VERY EASY EASY CHALLENGE!

1) Determine the number from these clues:
- the number is multiple of 5
- it is divisible by 3
- it is less than 200
-its tens digit equals the sum of its two digits

2)Students at schooll dance formed equal teams to play a game. When they formed teams of 3,4,5,or 6 there was always one person left. What is the smallest sumber of people that there could ahve been at the dance?

3) A carpenter can build one doghouse in one day. How many doghouses can 12 carpenters build in 20 days?

 :D  :lol:  :D  :lol:  :D  ;)

1. [hide]well, if it's divisiby by both 3 and 5, then it's divisiby by 15 (am i speling divisiby wrong?) and it has 2 digit. and if x = the tens digit and x+y = x, then y must be 0, meaning the number is 90... right?[/hide]

2. [hide]i'm guessing you use LCM of 3, 4, 5, and 6... which is 60.. then you add 1 and you get 61[/hide]

3. [hide]x*y=z so 12x*20y would equal to 240... unless i multiplied wrong...  :blush: [/hide]

hey Lucy
a little trick for number 1
read the problem again and properly!
 :D
mite help u

hm... when you said "-its tens digit equals the sum of its two digits" does it mean its OTHER two digits or its TWO digit?  other wise, i think my solution is right.... 90 does work for number 1....

rite Lucy
the other two when added must equal the tens digit.
that means the answer has to be bigger than _____
and smaller than 200.
this is a big clue!

ooo, I thought you meant that say ab is a number, and a+b = a... i didn't know that you meant there was 3 digits and when you said the tens digit is equal to the sum of its two digits, i didn't know that you meant the tenth digit wasn't part of them... so you mean that it's gonna say for a integer abc, b=a+c? I thought you mean ab, where a=a+b....  :blush:  :blush:  :blush:

ya Lucy it means that the other two digits other than tens digit's sum equals to its ten digits place!
so when u read that u no that it has to be a 3 digit number!
and than usin the clues u arrive the answer!

HE HE HE :D  :D  :D


find th value of $sin\frac{\pi}{16}(sin^4\frac{\pi}{16}-cos^4\frac{\pi}{16})cos\frac{\pi}{16}$


I am a pupil of one high school in HCMC. My friends say a lot of things about you. You know, I am very interested in examinations in Vietnam. So, would you mind sending me some problems from the exams of Ho Chi Minh city and Vietnam? Thank you very much. :)

I think that you should address a personal message to Namdung. He should be able to help you out more.

Or even ask a fellow Vietnamese user (you can see them by the flag ;) ).

[quote="Valentin Vornicu"]Or even ask a fellow Vietnamese user (you can see them by the flag ;) ).[/quote]
Hi xdragon,
If you want I'll send you some problems from exams of the Vietnam (TST) ;)


Given a positive real number $ x_1$, define a sequence recursively as

$ x_n\equal{}\left(x_{n\minus{}1}\right)^{x_{n\minus{}1}}$

I noticed that, if you define $ x_1\equal{}1\plus{}\frac{1}{a}$, it generally takes remarkably close to $ a$ terms before the sequence "blows up" (reaches somewhere around 2), given of course that $ x_1$ is reasonably close to 1 (1.1 or lower serves a decent estimate of "reasonably close to 1" in this case).

Why does this happen?

Just a warning: this question might be in completely the wrong section.

Look at $ y_n \equal{} \ln x_n$.  Are you familiar with the first-order approximation of the logarithm?

Is it $ x\minus{}1$?

I just used that and got $ x_n\approx\left(x_{n\minus{}1}\right)^{2}\minus{}x_{n\minus{}1}\plus{}1$, at which point I'm stuck.

Probably because that's not the correct recurrence relation I should've found for $ x_n$  :oops:

Maybe I'll look at this again tomorrow when I've caught up a little on ARML sleep deprivation...


It would be very interesting to investigate on the authors of old IMO problems.

[color=red]1959/1 proposed by Poland
1959/2 proposed by Constantin Ionescu-Tiu, Romania
1959/3 proposed by Hungary
1959/4 proposed by Hungary
1959/5 proposed by Cezar Cosnita, Romania
1959/6 proposed by Czechoslovakia

1960/1 proposed by Bulgaria
1960/2 proposed by Hungary
1960/3 proposed by Gheorghe D. Simionescu, Romania
1960/4 proposed by Hungary
1960/5 proposed by Czechoslovakia
1960/6 proposed by Bulgaria
1960/7 proposed by Eastern Germany

1961/1 proposed by Hungary
1961/2 proposed by Poland
1961/3 proposed by Bulgaria
1961/4 proposed by Eastern Germany
1961/5 proposed by Czechoslovakia
1961/6 proposed by Gheorghe D. Simionescu, Romania

1962/1 proposed by Poland
1962/2 proposed by Hungary
1962/3 proposed by Czechoslovakia
1962/4 proposed by Cezar Cosnita, Romania
1962/5 proposed by Bulgaria
1962/6 proposed by Eastern Germany
1962/7 proposed by USSR

1963/1 proposed by Czechoslovakia
1963/2 proposed by USSR
1963/3 proposed by Hungary
1963/4 proposed by USSR
1963/5 proposed by Wolfgang Engel, Eastern Germany
1963/6 proposed by Hungary

1964/1 proposed by Czechoslovakia
1964/2 proposed by Hungary
1964/3 proposed by Yugoslavia
1964/4 proposed by Hungary
1964/5 proposed by Gheorghe D. Simionescu, Romania
1964/6 proposed by Poland

1965/1 proposed by Yugoslavia
1965/2 proposed by Poland
1965/3 proposed by Czechoslovakia
1965/4 proposed by USSR
1965/5 proposed by Gheorghe D. Simionescu, Romania
1965/6 proposed by Poland

1966/1 proposed by USSR
1966/2 proposed by Hungary
1966/3 proposed by Bulgaria
1966/4 proposed by Yugoslavia
1966/5 proposed by Czechoslovakia
1966/6 proposed by Poland

1967/1 proposed by Poland
1967/2 proposed by Czechoslovakia
1967/3 proposed by United Kingdom
1967/4 proposed by Italy
1967/5 proposed by USSR
1967/6 proposed by Hungary

1968/1 proposed by Romania (apparently taken from Romanian problem book)
1968/2 proposed by Czechoslovakia
1968/3 proposed by Bulgaria
1968/4 proposed by Poland
1968/5 proposed by Eastern Germany
1968/6 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom

1969/1 proposed by Wolfgang Engel, Eastern Germany
1969/2 proposed by Hungary
1969/3 proposed by Poland
1969/4 proposed by Netherlands
1969/5 proposed by A. Mekei, Mongolia
1969/6 proposed by USSR

1970/1 proposed by Poland
1970/2 proposed by Dumitru Batinetzu, Romania
1970/3 proposed by Åke Samuelsson, Sweden
1970/4 proposed by Czechoslovakia
1970/5 proposed by Bulgaria
1970/6 proposed by USSR

1971/1 proposed by Hungary
1971/2 proposed by USSR
1971/3 proposed by Poland
1971/4 proposed by Netherlands
1971/5 proposed by Bulgaria
1971/6 proposed by Sweden

1972/1 proposed by USSR
1972/2 proposed by Netherlands
1972/3 proposed by United Kingdom
1972/4 proposed by Netherlands
1972/5 proposed by Bulgaria
1972/6 proposed by United Kingdom

1973/1 proposed by Czechoslovakia
1973/2 proposed by Poland
1973/3 proposed by Åke Samuelsson, Sweden
1973/4 proposed by Yugoslavia
1973/5 proposed by Poland
1973/6 proposed by Åke Samuelsson, Sweden

1974/1 proposed by Murray Klamkin, USA
1974/2 proposed by Matti Lehtinen, Finland
1974/3 proposed by Ciprian Borcea, Romania
1974/4 proposed by Bulgaria
1974/5 proposed by Jan van de Craats, Netherlands
1974/6 proposed by Sweden

1975/1 proposed by Czechoslovakia
1975/2 proposed by United Kingdom
1975/3 proposed by Jan van de Craats, Netherlands
1975/4 proposed by USSR
1975/5 proposed by USSR
1975/6 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom

1976/1 proposed by Czechoslovakia
1976/2 proposed by Matti Lehtinen, Finland
1976/3 proposed by Jan van de Craats, Netherlands
1976/4 proposed by Murray Klamkin, USA
1976/5 proposed by Jan van de Craats, Netherlands
1976/6 proposed by Robert Lyness and David Monk, United Kingdom

1977/1 proposed by Jan van de Craats, Netherlands
1977/2 proposed by Phan Đức Chính, Vietnam
1977/3 proposed by Jan van de Craats, Netherlands
1977/4 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1977/5 proposed by Hermann Frasch, Western Germany
1977/6 proposed by Ivan Prodanov, Bulgaria

1978/1 proposed by Cuba
1978/2 proposed by Murray Klamkin, USA
1978/3 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1978/4 proposed by Murray Klamkin, USA
1978/5 proposed by France
1978/6 proposed by Jan van de Craats, Netherlands

1979/1 proposed by Arthur Engel, Western Germany
1979/2 proposed by Nickolay Khadzhiivanov and Nedjalko Nenov, Bulgaria
1979/3 proposed by N, Vasil'ev & I/ Sharygin, USSR (also Kvant" 12-1979)
1979/4 proposed by Murray Klamkin, USA
1979/5 proposed by Joe Gillis, Israel
1979/6 proposed by Arthur Engel, Western Germany

1981/1 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1981/2 proposed by Arthur Engel, Western Germany
1981/3 proposed by Jan van de Craats, Netherlands
1981/4 proposed by Belgium
1981/5 proposed by USSR
1981/6 proposed by Juha Oikkonen, Finland

1982/1 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1982/2 proposed by Jan van de Craats, Netherlands
1982/3 proposed by Alexander N. Grishkov, USSR
1982/4 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1982/5 proposed by Jan van de Craats, Netherlands
1982/6 proposed by Văn Như Cương, Vietnam

1983/1 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1983/2 proposed by Igor F. Sharygin, USSR
1983/3 proposed by Arthur Engel, Western Germany
1983/4 proposed by Belgium
1983/5 proposed by Marcin Kuczma, Poland
1983/6 proposed by Murray Klamkin, USA

1984/1 proposed by Michael Stoll and Bruno Haible, Western Germany
1984/2 proposed by Aiko Tiggelaar, Netherlands
1984/3 proposed by Ioan Tomescu, Romania
1984/4 proposed by Laurentiu Panaitopol, Romania
1984/5 proposed by Tserendorj Dashdorj, Mongolia
1984/6 proposed by Poland

1985/1 proposed by Frank Budden, United Kingdom
1985/2 proposed by George Szekeres, Australia
1985/3 proposed by Jan van de Craats, Netherlands
1985/4 proposed by R. Gonchigdorj, Mongolia
1985/5 proposed by Igor Sharygin, USSR
1985/6 proposed by Åke Samuelsson, Sweden

1986/1 proposed by Arthur Engel, Western Germany
1986/2 proposed by Gengzhe Chang and Dongxu Qi, China
1986/3 proposed by Elias Wegert, Eastern Germany
1986/4 proposed by Sven Sigurðsson, Iceland
1986/5 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1986/6 proposed by Konrad Engel, Eastern Germany

1987/1 proposed by Horst Sewerin, Western Germany
1987/2 proposed by I.A.Kushnir, USSR
1987/3 proposed by Arthur Engel, Western Germany
1987/4 proposed by Nguyễn Minh Đức, Vietnam
1987/5 proposed by Hans-Dietrich Gronau, East Germany
1987/6 proposed by Vsevolod F. Lev, USSR

1988/1 proposed by Lucien Kieffer, Luxembourg
1988/2 proposed by Czechoslovakia
1988/3 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1988/4 proposed by Finbarr Holland, Ireland
1988/5 proposed by Dimitris Kontogiannis, Greece
1988/6 proposed by Stephan Beck, Western Germany

1989/1 proposed by Philippines
1989/2 proposed by Esther Szekeres, Australia
1989/3 proposed by Hendrikus Gerardus Derksen, Netherlands
1989/4 proposed by Eggert Briem, Iceland
1989/5 proposed by Bernt Lindström, Sweden
1989/6 proposed by Marcin Kuczma, (professor university of Warsaw,) Poland

1990/1 proposed by C.R. Pranesachar, India
1990/2 proposed by Pavol Černek, Czechoslovakia
1990/3 proposed by Laurentin Panaitopol, (University of Bucharest,) Romania
1990/4 proposed by Albert Erkip, Turkey
1990/5 proposed by Hagen von Eitzen, Germany
1990/6 proposed by Hendrikus Gerardus Derksen, Netherlands

1991/1 proposed by Arkadii Skopenkov, (Moscow State University,) USSR
1991/2 proposed by Laurentiu Panaitopol, Romania
1991/3 proposed by Chengzhang Li, China
1991/4 proposed by Laurentiu Panaitopol, USA
1991/5 proposed by Johan Yebbou, France
1991/6 proposed by Hendrikus Gerardus Derksen, Netherlands

1992/1 proposed by Achim Zulauf, New Zealand
1992/2 proposed by Belur Jana Venkatachala, India
1992/3 proposed by Chengzhang Li, China
1992/4 proposed by Johan Yebbou, France
1992/5 proposed by Stefano Mortola, Italy
1992/6 proposed by Tony Gardiner, United Kingdom

1993/1 proposed by Tom Laffey, Ireland
1993/2 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1993/3 proposed by Kerkko Luosto, Finland
1993/4 proposed by Donco Dimovski, F.Y.R. Macedonia
1993/5 proposed by Elias Wegert, Germany
1993/6 proposed by Nicholas de Bruijn, Netherlands

1994/1 proposed by Vincent Lafforgue, France
1994/2 proposed by Hans Lausch, (Monash University,) Australia and Armenia independently
1994/3 proposed by Gabriel Istrate, Romania
1994/4 proposed by Hans Lausch, Australia
1994/5 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1994/6 proposed by Kerkko Luosto, Finland

1995/1 proposed by Borislav Mihailov, Bulgaria
1995/2 proposed by Arkadii Skopenkov (from Moscow State University,) Russia
1995/3 proposed by Jaromir Simsa, Czech Republic
1995/4 proposed by Marcin Kuczma, (professor University of Warsaw,) Poland
1995/5 proposed by Alastair McNaughton, New Zealand
1995/6 proposed by Marcin Kuczma, (professor University of Warsaw,) Poland

1996/1 proposed by Kerkko Luosto, Finland
1996/2 proposed by Jeffrey Grossman, Canada
1996/3 proposed by Mircea Becheanu, Romania
1996/4 proposed by Alexander S. Golovanov, Russia
1996/5 proposed by Nairi M. Sedrakyan, Armenia
1996/6 proposed by Laurent Rosaz, France

1997/1 proposed by Igor Akylich and Igor Voronovich, Belarus
1997/2 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1997/3 proposed by Oleg Bogopolsky, Russia
1997/4 proposed by Ebadollah S. Mahmoodian and Mohammad Mahdian, Iran
1997/5 proposed by Petr Kanovsky, Czech Republic
1997/6 proposed by Giedrius Alkauskas, Lithuania

1998/1 proposed by Charles Leytem, high school Dierkirch, Luxembourg
1998/2 proposed by R.B. Bapat, India
1998/3 proposed by Igor Voronovich, Belarus
1998/4 proposed by David Monk, (formerly University of Edinburgh,) United Kingdom
1998/5 proposed by Vyacheslav Yasinsky, Ukraine
1998/6 proposed by Nikolai Nikolov and Oleg Mushkarov Institut Mathematiques Informatiques Academie des Sciences Sofia Bulgarie

1999/1 proposed by Jan Willemson (University of Tartu,) Estonia
1999/2 proposed by Marcin Kuczma, (professor University of Warsaw,) Poland
1999/3 proposed by Eugenii Barabanov and Igor Voronovich, Belarus
1999/4 proposed by Liang-Ju Chu, Taiwan
1999/5 proposed by Pavel Kozhevnikov, Russia
1999/6 proposed by Tetsuya Ando, Japan

2000/1 proposed by Serge Berlov, (Saint-Petersbourg,) Russia
2000/2 proposed by Titu Andreescu, USA
2000/3 proposed by Sergei Shikh and Igor Voronovich, Belarus
2000/4 proposed by Sándor Dobos, (Professor lyc Fazekas Mihaly, Budapest,) Hungary
2000/5 proposed by Valerii Senderov, (Moscow,) Russia
2000/6 proposed by Lev Emelianov and Tatiana Emelianov, (Kaluga,) Russia

2001/1 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2001/2 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2001/3 proposed by Christian Bey, (Dr. University of Rostock,) Germany
2001/4 proposed by Professor Bill Sands university of Calgary, Canada
2001/5 proposed by Shay Gueron, Israel (see also Toshio Seimiya, Kawasaki Japan: Problem 2501 in Crux Mathematicorum, February 2000)
2001/6 proposed by Alexander Ivanov, Bulgaria

2002/1 proposed by Federico Ardila, (University of Los Andes, Bogota,) Colombia
2002/2 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2002/3 proposed by Laurentin Panaitopol, (University of Bucharest,) Romania
2002/4 proposed by Mihai Manea, Galati Romania
2002/5 proposed by Belur Jana Venkatachala, (Professor University of Indian Institut of Sciences, Bangalore India
2002/6 proposed by Vyacheslav Yasinskiy, (Vinnitsa Pedagogical University,) Vinnitsa Ukraine

2003/1 proposed by Carlos Gustavo Moreira, Brasil
2003/2 proposed by Aleksander Ivanov, Bulgaria
2003/3 proposed by Waldemar Pompe, Poland
2003/4 proposed by Matti Lehtinen, Finland
2003/5 proposed by Finbarr Holland, Ireland
2003/6 proposed by Johan Yebbou (Professor Mathematiques Speciales MP*,) France

2004/1 proposed by Valentin Vornicu and Dinu Serbanescu, Romania
2004/2 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2004/3 proposed by Jan Villemson, Estonia
2004/4 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2004/5 proposed by Waldemar Pompe, Poland
2004/6 proposed by Mohsen Jamaali and Armin Morabbi, Iran

2005/1 proposed by Bogdan Enescu, Romania
2005/2 proposed by Nicholas de Bruijn, Netherlands
2005/3 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2005/4 proposed by Mariusz Skalba, Poland
2005/5 proposed by Waldemar Pompe, Poland
2005/6 proposed by Radu Gologan and Dan Schwarz, Romania

2006/1 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2006/2 proposed by Dušan Djukić, Serbia
2006/3 proposed by Finbarr Holland, Ireland
2006/4 proposed by Zuming Feng, USA
2006/5 proposed by Dan Schwarz, Romania
2006/6 proposed by Dušan Djukić, Serbia

2007/1 proposed by Michael Albert, New Zealand
2007/2 proposed by Charles Leytem, Luxembourg
2007/3 proposed by Vasily Astakhov, Russia
2007/4 proposed by Marek Pechal, Czech Republic
2007/5 proposed by Kevin Buzzard and Edward Crane, United Kingdom
2007/6 proposed by Gerhard Woeginger, Netherlands

2008/1 proposed by Andrey Gavrilyuk, Russia
2008/2 proposed by Walther Janous, Austria
2008/3 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=4906]Kęstutis Česnavičius, Lithuania[/url]
2008/4 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
2008/5 proposed by Bruno Le Floch and Ilia Smilga, France
2008/6 proposed by Vladimir Shmarov, Russia

2009/1 proposed by Ross Atkins, Australia
2009/2 proposed by Sergei Berlov, Russia
2009/3 proposed by Gabriel Carroll, USA
2009/4 proposed by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, South Korea[/url], [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=2639]Peter Vandendriessche, Belgium[/url] and [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=9797]Jan Vonk, Belgium[/url]
2009/5 proposed by Bruno Le Floch, France
2009/6 proposed by Dmitry Khramtsov, Russia

[/color]

[b]Last Edited: 19.09.2009[/b]

Well I have a few knowledge in that field, so let's see:

prof. dr. Laurentiu Panaitopol proposed in 1990, problem no. 3 (the one with n2 | 2n + 1) also another problem, in 2002, also problem no. 3

mihai manea (former IMO team member) is atributed with problem no. 4 in 2002

also one mathlink-er (member of the mathlinks community) has proposed two problems in the same day at some IMO ;)  he doesn't want his name public, so I willl not make it public just now.

well in fact he gave me permission to tell his name: it's hojoo lee. he is the author of IMO 2001 / pb. 1 and pb. 2 (!) and also IMO 2002 pb. 2, so I wonder anyone else than me had a quarter of him IMO problems created by hojoo ? :)

[quote="Valentin Vornicu"]well in fact he gave me permission to tell his name: it's hojoo lee.[/quote]

Aha, thanks, he is (or was) at Hyacinthos, too... What is, btw, his nickname on Mathlinks?

  dg

one of the questio in shortlist 2002 was created by DR naghshineh who is a leader of iranian mathematical olympiad ,here is problem:
  A is a non-empty set of positive integers. There are positive integers b1, b2, ... , bn, c1, c2, ... , cn such that each set biA + ci = {bia + ci : a belongs to A} is a subset of A, and the n sets biA + ci are pairwise disjoint. Prove that 1/b1 + 1/b2 + ... + 1/bn ≤ 1.
this realy hard problem or maybe the hardest problem thathas been ever proposed but the story of creatin of this problem is nice than itself :D
when Dr naghshineh was working on fractal dimention conjecture this problem maybe I clarify it later

Hi,

I think IMO 1986/6 is by Konrad Engel from Germany, IMO 2000/2 is by Titu Andreescu (but who can be sure  :D ) and there are two questions by Hans Lausch onto the IMO 1994 paper (but I don't know which ones yet, I have to consult Mircea Becheanu's IMO book first). Please have a look here: http://www.amt.canberra.edu.au/bhnlausc.html . I am interested into the process of creating problems. Please tell the story of the ISL problem, sam-n ! I am particularly interested in the history of IMO 1996/5 - I know there was account on this problem in one of the recent volumes of the WFNMC journals.

I found in the book some authors of IMO problems

Olympiades Internationales de Mathmatiques
Authors Mohamed Assila
Editor Ellipses

electronic mail of Mohamed Assila
aassilam@yahoo.fr

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

1998 problem 1 by Charles Leytem, high school Dierkirch, Luxembourg

1998 problem 4 by David Monk university Edimbourg

1998 problem 6 by Nikolai Nikolov and Oleg Mushkarov Institut Mathmatiques Informatiques Acadmie des Sciences Sofia Bulgarie

-----------------------------------------------------------------------------
1999 problem 1 by Jan Willemson university of Tartu, Estonia

1999 problem 2 by Professor Marcin Kuczma, university of Warsaw, Poland
from the same author problem 6 IMO 1983, problem 6 IMO 1989, problems 4 and 6 IMO 1995.

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

2000 problem 1 by Sergeï Berlov, Saint-Petersbourg Russia

2000 problem 2 by Titu Andreescu

2000 problem 4 by Sandor Dobos professor lyce Fazekas Mihaly, Budapest, Hungary

2000 problem 5 by Valerii Senderov, Moscou Russia

2000 problem 6 by Lev and Tatiana Emelianovii, Kaluga Russia
---------------------------------------------------------------------------

2001 problems 1 and 2 by Hojoo Lee university Kwangwoon, Seoul Korea

2001 problem 3 by Dr.Christian Bey university of Rostock, Germany

2001 problem 4 by Professor Bill Sands university of Calgary, Canada

2001 problem 5 by Toshio Seimiya, Kawasaki Japan
see cruxmathematicorum problem 2501

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

2002 problem 1 by Federico Ardila, university of Los Andes, Bogota Colombie

2002 problem 3 by Laurentin Panaitopol, university of Bucharest Romania

2002 problem 4 by Mihai Manea, Galati Romania

2002 problem 5 B.J.Venkatachala, professor university of Indian Institut
of Sciences, Bangalore India

2002 problem 6 Vyacheslav Yasinskiy, Vinnitsa Pedagogical University,
Vinnitsa Ukraine

Hi,

I would appreciate identifying more of the IMO problem authors. At least the proposer country could be discovered in the following book:

Mircea Bechanu, International Mathematical Olympiad 1959-2000, The Academic Distribution Center Freeland, Maryland (USA), 2001

There is not too much helpful information regarding this matter in these books I think:

Istvan Reiman, International Mathematical Olympiad 1959-1999, Anthem Press / Wimbledon Publishing Company, London, 2001

Greitzer, IMO 1959-1977, MAA 1978

M S Klamkin, IMO 1978-1985, MAA 1986, ISBN 088386631X

But does anyone know the following books:

Derek Holton, International Mathematical Olympiad Problems, MAA 1990, ISBN 0906588189

Marcin E. Kuczma, International Mathematical Olympiads 1986–1999, MAA 2003, ISBN 0883858118

I don't think the problem (2001/5, a geometry problem) was proposed by Japan. It was sent it by Israel.

Does anyone know better? (For example those who were at IMO 2001 in Washington)

I was there, but as a contestant, so I wouldn't  know :D

imo1995 ,problem 2 (that inequality which has innumerable solutions) was proposed by Nazar Agakhanov,russia

Problem 6, IMO  2003  by Johan Yebbou Professeur Mathmatiques Spciales MP* in France.

I made an update regarding the proposer countries.

According to the official book published by the C.M.S. for 1995 IMO :
Pb n 3 is due to Karel Horak, leader of the Czech Republic team.

Pierre.

[quote="gjw"]=============================================
Some statistics based on
http://www.artofproblemsolving.com/Wiki/index.php/IMO_problems_and_their_authors
=============================================
According to the currently available data, there are 24 authors who
have contributed more than one IMO problem:
[/quote]
=============================================
Some more statistics based on
http://www.artofproblemsolving.com/Wiki/index.php/IMO_problems_and_their_authors
=============================================

According to the currently available data, there are seven problem authors
who have contributed more than one problem to the same IMO:

Åke Samuelsson, Sweden: 2 problems for IMO'1973
Jan van de Craats, Netherlands: 2 problems for IMO'1976, IMO'1977, IMO'1982
Murray Klamkin, USA: 2 problems for IMO'1978
Hans Lausch, Australia: 1.5 problems for IMO'1994
Marcin Kuczma, Poland  : 2 problems for IMO'1995
Hojoo Lee, South Korea: 2 problems for IMO'2001, IMO'2004
Dusan Djukic, Serbia: 2 problems for IMO'2006

[quote="Virgil Nicula"]Know anyone and another authors of IMO's problems ?[/quote]

Does anybody know the author of the IMO Problems? I know the countries but except Charles Leytem from Luxembourg, author of problem 2, none of the authors. I also would like to have author information for IMO Shortlist 2007 (and previous versions). The efforts for ISL 2007 is shown in the  [url=http://www.mathlinks.ro/resources.php?c=1&cid=17&year=2007]Resources' section.[/url] So far we know the following:

ISL Algebra A1: Michael Albert, New Zealand

ISL Combinatorics C5: Radu Gologan and Dan Schwarz, Romania

ISL Geometry G4: Charles Leytem, Luxembourg
ISL Geometry G5: Christopher Bradley, United Kingdom
G7: Davoud Vakili, Iran // Thanks to Omid Hatami

N5: Mohsen Jamali and [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=4683]Nima Ahmadi Pour Anari[/url], Iran // Thanks to Omid Hatami
ISL Number Theory N6: Kevin Buzzard and Edward Crane, United Kingdom

All countries for IMO 2008 are known by know. And we got all authors except for IMO question 5 which was proposed by France.

The author of G8 from IMO SL 07 is Waldemar Pompe (he had already two geometry problems on IMO).

I think the original question which gave rise to IMO 2007 Q5 was proposed by Kevin Buzzard (Imperial College, London).

We made some progress on the list of IMO problem authors. I would like to thank Geoff Smith, Joseph S. Myers and David Monk from UK. Dmitrii Tereshin, Pavel Kozhevnikov and Nazar Agakhanov from Russia. And Michael Stoll from Germany. As of now David Monk is the author who got the most problems used in the IMO competition, followed by Jan van de Craats, Netherlands, with at least 11 problems. But currently many early Netherlands IMO problems have no proposing author assigned.

Here are the German authors of some IMO-problems:
====================================
69-1 Wolfgang Engel, East Germany
77-5 Hermann Frasch, West Germany
79-1 Arthur Engel, West Germany
79-6 Arthur Engel, West Germany
81-2 Arthur Engel, West Germany
83-3 Arthur Engel, West Germany
84-1 Michael Stoll and Bruno Haible, West Germany
86-1 Arthur Engel, West Germany
87-1 Horst Sewerin, West Germany
87-3 Arthur Engel, West Germany
87-5 Hans-Dietrich Gronau, East Germany
88-6 Stephan Beck, West Germany
90-5 Hagen von Eitzen, West Germany
====================================

The current list is also available at:
http://www.artofproblemsolving.com/Wiki/index.php/IMO_Problems_and_Solutions

The years 1978-2008 are almost complete now.
Missing are only the names of the following problem authors:
====================================
1978/1 proposed by Cuba
1978/5 proposed by France
1981/4 proposed by Belgium
1981/5 proposed by USSR
1983/4 proposed by Belgium
1984/6 proposed by Poland
1988/2 proposed by Czechoslovakia
1989/1 proposed by Philippines
====================================

Thanks, this was the list that Dieter Gronau gave me and apparently also forwarded to the IMO Compendium group. He also sent them some German authors of shortlisted problems:

77-5 Hermann Frasch, West Germany [b]// Mircea Becheanu's IMO book: http://www.mathpropress.com/imoBooks/ lists it as East Germany[/b]
79-1 Arthur Engel, West Germany
79-6 Arthur Engel, West Germany [b]// Mircea Becheanu's IMO book: http://www.mathpropress.com/imoBooks/ lists it as East Germany[/b]
81-2 Arthur Engel, West Germany
83-3 Arthur Engel, West Germany
84-1 Arthur Engel, West Germany [b]// -> according Prof. Stoll: 1984/1 proposed by Michael Stoll and Bruno Haible, Western Germany[/b]
86-1 Arthur Engel, West Germany
87-3 Arthur Engel, West Germany
87-1 Horst Sewerin, West Germany
88-6 Stephan Beck, West Germany
90-5 Hagen von Eitzen, West Germany

[b]1989-2007 UK shortlist problems figured out by Joseph S. Myers jsm &a& polyomino &.& org &.& uk:[/b]

SL92-21  Tony Gardiner (IMO problem 6)
SL93-21  Christopher Bradley
SL93-22  David Monk (IMO problem 2)
SL93-23  David Monk
SL94-A3  David Monk (IMO problem 5)
SL94-N7  David Monk
SL96-G1  David Monk
SL96-G3  David Monk
SL96-G7  Christopher Bradley
SL97-8   David Monk (IMO problem 2)
SL97-18  David Monk
SL97-23  Christopher Bradley
SL98-7   David Monk
SL98-14  David Monk (IMO problem 4)
SL98-17  David Monk
SL99-G4  David Monk
SL99-C3  David Monk
SL99-C4  Ben Green
SL99-C6  Mansur Boase
SL00-A2  Imre Leader and Peter Shiu
SL00-A4  David Monk
SL00-G5  David Monk
SL01-G3  Christopher Bradley
SL01-N3  Kevin Buzzard
SL04-G6  Ben Green and Edward Crane
SL07-G5  Christopher Bradley
SL07-N6  Kevin Buzzard and Edward Crane (IMO problem 5)

[quote="gjw"][quote="orl"]
77-5 Hermann Frasch, West Germany [b]// Mircea Becheanu's IMO book: http://www.mathpropress.com/imoBooks/ lists it as East Germany[/b]
79-6 Arthur Engel, West Germany [b]// Mircea Becheanu's IMO book: http://www.mathpropress.com/imoBooks/ lists it as East Germany[/b]
[/quote]
Becheanu's book contains many small mistakes.

All other sources agree that the IMO problems 1977/5 and 1979/6 have been proposed by West Germany.
For instance, the two old German books
*** Internationale Mathematik-Olympiaden II: 1969-1978, Manz-Verlag 1983, ISBN 3-7863-0359-2
*** Internationale Mathematik-Olympiaden III: 1979-1988, Manz-Verlag 1988, ISBN 3-7863-0820-9
attribute them to West Germany.[/quote]

I think I also have copies deep down in my piles by Hermann-Dietrich Hornschuh. :) This guy is publishing lots of school books if you check amazon.de, even for English!  :D

[quote="gjw"]
1989/1 proposed by Philippines
[/quote]

1989/1 was proposed by Jose Marasigan  :D

How about the authors for 2009? I know them for 2009/4, this was a joint work of [list][*][url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=310]Hojoo Lee, Seoul in South Korea[/url]
[*][url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=2639]Peter Vandendriessche, Ghent, Belgium[/url]
[*][url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=9797]Jan Vonk, Ghent, Belgium[/url][/list]
but I'm curious about the other questions.

They are all written in respective problem topics, and I wrote them on Wiki some time ago.
Problem 1 proposed by Ross Atkins, Australia
Problem 2 proposed by Sergei Berlov, Russia
Problem 3 proposed by Gabriel Carroll, USA
Problem 4 proposed by Hojoo Lee, Peter Vandendriessche and Jan Vonk, Belgium
Problem 5 proposed by Bruno Le Floch, France
Problem 6 proposed by Dmitry Khramtsov, Russia

http://www.artofproblemsolving.com/Wiki/index.php/2009_IMO_Problems

Well, I have a super tough math problem I want to propose to the IMO but was told that I have to go through the organization representing the country where I live.

I sent it to the organization and people there have no clue on how to solve it and ignore my problem. What should I do? Any idea on how to work around this?

It goes without saying that you should enclose your own proposed solution.


If $a_i\geq 1$, prove that

$2^{n-1}(a_1a_2\dots a_n+1)\geq (1+a_1)(1+a_2)\ldots (1+a_n)$

AM-GM probably won't help very much...

Induction on n.

Base case: the two sides are equal.

Assume $\displaystyle 2^{n-1}(a_1a_2\dots a_n+1)\geq (1+a_1)(1+a_2)\ldots (1+a_n)$.

Now:

$\displaystyle 2^{n-1}(a_1a_2\dots a_n+1)\geq (1+a_1)(1+a_2)\ldots (1+a_n)$

$\Rightarrow \displaystyle 2^{n-1}(a_1a_2\dots a_n+1)(1+a_{n+1})\geq (1+a_1)(1+a_2)\ldots (1+a_n)(1+a_{n+1})$

from whence it remains to prove $2(a_1a_2\dots a_na_{n+1} + 1) \geq (a_1a_2\dots a_n+1)(1+a_{n+1})$.

Let $r = a_1a_2\dots a_n$, and $y = a_{n+1}$ for convenience.

Then we are required to prove $2(ry + 1) \geq (r + 1)(y+1)$.  This is equivalent to $(r-1)(y-1) \geq 0$.  Clearly, $y = a_{n+1} \geq 1$ as given.  Also, any mean (such as $r$, the nth power of a geometric mean) cannot be less than the minimum of the elements involved, whereby $r^n \geq 1$ so that $r \geq 1$.

and the induction is complete.

Here was my solution.
[hide]
Write $f(a_n)=2^{n-1}(a_1a_2\ldots a_n+1)-(1+a_1)(1+a_2)\ldots (1+a_n)$

We calculate $f'(a_n)=2^{n-1}a_1a_2\ldots a_{n-1}-(1+a_1)(1+a_2)\ldots (1+a_{n-1})$

Clearly, this is $\geq 0$, because

$a_1a_2\ldots a_{n-1}\geq\left(\frac {1+a_1}{2}\right)\ldots\left(\frac {1+a_{n-1}}{2}\right)$

since $a_i\geq \frac {1}{2}+\frac {a_i}{2}\geq 1$

So $f(a_n)$ is strictly increasing in $a_n$.  Thus it's enough to show that the inequality holds for $a_n=1$.  By induction, we see it's enough to show that the inequality holds for all $a_i=1$

But then we just get $2^n=2^n$ which establishes the result.[/hide]


function
vector
calculus
integration
absolute value
real analysis
real analysis theorems

Hello everyone! My question is via the Weierstrass M-test how does one go about proving a [u]conditonally convergent[/u] series uniformly convergent?

For example

$ \forall{x\in[\minus{}1,1]}\quad\arctan(x)\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^nx^{2n\plus{}1}}{2n\plus{}1}$

So we know that the above series must be uniformally convergent on that interval since it is the interval of convergence and a series is uniformally convergent for all subsets of its interval of convergence. But how does one go about validating uniform convergence via the M-test for this series?

We can start by observing that

$ \forall{x}\in[\minus{}1,1]\quad{|x^{2n\plus{}1}|\leq{1}}$

So now here is where I am a little bit confused, we have that

$ {\color{red}|}u_n(x){\color{red}|}\leq{M_n}$

So the only statement I can make including that absolute value is that

$ \forall{x}\in[\minus{}1,1]\quad\left|\frac{(\minus{}1)^nx^{2n\plus{}1}}{2n\plus{}1}\right|\leq\frac{1}{2n\plus{}1}$

The only problem is that the series on the right is not convergent...so what do you do from here?

You cannot say that

$ \forall{x}\in[\minus{}1,1]\quad\left|\frac{(\minus{}1)^nx^{2n\plus{}1}}{2n\plus{}1}\right|\leq\frac{(\minus{}1)^n}{2n\plus{}1}$

Right?

So how do you use the M-test to show uniform convergence...what comparison series do you use?

My second question pertains similarly to

$ \sum_{n\equal{}1}^{\infty}\frac{x^n}{\sqrt{n}}\quad{x\in[\minus{}1,0]}$

We can say that

$ \forall{x}\in[\minus{}1,0]\quad{|x^n|\leq{1}}$

and then say that

$ \forall{x}\in[\minus{}1,0]\quad\left|\frac{x^n}{\sqrt{n}}\right|\leq\frac{1}{\sqrt{n}}$

Once again the problem being that the comparison series is divergent. My book says that this is uniformally convergent...and it must be considering once again that the interval in quetsion is a subset of the interval of covnergence...but how does one show it via the M-test?

Any help appreciated

Alex.

Alas, conditional uniform convergence is something the $ M$-test is completely useless for. So, the formal answer is "nohow". Still, sometimes you can use it in combination with some clever tricks. To give you some flavor of what can be done, here is a simple model
[b]Proposition:[/b]
If there are $ m_k>0$ and $ M_k>0$ such that $ m_k\to 0$, $ \sum M_k<\plus{}\infty$ and $ |f_k(x)|\le m_k$, $ |f_k(x)\plus{}f_{k\plus{}1}(x)|<M_k$ on $ E$, then $ \sum f_k(x)$ converges uniformly on $ E$.

This immediately takes care of both $ \sum\frac {x^n}{\sqrt n}$ on $ (\minus{}1,0]$ and the $ \arctan$ series on $ [\minus{}1,1]$.

In general you may need to combine more than 2 terms, to use Abel transform, or to do something else to reduce the problem to absolute convergence. The point is that some preliminary work has always to be done before you can apply the $ M$-test to a conditionally convergent series.

[color=green] Unnecessary quote removed by moderator [/color]

Hmm, that is very interetsting,but isnt $ M_k$ basically just as hard to find? For example in the second one the obvious choice for $ m_k$ would be $ \frac {1}{\sqrt {k}}\to_{\text{which}}{0}$ But now I need to find a convergent function greater than$ \frac {x^k(\sqrt {k + 1} + \sqrt {k})}{\sqrt {k^2 + k}}$ that converges...which by definition of the direct comparison test is impossible...sorry clearly I am misunderstanding you.  :oops:

Thanks for your time

Actually can't I apply Abel's test here

Since if we define $ \frac {( - 1)^nx^{2n + 1}}{2n + 1} = a_nf_n(x)$ where $ a_n = \frac {( - 1)^n}{2n + 1}$ and $ f_n(x) = x^{2n + 1}$..since $ \sum{a_n}$ converges and $ f_n(x)\in\,\downarrow$ and $ \forall{x}\in[ - 1,1]\quad{ - 1\leq{x^{2n + 1}\leq{1}}}$ this implies that the series is uniformly convergent?

You are forgetting about the sign change or making some other stupid error in algebra; it is $ \left|\frac {x^k}{\sqrt k} \plus{} \frac {x^{k \plus{} 1}}{\sqrt {k \plus{} 1}}\right|\le |x|^{k \plus{} 1} \left|\frac 1{\sqrt k} \minus{} \frac 1{\sqrt {k \plus{} 1}}\right| \plus{} \frac {|x^k(1 \plus{} x)|}{\sqrt k}$ ;). By the way, do not quote the post directly preceding yours entirely; it never makes sense.

Yes, the Abel test is fine also and it is an even better way of doing the problem . My goal was just to answer your question about the possibility to apply $ M$-test, not to give the best solution :)

[quote="fedja"]My goal was just to answer your question about the possibility to apply $ M$-test, not to give the best solution :)[/quote]

(I hope that that quote is ok)

Yeah, sorry I am only in eleventh grade and sometimes I get a little focused on something. This is so strange for me, because I have done all the way up to vector calc and now I am going back and solidifying by analytic background

It is strange for me to have to prove this because I always used to go

"Now switching the integral and the sum assuming uniform convergence"-Me

"How do you know it is uniformally convergent?"-Someone else

"...................................because I said so........................"  :blush:

Anyways thanks for all your help...you've been great!


blogs
inequalities
probability
summer program
Mathcamp
MATHCOUNTS
AMC

The rules of this marathon will be the same as the other marathons:

1. Post a problem/question about anything you like.
2. Someone answers it, and he/she will post a new problem.
3. Someone else answers that, and he/she continues.

Unless a good discussion is in effect, each question will be given 5 days. If it isn't answer by then, someone may choose to continue with a new question. If a person answers correctly but doesn't post another question for 2 days, someone else may post a question in his/her place. How does this sound?

[size=200]Question 1[/size] (easy)

Who was the United States' first president?

George Washington

 $Question$ $2$
Who was the president of the Confederacy?

Civil War? Jefferson Davis? I think that's who it was...
Confirmation need.

Yes it was Jefferson Davis
Should I post a new question or should you?

[quote="aznmonkey1992"]hey can i post a question?
what is minnesota's representative[/quote]

 :huh:

dude...read the instructions at the top ;)

jesse ventura is the only person I know vaguely related to minnesota politics. heck, he may not even be from minnesota. maybe he's from somewhere else.

sry, gone for too long  :P

[size=200]Question 3[/size]

Which country won the second most swimming medals at the 2004 Athens Olympics?

My guess:

Nothing. Maybe a hole. Or at least something along the same lines.

No. Here's a hint, everybody's born with it.

baby teeth? virginity? freckles? nose hair?

yea I'd have put that all by itself without the other random guesses (well after guessing godel's incompleteness theorem you HAVE to guess that first) but then I'd seem like a pervert

What is currently the longest, nonscientific English word?

Sir Edward Teapot (of the United Plates of America)

[quote]who was the 30th president?[/quote]

Of which country? And by president, do you mean an actual president, a figurehead (e.g. king/queen), political leader (e.g. prime minister), or something else?

Um, isn't it my turn to ask a question?

What is the name of the 30th president?

Who was the 10th president of the United States?

John Tyler(I think)

When did jigglykongisFUM16 leave AoPS?


I'm having some trouble understanding effective nuclear charge...
I understand that its a way to measure the energy of an electron by considering its interaction with the average electrical field created by the other electrons and the nucleus. I know that it increases as you go from left to right in a period, but how does it vary as you go down a group? Does shielding play a role here?

There's no set rule for going down a group, since the number of electron shells varies. But for an electron in the same energy level, going down the group, the effective nuclear charge increases since the nuclear charge increases while shielding remains constant.

there is no regular trend even then we can see  decrease in effective nuclear charge(ENC) .
as we go down across a group,one new principle quantum no. come into action.Thus The electron in the outermost orbits i.e. valence electron get a strong repulsive force  from the surrounding electrons in comparison
with attraction from photons.This causes the net force on electrons in outer shells to be significantly smaller in magnitude; therefore, these electrons are not as strongly bonded to the nucleus as electrons closer to the nucleus.
u probably knowing the formula for estimating ENC
i.e.    Z-eff=Z- s

Where Z is the number of protons in the nucleus or atomic no.and s, is the average number of electrons between the nucleus and the electron in question. s, can be found by using quantum chemistry and the Schrödinger equation,
but in d-block the situation is different .We observe d-block contraction.The cause of the d-block contraction is the poor shielding of the nuclear charge by the electrons in the d orbitals. The outer valence electrons are more strongly attracted by the nucleus causing the observed increase in ionization potentials.
to understand it more clearly I'll suggest to see the shape of s,p,d,f,orbitals.


I was trying different things with the equation: $3a^{2}+1=b^{2}$, and I was getting rather frustrated with integers not working out, so I rewrote it as:

$(b+\sqrt{3}a)(b-\sqrt{3}a)=1$.

Then I set: $x=b+\sqrt{3}a$, so that $\frac{1}{x}=b-\sqrt{3}a$.  Using that, I got:

$\frac{b+\sqrt{3}a}{b-\sqrt{3}a}=x^{2}$

Multiplying by the conjugate:

$\frac{3a^{2}+b^{2}+2ab\sqrt{3}}{b^{2}-3a^{2}}=x^{2}$

Using the original equation to substitute back in simplifies the thing to:

$(2b^{2}-1)+(2ab)\sqrt{3}=x^{2}$.

Now here's the weird thing: $(2ab)$ and $(2b^{2}-1)$ are also solutions to the original Pell's equation!  (For example, using 4 and 7 produces 56 and 97, which are also solutions.)

I don't know why this happens.  The explanation for this may be obvious, but it eludes me.

Plugging $(a,b)=(2ab,2b^{2}-1)$ is equivalent to multiplying the original equation by $4b^{2}$.

This is at the heart of the general method for solving Pell's equations.

We're working in a system called $\mathbb{Z}[ \sqrt{3}]$, which is the set of all numbers of the form $x+y \sqrt{3}, x, y \in \mathbb{Z}$.

Given that $z = x+y \sqrt{3}$, we define a few functions on $z$.

[b]Definition:[/b] $\bar{z}= x-y \sqrt{3}$ is called the conjugate of $z$.  (The conjugate is a multiplicative function; that is, $\overline{zw}= \bar{z}\cdot \bar{w}$.  This is not so hard to prove.)

[b]Definition:[/b]  $N(z) = z \cdot \bar{z}= (x+y \sqrt{3})(x-y \sqrt{3}) = x^{2}-3y^{2}$ is called the norm of $z$.  ([i]The norm is also a multiplicative function.  In other words, $N(zw) = N(z)\cdot N(w)$[/i].)

What does this mean?

The equation $b^{2}-3a^{2}= 1$ can be rewritten as $N(b+a \sqrt{3}) = 1$.  Because we know that $N(zw) = N(z) N(w)$, it follows that if $N(z) = 1$ then $N(z^{k}) = 1$.

This means that if we take $z = b+a \sqrt{3}$ to integer powers, we get new numbers $z^{k}= b_{k}+a_{k}\sqrt{3}$ such that $N(z^{k}) = 1$, in other words,

$b_{k}^{2}-3a_{k}^{2}= 1$

So we have generated another solution to the Pell's equation!

It turns out that the powers of a single "primitive" solution generate all solutions.  This technique can be used to solve any equation of the form $x^{2}-Dy^{2}= 1$ for squarefree $D$.


geometry
circumcircle
trapezoid
incenter
geometric transformation
homothety
ratio

Let $ P$  be a random point on the circle circumscribed to the trapezoid $ ABCD$ with $ AB \parallel CD$ whilst $ K, L, M$ and $ N$ respectively being the projections of $ P$  onto $ AC, BD, BC$ and $ AD$ respectively.
Prove that $ K, L, M$ and $ N$ lie on a circle; find the center of this circle.

Best regards,
sunken rock

Let $ Q,R$ be the projections of $ P$ onto $ DC,AB.$ $ (N,L,R),(M,K,R),(N,Q,K)$ are  collinear on the Simson lines with pole $ P$ with respect to the triangles $\triangle ABD,$ $ \triangle ABC$ and $\triangle ACD.$ From the cyclic pentagons $ DLQPN$ and $ CKQPM,$ it follows that $ \overline{PQR}$ is the radical axis of their circumcircles $ \Longrightarrow$ $ RK \cdot RM \equal{} RL \cdot RN$ $ \Longrightarrow$ Quadrilateral $ KLNM$ is cyclic. From the collinearity of $ N,Q,K$ and the cyclic $ LQPN,KQPM$ we have

$ \angle KMP \equal{} \angle RQK \equal{} \angle NQP \equal{} \angle NLP \Longrightarrow LRMP$ is cyclic.

Note that $ PR$ bisects $ \angle LPK,$ because of $ \angle BDC \equal{} \angle LPR \equal{} \angle ACD \equal{} \angle KPR.$ Hence, if $ O$ is the circumcenter of $KLMN,$ we have $ \angle LOK \equal{} 2\angle LMK \equal{} \angle LPK$ $ \Longrightarrow$ $ LKOP$ is cyclic. Thus, circumcenter of $ KLMN$ is the midpoint of the arc $ LPK.$

P.S. However, I cannot see O as an incenter of some triangle, could you clarify what you mean with "find the incenter" ?.

Let $ O_1$ be the projection of $ P$ onto the perpendicular bisector of $ AB$ (which is the same as the perpendicular bisector of $ CD$).  Let $ S \equal{} AD \cap BC$, and let $ T \equal{} AC \cap BD$.

It is not hard to see that $ S$, $ M$, $ N$, $ O_1$, and $ P$ are concyclic.  Furthermore, $ \angle NSO_1 \equal{} \angle MSO_1$, so $ O_1$ lies on the perpendicular bisector of $ MN$.  Similarly, $ T$, $ K$, $ L$, $ O_1$, and $ P$ are concyclic, and $ O_1$ lies on the perpendicular bisector of $ KL$.  Hence, $ O_1$ is the circumcenter of quadrilateral $ KLMN$.

To Luis:
Sorry for the wrong wording, I edited (the title).
For the first step, my proof was exactly as yours, for the 2nd far more complicated than [b]nsato[/b]'s, using a homothecy centered P and ratio 1/2, i.e. O' being on the circle diameter OP, that is, the midpoint of the chord PP' || AB.

[b]Remark:[/b] [i] The projections of P', the other end of the chord parallel to the trapezoid bases onto the same sides and diagonals lie on the same circle.[/i].

Best regards,
sunken rock


How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable and neither are the boxes?

EDIT: I should start reading the problem... :)

ernie, READ THE PROBLEM.

There are 5 possibilities:

5, 0, 0

4, 1, 0

3, 1, 1

3, 2, 0

2, 2, 1

so the answer is 5.

The ways to arrange indistinguishable balls into indistinguishable boxes only depends on the number of balls in the boxes. The ways to do this are (5,0,0), (4,1,0), (3,2,0), (3,1,1)
, (2,2,1). There are \boxed{5}ways.

How do I edit silu's post. If there isn't a way, then here is a message to silu: [hide]Add a dollars sign before \boxed and one after {5} It will make the latex work.[/hide]

[quote=dsmandera]I am just testing this out right now:
[hide=It should say "IT WORKED!!!!1323"]IT WORKED!!! 1323[/hide][/quote]

That would be better placed in Test Forum.

[quote=dsmandera][color=#888]THIS POST HAS BEEN MOVED[/color][/quote]

oops forgot you can put 0 balls in a box


Let $ p \geq 1$ to be a real number, and $ f: [a,b] \rightarrow \mathbb R$ an absolutely continuous function. Show that $ f^p$ is also absolutely continuous.

You must mean $ |f|^p$. Hint: $ x\rightarrow |x|^p$ is Lipschitz on every bounded interval.

thanks!

I found this exercice without the |.|. But I think you are right! we need the |.| to have it true

thank you again


linear algebra
matrix
superior algebra
superior algebra unsolved

1/ Prove that $rank(A)=rank(^{t}A)$ for all matrices $A$ over a division ring $D$ iff $D$ is a field

2/ Propose  $D$ division ring and a matrix $A$ such that $rank(A)\neq rank(^tA)$

What's your definition of rank in the non-commutative case? I'm a bit confused :?


30. Segment $ AB$ has midpoint $ C$, and segment $ BC$ has midpoint $ D$.
Semi-circles are constructed with diameters $ \bar{AB}$ and $ \bar{BC}$ to form the entire region shown.
Segment $ CP$ splits the region into two sections of equal area.
What is the degree measure of $ \angle{ACP}$? Express you answer as a decimal to the nearest tenth.

I think you use circumference to solve this problem, right??...

[quote="CircleSquared"][hide]
The small semicircle is one fourth the large semicircle.
Call the radius of the large semicircle r.
Then the area of the section on the left is $ \frac {5\pi r^2}{16}$.
By the sector area formula, the angle ACP is $ 112.5$ so BCP is $ 67.5$

[/hide][/quote]

What is the sector area formula?

It's A=(1/2)xr^2, where r is the radius and x is the angle measure of the sector.

[quote="i_like_pie"]Problem #93 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=104081&start=520]MC Marathon 2[/url].[/quote]

How did you get your answer so fast?

Can you explain the last part of Sonny's solution how Sonny divides $ \frac {5x^{2}\pi}{4}$ by the total area of the circle which is $ 4x^{2}\pi$ to get$ \dfrac{5}{16}$. Multiply that by 360 to get $ 112.5$

[quote="i_like_pie"]I remembered solving this problem even though it was almost a year ago, so I just had to search for a post by me, look up the date, and find it in the topic (unfortunately there's no way to directly link to a post from search, it is only possible to be directed to the topic).

Sonny divided $ \frac {5x^{2}\pi}{4}$, the total area of the sector, by $ 4x^{2}\pi$, the area of the large circle (not drawn), to get the ratio of the degrees in the sector to the degrees in the circle, $ 360$. This is then multiplied by $ 360$ to get the number of degrees in the sector.[/quote]
I think I get it.. Thank you

[quote="CircleSquared"][hide]
The small semicircle is one fourth the large semicircle.
Call the radius of the large semicircle r.
Then the area of the section on the left is $ \frac{5\pi r^2}{16}$.
By the sector area formula, the angle ACP is $ 112.5$ so BCP is $ 67.5$

[/hide][/quote]
Uh where do you get  $ \frac{5\pi r^2}{16}$
the fraction 5/16? No angles are given

[quote="DonkeyKong"]
Segment $ CP$ splits the region into two sections of equal area.
[/quote]

Still don't see it. I know they're equal in area but don't you need $x$ in the equation $\frac{1}{2}xr^2$?


Prove that there are no three nonzero integers a, b, c satisfying

$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=0$.

Particularly, a solution without using advanced lemmas would be appreciated...

  Darij

/10 years old bumpppppp

I managed by finding the discriminant that we need to prove that for integers b and c , $$(a-c)^4-12a^2c^2$$ couldn't be a square


If

$a-bx-cx=0$

$-ay+b-cy=0$

$-az-bz+c=0$

Find the value of $xy+yz+zx+2xyz$

 :)

[hide="Answer"]For $(a+b)(b+c)(c+a)\neq 0,$ the answer is $1.$[/hide]

Yes that's right.Could you hide your answer? :D

[quote="kunny"]The given system of equation reminds us the first cosine's law.
Thus we are to set $x=\cos A, y=\cos B,\ z=\cos C.$[/quote]

how do you justify that $-1<x,y,z<1$?

(i can see that you can homogenize a,b,c though)

but if this problem were on a contest, then assuming that a=b=c would solve it instantly...

You are right, Altheman. :P

As you say, we should consider in case of $(a+b)(b+c)(c+a)=0,$ right?

[hide="Heres how I did it"]By using crosas multiplication for solving for 1st and 2nd equation:-

$\frac{a}{xy+x}=\frac{b}{xy+y}=\frac{c}{1-xy}$

Substitute for $a,b,c$ the proportionate values in the 3rd equation

$-(xy+x)z-(xy+y)z+(1-xy)=0$

$\implies xy+yz+zx+2xyz=1$[/hide]

I have another tedious method of substituting $x,y,z$ in terms of $a,b,c$ but the above is better.

[quote="kunny"]Humm, it looks like right solution, but why the denominators are not $0$ ?[/quote]

Which step?The first one?Why should they be $0$?


Anyone knows of a good online reference covering transcendental numbers ?
I glanced through some online books on algebraic number theory, but they did not cover transcendental number.

What about that? :
http://bugs.planetmath.org/files/papers/323/e.pdf
http://people.cs.uchicago.edu/~laci/reu03/notes1.pdf
http://www.math.u-psud.fr/~fischler/inde/inde_fischlerpondichery.pdf
http://home.wlu.edu/~dresdeng/papers/lnzd2.pdf
http://www.math.nus.edu.sg/~urops/Projects/transcendental.pdf

:)


A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards.  Two of the suits ($ \heartsuit$ and $ \diamondsuit$, called `hearts' and `diamonds') are red, the other two ($ \spadesuit$ and $ \clubsuit$, called `spades' and `clubs') are black.  The cards in the deck are placed in random order (usually by a process called `shuffling').  What is the probability that the first two cards drawn from the deck are both red?

hello
probability=$ \frac{^{26}C_2}{^{52}C_2}$
thank u

Um, you should calculate that out.

Another solution:  This is $ \frac{26}{52} \times \frac{25}{51} \equal{} \frac{1}{2} \times \frac{25}{51}\equal{}\boxed{\frac{25}{102}}$.


function
logarithms
complex numbers
imaginary numbers

How would the following be expressed in standard form ($x+yi$, where $x$ and $y$ are real numbers and $i$ is the imaginary number, $\sqrt{-1}$)?

$(a+bi)^{c+di}$

($a$, $b$, $c$, and $d$ are real numbers and $i$ is the imaginary number, $\sqrt{-1}$)

I have found a long, tedious way that involves changing the base into $re^{\theta i}$ form.

Does anybody have ANY other ways?  Any suggestions to make the change in expression as quick, easy, and efficient as possible?

rawr

a^i is not a function

don't waste your time with problems like these...

but if you want to do it, obviously you use euler's formula for this

e^{ix}=cos x+i sin x

[quote="Altheman"]a^i is not a function[/quote]

To elaborate slightly: if you take two complex numbers $z$ and $\zeta \neq 0$ and you want to compute $\zeta^{z}$, there are several possibilities: if $z$ is an integer, there is a uniquely defined value for $\zeta^{z}$.  If $z$ is a rational number with denominator $q$, there are $q$ values for the expression $\zeta^{z}$.  If $z$ is any other complex number, there are infinitely many values for the expression $\zeta^{z}$.  This follows from the fact that $e^{2\pi i}= 1$, so that the logarithm becomes multi-valued.  Since $\zeta^{z}= e^{z}\log\zeta$, the multi-valued-ness of the logarithm results in multi-valued-ness for exponentials.

[quote="Altheman"]

don't waste your time with problems like these...[/quote]

There are no problems that inspired me to ask about this... I was just curious if there was a general way of expressing a complex number raised to a complex number in standard form (there may be many numbers that represent the same number in standard form).

So, is there any general formula?  The results may have to be more than one...


I'm NOT a new user, but somehow, my previous account (miyomiyo) got taken away from me.  When I try to log in, it says my name and password don't match.  I checked if I was in uppercase and everything but it still didn't work.  Is anybody else experiencing this?  Can an administrator please help me get my account back?  Any help is much appreciated.

Not a glitch; I've PM'ed you about the details.  Please respond via email.


geometry
3D geometry
algebra
polynomial
number theory solved
number theory

Here is a nice NT problem which will come in hand for the problem I posted at:
http://www.mathlinks.ro/phpBB/viewtopic.php?p=4527#4527

If a, b are integers then ax^2+b is a perfect square for any x natural>0, then one of the numbers (a or b) is a square and the other one is 0.

cheers! :D :D

Try first to show that det(A^2)=0.
The  inverse, that is, if det(A^3+X^3) is a cube for all X in M_2(Z) is more simpler. :)   :)  :)  :)
good luck

More generally :
If P(x) is a polynomial with integer coefficients such that, for each integer k  \geq 0, the number P(k) is the square of some integer, then there exists a polynomial Q with integer coefficients such that P = Q 2 .

Pierre.


[color=darkred]Se considera sirurile $ a_n$ , $ b_n$ unde $ n\in N^*$ , definite recurent inlantuit astfel :

$ \{\begin{array}{c} 1\le a_1\le b_1 \\
 \\
1 + a_{n + 1} = a_n + \frac {b_n}{a_n} \\
 \\
1 + b_{n + 1} = b_n + \frac {a_n}{b_n}\end{array}$ . Sa se arate ca $ \lim_{n\to\infty}a_n = \lim_{n\to\infty}b_n\in\mathcal R\ ,$

adica cele doua siruri sunt convergente si au aceeasi limita.

[b]Enunt echivalent.[/b] Se considera sirurile $ x_n$ , $ y_n$ unde $ n\in N^*$ ,

definite recurent inlantuit astfel : $ \{\begin{array}{c} 0\le x_1\le y_1\ ,\ 1\not\in(x_1,y_1)\\\\
x_{n + 1} =  \frac {x^2_n+y_n}{x_n+1}\ ,\ y_{n + 1} =\frac {y^2_n+x_n}{y_n+1}\end{array}$ .

Sa se arate ca cele doua siruri sunt convergente si au aceeasi limita.[/color]

${{ b_{2} - a_{2} = b_{1} - a_{1} + \frac {a_{1}}{b_1}} - \frac {b_{1}}{a_1}} \geq 0$ ( dupa calcule se ajunge ca expresia din partea dreapta este pozitiva daca si numai daca $ a_{1}b_{1} \geq a_{1} + b_{1}$ )
Deci $ b_{2} \geq a_{2}$ si din cele doua relatii din ipoteza rezulta ca $ a_{2} \geq a_{1}$ si $ b_{1} \geq b_{2}$
Prin inductie se arata ca $ b_{n} \geq a_{n} \geq 1$ si $ a_{n}$ este crescator si $ b_{n}$ este descrescator
;deci $ b_{n}$ este convergent fiind descrescator si marginit inferior.
$ 1 + a_{n + 1} = a_{n} + \frac {b_{n}}{a_{n}} \le b_{n} + b_{n} \le 2b_{1}$ deci si sirul $ a_{n}$ este convergent fiind crescator si marginit superior
Notand cele doua limite cu $ l_{1}$ respectiv $ l_{2}$ obtinem ca $ l_{1} = l_{2}$

Domnule Nicula daca se poate sa veniti cu mai multe probleme de analiza
Va multumesc !


LaTeX
inequalities unsolved
inequalities

1. Let a,b,c be positive real numbers.Prove that:
  SUM {(a+b)/(b+c)\* (a/2a+b+c)} >=3/4

2.Let a,b,c be positive real numbers such that a+b+c=1.Prove that
  SUM {b\sqrt(c)/a*[sqrt(3c)+sqrt(ab)]} >=(3sqrt3)/4

Please send your problem from $LATEX$.O.K. :?:

sorry for my stupid but i can't write LATEX

Ok don't worry,I wonted anderstend your problems,but I don't know what means this[color=red]\*[/color]or this[color=red]a*[/color]

OK
        * means multiply
         / means divide
         \ means nothing

can you help 2 problem tiks :D

[quote="bui_haiha2000"]1. Let a,b,c be positive real numbers.Prove that:
 $\sum\frac{a+b}{b+c}\cdot \frac{a}{2a+b+c} \geq \frac{3}{4}$

2.Let a,b,c be positive real numbers such that $a+b+c=1$.Prove that
 $\sum\frac{b\sqrt{c}}{a}\cdot (\sqrt{3c}+\sqrt{ab}) \geq \frac{3\sqrt{3}}{4}$[/quote]

Is it better now? ;)

1 is true but 2 is false
 2 must be {a*(sqrt(3c)+sqrt(ab))} in the denominator.now is ok

Ok then :) :

[quote="bui_haiha2000"]1. Let a,b,c be positive real numbers.Prove that:
 $\sum\frac{a+b}{b+c}\cdot \frac{a}{2a+b+c} \geq \frac{3}{4}$

2.Let a,b,c be positive real numbers such that $a+b+c=1$.Prove that
 $\sum\frac{b\sqrt{c}}{{a}\cdot (\sqrt{3c}+\sqrt{ab})} \geq \frac{3\sqrt{3}}{4}$[/quote]

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46116]here[/url] for a stronger form of number 1.

Both are [i]Cauchy-Schwarz-able[/i], but is there a smarter way ?

tµtµ can you post the detail solution :D


how to draw a circle or a square here ?
thnx

This picture of a circle and a square
$\begin{picture}(100,100) \put(0,0){\circle{40}}\put(0,0){\framebox(30,30)}\end{picture}$
is produced by
[code]$\begin{picture}(100,100)
\put(0,0){\circle{40}}
\put(0,0){\framebox(30,30)}
\end{picture}$[/code]
[i]\begin{picture}(100,100)[/i] draws a picture box of size 100x100 needed to contain the graphics;
[i]\put(0,0){\circle{40}}[/i] draws a circle of diameter 80 centred at (0,0);
[i]\put(0,0){\framebox(30,30)}[/i] draws a rectangle of size 30x30 with its bottom left at (0,0).

The numbers 100, 40, 30 are measured in points (pt) where 72.72pt = 1 in, so a point is very small.

Section 5.2 of [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]The Not So Short Introduction to LaTeX2e[/url] explains it in much more detail.

You'll find more pictures in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122562]Clever ideas in LaTeX[/url]


algebra
polynomial
real analysis
real analysis unsolved

For every $k \in \mathbb{N}$ prove that $\ds a_{k}=\sum_{j=1}^\infty{\frac{j^{k}}{j!}}\notin\mathbb{Q}.$

For every $k$, $a_{k}$ is an integer multiple of $e$.

Proof: We will prove the stronger fact that for every $k$ $f_{k}(x) = \sum_{j=1}^\infty{\frac{j^{k}}{j!}}x^{j}$ is equal to $p(x)e^{x}$ for some integer polynomial $p(x)$. It is clear that $f_{1}(x) = xe^{x}$.

Assume now that what is said above is true for $f_{n}$, so that $f_{n}(x) = p(x)e^{x}$ for some polynomial $p$ with integer coefficients. By differentiating the sum expression it is clear that $f_{n+1}(x) = xf_{n}'(x) = xp_{2}(x)e^{x}= p_{3}(x)e^{x}$, where $p_{2}(x)e^{x}= f_{n}'(x)$ and it is clear that $p_{2}$ is an integer polynomail. Then $p_{3}(x) = xp_{2}(x)$ is also clearly an integer polynomial.

edit: Of course for each $f_{k}$, $p$ is a different polynomial. I just didn't feel like creating complicated notation.


Hi,

let $ f_0$ and $ f_1$ be two paths in $ \mathbb{R}^n$ with the same endpoints $ x_0$ and $ x_1$. The linear homotopy is defined by $ f_t(s)\equal{}(1\minus{}t)f_0(s)\plus{}tf_1(s)$. But why is the associated map $ F: I\times I\to\mathbb{R}^n$ continuous? Intuitively it's clear, but how can I rigorously prove it?

Another question about homotopies in general: does the continuity of the associated map $ F: I\times I\to X,F(s,t)\equal{}f_t(s)$ imply the continuity of the map $ f_t: I\to X,s\mapsto F(s,t)$?

I'm not sure whether I made my problems clear.

Mark

[quote="laxton"]Hi,

But why is the associated map $ F: I\times I\to\mathbb{R}^n$ continuous? Intuitively it's clear, but how can I rigorously prove it?[/quote]

$ f_0$ and $ f_1$ are continuous maps also if you consider them $ I \times I \rightarrow \mathbb R^n$, and so are  their linear combinations.

[quote]Another question about homotopies in general: does the continuity of the associated map $ F: I\times I\to X,F(s,t) \equal{} f_t(s)$ imply the continuity of the map $ f_t: I\to X,s\mapsto F(s,t)$?[/quote]

You can see $ f_t \equal{} F \circ i_t$ where $ i_t : I \rightarrow I \times I$ is defined by $ s \mapsto (s,t)$ for fixed $ t \in I$. You can see that $ i_t$ is continuous (use definition), since also $ F$ is, also the path $ f_t$ is continuous.

bye

Hi,

I understood the answer to my second question ($ f_t$ is continuous because it equals $ F\circ i_t$).

But I don't understand your answer to my first question: how can you prove with the definition of continuity ("preimages of open sets are open" or the $ \epsilon$-$ \delta$-definition in metric spaces) that the linear homotopy $ F$ is continuous? My difficulty is that $ I\times I$ is equipped with the [b]product topology[/b], isn't it?

Mark

yes, it is equipped with the product topology, which is generated by the product of couples of open sets.

First of all the values of $ F$ are vectors in $ \mathbb R^n$ and it is continuous iff all the components are (definition!)
So we can suppose we have to prove the continuity of
$ G: I \times I \rightarrow \mathbb R$ defined by $ G(s,t) \equal{} (1\minus{}t)g(s)\plus{}th(s)$ where $ g,h: I \mathbb R$ are continuous functions.

Also the function $ L: t \mapsto (1\minus{}t)A\plus{}tB$ is continuous (it's linear!) in the variable t.

Fix $ t_0, s_0 \in I$ and $ \varepsilon >0$. Let $ z_0 \equal{} G(s_0,t_0)$. There exist $ \delta_g, \delta_h >0$ such that $ g( (s_0\minus{}\delta_g, s_0 \plus{} \delta_g) ) \subseteq (z_0\minus{}g(s_0), g(s_0)\plus{}\varepsilon)$ and also $ h( (s_0\minus{}\delta_h, s_0 \plus{} \delta_h) ) \subseteq (h(s_0)\minus{}\varepsilon, h(s_0)\plus{}\varepsilon)$.

There also exist $ delta >0$ such that $ L ( t_0\minus{}\delta, t_0\plus{}\delta) \subseteq (z_0\minus{}\varepsilon, z_0\plus{}\varepsilon)$.

Now let $ \eta < \min(\delta_g,\delta_h)$ and $ V \equal{} (s_0\minus{}\eta, s_0\plus{}\eta) \times (t_0\minus{}\delta, t_0 \plus{} \delta)$ and you can check that  $ G(V) \subseteq (z_0\minus{}\varepsilon, z_0\plus{}\varepsilon)$.

bye

If you are willing to accept that the
multiplication and addition maps $ \mathbb{R} \times \mathbb{R} \to \mathbb{R}$
are continuous, then you can do this without $ \epsilon,\delta$ mangling.

There is a "universal property" that product spaces satisfy, which is
very useful: a product space $ Y \times Z$ is equipped with projections
$ Y \times Z \to Y$, $ Y \times Z \to Z$. Any map $ X \to Y \times Z$
is continuous if and only if the composites $ X \to Y \times Z \to Y$,
$ X \to Y \times Z \to Z$ are both continuous.  This allows us to
easily work with maps into products, and this idea is something
you are probably already somewhat familiar with.

Proposition 1: If $ A,B,X,Y$ are spaces, and $ f : A \to X, g : B \to Y$
are continuous maps, then so is $ (f \times g) : A \times B \to X \times Y$
defined by $ (f \times g)(a,b) \equal{} (f(a), g(b))$.

Proof: We build $ (f \times g)$ from two maps $ A \times B \to X \times Y$:
firstly, the composite map $ A \times B \to A \stackrel{f}{\to} X$, and then
$ A \times B \to B \stackrel{g}{\to} Y$; giving a product map $ A \times B \to
X \times Y$; it is easy to check this is pointwise the same as $ f \times g$.

Proposition 2: The diagonal map $ \Delta : X \to X \times X$ is continuous, where
$ \Delta(x) \equal{} (x,x)$.
Proof: $ \Delta \equal{} id \times id$ where $ id : X \to X$ is the identity map.

Here are some lemmas that are easy to fill in using the above (without $ \epsilon,\delta$s!
Just use that $ \plus{},\times : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ are continuous:

Lemma 1: The map $ \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$ defined by
$ (t,(x_1,\dots,x_n)) \mapsto (tx_1,\dots,tx_n)$ is continuous.

Lemma 2: The addition map $ \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$
is continuous.

The result you want then follows from the following propositions (with a minimal
amount of work):

Proposition 3: If $ f,g : X \to \mathbb{R}^n$ are continuous maps, then so is
$ f\plus{}g$, defined by $ (f\plus{}g) (x) \equal{} f(x) \plus{} g(x)$.

Proof:
We can express $ f\plus{}g$ as a composition of continuous maps:
\[ (f\plus{}g) \equal{} \plus{} \circ (f \times g) \circ \Delta : X \to X \times X \to \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n.\]

Proposition 4: If $ f: I \to \mathbb{R}^n$ is continuous, then so is
$ g : I \times I \to \mathbb{R}^n$ defined by $ g(s,t) \equal{} sf(t)$.

Proof: Let $ i : I \to \mathbb{R}$ be the inclusion map (continuous
since $ I \subset \mathbb{R}$). Then $ g$ is the composition

$ I \times I \stackrel{i \times f}{\to} \mathbb{R} \times \mathbb{R}^n \stackrel{\mathrm{scaling}}{\to} \mathbb{R}^n.$

NOTE: Yes, this is a lot of machinery, but it allows the result to generalize to arbitrary topological vector spaces,
and relies a lot less on metric space notions and more on useful topological / category-theoretic notions.


geometry
trigonometry
inequalities
geometry unsolved

The trigonal ABC has feature is [img]http://www.fileden.com/files/2008/8/1/2029422/tgd.gif[/img]. Prove that $ a\equal{}b\equal{}c$

[quote="L_Euler"][img]http://www.fileden.com/files/2008/8/1/2029422/tgd.gif[/img][/quote]
then $ \sum (a\minus{}b)^2(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)\equal{}0$
then $ a\equal{}b\equal{}c$
:D

[quote="Mr Green"][quote="L_Euler"][img]http://www.fileden.com/files/2008/8/1/2029422/tgd.gif[/img][/quote]
then $ \sum (a \minus{} b)^2(a \plus{} b \plus{} c)(a \plus{} b \minus{} c) \equal{} 0$
then $ a \equal{} b \equal{} c$
:D[/quote]
Can you give me a detail solution, i need an other solution! :)

$ a \ge b \ge c \to \cos A \le \cos B \le \cos C$

Use Trebusep inequality, we have:

$ \begin{array}{l}(a \plus{} b \plus{} c)(\cos A \plus{} \cos B \plus{} \cos C) \ge 3\left( {a\cos A \plus{} b\cos B \plus{} c\cos C} \right) \\\leftrightarrow \cos A \plus{} \cos B \plus{} \cos C \ge \frac{{3\left( {a\cos A \plus{} b\cos B \plus{} c\cos C} \right)}}{{a \plus{} b \plus{} c}} \equal{} \frac{3}{2} \\\end{array}$

But $ \cos A \plus{} \cos B \plus{} \cos C \le \frac{3}{2}$

$ \Longrightarrow 2(a\cos{A} \plus{} b\cos{B} \plus{} c\cos{C}) \equal{} a \plus{} b \plus{} c$

$ \Longrightarrow \sin{2A} \plus{} \sin{2B} \plus{} \sin{2C} \equal{} \sin{A} \plus{} \sin{B} \plus{} \sin{C}$

$ \Longrightarrow 4\sin{A}\sin{B}\sin{C} \equal{} \sin{A} \plus{} \sin{B} \plus{} \sin{C}$

$ \Longrightarrow 4\cdot\frac {abc}{8R^{3}} \equal{} \frac {a \plus{} b \plus{} c}{2R}\Longrightarrow R \equal{} 2r\Longrightarrow a \equal{} b \equal{} c \ (\because OI \equal{} 0)$

guys here is my soln

after simplifying and putting in the values of cosA, cosB, cosC

we get our eqn reduced to following

{(a+b-c)(a-b+c)(-a+b+c)}=abc

which can be written like

8{(s-a)(s-b)(s-c)}=abc

where s={(a+b+c)/2}

but

abc>=8{(s-a)(s-b)(s-c)}              (a result of AM GM)

hence equality holds only when three nos are equal

hence a=b=c

guys i am jaswinder from delhi india

i am in class 10th

and interested in solving hard problems

so plz post a few and i am new to this forum so plz help me

???????????

this problem is damn simple and can be solved using atleast 5 different methods

i have posted the one which i liked the most

so plz reply and tell whether it is nice or not

hope u like it?????????????


Solve the equation below:
$ 7x^2\plus{}7x\equal{}\sqrt{\frac{4x\plus{}9}{28}}$, $ x\in \mathbb{R}$

hello, squaring both sides of your equation and factoring the terms we get
$ (98x^2 \plus{} 112x \plus{} 9)(14x^2 \plus{} 12x \minus{} 1) \equal{} 0$
Only $ x_1 \equal{} \minus{} \frac {4}{7} \minus{}\frac {1}{14}\sqrt {46}$
or $ x_2 \equal{} \minus{} \frac {3}{4} \plus{} \frac {5}{14}\sqrt {2}$
are solutions of our equation.
Sonnhard.

Dr Sonnhard Graubner, how did you get that factorization?

[quote="Farenhajt"][hide="Hint"]Subtract $ {1\over 2}$ from both sides and then prove that the two sides are mutually inverse functions.[/hide][/quote]
Could you explain it more clearly? :|

hello, i got this factorization by making the ansatz
$ 49x^4\plus{}98x^3\plus{}49x^2\minus{}\frac{1}{7}x\minus{}\frac{9}{28}\equal{}(ax^2\plus{}bx\plus{}c)(dx^2\plus{}ex\plus{}f)$
Sonnhard.


I am a macbook user, how do I install latex into my computer. I want to type my notes so I will really appriciate any help.
Thanks,
Akech

You would want to download MacTeX.  See [url=http://www.tug.org/mactex/]this website[/url].


linear algebra
matrix
induction
linear algebra unsolved

I don't know the proof of Birkoff theorm about doubly stochastic  matrices(Every stochastic matrice is convex combination of permutation matrices) Can any one help me?

assume that $A$ is doubly stochastic n by n matrix such that sum of entries of column and rows is $l$ then we prove the statment on induction of number of nonzero entries,note that we have at least $n$ non zero entries if there are exactly $n$ nonzero entries is obvious suppose otherwise .define $S_{i}$ be the set of index like $j$ such that $a_{ij}>0$,now easily you can check that it satisfies Hall condition.so there is a system of distinct representative $j_{i}\in S_{i}$ take first permutation matrix $P_{1}$  as follows $p_{ij}=1$ iff $j=j_{i}$ then easily you can find approprate $\lambda_{1}$ such that $A-\lambda_{1}.P_{1}$ has less non zero entries so by induction we are done.


integration
trigonometry
calculus
function
complex analysis

how to integrate (sinx/x)^2 over the real line
thank you

The nice method to follow is by Fourier Transforms.

Consider $ f(x)\equal{}1 , \forall \ , |x|<a$  , and 0 otherwise.

The Fourier transform is $ \frac {2}{ \sqrt{2 \pi }}$$ \frac {sin(as)}{s}$

Then using the inversion formula for $ |x|<1$ we get  $ \int _{0}^{\infty} \frac {sin(x)}{x}.dx\equal{}\frac{\pi}{2}$

Now by using Parseval's Identity we get,

$ \int_{\minus{}a}^{a} (1)^2.dx \equal{} \frac{2}{\pi}\int_{\minus{}\infty}^{\infty}\frac{sin^2(as)}{s^2}.ds$

Take $ a\equal{}1$ , we get $ \int_{\minus{}\infty}^{\infty}\frac{sin^2(x)}{x^2}.dx\equal{}\pi$

Using real variable techniques:

$ \int_{\minus{}\infty}^{\infty}\left(\frac{\sin x}x\right)^2\,dx\equal{}2\int_0^{\infty}\left(\frac{\sin x}x\right)^2\,dx\equal{}\int_0^{\infty}\frac{1\minus{}\cos 2x}{x^2}\,dx$

Now let $ \frac1{x^2}\equal{}\int_0^{\infty}te^{\minus{}tx}\,dt,$ insert that to make a double integral, and switch the order of summation.

er~but do you know how to integrate it by complex variable method,i mean contour integrals?

[quote="DineshManohar"]The nice method to follow is by Fourier Transforms.

Consider $ f(x) \equal{} 1 , \forall \ , |x| < a$  , and 0 otherwise.

The Fourier transform is $ \frac {2}{\sqrt {2 \pi }}$$ \frac {sin(as)}{s}$

Then using the inversion formula for $ |x| < 1$ we get  $ \int _{0}^{\infty} \frac {sin(x)}{x}.dx \equal{} \frac {\pi}{2}$

Now by using Parseval's Identity we get,

$ \int_{ \minus{} a}^{a} (1)^2.dx \equal{} \frac {2}{\pi}\int_{ \minus{} \infty}^{\infty}\frac {sin^2(as)}{s^2}.ds$

Take $ a \equal{} 1$ , we get $ \int_{ \minus{} \infty}^{\infty}\frac {sin^2(x)}{x^2}.dx \equal{} \pi$[/quote]

Well I am landing in a bad situation. I don't know whether I am right.

Consider $ f(z)\equal{}\frac{e^{2iz}\minus{}1\minus{}2iz}{z^2}$ and choose the contour C , which has $ (\minus{}R,R)$ and $ \gamma_{R}$ , R is chosen big enough to enclose the poles of the integrand.

Now, $ \int_{C}f(z).dz\equal{}0$, then,

0= $ \int_{\minus{}R}^{R} \frac{sin^2(x)}{x^2}.dx\plus{} \int_{\gamma R}\frac{e^{2iz}\minus{}1}{z^2}.dz\minus{}2i\int_{\gamma R}\frac{1}{z}.dz$

Now letting $ R \to \infty$.. I am landing at $ \int_{\minus{}\infty}^{\infty}\frac{sin^2(x)}{x^2}.dx \equal{} 2\pi$  :maybe:. Where am i wrong?? :(

I am  having the following concerns.. Since for the function $ f(z)\equal{}\frac{sin^2(z)}{z^2} , z\equal{}0$ occurs as a pole of order 2. Is it not possible to choose a contour $ (\minus{}\epsilon,\epsilon) ,\epsilon >0$ to avoid the pole 0 ?

The assumption which I made to compute the integral is given as a hint in Joesph Bak "Complex analysis", why is it so to choose?

Actually, $ \frac{\sin^2 z}{z^2}$ has no poles. That's a removable singularity at the origin, and the function you integrated also has a removable singularity, and no poles anywhere on the plane.

The problem is a simple one: on the real axis, $ \frac{e^{2iz}\minus{}1\minus{}2iz}{z^2}\equal{}\frac{\cos 2z\minus{}1}{z^2}\plus{}\frac{i\sin 2z\minus{}2iz}{z^2}\equal{}\minus{}2\frac{\sin^2 z}{z^2}\plus{}(\text{odd})$
As $ R\to\infty$, we have $ 0\equal{}\minus{}2\int_{\infty}^{\infty} \frac{\sin^2 x}{x^2}\,dx\plus{}0\plus{}2\pi$, and the integral we seek is $ \pi$. You just missed the $ 2$, and handled some signs wrong.

your trick is splendid, especially the integral you choose!  thanks! :)

[quote="DineshManohar"]Well I am landing in a bad situation. I don't know whether I am right.

Consider $ f(z) \equal{} \frac {e^{2iz} \minus{} 1 \minus{} 2iz}{z^2}$ and choose the contour C , which has $ ( \minus{} R,R)$ and $ \gamma_{R}$ , R is chosen big enough to enclose the poles of the integrand.

Now, $ \int_{C}f(z).dz \equal{} 0$, then,

0= $ \int_{ \minus{} R}^{R} \frac {sin^2(x)}{x^2}.dx \plus{} \int_{\gamma R}\frac {e^{2iz} \minus{} 1}{z^2}.dz \minus{} 2i\int_{\gamma R}\frac {1}{z}.dz$

Now letting $ R \to \infty$.. I am landing at $ \int_{ \minus{} \infty}^{\infty}\frac {sin^2(x)}{x^2}.dx \equal{} 2\pi$  :maybe:. Where am i wrong?? :(

I am  having the following concerns.. Since for the function $ f(z) \equal{} \frac {sin^2(z)}{z^2} , z \equal{} 0$ occurs as a pole of order 2. Is it not possible to choose a contour $ ( \minus{} \epsilon,\epsilon) ,\epsilon > 0$ to avoid the pole 0 ?

The assumption which I made to compute the integral is given as a hint in Joesph Bak "Complex analysis", why is it so to choose?[/quote]


inequalities
trigonometry
calculus
derivative
trig identities
tangent half angle
calculus computations

$ \cos x < \left( {\frac {{\sin x}}{x}} \right)^3$ for $ |x| \in ]0,\frac{\pi }{2}[$

Don't know which methods to solve it with though. Messy derivative tests..  :blush:

[hide="Brute force"]

Equivalent to, \[ x^3 < \tan x \minus{} \frac{1}{2} \sin 2x\]

Let $ f(x) \equal{} \tan x \minus{} \frac{1}{2} \sin 2x \minus{} x^3$

$ f'(0) \equal{} f''(0) \equal{} 0$ and so it suffices to show $ f'''(x) > 0$

$ \Leftrightarrow f'''(x) \equal{} 2\sec^2 x \plus{} 6 \tan^2 x\cdot \sec^2 x \plus{} 4\cos 2x \minus{} 6 > 0$

Using the tangent half angle formula to convert cos 2x into a combination of tan x, you get that the above inequality is equivalent to,

\[ 2(1\plus{}z^2)^2 \plus{} 6z^2(1\plus{}z^2)^2 \plus{} 4(1\minus{}z^2) \minus{} 6(1\plus{}z^2) \equal{} 6z^6 \plus{} 14z^4 > 0\]

where $ z \equal{} \tan x$ as required.

[/hide]

Why do you want $ f'''(x)>0$ instead of $ f'(x)$?

[quote="mathwizarddude"]Why do you want $ f'''(x) > 0$ instead of $ f'(x)$?[/quote]
Coz that was easier to get probably. (because the angle isn't outside anymore)
But you see
$ f'''(x)>0\text{ and }f''(0)\equal{}0\implies f''(x)>0,f''(x)>0\text{ and }f'(0)\equal{}0\implies f'(x)>0\cdots$
(for all positive $ x$)

Does anyone see the equivalence of the two inequalities?  :maybe:

hello, form $ \cos(x)<\frac{\sin(x)^3}{x^3}$ we get $ x^3<\frac{\sin(x)(1\minus{}\cos(x)^2)}{\cos(x)}\equal{}\frac{\sin(x)}{\cos(x)}\minus{}\frac{1}{2}(2\sin(x)\cos(x))$. Note that $ \sin(2x)\equal{}2\sin(x)\cos(x)$.
Sonnhard.


Find the sum of real roots of the equation
$ (3\minus{}x)^{5}\plus{}(x\minus{}5)^{5}\equal{}\minus{}32$

hello, all solutions are
$ x_1\equal{}5, x_2\equal{}3,x_3\equal{}4\plus{}i\sqrt{3}, x_4\equal{}4\minus{}i\sqrt{3}$ so their sum is $ 16$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, all solutions are
$ x_1 \equal{} 5, x_2 \equal{} 3,x_3 \equal{} 4 \plus{} i\sqrt {3}, x_4 \equal{} 4 \minus{} i\sqrt {3}$ so their sum is $ 16$.
Sonnhard.[/quote]

I don't think your solution to have good sense.

[b]My solution[/b]
Let, $ (3\minus{}x)^{5}\plus{}(x\minus{}5)^{5}\equal{}\minus{}32 \iff a_4x^4\plus{}a_3x^3\plus{}a_2x^2\plus{}a_1x\plus{}a_0\equal{}0$.
So, sum of real roots of the equation is $ \minus{}\frac{a_3}{a_4}\equal{}\minus{}\frac{\binom{5}{3}3^{5\minus{}3}(\minus{}1)^3\plus{}\binom{5}{3}1^3(\minus{}5)^{5\minus{}3}}{\binom{5}{4}3^1(\minus{}1)^{5\minus{}1}\plus{}\binom{5}{4}1^4(\minus{}5)^{5\minus{}4}}\equal{}\minus{}\frac{\minus{}90\plus{}250}{15\minus{}25}\equal{}\minus{}\frac{160}{\minus{}10}\equal{}\boxed{16}$.

[quote][b]My solution[/b]
Let, $ (3 \minus{} x)^{5} \plus{} (x \minus{} 5)^{5} \equal{} \minus{} 32 \iff a_4x^4 \plus{} a_3x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$.
So, sum of real roots of the equation is $ \minus{} \frac {a_3}{a_4} \equal{} \minus{} \frac {\binom{5}{3}3^{5 \minus{} 3}( \minus{} 1)^3 \plus{} \binom{5}{3}1^3( \minus{} 5)^{5 \minus{} 3}}{\binom{5}{4}3^1( \minus{} 1)^{5 \minus{} 1} \plus{} \binom{5}{4}1^4( \minus{} 5)^{5 \minus{} 4}} \equal{} \minus{} \frac { \minus{} 90 \plus{} 250}{15 \minus{} 25} \equal{} \minus{} \frac {160}{ \minus{} 10} \equal{} \boxed{16}$.[/quote]

Note the problem says finding the sum of [b]real[/b] roots.

[quote="timwu"]Note the problem says finding the sum of [b]real[/b] roots.[/quote]Good catch. Funny to see how people rush in to reply without paying much attention to what is asked for :P

so that means, sum of roots = 3 + 5 = 8? ... REAL roots indeed!

[quote="licu-an and Arrange your tan"]so that means, sum of REAL roots = 3 + 5 = 8? ...  [/quote]

I moved the modifier in your question for its importance and
deleted a few words.

The answer is "yes."

Because $ (4 - i\sqrt{3)}$ and $ (4 + i\sqrt{3)}$ are roots that
are not real, they cannot be used here for the sum.  Their
real parts cannot be used for the sum of the real roots,
because taken as units unto themselves, they are prevented
from being summed with the other individual real numbers,
3 and 5.  Hence, it is just 3 + 5 = 8, for the sum of the
real roots.

Thanks for the clarification . . . it pays to read the problem indeed.

I agree with the 8.  Everyone who said 16 really confused me.


AMC USA(J)MO USAMO AIME LaTeX geometry MATHCOUNTS

A silly question, but do you use a pen or a pencil when doing your math work?  :D

i like to use pen, and they're required on USAMO (faxing?)
but for bubbling, AMC, AIME, it has to be pencil...

Pencil, duh! Because you can control the lightness/darkness and have better diagrams. But actually I have to admit pens write better. I plan on using a pen on the USAMO this year (assuming I qualify, which I really should).

I have seen fusions of the two and have grown very fond of them:
The erasible pens are awsome pecause they are easier to read but u can still erase.
 :!: Warning: When using erasible pens be very carefull! One brand uses presurized ink that can erase because it is so condensed, so under no circumstances should u try to break one! I have seen people ruin their clothing with ink doing this  :!:   ;)

PEN!

feels better and helps you organize better

and I heard it's a very bad habit to erase your work even though it is wrong or something

Pencil and erasable pen both suffer from high smudginess.  I've always prefered pencil, but I like pen relatively more than I used to because it is clearer and fades and smudges less.

i prefer pencils, especially the mechanical ones, as they don't smudge that much but (i feel) are better for diagrams and such

I really like pens for general writing, but somehow pencils fit better for math.

Pen all the way except when pencil is absolutely necessary.

Pens all the way!!! They are a lot neater to write with and they last longer too

I've always used pencil, but now I use $\LaTeX$ more than either. It's much better for everyone if my handwriting doesn't appear at all.
Even more editing power than pencil, even clearer than pen- and all you have to do is do it in advance.

pencils will take over the world because they are smart and you can become smart if you use one...

i use pencils because they can erase.

I use mechanical pencils and fine-point markers.

Pen

While solving problems, I try different methods,one of which may be right.

So,I feel it's better not to erase things because you never know you can need them.

:?  I just use whatever's closest to me... I've found myself writing on paper plates with brown whiteboard-markers

In general though, I prefer pens. Erasing is a pain, it's a lot more fun to cross out things you don't like.

For me it's almost even.  I like pens, because for some reason, I can't stand having a dull pencil, even if it's only moderately dull.  On the other hand, you can erase with pencils.  I suppose mechanical pencils (or erasable pens) would be the best for me, but they've never really caught on with me.

A pencil just feels right when I'm doing math, for some reason.

[quote="biffanddoc"]Pencils are the best.
They can erase and pens cant.[/quote]

umm....there are erasable pens, but yeah, they suck

There should be mechanical pencils on this poll!, I think they are really the best for doing math, never gets dull, and ofcourse when making stupid mistakes, erasing is necessary.

Not to mention, when drawing diagrams for geometry problems, pencils are a necessity in order to get the figure just to scale

Always pencil, that's why I forget my pen for spanish tests 60% of the time.

Pencils on math competitions, especially the timed ones, even though I almost never erase work, just answers.

Crossing things out is funner!

^^ Amen

Pencils also smear if you do your homework with them and shove it in your backpack (which I, unfortunately, tend to do way to often)

I've made a complete reversal on this issue. In my MATHCOUNTS years, I always used pencil, usually mechanical. Then between 8th grade and 9th grade, I made a serious adjustment: I went away from mechanical pencils. I've never been one of those people that just gives up on something because they don't understand it. I took apart my mechanical pencils to see what was wrong, put them back together, and they still didn't work. Eventually, enough knowledge about their structure led me to realize the many things that could go wrong. One big thing was the lead breaking, and since small leads are all practically useless and often end up clogging up the slot thingy, I found that mechanical pencils can break down much more easily than other writing implements. They also cost more. So I stopped using them.

Over the last year and a half, I've enjoyed writing in both pen and normal pencil. But for math tests, pen far exceeds normal pencils. I just hate those soft pencils that if you press really hard so you can actually see the mark, they break. But usually, given a good eraser, pencils are my favorite. Especially in things where you don't get a lot of space (and thus must erase), like standardized tests, it's best to use pencil. But in all other circumstances, pen is best. Yes, it doesn't give you as much variety, but I find dependability more important. Plus, you don't have to sharpen. On all of the math tests (except bubbling on AMC), I always use pen. It's so much easier to just write without having to worry about turning the pencil over, erasing, and wiping the eraser stuff off, then realizing that it's all smudged and the background is 50% darker. That's one reason why I only write in pencil if they have good erasers, or with those block erasers nearby. With pen, you don't waste the time and effort, and you just do the same motions over, whereas with erasing, you have to turn the pencil around, use your other hand to hold the paper, basically "scribble" your answer out with an eraser, and use the back of your writing hand to wipe away the junk. The end result is the same, except with pencil, you have a little more room on your piece of paper. Paper's endless, so why bother?

Of course, with proofs, you might have to block out sections with X's, but this doesn't waste as much time as erasing. Suppose you just needed to remove one word from a sentence (like "distinct"). If you erase, it leaves a big, awkward gap. If you cross it out, it looks perfectly normal from a distance and the reader just skips over it all the same. Proofs look like they have holes if the text actually does. :P

So the moral of the story is, USE PENS!!! They don't allow you to draw the Mona Lisa, but WHO CARES?

Technically, crossing things out is possible with pencils too, in case some of you didn't notice...I do it a lot.

I've also made a reversal, but only within the category of "pencil". Up to summer of last year I had a phobia of mechanical pencils. They looked wrong, they felt wrong when I held them, the forever-thin line they wrote in was creepy, ugly, unnatural...I only wrote with regular pencils, though I was practically the only person in my school to do so. Then, since I didn't have a portable pencil sharpener, I had to bring mechanical pencils to summer camp, and I got used to them. Now I normally only use mechanical pencils, but I always bring some regular ones to tests as backup. So far I haven't needed them.

Wow, I don't see how people use regular pencils anymore. I hate them because you grab one, and you can never tell if it's one of the ones with the good soft eraser or the rubbery hard one that stinks. But I never use pen. It bleeds through so I can't see what I wrote on the other side of the page, it smears (I'm a leftie) terribly, you can't erase (unless you get erasables, and those would get ALL OVER my hands), and they mysteriously give out on you. Good mechanical pencils are the way to go. Just don't get the cheap ones or they will break in your backpack, and I would definitely recommend getting the twist erasers (I <3 twist erasers). I always keep a pen handy though, because you know what? You never know when you'll need to write something down and you won't have paper- and you can't use pencil on your hand!

The thing about using pens on math tests: you say paper doesn't run out, but it so does. When I took the AIME this year, I had to raise my hand twice for new sheets, I still ended up writing on the desk, AND I was using a pencil! That wasted SO much time... especially when I had to lean to the side to see what I had written on the shiny desk.

i'd rather use pencil for almost everything, but my english teachers don't seem to agree. :D

I'm left handed... so ink/non-ballpoint pens smudge abundantly. So... mechanical pencils ftw.

pens ftw! unless you're bubbling in for AMC, AIME, etc...


AMC
AMC 10
AIME
geometry
USA(J)MO
USAMO
trigonometry

Really, I think those books are a lot of practice for the AMC. If you want, you can find some old tests on this site. It seems you should study for higher-level competitions to me...

Do a lot of old AMC problems, and if you feel really comfortable with AMC stuff, then move onto more advanced things, as those will become a lot more important.

Do the AMC12.
AMC10 is for wimps.
:P
Meh, I'll probably have to do the two AMC 10's though because my school doesn't provide AMC 12.
Anyways, just do old tests and understand the concepts.

[quote="ZhangPeijin"]Do the AMC12.
AMC10 is for wimps.
:P[/quote]

Excuse me?

[quote="georgiamathdude"]LOL. However, in a way AMC 12 has slight advantage since they decide the floor score.[/quote]

Not if you want to pass AIME. AMC 10 has advantage on AIME.

AMC 12 is easier to pass though. I recommend doing practice AMC 10s + 12s. Aops books are good. What is you current score and your weaknesses, illusiat? Also, I recommend Volume 2, though it is possible to pass using just Volume 1. when you can get about 5 on AIME, take WOOT and AIME problem series classes if you are willing to spend money. Also, for a more advanced Geometry textbook, I recommend Geometry Revisited.

if you are prepared for AIME and USAMO then AMC 10 and 12 will be pieces of cake. just learn the basics before you start jumping into problems. like me, my first experience with trigonometry was on AIME and it stunk.  :(


$(B)$
[hide]$a=\log_{8}225\Leftrightarrow\ 15^2=2^{3a}$ and  $b=\log_{2}15 \Leftrightarrow\ 15=2^{b}\Leftrightarrow\ 15^2=2^{2b}$ So $3a=2b\Leftrightarrow\ a=\frac{2}{3}b$[/hide]

[hide]Changing $a$ to base 2 we get $\frac {\log_2 225}{3}$ the numerator being equal to $2 \log_2 15$ and since $\log_2 15$ is $b$, the answer is $a = \frac {2b}{3}$.[/hide]

[hide="solution"]B, just use a change of base formula [/hide]

[hide]We can solve the problem by writing $a=xb$, and finding the value of $x$.  Using substitution, $\log_8 225=x\log_2 15$.  We can simplify this as $\log_2 15^{\frac{2}{3}}=\log_2 15^{x}$.  From this we can see that $x=\frac{2}{3}$, and $a=\frac{2b}{3}$.  So the answer is [b]B[/b].[/hide]


AMC
AIME
trigonometry
algebra
partial fractions
geometric series
complex numbers

Help would be greatly appreciated here.

My question is: does anyone know any problems involving taking the sigma of products (i.e. sums of products)? Does anyone know a general method of approaching these types of problems?

This question stemmed from an old AMC or AIME question (can't remember which). I know that when I was taking the practice test I was so scared of the question that I didn't even want to tackle it.

One way is to look for a pattern or telescoping sum. Those obviously simplify the problem up a lot.

It depends.

When I see problems involving complicated sums, the first thing I do is watch out for commonly known forms. Identifying geometric, arithmetic, and other common sums is helpful.

After that, I look out for sums that are probably vulnerable to certain methods. Sums that fall to partial fractions are usually obvious. Stuff involving sine and cosine can often be reduced to geometric series using complex numbers and Euler's identity, and so on.

If that doesn't work, it sometimes helps to try to express the sum in terms of itself, or in different ways, and find a closed form in that way (like how the closed forms for arithmetic and geometric series are derived).


geometry
rectangle
circumcircle
geometry unsolved

Let ABC an acute triangle, build , two rectangles out side of ABC, ACKL and BCMN, suppose tha  (AC)(CK)= BC(CM),
prove that the circumcenter of ABC , C and the midpoint of KM are collinear..

Sorry because I have misunderstanded your problem before!
Now here is my solution:
$ \frac{S_{OCK}}{S_{OCM}}\equal{}\frac{CK.sin\angle OCK}{CM.sin\angle OCM}\equal{}\frac{CK.cos\angle OCA}{CM.cos\angle OCB}\equal{}\frac{CK.sin\angle B}{CM.sin\angle A}\equal{}\frac{CK.AC}{CM.BC}\equal{}1$
Then $ OC$ pass through midpoint of $ KM$

$ \triangle ABC$ does not have to be acute. Let OC meet KM at D and let parallel to OC through K meet CM at P.

$ AC \cdot CK \equal{} BC \cdot CM\ \Longrightarrow \frac {CM}{CK} \equal{} \frac {AC}{BC} \equal{} \frac {b}{a}.$

$ \angle CPK \equal{} \angle MCD \equal{} 90^\circ \minus{} \angle OCB \equal{} \angle A,\ \angle KCP \equal{} \angle C\ \Longrightarrow$
$ \triangle PKC \sim \triangle ABC\ \Longrightarrow \frac {CP}{CK} \equal{} \frac {b}{a}$

CM = CP, C is the midpoint of MP, $ OC \equiv CD \parallel PK$ is a midline of $ \triangle MPK,$ D is midpoint of KM.


[hide]
Just use symmetric polynomials.

$a^{2}+b^{2}+c^{2}= (a+b+c)^{2}-2(ab+bc+ac)$.

We know $LHS = 5$, and that $(a+b+c)^{2}=9$.

It follows that $ab+bc+ac=2$.

Express $a^{3}+b^{3}+c^{3}$ in terms of $a+b+c$, $ab+bc+ac$, and $abc$ to find the values of the three elementary symmetric polynomials.

Express $a^{4}+b^{4}+c^{4}$ in terms of these three and evaluate.[/hide]


Using the numbers $ 1, 2, 3, 4, 5$ and $ 6$ and addition, subtraction, multiplication, division and exponentiation, make the number $ 1234321$.

Thanks kanonkula. This gives $ 1234321 \equal{} ((5^{2} \minus{} 3) \plus{} (6^{2} \minus{} 3)^{2})^{2}$ which uses $ 2$ four times and $ 3$ twice: I think the problem is to use each number once.

We could use $ 4\minus{}1$ instead of one of the $ 3$s and write "squared", but I think there must be a way of doing it with just the numbers.


induction
Ring Theory
superior algebra
superior algebra unsolved

A commutative ring $ A$ is Boolean if $ x^2 \equal{} x$ for all $ x \in A$.
In a Boolean ring $ A$, show that every ﬁnitely generated ideal in $ A$ is principal.

I am pretty sure this proof is by induction because: $ (x,y)\equal{}(x\minus{}xy\plus{}y)$ The inclusion $ (\supseteq)$ is clear, and the identities $ x(x\minus{}xy\plus{}y)\equal{}x$ and $ y(x\minus{}xy\plus{}y)\equal{}y$ give $ (\subseteq)$. How do I prove this by induction?

I thinks you got it already
suppose that $ \{x_{1},...,x_{n\minus{}1}, x_{n}\}$ is the  set of generators of ring $ A$ (we can choose $ n$ to be the minimum) since we know that $ \{x_{1},...,x_{n\minus{}1}, x_{n}\}\equal{}\{x_{1},...,x_{n\minus{}1}\}\cup\{ x_{n}\}$
and $ I\equal{}<\{x_{1},...,x_{n\minus{}1}>$ is a maximal ideal of $ A$, it is also a subring of $ A$ so $ I$ is Boolean , therefore $ I$ is generated by $ y\in A$ and $ A\equal{}<\{x_{n}, I\}>\equal{}<\{x_{n},y\}>$. Thus by the case $ n\equal{}2$, it is done

correct me if I am wrong


induction
algorithm
Additive Number Theory

Prove that any positive integer can be represented as a sum of Fibonacci numbers, no two of which are consecutive.

How about this approach?

Lemma $ 1$ Lets prove by induction that every integer can be written as the sum of some different fibonacci numbers.
It is true for $ n\equal{}1\equal{}f_1$
Assume that it is true for $ n$ then
if $ f_1$ is not in the sum of $ n$ we are done, add it and we obtain a sum for $ n\plus{}1$
suppose that $ f_1,f_2, \cdots,f_{i\minus{}1},f_i$ are in the sum of $ n$ and $ f_{i\plus{}1}$ is not
then you can replace by pairs $ (f_{i\minus{}1},f_i) \rightarrow f_{i\plus{}1}$  $ (f_{i\minus{}3},f_{i\minus{}2}) \rightarrow f_{i\minus{}1}$ replacing either $ f_2$ or $ (f_2$ and $ f_1)$ depending on the parity of $ i$ and
 add $ f_2$ and we obtain a sum for $ n\plus{}1$

From the lemma $ 1$ we can applying the following rule to obtain the desired sum:
$ (f_{i\minus{}1},f_i) \rightarrow f_{i\plus{}1}$ with $ i$ maximum index such as $ (f_{i\minus{}1},f_i)$ are in the sum and $ f_{i\plus{}1}$ is not in the sum.

Assume that each number less than or equal to $ f_n$ satisfies the condition. We want to prove that this implies all numbers between $ f_n$ and $ f_{n\plus{}1}$ satisfy the condition. Each number in this interval can be written as: $ f_n\plus{}k$ where $ k<f_{n\plus{}1}\minus{}f_{n}$ and so $ k<f_{n\minus{}1}$. But by induction all the $ k$ can be represented as a sum of fibonaci numbers all stricly less than $ f_{n\minus{}1}$. Hence by writing $ k$ in this form, all the $ f_n\plus{}k$ satisfy the condition, and so does $ f_{n\plus{}1}$ trivially, completing the induction

As the small-print caption of the topic mentions, this is Zeckendorf's theorem, see [url]http://en.wikipedia.org/wiki/Zeckendorf%27s_theorem[/url].

The simplest proof is by the greedy algorithm, see also above posts.


A group representative at a company are chosen randomly to run a workshop.
One person is chosen as the speaker, one as the coordinator, and three to
help prepare the materials. If the names are chosen from a hat with the
names of 14 people in it, determine the probability that sharon is the
coordinator and tom is one of the three people preparing materials.

Your answer is right, but check your value for Tom.  :wink:

[quote="AstroPhys"][hide]
$\text{Probability of Sharon}\ \ \frac{1}{14}$
$\text{Probability of Tom}\ \ \frac{3}{14}$
$\text{Probability of Both Events}\ \ \frac{3}{14 \cdot 13}$.
[/hide][/quote]

Hmm... shouldn't Tom be $\frac{3}{13}$ because Sharon is already chosen?