[quote="hsiljak"]I've just read that a TV station in Belgium made a false report on seccession of the Flandern from Belgium (I don't have the link, 'cos I'm in a hurry, so I expect that fredbel6 will write something about this... Pretty interesting thing (just like Orson Welles radio drama War of the Worlds)...[/quote]
Indeed I will, in fact I had already notified deimos through private messages.

It was the public Frenchspeaking channel (La Une) that did that.  They interrupted their normals programs with a very realistic announcement that the Flemish parliament in Brussels had decided to make Flanders an independent nation.(well secession from Belgium, since we have an absolute majority, I don't think they would keep the name "Belgium")
At first, they acted as if it was real, later they said it was all fiction.

Then they had a heavy debate (in French) with Flemish and Frenchspeaking politicians.

[i]For those of you not familiar with Belgium : Belgium is a double federal state, where the Dutchspeaking Belgians in the north (the Flemish) have an absolute majority.  The capital Brussels is officially bilingual but the Flemings are heavily outnumbered there.  The southern part (Wallonia) speaks French.[/i]

[url]http://news.bbc.co.uk/2/hi/europe/6178671.stm[/url] is a BBC reaction in English.

[url=http://www.nos.nl/nosjournaal/artikelen/2006/12/14/141206_belgie.html]Dutch article on it with the original report[/url] (only in French)

[url=http://www.vrtnieuws.be/nieuwsnet_master/versie2/nieuws/details/061214RTBFheisa/index.shtml#]Flemish clip about it[/url]
This is from the Flemish public broadcasting, the counterpart of the channel who made this report.  It's in Dutch.  It shows a tram not being allowed to drive pass the border between bilingual Brussels and Flanders.  It also mentions the king leaving the country and Walloon police officers keeping an eye on the border.

The major issue is of course Brussels, and that was fiercely discussed in the ensuing debate.  Brussels is bilingual and in practice completely francophone.  Some Flemings propose simply dropping it.  Others propose a Washington DC model.
Francophone politicians claimed that communities in Flanders in the vicinity of Brussels, where Frenchspeakers (mostly from Brussels) have reached an absolute majority, can do their own referendum and join Brussels if they choose to.  Flemish politicians disagreed, because that majority is a consequence of massive immigration since the official definitive establishment of the language border in 1963.  When you move somewhere where another language is spoken, you cannot just enforce your own language once you have reached a majority in that particular community.  Anyway, the Flemish territory around Brussels is a sensitive region, and if it would not accept being part of the nation Flanders, it could become our own "Krajina" or "Srpska".

[quote="Kent Merryfield"]fredbel, is there any religious component to the linguistic divide? Francophone Catholics and Flemish Protestants? Or is it not that simple? Or not that important?[/quote]
No.  In "theory" all of Belgium is catholic.  I was baptised as a catholic, and I went to a catholic school( but so like half of all the kis here)
In practice catholicism isn't doing too well.  Most people I know don't go to church, except for weddings and funerals.  I myself am an atheist.  One could say that Wallonia is even less religious.  In the past, that was different.  In fact, the Yser Tower, the Flemish World War I monument that expresses a desire for peace, said :
   A
V V K
   V
Alles Voor Vlaanderen, Vlaanderen voor Kristus = Everything For Flanders, Flanders For Christ.

One shouldn't wonder why the country is divided, one should wonder why it exists in the first place.

[quote="solafidefarms"]I thought that Belgium was the seat of the EU; wouldn't it be ironic for the seat of the European [i]Union[/i] to divide? It's already amusing that Belgium has so much discord. :)[/quote]
Brussels is indeed the place where most of the institutions of the EU are.  The European Parliament, the Council of the European Union,.... are all there.
What you are pointing out there is extremely crucial!   Belgium is often seen as an example of an artificial construction housing several kinds of Europeans.  Europe is supposed to consider it as an example.  The fact that the very heart of Brussel is disputed shows that it cannot work.
There are two ways to look at Europe.  A confederation of nations doing things together and deciding on certain matters together.  It makes no sense to raise taxes on cigarettes in Luxemburg if people can go buy them in France at a lower price instead.  It's also a waste of time and money that a person with an educational degree has to go through all sorts of trouble to get his degree recognised in other nations.
These are good things, several groups of people who are still boss in their own country working together as Europeans.
But some people propose becoming one country, with a president and with a standard official language in every country,... This is not good in my opinion.  There is simply no sense of European unity.  That will be a Europe of the big nations and of the big languages (German, French and English).  Borders should be crossed, but that does not mean they should disappear.

So for a Europe of the second kind, a breakup of Belgium would make no sense.
But in the first case, it's in fact a natural consequence.  A Europe in which Belgium has no use, now that their is a single currency (the Euro) and agreements on all sorts of things.  This applies to Catalonia, Basque country, ... as well.

"Little" languages - it's an interesting topic. I was reading an article a couple of weeks ago about the resurgence of Catalan over the last 20 or 30 years, and how one risks disapproval or heckling if one dares to give a public lecture in Spanish in Barcelona.

What is Catalan, anyway? Following the collapse of the Roman Empire, the Latin language, no longer anchored to its written form, splintered into a hundred local dialects. With the eventual rise of centralized nation states, a few lucky ones of those hundred dialects, such as Castilian Spanish or Parisian French, became national languages that swept aside their local cousins. Catalan is a survivor of the local Romance languages, poised somewhere between Spanish and French.

A generation ago, Catalan was a rural, traditional language. Now its advocates are urban and educated, seizing it as an expression of identity.

There's a practical issue, of course. Catalan is the language of one city (Barcelona) and a modest region around it; Spanish is the language of a continent. (The continent in question is not Europe, but it does contain hundreds of millions of people.) You can't really take Catalan with you around the world; you can take Spanish. Catalan may be the language you own and that somehow defines you, but it would help if you also spoke something else.

Dutch has much more of a foothold than Catalan, but some of the same practical issues apply. Where can you go to speak Dutch? The Netherlands, of course, and Flanders, and a few Caribbean islands and small slices of South America. Are there many in Indonesia who speak Dutch? 40+ years after the end of colonial rule, and as a Moslem country, I doubt it. Then there's South Africa, where Afrikaans is widely spoken. After 300 years of isolation, it isn't exactly Dutch any more, but it's pretty close, isn't it? But Dutch won't take you to as many places as German, and German won't get you as far as French will, and then there's the Dutch language's long-lost first cousin, English.

[quote]Dutch has much more of a foothold than Catalan, but some of the same practical issues apply. Where can you go to speak Dutch? The Netherlands, of course, and Flanders, and a few Caribbean islands and small slices of South America. Are there many in Indonesia who speak Dutch? 40+ years after the end of colonial rule, and as a Moslem country, I doubt it. Then there's South Africa, where Afrikaans is widely spoken. After 300 years of isolation, it isn't exactly Dutch any more, but it's pretty close, isn't it? But Dutch won't take you to as many places as German, and German won't get you as far as French will, and then there's the Dutch language's long-lost first cousin, English.[/quote]
Mister Kent Merryfield, you are obviously a very well informed man and I have enjoyed many of your posts ( maths-related and non-maths-related).  I do hope you allow me to correct you here.

Where can you go with Dutch? You can get around in Flanders, the Netherlands (in Friesland as well), Surinam (they have a very distinctive accent but if you pay attention you can understand them).  Apart from that, Afrikaans, one of the languages in South-Africa, is mutually intelligible if you speak really slowly.  In fact, there was an interview on the Flemish public channel with a South African in Afrikaans, but they did subtitle him in Dutch.  And in French-Flanders, the area of West-Flanders that French annexed centuries ago, some still speak Dutch, but French is the official language and (as expected) the website of cities like Lille and Dunkerque is strictly in French.
In Indonesia you won't get around with Dutch.

This makes Dutch a medium-sized language (it's still way bigger than Basque, Slovenian, Danish, Georgian, Latvian, Icelandic,.....) but without much international appeal.

But this is irrelevant.  Both Flemings and Dutch have accepted the position of their language a long time ago.  Our region has a very unique position in Europe : it is surrounded by the three big European languages : German, English and French.
Practically all Flemings speak French and English (the young generation tends to be more fluent in English).  Apart from that, many also had some German in school (I had two years).
In the Netherlands it's the same, but their second language tends to be English, and the third is more likely to be German than French.

This is a strength but also a weakness. Frenchspeaking people who settle in Flanders often don't bother to learn any Dutch.  In short, Flemings moving elsewhere adapt to the environment, but the Francophones tend to make the environment adapt to them.

That is one of the reasons why Flemings want their own state.  This has nothing to do with making our language stronger. Making English the first language to be instructed in schools can be a consequence, but I'm pretty sure French will still be instructed as well.  Wallonia and France will still be there and there won't be a wall or something crazy like that.  Flemings can already find their way in most of the world with English. You can't do that with French.  The Flemings may speak a smaller language, but in fact they are the group with the most international approach (there are lots of Flemings on the English Wikipedia, Flemings are popular because of their willingness to adapt in companies worldwide...)
So I don't understand your reasoning.  Why would we need to stay united with a (smaller !)group just because they speak a stronger language?

There are two other things : in Europe,German will in fact get you further than French.  German is instructed in many places all over the world.  It is the second most important language on the internet.  Germany alone has 83 million inhabitants, and Austrians and the majority of Swiss people are Germanspeaking as well.  Many people in Sweden, Denmark,... or Eastern Europe learn it as well.
This is actually something that irks many Flemings, it would be nicer if we could communicate in English with each other instead, but there is still the idea that French is the big language, while in fact it has even found itself surpassed in Europe.

About English being a cousin... English is a Germanic language, just like Dutch, German, Danish....   French, Spanish, Catalan are all Romance languages.
English has adopted lots of Romance elements.  Dutch on the other hand is a lot more "pure Germanic".  It's also extremely similar to German, especially when it comes to word order.
This is also a major difference between our struggle and the Catalan issue, or language is completely different as it is a Germanic language, and French is Romance.

A few examples :
English-Dutch-German-French

I can't play - Ik kan niet spelen - Ich kann nicht spielen - Je ne peux pas jouer

The population doesn't understand - De bevolking verstaat het niet-Die Bevölkerung versteht es nicht- La population ne le comprend pas

She is a strong girl - Ze is een sterk meisje - Sie ist ein starkes Mädhcen- Elle est une fille forte

As you can see, I consider myself a mini-ambassador of Flanders and its language, and I leave no opportunities to explain this region, its language and its struggle to the international public (and to point out that not everyone who waves a Flemish flag and supports independence is a jew-hating, african-hating, muslim-hating violent skinhead  :o )

You might wish to ask something else or point out other things.  In that case, I will definitely read it and reply. :)

[quote="fredbel6"]Wallonia and France will still be there and there won't be a wall or something crazy like that.  [/quote]

In a way, that would be kind of cool. Our own Berlin wall!  :D

Anyway, I think this report has been overrated by some people. It was quite clear that it was fake, and even if it weren't for some viewers, they should have checked other tv stations (like RTL) which would surely have reported the same thing immediately.
I think it's too soon to see what has been its impact, but (besides the fact that RTBF won't be making a joke on April 1st  :P ) I don't think it will have any consequences: Flanders will never be independent, nor will there be any real debate on it soon, I think. But it's still pretty interesting to see how it COULD happen. I would never have thought of that king-leaving-the-country-scenario.

Hi Kurt, long time no read. :wink:

[quote="Kurt Gödel"]

Anyway, I think this report has been overrated by some people. It was quite clear that it was fake, and even if it weren't for some viewers, they should have checked other tv stations (like RTL) which would surely have reported the same thing immediately.
[/quote]
Which many didn't bother to do (apparently).
A tenth (underground) ball of the Atomium in Brussels... :huh:
Also, that guy who stopped the train when it crossed the border spoke extremely accented Dutch.  Any Fleming watching that report would have realized it was fake.

[quote]I think it's too soon to see what has been its impact, but (besides the fact that RTBF won't be making a joke on April 1st  :P ) I don't think it will have any consequences: Flanders will never be independent, nor will there be any real debate on it soon, I think. But it's still pretty interesting to see how it COULD happen. I would never have thought of that king-leaving-the-country-scenario.[/quote]
Well I guess that's where we disagree.  Maybe Flanders will be independent, maybe it won't, but I'm quite sure this form of Belgium is not working and eventually the number of Flemings opposing it will become too large.
Consequences of this thing?  Well, they managed to trick their people into thinking that our independence will be equivalent to being almost being deported and border conflicts. The fact that so many people believed and didn't cross the border in their heads and switch to a Flemish channel, pretty much shows how easy it is for the RTBF to influence them.
I think Flemings and Walloons can only get along once they have their own state, and everyone respects each other.  I can understand you think otherwise as you have some francophone roots (and so do I but really not much), but I repeat : Flemish independence is not a declaration of war. :|

[quote="fredbel6"]Well I guess that's where we disagree.  Maybe Flanders will be independent, maybe it won't, but I'm quite sure this form of Belgium is not working and eventually the number of Flemings opposing it will become too large.[/quote]

Do you really think that number will ever be that large that it would force the politicians to split Belgium up? You've got Vlaams Belang and NVA now, but I don't think the other parties will ever agree on that. I could be wrong, of course.
You could be right about the form of Belgium not working right now, I don't know enough about politics. Do you mean that Walloon and Flemish politicians are just working against each other, instead of together?

[quote="fredbel6"]Consequences of this thing?  Well, they managed to trick their people into thinking that our independence will be equivalent to being almost being deported and border conflicts. The fact that so many people believed and didn't cross the border in their heads and switch to a Flemish channel, pretty much shows how easy it is for the RTBF to influence them.[/quote]

Yeah, that's what everyone is saying. RTBF is the same channel that broadcasts a program about the holocaust, every evening before an election, and it seems to be working: Front National isn't really getting anywhere. However, I don't see what that has to do with Belgium breaking up. How would the fact that RTBF can influence a lot of Walloons contribute to the independence of Flanders?

[quote="fredbel6"]I think Flemings and Walloons can only get along once they have their own state, and everyone respects each other.  I can understand you think otherwise as you have some francophone roots (and so do I but really not much), but I repeat : Flemish independence is not a declaration of war. :|[/quote]

I guess you're right about my roots  :) But I'm a real [i]Flamingant[/i], you know  :P

[quote="Kurt Gödel"][quote="fredbel6"]Well I guess that's where we disagree.  Maybe Flanders will be independent, maybe it won't, but I'm quite sure this form of Belgium is not working and eventually the number of Flemings opposing it will become too large.[/quote]

Do you really think that number will ever be that large that it would force the politicians to split Belgium up? You've got Vlaams Belang and NVA now, but I don't think the other parties will ever agree on that. I could be wrong, of course.
[/quote]
It's not because people don't vote for them that they don't support the independence.  There has been no referendum.  Don't forget that when the Volksunie split up into Spirit and N-VA, lots of other Volksunie politicians like Vankrunkelsven and Sauwens went to parties like VLD ( the Flemish liberals, which means right-wing here) and CD&V (the Flemish christian democrats, centrum party).  Boudewijn Brouckaert and Dedecker, who were both VLD members until recently have also expressed support for the independence.
Flemish prime minister Leterme from CD&V has also expressed doubts about the future of Belgium, he says stuff like "Belgium can continue but then we should definitely...."

[quote]You could be right about the form of Belgium not working right now, I don't know enough about politics. Do you mean that Walloon and Flemish politicians are just working against each other, instead of together? [/quote]
There are several reasons why it's not working.
One of them is the unwillingness of the francophone parties to cooperate.  The unconstitutional federal electoral district Brussels Halle Vilvoorde, that they refuse to split even though all Flemish parties demand it, is an example.  Not executing and funding the facilities in Wallonia and the linguistic laws in Brussels, while fully taking advantage of them and demanding more in Flanders, is another example.
Another one is that Flemings think differently, more right-wing.  We would like to fight crime harder for instance.  Is it normal that someone like Onckelinkx, the federal minister of justice, decides what we should do with criminals here in Flanders?
Another thing is the fact that we are not homogenous.  I vote for Flemish parties, francophones for Frenchspeaking parties.  Francophone politicians like to let planes come down at night above Flemish regions, and the other way around.... Another thing is the difference in wealth.  When we continue to pay for the disastrous PS regime, the PS is only tying more and more voters to its party.  It is a very common thing for small homogenous nations to do better.  Slovakia was poorer than the Czech republic when they split, and now Slovakia is backing up too.

[quote="fredbel6"]Consequences of this thing?  Well, they managed to trick their people into thinking that our independence will be equivalent to being almost being deported and border conflicts. The fact that so many people believed and didn't cross the border in their heads and switch to a Flemish channel, pretty much shows how easy it is for the RTBF to influence them.[/quote]

[quote]Yeah, that's what everyone is saying. RTBF is the same channel that broadcasts a program about the holocaust, every evening before an election, and it seems to be working: Front National isn't really getting anywhere. [/quote]
For those of you not familiar with Front National, they are a francophone party in Belgium.  They are far-right, anti-immigration.  Unlike Vlaams Belang, they are pro-Belgian, but they definitely are francophone (their website is only in French, hypocrites  :roll: ).  They are poorly organised and not succesful at all.
But the suggestion that Flemings are more racist than Francophones is unfair.  It's just that there is much more poverty in Wallonia, forcing the population to vote PS instead of FN.

[quote]However, I don't see what that has to do with Belgium breaking up. How would the fact that RTBF can influence a lot of Walloons contribute to the independence of Flanders? [/quote]
If they keep acting as if Flanders wants to do nothing but mean things once we're independent, they will support politicians opposing any reform, saying "non" instead.

[quote="fredbel6"]I think Flemings and Walloons can only get along once they have their own state, and everyone respects each other.  I can understand you think otherwise as you have some francophone roots (and so do I but really not much), but I repeat : Flemish independence is not a declaration of war. :|[/quote]
[quote]
I guess you're right about my roots  :) But I'm a real [i]Flamingant[/i], you know  :P[/quote]
A Flamingant is, according to Wikipedia, a Flemish nationalist? :wink:
Besides, roots are less important than how you were raised and how you have been living, just think of Leterme who was even raised in French for a while and who has a Frenchspeaking niece, who is politically active (as member of another party!).

In the history of the world, secession has usually led to war - it's not a pretty picture. But that's not true 100% of the time: a "civil divorce" is possible, as shown by the example of Slovakia and the Czech Republic. I would assume that whatever would or could happen in Belgium would be civilized.

But talk of secession is far more abundant than the fact of secession. The possible independence of Quebec from Canada has been front and center in the rhetoric of Francophone political parties for several decades now. It hasn't happened yet, and I'm skeptical that it will ever happen. (Even less likely that the energy-exporting Canadian west would secede - which doesn't mean there hasn't been loose talk.) Even at a more local level: if you search the web you could easily find organizations devoted to taking some particular region of one of the 50 states of the U.S.A. and making a new state out of it - far northern California (maybe together with southern Oregon), the Upper Peninsula of Michigan, and so on. The chances that any of that will happen? Don't hold your breath.

OK, those movements aren't likely to be taken seriously. The cause of Quebec independence is certainly a serious movement, but as I said, it's been going on for decades now, and hasn't happened yet.

[quote="Kent Merryfield"]In the history of the world, secession has usually led to war - it's not a pretty picture. But that's not true 100% of the time: a "civil divorce" is possible, as shown by the example of Slovakia and the Czech Republic. I would assume that whatever would or could happen in Belgium would be civilized.
[/quote] All those who are not in favour of secession often forget that Belgium itself separated from the short-lived United Kingdom of the Netherlands (1815-1830). And that separation was not supported by a referendum or anything like that.  So who are they to tell us separation is bad.

[quote]But talk of secession is far more abundant than the fact of secession. The possible independence of Quebec from Canada has been front and center in the rhetoric of Francophone political parties for several decades now. It hasn't happened yet, and I'm skeptical that it will ever happen. (Even less likely that the energy-exporting Canadian west would secede - which doesn't mean there hasn't been loose talk.) Even at a more local level: if you search the web you could easily find organizations devoted to taking some particular region of one of the 50 states of the U.S.A. and making a new state out of it - far northern California (maybe together with southern Oregon), the Upper Peninsula of Michigan, and so on. The chances that any of that will happen? Don't hold your breath.[/quote]

Quebec is a different story, but those other movements are of a different kind.  The topic isn't as sensitive.  Are people calculating the flow of money from North California to South etc...? Do people from the Upper Peninsula complain about immigrants from other parts who don't adapt?

[quote]OK, those movements aren't likely to be taken seriously. The cause of Quebec independence is certainly a serious movement, but as I said, it's been going on for decades now, and hasn't happened yet.[/quote]
A common argument is that Belgium will hold forever because it's been here for more than 175 years.  But Belgium didn't start off as a democratic state.  It was only after the First World War that every male adult had exactly one vote, and only after the Second World War that women could vote as well.
What evolutions have been seeing?  Well :
-1930 : the university of Ghent is the first to become completely Dutchspeaking
- 1963 :permanent definition of the linguistic borders, seriously diminishing the chances of realization of the "La Belgique sera latine ou ne sera pas" dream(="Belgium will be Latin or it won't be")
-1974 : the last non-Flemish prime minister of Belgium steps down
- 1980 : regions and communities become more autonomous
-1993 : the official end of the unitary Belgium and the start of federalism

[quote="Kent Merryfield"]
But talk of secession is far more abundant than the fact of secession. The possible independence of Quebec from Canada has been front and center in the rhetoric of Francophone political parties ...[/quote]

Yes, and further examples all over Europe are many. Northern Ireland, Scotland and UK, Catalunya and Spain, etc. The thing is that I think it has proven to be important in all cases how the rest of the population of the original undivided country thinks about secession. In my opinion, this is very wrong, however sometimes there are economic facts that can't be overlooked. In a country the state budget is divided between states, provinces, communities, counties, cantons, regions or whatever the political division happens to be. In order for two portions of a country to divorce one must proceed to make counts about how the money has been distributed over the country in the years they were together and if the payment of some debt from one portion to the other proceeds. Nevertheless, just like real life divorces, the decision of splitting should be left with conditions to the citizens of each portion alone.

Apart from that, from an external point of view, one can be pro-splitting or one can be against. In general I don't understand the idea of splitting  a country when the portions share without adoubt a common past. This is because I tend to think that uniting is better than dividing. But for example Scotland and Ireland have at least since early middle age been confronted in different conflicts with England so one can speak about a long lasting controversy between different countries.

Ursula

I would like to say something else about Quebec, there are several differences between their advances and ours :

- they have actually had a referendum (like in 1995) when the "no" voters barely gained the upper hand
- as far as I know, Canada isn't really bilingual on an individual basis, like the Flemings.  Can the average adult man and woman in Quebec talk fluently to native anglophones?  I mean : do they have to fear a massive invasion near their borders of anglophones who refuse to learn any French as they don't have to?  In any case, it is a fact that they have much tougher language laws (like "Arrêt" signs instead op "stop") than ours, thus leaving room for more annoyance here.
- There are six million Flemings and about 7.6 million in Quebec, but both in South and North America, Quebec would find itself quite a loner.  We live in the vicinity of the United Kingdom (and we speak English), we border Germany (German could become the third language, it could be made optional after the split) and last but not least : 16 million Dutch people north of us, with whom we share many ties, like language.  Apart from that, we could still enjoy the EU collaboration.  Don't forget that even within the separatist parties, there are people who favor a re-unification with the Netherlands and people who don't.  There are also politicians in the Netherlands who have expressed support for this idea.

[quote="ursula"][quote="Kent Merryfield"]
But talk of secession is far more abundant than the fact of secession. The possible independence of Quebec from Canada has been front and center in the rhetoric of Francophone political parties ...[/quote]

Yes, and further examples all over Europe are many. Northern Ireland, Scotland and UK, Catalunya and Spain, etc. The thing is that I think it has proven to be important in all cases how the rest of the population of the original undivided country thinks about secession. In my opinion, this is very wrong, however sometimes there are economic facts that can't be overlooked. In a country the state budget is divided between states, provinces, communities, counties, cantons, regions or whatever the political division happens to be. In order for two portions of a country to divorce one must proceed to make counts about how the money has been distributed over the country in the years they were together and if the payment of some debt from one portion to the other proceeds. Nevertheless, just like real life divorces, the decision of splitting should be left with conditions to the citizens of each portion alone.[/quote]
Uhm, if we are going to have to wait until the Francophones (the minority!) let us go, we will never be free. In fact, there was a proposal in the federal parliament this year by a Vlaams Belang member to prepare the dissolution of the state.  Almost all Flemish members voted "yes" on taking it into consideration (just a formality) , causing outcries of indignation from the francophone members.

Flanders is richer, there are about six million Flemings and they pay Wallonia about $6\ 10^{9}$ Euros a year.  Critics like to point out that Wallonia used to be richer, but a recent inquiry showed that money has nevertheless been flowing south since 1830.

[quote]Apart from that, from an external point of view, one can be pro-splitting or one can be against. In general I don't understand the idea of splitting  a country when the portions share without adoubt a common past. This is because I tend to think that uniting is better than dividing. But for example Scotland and Ireland have at least since early middle age been confronted in different conflicts with England so one can speak about a long lasting controversy between different countries.[/quote]
You tend to think that uniting is better than dividing?  Well in that case you should definitely scroll up and read my replies to solafidefarms's comment about the European Union.

Indeed, what if the common history isn't a very good one?  I'm thinking of abolishing education in our own language, executions of innocent Flemings because their trial was in French, people not being able to do most basic administrative stuff because it was all in French no matter where they lived, officers during World War I who didn't know any Dutch and expected poor uneducated and sometimes immature boys to understand theirs instead which got them killed, massive repression of the Flemish movement after the World Wars.....

[quote="fredbel"]
You tend to think that uniting is better than dividing?  Well in that case you should definitely scroll up and read my replies to solafidefarms's comment about the European Union.[/quote]

The subject obviously affects you closely. However, you misqunderstood me, or maybe I didn't express myself well enough. I wasn't referring to the particular case of Flanders within Belgium with which I'm not familiar. I was speaking in general and I can point you out the Catalunya within Spain case which I'm familiar with and well I'm not going to extend on it but I'm not pro splitting because I think the relatively strong separatist movement there is a result of recent history particularly under Franco, where different nationalisms in Spain where reppressed cruelly. In my paragraph I also cited the UK case as one in which separatoin seems logical to me. You also send me to read your comments about the EU as if I haven't read them when in fact I did and with which I agree. I don't support Europe becoming one country.

[quote="fredbel"]Indeed, what if the common history isn't a very good one?...[/quote]

Now you are the one who should read my message because this I point out I believe. I repeat in my last words I wasn't making a statement about Belgium's particular case. Nevertheless this is not the 19th century, protection of distinct identities can be achieved within one country with our modern conceptions. In Catalunya the use of the catalan language is regulated by law and of mandatory use at schools and indeed there is the case of irish gaelic which might be in the brink of extinction and still there's a lot of effort to bring it back to live which by the way seems to be giving fruits.

My point is, do most of you in Flanders feel the differences between you and the french speaking wallonians are irreconcilable. If so, what [i]I[/i] would encourage would be separation.  Also what [i]I[/i] think is that it would take a great deal of mutual damage between two different portions of a country [i]I[/i] inhabit to make me think that our differences are irreconcilable -not saying this is not the belgian case-. But I understand other opinions, understand at the level of 'I see what you mean' not at the level of 'I would do it if I were you'.

Ursula

PD: Oh I forgot, you referred to the signs on the street. Do you mean your signs in Flanders are not in Dutch. That I would say is weird.

[quote="ursula"][quote="fredbel"]
You tend to think that uniting is better than dividing?  Well in that case you should definitely scroll up and read my replies to solafidefarms's comment about the European Union.[/quote]

The subject obviously affects you closely.[/quote]
Of course, it is my home.  And it does affect lots of other Flemings, including young ones, many of whom are using the internet to explain the relatively unknown situation (which is unique, I can't think of any other country where the majority wants to split)

[quote]You also send me to read your comments about the EU as if I haven't read them when in fact I did and with which I agree. I don't support Europe becoming one country.[/quote]
I wasn't questioning whether or not you read them, don't worry. :wink:

[quote]Now you are the one who should read my message because this I point out I believe. I repeat in my last words I wasn't making a statement about Belgium's particular case. Nevertheless this is not the 19th century, protection of distinct identities can be achieved within one country with our modern conceptions[/quote].
It doesn't matter which century this is.  Languages are still being crushed.  It's not because this new era expects us to be cilivized and cooperative, that we all are.

[quote]

My point is, do most of you in Flanders feel the differences between you and the french speaking wallonians are irreconcilable. If so, what [i]I[/i] would encourage would be separation.  Also what [i]I[/i] think is that it would take a great deal of mutual damage between two different portions of a country [i]I[/i] inhabit to make me think that our differences are irreconcilable -not saying this is not the belgian case-. But I understand other opinions, understand at the level of 'I see what you mean' not at the level of 'I would do it if I were you'.[/quote]
Belgium should not exist... because it exists.  That's a flawed reasoning, all too often used here.  Unity is not by default the best thing.
I can only speak for myself (and refer to those 900,000 people, out of less than 7 million votes in all Belgium, who voted for parties explicity striving for Flemish independence).
Are people willing to erase those irreconcilable differences?  That's the question.  In 2003 during the last federal elections, only 0.1% voted for a certain party in favor of a return to the unitary state. Flemings and Francophones are a separate kind of people, in some municipalities they say "the only thing we have in common is the weather report".  But what would make us less eager to split? Well, not claiming our territory would be a great start.  Them keeping their end of the bargains would be another thing.  Providing reliable healthcare in Dutch in bilingual Brussels would be nice too.  It's easy to say a little country should settle its differences when it's always the same group.  It's also not very normal that Flemish schools still teach French as first language, while Dutch isn't enforced at a proper level in Wallonia.  More financial autonomy so the scandal party PS loses power and Wallonia can rebuild itself, would be great too.
I've seen so many people complaining about the mean Flemings since the report, I don't see many people eager to learn Dutch.  On the contrary : why learn the language of those who "hate" them?

[quote]
Ursula

PD: Oh I forgot, you referred to the signs on the street. Do you mean your signs in Flanders are not in Dutch. That I would say is weird.[/quote]
No, I meant that while Francophones in our country promote a "linguistic liberty", those in Quebec enforce the toughest linguistic laws on earth.  You will find STOP signs in all of Belgium.
But coincidentally, yesterday the Flemish government decides to remove all signs with the name of a municipality in two languages on them.  Not a very spectacular measure, but still good : we have to do everything to preserve the Flemish character on our territory.


If $ a$ $ b$ and $ c$ are positive reals then prove the following inequality:

$ \frac{ab^{2}}{2a^{2}\plus{}b^{2}}\plus{}\frac{bc^{2}}{2b^{2}\plus{}c^{2}}\plus{}\frac{ca^{2}}{2c^{2}\plus{}a^{2}}$ $ \leq\frac{a\plus{}b\plus{}c}{3}$

[quote="canada"]If $ a$ $ b$ and $ c$ are positive reals then prove the following inequality:

$ \frac {ab^{2}}{2a^{2} \plus{} b^{2}} \plus{} \frac {bc^{2}}{2b^{2} \plus{} c^{2}} \plus{} \frac {ca^{2}}{2c^{2} \plus{} a^{2}}$ $ \leq\frac {a \plus{} b \plus{} c}{3}$[/quote]
See here the equivalent form for your inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937


calculus
integration
logarithms
inequalities
calculus computations

For a positive real number $ a > 1$, prove the following inequality.

$ \frac {1}{a \minus{} 1}\left(1 \minus{} \frac {\ln a}{a\minus{}1}\right) < \int_0^1 \frac {x}{a^x}\ dx < \frac {1}{\ln a}\left\{1 \minus{} \frac {\ln (\ln a \plus{} 1)}{\ln a}\right\}$

Why bother to establish an inequality while the exact value can be determined by elementary means?

$ \int_{0}^{1}\frac{x}{a^{x}}dx\equal{}\allowbreak \frac{a\minus{}\ln a\minus{}1}{a\left( \ln a\right) ^{2}}$

You are right, I do know that, but I dare to post the problem.


calculus
integration
function
trigonometry
calculus computations

This is unclear what is meant by $ M(a)$ since $ f(x)$ is a function of both $ a$ and $ x$.

$ \sin (x\minus{}t)\sin (2t\minus{}a)$ = $ \frac {1}{2} (\cos (x\plus{}a\minus{}3t) \minus{} \cos (x\minus{}a\plus{}t))$

integrating from $ 0$ to $ \pi$, we find that

$ f(x)$ = $ \frac {1}{2} [\frac {1}{3} (\sin (x\plus{}a)\plus{}\sin (x\plus{}a)) \plus{} \sin (x\minus{}a) \plus{} \sin (x\minus{}a)]$

      = $ \frac {1}{2} [\frac {2}{3} \sin (x\plus{}a) \plus{} 2\sin (x\minus{}a)]$

      = $ \frac {1}{3} \sin (x\plus{}a) \plus{} \sin (x\minus{}a)$.

I guess we have to maximise taking x as the variable. and then evaluate the integral.

Hhmm..its 28/27 i multiply by 2 and i shouldnt...

After get the Hun's result, just check that

$ \frac{1}{3}\sin(x\plus{}a)\plus{}\sin(x\minus{}a)\equal{}\frac{4}{3}\cos{a}\sin{x} \minus{}\frac{2}{3}\sin{a}\cos{x}$

 :)

[quote="kunny"]For $ a\in\mathbb{R}$, let $ M(a)$ be the maximum value of the function $ f(x)\equal{}\int_{0}^{\pi}\sin (x\minus{}t)\sin (2t\minus{}a)\ dt$.
Evaluate $ \int_{0}^{\frac{\pi}{2}}M(a)\sin (2a)\ da$.[/quote]
$ f(x)=\int_{0}^{\pi}\sin (x-t)\sin (2t-a) dt=\int_{0}^{\pi} \frac{1}{2}(cos (x+a-3t)-cos (x-a+t))dt $$=\frac{sin(x+a)}{3}+sin(x-a)$

$f(x)=\frac{sin(x+a)}{3}+sin(x-a)=\frac{4}{3}cosasinx-\frac{2}{3}sinacosx\leq{\frac{2}{3}\sqrt{1+3cos^{2}a}}$ [b]from C-S[/b]

NOW $ I=\int_{0}^{\frac{\pi}{2}}M(a)\sin (2a) da=\frac{2}{3}\int_{0}^{\frac{\pi}{2}}\sqrt{1+3cos^{2}a}.sin2ada$

LET $1+3cos^{2}a=t^2$,  then $I=\frac{4}{9}\int_{1}^{2} t^{2}dt=\frac{28}{27}$


Solve next system for all real solutions

       $e^{x}+e^{-y^{2}}= 1$                      (1)

        e^{2x} + sqrt( e^{- y^2}) = 1             (2)

Substitute $u=e^{x}, v=\sqrt{e^{-y^{2}}}$ to get

$u+v^{2}=1$
$u^{2}+v=1$

Subtracting the equations we get

$(u-v)-(u^{2}-v^{2})=0\iff (u-v)(1-u-v)=0$

Hence either $u=v$ or $u+v=1$

The first possibility yields $u^{2}+u-1=0\implies u=v={\sqrt{5}-1\over 2}$ (the negative solution is extraneous), hence $x=\ln{\sqrt{5}-1\over 2}, y=\pm\sqrt{-\ln{3-\sqrt{5}\over 2}}=\pm\sqrt{\ln{3+\sqrt{5}\over 2}}$

The second possibility, combined with the first equation, yields $v=v^{2}\implies v=1$, since $v=0$ is extraneous, but then $u=0$ which is impossible.

Therefore the above solutions are unique.


$ f(1) \equal{} 2$,$ f(n \plus{} 1)\ge f(n) \ge \frac {n}{n \plus{} 1}f(2n)$
Find the min value $ M\in N^{ \plus{} }$, which $ f(n) < M$ for every $ n\in N^{ \plus{} }$.

I'm sorry for missing a condition.
M is a integer.
And I ask for that I don't know how to calculate this limit

There is a small mistake:
$ f(2^{n\plus{}1})\le \prod_{i\equal{}1}^{n}(1\plus{}\frac{1}{2^i})f(1)\equal{}2\prod(1\plus{}\frac{1}{2^n})$
And the key is $ M\equal{}10$
anyway,thank you all the same

[quote="wxydx00"]There is a small mistake:
$ f(2^{n \plus{} 1})\le \prod_{i \equal{} 1}^{n}(1 \plus{} \frac {1}{2^i})f(1) \equal{} 2\prod(1 \plus{} \frac {1}{2^n})$
And the key is $ M \equal{} 10$
anyway,thank you all the same[/quote]
$ f(1)\equal{}(1\plus{}2^{\minus{}0})$ . Were is my mistake?

I tried again,and find you're right.
but why is the key is 10?
maybe I didn't copy the problem from my teacher.
And i'm quite sorry to waste you so much time.
Thank you very much.
and our teacher gave another solution to prove the M is 5
$ \rightleftarrows (1 + \frac {1}{2})(1 + \frac {1}{2^2})...(1 + \frac {1}{2^{k}}) < \frac {5}{2}(1 - \frac {1}{2^k})$
and then $ (1 + \frac {1}{2})(1 + \frac {1}{2^2})...(1 + \frac {1}{2^{k}})(1 + \frac {1}{2^{k + 1}}) < \frac {5}{2}(1 - \frac {1}{2^{k + 1}})$
$ M = 2*2.5 = 5$

another question here:
Find $ \lim_{n - > + \infty}\prod(1 + \frac {1}{9^n})$


trigonometry
limit
real analysis
real analysis unsolved

Let $x_{n}=\sqrt{n}\sin(\sin(\sin\ldots\sin x))\ldots)$, where n appears $n$ times $n\geq 1$. Prove that $\lim_{n\to\infty}x_{n}\sqrt{3}$.

You mean where $\sin$ appears $n$ times.  And I think you mean

$\lim_{n \to \infty}x_{n}= \sqrt{3}$

already posted I think. A great classic .. It suffices to find an asymptotic formula for $u_{n+1}= sin(u_{n})$, $u_{0}=1$. (And I'm not sure it works for every $x$  :lol: )

Hmm, could you give me the link allek? :)

Cesar, you really don't need a link for that: if $f_{n}$ is the $n$-th iterated of $sin (x)$ just observe that $\frac{1}{f_{n+1}(x)^{2}}-\frac{1}{f_{n}(x)^{2}}$ tends to $1/3$,using the obvious fact that $1/\sin(x)^{2}-1/x^{2}$ tends to $1/3$ as $x$ tends to $0$. And now just sum up these equivalents to deduce the result. And by the way, by changing $x$ in $-x$ and assuming your problem as you stated correct, we find the nice result: $2\cdot \sqrt{3}=0$, which shows that $R$ has characteristic $2$.  :P

seee also here

G.Polya--G.Szego , Problems and Theorems in
Analysis I, Springer--Verlag Ed. pag.38}]

[quote="Cezar Lupu"]Let $ x_{n} \equal{} \sqrt {n}\sin(\sin(\sin\ldots\sin x))\ldots)$,......[/quote]


Given a postive integer $ n$ ,  there exists a unique positive integer $ k$ such that $ \frac{8}{15 } \le \frac{n}{n\plus{}k }\le \frac{7}{13 }$, find the maximum value of $ n$ that satisfies such inequality?

[hide="Solution"]
The ineq is equivalent to $ \frac {6}{7}n\le k\le \frac {7}{8}n$.
The length of interval of this ineq is $ \frac {7}{8}n \minus{} \frac {6}{7}n \equal{} \frac {1}{56}n$

If this length $ \ge 1$,there are at least one such $ k$,and if  $ > 2$,there exist two such $ k$ so ruled out,

so let's check the case:$ 1\le \frac {1}{56}n\le 2 \Longleftrightarrow 56\le n\le 112$.

Now,in the ineq $ \frac {6}{7}n\le k\le \frac {7}{8}n$,and $ k \equal{} [\frac {7}{8}n]$ is one of integer solutions,so we have the condition so that there exist a unique $ k: [\frac {7}{8}n] \minus{} 1 < \frac {6}{7}n$.

Let $ m$ be an integer,when $ n \equal{} 8m \plus{} j (j \equal{} 0,1,2,\cdots ,7)$,the ineq is equivalent to

$ m\le 6j \plus{} 6 \minus{} 7[\frac {7}{8}j]$ ,for each $ j \equal{} 0,1,2,3,4,5,6,7$ we have the maximum $ m \equal{} 6,12,11,10,9,8,7,6$

so finally we get the maximum $ n \equal{} 8\cdot 12 \plus{} 1 \equal{} \boxed{97}$[/hide]


Some one can post all problems from Austrian-Polish 2003 (in english) ?

[quote="Solar Plexsus"]You can find the problems of APCM 2003 in english here:

[url]http://www.kalva.demon.co.uk/aus-pol/ap03.html[/url][/quote]

Thanks Solar Plexsus, I know the site you have pointed out, but I am searching the problems in original formulations and if possible in pdf format.


What is the units digit of the product of all the prime numbers less than 100?

[hide="Solution"]We notice that $ 2$ and $ 5$ are in the list of the first $ 100$ primes. Multiplying $ 2\times5$ gives $ 10$ which ends in a zero. Thus, multiplying the rest of the primes by $ 10$ results in the end product having a units digit of $ \boxed{0}$.[/hide]


Prove that $\displaystyle\binom{2n}{2i-1}$ is always even for all natural numbers n and $i=1,2,...,n$.

This is an immediate consequence of Lucas' theorem which states that if $p$ is any prime, and $n=n_tp^t + \cdots n_1p+n_0$ be the expansion of $n$ in base $p$ and
$m=m_t2^t + \cdots + m_1p+m_0$ is quite the same for $m$ (here itis allowed that some of the first digits are $0$, just to have the same number of digits as in the expansion of $n$), then :
$\binom {n} {m} = \binom {n_t} {m_t} \cdots \binom {n_1} {m_1} \cdot \binom {n_0} {m_0} \mod [p]$

In particular, $\binom {n} {m} = 0 \mod [p]$ if and only if $m_k > n_k$ for some $k$.

Here, just choose $p=2$ and note that the unit digit of $2n$ in base $2$ is $0$, whereas the unit digit of $2i-1$ is $1$, hence result.

Pierre.

More simply, we can use the relation : $2n\binom{2n-1}{2i-2}=(2i-1)\binom{2n}{2i-1}$

__________
blang

Why always doing things simply?  :P

Pierre.

Merci bouque Pierre and Blang.
Pierre, I also used Lucas, but I was dissatisfied with the complex solution, fortunetly, Blang came up with that nice identity.


Solve the equation:
$x^{3}-\sqrt[3]{6+\sqrt[3]{x+6}}=6$

Let $f(x)=x^{3}-6$, equation equavalent to
(1) $x=f(f(f(x))).$
$f(x)-x=(x-2)(x^{2}+2x+3).$ Therefore (1) had unique real solution $x=2$.

[quote="toan hoc"]I need find other solution!
Help me.please![/quote]
There are not other real solutions. You need complex solutions?

yes, our equation is $x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}=\sqrt[3]{6+f(x)}$(remember the definition of $f(x)$).
but this is equivalent to $((x^{3}-6)^{3}-6)^{3}-6=x$
that is $f(f(f(x)))=x$.
wait i think Rust only solved for $f(x)-x=0=(x-2)((x+1)^{2}+2)$ but not $x=f(f(f(x)))$


trigonometry
inequalities proposed
inequalities

Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$|\sin nA+\sin nB+\sin nC|\leq\frac{3\sqrt{3}}{2}$ for all $n\in\mathbb{Z},n\geq 2$

[color=purple]Hien are you Vietnamese ? If i don't remember false your problem was proposed in Math& youth [/color]


abstract algebra
geometric series
superior algebra
superior algebra unsolved

Prove that the intersection of all maximal left ideals of a ring is a (two-sided) ideal.

this is the jacobson radical $ jac(R)$. we have $ jac(R) \equal{} \{x \in R : \forall y \in R : 1 \minus{} yx \text{ is left invertible}\}$.[hide="proof"]if $ x \in jac(R)$ and $ y \in R$, $ 1\minus{}yx$ is left-invertible since otherwise it lies in some maximal left ideal, and then also $ 1 \equal{} (1\minus{}yx) \plus{} yx$, contradiction. conversely, assume that $ \mathfrak{m}$ is some maximal left ideal not containing $ x$. then we have $ Rx \plus{} \mathfrak{m} \equal{} R$ by maximality, i.e. there exists $ y \in R$ such that $ 1\minus{}yx \in \mathfrak{m}$. thus $ 1\minus{}yx$ is not left-invertible.[/hide]
thus it's enough to prove the more general result: if $ 1\minus{}yx$ is left invertible, then also $ 1\minus{}xy$ is left-invertible. the idea is the following formal calculation with geometric series:
$ \frac{1}{1\minus{}xy} \equal{} \sum_{k \geq 0} (xy)^k \equal{} 1 \plus{} x \left(\sum_{k \geq 0} (yx)^k\right) y \equal{} 1 \plus{} x \frac{1}{1\minus{}yx} y$
this motivates a rigorous proof: let $ u$ be a left-inverse of $ 1\minus{}yx$. then $ 1\plus{}xuy$ is a left-inverse of $ 1\minus{}xy$ since $ (1\plus{}xuy)(1\minus{}xy)\equal{}1\minus{}xy \plus{} xuy \minus{} xuyxy \equal{} 1\minus{}xy \plus{} xu(1 \minus{} yx)y \equal{} 1$.

The ideal in question (that is the Jacobson radical) can be easily proved to be equal to $ J(R)$={$ \cap_{M}Ann(M)$| $ M$ simple  $ R-$module},where $ Ann(M)$={$ r\in R$| $ rm=0$ for all $ m\in M$} is the annihilator of $ M$ and obviously a two sided ideal. Hence, so is any intersection of annihilators and particularly $ J(R)$.


[quote="Potato Theory"]I liked Phelpedo's problem so I made one:

$\sum_{n=1}^{\infty}\frac{2F_{n}}{F_{n-1}F_{n+2}}$[/quote]

that's the Fibonacci??

No, the Fibonacci numbers are defined recursively:

$F_{n}=F_{n-1}+F_{n-2}$

with $F_{1}= 1$ and $F_{2}= 1$.

The series has quite a few interesting properties, and that sum I gave utilizes some of them.

The numbers start:

$1,1,2,3,5,8,13,21,34,55$.

See [url]http://en.wikipedia.org/wiki/Fibonacci_Numbers[/url] for more about them.


linear algebra
matrix
algebra
polynomial
vector

Hi, could anyone help to show the following matrix is defective?

Show that the matrix $\begin{bmatrix} c & 1\\ &c&1 \\ &&&.\\&&&&.\\ &&&&&c&1\\&&&&&&c \end{bmatrix}$ is defective.
A matrix of this form is called a Jordan block.

well I had never heard the term defective, but if it means not having a basis of eigenvectors : well that's just the point of jordan blocks

You can  see easily that c is an eigenvalue and the only one.  The diagonal is exactly the spectrum for upper and lower diagonal matrices.  (If you doubt this work out the characteristic polynomial)

Now we need a basis of eigenvectors for the total space : that means a basis of eigenvectors of c for the entire space.  But all the eigenvectors of c (combined with zero) form a subspace (this is true in general : eigenvectors of an eigenvalue form a space, if you add the zero vector) .

---> every vector is zero or eigenvector of c

contradiction : the vector $\left[\begin{matrix}0\\1\\0\\ \vdots\\0\end{matrix}\right]$ is not an eigenvector


Circle $ C$ has radius 6 cm. How many square centimeters are in the area of the largest possible inscribed triangle having one side as a diameter of circle $ C$?

This happens when it is an isosceles right triangle. The legs have $ \frac{6}{\sqrt{2}}\equal{}3\sqrt{2}$. The area of the triangle is thus $ \frac{\left(3\sqrt{2}\right)^2}{2}\equal{}\frac{18}{2}\equal{}\boxed{9}$.

I didn't get how to do this at first ( :rotfl: ), but now I do.  The RADIUS is $ 6$, so the diameter is $ 12$.  Now, using what RoFl said, we have that the legs are $ \frac {12}{\sqrt{2}}={6\sqrt{2}}$.  We now have $ \frac{(6\sqrt{2})^2}{2}=\frac{72}{2}=\boxed{36}$

The largest area will be when the triangle is isosceles and we know it will also be right since the diameter is one side of it.

Since the hypotenuse of the triangle is the diameter (which is $6\cdot2=12$; RoFlLoLcOpT forgot to multiply by $2$ for the [b]diameter[/b]), one of the legs would be $\frac{12}{\sqrt{2}}=6\sqrt2$

The area of the triangle is $\frac{(6\sqrt2)^2}{2}=\frac{36\cdot2}{2}=\boxed{36}$


find all integer values that
((a + 1) / b) + ((b + 1) / a)
can assume and determine all pairs of integers a,b, giving these solutions.

Nice, but classical. If I'm not wrong, infinite descending shows that the only integer valus are 3 and 4. And if we plug these values, we will find to Pell equations, which aren't hard to be solved.


vector
geometry
geometric transformation
rotation
analytic geometry
complex numbers

a, b, c,  and d are the complex numbers coresponding to the vertices of a square ABCD in the complex plane.

Prove that a^2 + b^2  +c^2  +d^2 = 2(ac +bd )

Consider the vector representations of a,b,c,d with respect to the origin O.
Then (a-c) is one diagonal of the square while (b-d) is the other diagonal of the square.
Since the diagonals are of the same length, but intersect at right angles, we have $(a-c)=\pm i*(b-d)$ because $i$ is the rotation of $90^\circ$ in the complex plane.
Then $(a-c)^2=-(b-d)^2$ or $(a-c)^2+(b-d)^2=0$.
Finally $a^2+b^2+c^2+d^2=2(ac+bd)$.

[quote="pkothari13"]brute force always works too  :D   move one of the vertices to the origin and name the other vertices $0+ai$, $a+0i$, and $a+ai$.  then using coordinates, it becomes pretty straightforward[/quote]

How do you know that when you move it back, it still works, though?


geometry
Triangle
minimization
parallelogram
IMO Shortlist

Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

We can prove this one by this lemma :
     If P and Q are abitrary points then
         $\sum \sin{BQC}.PA \geq \sum \sin{BQC}QA$
    Plug Q = G.   :D

[quote="orl"]Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG+BP{\cdot}BG+CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.[/quote]

Here  is my solution  to that and  i would like  someone  to  tell me please if its ok :

We use complex numbers.
Put the  triagle  on  complex  plane  with  the  origin  0(0,0)  on  G. Also  let z, a , b , c  the  complex  numbers  for  the  points P, A , B , C.Then  we  have :

$AP\cdot AG+BP\cdot BG+CP\cdot CG=\mid(z-a)a\mid+\mid(z-b)b\mid+\mid(z-c)c\mid\geq\mid(a+b+c)z-(a^{2}+b^{2}+c^{2})\mid=\mid a^{2}+b^{2}+c^{2}\mid=\frac{4}{9}(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})=\frac{1}{3}(AB^{2}+BC^{2}+CA^{2})$

I've been using this problem to demonstrate techniques for solving geometric inequalities.  Here's another solution.  At the minimum, $P$ will be inside the triangle.  For any value of $AP$, let us write the expression under consideration as a function of $\theta=\angle PAB$ using the Cosine Rule:
\[ AG\times AP+BG\sqrt{AB^2+AP^2-2AB\times AP\cos\theta}+CG\sqrt{AC^2+AP^2-2AC\times AP\cos(\angle A-\theta)}.  \]
Differentiating, we find that when this is minimised
\[ BG\frac{2AB\times AP\sin\theta}{2\sqrt{AB^2+AP^2-2AB\times AP\cos\theta}}=CG\frac{2AC\times AP\sin(\angle A-\theta)}{2\sqrt{AC^2+AP^2-2AC\times AP\cos(\angle A-\theta)}}.  \]
Cancelling and simplifying back the denominators using the Cosine Rule,
\[ \frac{BG\times AB\sin\angle PAB}{BP}=\frac{CG\times AC\sin\angle PAC}{CP}  \]
or, simplifying using the Sine Rule in triangles $ABP$ and $ACP$,
\[ \frac{BG}{\sin \angle APC}=\frac{CG}{\sin \angle APB}.  \]
By symmetry, we find that the condition $\sin \angle APB: \sin \angle APC: \sin \angle BPC = CG: BG: AG$ must hold at the point $P$ which minimises the expression.  As the three angles sum to $360^o$, this condition in fact uniquely determines them, and hence the point $P$, so that this condition is not only necessary for $P$ to minimize the expression, but also sufficient.  It is therefore enough to prove that when $P$ is set to $G$ this condition is satisfied.

This last bit is not difficult.  Extending $AG$ to meet $BC$ gives the midpoint $M$ of $BC$.  Applying the Sine Rule twice, $BG\sin \angle BGA=BA\sin \angle BAG =BM\sin\angle BMA$.  As $BM=CM$, $\sin\angle BMA=\sin\angle CMA$, it follows that $BG\sin \angle BGA= CG\sin\angle CGA$.  By symmetry the whole set of ratio equalities holds.

Let $A_1$, $B_1$, and $C_1$ be the midpoints of $BC$, $AC$, and $AB$, respectively. Let the line through $C$ parallel to $BG$ hit $AA'$ at $X$. $BGCX$ is a parallelogram, so $CX = GB$ and $GX = 2 GA_1 = AG$.

Let $AG$ intersect the circumcircle of $\triangle BCG$ at $A'$. $\angle CBA' = \angle CGA' = \angle CGX$, and $\angle BA'C = 180 - \angle BGC = 180 - (180 - \angle GCX) = \angle GCX$, so $\triangle BCA'$ is similar to $\triangle GCX$. Hence, there exists a positive real number $k$ such that $BC = k GX = kGA$, $CA' = k GC$, and $A'B = k CX = kGB$.

By Ptolemy's inequality, $PB \cdot CA' + PC \cdot BA' \geq PA' \cdot BC$, so $k PB \cdot GC + k PC \cdot GB \geq k PA' \cdot GA$. Dividing by $k$ and adding $PA \cdot GA$ to both sides yields $PA \cdot GA + PB \cdot GC + PC \cdot GB \geq (PA' + AP) \cdot GA$. By the triangle inequality, $PA' + AP \geq AA'$, so $PA \cdot GA + PB \cdot GC + PC \cdot GB \geq AA' \cdot GA$, with equality holding if and only if $A$, $P$, and $A'$ are collinear and $PBA'C$ is cyclic, i.e., if and only if $P = G$. Hence, the minimum value of the $AP \cdot AG + BP \cdot BG + CP \cdot CG$ is $GA^2 + GB^2 + GC^2= \frac{a^2+b^2+c^2}{3}.$

By law of sines on $\triangle BGL$, we have $\frac{BG}{\sin\angle BLG}=\frac{BL}{\sin\angle BGK}$. Similarly, in $\triangle CGL$ we have $\frac{CG}{\sin\angle CLG}=\frac{CL}{\sin\angle CGK}$. However, $L$ is the midpoint of $Bc$ and we have $\sin\angle BLG=\sin\angle CLG$, so $\frac{BG}{CG}=\frac{\sin\angle CGK}{\sin\angle BGK}$.

Now, since $\angle BOK=2\angle BGK$, with $O$ the circle's center, we have $BK=2R\sin\angle BGK$, where $R$ is the radius of the circle. Similarly, $CK=2R\sin\angle CGK$, so $\frac{CK}{BK}=\frac{BG}{CG}$, and hence $\frac{BG}{CK}=\frac{CG}{BK}$.

Similarly we have $\frac{AG}{BG}=\frac{\sin\angle BGN}{\sin\angle AGN}=\frac{\sin\angle BGN}{\sin CGK}$ since they are opposite angles. Since quadrilateral $BGCK$ is cyclic, we have $\angle BKC=\angle BGN\implies BC=2R\sin\angle BKC=2R\sin\angle BGN$. It follows that $\frac{BC}{CK}=\frac{\sin\angle BGN}{\sin\angle CGK}=\frac{AG}{BG}$, so $\frac{BG}{CK}=\frac{AG}{BC}$ and hence $BC:CK:BK=AG:BG:CG$.

Now, by Ptolemy's inequality on quadrilateral $PBKC$, we have $(PK)(BC)\le (BP)(CK)+(CP)(BK)$. Hence $(PK)(AG)\le (BP)(BG)+(CP)(CG)$. Therefore, \[AG(AP+PK)\le (AP)(AG)+(BG)(BG)+(CP)(CG)\] and it follows that $(AK)(AG)\le (AP)(AG)+(BP)(BG)+(CP)(CG)$, and equality holds if and only if $P$ lies on the circle between $B$ and $C$, which will give Ptolemy's equality, and $P$ lies on $AK$, for triangle equality. Thus the minimum is achieved when $P=G$.

Finally, by the law of cosines, we have  \[\begin{cases}AB^{2}=AL^2+BL^2-2(BL)(AL)\cos\angle ALB \\ AC^2=AL^2+CL^2-2(CL)(AL)\cos\angle ALC\end{cases}\] Adding the two equations yields \[AB^2+AC^2=2AL^2+\frac{BC^2}{2},\] so \[AG^2=\frac{4}{9}AL^2 = \frac{2AB^2+2AC^2-BC^2}{9}\] It follows that \[\text{min}\{AG^2+BG^2+CG^2\}=\frac{AB^2+BC^2+CA^2}{3}\] $\Box$

Note that $|AP|\cdot |AG|\ge (A-P)\cdot (A-G)$, where the right hand side is a vector dot product. Thus if $S$ is the desired sum,
\begin{align*}
S&\ge (A-P)\cdot(A-G)+(B-P)\cdot(B-G)+(C-P)\cdot(C-G)\\
&=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-P\cdot (A+B+C-3G)\\
&=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-P\cdot 0\\
&=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-G\cdot 0\\
&=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-G\cdot (A+B+C-3G)\\
&=(A-G)\cdot(A-G)+(B-G)\cdot(B-G)+(C-G)\cdot(C-G)\\
&=|AG|^2+|BG|^2+|CG|^2
\end{align*}
So $S$ is minimized when $P=G$ (and equality can only hold if $A,P,G$, $B,P,G$, and $C,P,G$ are collinear triples, i.e. if $P=G$). By well known-formulas the minimum value of $S$ can also be expressed as $\frac{1}{3}(a^2+b^2+c^2)$.

Let $X,Y,Z$ be the midpoints of $BC,CA,AB$. By the Law of Cosines on $APX$, we have that $PX^2=AP^2+AX^2-2\cdot AP\cdot AX\cdot\cos\angle PAX$, so $AP\cdot AX=\frac{AP^2+AX^2-PX^2}{2\cdot\cos\angle PAX}\leq\frac{AP^2+AX^2-PX^2}{2}$. By the length of the median formula on $BPC$, we have that $PX^2=\frac{BP^2}{2}+\frac{CP^2}{2}-\frac{BC^2}{4}$, so $AP\cdot AX\leq\frac{AP^2+AX^2-\frac{BP^2}{2}-\frac{CP^2}{2}+\frac{BC^2}{4}}{2}$. Summing through the other vertices, we have that $AP\cdot AG+BP\cdot BG+CP\cdot CG=\frac{2}{3}(AP\cdot AX+BP\cdot BY+CP\cdot CZ)\leq \frac{AX^2+BY^2+CZ^2+\frac{BC^2+CA^2+AB^2}{4}}{2}=\frac13(AB^2+BC^2+CA^2)$ by well-known formulas. Equality holds iff $\angle PAX=\angle PBY=\angle PCZ=0^\circ$, or iff $P=G$.

The only problem is the possibility that (WLOG) $\angle PAX\geq90^\circ$. This is easily resolved by some weak inequalities that are too annoying to type.

This solution was found with prior knowledge of what the answer was, but oh well, I still think it's pretty nice.

We first present a lemma.

[b]LEMMA:[/b] In any triangle $\triangle ABC$, we have \[AP^2+BP^2+CP^2=AG^2+BG^2+CG^2+3GP^2.\]

[i]Proof.[/i] We use physics.  Consider the set of mass points $\Sigma = \{1A,1B,1C\}$.  Let $J_{P,\Sigma}$ denote the moment of inertia of $P$ about $\Sigma$.  Note that $G$ is trivially the center of mass of $\Sigma$.  Now recall that by the Parallel Axis Theorem \[J_{P,\Sigma} = J_{G,\Sigma}+3GP^2,\] where the $3$ is the total mass of the system $\Sigma$.  Hence by the definition of moment of inertia we get the desired equality.  $\blacksquare$

Now note that by Law of Cosines we can rewrite this as \[AP^2+BP^2+CP^2 = AG^2+BG^2+CG^2+\sum_{cyc}\left(AP^2+AG^2-2AP\cdot AG\cos\angle PAG\right).\] This rearranges to \[\sum_{cyc}AP\cdot AG\cos\angle PAG = AG^2+BG^2+CG^2.\] Finally, note that $AP\cdot AG\cos\angle PAG\leq AP\cdot AG$ and similar, so we end up with \[AP\cdot AG+BP\cdot BG+CP\cdot CG\geq AG^2+BG^2+CG^2,\] with equality occurring when $P\equiv G$.  As others have pointed out above, this can be rewritten as $\tfrac13(a^2+b^2+c^2)$.

[hide=Solution]
We claim that the minimum value is $\frac{a^2+b^2+c^2}{3}$ and it occurs when $P=G$. By Stewart one can easily check that $AG^2+BG^2+CG^2=\frac{a^2+b^2+c^2}{3}$.
Place $O$ at the origin and let $\vec{X}$ denote the vector of point $X$. By dot product definition, $|\vec{AP}| |\vec{AG}| \cos \angle PAG = \vec{AP} \cdot \vec{AG}$. Therefore,
\[ \sum AP \cdot AG \ge (\vec{P}-\vec{A})\cdot(\vec{G}-\vec{A})+(\vec{P}-\vec{B})\cdot(\vec{G}-\vec{B})+(\vec{P}-\vec{C})\cdot(\vec{G}-\vec{C}) \]
Observe that $\vec{P}(3\vec{G}-\vec{A}-\vec{B}-\vec{C})=0$ so the inequality simplifies to
\[ \sum AP \cdot AG \ge \vec{A}\cdot\vec{A}+\vec{B}\cdot\vec{B}+\vec{C}\cdot\vec{C}-\vec{G}(\vec{A}+\vec{B}+\vec{C}) \]
Now $\vec{A}\cdot\vec{B}=R^2-\frac{c^2}{2}$, which is easy to derive, so $(\vec{A}+\vec{B}+\vec{C})\cdot\vec{G}=\frac{1}{3}(9R^2-a^2-b^2-c^2)$ and $\sum AP \cdot AG \ge \frac{a^2+b^2+c^2}{3}$. Equality holds iff $\cos \angle PAG = \cos \angle PBG = \cos \angle PCG = 1$, that is $P=G$. We have already checked that $P=G$ yields this minimum value so we are done.
[/hide]

Any barybash ? By totally thoughtless barybash, reduced the problem to given $a,b,c$ sides of a triangle, find all triples $(x,y,z)$ of reals such that  the following is minimized, $$ \sum_{\sigma} \frac{xy}{(x+y+z)^2} c_{xy} + \sum_{\sigma} \frac{x}{x+y+z} d_x $$, where $$c_{xy} = 6a^4 + a^2 \sum_{\sigma} a^2 $$ $$ d_x = 6c^2b^2-2(b^2+c^2)(\sum_{\sigma} a^2)$$ (and cyclic permutations)

[quote=darij grinberg]The minimum of the expression $AP\cdot AG+BP\cdot BG+CP\cdot CG$ is achieved when P coincides with the centroid G of triangle ABC, and this minimum equals $AG^2+BG^2+CG^2=\frac13 \left( a^2+b^2+c^2\right) $.

[i]Proof.[/i] Using vectors and their scalar products, we have $UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$ for any three points U, V and W, and equality holds if and only if the point W lies on the ray UV. Thus,

$AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$,

with equality holding if and only if the point G lies on the ray AP, on the ray BP and on the ray CP simultaneously, i. e. if and only if the point P coincides with the point G.

Now,

$\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$
$=\left( \overrightarrow{AG}+\overrightarrow{GP}\right) \cdot \overrightarrow{AG}+\left( \overrightarrow{BG}+\overrightarrow{GP}\right) \cdot \overrightarrow{BG}+\left( \overrightarrow{CG}+\overrightarrow{GP}\right) \cdot \overrightarrow{CG}$
$=\left( \overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2\right) +\overrightarrow{GP}\cdot \left( \overrightarrow{AG}+\overrightarrow{BG}+\overrightarrow{CG}\right) $.

Now, since the point G is the centroid of triangle ABC, we have $\overrightarrow{AG}+\overrightarrow{BG}+\overrightarrow{CG}=\overrightarrow{0}$, and thus

$\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}=\left( \overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2\right) +\overrightarrow{GP}\cdot \overrightarrow{0}$
$=\overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2=AG^2+BG^2+CG^2$.

After the well-known formula for the median lengths in a triangle, the length $m_a$ of the median of triangle ABC issuing from the vertex A satisfies the equation $4m_a^2=2b^2+2c^2-a^2$. Since the point G is the centroid of triangle ABC and thus divides all medians in the ratio 2 : 1, we have $AG=\frac23 m_a$, and thus

$AG^2=\left( \frac23 m_a\right) ^2=\frac{4m_a^2}{9}=\frac{2b^2+2c^2-a^2}{9}$.

Similarly,

$BG^2=\frac{2c^2+2a^2-b^2}{9}$;
$CG^2=\frac{2a^2+2b^2-c^2}{9}$,

and thus

$AG^2+BG^2+CG^2=\frac{2b^2+2c^2-a^2}{9}+\frac{2c^2+2a^2-b^2}{9}+\frac{2a^2+2b^2-c^2}{9}=\frac13 \left( a^2+b^2+c^2\right) $.

Hence, we have

$\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}=\frac13 \left( a^2+b^2+c^2\right) $.

Now, since

$AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$,

it follows that

$AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \frac13 \left( a^2+b^2+c^2\right) $,

with equality if and only if the point P coincides with the point G. In other words, the minimum of the expression $AP\cdot AG+BP\cdot BG+CP\cdot CG$ is achieved for P = G and equals $\frac13 \left( a^2+b^2+c^2\right) $. Proof complete.

  Darij[/quote]

"$UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$" what is this inequality?

[i]A synthetic solution, the motivation came from [url=https://artofproblemsolving.com/community/c6h18479p124397]IMOSL 1998 G4[/url]:[/i]
[rule]
For any point $P$ in the plane define $f(P) = AP \cdot AG + BP \cdot BG + CP \cdot CG$. Pick $P$ such that $f(P)$ is minimal. We will prove $P \equiv G$.

Let $K$ be the isogonal conjugate of $G$ wrt $\triangle ABC$ and let $D,E,F$ be the projection of $K$ onto sides $BC,CA,AB$, respectively.

[b][color=#f00]Claim: [/color][/b] $P$ must be contained in $\triangle ABC$.

[b][color=#00f]Proof: [/color][/b] Let $\mathcal R_A$ be the region of all points that lie on the same side of line $BC$ as of point $A$ (including all points on line $BC$). Define $\mathcal R_B,\mathcal R_C$ similarly. If $P \notin \mathcal R_A$, then if we denote by $P'$ the reflection of $P$ in line $BC$, then we have $f(P') < f(P)$, which would contradict the minimality of $f(P)$. Thus $P \in \mathcal R_A$. Similarly, $P \in \mathcal R_B$ and $P \in \mathcal R_C$. Now it is not hard to see that $\mathcal R_A \cap \mathcal R_B \cap \mathcal R_C$ is precisely the set of all points contained in $\triangle ABC$, which proves our claim. $\square$

[b][color=#f00]Claim: [/color][/b] $AK \sin A : BK \sin B : CK \sin C = AG : BG : CG$

[b][color=#00f]Proof: [/color][/b] It suffices to $\frac{AK}{BK} = \frac{AG}{BG} \cdot \frac{\sin B}{\sin A}$ as other results follows analogously. Let $Z = \overline{CG} \cap \overline{AB}$. Then,
\begin{align*}
\frac{AK}{BK} &= \frac{\sin \angle ABK}{\sin \angle BAK} = \frac{\sin \angle CBG}{\sin CAG} = \frac{\sin \angle GCB}{\sin \angle GCA} \cdot \frac{ \sin \angle CBG / \sin \angle GCB}{\sin \angle CAG / \sin \angle GCA}  \\
&= \frac{\sin \angle BCZ / BZ}{\sin \angle ACZ / AZ} \cdot \frac{CG / CB}{CG / CA} = \frac{\sin B / CZ}{\sin A /CZ} \cdot \frac{AG}{BG} = \frac{AG}{BG} \cdot \frac{\sin B}{\sin A}
\end{align*}
$\square$

Now by Sine rule we have
$$EF = AK \cdot \sin A ~,~ FD = BK \cdot \sin B ~,~ DE = CK \cdot \sin C$$
So because of the previous claim, we obtain that there is positive real number $t$ such that
$$EF = t \cdot AG ~,~ FD = t \cdot BG ~,~ DE = t \cdot CG$$
Now denote by $[\lambda]$ the area of any figure $\lambda$. Then,
$$AP \cdot AG + BP \cdot BG + CP \cdot CG = AP \cdot \frac{EF}{t} + BP \cdot \frac{FD}{t} + CP \cdot \frac{DE}{t} \ge \frac{[AEPF]}{t/2} + \frac{[BFPD]}{t/2} + \frac{CDPE}{t/2} = \frac{[ABC]}{t/2}$$
With equality iff $\overline{AP} \perp \overline{EF}$ , $\overline{BP} \perp \overline{FD}$ , $\overline{CP} \perp \overline{DE}$ ; which (because of the isogonal conjugate thing) is true iff $P \equiv G$.

So we have finally proven $\boxed{P \equiv G}$.

Now we want to find the value of
$$f(G) = AG^2 + BG^2 + CG^2$$
in terms of $a = BC$ , $b = CA$ , $c = AB$. But using appolonius theorem and the fact $G$ divides each median in the ratio $2:1$, we directly obtain that
$$\boxed{f(G) = \frac{a^2 + b^2 + c^2}{3}}$$
$\blacksquare$

[b][color=#f0f]Comment: [/color][/b] I think we can most probably avoid proving the first claim, but I am not exactly sure how to handle the configuration issues.

Solution with [b]islander7[/b].

Let \(\hat\bullet\) denote the unit vector. Let \(f(P)=AP\cdot AG+BP\cdot BG+CP\cdot CG\), which is convex, so at the unique point \(P\) that minimizes \(f\), \[0=\nabla f(P)=AG\cdot\widehat{\overrightarrow{AP}}+BG\cdot\widehat{\overrightarrow{BP}}+CG\cdot\widehat{\overrightarrow{CP}}.\] But \(\nabla f(G)=\vec{AG}+\vec{BG}+\vec{CG}=0\), so \(P=G\).

[quote=hydo2332]

"$UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$" what is this inequality?[/quote]

Follows from $a \cdot b = |a||b|\cos\theta$ where $\theta$ is the angle between $a$ and $b$. Alternatively its Cauchy-Schwarz.

[quote=OlympusHero]Hi all. I'm trying to use vectors to solve this Olympiad Geometry problem (I've actually never done this before!) Here's what I have at the moment:

We use vectors. Firstly, note that $G = \frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})$. We can rewrite the desired expression as $|\overrightarrow{P} - \overrightarrow{A}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{A}| + |\overrightarrow{P} - \overrightarrow{B}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{B}| + |\overrightarrow{P} - \overrightarrow{C}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{C}|$.

My natural next step would be to expand that out, but it seems like it's going to get extremely ugly. Is there a better way or am I just destined to bash this out and simplify (or do my best to) what I get?[/quote]

[quote=OlympusHero]Bumping this. Thanks in advance for any help![/quote]

Note that
\begin{align*}
AP\cdot AG +BP\cdot BG+CP\cdot CG &\ge \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG} \\
&= (\overrightarrow{AG}+\overrightarrow{GP})\cdot (\overrightarrow{AG}) + (\overrightarrow{BG}+\overrightarrow{GP})\cdot (\overrightarrow{BG}) + (\overrightarrow{CG}+\overrightarrow{GP})\cdot (\overrightarrow{CG}) \\
&= AG^2 + BG^2 + CG^2
\end{align*}
with equality if and only if $P=G$.

In what follows all point names except for $P$, $O$, and $G$ are taken cyclically.
Let $D,E,F$ be the feet of the altitudes from $P$ to $AG$, $BG$, $CG$ respectively, and let $O$ be the midpoint of $PG$. We have
\begin{align*}
\sum_{\text{cyc}} AG \cdot AP &\ge \sum_{\text{cyc}} AG \cdot AD \\
&= \sum_{\text{cyc}} (AO^2 - OG^2) \\
&= \sum_{\text{cyc}} (AG^2 + AO^2 - OG^2 - AG^2) \\
&= \sum_{\text{cyc}} AG^2 - (\vec{GA} \cdot \vec{GO}) \\
&= \vec{0} \cdot \vec{GO} + \sum_{\text{cyc}}AG^2 \\
&= \sum_{\text{cyc}} AG^2 .
\end{align*}
Thethird to last equality follows from the law of cosines, equality holding if and only if $D=E=F=G.$


Ring Theory
superior algebra
superior algebra unsolved

Hello,

yet again I am confused, my book is not clear. :|

Let $R$ be a graded ring : $R_{0}\oplus R_{1}\oplus \ldots$
Let $p$ be a homogeneous prime ideal of $R$

1.  Show that $T=R (R\backslash p)^{-1}$ is a graded ring. (so this is localisation)
2. Show that $T_{0}$ is a ring too.

I don't get it. :huh:   How can I make $T$ graded?  It's not that straightforward as it seems.
And then of course,the second problem is particularly hard when I don't even know the book's definition of $T_{0}$.
All ideas are welcome?

[quote="owk"]Usually, you consider only the localization w.r.t. the set of homogeneous elements not in p. owk[/quote]
I'm sorry, but that answer is a bit too fast for me, can you elaborate a bit perhaps?

I mean, what do the components of the new graded ring look like? :!:

*If* you localize only w.r.t. homogeneous elements (which is not what you have written), just set $\deg(a/s)=\deg a-\deg s$ for homogeneous $a,s$, and every element is a sum of such fractions. owk

Thx, but I still don't get it, I don't even know what "with respect to homogeneous elements" really means in this context?
Is anyone familiar with this? :maybe:

[quote="owk"]Let $S$ be the set of homogeneous elements of $R$ which are not contained in $\mathfrak p$. Then the localization $R[S^{-1}]=S^{-1}R$ is a graded ring with the grading as explained above (which is also the unique grading which is compatible with the canonical map $R\to S^{-1}R$ in the obvious sense). The ring of degree-zero elements of $S^{-1}R$ is denotes $R_{(\mathfrak p)}$, it is the local ring of the projective scheme $\mathrm{Proj}\,R$ at $\mathfrak p$, cf. Hartshorne p.76. owk[/quote]
I'm still confused. :(
My book says :

Exercise : Let $R$ be a graded ring and $p$ a homogeneous prime ideal.  The ring $R (R\backslash p)^{-1}$ is then a graded ring, and $(R (R\backslash p)^{-1})_{0})$ is a local ring

That's what it says. :!:
And I'm afraid your terminology is way too advanced for me, for instance I don't know what a projective scheme is. :o


function
integration
probability
calculus
advanced fields
advanced fields unsolved

Suppose we have a complex-valued function $ \phi(t)$, let it be a "good" function, integrable and differentiable. Suppose it's "inverse" Fourier transform $ p(x) \equal{}\frac{1}{2\pi}\int\phi(t) e^{\minus{}itx}dx$ is a probability density, so it is positive and it's integral is 1. Is it true that $ \phi(t)$ must be the characteristic function of distribution with density $ p(x)$? Or we have to check this by computing the Fourier transform of $ p(x)$ every time? More generally, do the function with the same Fourier transform have to be the same?

The Fourier transform is a bijection on the set of tempered distributions. In particular, it is one-to-one on integrable functions, or on finite measures.


Since 40% of a number is 84, we can therefore set up a proportion as follows:

$ \frac{40}{100}$ = $ \frac{84}{x}$

To solve this proportion, cross multiply:

$ 40x$=$ 8400$
$ x\equal{}210$

$ x\equal{}\fbox{210}$

forum glitch?

btw
40% of a number is 84
makes $ \frac{40}{100}x\equal{}84$(same thing, but easier to solve)


[quote="ifai"]Find all functions $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y\in\mathbb R:$
\[ f(xf(y))=(1-y)f(xy)+x^2y^2f(y)  \][/quote]

[hide]If $y=1$, then $f(xf(1))=x^2f(1)$. Since $f(1)=k$ is constant, $f(xk)=x^2k$, and $f(xk/k)=(x/k)^2k=x^2/k$. Thus \[ f(x) = \frac{x^2}{k}  \] for any constant $k$. Substituting back into the original equation, we have \[ \frac{x^2[\frac{y^2}{k}]^2}{k}=(1-y)\frac{x^2y^2}{k} + x^2y^2\frac{y^2}{k},  \] or \[ \frac{x^2y^4}{k^2} = x^2y^2(1-y+y^2),  \] and \[ \frac{y^2}{k^2} = y^2-y+1,  \] which isn't true for all $k$, so there are no solutions.
[/hide]
Is it right :?: I probably might have made a mistake in substituting. It seemed too easy though... :huuh:

[quote="chess64"][quote="ifai"]...  so there are no solutions.

[/quote][/quote]
 :cool:  What about $f(x)=0$  ?? :cool:

[quote="chess64"]...so there are no solutions.[/quote]
There are TWO functions satisifying the equation.

[quote="chess64"]Since $f(1)=k$ is constant, $f(xk)=x^2k$, and $f(xk/k)=(x/k)^2k=x^2/k$. Thus \[ f(x) = \frac{x^2}{k}  \] for any constant $k$.[/quote]
$k=f(1)$, so $k$ is not arbitrary.

[quote="chess64"][quote="ifai"]Find all functions $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y\in\mathbb R:$
\[ f(xf(y))=(1-y)f(xy)+x^2y^2f(y)  \][/quote]

Okay, let me try again :D

[hide]
If $y=1$, then we have $f(xf(1))=f(x)+x^2f(1)$. Hence $f(xf(1)/f(1))=f(x)=f(x/f(1))+(x/f(1))^2f(1)=f(x/f(1))+x^2/f(1)$.

uhh....
[/hide]

Okay, somebody else give it a try then :D[/quote]

It's not that hard, is it? :huh:

[hide=" The answer should be..."]$f(x)\equiv0,\ f(x)=x-x^2$[/hide]
And who can give the complete solution?

Let us put f(x) =0 aside and focus on other solutions:
Trivial. Put y=1 and let f(1) =b
then f(bx) = b x^2
or that
f(y) = y^2 /b for all y. one can see that this doesn't work unless b=0. hence f(0) = f(1)=0.

Now assuming that xy =1 we get that
f(f(y)/y) = f(y)
Since f(f(y)) = (1+y^2 -y) f(y) we see that if f(x)=f(y) then
y^2 -y = x^2 - x and therefore either x=y or y = 1-x.

hence either f(y)/y = y or f(y)= y-y^2 for all y.
If f(y) /y = y then f(y) =y^2   and so f(y^2) = (1+y^2 -y) y^2  which is not equal to either of y^4 or y^2 -y^4 in general
This indeed implies after some simple considerations (there are exceptional values of y, maybe only one or two values for which we don't get the contradiction above, but since they are limited in number we can ignore them...) that
f(x) = x - x^2 is the only non-trivial solution to this problem.
-Ali

What a beautiful solution! Keep going! :first:  :10:


trigonometry
function
algebra
polynomial
complex numbers
algebra solved

Find all numbers a,b,c,d,e, in the interval [-2,2] satisfying the system:

\[
\left\{ \begin{array}{l}
 a + b + c + d + e = 0 \\
 a^3  + b^3  + c^3  + d^3 + e^3  = 0 \\
 a^5  + b^5  + c^5  + d^5 + e^5  = 10 \\
 \end{array} \right.
\]

Let $w=\frac{\sqrt{5}-1}{2}$, then the solution is $(-1-w,-1-w,w,w,2)$. At least, it is what I have managed to find numerically.
I guess, the proof is quite boring, isn't it?

Yes, I think it is :). I actually had this on the contest, since I was in $10$'th grade in 2002 :). I saw that the main thing was putting $a=2\cos\alpha_1$ and so on and then using complex numbers, but left it like that (I think I didn't have time to finish it).

Pretty much the same. They wrote $a,b,c,d,e$ as $2\cos(\hbox{something})$, and then used complex numbers, all sorts of ugly (not [b]that[/b] ugly, but boring anyway) computations (like the ormula for $\cos(5x)$ as a function of $\cos x$; that can't be nice :)).

Denote $a=2\cos x,$ etc. By standard formulae, $2\cos 5x=a^{5}-5a^{3}+5a,$
hence
\[
\sum 2\cos 5x=\sum a^{5}-5\sum a^{3}+5\sum a=10,
\]
therefore
\[
\cos 5x=\cos 5y=\ldots =1.
\]
We find that $x,y\ldots \in \left\{ \frac{2k\pi }{5}|k\in \mathbb{Z}\right\}
.$ But $\cos \frac{2\pi }{5}=\frac{\sqrt{5}-1}{4},$ $\cos \frac{4\pi }{5}=-%
\frac{\sqrt{5}+1}{4},$ $\cos 0=1,$ hence $a,b,c,d,e\in \left\{ \frac{\sqrt{5}%
-1}{2},-\frac{\sqrt{5}+1}{2},2\right\} .$ Since $\sum a=0,$ we deduce that one of the numbers equals $2,$ other two equal $-\frac{\sqrt{5}+1}{2}$ and the others equal $\frac{\sqrt{5}-1}{2}.$ One can check that $\sum a^{3}=0,$
as well.

I forgot to say that your solution is cool.

[quote="Myth"]Please, tell what you were talking about in http://www.mathlinks.ro/Forum/viewtopic.php?t=25127[/quote]

There is a downloadable file with problems of UNIREA 2005 competition.
Some of the problems look nice but long to type in English.  Two short ones:

1.  $ (2^n - 1)!! = 1$ mod $2^n$

2.  If a monic polynomial has all its coefficients in the interval $[0,1]$ then any root $z$ has
 Re $z < (1 + \sqrt{5}) / 2$

Thank you!

Bogdan Enescu have already translated  for me  the most interesting problems :)


geometry
geometric transformation
rotation
LaTeX
linear algebra
matrix
rotation matrix

Let $ (x,y)$ be a pair of real numbers satisfying \[ 56x \plus{} 33y \equal{} \frac{\minus{}y}{x^2\plus{}y^2}, \qquad \text{and} \qquad 33x\minus{}56y \equal{} \frac{x}{x^2\plus{}y^2}.
\]Determine the value of $ |x| \plus{} |y|$.

[hide="Hidden by mod"]For constants $ a,\ b\ (ab\neq 0)$,

$ ax \plus{} by \equal{} \minus{} \frac {y}{x^2 \plus{} y^2}\ \cdots [1]$.

$ bx \minus{} ay \equal{} \frac {x}{x^2 \plus{} y^2}\ \cdots [2]$.

$ [1]^2 \plus{} [2]^2$ gives $ (a^2 \plus{} b^2)(x^2 \plus{} y^2) \equal{} \frac {1}{x^2 \plus{} y^2}\Longleftrightarrow x^2 \plus{} y^2 \equal{} \frac {1}{\sqrt {a^2 \plus{} b^2}} > 0$.

The problem seems to related to an inversion of circle and a rotation of matrix.[/hide]

Multiply the first equation by $ i$ and add it to the other equation. Let $ z\equal{}x\plus{}yi$. Then \[ (z)^{\minus{}1}\equal{}z\cdot ( 33\plus{}56i)\]
\[ z^{\minus{}2}\equal{}(7\plus{}4i)^2\]

The rest is easy...

I solved the solution by solving both equations for $ \frac {1}{x^2 \plus{} y^2}$ and then finding out that $ 4x \equal{} \minus{} 7y$

Which gives, quite nicely, $ x \equal{} \frac {7}{65}$ and $ y \equal{} \minus{} \frac {4}{65}$

But, I admit, Altheman's solution is about 40x more elegant and probably easier if you know it.

Edit... Ahh, poor latex!

I think that the most natural solution is to do LHS_1*RHS_2=RHS_1*LHS_2 then factor to get what you got basically.

[hide="Following kunnys discussion..."] Write as

$ \left[ \begin{array}{cc} 33 & \minus{}56 \\ 56 & 33 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] \equal{} \frac{1}{x^2 \plus{} y^2} \left[ \begin{array}{c} x \\ \minus{}y \end{array} \right]$.

The LHS is a [url=http://mathworld.wolfram.com/RotationMatrix.html]rotation matrix[/url] scaled by $ \sqrt{33^2 \plus{} 56^2} \equal{} 65$.  It follows that $ x^2 \plus{} y^2 \equal{} 65$.  Now what we have is that a particular matrix rotates $ (x, y)$ to $ (x, \minus{}y)$; in other words, the angle through which it rotates tells us the angle $ \left< x, y \right>$ makes.

Computationally, this is easiest to handle using complex numbers.  In fact, the complex numbers can be represented using real (rotation) matrices. [/hide]


A CD can store 120 minutes of sound. How many cassette
tapes with 75 minutes of recorded sound can be stored on
10 CD's?

in 10 CD's, there can be $ 120\times 10$ minutes of sound, or 1200 minutes
each cassette tape can hold 75 minutes of sound, so it would take $ \frac{1200} {75}$ tapes to hold 1200 minutes of sound

solving we get, $ \frac{1200}{75} = 16$

our answer is 16


Find all triples of integer numbers $ (a,b,c)$ such that

$ a\plus{}b\plus{}c\equal{}1$ and $ a^3\plus{}b^3\plus{}c^3\equal{}1$

 :)

what about cauchy ?

$ (a^3\plus{}b^3\plus{}c^3)(a\plus{}b\plus{}c)\ge (a^2\plus{}b^2\plus{}c^2)^2$
$ 1\ge a^2\plus{}b^2\plus{}c^2$. So $ a,b,c\le 1$.
Then $ (0,0,1)$ and permutations are the unique solutions i think  :oops:

What about $ (1,\minus{}17,17)$ :?:

note that it says integers, not only positive integers

verify how you applied Cauchy...

 :)

[quote="ElChapin"]Find all triples of integer numbers $ (a,b,c)$ such that

$ a \plus{} b \plus{} c \equal{} 1$ and $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 1$

 :)[/quote]

aren't there an infinite number of solutions? a general form for a solution would be:
(1,b,-b), please correct me if wrong-i dont think i understand the problem....

[quote="pelao_malo"]what about cauchy ?

$ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)\ge (a^2 \plus{} b^2 \plus{} c^2)^2$
$ 1\ge a^2 \plus{} b^2 \plus{} c^2$. So $ a,b,c\le 1$.
Then $ (0,0,1)$ and permutations are the unique solutions i think  :oops:[/quote]

Here, you have used cauchy assuming that $ a,b,c$ are positive, because you let the sequences be $ \sqrt{a}, \sqrt{b}, \sqrt{c}$ and $ a^{\frac32}, b^{\frac32}, c^{\frac32}$, which are only real if $ a,b,c\geq0$.

Thus, you still have to consider the case when any of $ a,b,c$ is less than 0.

$ 3(x \plus{} y)(y \plus{} z)(z \plus{} x)\equal{}(x \plus{} y \plus{} z)^3 \minus{} (x^3 \plus{} y^3 \plus{} z^3) \equal{} 1^3 \minus{} 1 \equal{} 0$

So WLOG $ y \plus{} z \equal{} 0$, then $ x \equal{} 1$, and this satisfies both equations...so $ (1,k, \minus{} k)$ and permutation where $ k\in\mathbb{Z}$.

[quote="Altheman"]$ 3(x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} (x \plus{} y \plus{} z)^3 \minus{} (x^3 \plus{} y^3 \plus{} z^3) \equal{} 1^3 \minus{} 1 \equal{} 0$

So WLOG $ y \plus{} z \equal{} 0$, then $ x \equal{} 1$, and this satisfies both equations...so $ (1,k, \minus{} k)$ and permutation where $ k\in\mathbb{Z}$.[/quote]
$ k\in\mathbb{Z}$
what is Z standard for?

[quote="linj"]$ k\in\mathbb{Z}$
what is Z standard for?[/quote]
$ \mathbb{Z}$ is the set of all integers.

What about the identity

$ x^{3}\plus{}y^{3}\plus{}z^{3}\minus{}3xyz\equal{}(x\plus{}y\plus{}z)(x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}xy\minus{}yz\minus{}zx)$
$ 1 \minus{} 3xyz \equal{} 1 ( 1^{2} \minus{} 3(xy \plus{} yz \plus{} zx)$
$ xyz \equal{} xy \plus{} yz \plus{} zx$
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$

And... where do I go from there?

[quote="b-flat"][quote="TheHeinrich"]for those who used cauchy: i think it only works for positive rationals. :wink:[/quote]

No; Cauchy works for all real numbers.  The problem (as noted in my last post) in the Cauchy solutions was that the sequences that were used were non-real in some cases.[/quote]

It is okay to think of cauchy as

$ (a_1\plus{}a_2\plus{}...\plus{}a_n)(b_1\plus{}b_2\plus{}...\plus{}b_n)\ge (\sqrt{a_1b_1}\plus{}\sqrt{a_2b_2}\plus{}...\plus{}\sqrt{a_nb_n})^2$ where $ a_i,b_i\ge 0$ for $ 1,2,...,n$.

[quote="iniquitus"]What about the identity

$ x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} zx)$
$ 1 \minus{} 3xyz \equal{} 1 ( 1^{2} \minus{} 3(xy \plus{} yz \plus{} zx)$
$ xyz \equal{} xy \plus{} yz \plus{} zx$
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$

And... where do I go from there?[/quote]
Since you got the identity $ xyz\equal{}xy\plus{}yz\plus{}xz$ you can go like this :
we will prove that $ (x\minus{}1)(y\minus{}1)(z\minus{}1)\equal{}0$
it is equivalent to $ xyz\minus{}xy\minus{}yz\minus{}xz\plus{}x\plus{}y\plus{}z\equal{}1$ which is true since the first condition is $ x\plus{}y\plus{}z\equal{}1$
and now Wlog $ x\equal{}1$ we obtain $ y\equal{}k$ and $ z\equal{}\minus{}k$ for some k interger

Nice job  :lol:  i completely missed that factorization.

I was too busy thinking of the reciprocals


When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:

$\textbf{(A)}\ 5\text{ in} \qquad
\textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad
\textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad
\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad
\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}$

[hide="Solution"]
$ 2\pi r \equal{} 20 \rightarrow r \equal{} \frac{10}{\pi}$
$ 2\pi r \equal{} 25 \rightarrow r \equal{} \frac{25}{2\pi}$
$ \frac{25}{2\pi} \minus{} \frac{10}{\pi} \equal{} \frac{5}{2\pi}$
D
[/hide]


I figured that the other post would get cluttered if I put the scoreboards in there also.  So I just made a new post.

Here are the results for the first 5 days.

[size=150][b]5 points:[/b][/size]

dts

JGeneson

mathfanatic

white_horse_king88

[size=150][b]4 points[/b][/size]

Magnara

[size=150][b]3 points[/b][/size]

Rep123Max

h_s_potter2002

pascal012

Yrael

hmmm

And there are a bunch of ppl with scores of <3.  If you believe I have a wrong score for you or you should be on this list pm me and we ll see about it.

Is hmmm a screen name? If it is... hmm.... yeah. LOL. I was confused.  :D  Thanks for hosting this - It does help alot because I usually don't have time to do whole sets.

"hmmm"  is a screen name

I am aiming to update it at least weekly.

[size=150][b]Week 2 Update[/b][/size]
[b]
10  <sup>+</sup> Points
[/b]

13   dts

13   JGeneson

10   white_horse_king88

[b]Next 6 highest[/b]

8    Yrael

6    Magnara

5    Mathfanatic

4    Rep123Max

4    RC-7th

4    pascal012

[size=150][b]Week 3 Update[/b][/size]

dts  18

JGeneson 16

Yrael  13

white_horse_king88 12

h_s_potter2002  8

[size=150][b]Week 4 Update[/b][/size]

dts  24

Yrael  17

JGeneson 16

white_horse_king88 13

h_s_potter2002  11

mathfanatic  9


function
trigonometry
calculus
calculus computations

Find the period of the function ( | sin x | + | cos x | )  .

[hide]Actually graphically i get it to be  $ \frac{\pi}{2}$ , any algebraic solutions . Anyways please post ur solution .[/hide]

$ |\sin x|$ and $ |\cos x|$ are both periodic with period $ \pi$, so their sum is periodic with period $ \pi$.  $ \pi$ is also the smallest period, since $ |\sin x| \plus{} |\cos x| > 1$ for $ x\in (0,\pi)$.

no but is that always true ? Actually what i feel is  that it has a period of $ \frac{\pi}{2}$ when i tried drawing the graph , not sure whether its correct ..can somebody check ?

Let $ f(x)\equal{}|\sin x|\plus{}|\cos x|$, we are to find the positive mimum value of $ T$ such that $ f(x\plus{}T)\equal{}f(x)$.

vijaymenon is correct: the period is $ \frac{\pi}2.$

The key observation: $ \left|\sin\left(x\plus{}\frac{\pi}2\right)\right|\equal{}|\cos x|$ and $ \left|\cos\left(x\plus{}\frac{\pi}2\right)\right|\equal{}|\sin x|$

A Fourier analysis idea for you could have $ f$ and $ g$ functions with period $ P$ while $ f\plus{}g$ has period $ \frac{P}2$:

$ f$ and $ g$ could both have Fourier series (with period $ P$) with all coefficients nonzero, but all of the odd coefficients of $ f\plus{}g$ could cancel out, leaving only even coefficients.

Obviously, we could do other things as well. $ f\plus{}g$ could have period $ \frac{P}7.$ Or $ f\plus{}g$ could be constant.

Its easy to prove that $ f\left( x \right) \equal{} f\left( {x \plus{} \frac {\pi }{2}} \right) \equal{} f\left( {x \minus{} \frac {\pi }{2}} \right)$
. By definition of period,  $ {\frac {\pi }{2}}$ is a period. If there was another one T' smaller than $ {\frac {\pi }{2}}$
then  T' should be one of $ \frac {\pi }{4},\frac {\pi }{6},\frac {\pi }{8},\frac {\pi }{{10}},...$ since $ {\frac {\pi }{2}}$ is one period.
For every $ x \in \left[ {0,\frac {\pi }{2}} \right]$
 we have $ f\left( x \right) \equal{} \sin x \plus{} \cos x \equal{} \sqrt 2 \sin \left( {x \plus{} \frac {\pi }{4}} \right)$
From relation $ f\left( x \right) \equal{} f\left( {x \plus{} T'} \right)$ if we put $ x \equal{} \frac{\pi }{4}$, since $ \frac{\pi }{4} \plus{} T' \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right]$, we get $ 1 \equal{} \cos \left( {T'} \right)$, contradiction

Thanks Kent .

But suddenly now i checked with wolfram alpha .here is what it showed [url]http://www.wolframalpha.com/input/?i=|+sin+x|+%2B+|+cos+x+|[/url]

Vijay, when you know you know more than the machine does, believe in yourself.

Ok thanks kent , maybe i should learn to believe myself .  :(  .


quadratics
function
algebra
quadratic formula
special factorizations

Hello;
happy new year.
Use Complete the Square Method to solve

hello, at first we divide the given equation by $ 2$
$ x^2\plus{}3x\minus{}\frac{3}{2}\equal{}0$
$ x^2\plus{}2\cdot \frac{3}{2}x\plus{}\frac{9}{4}\minus{}\frac{3}{2}\minus{}\frac{9}{4}\equal{}0$
$ (x\plus{}\frac{3}{2})^2\minus{}\frac{15}{4}\equal{}0$
$ (x\plus{}\frac{3}{2})^2\minus{}(\sqrt{\frac{15}{4}})^2\equal{}0$
$ (x\plus{}\frac{3}{2}\minus{}\frac{\sqrt{15}}{2})(x\minus{}\frac{3}{2}\minus{}\frac{\sqrt{15}}{2})\equal{}0$
from here you get all solutions.
Sonnhard.

ok, thank you very much Dr Sonnhard Graubner

Completing the square certainly works, but isn't is easier to use the quadratic formula?

$ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$

Completing the square shows us the veretex at a glance and along with the uncompleted form is enough to sketch the graph.

[quote="snowpaw"]I think the quadratic formula would be easier in this case because they're asking you to solve the problem, not graph it.[/quote]

That's basically what I meant.

I agree: in this case it is easier. I was speaking in general.

if u guys have graphing calculators, it is relatively simple to create a program that will tell you pretty much anything u want to know about a quadratic function, including vertex cords and solutions and stuff like that


algebra
polynomial
linear algebra
matrix
linear algebra theorems

can you tell me how i'd find the minimal polynomial of the matrix

 1  2  3
 0  1  2
 0  0  1

[/code]

The characteristic polynomial of the matrix is $(x-1)^3$. The minimal polynomial is a factor of the characteristic polynomial, so it is $(x-1)^3$, $(x-1)^2$ or $x-1$. Now try it out yourself.

Let $A$ be an $n\times n$ matrix. Then the characteristic polynomial of $A$ with respect to a certain variable, say $t$, is the determinant of the matrix $A-tI_n$.

I've only seen the characteristic polynomial being used to chas after the eigenvalues/eigenspaces of a matrix, so does it  matter what sign the leading coefficient is?

[quote="towersfreak2006"]I've only seen the characteristic polynomial being used to chas after the eigenvalues/eigenspaces of a matrix, so does it  matter what sign the leading coefficient is?[/quote]
No, it doesn't matter.
In fact, it is $\lambda E-A$ in most of books here in China.

Computationally, the degree of the minimal polynomial of $A$ is the least $d$ such that $I, A, \ldots, A^d$ are linearly dependent, and the minimal polynomial is given by that linear dependence.  You can write down the successive powers of $A$, 'flattening' them into rows of a large matrix and doing row reduction as you enter each new row until you can reduce a row to all zero.