Israeli Team for SEEMOUS
Second Stage Solutions.
1. A graph is, by definition, a collection of vertices and a collection of edges
that connect pairs of vertices. Two vertices are called adjacent, if they share
an edge.
Given a graph, consider the function c(n) – the number of ways to color each
vertex with one of n given colors, so that no two adjacent vertices will have
the same color. Show, that c(n) is a polynomial of n.
First solution. Induction over number of vertices + number of edges.
The only graph of 1 vertex gives c(n) = n.
Of course, if graph is disconnected function c(n) is a product of functions,
corresponding to his connected components, and product of polynomials is a
polynomial.
Take two adjacent vertices A, B in a graph. Let us erase the edge AB.
Number of ways to color the new graph, c1(n) is a polynomial by induction
(same vertices, less edges). Of those, there are c(n) ways to color it so that A
and B will be of different color, and c2(n) ways to color it so that so that A
and B will having the same color. If we shall glue vertices A and B, the new
graph will have less edges and less vertices than the original graph, and it
can be colored in c2(n) ways. Hence c(n) = c1(n) – c2(n), so it is a difference
of two polynomials, hence it is itself polynomial.
Second solution. A way to split the vertices of given graph into certain
equivalence classes will be called configuration. Configuration is called
good if no to vertices of the same class are adjacent. There is only finite
number of configuration.
Each coloring corresponds to a specific configuration: vertices of the same
color are declared equivalent. Let us count, how many colorings correspond
to the same configuration. Take a configuration which has M classes.
First class can be colored in one of n colors, second in one of n-1 colors, and
so on, hence if M ≥ n it corresponds to n(n-1)(n-2)…(n-M+1)
If M < n then the product we wrote, as well as the number of colorings, is 0.
So, number of colorings corresponding to certain configuration is a
polynomial (which we wrote explicitly) and since we have finite number of
configurations, the total number of colorings is a sum of finite number of
polynomials, which is a polynomial.
2. A disc of radius 1/N is rolling inside the circular box of radius 1,
where N > 2. (The friction between the edge of the disc and the wall of the
box is very high so the disc doesn’t slip with respect to the box at the point
of tangency). A red point on the boundary of a small circle goes along a starshaped closed trajectory.
Compute the area, bounded by this trajectory (as a function of N).
First solution. Let us start by building a parametrical equation of the star.
1
The center of the disc goes in circles of radius 1 
so it can be described
N

1
1


as v=  1   cos t , 1   sin t  . The vector which goes from the center of
 N
 N 

1
the disc to the red point goes around a circle of radius
in the opposite
N
1
1

direction, so it can be described as u=  cos s,  sin s  . Both parameters
N
N

depend linearly on the length of the arc that we cover, t.
While the center goes around one time, the red point meets the boundary N
times. This means the small discs rotates around itself N  1 times, hence
1
1

u=  cos   N  1 t  ,  sin   N  1 t   .
N
N

The point on the star can be described as w = u + v, which is also a vector
function of t. A simple way to check we wrote it correctly – differentiating
vectors u, v shows that their velocities are equal in their absolute value and
that they cancel each other when the red point is near at the boundary (and
then its velocity should be 0, because of the friction).
Of course, since u looks always directly clockwise and v is of the same
absolute value the vector will always go clockwise so the star won’t have
self-intersections.
Integrating  ydx around the star should, as usual, give the area inside.
Minus sign is because the trajectory, the way we have parameterized it, goes
clockwise, so the upper boundary must be consider with plus and the lower
with minus. So we get the following integral:
2

1
1
 d 
1
1


1

sin
t

sin
N

1
t
1

cos
t

cos
N

1
t















 dt 
0   N 
N
dt
N
N


 

1
 2
N
2
   N  1 sin t  sin   N  1 t    N  1 sin t  sin   N  1 t   dt 
0
2
N 1

N  1 sin 2 t  sin 2   N  1 t    N  2  sin t sin   N  1 t  dt
2  
N 0
To finish this, it is useful to know the following exercises:
2
Exercise 1.
 sin
2
2
tdt 
 sin   N  1 t  dt  
2
0
0
(Hint: sin  cos  1 )
2
2
2
Exercise 2.
 sin t sin   N  1 t  dt  0
0
(Hint: 2sin a  sin b  cos  a  b   cos(a  b) )
So, the answer is
 N  1 N  2  .
N2
Second solution. Like before, we describe the position of red point as the
sum of two vectors w = u + v where u goes clockwise in a circle of radius
1
1
1  one time and v goes counter-clockwise in a circle of radius , N  1
N
N
times, but we don’t write the coordinates explicitly.
For any to vectors k   k x , k y  , m   mx , m y  denote the oriented area of the
parallelogram they form k   k x , k y  , m   mx , m y  . So, in time dt the vector
dw  w
(since its clockwise) which gives total area
2
2
2
2
1
1
0 dw  w  0 2  du  dv   u  v   2 0 du  u  dv  v  du  v  dv  u
w sweeps area
The integrals
2
2
0
0
 du  v,  dv  u are 0, since the angle between du and v, as
well as dv and u rotates uniformly around 0 and makes several full circles.
2
2
1
1
1


1

since
u
sweeps
one
circle
of
radius
du

u


1


.


N
2 0
N




2
2
1
1
dv

v


N

1



  since v sweeps N  1 circles of radius in the
2 0
N
opposite direction. So the total integral is:
2
2
2
2
  N  1 2
1
1

1
1 
du  u  dv  v  1     N  1       
 N  1    
  N  
2 0
 N
N
 N  

N 1
 N  1 N  2 

N  1  1  
2 
N
N2
3. A natural number k is considered good, if for each N the number
1k+2k+…+Nk is divisible by 1+2+…+N.
Describe the set of all good numbers.
Solution. If k is good, then it 1k+2k is divisible by 3. So 1+(-1)k = 0 (mod 3)
hence k can’t be even.
Suppose now k is odd. 1k+2k+…+Nk is divisible by 1+2+…+N if and only if
2(1k+2k+…+Nk) is divisible by 2(1k+2k+…+Nk)=N(N+1).
N and N+1 are co-prime, so it is sufficient to verify separately that it is
divisible by N and by N+1. It is enough to prove 2(1k+2k+…+Nk) is
divisible by N+1 for all N, then 2(1k+2k+…+(N-1)k) is divisible by N and
2(1k+2k+…+Nk) also. We shall use “Gauss trick”:
2(1k+2k+…+Nk) = 2((1k+Nk) +(2k+(N-1)k) +…+(Nk+1k)).
But this is definitely divisible by N+1 since ak+bk is always divisible by a+b
for odd k since ak+bk =(a+b)(ak-1 – ak-2b + ak-3b2 – . . . + ak-1).
4. Let A1, A2,…,AN be nonzero matrices M×M (a matrix is called nonzero if
at least one of its elements is nonzero). Prove that there exists a matrix B of
the same size such that BA1BA2B…BANB is a nonzero matrix.
Solution. The key is to consider the kernel and image spaces of matrices.
We shall construct projection matrix B of rank 1, which satisfies the
conditions. Projection matrix of rank 1 is defined by 2 linear subspaces:
kernel of codimension 1 and image of dimension 1, which doesn’t contain
kernel. Each vector can be uniquely decomposed into sum of two vectors –
one from the space of dimension 1 and second from the space of
codimension 1. So, the projection can be described as taking the first vector
of that decomposition.
For this product to be non-zero, all we need is that the image of B won’t be
sent into its kernel. So, we have to prove that we can choose a nonzero
vector v (or the one-dimensional space) and a space W of codimension that
neither Ai will send v into W.
To do this, we must achieve 2 things:
a) Find a vector v which don’t belong to kernel of Ai for all i.
b) Find a hyperplane (containing 0) which doesn’t contain Aiv for all i.
So, it remains to prove 2 lemmas:
Lemma 1. There exists a vector which is not contained in all given linear
subspaces, where number of subspaces is finite.
Lemma 2. There exists a hyperplane (containing 0), which doesn’t intersect
with a given finite set of points.
Since any subspace can be enlarged to hyperspace, lemma 1 is equivalent to
its special case:
Lemma 3. There exists a vector which is not contained in all given
hyperplanes, where number of hyperplanes is finite.
Lemma 2 is also follows from lemma 3, since if we replace a hyperplane
a1x1+a2x2+…+anxn = 0 by a vector (a1, a2, … , an) and vice versa, the
condition “a hyperplane contains the vector” turns into “a vector belongs to
the hyperplane”.
So, it is enough to prove lemma 3.
Remark. All this works only for infinite fields.
Proof of lemma 3. Apply the induction over dimension of the space.
The base of induction: space of dimension 1 can’t be covered by finite
number of points (field is infinite).
The step of induction: assume it is proven for spaces of dimension smaller
than n. So, we have finite number of hyperplanes, and we try to prove they
don’t cover the space. There is infinite number of hyperplanes in the space,
so we can choose a huperplane H which is different from all given
hyperplanes. Intersection of H with other hyperplanes are sub-hyperplanes in
H, so, by induction, they can’t cover it.
5. An infinite sequence of real numbers {xi} will be called nice if ∑xi2
converges. Let {ai} be a sequence, such that for each nice sequence {xi} the
series ∑aixi converges. Prove that the sequence {ai} is nice.
Solution. Assume {ai} isn’t nice. So ∑ai2 diverges. We can cut the sequence
{ai} into infinite number of segments, each of which is greater than 1.
(That is done by induction, simply sum up the numbers from the end of
segment number k until it exceeds 1, and that will be segment k+1.)
Let segment number k start at mk and have nk elements.
Then, by construction, bk 
mk
a 2j > 1.

j m n 1
k
Now, construct a sequence x j 
We shall use a nice lemma:
k
aj
, for each j which belongs to segment
k  bk
number k. Then



mk
mk
a 2j
ajxj    ajxj   

j 1
k 1 j m n 1
k 1 j m n 1 k  b
k
mk


a 2j
k
k
k

bk
1




k

b
k

b
k
k 1
k 1
k 1
k
k
But sequence {xi} is good because
mk





a 2j
bk
1
1
2
xj 




2
2
2
2
2
2
j 1
k 1 j mk nk 1 k  bk
k 1 k  bk
k 1 k  bk
k 1 k


j  mk  nk 1
 

k





