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==> competition/tests/math/putnam/putnam.1987.p <==
WILLIAM LOWELL PUTNAM MATHEMATICAL COMPETITION

FORTY EIGHTH ANNUAL   Saturday, December 5, 1987

Examination A;

Problem A-1
------- ---

Curves A, B, C, and D, are defined in the plane as follows:

A = { (x,y) : x^2 - y^2 = x/(x^2 + y^2) },

B = { (x,y) : 2xy + y/(x^2 + y^2) = 3 },

C = { (x,y) : x^3 - 3xy^2 + 3y = 1 },

D = { (x,y) : 3yx^2 - 3x - y^3 = 0 }.

Prove that the intersection of A and B is equal to the intersection of
C and D.

Problem A-2
------- ---

The sequence of digits

1 2 3 4 5 6 7 8 9 1 0 1 1 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 ...

is obtained by writing the positive integers in order.  If the 10^n th
digit in this sequence occurs in the part of the sequence in which the
m-digit numbers are placed, define f(n) to be m.  For example f(2) = 2
because the 100th digit enters the sequence in the placement of the
two-digit integer 55.  Find, with proof, f(1987).

Problem A-3
------- ---

For all real x, the real valued function y = f(x) satisfies

y'' - 2y' + y = 2e^x.

(a)  If f(x) > 0 for all real x, must f'(x) > 0 for all real x?  Explain.

(b)  If f'(x) > 0 for all real x, must f(x) > 0 for all real x?  Explain.

Problem A-4
------- ---

Let P be a polynomial, with real coefficients, in three variables and F
be a function of two variables such that

P(ux,uy,uz) = u^2*F(y-x,z-x) for all real x,y,z,u,

and such that P(1,0,0) = 4, P(0,1,0) = 5, and P(0,0,1) = 6.  Also let
A,B,C be complex numbers with P(A,B,C) = 0 and |B-A| = 10.  Find
|C-A|.

Problem A-5
------- ---

^
Let G(x,y) = ( -y/(x^2 + 4y^2) , x/(x^2 + 4y^2), 0 ).  Prove or disprove
that there is a vector-valued function

^
F(x,y,z) = ( M(x,y,z) , N(x,y,z) , P(x,y,z) )

with the following properties:

(1)  M,N,P have continuous partial derivatives for all
(x,y,z) <> (0,0,0) ;

^   ^
(2)  Curl F = 0 for all (x,y,z) <> (0,0,0) ;

^          ^
(3)  F(x,y,0) = G(x,y).

Problem A-6
------- ---

For each positive integer n, let a(n) be the number of zeros in the
base 3 representation of n.  For which positive real numbers x does
the series

inf
-----
\     x^a(n)
>    ------
/      n^3
-----
n = 1

converge?
--

Examination B;

Problem B-1
------- ---

4/        (ln(9-x))^(1/2)  dx
Evaluate  | --------------------------------- .
2/ (ln(9-x))^(1/2) + (ln(x+3))^(1/2)

Problem B-2
------- ---

Let r, s, and t be integers with 0 =< r, 0 =< s, and r+s =< t.

Prove that

( s )     ( s )     ( s )           ( s )
( 0 )     ( 1 )     ( 2 )           ( s )          t+1
----- +  ------- + ------- + ... + ------- = ---------------- .
( t )    (  t  )   (  t  )         (  t  )   ( t+1-s )( t-s )
( r )    ( r+1 )   ( r+2 )         ( r+s )            (  r  )

( n )                                  n(n-1)...(n+1-k)
( Note: ( k ) denotes the binomial coefficient ---------------- .)
k(k-1)...3*2*1

Problem B-3
------- ---

Let F be a field in which 1+1 <> 0.  Show that the set of solutions to
the equation x^2 + y^2 = 1 with x and y in F is given by

(x,y) = (1,0)

r^2 - 1     2r
and (x,y) = ( ------- , ------- ),
r^2 + 1   r^2 + 1

where r runs through the elements of F such that r^2 <> -1.

Problem B-4
------- ---

Let ( x(1), y(1) ) = (0.8,0.6) and let

x(n+1) = x(n)*cos(y(n)) - y(n)*sin(y(n))

and y(n+1) = x(n)*sin(y(n)) + y(n)*cos(y(n))

for n = 1,2,3,...  .  For each of the limits as n --> infinity of
x(n) and y(n), prove that the limit exists and find it or prove that
the limit does not exist.

Problem B-5
------- ---

Let O(n) be the n-dimensional zero vector (0,0,...,0).  Let M be a
2n x n matrix of complex numbers such that whenever
( z(1), z(2), ..., z(2n)*M = O(n), with complex z(i), not all zero,
then at least one of the z(i) is not real.  Prove that for arbitrary
real number r(1), r(2), ..., r(2n), there are complex numbers w(1),
w(2), ..., w(n) such that

(   ( w(1) ) )   ( r(1)  )
(   (  .   ) )   (   .   )
Re ( M*(  .   ) ) = (   .   )  .
(   (  .   ) )   (   .   )
(   ( w(n) ) )   ( r(2n) )

(Note:  If C is a matrix of complex numbers, Re(C) is the matrix whose
entries are the real parts of entries of C.)

Problem B-6
------- ---

Let F be the field of p^2 elements where p is an odd prime.  Suppose S
is a set of (p^2-1)/2 distinct nonzero elements of F with the property
that for each a <> 0 in F, exactly one of a and -a is in S.  Let N be
the number of elements in the intersection of S with { 2a : a e S }.
Prove that N is even.
--

==> competition/tests/math/putnam/putnam.1987.s <==
Problem A-1
------- ---

Let z=x+i*y.  Then A and B are the real and imaginary parts of
z^2=3i+1/z, and C, D are likewise Re and Im of z^3-3iz=1, and
the two equations are plainly equivalent.  Alternatively, having
seen this, we can formulate a solution that avoids explicitly
invoking the complex numbers, starting with C=xA-yB, D=yA+xB.

Problem A-2
------- ---

Counting, we see that the numbers from 1 to n digits take
up (10^n*(9n-1)+1)/9 spaces in the above sequence.  Hence we need
to find the least n for which 10^n*(9n-1)+1 > 9*10^1987, but it
is easy to see that n = 1984 is the minimum such.  Therefore
f(1987) = 1984.

In general, I believe, f(n) = n + 1 - g(n), where g(n) equals
the largest value of m such that (10^m-1)/9 + 1 =< n if n>1,
and g(0) = g(1) is defined to be 0.

Hence, of course, g(n) = [log(9n-8)] if n>0.  Therefore

f(0) = 1,

f(n) = n + 1 - [log(9n-8)] for n>0.
Q.E.D.

Problem A-3
------- ---

We have a differential equation, solve it.  The general solution is

y = f(x) = e^x*(x^2 + a*x + b),

where a and b are arbitrary real constants.  Now use completing the
square and the fact that e^x > 0 for all real x to deduce that

(1)  f(x) > 0 for all real x iff 4b > a^2.

(2)  f'(x) > 0 for all real x iff 4b > a^2 + 4.

It is now obvious that (2) ==> (1) but (1) /==> (2).

Q.E.D.

Problem A-4
------- ---

Setting x=0, u=1 we find F(y,z)=P(0,y,z) so F is a polynomial; keeping
x=0 but varying u we find F(uy,uz)=u^2*F(y,z), so F is homogeneous of
degree 2, i.e. of the form Ay^2+Byz+Cz^2, so
P(x,y,z)=R(y-x)^2+S(y-x)(z-x)+T(z-x)^2
for some real R,S,T.  The three given values of P now specify three
linear equations on R,S,T, easily solved to give (A,B,C)=(5,-7,6).
If now P(A,B,C)=0 then (C-A)=r(B-A), r one of the two roots of
5-7r+6r^2.  This quadratic has negative discrminant (=-71) so its
roots are complex conjugates; since their product is 5/6, each
one has absolute value sqrt(5/6).  (Yes, you can also use the
Quadratic Equation.)  So if B-A has absolute value 10, C-A must
have absolute value 10*sqrt(5/6)=5*sqrt(30)/3.

Problem A-5
------- ---

There is no such F.  Proof: assume on the contrary that G extends
to a curl-free vector field on R^3-{0}.  Then the integral of G
around any closed path in R^3-{0} vanishes because such a path
bounds some surface [algebraic topologists, read: because
H^2(R^3-{0},Z)=0 :-) ].  But we easily see that the integral
of F around the closed path z=0, x^2+4y^2=1 (any closed path
in the xy-plane that goes around the origin will do) is nonzero---

Problem A-6
------- ---

For n>0 let

T(n) = x^a(n)/n^3	and 	U(n) = T(3n) + T(3n+1) + T(3n+2)

and for k>=0 let

Z(k) = sum {n=3^k to 3^(k+1)-1} T(n)

We have

Z(k+1)	= sum {n=3^(k+1) to 3^(k+2)-1} T(n)
= sum {n=3^k to 3^(k+1)-1} [T(3n) + T(3n+1) + T(3n+2)]
= sum {n=3^k to 3^(k+1)-1} U(n)

Let us compare U(n) to T(n). We have a(3n)=a(n)+1 and a(3n+1)=a(3n+2)=a(n).
Thus

U(n) = x^[a(n)+1]/(3n)^3 + x^a(n)/(3n+1)^3 + x^a(n)/(3n+2)^3

and so U(n) has as upper bound

x^a(n) * (x+2)/(3n)^3 = T(n) * (x+2)/27

and as lower bound

x^a(n) * (x+2)/(3n+2)^3 = T(n) * (x+2)/(3+2/n)^3

in other words U(n) = T(n) * (x+2)/(27+e(n)), where e(n)<(3+2/n)^3-27 tends to
0 when n tends to infinity. It follows then that

Z(k+1)= Z(k)*(x+2)/(27+f(k))

where f(k)<(3+2/3^k)^3-27 tends to 0 for n tending to infinity.

Now the series is the sum of all Z(k). Thus for x>25 we have Z(k+1)>Z(k) for k
large enough, and the series diverges; for x<25 we have Z(k+1)< r * Z(k) (with
r=(x+2)/27<1) for every k, and the series converges. For x=25 the series
diverges too (I think so), because Z(k+1)/Z(k) tends to 1 for k tending to
infinity.

Another proof:

I would say,for x<25.  Let S(m) be the sum above taken over 3^m <= n < 3^(m+1).
Then for the n's in S(m), the base 3 representation of n has m+1 digits.
Hence we can count the number of n's with a(n)=k as being the number
of ways to choose a leading nonzero digit, times the number of ways
to choose k positions out of the m other digits for the k zeroes, times
the number of ways to choose nonzero digits for the m-k remaining positions,
namely

( m )  m-k
2 (   ) 2   .
( k )

Hence we have

3^(m+1)-1                m
-----                  -----
\            a(n)      \        ( m )  m-k k
>          x      =    >     2 (   ) 2   x
/                      /        ( k )
-----                  -----
n=3^m                   k=0

m
= 2 (x+2)  .
m  -3m             m -3(m+1)
Hence we can bound S(m) between 2 (x+2)  3     and 2 (x+2) 3       .
It is then clear that the original sum converges just when

inf
-----
\           m -3m
>     (x+2) 3                converges, or when x<25.
/
-----
m=0

Problem B-1
------- ---

Write the integrand as

(ln(x+3))^(1/2)
1 - --------------------------------- .
(ln(9-x))^(1/2) + (ln(x+3))^(1/2)

Use the change of variables x = 6-t on the above and the fact that
the two are equal to deduce that the original is equal to 1.

QED.

Problem B-3
------- ---

First note that the above values for x and y imply that
x^2 + y^2 = 1.  On the other foot note that if x<>1 ,x^2 + y^2 = 1,
and 2 <> 0, then (x,y) is of the required form, with r = y/(1-x).
Also note that r^2 <> -1, since this would imply x = 1.

Derivation of r:  We want x = (r^2-1)/(r^2+1) ==> 1-x = 2/(r^2+1),
and also y = 2r/(r^2+1) ==> 1-x = (2y)/(2r) if 2<>0.  Hence if
2<>0, r = y/(1-x).

The above statement is false in some cases if 1+1 = 0 in F.  For
example, in Z(2) the solution (0,1) is not represented.

QED.

Problem B-4
------- ---

Observe that the vector (x(n+1), y(n+1)) is obtained from (x(n), y(n))
by a rotation through an angle of y(n).  So if Theta(n) is the inclination
of (x(n), y(n)), then for all n,

Theta(n+1) = Theta(n) + y(n)

Furthermore, all vectors have the same length, namely that of (x1, y1),
which is 1.  Therefore y(n) = sin (Theta(n)) and x(n) = cos (Theta(n)).

Thus the recursion formula becomes

(*)	Theta(n+1) = Theta(n) + sin (Theta(n))

Now 0 < Theta(1) < pi.  By induction 0 < sin(Theta(n)) = sin(pi - Theta(n))
< pi - Theta(n), so 0 < Theta(n+1) < Theta(n) + (pi - Theta(n)) = pi.

Consequently, {Theta(n)} is an increasing sequence bounded above by pi, so
it has a limit, Theta.  From (*) we get Theta = Theta + sin(Theta),
so with Theta in the interval (0,pi], the solution is Theta = pi.

Thus lim (x(n),y(n)) = (cos(Theta), sin(Theta)) = (-1, 0).

Problem B-5
------- ---

First note that M has rank n, else its left nullspace N has C-dimension >n
and so R-dimension >2n, and thus nontrivially intersects the R-codimension
2n subspace of vectors all of whose coordinates are real.  Thus the
subspace V of C^(2n) spanned by M's columns has C-dimsension n and so
R-dimension 2n, and to prove the R-linear map Re: V-->R^(2n) surjective,
we need only prove it injective.  So assume on the contrary that v is
a nonzero vector in V all of whose coordinates are purely imaginary,
and let W be the orthogonal complement of <v>; this is a subspace of
C^(2n) of C-dim. 2n-1 and R-dim. 4n-2 .  W contains N,
which we've seen has R-dimension 2n; it also contains the
orthogonal complement of <i*v> in R^(2n), which has R-dimension 2n-1.
Since (2n)+(2n-1) > (4n-2), these two real subspaces of W intersect
nontrivially, producing a nonzero real vector in N---contradiction.
So Re: V-->R^(2n) indeed has zero kernel and cokernel, Q.E.D. .

Problem B-6
------- ---

Let P be the product of elements of S; then P'=2^|S|*P, the product of
the elements of 2S, is either P or -P according to whether |2S-S| is
even or odd.  (each element of 2S is either in S or in -S, so match
the factors in the products for P and P'.)  But by Fermat's little
theorem, 2^(p-1)=1, and since |S|=(p^2-1)/2=(p-1)*(p+1)/2 is a multiple
of p-1, also 2^|S|=1 and P=P', Q.E.D. .

This solution--analogous to one of Gauss' proof of Quadratic
it been the only solution, B-6 would be a difficult problem on a par
with B-6 problems of previous years.  Unfortunately, just knowing
that F* is a cyclic group of order |F|-1 for any finite field F,
one can split F* into cosets of the subgroup generated by 2 and -1
and produce a straightforward, albeit plodding and uninspired, proof.
I wonder how many of the contestants' answers to B-6 went this way
and how many found the intended solution.

Another proof:

Given such a set S, it is immediate to verify that for any a in S, if
one deletes a and adjoins -a to obtain a new set S' then the number
of elements in the intersection of S' and 2S' is congruent (modulo 2)
to the number of elements in the intersection of S and 2S.  If S and
S' are any two sets meeting the condition of this problem, then S can
be changed to S' by repeating this operation several times.  So, it
suffices to prove the result for any one set S meeting the condition of
the problem.  A simple candidate for such an S is obtained by letting
(u, v) be a basis for F over the field of p elements and letting S
be the unions of the sets {au + bv: 1 <= u <= (p-1)/2, 0 <= b < p} and
{bv: 0 <= b < (p-1)/2}.  An elementary counting argument completes the
proof.

==> competition/tests/math/putnam/putnam.1988.p <==
Problem A-1: Let R be the region consisting of the points (x,y) of the
cartesian plane satisfying both |x| - |y| <= 1 and |y| <= 1.  Sketch
the region R and find its area.

Problem A-2: A not uncommon calculus mistake is to believe that the
product rule for derivatives says that (fg)' = f'g'.  If f(x) =
exp(x^2), determine, with proof, whether there exists an open interval
(a,b) and a non-zero function g defined on (a,b) such that the wrong
product rule is true for x in (a,b).

Problem A-3: Determine, with proof, the set of real numbers x for
which  sum from n=1 to infinity of ((1/n)csc(1/n) - 1)^x  converges.

Problem A-4:
(a) If every point on the plane is painted one of three colors, do
there necessarily exist two points of the same color exactly one inch
apart?
(b) What if "three" is replaced by "nine"?

Problem A-5: Prove that there exists a *unique* function f from the
set R+ of positive real numbers to R+ such that
f(f(x)) = 6x - f(x)  and  f(x) > 0  for all x > 0.

Problem A-6: If a linear transformation A on an n-dimensional vector
space has n + 1 eigenvectors such that any n of them are linearly
independent, does it follow that A is a scalar multiple of the

---------------------------------------------------------------------------

Problem B-1: A *composite* (positive integer) is a product ab with a
and b not necessarily distinct integers in {2,3,4,...}.  Show that
every composite is expressible as xy + xz + yz + 1, with x, y, and z
positive integers.

Problem B-2: Prove or disprove: If x and y are real numbers with y >= 0
and y(y+1) <= (x+1)^2, then y(y-1) <= x^2.

Problem B-3: For every n in the set Z+ = {1,2,...} of positive
integers, let r(n) be the minimum value of |c-d sqrt(3)| for all
nonnegative integers c and d with c + d = n.  Find, with proof, the
smallest positive real number g with r(n) <= g for all n in Z+.

Problem B-4: Prove that if  sum from n=1 to infinity a(n)  is a
convergent series of positive real numbers, then so is
sum from n=1 to infinity (a(n))^(n/(n+1)).

Problem B-5: For positive integers n, let M(n) be the 2n + 1 by 2n + 1
skew-symmetric matrix for which each entry in the first n subdiagonals
below the main diagonal is 1 and each of the remaining entries below
the main diagonal is -1.  Find, with proof, the rank of M(n).
(According to the definition the rank of a matrix is the largest k
such that there is a k x k submatrix with non-zero determinant.)

One may note that M(1) = (  0 -1  1 ) and M(2) = (  0 -1 -1  1  1 )
(  1  0 -1 )            (  1  0 -1 -1  1 )
( -1  1  0 )            (  1  1  0 -1 -1 )
( -1  1  1  0 -1 )
( -1 -1  1  1  0 )

Problem B-6: Prove that there exist an infinite number of ordered
pairs (a,b) of integers such that for every positive integer t the
number at + b is a triangular number if and only if t is a triangular
number.  (The triangular numbers are the t(n) = n(n + 1)/2 with n in
{0,1,2,...}.)

==> competition/tests/math/putnam/putnam.1988.s <==
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Chris Long, Rutgers University \hfill January 22, 1989}
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%
%  Body of article
%
\problem {A-1:}
Let $R$ be the region consisting of the points $(x,y)$ of the cartesian
plane satisfying both $|x| - |y| \le 1$ and $|y| \le 1$.  Sketch the
region $R$ and find its area.

\solution {Solution:}
The area is $6$; the graph I leave to the reader.

\problem {A-2:}
A not uncommon calculus mistake is to believe that the product rule for
derivatives says that $(fg)' = f'g'$.  If $f(x) = e^{x^{2}}$, determine,
with proof, whether there exists an open interval $(a,b)$ and a non-zero
function $g$ defined on $(a,b)$ such that the wrong product rule
is true for $x$ in $(a,b)$.

\solution {Solution:}
We find all such functions.  Note that $(fg)' = f'g' \Rightarrow f'g' = f'g + fg'$ hence if $g(x), f'(x) - f(x) \neq 0$ we get that $g'(x)/g(x) = f'(x)/(f'(x) - f(x))$.  For the particular $f$ given, we then get that
$g'(x)/g(x) = (2x)e^{x^2}/( (2x-1) (e^{x^2}) ) \Rightarrow g'(x)/g(x) = 2x/(2x-1)$ (since $e^{x^2} > 0$).  Integrating, we deduce that $\ln{|g(x)|} = x + (1/2)\ln{|2x-1|} + c$ (an arbitrary constant) $\Rightarrow |g(x)| = e^{c} \sqrt{|2x-1|} e^{x} \Rightarrow g(x) = C \sqrt{|2x-1|} e^{x}, C$
arbitrary $\neq 0$.  We finish by noting that any $g(x)$ so defined is
differentiable on any open interval that does not contain $1/2$.

Q.E.D.

\problem {A-3:}
Determine, with proof, the set of real numbers $x$ for which
$\sum_{n=1}^{\infty} {( {1\over n} \csc ({1\over n}) - 1)}^{x}$
converges.

\solution {Solution:}
The answer is $x > {1\over 2}$.  To see this, note that by Taylor's
theorem with remainder $\sin( {1\over{n}} ) = \sum_{i=1}^{k-1} { {(-1)}^{i-1} {n}^{-(2i+1)} } + c { {(-1)}^{k-1} {n}^{-(2k+1)} }$,
where $0 \leq c \leq {1\over n}$.  Hence for $n\geq 1 ( 1/n )/( 1/n - 1/(3! n^{3}) + 1/(5! n^{5}) - 1 < (1/n) \csc(1/n) - 1 < ( 1/n )/( 1/n - 1/(3! n^{3}) ) - 1 \Rightarrow$ for $n$ large enough, $(1/2) 1/(3! n^{2}) < (1/n) \csc(1/n) - 1 < 2\cdot 1/(3! n^{2})$.  Applying the p-test and the
comparison test, we see that $\sum_{n=1}^{\infty} {( {1\over n} \csc({1\over n}) - 1)}^{x}$ converges iff $x > {1\over 2}$.

Q.E.D.

\problem {A-4:}

\subproblem {(a)}
If every point on the plane is painted one of three colors, do there
necessarily exist two points of the same color exactly one inch apart?

\solution {Solution:}
The answer is yes.  Assume not and consider two equilateral
triangles with side one that have exactly one common face $\Rightarrow$
all points a distance of $\sqrt{3}$ apart are the same color; now
considering a triangle with sides $\sqrt{3}, \sqrt{3}, 1$ we reach the

Here is a pretty good list of references for the chromatic number of
the plane (i.e., how many colors do you need so that no two points 1
away are the same color) up to around 1982 (though the publication
dates are up to 1985). This asks for the chromatic number of the graph
where two points in R^2 are connected if they are distance 1 apart.
Let this chromatic number be chi(2) and in general let chi(n) be the
chromatic number of R^n. By a theorem in  this is equivalent to
finding what the maximum chromatic number of a finite subgraph of this
infinite graph is.

  H. Hadwiger, Ein Ueberdeckungssatz f\"ur den
Euklidischen Raum,'' Portugal. Math. #4 (1944), p.140-144

This seems to be the original reference for the problem

 N.G. de Bruijn and P. Erd\"os, A Color Problem for Infinite
Graphs and a Problem in the Theory of Relations,'' Nederl. Akad.
Wetensch. (Indag Math) #13 (1951), p. 371-373.

 H. Hadwiger, Ungel\"oste Probleme No. 40,'' Elemente der Math.
#16 (1961), p. 103-104.

Gives the upper bound of 7 with the hexagonal tiling and also
a reference to a Portugese journal where it appeared.

 L. Moser and W. Moser, Solution to Problem 10,'' Canad. Math.
Bull. #4 (1961), p. 187-189.

Shows that any 6 points in the plane only need 3 colors but
gives 7 points that require 4 (the  Moser Graph'' see ).

 Paul Erd\"os, Frank Harary, and William T. Tutte, On the
Dimension of a Graph,'' Mathematika #12 (1965), p. 118-122.

States that 3<chi(2)<8. Proves that chi(n) is finite for all n.

 P. Erd\"os, Problems and Results in Combinatorial Geometry,''
in Discrete Geometry and Convexity,'' Edited by Jacob E. Goodman,
Erwin Lutwak, Joseph Malkevitch, and Richard Pollack, Annals of
the New York Academy of Sciences Vol. 440, New York Academy of
Sciences 1985, Pages 1-11.

States that 3<chi(n)<8 and I am almost sure that chi(2)>4.''
States a question of L. Moser: Let R be large and S a measurable
set in the circle of radius R so that no two points of S have
distance 1. Denote by m(S) the measure of S. Determine

Lim_{R => infty} max m(S)/R^2.

Erd\"os conjectures that this limit is less than 1/4.

Erd\"os asks the following: Let S be a subset of the plane. Join
two points of S if their distances is 1. This gives a graph G(S).
Assume that the girth (shortest circuit) of G(S) is k. Can its
chromatic number be greater than 3? Wormald proved that such
a graph exists for k<6. The problem is open for k>5. Wormald
suggested that this method may work for k=6, but probably a
new idea is needed for k>6. A related (perhaps identical)
question is: Does G(S) have a subgraph that has girth k and
chromatic number 4?' ''

 N. Wormald, A 4-chromatic graph with a special plane drawing,''
J. Austr.  Math. Soc. Ser. A #28 (1970), p. 1-8.

The reference for the above question.

 R.L. Graham, Old and New Euclidean Ramsey Theorems,''
in Discrete Geometry and Convexity,'' Edited by Jacob E. Goodman,
Erwin Lutwak, Joseph Malkevitch, and Richard Pollack, Annals of
the New York Academy of Sciences Vol. 440, New York Academy of
Sciences 1985, Pages 20-30.

States that the best current bounds are 3<chi(2)<8. Calls the
graph in  the Moser graph. Quotes the result of Frankl and
Wilson  that chi(n) grows exponentially in n settling an
earlier conjecture of Erd\"os (I don't know the reference for
this). The best available bounds for this are

(1+ o(1)) (1.2)^n \le chi(n) \le (3+ o(1))^n.

 P. Frankl and R.M. Wilson, Intersection Theorems with Geometric
Consequences,'' Combinatorica #1 (1981), p. 357-368.

  H. Hadwiger, H. Debrunner, and V.L. Klee, Combinatorial
Geometry in the Plane,'' Holt, Rinehart & Winston, New York
(English edition, 1964).

  D.R. Woodall, Distances Realized by Sets Covering the Plane,''
Journal of Combinatorial Theory (A) #14 (1973), p. 187-200.

Among other things, shows that rational points in the plane can
be two colored.

  L. A. Sz\'ekely, Measurable Chromatic Number of Geometric
Graphs and Sets without some Distances in Euclidean Space,''
Combinatorica #4 (1984), p.213-218.

Considers \chi_m(R^2), the measurable chromatic number,
where sets of one color must be Lebesgue measurable.
He conjectures that \chi_m(R^2) is not equal to
\chi(R^2) (if the Axiom of Choice is false).

  Martin Gardner, Scientific American,'' October 1960, p. 160.

  Martin Gardner, Wheels, Life and other Mathematical Amusements,''
W.H. Freeman and Co., New York 1983, pages 195-196.

This occurs in a chapter on mathematical problems including
the 3x+1 problem. I think that his references are wrong, including
attributing the problem to Erd\"os and claiming that Charles Trigg
had original solutions in Problem 133,'' Crux Mathematicorum,
Vol. 2, 1976, pages 144-150.

Q.E.D.

\subproblem {(b)}
What if "three" is replaced by "nine"?

In this case, there does not necessarily exist two points of the same
color exactly one inch apart; this can be demonstrated by considering
a tessellation of the plane by a $3\times 3$ checkboard with side $2$,
with each component square a different color (color of boundary points
chosen in an obvious manner).

Q.E.D.

The length of the side of the checkerboard is not critical (the reader
my enjoy showing that $3/2 <$ side $< 3\sqrt{2}/2$ works).

\problem {A-5:}
Prove that there exists a {\it unique} function $f$ from the set
${\R}^{+}$ of positive real numbers to ${\R}^{+}$ such that $f(f(x)) = 6x - f(x)$ and $f(x) > 0$ for all $x > 0$.

\solution {Solution 1:}

Clearly $f(x) = 2x$ is one such solution; we need to show that it is the
{\it only} solution.  Let $f^{1}(x) = f(x), f^{n}(x) = f(f^{n-1}(x))$ and
notice that $f^{n}(x)$ is defined for all $x>0$.  An easy induction
establishes that for $n>0 f^{n}(x) = a_{n} x + b_{n} f(x)$, where $a_{0} = 0, b_{0} = 1$ and $a_{n+1} = 6 b_{n}, b_{n+1} = a_{n} - b_{n} \Rightarrow b_{n+1} = 6 b_{n-1} - b_{n}$.  Solving this latter equation
in the standard manner, we deduce that $\lim_{n\to \infty} a_{n}/b_{n} = -2$, and since we have that $f^{n}(x) > 0$ and since $b_{n}$ is
alternately negative and positive; we conclude that $2x \leq f(x) \leq 2x$
by letting $n \rightarrow \infty$.

Q.E.D.

\solution {Solution 2:}
(Dan Bernstein, Princeton)

As before, $f(x) = 2x$ works.  We must show that if $f(x) = 2x + g(x)$
and $f$ satisfies the conditions then $g(x) = 0$ on ${\R}^{+}$.
Now $f(f(x)) = 6x - f(x)$ means that $2f(x) + g(f(x)) = 6x - 2x - g(x)$,
i.e., $4x + 2g(x) + g(f(x)) = 4x - g(x)$, i.e., $3g(x) + g(f(x)) = 0$.
This then implies $g(f(f(x))) = 9g(x)$.  Also note that $f(x) > 0$
implies $g(x) > -2x$.  Suppose $g(x)$ is not $0$ everywhere.  Pick $y$
at which $g(y) \neq 0$.  If $g(y) > 0$, observe $g(f(y)) = -3g(y) < 0$,
so in any case there is a $y_{0}$ with $g(y_{0}) < 0$.  Now define $y_{1} = f(f(y_{0})), y_{2} = f(f(y_{1}))$, etc.  We know $g(y_{n+1})$ equals
$g(f(f(y_{n}))) = 9g(y_{n})$.  But $y(n+1) = f(f(y_{n})) = 6y_{n} - f(y_{n}) < 6y_{n}$ since $f>0$.  Hence for each $n$ there exists
$y_{n} < 6^{n} y_{0}$ such that $g(y_{n}) = 9^{n} g(y_{0})$.
The rest is obvious: $0 > g(y_{0}) = 9^{-n} g(y_{n}) > -2\cdot 9^{-n} y_{n} > -2 (6/9)^{n} y_{0}$, and we observe that as $n$ goes to infinity

Q.E.D.

\problem {A-6:}
If a linear transformation $A$ on an $n$-dimensional vector space has
$n+1$ eigenvectors such that any $n$ of them are linearly independent,
does it follow that $A$ is a scalar multiple of the identity?  Prove your

\solution {Solution:}
The answer is yes.  First note that if $x_{1}, \ldots, x_{n+1}$ are the
eigenvectors, then we must have that $a_{n+1} x_{n+1} = a_{1} x_{1} + \cdots + a_{n} x_{n}$ for some non-zero scalars $a_{1}, \ldots, a_{n+1}$.
Multiplying by $A$ on the left we see that $\lambda_{n+1} a_{n+1} x_{n+1} = \lambda_{1} a_{1} x_{1} + \cdots + \lambda_{n} a_{n} x_{n}$, where
$\lambda_{i}$ is the eigenvalue corresponding to the eigenvectors $x_{i}$.
But since we also have that $\lambda_{n+1} a_{n+1} x_{n+1} = \lambda_{n+1} a_{1} x_{1} + \cdots + \lambda_{n+1} a_{n} x_{n}$ we conclude that
$\lambda_{1} a_{1} x_{1} + \cdots + \lambda_{n} a_{n} x_{n} = \lambda_{n+1} a_{1} x_{1} + \cdots + \lambda_{n+1} a_{n} x_{n} \Rightarrow a_{1} (\lambda_{1} - \lambda_{n+1}) x_{1} + \cdots + a_{n} (\lambda_{n} - \lambda_{n+1}) x_{1} = 0 \Rightarrow \lambda_{1} = \cdots = \lambda_{n+1} = \lambda$ since $x_{1}, \ldots, x_{n}$ are linearly independent.  To
finish, note that the dimension of the eigenspace of $\lambda$ is equal
to $n$, and since this equals the dimension of the nullspace of $A - \lambda I$ we conclude that the rank of $A - \lambda I$ equals $n - n = 0 \Rightarrow A - \lambda I = 0$.

Q.E.D.

\problem {B-1:}
A {\it composite} (positive integer) is a product $ab$ with $a$ and $b$
not necessarily distinct integers in $\{ 2,3,4,\ldots \}$.  Show that
every composite is expressible as $xy + xz + yz + 1$, with $x, y$, and
$z$ positive integers.

\solution {Solution:}
Let $x = a-1, y = b-1, z = 1$; we then get that $xy + xz + yz + 1 = (a-1)(b-1) + a-1 + b-1 + 1 = ab$.

Q.E.D.

\problem {B-2:}
Prove or disprove:  If $x$ and $y$ are real numbers with $y \geq 0$
and $y(y+1) \leq {(x+1)}^2$, then $y(y-1) \leq x^2$.

\solution {Solution:}
The statement is true.  If $x+1 \geq 0$ we have that $\sqrt{y(y+1)} - 1 \leq x \Rightarrow x^{2} \geq y^{2} + y + 1 - 2 \sqrt{y^{2}+y} \geq y^{2} - y$ since $2y + 1 \geq 2 \sqrt{y^{2}+y}$ since ${(2y + 1)}^{2} \geq 4 (y^{2}+y)$ if $y\geq 0$.  If $x+1 < 0$, we see that $\sqrt{y(y+1)} \leq -x - 1 \Rightarrow x^{2} \geq y^{2} + y + 1 + 2 \sqrt{y^{2}+y} \geq y^{2} - y$.

Q.E.D.

\problem {B-3:}
For every $n$ in the set ${\Z}^{+} = \{ 1,2,\ldots \}$ of positive
integers, let $r(n)$ be the minimum value of $|c-d \sqrt{3}|$ for all
nonnegative integers $c$ and $d$ with $c + d = n$.  Find, with proof,
the smallest positive real number $g$ with $r(n) \leq g$ for all $n$
in ${\Z}^{+}$.

\solution  {Solution:}
The answer is $(1 + \sqrt{3})/2$.  We write $|c-d\sqrt{3}|$ as
$|(n-d) - d\sqrt{3}|$; I claim that the minimum over all $d, 0 \leq d \leq n$, occurs when $d = e = \floor{n/(1+\sqrt{3})}$ or when $d = f = e+1 = \floor{n/(1+\sqrt{3})} + 1$.  To see this, note that $(n-e) - e \sqrt{3} > 0$ and if $e'<e$, then $(n-e') - e' \sqrt{3} > (n-e) - e \sqrt{3}$, and similarly for $f'>f$.  Now let $r = n/(1+\sqrt{3}) - \floor{n/(1+\sqrt{3})}$ and note that $|(n-e) - e \sqrt{3}| = r (1+\sqrt{3})$ and $|(n-f) - f \sqrt{3}| = (1-r) (1+\sqrt{3})$.  Clearly
one of these will be $\leq (1+\sqrt{3})/2$.  To see that $(1+\leq{3})/2$
cannot be lowered, note that since $1+\sqrt{3}$ is irrational, $r$ is
uniformly distributed $\mod{1}$.

Q.E.D.

\notes {Notes:}
We do not really need the result that $x$ irrational $\Rightarrow x n - \floor{x n}$ u. d. $\mod{1}$, it would suffice to show that $x$
irrational $\Rightarrow x n - \floor{x n}$ is dense in $(0,1)$.  But
this is obvious, since if $x$ is irrational there exists arbitrarily
large $q$ such that there exists $p$ with $(p,q) = 1$ such that $p/q < x < (p+1)/q$.  The nifty thing about the u. d. result is that it answers
the question:  what number $x$ should we choose such that the density
of $\{ n : r(n) < x \}$ equals $t, 0 < t < 1$?  The u. d. result implies
that the answer is $t (1+\sqrt{3})/2$.  The u. d. result also provides
the key to the question:  what is the average value of $r(n)$?  The
answer is $(1+\sqrt{3})/4$.

\problem {B-4:}
Prove that if $\sum_{n=1}^{\infty} a(n)$ is a convergent series of
positive real numbers, then so is $\sum_{n=1}^{\infty} {(a(n))}^{n/(n+1)}$.

\solution {Solution:}
Note that the subseries of terms ${a(n)}^{n\over{n+1}}$ with
${a(n)}^{1\over{n+1}} \leq {1\over 2}$ converges since then
${a(n)}^{n\over{n+1}}$ is dominated by $1/2^{n}$, the subseries of
terms ${a(n)}^{n\over{n+1}}$ with ${a(n)}^{1\over{n+1}} > {1\over 2}$
converges since then ${a(n)}^{n\over{n+1}}$ is dominated by $2 a(n)$,
hence $\sum_{n=1}^{\infty} {a(n)}^{n\over{n+1}}$ converges.

Q.E.D.

\problem {B-5:}
For positive integers $n$, let $M(n)$ be the $2n + 1$ by $2n + 1$
skew-symmetric matrix for which each entry in the first $n$ subdiagonals
below the main diagonal is $1$ and each of the remaining entries below
the main diagonal is $-1$.  Find, with proof, the rank of $M(n)$.
(According to the definition the rank of a matrix is the largest $k$
such that there is a $k \times k$ submatrix with non-zero determinant.)

One may note that \break\hfill
$M(1) = \left( \matrix{0&-1&1 \cr 1&0&-1 \cr -1&1&0} \right)$ and $M(2) = \left( \matrix{0&-1&-1&1&1 \cr 1&0&-1&-1&1 \cr 1&1&0&-1&-1 \cr -1&1&1&0&-1 \cr -1&-1&1&1&0} \right)$.

\solution {Solution 1:}
Since $M(n)$ is skew-symmetric, $M(n)$ is singular for all $n$, hence
the rank can be at most $2n$.  To see that this is indeed the answer,
consider the submatrix $M_{i}(n)$ obtained by deleting row $i$ and column
$i$ from $M(n)$.  From the definition of the determinant we have that
$\det(M_{i}(n)) = \sum {(-1)}^{\delta(k)} a_{1 k(1)} \cdots a_{(2n) k(2n)}$,
where $k$ is member of $S_{2n}$ (the group of permutations on
$\{1,\ldots,2n\}$) and $\delta(k)$ is $0$ if $k$ is an even permutation or
$1$ if $k$ is an odd permutation.  Now note that ${(-1)}^{\delta(k)} a_{1 k(1)} \cdots a_{(2n) k(2n)}$ equals either $0$ or $\pm 1$, and is
non-zero iff $k(i) \neq i$ for all $i$, i.e. iff $k$ has no fixed points.
If we can now show that the set of all elements $k$ of $S_{2n}$, with
$k(i) \neq i$ for all $i$, has odd order, we win since this would imply that
$\det(M_{i}(n))$ is odd $\Rightarrow \det(M_{i}) \neq 0$.  To show this,
let $f(n)$ equal the set of all elements $k$ of $S_n$ with $k(i) \neq i$ for all $i$.  We have that $f(1) = 0, f(2) = 1$ and we see that
$f(n) = (n-1) ( f(n-1) + f(n-2) )$ by considering the possible values of
$f(1)$ and whether or not $f(f(1)) = 1$; an easy induction now establishes
that $f(2n)$ is odd for all $n$.

Q.E.D.

\notes {Notes:}
In fact, it is a well-known result that $f(n) = n! ( 1/2! - 1/3! + \cdots + {(-1)}^{n}/n! )$.

\solution {Solution 2:}
As before, since $M(n)$ is skew-symmetric $M(n)$ is singular for all $n$
and hence can have rank at most $2n$.  To see that this is the rank,
let $M_{i}(n)$ be the submatrix obtained by deleting row $i$ and column
$i$ from $M(n)$.  We finish by noting that ${M_{i}(n)}^{2} \equiv I_{2n} \Mod{2}$, hence $M_{i}(n)$ is nonsingular.

Q.E.D.

\problem {B-6:}
Prove that there exist an infinite number of ordered pairs $(a,b)$ of
integers such that for every positive integer $t$ the number $at + b$
is a triangular number if and only if $t$ is a triangular number.
(The triangular numbers are the $t(n) = n(n + 1)/2$ with $n$ in
$\{ 0,1,2,\ldots \}$ ).

\solution {Solution:}
Call a pair of integers $(a,b)$ a {\it triangular pair} if $at + b$
is a triangular number iff $t$ is a triangular number.  I claim that
$(9,1)$ is a triangular pair.  Note that $9 (n(n+1)/2) + 1 = (3n+1)(3n+2)/2$ hence $9t+1$ is triangular if $t$ is.  For the other
direction, note that if $9t+1 = n(n+1)/2 \Rightarrow n = 3k+1$
hence $9t+1 = n(n+1)/2 = 9(k(k+1)/2)+1 \Rightarrow t = k(k+1)/2$,
therefore $t$ is triangular.  Now note that if $(a,b)$ is a triangular
pair then so is $(a^{2},(a+1)b)$, hence we can generate an infinite
number of triangular pairs starting with $(9,1)$.

Q.E.D.

\notes {Notes:}
The following is a proof of necessary and sufficient conditions for $(a,b)$
to be a triangular pair.

I claim that $(a,b)$ is a triangular pair iff for some odd integer $o$
we have that $a = o^{2}, b = (o^{2}-1)/8$.  I will first prove the
direction $\Leftarrow$.  Assume we have $a = o^{2}, b = (o^{2}-1)/8$.
If $t = n(n+1)/2$ is any triangular number, then the identity $o^{2} n(n+1)/2 + (o^{2}-1)/8 = (on + (o-1)/2) (on + (o+1)/2)/2$ shows that
$at + b$ is also a triangular number.  On the other hand if $o^{2} t + (o^{2}-1)/8 = n(n+1)/2$, the above identity implies we win if we can show
that $( n - (o-1)/2 )/o$ is an integer, but this is true since $o^{2} t + (o^{2}-1)/8 \equiv n(n+1)/2 \Mod{o^{2}} \Rightarrow 4n^{2} + 4n \equiv -1 \Mod{o^{2}} \Rightarrow {(2n + 1)}^{2} \equiv 0 \Mod{o^{2}} \Rightarrow 2n + 1 \equiv 0 \Mod{o} \Rightarrow n \equiv (o-1)/2 \Mod{o}$.  For the direction
$\Rightarrow$ assume that $(a,b)$ and $(a,c), c\ge b$, are both triangular
pairs; to see that $b = c$ notice that if $at + b$ is triangular for all
triangular numbers $t$, then we can choose $t$ so large that if $c>b$ then
$at + c$ falls between two consecutive triangular numbers; contradiction hence
$b = c$.  Now assume that $(a,c)$ and $(b,c)$ are both triangular pairs;
I claim that $a = b$.  But this is clear since if $(a,c)$ and $(b,c)$
are triangular pairs $\Rightarrow (ab,bc+c)$ and $(ab,ac+c)$ are
triangular pairs $\Rightarrow bc+c = ac+c$ by the above reasoning
$\Rightarrow bc=ac \Rightarrow$ either $a=b$ or $c=0 \Rightarrow a=b$
since $c=0 \Rightarrow a=b=1$.  For a proof of this last assertion,
assume $(a,0), a>1$, is a triangular pair; to see that this gives a
contradiction note that if $(a,0)$ is a triangular pair $\Rightarrow (a^{2},0)$ is also triangular pair, but this is impossible since then
we must have that $a(a^{3}+1)/2$ is triangular (since $a^{2} a (a^{3}+1)/2$
is triangular) but $(a^{2}-1)a^{2}/2 < a(a^{3}+1)/2 < a^{2}(a^{2}+1)/2$
(if $a>1$).  We are now done, since if $(a,b)$ is a triangular pair
$\Rightarrow a 0 + b = n(n+1)/2$ for some $n\geq 0 \Rightarrow b = ({(2n+1)}^{2} - 1)/8$.

Q.E.D.

\bye

==> competition/tests/math/putnam/putnam.1990.p <==
Problem A-1
How many primes among the positive integers, written as usual in base
10, are such that their digits are alternating 1's and 0's, beginning
and ending with 1?

Problem A-2
Evaluate \int_0^a \int_0^b \exp(\max{b^2 x^2, a^2 y^2}) dy dx where a and
b are positive.

Problem A-3
Prove that if 11z^10 + 10iz^9 + 10iz - 11 = 0, then |z| = 1. (Here z is
a complex number and i^2 = -1.)

Problem A-4
If \alpha is an irrational number, 0 < \alpha < 1, is there a finite
game with an honest coin such that the probability of one player winning
the game is \alpha? (An honest coin is one for which the probability of
heads and the probability of tails are both 1/2. A game is finite if
with probability 1 it must end in a finite number of moves.)

Problem A-5
Let m be a positive integer and let G be a regular (2m + 1)-gon
inscribed in the unit circle. Show that there is a positive constant A,
independent of m, with the following property. For any point p inside G
there are two distinct vertices v_1 and v_2 of G such that
1     A
| |p - v_1| - |p - v_2| | < --- - ---.
m    m^3
Here |s - t| denotes the distance between the points s and t.

Problem A-6
Let \alpha = 1 + a_1 x + a_2 x^2 + ... be a formal power series with
coefficients in the field of two elements. Let
/ 1 if every block of zeros in the binary expansion of n
/    has an even number of zeros in the block,
a_n = {
\ 0 otherwise.
(For example, a_36 = 1 because 36 = 100100_2 and a_20 = 0 because 20 =
10100_2.) Prove that \alpha^3 + x \alpha + 1 = 0.

Problem B-1
A dart, thrown at random, hits a square target. Assuming that any two
points of the target of equal area are equally likely to be hit, find
the probability that the point hit is nearer to the center than to any
edge. Express your answer in the form (a \sqrt(b) + c) / d, where a, b,
c, d are integers.

Problem B-2
Let S be a non-empty set with an associative operation that is left and
right cancellative (xy = xz implies y = z, and yx = zx implies y = z).
Assume that for every a in S the set { a^n : n = 1, 2, 3, ... } is
finite. Must S be a group?

Problem B-3
Let f be a function on [0,\infty), differentiable and satisfying
f'(x) = -3 f(x) + 6 f(2x)
for x > 0. Assume that |f(x)| <= \exp(-\sqrt(x)) for x >= 0 (so that
f(x) tends rapidly to 0 as x increases). For n a non-negative integer,
define
\mu_n = \int_0^\infty x^n f(x) dx
(sometimes called the nth moment of f).
a. Express \mu_n in terms of \mu_0.
b. Prove that the sequence { \mu_n 3^n / n! } always converges, and that
the limit is 0 only if \mu_0 = 0.

Problem B-4
Can a countably infinite set have an uncountable collection of non-empty
subsets such that the intersection of any two of them is finite?

Problem B-5
Label the vertices of a trapezoid T (quadrilateral with two parallel
sides) inscribed in the unit circle as A, B, C, D so that AB is parallel
to CD and A, B, C, D are in counterclockwise order. Let s_1, s_2, and d
denote the lengths of the line segments AB, CD, and OE, where E is the
point of intersection of the diagonals of T, and O is the center of the
circle. Determine the least upper bound of (s_1 - s_2) / d over all such
T for which d \ne 0, and describe all cases, if any, in which it is
attained.

Problem B-6
Let (x_1, x_2, ..., x_n) be a point chosen at random from the
n-dimensional region defined by 0 < x_1 < x_2 < ... < x_n < 1.
Let f be a continuous function on [0, 1] with f(1) = 0. Set x_0 = 0 and
x_{n+1} = 1. Show that the expected value of the Riemann sum
\sum_{i = 0}^n (x_{i+1} - x_i) f(x_{i+1})
is \int_0^1 f(t)P(t) dt, where P is a polynomial of degree n,
independent of f, with 0 \le P(t) \le 1 for 0 \le t \le 1.

==> competition/tests/math/putnam/putnam.1990.s <==
Problem A-1
How many primes among the positive integers, written as usual in base
10, are such that their digits are alternating 1's and 0's, beginning
and ending with 1?

Solution:
Exactly one, namely 101. 1 is not prime; 101 is prime. The sum
100^n + 100^{n - 1} + ... + 1 is divisible by 101 if n is odd,
10^n + 10^{n - 1} + ... + 1 if n is even. (To see the second part,
think about 101010101 = 102030201 - 1020100 = 10101^2 - 1010^2.)

Problem A-2
Evaluate \int_0^a \int_0^b \exp(\max{b^2 x^2, a^2 y^2}) dy dx where a and
b are positive.

Solution:
Split the inner integral according to the max{}. The easy term becomes
an integral of t e^{t^2}. The other term becomes an easy term after you

Problem A-3
Prove that if 11z^10 + 10iz^9 + 10iz - 11 = 0, then |z| = 1. (Here z is
a complex number and i^2 = -1.)

Solution:
z is not zero, so divide by z^5 to make things a bit more symmetric.
Now write z = e^{i \theta} and watch the formula dissolve into a simple
trigonometric sum. The 11 sin 5 \theta term dominates the sum when that
sine is at its maximum; by this and similar considerations, just *write
down* enough maxima and minima of the function that it must have ten
real roots for \theta. (This cute solution is due to Melvin Hausner,
an NYU math professor.)

Problem A-4
If \alpha is an irrational number, 0 < \alpha < 1, is there a finite
game with an honest coin such that the probability of one player winning
the game is \alpha? (An honest coin is one for which the probability of
heads and the probability of tails are both 1/2. A game is finite if
with probability 1 it must end in a finite number of moves.)

Solution:
Yes. Write \alpha in binary---there's no ambiguity since it's irrational.
At the nth step (n >= 0), flip the coin. If it comes up heads, go to the
next step. If it comes up tails, you win if the nth bit of \alpha is 1.
Otherwise you lose. The probability of continuing forever is zero. The
probability of winning is \alpha.

This problem could have been better stated. Repeated flips of the coin
must produce independent results. The note that finite'' means only
finite with probability 1'' is hidden inside parentheses, even though
it is crucial to the result. In any case, this problem is not very
original: I know I've seen similar problems many times, and no serious
student of probability can take more than ten minutes on the question.

Problem A-5
Let m be a positive integer and let G be a regular (2m + 1)-gon
inscribed in the unit circle. Show that there is a positive constant A,
independent of m, with the following property. For any point p inside G
there are two distinct vertices v_1 and v_2 of G such that
1     A
| |p - v_1| - |p - v_2| | < --- - ---.
m    m^3
Here |s - t| denotes the distance between the points s and t.

Solution:
Place G at the usual roots of unity. Without loss of generality assume
that p = re^{i\theta} is as close to 1 as to any other vertex; in other
words, assume |\theta| <= 2\pi / (4m + 2) = \pi / (2m + 1). Now take the
distance between p and the two farthest (not closest!) vertices. Make
sure to write | |X| - |Y| | as the ratio of | |X|^2 - |Y|^2 | to |X| + |Y|.
I may have miscalculated, but I get a final result inversely proportional
to (4m + 2)^2, from which the given inequality follows easily with, say,
A = 0.01.

Alternate solution:
The maximum distance between p and a point of G is achieved between two
almost-opposite corners, with a distance squared of double 1 + \cos\theta
for an appropriately small \theta, or something smaller than 2 - A/m^2
for an appropriate A. Now consider the set of distances between p and
the vertices; this set is 2m + 1 values >= 0 and < 2 - A/m^2, so that
there are two values at distance less than 1/m - A/m^3 as desired.

Problem A-6
Let \alpha = 1 + a_1 x + a_2 x^2 + ... be a formal power series with
coefficients in the field of two elements. Let
/ 1 if every block of zeros in the binary expansion of n
/    has an even number of zeros in the block,
a_n = {
\ 0 otherwise.
(For example, a_36 = 1 because 36 = 100100_2 and a_20 = 0 because 20 =
10100_2.) Prove that \alpha^3 + x \alpha + 1 = 0.

Solution:
(Put a_0 = 1, of course.) Observe that a_{4n} = a_n since adding two zeros
on the right end does not affect the defining property; a_{4n + 2} = 0
since the rightmost zero is isolated; and a_{2n + 1} = a_n since adding
a one on the right does not affect the defining property. Now work in the
formal power series ring Z_2[[x]]. For any z in that ring that is a
multiple of x, define f(z) as a_0 + a_1 z + a_2 z^2 + ... . Clearly
f(z) = f(z^4) + z f(z^2) by the relations between a's. Now over Z_2,
(a + b)^2 = a^2 + b^2, so f(z) = f(z)^4 + z f(z)^2. Plug in x for z and
cancel the f(x) to get 1 = \alpha^3 + x \alpha as desired.

Problem B-1
A dart, thrown at random, hits a square target. Assuming that any two
points of the target of equal area are equally likely to be hit, find
the probability that the point hit is nearer to the center than to any
edge. Express your answer in the form (a \sqrt(b) + c) / d, where a, b,
c, d are integers.

Solution:
This is straightforward. The closer-to-the-center region is centered on
a square of side length \sqrt 2 - 1; surrounding the square and meeting
it at its corners are parabolic sections extending out halfway to the
edge. b is 2 and d is 6; have fun.

Problem B-2
Let S be a non-empty set with an associative operation that is left and
right cancellative (xy = xz implies y = z, and yx = zx implies y = z).
Assume that for every a in S the set { a^n : n = 1, 2, 3, ... } is
finite. Must S be a group?

Solution:
Yes. There is a minimal m >= 1 for which a^m = a^n for some n with n > m;
by cancellation, m must be 1. We claim that a^{n-1} is an identity in S.
For ba = ba^n = ba^{n-1}a, so by cancellation b = ba^{n-1}, and similarly
on the other side. Now a has an inverse, a^{n-2}. This problem is not new.

Problem B-3
Let f be a function on [0,\infty), differentiable and satisfying
f'(x) = -3 f(x) + 6 f(2x)
for x > 0. Assume that |f(x)| <= \exp(-\sqrt(x)) for x >= 0 (so that
f(x) tends rapidly to 0 as x increases). For n a non-negative integer,
define
\mu_n = \int_0^\infty x^n f(x) dx
(sometimes called the nth moment of f).
a. Express \mu_n in terms of \mu_0.
b. Prove that the sequence { \mu_n 3^n / n! } always converges, and that
the limit is 0 only if \mu_0 = 0.

Solution:
The only trick here is to integrate \mu_n by parts the wrong way,''
towards a higher power of x. A bit of manipulation gives the formula for
\mu_n as \mu_0 times n! / 3^n times the product of 2^k / (2^k - 1) for
1 <= k <= n. Part b is straightforward; the product converges since the
sum of 1 / (2^k - 1) converges (absolutely---it's positive).

Problem B-4
Can a countably infinite set have an uncountable collection of non-empty
subsets such that the intersection of any two of them is finite?

Solution:
Yes. A common example for this very well-known problem is the set of
rationals embedded in the set of reals. For each real take a Cauchy
sequence converging to that real; those sequences form the subsets of
the countably infinite rationals, and the intersection of any two of
them had better be finite since the reals are Archimedian. Another
example, from p-adics: Consider all binary sequences. With sequence
a_0 a_1 a_2 ... associate the set a_0, a_0 + 2a_1, a_0 + 2a_1 + 4a_2,
etc.; or stick 1 bits in all the odd positions to simplify housekeeping
(most importantly, to make the set infinite). Certainly different
sequences give different sets, and the intersection of two such sets
is finite.

Alternative solution:
Let C be a countable collection of non-empty subsets of A with the property
that any two subsets have finite intersection (from now
on we call this property, countable intersection property). Clearly
such a collection exists. We will show that C is not maximal, that is,
there exists a set which does not belong to C and it intersects finitely
with any set in C. Hence by Zorn's lemma, C can be extended to an
uncountable collection.

Let A1, A2, .... be an enumeration of sets in C. Then by axiom of choice,
pick an element b sub i from each of A sub i - Union {from j=1 to i-1} of
A sub j. It is easy to see that each such set is non-empty. Let B be the
set of all b sub i's. Then clearly B is different from each of the A sub i's
and its intersection with each A sub i is finite.

Yet another alternative solution:
Let the countable set be the lattice points of the plane.  For each t in
[0,pi) let  s(t) be the lattice points in a strip with angle of inclination
t and width greater than 1.  Then the set of these strips is uncountable.
The intersection of any two is bounded, hence finite.

More solutions:
The problem (in effect) asks for an uncountable collection of
sets of natural numbers that are "almost disjoint," i.e., any two
have a finite intersection.  Here are two elementary ways to
get such a collection.

1. For any set A={a, b, c, ...} of primes, let A'={a, ab, abc, ...}.
If A differs from B then A' has only a finite intersection with B'.

2.  For each real number, e.g. x=0.3488012... form the set
S_x={3, 34, 348, 3488, ...}.  Different reals give almost disjoint sets.

Problem B-5
Label the vertices of a trapezoid T (quadrilateral with two parallel
sides) inscribed in the unit circle as A, B, C, D so that AB is parallel
to CD and A, B, C, D are in counterclockwise order. Let s_1, s_2, and d
denote the lengths of the line segments AB, CD, and OE, where E is the
point of intersection of the diagonals of T, and O is the center of the
circle. Determine the least upper bound of (s_1 - s_2) / d over all such
T for which d \ne 0, and describe all cases, if any, in which it is
attained.

Solution:
Center the circle at the origin and rotate the trapezoid so that AB and
CD are horizontal. Assign coordinates to A and D, polar or rectangular
depending on your taste. Now play with s_1 - s_2 / d for a while;
eventually you'll find the simple form, after which maximization is
easy. The answer, if I've calculated right, is 2, achieved when rotating
the trapezoid by 90 degrees around the circle would take one vertex into
another. (A right triangle, with the hypoteneuse the length-two diamater
and d = 1, is a degenerate example.)

Alternative solution:
Let a be the distance from O (the center of the circle) to AB (that is
the side with length s1), and b the distance from O to CD. Clearly,
a = sqrt(1-s1*s1/4) and b = sqrt(1-s2*s2/4). Then with some mathematical
jugglery, one can show that (s1-s2)/d = (s1*s1-s2*s2)/(b*s1-a*s2).
Then differentiating this with respect to s1 and s2 and equating to
0 yields s1*s1+s2*s2=4, and hence s1=2*b and s2=2*a. The value of (s1-s2)/d
for these values is then 2. Hence (s1-s1)/d achieves its extremeum when
s1*s1+s2*s2=4 (that this value is actually a maximum is then easily seen),
and the lub is 2.

Problem B-6
Let (x_1, x_2, ..., x_n) be a point chosen at random from the
n-dimensional region defined by 0 < x_1 < x_2 < ... < x_n < 1.
Let f be a continuous function on [0, 1] with f(1) = 0. Set x_0 = 0 and
x_{n+1} = 1. Show that the expected value of the Riemann sum
\sum_{i = 0}^n (x_{i+1} - x_i) f(x_{i+1})
is \int_0^1 f(t)P(t) dt, where P is a polynomial of degree n,
independent of f, with 0 \le P(t) \le 1 for 0 \le t \le 1.

Solution:
Induct right to left. Show that for each k, given x_{k-1}, the
expected value at a point chosen with x_{k-1} < x_k < ... < x_n < 1
is a polynomial of the right type with the right degree. It's pretty
easy once you find the right direction. 0 \le P(t) \le 1 comes for
free: if P(t) is out of range at a point, it is out of range on an
open interval, and setting f to the characteristic function of that

`

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