Internet Amateur Mathematics Society
Newsletter 3

IAMS@quack.kfu.com 




>From The Editor

Greetings.  This is IAMS's third newsletter, but for a lot of you,
this is new.  The old newsletters are kept at princeton.edu, organized
by David Wagner (dawagner@phoenix.princeton.edu).  Special thanks to
David.

Our first contest was a success.  Thank you all for supports.  I will
include the contest questions and results in this newsletter, answers
and solutions will be available upon request.




IAMS Mathematics Contest, 1

0. No person has more than 300,000 hairs on his head.  The Capital of
Sikinia has a population of 300,001.  Can you assert that there are
two persons in the city having the same number of hairs on their
heads?

1. Let C1 and C2 be circles of radius 1 that are in the same plane and
tangent to each other.  How many circles of radius 3 are in this plane
and tangent to both C1 and C2?
(A) 2		(B) 4		(C) 5		(D) 6		(E) 8

2. Which of the following sets could NOT be the lengths of the
external diagonals of a right rectangular prism [a "box"]? (An
external diagonal is a diagonal of one of the rectangular faces of the
box.)
(A) {4,5,6}		(B) {4,5,7}		(C) {4,6,7}
(D) {5,6,7}		(E) {5,7,8}

3. If the limit of (sqrt{n^2 + an + 2}-sqrt{n^2 + 2n + 3}) as n
approaches infinity is 3, what is a?

4. One thousand unit cubes are fastened together to form a large cube
with edge length 10 units.  This is painted and then separated into
the original cubs.  How many of these unit cubes have at least one
face painted?

5. If x>y, what is the ordered pair of real numbers (x,y) for which
 64^{2x} + 64^{2y} = 12  and  64^{x+y} = 4sqrt{2} ?

6. If (x+y)^2 = 2 and (1/x + 1/y)^2 = 8, then what
would xy be?

7. A triangle has sides 19, 92 and 93.  Find the area of the circle
inscribed in this triangle.

8. Find the smallest positive integer x such that 1992\times x has exactly
105 factors.

9. Let n be a nature number, prove 
120 | n(n^{2} -1)(n^{2} -5n+26)

10. The curve traced by a fixed point P on the circumference of a
circle as the circle rolls along a line in a plane is called a
cycloid.  Find the parametric equations for a cycloid.  (Parametric
equations of a curve is the set of equations in term of t which
describes the x and the y of the equation of the curve)

11. Find the sum
1/(1*3) + 1/(3*5)+ ... + 1/((2n-1)(2n+1)) + ... + 1/(255*257)

12. The base three representation of x is 12112211122211112222.  The
first digit on the left of the base nine representation of x is what?

13. If a, b, c, d are the solutions of the equation x^4 - bx - 3 = 0,
then an equation whose solutions are (a+b+c)/(d^2), 
(a+b+d)/(c^2), (a+c+d)/(b^2), (b+c+d)/(a^2) is ?

14. There is a famous story which runs as follows: a traveller with no
money but a golden chain with n links comes to an inn.  The innkeeper
agrees to take one link for each day.  Since the innkeeper has a bad
reputation it is not safe to give him advance payment.  If the
traveller opened two links and was able to pay one link per day, what
is the maximum number of links in the chain?

15. Find the smallest natural number with the following property: if
you transfer the left-most digit to the right hand end, the new number
will be 1.5 times the old number.

16. Two checkers are placed on the 1 x n board in figure.  A and B may
move their checkers any number of squares forward or backward.  The
white checker must stay to the left of the black checker.  The winner
is the one who makes the last move blocking his opponent.  Who wins?
--------------------------------------------------------------------
|     |     | white |      |      |      |     | black |     |     |
--------------------------------------------------------------------

17. If you are condemned to die in Sikinia, you are put into Death Row
until the last day of the year.  Then all prisoners from Death Row are
arranged in a circle and numbered 1,2,...,n.  Starting with number 2
every second one is shot until only one remains who is immediately set
free.  How do you find the place number of the sole survivor?

18. (3 points) Describe a dense order on the set of natural numbers.
(A dense order on a set is a special arrangement of elements of the
set such that between every two elements, there's another element.
For example, the counting order of set of real numbers is a dense
order because between every two real numbers, there's at least one
other real number in between.)




Mathematics Challenger

Who said there's no math involved in basketball?  The National
Basketball Association (American basketball league) is having their
annual lottery draft in the near future, and 11 teams will have their
chances at the 1st, 2nd and 3rd picks.  It works like this: The best
team out of the 11 teams gets 1 ping-pong ball, the second best team
gets 2 balls, the third best team gets 3 balls ... and the worst team
gets 11 ping-pong balls.  In the 66 ping-pong balls, 3 of them have a
tag in it.  One of the tag says 1st, one says 2nd and the other one
says 3rd.  The team who gets a tag will end up picking in that place.
Assume that a team will never pick two or more tags together, then
what are the possibilities of each team picking the tag says 1st? How
about 2nd?  and 3rd?




Historic Note
(some of the stuff below are copied from Book 0 Chapter 10 of the
Elements of Mathematics series)

A famous story runs like this: "a boy was sentenced to life. The
jailer, a math wiz told the boy `I have a fixed positive number for
you to guess, it could be an integer, or a fraction.  If you can guess
it right, I will let you go.'  The boy thought about it.  He knew that
if he start from 1, then 2, then 3, he will never get to fractions.
If he go from 0.1, then 0.11, he will never get to 0.2.  There must be
an order that will allow him to guess every number.  But what is it?"

This problem had bothered a lot of mathematicians, but the great
German mathematician Georg Cantor (1845-1918) showed that there is at
least one such order: a order that is not dense (there's no number
between two numbers) on the set of real numbers.  Let us denote the
"Cantorian order" on Q+ (positive real numbers) by "<c".  The
Cantorian order may be described as follows.  Arrange all the
fractions that name positive rational numbers in an array as indicated
below.  All the fractions with the same denominator are in the same
row.  Within each row the fractions are ordered from left to right
according to the natural order of their numerators.  Within each
column the fractions are ordered from top to bottom according to the
natural order of their denominators.  Obviously, each positive
rational number is named infinitely often in this array.  Circle the
repeated appearance of those number, and we will pretent that they
don't exist.  Let us start from 1/1, then to 2 horizontally, then to
1/2 down and left, then to 1/3, then to 3, and so on (It may help if
you sketch the graph).  In this way we obtain a linear order on Q+
with respect to which 1 is the first positive number, 2 is the second,
1/2 is the third, and so on.  This linear order is the Cantorian order
"<c" on Q+, and it should be clear that <c is not a dense order on Q+.
In fact, 1 has one immediate neighbor, namely 2, with respect to <c.

Historically, the Cantorian order on Q+ was not primarily of interest
to mathematicians because it exemplified a nondense order on Q+.  It
was the deductions that could be made on the basis of the existence of
such an order that stimulated most interest.  Notice that, relative to
the Cantorian order, we can speak of the first positive rational
number, the second positive rational number, and so on.  That is, this
order allows us to assign to each positive rational number a natural
number indicating its position in the Cantorian order.  This
assignment is a one-to-one mapping from Q+ to N because two different
positive rational numbers cannot aviod being in the two different
positions in the Cantorian order.  We conclude that there can be no
more positive rational numbers than there are natural numbers.  If you
think of the apparent abundance of positive rational numbers relative
to the seeming sparseness of natural numbers on the number line, you
can see why this result caught the imagination of Cantor's
contemporaries.  Of course, it also aroused a great deal of disbelief
and unkind criticism.  Moreover, once you have managed to swallow the
fact that the positive rational numbers can be counted with the
natural numbers, there are still surprises in store for you.  Cantor
also showed that there are infinite sets whose elements cannot be
counted with the natural numbers.  Among these sets are those on which
linear orders cannot be represented by linear arrangements because
they have too many elements to fit on a line.  Sadly, I don't have a
lot of information on these sets.  If you do, please post them.

The solution (to number 18 of the contest questions) that I was
looking for had something to do with Cantorian order.  The only one
who summitted that solution was cpf@TC.Cornell.Edu, and here is his
solution:

Make a one-to-one mapping between the set of positive rationals and
the set of positive integers by the traditional triangular method:
Make an infinite square array of the rationals, ordered by numerator
over denominator.  That is, Aij = i/j; i, j >= 1. Then map the
rationals to the integers by starting at the upper left, and moving in
diagonal lines to the upper right.  If a rational in the array is not
in simplest form, (ie numerator and denominator share a common factor
other than one), we have seen it already; skip it and go onto the
next.  Thus: 1/1->1; 2/1->2; 1/2->3; 3/1->4; 2/2 = 1/1 seen already,
skip; 1/3->5; etc.

(If you want zero, map zero to zero; if you want negative integers,
map them to the negative rationals as above.)

(A diagram would be really useful here; but I can't do one.  Here is
the array at least-- imagine sweeping from 1, up and to the right
until one hits the top, then go to the left side and start again.)

1/1  2/1  3/1  4/1  5/1   
1/2  2/2  3/2  4/2  5/2
1/3  2/3  3/3  4/3  5/3
1/4  2/4  3/4  4/4  5/4
1/5  2/5  3/5  4/5  5/5

Now order the set of integers by the order of their corresponding
rationals.  Since the rationals are dense, this order will be dense.




Last Words

Thank you to all who participated in our contest.  I am looking for
problems for future contests.  If you have some really good ones, send
it to me.  I am also looking for articles for future newsletters.  If
you like writing and/or have good articles, send them to me.  

As always, comments and suggestions are welcome.

Enjoy and happy mathing,

Benjie