Covering families of triangles

Otfried Cheong, Olivier Devillers, Ji-Won Park, Marc Glisse

► To cite this version:

Otfried Cheong, Olivier Devillers, Ji-Won Park, Marc Glisse. Covering families of triangles. Periodica Mathematica Hungarica, 2023, 87, pp.86–109. 10.1007/s10998-022-00503-4. hal-03662311

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Distributed under a Creative Commons Attribution 4.0 International License
Covering families of triangles*

Otfried Cheong†

Olivier Devillers‡

Marc Glisse§

Ji-won Park¶

**Abstract**

A cover for a family $\mathcal{F}$ of sets in the plane is a set into which every set in $\mathcal{F}$ 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 $\mathcal{X}$ there is a triangle $\mathcal{Z}$ whose area is not larger than the area of $\mathcal{X}$, such that $\mathcal{Z}$ covers the family of triangles contained in $\mathcal{X}$. We prove this claim for the case where a diameter of $\mathcal{X}$ 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.

<footnote>*ANR-17-CE40-0017 of the French National Research Agency (Aspag)</footnote>

<footnote>†School of Computing, KAIST, Daejeon, Korea. otfried@kaist.airpost.net.
https://orcid.org/0000-0003-4467-7075</footnote>

<footnote>‡Université de Lorraine, CNRS, Inria, LORIA, F-54000 Nancy, France.
Olivier.Devillers@inria.fr. https://orcid.org/0000-0003-4275-5068.</footnote>

<footnote>§Université Paris-Saclay, CNRS, Inria, Laboratoire de Mathématiques d'Orsay, 91405, Orsay, France. Marc.Glisse@inria.fr. https://orcid.org/0000-0001-6914-1651.</footnote>

<footnote>¶Université de Lorraine, CNRS, Inria, LORIA, F-54000 Nancy, France. Ji-won.Park@inria.fr. https://orcid.org/0000-0003-2490-1359.</footnote>
# 1 Introduction

A cover for a family $\mathcal{F}$ of sets in the plane is a set into which every set in $\mathcal{F}$ 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 $\mathcal{T}$ of triangles there is a triangle $Z$ that is a smallest convex cover for $\mathcal{T}$.

It is easy to see that this is equivalent to the following conjecture:

**Conjecture 2 ([8]).** Let $\mathcal{X}$ be a convex set. Then there is a triangle $Z$ whose area is at most the area of $\mathcal{X}$, such that $Z$ is a convex cover for the family of triangles contained in $\mathcal{X}$.

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.
<figure><img src="image_2.png" /><figcaption>Fig. 1: The smallest convex covers (blue triangles) for triangles in a half-disk.</figcaption></figure>

**Theorem 3.** The triangle with sides $\sqrt{2}$, $1+\sqrt{2}$, and $\sqrt{3}$ is a smallest convex cover for the family of triangles contained in the half-disk of radius one.

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 $\sqrt{2}$ that covers “most” triangles contained in the unit square. However, some skinny triangles—the worst case being the isosceles triangle with apex angle $\approx 5.6^\circ$—do not fit into $C^*$, and the actual smallest convex cover is a triangle $C$ whose longest edge has length about $\sqrt{2} + 0.005$. We prove that $C$ indeed covers all triangles contained in the unit square in Section 6; see Figure 2.

**Theorem 4.** The unique smallest convex cover for the family of triangles contained in the unit square is the triangle $\triangle XYZ$ with $\angle XZY = \frac{\pi}{3}$, $|ZY| = \frac{1}{\cos \frac{\pi}{12}}$, and

$$|XZ| = \frac{\sin(\frac{\pi}{3} + 2\theta_0)}{\cos(\frac{\pi}{4} - \theta_0) \sin(\frac{\pi}{3})} \approx 1.4195,$$ \\
$$\text{where } \theta_0 = \tan^{-1} \left( \frac{1}{\sqrt{2} + \sqrt{3}} \right) - \frac{\pi}{6} \approx 0.049 \approx 2.81^\circ.$$

In our second main result, we consider Conjecture 2. It is known to hold when $\mathcal{X}$ 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 $\mathcal{X}$:

<figure><img src="image_3.png" /><figcaption>Fig. 2: The smallest convex covers (blue triangles) for triangles in a square.</figcaption></figure>
**Theorem 5.** Let $\mathcal{X}$ 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 $\mathcal{X}$, such that $Z$ is a convex cover for the family of triangles contained in $\mathcal{X}$.

Note that we do not claim that the triangle $Z$ is a smallest cover for the family of triangles contained in $\mathcal{X}$. 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.

## 2 Notation

For three points $A, B, C \in \mathbb{R}^2$, we let $AB$ denote the line through $A$ and $B$, let $\overline{AB}$ denote the line segment connecting $A$ and $B$, and let $\triangle ABC$ 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 $P \in H_{AB}$, we define $\zeta_{AB}(P)$ as the horizontal distance between $P$ and the line $AB$. Formally, $\zeta_{AB}(P) = |PX|$, where $X$ is the intersection point of $AB$ with the horizontal line through $P$.

For a point $P$ and a distance $t \ge 0$, we define points $P \ominus t = P - (t, 0)$ and $P \oplus t = P + (t, 0)$. In other words, $P \ominus t$ and $P \oplus t$ 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 $\mathcal{X}$ if there is a triangle $T' \subset \mathcal{X}$ 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 $\mathcal{X}$ if $T$ fits into $\mathcal{X}$, but there is no triangle $T' \supseteq T$ that fits into $\mathcal{X}$.

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 $\mathcal{X}$, let $|\mathcal{X}|$ denote the area of $\mathcal{X}$.

## 3 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 $\mathcal{X}$ be a crescent with diameter $AB \subset \partial\mathcal{X}$. We assume that $AB$ is horizontal and $A$ is to the left of $B$. Let $C$ be a highest point on $\partial\mathcal{X}$, that is, a point maximizing the distance from $AB$, let $D$ be a point on the curve $AC \subset \partial\mathcal{X}$ maximizing the horizontal distance from $AC$, and let $E$ be a point on the curve $BC \subset \partial\mathcal{X}$ maximizing the horizontal distance from $BC$. In other words, $\mathcal{X}$ has a horizontal tangent in $C$, a tangent parallel to $AC$ in $D$, and a tangent parallel to $BC$ in $E$. Let
<figure><img src="image_4.png" /><figcaption>Fig. 3: Construction of a triangular cover from a crescent.</figcaption></figure>

$A' = A \ominus \zeta_{BC}(E)$ and $B' = B \oplus \zeta_{AC}(D)$. We claim that $\triangle A'B'C$ is indeed a cover for the set of triangles in $\mathcal{X}$, and that $|\triangle A'B'C| \le |\mathcal{X}|$.

**Theorem 6.** If a triangle fits into the crescent $\mathcal{X}$, then it fits into the triangle $\triangle A'B'C$.

Before we prove Theorem 6, we show how it implies the result stated in the introduction.

of Theorem 5. It suffices to observe that for the triangle $Z = \triangle A'B'C$ constructed in Theorem 6 we have $|Z| \le |\mathcal{X}|$ since

$$Z = \triangle A'AC \cup \triangle ABC \cup \triangle BB'C, \\ \mathcal{X} \supset \triangle ADC \cup \triangle ABC \cup \triangle BEC,$$

$$\text{and } |\triangle A'AC| = |\triangle BEC| \text{ and } |\triangle BB'C| = |\triangle ADC|. \quad \square \quad \square$$

<figure><img src="image_5.png" /></figure>

<figure><img src="image_6.png" /></figure>

To prove Theorem 6, we first need a few lemmas. The first one characterizes triangles that maximally fit into a crescent.

**Lemma 7.** Let $\mathcal{X}$ be a crescent with horizontal diameter $\overline{AB}$, A left of B, contained in the upper halfplane bounded by AB. If a triangle $\triangle PQR$ fits maximally into $\mathcal{X}$, then it is of one of the following three forms:

(i) $P = A$ and $Q = B$, and $R \in \partial \mathcal{X} \setminus \overline{AB}$;

(ii) $P = A$ and $R$, $Q \in \partial \mathcal{X} \setminus \overline{AB}$, with $R$ to the left of and strictly above $Q$;

(iii) $P = B$ and $R$, $Q \in \partial \mathcal{X} \setminus \overline{AB}$, with $R$ to the left of and strictly below $Q$.

*Proof.* Since $\triangle PQR$ maximally fits into $\mathcal{X}$, we can assume that $P, Q, R$ all lie on the boundary $\partial \mathcal{X}$. If no vertex lies on $\overline{AB}$, we can translate the triangle downwards until it touches $\overline{AB}$, so we can assume that $P \in \overline{AB}$. If $Q \in \overline{AB}$,
<figure><img src="image_7.png" /><figcaption>Fig. 4: Proof of Lemma 7.</figcaption></figure>

then $\triangle PQR \subset \triangle ABR$, so the maximality implies that $\triangle PQR = \triangle ABR$ and we are in case (i). It remains to consider the case where $P \in \overline{AB}$, while $Q$ and $R$ lie on the upper chain $\partial X \setminus \overline{AB}$, 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 $\ell$ be the bisector of the angle $\angle AKR$; see Figure 4. We reflect the points $R$ and $Q$ about $\ell$ to obtain points $R^*$ and $Q^*$ on the line $AB$. Since $|KR^*| = |KR| < |KB| + |BR| \le |KB| + |BA| = |KA|$, we have $R^* \in \overline{AB}$ but is not equal to $A$. We also note that $Q^*$ lies strictly between $R^*$ and $B$ and thus $Q^* \in \overline{AB}$.

If $P$ lies between $R^*$ and $Q^*$, then we can reflect it about $\ell$ to obtain a point $P^*$ on the segment $RQ$ so that $\triangle P^*Q^*R^*$ is congruent to $\triangle PQR$; see Figure 4(left). Since $\triangle P^*Q^*R^* \subsetneq \triangle ABP^*$, 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 $\triangle PQR$ clockwise around $R$. This moves $Q$ and $P$ into the interior of $X$, so $\triangle PQR$ does not maximally fit into $X$.

If $P$ lies to the right of $Q^*$, then we rotate $Q$ by $180^\circ$ about the midpoint of $\overline{PR}$ to obtain $Q'$, see Figure 4(right). The quadrilateral $PQRQ'$ is a parallelogram, and $\triangle PRQ'$ is congruent to $\triangle PQR$. Then $Q' \in \triangle PRR^*$ since $R$ above $Q$ implies $Q'$ above $P$ and $P$ right of $QQ^*$ implies $Q'$ right of $RR^*$. Since $\triangle PRQ' \subsetneq \triangle APR$, $\triangle PQR$ 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 $\ell$ be the horizontal line equidistant from $AB$ and $RQ$. Again we mirror $R$ and $Q$ about $\ell$ to obtain $R^*$ and $Q^*$ on $\overline{AB}$. 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 $\triangle PQR$ does not maximally fit into $X$. $\square \square$
We now state two lemmas, postponing their proofs to Section 4.

<figure><img src="image_8.png" /><figcaption>Fig. 5: Lemma 8.</figcaption></figure>

**Lemma 8.** Let $\triangle ABC$ be a triangle with the longest edge $|AB|$. Let $AB$ be horizontal, A left of B, and let $Q \in \mathcal{H}_{AC}$ lie to the left of AC with $|QB| \le |AB|$. Then $\triangle BCQ$ fits into $\triangle AB'C$, where $B' = B \oplus \zeta_{AC}(Q)$ (see Figure 5).

**Lemma 9.** Let $\triangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $AB$ be horizontal, A left of B, let $AH$ be the height of $\triangle ABC$ with respect to $BC$, and let $R \in \mathcal{H}_{AC}$ lie to the left of AC with $|AR| \ge |AH|$ and $|BR| \le |AB|$ (that is, R lies in the green area of Figure 6). Let $A' = A \ominus \mu$ for some $\mu \ge 0$ such that $|\frac{A'B}{|AB|}| \le |\frac{AB}{|AH|}|$, let $B' = B \oplus \zeta_{AC}(R)$, 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 $\angle CAR$, obtaining points $B''$ and $H''$, respectively. Then $B''$ lies in $\triangle BB'H$ and $H'' \in \triangle A'H'R^*$.

<figure><img src="image_9.png" /><figcaption>Fig. 6: Lemma 9.</figcaption></figure>
*Proof of Theorem 6.* It suffices to prove the statement for an arbitrary triangle $\triangle PQR$ that maximally fits into $\mathcal{X}$. By Lemma 7, this implies that $\triangle PQR$ is one of the three types in the lemma.

**Case $\triangle PQR$ of type(i).**

If $\triangle PQR$ is of type (i) with $P = A$ and $Q = B$, then, depending on the location of $R$, we translate it leftwards by $\zeta_{BC}(R)$ or rightwards by $\zeta_{AB}(R)$ to place it in $\triangle A'BC$ or $\triangle AB'C$, which are both included in $\triangle A'B'C$.

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 $C$**

If $R$ does not lie to the left of $C$, we translate $\triangle PQR$ leftwards by $\zeta_{BC}(R)$ to obtain a triangle $\triangle P'Q'R'$ with $R'$ on $\overline{BC}$; see Figure 7. If $Q' \in \triangle ABC$, we are done since $\triangle P'Q'R' \subset \triangle A'BC$. Otherwise we apply Lemma 8 (after symmetry) to the triangle $\triangle BP'R'$ and the point $Q'$ and obtain that $\triangle PQR$ fits in the triangle $\triangle (P' \ominus \zeta_{BC}(Q'))BR' \subset \triangle A'BC$.

Now we are left with the case where $R$ lies strictly to the left of $C$. Since $Q$ lies below $R$, $Q$ lies on the right chain from $B$ up to but not including $C$. Let $H$ be the foot of the height of $\triangle ABC$ with respect to $\overline{BC}$ and let $B^*$ be the mirror image of $B$ about $AH$.

<figure><img src="image_10.png" /><figcaption>Fig. 7: Proof of Theorem 6: $R$ not left of $C$.</figcaption></figure>
<figure><img src="image_11.png" /><figcaption>Fig. 8: Proof of Theorem 6: $Q$ below $AH$.</figcaption></figure>

**Case Q below AH.**

If $\angle BAQ \le \angle BAH$, then we rotate $\triangle PQR$ clockwise around $A$. During the rotation, both $\zeta_{AC}(R)$ and $\zeta_{BC}(Q)$ are decreasing. We continue the rotation to obtain $\triangle PQ'R'$ until either $\zeta_{AC}(R') = 0$ or $Q' \in \overline{AB}$; see Figure 8. If $\zeta_{AC}(R') > 0$, then $Q' \in \overline{AB}$. We then translate the triangle rightwards by $\zeta_{AC}(R') < |BB'|$ to place it in $\triangle AB'C$. Otherwise, $R' \in \overline{AC}$. Since $\triangle AR'Q' \subset \triangle ACQ'$ and $\zeta_{BC}(Q') < |AA'|$, we apply Lemma 8 to $\triangle ABC$ and $Q'$ to conclude that $\triangle AR'Q'$ fits in $\triangle A'BC$.

**Case $R \in \triangle AHB^*$:**

We now have that $\angle BAQ > \angle BAH$. If $R \in \triangle AHB^*$, then we mirror $\triangle PQR$ about $AH$ to obtain a new triangle $\triangle PR'Q'$ with $R' \in \triangle ABH \subset \triangle ABC$ and $\zeta_{BC}(Q') = \zeta_{BC}(Q)$; see Figure 9. We can then rotate the triangle clockwise. The rotation decreases $\zeta_{BC}(Q')$. We stop when either $Q' \in \overline{BC}$ or $R' \in \overline{AB}$ and denote by $\triangle PQ''R''$ the triangle in the new position. In the first case, we have $\triangle PR''Q'' \subset \triangle ABC$, in the second case we can translate leftwards by $\zeta_{BC}(Q'') < \zeta_{BC}(Q) \le |AA'|$ to place the triangle in $\triangle A'BC$.

<figure><img src="image_12.png" /><figcaption>Fig. 9: Proof of Theorem 6: $R \in \triangle AHB^*$.</figcaption></figure>
<figure><img src="image_13.png" /><figcaption>Fig. 10: Proof of Theorem 6: $|AR| \le |AH|$.</figcaption></figure>

**Case** $|AR| \le |AH|$.

Consider now the case where $|AR| \le |AH|$. We can rotate $\triangle PQR$ clockwise around $A$ to obtain a new triangle $\triangle PQ'R'$ with $Q' \in AB$; see Figure 10. Since $R'$ lies in the interior of $X$, the triangle $\triangle PQ'R'$ does not maximally fit into $X$.
<figure><img src="image_14.png" /><figcaption>Fig. 11: Proof of Theorem 6: the final case.</figcaption></figure>

**Final Case.**

We are now left with the final case where $R \in \mathcal{H}_{AC} \subset \mathcal{H}_{AB^*}$ lying to the left of $AB^*$, $|AR| \ge |AH|$, $|BR| \le |AB|$, $\angle BAQ > \angle BAH$, and $Q \in \mathcal{H}_{AC} \subset \mathcal{H}_{AB^*}$ lying to the right of $BC$; see Figure 11. We let $\rho = \angle B^*AR$. We first mirror $\triangle PQR$ about $AH$ to obtain $\triangle PR'Q'$, with $R'$ below $AB$, $\angle R'AB = \rho$, and $\zeta_{BC}(Q') = \zeta_{BC}(Q)$. We then translate $\triangle PR'Q'$ leftwards by $\zeta_{BC}(Q')$ to obtain $\triangle P''Q''R''$ and finally rotate counter-clockwise about $P''$ by angle $\rho$ to obtain $\triangle P''Q'''R''$. To see that $\triangle P''R'''Q'''$ fits into $\triangle A'B'C$, it remains to show that $Q''' \in \triangle A'B'C$, since $P'' \in \overline{A'A}$ and $R''' \in \overline{AB}$.

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 $PH$ by $\zeta_{BC}(Q')$, and $Q'$ lies below the line $AH$, we have $Q'' \in \overline{H'B}$. We rotate $H'$ and $B$ about $P''$ by angle $\rho$ to obtain $H''$ and $B''$, respectively, so that $Q''' \in \overline{H''B''}$.

We now apply Lemma 9 to $\triangle ABB^*$ and $R$, with $\mu = \zeta_{BC}(Q) = \zeta_{BB^*}(Q)$. 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 $B^\oplus = B \oplus \zeta_{AB^*}(R) \in \overline{BB'}$ since $\zeta_{AB^*}(R) \le \zeta_{AC}(R) \le \zeta_{AC}(D)$. To check the lemma's condition on $\mu$, let $Q_0$ be the point on the line $AH$ at distance $|AB|$ from $A$ and let $A_0 = A \ominus \zeta_{BB^*}(Q_0)$; see Figure 12. Since $\mu \le \zeta_{BB^*}(Q_0)$ under the constraint $|AQ| \le |AB|$, we have that $\frac{\mu + |AB|}{|AB|} \le \frac{|A_0B|}{|AB|}$. Let $Y = Q_0 \ominus \zeta_{BB^*}(Q_0)$ and let $Z = Q_0 \oplus |AB|$. The
<figure><img src="image_15.png" /><figcaption>Fig. 12: Proof of Theorem 6: verifying the condition of Lemma 9.</figcaption></figure>

quadrilateral $ABZQ_0$ is a rhombus, and the triangle $\triangle BZY$ is right-angled at $B$, and therefore similar to $\triangle ABH$. It follows that $\frac{|A_0B|}{|AB|} = \frac{|YZ|}{|AB|} = \frac{|AB|}{|AH|}$, and the condition in Lemma 9 is satisfied. The lemma implies that $B'' \in \triangle BB^{\oplus}H$, and that $H'' \in \triangle H'P''R^*$, where $R^*$ is the horizontal projection of $R$ on the line $BB^*$. Since $R \in \mathcal{H}_{AC}$, the point $C$ must lie on the segment $R^*B^*$, and thus $H'' \in \triangle H'P''R^* \subset \triangle H'P''C \subset \triangle A'BC$.

Since $Q''' \in \overline{H''B''}$, $H'' \in \triangle A'BC \subset \triangle A'B'C$ and $B'' \in \triangle BB^{\oplus}H \subset \triangle A'B'C$, convexity of $\triangle A'B'C$ implies that $Q''' \in \triangle A'B'C$, completing the proof. $\square$

<figure><img src="image_16.png" /></figure>

<figure><img src="image_17.png" /></figure>

# 4 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 $\alpha + \phi = \psi + \theta$. We claim that $\triangle BCQ$ fits into $\triangle AB'C$. We distinguish two cases.

**Case 1:** $\theta \ge \phi$

In this case, we rotate $\triangle BCQ$ by angle $\phi$ around $B$, resulting in $\triangle BRS$ with $S$ on the line $AB$; see Figure 13(right). Since $|BS| = |BQ| \le |AB|$, $S$ lies on $\overline{AB}$.

<figure><img src="image_18.png" /><figcaption>Fig. 13: Lemma 8. Angle notation (left); Case 1 (right).</figcaption></figure>
<figure><img src="image_19.png" /><figcaption>Fig. 14: Lemma 8. Case 2.</figcaption></figure>

On the other hand $\psi = \alpha + (\phi - \theta) \le \alpha$, so $R$ lies below $AC$, and therefore in $\triangle ABC$.

**Case 2: $\theta < \phi$**

In this case, we rotate $\triangle BCQ$ by angle $\theta$ around $B$, resulting in $\triangle BRS$ with $RS$ parallel to $AC$; see Figure 14(left). We let $\triangle BS'R'$ be the image of $\triangle BRS$ mirrored about the angular bisector of $\angle ABR$, which means that $R'$ lies on $\overline{AB}$; see Figure 14(right).

We claim that $\min\{\zeta_{AC}(S), \zeta_{AC}(S')\} \le \zeta_{AC}(Q) = |BB'|$, which implies that at least one of $\triangle BRS$ or $\triangle BS'R'$ can be translated rightward to fit into $\triangle AB'C$. By the law of sines, applied to triangles $\triangle A(Q \oplus \zeta_{AC}(Q))(B \oplus \zeta_{AC}(Q))$, $\triangle A(S \oplus \zeta_{AC}(S))(B \oplus \zeta_{AC}(S))$, and $\triangle A(S' \oplus \zeta_{AC}(S'))(B \oplus \zeta_{AC}(S'))$ (see blue shaded triangles in Figures 13 and 14), we have

$$ \frac{|BQ|}{\sin \alpha} = \frac{\zeta_{AC}(Q) + |AB|}{\sin(\alpha + \phi)} = \frac{\zeta_{AC}(S) + |AB|}{\sin \psi} = \frac{\zeta_{AC}(S') + |AB|}{\sin(\alpha + \beta - \phi)}, $$

so we need to prove $\min\{\sin(\psi), \sin(\alpha + \beta - \phi)\} \le \sin(\alpha + \phi)$. Suppose this is not the case. Then we have $\sin(\alpha + \phi) < \sin \psi$. Since $\alpha + \phi = \psi + \theta > \psi$ and $x \mapsto \sin x$ is monotonously increasing on $[0, \frac{\pi}{2}]$, we must have $\alpha + \phi > \frac{\pi}{2}$ and $\pi - (\alpha + \phi) < \frac{\pi}{2}$. On the other hand, we also have $\sin(\alpha + \beta - \phi) > \sin(\alpha + \phi) = \sin(\pi - (\alpha + \phi))$, which implies $\alpha + \beta - \phi > \pi - (\alpha + \phi)$ and therefore $(\alpha + \phi) + (\alpha + \beta - \phi) > \pi$. However, since $\gamma$ is the largest angle in $\triangle ABC$, $(\alpha + \phi) + (\alpha + \beta - \phi) = 2\alpha + \beta \le \alpha + \beta + \gamma = \pi$, a contradiction. $\square$

of Lemma 9. We scale the problem such that $|AB| = |AC| = 1$ and place $A$ at the origin, so that $B = (1,0)$. Let $\beta = \angle ABC$, $\alpha = \angle BAC = \pi - 2\beta$, and $\rho = \angle CAR$. We have $|AH| = \sin \beta$, and $|A'B| \le \frac{1}{\sin \beta}$; see Figure 15.

We first observe that we can replace $R$ by the point at distance $\sin \beta$ from $A$ on $\overline{AR}$. This keeps $\rho$ unchanged, decreases $|BR|$, decreases $\zeta_{AC}(R)$, and decreases $|H'R^*|$ so that $\triangle A'H'R^*$ becomes strictly smaller. So in the following, $|AR| = |AH| = \sin \beta$.
<figure><img src="image_20.png" /><figcaption>Fig. 15: Lemma 9: Notation.</figcaption></figure>
<figure><img src="image_21.png" /><figcaption>Fig. 16: Lemma 9: definitions of $\beta_0$ and $\beta_1$.</figcaption></figure>

Let next $\delta = \zeta_{AC}(R)$, and let $X \in \overline{AC}$ be the point $R \oplus \delta$. Applying the law of sines to $\triangle AXR$, we have

$$ \frac{\delta}{\sin \rho} = \frac{\sin \beta}{\sin \alpha} = \frac{\sin \beta}{\sin 2\beta} = \frac{\sin \beta}{2 \sin \beta \cos \beta} = \frac{1}{2 \cos \beta} \quad \text{so} \quad \delta = \frac{\sin \rho}{2 \cos \beta}. \quad (1) $$

We now analyse the interval of angles $\beta$ for which the conditions of the lemma can be satisfied. Consider the point $R_0 = (-\sin \beta \cos 2\beta, \sin \beta \sin 2\beta)$ on $\overline{AC}$ with $|AR_0| = \sin \beta$, and let

$$ \phi(\beta) := |R_0B|^2 = (1 + \sin \beta \cos 2\beta)^2 + (\sin \beta \sin 2\beta)^2. $$

Notice<fnref n="1" /> that $\frac{d}{d\beta}\phi(\beta) = 2\cos\beta(6\cos^2\beta + \sin\beta - 5)$, which is negative on $[\frac{\pi}{4}, \frac{\pi}{2}]$ since $6\cos^2\beta + \sin\beta - 5 \le 6\cos^2\frac{\pi}{4} - 4 = -1$. Thus, as $\beta$ increases from $\frac{\pi}{4}$ to $\frac{\pi}{2}$, $\phi(\beta)$ decreases monotonously from $\phi(\frac{\pi}{4}) = \frac{3}{2}$ to $\phi(\frac{\pi}{2}) = 0$, so there is a $\beta_0 \in (\frac{\pi}{4}, \frac{\pi}{2})$ with $\phi(\beta) = 1$. For $\beta < \beta_0 \approx 1.003 \approx 57.47^\circ$, no point R left of AC at distance smaller than one from B can exist; see Figure 16. In the following we therefore have $\beta \ge \beta_0$.

The point R lies on an arc of circle around A with right endpoint $R_0$. There are two other critical points on this circle: let $R_1$ be the point with $|AR_1| = \sin \beta$ and $|BR_1| = 1$, and let $R_2$ be the point on the horizontal line through C with $|AR_2| = \sin \beta$ and a positive x-coordinate. A point R satisfying the conditions of the lemma cannot lie to the left of $R_1$ because $|BR| \le 1$, and cannot lie to the left of $R_2$ since $R \in \mathcal{H}_{AC}$.

The triangle $\triangle ABR_1$ is isosceles with two sides of length one and a short side $\overline{AR_1}$ of length $\sin \beta$, so $\angle ABR_1 = 2\sin^{-1}(\frac{1}{2} \sin \beta)$. The law of sines applied to triangle $\triangle (A \ominus \zeta_{AC}(R_1))R_1B$ now shows that

$$ \frac{\zeta_{AC}(R_1) + 1}{\sin(\pi - (\pi - 2\beta) - 2\sin^{-1}(\frac{1}{2} \sin \beta))} = \frac{1}{\sin 2\beta}. $$

<footnote n="1">¹ The computations in Maple can be found in the appendix.</footnote>
We set

$$h(\beta) := \zeta_{AC}(R_1) = \frac{\sin(2\beta - 2\sin^{-1}(\frac{1}{2}\sin\beta))}{\sin 2\beta} - 1. \quad (2)$$

Since $C = (-\cos 2\beta, \sin 2\beta)$, the x-coordinate of $R_2$ is $\sqrt{\sin^2\beta - \sin^2 2\beta}$. For $\beta = \beta_1 := \cos^{-1}(\frac{1}{\sqrt{5}}) \approx 1.107 \approx 63.4^{\circ}$, we have $C = (\frac{3}{5}, \frac{4}{5})$, $R_2 = (\frac{2}{5}, \frac{4}{5})$, implying $|BR_2| = 1$, that is, $R_1 = R_2$; see Figure 16. We set

$$g(\beta) := \zeta_{AC}(R_2) = |R_2C| = -\cos 2\beta - \sqrt{\sin^2\beta - \sin^2 2\beta}. \quad (3)$$

To summarize:

* For $\beta_0 \le \beta \le \beta_1$, $R$ lies on the arc between $R_0$ and $R_1$. The angle $\rho$ is maximized when $R = R_1$. For $\beta = \beta_0$, we have $R_1 = R_0$ (so there is only a single choice for $R$), for $\beta = \beta_1$ we have $R_1 = R_2 = (\frac{2}{5}, \frac{4}{5})$. Since $R$ cannot lie to the left of $R_1$, we have $\delta \le h(\beta)$.

* For $\beta_1 \le \beta < \frac{\pi}{2}$, $R$ lies on the arc between $R_0$ and $R_2$, with $\rho$ maximized when $R = R_2$. Since $R$ cannot lie to the left of $R_2$, we have $\delta \le g(\beta)$.

### B'' position.

Consider now the point $B''$. Since $|A'B| \le \frac{1}{\sin\beta}$, it has y-coordinate at most $\frac{\sin\rho}{\sin\beta}$. We will prove that $HB'$ intersects the vertical line $x = 1$ through $B$ at y-coordinate at least $\frac{\sin\rho}{\sin\beta}$, implying that $B''$ lies below $HB'$, and therefore is in $\triangle BB'H$.

Since $H = (\sin^2\beta, \sin\beta\cos\beta)$ and $B' = (1 + \delta, 0)$, the line $x = 1$ intersects $HB'$ at y-coordinate

$$\delta \cdot \frac{\sin \beta \cos \beta}{1 + \delta - \sin^2 \beta} = \frac{\sin \rho}{2 \cos \beta} \cdot \frac{\sin \beta \cos \beta}{\cos^2 \beta + \delta} = \frac{\sin \rho \sin \beta}{2(\cos^2 \beta + \delta)}.$$

This is at least $\frac{\sin\rho}{\sin\beta}$ if and only if

$$\frac{1}{\cos^2 \beta + \delta} \ge \frac{2}{\sin^2 \beta},$$

which is equivalent to

$$\delta \le \frac{1}{2} \sin^2 \beta - \cos^2 \beta = \frac{3}{2} \sin^2 \beta - 1.$$

Setting $f(\beta) = \frac{3}{2} \sin^2 \beta - 1$, it remains to show that $\delta = \zeta_{AC}(R) \le f(\beta)$ under the conditions of the lemma.

We first consider the case $\beta \ge \beta_1$, where $\delta \le g(\beta)$. Since $g(\beta)$ is a decreasing function, while $f(\beta)$ is an increasing function, this implies that $\delta \le g(\beta) \le g(\beta_1) = \frac{1}{5} = f(\beta_1) \le f(\beta)$.
We next consider $\beta_0 \le \beta \le \beta_1$. For $\beta = \beta_1$, $R_2 = R_1$, so $h(\beta_1) = g(\beta_1) = \frac{1}{5} = f(\beta_1)$. We consider the function $\beta \mapsto f(\beta) - h(\beta)$. Plotting its derivative on the interval $[\beta_0, \beta_1]$ shows that it is smaller than $-0.2$, so $f(\beta) - h(\beta)$ is decreasing on the interval. This implies that $\delta \le h(\beta) \le f(\beta)$ for $\beta \in [\beta_0, \beta_1]$, completing the proof of $B'' \in \Delta BB'H$.

### $H''$ position.

We now turn to the point $H''$. It is obtained by rotating $H'$ counter-clockwise around $A'$ by angle $\rho$. Since $A'H'$ is orthogonal to $BC$, $H''$ lies below the line $BC$. Since $\rho < \pi$, $H''$ lies above the line $A'H'$. To show that $H'' \in \Delta A'H'R^*$, it remains to prove that $H''$ lies below the line $A'R^*$. This is equivalent to proving $\rho \le \angle H'A'R^*$.

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 $\sin \beta \sin 2\beta$, we have $R_0^* = (1 - \cos \beta \sin 2\beta, \sin \beta \sin 2\beta)$. We have

$$ \begin{aligned} \angle H'A'R^* &= \angle BA'R^* - \angle BA'H' = \angle BA'R^* - \angle BAH \\ &= \angle BA'R^* - (\frac{\pi}{2} - \beta) \ge \angle BA'R_0^* - (\frac{\pi}{2} - \beta) \end{aligned} $$

Since $|A'B| \le \frac{1}{\sin \beta}$, we can therefore bound from below $\angle H'A'R^*$ by $r(\beta)$, where

$$ r(\beta) := \tan^{-1} \left( \frac{\sin \beta \sin 2\beta}{(1 - \cos \beta \sin 2\beta) + \left(\frac{1}{\sin \beta} - 1\right)} \right) - \frac{\pi}{2} + \beta. $$

Plotting $r(\beta)$ shows that it is larger than 0.25 on the interval $[\beta_0, \frac{2\pi}{5}]$.

We consider the case $\beta_0 \le \beta \le \beta_1$. This implies that $\rho$ is maximized when $R = R_1$. Combining (1) and (2), this gives us $\sin \rho \le 2h(\beta) \cos \beta$. Plotting $\sin^{-1}(2h(\beta) \cos \beta)$ on the interval $[\beta_0, \beta_1]$ shows that $\rho < 0.2 < 0.25 < r(\beta)$.

Finally, we turn to the case $\beta_1 \le \beta < \frac{\pi}{2}$. Here, $\rho$ is maximized when $R = R_2$. Combining (1) and (3), this gives us $\sin \rho \le 2g(\beta) \cos \beta$. Plotting the function $\beta \mapsto \sin^{-1}(2g(\beta) \cos \beta)$ on the interval $[\beta_1, \frac{\pi}{2}]$ shows that $\rho < 0.2$ on that interval. For $\beta \le \frac{2\pi}{5}$, this implies $\rho < 0.25 < r(\beta)$. For $\beta \ge \frac{2\pi}{5}$, we consider the function $t(\beta) = r(\beta) - \sin^{-1}(2g(\beta) \cos \beta)$. We plot the derivative of $t(\beta)$ on the interval $[\frac{2\pi}{5}, \frac{\pi}{2}]$ to show that it is smaller than $-0.05$, so $t(\beta)$ is a decreasing function on that interval. Since $t(\frac{\pi}{2}) = 0$, this implies that $t(\beta) \ge 0$, and therefore $\rho \le r(\beta)$ for $\frac{2\pi}{5} \le \beta < \frac{\pi}{2}$. To summarize, we have shown $\rho \le r(\beta) \le \angle H'A'R^*$, so $H'' \in \Delta A'H'R^*$ for all values of $\beta$. $\square$ $\square$
<figure><img src="image_22.png" /><figcaption>Fig. 17: Right triangles of diameter two.</figcaption></figure>

# 5 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 $y \ge 0$. 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 $(-1, 0)$, $(1, 0)$, $(-x, \sqrt{1-x^2})$, for $x \in [0, 1]$; 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 $(x + \sqrt{1-x^2}, 0)$. The $x$-coordinate of this point is maximized for $x = \frac{1}{\sqrt{2}}$, so the triangle $Z$ with vertices $(-1, 0)$, $(\sqrt{2}, 0)$, 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 $T_{\frac{1}{\sqrt{2}}}$, as can be seen using Corollary 10 of Park and Cheong [8]. $\square$ $\square$
<figure><img src="image_23.png" /><figcaption>Fig. 18: Triangles that fit in a square.</figcaption></figure>

# 6 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 $\mathcal{X} = ABCD$ be a square. If a triangle $T = \triangle PQR$ fits maximally into $\mathcal{X}$, 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 $\mathcal{X}$, we can assume that $P, Q, R$ all lie on the boundary $\partial\mathcal{X}$. Suppose two vertices of $T$ lie on the same side of $\mathcal{X}$, say, $P, Q$ lie on $AB$. Then $T \subset \triangle ABR \subset \mathcal{X}$ as in Figure 18(left). Since $T$ maximally fits into $\mathcal{X}$, this implies $P = A$, $Q = B$. Suppose next that no vertex of $T$ coincides with a vertex of $\mathcal{X}$. Then $P, Q, R$ lie on three different sides of $\mathcal{X}$, 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 $\mathcal{X}$; see Figure 18(middle). It follows that we can assume that $P = A$ and that $Q, R$ lie on the sides $BC$ and $CD$. $\square$ $\square$

By Lemma 10, it suffices to study the triangles with $P = A$, $Q \in \overline{BC}$, and $R \in \overline{CD}$. We parameterize these triangles $\triangle PQR$ by the two angles $\theta$ and $\theta'$ made by the edges $\overline{PQ}$ and $\overline{PR}$ with the diagonal $\overline{AC}$ of the square. We denote this triangle $T_{\theta,\theta'}$; see Figure 18(right). These parameters range in $[0, \frac{\pi}{4}]$ and the case $\theta = \theta' = \frac{\pi}{6}$ corresponds to the largest equilateral triangle that can fit in the square. We denote this equilateral triangle as $T_0 = \triangle P_0Q_0R_0 = T_{\frac{\pi}{6}, \frac{\pi}{6}}$; see the red triangle in Figure 18(right).

## 6.1 The isosceles case: construction of the cover

We first consider the isosceles triangle $T_{\theta,\theta}$ with $\theta \le \frac{\pi}{6}$. A convex cover $C_\theta = \triangle X_\theta Y_\theta Z_\theta$ for $T_{\theta,\theta}$ and $T_0$ is obtained when $P'R'$ is aligned with $P_0R_0$, and $Q'$ is on $Q_0R_0$; see Figure 19. We have $|P'Q'| = |P_0Q| = \frac{1}{\cos(\frac{\pi}{4}-\theta)}$. Hence we
<figure><img src="image_24.png" /><figcaption>Fig. 19: Construction of the smallest convex cover for the equilateral triangle $T_0$ and the isosceles triangle $T_{\theta,\theta}$.</figcaption></figure>

compute the distance $\ell(\theta)$ between $P'$ and $R_0$ by the law of sines in $\triangle X_{\theta}Q'Z_{\theta}$:

$$ \ell(\theta) = \frac{|P'Q'| \cdot \sin \angle Z_{\theta}Q'X_{\theta}}{\sin \angle X_{\theta}Z_{\theta}Q'} = \frac{\sin(\frac{\pi}{3} + 2\theta)}{\cos(\frac{\pi}{4} - \theta) \sin(\frac{\pi}{3})} $$

When $\theta = 0$, $T_{0,0}$ degenerates to the diagonal of the square and $\ell(0) = \sqrt{2}$. As $\theta$ increases from zero, $\ell(\theta)$ increases to a maximum<fnref n="2" /> at

$$ \theta_0 = \tan^{-1} \left( \frac{1}{\sqrt[3]{2 + \sqrt{3}}} \right) - \frac{\pi}{6} \approx 0.049 \approx 2.81^{\circ}, $$

then decreases to $\ell(\frac{\pi}{6}) = 1/\cos \frac{\pi}{12}$. We have $\ell(\theta_0) \approx 1.4195$.

It follows that the triangle $C = \triangle XYZ$, where $X = X_{\theta_0}$, $Y = Y_{\theta_0}$, $Z = Z_{\theta_0}$, is a cover for the family of all isosceles triangles $T_{\theta,\theta}$ for $0 < \theta \le \frac{\pi}{6}$. We note that $\angle XZY = \frac{\pi}{3}$, $|XZ| = \ell(\theta_0)$, and $|ZY| = \ell(\frac{\pi}{6})$.

It is intriguing that $C$ is just slightly larger than the much “nicer” triangle $\triangle X_0Y_0Z_0$ obtained for $\theta = 0$. We will denote this triangle as $C^* = \triangle X^*Y^*Z^*$. The angles of $C^*$ are $\frac{\pi}{4} = 45^{\circ}$, $\frac{5\pi}{12} = 75^{\circ}$, and $\frac{\pi}{3} = 60^{\circ}$. The longest side is $|X^*Z^*| = \sqrt{2}$, and, by construction, we have $C^* \subset C$.

We have $\ell(\theta) \ge \sqrt{2}$ when $\theta \in [0, \theta_1]$ and $\ell(\theta) \le \sqrt{2}$ when $\theta \in [\theta_1, \frac{\pi}{6}]$ with

$$ \theta_1 = \tan^{-1} \left( \frac{4 \sin^2 \frac{\pi}{12} + 1}{8 \sin^2 \frac{\pi}{12} - 6 + \sqrt{16 \sin^4 \frac{\pi}{12} - 72 \sin^2 \frac{\pi}{12} + 57}} \right) - \frac{\pi}{6} \approx 0.0996 \approx 5.7^{\circ}, $$

so the triangle $T_{\theta,\theta}$ actually fits into $C^*$ for $\theta_1 \le \theta \le \frac{\pi}{6}$.

<footnote n="2">² The computations in Maple can be found in the appendix.</footnote>
<figure><img src="image_25.png" /><figcaption>Fig. 20: Six cases cover all possible triangles. All cases but Case A fit in $C^*$.</figcaption></figure>

In the following six sections, we discuss why each triangle $T_{\theta,\theta'}$ indeed fits into $\mathcal{C}$. Figure 20 shows how the six cases cover the entire domain of $(\theta, \theta')$. It turns out that only case A requires the cover $\mathcal{C}$, in all other cases $T_{\theta,\theta'}$ fits into the nicer triangle $\mathcal{C}^*$—so in a sense $\mathcal{C}^*$ 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 $\mathcal{C}$ is a smallest convex cover for $T_0$ and $T_{\theta_0, \theta_0}$. This makes $\mathcal{C}$ a smallest convex cover for the family of all triangles contained in the unit square.

Moreover, $\mathcal{C}$ 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 $T_{\theta_0, \theta_0}$ that is different from $\mathcal{C}$ is a quadrilateral, and that this quadrilateral does not cover either $T_{\theta_0+\varepsilon, \theta_0+\varepsilon}$ or $T_{\theta_0-\varepsilon, \theta_0-\varepsilon}$ for small enough $\varepsilon$.

We now turn to the six cases. Without loss of generality we will always assume that $\theta \le \theta'$.
<figure><img src="image_26.png" /><figcaption>Fig. 21: Case A: covering $\triangle PQR$ with $\theta \le \theta_1$ and $\theta \le \theta' \le \pi/12$.</figcaption></figure>

## 6.2 Case A

We start with the triangles where $\theta \le \theta_1 \approx 5.7^\circ$ and $\theta' \le \frac{\pi}{12}$. This is the only case where we need to use the cover $\mathcal{C}$—that should not come as a surprise, since $T_{\theta_0, \theta_0}$ falls into this case.

Let $\triangle PQR = T_{\theta, \theta'}$ be a triangle with $\theta \le \theta_1$. Let $Q' \in \overline{BC}$ be such that $\triangle PQ'R = T_{\theta, \theta}$. We have seen in Section 6.1 that $\mathcal{C}_{\theta} \subset \mathcal{C}$ covers $\triangle PQ'R$ as in Figure 21. The point $Q$ lies on the segment $\overline{BQ'}$, so $\triangle PQR \subset \mathcal{C}_{\theta}$ as long as $\theta + \theta' \le \angle Y_{\theta}X_{\theta}Z_{\theta}$. Since the angle $\angle Y_{\theta}X_{\theta}Z_{\theta}$ is minimized by $\angle Y_{\theta_0}X_{\theta_0}Z_{\theta_0} > 44.8^\circ$, this holds by $\theta + \theta' \le \theta_1 + \frac{\pi}{12} < 21^\circ$.
<figure><img src="image_27.png" /><figcaption>Fig. 22: Case B: covering $\triangle PQR$ when $\theta + \theta' \le \frac{\pi}{4}$ and $\theta_1 \le \theta \le \theta'$.</figcaption></figure>

## 6.3 Case B

Case B covers those triangles where $\theta + \theta' \le \frac{\pi}{4}$, 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 $\overline{X^*Z^*}$.

For $\theta \ge \theta_1$, we again let $Q' \in \overline{BC}$ be such that $\triangle PQ'R = T_{\theta,\theta}$. We argued in Section 6.1 that $C^*$ covers $\triangle PQ'R$ as in Figure 22. Since $\angle RAQ \le \frac{\pi}{4} = \angle Z^*X^*Y^*$, $Q \in C^*$ and so $\triangle PQR \subset C^*$.

It remains to consider the situation where $\theta < \theta_1$ and $\theta' \ge \frac{\pi}{12}$. Let $Q''$ be the point on $BC$ with $\angle BAQ'' = \frac{\pi}{6}$. Since $|AQ''| = \frac{2}{3}\sqrt{3}$ is less than the height of $X^*$ in $C^*$, the point $Q''$ lies in $C^*$. From $\theta' \ge \frac{\pi}{12}$ follows that $Q$ lies on $\overline{BQ''}$, and since $\theta + \theta' \le \frac{\pi}{4}$, it therefore lies in $C^*$ as in Figure 23.

<figure><img src="image_28.png" /><figcaption>Fig. 23: Case B: covering $\triangle PQR$ when $\theta + \theta' \le \frac{\pi}{4}$ and $\theta < \theta_1 < \frac{\pi}{12} \le \theta'$.</figcaption></figure>
<figure><img src="image_29.png" /><figcaption>Fig. 24: Case C: covering $\triangle PQR$ when $\frac{\pi}{6} \le \theta \le \theta'$.</figcaption></figure>

## 6.4 Case C

We now consider the triangles where $\theta, \theta' \ge \frac{\pi}{6}$. In other words, $Q \in \overline{BQ_0}$, $R \in \overline{R_0D}$.

We first observe that $C^*$ can be placed such that $\overline{X^*Y^*}$ is vertical and lies on the line $AB$, while $Z^*$ lies on the line $CD$ (recall that $\angle Y^*X^*Z^* = \frac{\pi}{4}$ while $|X^*Z^| = \sqrt{2}$). When $Z^* \in \overline{R_0D}$, then the side $\overline{X^*Y^*}$ covers the entire square edge $AB$. Figure 24 shows the two extreme cases where $Z^* = R_0$ (top left) and where $Z^* = D$ (top right).

Consider now our triangle $\triangle PQR$. We place $C^*$ such that $Z^* = R$; see Figure 24(bottom). Since the line $Z^*X^*$ has slope $-1$, it intersects $\overline{BC}$ in a point $Q'$ such that $\triangle PQ'R = T_{\theta,\theta}$. Since $\theta' \ge \theta$, we have $Q \in \overline{BQ'} \subset C^*$ and thus $\triangle PQR \subset C^*$.
<figure><img src="image_30.png" /><figcaption>Fig. 25: Case D: covering $\triangle PQR$ when $\frac{\pi}{12} < \theta \le \theta' \le \frac{\pi}{6}$.</figcaption></figure>

## 6.5 Case D

We now look at the situation where we have $\frac{\pi}{12} = 15^\circ < \theta \le \theta' \le \frac{\pi}{6}$. In other words, we have $Q \in \overline{Q_0Q_1}$ and $R \in \overline{R_1R_0}$ where $\angle Q_1AC = \angle R_1AC = \frac{\pi}{12}$ as in Figure 25(top left).

We observe that $C^*$ can be placed to cover $T_0 = \triangle PQ_0R_0$ as in Figure 25(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 $Y^* = R_1$; see Figure 25(bottom right). Since the line $Y^*Z^*$ has slope $-1$, it intersects $BC$ in a point $Q'$ such that $\triangle PQ'R = T_{\theta,\theta}$. Since $\theta' \ge \theta$, we have $Q \in \overline{Q_0Q'} \subset C^*$ and thus $\triangle PQR \subset C^*$.
<figure><img src="image_31.png" /><figcaption>Fig. 26: Case E: covering $\triangle PQR$ when $\theta \le \frac{\pi}{6} \le \theta'$ and $\frac{\pi}{4} - \theta \le \theta' \le \frac{\pi}{3} - \theta$.</figcaption></figure>

## 6.6 Case E

We consider the situation where $\theta \le \frac{\pi}{6} \le \theta'$, with the constraints $\frac{\pi}{4} - \theta \le \theta' \le \frac{\pi}{3} - \theta$. In other words, $R$ lies on $\overline{CR_0}$, while $Q$ lies on $\overline{BQ_0}$ in Figure 26, with $\frac{\pi}{4} \le \angle RAQ \le \frac{\pi}{3}$. 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 $\angle R'AQ' = \frac{\pi}{4}$. To see this, consider the point $H \in \overline{X^*Y^*}$ such that $\overline{Z^*H}$ is a height of $C^*$. Since the height $|AH| = 1$, we have $\triangle ADR' \equiv \triangle AHR'$ and $\triangle ABQ' \equiv \triangle AHQ'$.

We continue rotating $C^*$ until either $R$ lies on $\overline{X^*Y^*}$ or $Q$ lies on $\overline{Y^*Z^*}$.

In the first case, $R = R'$; see Figure 26(bottom left). Then $\angle RAQ \ge \frac{\pi}{4} = \angle R'AQ'$ 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 $C^*$.

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 $\angle QAR \le \frac{\pi}{3} = \angle Y^*Z^*X^*$, $R$ lies below the line $X^*Z^*$, and therefore on the highlighted segment in $C^*$.
## 6.7 Case F

In the final case we consider the angles $\frac{\pi}{12} = 15^\circ \le \theta \le \frac{\pi}{6}$ and $\frac{\pi}{6} \le \theta' \le \frac{\pi}{4}$. In other words, $Q$ lies on $\overline{BQ_0}$, while $R \in \overline{R_1R_0}$; see Figure 27(top left).

We again start by covering $T_0 = \triangle PQ_0R_0$ with $C^*$, but this time we need to cover it in two different ways; see Figure 27(top right). The first copy $C_1^*$ has $Y_1^* = Q_0$ and $Z_1^* = A$ and the second copy $C_2^*$ has $Z_2^* = Q_0$ and $Y_2^* = A$. Note that $\overline{X_1^*Z_1^*}$ and $\overline{X_2^*Z_2^*}$ intersect exactly at $R_0$.

Consider now the point $Q \in \overline{BQ_0}$. We rotate $C_2^*$ counter-clockwise around $A$ until $Q \in \overline{X_2^*Z_2^*}$ and translate $C_1^*$ to the right until $Y_1^* = Q$. This places $A$ outside of $C_1^*$, so we then rotate $C_1^*$ counter-clockwise around $Q$ until $A \in \overline{X_1^*Z_1^*}$. Figure 27 depicts the situation for different positions of $Q$.

Let $M_1$ be the intersection of $\overline{X_1^*Y_1^*}$ and $CD$ and let $M_2$ be the intersection of $\overline{X_2^*Y_2^*}$ and $CD$; see Figure 27(middle right). When $Q$ moves from $Q_0$ to $B$, the line $AM_2$ rotates around $A$ and thus $M_2$ moves downwards monotonously. We let $Q'$ be the position of $Q$ when $B \in \overline{Y_2^*Z_2^*}$ and $M_2 = R_0$; see Figure 27(bottom left).

Let $N$ be the intersection of $\overline{X_1^*Z_1^*}$ and $\overline{X_2^*Z_2^*}$. We will show below that for $Q \in \overline{BQ_0}$, the point $N$ always lies on or to the left of $CD$. This will imply that the segment $\overline{M_1M_2}$ lies entirely in $C_1^* \cup C_2^*$, so as long as $R \in \overline{M_1M_2}$, we have $\triangle PQR \subset C_1^*$ or $\triangle PQR \subset C_2^*$.

Assume now that $R$ lies above $M_2$, that is on $\overline{M_2R_0}$. This implies that $R \in \overline{R_2R_0}$, where $R_2$ is the position of $M_2$ when $Q = B$ as in Figure 27(bottom right). Such a triangle $\triangle PQR$ is covered by $C_2^*$ in its position when $Q = Q'$, as illustrated in Figure 27(bottom left).

Otherwise $R$ lies below $M_1$, that is on $\overline{R_1M_1}$. (This is indeed possible: while $Q$ moves from $Q_0$ to $B$, $M_1$ initially moves slightly upwards above $R_1$ before starting a monotone movement downwards.) In this case we rotate $C_1^*$ further counter-clockwise until $R \in X_1^*Y_1^*$. Since $\angle RQA \le \angle R_1QA \le \angle R_1Q_0A = 75^\circ = \angle X^*Y^*Z^*$, we then have $P = A \in C_1^*$ and $\triangle PQR \subset C_1^*$.

It remains to prove the claim that the point $N$ lies on or to the left of the line $CD$. We will compute the $x$-coordinate of $N$ as a function of $q := |AQ|$. As $Q$ ranges from $B$ to $Q_0$, $q$ ranges from $1/1/\cos \frac{\pi}{12} \approx 1.035$. Let $h$ be the height of $Y^*$ in $\triangle X^*Y^*Z^*$. We have

$$h = |Z^*Y^*| \sin \frac{\pi}{3} = |AQ_0| \frac{1}{2} \sqrt{3} = \frac{1}{\cos \frac{\pi}{12}} \frac{1}{2} \sqrt{3} = \sqrt{6 - 3\sqrt{3}}.$$

We next observe that $Z_1^*X_1^*$ is the line at distance $h$ from $Q$ through $A$, while $Z_2^*X_2^*$ is the line at distance $h$ from $A$ through $Q$. This implies that $\triangle AQN$ is isosceles, with two equal heights of length $h$; see Figure 28. Let $\alpha := \angle QAN = \angle AQN$ and $d := |AN| = |QN|$. We have $\sin \alpha = \frac{h}{q}$ and $\cos \alpha = \frac{q}{2d}$. Let $\beta := \angle BAQ$. Then, $\cos \beta = \frac{1}{q}$.
<figure><img src="image_32.png" /><figcaption>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**: $Q = B$.</figcaption></figure>
<figure><img src="image_33.png" /><figcaption>Fig. 28: Case F: computing the x-coordinate of N.</figcaption></figure>

Now we compute the horizontal distance $f(q)$ between A and N:

$$ 
\begin{aligned}
f(q) &= d \cos(\frac{\pi}{2} - \alpha - \beta) = d \sin(\alpha + \beta) = d \sin \alpha \cos \beta + d \cos \alpha \sin \beta \\
&= \frac{q}{2 \cos \alpha} \frac{h}{q} \frac{1}{q} + \frac{q}{2} \sin \beta = \frac{h}{2q\sqrt{1 - \frac{h^2}{q^2}}} + \frac{q}{2} \sqrt{1 - \frac{1}{q^2}} \\
&= \frac{1}{2} \left( \frac{h}{\sqrt{q^2 - h^2}} + \sqrt{q^2 - 1} \right).
\end{aligned}
$$

Plotting the function $f(q)$ shows that $f(q) > 1.01$ on the interval $1 \le q \le 1.02$. Plotting the derivative $f'(q)$ shows that $f'(q) < -0.9$ on the interval $1.01 \le q \le 1.05$, so $f(q)$ is decreasing on this interval. We also know that $f(|AQ_0|) = 1$ since then $N = R_0$. This implies that $f(q) \ge 1$ for any $Q \in \overline{BQ_0}$. It follows that $N$ lies on or to the left of $CD$, completing this case and the entire proof.

## Acknowledgement

This work was initiated at the 18th INRIA-McGill-Victoria Workshop on Computational Geometry at the Bellairs Research Institute.

## References

[1] John C. Baez, Karine Bagdasaryan, and Philip Gibbs. The Lebesgue universal covering problem. *Journal of Computational Geometry*, 6:288–299, 2015. doi:10.20382/jocg.v6i1a12.

[2] Peter Brass, William Moser, and János Pach. *Research Problems in Discrete Geometry*. Springer-Verlag, 2005.
[3] Peter Brass and Mehrbod Sharifi. A lower bound for Lebesgue's universal cover problem. *International Journal of Computational Geometry & Applications*, 15:537–544, 2005. doi:10.1142/S0218195905001828.

[4] Zoltan Füredi and John E. Wetzel. The smallest convex cover for triangles of perimeter two. *Geometriae Dedicata*, 81:285–293, 2000. doi:10.1023/A:1005298816467.

[5] Philip Gibbs. An upper bound for Lebesgue's covering problem. *arXiv preprint arXiv:1810.10089*, 2018.

[6] Tirasan Khandhawit, Dimitrios Pagonakis, and Sira Sriswasdi. Lower bound for convex hull area and universal cover problems. *International Journal of Computational Geometry & Applications*, 23(03):197–212, 2013. doi:10.1142/S0218195913500076.

[7] Mikhail D. Kovalev. A minimal convex covering for triangles (in Russian). *Ukrain. Geom. Sb.*, 26:63–68, 1983.

[8] Ji-won Park and Otfried Cheong. Smallest universal covers for families of triangles. *Computational Geometry, Theory and Applications*, 92:101686, 2021. doi:10.1016/j.comgeo.2020.101686.

[9] W Wang. An improved upper bound for worm problem. *Acta Mathematica Sinica-Chinese Series*, 49(4):835–846, 2006.

[10] John E. Wetzel. Fits and covers. *Mathematics Magazine*, 76:349–363, 2003. URL: http://www.jstor.org/stable/3654881.

[11] John E. Wetzel. Bounds for covers of unit arcs. *Geombinatorics*, 3:116–122, 01 2013. URL: https://www.researchgate.net/publication/266915918_Bounds_for_covers_of_unit_arcs.
A Computations

Here we perform the computations needed in the paper using Maple. The Maple
source file is available as an auxiliary file with this paper

Proof of Lemma 9

$$
\phi := \beta \mapsto (1 + \sin(\beta) \cos(2\beta))^{2+\left(\frac{\sin(\beta)}{\sin(2\beta)}\right)^2}
$$

$$
\phi := \beta \mapsto \left(1 + \sin(\beta) \cos(2\beta)\right)^2 + \sin(\beta)^2 \sin(2\beta)^2 \quad (1)
$$

$$
\texttt{ simplify(diff(\phi(\beta),\beta));}
$$

$$
2 \cos(\beta) (6 \cos(\beta)^2 + \sin(\beta) - 5) \quad (2)
$$

$$
\texttt{beta0:=solve(\phi(\beta)=1)[1];evalf(\beta0);evalf(\beta0*180/Pi);}
$$

$$
\beta0 := \arctan \left( \frac{8 \left( \frac{1}{8} + \frac{\sqrt{33}}{8} \right)}{\sqrt{30 - 2\sqrt{33}}} \right)
$$
$$
1.002966954
$$
$$
57.46577344
$$

$$
\texttt{h:= beta -> sin(2*beta-2*arcsin(sin(\beta)/2))/sin(2*beta)-1;}
$$

$$
h := \beta \mapsto \frac{\sin\left(2\beta - 2\arcsin\left(\frac{\sin(\beta)}{2}\right)\right)}{\sin(2\beta)} - 1 \quad (4)
$$

$$
\texttt{beta1:=arccos(1/sqrt(5));evalf(\beta1);evalf(\beta1*180/Pi);
}$$

$$
\beta1 := \arccos\left(\frac{\sqrt{5}}{5}\right)
$$
$$
1.107148718
$$
$$
63.43494883
$$

$$
\texttt{g := beta-> -cos(2*beta)-sqrt(sin(\beta)^2-sin(2*beta)^2);
}$$

$$
g := \beta \mapsto -\cos(2\beta) - \sqrt{\sin(\beta)^2 - \sin(2\beta)^2} \quad (6)
$$

$$
\texttt{f := beta-> 3/2*sin(\beta)^2-1;
}$$

$$
\arcsin(\lambda)^2
$$
B"position

> plot([f beta],h beta],beta=beta0..beta1); plot([f beta],g beta],
beta=beta1..Pi/2);

<figure><img src="image_34.png" /></figure>

<figure><img src="image_35.png" /></figure>
H''position

$$
> r := \text{beta} \rightarrow \arctan\left(\frac{\sin(\text{beta})\sin(2*\text{beta})}{((1-\cos(\text{beta})\sin(2*\text{beta})) - (1-1/\sin(\text{beta}))))} - \frac{\pi}{2} + \text{beta}\right;
$$

$$
r := \beta \mapsto \arctan \left( \frac{\frac{\sin(\beta) \sin(2\beta)}{-\cos(\beta) \sin(2\beta)} + \frac{1}{\sin(\beta)}}{\frac{\pi}{2} + \beta} \right) \quad (8)
$$

$$
> plot([r(beta), 1/4], beta=beta0..2*Pi/5);
$$

<figure><img src="image_36.png" /></figure>

$$
> plot(arcsin(2*h(beta)*\cos(beta)), beta=beta0..beta1);
$$

<figure><img src="image_37.png" /></figure>

$$
> plot(arcsin(2*g(beta)*\cos(beta)), beta=beta1..Pi/2);
$$

<figure><img src="image_38.png" /></figure>

$$
> t := \text{beta} \rightarrow r(\text{beta}) - \arcsin(2 \cdot g(\text{beta}) \cdot \cos(\text{beta})); \quad \text{dt} := \text{diff}(t(\text{beta}), \text{beta});
$$

$$
t := \beta \mapsto r(\beta) - \arcsin(2 g(\beta) \cos(\beta)) \quad (9)
$$

$$
> plot([t(beta), dt(beta)], beta=2*Pi/5..Pi/2);
$$

<figure><img src="image_39.png" /></figure>
## Triangles fitting in a square

### Isosceles triangles

```R
> l:= theta-> sin(Pi/3+2*theta)/(cos(Pi/4-theta)*sin(Pi/3));
```

$$l := \theta \mapsto \frac{\sin\left(\frac{\pi}{3} + 2\theta\right)}{\cos\left(\frac{\pi}{4} - \theta\right) \sin\left(\frac{\pi}{3}\right)} \quad (10)$$

```R
> plot([l theta], sqrt(2)], theta=0..Pi/6);
```

<figure><img src="image_40.png" /></figure>

```R
> theta0:=simplify(solve(diff(l(t),t),t)[1]); evalf([theta0,theta0*180/Pi]);
maxl:=simplify(l theta0); evalf(maxl);
theta1:=solve(l(t)=sqrt(2),t)[3]; evalf([theta1,theta1*180/Pi]);
```

$$\theta 0 := \arctan\left(\frac{1}{(2 + \sqrt{3})^{1/3}}\right) - \frac{\pi}{6} \quad [0.0490343963, 2.809463958]$$

$$maxl := \frac{2 \sin\left(2 \arctan\left(\frac{1}{(2 + \sqrt{3})^{1/3}}\right)\right) \sqrt{3}}{3 \sin\left(\frac{\pi}{12} + \arctan\left(\frac{1}{(2 + \sqrt{3})^{1/3}}\right)\right)} \\ 1.419493024$$

$$\theta 1 := \arctan\left(\frac{4 \sin\left(\frac{\pi}{12}\right)^2 + 1}{8 \sin\left(\frac{\pi}{12}\right)^2 - 6 + \sqrt{16 \sin\left(\frac{\pi}{12}\right)^4 - 72 \sin\left(\frac{\pi}{12}\right)^2 + 57}}\right) - \frac{\pi}{6} \quad [0.0995677357, 5.704811030] \quad (11)$$
X angle of the cover

$$
> xxx:=simplify(solve(sin(xx)/1(Pi/6)=sin(2*Pi/3-xx)/1 theta0,xx));
evalf([xxx,xxx*180/Pi]);
$$

$$
\begin{align*}
xxx &:= -\frac{\pi}{6} - \arctan\left(\left(2 \sin\left(\frac{\pi}{12}\right) \cos\left(\frac{\pi}{12}\right)\right) \left(-2 \sin\left(2 \arctan\left((2 + \sqrt{3})^{2/3} (\sqrt{3} - 2)\right)\right) \cos\left(\frac{\pi}{12}\right) \sqrt{(2 + \sqrt{3})^{2/3} + 1} + \sqrt{3} (2 + \sqrt{3})^{1/3} \sin\left(\frac{\pi}{12}\right) \right.\right. \\
&\qquad \left. + \sqrt{3} \cos\left(\frac{\pi}{12}\right)\right)\right) / \left(\left(2 \cos\left(\frac{\pi}{12}\right)^2 - 1\right) \left(2 \sin\left(2 \arctan\left((2 + \sqrt{3})^{2/3} (\sqrt{3} - 2)\right)\right) \cos\left(\frac{\pi}{12}\right) \sqrt{(2 + \sqrt{3})^{2/3} + 1} + \sqrt{3} (2 + \sqrt{3})^{1/3} \sin\left(\frac{\pi}{12}\right) \right.\right. \\
&\qquad \left. + \sqrt{3} \cos\left(\frac{\pi}{12}\right)\right)\right)
\end{align*}
$$

[0.7824625782, 44.83180336] (12)

Position of N, x coordinate

$$
> ff:=q-> 1/2*(sqrt(6-3*sqrt(3))/sqrt(q^2-6+3*sqrt(3))+sqrt(q^2-1));
$$

$$
ff := q \mapsto \frac{\sqrt{6 - 3\sqrt{3}}}{2\sqrt{q^2 - 6 + 3\sqrt{3}}} + \frac{\sqrt{q^2 - 1}}{2} \quad (13)
$$

> plot(ff(q),q=1..1.05);

<figure><img src="image_41.png" /></figure>

> ffd:= diff(ff(q),q);

$$
ffd := - \frac{\left( \frac{3\sqrt{2}}{2} - \frac{\sqrt{6}}{2} \right) q}{2\left(q^2 - 6 + 3\sqrt{3}\right)^{3/2}} + \frac{q}{2\sqrt{q^2 - 1}} \quad (14)
$$

> plot(ffd(q),q=1.01..1.05);

<figure><img src="image_42.png" /></figure>
Position of M1, y coordinate

$$
> plot((1-tan(x)) * tan(arcsinovernode(sqrt(6 - 3*sqrt(3))) * cos(x)) + x-Pi/4), x=0..Pi/12);
$$

<figure><img src="image_43.png" /></figure>

x

$$
> plot((1-tan(x))*tan(arcsinovernode(sqrt(6 - 3*sqrt(3)))*cos(x)) + x-Pi/4), x=Pi/16..Pi/12);
$$

<figure><img src="image_44.png" /></figure>

x