Ugrás a tartalomhoz

## Convex Geometry

Csaba Vincze (2013)

University of Debrecen

4.2 Universal covers and approximately central symmetry

## 4.2 Universal covers and approximately central symmetry

Definition A compact subset K in the coordinate plane is a universal cover if any compact convex subset having diameterone can be covered by a congruent copy of K.

The problem of finding the smallest universal cover of a given class of objects is very natural and important. Here we illustrate only the cases of disks (as an excercise) and squares because they are directly in the competence of Helly's theorem.

Definition Consider an equilateral triangle in the coordinate plane. The Reuleaux triangle is formed by three circular arcs lyingon the sides of the triangle with centers running through the vertices.

Figure 29: The Reuleaux triangle.

The problem we are going to discuss here is how a square can be rotated around the Reuleaux triangle?

Figure 30: Tangent lines.

Let

be the vertices of the regular triangle (the common length of the sides is 2) and consider the one-parameter family

 $x\mathrm{s}\mathrm{i}\mathrm{n}t-y\mathrm{c}\mathrm{o}\mathrm{s}t=c\left(t\right)$ (4.5)

of lines, where c(t)=sin t - 2 is given in such a way that the line at t is tangent to the arc AC of the circle centered at B with radius 2. The one-parameter family of lines parallel to 4.5 through the point B is

 $x\mathrm{s}\mathrm{i}\mathrm{n}t-y\mathrm{c}\mathrm{o}\mathrm{s}t=\mathrm{s}\mathrm{i}\mathrm{n}t.$

In a similar way

 $x\mathrm{c}\mathrm{o}\mathrm{s}t+y\mathrm{s}\mathrm{i}\mathrm{n}t=c\left(t\right)$ (4.6)

is the one-parameter family of lines, where

 $c\left(t\right)=\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}t-2$

is given in such a way that the line at t is tangent to the arc AB of the circle centered at C with radius 2. Lines 4.5 and 4.6 give adjacent sides of the circumscribed square around the Reuleaux triangle when the parameter is between 60 and 90; see figure 30. The corresponding vertex[5] moves along the path K(t) with coordinate functions

 ${K}^{1}\left(t\right)=\mathrm{s}\mathrm{i}\mathrm{n}t\left(\mathrm{s}\mathrm{i}\mathrm{n}t-2\right)+\mathrm{c}\mathrm{o}\mathrm{s}t\left(\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}t-2\right),$

 ${K}^{2}\left(t\right)=\mathrm{s}\mathrm{i}\mathrm{n}t\left(\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}t-2\right)-\mathrm{c}\mathrm{o}\mathrm{s}t\left(\mathrm{s}\mathrm{i}\mathrm{n}t-2\right).$

The opposite vertex moves along M(t) with coordinate functions

 ${M}^{1}\left(t\right)={\mathrm{s}\mathrm{i}\mathrm{n}}^{2}t+\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}t\mathrm{c}\mathrm{o}\mathrm{s}t,$

 ${M}^{2}\left(t\right)=\sqrt{3}{\mathrm{s}\mathrm{i}\mathrm{n}}^{2}t-\mathrm{s}\mathrm{i}\mathrm{n}t\mathrm{c}\mathrm{o}\mathrm{s}t.$

Finally (the missing vertices)

 $L\left(t\right)=M\left(t\right)-2\left(\mathrm{c}\mathrm{o}\mathrm{s}t,\mathrm{s}\mathrm{i}\mathrm{n}t\right),N\left(t\right)=K\left(t\right)+2\left(\mathrm{c}\mathrm{o}\mathrm{s}t,\mathrm{s}\mathrm{i}\mathrm{n}t\right)$

and the motion of the center can be described as

 $T\left(t\right)=\frac{1}{2}\left(K\left(t\right)+M\left(t\right)\right)$

with coordinate functions

 ${T}^{1}\left(t\right)={\mathrm{s}\mathrm{i}\mathrm{n}}^{2}t-\mathrm{s}\mathrm{i}\mathrm{n}t+\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}t\mathrm{c}\mathrm{o}\mathrm{s}t-\mathrm{c}\mathrm{o}\mathrm{s}t,$

 ${T}^{2}\left(t\right)=\sqrt{3}{\mathrm{s}\mathrm{i}\mathrm{n}}^{2}t-\mathrm{s}\mathrm{i}\mathrm{n}t-\mathrm{s}\mathrm{i}\mathrm{n}t\mathrm{c}\mathrm{o}\mathrm{s}t+\mathrm{c}\mathrm{o}\mathrm{s}t,$

where t is between 60 and 90 degree's (a routine calculation shows that T(60) and T(90) can be seen under the angle of measure 120 degree from the center of the triangle). If R is the rotation about the center of the triangle with magnitude +120 then the center of the circumscribed squares moves along the path

 $T\cup R\left(T\right)\cup {R}^{-1}\left(T\right);$

the figure 31 shows the curve T(t) with a full period t=0 ... 360. An animation can be available at zeus.nyf.hu/ kovacsz/Csaba.

Figure 31: The curve T(t) with a full period: the limacon.

Remark Conversely the Reuleaux triangle can be rotated through 360 degree inside a square although the center of the rotation moves along an excentric path.

Theorem 4.2.1 The smallest universal cover among squaresin the coordinate plane is the square having sides of length one.

Proof Let K be a square having sides of lenght one and consider a compact convex set F with diameter 1 in the coordinate plane. In the sense of theorem 4.1.4 it is enough to prove that for any three points in F there exists a translate of K covering them. Let p(1), p(2) and p(3) be three points in F. Since its diameter is one there exists a Reuleaux triangle constructed from an equilateral triangle with sides of length one such that it contains p(1), p(2) and p(3). But such a shape can be rotated through 360 degree in a square of side 1. This means that there exists a translate of K covering the Reuleaux triangle (together with p(1), p(2) and p(3)) independently of its orientation. Therefore 4.1.4 says that there exists a translate of K covering F. To cover a disk of diameter one we obviously need a square having sides of length at least one.

Theorem 4.2.2 Let K be a two-dimensional compact convex subset in the plane. There exists a point in the interior of K such that it belongs to the middle part of the trisection of any chord passing through this point.

Proof Consider a triangle Δ formed by the boundary points of K and use central similarities from the vertices of the triangle with ratio 2/3. The barycenter of Δ is a common point of the images of K under the three similarities. Using the general version 4.1.2 of Helly's theorem there exists a point p* in the intersection of the images of K under the similarities as the center runs through the boundary of K. If a chord contains this common point then we can use the central similarities relative to the endpoints of such a chord with ratio 2/3 to prove that p* must be in the middle part of the trisection.

### 4.2.1 The Brunn-Minkowski inequality

The most famous problem related to universal covers was posed by H. Lebesgue: what is the minimum area that a universal cover in the coordinate plane can have? This is related to a whole class of extremum problems wherein one quantity is to be minimized or maximized subject to certain restraining conditions. The best known of all the extremum problems is the classical isoperimetric problem; which simple closed curve of given perimeter encloses the greatest area? It is not hard to prove that among all rectangles of perimeter 1 the square has the greatest area. Likewise among triangles of perimeter 1 the equilateral triangle has the greatest area. In general the answer is again the most symmetrical shape, which in this case is the circle. Rigorous proofs are hard to find. In what follows we present a theoretical way of the solution via the Brunn-Minkowski inequality for convex bodies in the space. It is not at all unreasonable to expect that a solution to the isoperimetric problem is a convex body. If we accept, for the moment, an intuitive notion of the area and perimeter of a non-convex set then the convex hull of this set has expectedly a shorter perimeter and a larger area.

Definition Convex bodies mean compact convex sets with non-empty interiors in the coordinate space.

Theorem 4.2.3 (Brunn-Minkowski inequality) For convex bodies in the coordinate space of dimension n we have

 $V\left(K+L{\right)}^{1/n}\ge V\left(K{\right)}^{1/n}+V\left(L{\right)}^{1/n},$ (4.7)

where V refers to the volume of the bodies.

Proof In what follows we sketch the steps of the proof due to W. Blaschke; for more details and historical remarks see [27]. It is based on the so-called Steiner symmetrization process. Let u be a unit vector in the space. The Steiner symmetral S(u)K of K in the direction u is a convex body obtained from K by sliding eachof its chords parallel to u so that they are bisected by its orthogonal complement. By Cavaliéri's principle K and S(u)K have the same area. On the other hand the Steiner symmetral of the sum K+L contains the sum of the Steiner symmetrals in any given direction. Therefore

 $V\left(K+L\right)=V\left({S}_{u}\left(K+L\right)\right)\ge V\left({S}_{u}K+{S}_{u}L\right).$ (4.8)

Let B be the unit ball centered at the origin. One can also prove that there is a sequence of directions u(m) such that the iteration

 ${K}_{m}:={S}_{{u}_{m}}{K}_{m-1},{K}_{0}:=K$

tends to the Euclidean ball r(K)B with respect to the Hausdorff metric. It is clear that the constant r(K) is just the nth root of the ratio between the volumes of K and B. Since r(K)B+r(L)B=(r(K)+r(L))B we have that

 $V\left(K+L\right)\ge V\left({r}_{K}B+{r}_{L}B\right)=\left({r}_{K}+{r}_{L}{\right)}^{n}V\left(B\right)$ (4.9)

which is just the Brunn-Minkowski inequality 4.7 as was to be proved.

Minkowski's definition of the surface area A(K) of a convex body is

 $A\left(K\right):=\underset{\epsilon \to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{V\left(K+\epsilon B\right)-V\left(K\right)}{\epsilon }.$ (4.10)

Excercise 4.2.4 Compute the surface area of a square in the plane by Minkowski's definition.

Remark Let K be a convex body in the plane bounded by a smooth curve c. Since tangent lines of c are working as supporting hyperplanes it follows that c+ε n is the parameterization of the boundary of K+εB, where n is the outer pointing unit normal vector field along c. Under the choice of the arclenght parameter into the counterclockwise direction

see section 11.2 for the elements of differential geometry. We have

 $V\left(K+\epsilon B\right)=\frac{1}{2}{\int }_{0}^{P}\mathrm{d}\mathrm{e}\mathrm{t}\left(c+\epsilon n,c\mathrm{\text{'}}+\epsilon n\mathrm{\text{'}}\right)=\frac{1}{2}{\int }_{0}^{P}\left(1+{\kappa }_{s}\epsilon \right)\mathrm{d}\mathrm{e}\mathrm{t}\left(c+\epsilon n,c\mathrm{\text{'}}\right),$

where P is the arclength of c. Therefore

 $A\left(K\right)=\frac{1}{2}{\int }_{0}^{P}\mathrm{d}\mathrm{e}\mathrm{t}\left(n,c\mathrm{\text{'}}\right)+{\kappa }_{s}\mathrm{d}\mathrm{e}\mathrm{t}\left(c,c\mathrm{\text{'}}\right).$

Since n and c' are orthogonal unit vectors det( n, c')=1. On the other hand

 ${\kappa }_{s}\mathrm{d}\mathrm{e}\mathrm{t}\left(c,c\mathrm{\text{'}}\right)=\mathrm{d}\mathrm{e}\mathrm{t}\left(c,n\mathrm{\text{'}}\right)=\mathrm{d}\mathrm{e}\mathrm{t}\left(\begin{array}{ll}x& y\mathrm{\text{'}}\mathrm{\text{'}}\\ y& -x\mathrm{\text{'}}\mathrm{\text{'}}\end{array}\right)=-\left(xx\mathrm{\text{'}}\mathrm{\text{'}}+yy\mathrm{\text{'}}\mathrm{\text{'}}\right).$

Using the rule of partial integration and the periodocity of the curve c it follows that

 $A\left(K\right)=\frac{1}{2}{\int }_{0}^{P}1+\left(x\mathrm{\text{'}}{\right)}^{2}+\left(y\mathrm{\text{'}}{\right)}^{2}=\frac{1}{2}{\int }_{0}^{P}1+1=P$

because of the arclenght parameter.

Excercise 4.2.5 Prove that the ratio between the surface area and the volume of the unit ball is just the dimension of the space.

The last step of the derivation of the isoperimetric inequality for convex bodies is

 $A\left(K\right):=\underset{\epsilon \to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{V\left(K+\epsilon B\right)-V\left(K\right)}{\epsilon }$

 $\ge \underset{\epsilon \to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{{\left(V\left(K{\right)}^{1/n}+\epsilon V\left(B{\right)}^{1/n}\right)}^{n}-V\left(K\right)}{\epsilon }=nV\left(K{\right)}^{\left(n-1\right)/n}V\left(B{\right)}^{1/n}$

which implies by the ratio A(B) : V(B)=n, that

 ${\left(\frac{V\left(K\right)}{V\left(B\right)}\right)}^{1/n}\le {\left(\frac{A\left(K\right)}{A\left(B\right)}\right)}^{1/\left(n-1\right)}.$ (4.11)

Remark Equality holds in the isoperimetric inequality 4.11 if and only if K is a ball. For the Brunn-Minkowski inequality and its relatives see [27], see also [52]. A nice presentation of a differential geometric proof in the plane can be found in [51].