Download Targil 3. Reminder: a set is convex, if for any two points inside the

Targil 3.
Reminder: a set is convex, if for any two points inside the set, it contains the
interval connecting these two points.
1. Consider a subset in K. At each step, we add to our subset all the interior points
of all intervals having both ends in that subset. The process is stopped when the set
is already convex. What is the maximal possible number of steps in this process?
Answer. log 2  K  1 , where  x  denote ceiling of x, which is the least integer
number not smaller than x.
Solution.
Definition. For vectors v1, v2, …, vn a convex combination is a sort of linear
combination, a1v1 + a2v2 + … + anvn with two additional conditions: the
coefficients are nonnegative and sum of coefficients is 1.
The physical meaning of convex combination is the mass center with some masses.
Definition. The minimal convex set containing the original set is called a convex
hull (in Hebrew ‫ )קמור‬of the original set.
So, first of all, we had some original set A. Take all (finite) convex combinations
of elements of A.
Lemma 1. All convex combinations form a convex set.
Lemma 2. WLOG any convex combination is of no more than K+1 points. If there
are more points, we can write the same point as a convex combination of fewer
points from A.
Now denote A1 union of all intervals with endpoints at A. Similarly Am+1 union of
all intervals with endpoints at Am.
Lemma 3. Am is a set of all convex combinations of no more than 2m points.
From the lemmas we get our claim in one direction. That is if m  log2  K  1 ,
in other words m is the smallest integer such that 2m ≥ K + 1, then by lemmas 2 and
3, Am contains all convex combinations of points from A. That is a convex set by
lemma 1, so the process will stop.
For any K we can construct many examples in which the process won’t stop
earlier. Indeed, assume the original set A was K + 1 points v0, v1, …, vK in general
position. That is, they were not on a hyperplane.
The center of mass (v0 + v1 + … + vK) / (K + 1) is a convex combination of K + 1
points but not of K points, otherwise a point of A would belong to the hyperplane
spanned by the other points. Therefore Am-1 doesn’t contain it, since it contains
only convex combinations of at most 2m-1 points, and by definition m is the
smallest integer such as … .
It remains to prove the lemmas. All of them are based on the expression from
analytic geometry to the interval connecting points P and Q. The interval PQ is
described as  P   Q     1 , 0   ,   .
That formula is an exercise to the reader. Direct application of this formula makes
lemmas 1 and 3 obvious,
Proof of lemma 2. Consider convex combination of N vectors, N > K + 1.
C = a1v1 + a2v2 + … + aNvN
We want to prove that we can get C as a convex combination of fewer points.
Construct vectors wm in K+1 : first K coordinates of wm coincide with those of vm,
and the last coordinate of wm is 1. Now we have N vectors in K+1, hence they are
linearly dependent. Hence there is a nontrivial zero linear combination of those
N
which is
b w
m 1
m
m
 0 . This condition can be written as two conditions: when you
N
look at the last coordinate, you see that
b
m 1
N
coordinates, you get
b v
m 1
m m
m
 0 , and when you look at the first K
 0 . We have both positive and negative coefficients,
since the combination is non-trivial, and sum of coefficients is zero. Let i1, i2, i3, …
be the indexes of positive coefficients, j1, j2, j3, … be the indexes of negative
coefficients, and cm = |bm|, Then after moving negative coefficients to the RHS
(right hand side), we get  cik vik   c jk v jk , and here all the coefficients are
positive. Take q  max  ak ck  . WLOG, we may assume that the maximal value
ck 0
was achieved on the RHS (otherwise we shall revert the inequality).
We shall take the original convex combination a1v1 + a2v2 + … + aNvN and add to it
the expression q 
 c v  c
ik ik
jk

v jk . Sum of coefficients will remain the same
(i.e. 1), one coefficient will cancel out, the others will remain positive. So we get a
convex combination with fewer points. QED.
2. A family of N convex sets in K is given, N > K. Each K + 1 sets of the family
have a common point. Prove that all sets have a common point.
Proof. The proof goes by induction. Assume we have proven that each M sets have
a common point, M > K + 1. It remains to prove that each M + 1 sets, for instance
A0, A1, …, AM, have a common point.
Consider sets Bi  Ai A0 , for i = 1, 2, …, M. It is easy to see that these sets are
convex, since an intersection of two convex set is obviously convex.
Any M – 1 of sets Bi have a common point (since their intersection is actually an
intersection of M sets Ai). So, intersection of any K + 1 among sets Bi is nonempty, and by induction assumption the intersection of M of them is non-empty.
Therefore B1 B2 ... BM  A0 A1 A2 ... AM is nonempty.
It remains to build a base for this induction. It has to be based on a different idea.
Because, for example, intersection of any two sides of triangle is nonempty, but
intersection of all 3 is empty.
During the proof of lemma 2 in the solution of the previous problem we have
noticed the following fact: for more than K + 1 points v1, …, vN in K – dimensional
space, we can choose some two subsets of indexes and positive coefficients ck,
such that  cik vik   c jk v jk and  cik   c jk . If we divide all coefficients by
s   cik   c jk we arrive to the following conclusion: there is a point which is a
convex combination of any of the two disjoint subsets of given points (assuming
the number of points is at least K+2).
The base of induction, that we have to prove, is the following: if any K+1 sets
intersect, then any K+2 sets intersect. Take K+2 sets A1, A2, …, AK+2. Intersection
of all those sets except Am contains at least one point Pm.
Now we have K+2 points. We can choose two disjoint subsets of indexes i1, i2, …
and j1, j2, … such that a certain point P in space is a convex combination both of
Pi1 , Pi2 ,... and of Pj1 , Pj2 ,... . We shall prove that the point P belongs to any Am,
therefore it belongs to their intersection and hence it is nonempty.
Indeed, assume that m is not one of i1, i2, i3, …. Then Am contains Pi1 , Pi2 ,... and
because it is convex, it contains all their convex combinations, and P in particular.
If is not one of i1, i2, i3, …, then it is not one of the j1, j2, …, then Am contains
Pj1 , Pj2 ,... and all their convex combinations, hence it contains P.
Remarks. I think problem 2 was also called Haley theorem, but I can’t find it in
google. In order to solve such a problem, a good approach is to solve 2- or 3dimensional case first (worked for me), even if when you write it down you find
out you don’t use
3. Matrix S is square and has the following 3 properties:
(a) All entries are nonnegative.
(b) Sum of numbers in any row is 1.
(c) Sum of numbers in any column is 1.
Prove that this matrix is a weighted average of permutation matrixes.
Remark. The matrixes described in the question are called bistochastic or doubly
stochastic. Stochastic is a scientific word for probabilistic. Nonnegative numbers
with sum 1 can mean probabilities; bistochastic means both rows and columns can
be probabilities.
Solution. The cells of the matrix which are zeroes or ones will be called good
cells. The other cells will be called bad cells.
Lemma. If we have a matrix which has some bad cells, it is a convex combination
(weighted averaged) of matrixes with fewer bad cells.
Proof of lemma. Choose a bad cell and mark it with black. There is another bad
cell in the same row, mark it with white and move to that cell. There is another bad
cell in the same column, mark it with black and move to that cell and so on. At
some moment, we shall be able to move to the cell which was already marked with
the opposite color.
In that way we shall create a closed loop of even length, all even steps are vertical
from a black cell to a white cell, and all odd steps are horizontal from a white cell
to the left cell. If we add ε to all white cells of that loop, and subtract ε from all
white cells of that loop, then the sum in every row and column remains the same.
So, when ε is close to 0, the matrix remains bistochastic, when it is far enough for
0, the matrix stops being bistochastic, because some numbers in the matrix become
negative. When we substitute different values of ε we get a line in the linear space
of all matrixes. When ε is 0, we get the original matrix A. When ε is minimal
possible/maximal possible so that matrix is still bi-stochastic, we get matrixes A0
and A1 which have fewer bad cells than A, and A is on the interval connecting A0
to A1.
So, we can replace each matrix by a convex combination of matrixes with less bad
cells; those can be replaced by convex combinations of matrixes with even smaller
number of bad cells etc. With each step in our convex combination maximal
number of bad cells in a matrix will be reduced, and after n2 steps at most we get
the same matrix as convex combination of matrixes with no bad cells.
Matrixes which have only zeroes and ones and sum in each column and in each
row is 1 are precisely the permutation matrixes.
Remark. There’s a more general statement called Krein-Milman theorem. In finite
dimensional case, it states that a compact convex set is a convex hull of its extreme
points (where extreme points are such points of the set that are not interior points
of the intervals joining points of that set). In the infinite dimensional case, convex
hull is replaced by the closure of convex hull.
4. Given k balls of radius 1 in 3 a point on the boundary of a ball is called
"isolated" if it doesn't see any other ball (the balls are not transparent).
What is the area of the set of isolated points?
Answer. The area of the unit sphere, that is 4π.
Solution. For every isolated point of any sphere, consider a unit normal vector to
its sphere, looking outside the sphere.
That defines a map from the set of isolated points to the unit sphere (since a unit
vector always belongs to the unit sphere). We shall prove that this map, up to a set
of measure zero, is bijective (‫)חח"ע ועל‬.
Injectivity (‫)חח"ע‬: at any isolated point its sphere has a tangent plane. By definition
of isolated point, all the other points of all the spheres are on one side of that plane.
Therefore, scalar product of corresponding normal vector with the given isolated
point is greater than for any other point. Hence it is unique.
Surjectivity (‫ )על‬up to measure 0: consider any unit vector v, and consider of all
points on given spheres point p which has highest scalar product with v.
It might happen that we have two or more points of that kind, from different
spheres, but probability of it is 0. It would mean that scalar product of v with two
centers of different spheres is 0, and that means that v belongs to one of the finite
number of arcs, defined by orthogonality to interval connecting two centers of
spheres. There is a finite number of those intervals.
Outside those cases with probability 0, we get just one point with highest scalar
product, and that is the isolated point with given normal. QED.
5. Consider a bounded convex shape of area S in plane with smooth boundary of
length l. Find the area of R-neighborhood given R, S and l. (By R-neighborhood we
mean the set of all points that have distance at most R from the original shape.)
Solution. Consider a convex polygon, that is inscribed into the shape and
approximates it (it can be done by walking around along the boundary in small
steps). The R neighbourhood of the polygon consists of:
a. Rectangles of height R and bases = sides of the polygon.
b. Sectors of disc of radius R (angles are 180° – internal angle of polygons).
The rectangles can be glued together into a rectangle of area R × perimeter.
The sectors can be glued together into a disc of radius R and are πR2.
In addition, we have the internal area of the polygon.
When the polygon is close to the shape, the R-neighborhood of the polygon is close
to (but slightly smaller than) R-neighborhood of the original shape.
So the limit area is πR2 + lR + S.
Remark. Of course, similar thing happens in higher dimensions (for almost the
same reason): the answer is a polynomial in R of degree = dimension, the first
coefficient = volume of the unit ball, linear coefficient = area of surface (n-1
dimensional) and the free coefficient is the volume. Other coefficients are more
complicated.
6. We are given N > 2 circles of radius 1. Every straight line meets less than 3
circles. Their centers are O1, O2, … ON. Prove that
1
n
a.


4
i  j Oi O j
1
OO
b.*
i j
i

 n  1
4
j
Solution. a. Consider angles of view of all circles other than number j from the
point Oj. By condition, those angles don’t intersect. Even if we add to each angle
the opposite angle, the angles still don’t intersect (otherwise there would be a line
through Oj cutting circle j and two more circles). So their sum is less than 2π.
Denote by  i , j half the angle of view of circle number i from point Oj.
We actually proved that

i j
i, j


2
for any j. But
1
 sin  i , j   i , j .
OiO j
1

Therefore 
 for any j. Summing over j gives
2
i  j Oi O j
Hence
1
 O O
i j
i

j
n
1
j 1 i  j
i
 O O

j
n .
2
n
.
4
b. Consider angle OjOkOi. The distance from Oi to the line OjOk is at least 2.
2
Therefore OiOk  sin  O j Ok Oi   2 . Hence the angle sin  O j Ok Oi  
.
OiOk
Hence both the angle OjOkOi and 180° – OjOkOi are greater than
2
for the same reason).
O jOk
2
(or than
OiOk
Let n be the total number of points. For given k, consider lines via Ok and all other
Oj this line split plane into n – 1 pairs of symmetric angles. Angle bounded (from
2
the clockwise direction) by OjOk is greater than
. There are two symmetric
O jOk
2
 2 .
j  k O j Ok
angles of that kind, so for each k we get 2
Summing these things will lead to the same conclusion once again, so we need yet
another idea. The idea is: consider the convex hull of all Oj. Not all Oj are the
vertexes of the convex hull, just m of them.
Sum of all angles of the convex hull is not m but  m  2  .
Fix vertex of the convex hull Ok and denote the angle of the convex hull at it  k .
Then rays OkOj split the angle  k into n – 2 parts. The part OiOkOj is greater than
by both
2
2
and
.
OiOk
O jOk
Summing inequalities of that kind we can write
2
OO
j i , k
j
  k . In this sum
k
2
OiOk
is omitted. We can omit any we choose, so we shall omit the longest OiOk .
Therefore
2
O O
j k
j
k

n 1
2
n 1

k .

n  2 j i ,k O jOk n  2
Sum it over all vertexes of the convex hull:
2
n 1
n 1
n 1
 
k 
k 
 m  2   m  1



n  2 Ok hull
n2
Ok hull j  k O j Ok
Ok hull n  2
Hence the sum over all vertexes
2
2
2
  
  
  m  1   n  m    n  1

k j  k O j Ok
Ok hull j  k O j Ok
Ok  hull j  k O j Ok
Each pair here appears twice. Therefore sum over pairs 2
i j
2
  n  1 . QED.
OiO j