Universal convex covering problems under translation and discrete rotations

We consider the smallest-area universal covering of planar objects of perimeter 2 (or equivalently closed curves of length 2) allowing translation and discrete rotations. In particular, we show that the solution is an equilateral triangle of height 1 when translation and discrete rotation of $\pi$ are allowed. Our proof is purely geometric and elementary. We also give convex coverings of closed curves of length 2 under translation and discrete rotations of multiples of $\pi/2$ and $2\pi/3$. We show a minimality of the covering for discrete rotation of multiples of $\pi/2$, which is an equilateral triangle of height smaller than 1, and conjecture that the covering is the smallest-area convex covering. Finally, we give the smallest-area convex coverings of all unit segments under translation and discrete rotations $2\pi/k$ for all integers $k\ge 3$.


Introduction
Given a (possibly infinite) set S of planar objects and a group G of geometric transformations, a Gcovering K of S is a region such that every object in S can be contained in K by transforming the object with a suitable transformation g ∈ G. Equivalently, every object of S is contained in g −1 K for a suitable transformation g ∈ G.That is, ∀γ ∈ S, ∃g ∈ G such that gγ ⊆ K.
We denote the group of planar translation by T and that of planar translation and rotation by T R. Mathematically, T R = T R is the semidirect product of T and the rotation group R = SO(2, R).We often call coverings for G-coverings if 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 = T R have been widely studied.
The universal covering problem has attracted many mathematicians.Henri Lebesgue (in his letter to J. Pál in 1914) proposed a problem to find the smallest-area convex T R-covering of all objects of unit diameter (see [7,4,10] for its history).Soichi Kakeya considered in 1917 the T -covering of the set S seg of all unit line segments (called needles) [13].Precisely, his formulation is to find the smallest-area region in which a unit-length needle can be turned round, but it is equivalent to the covering problem if the covering is convex [3].Originally, Kakeya considered the convex covering, and Fujiwara conjectured that the equilateral triangle of height 1 is the solution.The conjecture was affirmatively solved by Pál in 1920 [16].For the nonconvex covering, Besicovitch [5] gave a construction such that the area can be arbitrarily small.Generalizing Pál's result, for any set of n segments, there is a triangle to be the smallest-area convex T -covering of the set, and the triangle can be computed efficiently in O(n log n) time [1].It is further conjectured that the smallest-area convex T R-covering of a family of triangles is a triangle, which is shown to be true for some families [18].
The problem of finding the smallest-area covering and convex covering of the set of all curves of unit length and G = T R was given 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 slightly larger than 0.23, while the best upper bound was about 0.27 for long time [8,7].Wetzel informally conjectured (formally published in [19]) in 1970 that the 30 • circular fan of unit radius, which has an area π/12 ≈ 0.2618, is a convex T R-covering of all unit-length curves, and it is recently proved by Paraksa and Wichiramala [17].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 the case of considering the unit line segments.Corollary 4 is folklore and the authors cannot find its concrete proof in the literature.The fact can be confirmed analytically, since it suffices to consider polyline worms with two legs.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 the topic can be found in an 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 S c 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 degenerate 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 T R despite of substantial efforts in the literature [9,19,20,8,7].As far as the authors know, the smallest-area convex T R-covering known so far is a hexagon obtained by clipping two corners of a rectangle given by Wichiramala, and its area is slightly less than 0.441 [22].It is also shown that the smallest area is at least 0.39 [11], which has been recently improved to 0.4 [12] with help of computer programs.The smallest area of the convex T -covering is known to be between 0.620 and 0.657 [7].
There are some works on restricted shapes of covering.Especially, if we consider triangular coverings, Wetzel [20,21] gave a complete description, and it is shown that an acute triangle with side lengths a, b, c and area X becomes a T -covering (resp.T R-covering) of S c if and only if 2 ≤ 8X 2 abc (resp. 2 ≤ 2πX a+b+c ).As a consequence, the equilateral triangle of side length 4/3 (resp. 2 √ 3 π ) is the smallest triangular T -covering (resp.T R-covering) of S c .Unfortunately, their areas are larger than those of the known smallest-area convex coverings.
If H is a subgroup of G, an H-covering is a G-covering.Since T ⊂ T R, it is quite reasonable to consider groups G lying between them, that is T ⊂ G ⊂ T R. The group R = SO(2, R) is an abelian group, and its finite subgroups are 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 G 2 -covering of S c 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 ≈ 0.538675, and we conjecture that it is the smallest-area convex G 4 -covering.It is a pleasant surprise that the equilateral triangular covering becomes the optimal convex covering if we consider rotation by π (G 2 -covering), and it also seems to be true if we consider rotations by π/2 (G 4 -covering).
However, the minimum area convex G 3 -covering is no longer an equilateral triangle.We give a convex G 3 -covering of S c which has area not larger than 0.568.Our G 3 -covering has areas strictly smaller than the area of the G 2 -covering 1 , and a bit larger than the area of the G 4 -covering β .Unlike the G 2 -and G 4 -coverings, the G 3 -covering is not regular under rotation.We show that any convex G 3 -covering of S c which is regular under rotation by π/2 or 2π/3 has area strictly larger than the area of our G 3 -covering.For triangles of perimeter 2, we give a smaller convex G 3 -covering of them that has area 0.563.
We also determine the set of all smallest-area convex G k -coverings of S seg , which are all triangles.
We use (C) to denote the perimeter of a compact set C, and (γ) to denote the length of a curve γ.The slope of a line is the angle swept from the x-axis in a counterclockwise 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 use pq to denote the line segment connecting p and q, and by |pq| the length of pq.We use pq to denote the line passing through p and q.

Covering under rotation by 180 degrees
In this section, we show that the smallest-area convex G 2 -covering of the set S c of all closed curves of length 2 is the equilateral triangle of height 1, denoted by 1 , whose area is √ 3/3.

The smallest-area covering and related results
First, we recall a famous result mentioned in the introduction.
Theorem 1 (Pal's theorem for the convex Kakeya problem).The equilateral triangle 1 is the smallestarea convex T -covering of the set of all unit line segments.
Proof.Observe that all unit line segments are in S c , and line segments are stable under the action of rotation by π.Thus, any convex G 2 -covering of S c must be a T -covering of all unit line segments, and the corollary follows from Theorem 1 (Pal's theorem).
From Corollary 1, it suffices to show that any closed curve γ in S c can be contained in 1 by applying an action of G 2 to prove the following main theorem in this section.
Theorem 2. The equilateral triangle 1 is the smallest-area convex G 2 -covering of S c .
Before proving the theorem, we give some direct implications of it.An object P is centrally symmetric if −P = P , where −P = {−x | x ∈ P }.Let S sym be the set of all centrally symmetric closed curves of length 2.
Corollary 2. The equilateral triangle 1 is the smallest-area convex T -covering of S sym .
Proof.S sym contains all unit segments.Thus from Theorem 1 (Pal's theorem), the smallest convex T -covering of S sym has area at least the area of 1 as in the proof of Corollary 1.On the other hand, S sym ⊂ S c , and by Theorem 2, any curve in S sym can be contained in 1 by applying a suitable transformation in G 2 .However, a centrally symmetric object is stable under the action of Z 2 , and hence it can be contained in 1 by applying a transformation in T .Thus, 1 is the smallest-area convex T -covering of S sym .
We can consider two special cases shown below.Corollary 2 implies that 1 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 1 is the smallest-area convex T -covering of the set of all rectangles of perimeter 2, and also of the set of all parallelograms of perimeter 2.
We can also obtain the following well-known result about the covering of the set S worm of all curves of length 1 (often called worms) mentioned in the introduction as a corollary.Proof.Given a curve ζ in S worm , we consider the rotated copy −ζ by angle π.Suitably translated, we can form a closed curve γ(ζ) by connecting them at their endpoints.Then γ(ζ) ∈ S sym , and can be contained in 1 under translation.Therefore, ζ is also can be contained there.The corollary follows from Theorem 1 (Pal's theorem).
We can prove Corollary 4 directly by applying a reflecting argument similar to the proof of Theorem 2, which is given in Appendix.

Proof of Theorem 2
A slab is the region bounded by two parallel lines in the plane, and its width is the distance between the two bounding lines.For a closed curve γ of S c , let L θ be the minimum-width slab of orientation θ with 0 ≤ θ < π containing γ.We denote the width of L θ by d θ .
Proof.Let H γ be the hexagon which is the intersection of L 0 , L π/3 and L 2π/3 of γ.See Figure 1(a).Let e 1 , . . ., e 6 be the edges of H γ in counterclockwise 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 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 d 0 + 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 from the definition of S c , the lemma follows.Proof.For a closed curve γ of S c , let P γ be the parallelogram which is the intersection of L π/3 and L 2π/3 of γ.Then, the height h of P γ is d π/3 + d 2π/3 (Figure 2(a)).
In case of h ≤ 1, P γ can be contained in both 1 and − 1 under translation.
In case of h > 1, consider two horizontal lines t and b such that t lies above the bottom corner of P γ at distance 1 and b lies below the top corner of P γ at distance 1.For the distance d between t and b , d + h equals 2 (Figure 2(b)).With Lemma 1, we have This implies that L 0 does not contain both of the lines b and t , so L 0 lies above b or below t .If L 0 lies above b , then γ can be contained in 1 under translation, and if L 0 lies below t , then γ can be contained in − 1 under translation.
For a closed curve γ, either γ or −γ can be contained in 1 under translation by Lemma 2, so we get Theorem 2.

Covering under rotation by 90 degrees
In this section, we consider the convex G 4 -coverings of the sets S seg and S c , where G 4 = T Z 4 .We show that the smallest-area convex G 4 -covering of S seg is the isosceles right triangle with legs (the two equal-length sides) of length 1.For S c , we give an equilateral triangle of height slightly smaller than 1 as a minimal convex G 4 -covering.

Covering of unit line segments
First, we consider the set S seg of all unit segments.Theorem 3. The smallest-area convex G 4 -covering of all unit segments has area 1/2, and it is uniquely attained by the isosceles right triangle with legs of length 1 under closure.
Proof.Let 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 for 0 ≤ θ ≤ π/2.Thus, is a G 4 -covering of S seg , and its area is 1/2.Now we show that any convex G 4 -covering of S seg has area at least 1/2.Let X be a smallest-area convex G 4 -covering of S seg , 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.
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 Since X contains a unit segment s(π/2) and a unit segment 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, ) be the open interval in [0, π/2) centered at an angle x with radius , and let θ be an angle in A − .Since A − contains no interval, there is a sequence {θ k } ∞ k=1 such that θ k ∈ I(θ , 1/k) ∩ A + for each k, and lim k→∞ θ k = θ .Since X contains a unit segment s(θ ) and a unit segment s(θ k + π/2) for any k, X contains their convex hull.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 θ, other than 0 and π/2.Then there is a sequence for each n, and lim n→∞ θ n = θ.Since X contains unit segments s( θk ) and s( θ n + 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 convex G 4 -covering of S seg under its closure.We assume that X is compact.First, we prove that if there is a sequence of angles in p=1 that converges 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 unit segments s( θ) and s( θ + π/2), orthogonal to each other.Moreover, if X is a covering of area 1/2, X is the convex hull Φ of s( θ) and s( θ + π/2).There are two cases of Φ : it is either a convex quadrilateral or a triangle of height 1 and base length 1.
Consider the case that Φ is a convex quadrilateral.Then both θ and θ + π/2 are isolated points in A. Thus, Φ is not a G 4 -covering of S seg .Consider the case that Φ is a triangle of height 1 and base length 1.Without loss of generality, assume that s( θ) is the base of Φ and s( θ + π/2) corresponds to the height of Φ. Suppose that Φ is an acute triangle.Since s( θ) is the base of Φ, θ is an isolated point in A. Observe that s( θ + π/2) can be rotated infinitesimally around the corner opposite to the base in both directions while it is still contained in Φ.Thus, θ + π/2 is an interior point of an interval I in A. So for an endpoint θ of I other than θ, Φ contains unit segments s( θ) and s( θ + π/2).The convex hull Φ of the two segments is also a convex quadrilateral or a non-obtuse triangle of area at least 1/2.Since Φ is contained in Φ of area 1/2, Φ = Φ, that is, Φ must be an acute triangle with base b of length 1.Since θ = θ, b is not the base of Φ, and b must be one of the other two sides of Φ.But clearly both the sides are longer than 1, the height of Φ, a contradiction.Hence Φ is not an acute triangle.Thus, Φ is the isosceles right triangle, and it is the unique G 4 -covering of S seg under closure.

An equilateral triangle covering of closed curves
Unfortunately, the isosceles right triangle with legs of length 1 is not a G 4 -coverings of S c , 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 [2], it can be shown that the area of the convex hull of the union of the isosceles right triangle and any translated copy of the disk has area at least 0.543.
Naturally, the equilateral triangle of height 1 is a G 4 -covering of S c , since it is a G 2 -covering of S c .However, a smaller equilateral triangle can be a G 4 -covering of S c , and we seek for the smallest one, which is conjectured to be the smallest-area G 4 -covering of S c .Consider an equilateral triangle that is a G 4 -covering of S c .Since it is a G 4 -covering of S seg , 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 ≈ 0.538675.We denote this triangle by β .
The smallest equilateral triangle ∆ β containing a pair of orthogonal unit segments.
Theorem 4. The equilateral triangle β is a convex G 4 -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 β is at least 2.
The claim implies that any closed curve of length smaller than 2 can be placed in β under G 4 transformation, but it is not G 4 -brimful for β .Thus, if the claim is true, then every closed curve of length 2 can be placed in β under G 4 transformation, so Theorem 4 holds.Thus, we will prove the claim.
Let γ be a G 4 -brimful curve for β .Let g i = e 2iπ √ −1/4 ∈ Z 4 for i = 0, . . ., 3. We consider the rotated copy g i β of β , and find the smallest scaled copy Υ i of g i β which circumscribes γ with a proper translation.Let a i be the scaling factor, and 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 intersection H = H(γ) = 3 i=0 Υ i 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 degenerated edges.Note that H consists of edges of slopes 2kπ/12 (mod π) for k = 0, . . ., 11.
The following lemma states that the perimeter of H only depends on a i (i = 0, 1, 2, 3).
) \ H consists of 12 triangles (gray), each of which is an isosceles triangle with apex angle 2π/3.

Lemma 3. The perimeter (H) equals a0+a1+a2+a3
2 cos(π/12) .Proof.Let δ be the side length of ∆ β .The intersection X 0,2 = Υ 0 ∩ Υ 2 is a hexagon where γ touches all of its six (possibly degenerated) edges.Then, (Υ 0 ∪ Υ 2 ) \ X 0,2 consists of six equilateral triangles, and the total sum of the perimeters of them equals (Υ 0 ) + (Υ 2 ) = 3(a 0 + a 2 )δ.On the other hand, one edge of each equilateral triangle contributes to the boundary polygon of X 0,2 .Hence, (X 0,2 ) = (a 0 + a 2 )δ.Similarly, the perimeter of . 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 equal (X 0,2 ) + (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 The third equality comes from δ = 2  Proof.Let p k be the point on e k for k = 0, . . ., 11.The circuit C connects the points from p 0 to p 12 in increasing order of their indices, where p 12 is a duplicate of p 0 .We also assume that p 12 is on e 12 which is a duplicate of e 0 .While incrementing k by 1 from 1 to 11, we reflect the edges e k+1 , . . ., e 12 about the edge e k .Then, the edges e 0 ,. . .,e 12 are transformed to a zigzag path alternating a horizontal edge and an edge of slope 2π/12, and C becomes the path connecting p 0 , p 1 , p 2 , . . ., p 12 , where p k is the location of p k after the series of reflections for k = 2, . . ., 12. Thus, (C) is at least the distance between p 0 to p 12 .See Figure 5 for the illustration.The vector x = p 0 p 12 is the sum of a horizontal vector a and another vector b of slope π/6 such that |a| + |b| = (P ).The size |x| of x is minimized when |a| = |b| to attain the value (|a| + |b|) cos(π/12).Thus, |x| ≥ (P ) cos(π/12) and (C) ≥ (P ) cos(π/12).
From Lemma 4, the following corollary is immediate.
Combining Lemma 3 and Corollary 5, we have (γ) The claim is proved.

Minimality of the covering
There are many G 4 -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.They suggest that β is the smallest-area convex G 4 -covering of S c .Conjecture 1. β is the smallest-area convex G 4 -covering of S c .

Although we
have not yet proven it rigorously, we can prove that it is minimal in the set theoretic sense.That is, no proper closed subset of β can be a G 4 -covering of S c .
Ahn et al. [1] proved there exists a triangle that is the smallest-area convex T -covering of any given set of segments, and showed an algorithm to construct the triangle.For the six unit segments, their algorithm computes the smallest regular hexagon that contains them under translation.Since the hexagon is the Minkowski symmetrization of β , which is { 1 2 (x − y) | x, y ∈ β }, the algorithm returns β as the smallest-area closed convex T -covering of the six unit segments.Thus, the lemma follows.
Proposition 1. β is a minimal closed convex G 4 -covering of S c .
Proof.Suppose P ⊆ β is a closed subset and a G 4 -covering of S c .Observe that each angle θ in A has no other angle in A that is equivalent to θ under the Z 4 action.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 β .Therefore, P must be β itself.

G k -covering of unit line segments
Consider the smallest-area convex G k -covering of the set S seg of all unit line segments.In general, a smallest-area convex T -covering of any given set of segments is attained by a triangle [1].This implies that there is a triangle that is a smallest-area convex G k -covering of S seg .The following theorem determines the set of all smallest-area convex G k -coverings.
Theorem 5.If k ≥ 3 is odd, the smallest area of convex G k -covering of S seg is 1  2 sin(π/k), and it is attained by any triangle XY Z with bottom side XY of length 1 and height sin(π/k) such that π/2 ≤ ∠X ≤ (k − 1)π/k.If k ≥ 4 is even, the smallest area of convex G k -covering of S seg is 1  2 sin(2π/k), and it is attained by any triangle XY Z with bottom side XY of length 1 and height sin(2π/k) such that π/2 ≤ ∠X ≤ (k − 2)π/k.
Proof.We have already seen a proof for k = 4, and it can be generalized as follows.Let Λ be a smallest-area convex G k -covering of S seg .
First, let us consider the case that 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/k (mod π) must be contained in Λ under translation for each angle θ with 0 ≤ θ < 2π/k.We define the smallest one of such angles to be f (θ).As before, we can assume there exists an angle θ such that f ( θ) = θ.
Let A = {f (θ) | 0 ≤ θ < 2π/k} be the set of angles, let Ā be the complement of A, and let s(θ) be a unit segment of slope θ.There exists an angle θ with 0 ≤ θ < 2π/k such that θ is contained in both A and the closure of Ā.There is a sequence { θn } ∞ n=1 ⊆ A such that lim n→∞ θn = θ + 2iπ/k for some i.Then, Λ contains two segments s( θ) and s( θn ) for any n, and their convex hull Φ n .Since lim n→∞ θn = θ + 2iπ/k, , the area of Λ is at least 1  2 sin(π/k).On the other hand, let XY Z 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 XY Z. Any unit line segment of slope θ 1 for 0 ≤ θ 1 ≤ ∠Y can be placed within XY Z with one endpoint at Y , and any unit line segment of slope θ 2 for ∠Y ≤ θ 2 ≤ max{∠Y, π/k} can be placed within XY Z with one endpoint at Z. Thus, any segment of slope θ with 0 ≤ θ ≤ π/k can be placed within XY Z.
The case of even k can be proven analogously.The only difference is that i = k/2, and the minimum area is attained at i = k/2 − 1. Analogously to the odd k case, the area can be minimum only if the convex hull of s(θ) ∪ s(θ + 2iπ/k) is a triangle, and it is routine to derive the conditions that the triangle is a G k -covering.

Covering under rotation by 120 degrees
We construct a convex G 3 -covering of S c , denoted by Γ 3 , as follows.Let Γ be the convex region bounded by y 2 = 1 + 2x and y 2 = 1 − 2x, and containing the origin O. Then Γ 3 is the convex subregion of Γ bounded by the x-axis and the line y = 2/3.The area of Γ 3 is 81 , which is smaller than 0.5680.See Figure 7(a).
We show that Γ 3 is a G 3 -covering of S c .We first show a few properties that we use in the course.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 + .Γ + is called a church window, which is a T -covering of S c [6].The following lemma gives a lower bound on the length of a T -brimful curve for Γ + .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 always be translated to lie in the interior of Γ + .Let XY Z be a triangle for the touching points X, Y, Z of the curve with the boundary of Γ + .Since (γ) ≥ ( XY Z), γ is XY Z.
Without loss of generality, 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), γ becomes a line segment and the length of the line segment is always larger than or equal to 1. Thus, (γ) ≥ 2. Now we assume that none of X, Y , and Z is on a corner of Γ + .Let 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 Z be the point symmetric to Z with respect to X .See Figure 8(a).
We claim that Z, X, Y , and Z are collinear.If X, Y , and Z are not collinear, consider the intersection For the point X where X Y intersects the left side of Γ + , we have X Y Z ⊂ X Y Z. Therefore, ( X Y Z) < ( X Y Z) < (γ), and this contradicts the assumption on γ.Suppose that XZ is not horizontal.From the collinearity of X, Z, and Z , the reflection of XZ in the tangent line X is 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 .From the geometry of the parabola, the reflection of XXr in the tangent line X is XX r .Therefore, ∠Z X X r = ∠Z XX r , implying that 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.This contradicts that XZ is not horizontal.
Since XZ is horizontal, Y is at the origin.Thus, XY Z is an isosceles triangle with base XZ, and ( XY Z) = 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]).For k compact convex figures C i for i = 1, . . ., k in the plane, the perimeter function of their convex hull of C i (r) is convex, where C i (r) for a vector r = (r 1 , . .
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.For k compact convex figures C i for i = 1, . . ., k that have point symmetry in the plane, the perimeter function of their convex hull of C i (r) is minimized when their centers (of the symmetry) meet at a point, where C i (r) for a vector r = (r 1 , . . Proof.Without loss of generality, assume that the k compact convex figures are given with centers all lying at the origin.Let r = (r 1 , . . ., 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 = (−r 1 , . . ., −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 C i (0) also has the minimum perimeter, where 0 is the zero vector (a vector of length zero).
We are now ready to have a main result.Theorem 6. Γ 3 is a convex G 3 -covering of all closed curves of length 2.
Proof.By Lemma 6, any closed curve of length 2 can be contained in Γ + under translation.Without loss of generality, assume that Γ 3 is given as a part of Γ + .Let C be a closed curve of length 2 that is contained in Γ + and touches its bottom side, and let C be the convex hull of C.
Suppose that C crosses the top side of Γ 3 .Let s be a segment contained in C and connecting the top side and the bottom side of Γ 3 such that the upper endpoint of s lies in the interior of C. For each 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 Γ Assume to the contrary that neither C 1 nor C 2 is contained in Γ 3 .Then both curves cross the top side of Γ 3 .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 Γ 3 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 C. Let Φ be the convex hull of s, s 1 , and s 2 .Since s, s 1 , s 2 ⊂ C and the upper endpoint of s lies in the interior of C, (Φ) Now consider a translation of these three segments such that their midpoints meet at a point, and let Φ m be the convex hull of the three translated segments.By Lemma 8, (Φ m ) ≤ (Φ).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 L 0 , L π/3 , and L 2π/3 of Φ m .Observe that d θ for θ = 0, π/3, 2π/3 is at least height Recall that the smallest-area convex G 2 -covering 1 and G 4 -covering β of S c are equilateral triangles.Our G 3 -covering, Γ 3 , has area smaller than the area of 1 , but a bit larger than the area of β , which sounds reasonable.
However, it may look odd that Γ 3 is not regular under any discrete rotation while 1 and β are regular under rotation by 2π/3.We show that any convex G 3 -covering regular under rotation by 2π/3 or π/2 has area strictly larger than the area of Γ 3 .
Let Λ be a convex G 3 -covering which is regular under rotation by 2π/3.Then Λ is a T -covering of all unit segments of any slope θ for 0 ≤ θ < π.Since 1 is the smallest-area convex T -covering of the set of all unit segments by Theorem 1, the area of Λ is at least the area of 1 , which is strictly larger than the area of Γ 3 .
Let Λ be a convex G 3 -covering which is regular under rotation by π/2.Assume that a unit segment s of slope π/4 is contained in Λ.Since Λ is a G 3 -covering, there is a unit segment s 1 of one of slopes {0, π/3, 2π/3} contained in Λ. Assume that s 1 of slope π/3 is contained in Λ.Since Λ is regular under rotation by π/2, there are unit segments s of slope 3π/4 = π/4 + π/2 and s 1 of slope 5π/6 = π/3 + π/2 contained in Λ.Thus, the four segments s, s , s 1 , s 1 are contained in Λ.
Let c be the point of symmetry of Λ.Let Φ be the convex hull of the translated copies of s, s , s 1 , s 1 such that their midpoints are all at c.We will prove that the area Λ of Λ is at least the area Φ of Φ.Then Λ ≥ Φ = √ 6/4 > Γ 3 , where Γ 3 is the area of Γ 3 .Suppose that the midpoint of s is not at c. Let s iπ/2 be the copy obtained by rotating s around c by iπ/2 for i = 1, 2, 3. Since Λ is regular under the rotation by π/2, s iπ/2 is contained in Λ for every i = 1, 2, 3. Since Λ is convex, the convex hull Φ 1 of s and the segments s iπ/2 for all i = 1, 2, 3 is contained in Λ, and thus Λ ≥ Φ 1 .See Figure 9(a).
Let s 1 be the translated copy of s such that the midpoint of s 1 is at c, and s 2 be the copy of s 1 rotated by π/2 around c. Let Φ 2 be the convex hull of s 1 and s 2 .Since s 1 is contained in the convex hull of s and s π , and s 2 is contained in the convex hull of s π/2 and s 3π/2 , Φ 2 ⊂ Φ 1 .See Figure 9(b).Similarly, the convex hull of the translated copy s of s 1 with midpoint lying at c and the rotated copy of s by π/2 around c is contained in Λ.Thus, we conclude that Λ ≥ Φ ≥ √ 6/4 > 0.6 > Γ 3 .We can show this for s 1 of slopes 0 and 2π/3 contained in Λ in a similar way.
6 Covering of triangles under rotation by 120 degrees

Construction
Let S t be the set of all triangles of perimeter 2. We construct a convex G 3 -covering of S t , denoted by Γ t , from Γ 3 by shaving off some regions around the top corners.Consider an equilateral triangle for t varying from −4/9 to −1/3.The trajectory of Z forms the top-right boundary of Γ t that connects the top side and the right side of Γ 3 .Thus, the region of Γ 3 lying above the trajectory is shaved off.The top-left boundary of Γ t can be obtained similarly.Figure 10(c .
We obtain where f 1 (t) = √ −18t − 5 and f 2 (t) = √ 2t + 1.Since f 1 (t) > 0 and f 2 (t) > 0, the denominator of f (t) is positive.Since the numerator of f (t) is negative, d dt ( dy dx )/ dx dt ≤ 0. At t = −4/9, dy dx = 0, which is the slope of the top side of Γ 3 .At t = −1/3, dy dx = − √ 3, which is the slope of the tangent to the right side of Γ 3 at the same point.Thus, Γ t is convex.Now we show the area of Γ t .The area that is shaved off from Γ 3 is 2 2 3 where f (x) is the function of γ DE such that Thus, Γ t has area smaller than 0.5634.

Covering of triangles of perimeter 2
We show that Γ t is a G 3 -covering of all triangles of perimeter 2. To do this, we first show a few properties that we use in the course.Let A, B, C, D, E, and F be the boundary points of Γ t as shown in Figure 10.We denote the boundary curve of Γ t from a point a to a point b in clockwise direction along the boundary by γ ab .Proof.If XY Z lies in the left of OE (including the line), XY Z lies in between OE and the line tangent to Γ t at B. The two lines are of slope π/3 and they are at distance √ 3/3.Thus there are copies of XY Z rotated by 2π/3 and lying in between BE and AF .Among such copies, let be the one that touches γ F A from above and γ AB from right.Since ( ) = 2, is contained in Γ t by Lemma 6. See Figure 11(a).The case of XY Z lying in the right of OB can be shown by a copy of the triangle rotated by −2π/3.
In the following, we assume that XY Z is contained in Γ but it is not contained in Γ t .If there is no corner of XY Z lying in the left of OB or in the right of OE (including the lines), Γ t is a G 3 -covering of XY Z by Lemma 9. Thus, we assume that X is in the left of OB and Y is in the right of OE .Since Γ t is convex, the remaining corner Z of XY Z lies in Γ 3 \ Γ t .Let R 1 and R 2 denote the left and right regions of Γ 3 \ Γ t , respectively, as shown in Figure 11(a).Translate XY Z leftwards horizontally until X or Z hits the left side of Γ 3 .If the triangle lies in the left of OE (including the line), we are done by Lemma 9. Thus, we assume that Z is in R 1 ∪ R 2 and Y lies in the right of OE .There are two cases, either Z ∈ R 1 or Z ∈ R 2 .See Figure 11(b) and (c).If Z is in R 2 , then X is at A. Lemma 10.Let XY Z be a triangle contained in Γ 3 such that X is at A, Y is in the right of OE , and Z ∈ R 2 .Then Γ t is a convex G 3 -covering of XY Z.
Proof.Let = X Y Z be the copy of XY Z rotated by 2π/3 such that X lies at F .We show that is contained in Γ t .Assume to the contrary that is not contained in Γ t .Since ∠ZAF > π/6 and F is at distance at least 1 from any point on γ AB , Z must be contained in Γ t .See Figure 12(a) and (b).
If Y lies on or below BE , is contained in Γ t by Lemma 6.So assume that Y lies above BE .Then Y must lie to the right of the line of slope π/3 and passing through the point of γ F O at distance 1/3 from F .See Figure 12(c).Let H be the point at (0, 2/3).Then there is a triangle XV W with V ∈ , and W ∈ HE such that ( XV W ) < ( XY Z).Thus, to show a contradiction, it suffices to show ( XV W ) ≥ 2. For a point p, let p denote the point symmetric to p along .Then Since X is at distance at least 7/6 to any point in H Ē and at distance at least 5/6 to any point in HE, We can also show that Γ t is a G 3 -covering of XY Z for the remaining case of Z ∈ R 1 .
Lemma 11.Let XY Z be a triangle contained in Γ 3 such that X is on γ OA , Y is in the right of OE , and Z ∈ R 1 .Then Γ t is a convex G 3 -covering of XY Z.
Before proving the lemma, we need a few technical lemmas.
Lemma 12. Let XY Z be an isosceles triangle of perimeter 2 such that its base Y Z is of length ≥ 2/3 and parallel to the bottom side of Γ t , and X lies at O. Then XY Z can be rotated in a clockwise direction within Γ t such that X moves along γ OA and Y moves along γ EF until Y meets F .
Proof.Let XY Z be an equilateral triangle satisfying the conditions in the lemma statement.Then |Y Z| = 2/3 and ∠Y OF = π/3.Let p be a point in Γ t lying below Y O , and let Q be the convex hull of p and XY Z.Let θ be the angle ∠pOF .See Figure 13(a).We claim that Q can be rotated within Γ t in a clockwise direction such that Y moves along γ EF and X moves along γ OA until p reaches AF .Let Q(θ) = X Y Z p be the rotated copy of Q by θ with 0 ≤ θ ≤ θ 0 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 Q T (θ) = X * Y * Z * p * be the translated copy of Q(θ) such that Y * ∈ γ EF , X * ∈ γ OA , and each corner κ * of Q T (θ) corresponds to the translated point of κ for κ ∈ {X , Y , Z , p }. See Figure 13(b) and (c) for an illustration.
Let Φ be the convex hull of XY Z and X Ȳ Z. See Figure 14(b).Imagine that Φ rotates in a clockwise direction such that X moves along γ OA and Y moves along γ EF until Ȳ reaches γ F A .Then Z moves along γ BC , and then moves into the interior of Γ t during the rotation because XY Z is an equilateral triangle with |XZ| = 2/3.Let Φ(θ ) denote the rotated copy of Φ by θ − π/3, where θ = ∠ZXA.Then θ increases monotonically from π/3 during the rotation.See Figure 14(c).
Consider the rotation for θ from π/3 to π/2.Observe that Ȳ is contained in Γ t during the rotation by the argument in the first paragraph of the proof.Since Z = Z and Z ∈ Γ t during the rotation (shown in Section 6.1), Z is also in Γ t .Now we claim that X is contained in Γ t during the rotation.For any θ with π/3 < θ ≤ π/2, X lies above OA , but it lies below BE because ∠ XF A ≤ π/6 for θ ≤ π 2, . . ., where e θ √ −1 means the rotation of angle θ.

Corollary 4 .
The equilateral triangle 1 is the smallest convex T -covering of S worm .

2 Figure 5 :
Figure 5: Illustration of the proof of Lemma 4.

Lemma 5 .
β is the smallest-area closed convex T -covering of S A .

Figure 8 :
Figure 8: Illustration of the proof of Lemma 6.

2 OFigure 11 :
Figure 11: (a) XY Z lying in the left of OE .There is a copy that is contained in Γ t .(b) XY Z contained in Γ 3 such that X is at A, Y is in the right of OE , and Z ∈ R 2 .(c) XY Z contained in Γ 3 such that X is on γ OA , Y is in the right of OE , and Z ∈ R 1 .
Figure 12: (a) XY Z with Z ∈ R 2 .(b) A rotated copy X Y Z of XY Z by 2π/3 with X lying at F .(c) If Y lies in the right of , then ( XY Z) > ( XV W ) ≥ 2.

Figure 13 :Figure 14 :
Figure 13: (a) The convex hull Q of an equilateral triangle XY Z and p.(b) Rotated copy Q(θ) of Q by θ around O. (c) Translated copy Q T (θ) of Q(θ) such that Y * ∈ γ EF and X * ∈ γ OA

/ 2 . 3 Figure 15 :
Figure 15: (a) X lies below B or X = B .(b) X lies below or on X .

Figure 16 :
Figure 16: (a) Z ∈ R 1 and ZX is the longest side.(b) A copy X Ȳ Z of XY Z rotated by 2π/3.(c) A copy X Ȳ Z of XY Z rotated by −2π/3.
) shows Γ t .We show that Γ t is convex by showing that d dx ( dy dx ) = d dt ( dy dx )/ dx dt ≤ 0 for Z = (x(t), y(t)), and the boundary of Γ t has a unique tangent at t = −4/9 and −1/3.Since the x-coordinate of Z increases as t increases, dx dt > 0. Thus, it suffices to show that d dt ( dy dx ) ≤ 0 for t with −4/9 ≤ t ≤ −1/3.Observe that