Integral geometry and Mizel’s problem

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INTEGRAL GEOMETRY AND MIZEL’S PROBLEM
Yu.B. Zelinskii, M.V. Tkachuk, B.A. Klishchuk
Institute of Mathematics of the National Academy of Sciences of Ukraine
e-mail: zel@imath.kiev.ua; max@imath.kiev.ua; bogdanklishchuk@mail.ru
Key words and phrases: Jordan curve, circle, sphere, convex set, rectangle
In memory of professor Promarz Tamrazov
Abstract. The solution of a Zamfirescu’s problem was obtained. The unsolved questions
related to Mizel’s problem were discussed.
1. Introduction
The subject of this article is to combine in one bundle some questions of complex analysis,
geometry and probability theory. First investigations of geometric probability start with the
well-known Buffoon’s needle problem and related paradoxes. A needle is considered as the
real line and then the problem reduces to finding an invariant measure of a set relative to a
movement (R. V. Ambartsumyan, G. Matheron, L. Santalo [1, 9, 13]).
Many questions of integral geometry are reduced to the estimation of a measure of linear
spaces crossing a convex set. Finding such a measure we obtain probabilistic estimations.
Other problems, more close to geometry, are the estimations of properties of the set under
investigation if properties of its intersections with families of some sets are well known:
1) with planes of fixed dimension:
a) the real case (G. Aumann, A. Kosi´nski, E. Shchepin [2, 7, 14]);
b) the complex case (Yu. Zelinskii [19]);
2) with a set of vertices of an arbitrary rectangle (A.S. Besicovitch, L. Danzer, M. Tkachuk,
T. Zamfirescu [3, 4, 15, 18]).
The first problem is connected with the well-known Ulam’s problem from the [10].
Ulam’s problem. Let Mnbe an n-dimensional manifold and every section of Mnby
arbitrary hyperplane Lbe homeomorphic to an (n−1)-dimensional sphere Sn−1. Is it true that
Mnis an n-dimensional sphere?
In the real case A. Kosi´nski solved this problem in 1962 [7]. L. Montejano obtained the
same result in 1990 [11]. In the complex case Yu. Zelinskii got a similar result in 1993 [19].
The second problem is known in literature as Mizel’s problem(A characterization of the
circle): A closed convex curve such that, if three vertices of any rectangle lie on it, so does the
fourth, must be a circle.
In 1961 A.S. Besicovitch [3] solved this problem. Later, a modified proof of this statement
was presented by L.W. Danzer, W.H. Koenen, C. St. J. A. Nash-Williams, A.G.D. Watson [4,
6, 12, 17].
In 1989 T. Zamfirescu [18] proved a similar result for a Jordan curve (not convex a priory)
and for a rectangle with the infinitesimal relation between its sides:
¯
¯
¯
a
b
¯
¯
¯≤ε > 0,
where aand bare sidelengths of a rectangle.
In 2006 M. Tkachuk [15] obtained the most general result in this area for any arbitrary
compact set C⊂R2, where the complement R2\Cis not connected.
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It is obvious that the requirement of the compactness is necessary, otherwise the straight
line and some other sets noted below will satisfy the stated requirement. But if the partition
of the plane is not required, the ensembles below will satisfy, for instance, that condition of
a rectangle: an ensemble from three points of a plane such that the triangle with vertices in
these points will not be rectangular, a proper arc of a semicircle, a set of points of a plane with
rational (irrational) coordinates.
T. Zamfirescu proved that every analytic curve of constant width satisfying the infinitesimal
rectangle property is a circle. Our aim is to prove the theorem for a convex curve without
the analyticity condition(Theorem 1) and to discuss new unsolved problems concerning Mizel’s
problem.
2. The main theorem
Definition 2.1. A set in R2is called convex if it contains, with any two distinct points x
and y, the (closed) line segment between xand y.
Definition 2.2. A width of a convex curve in the direction is the distance between two
parallel support lines of this curve perpendicular to this direction. A curve has a constant width
if its width is equal in all directions.
Definition 2.3. A set in R2has the infinitesimal rectangle property if there is some ε > 0
such that no rectangle with sidelengths ratio at most εhas exactly three vertices in the set( if
three vertices of any such rectangle lie on the set, so does the fourth).
We introduce the following notation: let Γ be a convex curve of constant width satisfying
the infinitesimal rectangle property and for each point x∈Γ is denoted by ∂∆xthe circle
having with the curve Γ a common tangent at the point x; in some neighborhood of a point
of the curve Γ we assume that the upward direction is the direction along the inner normal
and thus we consider the right direction and the left direction along the curve Γ ; Uε(x, Γ) is
the ε-neighborhood of xon the curve Γ, Ul
ε(x, Γ) is the left ε-neighborhood of xon the curve
Γ, i.e. the subset of Uε(x, Γ) each point of which lies to the left of x;Ur
ε(x, Γ) is the right
ε-neighborhood of xon Γ.
Definition 2.4. A point y∈Γ is called opposite to a point x∈Γ if the straight line xy is
perpendicular to the straight line Txthat is supporting for Γ and passes through the point x.
The straight line xy is called a normal of a convex curve at the point x. A straight line that is
a normal of a convex curve at two points of their intersection is called a binormal.
Theorem 2.1. Any convex curve of constant width satisfying the infinitesimal rectangular
condition is a circle.
Proof. We are only interested in rectangles with the diagonal length d. Let aand bbe
sides of the rectangle. By the infinitesimal rectangular property:
a
b< δ.
From the Pythagorean theorem we have:
b=√d2−a2.
Hence
a
√d2−a2< δ
a2< δ2(d2−a2)
2

a2<δ2d2
1 + δ2.
Let δd
√1+δ2=εthen a < ε,ε > 0. So the infinitesimal condition may be replaced by the
requirement that the smaller rectangle side has a length less than some εfor the fixed diagonal
length d.
In [15] it was proved that the curve Γ has a continuous tangent.
We prove that Γ does not contain the points through which several different straight lines
supporting Γ may pass. In what follows, we call these points angular and the other points
regular. The proof of the statement that every normal of a convex curve that satisfies rectangle
condition( if three vertices of a rectangle belong to Γ then its fourth vertex also belongs to Γ)
is a binormal is given in [3] and can also be applied to Γ in the case considered (this was proved
by T. Zamfirescu).
Let xbe an angular point of Γ. Consider the set λof points z∈Γ opposite to the point x.
Consider a mapping that associates every point z∈λwith the straight line pperpendicular to
xz,z∈pthat is supporting for Γ at the point z. We obtain a continuous mapping of the arc
λonto the set of elements supporting for the arc λof the convex curve Γ [8]. Thus the arc λ
cannot contain angular points because the mapping p(z) is continuous.
Let us prove that the supporting straight line p(z) is tangent to λat a point zin the
classical sense of this word (the limit location of a secant). It is known that one can represent a
convex curve in the neighborhood of an arbitrary point of it as the graph of a convex function
by choosing corresponding coordinate axes. A convex function has right-hand and left-hand
derivatives everywhere but they may be different at certain points [8]. We prove that the
tangents to λfrom the right and from the left at all points coincide with supporting straight
lines. For an arbitrary point z0∈λwe take a sequence of points zn∈λsuch that zn→z0,
from the right with respect to the chosen orientation of axes. The straight line znz0tends to
the right tangent at the point z0. Since the mapping p(z) is continuous, there exists a point
tn∈Sz0znsuch that p(tn) parallel to z0zn. Since zn→z0and tn∈Sz0zn, we have tn→z0,
and, hence, p(tn) tends to the right tangent at the point z0that coincides with p(z0). For the
left tangent the arguments are analogous.
Thus, the arc λis a smooth arc, i.e., a diffeomorphism of an open segment onto the plane.
Using differential calculus and taking into account the condition p(z) perpendicular to xz, we
prove that λis an arc of a circle. For an arbitrary point of a circle, there exists its neighborhood
on the arc symmetric relative to the normal to the arc at this point, and the neighborhood
of the angular point xcan be symmetric only relative to the bisector of the supporting angle
(the supporting straight lines must also be symmetric). We draw a binormal different from the
bisector of the supporting angle through the point xand an interior point of the arc of the
circle and construct a rectangle with two points of the arc of the circle symmetric relative to
the binormal and the third vertex in the neighborhood of the point x. The fourth vertex, which
is symmetric to the third one, belongs to Γ. Thus, there exists a neighborhood of the angular
point xthat is symmetric relative to this binormal, which is impossible. Therefore Γ does not
contain angular points and is a smooth curve.
Let A⊂R2be some set of points and ΛAlinear measure of A.
A.S. Besicovitch proved the following four lemmas:
Lemma 2.1. If E,ΛE= 0, is the set of the end points of a set of normals of Γand Uthe
set of mid-points of the normals, then ΛU= 0.
Lemma 2.2. The set of centres of curvature of Γat the points, at which the curvature is
equal to 2
d, is of linear measure 0.
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Lemma 2.3. Given a set E,ΛE > 0, of points of Γ, at which the radius of curvature is
different from 1
2d. Then the set Uof mid-points of the normals at Eis also of positive measure
and at almost all points of Uthe tangent to Uexists and is perpendicular to the normal to
which the point belongs.
Lemma 2.4. If at almost all points of an arc ^ AB of Γ, and of the opposite arc ^ A0B0
the radius of curvature is 1
2dthen ^ AB and ^ A0B0are circular arcs.
Notice that these lemmas are valid for the infinitesimal rectangle property since they have
a local character. But the result following from these lemmas changes: if at some point x∈Γ
the circle ∂∆xintersects Γ at yand the distance between xand yis less than ε, then in some
neighborhood of x(and also of the opposite point x∗) the curve Γ is an arc of the circle ∂∆x.
Consequently, all points of the curve Γ can be divided into five disjoint sets:
A={x∈Γ|Uε(x, Γ)\{x} ⊂ ∆x}is the set of points of the curve Γ such that the curve Γ
lies in the open circular disk ∆xin ε-neighborhood of x∈Aexcept the point xitself;
B={x∈Γ|Uε(x, Γ)\{x} ⊂ R2\∆x}is the set of points of the curve Γ such that in their
ε−neighborhood the curve Γ is situated outside of the closed circular disk 4xexcept the point
xitself;
AB ={x∈Γ|Ul
ε(x, Γ) ⊂∆x, Ur
ε(x, Γ) ⊂R2\∆x}is the set of points of the curve Γ such
that in their ε−neighborhood the curve Γ lies in the open circular disk 4xto the left of xand
outside of the closed circular disk 4xto the right of xexcept the point xitself;
BA ={x∈Γ|Ul
ε(x, Γ) ⊂R2\∆x, U r
ε(x, Γ) ⊂∆x}is the set of points of the curve Γ such
that in their ε−neighborhood the curve Γ lies in the open circular disk 4xto the right of x
and outside of the closed circular disk 4xto the left of xexcept the point xitself;
Cis the set of points of the curve Γ each of which has the neighborhood where the curve
coincides with an arc of the circle ∂∆x.
Hence Γ = ASBSAB SBA SC.
Suppose that xn→x,xn∈Γ and let all xnbe in ε-neighborhood of xto the left of it.
Assume that xn∈ASBA and x∈AB then in some neighborhood of xthe curve Γ lies above
each circle ∂∆xn. Γ is inside the circular disk 4x. This fact follows from the continuity of a
tangent to Γ. We obtain a contradiction with x∈AB. So in some neighborhood of xto the
left of it there are no points of the sets Aand BA. In this neighborhood we consider a similar
sequence of points xn∈BSAB and as a consequence we find that in some neighborhood of x
the curve Γ lies under every circle ∂∆xnand therefore outside of the circular disk ∆xwhich is
impossible. So in some left half-neighborhood of xour curve is an arc of a circle ∂∆xwhich is
also not compatible with the condition x∈AB. We conclude that AB =∅. Similarly, BA =∅.
Thus Γ = ASBSC. Suppose that C=∅, Γ = ASB. Consider a sequence of points
xn∈A,xn→x. If x∈Bthen as we know the contradiction is obtained. Hence xdoes belong
to Aand Ais a closed set. Similarly, Bis closed. The curve Γ is connected and therefore either
Aor Bis empty. It means that Γ = Aor Γ = B, which implies that the curve Γ either contains
a circular disk ∆xor is contained in a circular disk ∆x, which contradicts the fact that Γ is the
convex curve of constant width and has a length πd.
Hence C6=∅and there exists a point x∈C. Some neighborhood of xis also contained in
the set C. We shall move along the curve Γ to the left from the point xto the first point y
that does not belong to C. It is obvious that ycan belong neither to Anor to Band therefore
Γ = C. Then by applying the Heine-Borel lemma we conclude that Γ is a circle and theorem 1
is proved.
Combining this result with the result of Zamfirescu [18], we obtain:
Corollary 2.1. Every Jordan curve satisfying the infinitesimal rectangle condition is a
circle.
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3. Open Problems
A row of similar open problems in the plane and in an n-dimensional case appears in
connection with Mizel’s problem.
Problem 3.1. Let Cbe a closed Jordan curve in R2, and for an arbitrary algebraic closed
curve Lof the order nfrom the property that the intersection C∩Lcontains mpoints follows
that C∩Lcontains no less than m+ 1 points. Does there exist a number msuch that from
the property above follows that Cis an algebraic curve of the order n?
Problem 3.2. In the previous question let Lbe a circle and m= 3. Is it true that Calso
is a circle?
Problem 3.3. In Problem 3.1 let Lbe an ellipse and m= 5. Is it true that Calso is an
ellipse?
Problem 3.4. Will a compact Cbe a sphere in Rn, if Cdivides the space, and if from the
belonging n+ 1 tops of the arbitrary rectangular parallelepiped to a compact C, it follows that
one more top lies in Ctoo?
The last question is interesting even if Cis an (n−1)-dimensional manifold or a boundary
of a convex set.
Problem 3.5. Let Cbe an (n−1)-dimensional manifold (or a boundary of a convex
domain) in Rnand there does not exist an (n−1)-dimensional sphere Sn−1such that the
intersection C∩Sn−1contains n+ 1 points exactly. Is it true that Cis an (n−1)-dimensional
sphere?
Problem 3.6. Does the result of [18] remain valid, if we consider a compact set C⊂R2
where the complement R2\Cis not connected?
Problem 3.7. Are the cited results and Problems 3.1-3.5 true, if one point (top) on Cis
fixed?
Problem 3.8. Will the problems 3.1-3.6 be true if one of the intersection points must belong
to some selected subset E⊂Cthat is not everywhere dense in C? What is the minimum power
of this set for the correctness of problems?
Example 3.1. Let C=F∪S1, where S1is a unit circle in the plane and Fis any compact
subset of y-axis. Let all points of Fbe at the distance more than one unit from the origin. If
the selected set consists of the pair of points E= (0,1) ∪(0,−1), it is obvious that the compact
Csatisfies the Mizel’s problem for the vertices of a rectangle which one vertex lies on the set
E, but Cis not a circle.
This example shows that the selected set of Problem 3.8 should contain at least three points
or we restrict ourselves to the study compacts that are Jordan curves.
The example below will show that in Problem 3.2 it is impossible instead of a curve, in
analogy with Tkachuk’s result, to consider a compact set dividing the plane.
Example 3.2. Consider the domain Don the plane bounded by the circle S1. Next, remove
from the interior of the domain Deverywhere dense in Dinfinite ensemble of opened balls Di
which do not intersect pairwise even on the border and also do not intersect the circle S1.
Then we receive the fractal compact set K=¯
D\ ∪ Diwithout interior points that divides
the plane into the countable set of components. But it is easy to see that the intersection of
an arbitrary circle and Kis only one point or an infinite number of points (see Figure 1).
We receive other examples if domains D(Di) are the domains bounded by ellipses or squares.
In this case intersections with a circle can contain one, two, four or infinitely many points but
never three points. These examples give a negative answer to Problem 8 from [20].
Corollary 3.1. On the two-dimensional plane R2there exist compact sets that divide the
plane and such that there are no circles having with them in accuracy three cross points.
In particular, a class of similar sets includes the Shottka sets and Sierpi´nski’s carpet.
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Figure 1
Other similar problems can be found in the work of Gr¨unbaum [5].
This investigation was partially supported by Tubitek-NASU grant number 110T558.
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