Universal convex covering problems under translations and discrete rotations

Mook Kwon Jung; Sang Duk Yoon; Hee-Kap Ahn; Takeshi Tokuyama

doi:10.1515/advgeom-2023-0021

Article Open Access

Universal convex covering problems under translations and discrete rotations

Mook Kwon Jung , Sang Duk Yoon , Hee-Kap Ahn and Takeshi Tokuyama

Published/Copyright: October 13, 2023

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From the journal Advances in Geometry Volume 23 Issue 4

Abstract

We consider the smallest-area universal covering of planar objects of perimeter 2 (or equivalently, closed curves of length 2) allowing translations and discrete rotations. In particular, we show that the solution is an equilateral triangle of height 1 when translations and discrete rotations of π are allowed. We also give convex coverings of closed curves of length 2 under translations and discrete rotations of multiples of π/2 and of 2π/3. We show that no proper closed subset of that covering is a covering for discrete rotations of multiples of π/2, which is an equilateral triangle of height smaller than 1, and conjecture that such a covering is the smallest-area convex covering. Finally, we give the smallest-area convex coverings of all unit segments under translations and discrete rotations of 2π/k for all integers k=3.

Keywords: Universal covering; Kakeya problem; discrete rotation

MSC 2010: 52C15; 52B55

1 Introduction

Given a (possibly infinite) set S of planar objects and a group G of geometric transformations, a G-covering K of S is a region such that every object in S can be included in K by transforming the object with a suitable transformation g ∈ G. Equivalently, every object of S is contained in g⁻¹K for a suitable transformation g ∈ G; that is, ∀γ∈S∃g∈G:gγ⊆K.

We denote the group of planar translations by T and that of planar translations and rotations by TR. Group-theoretically, TR=T⋊ R is the semidirect product of T and the rotation group R=SO(2,R). We often omit G and use coverings for G-coverings if the group G is known from the context.The problem of finding a smallest-area covering is a classical problem in mathematics, and such a covering is often called a universal covering. In the literature, the cases where G=T or G=TR have been widely studied.

The universal covering problem has attracted many mathematicians. Henri Lebesgue (in his letter to J. Pál in 1914) proposed the problem to find the smallest-area convex TR-covering of all objects of unit diameter; see [7,4,10] for the history. Kakeya considered in 1917 the T-covering of the set Sseg of all unit line segments (called needles); see [13]. Precisely, his formulation was to find the smallest-area region in which a unit-length needle can be turned around, but this is equivalent to the universal convex covering problem if the region is convex; see [3]. Originally, Kakeya considered convex coverings, and Fujiwara conjectured that the equilateral triangle of height 1 is the solution. This conjecture was proved affirmatively solved by Pál in 1920; see [17]. For the case where the region is not necessarily convex, Besicovitch [5] gave a construction such which shows that the area can be arbitrarily small.

Generalizing Pál’s result, for any set of n segments, there is a a triangle is the smallest-area convex T-covering of the set, and that triangle can be computed efficiently in O(nlogn) time; see [1]. It is further conjectured that the smallest-area convex TR-covering of a family of triangles is a triangle, which is shown to be true for some families in [19].

The problem of finding the smallest-area covering and convex covering of the set of all curves of unit length for G=TR was listed by Leo Moser as an open problem in 1966 [14]. The problem is still unsolved, and the best lower bound of the smallest area of the convex covering is 0.21946; see [22]. For the best upper bound, Wetzel informally conjectured in 1970 (and published in [23]) that the 30° circular fan of unit radius, which has the area π/12 ≈ 0.2618, is a convex TR-covering of all unit-length curves; this was proved by Panraksa and Wichiramala in [18]. Recently, the upper bound was improved to 0.260437 in [16], but there still remains a substantial gap between the lower and upper bounds.

However, when only translations are allowed, the equilateral triangle of height 1 is the smallest-area convex covering (Corollary 4), which is the same as in the case of considering the unit line segments. There has been no concrete proof of Corollary 4 in the literature. It can be confirmed analytically, since it suffices to consider a polyline consisting of two segments. Also, we can directly prove it by applying a reflecting argument (similar to the proof of Theorem 2) which is given in Appendix.

There are many variants of Moser’s problem, and they are called Moser’s worm problem. The history of progress on this topic can be found in the article [15] by William Moser (Leo’s younger brother), in Chapter D18 in [8], and in Chapter 11.4 in [7]. It is interesting to find a new case of Moser’s worm problem with a clean mathematical solution.

From now on, we focus on the convex G-covering of the set Sc of all closed curves γ of length 2. Here, We follow the tradition of previous works on this problem to consider length 2 instead of 1, since a unit line segment can be considered as a convex closed curve of length 2. Since the boundary curve of the convex hull C(γ) of any closed curve γ is not longer than γ, it suffices to consider only convex curves.

This problem is known to be an interesting but hard variant of Moser’s worm problem, and remains unsolved for T and TR despite of substantial efforts in the literature; see [9,23,20,8,7]. As far as the authors know, the smallest-area convex TR-covering known today is a hexagon obtained by clipping two corners of a rectangle, see given by Wichiramala [24], and its area is slightly less than 0.441. [24].By [11] the smallest area is at least 0.39, which has been recently improved to 0.4 in [12] using computer programs. The smallest area of the convex T-covering is known to be between 0.620 and 0.657 by [7].

There are some results on restricted shapes of covering. Especially, ifFor triangular coverings, Wetzel [20,21] gave a complete description, and he it is showed that an acute triangle with side lengths a, b, c and area X becomes a T-covering (respectively TR-covering) of Sc if and only if abc ⩽ 4X² (respectively a + b + c ⩽ π X). As a consequence, the equilateral triangle of side length 4/3 (respectively 23/π) is the smallest triangular T-covering (respectively TR-covering) of Sc. Unfortunately, the areas of these triangles are larger than those of the known smallest-area convex coverings.

If H is a subgroup of G, then an H-covering is a G-covering. Since T⊂ TR, it is quite reasonable to consider groups G lying in between, that is T ⊂ G⊂ TR. The groupR=SO(2,R) is abelian, and its finite subgroups are the groups Zk={e2iπ−1/k∣0≤i≤k−1} fork=1,2,…, where eθ−1 denotes the rotation of angle θ.

In this paper, we consider the coverings under the action of the group G_k=T⋊ Z_k. We show that the smallest-area convex G₂-covering of Sc is the equilateral triangle of height 1, whose area is 3/3≈0.577. A nice feature is that the proof is purely geometric and elementary, assuming Pal’s result mentioned above.Then we show that the equilateral triangle with height β= cos (π/12)≈0.966 is a G₄-covering of Sc. Its area is (23+3)/12≈0.538675, and we conjecture that it is the smallest-area convex G₄-covering.

However, the smallest-area convex G₃-covering is no longer an equilateral triangle. We give a convex G₃-covering of Sc which has area not larger than 0.568. The area of our G₃-covering is strictly smaller than the area of the G₂-covering Δ₁, and a bit larger than the area of the G₄-covering Δ_β. Our G₃-covering is not invariant under any nontrivial rotation, while Δ₁ and Δ_β are invariant under rotation by 2π/3. For triangles of perimeter 2, we give a smaller convex G₃-covering of them that has area 0.563.We also determine the set of all smallest-area convex G_k-coverings of Sseg, which are all triangles; see Table 1 for the results.

Table 1

A table listing our results

	All closed curves with length 2	All unit segments
G₂-covering	Equilateral triangle with height 1 (smallest area)	-
G₃-covering	Part of a church window	-
G₄-covering	Equilateral triangle with height cos(π/12) (minimal)	Isosceles right triangle with legs of length 1 (smallest area)
G_k-covering for odd k ≥ 3	-	Triangle with base length 1, height sin(π/k), and π/2 ≤ θ ≤ (k − 1)π/k for a corner angle θ incident to the base (smallest area)
G_k-covering for even k ≥ 4	-	Triangle with base length 1, height sin(2π/k), and π/2 ≤ θ' ≤ (k − 2)π/k for a corner angle θ' incident to the base (smallest area)

We denote by ℓ(C) the perimeter of a compact set C, and by ℓ(γ) the length of a curve γ. The slope of a line is the angle swept from the x-axis in a counter-clockwise direction to the line, and it is thus in [0, π). The slope of a segment is the slope of the line that extends it. For two points p and q, we denote by pq the line segment connecting p and q, and by |pq| the length of pq.We denote by ℓ_pq the line passing through p and q.

Covering under rotation by 180 degrees

In this section, we show that the smallest-area convex G₂-covering of the set Sc of all closed curves of length 2 is the equilateral triangle of height 1, denoted by Δ₁, whose area is 3/3.

2.1 The smallest-area covering and related results

We recall a famous result mentioned in the introduction.

Theorem 1

(Pal’s theorem for the convex Kakeya problem). The equilateral triangle Δ₁ is the smallest-area convex T-covering of the set of all unit line segments.

Corollary 1

The area of a convex G ₂-covering ofScis at least3/3.

Proof. Observe that all unit line segments are in Sc, and that line segments are stable under the action of rotation by π. Thus any convex G₂-covering of Sc must be a T-covering of all unit line segments, and the corollary follows from Theorem 1 (Pal’s theorem). □

By Corollary 1, it suffices to show that any closed curve γ in Sc can be included in Δ₁ by applying an action of G₂ to prove the following main theorem in this section.

Theorem 2

The equilateral triangle Δ₁is the smallest-area convexG₂-covering ofSc.

Before proving the theorem, we give some direct implications of it. An object P is centrally symmetric with respect to the origin if −P=P, where −P={−x∣ x ∈ P}. Let Ssym be the set of all centrally symmetric closed curves of length 2.

Corollary 2

The equilateral triangle Δ₁is the smallest-area convexT-covering ofSsym.

Proof. Ssym contains all unit segments. Thus by from Theorem 1 (Pal’s theorem), the smallest-area convex T-covering of Ssym has area at least the area of Δ₁ as in the proof of Corollary 1. On the other hand, Ssym⊂Sc and by Theorem 2 any curve in Ssym can be included in Δ₁ by applying a suitable transformation in G₂. However, a centrally symmetric object is stable under the action of Z₂, and hence it can be included in Δ₁ by applying a transformation in T. Thus Δ₁ is the smallest-area convex T-covering of Ssym. □

We can consider two special cases shown below. Corollary 2 implies that Δ₁ is a T-covering of all rectangles of perimeter 2, and also of all parallelograms of perimeter 2. Since a unit line segment is a degenerate rectangle and a degenerate parallelogram of perimeter 2, we have the following corollary.

Corollary 3

The equilateral triangle Δ₁ is the smallest-area convex T-covering of the set of all parallelograms of perimeter 2. This also holds for the set of all rectangles of perimeter 2.

We also obtain the following well-known result about the covering of the set Sworm of all curves of length 1 (often called worms) mentioned in the introduction as a corollary.

Corollary 4

The equilateral triangle Δ₁is the smallest-area convexT-covering ofSworm.

Proof. Given a curve ζ in Sworm, we consider the copy ζ′ obtained by rotating ζ around the midpoint of the two endpoints by the angle π. Then γ(ζ)=ζ∪ζ′ is a closed curve. We have γ(ζ)∈Ssym, and it can be included in Δ₁ by a translation. Therefore, ζ is also can be included there. The corollary follows from Corollary 2. □

We can prove Corollary 4 directly by applying a reflecting argument (similar to the proof of Theorem 2) which is given in Appendix.

Figure 1

Tiling argument. (a) H_γ=L₀∩ L_π/3∩ L_2π/3. (b) A tiling of six copies H_γ(1), ⋅, H_γ(6) of H_γ. Any closed curve that touches every side of H_γ is not shorter than the dashed (blue) segment of length at least d₀+d_π/3+d_2π/3.

2.2 Proof of Theorem 2

A slab is the region bounded by two parallel lines in the plane; its width is the distance between the two bounding lines. For a closed curve γ of Sc, let L_θ for 0 ⩽ θ<π be the minimum-width slab of orientation θ containing γ. We denote the width of L_θ by d_θ.

Lemma 1

For a closed curve γ of S c we have d₀+d_π/3+d_2π/3⩽2 for slabs L₀, L_π/3 and L_2π/3 of γ.

Proof. Let H_γ be the hexagon which is the intersection of L₀, L_π/3 and L_2π/3 of γ; see Figure 1(a). Let e₁, ⋅, e₆ be the edges of H_γ in counter-clockwise order. We can obtain a tiling of six copies H_γ(1), ⋅, H_γ(6) of H_γ such that H_γ(1)=H_γ and H_γ(k + 1) is the copy of H_γ(k) reflected about e_k of H_γ(k) for k=1, ⋅, 5 from 1 to 5 as shown in Figure 1(b). We observe that the length of a closed curve that touches every side of H_γ is at least d₀+d_π/3+d_2π/3; see Figure 1(b). Since γ touches every side of H_γ and the length of γ is 2 by the definition ofSc, the lemma follows. □

Figure 2

(a) P_γ=L_π/3∩ L_2π/3, andh=d_π/3+d_2π/3. (b) h+d=2.

Lemma 2

Any closed curve γ of Sccan be included in Δ₁or in −Δ₁under translation.

Proof. Let P_γ be the parallelogram which is the intersection of L_π/3 and L_2π/3 of γ. The height h of P_γ is h=d_π/3+d_2π/3; see Figure 2(a). If h ⩽ 1, then P_γ can be included in both Δ₁ and −Δ₁ under translation.

For h > 1 we consider two horizontal lines ℓ_t and ℓ_b such that they intersect P_γ, and ℓ_t lies above the bottom corner of P_γ at distance 1 and ℓ_b lies below the top corner of P_γ at distance 1. If d is the distance between ℓ_t and ℓ_b, then d+h=2; equals 2 (see Figure 2(b). By Lemma 1 we have d₀+d_π/3+d_2π/3=d₀+h ⩽ 2, so d=d₀. This implies that ℓ_b or ℓ_t is not contained in the interior of L₀. If ℓ_b lies below the interior of L₀, then γ can be included in Δ₁ under translation. If ℓ_t lies above the interior of L₀, then γ can be included in −Δ₁ under translation. □

For a closed curve γ, either γ or −γ can be included in Δ₁ under translation by Lemma 2, so we obtain Theorem 2.

3 Covering under rotation by 90 degrees

In this section we consider the convex G₄-coverings of the setsSseg and Sc where G₄=T⋊ Z₄. We show that the smallest-area convex G₄-covering of Sseg is the isosceles right triangle with legs (the two equal-length sides) of length 1. A G-covering K of S is minimal if no proper closed subset of K is a G-covering of S. For Sc we give an equilateral triangle of height slightly smaller than 1 as a minimal convex G₄-covering.

3.1 Covering of unit line segments

First, we consider the set Sseg$ of all unit segments.

Theorem 3

The smallest-area convex G ₄-covering of all unit segments has area 1/2, and the unique compact convexG₄-covering of all unit segments with smallest area is the isosceles right triangle with legs of length 1.

Proof. Let Δ be the isosceles right triangle with legs of length 1 and base of slope π/4. Any unit segment of slope θ can be placed in Δ for 0 ⩽ θ ⩽ π/2. Thus, Δ is a G₄-covering of Sseg, and its area is 1/2.

Now we show that any convex G₄-covering of Sseg has area at least 1/2. Let X be a smallest-area convex G₄-covering ofSseg, and let s(θ) be a unit segment of slope θ. Let A be the set of angles θ such that s(θ) can be placed in X under translation with 0 ⩽ θ <π. Let A⁻=A∩[0, π/2) and A⁺={θ−π/2∣θ ∈ A∩[π/2, π)}. If A⁻∩ A⁺≠∅, then X contains two unit segments which are orthogonal to each other and also contains their convex hull. Thus the area of X is at least 1/2. So we assume that A⁻∩ A⁺=∅. Since X is a G₄-covering of Sseg, A⁻∪ A⁺=[0, π/2). If A⁻=∅, then A⁺=[0, π/2) and there is a sequence {θk}k=1∞ in A⁺ such that lim_{k → ∞}θ_k=π/2. Since X contains a translated copy of s(π/2) and a translated copy of s(θ_k+π/2) for any k, X contains their convex hull. Since lim_{k → ∞}θ_k=π/2, the area of X is at least 1/2. Similarly, we can prove this for the case A⁺=∅. So we assume that A⁻≠∅ and A⁺≠∅. Then there are two cases : A⁻ contains no interval or A⁻ contains an interval.

Consider the case that A⁻ contains no interval. Let I(x, ε) be the open interval in [0, π/2) centered at an angle x with radius ε, and let θ′ be an angle in A⁻. Observe that I(θ′, 1/k) ∩ A⁺≠∅ since A⁻ contains no interval and A⁻∪ A⁺=[0, π/2). So there is a sequence{θk′}k=1∞ such that θ′_k ∈ I(θ′, 1/k) ∩ A⁺ for each k, and lim_{k → ∞}θ′_k=θ′. Since X contains a translated copy of s(θ′) and a translated copy of s(θ′_k+π/2) for any k, their convex hull is contained in X. Thus the area of X is at least 1/2.

Consider now the case that A⁻ contains an interval. Since A⁺≠∅, there is an interval in A⁻ with an endpoint θˉ other than 0 and π/2. Then there is a sequence {θˉk}k=1∞ in A⁻ such thatlimk→∞θˉk=θˉ. Sinceθˉ is an endpoint of the interval in A⁻,I(θˉ,1/n)∩A+≠∅ for any n. So there is a sequence {θˉn′}n=1∞ such thatθˉn′∈I(θˉ,1/n)∩A+ for each n, and limn→∞θˉn′=θˉ. Since X contains a translated copy of s(θˉk) and a translated copy of s(θˉn′+π/2) for any k and any n, X contains their convex hull. Thus the area of X is at least 1/2.

We show the uniqueness of the smallest-area compact convex G₄-covering of Sseg. We assume that X is compact. SinceΔ is a G₄-covering of Sseg with area 1/2, X is a G₄-covering of Sseg with area 1/2. First, we prove that if there is a sequence of angles in A converging to θ ∈ [0, π), then X contains a translated copy of s(θ). Consider a sequence {θk}k=1∞ in A such that lim_{k → ∞}θ_k=θ. Then there is a sequence {s(θk)}k=1∞ of unit segments s(θ_k) contained in X. Since X is compact, there is a subsequence {s(θkp)}p=1∞ converging to a unit segment s(θ). Since X is closed, s(θ) is contained in X. By the argument in the previous paragraphs, there is an angle θ˜∈[0,π/2) such that X contains a translated copy of s(θ˜) and a translated copy ofs(θ˜+π/2), orthogonal to each other. Since X is a covering of area 1/2, X is the convex hull of the two segments. There are two cases for X: it is either a convex quadrilateral or a triangle of height 1 and base length 1.

Let X be a convex quadrilateral. Then bothθ˜ and θ˜+π/2 are isolated points in A, hence X is not a G₄-covering of Sseg. Consider the case that X is a triangle of height 1 and base length 1. Without loss of generality, we assume that s(θ˜) is the base of X and s(θ˜+π/2) corresponds to the height of X. Note Observe that the corner opposite to the base of X is acute. If it is not acute, the base has length strictly larger than 1, a contradiction. Suppose that both corners of X incident to the base are acute. Since s(θ˜) is the base of X, θ˜ is an isolated point in A. Note Observe thats(θ˜+π/2) can be rotated infinitesimally around the corner opposite to the base in both directions while it is still contained in X. Thus, θ˜+π/2 is an interior point of an interval I in A. So for an endpoint θˆ of I other than θ˜, X contains a translated copy ofs(θˆ) and a translated copy of s(θˆ+π/2). The convex hull X′ of the two segments is also a convex quadrilateral or a non-obtuse triangle of area at least 1/2. Since X′ is contained in X of area 1/2, X′=X, that is, X′ is must be an acute triangle with base b of length 1. Sinceθˆ≠θ˜, b is not the base of X, b must be one of the other two sides of X. But clearly both these sides are longer than 1, the height of X, a contradiction. Hence X is not an acute triangle. Thus X is an isosceles right triangle with legs of length 1, and it is the unique compact convex G₄-covering of Sseg. □

3.2 An equilateral triangle covering of closed curves

Unfortunately, the isosceles right triangle with legs of length 1 is not a G₄-coverings of Sc, since it does not contain a circle of perimeter 2. Using the convexity of the area function on the convex hull of two translated convex objects, see [2], it can be shown that the convex hull of an isosceles right triangle and a circle of perimeter 2 has area at least 0.543.

Naturally, the equilateral triangle of height 1 is a G₄-covering of Sc, since it is a G₂-covering of Sc. However, a smaller equilateral triangle can be a G₄-covering of Sc, and we seek for the smallest one, which is conjectured to be the smallest-area G₄-covering of Sc. Consider an equilateral triangle that is a G₄-covering of Sc. Since it is a G₄-covering of Sseg, it must contain a pair of orthogonally crossing unit segments as shown in the proof of Theorem 3. Figure 3 shows a smallest equilateral triangle containing such a pair. The side length of the triangle is 1+36≈1.115 and the height isβ=cos(π/12)=1+322≈0.966. Its area is 23+312≈0.538675. We denote such a triangle with a horizontal base and centered at the origin by Δ_β.

Figure 3

The smallest equilateral triangle Δ_β containing a pair of orthogonal unit segments

Theorem 4

The equilateral triangle Δ_βis a convexG₄-covering of all closed curves of length 2.

Definition 1

A closed curve γ is called G-brimful for a set Λ if it can be placed in Λ but cannot be placed in any smaller scaled copy of Λ under G transformation.

To prove Theorem 4, we consider the following claim.

Claim 1

The length of any G ₄-brimful curve for Δ_βis at least 2.

The claim implies that any closed curve of length smaller than 2 can be placed in Δ_β under G₄ transformation, but it is not G₄-brimful for Δ_β. Thus, if the claim is true, then every closed curve of length 2 can be placed in Δ_β under G₄ transformation, so Theorem 4 holds. Thus, we will prove the claim.

Let γ be a G₄-brimful curve for Δ_β. Letgi=e2iπ−1/4∈Z4 for i=0, ⋅, 3. We consider the rotated copy gⁱΔ_β of Δ_β, and find the smallest scaled copy ϒ_i of gⁱΔ_β which circumscribes γ with a proper translation. Let a_i be the scaling factor; the brimful property implies that a_i⩾1 for i=0, 1, 2, 3 and the minimum of them equals 1. Also, γ touches all three sides of ϒ_i for each i, where we allow it to touch two sides simultaneously at the vertex shared by them.

As illustrated in Figure 4(a), the intersectionH=H(γ)=⋂i=03Υi contains γ, and H is a (possibly degenerate) convex 12-gon such that γ touches all 12 edges of H. The 12-gon can be degenerate as shown in Figure 4(b), where γ touches multiple edges simultaneously at vertices corresponding to degenerate edges. Note that H consists of edges of slopes 2kπ/12modπ for k=0, ⋅, 11.The following lemma shows that the perimeter of H only depends on a_i(i=0, 1, 2, 3).

$Figure 4 (a) H=⋂i=03Υi$ H=\bigcap_{i=0}^3 \Upsilon_i $ is a (possibly degenerate) convex 12-gon, and γ touches every edge of H. (b) An example of a degenerate convex 12-gon. The degenerate edges are denoted by black disks.(c) For X0,2=ϒ0∩ϒ2 (red) and X1,3=ϒ1∩ϒ3 (blue), the set (X0,2∪ X1,3)∖ H consists of 12 triangles (gray), each of which is an isosceles triangle with apex angle 2π/3.$

Figure 4

(a) H=⋂i=03Υi is a (possibly degenerate) convex 12-gon, and γ touches every edge of H. (b) An example of a degenerate convex 12-gon. The degenerate edges are denoted by black disks.(c) For X_0,2=ϒ₀∩ϒ₂ (red) and X_1,3=ϒ₁∩ϒ₃ (blue), the set (X_0,2∪ X_1,3)∖ H consists of 12 triangles (gray), each of which is an isosceles triangle with apex angle 2π/3.

Lemma 3

The perimeter ℓ(H) equalsa0+a1+a2+a32cos(π/12).

Proof. Let δ be the side length of Δ_β. The intersection X_0,2=ϒ₀∩ϒ₂ is a hexagon where γ touches all of its six (possibly degenerate) edges. Then, (ϒ₀∪ϒ₂)∖ X_0,2 consists of six equilateral triangles, and the total sum of the perimeters of them is ℓ(ϒ₀) + ℓ(ϒ₂)=3(a₀ + a₂) δ. On the other hand, one edge of each equilateral triangle contributes to the boundary polygon of X_0,2, hence ℓ(X_0,2)=(a₀ + a₂)δ. Similarly, the perimeter of X_1,3=ϒ₁∩ϒ₃ is (a₁ + a₃)δ.

Let H=⋂i=03Υi=X0,2∩X1,3 ; see Figure 4(c). The set (X_0,2∪ X_1,3)∖ H consists of 12 triangles, each of which is an isosceles triangle with apex angle 2π/3. The total sum of the perimeters of these triangles is equals ℓ(X_0,2) + ℓ(X_1,3), while the bottom side of each isosceles triangle contributes to the 12-gon H. Since the ratio of the bottom side length to the perimeter of the isosceles triangle is 32+3, we have

ℓ ( H ) = 3 2 + 3 ( ℓ ( X 0 , 2 ) + ℓ ( X 1 , 3 ) ) = 3 2 + 3 ( a 0 + a 1 + a 2 + a 3 ) δ = 2 2 + 3 cos ⁡ ( π / 12 ) ( a 0 + a 1 + a 2 + a 3 ) = a 0 + a 1 + a 2 + a 3 2 cos ⁡ ( π / 12 ) .

The third equality comes fromδ=23β=23cos(π/12), and the last equality comes fromcos(π/12)=1+322 and hencecos2(π/12)=2+34. □

The following lemma states the relation between the lengths of γ and H.

Figure 5

Illustration of the proof of Lemma 4

Lemma 4

Let P be a (possibly degenerate) convex 12-gon with edges e _k of slopes 2 k π / 12 m o d π for k=0, ⋅, 11. LetCbe a convex 12-gon such that one vertex lies on each edge ofP. Then ℓ(C) ⩾ ℓ(P) cos (π/12).

Proof. Let C be a convex 12-gon such that one vertex lies on each edge of P, and let Let p_k be the vertex of C on e_k for k=0, ⋅, 11. The boundary of C connects the points from p₀, ⋅, p₁₂ in counter-clockwise order along the boundary of P, where p₁₂ is a duplicate of p₀. We also assume that p₁₂ is on e₁₂ which is a duplicate of e₀. For k=1, ⋅, 11 we reflect the edges e_k+1, ⋅, e₁₂ about the edge e_k. Then, the edges e₀, …,e₁₂ are transformed to a zigzag path alternating between a horizontal edge and an edge of slope 2π/12, and C becomes the path connecting p₀, p₁, p′₂, ⋅, p′₁₂ where p′_k is the location of p_k after the series of reflections for k=2, ⋅, 12. Thus, ℓ(C) is at least the distance between p₀ to p′₁₂. See Figure 5 for an illustration. The vector x=p₀p′₁₂ is the sum of a horizontal vector a and another vector b of slope π/6 such that ∣a∣ + ∣b∣=ℓ(P). The size ∣x∣ of x is minimized when ∣a∣=∣b∣ attains the value (∣a∣+∣b∣) cos (π/12). Thus ∣x∣ ⩾ ℓ(P) cos (π/12) and ℓ(C) ⩾ ℓ(P) cos (π/12). □

From Lemma 4, the following corollary is immediate.

Corollary 5

For any G ₄-brimful curve γ for Δ_β, we have ℓ(γ) ⩾ ℓ(H) cos (π/12) whereH=H(γ).

Lemma 3 and Corollary 5 imply we have ℓ(γ) ⩾ ℓ(H) cos (π/12)=(a₀+ a₁ + a₂+ a₃)/2=2, which proves Claim 1. . The claim is proved.

3.3 Minimality of the covering

Figure 6

Examples of G₄-brimful curves of length 2

There are many G₄-brimful curves of length 2, and some examples are illustrated as red curves (the line segment is considered as a degenerate closed curve) in Figure 6 together with the (possibly degenerate) 12-gons they inscribe. Based on them, we conjecture that Δ_β is the smallest-area convex G₄-covering of S_C.

Conjecture 1

Δ_βis the smallest-area convexG₄-covering ofS_C.

Although we have not yet proven the conjecture rigorously, we can prove that Δ_β is minimal.Recall that Δ_β has a horizontal base. Let S be the set of unit line segments that can be included in Δ_β under translation (without considering the Z₄-action). Then A=[0, π/12]∪[3π/12, 5π/12]∪[7π/12, 9π/12]∪[11π/12, π] is the set of slope angles of the line segments of S. Let A′=(0, π/12)∪(3π/12, 5π/12)∪(7π/12, 9π/12)∪(11π/12, π) and let S_A′ be the set of all unit line segments of slopes in A′.

Lemma 5

Δ_βis the smallest-area closed convexT-covering of S_A′.

Proof. Consider six unit segments of slopes π/12+i·π/6 for i=0, 1, ⋅, 5. Then any compact convex T-covering of S_A′ contains these six unit segments under translation.

By [1, Theorem 9] there exists a triangle that is the smallest-area convex T-covering of any given set of segments. More specifically, every segment of the set is placed with its center at the origin, and the smallest-area affine-regular hexagon that contains the segments is considered. It is proved that a triangle Δ such that the hexagon is the Minkowski symmetrization of Δ, which is {12(x−y)∣x,y∈Δ}, is the smallest-area closed convex T-covering.

Observe that the smallest-area affine-regular hexagon containing the six unit segments that we consider is the Minkowski symmetrization of Δ_β, which is{12(x−y)∣x,y∈△β}. Thus Δ_β is the smallest-area closed convex T-covering of the six unit segments by the proof of Theorem 9 in [1]. Thus the lemma follows. □

Proposition 1

Δ_βis a minimal closed convexG₄-covering ofS_C.

Proof. Suppose that P⊆Δ_β is a closed subset and a G₄-covering of S_C. Observe that, for each angle θ in A′, A does not contain θ+π/2modπ. Thus P must contain all line segments of S_A′ under translation. Lemma 5 implies that the area of P is the same as that of Δ_β. Therefore, P must be Δ_β itself. □

4 G _k-covering of unit line segments

Consider the smallest-area convex G_k-covering of the set Sseg of all unit line segments. In general, a smallest-area convex T-covering of any given set of segments is attained by a triangle; see [1]. This implies that there is a triangle that is a smallest-area convex G_k-covering of Sseg. The following theorem determines the set of all smallest-area convex G_k-coverings.

Theorem 5

If k=3 is odd, then the smallest area of a convexG_k-covering ofSseg is 12sin(π/k), and it is attained by any triangle Δ XYZwith bottom sideXYof length 1 and height sin (π/k) such thatπ/2 ⩽ ∠ X ⩽ (k−1)π/k. Ifk=4 is even, then the smallest area of a convexG_k-covering ofSseg is12sin(2π/k), and it is attained by any triangle Δ XYZwith bottom sideXYof length 1 and height sin (2π/k) such thatπ/2 ⩽ ∠ X ⩽ (k−2)π/k.

Proof. We have already seen a proof for k=4, and it can be generalized asfollows. Let Λ be a smallest-area convex G_k-covering ofSseg.

First we consider the case where k is odd. Since Z_k consists of 2iπ/k rotations for i=0, 1, 2, ⋅, k − 1, one of the unit segments of slopes θ+2iπ/kmodπ must be contained in Λ under translation for each angle θ with 0 ⩽ θ<2 π/k. We define the smallest one of such angles to be f(θ). As before, we can assume that there exists an angle θˉ such that f(θˉ)=θˉ.

Let A={f(θ)∣ 0 ⩽ θ<2π/k } be the set of angles, let A^C denote the complement of A, and let s(θ) be a unit segment of slope θ. There exists an angleθ˜ with 0≤θ˜<2π/k such thatθ˜ is contained in both A and the closure of A^C. There is a sequence {θ˜n}n=1∞⊆A such thatlimn→∞θ˜n=θ˜+2iπ/k for some i. Then for any integer n > 0, the set Λ contains a translated copy of s(θ˜) and a translated copy ofs(θ˜n), so the area of Λ is at least12|sin(θ˜n−θ˜)|. Sincelimn→∞(θ˜n−θ˜)=2iπ/k, the area of Λ is at least 12sin(π/k).

On the other hand, let Δ XYZ be a triangle with bottom side XY of length 1 and height sin (π/k) such that π/2 ⩽ ∠ X ⩽ (k−1)π/k. Then ∠ Y+∠ Z ⩾ π/k. We show that any segment of slope θ with 0 ⩽ θ ⩽ π/k can be placed within Δ XYZ. Any unit line segment of slope θ₁ with 0 ⩽ θ₁ ⩽ ∠ Y can be placed within Δ XYZ with one endpoint at Y, and any unit line segment of slope θ₂ with ∠ Y ⩽ θ₂ ⩽ max{∠ Y, π/k} can be placed within Δ XYZ with one endpoint at Z. Thus, any segment of slope θ with 0 ⩽ θ ⩽ π/k can be placed within Δ XYZ.

The case where k is even can be handled analogously; the only difference is that the area is minimized at i=k/2 − 1. As in the case of odd k, a triangle satisfying the conditions stated in Theorem 5 is a G_k-covering. □

5 Covering under rotation by 120 degrees

We construct a convex G₃-covering Γ₃ of S_C as follows. Let Γ be the convex region bounded by y²=1 + 2x and y²=1 − 2x and containing the origin O. Then Γ₃ is the convex subregion of Γ bounded by the x-axis and the line y=2/3. The area of Γ₃ is |Γ3|=2(527+∫5/181/21−2xdx)=46/81<0.5680. See Figure 7(a).

Figure 7

(a) Illustration of Γ₃ (b)H=L₀∩ L_π/3∩ L_2π/3

We show that Γ₃ is a G₃-covering of S_C. We first show a few properties that we use for proving Theorem 6. Let Γ⁺ be the region of Γ above the x-axis. We call the boundary segment on the x-axis the bottom side, the boundary curve on y=1+2x the left side, and the boundary curve on y=1−2x the right side of Γ⁺. The region Γ⁺ is called a \textitchurch window, which is a T-covering of S_C ; see [6]. The following lemma gives a lower bound on the length of a T-brimful curve for Γ⁺. It is stated in [6] without proof. We give a proof of the lemma.

Lemma 6

Any closed T-brimful curve for Γ⁺has length at least 2.

Proof. Recall the definition of a G-brimful curve in Definition 1. Consider a closed T-brimful curve γ of minimum length for Γ⁺. Observe that γ touches every side of Γ⁺; otherwise γ can be translated to lie in the interior of Γ⁺. Let ΔXYZ be a triangle for the touching points X, Y, Z of the curve with the boundary of Γ⁺. Since ℓ(γ) ⩾ ℓ(ΔXYZ), γ is ΔXYZ.

Figure 8

Illustration of the proof of Lemma 6

Without loss of generality, we assume that X is on the left side, Z is on the right side, and Y is on the bottom side of Γ⁺. If X and Y are at (−1/2, 0), or X and Z are at (0, 1), or Y and Z are at (1/2, 0), then γ becomes a line segment and the length of the line segment is larger than or equal to 1. Thus ℓ(γ)⩾2.

Now we assume that none of X, Y and Z is on a corner of Γ⁺. Let ℓ_X be the line tangent to the left side of Γ⁺ at X. Let Z′ be the point symmetric to Z with respect to the x-axis, and let Zˉ be the point symmetric to Z with respect to ℓ_X. See Figure 8(a).

We claim that Zˉ,X,Y and Z′ are collinear. If X, Y and Z′ are not collinear, consider the intersection point Y′ of XZ′ and the bottom side of Γ⁺. Then ℓ(ΔXY′Z)<ℓ(ΔXYZ), since ∣XY′∣ + ∣Y′Z∣=∣XZ′∣ < ∣XY∣+∣YZ′∣=∣XY∣+∣YZ∣. This contradicts the assumption on γ. Thus X, Y and Z′ are collinear. If Zˉ is not on the line through X and Y, consider the point X′ where YZˉ intersects ℓ_X. Then we have ℓ(ΔX′YZ)<ℓ(ΔXYZ) because|YX′|+|X′Z|=|YZˉ|<|YX|+|XZˉ|=|YX|+|XZ|. For the point X″ where X′Y intersects the left side of Γ⁺, we have ΔX″YZ⊂ΔX′YZ. Therefore ℓ(ΔX″YZ)<ℓ(ΔX′YZ)<ℓ(γ), and this contradicts the assumption on γ.

Suppose that XZ is not horizontal. By the collinearity ofX,Zˉ and Z′, the reflection of ℓ_XZ over ℓ_X is ℓ_XZ′. Let X_r be the point symmetric to X with respect to the y-axis, and let X′ and X′_r be the points symmetric to X and X_r with respect to the x-axis. By the symmetry, ∠ ZXX_r=∠ Z′X′X′_r. Observe that for any line ℓ parallel to the axis of a parabola V, the reflection of ℓ over the line tangent to V at ℓ∩ V passes through the focus of V. So the reflection of ℓXXr over ℓ_X is ℓXXr′. Therefore ∠ Z′X′X′_r=∠ Z′XX′_r, implying that X, X′, X′_r and Z′ are on a circle C; see Figure 8(b). Since C passes through X, X′ and X′_r, the center of C is at the origin. This implies that X′Z′ is horizontal, and thus XZ is also horizontal, contrary to the assumption that XZ is not horizontal.

Since XZ is horizontal, Y is at the origin. Thus ΔXYZ is an isosceles triangle with base XZ, and ℓ(ΔXYZ)=2 by the construction of Γ⁺. □

The following lemma shows the convexity of the perimeter function on the convex hull of planar figures under translation.

Lemma 7

(Theorem 2 of [2]. Givenkcompact convex figuresC₁, C₂, ⋅, C_kC_i for i=1, ⋅, kin the plane, the perimeter functionψ:R2k→R of the convex hull conv(⋃iCi(r))is convex, whereC_i(r)=C_i+r_ifor a vectorr=(r₁, ⋅, r_k) ∈ ℝ^2kand isC_i+r_iforr_i ∈ ℝ².

For compact convex figures that have point symmetry in the plane, we can show an optimal translation of them using the convexity of the perimeter function in Lemma 7.

Lemma 8

Given k compact convex figures C ₁, C₂, ⋅, C_kForkcompact convex figuresC_i for i=1, ⋅, kthat have point symmetry in the plane, the perimeter functionψ:R2k→Rof the convex hullconv(⋃iCi(r))is minimized when their centers of symmetry coincide, (of the symmetry) meet at a point, whereC_i(r)=C_i+r_ifor a vectorr=(r₁, ⋅, r_k) ∈ ℝ^2kisC_i+r_i for and r_i ∈ ℝ².

Proof. Without loss of generality, we assume that the k compact convex figures are given with centers all lying at the origin. Let r=(r₁, ⋅, r_k) ∈ ℝ^2k be a vector such that the perimeter of their convex hull of C_i(r) is minimized among all translation vectors in ℝ^2k. Then −r=(−r₁, ⋅, −r_k) is also a vector such that the convex hull of C_i(−r) has the minimum perimeter. This is because the two convex hulls are symmetric to the origin. Since the perimeter function is convex by Lemma 7, the convex hull of Ci(0) also has the minimum perimeter, where0 is the zero vector (a vector of length zero). □

We are now ready to prove a main result.

Theorem 6

Γ₃is a convexG₃-covering of all closed curves of length 2.

Proof. By Lemma 6, any closed curve of length 2 can be included in Γ⁺ under translation. Let C be a closed curve of length 2 that is contained in Γ⁺ and touches its bottom side, and let Cˉ be the convex hull of C.

Suppose that C crosses the top side of Γ₃. Let s be a segment contained in Cˉ and connecting the top side and the bottom side of Γ₃ such that the upper endpoint of s lies in the interior of Cˉ. For i=1, 2 let C_i be a rotated and translated copy of C by 2iπ/3 such that they are contained in Γ⁺ (by Lemma 6) and touch the bottom side of Γ₃. If C₁ or C₂ is contained in Γ₃, then Γ₃ is a convex G₃-covering of C and we are done.

Assume now to the contrary that neither C₁ nor C₂ is contained in Γ₃. Then both curves cross the top side of Γ₃. For i=1, 2 let s_i be a line segment contained in the convex hull of C_i and connecting the top side and the bottom side of Γ₃ such that the upper endpoint of s_i lies in the interior of the convex hull of C_i; see Figure 7(a). Then there is a rotated and translated copy s˜i of s_i by −2iπ/3 such that s˜i is contained in Cˉ. Let Φ be the convex hull of s,s˜1 and s˜2. Since s,s˜1,s˜2⊂Cˉ and the upper endpoint of s lies in the interior of Cˉ, we haveℓ(Φ)<ℓ(Cˉ)≤ℓ(C)=2.

Now consider a translation of these three segments such that their midpoints coincide, meet ???? at a point, and let Φ_m be the convex hull of the three translated segments. By Lemma 8, ℓ(Φ_m) ⩽ ℓ(Φ). For 0 ⩽ θ < π let L_θ denote the slab of minimum width at orientation θ for 0 ⩽ θ < π that contains Φ_m. Let d_θ be the width of L_θ. Consider the three slabs L₀, L_π/3 and L_2π/3 of Φ_m. Observe that d_θ for θ=0, π/3, 2π/3 is at least the height of Γ₃, which is 2/3. Let H=L₀∩ L_π/3∩ L_2π/3 as shown in Figure 7(b). Then Φ_m is contained in H and touches every side of H. Since H is a (possibly degenerate) hexagon, ℓ(Φ_m) ⩾ d₀+d_π/3+d_2π/3⩾2, which can be shown by a proper tiling of copies of H as in Lemma 1. Thus ℓ(Φ_m)⩾2, which is a contradiction to ℓ(Φ_m) ⩽ ℓ(Φ)<2. □

Recall that the smallest-area convex G₂-covering Δ₁ and the G₄-covering Δ_β of S_C are equilateral triangles. Our G₃-covering Γ₃ has area smaller than the area of Δ₁, but a bit larger than the area of Δ_β, which sounds reasonable. However, it may look odd that Γ₃ is not invariant under any discrete rotation while Δ₁ and Δ_β are invariant under rotation by 2π/3.

Theorem 7

Any convex G ₃-covering invariant under rotation by 2π/3 orπ/2 has area strictly larger than the area of Γ₃.

Proof. Let Λ_2π/3 be a convex G₃-covering which is invariant under rotation by 2π/3. Then Λ_2π/3 is a T-covering of all unit segments of any slope θ with 0 ⩽ θ<π. As Since Δ₁ is the smallest-area convex T-covering of the set of all unit segments by Theorem 1, the area of Λ_2π/3 is at least the area of Δ₁, which is strictly larger than the area of Γ₃.

$Figure 9 (a) siπ/2 is contained in Λπ/2 for i=1, 2, 3. (b) The convex hull of sˉ$ \bar{s} $ ands˜$ \tilde{s} $ is contained in the convex hull Φ1 of s, sπ/2, sπ, s3π/2.$

Figure 9

(a) s_iπ/2 is contained in Λ_π/2 for i=1, 2, 3. (b) The convex hull of sˉ ands˜ is contained in the convex hull Φ₁ of s, s_π/2, s_π, s_3π/2.

Let Λ_π/2 be a convex G₃-covering which is invariant under rotation by π/2. Assume that a unit segment s of slope π/4 is contained in Λ_π/2. Since Λ_π/2 is a G₃-covering, there is a unit segment of one of slopes slope in {0, π/3, 2π/3} contained in Λ_π/2. Assume that a unit segment t of slope π/3 is contained in Λ_π/2. Since Λ_π/2 is invariant under rotation by π/2, there are a unit segment s′ of slope 3π/4=π/4+π/2 and a unit segment t′ of slope 5π/6=π/3+π/2 contained in Λ_π/2. Thus, the four segments s, s′, t, t′ are contained in Λ_π/2.

Let c be the point of symmetry of Λ_π/2. Let Φ be the convex hull of the translated copies of s, s′, t, t′ such that their midpoints are all at c. We will prove that the area ∣Λ_π/2∣ of Λ_π/2 is at least the area ∣Φ∣ of Φ.

Suppose that the midpoint of s is not at c. Let s_iπ/2 beobtained by rotating s around c by iπ/2 for i=1, 2, 3. Since Λ_π/2 is regular invariant under the rotation by π/2, s_iπ/2 is contained in Λ_π/2 for every i=1, 2, 3. Since Λ_π/2 is convex, the convex hull Φ₁ of s and the segments s_iπ/2 for i=1, 2, 3 is contained in Λ_π/2, and thus ∣Λ_π/2∣ ⩾ ∣Φ₁∣. See Figure 9(a).

Let sˉ be the translated copy of s such that the midpoint ofsˉ is at c, and s˜ be the copy of sˉ rotated by π/2 around c. Let Φ₂ be the convex hull of sˉ and s˜. Since sˉ is contained in the convex hull of s and s_π, and s˜ is contained in the convex hull of s_π/2 and s_3π/2, we have Φ₂⊂Φ₁. See Figure 9(b). Similarly, the convex hull of the translated copy tˉ of t with midpoint lying at c and the rotated copy of tˉ by π/2 around c is contained in Λ_π/2. Thus we conclude that ∣Λ_π/2∣ ⩾ ∣Φ∣ ⩾ \sqrt6/4 > 0.6>∣Γ₃∣, where ∣Γ₃∣ is the area of Γ₃.

We can show this for unit segments of slopes 0 and 2π/3 contained in Λ_π/2 in a similar way. □

6 Covering of triangles under rotation by 120 degrees

6.1 Construction

Let S_t be the set of all triangles of perimeter 2. We construct a convex G₃-covering of S_t, denoted by Γt, from Γ₃ by shaving off some regions around the top corners. Consider an equilateral triangle Δ=Δ XYZ of perimeter 2 such that the side YZ is vertical, Y lies on the bottom side of Γ₃ and X lies on the left side of Γ₃. See Figure 10(a-b) for an illustration.

$Figure 10 Construction: (a) The covering Γ3 (b) The trajectories of X, Y and Z (c) The G3-covering Γt$ {\Gamma_{\text{t}}} $ of St$

Figure 10

Construction: (a) The covering Γ₃ (b) The trajectories of X, Y and Z (c) The G₃-covering Γt of S_t

Imagine that Δ rotates in a clockwise direction such that X moves along the left side and Y moves along the bottom side of Γ₃. Let t denote the x-coordinate of X and θ=∠XYO. Thentanθ=(2t+1)/(−2t−59) andZ=((6t+3+−2t−59+2t)/2,(−6t−53+2t+1)/2) for t varying from −4/9 to −1/3. The trajectory of Z forms the top-right boundary of Γt that connects the top side and the right side of Γ₃. Thus the region of Γ₃ lying above the trajectory is shaved off. The top-left boundary of Γt can be obtained similarly. Figure 10(c) shows Γt.

We show that Γt is convex by showing thatddx(dydx)=ddt(dydx)/dxdt≤0 for Z=(x(t), y(t)), and that the boundary of Γt has a unique tangent at t=−4/9 and t=−1/3. Since the x-coordinate of Z increases as t increases, we have dxdt>0. Thus it suffices to show thatddt(dydx)≤0 for t with −4/9 ⩽ t ⩽ −1/3. Observe that

d y d x = − 3 ( − 6 t − 5 3 ) − 1 / 2 + ( 2 t + 1 ) − 1 / 2 3 ( 6 t + 3 ) − 1 / 2 − ( − 2 t − 5 9 ) − 1 / 2 + 2 .

We obtain

f ( t ) := d d t ( d y d x ) = ( 36 t + 10 ) f 1 ( t ) − 3 ( 54 + 108 t ) f 2 ( t ) − 24 f 1 ( t ) f 2 ( t ) { 2 f 1 ( t ) f 2 ( t ) + 3 f 1 ( t ) − 3 f 2 ( t ) } 2

where f1(t)=−18t−5 and f2(t)=2t+1. Since f₁(t) > 0 and f₂(t) > 0, the denominator of f(t) is positive. The numerator of f(t) is negative, hence ddt(dydx)/dxdt≤0. At t=−4/9 we have dydx=0, which is the slope of the top side of Γ₃. At t=−1/3 we have dydx=−3, which is the slope of the tangent to the right side of Γ₃ at the same point. Thus Γt is convex, and the boundary of Γt, except the bottom side, is a C¹ curve.

Now we estimate the area of Γt. The area that is shaved off from Γ₃ is

2 ( 2 3 ( 13 18 − 1 3 ) + ∫ 5 18 1 3 1 − 2 x d x − ∫ 1 3 − 4 9 1 3 f ( x ) d x ) = 1 81 ( 86 − 44 3 − 3 π ) ,

where f(x) is the function of γ_DE such that

∫ 1 3 − 4 9 1 3 f ( x ) d x = 1 4 ∫ − 4 9 − 1 3 ( − 6 x − 5 3 + 2 x + 1 ) ( 3 6 x + 3 − 1 − 2 x − 5 9 + 2 ) d x .

Thus the area of Γt is smaller than 0.5634.

6.2 Covering of triangles of perimeter 2

We show that Γt is a G₃-covering of all triangles of perimeter 2. To do this, we first show a few properties of Γt. Let A, B, C, D, E and F be the boundary points of Γt as shown in Figure 10. We denote the boundary curve of Γt from a point a to a point b in clockwise direction along the boundary by γ_ab.

Lemma 9

Let ΔXYZbe a triangle of perimeter 2 contained in Γ⁺. If it is on the left of ℓ_OEor on the right of ℓ_OB(including the lines), thenΓtis a convexG₃-covering of ΔXYZ.

Proof. If ΔXYZ lies on the left of ℓ_OE (including the line), then ΔXYZ lies in between ℓ_OE and the line ℓ tangent to Γ_t at B. The two lines have slope π/3 and are at distance 3/3. Thus there are copies of ΔXYZ rotated by 2π/3 and lying in between ℓ_BE and ℓ_AF. Among such copies, let Δ be the one that touches γ_FA from above and γ_AB from the right. Since ℓ(Δ)=2, Δ is contained in Γt by Lemma 6. See Figure 11(a). The case of ΔXYZ lying on the right of ℓ_OB can be shown by a copy of the triangle rotated by −2π/3. □

$Figure 11 (a) ΔXYZ lying on the left of ℓOE. There is a copy Δ that is contained in Γt$ {\Gamma_{\text{t}}} $. (b) ΔXYZ contained in Γ3 such that X is at A, Y is on the right of ℓOE, and Z ∈ R2. (c) ΔXYZ contained in Γ3 such that X is on γOA, Y is on the right of ℓOE, and Z ∈ R1.$

Figure 11

(a) ΔXYZ lying on the left of ℓ_OE. There is a copy Δ that is contained in Γt. (b) ΔXYZ contained in Γ₃ such that X is at A, Y is on the right of ℓ_OE, and Z ∈ R₂. (c) ΔXYZ contained in Γ₃ such that X is on γ_OA, Y is on the right of ℓ_OE, and Z ∈ R₁.

In the following, we assume that ΔXYZ is contained in Γ₃ but it is not contained in Γt. If there is no corner of ΔXYZ lying on the left of ℓ_OB or on the right of ℓ_OE (including the lines), then Γt is a G₃-covering of ΔXYZ by Lemma 9. Thus we assume that X is on the left of ℓ_OB and Y is on the right of ℓ_OE. Since Γt is convex, the remaining corner Z of ΔXYZ lies in Γ3∖Γt. Let R₁ and R₂ denote the left and right regions of Γ3∖Γt, respectively, as shown in Figure 11(a). Translate ΔXYZ leftwards horizontally until X or Z hits the left side of Γ₃. If the triangle lies on the left of ℓ_OE (including the line), we are done by Lemma 9. Thus we assume that Z is in R₁∪ R₂ and Y lies on the right of ℓ_OE. There are two cases, either Z ∈ R₁ or Z ∈ R₂; see Figure 11(b) and (c). If Z is in R₂ then X is at A.

Lemma 10

Let ΔXYZbe a triangle contained in Γ₃such thatXis atA, Yis on the right of ℓ_OE, andZ ∈ R₂. ThenΓtis a convexG₃-covering of ΔXYZ.

Proof. Let △=△X˜Y˜Z˜ be the copy of ΔXYZ rotated by 2π/3 such that X˜ lies at F. We show that Δ is contained in Γt. Assume to the contrary that Δ is not contained in Γt. Since ∠ ZAF>π/6 and F is at distance at least 1 from any point on γ_AB, Z˜ must be contained in Γt. See Figure 12(a) and (b).

$Figure 12 (a) ΔXYZ with Z ∈ R2. (b) A rotated copy △X˜Y˜Z˜$ \triangle{\tilde{X}\tilde{Y}\tilde{Z}} $ of ΔXYZ by 2π/3 with X˜$ \tilde{X} $ lying at F. (c) If Y lies on the right of ℓ, then ℓ(ΔXYZ)>ℓ(ΔXVW)⩾2.$

Figure 12

(a) ΔXYZ with Z ∈ R₂. (b) A rotated copy △X˜Y˜Z˜ of ΔXYZ by 2π/3 with X˜ lying at F. (c) If Y lies on the right of ℓ, then ℓ(ΔXYZ)>ℓ(ΔXVW)⩾2.

If Y˜ lies on or below ℓ_BE, then Δ is contained in Γt by Lemma 6. So we assume that Y˜ lies above ℓ_BE. Then Y must lie in the right of the line ℓ of slope π/3 passing through the point of γ_FO at distance 1/3 from F; see Figure 12(c). Let H be the point at (0, 2/3). Then there is a triangle ΔXVW with V ∈ ℓ and W ∈ HE such that ℓ(ΔXVW)<ℓ(ΔXYZ). To produce a contradiction, it suffices to show that ℓ(ΔXVW)⩾2. For a point p let p′ denote the point symmetric to p along ℓ. Then ∣VW∣=∣VW′∣. Since X is at distance at least 7/6 to any point in HˉEˉ and at distance at least 5/6 to any point in HE, we obtain ℓ(ΔXVW)=∣XV∣+∣VW∣+∣WX∣=∣XV∣+∣VW′∣+∣WX∣⩾7/6 + 5/6=2. □

We can also show that Γt is a G₃-covering of ΔXYZ for the remaining case of Z ∈ R₁.

Lemma 11

Let ΔXYZbe a triangle contained in Γ₃such thatXis onγ_OA, Yis on the right of ℓ_OEandZ ∈ R₁. ThenΓtis a convexG₃-covering of ΔXYZ.

Before proving this, we need a few technical lemmas.

Lemma 12

Let ΔXYZbe an isosceles triangle of perimeter 2 such that its baseYZis of length ⩾2/3, at least 2/3, and letX ∈ γ_OA, Y ∈ γ_EFandZ ∈ ℓ_BE. Then ΔXYZcan be rotated in clockwise direction withinΓtsuch thatX moves along γ_OAandYmoves alongγ_EFuntilYmeetsF.

Proof. By Lemma 6, Z lies on BE. Let △XˉYˉZˉ be a translated copy of ΔXYZ such that Zˉ lies at B; then Xˉ∈γOA. Observe that Yˉ is contained in Γt. See Figure 13(a).

$Figure 13 (a) An isosceles triangle Δ XYZ and a translated copy △XˉYˉZˉ$ \triangle \bar{X}\bar{Y}\bar{Z} $ of Δ XYZ (b) Convex hull Φ of Δ UVW and △XˉYˉZˉ$ \triangle \bar{X}\bar{Y}\bar{Z} $ (c) The rotated copy Φ(θ′) of Φ by θ′−π/3$

Figure 13

(a) An isosceles triangle Δ XYZ and a translated copy △XˉYˉZˉ of Δ XYZ (b) Convex hull Φ of Δ UVW and △XˉYˉZˉ (c) The rotated copy Φ(θ′) of Φ by θ′−π/3

Let ΔUVW be an equilateral triangle with base VW of length ∣VW∣=2/3, W lying at B, and U lying at O. Observe that |XˉU|≤1/3 : otherwise, Yˉ is not contained in Γt since △XˉYˉZˉ is an isosceles triangle of perimeter 2 with base length |YˉZˉ|≥2/3. We will show below that ΔXYZ can be rotated as stated in the lemma by using the rotation of ΔUVW and △XˉYˉZˉ within Γt.

Let p be a point in Γt lying below ℓ_YO, let Q be the convex hull of p and ΔUVW, and let θ₀ be the angle ∠ pOF. See Figure 14(a). We claim that Q can be rotated within Γt in a clockwise direction such that V moves along γ_EF and U moves along γ_OA until p reaches ℓ_AF. Let Q(θ)=U˜V˜W˜p˜ be the rotated copy of Q by θ with 0 ⩽ θ ⩽ θ₀ in clockwise direction around O such that each corner κ˜ of Q(θ) corresponds to the rotated point of κ for κ ∈ {X, Y, Z, p}. Let Qˉ(θ)=UˉVˉWˉpˉ be the translated copy of Q(θ) such that Vˉ∈γEF, Uˉ∈γOA and each corner κˉ of Qˉ(θ) corresponds to the translated point of κ˜ for κ˜∈{U˜,V˜,W˜,p˜}. See Figure 14(b) and (c).

$Figure 14 (a) The convex hull Q of an equilateral triangle Δ UVW and p (b) Rotated copy Q(θ) of Q by θ around O (c) Translated copy Qˉ(θ)$ \bar{Q}(\theta) $ of Q(θ) such that Vˉ∈γEF$ \bar{V}\in\gamma_{EF} $ and Uˉ∈γOA$ \bar{U}\in\gamma_{OA} $$

Figure 14

(a) The convex hull Q of an equilateral triangle Δ UVW and p (b) Rotated copy Q(θ) of Q by θ around O (c) Translated copy Qˉ(θ) of Q(θ) such that Vˉ∈γEF and Uˉ∈γOA

Since ΔUVW is an equilateral triangle, Wˉ is contained in Γt as shown in Subsection 6.1. Thus we show that pˉ∈Γt. We may assume that p ∈ γ_EF. Let p′ be the intersection between the horizontal line through pˉ and γ_EF. We show that f(θ)=x(p′)−x(pˉ)≥0 with 0 ⩽ θ ⩽ θ₀, implying pˉ∈Γt. Then f(θ)=x(p′)−x(pˉ)=x(p′)−x(p˜)+x(p˜)−x(pˉ)=x(p′)−x(p˜)+x(V˜)−x(Vˉ). Let g(θ0,θ)=x(p˜)−x(p′) for 0 ⩽ θ ⩽ θ₀ ⩽ π/3. Observe that g(θ0,θ)=x(p˜)−x(p′)=h(θ0)cos(θ0−θ)−(12−12(h(θ0)sin(θ0−θ))2), where h(θ₀)=1/(1 + cos(θ₀)). For θ₀=π/3, p˜ is at V˜ and p′ is at Vˉ, and thus x(V˜)−x(Vˉ)=g(π/3,θ). From this, f(θ)=g(π/3, θ)−g(θ₀, θ). Since pˉ is on γ_EF at θ=0, we have f (0)=0. Thus it suffices to show that g(θ₀, θ) is not decreasing for θ₀ increasing from θ to π/3 for any fixed θ with 0<θ ⩽ π/3. We have

∂ g ∂ θ 0 = 1 ( 1 + cos θ 0 ) 3 ( g 1 ( θ 0 ) + g 2 ( θ 0 ) ) > 0 where g 1 ( θ 0 ) = sin θ 0 sin 2 ( θ − θ 0 ) + ( sin ( θ − θ 0 ) + sin θ 1 cos ( θ − θ 0 ) ) ( 1 + cos θ 0 ) = sin θ 0 sin 2 ( θ − θ 0 ) + ( cos θ 0 sin ( θ − θ 0 ) + sin θ 0 cos ( θ − θ 0 ) ) ( 1 + cos θ 0 ) = + sin ( θ − θ 0 ) ( 1 − cos θ 0 ) ( 1 + cos θ 0 ) = sin θ 0 sin 2 ( θ − θ 0 ) + sin θ ( 1 + cos θ 0 ) + sin ( θ − θ 0 ) sin 2 θ 0 = ( sin ( θ 0 − θ ) − sin θ 0 ) sin θ 0 sin ( θ 0 − θ ) + sin θ + sin θ cos θ 0 and g 2 ( θ 0 ) = sin ( θ 0 − θ ) ( 1 + cos θ 0 ) ( cos ( θ 0 − θ ) − cos θ 0 ) .

Observe that sin θ cos θ₀⩾0. Since 0< sin θ₀ sin (θ₀−θ)<1, it suffices to show that (sin (θ₀−θ)− sin θ₀)+ sin θ>0 to show g₁(θ₀)>0. Let gˉ(θ0)=sin(θ0−θ)−sinθ0+sinθ. Since ∂gˉ∂θ0=cos(θ0−θ)−cosθ0>0 and gˉ(θ)=0, we have gˉ(θ0)>0, implying g₁(θ₀)>0. Observe that 1+ cos θ₀ > 0 and (cos (θ₀−θ)− cos θ₀)>0, so we have g₂(θ₀)>0. Thus ∂g∂θ0>0. Since pˉ is contained in Γt at θ=θ₀, pˉ is contained in Γt.

Let Φ be the convex hull of △XˉYˉZˉ and ΔUVW. See Figure 13(b). Imagine that Φ rotates in a clockwise direction such that U moves along γ_OA and V moves along γ_EF until Yˉ reaches γ_FA. Then W moves along γ_BC, and then moves into the interior of Γt during the rotation because ΔUVW is an equilateral triangle with ∣UW∣=2/3. Let Φ(θ′) denote the rotated copy of Φ by θ′−π/3, where θ′=∠WUA. Then θ′ increases monotonically from π/3 during the rotation. See Figure 13(c).

Consider the rotation for θ′ from π/3 to π/2. Observe that Yˉ is contained in Γt during the rotation by the argument in the second paragraph of the proof. Since Zˉ=W and W∈Γt during the rotation (as shown in Subsection 6.1), Zˉ is also in Γt.

Now we claim that Xˉ is contained in Γt during the rotation. For any θ′ with π/3<θ′ ⩽ π/2, Xˉ lies above ℓ_OA, but it lies below ℓ_BE because ∠XˉFA≤π/6 for θ′ ⩽ π/2. Observe that ∣UF∣⩽2/3. Since |UXˉ|≤1/3, we have |XˉF|≤|XˉU|+|UF|≤1. Since the distance between F and γ_AB is 1, Xˉ is contained in Γt. Thus △XˉYˉZˉ is contained in Γt during the rotation.

For a fixed θ′ with π/3 ⩽ θ′ ⩽ π/2, we can translate △XˉYˉZˉ in Φ(θ′) within Γt such that Xˉ is on γ_OA and Yˉ is on γ_EF. This implies that ΔXYZ can be rotated as stated in the lemma. □

Lemma 13

Let ΔXYZbe an isosceles triangle of perimeter 2 such that its baseYZis of length at least 2/3, and letY ∈ γ_OAandZ ∈ γ_EF. Then the line throughXand parallel toYZintersectsγ_OB.

Proof. Let ℓ_X and ℓ_B be the lines parallel to YZ such that ℓ_X passes through X and ℓ_B passes through B. See Figure 15(a). Let θ=∠ZYF and let p denote the y-coordinate of Z. Then 0≤p≤3/3, sin ⁻¹p ⩽ θ ⩽ sin ⁻¹(3p/2) and Z=(1−p22,p). Let f(p, θ) denote the distance between ℓ_B and X. Observe that ℓ_YZ has the equation y=tanθ(x−1−p22)+p. The distance between B=(−13,33) and ℓ_YZ is (33−p)cosθ+(5−3p26)sinθ. Observe that |YZ|=psinθ. The perimeter of ΔXYZ is 2, hence |XZ|=1−p2sinθ. Since ΔXYZ is an isosceles triangle, the distance between X and ℓ_YZ is 1−psinθ. Thus we have

f ( p , θ ) = ( 3 3 − p ) cos θ + ( 5 − 3 p 2 6 ) sin θ − 1 − p sin θ .

We will show that for 0≤p≤3/3 and for θ with sin ⁻¹p ⩽ θ ⩽ sin ⁻¹\bigl(3p/2\bigr) (or equivalently for 23sinθ≤p≤min{3/3,sinθ} and 0 ⩽ θ ⩽ π/3) the following holds:

∂ f ∂ p = − p sin θ − cos θ + 1 2 sin 2 θ − p sin θ ≥ 0.

We have ∂2f∂p2=sinθ(14(sin2θ−psinθ)3/2−1). As 0≤sinθ−p≤sinθ3, we have 0≤4(sin2θ−psinθ)3/2≤4sin3θ33≤12. This implies that ∂2f∂p2≥0 ; so it suffices to show that ∂f∂p≥0 for p=23sinθ. We compute

∂ f ∂ p | p = 2 3 sin θ = − 2 sin 2 θ 3 − cos θ + 3 2 sin θ = 1 sin θ ( − 2 sin 3 θ 3 − sin 2 θ 2 + 3 2 )

and with g(θ)=−23sin3θ−12sin2θ+123 we obtain

(1) ∂ g ∂ θ = − cos 2 θ − 2 sin 2 θ cos θ

(2) = − cos 2 θ + 2 cos 3 θ − 2 cos θ

(3) = − cos 2 θ − cos θ + cos 2 θ cos θ .

For θ≤π/4 we have ∂g∂θ≤0 by Equation (1). For θ with π/4<θ≤π/3 we have −cos2θ−cosθ≤0 and cos2θcosθ≤0. So ∂g∂θ≤0 by Equation (3). Therefore g(θ)≥g(π/3)=0, which implies that ∂f∂p≥0.

Now we show that f(23sinθ,θ)≥0 for θ with 0 ⩽ θ ⩽ π/3. Writing h(θ):=f(23sinθ,θ) we have

h ( θ ) = 5 sin θ 6 − 2 sin 3 θ 9 + cos θ 3 − 2 sin θ cos θ 3 − 1 3 .

Let t= sin θ. Then 1−t2=cosθ and 0≤t≤3/2. We determine the roots of h(θ): if h(θ)=0, then

⟹ − 2 9 t 3 + 5 6 t − 1 3 = ( 2 3 t − 1 3 ) 1 − t 2 ⟹ ( 2 3 t − 1 3 ) ( − 1 3 t 2 − 1 2 3 t + 1 ) = ( 2 3 t − 1 3 ) 1 − t 2 ⟹ ( − 1 3 t 2 − 1 2 3 t + 1 ) 2 = 1 − t 2 or t = 3 2 ⟹ 1 9 t 4 + 1 3 3 t 3 + 5 12 t 2 − 1 3 t = 0 or t = 3 2 ⟹ t ( t − 3 2 ) ( 1 9 t 2 + 1 2 3 t + 2 3 ) = 0 or t = 3 2 ⟹ t = 0 or t = 3 / 2 ⟺ θ = 0 or θ = π / 3 .

Since h(π/6)=8/9−3/2>0 and h is continuous with roots only at 0 and π/3, we have h(θ)⩾0 for θ with 0 ⩽ θ ⩽ π/3.

Observe that f(p, θ) is minimized at p=23sinθ for a fixed θ. In other words, f(p, θ) is minimized when ΔXYZ is an equilateral triangle. Since f(p, θ)⩾0 for p=23sinθ, any isosceles triangle Δ XYZ such that ℓ_X does not intersect γ_OB has height larger than 3/3. Thus ℓ_X intersects γ_OB. □

Lemma 14

Let ΔXYZbe an isosceles triangle of perimeter 2 such that its baseYZis of length at least 2/3, Xlies above or on ℓ_YZ, Y ∈ γ_OAandZ ∈ γ_EF. Then ΔXYZis contained inΓt.

Proof. Let ΔXYZ be an isosceles triangle satisfying these conditions in the lemma statement with φ=∠XYZ and θ=∠ZYF. Observe that 0 ⩽ φ, θ ⩽ π/3. Let f(φ) be the y-coordinate of X. Then f(φ)=sin(φ+θ)1+cosφ. Since f′(φ)⩾0, f(φ) is maximized at φ=π/3, implying that f(φ) is maximized when ΔXYZ is the equilateral triangle.

Let ΔUVW be the equilateral triangle such that V ∈ γ_OA, W ∈ γ_EF and VW is parallel to YZ. Let ℓ_U be the line parallel to VW and passing through U. See Figure 15(b). Then X lies below or on ℓ_U by the proof of Lemma 13, and its y-coordinate f(φ) is smaller than or equal to the y-coordinate of U by the argument in the previous paragraph. Since U is on the boundary of Γt, X is contained in Γt. □

Now we are ready to prove Lemma 11.

Proof. Translate ΔXYZ to the right until Y meets γ_EF; see Figure 16(a). If Z∈Γt then △XYZ⊂Γt and we are done. Suppose that We have Z ∈ R₁. If X ∈ γ_FO then ΔXYZ lies on the right of ℓ_OB (including the line), and Γt is a G₃-covering of ΔXYZ by Lemma 9. So we assume that X ∈ γ_OA, and thus ∣XY∣⩾1/2. There are three cases: the longest side of ΔXYZ is either (1) XZ or (2) YZ or (3) XY. For each case, we show that there is a rotated copy of ΔXYZ that is contained in Γt.

Consider Case (1) that XZ is the longest side. There are two subcases, ∣XY∣ ⩾ ∣YZ∣ or ∣XY∣ < ∣YZ∣. Suppose that ∣XY∣ ⩾ ∣YZ∣. Let △X˜Y˜Z˜ be the copy of ΔXYZ rotated by 2π/3 such that Z˜∈γFA and X˜ lies on the right side of Γ. Since ∠ZXF > π/3, Z˜ is the lowest corner. So X˜ lies on the right side of Γ⁺. If △X˜Y˜Z˜ lies on the right of ℓ_OB, on the left of ℓ_OE, or in the below ℓ_BE (including the line), then Γt is a G₃-covering of ΔXYZ by Lemma 9. Then △X˜Y˜Z˜ has one corner lying on the left of ℓ_OB, one corner lying on the right of ℓ_OE, and one corner lying above ℓ_BE. Since ∠ZXF⩽2π/3, we have ∠X˜Z˜F≤π/3. If X˜∉γEF, then Z˜∈γFO and thus ∠X˜Z˜F>π/3, which is a contradiction. Thus X˜∈γEF. Consequently, Y˜ lies above ℓ_BE and Z˜∈γOA. See Figure 16(b).

Figure 15

(a) ℓ_X lies below ℓ_B or ℓ_X=ℓ_B. (b) X lies below or on ℓ_U.

$Figure 16 (a) Z ∈ R1 and ZX is the longest side. ΔXYZ is translated to the right until Y meets γEF. (b) A copy △X˜Y˜Z˜$ \triangle{\tilde{X}\tilde{Y}\tilde{Z}} $ of ΔXYZ rotated by 2π/3. (c) A copy △X˜Y˜Z˜$ \triangle{\tilde{X}\tilde{Y}\tilde{Z}} $ of ΔXYZ rotated by −2π/3. \labelfig:G3-tri-cases-Z$

Figure 16

(a) Z ∈ R₁ and ZX is the longest side. ΔXYZ is translated to the right until Y meets γ_EF. (b) A copy △X˜Y˜Z˜ of ΔXYZ rotated by 2π/3. (c) A copy △X˜Y˜Z˜ of ΔXYZ rotated by −2π/3. \labelfig:G3-tri-cases-Z

Since |Y˜Z˜|≤2/3, Y˜ does not lie above ℓ_CD. If Y˜∈R2 then Γt is a convex G₃-covering of ΔXYZ by Lemma 6. So we assume that Y˜∈R1. Let Y* be a point lying on the left of ℓZ˜X˜ (including the line) such that △X˜Y∗Z˜ is an isosceles triangle of perimeter 2 with base Z˜X˜. Then both Y˜ and Y* are on the same ellipse with foci X˜ and Z˜. Since Z˜ is the lowest corner and |X˜Y˜|≥|Y˜Z˜|, Y˜ lies on or below ℓY∗, where ℓY∗ is the horizontal line through Y*. By Lemma 14 we have Y∗∈Γt. Since |XˉY˜|≥|Y˜Z˜|, Y˜ lies on or below the axis line of the ellipse through Y*. Since Y˜ lies on or below the tangent line of the ellipse at Y* and Y˜ lies on or below ℓY∗, we obtain Y˜∉R1 by Lemma 13, which is contradiction. Thus Y˜∈Γt, implying △X˜Y˜Z˜⊂Γt.

Suppose now that ∣XY∣ < ∣YZ∣. Let △X˜Y˜Z˜ be the copy of ΔXYZ rotated by −2π/3 such that X˜∈γBD and Z˜∈γEF. Since ∠ZXF<2π/3, X˜ is the highest corner. Let Y* be a point such that △X˜Y∗Z˜ is an isosceles triangle with base Z˜X˜ and lies below ℓZ˜X˜. Then both Y˜ and Y* are on the same ellipse with foci X˜ and Z˜. Since |X˜Y˜|<|Y˜Z˜| and X˜ is the highest corner, Y˜ lies on or above ℓY∗, where ℓY∗ is the horizontal line through Y*. By Lemma 12 we obtain Y∗∈Γt and thus Y˜ lies above ℓ_FA. So we have Y˜∈Γt and △X˜Y˜Z˜⊂Γt.

Now consider Case (2) that YZ is the longest side. Translate ΔXYZ such that Y ∈ γ_EF and Z ∈ γ_BC. If ∣XY∣ ⩾ ∣ZX∣, then X∈Γt by the argument for Case (1). So we assume that ∣XY∣< ∣ZX∣. Let △X˜Y˜Z˜ be the copy of ΔXYZ rotated by −2π/3 such that Y˜∈ℓFA and Z˜ lies on the right side of Γ⁺. Then ∠Z˜Y˜F≤π/3. If △X˜Y˜Z˜ lies to the right of ℓ_OB or to the left of ℓ_OE (including the lines), then Γt is a G₃-covering of the triangle by Lemma 9. then there is a copy of ΔXYZ rotated by ± 120° contained in Γt. So we assume that Z˜∈γEF and Y˜∈γOA. Since ∣XY∣<∣ZX∣, we obtain X˜∈Γt by the argument for Case (1), and thus △X˜Y˜Z˜⊂Γt.

Finally, consider Case (3) that XY is the longest side. Translate ΔXYZ such that X ∈ γ_OA and Y ∈ γ_EF. If ∣XZ∣ ⩽ ∣YZ∣, then Z∈Γt by the argument for Case (1). So we assume that ∣XZ∣>∣YZ∣. Let △X˜Y˜Z˜ be the copy of ΔXYZ rotated by 2π/3 such that X˜ lies on the right side of Γ⁺ and Z˜∈γFA. Since ∠ZXF > π/3, Z˜ is the lowest corner and ∠X˜Z˜F≤π/3. If △X˜Y˜Z˜ lies to the right of ℓ_OB or to the left of ℓ_OE (including the lines), then Γt is a G₃-covering of the triangle by Lemma 9. then there is a copy of ΔXYZ rotated by ± 120° contained in Γt. So we assume that X˜∈γEF and Z˜∈γOA. Since Y˜Z˜ is the shortest side, Y˜ lies below ℓ_CD. Thus it belongs to Case (2). □

Combining Lemmas 9, 10 and 11, we have the following result.

Theorem 8

Γ t is a convex G ₃-covering of all triangles of perimeter 2.

Observe that no proper subset of Γt is a convex G₃-covering of all triangles of perimeter 2. For any boundary point q of Γt there is a triangle of perimeter 2 that cannot be placed in Γt∖q under G₃ transformations. Therefore, Γt is a minimal convex G₃-covering of all triangles of perimeter 2.

7 Conclusion

We considered the smallest-area covering of planar objects of perimeter 2 allowing translations and discrete rotations of multiples of π, π/2 and 2π/3. We gave a geometric and elementary proof of the smallest-area convex coverings for translations and rotation of π while showing convex coverings for the other discrete rotations. We also gave the smallest-area convex coverings of all unit segments under translations and discrete rotations π/k for all integers k=3. Open problems include the optimality proof of the equilateral triangle covering for the rotation of multiples of π/2, and the smallest-area coverings allowing other discrete rotations with clean mathematical solutions.

Funding statement: Work by M. K. Jung and H.-K. Ahn was supported by the Institute of Information & Communications Technology Planning & Evaluation (IITP) grant funded by the Korea government (MSIT) (No. 2017-0-00905, Software Star Lab (Optimal Data Structure and Algorithmic Applications in Dynamic Geometric Environment)) and (No. 2019-0-01906, Artificial Intelligence Graduate School Program (POSTECH)). S. D. Yoon was supported by “Cooperative Research Program for Agriculture Science & Technology Development (Project No. PJ015269032022)”, Rural Development Administration, Republic of Korea. Work by T. Tokuyama was partially supported by MEXT JSPS Kakenhi 20H04143.

8 Appendix

We give a proof of Corollary 4 using a reflecting argument similar to the proof of Theorem 2.

Proof. Given a worm γ, let γ′ be its largest scaled copy contained in Δ₁. It suffices to show that the length of γ′ is not shorter than 1.

Observe that γ′ touches all three edges of Δ₁, possibly touching two edges simultaneously at a vertex, since otherwise we can translate γ′ into the interior of Δ₁ and get a larger scaled copy of γ′ contained in Δ₁. If γ′ touches a vertex of Δ₁, then it connects the vertex and the edge opposite to the vertex. Thus its length is not shorter than the height 1 of Δ₁.

Consider the case that γ′ touches the three edges of Δ₁ at three distinct points a, b, c along γ′ in that order. Then the length of γ′ is not shorter than the two-leg polyline abc with the joint point b. Let ℓ be the line containing the edge of Δ₁ that contains b. Now, abc has the same length as abc′, where c′ is the reflection of c over ℓ. Consider the union of Δ₁ and the reflected copy of Δ₁ over ℓ. The union is a rhombus of height 1 containing a and c′ on opposite parallel edges. Thus the length of abc′ is at least 1, and we complete the proof. □

References

[1] H.-K. Ahn, S. W. Bae, O. Cheong, J. Gudmundsson, T. Tokuyama, A. Vigneron, A generalization of the convex Kakeya problem. Algorithmica 70 (2014), 152–170. MR3239787 Zbl 1314.5200210.1007/s00453-013-9831-ySearch in Google Scholar

[2] H.-K. Ahn, O. Cheong, Aligning two convex figures to minimize area or perimeter. Algorithmica 62 (2012), 464–479. MR2886052 Zbl 1311.6816110.1007/s00453-010-9466-1Search in Google Scholar

[3] S. W. Bae, S. Cabello, O. Cheong, Y. Choi, F. Stehn, S. D. Yoon, The reverse Kakeya problem. Adv. Geom. 21 (2021), 75–84. MR4203286 Zbl 1479.5202810.1515/advgeom-2020-0030Search in Google Scholar

[4] J. C. Baez, K. Bagdasaryan, P. Gibbs, The Lebesgue universal covering problem. J. Comput. Geom. 6 (2015), 288–299. MR3400942 Zbl 1408.52029Search in Google Scholar

[5] A. S. Besicovitch, On Kakeya's problem and a similar one. Math. Z. 27 (1928), 312–320. MR1544912 Zbl 0258375110.1007/BF01171101Search in Google Scholar

[6] K. Bezdek, R. Connelly, Covering curves by translates of a convex set. Amer. Math. Monthly 96 (1989), 789–806. MR1033346 Zbl 0706.5200610.1080/00029890.1989.11972283Search in Google Scholar

[7] P. Brass, W. Moser, J. Pach, Research problems in discrete geometry. Springer 2005. MR2163782 Zbl 1086.52001Search in Google Scholar

[8] H. T. Croft, K. J. Falconer, R. K. Guy, Unsolved problems in geometry. Springer 1991. MR1107516 Zbl 0748.5200110.1007/978-1-4612-0963-8Search in Google Scholar

[9] Z. Füredi, J. E. Wetzel, Covers for closed curves of length two. Period. Math. Hungar. 63 (2011), 1–17. MR2853167 Zbl 1265.5202010.1007/s10998-011-7001-zSearch in Google Scholar

[10] P. Gibbs, An Upper Bound for Lebesgue's Covering Problem. Preprint 2018, arXiv:1810.10089Search in Google Scholar

[11] B. Grechuk, S. Som-am, A convex cover for closed unit curves has area at least 0.0975. Internat. J. Comput. Geom. Appl. 30 (2020), 121–139. MR4224201 Zbl 1508.6838610.1142/S0218195920500065Search in Google Scholar

[12] B. Grechuk, S. Som-am, A convex cover for closed unit curves has area at least 0.1. Discrete Optim. 38 (2020), Article ID 100608, 15 pages. MR4143497 Zbl 1506.5203010.1016/j.disopt.2020.100608Search in Google Scholar

[13] S. Kakeya, Some problems on maxima and minima regarding ovals. Tohoku Science Reports 6 (1917), 71–88. Zbl 46.1119.02Search in Google Scholar

[14] L. Moser, Poorly formulated unsolved problems of combinatorial geometry. Mimeographed, 1966.Search in Google Scholar

[15] W. O. J. Moser, Problems, problems, problems. Discrete Appl. Math. 31 (1991, 201–225. MR1106701 Zbl 0817.5200210.1016/0166-218X(91)90071-4Search in Google Scholar

[16] R. Norwood, G. Poole, An improved upper bound for Leo Moser's worm problem. Discrete Comput. Geom. 29 (2003), 409–417. MR1961007 Zbl 1028.5200110.1007/s00454-002-0774-3Search in Google Scholar

[17] J. Pál, Ein Minimumproblem für Ovale. Math. Ann. 83 (1921), 311–319. MR1512015 Zbl 0260303010.1007/BF01458387Search in Google Scholar

[18] C. Panraksa, W. Wichiramala, Wetzel's sector covers unit arcs. Period. Math. Hungar. 82 (2021), 213–222. MR4260141 Zbl 1488.5200910.1007/s10998-020-00354-xSearch in Google Scholar

[19] J.-w. Park, O. Cheong, Smallest universal covers for families of triangles. Comput. Geom. 92 (2021), Paper No. 101686, 10 pages. MR4126371 Zbl 1471.5201710.1016/j.comgeo.2020.101686Search in Google Scholar

[20] J. E. Wetzel, Triangular covers for closed curves of constant length. Elem. Math. 25 (1970), 78–82. MR266051 Zbl 0199.57201Search in Google Scholar

[21] J. E. Wetzel, On Moser's problem of accommodating closed curves in triangles. Elem. Math. 27 (1972), 35–36. MR295214 Zbl 0228.50013Search in Google Scholar

[22] J. E. Wetzel, Sectorial covers for curves of constant length. Canad. Math. Bull. 16 (1973), 367–375. MR362009 Zbl 0271.510.4153/CMB-1973-058-8Search in Google Scholar

[23] J. E. Wetzel, Fits and covers. Math. Mag. 76 (2003), 349–363. MR2085394 Zbl 1059.5202210.1080/0025570X.2003.11953209Search in Google Scholar

[24] W. Wichiramala, A smaller cover for closed unit curves. Miskolc Math. Notes 19 (2018), 691–698. MR3895606 Zbl 1463.5200710.18514/MMN.2018.2485Search in Google Scholar

Received: 2022-03-31

Revised: 2023-03-26

Published Online: 2023-10-13

Published in Print: 2023-10-26

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/advgeom-2023-0021

Keywords for this article

Universal covering; Kakeya problem; discrete rotation

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BY 4.0

Universal convex covering problems under translations and discrete rotations

Article

Abstract

1 Introduction

Covering under rotation by 180 degrees

2.1 The smallest-area covering and related results

Theorem 1

Corollary 1

Theorem 2

Corollary 2

Corollary 3

Corollary 4

2.2 Proof of Theorem 2

Lemma 1

Lemma 2

3 Covering under rotation by 90 degrees

3.1 Covering of unit line segments

Theorem 3

3.2 An equilateral triangle covering of closed curves

Theorem 4

Definition 1

Claim 1

Lemma 3

Lemma 4

Corollary 5

3.3 Minimality of the covering

Conjecture 1

Lemma 5

Proposition 1

4 G k -covering of unit line segments

Theorem 5

5 Covering under rotation by 120 degrees

Lemma 6

Lemma 7

Lemma 8

Theorem 6

Theorem 7

6 Covering of triangles under rotation by 120 degrees

6.1 Construction

6.2 Covering of triangles of perimeter 2

Lemma 9

Lemma 10

Lemma 11

Lemma 12

Lemma 13

Lemma 14

Theorem 8

7 Conclusion

8 Appendix

References

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4 G _k-covering of unit line segments