Triangulations and a discrete Brunn-Minkowski inequality in the plane

TL;DR

This paper explores a discrete analogue of the Brunn-Minkowski inequality in the plane, relating the number of triangles in triangulations of point sets and proving the conjecture in specific cases.

Contribution

The paper introduces a conjecture extending the Brunn-Minkowski inequality to triangulation counts and proves it in several special scenarios.

Findings

Proved the conjecture when sets have the same convex hull.

Established the inequality for sets where one is a union with a single point.

Confirmed the inequality for sets with no interior points.

Abstract

For a set $A$ of points in the plane, not all collinear, we denote by ${\rm tr}(A)$ the number of triangles in any triangulation of $A$; that is, ${\rm tr}(A) = 2i+b-2$ where $b$ and $i$ are the numbers of points of $A$ in the boundary and the interior of $[A]$ (we use $[A]$ to denote "convex hull of $A$"). We conjecture the following analogue of the Brunn-Minkowski inequality: for any two point sets $A,B \subset {\mathbb R}^2$ one has \[ {\rm tr}(A+B)^{\frac12}\geq {\rm tr}(A)^{\frac12}+{\rm tr}(B)^{\frac12}. \] We prove this conjecture in several cases: if $[A]=[B]$, if $B=A\cup\{b\}$, if $|B|=3$, or if none of $A$ or $B$ has interior points.

Equations

$$|A+B|\ge |A|+|B|-1,$$

$$\lambda (X+Y{)}^{\frac{1}{d}}\ge \lambda (X{)}^{\frac{1}{d}}+\lambda (Y{)}^{\frac{1}{d}}$$

$$|A+B|\ge (d!{)}^{-\frac{1}{d}}(|A|-d{)}^{\frac{1}{d}}+|B{|}^{\frac{1}{d}}.$$

$$|A+A|\ge (d+1)|A|-\left(\genfrac{}{}{0ex}{}{d+1}{2}\right).$$

$$\mathrm{tr}(A)={\Delta }_A+2{\Omega }_A-2=2|A|-{\Delta }_A-2.$$

$$\mathrm{tr}(A+B{)}^{\frac{1}{2}}\ge \mathrm{tr}(A{)}^{\frac{1}{2}}+\mathrm{tr}(B{)}^{\frac{1}{2}}.$$

$$|A+B|\ge |A|+|B|+(2|A|-{\Delta }_A-2{)}^{1/2}(2|B|-{\Delta }_B-2)-1.$$

$$|A+A|\le (4-\epsilon )|A|,$$

$$\frac{2}{\epsilon }\cdot (1+\frac{32}{|A|{\epsilon }^2})$$

$$|A+A|\le (4-\epsilon )|A|,$$

$$\Vert M{\Vert }^{\frac{1}{d}}\ge |T_A{|}^{\frac{1}{d}}+|T_B{|}^{\frac{1}{d}}.$$

$$|M_{11}|\ge \sqrt{|T_A|\cdot |T_B|}.$$

$$A=\{(0,0),(-1,-2),(2,1)\}.$$

$$B=\{p,q,l_0,\dots ,l_k,r_0,\dots ,r_{k-1}\},$$

$$[p,l_i,r_i],[q,l_i,r_i],i=0,\dots ,k-1\text{and }[p,l_i,r_{i-1}],[q,l_i,r_{i-1}],i=1,\dots ,k.$$

$$\Vert M\Vert =2^d|T_A|.$$

$$\mathrm{tr}(A+B)\ge 2\mathrm{tr}(A)+2\mathrm{tr}(B),$$

$$f_{X}(z)=\left\{\begin{array}[]{rl}1&\mbox{ \ if $z\in\partial[X]$}\\[2.15277pt] 2&\mbox{ \ if $z\in{\rm int}\,[X]$}\end{array}\right.,$$

$$\mathrm{tr}(X)=\left(\sum _{z\in X}{f}_X(z)\right)-2.$$

$$\mathrm{tr}(A_t+B_t)\ge 2\mathrm{tr}(A_t)+2\mathrm{tr}(B_t),$$

$$\sum _{p\in A_t+B_t}{f}_{A_t+B_t}(p)=\left(\sum _{p\in A_t}{f}_{A_t}(p)\right)+\left(\sum _{p\in B_t}{f}_{B_t}(p)\right)-6.$$

$$v_i+p=v_j+q\text{mbox}yields{v}_i=v_j\text{mbox}andp=q.$$

$$\left(\sum _{i,j=1,2,3}{f}_{A_t+B_t}(v_i+v_j)\right)=\left(\sum _{i=1}^3{f}_{A_t}(v_i)\right)+\left(\sum _{j=1}^3{f}_{B_t}(v_j)\right)=6.$$

$$\begin{array}[]{rcl}\sum_{j=1}^{3}f_{A_{t}+B_{t}}(p+v_{j})&>&2f_{A_{t}}(p)\\ \sum_{i=1}^{3}f_{A_{t}+B_{t}}(v_{i}+q)&>&2f_{B_{t}}(q).\end{array}$$

$$\mathrm{tr}(A+A)=4\cdot \mathrm{tr}(A)$$

$${\rm tr}(A+A)={\rm tr}(\mbox{$\frac{1}{2}$}(A+A))\geq{\rm tr}(A_{T})=4\cdot{\rm tr}(A).$$

$$A_T=\frac{1}{2}(A+A).$$

$$A_T=\frac{1}{2}(A+A),$$

$$|A_{\ell }+A_{\ell }|=|(A+A)\cap (\ell +\ell )|=|A_T\cap \ell |=2|A_{\ell }|-1,$$

$$(4a+2k{)}^{\frac{1}{2}}>a^{\frac{1}{2}}+(a+k{)}^{\frac{1}{2}}.$$

Full text

Triangulations and a discrete Brunn-Minkowski inequality in the plane

Károly J. Böröczky

Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Reltanoda u. 13-15, H-1053 Budapest, Hungary, and Department of Mathematics, Central European University, Nador u 9, H-1051, Budapest, Hungary, E-mail: boroczky.karoly.j@renyi.mta.hu, supported by NKFIH grants 116451, 121649 and 129630   

Máté Matolcsi Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Reltanoda u. 13-15, H-1053 Budapest, Hungary, E-mail: matolcsi.mate@renyi.mta.hu, and Technical University of Budapest, Egry J. u. 1., H-1111 Budapest, supported by NKFIH grant 109789   

Imre Z. Ruzsa Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Reltanoda u. 13-15, H-1053 Budapest, Hungary, E-mail: ruzsa@renyi.hu, supported by NKFIH grant 109789   

Francisco Santos Depto. de Matemáticas, Estadística y Computación, Universidad de Cantabria, 39012 Santander, SPAIN. E-mail: francisco.santos@unican.es. Supported by grant MTM2017-83750-P of the Spanish Ministry of Science and grant EVF-2015-230 of the Einstein Foundation Berlin   

Oriol Serra Department of Mathematics, Universitat Politècnica de Catalunya, and Barcelona Graduate School of Mathematics, Barcelona, Spain. E-mail: oriol.serra@upc.edu. supported by grants MTM2017-82166-P and MDM-2014-0445 of the Spanish Ministry of Science

1 Introduction

In this paper we write $A,B$ to denote finite subsets of ${\mathbb{R}}^{d}$, and $|\cdot|$ stands for their cardinality. We say that $A\subset{\mathbb{R}}^{d}$ is $d$–dimensional if it is not contained in any affine hyperplane of ${\mathbb{R}}^{d}$. Equivalently, the real affine span of $A$ is ${\mathbb{R}}^{d}$. For objects $X_{1},\ldots,X_{k}$ in ${\mathbb{R}}^{2}$, $[X_{1},\ldots,X_{k}]$ denotes their convex hull. The lattice generated by $A$ is the additive subgroup $\Lambda=\Lambda(A)\subset{\mathbb{R}}^{d}$ generated by $A-A$, and $A$ is called saturated if it satisfies $A=[A]\cap\Lambda(A)$.

Our starting point are two classical results. The first one is from the 1950’s, due to Kemperman [10], and popularized by Freiman [4]: if $A$ and $B$ are finite nonempty subsets of ${\mathbb{R}}$, then

[TABLE]

with equality if and only if $A$ and $B$ are arithmetic progressions of the same difference. The other result, the Brunn-Minkowski inequality, dates back to the 19th century. It says that if $X,Y\subset{\mathbb{R}}^{d}$ are compact nonempty sets then

[TABLE]

where $\lambda$ stands for the Lebesgue measure. Moreover, provided that $\lambda(X)\lambda(Y)>0$, equality holds if and only if $X$ and $Y$ are convex homothetic sets.

Various discrete analogues of the Brunn-Minkowski inequality have been established in Bollobás, Leader [1], Gardner, Gronchi [5], Green, Tao [6], González-Merino, Henze [11], Hernández, Iglesias and Yepes [8], Huicochea [9] in any dimension, and Grynkiewicz, Serra [7] in the planar case. Most of these papers use the method of compression, which changes a finite set into a set better suited for sumset estimates, but does not control the convex hull.

Unfortunately the known analogues are not as simple in their form as the original Brunn–Minkowski inequality. For instance, a formula due to Gardner and Gronchi [5] says that, if $A$ is $d$–dimensional, then

[TABLE]

Concerning the case $A=B$, Freiman [4] proved that if the dimension of $A$ is $d$, then

[TABLE]

Both estimates are optimal. In particular, we can not expect a true discrete analogue of the Brunn–Minkowski inequality if the notion of volume is replaced by cardinality.

We here conjecture and discuss a more direct version of the Brunn–Minkowski inequality where the notion of volume is replaced by the number of full dimensional simplices in a triangulation of the convex hull of the finite set.

For any finite $d$–dimensional set $A\subset{\mathbb{R}}^{d}$ we write $T_{A}$ to denote some triangulation of $A$, by which we mean a triangulation of $[A]$ using $A$ as the set of vertices. We denote $|T_{A}|$ the number of $d$-dimensional simplices in $T_{A}$.

In dimension two the number $|T_{A}|$ is the same for all triangulations of $A$, so we denote it ${\rm tr}(A)$. More precisely, if $\Delta_{A}$ and $\Omega_{A}$ denote the number of points of $A$ in the boundary $\partial[A]$ and in the interior ${\rm int}[A]$, respectively, then

[TABLE]

Therefore in dimension two we can formulate the following discrete analogue of the Brunn–Minkowski inequality.

Conjecture 1

If finite $A,B\subset{\mathbb{R}}^{2}$ in the plane are not collinear, then

[TABLE]

One case where Conjecture 1 holds with equality is when $A$ and $B$ are homothetic saturated sets with respect to the same lattice; namely, $A=\Lambda\cap k\cdot P$ and $B=\Lambda\cap m\cdot P$ for a lattice $\Lambda$, polygon $P$ and integers $k,m\geq 1$. This follows from the original Brunn-Minkowski equality, since $A+B=\Lambda\cap(k+m)\cdot P$ and the area of any triangle in a suitable triangulation is $\frac{1}{2}\det\Lambda$.

We also note that Conjecture 1, together with the equality (3) and the fact that $\Delta_{A+B}\geq\Delta_{A}+\Delta_{B}$, would imply the following inequality of Gardner and Gronchi [5, Theorem 7.2] for sets $A$ and $B$ saturated with respect to the same lattice:

[TABLE]

Unfortunately we have not been able to prove Conjecture 1 in full generality. Our main results are the following four cases of it: if $[A]=[B]$ (Theorem 2), in which case we also determine the conditions for equality in Conjecture 1; if $A$ and $B$ differ by one element (Theorem 4); if either $|A|=3$ or $|B|=3$ (Theorem 7); and if none of $A$ and $B$ have interior points (Theorem 8). Actually, the last two theorems verify a stronger conjecture (Conjecture 5) discussed below.

We start with the case $[A]=[B]$, which naturally include the case $A=B$.

Theorem 2

Let $A,B\subset{\mathbb{R}}^{2}$ be finite two dimensional sets. If $[A]=[B]$ then Conjecture 1 holds. Moreover equality holds if and only if $A=B$, and

(a) either $A$ is a saturated set, or

(b) $A=\{z_{1},\ldots,z_{k}\}$ for $k\geq 4$, where $z_{1},\ldots,z_{k-3}\in{\rm int}[z_{k-2},z_{k-1},z_{k}]$, and $z_{1},\ldots,z_{k-2}$ are collinear and equally spaced in this order (see Figure 1).

Let us mention that Theorem 2 (in fact, its particular case $A=B$) gives a simple proof of the following structure theorem of Freiman [4] for a planar set with small doubling. We recall that according to (2), if finite $A\subset{\mathbb{R}}^{2}$ is two dimensional, then $|A+A|\geq 3|A|-3$ and, if the dimension of $A$ is at least $3$, then $|A+A|\geq 4|A|-6$.

Corollary 3** (Freiman)**

Let $A\subset{\mathbb{R}}^{2}$ be a fnite two dimensional set and $\varepsilon\in(0,1)$. If $|A|\geq 48/\varepsilon^{2}$ and

[TABLE]

then there exists a line $l$ such that $A$ is covered by at most

[TABLE]

lines parallel to $l$.

We note that, for $A$ the grid $\{1,\ldots,k\}\times\{1,\ldots,k^{2}\}$ and large $k$,

[TABLE]

with $\varepsilon=\varepsilon_{k}=\frac{2}{k}$ and $A$ can not be covered by less than $k$ parallel lines. Therefore the constant $2$ in the numerator of $\frac{2}{\varepsilon}$ is asymptotically optimal in Corollary 3.

The next case w address is when $A$ and $B$ differ by one element.

Theorem 4

*Let $A\subset{\mathbb{R}}^{2}$ be a finite two dimensional set. If $B=A\cup\{b\}$ for some $b\not\in A$ then Conjecture 1 holds. *

For our next results we need the notion of mixed subdivision (see De Loera, Rambau, Santos [3] for details). For finite $d$–dimensional sets $A,B\subset{\mathbb{R}}^{d}$ and triangulations $T_{A}$ and $T_{B}$ of $[A]$ and $[B]$, we call a polytopal subdivision $M$ of $[A+B]$ a mixed subdivision corresponding to $T_{A}$ and $T_{B}$ if

(i) every $k$-cell of $M$ is of the form $F+G$ where $F$ is an $i$-simplex of $T_{A}$ and $G$ is a $j$-simplex of $T_{B}$ with $i+j=k$;

(i) for any $d$-simplices $F$ of $T_{A}$ and $G$ of $T_{B}$, there is a unique $b\in B$ and a unique $a\in A$ such that $F+b\in M$ and $a+G\in M$.

We write $\|M\|$ to denote the weighted number of $d$-polytopes, where $F+G$ has weight ${i+j\choose i}$ if $F$ is an $i$-simplex of $T_{A}$, and $G$ is a $j$-simplex of $T_{B}$ with $i+j=d$. In particular, all vertices of $M$ are in $A+B$, and the number of $d$-simplices is $\|M\|$ for any triangulation of $M$ with the same set of vertices (see e.g. [3, Proposition 6.2.11]).

The main goal of this paper is to investigate the following problem: For which triangulations $T_{A}$ and $T_{B}$ there exists a corresponding mixed subdivision $M$ for $[A+B]$ such that

[TABLE]

In ${\mathbb{R}}^{2}$, we write $M_{11}$ to denote the set of parallelograms in a mixed subdivision $M$. In this case (5) is equivalent to the following stronger version of Conjecture 1.

Conjecture 5

For every finite two dimensional sets $A,B\subset{\mathbb{R}}^{2}$ there exist triangulations $T_{A}$ and $T_{B}$ of $[A]$ and $[B]$ using $A$ and $B$, respectively, as the set of vertices, and a corresponding mixed subdivision $M$ of $[A+B]$ such that

[TABLE]

Conjecture 5 offers a geometric and algorithmic approach to prove Conjecture 1.

The following example shows that one cannot a priori fix the triangulations $T_{A}$ and $T_{B}$ in Conjecture 5:

Proposition 6

Let

[TABLE]

For $k\geq 145$, let

[TABLE]

where $p=(-1,k+1)$, $q=(k+1,-1)$, $l_{i}=(i,i)$ for $i=0,\ldots,k$ and $r_{i}=(i,i+1)$ for $i=0,\ldots,k-1$.

Let $T_{B}$ be the triangulation of $B$ consisting of the triangles

[TABLE]

Then, no mixed subdivision of $A+B$ corresponding to $T_{B}$ and any triangulation $T_{A}$ of $A$ satisfies (5) for $d=2$.

Now Conjecture 5 is verified if either $A$ or $B$ has only three elements.

Theorem 7

If $|B|=3$, then Conjecture 5 holds for any finite two dimensional set $A\subset{\mathbb{R}}^{2}$.

**Remark ** It follows that if $B$ is the sum of sets of cardinality three, then Conjecture 1 holds for any finite two dimensional set $A\subset{\mathbb{R}}^{2}$. For example, if $m\geq 1$ is an integer, and $B=\{(t,s)\in{\mathbb{Z}}^{2}:t,s\geq 0\mbox{ and }t+s\leq m\}$, or $B=\{(t,s)\in{\mathbb{Z}}^{2}:|t|,|s|\leq m\mbox{ and }|t+s|\leq m\}$.

Conjecture 1 was verified by Böröczky, Hoffman [2] if $A$ and $B$ are in convex position; namely, $A\subset\partial[A]$ and $B\subset\partial[B]$. Here we even verify Conjecture 5 under these conditions.

Theorem 8

Let $A,B\subset{\mathbb{R}}^{2}$ be finite two dimensional sets. If $A\subset\partial[A]$ and $B\subset\partial[B]$ then Conjecture 5 holds.

Part of the reason why we could not verify Conjecture 1 in general is that, except for Theorem 7, our arguments actually prove the inequality ${\rm tr}(A+B)\geq 2({\rm tr}(A)+{\rm tr}(B))$, which is stronger than Conjecture 1, but which does not hold for all pairs with $A\subset B$. For example, if $A$ are the nonnegative integer points with sum of coordinates at most $k$ and $B$ is the same with sum of coordinates at most $l$, we have ${\rm tr}(A+B)=(k+l)^{2}$, ${\rm tr}(A)=k^{2}$ and ${\rm tr}(B)=l^{2}$. So we have ${\rm tr}(A+B)<2({\rm tr}(A)+{\rm tr}(B))$ if $k\neq l$.

Turning to higher dimensions, we note that if $T_{A}=T_{B}$, then a mixed subdivision satisfying (5) does exist.

Theorem 9

For a finite $d$–dimensional set $A\subset{\mathbb{R}}^{d}$ and for any triangulation $T_{A}$ of $[A]$ using $A$ as the set of vertices there exists a corresponding mixed subdivision $M$ of $[A+A]$ such that

[TABLE]

Therefore in certain cases, mixed subdivisions point to a higher dimensional generalization of Conjecture 1. This is specially welcome knowing that, if $d\geq 3$, then the order of the number of $d$-simplices in a triangulation of the convex hull of a finite $A\subset{\mathbb{R}}^{d}$ spanning ${\mathbb{R}}^{d}$ might be as low as $|A|$ and as high as $|A|^{\lfloor d/2\rfloor}$ for the same $A$. In particular, one can not assign the number of $d$-simplices as a natural notion of discrete volume if $d\geq 3$.

2 Proof of Theorem 2

We will actually prove that

[TABLE]

a stronger inequality than Conjecture 1.

For a finite two dimensional set $X\subset{\mathbb{R}}^{2}$, we define

[TABLE]

so that

[TABLE]

Lemma 10

Let $A,B\subset{\mathbb{R}}^{2}$ satisfy $[A]=[B]$. Then inequality (7) holds. Moreover, equality in (7) yields $A=B$.

*Proof: * Let $T$ be a triangulation of $[A]=[B]$ using the points in $A\cap B$ as vertices. One nice thing about inequality (7) is that, since it is linear, it is additive over the triangles of $T$. Therefore, it suffices to show that, for each triangle $t$ of $T$, if $A_{t}=A\cap t$ and $B_{t}=B\cap t$, then

[TABLE]

and that equality in (8) implies that $A_{t}=B_{t}$ consists of the three vertices of $t$ alone. Moreover, inequality (8) is equivalent to

[TABLE]

Let $A_{t}\cap B_{t}=\{v_{1},v_{2},v_{3}\}$ be the three vertices of the triangle $t=[A_{t}]=[B_{t}]$. We claim that if $i,j\in\{1,2,3\}$, $p\in(A_{t}\cup B_{t})\backslash\{v_{1},v_{2},v_{3}\}$ and $q\in A_{t}\cup B_{t}$, then

[TABLE]

We may assume that $v_{i}$ is the origin and, to get a contradiction, $v_{i}\neq v_{j}$. Then the line $l$ passing through $v_{j}$ and parallel to the side of $t$ opposite to $v_{j}$ separates $t$ and $v_{j}+t$, and intersects $t$ only in $v_{j}\neq p$. Since $v_{j}+q\in v_{j}+t$, we get the desired contradiction.

It follows from (10) that the six points $v_{i}+v_{j}$, $1\leq i\leq j\leq 3$, and the points of the form $v_{i}+p$, $i=1,2,3$ and $p\in(A_{t}\cup B_{t})\backslash\{v_{1},v_{2},v_{3}\}$ are all different. Since the six points $v_{i}+v_{j}$, $1\leq i\leq j\leq 3$, belong to $\partial(A_{t}+B_{t})$, we have

[TABLE]

On the other hand, we claim that, if $p\in A_{t}\backslash\{v_{1},v_{2},v_{3}\}$ and $q\in B_{t}\backslash\{v_{1},v_{2},v_{3}\}$, then

[TABLE]

Indeed, the inequality readily holds if $p\in\partial[A_{t}]$ and, if $p\in{\rm int}\,[A_{t}]$, then $p+v_{j}\in{\rm int}\,[A_{t}+B_{t}]$ for $j=1,2,3$, as well, yielding (12).

By combining (11) and (12) we get (9) and in turn (7). Moreover, (12) shows that if equality holds in (8) then $A_{t}=B_{t}$ and, therefore, if equality holds in (7), then $A=B$. $\Box$

For a finite two dimensional set $A\subset{\mathbb{R}}^{2}$ and a triangulation $T$ of $A$ we denote by $A_{T}$ the union of $A$ and the set of midpoints of the edges of $T$ (see Figure 3).

Lemma 11

Let $A\subset{\mathbb{R}}^{2}$ be a finite a finite two dimensional set. The equality

[TABLE]

holds if, and only if, for every triangulation $T$ of $[A]$, we have $A_{T}=\frac{1}{2}(A+A)$.

*Proof: * Divide each triangle $t$ of $T$ into four triangles using the vertices of $t$ and the midpoints of the sides of $t$. This way we have obtained a triangulation of $[A]=[A_{T}]$ using $A_{T}$ as the vertex set. Therefore

[TABLE]

Moreover, there is equality if and only if $A_{T}=\frac{1}{2}(A+A)$. $\Box$

We observe that the equation in Lemma 11 is equivalent to Conjecture 1 for the case $A=B$. Therefore all we have left to prove is that ${\rm tr}(A+A)=4\cdot{\rm tr}(A)$ if and only if $A$ is of the form either (a) or (b) in Theorem 2. The if part is simple.

Lemma 12

Suppose that either (a) or (b) in Theorem 2 hold for the finite set $A$. Then

[TABLE]

*Proof: * Suppose first that $A=[A]\cap\Lambda$ for a lattice $\Lambda$. We may assume $\Lambda={\mathbb{Z}}^{2}$. Then clearly the midpoints of sides of every triangulation $T$ of $[A]$ using $A$ as vertex set are precisely the points of $\frac{1}{2}(A+A)$.

Next, if we have property (b), then there is a unique triangulation $T$ of $[A]$ using $A$ as vertex set. For $1\leq i<j\leq k$, $[z_{i},z_{j}]$ is an edge of $T$, unless $j\leq k-2$, an hence we have $A_{T}=\frac{1}{2}(A+A)$ again. $\Box$

The next Lemma shows the reverse direction and concludes the proof of Theorem 2.

Lemma 13

Let $A\subset{\mathbb{R}}^{2}$ be a finite two dimensional set. If for every triangulation $T$ of $A$ it holds that

[TABLE]

then either (a) or (b) from Theorem 2 hold.

*Proof: * We first prove two simple claims. All throughout we assume that $A_{T}=\frac{1}{2}(A+A)$ for every triangulation $T$ of $A$.

Claim 14

Let $\ell$ be a line intersecting $A$ in at least two points and $A_{\ell}=A\cap\ell$. If $A_{\ell}+A_{\ell}=(A+A)\cap(\ell+\ell)$ then the points in $A_{\ell}$ form an arithmetic progression. In particular, the points on each side of the convex hull of $A$ form an arithmetic progression.

*Proof: * There is a triangulation $T$ of $A$ which contains the edges defined by consecutive points in $A_{\ell}$. Since there are $|A_{\ell}|-1$ midpoints of $T$ on $A_{\ell}$, by the hypothesis of the Lemma and of the Claim, we have

[TABLE]

which implies that $A_{\ell}$ consists of an arithmetic progression. $\Box$

Call a set of four points of $A$ no three of which collinear an empty quadrangle of $A$ if their convex hull contains no further points of $A$.

Claim 15

Let $x_{1},x_{2},x_{3},x_{4}\in A$ form an empty quadrangle of $A$. If they are in convex position then the four points form a parallelogram. That is, assuming they are listed in clockwise order, we have $x_{1}+x_{3}=x_{2}+x_{4}$.

*Proof: * There are two triangulations of $A$ containing the edges of the convex quadrangle, one of them containing the edge $x_{1}x_{3}$ and the other one containing $x_{2}x_{4}$. Since $A_{T}$ cannot depend on the triangulation, the midpoints of these two edges must coincide and therefore $x_{1}+x_{3}=x_{2}+x_{4}$. $\Box$

The proof of the Lemma is by induction on $k=|A|$. The Lemma clearly holds if $k=3$.

Suppose $k=4$. If three of the points are collinear then they are on an edge of the convex hull of $A$ and, by Claim 14, they form an arithmetic progression. With the fourth one they form a saturated set. If no three of the points are collinear then the four points form an empty quadrangle. If they are in convex position then by Claim 15 they form a saturated set, otherwise case (b) holds.

Let $k>4$. Choose a vertex $v$ of the convex hull of $A$ and let $A^{\prime}=A\setminus\{v\}$. If all points of $A^{\prime}$ are collinear then by Claim 14 they are in a progresion and, with $v$, they form a saturated set. Suppose that $A^{\prime}$ is not on o a line. For every triangulation $T^{\prime}$ of $A$ there is a triangulation $T$ of $A$ containing $T^{\prime}$. The points in $\frac{1}{2}(A^{\prime}+A^{\prime})$ are contained in the convex hull of $A^{\prime}$ and, by the condition of the Lemma, coincide with $A^{\prime}_{T^{\prime}}$. By induction either (a) or (b) hold for $A^{\prime}$. We consider the two cases.

Case 1. $A^{\prime}$ is a saturated set.

Case 1.1. There is a convex empty quadrangle formed by $v$ and three points of $A^{\prime}$. Then, by Claim 15, $v$ belongs to the lattice generated by $A^{\prime}$ as well. Moreover, since $A^{\prime}$ is convex, $A$ is also convex and case (a) holds.

Case 1.2. There is no convex empty quadrangle involving $v$ and three points of $A^{\prime}$. Then it is easily checked that $A^{\prime}$ has at most one empty convex quadrangle.

If there is none in $A^{\prime}$ then, up to an affine transformation, $A^{\prime}$ consists of the point $(0,1)$ or the two points $(0,\pm 1)$, and the remaining points on the line $y=0$. Then either (i) $v$ belongs to the same line $y=0$, which satisfies the condition of Claim 14, and all points on that line in $A$ are in arithmetic progression, so that $A$ is a saturated set, or (ii) $A^{\prime}$ contains only the point $(0,1)$ and $v$ is on the line $x=0$, in which case Claim 14 yields that the three points of $A$ on that line are in arithmetic progression and $A$ is a saturated set again, or (iii) $A^{\prime}$ contains only the point $(0,1)$ and $v$ belongs to none of the two lines containing $A^{\prime}$ and case (b) holds (see Figure 4).

If $A^{\prime}$ contains one convex empty quadrangle then, up to affinities, $A^{\prime}$ consists of the four points $(0,0),(1,0),(1,1),(0,1)$ and the remaining ones are on the line $x=y$. Moreover $v$ must belong to the latter line as well and Claim 14 yields that the points on that line are in arithmetic progression and $A$ is a saturated set (see Figure 4).

Case 2. $A^{\prime}$ is as in (b). We may assume that the progression of points of $A^{\prime}$ lies on the line $x=0$. If $v$ is not on this line then it forms a convex empty quadrangle with two extreme points of the progression and one of the vertices $w$ of the triangle. By Claim 15, $v$ must be the point $w+(\pm 1,0)$, which gives a configuration not satisfying the condition of the Lemma. Therefore $v$ lies on the line $x=0$ which satisfies the condition of Claim 14, so that $v$ belongs to the progression on that line yielding case (b). $\Box$

3 Proof of Theorem 4

The inequality between the quadratic and arithmetic means gives that, if $a,k>0$, then

[TABLE]

Therefore to prove Theorem 4, it is sufficient the verify the following: Let $B=A\cup\{b\}$ for $b\not\in A$.

(*)

If ${\rm tr}(A)=a$ and ${\rm tr}(B)=a+k$, then ${\rm tr}(A+B)\geq 4a+2k$.

We fix a triangulation $T$ of $A$, and let $A_{T}$ be the union of $A$ and the family of midpoints of the edges of $T$. It follows by (3) that

[TABLE]

To estimate {\rm tr}(A+B)={\rm tr}(\mbox{\frac{1}{2}}(A+B)), we isolate certain subset $V$ of $A$ in a way such that

[TABLE]

Therefore

[TABLE]

We distinguish two cases depending on how to define $V$.

**Case 1 ** $b\not\in[A]$

We say that $x\in[A]$ is visible if $[b,x]\cap[A]=\{x\}$. In this case $x\in\partial A$. We note that there are exactly two visible points on $\partial[B]$, which are on the two supporting lines to $[A]$ passing through $b$ (see Figure 5). Let $k+1$ be the number of visible points of $A$, and hence $k\geq 1$. Now $k-1$ visible points of $A$ lie in ${\rm int}[B]$, thus (3) yields that ${\rm tr}(B)=a+k$.

Let $V$ be the set of visible points of $A$. The condition (13) is satisfied because [A]\cap(\mbox{\frac{1}{2}}(V+\{b\}))=\emptyset. We have |\mbox{\frac{1}{2}}(V+\{b\})|=k+1, and $2k-1$ visible points of $A_{T}$ lie in ${\rm int}[B]$. In particular, $(*)$ follows as (14) yields

[TABLE]

**Case 2 ** $b\in[A]$

In this case ${\rm tr}(B)=a+k$ for $k\leq 2$ by (3), and $b$ is contained in a triangle $T=[p,q,r]$ of $T$ (see Figure 6).

We may assume that $b$ is not contained in the sides $[r,p]$ and $[r,q]$ of $T$. We take $V=\{p,q,r\}$, which satisfies (13). Since $\frac{1}{2}(b+q)\in{\rm int}T\subset{\rm int}[A]$, (14) yields ${\rm tr}(A+B)\geq 4a+4$. In turn, we conclude Theorem 4.

**Remark: ** The argument does not work if we only asssume that $A\subset B$, because we may have equality in Conjecture 1 in this case.

4 Proof of Theorem 7

Let $A\subset{\mathbb{R}}^{2}$ be finite and not contained in any line. By a path $\sigma$ on $A$ we mean a piecewise linear simple path whose vertices are in $A$, and every point of $A$ in the support of $\sigma$ is a vertex of the path. We write $|\sigma|$ to denote the number of segments forming $\sigma$. We allow the case that $\sigma$ is a point, and in this case $|\sigma|=0$. We say that $\sigma$ is transversal to a non-zero vector $u$ if every line parallel to $u$ intersects $\sigma$ in at most one point. In this case, the segments in $\sigma$ induce a subdivision of $\sigma+[o,u]$ into $|\sigma|$ parallelograms if $|\sigma|\geq 1$. For the proof of Theorem 7 the idea is to find an appropriate set of paths on $A$ with total length at least $\sqrt{|T_{A}|}$.

First, we explore the possibilities using only one or two paths. We will see in Remark 16 that one path is not enough, but Proposition 17 shows that using two paths $\sigma_{1},\sigma_{2}$ almost does the job.

Observe that for any given non-zero vector $w$, the length of the longest path on $A$ transversal to $w$ equals the number of lines parallel to $w$ intersecting $A$, minus one.

Remark 16

Given pairwise independent vectors $w_{1},\ldots,w_{n}$ let $f(w_{1},\ldots,w_{n},s)$ be the minimal number such that, for every finite set $A\subset{\mathbb{R}}^{2}$ with ${\rm tr}(A)=s$, there is a $w_{i}$ and a path on $A$ transversal to $w_{i}$ of length $f(w_{1},\ldots,w_{n},s)$.

For $n=2$, $f(w_{1},w_{2},s)\geq\sqrt{s/2}$, with equality provided that $k:=\sqrt{s/2}$ is an integer. An extremal configuration consists of the points $\{iw_{1}+jw_{2}:i,j\in\{0,\dots,k\}\}$.

For $n=3$, $f(w_{1},w_{2},w_{3},s)\geq\sqrt{2s/3}$ and equality holds provided that $s=6k^{2}$. Assuming without loss of generality that $w_{1}+w_{2}+w_{3}=0$, an extremal configuration is given by the points of the lattice generated by $w_{1},w_{2}$ in the affine regular hexagon $[\pm kw_{1},\pm kw_{2},\pm kw_{3}]$.

Let $e_{1}=(1,0)$ and $e_{2}=(0,1)$, and let $\sigma_{1},\sigma_{2}$ be piecewise linear paths whose vertices are among the vertices of $A$. We say that the ordered pair $(\sigma_{1},\sigma_{2})$ is a horizontal-vertical path if

(i’) $\sigma_{i}$ is transversal with respect $e_{3-i}$ (possibly a point), $i=1,2$;

(ii’) the right endpoint $a$ of $\sigma_{1}$ is the upper endpoint of $\sigma_{2}$

(iii’) writing ${\mathbb{R}}_{+}=\{t\in{\mathbb{R}}:\,t>0\}$, if $|\sigma_{1}|,|\sigma_{2}|>0$, then

[TABLE]

We call $\sigma_{1}$ the horizontal branch, and $\sigma_{2}$ the vertical branch, and $a$ the center.

We observe that if $\sigma^{\prime}_{i}$ is the image of $\sigma_{i}$ by reflection through the line ${\mathbb{R}}(e_{1}+e_{2})$, then the ordered pair $(\sigma^{\prime}_{2},\sigma^{\prime}_{1})$ is also a horizontal-vertical path.

For any polygon $P$ and non-zero vector $u$, we write $F(P,u)$ to denote the face of $P$ with exterior normal $u$. In particular, $F(P,u)$ is either a side or a vertex.

Proposition 17

For every finite $A\subset{\mathbb{R}}^{2}$ not contained in a line, and for every triangulation $T$ of $[A]$ using $A$ as a vertex set, there exists a horizontal-vertical path $(\sigma_{1},\sigma_{2})$ whose vertices belong to $A$, and satisfies

[TABLE]

*Proof: *Let us write

[TABLE]

By the invariance with respect to reflection through the line ${\mathbb{R}}(e_{1}+e_{2})$, we may assume that

[TABLE]

We set $\{\langle e_{1},p\rangle:\,p\in A\}=\{\alpha_{0},\ldots,\alpha_{k}\}$ with $\alpha_{0}<\ldots<\alpha_{k}$, $k\geq 1$. For $i=0,\ldots,k$, let $A_{i}=\{p\in A:\,\langle e_{1},p\rangle=\alpha_{i}\}$, let $x_{i}=|A_{i}|$, and let $a_{i}$ be the top most point of $A_{i}$; namely, $\langle e_{2},a_{i}\rangle$ is maximal. In particular, $x_{0}=|F([A],-e_{1})\cap A|$. For each $i=1,\ldots,k$, we consider the horizontal-vertical path $(\sigma_{1i},\sigma_{2i})$ where

[TABLE]

and the vertex set of $\sigma_{2i}$ is $A_{i}$. In particular, the total length of the horizontal-vertical path is $(\sigma_{1i},\sigma_{2i})$ is

[TABLE]

The average length of these paths for $i=1,\ldots,k$ is

[TABLE]

We observe that $2|A|=|T|+\Delta_{A}+2$, according to (3), and (15) yields

[TABLE]

Therefore we deduce from the inequality between the arithmetic and geometric mean that

[TABLE]

Therefore there exists some horizontal-vertical path $(\sigma_{1i},\sigma_{2i})$ satisfying (17). $\Box$

The estimate of Proposition 17 is close to be optimal according to the following example.

Example 18

Let $k\geq 2$ and $t>0$. Let $A^{\prime}$ be the saturated set with $[A^{\prime}]$ having vertices $(0,0),(0,k),(k-1,0)$ and $(k-1,1)$, and let $A=A^{\prime}\cup\{(k+t,0)\}$. A triangulation of $A$ has $k^{2}+k-1$ triangles and every horizontal–vertical path $(\sigma_{1},\sigma_{2})$ on $A$ has total length

[TABLE]

$\Box$

We next proceed to the proof of Theorem 7 by a similar strategy using three paths. Let $B=\{v_{1},v_{2},v_{3}\}$ and, for $\{i,j,k\}=\{1,2,3\}$ denote by $u_{i}$ the exterior unit normal to the side $[v_{j},v_{k}]$ of $B$. A set of three paths $(\sigma_{1},\sigma_{2},\sigma_{3})$ meeting at some point $a\in A$ and using the edges of a triangulation $T$ of $A$ is called a proper star if the following conditions hold:

(i) $\sigma_{i}$ is transversal with respect $v_{j}-v_{k}$ (possibly $\sigma_{i}=\{a\}$);

(ii) $\sigma_{i}$ has an end point $b_{i}\in\partial[A]$ such that $u_{i}$ is an exterior unit normal to $[A]$ at $b_{i}$, and

[TABLE]

(iii) writing ${\mathbb{R}}_{+}=\{t\in{\mathbb{R}}:\,t>0\}$, if $|\sigma_{j}|,|\sigma_{k}|>0$, then

[TABLE]

If the semi-open paths $\sigma_{i}\backslash\{a\}$, $i=1,2,3$, are all non-empty and pairwise disjoint, then (iii) means that they come around $a$ in the same order as the orientation of the triangle $[v_{1},v_{2},v_{3}]$ (see Figure 7 for an illustration).

The next Lemma shows how to construct an appropriate mixed subdivision of $A+B$ using a proper star.

Lemma 19

Given a proper star with rays $\sigma_{1},\sigma_{2},\sigma_{3}$ such that $|\sigma_{1}|+|\sigma_{2}|+|\sigma_{3}|>0$, there exists a mixed subdivision $M$ for $A+B$ satisfying

[TABLE]

*Proof: *We may assume that $|\sigma_{1}|>0$ and $v_{3}=o$. We partition the triangles of $T_{A}$ into three subsets $\Sigma_{1},\Sigma_{2},\Sigma_{3}$ (some of them might be empty). The idea is that if the semi-open paths $\sigma_{i}\backslash\{a\}$, $i=1,2,3$, are all non-empty and pairwise disjoint and $\{i,j,k\}=\{1,2,3\}$, then $\Sigma_{i}$ consists of the triangles cut off by $\sigma_{j}\cup\sigma_{k}$.

A triangle $\tau$ of $T_{A}$ is in $\Sigma_{1}$ if and only if there exists a $p\in({\rm int}\,\tau)\backslash(a+{\mathbb{R}}v_{1})$ such that

[TABLE]

is finite and odd. Similarly, $\tau\in T_{A}$ is in $\Sigma_{2}$ if and only if there exists a $p\in{\rm int}\,\tau$ such that

[TABLE]

is finite and odd. The rest of the triangles of $T_{A}$ form $\Sigma_{3}$.

The triangles of the mixed subdivision $M$ are as follows. If $\tau\in\Sigma_{i}$, then the corresponding triangle in $M$ is $\tau+v_{i}$. In addition, $[B]+a$ is in $M$. For the parallelograms, let $\{i,j,k\}=\{1,2,3\}$. If $e$ is an edge of $\sigma_{i}$, then $e+[v_{j},v_{k}]$ is in $M$. $\Box$

For the rest of the section, we fix finite $A\subset{\mathbb{R}}^{2}$ and $B=\{v_{1},v_{2},v_{3}\}\subset{\mathbb{R}}^{2}$ such that both of them spans ${\mathbb{R}}^{2}$ affinely, and confirm Conjecture 5 in this case.

The following statement is a simple consequence of the definition of a proper star.

Lemma 20

Assuming $B=\{v_{1},v_{2},v_{3}\}$ with $v_{1}=(1,0)=-u_{1}$, $v_{2}=(0,1)=-u_{2}$ and $v_{3}=(0,0)$, and hence $u_{3}=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$, if $(\sigma_{1},\sigma_{2})$ is a horizontal-vertical path for $A$ centered at $a\in A$, then

• •

there exists a proper star $(\sigma^{\prime}_{1},\sigma^{\prime}_{2},\sigma^{\prime}_{3})$ centered at $a$ such that $\sigma_{1}\subset\sigma^{\prime}_{1}$, $\sigma_{2}\subset\sigma^{\prime}_{2}$,

• •

if in addition $a\not\in F([A],u_{3})$, then $|\sigma^{\prime}_{3}|\geq 1$.

**Proof of Theorem 7 ** We may assume that $B=\{v_{1},v_{2},v_{3}\}$ with $v_{1}=(1,0)=-u_{1}$, $v_{2}=(0,1)=-u_{2}$ and $v_{3}=(0,0)$, and hence $u_{3}=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. In addition, we may assume that

[TABLE]

Using the notation of the proof of (16), we set $\{\langle u_{1},p\rangle:\,p\in A\}=\{\alpha_{0},\ldots,\alpha_{k}\}$ with $\alpha_{0}<\ldots<\alpha_{k}$, and $\Delta_{A}^{\prime}=|(A\cap\partial[A])\backslash(F([A],-u_{1})\cup F([A],-u_{2}))|$. For $i=0,\ldots,k$, let $A_{i}=\{p\in A:\,\langle u_{1},p\rangle=\alpha_{i}\}$, let $x_{i}=|A_{i}|$ and let $a_{i}$ be the top most point of $A_{i}$; namely, $\langle u_{2},a_{i}\rangle$ is maximal. According to (16) and (17), we have

[TABLE]

Let $I$ be the set of all $i\in\{1,\ldots,k\}$ such that

[TABLE]

Since $\xi\geq\sqrt{|T_{A}|+1}-\frac{1}{2}$, if strict inequality holds for some $i$ in (19), then we have a required proper star by Lemma 20. Thus we assume that $i+x_{i}-1=\xi$ for $i\in I$.

Let $\theta=|I|$. Since $i+x_{i}-1\leq\xi-1$ if $i\not\in I$, we have

[TABLE]

We deduce from (18) that if $i\in I$, then

[TABLE]

If $i\in I$ and $a_{i}\not\in F([A],u_{3})$, then $\xi\geq\sqrt{|T_{A}|+1}-\frac{1}{2}$ and Lemma 20 yields the existence of a required proper star. Therefore we may assume that $a_{i}\in F([A],u_{3})$ for $i\in I$. Since $|F([A],u_{3})\cap F([A],-u_{2}))|\leq 1$, we deduce that

[TABLE]

Therefore if $i\in I$, then we conclude using the inequality betwen the arightmetic and the geometric mean at the last inequality that

[TABLE]

5 Proof of Theorem 8

We assume in this section that there are no points of $A$ (resp. $B$) in the interior of $[A]$, (resp. $[B]$).

Recall that $\Delta_{X}$ denotes the number of points of $X$ in the boundary of $[X]$. It is easy to check that $\Delta_{A+B}$ has at least as many points as $\Delta_{A}$ and $\Delta_{B}$ together, that is:

[TABLE]

As a motivation for the proof, we note that Conjecture 1 follows if the number $\Omega_{A+B}$ of points of $A+B$ in ${\rm int}([A+B])$ is at least

[TABLE]

Naturally we aim at the stronger Conjecture 5. Given Theorem 7, Theorem 8 follows if $A$ and $B$ being in convex position and $|A|,|B|\geq 4$ yield that there exists a mixed subdivision of $A+B$ satisfying

[TABLE]

Throughout the proof we assume that $[B]$ has at most as many vertices as $[A]$ and $v$ denotes a unit vector (which we assume pointing upwards) not parallel to any side of $[A+B]$. We denote by $a_{0}$ and $a_{1}$ the leftmost and rightmost vertex of $[A]$ and by $b_{0}$ and $b_{1}$ the leftmost and rightmost vertex of $[B]$.

To prove (21), we say that $A$ and $B$ form a strange pair if $[B]$ is a triangle and the three exterior normals to $[B]$ are also exterior normals of edges of $[A]$.

We will use that, for $t,s\geq 1$,

[TABLE]

**Case 1 ** $A$ and $B$ are not strange pairs.

We choose a unit vector $v$ as above in the following way: if $B$ is a triangle, then the upper arc of $\partial[B]$ is a side such that $[A]$ has no side with same exterior unit normal; if $[B]$ has at least four sides, then the two supporting lines of $[B]$ parallel to $v$ touch at non-consecutive vertices of $[B]$. For the existence of the latter pair of supporting lines, we note that while continuously rotating $[B]$, the number of upper - lower vertices changes by either zero or two units at a time when a side of $[B]$ is parallel to $v$, and after rotation by $\pi$ it changes to its opposite. Hence, at some position that difference is zero or one which implies, since $[B]$ has at least four vertices, that at that position there is at least one upper and one lower vertex, as required.

Claim 21

One of the two following statements hold:

[TABLE]

*Proof: * We may assume that $b_{1}=a_{0}=o$ (see Fig. 8). Observe first that the only repetitions $x+b_{0}=a_{1}+y$ or $x+b_{1}=a_{0}+y$ in these configurations are the points $a_{1}+b_{0}$ and $a_{0}+b_{1}$ (which are interior to $[A+B]$ by our hypothesis). To prove (23), we verify first that

(i) for every $x\in A\setminus\{a_{0},a_{1}\}$ except perhaps two of them, at least one of $x+b_{0}$ or $x+b_{1}$ is interior in $A+B$,

(ii) for every $y\in B\setminus\{b_{0},b_{1}\}$ except perhaps two of them, at least one of $a_{0}+y$ or $a_{1}+y$ is interior in $A+B$.

For (i), we note that if both $x+b_{0}$ or $x+b_{1}$ are in $\partial[A+B]$, then they are the end points of a segment translated from $[b_{0},b_{1}]$ and only two such translations have their end-points in $\partial[A+B]$ because $A$ and $B$ are not a strange pair. The argument for (ii) is similar.

Now (i) and (ii) say that counting the interior points of $(A+b_{0})\cup(a_{1}+B)$ and $(a_{0}+B)\cup(A+b_{1})$ except $a_{0}+b_{1}$ and $a_{1}+b_{0}$ we have altogether at least $|\Delta_{A}|+|\Delta_{B}|-8$ of them. Including the latter we have at least $|\Delta_{A}|+|\Delta_{B}|-6$ of them and at least half of these in either $(A+b_{0})\cup(a_{1}+B)$ or $(a_{0}+B)\cup(A+b_{1})$, which yields (23). $\Box$

Let us construct the suitable mixed triangulation of $[A+B]$. For every path $\sigma$ in $\partial A$, we assume that every point of $A$ in $\sigma$ is a vertex of $\sigma$. According to (23), we may assume that

[TABLE]

Let $a_{\rm upp}$ ($a_{\rm low}$) be the neighboring vertex of $[A]$ to $o$ on the upper (lower) arc of $\partial A$, and let $b_{\rm upp}$ ($b_{\rm low}$) be the neighboring vertex of $[B]$ to $o$ on the upper (lower) arc of $\partial B$. We write $\omega^{A}_{\rm upp}$ and $\omega^{A}_{\rm low}$ to denote the paths determined by $[o,a_{\rm upp}]$ and $[o,a_{\rm low}]$ and $\omega^{B}_{\rm upp}$ and $\omega^{B}_{\rm low}$ to denote the paths determined by $[o,b_{\rm upp}]$ and $[o,b_{\rm low}]$. Next let $\sigma^{A}_{\rm upp}$ ($\sigma^{A}_{\rm low}$) be the longest path on the upper (lower) arc of $\partial[A]$ starting from $o$ such that every segment $s$ of $\sigma^{A}_{\rm upp}$ ($\sigma^{A}_{\rm low}$) satisfies that $s+[o,b_{\rm upp}]$ ($s+[o,b_{\rm low}]$) is a parallelogram that does not intersect ${\rm int}[A]$. Similarly, let $\sigma^{B}_{\rm upp}$ ($\sigma^{B}_{\rm low}$) be the longest path on the upper (lower) arc of $\partial[B]$ starting from $o$ such that every segment $s$ of $\sigma^{B}_{\rm upp}$ ($\sigma^{B}_{\rm low}$) satisfies that $s+[o,a_{\rm upp}]$ ($s+[o,a_{\rm low}]$) is a parallelogram that does not intersect ${\rm int}[B]$. Since $a_{1}=b_{0}=o$ is a common point of $\sigma^{A}_{\rm upp}$, $\sigma^{A}_{\rm low}$, $\sigma^{B}_{\rm upp}$, $\sigma^{B}_{\rm low}$, we deduce from (24) that

[TABLE]

equivalently,

[TABLE]

We construct the mixed subdivision by considering the subdivisions into suitable paralleograms of $\sigma^{A}_{\rm upp}+\omega^{B}_{\rm upp}$ and $\sigma^{B}_{\rm upp}+\omega^{A}_{\rm upp}$ that have $\omega^{A}_{\rm upp}+\omega^{B}_{\rm upp}$ in common, and the subdivisions into suitable parallelograms of $\sigma^{A}_{\rm low}+\omega^{B}_{\rm low}$ and $\sigma^{B}_{\rm low}+\omega^{A}_{\rm low}$ that have $\omega^{A}_{\rm low}+\omega^{B}_{\rm low}$ in common (see Figure 9).

In particular, $|\omega^{A}_{\rm upp}|,|\omega^{B}_{\rm upp}|\geq 1$, (22) and (25) yield that

[TABLE]

proving (21) in Case 1.

**Case 2 ** $A$ and $B$ form a strange pair with $|A|,|B|\geq 4$, and $[A]$ and $[B]$ are not similar triangles

We write $\alpha_{\rm upp}$ ($\alpha_{\rm low}$) to denote the number of segments that the points of $A$ divide the upper (lower) arc of $\partial[A]$. We denote by $b_{2}$ the third vertex of $[B]$ and by $[x_{0},x_{1}]$ the side of $A$ with $x_{1}-x_{0}=t(b_{1}-b_{0})$ for $t>0$. For $i=0,1,2$, let $s_{i}$ be the number of segments that the points of $B$ divide the side of $[B]$ opposite to $b_{i}$.

Claim 22

There exists a $v$ such that one of the following holds:

[TABLE]

*Proof: * Since $\alpha_{\rm upp}+\alpha_{\rm low}=\Delta_{A}$ and $s_{0}+s_{1}+s_{2}=\Delta_{B}$, the claim easily follows if there is a $v$ such that, for each the sets $A$ and $B$, both the upper arc and the lower arc contain a point of the set strictly between the two supporting lines parallel to $v$.

Otherwise, choose a $v$ such that the side $[b_{0},b_{1}]$ of $[B]$ contains at least $3$ points of $B$ (this is possible since $|B|\geq 4$). Then $[x_{0},x_{1}]$ has no other point of $A$ than $x_{0},x_{1}$ and the other side of $[A]$ at $x_{i}$, $i=0,1$ is parallel to $[b_{i},b_{2}]$. As $[A]$ and $[B]$ are not similar triangles , $[A]$ has some more sides, which in turn yields that $[b_{i},b_{2}]\cap B=\{b_{i},b_{2}\}$ for $i=0,1$. In summary, we have $\alpha_{\rm upp}=s_{0}=s_{1}=1$ and $\alpha_{\rm low},s_{2}\geq 2$. Since $\alpha_{\rm low}+s_{2}>\alpha_{\rm upp}+s_{0}+s_{1}$, we conclude (27). $\Box$

To prove (21) based on (26) and (27), we introduce some further notation. After a linear transformation, we may assume that $v$ is an exterior normal to the side $[b_{0},b_{1}]$ of $[B]$. We say that $p,q\in\partial[A]$ are opposite if there exists a unit vector $w$ such that $w$ is an exterior normal at $p$ and $-w$ is an exterior normal at $q$. If $p,q\in\partial[A]$ are not opposite, then we write $\overline{pq}$ the arc of $\partial[A]$ connecting $p$ and $q$ and not containing opposite pair of points.

First we assume that (26) holds and $b_{2}=o$. Since $[x_{0},x_{1}]$ has exterior normal $v$ and $\alpha_{\rm upp}\geq 2$, there exists $a\in A\backslash\{x_{0},x_{1}\}$ such that $v$ is an exterior normal to $\partial[A]$ at $a$. We write $l_{\rm upp}$ and $r_{\rm upp}$ to denote the number of segments the points of $A$ divide the arcs $\overline{ax_{0}}$ and $\overline{ax_{1}}$, respectively. To construct a mixed subdivision, we observe that every exterior normal $u$ to a side of $[A]$ in $\overline{ax_{0}}$ satisfies $\langle u,b_{0}\rangle>0$, and every exterior normal $w$ to a side of $[A]$ in $\overline{ax_{1}}$ satisfies $\langle w,b_{1}\rangle>0$. We divide $\overline{ax_{0}}+[o,b_{0}]$ into suitable $s_{1}l_{\rm upp}$ parallelograms, and $\overline{ax_{1}}+[o,b_{1}]$ into suitable $s_{0}r_{\rm upp}$ parallelograms. It follows from (22) that

[TABLE]

Secondly we assume that (27) holds. Since $s_{2}\geq 2$, we may assume that $o\in([b_{0},b_{1}]\backslash\{b_{0},b_{1}\})\cap B$. For $i=0,1$, we write $s_{2i}$ to denote the number of segments the points of $B$ divide $[o,b_{i}]$. Let $\tilde{x}_{0}$ and $\tilde{x}_{1}$ be the leftmost and rightmost points of $A$ such that $-v$ is an exterior normal to $\partial[A]$, where possibly $\tilde{x}_{0}=\tilde{x}_{1}$. Since $[A]$ has sides parallel to the sides $[b_{2},b_{0}]$ and $[b_{2},b_{1}]$ of $[B]$, we deduce that $\tilde{x}_{0}\neq x_{0}$ and $\tilde{x}_{1}\neq x_{1}$. To construct a mixed subdivision, we set $m_{\rm low}=0$ if $\tilde{x}_{0}=\tilde{x}_{1}$, and $m_{\rm low}$ to be the number of segments the points of $A$ divide $\overline{\tilde{x}_{0},\tilde{x}_{1}}$ if $\tilde{x}_{0}\neq\tilde{x}_{1}$. In addition, we write $l_{\rm low}\geq 1$ and $r_{\rm low}\geq 1$ to denote the number of segments the points of $A$ divide the arcs $\overline{\tilde{x}_{0}x_{0}}$ and $\overline{\tilde{x}_{1}x_{1}}$, respectively. We divide $\overline{\tilde{x}_{0}x_{0}}+[o,b_{0}]$ into suitable $l_{\rm low}s_{20}$ parallelograms, and $\overline{\tilde{x}_{1}x_{1}}+[o,b_{1}]$ into suitable $r_{\rm upp}s_{21}$ parallelograms. In addition, if $\tilde{x}_{0}\neq\tilde{x}_{1}$, then we divide $[\tilde{x}_{0}\tilde{x}_{1}]+[o,b_{2}]$ into suitable $m_{\rm low}$ parallelograms. It follows from (22) that

[TABLE]

finishing the proof of (21) in Case 2.

**Case 3 ** $[A]$ and $[B]$ are similar triangles and $|A|,|B|\geq 4$

We recall that $s_{1},s_{2}$ and $s_{3}$ denote the number of segments the points of $B$ divide the sides of $[B]$ and let $s^{\prime}_{1},s^{\prime}_{2},s^{\prime}_{3}$ be the number of segments the points of $A$ divide the corresponding sides of $[A]$. We have ${\rm tr}(A)=s_{1}^{\prime}+s^{\prime}_{2}+s^{\prime}_{3}-2$ and ${\rm tr}(B)=s_{1}+s_{2}+s_{3}-2$. We may assume that $s_{1}$ is the largest among the six numbers and that $s^{\prime}_{2}\geq s^{\prime}_{3}$. Readily

[TABLE]

If $s_{2}^{\prime}\geq 3$, then

[TABLE]

If $s_{2}^{\prime}=2$, then $s_{3}^{\prime}\leq 2$ and

[TABLE]

Therefore we assume that $s_{2}^{\prime}=s_{3}^{\prime}=1$. In particular, we may also assume that $s_{2}\geq s_{3}$. Since $s^{\prime}_{1}\geq 2$ and $s_{2}\geq 1$ we have $s^{\prime}_{1}s_{2}\geq s^{\prime}_{1}+2s_{2}-2$. Therefore,

[TABLE]

and we conclude (21) in Case 3, as well. $\Box$

6 Proof of Theorem 9

Let $A=\{a_{1},\ldots,a_{n}\}$. Naturally, $[A+A]$ has a triangulation $\{F+F:\,F\in T_{A}\}$, which we subdive in order to obtain $M$. We define $M$ to be the collection of the sums of the form

[TABLE]

where $k\geq 0$, $0\leq m\leq k$, $i_{j}<i_{l}$ for $j<l$, and $[a_{i_{0}},\ldots,a_{i_{k}}]\in T_{A}$.

To show that we obtain a cell decomposition, let

[TABLE]

be a $k$-simplex with $k>0$ where $i_{j}<i_{l}$ for $j<l$, and hence

[TABLE]

We write ${\rm relint}\,C$ to denote the relative interior of a compact convex set $C$. For some $0\leq m\leq k$, $\alpha_{0},\ldots,\alpha_{k}\geq 0$ with $\sum_{i=0}^{k}\alpha_{j}=2$, we have

[TABLE]

if and only if $\sum_{j<m}\alpha_{j}<1$ and $\sum_{i=0}^{m}\alpha_{j}>1$ where we set $\sum_{j<0}\alpha_{j}=0$. We conclude that $M$ forms a cell decomposition of $[A+A]$.

For any $d$-simplex $F\in T_{A}$, and for any $m=0,\ldots,d$, we have constructed one $d$-cell of $M$ that is the sum of an $m$-simplex and a $(d-m)$-simplex. Therefore

[TABLE]

7 Proof of Corollary 3

In this section, let $A\subset{\mathbb{R}}^{2}$ be finite and not collinear. We prove four auxiliary statements about $A$. The first is an application of the case $A=B$ of Conjecture 1 (see Theorem 2).

Lemma 23

[TABLE]

*Proof: *We have readily $\Delta_{A+A}\geq 2\Delta_{A}$. Thus (3) and Theorem 2 yield

[TABLE]

We note that the estimate of Lemma 23 is optimal, the configuration of Theorem 2 (b) being an extremal set.

Next we provide the well-known elementary estimate for $|A+A|$ only in terms of boundary points.

Lemma 24

Let $m_{A}$ denote the maximal number of points of $A$ contained in a side of $[A]$. We have,

[TABLE]

*Proof: *We choose a line $l$ not parallel to any side of $[A]$, that we may assume to be a vertical line, and denote by $s_{1},\ldots,s_{k}$ the sides of $[A]$ on the upper chain of $[A]$ in left to right order. Let $A_{i}$ be the set obtained from $A\cap s_{i}$ by removing its rightmost point. We may assume that

[TABLE]

We observe that, for $1\leq i<j\leq k$, we have

[TABLE]

It follows that

[TABLE]

The following Lemma can be found in Freiman [4].

Lemma 25

Let $\ell$ be a line intersecting $[A]$ in $m$ points of $A$. If $A$ is covered by exactly $s$ lines parallel to $\ell$, then

[TABLE]

Moreover,

[TABLE]

*Proof: * We may assume that $\ell$ is the vertical line through the origin, that $a_{1},\ldots,a_{s}$ are $s$ points of $A$ ordered left to right such that $A=\cup_{i=1}^{s}(A\cap(\ell+a_{i}))$ and that $|A\cap(\ell+a_{1})|=m$. Let $A_{i}=A\cap(a_{i}+\ell)$. Then,

[TABLE]

from which (29) follows. On the other hand,

[TABLE]

If the latter estimate is larger than the former one we obtain (30), otherwise we get the stronger inequality $|A+A|\geq(4-2/s^{2})|A|-(2s-1)$. $\Box$

**Proof of Corollary 3 ** Let $|A+A|\leq(4-\varepsilon)|A|$ where $\varepsilon\in(0,1)$ and $\varepsilon^{2}|A|\geq 48$. To simply formulae, we set $\Delta=\Delta_{A}$ and $m=m_{A}$.

We deduce from Lemma 23 that $\Delta\geq\varepsilon|A|-3$. Substituting this into Lemma 24 yields

[TABLE]

Therefore

[TABLE]

as $\varepsilon|A|-3\geq\frac{48}{\varepsilon}-3>\frac{12}{\varepsilon}$. In particular, $m-1>\frac{1}{2}\varepsilon|A|-\frac{12}{\varepsilon}$.

Next let $l$ be the line determined by a side of $[A]$ containing $m=m_{A}$ point of $A$, and let $s$ be the number of lines parallel to $l$ intersecting $A$. According to (29),

[TABLE]

thus first rearranging, and then applying $\varepsilon^{2}|A|\geq 48$ yield

[TABLE]

Therefore $s<\frac{8}{\varepsilon}$.

We deduce from (30) and $s<\frac{8}{\varepsilon}$ that

[TABLE]

Rearranging, and then applying $\varepsilon^{2}|A|\geq 48$ imply

[TABLE]

8 Proof of Proposition 6

We call the points of $A$,

[TABLE]

If $k\geq 2$, then we show that every mixed subdivision $M$ corresponding to $T_{A}$ and $T_{B}$ satisfies

[TABLE]

We prove (31) in several steps. First we verify

[TABLE]

For (32), we observe that $a_{1}+l_{i+1}$ if $i\leq k-1$ or $a_{1}+l_{i-1}$ if $i\geq 1$ is a point of $A+B$ in $[a_{1},a_{2}]+l_{i}$ different from the endpoints. Similarly, for (33), we observe that $a_{1}+r_{i+1}$ if $i\leq k-2$ or $a_{1}+r_{i-1}$ if $i\geq 1$ is a point of $A+B$ in $[a_{1},a_{2}]+r_{i}$ different from the endpoints.

Next, we have

[TABLE]

as $l_{i+1}\in{\rm int}\,[a_{0},a_{2}]+[l_{i},r_{i}]$ and $l_{i}\in{\rm int}\,[a_{0},a_{1}]+[r_{i},l_{i+1}]$.

Let us call the edges of $T_{B}$ of the form either $[l_{i},r_{i}]$ or $[r_{i},l_{i+1}]$ for $i=0,\ldots,k-1$ small edges, and the edges of $T_{B}$ of the form either $[p,l_{i}]$, $[q,l_{i}]$ for $i=0,\ldots,k$, or $[p,r_{i}]$, $[q,r_{i}]$ for $i=0,\ldots,k-1$ long edges. In other words, long edges of $T_{B}$ contain either $p$ or $q$, while small edges of $T_{B}$ contain neither.

Concerning long edges, we prove that that the number of parallelograms of $M$ of the form

[TABLE]

If $e_{A}$ is an edge of $T_{A}$, then there exist at most two cells of $M$ whose side are $p+e_{A}$. Since $T_{A}$ has three edges, there are at most six of parallelograms of $M$ of the form $e_{A}+e_{B}$ where $e_{A}$ is an edge of $T_{A}$ and $e_{B}$ is an edge of $T_{B}$ with $p\in e_{B}$. Since the same estimate holds if $q\in e_{B}$, we conclude (36).

Finally, we prove that that the number of parallelograms of $M$ of the form

[TABLE]

The argument for (37) is based on the claim that if $e_{A}+e_{B}$ is a parallelogram of $M$ for an edge $e_{A}$ of $T_{A}$ and a small edge $e_{B}$ of $T_{B}$, then there is a long edge $e^{\prime}_{B}$ of $T_{B}$ such that

[TABLE]

We have $e_{A}\neq[a_{1},a_{2}]$ according to (32) and (33). If $e_{A}=[a_{0},a_{1}]$, then $e_{B}=[l_{i},r_{i}]$ for some $i\in\{1,\ldots,k-1\}$ according to (35). Now $r_{i}+e_{A}$ intersects the interior of $[A+B]$ as $r_{i}\in{\rm int}\,[A]$, thus it is the edge of another cell of $M$, as well. This other cell is either a translate of $[A]$, which is impossible by (32), (33), and as $r_{i}\not\in p+[A],q+[A]$, or of the form $e_{A}+e^{\prime}_{B}$ for an edge $e^{\prime}_{B}\neq e_{B}$ of $T_{B}$ containing $r_{i}$. However, $e^{\prime}_{B}\neq[r_{i},l_{i+1}]$ by (35), therefore $e^{\prime}_{B}$ is a long edge.

On the other hand, if $e_{A}=[a_{0},a_{2}]$, then $e_{B}=[r_{i},l_{i+1}]$ for some $i\in\{1,\ldots,k-1\}$ according to (34), and (38) follows as above.

Now if $e_{A}+e^{\prime}_{B}$ is a parallelogram of $M$ for an edge $e_{A}$ of $T_{A}$ and a long edge $e^{\prime}_{B}$ of $T_{B}$, then there is at most one neighboring paralellogram of the form $e_{A}+e_{B}$ for a small edge $e_{B}$ of $T_{B}$ because $e_{A}+e_{B}$ does not intersect $e_{A}+p$ and $e_{A}+q$. In turn, (37) follows from (36) and (38). Moreover, we conclude (31) from (36) and (37).

Finally, it follows from (31) that if $k\geq 145$, then

[TABLE]