ratio videos articles Princeton college Stanford geometry

http://www.universalseed.org

There have been many questions concerning the attacks of 9/11. I was wondering if any of you guys were into it.

Do you believe what the government has told you? The planes crashed into the World Trade Centers and they simply collapsed and a plane flew head on into the Pentagon.

Conspiracy theorists such as myself believe that the fire from towers could not have reached melting point temperatures. There must have been some other factor such as explosives used to bring down the towers in close to freefalling speeds. There must have been some other object that hit the Pentagon without leaving any plane wreckage on the site of the crash.

Many respected physicists and engineers have interesting hypotheses at http://www.scholarsfor911truth.org/ and http://www.physics911.net.

There has also been well illustrated documentaries such as Loose Change 2nd Edition produced by Dylan Avery that are available for free to the public. http://video.google.com/videoplay?docid=-8260059923762628848&q=loose+change+2nd+edition

I think USA sisn't know about what was going to happen. They thought they were secure enough, but terrorism is terrorism, and they defeated their security... however, what happened was horrible, a really bad thing

[quote="SuperNova"][quote="Seekeroftruth"]Conspiracy theorists such as myself believe that the fire from towers could not have reached melting point temperatures.[/quote]
The fires wouldn't have to reach melting point temperatures - if you heat metal, it weakens long before it melts.[/quote]

Prior to September 11th, no steel frame building has ever collapsed due to fire. The towers were designed so they could hold 6 times their own weight.

One of your links (physics911.net) say this:

"The official story maintains that fires weakened the buildings. Jet fuel supposedly burned so hot it began to melt the steel columns supporting the towers. But steel-framed skyscrapers have never collapsed from fire, since they're built from steel that doesn't melt below 2750 degrees Fahrenheit. No fuel, not even jet fuel, which is really just refined kerosene, will burn hotter than [b]1500[/b] degrees Fahrenheit."

If you look at this website, [url=http://www.cwc.ca/design/fire/index.php]http://www.cwc.ca/design/fire/index.php[/url], there's a graph showing the ratio of the strength of steel at temperatures of up to 1000 C to its strength at 20 C (room temperature). 1500 F is approximately equal to 815 C. If you look at 815 C on the chart, you'll see that the ratio is about .15 - less than one-sixth. Now, if the towers were designed to hold six times their weight...

[quote="seekeroftruth"]Prior to September 11th, no steel frame building has ever collapsed due to fire.[/quote]
I don't see how this proves anything. Just because it's never happened before doesn't mean it couldn't happen, especially considering that 9/11 was intentional and not an accident.

That is 6 times from the ground up. Considering the steel from the bottom of the towers were not burning and on fire and that 1500 degrees fahrenheit would be a maximum at the locations of the jet fuel, the whole towers should not have collapsed but maybe something like the building would have tipped over, right?

And well... you're right about the possibilities of it happening.

This here looks the same like all those claims that humans never landed on the moon/everything about that was faked: they just look true for someone who doesn't know much about the subject; but for experts, it is just stupid nonsense  (note that I just call the attempts stupid, I want not to clear if it happened or not ;) ).

Same here, I think (I know nothing about statics etc., but):
- do towers of that type tip over that easy¿
- is it really six times their wheight (we speak of real static load, not of idealistic ones!)
- there was more damage than just that one by fire, e.g. the hit itself (and the towers were not the newest buildings)
- how else then¿ explosives can be excluded I think since it hould be impossible to remove tracks of them and to let it look naturall...
- but the real question: what's the sense of doing that¿ (even if you claim e.g. to start a war, the hit itself would have sufficed)

So in short:
Can it really happen, based on facts from people having nothing to do with both viewpoints¿ And why¿ And how¿

who would be heartless enough to kill thousands of their own citizens? we're talking about the melting point of steel, but if you think about it, killing people in the U. S. in order to kill people in Iraq doesn't make much sense.

Yeah, that's another reason why I find this whole conspiracy theory unlikely.

[quote="Totally Zealous"]I believe the attacks were carefully planned for a long time. I read a whole big huge long list of "coincidences" that occured with 9/11 on a myspace bulliten. Such as it happened on 9/11, 911=emergency number. There was 254 passengers on the plane. 2+5+4=9+1+1=11. The madrid bombing happened 254 days after 9/11. 9/11 is the 254th day of the year. Stuff like that. And when you type up the flight number or something and change the font to windings (one of those picture fonts), a bunch of pictures shows up with plane, bomb, star of david, letter, or something. Anyway I will try and find the list for you.
I don't think all these "coincidences" were "coincidences".[/quote]
 [url=http://www.rotten.com/library/occult/numerology/]Numerology![/url]OMG!

I believe these were secretly planned by evil Pi nazi terroristz. The evidence is obvious!

9/11=$3^{2}/11$! Carrying over the extra three, we get 3/14! Coincidence? I think not!

On a more serious note, [url=http://www.rotten.com/library/conspiracy/september_11_conspiracy_theories/]conspiracy theories are a load of ----.[/url]

I highly doubt that US would kill its citizens.

But...I agree, it was most certainly planned. By terroists.

But, as they say, the impossible can happen.
There's always a first....

...but, seriously, I doubt that

I don't know, remember the burning of the Reichstag or the Mukden incident?  It wouldn't be the first time a nation damnages its own property in order to consolidate their regime and/or justify military action.

However, the number of casualties is enormous when talking about 9/11.

There is a video on the web giving numerous arguments : Bin Laden does not look like the man in the video released in december 2001 (which seemed to be evidence for the White House that Osama Bin Laden is involved in the 9/11 attacks, even though the USA didn't bother to begin the war anyway).  The man in the video also writes with the wrong hand and wears a ring (which is not allowed by his religion).  All his arguments however are not that strong.
It also claims the USA government published names very shortly after the attacks of hijackers that actually turned out to be alive.

There are other things that I have heard that I find weird : why is it not obvious from Pentagon images that it was a plane, why did the hijackers decide to fly around  the Pentagon (which did cost them more time) to hit the western wall?  The Pentagon is special as it is the only low target that was succesfully hit that day, some claim no pilot is good enough to do that job.

But then again : George Bush really acted confused and didn't know what to do when you look at those images in 9/11.  Would he have reacted that way if he knew about it?  Is he smart enough to play that dumb?  Is he smart enough to keep up with these lies for all these years?

I don't think the attack was staged by America. But the terrorists like Osama bin Laden have most definitely been raised by the US Government. Most probably they were meant to fight the eastern bloc, but the cold war was suddenly over. Driven by Islamic fundamentalism, these terrorists have turned against their first patrons. What a pity!

of course [quote]But then again : George Bush really acted confused and didn't know what to do when you look at those images in 9/11. Would he have reacted that way if he knew about it? Is he smart enough to play that dumb? Is he smart enough to keep up with these lies for all these years?
[/quote]

And little kid could pretend they are confused when they hit their sibling...

Okay, perhaps, but still he has had to keep up with these lies all this time.  And compared to his intelligence as we know it (read this [url]http://en.wikipedia.org/wiki/Bushism[/url] ) I find that hard to believe.  Or are you gonna tell me he has been playing dumb since birth?

The thing with 9/11 conspiracists  is :

they always point out usual things (like the way the Pentagon was hit, the refusal to release all of the recorded data,... ) but they don't seem to present a complete alternative version of what happened.

Who shot those rockets?  Who knew of this? Who had dealings with muslim fundamentalists willing to do this?  How many of the hijackers knew that they were being used?  ...

Read the book crossing the rubicon...and you will think that something is fishy...

Perhaps I should.

But here is yet another question :

After 9/11 the USA presented questionable about the identities of the hijackers. Proof that Bin Laden was behind the attacks when they were already in Afghanistan,and the proof was questionable.  And even so, how can we be so sure the Taleban government was behind the attacks or that Bin Laden has even been in Afghanistan since 9/11?

But even more important is this :

the 'case for Iraq' was laughable.  The WMD were a joke, and the ties between Zarqawi and Saddam had to confirm some sort of tie between Al Qaeda and the Baath party.  In my opinion there wasn't any more reason to attack Iraq than on 10 september 2001.

So : if this was orchestrated to justify interventions in the middle east, don't you they would have made it a bit more convincing.  Even when one completely believes that Bin Laden is behind 9/11 and the USA wasn't all, it is already hard to justify the Iraq war.  So if the conspiracists are correct, the USA government is either very twisted or just plain stupid.

When I saw that 'documentary' a few weeks ago, I was amazed by the Pentagon-story. Before, I never wondered how a plane could fly that low. I'm 99% sure it wasn't a plane. Ever seen a plane's engine blow things away (America's Funniest Home Videos for example  :D ) ? Cars would have been blown away etc. As Frédéric stated, "some claim no pilot is good enough to do that job". And then there's that story about the five images released. It must have been a missile. But that doesn't mean it was an American missile  :P

The story about the WTC was less convincing. Okay, they were built to resist such attacks, the melting point temperatures weren't reached and there were a lot of "second explosions" (like the one on the ground flour of the North Tower, but wasn't that because of the elevators?). But I guess those coincidences happen.

However - and this is quite off-topic - I don't think an imbicile could become president of the USA. Things like Bushisms are really ridiculous to use as argument againt Bush. That's like saying that he invaded Iraq because of oil. He's against non-democratic regimes. Yup. Yup. Nothing else. I saw a documentary on Belgian television which provided interviews with revolutionairies that succeeded. (Unfortunately, I don't remember the name of the countries.) They all admit that America helped them. In fact, there were even people talking about Americans teaching revolutionairies how to get attention from the media, how to win people for their ideas, how to organize their protest etc. There were a few other programs about America on that network which showed that Bush isn't thát stupid. (At least, that's how I felt after seeing them.) I guess you didn't see them, Frédéric? Really, I don't think he's as dumb as people think he is. Altough, I must admit, http://www.youtube.com/watch?v=FOomGp8AnH8&search=bush%20G8 is really funny.
So, I guess they are capable of faking those attacks. But would they really do that? Let's say I'm between the two opinions: I think the government knew about it, but I don't think they planned it.

By the way, PenguinIntegral, I agree. Evil Pi nazi terroristz díd attack the USA! That 9/11 = 3²/11 --> 3/14 is so marvellous  :omighty:

I think that none of these "conspiracy" theories have any credibility whatsoever. On the other hand, I do believe that this administration has done a very bad job on terrorism, and there is evidence that we could have been able to prevent these attacks (apparently the CIA had tapes in Arabic planning these attacks, but of course giving money to very wealthy people is a much more important use of it than paying for Arabic translators to protect against terrorism  :D ).

[quote="calc rulz"]I think that none of these "conspiracy" theories have any credibility whatsoever.[/quote]

Actually the same could be said of the official 9/11 commision report. All the steel from WTC 1,2 [i]and 7[/i](which "collapsed"  due to "structural damage and fire" later that day) was shipped off to China to be melted and recycled before any true investigation could take place.

You may also take into account that [b][u]many[/u] [/b]security tapes from local buildings such as a gas station and hotel displaying the plane crashing into the Pentagon have been seized by the FBI and are not able to be viewed by the general public.

http://www.gpoaccess.gov/911/index.html <-- 9/11 commision report. [i]Interesting [/i]read especially the part about telephone calls being made at the supposed altitudes.

I repeat my objection :

if 9/11 was orchestrated to gain influence in the Middle East : Afghanistan and Iraq they could have done it a bit better, even if one believes Osama Bin Laden is the main culprit and that he was in Afghanistan, Iraq is still hard to buy.

I personally do not believe in the theory, but this is an article I found on [url]http://www.comcast.net[/url]

9/11 Conspiracy Theorists Thriving
By JUSTIN POPE, AP Education Writer
3 hours ago

Kevin Barrett believes the U.S government might have destroyed the World Trade Center. Steven Jones is researching what he calls evidence that the twin towers were brought down by explosives detonated inside them, not by hijacked airliners.

These men aren't uneducated junk scientists: Barrett will teach a class on Islam at the University of Wisconsin this fall, over the protests of more than 60 state legislators. Jones is a tenured physicist at Brigham Young University whose mainstream academic job has made him a hero to conspiracy theorists.

Five years after the terrorist attacks, a community that believes widely discredited ideas about what happened on Sept. 11, 2001, persists and even thrives. Members trade their ideas on the Internet and in self-published papers and in books. About 500 of them attended a recent conference in Chicago.

The movement claims to be drawing fresh energy and credibility from a recently formed group called Scholars for 9/11 Truth.

The organization says publicity over Barrett's case has helped boost membership to about 75 academics. They are a tiny minority of the 1 million part- and full-time faculty nationwide, and some have no university affiliation. Most aren't experts in relevant fields. But some are well educated, with degrees from elite universities such as Princeton and Stanford and jobs at schools including Rice, Indiana and the University of Texas.

"Things are happening," said co-founder James Fetzer, a retired philosophy professor at the University of Minnesota Duluth, who maintains, among other claims, that some of the hijackers are still alive. "We're going to continue to do this. Our role is to establish what really happened on 9/11."

What really happened, the national Sept. 11 Commission concluded after 1,200 interviews, was that hijackers crashed planes into the twin towers. The National Institute of Standards and Technology, a government agency, filed 10,000 pages of reports that found fires caused by the crashing planes were more than sufficient to collapse the buildings.

The scholars' group rejects those conclusions. Their Web site contends the government has been dishonest. It adds: the "World Trade Center was almost certainly brought down by controlled demolitions" and "the government not only permitted 9/11 to occur but may even have orchestrated these events to facilitate its political agenda."

The standards and technology institute, and many mainstream scientists, won't debate conspiracy theorists, saying they don't want to lend them unwarranted credibility.

But some worry the academic background of the group could do that anyway.

Members of the conspiracy community "practically worship the ground (Jones) walks on because he's seen as a scientist who is preaching to their side," said FR Greening, a Canadian chemist who has written several papers rebutting the science used by Sept. 11 conspiracy theorists. "It's science, but it's politically motivated. It's science with an ax to grind, and therefore it's not really science."

Faculty can express any opinion outside the classroom, said Roger Bowen, general secretary of the American Association of University Professors. However, "with academic freedom comes academic responsibility. And that requires them to teach the truth of their discipline, and the truth does not include conspiracy theories, or flat Earth theories, or Holocaust denial theories."

Members of the group don't consider themselves extremists. They simply believe the government's investigation was inadequate, and maintain that questioning widely held assumptions has been part of the job of scholars for centuries.

"Tenure gives you a secure position where you can engage in controversial issues," Fetzer said. "That's what you should be doing."

But when asked what did happen in 2001, members often step outside the rigorous, data-based culture of the academy and defer to their own instincts.

Daniel Orr, a Princeton Ph.D. and widely published retired economics chair at the University of Illinois, said he knew instantly from watching the towers fall that they had been blown apart by explosives. He was reminded of watching an old housing project being destroyed in St. Louis.

David Gabbard, an East Carolina education professor, acknowledges this isn't his field, but says "I'm smart enough to know ... that fire from airplanes can't melt steel."

When they do cite evidence, critics such as Greening contend it's junk science from fellow conspiracy theorists, dressed up in the language and format of real research to give it a sense of credibility.

Jones focuses on the relatively narrow question of whether molten metal present at the World Trade Center site after the attacks is evidence that a high-temperature incendiary called thermite, which can be used to weld or cut metal, was involved in the towers' destruction. He concludes thermite was present, throwing the government's entire explanation into question and suggesting someone might have used explosives to bring down the towers.

"I have not run into many who have read my paper and said it's just all hogwash," Jones said.

Judy Wood, until recently an assistant professor of mechanical engineering at Clemson University, has been cited by conspiracy theorists for her arguments the buildings could not have collapsed as quickly as they did unless explosives were used.

"If the U.S. government is lying about how the buildings came down, anything else they say cannot be believed," she said. "So why would they want to tell us an incorrect story if they weren't part of it?"

In fact, say Greening and other experts, the molten metal Jones cites was most likely aluminum from the planes, and any number of explanations are more likely than thermite.

And the National Institute of Standards and Technology's report describes how the buildings collapsed from the inside in a chain reaction once the floors began falling.

"We respect the opinions of others, but we just didn't see any evidence of what people are claiming," institute spokesman Michael Newman said.

Wisconsin officials say they do not endorse the views of Barrett, an adjunct, but after investigating concluded he would handle the material responsibly in the classroom.

That didn't mollify many state legislators.

"The general public from Maine to Oregon knows why the trade towers went down," said state Rep. Stephen Nass, a Republican. "It's not a matter of unpopular ideas; it's a matter of quality education and giving students their money's worth in the classroom."

In a July 20 letter obtained by The Associated Press in an open records request, Wisconsin Provost Patrick Farrell warned Barrett to tone down his publicity seeking, and said he would reconsider allowing Barrett to teach if he continued to identify himself with the university in his political messages.

BYU's physics department and engineering school have issued statements distancing themselves from Jones' work, but he says they have not interfered.

At Clemson, Wood did not receive tenure last year, but her former department chair, Imtiaz ul Haque, denies her accusation that it was at least partly because of her Sept. 11 views.

"Are you blackballed for delving into this topic? Oh yes," Wood said. "And that is why there are so few who do. Most contracts have something to do with some government research lab. So what would that do to you? The consequences are too great for a career. But I made the choice that truth was more important."

"If we're in higher education to be trying to encourage critical thinking," Wood says, "why would we say 'believe this because everybody else does?'"

___

On the Net:

Scholars for Truth: http://www.scholarsfor911truth.org/

Nat'l Institute of Standards: http://wtc.nist.gov/

Debunking Conspiracy Theories: http://www.debunking911.com/

Copyright 2006 The Associated Press. All rights reserved. This material may not be published, broadcast, rewritten or redistributed.

[color=red]note this is not plagiarism because I stated the website I got it at and that the AP wrote it.

I personally do not believe in this.[/color]

http://www.telegraph.co.uk/news/main.jhtml?xml=/news/2001/11/11/wbin11.xml

Enough said to be honest.

These conspiracy theories are highly unlikely most people seem to go along with them as thanks to individuals such as Michael Moore, being anti-US is now fashionable.

[quote="fredbel6"]Very relevant link, but it doesn't completely exclude the possibility of USA's involvement :

- have we seen this video?
- is he telling the truth and the whole truth?
- does he know the whole truth

Remember, there have been terrorist attacks that were claimed by dozens of groups.  It's easy to claim a terrorist attack, it's harder to do it.[/quote]

The thing is, you have to remember that these people are extremely anti-western, Im sure if they were asked to blow up a target int he US by the US government they would refuse to do so as an act of defiance against them.

The question in my opinion is this:

Osama Bin Laden when he ordered this attack would know that all the major powers in the world would be hunting him down for the rest of his life, would he really be willing to make this sacrifice for doing a favour to the United States? No.

Fair enough - I accept the whole steel melting theory. Perhaps we dont know what actually happened within the buildings. What I find difficulty accepting is the fact that as soon as there is any grey area or any doubt they automatically pin the situation on the government. People like Michael Moore who believes he knows better than a government which consists of thousands of hard working and highly trained people.

Just my opinion on the situation,

Some of my first posts here hopefully the first of many!

-Malcolm

I'm only saying there are so many variations possible:

- US had nothing do with it, bin Laden planned it all
- without bin Laden's knowledge, USA allowed them to continue
- without bin Laden's knowledge, CIA or another agency infiltrated their organisation during preparations
- bin Laden made plans with certain people in the USA government
- bin Laden was not at all involved, except that his organisation was to take the blame
....

Well, does the "official story" make much sense.  bin Laden delivers a hard blow to the USA.  How did he think that would play out? Maybe he figured there would be considerable doubt about his location and thus no action against Afghanistan?  I don't think so however, and he should have known that the Taleban, already unable to control more than two thirds of Afghanistan, would stand no chance.  So all he accomplished was destroying the "only real Islamic country" in a few months?!

[quote="chess64"]You all need to watch [url=http://www.loosechange911.com/]Loose Change[/url]. After watching that video, it is impossible for a sane person to disagree that the government was responsible for the attacks.[/quote]

I am interested :
1. Can I download that film( I don't like streaming)
2. Have you done it and can you confirm that the site is safe?

fredbel-

I'm not sure if you can download it or not, but the site is safe. The link on the website takes you to google videos. I recommend this video to everyone.

That stream doesn't seem to work that is why I  asked.
 :(

http://video.google.com/videoplay?docid=1519312457137943386

Google video so I'm sure its a pretty safe site.

Enjoy. (Not that I did)

As far as the US staging attacks to gain support for war, [url=http://en.wikipedia.org/wiki/Operation_Northwoods]Never say never[/url] (though this never came to fruition).  Furthermore, parts of the Gulf of Tonkin Incident were lied about, but in that case, there was an actual attack (it was just embellished).

Three people who say you're wrong and make some interesting arguments.  (One of them is me.)[quote]Documentaries can make Hitler look like a nice guy.  They really can.

I mean, you have to accept the fact that the United States is not out to kill you. Look, the people that make these documentaries are alive. They are not in Guantanamo Bay, they are not being tortured in Iran, they are not being tortured in Syria, because they're not freedom fighters, they're crackpots.

And there's a reason nobody bothers to debunk conspiracy theories. Nobody really tries to disprove that Hitler isn't on a moon base on Mars, or that there aren't aliens in some English woods.

The World Trade Centers were designed to withstand unbelievable amounts of strength. Unfortunately, the WTC and pretty much every single skyscraper ever were built completely differently. Every other skyscraper ever was built with a skeleton of supporting interior structures designed to keep it up, and mostly out of concrete, which spans the whole thing. The World Trade Centers were built with a core and load-bearing walls. No other skyscraper features load-bearing exterior walls.

When a plane goes through load-bearing walls, they cease to bear weight.

There's a problem with some of the comparisons, too: Concrete does not burn. No amount of heat will stop concrete. (short of melting it into lava, I suppose.)  However, steel does, in fact, weaken.

The most ridiculous comparisons. The statement, "Two thousand degrees of heat weakened the steel." They made it into, "Two thousand degrees of heat melted the steel and collapsed the towers." Unfortunately, weakened steel does not have to melt to structurally fail.

And this is on top of the core of the center being gutted and approximately 1/3 of the load-bearing exterior walls demolished.

Here's some factual evidence:
http://www.popularmechanics.com/science/defense/1227842.html
http://vincentdunn.com/wtc.html[/quote]
[quote]I don't think I've seen that particular documentary (didn't click the link) but without going into specifics, I can address the problems with the Gov being directly involved in the 9/11 attacks.

The problem with conspiracy theories is that they assume people can keep their mouths shut. Imagine the number of people involved in a massive conspiracy to kill three thousand Americans. Several hundred at a bare minimum, several thousand, more likely. With every additional person who's added to the conspiracy, there's a whole set of things that can go wrong. Someone could be mistaken as immoral, and once their told about it, they go to the newspapers; someone could have a change of heart during the process; someone could even just let it slip to a family member. There is just no feasible way something that massive could be hidden.

Furthermore, do you know Occam's Razor? Basically, it states that the simplest explanation is probably the right one. What's simpler? A group of terrorists in the middle east were able to commandeer several airliners, and crash them into buildings; vs. a group of terrorists, inspired, funded, and aided by the country they were attacking, commandeered several airliners, and crashed them into buildings.

Think of all the extra work that would need to go unnoticed. Someone trustworthy (read: high ranking) would have to visit Afghanistan, or at least someone from al-qaeda, and persuade them, not only that they're serious, but that they are trustworthy. Then funding would have to go unnoticed by the general public, and more importantly the media. All sorts of stuff would have to work perfectly if this was to work; and them not get caught.

Look at Moussaoui (sp), he got caught alive, and was supposedly involved, if not a central organizer. Yet he didn't even claim the administration was involved. We haven't caught Osama, but we've caught many other high up Al-Qaeda peoples, none of whom claim Bush was involved.

Like I said, I haven't looked at the documentary; but this is just a generic logical argument against conspiracy theories, that I've almost never seen successfully countered.[/quote]
[quote]Lots of stuff. Mike covered most of the personal aspects, and too_late_again got a good start on the factual ones. I'm here to give some sociological explanations:

Look at it from the conspiracy theorist's POV. They have every motivation to make these outlandish claims---which, as too_late_again showed pretty well, they can justify by basically sounding convincing while pulling bullshit out of their collective butts. They get fame, they get coverage, they get their political goals advanced; they get to feel special, they get to feel like people pay attention to them... it's much easier not to fact-check when you have a political motivation for listening to those who are feeding you the facts. There's no reason for them to attempt truthfulness at all, whereas there's a lot of reasons why they'd want to B.S. stuff.

Furthermore, the most convincing quote I found somewhere went something like this:
[b]Why would a government able to pull off a massive real-time deception of the American public, on national TV, without any leaks, and get a bunch of Middle Easterners to go along with it.. NOT BE ABLE TO PLANT ANY WMD IN IRAQ?[/b]

So, they got the reaction they wanted from you... they also probably got a decently large amount of AdSense impressions... they make out great, you come away confused, all a day in the life of a conspiracy theorist with a paycheck to take home.

If you're really interested in details about the 9/11 attacks, the 9/11 Comission Report is a good place to start. You can't go about pretending to be objective if you look at stuff cobbled together by conspiracy theorists with Windows Movie Maker and an internet connection, without first looking at the data and investigations of an independent comission.[/quote]

What is so complicated in the matter.

The US Government are not staging the attracks. It is quite plain. But those that are came to power under US Patronization. That is the simple fact.

Okay this is on a comical side note (yes Bubka I'm not always serious  :wink: ):

[hide]How are we so sure 9/11 happened in 2001?  Conspiracists believe the USA government is consistently pushing it back to justify all their wars in the end, there is no doubt that there is considerable room for doubt,according to this article:
[url=http://news.yahoo.com/s/afp/20060809/od_afp/usattackspolloffbeat_060809145351;_ylt=ArnrtaXH3JkyylylPgoybNSs0NUE;_ylu=X3oDMTA3NW1oMDRpBHNlYwM3NTc]link[/url][/hide]

Okay, I was just watching V for Vendetta (for like the 16th time) and saw this thread and paused it right after Rockwood finishes telling his story. Imagine the irony.

But seriously, someone's been watching too many movies. :P

When even some everyday crimes such as arson and rape are published by national papers before they're proven, you must wonder why conspiracy theories on 9/11 have gained relatively little attention in the industry. Let me give you a hint: it's not because they're afraid of being bagged by a guy named Creedy.

Loose change is complete bunk. You can find refutations if you google them.

For those interested and who can receive it :

Flemish public channel Canvas will broadcast a shorter version of "Loose Change 2" on Monday 21 august 20.00 ( 8 pm),
and an anti-complot-theory documentary "9/11 did not take place" on Tuesday 22 august 20.00 (8 pm)

Both of them will be repeated all night long, starting a few hours after that.

Belgian magazine HUMO interviewed Avery and anticomplot people this week.

[quote="bubka"]Curbing the freedom of press might help.  :wink:[/quote] :|? I don't get it?

NVM!!! :blush:

And no , i pretty much don't believe in conpiracy theories, any kind of them

[quote="alex__ib"]Hmmm, i found something interesting.The "number" of that flight was Q33 NY...Now, write that in a word document and change the font to windlings  :ninja:

ANd no , i pretty much don't believe in conpiracy theories.[/quote]

Not again!..... :(

The flight numbers were AA11, AA77, UA93, UA175.
It's funny how most people immediately check that in Word, but don't check whether or not that was actually the flight number :huh:


What are the most important properties of complex numbers? For example, conj(w+z) = conj(w) + conj(z). Proofs and when they are used would be helpful too.

One of my favorite theorems in all of mathematics.

DeMoivre's Theorem:
eit = cos t + i*sin t

Or equivalently,
a + ib = r*eit, where r = :sqrt:(a2 + b2) and t = arctan (b/a)

The exact proof involves calculus. But it has some nice geometrical representations.


if you mix 1/3 of x degrees celcius of water and 2/3 of y degrees celcius of water, the result would be

(1/3)x + (2/3)y......?

p.s. or do you have to convert to kelvins??????

$\frac{2}{3}(y-t) = \frac{1}{3}(t-x)$

 $t = \frac{2}{3}y+\frac{1}{3}x$

 For the second question, $No$. it always cancels out,
 because we are not concerned with absolute temperatures
 but with the diference between them, which fortunately
 happens to be the same in the Centigrade and Kelvin scales.

omg..... thank you so much!! seriously..


We know that in nuclear reactions , mass is converted into energy which appears as Binding Energy. Now, when we consider electrons coming close to the nucleus during nuclear/atomic phenomenon, can we consider that the [b]energy present in the electric and Nuclear fields is converted into mass[/b]?? Can this be generalised to all forces and energy fields..i.e. If a [b]body moves in a certain force field associated with a vector field, does energy change to mass[/b]??? When do such changes take place? [b]Why/Why not?[/b]

I think that I can't really imagine it :) . The only example of this phenomenon that I'm aware of is the nuclear field - mass conversion during nuclear fission / fusion. Could you present some other example that you consider or that actually exists? I don't really understand the "electron close to nucleus" one... sorry :)

My idea is just sort of an imagination. I'm just asking this question from a theoretical point of view - when are we allowed to interchange energy and mass. Does it play a role everywhere????

In radioactive decay, the daughter nucleus is most often created in an excited state. Sometimes, the angular momentum difference between the excited and ground states is large, which makes the deexcitation by electromagnetic interaction "highly forbidden", because angular momentum difference has to be carried away by the emitted $\gamma$-ray or conversion electron and the emission probabilities of photons or electrons with high angular momentum are low. The excited state then can have considerable half-life (many days) and it is called isomer of the ground state. With a long half-life, it is not difficult to measure that the isomer nucleus has greater mass than the ground state nucleus. When a $\gamma$-ray is emitted, the excess mass of the nucleus $\triangle M$ is converted  by electromagnetic interaction into the $\gamma$-ray energy $E_\gamma = \triangle M c^2$, but no electrons are involved. When a conversion electron is emitted, the excess mass $\triangle M$ of the nucleus is converted by electromagnetic interaction directly into the energy of this electron, most likely one of the K-shell electrons. This is not strange, because the spatial probability distribution for the K-shell electrons has a maximum in the nucleus. The conversion electron kinetic energy is $E_e = \triangle M c^2 - E_K$ (usually hundreds of keV), where $E_K$ is the K-shell electron binding energy (usually tens of keV). So the excess mass of the nucleus is converted by electromagnetic interaction into the electron kinetic energy plus the increased elecric field energy. As an example, in the decay of the isomer $^\text{99m}\text{Tc}$ with the half-life $T_{1/2} \doteq 6\ \text{hr}$, $E_\gamma \doteq 140.5\ \text{keV}$, $E_e \doteq 119.5\ \text{keV}$, $E_K \doteq 21.0\ \text{keV}$.

I'm sorry, but even though I havent been able to understand much of your post yett - too high for me!  :lol: - , you have basiucally described the actual process of conversion of mass to energy in a nuclear phenomenon - thanks!  :lol:

What do you think of my idea - that motion of 'mass' in a 'force field' chanes energy to mass etc.??? When do such energy to mass and viceversa changes take place? How can we predict which will occur when?

I'm not sure if this is correct, but this is what I think:

Energy and mass are both intrisic properties of an object (planet, cat, electron, photon, etc.), which are always related by the relation $E=mc^2$. Therefore, strictly speaking, it is inaccurate to say that energy and mass can be interconverted, though it is often convenient to think of it as such. Any increase in the energy of a particle is always accompanied by a corresponding increase in mass, and vice versa. For example, in a nuclear reaction, the mass lost by the reacting nuclei is equal to the mass gained by the neutrons, electrons, and/or photons released during the reaction. The rest energy of the reacting nuclei is also converted to the kinetic energy of the neutrons, electrons, and the photons. However, all of these particles obey the equation $E=mc^2$. Moreover, all of the energy changes usually seen in classical physics and chemistry: changes in kinetic energy, energy changes in chemical reactions, etc: they are all accompanied by a corresponding mass change that is too small to detect and is usually neglected.

Nuclear fussion or nuclear fission are nuclear (strong) interactions. So is radioactive $\alpha$-decay (the formation of the alpha particle in the nucleus, the rest is just tunelling through the coulombic potential barier). Radioactive $\beta$-decay or electron capture are weak interactions. But the deexcitation afterwards (which I roughly described) is electromagnetic interaction, not nuclear, and it converts rest mass into energy.

Motion in a force field itself does not change rest mass, it has to be a transition between different energy levels. When the transition is from a higher to a lower energy level, something is emitted from the system, virtually always with non-zero kinetic energy, sometimes with non-zero rest mass as well. So the rest mass of the deexcited system decreases.  When the transition is from a lower to a higher energy level, something with non-zero kinetic energy and/or rest mass is absorbed by the system, so the rest mass of the excited system increases.

SO such 'energy to mass' conversions take place ONLY when electrons etc. undergo transitions from one enrgy level to another. SO basically, such 'energy-mass' interchanges occur only at the atomic level . Two questions.

1) When does energy change to mass - an eg please ?
2) DO the equations tell us that energy-mass conversions can take place only at the atomic level during nuclear phenomenon or it it just observed? What 'physics' is used to carry out these calculations?

P.S. Yetti - please answer the questions I asked in my PM

When $\gamma$-rays pass through matter, they interact in 3 possible ways: photoeffect, Compton effect and the pair creation. In photoeffect, the $\gamma$-ray disappears and its energy is fully transferred to an electron, which takes of with the energy $E_\gamma - E_B$, where $E_B$ is the electron binding energy in the atom. It has to be one of the inner well bound electrons, because some substantial momentum (and some very little energy) has to be transferred to the nucleus. Otherwise, it would not be possible to satisfy both the energy and momentum conservation laws. The Compton effect is essentially scattering of $\gamma$-rays on free electrons (the almost free electrons in outer shells). The energy of the scattered $\gamma$-ray is substantially reduced and the difference transferred to an electron. The 3rd interaction occurs only for $\gamma$-rays with energy greater than $E_\gamma > 2m_e \doteq 1022\ \text{keV}$. In this interaction, energy ($\gamma$-ray) is transformed into the rest mass of an electron-positron pair and they take off with whatever energy is left. The $\gamma$-ray cannot do this by itself, because again, the energy and momentum conservation laws could not be satisfied simultaneously. If has to interact with the electromagnetic field of a nucleus and transfer to it some momentum.

If a proton beam of sufficient energy hits a target, it creates a beam of $\pi$-mesons (pions), then the heavier K-mesons (kaons, the strange ones) and eventually proton-antiproton pairs and other strange particles, some with considerable half-lives (that's why they were called strange in the first place).  What I mean is that a proton hits another proton and the result are 3 protons and one antiproton. The pion creation occurs also in Earth's atmosphere hit by cosmic rays and what reaches the ground are mostly $\mu$-mesons (muons), the decay product of pions.

The physics to measure these effects is usually just application of the conservation laws. The physics to perform calculations is the quantum field theory and Feynman diagrams. Yesterday, I tried to explain to my son why the 1st Newton law is necessary and the inertial systems. He was perplexed, so I leafed through his high school textbook to see if it was there (it was). And somewhere at the end of the book, there was a Feynman diagram.

I'm sorry I'm being fastidious, but all this comes from solving the equations or is it just observed!?

[quote="yetti"]Yesterday, I tried to explain to my son why the 1st Newton law is necessary and the inertial systems. [/quote]

There is a way to derive Newton's 1st law, rather than postulating it?

By the way: your physics postings are highly informative.  Thank you.

No, the 1st law is a definition of inertial systems, in which the 2nd law is stated. Otherwise, the 1st law might seem unnecessary - according to the 2nd law, if no forces act on a body, the acceleration is zero and the body moves on a straight line with a constant velocity, which is what the 1st law states.


Let be $ x,y,z\in \mathbb{R}_\plus{}$ . Show that :

$ \left(x^2\plus{}\frac{3}{4}\right)\left(y^2\plus{}\frac{3}{4}\right)\left(z^2\plus{}\frac{3}{4}\right)\ge \sqrt{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$

[quote="alex2008"]Let be $ x,y,z\in \mathbb{R}_ \plus{}$ . Show that :

$ \left(x^2 \plus{} \frac {3}{4}\right)\left(y^2 \plus{} \frac {3}{4}\right)\left(z^2 \plus{} \frac {3}{4}\right)\ge \sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}$[/quote]
Because $ \left(x^2 \plus{} \frac {3}{4}\right)\left(y^2 \plus{} \frac {3}{4}\right)\geq x\plus{}y.$ :wink:

what is the best constant k that satisfy this inequality?
$ (x^2+k)(y^2+k)(z^2+k)\ge \sqrt{(x+y)(y+z)(z+x)}$

[quote="tuandokim"]what is the best constant k that satisfy this inequality?
$ (x^2 + k)(y^2 + k)(z^2 + k)\ge \sqrt {(x + y)(y + z)(z + x)}$[/quote]
$ k_{best}=\max_{[0,+\infty)}(\sqrt{2t}-t^2)=\frac{3}{4}.$ :wink:

[quote="alex2008"]Let be $ x,y,z\in \mathbb{R}_ \plus{}$ . Show that :

$ \left(x^2 \plus{} \frac {3}{4}\right)\left(y^2 \plus{} \frac {3}{4}\right)\left(z^2 \plus{} \frac {3}{4}\right)\ge \sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}$[/quote]
We have $ x^2  \plus{} \frac{3}{4} \equal{} \left( {x^2  \plus{} \frac{1}{4}} \right) \plus{} \frac{1}{2} \ge 2\sqrt {\frac{1}{2}\left( {x^2  \plus{} \frac{1}{4}} \right)}$(AM-GM)
Similarly we obtains $ \left( {x^2  \plus{} \frac{3}{4}} \right)\left( {y^2  \plus{} \frac{3}{4}} \right)\left( {z^2  \plus{} \frac{3}{4}} \right) \ge 8\sqrt {\frac{1}{8}\left( {x^2  \plus{} \frac{1}{4}} \right)\left( {y^2  \plus{} \frac{1}{4}} \right)\left( {z^2  \plus{} \frac{1}{4}} \right)}$
So ineq at first equivalent to:
          $ \prod {\left( {4x^2  \plus{} 1} \right)}  \ge 8\prod {(x \plus{} y)}$
          Then apply Cauchy-Shwar we have $ \left( {4x^2  \plus{} 1} \right)\left( {1 \plus{} 4y^2 } \right) \ge 4\left( {x \plus{} y} \right)^2$
similarly and multiply we have finished.

We have that $ x^2\left(y^2\plus{}\frac{3}{4}\right)\minus{}x\plus{}\frac{3}{4}\left(y^2\plus{}\frac{3}{4}\right)\minus{}y\ge 0$

We should prove that $ \Delta \le 0$

$ \Delta \equal{}1\minus{}3\left(y^2\plus{}\frac{3}{4}\right)^2\plus{}4y\left(y^2\plus{}\frac{3}{4}\right)\le 0\Leftrightarrow 3\left(y^2\plus{}\frac{3}{4}\right)^2\minus{}4y\left(y^2\plus{}\frac{3}{4}\right)\minus{}1\ge 0$

$ \Leftrightarrow \left(y\minus{}\frac{1}{2}\right)^2\left(y\minus{}\frac{3}{2}\right)^2\plus{}2\left(y\minus{}\frac{1}{2}\right)^2\left(y\plus{}\frac{1}{2}\right)^2\ge 0$

[quote="peine"][quote="arqady"][quote="alex2008"]Let be $ x,y,z\in \mathbb{R}_ \plus{}$ . Show that :

$ \left(x^2 \plus{} \frac {3}{4}\right)\left(y^2 \plus{} \frac {3}{4}\right)\left(z^2 \plus{} \frac {3}{4}\right)\ge \sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}$[/quote]
Because $ \left(x^2 \plus{} \frac {3}{4}\right)\left(y^2 \plus{} \frac {3}{4}\right)\geq x \plus{} y.$ :wink:[/quote]
Prove it please!  :wink:[/quote]
$ \left(x^2 \plus{} \frac {3}{4}\right)\left(y^2 \plus{} \frac {3}{4}\right) \minus{} (x \plus{} y) \equal{}$ $ \frac{(4xy\minus{}1)^2 \plus{} 4(x\plus{}y\minus{}1)^2\plus{}8 \left (x\minus{}\frac{1}{2} \right )^2 \plus{} 8 \left (y\minus{}\frac{1}{2} \right )^2 }{16}$ :P

[quote="peine"] Prove it please!  :wink:[/quote]

First CBS then AMGM:

$ (x^2\plus{}0.75)(y^2\plus{}0.75)\equal{}(x^2\plus{}0.25\plus{}0.5)(0.25\plus{}y^2\plus{}0.5)\ge$

$ (0.5x\plus{}0.5y\plus{}0.5)^2\equal{}0.25(x\plus{}y\plus{}1)^2\ge x\plus{}y$.

-----

I was too lazy for writing fractions :D

:)


This is going to sound weird but i am desperate to get some thoughts on the following question:

 How do we become better problem solvers?

 The obvious reply is of course "Do More Problems!", well that of course is valid but i am looking for information about the attitude and mental state necessary to improve ability. The list of questions that follow is just some of the things that i am curious about.

What are some personal qualities that good problem solvers have?

What does it take emotionally to become better at math?

How should we view the challenges we face during a problem solving session?

What does a good problem solver tell himself/herself when a solution is nowhere in sight?

What are some good reasons to solve problems?

Are most problem solvers at the more advanced level (Putnam Fellows etc) Competetive at heart? Or is it just about the math?

 Looking forward to replies.

p.s.- by the way, this is my first post, so...Hi everyone!

Hello,

This is a very frequent question.  First off, become a better problem solver takes practice.  However, I do believe that some of it is innate.  If you are born with curiosity there is a good chance that will become a better problem solver than someone who isn't.

[quote]What are some personal qualities that good problem solvers have? [/quote]

I don't think that there are too many qualities that all good problem solvers have in common.  Thinking all the time is probably one that they share.  Curiosity could be another one along with desire to become great.

[quote]What are some good reasons to solve problems?[/quote]

1.  They are fun.

2.  You feel a great sense of accomplishment.

3.  Real life problems have serious repercussions that need to be dealt with.

I think the answer to almost all of your questions would be: enjoyment. I think the main thing that you need to be a good problem solver is to actually enjoy doing it.

Thanks for the replies. Yeah i was thinking along those lines, just doing a reality check.

 Alot of great problem solvers gave me that advice: "just enjoy it!"

 Sometimes i am worried that i dont really like math and i begin think i am forcing myself to like it.

Hi
this problem is faced by maany of us doing maths.our basic problem is attitude.we should think maths as easy and start doing problems with posive attitude.of course,practice is a major factor but attitude helps alot.when u do a problem keep ur mind open.dont just think on one line.even after completing the problem try doing it by another method.initially do simple problems to boost ur moral and then go for the tougher problems.i hope this will help u.

Sriram

On harder problems, focus more on the method you used to find the solution rather than the solution. This may make the question easier to solve, or at least you'll enjoy more since you can achieve something out of it, instead of just worrying about the answer.

Always think outside the box; there are lots of solutions to be found that way. And always be determined to solve the problem; if you get sick of it, keep it in the back of your head, then the solution will pop out of the blue.

Yes things work out when u keep it at the back of ur mind.may be while having dinner or lunch the solution  might strike u.take it as a challenge.Dont leave the question.even talking to somebody else who is at ur level of understanding ,helps.They might think of a different method than urs and help u in finding the soln.

Sriram  :)

I think what has been keeping my from progressing is my impatience with not knowing the solution. I will follow this advice and see what happens.

Well, I learn to solve Maths problem since I was 5 yrs old. I found out that I am not learning it the hard way. Some how I take it as my hobby, rather than practicing it. I usually love to read reference books related to maths.
Is there any problem to think that maths is a type of hobby?( I mean frequent reading and not practising)

Also, persistence is necessary. Don't be like me - giving up in disgust once you cannot find the solution to the problem in several minutes.


$ x,y$ integer number . solve:
$ \mid x\minus{}4y\plus{}1 \mid\minus{}\mid y\minus{}3x\minus{}2 \mid\plus{}\mid x\plus{}y\plus{}2 \mid\plus{}\mid x\plus{}2y\plus{}3 \mid\equal{}4$

[quote="zaya_yc"]$ x,y$ integer number . solve:
$ \mid x \minus{} 4y \plus{} 1 \mid \minus{} \mid y \minus{} 3x \minus{} 2 \mid \plus{} \mid x \plus{} y \plus{} 2 \mid \plus{} \mid x \plus{} 2y \plus{} 3 \mid \equal{} 4$[/quote]
right  equation $ \mid x \minus{} 4y \plus{} 1 \mid \plus{}\mid x \plus{} y \plus{} 2 \mid\plus{} \mid x \plus{} 2y \plus{} 3 \mid \geq \mid 3x \minus{}y \plus{}6   \mid$
   left of equation :
 $ 4 \plus{}  \mid y \minus{}3x \minus{}2 \mid$

   maybe  :
$ 3x\plus{}2\minus{}y >0 , x\minus{}4y\plus{}1 , x\plus{}y \plus{}2>0$  and $ x\plus{}2y \plus{}3 >0$
  use  graphis.  maybe   find  value  $ x,y$


How do you do those problems where they give you 2+ circles intersecting and you have to find the area of the intersection? I can never get those, even though I've seen them so many times. :blush:

[quote="breez"]How would you find the area of the circular region?[/quote]

Do you mean the area enclosed by 2 radii and the circumference? Find the measure of the angle made by the 2 radii. Then divide the angle measure by 360 and multiply by the area of the circle.

[quote="chess64"][quote="Scrambled"]connect the two points of intersection (call them A and B). Then connnect the two points to the centers of both circles(call them O and O'). Find the area of the circular region AOB and AO'B and triangle AOB and AO'B and subtract the circular regions from the triangles[/quote]

wait can you draw a picture, i don't really understand because in my diagram A, O, and B are collinear[/quote]

well if your points are collinear, then that means that you have a degenerate triangle AOB and your slice AOB is half of the circle with center O.

im not quite sure what a picture would do, but w/e

Oops, sorry, I misread your definition of A and B :blush:

Thanks! :)

[quote="breez"]How would you find the area of the circular region?[/quote]

Let $x$ be the area of the sector with angle $\theta$, then $\frac{\theta}{360^\circ}=\frac{x}{\pi r^2}$ where $r$ is the radius of the circle.

Drawing triangles is always the key.  In the picture above, to find the right half of the area, notice it's part of circle O cut off by AB.  So the area of the sector OAB is $\frac{\theta}{2\pi} \pi r^2 = \frac{\theta r^2}{2}$  Then the area of triangle OAB is $\frac12 r^2 \sin \theta$.  So the area of the cut off part is merely the difference between the two.  Then do the same thing for the part cut off in circle O', and add the two sections.

Here's a fun one - take a square, and at each vertex, draw a circle with radius equal to the side length of the square.  Find the area of the union of all four circles.


Define the sequence $ \{a_n\}$ as:

$ a_1\equal{}1, a_n\equal{}a_{n\minus{}1}\plus{}\frac{1}{a_{n\minus{}1}}, a_n>0(n \in \mathbb{N}_\plus{})$

Does the sequence has a general term formula? If it do has, please find the formula :blush:

I think this can't be called the formula of the sequence.


integration
inequalities
calculus
real analysis
real analysis unsolved

[quote="N.T.TUAN"]Let $a,b$ be positive reals and let $f\in C^{2}[0,a]$ such that $f(0)=f(a)=b,f(x)\geq 0,f''(x)\geq 0\forall x\in [0,a]$. Prove that $2\int_{0}^{a}f(x)\sqrt{1+(f'(x))^{2}}dx\leq b\sqrt{a^{2}+b^{2}}.$[/quote]
one holder's and one jensens integral inequality .


search
superior algebra
superior algebra unsolved

Let $ I$ be an ideal of the ring $ R$ and $ f$ a nonidentity permutation of the set $ \{ 1,2,\ldots, k \}$ for some $ k$. Suppose that for every $ 0 \not\equal{} a \in R, \;aI \not\equal{} 0$ and $ Ia \not\equal{}0$ hold; furthermore, for any elements $ x_1,x_2,\ldots ,x_k \in I$, \[ x_1x_2\ldots x_k\equal{}x_{1f}x_{2f}\ldots x_{kf}\] holds. Prove that $ R$ is commutative.

[i]R. Wiegandt[/i]

I posted a proof in [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1895161787&t=29349]another topic[/url]


Let ${a_{n}}$ be a sequence of natural numbers such that for each  $n\geqq1$ , $a_{n+1}=\sum_{k=1}^{n}a_{k}^{2}$and $a_{2006}$ is divisible by $2006$. Find the smallest possible value of $a_{1}$.

[quote="reptile"]Rust are there a typing wrong? I think  If $a_{n+1}=0 (modp )$ then $a_{n}=0$ or $a_{n}=-1 mod (p)$ not $a_{n}=0$ or $a_{n}=1 mod (p)$ . And solution changes.[/quote]
I wrote: if $a_{n+1}=0$, then $a_{n}=0$ or $a_{n}=-1$.

Rust I am sorry . I corrected my post. Can you explain your solution?  Thanks.

{$2006=2*17*59$ and $a_{n+1}=a_{n}(a_{n}+1)$.  If $2006$ divides $a_{n+1}$ then $a_{n}\equiv 0 \textrm{ or }a_{n}\equiv-1(modp)$. $(p=2,17,59)$ But If  $a_{n}\equiv-1(mod17)$ then it can not hold by any natural number so that $a_{n}\equiv 0(mod17)$.  As a result $a_{n}\equiv 0(mod2,17,59)$ Furtermore, because of $a_{2}=a_{1}^{2}\equiv 0 (mod2,17,59)$ it must be $a_{1}^{2}\equiv 0 (mod2,17,59)$  and for this reason $a_{1}=2*17*19*k$.($k$ is a natural number) . Minimum $a_{1}$ is equal to $a_{1}=2*17*19*1=2006$.

Rust, yes you are right but what is the problem in my solution. I can not understand. :(


[quote="April"]Let $ ABC$ be acute triangle. The circle with diameter $ AB$ intersects $ CA,\, CB$ at $ M,\, N,$ respectively. Draw $ CT\perp AB$ and intersects above circle at $ T$, where $ C$ and $ T$ lie on the same side of $ AB$. $ S$ is a point on $ AN$ such that $ BT = BS$. Prove that $ BS\perp SC$.[/quote]

Let $ CT$ and $ AB$ intersect at $ K$.

Since $ AB$ is diameter of the circle and $ N$ lies on that circle, so $ \angle{BNA}=90$. Also we have that $ \angle{BKC}=90$ so triangles $ BKC$ and $ BNA$ are similar. Hence, $ \angle{BCK}=\angle{BAN}$.

We also have that $ \angle{BTA}=90$, because $ T$ lies on the circle. Hence we have that triangles $ BKT$ and $ BTA$ are similar thus $ \frac{BT}{BK}=\frac{BA}{BT}$ or $ BT^{2}=BA.BK$. Since $ BT=BS$ we have $ BA.BK=BS^{2}$ hence $ BS$ is tangent to circle circumscribed around $ KSA$. It follows that $ \angle{BSK}=\angle{KAS}=\angle{BAN}=\angle{BCK}$. From here we get that $ BCSK$ is cyclic and thus $ \angle{BSC}=\angle{BKC}=90$, in other words $ BS\perp SC$.

let$ H$ be feet of $ C$. $ \angle T=90^{\circ}$ hence $ BS^{2}=TB^{2}=BH.BA=BN.BC$ and cause $ \angle N=90^{\circ}$ hence $ \angle S=90^{\circ}$.

Just as $T$ is the intersection of the circle with diameter $AB$ with the height from $C$, we wish to show that $S$ is the intersection of the circle on $BC$ with the height from $A$. Let's rather redefine $S$ and $T$ as the intersection of a height and a Thales circle of a side, and show that $BS=BT$.

Now note that if $X$ is the midpoint of $BC$, and $BN=n=c\cos \beta$, $XB=XS=r=\frac a2$, then we have
$$BS^2=BN^2+NS^2=BN^2+(SX^2-NX^2)=n^2+r^2-(n-r)^2=2nr=ac\cos \beta,$$
which is symmetric in $a,c$, so $BT^2$ will be the same expression. This verifies $BS=BT$.

Hi! Here's my solution
[hide=Solution]
Let H=orthocenter, CP be the C-altitude, S' be the intersection of the semicircle with diameter BC and AN
Clearly, angles AMB, ANB, ATB, and BS'C are all right angles
In right triangle ABT with altitude TP, we have $BT^2=BP\cdot BA$ (As triangles BPT and TPA are similar)
Similarly, in right triangle BCS' with altitude S'N, $BS'^2=BN \cdot BC$ (As triangles BNS' and S'NC are similar)
Note that
$$\frac{BP}{BN}=\frac{ \frac{BP}{BH} }{\frac{BN}{BH} }=\frac{cos(HBP)}{cos(HBN)}=\frac{cos(90-A)}{cos(90-C)}=\frac{sin(A)}{sin(C)}=\frac{BC}{BA}$$
Hence, $BP\cdot BA=BN\cdot BC$, so $BT=BS'$
Therefore, S=S' and angle BSC is 90 degrees

[/hide]

Can't believe this problem is Chinese. :ninja:
First notice that $\angle BTN=\angle BAN=\angle TCN$

Thus ,we get $BT$ is tangent to $\odot(NTC)$,
So by power of a point
$BT^2=BN.BC=BS^2$
Thus
$\triangle BSC$ and$\triangle BNH$ are similar,
so we are done.

Let CT meet AB at K. ∠ATB = ∠ANB = 90 so BT^2 = BK.BA = BN.BC = BS^2 so ∠BSC = ∠BNS = 90.
we're Done.


logarithms
real analysis
real analysis unsolved

Let $x_0>0$ and $ x_n=x_{[\frac{n}{2}]}+x_{[\frac{n}{3}]}+\frac{n}{6}$. Prove that for any $k>log_3{ \frac{5}{2}}$ we have $ x_n=n+o(n^k)$. Can you give an asymptotic development with 2 terms? (the last question is unsolved to me).

I wonder where $\log_3 \frac52$ came from. I got $k>k_0$ where $k_0$ is the solution of the equation $2^{-k}+3^{-k}=1$. I suspect that $x_n=n+C(x_0)n^{k_0}+o(n^{k_0})$ but do not have a proof at the moment (the problem is very similar to but a bit more difficult than that with $x_{[\frac n6]}$ instead of $\frac n6$).


Find all order pairs of integers $\ (x,y)$ such that $\ 1 + 2x +3y=xy$

[hide]add a constant amount then you can factor it to $(x-3)(y-2)=7$

then it is obvious as you can just look at the factors of the other side and compare them with the other factors

these being $(-1)(-7)$, $(-7)(-1)$, $(1,7)$, $(7,1)$

and that makes the ordered pairs $(x,y) \rightarrow (2,-5) \, (-4,1) \, (4,9) \, (10,3)$

[/hide]


geometry
circumcircle
inequalities
inequalities solved

Let ABC be an acute triangle. Prove that
     a^6+b^6+c^6>=(a^2+b^2+c^2)^2*R^2
   R is the circumcircle radius.
     If ABC is obtuse ,it may be wrong. For example 6,9,14.

Well, since triangle ABC is acute, we have b^2 + c^2 > a^2, c^2 + a^2 > b^2 and a^2 + b^2 > c^2. Hence, if x = a^2, y = b^2 and z = c^2, then there exists a triangle with sides x, y, z.

Now, the inequality we have to prove

  a^6 + b^6 + c^6 >= (a^2 + b^2 + c^2)^2 * R^2

can be rewritten as

  x^3 + y^3 + z^3 >= (x + y + z)^2 * R^2.

But the area S of triangle ABC fulfills

  16 S^2 = 2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4,

i. e.

  16 S^2 = 2yz + 2zx + 2xy - x^2 - y^2 - z^2,

so that

  R^2 = (abc / 4S)^2 = a^2 b^2 c^2 / (16 S^2)
      = xyz / (2yz + 2zx + 2xy - x^2 - y^2 - z^2).

So we have to prove

  x^3 + y^3 + z^3 >= (x + y + z)^2 * xyz / (2yz + 2zx + 2xy - x^2 - y^2 - z^2),

i. e.

  (x^3 + y^3 + z^3) * (2yz + 2zx + 2xy - x^2 - y^2 - z^2) >= (x + y + z)^2 * xyz.

But

  (x^3 + y^3 + z^3) * (2yz + 2zx + 2xy - x^2 - y^2 - z^2) - (x + y + z)^2 * xyz
= (2xyz + y^2 z + z^2 y + z^2 x + x^2 z + x^2 y + y^2 x - x^3 - y^3 - z^3) * (x^2 + y^2 + z^2 - yz - zx - xy).

The second factor is >= 0, since

  x^2 + y^2 + z^2 - yz - zx - xy = 1/2 * ((y-z)^2 + (z-x)^2 + (x-y)^2) >= 0;

the first factor can be rewritten as

  2 xyz + y^2 z + z^2 y + z^2 x + x^2 z + x^2 y + y^2 x - x^3 - y^3 - z^3
= 4 * (4 XYZ + Y^2 Z + Z^2 Y + Z^2 X + X^2 Z + X^2 Y + Y^2 X),

where X = (y+z-x)/2, Y = (z+x-y)/2 and Z = (x+y-z)/2; hence, it is >= 0, too. Therefore, we have

  (x^3 + y^3 + z^3) * (2yz + 2zx + 2xy - x^2 - y^2 - z^2) - (x + y + z)^2 * xyz >= 0,

and

  (x^3 + y^3 + z^3) * (2yz + 2zx + 2xy - x^2 - y^2 - z^2) >= (x + y + z)^2 * xyz.

Qed..

If I would have found this solution at an IMO, without a PC at hand, (and if I would find the time to write up the complete calculation,) I guess it would give me a special prize...

  Darij

Your solution is very beautiful.
  Thanks to Schur's inequality, we have
    a^6+b^6+c^6>= \sum a^4*b^2+b^4*a^2 -3a^2*b^2*c^2 = M.
   I 'd be very grateful if you spend your valuable time on my original inequality
       M>=(a^2+b^2+c^2)^2*R^2.
   Thanks.


find $f:N*-->N*$ that satisfied
$f(n)>f(f(n-1)$$

f(n)=n satisfies. Do you mean find all f?

I think that's the actual question. This problem is folklore by now :).

There is an $x_1$ s.t. $f(x_1)$ is minimal. Since $f(x_1)>f(f(x_1-1))$ if $x_\ge 2$, it means that $x_1=1$. There must be an $x_2\ge 2$ s.t. $f(x_2)$ is minimal except for $f(1)$. Then $f(x_2)>f(f(x_2-1))$, which means that $f(x_2-1)=1$, and since $1$ must then be the minimal value of $f$, we get $x_2-1=1\Rightarrow x_2=2$. Now take an $x_3$ s.t. $f(x_3)$ comes right after $f(x_1)=1,f(x_2)=2$ (i.e. the third number in the image of $f$ in increasing order), and repeat the process used above. We get $f(n)=n$.

congratulation grobber!
and what do you think about this one
$f(n)>f(f(f(f.........(f(n-1)))..) $ (n times)
i think it will be more complex$

sorry , everyone , i think this one is impossible to prove
i have a big mistake (k times not n times) :(


Let $A_1A_2A_3$ be a triangle and, for $1 \leq i \leq 3$, let $B_i$ be an interior point of edge opposite $A_i$. Prove that the perpendicular bisectors of $A_iB_i$ for $1 \leq i \leq 3$ are not concurrent.

Suppose that the perpendicular bisectors meet at M.
WLOG Suppose $A_1M\geq A_2M \geq A_3M$.

Suppose $A_1M>A_2M$. Then $A_1M>A_2M\geq A_3M$. For all $X\in (A_2A_3)$ we have $MX \leq A_2M$. But M lies on the perpendicular bisector of $A_1B_1$ and thus $MB_1=MA_1$. $MA_1=MB_1\leq MA_2<MA_1$ is a contradiction.

So $A_1M=A_2M$. Suppose $A_2M>A_3M$.
We have $MA_2=MB_2$. Because $MA_1>MA_3$, on $(A_1A_3)$ there is no point X such as $MX=MA_1$. This leads to a contradiction.

So $MA_1=MA_2=MA_3$. On $(A_2A_3)$ there is no point X such as $MX=MA_1$.
Again a contradiction.

If you can find your way through all the assumptions you can see that now the problem is solved.

I also suppose that the perpendicular bisectors of these segments are concurrent in a point $ O $.
we have that in a triangle $ ABC $ with $ X\in (BC) $ the inequality $ AX<max(AB,AC) $.
we see that using this we obtain a contradiction immediately.


geometry
inradius
trigonometry
trig identities
Law of Cosines
Law of Sines
area of a triangle

If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

[b]EDIT:[/b] Whoops, made mistakes on the area and the problem itself... :oops:

[quote="JesusFreak197"][hide="Answer"]By Heron's Formula, the area is $\sqrt{(0.5)(1.5)(2.5)}=\frac{\sqrt{30}}{4}$, and the semi-perimeter is $\frac{9}{2}$, so we have $r=\frac{A}{s}=\frac{\frac{\sqrt{30}}{4}}{\frac{9}{2}}=\boxed{\frac{\sqrt{30}}{18}}$.[/hide][/quote]

Well, that would give the [i]inradius[/i], but i think it's asking for the circumradius...

[hide]If $\theta$ is the angle opposite the side with length 4, then by Law of Cosines, $\cos \theta=\frac{2^2+3^2-4^2}{2*2*3}=-\frac{1}{4}\Rightarrow \sin \theta=\frac{\sqrt{15}}{4}$.

Use Extended Law of Sines to find the circumradius. $2R=\frac{4}{\sin \theta}\Rightarrow R=\frac{2}{\sin \theta}=\boxed{\frac{8\sqrt{15}}{15}}$.[/hide]

[hide]
By Heron's formula, the area of the triangle is $\sqrt{\frac{9}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}}=\frac{3\sqrt{15}}{4}$

Using $R=\frac{abc}{4K}$, where $K$ is the area of the triangle, we obtain $R=\frac{2\cdot 3\cdot 4}{\frac{3\sqrt{15}}{4}}= \boxed{\frac{8\sqrt{15}}{15}}$
[/hide]


Πολυ ωραια θεματα.....
Πως τα πηγαν οι συμμετεχοντες....?

 :)

????

1on: Egine telika simera o diagwnismos tis EME?
Stin istoselida lene oti metatethike o diagwnismos gia tis 9 Dekemvri!!
Ti paixtike telika? den katalava...

2on: Afou pinakes den didaskontai katholou sto Likeio, pws kai evalan tetoia themata?

Alexandros

re paidia min trelenaiste apokleietai na dothike o thalis simera
9/12 ginetai
episimi anakoinosi tis eme

Χαβαλέ πρέπει να μας κάνει  :P

Αν εννοείς εμένα φίλε μου όχι αν εννοείς τον Σιλουανό πραγματικά τα έχει χαμένα

Ambros: Δεν νομίζω να τα έχει χαμένα, αυτό που αποσκοπούσε το πέτυχε. Τουλαχιστόν εγώ για κάποια στιγμή το πίστεψα και τρομοκρατήθηκα ( :P ) τρέχοντας κατευθείαν στην ιστοσελίδα της ΕΜΕ για να επικυρώσω ότι όντως ο διαγωνισμός θα διεξαχθεί στις [b]09/12/2006[/b].

--

Σταματήστε τις πλάκες, θα μας κάνετε καρδιακούς...  :mad:

Λοιπόν παιδιά οκ .Μια πλάκα ήταν (λίγο χοντρή όμως) και σόρρυ σε αυτούς που τρομοκρατήθηκαν .Πράγματι νομίζω ότι αυτά τα θέματα δεν θα μπορούσαν να δωθούν σε Θαλή (αν και για να διευκρινήσω κάτι για να μην υπάρχουν παρεξηγήσεις συγκεκριμένη ύλη δεν υπάρχει μπορούν να βάλλουν ότι θέλουν )
Νομίζω ότι είναι μια καλή προπόνηση όμως τα θέματα γιαυτό προσπαθήστε τα . Καλή διασκέδαση .

@Αλέξανδο : Αλέξανδρε και η ανισότητα Chebychev ή και η Schur δεν διδάσκονται αλλά πρέπει να τις ξέρεις .

Silouan den diafwnw oti prepei na tis gnwrizeis gia na sinexiseis kai stis epomenes faseis twn mathimatikwn olimpiadwn, alla toulaxiston stin 1h fasi den mporei na tethoun themata ta opoia toulaxiston den einai stin sxoliki ili! Den xreiazetai na valoun ena thema ektos ilis gia na mi to lisoun oi mathites! Stin 1h fasi paratirise oti panta mpainoun themata pou apla apaitoun poli kali gnwsi twn oswn exoun didaxthei 'h oswn epontai meta apo merika mathimata kai kali kritiki skepsi... Den tha deis oute paragwgous stin B likeiou, oute dianismata stin A' Likeiou. (Allo an kapoios exei perissoteres gnwseis kai linei me eukolo tropo ta provlimata auta).

Apo ekei kai pera vevaia, an kapoios mathitis thelei na sinexisei kai na diakrithei einai epivevlimeno na diavasei poli perissotero kai na kserei stoixeiwdi pragmata apo Algevra, Anisotites, Sindiastiki ktl ktl...

Alexandros

Y.G. Asxeto... alla an thimamai kala pinakes eixan mpei teleutaia fora stin 3h likeiou to etos 198? an thimamai kala(tha to psaksw kalitera molis anoiksw tin sillogi mou).

[quote="silouan"]Παραθέτω τα σημερινά θέματα του 67ου πανελληνίου μαθηματικού διαγωνισμού ο Θαλής

Θέμα 1ο
Αν Α,Β είναι πίνακες $n*n$  να δείξετε ότι
αν $A^{2}=0$ και $AB+BA=0$ νδο $AB=BA=0$

Θέμα 2ο
Να λύσετε στους πραγματικούς το σύστημα
$x+y+\frac{1}{z}=3$
$y+z+\frac{1}{x}=3$
$z+x+\frac{1}{y}=3$

Θέμα 3ο
Να δείξετε ότι σε κάθε τετράπλευρο οι τρεις ευθείες που συνδέουν τα μέσα των ζευγών των απέναντι πλευρών και διαγωνίων του αλληλοδιχοτομούνται

Θεμα 4ο
Δείξτε ότι $\cos\frac{8\pi}{35}+\cos\frac{12\pi}{35}+\cos\frac{18\pi}{35}=\frac{1}{2}\cdot\left(\cos\frac{\pi}{5}+\sqrt7\cdot\sin\frac{\pi}{5}\right).$

Διάρκεια εξέτασης 3 ώρες[/quote]

 :rotfl:  :rotfl: .ωραιο siluan...φαινεται οτι πολλοι .... :stretcher: +θα ειχε μεγαλυτερη<επιτυχια>αν τα 8εματα δεν εκαναν τοσο....:bomb:  οτι δεν κανουν για Θαλη...

Παντως,εδω που τα λεμε,δεν +ασχημα θεματακια...
μια διευκρινηση¨...στο 1ο θεμα ειναι σιγουρα Α^2=0 ΚΑΙ ΟΧΙΑ^2=Α???(ΕΙΝΑΙ ΓΝΩΣΤΗ ΣΧΟΛΙΚΗ ΑΣΚΣΗ,ΑΣΚΗΣΗ ΒΟΗΘΗΜΑΤΟΣ,+ΜΑΣ ΤΗΝ ΕΔΩΣΑΝ ΣΑΝ ΑΣΚΣΗΣΗ +ΣΤΗΝ ΓΡΑΜΜΙΚΗ ΑΛΓΕΒΡΑ,..)
ΓΙΑ ΤΟ 2Ο...καλο απλα θελει λιγο προσοχη μην χαθει καμια λυση...
βρηκα (3,3,3),(1/2,1/2,1/2), και η(-1/3,3,3) με κυκλικη εναλλαγη...
Για το 3ο...επισης ασκηση σε σχολικο βοηθημα,αποδεικνυεται με αναλυτικη...4 τυχαι σημεια στο καρτεσιανο(8 μεταβλητες δηλαδη!!!!)
και εφαρμοζοντας τον τυπο που δινει τισ συντεταγμενες του μεσου ευθ.τμηματος με δοσμενα ακρα,προκυπτει ευκολα το ζητουμενο...
+ΓιΑΤΟ 4ο...δεν θελει τιποτα τρομερο..πραξεις...,απλως....<εξυπνες>πραξεις...
Μου φαινεται οτι τελικα τα θεματα του φετινου Θαλη ηταν πολυ.... :icecream:  :juggle:  :harhar:

 :D

ενταξει ο Ραλλης τα  ξεπεταξε στα γρηγορα αλλα μιλαμε για ΘΑΛΗ β λυκειου,ειναι σχετικα καλα.παντως σε μερικους Θαληδες υπαρχουν θεματα που πεφτουν  στανταρ σε ευκλειδη

Αμα δεν ήσουν πολύ φίλος μου (κι το εννοώ αυτό) κι αν δεν ήσουν κι στυο φόρουμ μας οπου επιβάλλεται η κόσμια έκφραση θα σε είχα λούσει στ α Γαλλικά ευτυχώς που εγώ δεν τρώω κουτόχορτο (αν κι για να είμαι ελικρινής λίγο τρομοκραττήθηκα ΛΙΓΟ είπα μην παρεξηγηθώ) οκ??? όπως κι να έχει τα θέματα  αυτά είναι ωραία αν κι 2/4 (2,3) ήταν για babies
Καλή η πλάκα Σιλουανέ
Μπράβο :rotfl:

PS ο πατέρας του κα΄λύτερου φίλου μου είναι καρδιολόγος αν θέλετε τίποτα  (δηλαδή οαν ο Σιλουάν ξανακάνει τέτοια πλάκα ) :rotfl:  :D

[b]Ραλλη,[/b]
εχεις δικιο. Ειναι $A^{2}=A$....
H ασκηση αυτη ειχε μπει και σε παννεληνιες, το '96 αν θυμαμαι καλα...
[b]Σιλουανε[/b], κανε τον κοπο να το διορθωσεις....θενκς

'Οπως καταλαβατε, ημουν και εγω στο κολπο....
και συγκεκριμενα εδωσα το εναυσμα.Η ιδεα ειναι του σιλουανου ομως.... :D

Τελος, [b]Ραλλη [/b] μπορεις να δωσεις (εστω εν ολιγοις) τη λυση σου στο 4?

 :)

Ναι πράγματι έτσι είναι αλλά δεν μπορώ να κάνω edit  :maybe:  δεν ξέρω γιατί .Ράλλη βάλε αν θες τη λύση σου στο 4 .

olokliri i lisis ine fisi adinaton na tin stilo (8a grafo kana 2 meres...)
vsk ena pragma xrisimopiisa...pos metatrepume se ginomeno mia parastasi tis morfis asinx+bcosx
+edo fenete oto ayti i askisi den kani gia diagonismo giati ...(ego tulaxiston) xrisimopiisa prota kompiyteraki gia na prosdioriso piramatika mia gonia kai meta to apediksa....
(tora an kapios ixe tosi empneysi  tin ora tu diagonismu+tu erxotan etsi uranokatevata i gonia.... :P )
meta tin metatropi ayti tu deyteru melus  ine eykolo..
vasika stin arxi ixa ipoyies oti den isxii,alapira kompiyteraki +to epali8eysa +pezontas me to kompiyteraki...brika katala8os tin goniapu eyaxna..... :lol:  :lol:

 :D

[quote="cretanman"]

Y.G. Asxeto... alla an thimamai kala pinakes eixan mpei teleutaia fora stin 3h likeiou to etos 198?quote]

ως γνωστό έχουμε σπουδαίο παρελθόν...
υπήρχαν τότε πίνακες?

ξέχασες τη γεωμετρία :o

Mexri KAI to 1999 didaskotan sta sxoleia (kai sigekrimena stin A' Desmi kai ligo stin 4h desmi) to 1o kefalaio tis Algevras tou vivliou twn Mathimatikwn Kateuthinsis tis 3hs Likeiou opote emeis mathainame kanonika pinakes, grammika sistimata kai epilisi tous me pinakes, orizouses, methodos Crammer, g(r)ammoprakseis ktl ktl...

Den kserw gia pio logo exoun afairethei apo tin sxoliki ili tis 3hs Likeiou. Apotelouse (eidika oi orizouses kai oi pinakes se thewritiki morfi) ena apo ta pio omorfa kefalaia, me polla texnasmata kai krimmenes omorfies... Krima!!

Oso pame meiwnoume tin ili twn mathimatikwn... Distixws den kserw pou tha ftasei auti i katastasi! Den lew oti kai tote itan kala ta pragmata afou ipirxe ipervoliko forto se ola ta mathimata akoma kai sta mathimatika (fantasteite oti h A' Desmi didaskotan mathimatika 8 wres tin evdomada (4 wres Algevra, 4 wres Analisi) sto sxoleio, eixe 2 vivlia (Algevras & Analisis) kai eixe stin ili tis OLA osa didaskontai sta simerina mathimatika kateuthinsis tis B' Likeiou (ektos Thewrias Arithmwn) kai sta mathimatika Kateuthinsis tis 3hs Likeiou (mazi me to 1o kefalaio algevras pou twra einai ektos ilis). Goustarame na kanoume mathimatika...Pio polles wres vlepame ton mathimatiko para tin oikogeneia mas! :) ).

Tha mou peite oti twra anti gia 4 mathimata pou kaname emeis tote, oti twra kanoun 6. Alla siggnwmi... den kserw se ti wfelei ena paidi o programmatismos 'h h Viologia an den perasei Programmatistis 'h viologos antistoixa... Einai merika pragmata pou den tha ksanadei sti zwi tou! Mathimatika omws, opou ki an perasei [akoma kai stin filosofiki sxoli exoun valei statistiki (ta exoun paiksei vevaia oi foitites alla ok)] tha ta ksanadei. Idiaitera stis thetikes epistimes, politexneia, iatrikes, oikomika ktl... Ola omws ginontai gia kapoious xrisimothirikous skopous kai gia na valoun xrimata stin tsepi kapoioi epitideioi...

Akouste erwtisi apo mathiti tis B' Likeiou (ipopsifio Texnologikis Kateuthinsis) pou mou metefere enas sinadelfos mathimatikos apo ena sxoleio:

 "Ma kirie, auto to x tha pairnei kathe mera diaforetiki timi????"

Em... den ftaiei to paidi... Emeis ftaime pou to diapaidagwgisame me auto ton tropo, pou tou ta dinoume ola etoima (oi goneis) kai den to afinoume na parei prwtovoulies gia tipota. Pou tou exoume sinexws ton H/Y  kai to internet mipws kai pathei tipota. To paidi den vgainei eksw na paiksei alla paizei me to Playstation sto spiti. Etsi ta paidia apoxaunwnontai/pathitikopoiountai kai meta pws na exoun oreksi gia ta mathimatika, pou theloun psaksimo anazitisi,ipopsia,kaxipopsia mipws simvainei to ena 'h to allo? Den mporeis na exeis ta mathimatika etoima sto piato sou oute na valeis ena "password spasmeno/hint/tip" kai na pas stin "epomeni pista" opws sto Playstation. Alla den to katalavainoun kapoioi... Auta omws se ligo kairo tha ta plirwsoume!! Hdi fetos meinane eksw apo ta AEI kai TEI poses xiliades paidiwn.... Kai ta xeirotera erxontai...

Ouf... Ta eipa kai ksalafrwsa ligaki...

Alexandros

Y.G.1: Sorry pou apo allou ksekinisa kai allou kateliksa... Vrika aformi kai ksespasa!
Y.G.2: Kathe persi kai kalitera! AMHN!


Define an integer n to be k-abundant if $\sigma(n)>(k+1)n$. Show that for each positive integer k there are an infinite number of k-abundant integers.

The solution that a friend and I worked out was extremely nice, but long. Any and all thoughts welcome.

Partial answer: It is enough to show that there is at least [u]one[/u] $k$-abundant number for any $k$, since if a number is $k$-abundant then it is $l$-abundant for any $l<k$.

Now if $n$ is a squarefree number then $\sigma(n)=\prod_{p\vert n}(1+p)$, so find a set $S$ of primes such that $\sum_{p\in S}\frac{1}{p}>k$ ...


Has Indian IMO selection test for this year begun? Are there problems from past years? I'd be interested.

no they havent begun. As for the probs, see the resource section :)

http://www.mathlinks.ro/resources.php?c=78&cid=49

Sadly it has only for 2 years. You may have to search the forum for more. But don't all countries use the IMO Shortlist for their selection test?

The IMO Training Camp will begin in May.


calculus
integration
derivative
calculus computations

Hi, I'm having trouble with the integral of 5^sqrt(x) * sqrt(x)

I tried two ways, integration by parts and u-substitution. With U-substitution I subbed in sqrt(x) for u and got:

2du = dx/sqrt(x), which doesn't work out.

Integration by parts I got (5/2 * 1/ln 5) - (5/2 * x/ln 5) Which doesn't seem right either.... any input would be much appreciated!!!

The answer is probably really simple but im just missing it...

The substitution does, in fact, work; when $u = \sqrt{x}$ we have $2 \, du = x^{-\frac{1}{2}}\, dx$ and

$5^{ \sqrt{x}}\sqrt{x}\, dx = 2u^{2}\cdot 5^{u}\, du$

Okay I guess i'm confused as to how you could get u^2 from x^-1/2 doing the substitution....  What happens to the negative sign?  Thank you

Try writing the substitution backwards as $x=u^{2}$. It's clearer that way, and this is a very broadly applicable technique.

Continue from how t0rajir0u set it up:
$\int 5^{u}\cdot 2u^{2}du$.  Now use integration by parts (I prefer the tabular method).  Take the derivatives of $2u^{2}$ and integrate $5^{u}$:
$= 2u^{2}\cdot\frac{5^{u}}{\ln5}-4u\cdot\frac{5^{u}}{\ln5^{2}}+4\cdot\frac{5^{u}}{\ln5^{3}}$
$= 5^{u}\left(\frac{2u^{2}}{\ln5}-\frac{4u}{\ln5^{2}}+\frac{4}{\ln5^{3}}\right)$
$= 5^{\sqrt{x}}\left(\frac{2x}{\ln5}-\frac{4\sqrt{x}}{\ln5^{2}}+\frac{4}{\ln5^{3}}\right)$

[quote="Polaris878"]Okay so

u^2 = x
2u = dx

After that I'm still lost lol... sorry this is very difficult for me.  Using u=sqrt(x) I came up with 2du = dx/srqt(x)  Could someone walk me through how to manipulate 2du = dx/sqrt(x) so i could sub in for sqrt(x) dx?[/quote]

Certainly.  Our desired integrand is

$5^{ \sqrt{x}}\cdot \sqrt{x}\, dx$

Now, we make the substitution $u = \sqrt{x}\implies u^{2}= x \implies 2u \, du = dx$.  Then

$5^{ \sqrt{x}}\cdot \sqrt{x}\, dx = 5^{u}\cdot u (2u \, du) = 2u^{2}\cdot 5^{u}\, du$

[quote="bos1234"]jmerry can you pelase show your method??[/quote]

T0rajir0u just did.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="dhdung"]If $S$ is the group of all finitary of the set $\{1,2,3,..\}$ and $A$ is the alternating subgroups, prove that $S$ has only three normal subgroup, they are $1; A$ and $S$.[/quote]

Look at the composition series for $S_{n}$

$1< A_{n}< S_{n}$
For $n\geq 5$ we have $A_{n}$ is simple.

And nothing can be in between $A_{n}<S_{n}$
because $(S_{n}: A_{n})=2$

It's right. But you have to prove that the composition of $S_{n}$ is only $1\triangleleft A_{n}\triangleleft S_{n}$.
And note that, $A$ defined above is infinitive alternating subgroup.
If we retrict attention to those elements of  $A$ for every $m>t$ yields a subgroup $A(t)$ which is a copy of $A_{t}$.
So we have \[A=\bigcup_{t\in\mathbb{N}}A(t)\]
Hence $A$ is simple.
Can we show that $[S: A]=2$
From that show $S$ has only one composition, that is $1\triangleleft A\triangleleft S$.

[quote="dhdung"]It's right. But you have to prove that the composition of $S_{n}$ is only $1\triangleleft A_{n}\triangleleft S_{n}$.
[/quote]
Hmm, ... , it looks like the job of the Jordan-Holder theorem.


Latex processor is broken at this time. Nothing works even the simplest formula. Admin, can you fix it?

Should be fixed now.
[tex]x^2 + y^2 = z^3[/tex]


geometry
circumcircle
incenter
trigonometry
inequalities
inradius
3D geometry

[b]3nd homework problem set of the IMO training 2004[/b]

Here are the problems from the third homework problem set of our IMO training. The problems are on geometry and graph theory.

[b]1.[/b] Let the equilateral triangle A'B'C' be inscribed in a triangle ABC such that the points A', B', C' lie on the segments BC, CA, AB, respectively. Let S' and S, O' and O, I' and I, and H' and H be the centroids, circumcenters, incenters and orthocenters of triangles A'B'C' and ABC, respectively.

In which of the following cases we can then conclude that triangle ABC is equilateral?

[b](a)[/b]   S = S';
[b](b)[/b]   O = O';
[b](c)[/b]   I = I';
[b](d)[/b]   H = H'.

[b]2.[/b] Let H be the orthocenter of a triangle ABC, and let A', B', C' be the midpoints of the sides BC, CA, AB, respectively. A circle with center H meets the line B'C' at $D_{1}$ and $D_{2}$, the line C'A' at $E_{1}$ and $E_{2}$, and the line A'B' at $F_{1}$ and $F_{2}$.

Prove $AD_{1}=AD_{2}=BE_{1}=BE_{2}=CF_{1}=CF_{2}$.

[b]3.[/b] In the interior of a triangle ABC with BC > BA, consider a point P such that angle PAC = angle BCA. Construct the point Q exterior to triangle ABC such that PQ || AB and BQ || AC. Finally, let R be a point on the segment BC such that the points Q and R are on different sides of the line AP, and that angle PRQ = angle BCA.

Prove that the circumcircles of triangles ABC and PQR touch each other.

[b]4.[/b] [b](a)[/b] Let V be a convex quadrilateral inscribed in a circle with center O. Prove:

[b]i.[/b] If the distances from O to the sides of V equal one half of the respective opposite sides, then the diagonals of V are perpendicular to each other.

[b]ii.[/b] If the diagonals of V are perpendicular to each other, then the distances from O to the sides of V equal one half of the respective opposite sides.

[b](b)[/b] Now let G be a region symmetric to itself with respect to a point O. Assume that the assertion [b]i.[/b], or the assertion [b]ii.[/b] is valid for all quadrilaterals V inscribed in G. Does it then follow that G is a circle?

[b]5.[/b] Consider a convex quadrilateral ABCD with circumcircle k, where A, C and k are fixed, while B and D can be moved simultaneously in such a manner that the equation BC = CD always holds. Let the diagonals AC and BD of the quadrilateral ABCD meet at M, and let F be the circumcenter of triangle ABM.

Show that the locus of F is a circular arc.

[b]6.[/b] Let ABC be a triangle with an angle of 30° at C. Call O and I the circumcenter and incenter of triangle ABC, respectively; further, let D and E be points on the segments BC and AC such that BD = AE = AB.

Prove that the segments DE and OI are perpendicular and equal in length.

[Notably, this can be generalized: If ABC is an ARBITRARY triangle (not necessarily with an angle of 30°), and D and E be points on the rays BC and AC such that BD = AE = AB, then the segments DE and OI are perpendicular, and $DE=2\sin C\cdot OI$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=14655 for this problem.]

[b]7.[/b] Let ABC be an arbitrary triangle, and D, E, F any three points on the segments BC, CA, AB. Denote by P, Q, R the intersections of the lines AD, BE, CF with the circumcircle of triangle ABC (different from A, B, C).

Show the inequality $\frac{AD}{DP}+\frac{BE}{EQ}+\frac{CF}{FR}\geq 9$.
When does equality hold?

[b]8.[/b] Let ABC be a triangle with circumradius R and inradius r. It is well-known that there exist infinitely many triangles having the same circumcircle and the same incircle as triangle ABC.

[b](a)[/b] Show that the orthocenters of all of these triangles lie on a common circle.

[b](b)[/b] What can be said about the locus of the nine-point centers of all of these triangles in the case R = 4r ?

[b]9.[/b] Let G be a graph without double edges (i. e. two edges with the same startpoints and the same endpoints) and without loops (i. e. edges where the startpoint coincides with the endpoint). Moreover, assume at from every vertex of G, at least 3 edges issue. Finally, assume that G does not contain any two disjunct cycles. (A "cycle" is a graph looking like the graph consisting of the vertices and sides of a convex polygon.) Show that in this case, one of the following cases must be in existence:

- G is a complete graph with 5 vertices. ("Complete" means that any two vertices are joined with each other.)

- G is a "wheel", i. e. the graph consisting of the vertices and edges of a pyramid based on an n-gon (with $n\geq 3$).

- The vertices of G can be subdivided into two subsets E and F such that each vertex in E is joined with every one in F, no two vertices from E are joined with each other, and F has exactly 3 vertices.

[I guess I have got all graph theory notions wrongly translated, hence the explanations in brackets.]

[b]10.[/b] Given a polyhedron such that any vertex is an endpoint of exactly three edges. Show that we can attach to each vertex a complex number not equal 1, such that for each face-polygon of the polyhedron, the product of the numbers at the vertices of the side-polygon is 1.

  Darij

I saw problem 10 in the Chinese olympiad in 1991 (a PDF taken from Kalva) and I noticed it had an extra condition there (it always looked kind of strange to me): The sides can be colored with 3 colors s.t. each vertex has exactly one edge colored with each color. Is this condition redundant? Does it follow from the fact that the polyhedron is 3-regular?

I would point you to the file, but the strange thing is that there do not seem to be any Chinese problems on John Scholes' site anymore!?.

Thanks, grobber, for the information. I don't know if this extra condition is necessary, I have not solved the problem (as I have not solved problems 9, 1b, 1d and 4b yet). Maybe it's really redundant, but maybe it's necessary and was forgotten on our working sheet.

  Darij

Well, since it's not available on the Kalva page, I've uploaded the file so you can see the text and judge for yourself.


Let be $ a,b,c\in \mathbb{N}^*$ such that $ (\exists) u,v\in \mathbb{Z}$ that satisfy $ ua\plus{}vb\equal{}1$ . Let be $ S\equal{}\{(x,y)\in \mathbb{N}\times \mathbb{N}\ |\ ax\plus{}by\equal{}c\}$ . Prove that :

i)$ \sqrt{(x_1\minus{}x_2)^2\plus{}(y_1\minus{}y_2)^2}\ge \sqrt{a^2\plus{}b^2}\ ,\ (\forall)(x_1,y_1)\neq (x_2,y_2)\in S$;

ii)$ \{(x,y)\in S\ |\ x\ge 0,y\ge 0\}$ can't have more than $ 1\plus{}\frac{c}{ab}$ elements .

Suppose that we take some particular $ u,v$ such that $ ua\plus{}vb\equal{}1$ tells us that $ (a,b)\equal{}1$.

Now $ a(uc)\plus{}b(vc)\equal{}c$. Suppose $ (x,y)\in S$. Then $ ax\plus{}by\equal{}c$ or
$ a(uc\minus{}x)\equal{}\minus{}b(vc\minus{}y)$ Since $ (a,b)\equal{}1$, $ a|(vc\minus{}y)$ and likewise, $ b|uc\minus{}x$.

So $ vc\minus{}y\equal{}ka$ and $ uc\minus{}x\equal{}mb$

$ a(mb)\equal{}\minus{}b(ka)$ so $ k\equal{}\minus{}m$. So

$ x\equal{}uc\minus{}mb$ and $ y\equal{}vc\minus{}ka\equal{}vc\plus{}ma$

$ f(m)\equal{}(x,y)\equal{}(uc\minus{}mb,vc\plus{}ma)$

It is easy to check that $ f(m)\in S$ for all $ m\in\mathbb{Z}$. So we have an explicit characterization of $ S$ (which in retrospect was more complicated than it needed to be...).

Now i) is obvious as $ |x_1\minus{}x_2|\ge b$ and $ |y_1\minus{}y_2|\ge a$ for distinct elements of $ S$ since they differ by at least $ mb$ ($ m\ne 0$).

I am not sure about ii) but I think that it should follow relatively quickly.


Έστω $ ABC$ οξυγώνιο τρίγωνο και $ G$ το βαρύκεντρό του. Έστω $ M,N,P$ είναι τα ίχνη των καθέτων που φέρνουμε από το $ G$ στις πλευρές $ AB,BC,CA$ αντίστοιχα. Να αποδείξετε ότι $ \frac {4}{27} < \frac {(MNP)}{(ABC)}\leq\frac {1}{4}$ :)

Θα δώσω τα βήματα που λύνουν το πρόβλημα:

Έστω ορθοκανονικό σύστημα αξόνων και έστω, χωρίς βλάβη της γενικότητας, ότι  $ B(0,0), C(2,0), A(2p, 2q)$ με $ p, q > 0$ τότε

Έστω $ K, L, O$ τα μέσα των πλευρών $ AB, BC, CA$ τότε $ K(p, q), O(p \minus{} 1, q \minus{} 1), L(1, 0)$

βρίσκουμε την εξίσωση ευθείας του $ CK$
και την εξίσωση ευθείας του $ BO$

$ G \equal{}$ το σημείο τομής των $ CK, BO$ και συναληθεύοντας τις εξισώσεις ευθειών τους, βρίσκουμε το G

μετά βρίσκουμε τις προβολές του $ G$ στις πλευρές (ευθείες με συντελεστές διεύθυνσης που να έχουν γινόμενο -1 με αυτούς των πλευρών και να περνάνε από το G και μετά τα σημεία τομής αυτών των ευθειών με τις αντίστοιχες πλευρές)

$ (MNP) \equal{} \frac {1}{2}|det(\vec{MN}, \vec{NP})|$

$ (ABC) \equal{} \frac {1}{2}|det(\vec{AB}, \vec{BC})|$

παίρνουμε τους λόγους και τελικά θα βγάλουμε μια παράσταση με αριθμητή και παρανομαστή γραμμικούς συνδυασμούς των $ p, q$. (Σε περίπτωση που ακόμα δε βγαίνει συμπέρασμα, εφαρμόζουμε και την τριγωνική ανισότητα και φράζουμε τα p, q και με άνω τιμές. Είναι βέβαια είδη φραγμένα από κάτω από το 0.)

Δεν είμαι όμως ικανοποιημένος. :mad: Είμαι σίγουρος ότι βγαίνει και γεωμετρικά και σίγουρα πιο απλά:

Για το δεξί, αρκεί να δείξουμε ότι είναι μέγιστος ο λόγος όταν το (ABC) είναι ισόπλευρο (πράγματι, PN = μέγιστο για σταθερή τη γωνία Α  και το AC, όταν BC = AC και το ύψος στο MNP που διέρχεται από το M είναι μέγιστο όταν AB = AC). :lol:  Αλλά δεν μπορώ να δείξω το αριστερό μέλος. :(

P.S. Με μια δεύτερη ματιά παρατηρώ ότι υπάρχει και άλλος τρόπος να βρω το G:
$ BG \equal{} \frac {2}{3}BO \Leftrightarrow G(\frac {2}{3}(p \minus{} 1), \frac {2}{3}(q \minus{} 1))$
Αυτό κάνει πιο έυκολη την όλη διαδικασία.

Να η λύση μου .

Έχουμε $ \frac{(GMP)}{(ABC)}\equal{}\frac{\frac{AH}{3}\cdot\frac{CH}{3}}{AB\cdot BC}\equal{}\frac{(\sin B)^2}{9}$

όπου Η το ορθόκεντρο .

όμοια και για τα άλλα και μας μένει νδο $ \frac{4}{3}<(\sin A)^2\plus{}(\sin B)^2\plus{}(\sin C)^2\leq\frac{9}{4}$

το οποίο είναι αρκετά εύκολο αν δει κανείς ότι

$ (\sin A)^2\plus{}(\sin B)^2\plus{}(\sin C)^2\equal{}2\plus{}\cos A\cdot\cos B\cdot\cos C$ kai μετά Jensen ή οτιδήποτε άλλο . :wink:

Παραπάνω υπήρχαν κάποια λαθάκια αλλά δεν μπορώ να τα διορθώσω γιατί λέει ask a moderator opote na ek neou i lysi mou
[quote="silouan"]Να η λύση μου .

Έχουμε $ \frac {(GMP)}{(ABC)} \equal{} \frac {\frac {AD}{3}\cdot\frac {CF}{3}}{AB\cdot BC} \equal{} \frac {(\sin B)^2}{9}$

όπου D,F ta ίχνη των υψών από τα A,C  .

όμοια και για τα άλλα και μας μένει νδο $ \frac {4}{3} < (\sin A)^2 \plus{} (\sin B)^2 \plus{} (\sin C)^2\leq\frac {9}{4}$

το οποίο είναι αρκετά εύκολο αν δει κανείς ότι

$ (\sin A)^2 \plus{} (\sin B)^2 \plus{} (\sin C)^2 \equal{} 2 \plus{}2 \cos A\cdot\cos B\cdot\cos C$ kai μετά Jensen ή οτιδήποτε άλλο . :wink:[/quote]


Let $a, b$ and $c$ be positive integers, no two of which have a common divisor greater than $1$. Show that $2abc-ab-bc-ca$ is the largest integer which cannot be expressed in the form $xbc+yca+zab$, where $x, y, z \in \mathbb{N}_{0}$

The first we prove that the equation :
$ bcx \plus{} cay \plus{} abz \equal{} 2abc \minus{} ab \minus{} ac \minus{} bc$ has no solution on $ N_0$
Obivious $ a|x \plus{} 1,b|y \plus{} 1,c|z \plus{} 1$
so $ bcx \plus{} abz \plus{} cay\geq bc(a \minus{} 1) \plus{} ca(b \minus{} 1) \plus{} ab(c \minus{} 1)> 2abc \minus{} ab \minus{} ac \minus{} bc$ (contradition)
Now we prove that each $ n > 2abc \minus{} ab \minus{} ac \minus{} bc$ can represent in this form.
$ b(cx \plus{} az) \equal{} 2abc \minus{} ab \minus{} bc \minus{} ca \minus{} cay \plus{} k$
$ \{2abc \minus{} ab \minus{} ac \minus{} bc \minus{} cay\}_{y \equal{} 0}^b$ is a complete reside mod $ b$ so we can chose $ y_0$ such that:
$ cx \plus{} az \equal{} \frac {2abc \minus{} ab \minus{} ac \minus{} bc \minus{} cay_0 \plus{} k}{b}$
$ az \plus{} cx \equal{} \frac { 2abc \minus{} ba \minus{} bc \minus{} cay \minus{} ac \plus{} k}{b} > ac \minus{} a \minus{} c (1)$
[u]Result[/u] we know that for each $ n \geq (a \minus{} 1)(b \minus{} 1)$ then the equation $ ax \plus{} by \equal{} n$ has non negative integer solution.
So $ (1)$ has solution.
The problem has been proven.


[quote="Cauchy_Schwartz"]$A_1,B_1,C_1,D_1 \in (AB,(BC,(CD,(CA$;[/quote]

Do you happen to mean $A_1,B_1,C_1,D_1 \in (AB,(BC,(CD,(DA$ ?

And also, please use \in ($\in$) instead of \epsilon ($\epsilon$) for "is element of". (I have corrected this in your posting.)

  Darij

THank you, I couldn't find \[\in\], that's why I used \[\epsilon\]...Sorry, And thank for your support....but still, do you find it interesting?

Well, not bad but still not the kind of problems I really love... but I'll try.

  Darij

Well, is it so difficult or it's simply boring?


Compute the following sum

$ \sum_{k\equal{}0}^n(4^{k\plus{}1} C_{n\plus{}2}^{k\plus{}1}\minus{}4^k C_{n\plus{}1}^{k})$

where $ C_n^k$ are the binomial numbers

I'm assuming you meant
\[ \sum_{k \equal{} 0}^n \left( 4^{k \plus{} 1} \binom{n \plus{} 2}{k \plus{} 1} \minus{} 4^k \binom{n \plus{} 1}{k} \right)
\]
since the way you have written, it would be trivially zero since by definition, whenever the bottom part of the binomial coefficient is greater than the top part, it's 0.

[hide="solution"]
Expanding out, we have
\[ \left( 4\binom{n \plus{} 2}{1} \minus{} \binom{n \plus{} 1}{0} \right) \plus{} \left( 4^2\binom{n \plus{} 2}{2} \minus{} 4\binom{n \plus{} 1}{1} \right) \plus{} \cdots \plus{} \left( 4^{n \plus{} 1}\binom{n \plus{} 2}{n \plus{} 1} \minus{} 4^n\binom{n \plus{} 1}{n} \right)
\]
which is a telescoping sum equal to
\[ 4^{n \plus{} 1}\binom{n \plus{} 2}{n \plus{} 1} \minus{} \binom{n \plus{} 1}{0} \equal{} 4^{n \plus{} 1}(n \plus{} 2) \minus{} 1
\]
[/hide]

[hide="correct (hopefully) solution"]
\[ \sum_{k\equal{}0}^n \left( 4^{k\plus{}1} \binom{n\plus{}2}{k\plus{}1} \minus{} 4^k \binom{n\plus{}1}{k} \right)\equal{}\sum_{k\equal{}1}^n \left( 4^{k} \binom{n\plus{}2}{k}\right) \minus{} \sum_{k\equal{}0}^n \left( 4^k \binom{n\plus{}1}{k}\right) \equal{} (1\plus{}4)^{n\plus{}2} \minus{} 4^{n\plus{}2} \binom{n\plus{}2}{n\plus{}2}\minus{}4^{n\plus{}1} \binom{n\plus{}2}{n\plus{}1}\minus{}4^{0} \binom{n\plus{}2}{0} \plus{} (1\plus{}4)^{n\plus{}1} \minus{} 4^{n\plus{}1} \binom{n\plus{}1}{n\plus{}1}\]
which simplifies to
\[ 5^{n\plus{}2} \minus{} 4^{n\plus{}2} \minus{} 4^{n\plus{}1}(n\plus{}2) \minus{} 1 \plus{} 5^{n\plus{}1} \minus{} 4^{n\plus{}1}\]
and then to
\[ 5^{n\plus{}2} \plus{} 5^{n\plus{}1} \minus{} 4^{n\plus{}1}(n\plus{}3) \minus{} 1\]
[/hide]

p.s. If anyone knows why the spacing in my LaTeX is so wierd, please tell me!

How come it becomes a telescoping sum ? The very first term, $ 4 \binom{n \plus{} 2}{1}$ cannot be canceled out with any thing. Note that the similar one in the second term is $ 4 \binom{n \plus{} 1}{1}$.

I think that there's no easy formula for this sum.


I have an Olympiad Geometry class tonight starting in about 15 minutes, but when I click on "Classroom," it goes straight to the WOOT classroom instead of allowing me to chose which one...

Classrooms open approximately 10 minutes before class starts. Since it is still 15 minutes before class, the Olympiad Geometry classroom is not yet open.


Suppose it is now the month of November.  What month would it be 100 months from now?

[hide]  100/12= 8 remainder 4

So we will revolve around november for 96 months but then must move on 4 more months.  That would be december, january, februay, [b] march [/b][/hide]

[quote="236factorial"][hide]100 divided by 12 is 8r4, and adding 4 months to December is April. [/hide][/quote]
you are supposed to add 4 months to November not December.


calculus
integration
trigonometry
function
conics
ellipse
complex analysis

Evaluate
\[\int_{-1}^{1}\sqrt{\frac{1+x^{2}}{1-x^{2}}}\ dx \]
I came across this integral when a friend (as a joke) asked me to find the arclength of $\cos x$ from $0$ to $\pi$.  I tried using the "standard" complex variables substitution ($z=e^{i\theta})$ on $\int_{0}^{2\pi}\sqrt{1+\cos^{2}\theta}\ d\theta$ but didn't get anywhere (I think it's because the funtion has a singularity and a branch point at the origin). I'm not really sure if it's even possible to find a closed form for this integral, but any help would be appreciated.

Thanks in advance.

[quote="mlok"]First, carefully draw the graph of $y=\cos x$ for $0\le x\le 2\pi$ on a sheet of paper. Then wrap the paper into a cylinder of radius $1$ by gluing the vertical lines $x=0$ and $x=2\pi$ together. You will notice that the graph is nothing but the intersection of the cylinder with a plane slanted at $45$ degrees. This shape is an [b]ellipse[/b]. Thus, the length of the graph is equal to the length of an ellipse with semiaxes $1$ and $\sqrt{2}$.

The length of an ellipse (other than a circle) is impossible to express in terms of elementary functions of the semi-axes. This is why people invented [b]elliptic[/b] integrals. [url=http://mathworld.wolfram.com/Ellipse.html]Read more here[/url].[/quote]

Elegant; very nice.

[quote="mlok"]
The length of an ellipse (other than a circle) is impossible to express in terms of elementary functions of the semi-axes.[/quote]
This is crazy. Why is this so?

In the "real" world, integrals that aren't elementary are extremely common. If you see an integral that didn't come from a calculus textbook and you don't see a method right away, you should assume that you probably can't do it in closed form. Elliptic integrals are merely one of the more famous examples.

The complete arclength of an ellipse can be calculated very accurately with little effort; the simplest numerical integration methods work extraordinarily well on $\int_{0}^{2\pi}\sqrt{a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}\,d\theta$.

Well, ellipses seem to be such simple shapes that it should be possible to express their lengths in an easy way.

Anyway, in what kind of mathematics do you actually discuss issues such as if a number can be expressed with elementary functions? Is it some type of esoteric algebra?

[quote="Kalle"]Well, ellipses seem to be such simple shapes that it should be possible to express their lengths in an easy way.

Anyway, in what kind of mathematics do you actually discuss issues such as if a number can be expressed with elementary functions? Is it some type of [b]esoteric algebra[/b]?[/quote]

What exactly is this?


limit
function
calculus
calculus computations

how does one prove that

$ \lim_{n\to\infty}(1+\frac{1}{n})^{n}=\lim_{x\to\infty}(1+\frac{1}{n})^{n}$

where $ [x]=n$ after proving that the latter limit exists and knowing that the former exists

The second limit doesn't seem to make sense, because there is no "x" in the expression you want to find the limit.

As written, it is [b]exactly[/b] the same limit, so there is nothing to do. If you have any function $ f(x)$, then $ \lim_{x\to\infty}f([x])=\lim_{n\to\infty}f(n)$ (just use the $ \varepsilon-N$ definition of the limit and you'll see it in a few seconds :))


inequalities
integration
algebra
function
domain
logarithms
real analysis

$ f: [1,2]\rightarrow(0,\infty)$ f strictly ascending, with $ f^\prime(x)$ continous. If $ f(2) = 2f(1)$ prove the following ineq:

${ \int_1^2\frac {f^\prime(x)}{xf^\prime(x) + f(x)}}dx\leq\frac {1}{2}\ln2$

yes it is [1,2], sorry:( i was in a hurry and i didn`t notice the mistake

deleted because of mistakes   :oops:

but is $ f : [1,2]\rightarrow (0,\infty)$    ?

I assume the domain is meant to be $ [1,2]$.
$ \frac {1}{a + b}\leq \frac {1}{4a} + \frac {1}{4b}$ (AM-HM or just multiply by $ 4ab(a + b)$) so,
\begin{eqnarray*} \int_1^2 \frac {f'(x)}{xf'(x) + f(x)}dx & \leq & \frac {1}{4}\int_1^2\frac {1}{x}dx + \frac {1}{4}\int_1^2\frac {f'(x)}{f(x)}dx \\
& = & \frac {1}{4}\log 2 + \frac {1}{4}\log \left( f(2)\right) - \frac {1}{4}\log \left( f(1) \right) \\
& = & \frac {1}{2}\log 2\end{eqnarray*}


probability
function
calculus
derivative
algebra
polynomial
probability and stats

Prove that:
There exist random variables $ X_1$ and $ X_2$ taking values in $ \mathbb{R}$ which have the same moments but different probability densities.

I was thinking along those lines, but one thing worries me: sure, we can find two different functions whose Fourier transforms agree in all derivatives at the origin - but can we assure that both of those functions are nonnegative? We need that last condition to have probability distributions.

Just add some positive function decaying much slower than both of them (but still faster than any polynomial) with a big coefficient to both.