}{2}-u)+\frac{1}{2}(\frac{1}{2}-u)((y_{2}+\frac{\sqrt{3}}{18})+r-u)>0.0975$, and the claim follows.

Now, assume that $y_{2}\geq 0.13$. Let $C_{1}$ be the same circle as above, see Figure 2, and let $P^{\prime}$ be a point on $C_{1}$ with coordinates $(x_{2},y_{2}+\frac{\sqrt{3}}{18})$. Then $P^{\prime}\in T$, and $f(z)>0.0975$ by the claim.

The cases $x_{2}\leq-0.17$ and $y_{2}\leq-0.13$ are considered similarly. ∎

Corollary 5**.**

Either $f(z)>0.0975$, or $F\cup T\cup R$ is a subset of a rectangle with side lengths $0.386\times 0.644$.

Proof.

By Lemma 4, we can assume that $z=(x_{1},y_{1},x_{2},y_{2},\theta)\in Z$. Let $Y_{1}$ and $Y_{2}$ be the points of configuration $X=F\cup T\cup R$ with the lowest and highest $y$-coordinates $y^{*}_{1}$ and $y^{*}_{2}$, respectively. Because $0\leq y_{1}\leq 0.04$, $Y_{1}$ and $Y_{2}$ cannot belong to the rectangle $R$. If they both belong to $F$, then $y^{*}_{2}-y^{*}_{1}=\frac{1}{\pi}<0.386$. If they both belong to the triangle, then $y^{*}_{2}-y^{*}_{1}\leq\frac{1}{3}<0.386$. If $Y_{1}\in F$ and $Y_{2}\in T$, then $y^{*}_{1}=-\frac{1}{2\pi}$, and the inequality $y^{*}_{2}\leq 0.13+\frac{\sqrt{3}}{18}$ follows form the Claim 1 in the proof of Lemma 4. Then $y^{*}_{2}-y^{*}_{1}\leq 0.13+\frac{\sqrt{3}}{18}+\frac{1}{2\pi}<0.386$.

Similarly, Let $X_{1}$ and $X_{2}$ be the points of configuration $X=F\cup T\cup R$ with the lowest and highest $x$-coordinates $x^{*}_{1}$ and $x^{*}_{2}$, respectively. $z\in Z$ implies that neither $X_{1}$ nor $X_{2}$ belongs to $F$. If $X_{1}\in T$ and $X_{2}\in R$, then, by Lemma 4 $x^{*}_{1}\geq-0.17-\frac{\sqrt{3}}{9}$, and $x^{*}_{2}\leq 0.05+\frac{0.4625}{2}$, hence $x^{*}_{2}-x^{*}_{1}\leq 0.05+\frac{0.4625}{2}+0.17+\frac{\sqrt{3}}{9}<0.644$. ∎

The following lemma established Lipschitz continuity of $f$ in $Z$.

Lemma 6**.**

For every $(x_{1},y_{1},x_{2},y_{2},\theta)\in Z$, and any $\epsilon_{i}\geq 0$, $i=1,\dots,5$,

[TABLE]

with constants $C_{1}=0.212$, $C_{2}=0.322$ , $C_{3}=0.326$ , $C_{4}=0.398$, and $C_{5}=0.134$.

Proof.

If function $g:{\mathbb{R}}\to{\mathbb{R}}$ is convex on ${\mathbb{R}}$ and

[TABLE]

then

[TABLE]

Indeed, inequality $C^{\prime}:=(g(t_{0}+\epsilon)-g(t_{0}))/\epsilon>C$ for some $t_{0}$ and $\epsilon$, would, by convexity of $g$, imply that $g(t_{0}+2\epsilon)>g(t_{0})+2C^{\prime}\epsilon$, and, by induction, $g(t_{0}+2^{n}\epsilon)>g(t_{0})+2^{n}C^{\prime}\epsilon$, a contradiction with $\lim\limits_{t\to+\infty}\frac{g(t)}{t}\leq C<C^{\prime}$. Inequality $(g(t_{0}+\epsilon)-g(t_{0}))/\epsilon<-C$ leads to a contradiction for similar reasons.

Let us apply this result to convex function $g(x_{1})=f(x_{1},y_{1},x_{2},y_{2},\theta)$, where $y_{1},x_{2},y_{2},\theta$ are fixed. In this case,

[TABLE]

see Figure 4.

Hence,

[TABLE]

Similarly, with $g(x_{2})=f(x_{1},y_{1},x_{2},y_{2},\theta)$ for fixed $x_{1},y_{1},y_{2},\theta$,

[TABLE]

see Figure 5, while with $g(y_{2})=f(x_{1},y_{1},x_{2},y_{2},\theta)$,

[TABLE]

see Figure 6. This implies that

[TABLE]

and

[TABLE]

The proof of similar bounds for the second and the fifth coordinates requires a different approach. For the second coordinate, we need to prove that

[TABLE]

We claim that it is sufficient to prove (1) for $\epsilon_{2}\in(0,\epsilon)$ for some $\epsilon>0$, which can depend on $x_{1},y_{1},x_{2},y_{2}$, and $\theta$. Indeed, assume, by contradiction, that, for some $y_{1}$, (1) holds for $\epsilon_{2}\in(0,\epsilon)$ but not for all $\epsilon_{2}>0$. Let $\epsilon^{*}$ be the supremum of all $\epsilon$ such that (1) holds for $\epsilon_{2}\in(0,\epsilon)$. Then, by continuity of $f$, (1) also holds for $\epsilon_{2}=\epsilon^{*}$, that is,

[TABLE]

Applying (1) to $y^{\prime}_{1}=y_{1}+\epsilon^{*}$, we find that

[TABLE]

holds for all $\delta_{2}\in(0,\delta)$ for some $\delta>0$. But the last two inequalities imply that (1) holds for all $\epsilon_{2}\in(0,\epsilon^{*}+\delta)$, a contradiction with the definition of $\epsilon^{*}$.

We next prove (1) for $\epsilon_{2}\in(0,\epsilon)$. Let $R^{\prime}$ with vertices $R^{\prime}_{1}R^{\prime}_{2}R^{\prime}_{3}R^{\prime}_{4}$ be the rectangle $R$ which moved up by $\epsilon_{2}$ in Y-axis’s direction. Convex hulls $\mathcal{H}(R,F,T)$ and $\mathcal{H}(R^{\prime},F,T)$ are polygons, and, by selecting $\epsilon$ sufficiently small, we can assume that all vertices of these polygons, which are not vertices of $R$ and $R^{\prime}$, coincides. Then $\mathcal{A}(R^{\prime},F,T)-\mathcal{A}(R,F,T)$ is bounded by the total area of three triangles, say $R^{\prime}_{1}R_{1}D_{1},R^{\prime}_{2}R_{2}D_{2}$, and $R^{\prime}_{2}R_{2}T^{*}$ which $D_{1},D_{2}\in F$ and $T^{*}\in T$ see Figure 7. Let $h_{1},h_{2},h_{3}$ be the height of $R^{\prime}_{1}R_{1}D_{1},R^{\prime}_{2}R_{2}D_{2}$, $R^{\prime}_{2}R_{2}T^{*}$, respectively. By Corollary 5, $|f(x_{1},y_{1}+\epsilon_{2},x_{2},y_{2},\theta)-f(x_{1},y_{1},x_{2},y_{2},\theta)|\leq\frac{1}{2}\epsilon_{2}(h_{1}+h_{2}+h_{3})\leq\frac{1}{2}\epsilon_{2}(0.644)=0.322\epsilon_{2}=C_{2}\epsilon_{2}$.

Finally, we need to prove that

[TABLE]

To prove the bound for $C_{5}$, we need the following claim.

Claim 2. The diameter $d(\mathcal{F\cup R})$ of $\mathcal{F\cup R}$ is less than $0.46402$.

Indeed, let $R_{0}=(0.05,0.04)$. We get $R_{2}=(0.28125,0.05875)$. Let $F_{1}\in F$ be a point which $d(x,R_{2})$ is maximum for all $x\in F$ see Figure 8. By direct calculation, $R_{2}F_{1}=0.4465<|R_{2}R_{4}|$. Hence, the diameter of $\mathcal{F\cup R}$ is $|R_{2}R_{4}|<0.46402$.

Next, we will prove (2).

Let $T^{\prime}$ with vertices $T^{\prime}_{1},T^{\prime}_{2},T^{\prime}_{3}$ be the triangle $T$ rotated around $T_{0}$ by angle $\epsilon_{5}$. Then $|T_{1}T^{\prime}_{1}|=2|T_{0}T_{1}|\sin(\epsilon_{5}/2)<2|T_{0}T_{1}|(\epsilon_{5}/2)=|T_{0}T_{1}|\epsilon_{5}=\frac{\sqrt{3}}{9}\epsilon_{5}$. Similarly, $|T_{2}T^{\prime}_{2}|=|T_{3}T^{\prime}_{3}|<\frac{\sqrt{3}}{9}\epsilon_{5}$.

By selecting $\epsilon_{5}$ sufficiently small, we can ensure that all vertices of polygons $\mathcal{H}(R,F,T)$ and $\mathcal{H}(R,F,T^{\prime})$ coincides, except possibly the vertices of $T$ and $T^{\prime}$.

We assume that no vertices of the triangle $T$ are adjacent in the convex hull of $\mathcal{H}(R,F,T)$. Let $X_{1}$ and $X_{2}$ be the vertices of $\mathcal{H}(R,F,T)$ adjacent to $T_{1}$, $X_{3}$ and $X_{4}$ - vertices of $\mathcal{H}(R,F,T)$ adjacent to $T_{2}$, and $X_{5}$ and $X_{6}$ - vertices of $\mathcal{H}(R,F,T)$ adjacent to $T_{3}$, see Figure 9. Let us denote $S(ABC)$ the area of any triangle $ABC$.

Then area difference $|\mathcal{A}(R,F,T^{\prime})-\mathcal{A}(R,F,T)|$ is equal to

[TABLE]

But $|(S(T_{1}X_{1}X_{2})-S(T^{\prime}_{1}X_{1}X_{2})|=|\frac{1}{2}h_{1}|X_{1}X_{2}|-\frac{1}{2}h_{2}|X_{1}X_{2}||=\frac{1}{2}|X_{1}X_{2}|\cdot|h_{1}-h_{2}|$, where $h_{1}$ and $h_{2}$ are heights of triangles $T_{1}X_{1}X_{2}$ and $T^{\prime}_{1}X_{1}X_{2}$, respectively, see Figure 10. But $|X_{1}X_{2}|<0.46402$ by Claim 2, and $|h_{1}-h_{2}|\leq|T_{1}T^{\prime}_{1}|<\frac{\sqrt{3}}{9}\epsilon_{5}$, hence $|(S(T_{1}X_{1}X_{2})-S(T^{\prime}_{1}X_{1}X_{2})|<0.46402\cdot\frac{\sqrt{3}}{18}\epsilon_{5}$. The same bound holds for $|(S(T_{2}X_{3}X_{4})-S(T^{\prime}_{2}X_{3}X_{4})|$ and $|(S(T_{3}X_{5}X_{6})-S(T^{\prime}_{3}X_{5}X_{6})|$. Hence,

[TABLE]

∎

3. Computational results

In this section, we use box-search algorithm to prove that the minimal value of function $f(z)=f(x_{1},y_{1},x_{2},y_{2},\theta)$ in region $Z$ defined in Lemma 4 is grater than $0.0975$.

In general, let $B$ be a box $[a_{1},b_{1}]\times[a_{2},b_{2}]\times[a_{3},b_{3}]\times[a_{4},b_{4}]\times[a_{5},b_{5}]$ (that is, set of points $z=(x_{1},y_{1},x_{2},y_{2},\theta)$ such that $a_{1}\leq x_{1}\leq b_{1}$, $a_{2}\leq y_{1}\leq b_{2}$, $a_{3}\leq x_{2}\leq b_{3}$, $a_{4}\leq y_{2}\leq b_{4}$, $a_{5}\leq\theta\leq b_{5}$. Let $z^{*}=(\frac{a_{1}+b_{1}}{2},\frac{a_{2}+b_{2}}{2},\frac{a_{3}+b_{3}}{2},\frac{a_{4}+b_{4}}{2},\frac{a_{5}+b_{5}}{2})$ be the center of the box). Then, if

[TABLE]

where $d_{i}=\frac{b_{i}-a_{i}}{2},\,i=1,2,3,4,5$, then, by Lemma 6, $f(z)\geq 0.0975$ for all $z\in B$.

If the condition (3) does not hold for $B$, we will divide $B$ into two sub-boxes $B_{1}$ and $B_{2}$, by dividing its maximal edge by half. For example, if $b_{2}-a_{2}\geq b_{i}-a_{i}$, $i=1,3,4,5$, then edge with length $b_{2}-a_{2}$ is the maximal one, and we divide $B$ into $B_{1}=[a_{1},b_{1}]\times[a_{2},(a_{2}+b_{2})/2]\times[a_{3},b_{3}]\times[a_{4},b_{4}]\times[a_{5},b_{5}]$ and $B_{2}=[a_{1},b_{1}]\times[(a_{2}+b_{2})/2,b_{2}]\times[a_{3},b_{3}]\times[a_{4},b_{4}]\times[a_{5},b_{5}]$.

We then check (3) for $B_{1}$ and $B_{2}$. If it holds in both cases, then $f(z)\geq 0.0975$ for all $z\in B_{1}$ and for all $z\in B_{2}$, hence $f(z)\geq 0.0975$ for all $z\in B$. If (3) does not hold for $B_{1}$ (or for $B_{2}$, of for both), we divide the corresponding box by two sub-boxes, and proceed iteratively.

$a_{1}=0,b_{1}=0.05,a_{2}=0,b_{2}=0.04,a_{3}=-0.17,b_{3}=0.17,a_{4}=-0.13,b_{4}=0.13,a_{5}=0,b_{5}=\frac{2\pi}{3}$. Then we evaluate $f$ in the box center $z^{*}=(0.025,0.02,0,0,\pi/3)$ to check whether (3) holds. In this case, (3) reduces to

[TABLE]

or equivalently, to $f(z^{*})>0.3567$. However, the computation show that $f(z^{*})\approx 0.10605<0.3567$, hence (3) does not hold. Hence, we need to subdivide $B$ into $B_{1}$ and $B_{2}$. In this case, $b_{1}-a_{1}=0.05$, $b_{2}-a_{2}=0.04$, $b_{3}-a_{3}=0.34$, $b_{4}-a_{4}=0.26$, and $b_{5}-a_{5}=2\pi/3\approx 1.047$. Hence, $b_{5}-a_{5}>b_{i}-a_{i}$, $i=1,2,3,4$, and we divide $B$ into $B_{1}=[a_{1},b_{1}]\times[a_{1},b_{2}]\times[a_{3},b_{3}]\times[a_{4},b_{4}]\times[a_{5},(a_{5}+b_{5})/2]$ and $B_{2}=[a_{1},b_{1}]\times[a_{1},b_{2}]\times[a_{3},b_{3}]\times[a_{4},b_{4}]\times[(a_{5}+b_{5})/2,b_{5}]$. Then we repeat the above procedure for $B_{1}$ and for $B_{2}$, and proceed iteratively.

We use Matlab$\circledR$ R2016a to implement this algorithm, see Algorithm 1. The actual Matlab code is presented in the Appendix.

The programme successfully verified the inequality $f(z)>0.0975$ for all $z\in Z$ after $n=7,180,439,126$ iterations. The program actually returned the minimal area $0.09762$ for the optimal configuration with $x_{1}=0.0251,\,y_{1}=0.00258,\,x_{2}=0.0653,\,y_{2}=0.00542,\,\theta=0.07989$ see Figure 11 and Table 14.

Repeating the calculation for this particular configuration in Mathematica with actual circle instead of $500$-gon shows that the optimal convex hull area is about

[TABLE]

4. Main Theorems

Theorem**.**

(Theorem 1 in the Introduction) Any convex set $S$ on the plane which can cover circle of perimeter $1$, equilateral triangle of perimeter $1$, and rectangle of size $0.0375\times 0.4625$ (and perimeter $1$) has area at least $0.0975$.

Proof.

By numerical results, $f(z)>0.0975$ for all $z\in Z$, where $Z$ is defined in Lemma 4. By Lemma 4, this implies that $f(z)>0.0975$ for all $z\in{\mathbb{R}}^{5}$, hence $\mathcal{A}(F,R,T)>0.0975$. Because $500$-gon $F$ is the subset of circle $C$, $\mathcal{A}(C,R,T)\geq\mathcal{A}(F,R,T)>0.0975$. ∎

Corollary 7**.**

Any convex cover for closed unit curves has area of at least $0.0975$.

Proof.

Because every convex cover for closed unit curves should cover the circle $C$, equilateral triangle $T$, and rectangle $R$ of size $0.0375\times 0.4625$, the claim follows from Theorem 1. ∎

The following Theorem implies that this method (with circle, equilateral triangle, and rectangle of perimeter $1$) cannot be used to improve the lower bound in Corollary 7 beyond $0.09763$.

Theorem 8**.**

For any rectangle $R^{\prime}$ with perimeter $1$, there is a convex cover of $R^{\prime}$, $C$, and $T$ with area at most $0.09763$.

Proof.

Let $l,w$ be the length and width of rectangle $R^{\prime}$ such that $l+w=\frac{1}{2}$ and $w\in[0,0.25]$.

Let $F^{\prime}$ be the regular 500-gon inscribed into the circle with $r^{\prime}=\frac{sec(\frac{\pi}{500})}{2\pi}$. Then $C\subset F^{\prime}$, and $\mathcal{H}(X)=\mathcal{H}(R,C,T)\subset\mathcal{H}(R,F^{\prime},T)$. Thus, $\mathcal{A}(X)\leq\mathcal{A}(R,F^{\prime},T)$.

Let $f(w)$ denotes the minimal area of convex cover $R,F^{\prime},T$.

Claim For any $\epsilon>0$, $|f(w+\epsilon)-f(w)|\leq 0.318\epsilon$.

It suffices to prove the claim only for “small” $\epsilon$. We will prove that $f(w)-f(w+\epsilon)\leq 0.318\epsilon$, the proof for inequality $f(w+\epsilon)-f(w)\leq 0.318\epsilon$ is similar. Let $R^{\prime\prime}$ be the rectangle with width $w+\epsilon$ and perimeter $1$. Consider optimal configuration of $R^{\prime\prime},F^{\prime},T$, so that $f(\omega+\epsilon)=\mathcal{A}(R^{\prime\prime},F^{\prime},T)$. Let us put $R^{\prime}$ parallel to $R^{\prime\prime}$ as shown on Figure 12. This configuration is not necessary optimal, and, because $f$ denotes the area of the optimal configuration, $f(w)\leq\mathcal{A}(R^{\prime},F^{\prime},T)$. Hence, $f(w)-f(w+\epsilon)\leq\mathcal{A}(R^{\prime},F^{\prime},T)-\mathcal{A}(R^{\prime\prime},F^{\prime},T)$.

Convex hulls $\mathcal{H}(R^{\prime},F^{\prime},T)$ and $\mathcal{H}(R^{\prime\prime},F^{\prime},T)$ are polygons, and, by selecting $\epsilon$ sufficiently small, we can assume that all vertices of these polygons, which are not vertices of $R^{\prime}$ and $R^{\prime\prime}$, coincides. Then $\mathcal{A}(R^{\prime},F^{\prime},T)-\mathcal{A}(R^{\prime\prime},F^{\prime},T)$ is bounded by the total area of triangles $XQ_{1}R_{2}$, $YQ_{2}R_{3}$, and rectangle $Q_{1}R_{2}R_{3}Q_{2}$, which is

[TABLE]

We have $Q_{1}Q_{2}=w\leq 0.25$, and, by Corollary 5, $h_{1}+h_{2}+Q_{1}Q_{2}\leq 0.386$. Hence,

[TABLE]

which proves the claim.

To verify inequality $f(w)<0.09763$ at some specific point $w$, it is not necessary to find the optimal configuration of $R^{\prime}$, $F^{\prime}$, and $T$. In suffices just to find some configuration with $\mathcal{A}(R^{\prime},F^{\prime},T)<0.09763$, and then conclude that $f(w)\leq\mathcal{A}(R^{\prime},F^{\prime},T)<0.09763$. This makes the numerical verification simple.

We will verify inequality $f(w)<0.09763$ for $w$ belonging to some finite set $W=\{w_{1},w_{2},\dots,w_{N}\}$, where $0\leq w_{1}\leq w_{2}\leq\dots\leq w_{N}\leq 0.25$ are points to be specified below. By the claim, inequality $f(w_{i})<0.09763$ implies that $f(w)\leq 0.09763$ in the whole interval $w\in[w_{i}-d_{i},w_{i}+d_{i}]$, where $d_{i}=(0.09763-f(w_{i}))/0.318$.

We will select set $W$ in such a way that intervals $[w_{i}-d_{i},w_{i}+d_{i}]$, $i=1,2,\dots,N$ cover the whole interval $[0,0.25]$. In other words, $w_{1}-d_{1}<0$, $w_{N}+d_{N}>0.25$, and

[TABLE]

Set $W$ with $N=772$ points with this property is presented in the Appendix. For example, $w_{1}=0.00020$, $w_{2}=0.0034$, $w_{3}=0.0086$, and so on, $w_{772}=0.2415$.

∎

5. Conclusion and future work

Using the method proposed in [12], we improved the lower bound of convex covers for closed unit curves from $0.096694$ to $0.0975$. The method is based on studying configurations of the regular 500-gon, rectangle, and equilateral triangle. The use geometric methods to prove Lipschitz bound for the corresponding function, and then use numerical box-search algorithm to finish the proof. Based on the numerical results, we conjecture that $T_{1}$ and $R_{2}$ coincides in the optimal configuration, see Figure 13. Our numerical results actually imply lower bound $0.09762742$, corresponding to the optimal configuration with parameters $x_{1}=0.0255904,y_{1}=0.0013503,x_{2}=0.0653055,y_{2}=0.0050124$ and $\theta=0.0766554$. However, we can formally prove only the weaker bound $0.0975$, using the algorithm which require $n=7,180,439,126$ iterations. By using more powerful computers, the same methods can lead to, for example, bound $0.0976$. However, we have proved that one cannot achieve the bound $0.09763$ this way. An obvious way to improve the bound would be studying configurations with $4$ objects, but this would increase the dimension of the parameter space.

Appendix A Appendix

A.1. Box’s search method

Function $cvh(x_{1},x_{2},y_{1},y_{2},\alpha)$ below calculates the area of the convex hull for the configuration defined by parameters $x_{1},x_{2},y_{1},y_{2},\alpha$. Function $checkminNB4$ is used to check the main inequality 3 in a box domain defined by its parameters $a_{1},b_{1},\dots,a_{5},b_{5}$.

A.1.1. The area of convex hull fucntion

A.1.2. The box’s search method

A.1.3. The box’s search results

The Box search program displayed a message every $1,000,000$ steps. Figure 14 presents the output of these messages for (approximately) every $100,000,000$ steps. Here, the first column represents progress, in terms of the percentage of the area of the initial box for which the inequality 3 is verified. The second column is the iteration number. Figure 15 presents the graphical illustration how progress depends on the number of iterations.

When the program finished, it displayed the result:

r = 3.680103453346938e-04 (this is the area of the initial box, $100\%$ covered)

n = 7.180439126000000e+09 (total number of iterations needed)

min = 0.097626517574902 (the minimal area convex hull)

xx1 = 0.025097656250000

yy1 = 0.002578125000000

xx2 = 0.065327148437500

yy2 = 0.005422070312500

app = 0.079894832702377 (the 5 coordinates for the optimal configuration)

A.2. Matlab program in Theorem 8

Program cvhull2 below calculates the convex hull of the given configuration.

A.2.1. the area of convex hull fucntion

A.2.2. the main program

The function below finds set $W$ which satisfy the condition 4 in the proof of Theorem 8.

A.2.3. the results

Set $W$ satisfying the condition 4 in the proof of Theorem 8.

See pages - of maincodethm2