Covering families of triangles

A cover for a family F\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathcal {F}}}$$\end{document} of sets in the plane is a set into which every set in F\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathcal {F}}}$$\end{document} can be isometrically moved. We are interested in the convex cover of smallest area for a given family of triangles. Park and Cheong conjectured that any family of triangles of bounded diameter has a smallest convex cover that is itself a triangle. The conjecture is equivalent to the claim that for every convex set X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathcal {X}}}$$\end{document} there is a triangle Z whose area is not larger than the area of X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathcal {X}}}$$\end{document}, such that Z covers the family of triangles contained in X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathcal {X}}}$$\end{document}. We prove this claim for the case where a diameter of X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathcal {X}}}$$\end{document} lies on its boundary. We also give a complete characterization of the smallest convex cover for the family of triangles contained in a half-disk, and for the family of triangles contained in a square. In both cases, this cover is a triangle.


Introduction
A cover for a family F of sets in the plane is a set into which every set in F can be isometrically moved.We call a cover smallest if it has smallest possible area.Smallest convex covers have been studied for various families of planar shapes.In 1914, Lebesgue asked for the smallest convex cover for the family of all sets of diameter one.The problem is still open, with the best known upper bound on the area being 0.845 [1,5] and the best known lower bound being 0.832 [3].Moser's worm problem asks for the smallest convex cover for the family of all curves of length one, with the best known upper bound of 0.271 [9,11] and the best known lower bound of 0.232 [6].More variants can be found in [2,10].
These problems appear to be hard because we do not even have a conjecture on the shape of a smallest convex cover.The best lower bound for Lebesgue's problem, for instance, is based on an approximation to the optimal placement of a circle, a triangle, and a pentagon obtained through an exhaustive computer search [3].
While smallest convex covers have remained elusive for most families, we have a complete answer for some families of triangles.Kovalev showed that the smallest convex cover for the family of all triangles of perimeter one is a uniquely determined triangle [7,4].Füredi and Wetzel showed that the same holds for the family of all triangles of diameter one [4], and Park and Cheong showed the same for the family of triangles of circumradius one, as well as for any family of two triangles [8].These known results led Park and Cheong to make the following conjecture: Conjecture 1 ( [8]).For any bounded family T of triangles there is a triangle Z that is a smallest convex cover for T .
It is easy to see that this is equivalent to the following conjecture: Conjecture 2 ( [8]).Let X be a convex set.Then there is a triangle Z whose area is at most the area of X , such that Z is a convex cover for the family of triangles contained in X .
In this paper, we add to the existing evidence motivating these conjectures.In particular, we prove that Conjecture 1 is true for the family of triangles contained in a given half-disk, and for the family of triangles contained in a given square.The half-disk result is a rather easy warm-up exercise, proven in Section 5; see Figure 1.The family of triangles contained in the unit square turns out to be much harder.Intriguingly, there is a "nice" triangle C with angles 60 • , 75 • , and 45 • and a longest edge of length √ 2 that covers "most" triangles contained in the unit square.However, some skinny triangles-the worst case being the isosceles triangle with apex angle ≈ 5.6 • -do not fit into C , and the actual smallest convex cover is a triangle C whose longest edge has length about √ 2 + 0.005.We prove that C indeed covers all triangles contained in the unit square in Section 6; see Figure 2. , and In our second main result, we consider Conjecture 2. It is known to hold when X is a disk [8], a half-disk (Theorem 3), or a square (Theorem 4).In Section 3, we prove the following theorem, which extends this to a much larger family of shapes X : Fig. 2: The smallest convex covers (blue triangles) for triangles in a square.Theorem 5. Let X be a crescent, that is, a convex set that contains a diameter on its boundary.Then there is a triangle Z whose area is at most the area of X , such that Z is a convex cover for the family of triangles contained in X .
Note that we do not claim that the triangle Z is a smallest cover for the family of triangles contained in X .For instance, a half-disk is a crescent, but the triangle Z constructed in the proof of Theorem 5 is larger than the optimal triangle cover of Theorem 3.While proving Conjecture 2 would imply Conjecture 1, the special case of Theorem 5 does therefore not seem to imply any special case of Conjecture 1.In particular, it does not allow us to claim that the family of triangles contained in a given crescent has a triangular smallest convex cover.
The proofs of the three theorems are independent, we start with Theorem 5.

Notation
For three points A, B, C ∈ R 2 , we let AB denote the line through A and B, let AB denote the line segment connecting A and B, and let ABC denote the triangle ABC.When AB is not horizontal, then we let H AB denote the horizontal strip bounded by the horizontal lines through A and through B. For a point P ∈ H AB , we define ζ AB (P ) as the horizontal distance between P and the line AB.Formally, ζ AB (P ) = |P X|, where X is the intersection point of AB with the horizontal line through P .For a point P and a distance t 0, we define points P t = P − (t, 0) and P ⊕ t = P + (t, 0).In other words, P t and P ⊕ t lie on the horizontal line through P at distance t to the left and to the right of P .
We say that a triangle T fits into a convex planar set X if there is a triangle T ⊂ X such that T and T are congruent, that is, T is the image of T under a combination of translations, rotations, and reflections.We say that T maximally fits into X if T fits into X , but there is no triangle T T that fits into X .
We define a crescent to be a convex shape whose diameter lies on its boundary.Any triangle is itself a crescent.For a convex planar set X , let |X | denote the area of X .

Every crescent has a triangular cover
We start by describing how to construct a triangular cover for the family of all triangles in a given crescent.See Figure 3 for illustration.Let X be a crescent with diameter AB ⊂ ∂X .We assume that AB is horizontal and A is to the left of B. Let C be a highest point on ∂X , that is, a point maximizing the distance from AB, let D be a point on the curve AC ⊂ ∂X maximizing the horizontal distance from AC, and let E be a point on the curve BC ⊂ ∂X maximizing the horizontal distance from BC.In other words, X has a horizontal tangent in C, a tangent parallel to AC in D, and a tangent parallel to BC in E. Let To prove Theorem 6, we first need a few lemmas.The first one characterizes triangles that maximally fit into a crescent.Lemma 7. Let X be a crescent with horizontal diameter AB, A left of B, contained in the upper halfplane bounded by AB.If a triangle P QR fits maximally into X , then it is of one of the following three forms: (i) P = A and Q = B, and R ∈ ∂X \ AB; (ii) P = A and R, Q ∈ ∂X \ AB, with R to the left of and strictly above Q; (iii) P = B and R, Q ∈ ∂X \ AB, with R to the left of and strictly below Q.
Proof.Since P QR maximally fits into X , we can assume that P, Q, R all lie on the boundary ∂X .If no vertex lies on AB, we can translate the triangle downwards until it touches AB, so we can assume that P ∈ AB.
then P QR ⊂ ABR, so the maximality implies that P QR = ABR and we are in case (i).It remains to consider the case where P ∈ AB, while Q and R lie on the upper chain ∂X \ AB, so we can assume that R lies to the left of Q.
Let us first assume that R lies above Q.Let K be the intersection point of AB and RQ, and let be the bisector of the angle ∠AKR; see Figure 4. We reflect the points R and Q about to obtain points R * and Q * on the line AB.
We also note that Q * lies strictly between R * and B and thus Q * ∈ AB.
If P lies between R * and Q * , then we can reflect it about to obtain a point P * on the segment RQ so that P * Q * R * is congruent to P QR; see Figure 4(left).Since P * Q * R * ABP * , it does not maximally fit into X .If P lies to the left of R * but is not equal to A, then we can slightly rotate P QR clockwise around R. This moves Q and P into the interior of X , so P QR does not maximally fit into X .
If P lies to the right of Q * , then we rotate Q by 180 • about the midpoint of P R to obtain Q , see Figure 4(right).The quadrilateral P QRQ is a parallelogram, and P RQ is congruent to P QR.Then Q ∈ P RR * since R above Q implies Q above P and P right of QQ * implies Q right of RR * .Since P RQ AP R, P QR does not maximally fit into X .It follows that whenever R lies above Q, then P = A and we are in case (ii).By symmetry, whenever R lies below Q, then P = B and we are in case (iii).
Finally, when RQ is horizontal, we let be the horizontal line equidistant from AB and RQ.Again we mirror R and Q about to obtain R * and Q * on AB.The arguments above apply literally, and we conclude that P = A. By symmetry, however, we can also conclude that P = B, a contradiction.It follows that when RQ is horizontal, then P QR does not maximally fit into X .
We now state two lemmas, postponing their proofs to Section 4.   Proof of Theorem 6.It suffices to prove the statement for an arbitrary triangle P QR that maximally fits into X .By Lemma 7, this implies that P QR is of one of the three types in the lemma.
Case P QR of type(i).
If P QR is of type (i) with P = A and Q = B, then, depending on the location of R, we translate it leftwards by ζ BC (R) or rightwards by ζ AB (R) to place it in A BC or AB C, which are both included in A B C.
Types (ii) and (iii) are symmetric, so we break the symmetry and assume that P = A and R lies to the left and above Q.
Case R not to the left of C. Case Q below AH.
If BAQ BAH, then we rotate P QR clockwise around A. During the rotation, both ζ AC (R) and ζ BC (Q) are decreasing.We continue the rotation to obtain Case R ∈ AHB * . We

Case |AR| |AH|.
Consider now the case where |AR| |AH|.We can rotate P QR clockwise around A to obtain a new triangle P Q R with Q ∈ AB; see Figure 10.Since R lies in the interior of X , the triangle P Q R does not maximally fit into X .
Let H be the foot of the perpendicular from P to BC.Since the line P H is the image of a leftward translation of the line P H by ζ BC (Q ), and Q lies below the line AH, we have Q ∈ H B. We rotate H and B about P by angle ρ to obtain H and B , respectively, so that Q ∈ H B .
We now apply Lemma 9 to ABB * and R, with Note that C in the lemma is our B * , A in the lemma is our P .The B in the lemma will be denoted here To check the lemma's condition on µ, let Q 0 be the point on the line AH at distance |AB| from A and let

Proofs with trigonometry and calculus
In this section we provide the postponed proofs of Lemma 8 and Lemma 9.
of Lemma 8. We denote angles as in Figure 13(left).Note that α + φ = ψ + θ.We claim that BCQ fits into AB C. We distinguish two cases.On the other hand ψ = α + (φ − θ) α, so R lies below AC, and therefore in ABC.

Case 2: θ < φ
In this case, we rotate BCQ by angle θ around B, resulting in BRS with RS parallel to AC; see Figure 14(left).We let BS R be the image of BRS mirrored about the angular bisector of ∠ABR, which means that R lies on AB; see Figure 14 so we need to prove min {sin(ψ), sin(α + β − φ)} sin(α + φ).Suppose this is not the case.Then we have sin(α + φ) < sin ψ.Since α + φ = ψ + θ > ψ and x → sin x is monotonously increasing on [0, π 2 ], we must have α + φ > π 2 and π − (α + φ) < π 2 .On the other hand, we also have sin  Let next δ = ζ AC (R), and let X ∈ AC be the point R ⊕ δ.Applying the law of sines to AXR, we have We now analyse the interval of angles β for which the conditions of the lemma can be satisfied.The triangle ABR 1 is isosceles with two sides of length one and a short side AR 1 of length sin β, so ABR 1 = 2 sin −1 ( 1 2 sin β).The law of sines applied to triangle We set Since C = (− cos 2β, sin 2β), the x-coordinate of R 2 is sin To summarize: • For β 0 β β 1 , R lies on the arc between R 0 and R 1 .The angle ρ is maximized when R = R 1 .For β = β 0 , we have R 1 = R 0 (so there is only a single choice for R), ).Since R cannot lie to the left of R 1 , we have δ h(β).
• For β 1 β < π 2 , R lies on the arc between R 0 and R 2 , with ρ maximized when R = R 2 .Since R cannot lie to the left of R 2 , we have δ g(β).

B position.
Consider now the point B .Since |A B| 1 sin β , it has y-coordinate at most sin ρ sin β .We will prove that HB intersects the vertical line x = 1 through B at ycoordinate at least sin ρ sin β , implying that B lies below HB , and therefore is in BB H.
This is at least sin ρ sin β if and only if Setting f (β) = 3 2 sin 2 β − 1, it remains to show that δ = ζ AC (R) f (β) under the conditions of the lemma.

H position.
We now turn to the point H .It is obtained by rotating H counter-clockwise around A by angle ρ.Since A H is orthogonal to BC, H lies below the line BC.Since ρ < π, H lies above the line A H .To show that H ∈ A H R , it remains to prove that H lies below the line A R .This is equivalent to proving ρ H A R .Let R 0 be the horizontal projection of R 0 on the line BC; see Figure 15.Since the y-coordinate of R 0 is sin β sin 2β, we have R 0 = (1−cos β sin 2β, sin β sin 2β).We have Since |A B| Plotting r(β) shows that it is larger than 0.25 on the interval [β 0 , 2π 5 ].We consider the case β 0 β β 1 .This implies that ρ is maximized when R = R 1 .Combining (1) and ( 2), this gives us sin ρ 2h(β) cos β.Plotting sin −1 (2h(β) cos β) on the interval [β 0 , β 1 ] shows that ρ < 0.2 < 0.25 < r(β).
Fig. 17: Right triangles of diameter two.

Triangles contained in a half-disk
As a warm-up exercise to the square case, we determine the smallest convex cover for the family of triangles contained in the half-disk that is the intersection of the unit disk with the halfplane y 0. The half-disk is a crescent, but the triangular cover constructed in Theorem 6 is in this case not the smallest one.
of Theorem 3. Let T be a triangle that maximally fits into the half-disk, and so T falls into one of the three cases of Lemma 7. Cases (ii) and (iii) cannot occur, since such a triangle can rotate around its bottom vertex.Thus we are in case (i): T is a right-angled triangle whose hypotenuse is the diameter of the half-disk.
By symmetry, we in fact only have to consider the triangles T x whose vertices are (−1, 0), (1, 0), (−x, √ 1 − x 2 ), for x ∈ [0, 1]; see Figure 17(left).When translating T x horizontally so that its upper vertex is on the line segment (−1, 0)(0, 1), the right vertex of the translation of T x is at coordinate (x + √ 1 − x 2 , 0).The x-coordinate of this point is maximized for x = 1 √ 2 , so the triangle Z with vertices (−1, 0), ( √ 2, 0), and (0, 1) is a cover for all T x ; see the blue triangle in Figure 17(right).To complete the proof of Theorem 3, we need to argue that Z is a smallest cover for the family T x .This is true since it is already a smallest cover for the two triangles T 0 and T 1 √ 2 , as can be seen using Corollary 10 of Park and Cheong [8].18: Triangles that fit in a square.

Triangles contained in the unit square
In this section, we prove Theorem 4. We start again by characterizing triangles that maximally fit into the square.
Lemma 10.Let X = ABCD be a square.If a triangle T = P QR fits maximally into X , then without loss of generality, we can assume that P = A, Q lies on BC, and R lies on CD.
Proof.Since T maximally fits into X , we can assume that P, Q, R all lie on the boundary ∂X .Suppose two vertices of T lie on the same side of X , say, P, Q lie on AB.Then T ⊂ ABR ⊂ X as in Figure 18(left).Since T maximally fits into X , this implies P = A, Q = B. Suppose next that no vertex of T coincides with a vertex of X .Then P, Q, R lie on three different sides of X , so we can assume that no vertex lies on AD.We can then translate T upwards so that it no longer touches BC, which implies that T does not maximally fit into X ; see Figure 18(middle).It follows that we can assume that P = A and that Q, R lie on the sides BC and CD.
By Lemma 10, it suffices to study the triangles with P = A, Q ∈ BC, and R ∈ CD.We parameterize these triangles P QR by the two angles θ and θ made by the edges P Q and P R with the diagonal AC of the square.We denote this triangle T θ,θ ; see Figure 18(right).These parameters range in [0, π 4 ] and the case θ = θ = π 6 corresponds to the largest equilateral triangle that can fit in the square.We denote this equilateral triangle as ; see the red triangle in Figure 18(right).

The isosceles case: construction of the cover
We first consider the isosceles triangle T θ,θ with θ π 6 .A convex cover C θ = X θ Y θ Z θ for T θ,θ and T 0 is obtained when P R is aligned with P 0 R 0 , and In the following six sections, we discuss why each triangle T θ,θ indeed fits into C. Figure 20 shows how the six cases cover the entire domain of (θ, θ ).It turns out that only case A requires the cover C, in all other cases T θ,θ fits into the nicer triangle C -so in a sense C is a cover for "most" triangles contained in the unit square.
It follows from the complete characterization of the smallest convex cover for two given triangles by Park and Cheong [8] that C is a smallest convex cover for T 0 and T θ0,θ0 .This makes C a smallest convex cover for the family of all triangles contained in the unit square.
Moreover, C is indeed the unique smallest cover for this family.To show this, we can directly adapt the proof of Lemma 13 by Park and Cheong [8] to argue that a smallest cover for T 0 and T θ0,θ0 that is different from C is a quadrilateral, and that this quadrilateral does not cover either T θ0+ε,θ0+ε or T θ0−ε,θ0−ε for small enough ε.
We now turn to the six cases.Without loss of generality we will always assume that θ θ .θ

Case A
We start with the triangles where θ θ 1 ≈ 5.7 • and θ π 12 .This is the only case where we need to use the cover C-that should not come as a surprise, since T θ0,θ0 falls into this case.
Let P QR = T θ,θ be a triangle with θ θ 1 .Let Q ∈ BC be such that P Q R = T θ,θ .We have seen in Section 6.1 that C θ ⊂ C covers P Q R as in Figure 21.The point Q lies on the segment BQ , so P QR ⊂ C θ as long

Case B
Case B covers those triangles where θ + θ π 4 , except for those triangles we treated in case A. It is nearly identical to case A, but now we can use our "nice" cover C .We place C with X = A and such that R is on X Z .
For θ θ 1 , we again let Q ∈ BC be such that P Q R = T θ,θ .We argued in Section 6.1 that C covers P Q R as in Figure 22.Since RAQ

Case C
We now consider the triangles where θ, θ π 6 .In other words, We first observe that C can be placed such that X Y is vertical and lies on the line AB, while Z lies on the line CD (recall that Y X Z = π Consider now our triangle P QR.We place C such that Z = R; see Figure 24

Case D
We now look at the situation where we have π 12 = 15 • < θ θ π 6 .In other words, we have 12 as in Figure 25(top left).
We observe that C can be placed to cover T 0 = P Q 0 R 0 as in Figure 25

Case E
We consider the situation where θ The second case is illustrated in Figure 26(bottom right).The line X Y has not yet reached R, so R lies above that line.Since QAR π 3 = Y Z X , R lies below the line X Z , and therefore on the highlighted segment in C .Now we compute the horizontal distance f (q) between A and N : Plotting the function f (q) shows that f (q) > 1.01 on the interval 1 q 1.02.Plotting the derivative f (q) shows that f (q) < −0.9 on the interval 1.01 q 1.05, so f (q) is decreasing on this interval.We also know that f (|AQ 0 |) = 1 since then N = R 0 .This implies that f (q) 1 for any Q ∈ BQ 0 .It follows that N lies on or to the left of CD, completing this case and the entire proof.

Theorem 4 .
The unique smallest convex cover for the family of triangles contained in the unit square is the triangleXY Z with XZY = π 3 , |ZY | = 1 cos π 12

Lemma 9 .
Let ABC be an isosceles triangle with |AB| = |AC|.Let AB be horizontal, A left of B, let AH be the height of ABC with respect to BC, and let R ∈ H AC lie to the left of AC with |AR| |AH| and |BR| |AB| (that is, R lies in the green area of Figure 6).Let A = A µ for some µ 0 such that |A B| |AB| |AB| |AH| , let B = B ⊕ ζ AC (R), let H be the orthogonal projection of A on BC, and let R be the horizontal projection of R on BC.We rotate B and H around A by angle CAR, obtaining points B and H , respectively.Then B lies in BB H and H ∈ A H R .

a contradiction. of Lemma 9 .
We scale the problem such that |AB| = |AC| = 1 and place A at the origin, so that B = (1, 0).Let β = ABC, α = BAC = π − 2β, and ρ = CAR.We have |AH| = sin β, and |A B| 1 sin β ; see Figure 15.We first observe that we can replace R by the point at distance sin β from A on AR.This keeps ρ unchanged, decreases |BR|, decreases ζ AC (R), and decreases |H R | so that A H R becomes strictly smaller.So in the following, |AR| = |AH| = sin β.

Fig. 20 :
Fig. 20: Six cases cover all possible triangles.All cases but Case A fit in C .

4 while |X Z | = √ 2 )
. When Z ∈ R 0 D, then the side X Y covers the entire square edge AB.Figure24shows the two extreme cases where Z = R 0 (top left) and where Z = D (top right).

Fig. 25 :
Fig. 25: Case D: covering P QR when π 12 < θ θ (top right).Starting in this position, we can translate C downwards until Y = R 1 .Since X Y is parallel to AR 1 , A lies in C during the entire translation; see Figure 25(bottom left).Among these positions for C , we choose the one where Y = R; see Figure 25(bottom right).Since the line Y

π 6 θ , with the constraints π 4 − θ θ π 3 − 4 RAQ π 3 . 4 =
θ.In other words, R lies on CR 0 , while Q lies on BQ 0 in Figure26, with π We place C with Z = A and X = C. Rotating C clockwise around A, the line X Y intersects BC and CD in two points Q and R , respectively; see Figure 26(top right).We claim that R AQ = π 4 .To see this, consider the point H ∈ X Y such that Z H is a height of C .Since the height |AH| = 1, we have ADR ≡ AHR and ABQ ≡ AHQ .We continue rotating C until either R lies on X Y or Q lies on Y Z .In the first case, R = R ; see Figure 26(bottom left).Then RAQ π R AQ implies that Q lies to the right of Q in C .Since the line Y Z has not yet passed the point Q, Q lies on the highlighted segment in C .

Fig. 27 :Fig. 28 :
Fig. 27: Case F: top left: locations of Q and R; top right: double-covering of T 0 ; middle left: Q = Q 0 ; middle right: Q moving right, M 2 moving down, bottom left: when M 2 reaches R 0 , let Q be the position of Q, bottom right: Q = B.