| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1997.1 | Some Romanian Olympics | EVTSG8::ESANU | Au temps pour moi | Tue Sep 12 1995 08:38 | 47 |
| 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.
|
| 1997.2 | | AUSSIE::GARSON | achtentachtig kacheltjes | Tue Sep 12 1995 18:59 | 8 |
| 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.)
|
| 1997.3 | Notes related to this topic | EVTSG8::ESANU | Au temps pour moi | Wed Sep 13 1995 04:32 | 11 |
| ------------------------------------------------------------------------------------------------------------------------------------
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
|
| 1997.4 | Putnam exams vs. IMO | EVTSG8::ESANU | Au temps pour moi | Wed Sep 13 1995 04:43 | 19 |
| 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.
|
| 1997.5 | Re: .4 | DECADA::YODER | MFY | Wed Sep 13 1995 13:57 | 30 |
| 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.
|
| 1997.6 | some recent data | AUSSIE::GARSON | achtentachtig kacheltjes | Wed Sep 13 1995 18:26 | 129 |
| IMO Year Location Countries Competitors
22 1981 US, Washington D.C.
23 1982 Hungary, Budapest 30 119
24 1983 France, Paris 32
25 1984 Czechoslovakia, Prague 34
26 1985 Finland, Joutsa 38?42? 209
27 1986 Poland, Warsaw 37
28 1987 Cuba, Havana 42
29 1988 Australia, Canberra 49 260
30 1989 W. Germany, Brunswick� 50 291
31 1990 China, Peking ?
32 1991 Sweden, Sigtuna 55
33 1992 Russia, Moscow 56
34 1993 Turkey, Istanbul ?
35 1994 Hong Kong 68
�Braunschweig
I believe that the IMO formally began in 1959 (which means that one year must
have been skipped somewhere along the way).
Apparently the host country pays for the living costs for all competitors
during the competition but each country pays for its own travel costs to and
from the host country.
Placings
1983:
1. West Germany
2. USA
3. Hungary
4. USSR
5. Romania
6. Vietnam
1984:
1. USSR 18. Brazil
2. Bulgaria 19. Greece
3. Romania 20. Canada
4. USA 21. Colombia
5. Hungary 22. Cuba
6. UK 23. Belgium
7. Vietnam 24. Morocco
8. East Germany 25. Sweden
9. West Germany 26. Cyprus
10. Mongolia 27. Spain
11. Poland 28. Algeria
12. France 29. Finland
13. Czechoslovakia 30. Tunisia
14. Yugoslavia 31. Norway
15. Australia 32. Luxemburg
16. Austria 33. Kuwait
17. The Netherlands 34. Italy
1987:
1. Romania
2. West Germany
3. Hungary
1988:
1. USSR
2. China
2. Romania
4. West Germany
5. Vietnam
6. USA
7. Bulgaria
1989:
1. China
2. Romania
3. USSR
4. East Germany
5. USA
1991:
1. USSR
2. China
3. Romania
4. Germany
5. USA
6. Hungary
7. Bulgaria
8. Iran
8. Vietnam
10. India
1992:
1. China
2. USA
3. Romania
4. CIS
5. UK
6. Russia
7. Germany
8. Hungary
8. Japan
10. France
10. Vietnam
1993:
1. China
2. Germany
3. Bulgaria
4. Russia
5. Taiwan
6. Iran
7. USA
8. Hungary
9. Vietnam
10. Czech Republic
11. Romania
12. Slovakia
13. Australia
1994:
1. USA
2. China
3. Russia
4. Bulgaria
5. Hungary
6. Vietnam
7. UK
8. Iran
9. Romania
10. Japan
11. Germany
12. Australia
|
| 1997.7 | | AUSSIE::GARSON | achtentachtig kacheltjes | Fri Sep 15 1995 22:02 | 9 |
| 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
|
| 1997.9 | Is it really this easy? | WIBBIN::NOYCE | EV5 issues 4 instructions per meter | Mon Sep 18 1995 15:28 | 6 |
| 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.
|
| 1997.10 | | AUSSIE::GARSON | achtentachtig kacheltjes | Mon Sep 18 1995 18:57 | 4 |
| re .9
Yes, it was really that easy. As I said, the problems got harder from
there.
|
| 1997.11 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Tue Sep 19 1995 10:42 | 12 |
|
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
|
| 1997.12 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Tue Sep 19 1995 11:15 | 15 |
| 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
|
| 1997.13 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Wed Sep 20 1995 10:38 | 1 |
| ok got it now. thanks for the details
|