Conference rusure::math

Romania had very strong IMO teams, even more so in the 60's and 70's then
now. As a matter of fact, IMO was in great part the creation of Grigore C.
Moisil, great Romanian mathematician who left his mark on the Romanian
school of mathematics, and who was renowned for his wit.

Here is a (very incomplete) list of Romanian IMO prize winners that I know,
in chronological and alphabetic order:

Dan Voiculescu

Marius Radulescu
Sorin Radulescu

Radu Gologan
Dan Timotin

Alexandru Buium
Adrian Ocneanu
Mihai Putinar

Gruia Arsu
Iustin Coanda
Ioana-Anca Draghicescu
Nicolae Mihalache
Adrian Pasculescu

There were plenty of IMO participants and winners among my colleagues and
friends, including my ex-wife. The most stunning was Dan Voiculescu, 3
times in-a-row 1st prize (in Romania 3 was the maximum number of
participations to IMO, due to age restrictions), and he obtained each time
the maximum rate. Afterwards he was on the candidate list for the Fields
medal in the early 80's (but didn't get it...). He's a full professor at
Indiana University, Bloomington, with other great Romanian mathematicians,
like Ciprian Foias and Harry Bercovici. Tom Apostol also came from
Bucharest. Other ex-IMO participants are now mathematical researchers,
associate or full professors. Adrian Ocneanu, Mihai Putinar and Tibi
Constatinescu are full professors in the US (Pennsylvania, California, New
Mexico).

My brother-in-law, Mugur Abulius, was an IMO substitute, but quit
mathematics altogether and dedicated himself to VMS, and now OpenVMS - he
works for Digital Equipment France since 1981.

There is a French saying: "Si jeunesse savait, si vieillesse pouvait...",
which I'd translate by "If youth knew, if old age could...".

Mihai.

re .1
    
>There is a French saying: "Si jeunesse savait, si vieillesse pouvait...",
    
    Hey, I already used that one in note 1974.3. (-:
    
    (Your French spelling is better than mine so I corrected the title of
     the aforementioned note.)

------------------------------------------------------------------------------------------------------------------------------------
                                                         Mathematics at DEC
Created:  3-FEB-1986 11:36                                   1997 topics                                  Updated: 13-SEP-1995 04:26
 Topic  Author               Date         Repl  Title
------------------------------------------------------------------------------------------------------------------------------------
        EVTSG8::ESANU         5-SEP-1995  1992.5  re. .4
        DECADA::YODER         5-SEP-1995  1992.6  
        JOBURG::BUCHANAN      7-SEP-1995  1992.7  takes me back
         CSC32::D_DERAMO      7-SEP-1995  1992.8  ages ago ...
        DECADA::YODER         7-SEP-1995  1992.9  old Putnam exams
        EVTSG8::ESANU         8-SEP-1995  1992.10  Old dreams

The IMO was born almost 50 years ago as a Russian block contest. In the
beginning the only countries who participated were those from Eastern
Europe, namely Bulgaria, Czechoslovakia, Hungary, Poland, Romania, Soviet
Union.

The Russian and the Hungarians teams were constantly very strong, I think
that in average they were stronger than the Romanians.

Since the Americans have joined the IMO, they proved to be very strong. I
saw somewhere the results of last year IMO, and USA performed amazingly
well.

I don't know much about Putnam exams, but they seem to be very close to the
Olympiad system - prolonged in the IMO - and inherited from Communist
Russia.

Putnam / IMO - was this a political issue facing the East / West conflict?

Mihai.

If Putnam vs. IMO was a political issue, it didn't filter down to me; but I
couldn't swear one way or the other.  I believe George Putnam started the exam
because he thought mathematics needed to be encouraged in much the same way that
(say) football is encouraged, with competitions and scholarships.  Of course, he
might have been thinking this within the framework of US-USSR competition.

As I recollect, only colleges in the U.S. or Canada participated in the Putnams;
I don't know whether this was by design or accident.

(Those uninterested in details may skip the rest.)

The Putnam exam was done yearly; it comprised two halves, a morning test and an
afternoon test (on the same day), each 3 hrs. duration, each containing 6
questions.  The first 2 in each set of 6 were generally easier than the others. 
After scoring the tests, the numerical standing of each participant was
announced, except for the top 5.  These became "Putnam fellows," and their exact
ranking was kept secret.  A Putnam fellowship came with a $1000 (U.S.)
scholarship.

In addition, each college or university was allowed to name a team of 3 which
would represent that institution in a team competition.  (The team had to be
named before the exam was taken.)  Team scoring was done by summing the
numerical rankings of the members, e.g. 1+2+7 if the members came in 1st, 2nd,
and 7th.  The team with the lowest sum won (note that announcing the team winner
can theoretically interfere with the desire to keep the top 5 ranks secret).  I
don't remember whether the winning school or its team got any money for winning.

I'm pretty sure that the Putnam hasn't changed form since its inception, but I
once heard a rumor (false, I think) that they had changed so as to have 10
rather than 5 Putnam fellows per year.

    re .*
    
    For the record, here is the very *first* IMO problem i.e. IMO I, Q1. It
    is a little easy by IMO standards and is posted here for historical
    interest.
    
                                      21n+4
    Show that for all n, the fraction ----- cannot be cancelled.
                                      14n+3

Using the fact that  GCD(a,b) = GCD(b,a-b) = GCD(b,a)

	GCD(21n+4,14n+3) = GCD(14n+3,7n+1) = GCD(7n+1,7n+2) = GCD(7n+1,1) = 1

This is Euclid's Algorithm using subtraction instead of division,
as found near the beginning of Knuth Chapter 1.

    re .9
    
    Yes, it was really that easy. As I said, the problems got harder from
    there.

Well, easy for some of you.  I got lost on the first step...

	GCD(21n+4,14n+3) = GCD(14n+3,7n+1)...

I know that GCD is the greatest common divisor but there seems to be alot
of explanation left out of this first step.  It looks like the 14n+3 went
from the right parameter to the left, and the 21n+4 became 7n+1.

What's being done in this step ?

/Eric

        The greatest common divisor function (GCD) follows the
        following rules: for all positive integers n,m with n>m
        
        	GCD(n,m) = GCD(m,n)
        	GCD(n,m) = GCD(n-m,m)
        
        Or just consider any divisor d of both x and y: then d must
        also divide their sum and their difference.  So if d divides
        both 21n+4 and 14n+3 then d must also divide (21n+4)-(14n+3) =
        7n+1. Since d divides both 14n+3 and 7n+1 it also divides
        (14n+3)-(7n+1) = 7n+2.  Since d divides both 7n+2 and 7n+1 it
        also divides their difference, 1.  So any common divisor of
        21n+4 and 14n+3 must divide evenly into 1.
        
        Dan

ok got it now.  thanks for the details