Seems like US want a 3rd World War, phase No.1, to fight the "Islamic fascist".
http://www.freep.com/apps/pbcs.dll/article?AID=/20060831/NEWS07/608310334/1001/NEWS

sorry. url problem fixed

[quote="Buffalo"]Seems like US want a 3rd World War, phase No.1, to fight the "Islamic fascist".
[url]http://www.freep.com/apps/pbcs.dll/article?
AID=/20060831/NEWS07/608310334/1001/NEWS[/url][/quote]

It's Bush.  :o

What's wrong with the URL?  :maybe:

Anyway, there will be a third world war - the greatest of 'em all - and I am still unsure how many factions will join it ... It will be the war to end all wars - that's certain though!

A large percentage of the population of the USA is pretty dumb. In fact, half of them are below the average intelligence! Bush only needs a little more than half to win.

You know, the reason the Muslims in the Middle East hate the West is for the same reason that people in the West hate them -- the ones who control information and send cultural and political messages tell them to! In that abstract way, I don't consider either side to be worse. And in a sense it isn't the average citizen's fault -- it's when he realizes that what he's told might be wrong (or at least biased) and still doesn't care that he's really doing something wrong himself.

[url=http://youtube.com/watch?v=ugF4WyG23PM]Fox video[/url]
What to do when you're trying to make "American television" and some pesky Arab starts accusing the USA of abandoning its Arab citizens:
[hide]pull the Islamic fascism card! :clap2: [/hide]
[url]http://youtube.com/watch?v=ugF4WyG23PM[/url]

From [url]http://www.idleworm.com/arc/006/07jul.shtml[/url]
[quote]Remember Dermot's #1 rule: you can't have a World War without inviting the Russians.

So here's my easy-to-understand plan for the new World War naming convention:
7 Years War (1756-1763) = World War 1
Napoleonic Wars (~1799-1815) = World War 2
World War 1 (1914-1918) = World War 3
World War 2 (1937-1945) = World War 4
Cold War (1945-1990) = World War 5
War on Terror (2001-the rapture) = World War 6 [/quote]

[quote="Buffalo"]Do you mean it's Bush who want the war, not US?
I don't see the difference, US has no choice but going to whatever war that Bush wants, like what happened in Iraq.
Americans don't want the war? Then WHY did they make Bush president again?[/quote]

Well, there certainly are very many dumb Americans who've been brainwashed by the news media. That doesn't mean, however, that there aren't also many people who are against this war.

And who ever said that democracy was perfect  :wink: [url]http://www.rollingstone.com/news/story/10432334/was_the_2004_election_stolen/print[/url]

[quote="Buffalo"]Do you mean it's Bush who want the war, not US?
I don't see the difference, US has no choice but going to whatever war that Bush wants, like what happened in Iraq.
Americans don't want the war? Then WHY did they make Bush president again?[/quote]

I meant :" It's Bush. He says things like that all the time."
Of course that US citizens don't want war. The military industry and oil ones want war.
Why did they make Bush president again? Everybody can make a mistake.

I would like to point out that the decision to go to war had to be backed by the US House of Representatives and US Senate.  Representatives are elected every two years, senators every six, and not all states elect a new senator at the same time.

Whether or not USA citizens are actually to blame, that is hard to say...
But we are talking about a country where most of the people don't know where Israel is, Taiwan, Iraq,etc.... , where lots of people don't remember when 9/11 happened,...
Media has a role in this, but there is free speech and internet etc.. so it's not hard to inform yourself.  But maybe it's just the culture that promotes self-centeredness.
Perhaps we should all follow the teachings of this fine young lady :
[img]http://upload.wikimedia.org/wikipedia/commons/thumb/8/84/Britney_Spears.jpg/220px-Britney_Spears.jpg[/img]
[quote]I think we should just trust our president in every decision that he makes and we should just support that, you know, and be faithful in what happens. [/quote]

[quote="fredbel6"]
Perhaps we should all follow the teachings of this fine young lady :
[img]http://upload.wikimedia.org/wikipedia/commons/thumb/8/84/Britney_Spears.jpg/220px-Britney_Spears.jpg[/img]
[/quote]

lol  :rotfl:

[quote="Xevarion"]A large percentage of the population of the USA is pretty dumb. In fact, half of them are below the average intelligence! Bush only needs a little more than half to win. [/quote]
I don't think you've stated a solid reasoning. People could have all kind of different motivation to vote for Bush, for example, he's a republican, his domestic policies are attractive, Bush is charming--compared with Kelly, at least.
On the other hand, not every elector will vote, in 2004, there are only 60%, it's already the highest in the history. If most of the anti-Bush's stay home, but almost all the Bush fans vote, you can expect the result. Further more, US has a indirect election, this change things too.
  For those who voted Bush, either they don't realize this will bring disaster to the rest of the world(do you think US can always keep out of it?), or they think it's relatively unimportant, they're like some starving killers, you give them a piece of bread, they will kill someone for you, or at least be a stander-by when you kill someone. Being a citizen in the most powerful country, whose power is comparable to the sum of the rest of the world, voting this way, this is the real disater.
 BTW, although Bush's IQ is a little bit under the average, and he often behaves like a big-mouth, I don't think he is dumb. He's not alone, the group behind him--the military and oil industry, etc--decides what he should say, what he should do.

@[b]Buffalo[/b]: I once heard that Bush's IQ is 87. I'd guess that's not accurate, and I agree that the fact that he says stupid-sounding stuff when his speeches aren't prepared isn't sufficient to prove that he actually is stupid. People mock him because they don't know a better way to protest against his administration (or because mocking the current president is a time-honored tradition).

You say that people who vote for Bush either don't realize what negative effect his administration can have on the rest of the world or they don't care. This is exactly what I meant! Not all voters in the USA think like this (I don't!!!) but way too many of them do! A lot of those who vote against Bush probably still feel that way, and just don't like Bush for other reasons! It is (as [b]fredbel6[/b] mentions) largely due to the media and education here. However on one point I think [b]fredbel6[/b] is wrong: if a person firmly believes something like "all Muslims are evil" or "the Holocaust didn't happen" or "the Bible is literally exact and true" (or, perhaps more relevant [though not as wildly wrong], "Bush is a good president all around") they will not be easily influenced by evidence to the contrary, even when it is overwhelming in magnitude. [b]That[/b] is the barrier that keeps people like me from doing anything useful for the nation. We can be well-informed voters but it's really hard for us to create any more by informing others when they don't want to be informed!

[quote="Buffalo"]
I don't think you've stated a solid reasoning. People could have all kind of different motivation to vote for Bush, for example, he's a republican, his domestic policies are attractive, Bush is charming--compared with Kerry, at least.[/quote]

You've made a good point - although people may hear the newsmedia go on and on about the failure in Iraq, people still may have many reasons to vote for Bush or even Republican Congressmen (as fredbel6 pointed out that the Congress needs to approve the war). One important thing to note is that many people living in the current America believe that Democrats are weak on security - Americans constantly hear about the "cut and run" Democrats, the Democrats who are unpatriotic and don't "support our troops," or the Democrats who don't take Guantanamo "seriously." Therefore, they end up trusting Republicans on National Security. As an NYT Columnist, "it's better to have the wrong convictions than to have little or no convictions."

As well, Americans tend to prefer to hear optimistic news - nobody is going to sit there and watch more and more about how much of a failure Iraq is. Instead of actually trying to deal with the conflict, many Americans simply attempt to avoid it. A good assessment of the situation is:
[quote="Frank Rich"]Americans want this war canceled, and first- and last-place networks alike are more than happy to oblige.[/quote]

[quote="mustafa"][url]http://en.wikipedia.org/wiki/Image:Bush_approval_ratings_line_graph.png[/url]
His approval ratings have been below 50% for at least six or seven months, and he dropped below 33% for a little while. Not that many Americans really think that he is doing the right thing.[/quote]
The point is not how many people think he is wrong. It's how many people could do something to stop him that really matters.
Now the situation is just like the beginning of WWII, many people knew that Hitler was doing bad, but no one with necessary power stood out saying "No!", instead, they tried to share some residue of Hitler's prey, finally they became the next prey themselves.

[quote="JBL"]Buffalo, you're so [i]boring[/i].  The United States is a democracy (more or less), and we're 2 months away from a national election which is seeming increasingly likely to turn one (and, substantially less likely, both) houses of congress over to the opposition party.  I realize that you come from a country where the leaders [i]are[/i] the government, there are no opposition parties, there are no free elections, and the media is tightly controlled by the government, but that doesn't mean that you have to act like everyone else's countries are the same.[/quote]
Wow JBL, that is quite an explicit response, rather unlike your others. :o

But seriously, what country is Buffalo from then?

[quote="fredbel6"]Wow JBL, that is quite an explicit response, rather unlike your others. :o

But seriously, what country is Buffalo from then?[/quote]

Well, I get tired of it.  He doesn't come here to have a conversation, to think, he comes here to preach.  I have many of my own views about the world, but for some reason, every time I find a piece of information which re-enforces one of them, I don't feel a need to come and post it on the forum.  Even when I agree with him, I find the basic presentation of ideas tiresome.

I'm reasonably certain Buffalo is from China, from earlier threads.  I don't remember if he specifically said so, or if I picked it up from some of his sermons.  I suppose the hypocricy of it bothers me as well (as someone who finds the current regimes in both the U.S. and China to be highly distasteful).

I have to admit in a way I agree with [b]JBL[/b]. What are you trying to do? Most of us Americans here on AoPS already don't like our administration. Most of the people here are under 18 and therefore can't really do anything in our system of government. Unless you're hoping to spark a violent revolution led by the mathematics students of the USA or something.... :P

Anyway you are surely accusing [b]JBL[/b] of something he didn't say now! His complaint is against the kind of things you say and the way you present them. He never said there's anything wrong with Chinese in general.

In fact I really hope we don't ban Chinese because there are some very good contributors to this forum like [b]zhaobin[/b] and [b]shobber[/b] who are from China.

Xevarion :
The idea that all Muslims are evil, isn't that still somewhat the same?  Where does it come from?  Was it written on your arms when you were born?   You would have gotten it from your parents, teachers,etc....

I have a question : [url]http://www.overspun.com/video/Countdown.RummyOnTheRopes.wmv[/url]
In this video, a reporter is questioning Rumsfeld about his own claims in the past concerning Zarqawi and weapons of mass destruction.  (Interesting is Rumsfeld's reference to protection against chemical weapons, let us call this "proof by being afraid of it")

Anyway my question is: when the reporter asks him if he was lying, Rumsfeld says he wasn't, and the people start clapping, even before he has given any arguments?  How representative are these people for the entire USA population?  Applauding your leaders...because they say they are not lying :huh:

[quote="Buffalo"]Sorry for stepping on your tail, Am I making you feel uncomfortable?
You should put signs here saying "Chinese and dogs forbidden" etc.[/quote]

This is a classic falacy -- "you objected so strongly, so something I said must be right."

Explain to me again how telling [i]you[/i] that I find [i]you[/i] annoying and hypocritical is equivalent to my wanting to ban all Chinese from the forum?  (Also, I happen to like Gyan's dog who visits here (usually the intermediate forum?) from time to time.)

[quote="fredbel6"]Anyway my question is: when the reporter asks him if he was lying, Rumsfeld says he wasn't, and the people start clapping, even before he has given any arguments?  How representative are these people for the entire USA population?  Applauding your leaders...because they say they are not lying :huh:[/quote]
This is the kind of thing that discourages me so much that I'm not motivated to be politically active.

It doesn't matter what country someone is from, their personal views are not dependent on their nations acts or what type of government one has.

There are many people in the US who believe that the war against terrorists in Iraq was and is proper and appropriate.  Have any of you talked to any Iraqi citizens or soldiers who actually were in Iraq and have some potentail of knowing what is actually going on, or do you rely on the press and the stories that they print trying, not to represent the truth, but to sell papers and air time to their readers and viewers?

Have you gone back to see how the press reported Nazi Germany and Japan before the US was attacked and entered the war?  How about before the first world war (called the Great War before WW II)?  History books in class make this very clear, but it never is when the decisions have to be made.

People are decrying our losses in Iraq saying that we can't afford the cost in manpower.  While each death and injury is terrible and I don't want any to occur. (Many more Iraqi's are dying now than foreign soldiers and will die if we "cut and run" leaving a civil war to break out.)  Look up the death toll at just one civil war battle, say Getteyburg to see what a real civil war can do.

In Vietnam we lost how many men??

In WWII the US had over 30,000 men per month killed over the entire US participation - 5 years.  We are now approximately twice the size and have lost 2,000 in five years.

We can leave, but we shouldn't leave because someone is mislead to say we can't stand the loss of life.

We will leave when the US political sentiment reaches this decision and not a minute before.  The President is important, but the money is voted by Congress and they can pass a law requiring troops to come home.

Why haven't the Democrats stopped the funding bills and caused the financing to "end the war?"  They have the current political power to do this, despite their complaining on the news.  The real reason is that this isn't a moral issue to them, it's just politics as usual.  They know if they don't support the war the unvoting folks will turn up and they will have to find a "real job."

Don't be mislead.  Folks on the news may not support the war, but America as a whole does.

You folks claim that you are smart.  What is likely to happen if the US says, "Ok, you terrorists win" and then we leave?

Will any Iraqi be safer?  Better off?  Will the US be safer?  Will the other countries be safer?  Will China be safer?

I don't like war, but the terrorists haven't given us any choice.

Look, if they really wanted us out of Iraq, the smartest thing (I know discussing a smart thing to do when talking about terrorists is not working with the best correlation I've ever seen.) would be to have stopped all violence, have a government installed and ask the US to leave.  We would have been out of Iraq in less than a year.  We are only still there because of the terrorists that are mainly killing Iraqis and telling them that they are "fighting" to get the US out.

Some actually believe them.

Amazing.

[quote]People are decrying our losses in Iraq saying that we can't afford the cost in manpower. While each death and injury is terrible and I don't want any to occur. (Many more Iraqi's are dying now than foreign soldiers and will die if we "cut and run" leaving a civil war to break out.) Look up the death toll at just one civil war battle, say Getteyburg to see what a real civil war can do. [/quote]
I think I discovered the higher death toll among Iraqi civilians some time ago.  Uhm, when I decry the terrible situation,not  only am I talking about the civilians too, but also especially about them.

[quote]In Vietnam we lost how many men?? [/quote]
If "we" means Belgium or the Netherlands : zero :) . If you're from Denmark it's okay too (we are not by default from the USA)

[quote]You folks claim that you are smart.[/quote]
Me?  I'd remember such a claim.

[quote]What is likely to happen if the US says, "Ok, you terrorists win" and then we leave? [/quote]
If you mean by that retreating from Iraq : civil war and instability in the entire region.

[quote]I don't like war, but the terrorists haven't given us any choice. [/quote]
What are you referring to now?  The decision to invade in 2003 or the decision to stay in Iraq?
But it's okay, I'm pretty sure you are in the continental USA so you don't have like war.

[quote]We are only still there because of the terrorists that are mainly killing Iraqis and telling them that they are "fighting" to get the US out. [/quote]
Many terrorist, like Zarqawi, are from Jordan, Saudi Arabia or Egypt.  They opened a Pandora's Box, and the toll is on the Iraqis now.

As I don't fully understand your depiction of the situation, I'll make some comments. Feel free to respond :D

[quote="wizard2378"]It doesn't matter what country someone is from, their personal views are not dependent on their nations acts or what type of government one has.[/quote]
How can you know it? Under how many types of government have you been under? (I deduced that you're from the US, correct me if I'm wrong.)

[quote="wizard2378"]
People are decrying our losses in Iraq saying that we can't afford the cost in manpower...
In Vietnam we lost how many men??
In WWII the US had over 30,000 men per month killed over the entire US participation - 5 years.  We are now approximately twice the size and have lost 2,000 in five years.
We can leave, but we shouldn't leave because someone is mislead to say we can't stand the loss of life.[/quote]
You're waving some numbers in front of me. You're saying that because the US is big and [b]only[/b] a couple of thousands have died, they should continue the war?
Anyway, I agree with the other part, leaving would probably do no good as well.

[quote="wizard2378"]
Don't be mislead.  Folks on the news may not support the war, but America as a whole does.
[/quote]
You're talking about politicians just a minute ago (Democrats, Republicans), now you say that America as a whole supports it? How did the logical deduction go?

[quote="wizard2378"]
You folks claim that you are smart.  What is likely to happen if the US says, "Ok, you terrorists win" and then we leave?
Will any Iraqi be safer?  Better off?  Will the US be safer?  Will the other countries be safer?  Will China be safer?
I don't like war, but the terrorists haven't given us any choice.
[/quote]
You're right on one part. The terrorists shouldn't "win".

[quote="wizard2378"]
Look, if they really wanted us out of Iraq, the smartest thing (I know discussing a smart thing to do when talking about terrorists is not working with the best correlation I've ever seen.) would be to have stopped all violence, have a government installed and ask the US to leave.  [b]We would have been out of Iraq in less than a year.[/b]
[/quote]
That's a very naive idea (the one in bold and what's before it). The Americans would have to be REALLY stupid to do that, no matter what the real purpose of the war is.

Some good points, wizard2378.

We simply can't quit in Iraq now, regardless of whether it was good or bad to go in. First of all, it would break our trust with the Iraqis.  What would it be like if we rushed in there, removed a dictator and then just left the people alone to fend for themselves? Second, the terrorists will keep fighting until there are defeated. We can pull out of Iraq, but the terrorists will still fight. Perhaps by planning another 9/11 type attack, or perhaps by attacking Iraqis while we are gone. Hey, if they attacked us when we weren't "fighting" them on 9/11, you can expect them to fight us any other time until they are subdued.

Both of these things would make the US look bad. However, the loss of a reputation is worse because it will embolden the terrorists. If they have success with a 9/11 attack, and go unchallenged, can you imagine what they would do next?

This is exactly what the terrorists want. The know they can't defeat the U.S. in a conventional war. The want us all dead, 9/11 is exceedingly good evidence for that, but they can't defeat us by invading us with an army. So they want us to lose heart, by continuously hacking away at morale. The US media only seems to encourage it.

The more we think we are losing, the more they are winning. But we don't need to think we are losing. We invaded Iraq with exceeding effectiveness. We have removed Saddum Hussein from power. We have had a low number of casualties, considering the difficulty of defeating people who will blow themselves up while killing you. We have killed Al Zarqawi, the leader of the Iraq insurgents. People in Iraq have voted. Terrorist attacks on planes have been thwarted.

And people, who says that you can establish peace within a country in five years? Over 25 years after their defeat of the British, the U.S. still had trouble with Britain. Not to mention events such as the Whisky rebellion.

[quote]We simply can't quit in Iraq now, regardless of whether it was good or bad to go in. First of all, it would break our trust with the Iraqis. What would it be like if we rushed in there, removed a dictator and then just left the people alone to fend for themselves?[/quote]
I'm sorry but that's like using someone else's restroom without asking and expecting a praise for flushing :|

[quote]Both of these things would make the US look bad. However, the loss of a reputation is worse because it will embolden the terrorists. If they have success with a 9/11 attack, and go unchallenged, can you imagine what they would do next? [/quote]
I second Perfectradio on this, who is "they", it's suspiciously vague.

[quote]We have had a low number of casualties, considering the difficulty of defeating people who will blow themselves up while killing you.[/quote]
Number of people killed on 9/11 :2973 (excluding the hijackers)
Number of US soldiers killed in Iraq : over 2600

But the most important thing : Iraqi civilians killed since start war :definitely over 40.000 (some estimas go as far as 100.000)
Either you were simply not referring to Iraqi casualties at all, or I need to grab my US-Iraqi convertor again

[quote]Terrorist attacks on planes have been thwarted. [/quote]
IF we believe the official story, then we can still doubt whether or not the Iraq war didn't cause it, rather than helping to prevent it.
Just a moment ago I heard someone talk about two phases
1. many radical young muslims travel from countries like UK, Belgium,... to Iraq etc.. to receive training and/or blow themselves up or something like that.
2. those who didn't blow themselves up return to the west

I repeat again that Iraq isn't just full of Iraqi terrorist, it's crawling with insurgents from all around.  What was allegedly a training ground for muslims extremists is now certainly such a thing, thanks to the invasion.


integration
trigonometry
calculus
geometry
function
perimeter
analytic geometry

Let $ y \equal{} \sqrt {x^2 \minus{} r^2}$
and $ \displaystyle \alpha_x \equal{} \arctan{\left( \minus{} \frac {\sqrt {x^2 \minus{} r^2}}{x}\right)}$

$ \int^1_0\sqrt {2y^2 \plus{} 2y\sin{(x \plus{} \alpha_x)} \plus{} \cos^2{x}}\,dx \equal{} ?$

Bit over my head here...thanks!

What was the motivation for this integral?

Maybe if we integrate it some other way, it will be nicer

Well that is, to the best of my abilities, the area between a circle and a cosine function who has had it's x-axis redefined as path of said circle...after simplification.

By, 'some other way', what do you have in mind?

I'm sure we'd love to help (as soon as we're convinced that this isn't just a bunch of completely random functions thrown together) but what on Earth does it mean to "redefine" the x-axis?  I have never heard this expression.

Are you talking about bending a cosine graph along a circle?  What are the details of this deformation?  Are you maintaining distance to the axis when you bend?  Is it possible that as you bend the graph, it intersects with itself?  If so, are you more interested in the actual plane area, or of the value of this integral?

In short, if you want us to do something "gross", please tell us [i]WHY[/i].  I can already tell you what Mathematica says: that it doesn't know.

I apologize; when I posted this question I was very new to calculus.

The concept here is that the perimeter of a semicircle is $ \pi \equal{} |\beta \minus{} \alpha|$ represents the angular interval $ (\alpha,\beta) \equal{} (\pi, 0)$ upon which cosine is redefined. In polar coordinates, we set $ \alpha \minus{} \theta \equal{} \varphi$, which will increase in a clockwise rotation.

You extend the radius of the semicircle by $ \cos{(\theta \minus{} \pi/2)}$ in the direction of the unit vector corresponding to each point on the circle. I computed the area for a circle of radius $ r$, and found that
\[ A(r) \equal{} 2r \plus{} \pi/4
\]
in only three minutes. My, how far I've come.  :D


find all numbers (in base 10) whose base 5 representation is the reverse of its base 8 representation.

Could you stop posting problems from camp admission tests(mathcamp in this case)? Thanks.

Please lock/delete this.

Pm an admin please. I'm too tired to send a message tonight.

I will lock this though.


induction
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer.Consider an $ 2*n$ chessboard .In some cells there are some coins.In each step we are allowed
to choose a cell containning more than 2 coins,remove two of the coins in the cell, put one back into either the cell upper to the choosen cell or to the cell to the right of the choosen cell.Assumme there are at least $ 2^n$ coins on the cheesboard .prove that after finite number of moves ,it is possible for the upper right corner cell to contain a coin.

this problem is from iran 2005 and the solution I know uses induction on $ n$.


The lines 4x+3y=6, 2x-5y=16 and y=-3x+b intersect at the same point...find b.  how do i start this??

[quote="tim1234133"][hide="Method"]
Solve the first two, and find the unique point where they meet. Then put these $x$ and $y$ values into the third equation, and you get $b$.
[hide="Solution"]
The first two lines meet at $(3,-2)$: put these values in the third equation and get $b=7$[/hide][/hide][/quote]

thx this helped a ton  :wink:


The new game has started! Hopefully it works better this time :oops:
The game will be pretty much the same as the previous attempt, but this time there are more civilians, so the role-claiming won't work again.

[b]Players:[/b]

1. bubka
2. Klebian
3. Ars
4. mustafa
5. kheg_k
6. Ignite168
7. junggi
8. wobster109
9. tim1234133
10. Peter VDD
11. BoesFX
12. akalra1
13. ragnarok23
14. dts
15. LadyKn1ght
16. daine
17. perfect628
18. laughinghead505
19. mad_skillz_aops
20. pianoforte

[b]Roles:[/b]

--- 4 mafia ---
No day communication allowed, except during day 1. The mafia also get the corpseripping ability, which means that nobody but them get to know who they killed.

--- 2 angels ---
No day communication allowed, except during day 1.

--- 1 mathematician ---
This innocent role can, once during the game, during the day, reveal him/herself and pick a group of 2 or more people (not including him/herself) to "inspect". I will then tell everybody how many mafia there are in the group, but not which of them is mafia. It is illegal to claim mathematician if you are not really mathematician. Once the mathematician has used his/her power, (s)he becomes the same as a normal civilian.

--- 1 magician ---
This innocent role gets to choose 2 people each night and whatever would have happened to them, is swapped round. For example, if A would have been killed by the mafia and B would have been inspected, B would be killed and A would be inspected. I will tell the mafia that they have killed B and the role of B, and the inspector will be told that he inspected A, and if A is guilty or not.

--- 1 inspector ---
--- 1 suicide bomber ---
--- 1 priest ---
--- 1 devil ---
--- 2 masons ---
--- 5 civilians ---
--- 1 miller (who thinks (s)he is a civilian) ---

Other info:

- Everybody has to post at least once every 72 hours (unless they said they would be away), or else they might be replaced.

- If a day lasts longer than a week, the town will go to sleep without a lynch.

- One person has a fake ID. If this person is killed, it will only be revealed that (s)he had a fake ID.

Wow that was fast why do you accuse him Ignite

Alive and kickin'.

And it seems Ignite thinks we will need a random lynch and already makes a proposal?  :roll:

Back in the game!

Also, for the record, I was the Devil last time.

But not this time!

And Ignite168, why so fast, and why perfect628?

I'm in not as mafia :D

Did I say I am mafia in the last game(previous mafia 22)?

Oh well, just concentrate on this game now.

I am here :D

Last game, I was mafia, but not this game. I promise.

What did perfect628 do to offend you ignite? Just a random killing?

[quote="BoesFX"][quote="pianoforte"]That's what I thought earlier too, but upon checking, weiquan said 7 posts ago that "I am innocent", though it was easy too miss.

Edited:
[quote="BoesFX"]Bubka - Civilian
Jashper - Miller
Ars - Civilian
Tim - Civilian (likely)

I'm a civilian.[/quote]

omg, I was so sure this post said "I accuse PeterVDD". It must have been edited. Seems suspicious...[/quote]

Sorry about editing my post.

Previously I don't see that there are 5 civilian and 1 miller, what I see is that there are 4 civilian and 1 miller and because I am civilian, I think that all the last civilian-claims are fake so I start with PeterVDD. However, when I check the first page I am wrong. So I edited my post.

About voting Aye on Ars, maybe I am a bit emotional since we get no lynch 2 days ago without reason.[/quote]

Koda is innocent, laughinghead505 is innocent, I am innocent. If there are 2 more civilians, then either you, pianoforte or Peter VDD is guilty. Now I have just reread the posts in this game again and based on that I think you are more probably guilty. Well, if you say you are innocent then you have to prove sooner or later that either pianoforte or Peter VDD is guilty, OR that all 3 of you are innocent, which is [b]very unlikely[/b].

However we have to lynch a guilty today otherwise we lose, and since there are some guilty people whom we agree on (unless you are mafia), like Ignite168 and dts, we could lynch those first. So I [color=red]second both[/color] Ignite168 and dts. If anyone think either is innocent please state your reasons quickly, because we have to make the correct lynch by today.

And Koda, I suggest you save yourself at random with probability about 2/3, so the mafia can't kill you 2/3 of the time and 1/3 of the time you can save someone you think the mafia are likely to kill. But if you think you should do otherwise then just ignore what I say.

Anyway when exactly will the day end?

[quote="Melissa"]All right, this game is over, since we are indeed in a 4-4 situation. Sorry that I did not end it earlier... I was away for a week, and was in quite a hurry when I started this day, so I did not notice that the guilties had won.

[hide="Congratulations to..."]Peter VDD, Ignite168, dts, pianoforte (the devil) and (although he was lynched) me@home[/hide][/quote]

Congratulations :)

[quote="Peter VDD"]The day is not deadlined yet, so stop trying to push us. We don't need to be quick and the more we talk before acting, the more sure we can be of what we do. And especially, the more we can link players together in packs, so that if we lynch correct once, the others will follow. It's worthless to take the sure picks now if it leaves us the same doubt tomorrow.

Also you don't know laughinghead is innocent, she's replacing an unconfirmed player for the moment.[/quote]

Haha funny now that we know that I was right.

[quote="Ignite168"]Since when does everybody agree on the fact that I am mafia?

For one, I was NOT on the Ars lynch yesterday, and if you noticed, Ars posts started saying I was town before he died.

weiquan, how are you innocent? You replaced someone who made little to no contributions all game, and only AYEd people, and you were also responsible for the death of Ars.

I am almost 100% sure all of our 3 mafia were on the lynch of Ars or else it would not have been so rushed. That is

Peter VDD
weiquan
BoesFX
pianoforte
dts

The mafia are probably weiquan, BoesFX, and pianoforte.

Lastly, I am the Magician, so I really can't be guilty now, can I?[/quote]

This is even more amusing, when did "Magician" sound like "Mafia"? :P
Anyway it was fun figuring out that dts and you were guilty, and I realised that BoesFX was innocent sometime after my suspicion so I wanted to see what would happen when I accused the two of you instead of him :lol:

[quote="Really early in the game, Ars"]And by a pattern at the time I needed sleep so I just noticed that 6. Ignite, 10. Peter, 14. Dts, 18. Me at Home it goes up in a even interval.[/quote] I still don't know how you [i]really[/i] figured it out, but congratulations to Ars for this remarkable guess.  :lol:

And btw I won the bet with dts on who was the devil. :P

And yes, I said I had a winning strategy for 'us' if some technical detail worked, right?  :D

Sigh... for now I retire from playing mafia. Next time when I got another free time, I will sign up again.
Anyway, why is miller looked as mafia when inspected by mathematician?

[quote="Melissa"]
--- 1 mathematician ---
This innocent role can, once during the game, during the day, reveal him/herself and pick a group of 2 or more people (not including him/herself) to "inspect". I will then tell everybody how many mafia there are in the group, but not which of them is mafia. It is illegal to claim mathematician if you are not really mathematician. Once the mathematician has used his/her power, (s)he becomes the same as a normal civilian.
[/quote]

Should it be that miller is confirmed as not mafia and devil also as not mafia :roll:

About killing claimed inspector it is an irony. If Ignite168 told us earlier about that not when in the mafia 23 :roll: (although in mafia 23 the inspector is still lynched :mad:) That is one of the reasons why I start to second PeterVDD as my first suspect. :wink:

Mathematician and suicide bomber use the power only to get rid the miller, lynching inspector(when he/she would be most effective when miller is gone), and no lynch (due to inactivity). Great!

As I said

pianoforte started screwing with everything when I was getting lynched making me believe he was mafia

Ignite I gave up on because I was sure Peter was mafia so needed him dead first

Dts in my review just revealed himself entirely

Peter you were scum for pushing the inspector lynch and mine if we more active you should have died

And I noticed the pattern after Peter started acting way to mafia as did Ignite

That left the third person on my lynch Dts

I had all 3 mafia 3 days before I died but was screwed because they were all trying to lynch me

Ignite again pointed himself out on the last day when he accused 5 people you can never do that

Also the pattern was noticed because the masons were oppisite each other and I already suspected you 3.

yay,  :D

Anyways, I, inspected
[hide]Koda(angel)
Ladyknight/weiquan(preist)
Ars
ragnarok/laughinghead

I remember I accused ladyknight/weiquan earlier for no apparent reason, yet everyone just ignored it...
[/hide]

I did in my analysis but did not suspect you as the mafia jumped on you then

Later I figured it out but it really did not matter

We didn't ignore it, pianoforte. We didn't know if you were the devil or just trigger happy, but it did make us not kill you (which makes you succeed for your number 1 task: subtly hint us not to kill you).

And by the way I still find it funny how we just sacrificed me@home on turn 1, as if we wanted to put ourselves safe from his possible stupid moves.

"O God that madest this beautiful earth , when will it be ready to receive Thy saints (eg me, instead of eg Ignite)? How long, O Lord, how long?"

(you can't believe that though i hadn't opened saint joan for ten months, it was by some inexplicable coincidence right within my reach)


Could anyone post reading lists for SAT prep.?
i need to keep reading during the summer, but don't know what books to read........
Thanks,

[quote="Ignite168"]Umm, do you mean specific study books? If you don't mean that, then there isn't really any book that will help you with the SAT. If you read a lot of challenging books normally and look up the words you don't know, it should expand your vocabulary gradually.[/quote]
sorry, my post wasn't clear...
i meant those challenging books


modular arithmetic
number theory unsolved
number theory

Prove that for any positive integer s, there is a positive integer n whose sum of digits is s and s|n (s divdes n).

Take $ n$ to be a sum of $ s$ distinct powers of ten.

That wasn't intended to be a full solution.  How would you go about shuffling powers of ten around to ensure divisibility by $ s$?

Note that the sequence $ 10^n\pmod s$ must be eventually periodic. Take $ s$ different values of $ n$ which have the same value of $ 10^n\pmod s$, and there is your number.


calculus
integration
derivative
logarithms
real analysis
real analysis unsolved

I am stuck with the following integral.

$ I (1/x)*2*exp((\minus{}3*y)/x) dx$

Have tried substitution but it does not seem to work.

Have also looked through several calculus books, but could not find anything for exp with 2 variables.

Has anyone come across a similar problem?

Many thanx.

This looks like a multivariable calculus problem $ \int\int \frac {2}{x}\exp( \minus{} 3y/x)\,dx\,dy$ that is best approached by [url=http://www.math.umn.edu/~nykamp/m2374/readings/doubleintchange/]changing the order of integration[/url] to $ dy\,dx$.

Thanx a lot, however, it's not a double integral. I have to integrate it w.r.t. x only and that's where I am stuck...

Thanx for your help and the link though!

Substitution $ u\equal{}\minus{}3y/x$ (treat $ y$ as a constant) reduces this to the [url=http://en.wikipedia.org/wiki/Exponential_integral]exponential integral[/url], which is a non-elementary function. Do you have a definite integral by any chance? Also, the presence of $ y$ still puzzles me.

Thanx for this.

This integral is part of the solution of the differential equation:

$ (diff(y(x),x)*x \equal{} y \plus{} 2*x*exp(\minus{}3*y/x)$

when $ x \equal{} 1, y \equal{} 0$

Having re-arranged this (including dividing the whole expression by x) I got:

$ (diff(y(x),x) \minus{} (y/x) \equal{} 2*exp(\minus{}3*y/x)$

$ P(x)\equal{} (\minus{}1/x)$

$ Q(x) \equal{} 2*exp(\minus{}3*y/x)$

Don't know the symbol for mju(x) = E(x)

$ mju(x) \equal{} exp(I (P(x) dx) \equal{} exp (I (\minus{}1/x) dx) \equal{} exp(\minus{}ln(x)) \equal{} 1/x$

So, $ y(x) \equal{} 1/(1/x)* (I (1/x) * 2*exp(\minus{}3*y/x)$; $ y(x)\equal{} x*(I (1/x) *2*exp(\minus{}3*y/x)$

So, that's where the integral $ I (2/x)*exp(\minus{}3*y/x)$ is derived from...

Your method is nonsense- this integral is not part of what you need to do to solve the differential equation. You cannot integrate an expression with both $ x$ and $ y$ in it- you'll have to make a substitution, and this is not a linear differential equation.

hello, write your equation in the form
$ y^'=\frac{y}{x}-2\exp\left(-3\frac{y}{x}}\right)$ and let $ u=\frac{y}{x}$.
Sonnhard.

Hello, thanx for your help!

I have tried substitution $ u \equal{} y/x$

and got the following:

$ dy/dx \minus{} y/x \equal{} 2*exp( \minus{} 3*y/x)$

$ u \equal{} y/x$

$ E(u) \equal{} exp(Int ( \minus{} u)) du \equal{} exp( \minus{} u^2/2)$

$ Q(u) \equal{} 2*exp( \minus{} 3*u)$

$ y(u) \equal{} 1/E(u) * Int ( E(u) * Q(u)) du$

$ y(u) \equal{} (1/exp( \minus{} u^2/2) * I exp( \minus{} u^2/2) * 2 * exp( \minus{} 3*u) du$

$ y(u) \equal{} 2*exp(u^2/2) * I exp(( \minus{} u^2/2) \minus{} 3*u) du$

Not quite sure how to integrate this either. And the only method of solving this type of equations at the class was through integration...

The solution in Maple is:

$ ode : \equal{} (diff(y(x), x)) \minus{} y(x)/x \equal{} 2*exp( \minus{} 3*y(x)/x)$

$ dsolve(ode)$: $ y(x) \equal{} (1/3)* ln (6 ln(x) \plus{} 6*C(1)) x$

Is there any other method to solve this?

Thanx a lot. Antonio.

If $ u\equal{}\frac{y}{x}$, then $ y^\prime$ is not equal to $ u^\prime$ (where the derivatives are with respect to $ x$); instead, we have
\[ y\equal{}ux\Rightarrow y^\prime \equal{}u^\prime x\plus{}u\]
Now plug into the equation to get
\[ (u^\prime x\plus{}u)\minus{}u\equal{}2e^{\minus{}3u}\Rightarrow u^\prime x\equal{}2e^{\minus{}3u}\]
This is separable, so it can be integrated:
\[ e^{3u}\ du\equal{}\frac{2}{x}\ dx\Rightarrow \frac{e^{3u}}{3}\equal{}C\plus{}2\ln x\]
where $ C$ is some constant.

Now when $ x\equal{}1$, $ y\equal{}0$ so $ u\equal{}0$. This gives $ C\equal{}\frac{1}{3}$. So $ e^{3u}\equal{}1\plus{}6\ln x$, and isolating $ u$ gives
\[ u\equal{}\frac{\ln (1\plus{}6\ln x)}{3}\]
\[ y\equal{}\boxed{\frac{x\ln (1\plus{}6\ln x)}{3}}\]
(Using the substitution $ u\equal{}y/x$, any differential equation of the form $ y^\prime \equal{}f(y/x)$ can be reduced to a separable equation.)


trigonometry
logarithms
limit
calculus
calculus computations

investigate the convergence or divergence of the following series: :D

$ \textbf{(i)}\;\sum_{k=1}^\infty\;\sin k$

$ \textbf{(ii)}\;\sum_{k=1}^\infty\;\sin \frac{k\pi}{2}$

$ \textbf{(iii)}\;\sum_{k=1}^\infty\;\frac{\sin \frac{k\pi}{2}}{k}$

$ \textbf{(iv)}\;\sum_{k=3}^\infty\;\frac{\sin \frac{k\pi}{2}}{\ln k}$

$ \textbf{(v)}\;\sum_{k=1}^\infty\;\sin \frac{1}{k}$

$ \textbf{(vi)}\;\sum_{k=1}^\infty\;\sin \frac{\pi}{2k}$

$ \textbf{(vii)}\;\sum_{k=1}^\infty\;\frac{\sin \frac{\pi}{2k}}{k}$

$ \textbf{(viii)}\;\sum_{k=2}^\infty\;\frac{\sin \frac{\pi}{2k}}{\ln k}$

hello, $ \sum_{k=1}^\infty\sin(k)$is divergent, because $ \lim_{k\to \infty}\sin(k)$ ist not zero.

(i) Diverges, terms don't tend to zero.
(ii) Diverges, terms don't tend to zero.
Both (iii) and (iv) might as well be restricted to the terms with $ k$ odd. Then they both satisfy the conditions of the alternating series test, and both converge, but merely conditionally.
(v) Diverges by limit comparison to the harmonic series.
(vi) Diverges by limit comparison to the harmonic series.
(vii) Converges by limit comparison to $ \sum\frac1{k^{2}}.$
(viii) Diverges by limit comparison to $ \sum\frac1{k\ln k}.$

With experience, questions like these may be answered very quickly.


function
real analysis
real analysis solved

Let f be a real continuous function such that {f(x)}=f({x}) , where {} is the fractional part. Then f or g(x)=f(x)-x is periodical.
  Bucharest Olympiad, 1992, Mihai Baluna

For x in [0,1) we have {x}=x, so f(x)={f(x)} for all x in [0,1), so f(x) is in [0,1) for all x in the same interval (*). From the fact that {f(0)}=f(0)={f(1)} we find that f(1)-f(0) is in Z, and by combining this with (*) we get f(1)-f(0) = 0 (a) or 1 (b).

I don't think it's that hard now. Just try to prove that for the case (a) f(x+1)=f(x) and for (b) f(x+1)-(x+1)=f(x)-x.

Thank you, Grobber, now I realize how stupid was. I had already found that f(x+1)-f(x) is constant (since it is a continuous function that takes only integer values), but I couldn't believe that the argument with f(0) and f(1) could work. I was wrong. Thank you again.

What round is this problem from? And which grade was it proposed to? Do you, by any chance know this?

Mihai Baluna proposes e\really nice problems. This wasn't hard, but I think it was kind of nice  :)

It was " Etapa Judeteana" (I hope I've written correctly) and it was proposed for the 11 form. You are right, it was easy and I was stupid. And indeed, all the problems proposed by Mihai Baluna are nice.

Oh come on! What's up with this "I'm stupid" attitude? :D That's NOT what I meant by saying it was easy... :D

Btw, "Etapa judeteana" is the second round of the Romanian NMO. Judet is something like county.


[quote="Heather-M"]Solve this equation: 2s + s + 12 =132.

A.  s = 120

B.  s = 40

C.  s = -30

D.  s = 9[/quote]

[hide]add the like terms together.

3s=120
s=40[/hide]

You posted a problem that is very similar to this one, and you can use the exact same idea for both problems.


calculus
integration
conics
parabola
hyperbola
geometry
calculus computations

When the parabola which has the axis parallel to $y$ -axis and passes through the origin touch to the rectangular hyperbola $xy=1$ in the first quadrant moves,
prove that  the area of the figure sorrounded by the parabola and the $x$-axis is constant.

Just a note $xy=1$ is a rectangular hyperbola, not parabola Kunny.

Yes the area is a constant and comes out to be [b]4/3
[/b]. Sq. Units


[b]Using any digit in the set {1,2,3,4,5,6,7,8,9} only once, form two fractions with the closest sum to 1 without exceeding it. Express your answer as a common fraction.[/b]

[hide]
You can have $ \dfrac{1}{3}$ and $ \dfrac{6}{9}$. So the sum is $ 1$.
[/hide]
Or do you want two [b]common[/b] fractions?

yea, im sure it wasnt 1 b/c the numerator/denominator coudnt share a factor
SOO the maximum denominator possible when finding a least common denominator is 72 (8&9) so u test if u can get close to 1 by adding fractions w/ denominators 8 and 9
you might notice that $ \frac{1}{9}\equal{}\frac{1}{8}\minus{}\frac{1}{72}$
so if you have $ \frac{8}{8}$ and you take $ \frac{1}{8}$ away while adding a $ \frac{1}{9}$
you get $ \frac{7}{8}\plus{}\frac{1}{9}\equal{}\frac{71}{72}$

[hide]Since 1/9<1/x where x<9, we should have that as one, and then have the biggest other one possible, which can be found to be 7/8. The sum is 71/72.[/hide]

[quote="stevenmeow"]yea, im sure it wasnt 1 b/c the numerator/denominator coudnt share a factor
[/quote]
That's true, but the statement of the problem is a bit weird.
[quote="mihail911"]Using any digit in the set {1,2,3,4,5,6,7,8,9} only once, form two fractions with the closest sum to 1 without exceeding it. Express your answer as a common fraction.[/quote]
We never really know what exactly our answer is. We could either assume:

The sum is the answer and must be a common fraction, so my first answer is technically correct, or

The two fractions are the answer, but it says [b]...[i]a[/i] common fraction[/b]?!


If P is a point inside a triangle ABC and D,E,F are the feet of perpendiculars to BC,CA,AB from P . For which point P is the expression,
$ \frac{BC}{PD} \plus{} \frac{CA}{PE} \plus{}\frac{AB}{PF}$  minimal

[hide="Solution(?)"]
I'm not certain as to whether the solution depends on the type of triangle, but I believe that P would be the centroid, because we are looking to maximize each of PD, PE, and PF in order to minimize the final answer. This means that the point needs to be the furthest distance possible from each side without being too close to another side, upon which the term concerning the "other side" becomes larger. The centroid is equidistant from the sides, so it keeps (to my view) the expression given to a minimum. Then again, I haven't looked at the circumcenter or incenter yet...
[/hide]

[hide="Solution"]Note that
\[ 2[ABC]\left(\frac{BC}{PD}\plus{}\frac{CA}{PE}\plus{}\frac{AB}{PF}\right)\\
\equal{}(BC\cdot PD\plus{}CA\cdot PE\plus{}AB\cdot PF)\left(\frac{BC}{PD}\plus{}\frac{CA}{PE}\plus{}\frac{AB}{PF}\right)\\
\ge(BC\plus{}CA\plus{}AB)^2,\]with equality when $ PD^2\equal{}PE^2\equal{}PF^2$, or equivalently, when $ P$ is the incenter.[/hide]


projective geometry
geometry proposed
geometry

Dear Mathlinkers,
let ABC be a triangle,
P, Q two points not on the sides of ABC,
and Ab, Ba, Ac, Ca, Bc, Cb the points of intersection of PA and QB, of PB and QA, of PA and QC, of PC and QA, of PB and QC, of PC and QB.
Prove that AbBa, BcCb and CaAc are concurrent.
Sincerely
Jean-Louis

$ Let \ \{R\} \equal{} A_cC_a \cap A_bB_a.$
$ The \ points \ A_c, P, A_b \ colinear (\in AP) \ and \ the \ points \ B_a, Q, C_a \ colinear ( \in AQ) \\
and \ apply \ Pappus \ Theorem \Rightarrow \ the \ points \ B_c, R, C_b \ colinear \Rightarrow \\
\Rightarrow A_bB_a \cap A_cC_a \cap B_cC_b \equal{} \{ R \}.$

Dear Mihai miculita and Mathlinkers,
thank you for your nice contribution.
Here is my proof:
AbBcCa and AcBaCb are perspective with center P
AbBcCa and CbAcBa are perspective with center Q.
Accordling to "The double perspective theorem", AbBcCa and BaCbAc are perspective…
Sincerely
Jean-Louis


Prove that in an acute angled triangle $ \Delta ABC$,
\[ (a \plus{} b \plus{} c) \leqslant \frac {{4\Delta }} {R}\]

$ \Delta$= Area of the triangle.

$ a\plus{}b\plus{}c\le \frac{4rs}{R}$

$ 1\le \frac{2r}{R}$, which is not true (Euler formula)

True would be: $ 1\ge \frac{2r}{R}$

$ a\plus{}b\plus{}c \ge \frac{4rs}{R}$

$ a\plus{}b\plus{}c\ge \frac{4\Delta}{R}$


limit
function
Functional Analysis
real analysis
real analysis unsolved

Here's something I wondered:

If $ x_n$ is a sequence in a Banach space $ X$ such that for every $ \ell$ in $ X^*$, $ \ell(x_n)$ converges, does $ x_n$ converge weakly?

If $ x_n$ is seen as a sequence in $ X^{**}$ then this sequence converges weak* to some element in $ X^{**}$ (the linear functional on $ X^*$ defined by $ u(\ell) \equal{} \lim \ell(x_n)$ is bounded because it follows by the uniform boundedness principle that $ x_n$ is a bounded sequence). So if $ X$ is reflexive the result follows.

Is it true iff $ X$ is reflexive? Would it be correct to say that I'm asking what the sequential weak* closure of $ X$ in $ X^{**}$ is?

The image of $ X$ under the canonical embedding $ X\to X^{**}$ is weak*-dense in $ X^{**}$. I doubt that it is always sequentially weak* dense (consider something like $ c_0$ but with uncountable set of indices). However, it should be true that for any nonreflexive $ X$ the sequential weak* closure of $ X$ in $ X^{**}$ is a proper superset of $ X$. Possibly relevant fact: reflexivity is determined by separable subspaces [i.e. $ X$ is reflexive iff every separable subspace is reflexive].

Here is an obvious counterexample: consider the space of continuous functions on $ [0,1]$ that vanish at the endpoints with the usual uniform norm. Now, the dual space is the space of all finite complex valued Borel measures on $ (0,1)$. It remains to take any sequence $ f_n$ that is uniformly bounded and converges to $ 1$ pointwise on $ (0,1)$. The only weak limit can be $ 1$ but it is not in the space.


I would like to write my CV like in http://kompassi.lasipalatsi.fi/temporary/image.asp/data_id=1127/image001.gif . Is there any way to take the indentation away as default? My code looks now as

[code]
$\,$

\begin{flushleft}
\textbf{Name} Meknowsnothing
\end{flushleft}

$\,$

\begin{flushleft}
\textbf{Address} Street
\end{flushleft}
[/code]

And how can I write text in two columns?

If you use a table then it will solve both your problems. See [[LaTeX:Layout#Tables]]


linear algebra
matrix
search
graph theory

Bob really LOVES chocolate. He thinks he never gets enough. Imagine his joy when his parents told him that they would buy him many rectangular chocolate pieces for his birthday. A piece of chocolate is a 2x1 or 1x2 rectangle. Bob's parents also bought him a nice birthday cake, which can be imagined as a matrix having M(1<=M<=70) rows and N(1<=N<=7) columns. Some positions on the cake are occupied by candles, the others are empty. Bob's parents asked their son to place as many chocolate pieces as he can on the empty squares on the cake, in such a manner that no two chocolate pieces overlap. However, he would like to keep the chocolate pieces to himself. That's why, he wants to place only a minimal amount of them on the cake and keep the rest. In order not to make Mon and Dad suspicious, Bob wants to place the chocolate pieces in such a way, that no other piece may be placed on the cake (that is, there won't exist any two adjacent empty squares). Find the minimal number of pieces which need to be placed on the cake, so that they do not overlap and no extra piece may be added.

Isn't this a simple breadth-first search?

Observation: At each empty square, a piece of chocolate can be placed either:
a) on the square and the square above (if both are empty)
b) on the square and the square below (if both are empty)
c) on the square and the square to the right (if both are empty)
d) on the square and the square to the left (if both are empty)

Since the bounds are so low (7 x 70), you can breadth-first search it. This is a graph theory problem. The vertices are the possible states of the cake with different chocolate arrangements, the edges are the actions of placing the next piece of chocolate in one of four positions.

The end point in this graph is the state in which no square has an empty square above, below, left or right of it. As soon as we hit this state, exit and output the number of chocolate pieces we placed.

This may not be a 100% perfect solution, but by my calculations, it should come under reasonable time & memory limits because the problem has such low bounds.

I don't quite see why you are saying the bounds are low.. in the worst case (0 candles, 70x7), if you consider each square like you have; theres 5 ways of choosing a domino for that square (4 directions, or none there at all), and you are approximately cutting down the number of squares by half each time, leaving the BFS with a size of about 5^245.. thats quite large, to me :)

You can use a combination of dynamic programming and backtracking (see IOI 2001 booklet).
There are 3^7 states for each line and you do a DP on then.

domino plz tell us how it is done... actually i am not understanding that soln! so if u describe it......... ;)

You do a dynamic programming algorithm... all you need to know is the state of the last line.. each cell in the last line cam be of 3 types:
0 - empty
1 - empty and must be filled on the next line , otherwise you will be able to put a chocolate if you leave it empty
2 - filled

So.. 3^7 states.. you can do a backtracking and compute for each such state S the states S' which you can obtain by putting chocolates. Then , the DP is clear I believe. The IOI 2001 booklet explains it much better than me :)

thanks. well... is the booklet available over internet? if so then where?

http://olympiads.win.tue.nl/ioi/ioi2001/contest/A-2001-7.pdf

The particular problem using this style of solution is 'Pavement'.

Is this solution correct?
Let squares in the matrix be vertex of the graph. If you can put chocolate on the two adjanced sqares then the weight of that edge is 1 else it is 0. After this we aply minimum weighted matching.

it seems to me correct. but it is reminding me of: Vertex cover problem. can this problem be converted to vertex cover problem?


[quote="boxedexe"]Hello,  I do not know what to do.  I am going to buy a book, but I cannot make a choice due to the fact that I do not know much about the affectiveness of the books.  Would you suggest I buy the book IMO Compendium or a book on Algebraic Number Theory/Modern Graph Theory by Springer?[/quote]
[u]
If you want to buy a book: [/u]

A book on Algebraic Number Theory/Modern Graph Theory  from Springer Verlag ? My [b]answer[/b], taking into account your age,  is [b]negative ! [/b]

I suggest you to buy following [b] good book (indeed)[/b] :
G.Polya and Gabor Szego "[i] Aufgaben und Lehrsaetze aus der Analysis [/i]", vol.I-II, Vierte Auflage(=fourth edition) Heidelberger  Taschenbuecher,Springer-Verlag,Band 73-74 (1971), ...but [b] there is a most recently English version in one volume ,
 also in Springer Verlag. [/b] Do you have seen this book ?

Well, I do not consider learning mathematics relavent to age.  I've been studying Algebraic Number Theory and Graph Theory on my own, so I was willing to buy an introductory book on the subjects.  I've encountered a book with full of problems, i.e., Problems in Algebraic Number Theory.  The book called Modern Graph Theory published by Springer is a book I find quite interesting and highly understandable.  On the other hand, IMO Compendium seems to be filled with quite a lot of problems that would both increase my problem-solving abilities and give me a nice read throughout the year by solving several problems a day.  Your book suggections seem tempting to seek; I will take your suggestions into consideration.  Thank you.

Masoud Zargar

EDIT: No I have not seen the book. :( I'll be thankful if you translate the title of that book for me. :)

[quote="boxedexe"]......EDIT: No I have not seen the book.
( I'll be thankful if you translate the title of that book for me. :)[/quote]

[1] G. Polya and G. Szego, Problems and Theorems in Analysis, volume I, Springer Verlag (1972).
[2] G. Polya and G. Szego, Problems and Theorems in Analysis, Volume II, Springer-Verlag, New York (1976).

This book this very good one I recommend to you is called: [url=http://www.ofertondelibros.com/default.asp?R=0387946063B&pi=newyork&cam=books]"Basic Calculus"[/url]author is. Hahn, Alexander J.
Publisher: Springer-Verlag Telos
see you soon!!!!! :)


[quote="jli"]visit this website:

[url]http://www.thestatenislandboys.com/All_da_Crap_is_here/Sony%20Pic%20Taking.swf[/url][/quote]

I've seen this before in so many websites. Its still pretty funny though

[quote="aznphatso"]the original ones from like 2000 are better.

They're like

"look at the car advertisement, something weird happens after the car goes under the cloud"

and

"Is there something wrong with this picture?"[/quote]

what website is that on?

you almost got me... :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  however i knew it when it said view pic in a really really big icon.

[quote="stupidkid"]didn't know that I looked that nice...[/quote]

i know, i looked so hotandsexy


function
inequalities
inequalities theorems

what is a continous function??

im new at inequalities how much time do i need to get the feel to it?

where do i study? how?

find any analysis book and you should see clearly the definition of a continuous function...
Anyway, it's like, f(x) is continuous at a if
(1) lim x->a f(x) exists
(2) f(x) is well defined at a
(3) lim x->a f(x) = f(a)


Given a cube of cheese, what is the maximum number of pieces you can cut from the cube if you are only allowed 4 straight cuts?  (The pieces of cheese have flat faces.)

I don't know how hard this problem is, but I guess it belongs here. Does anyone have a solution?

Thanks.

[quote="Fierytycoon"]Actually, I think I just solved it...

[hide]Consider the cube with vertices (x,y,z) where x,y,z are either 0 or 10. Then consider the cuts alone the planes x=5, y=5, z=5, and (10/3)(x+y) - z = 30.[/hide][/quote]

Can you prove that this is the maximum number of pieces?

[hide]Hmm, yeah, that's a problem. I can show that 15 pieces can be made (with the example in my last post), but I can't prove that making 16 or more pieces is impossible.[/hide]

Any ideas?

[quote="Fierytycoon"][hide]Hmm, yeah, that's a problem. I can show that 15 pieces can be made, but I can't prove that making 16 pieces is impossible.[/hide]

Any ideas?[/quote]

I would say that 16 pieces are impossible. I have no idea though.

obviously, the cube doesnt matter.  What we want to know is this;  given n planes, what is the maximum number of parts it can divide the space into? (n=4)?

http://www.cs.colostate.edu/~rmm/mathChallenge/cairoliBinom.pdf

I give a source, where they have proved the result for general n.

[quote="Fierytycoon"][quote="lightrhee"]No one has comment on my proof?
Come on. This proof is beautiful. Admit it :D[/quote]

I would, but I don't quite understand it, heh.

So for the 2D case, what are the "pieces" that we are putting the beads on? The lines? The intersections? Also, how exactly are some of the beads dropping and some moving to the intersections? I am a bit confused...[/quote]

Pieces meant the each spaces that are divided. I'll try to explain this in more detail tomorrow(I need to sleep).

How do they "fall down because of gravity"?  That part doesn't appear to make sense...not to me at least...
In any case, the way i would see it for two dimensions is
You begin with 1 piece
The first line cuts it into 2 pieces
The second line can travel through 2 regions maximum by intersecting the first line, so then we have 4
The third line can go through both of the other two, so then we get 4+3=7
The fourth goes through all three lines and adds 4 more.  therefore, we have 11.

Apply it to 3-d and...
1st plane: 2 sections
2nd plane: 2+2=4 sections
3rd plane: cuts through 4 sections, so we have 8
4th: It obviously can't leave you with more than 16 pieces: in order for that, at least one of the pieces would have to be cut twice.  Therefore, we know that the maximum cannot exceed 16.  Then, we know that it can't be 16, because in order for that, it would have to cut every one of the 8 pieces.  In order to have this, it would have to contain two points exactly on opposite sides of the point of intersection of the three other planes while not going through the point of intersection.  That's not possible.  Therefore, the maximum is 15.

But, if you were allowed to pick up the pieces of the cheese between cuts and stack them on top of each other, you would easily be able to get 16 pieces.

Well...depends on what the word "only" modifies.  If it's that "you are allowed only to make four straight cuts" then you aren't allowed to stack.  If it's as Fierytycoon said "you are only allowed four straight cuts" then you aren't forced to make four straight cuts...you're only allowed to, not forced to.  Then you can do whatever you want to.  But we're assuming that the meaning meant was that of "you are allowed ONLY to make four straight cuts".  You aren't allowed to make five cuts, three cuts, stack the pieces, or eat them.

Well, technically, yeah, ok, but I'm sure everyone understands that the pieces of cheese cannot be moved. Otherwise the problem would be trivial.

[quote="lightrhee"]This is a proof that I came up with right after I turned off my computer yesterday:D . I have seen a proof for two dimemsion and I thought this might work for three dimemsion too. This proof is pretty.

First, we need to solve this in two-dimemsion. Let's say that there are four lines that cuts a plane. We can make sure that no line is horizental by slightly tilting it(this is possible since there are only finite lines). Now place a bead (yes a bead!) on each pieces. They fall down because of gravity. 4+1 ofthem will drop eternally, and each intersections of pair of lines will have a bead. Therefore, the maximum number of pieces in two dimemsion is $4+1+_4C_2=11$.

Now, we can solve for three dimemsion. Let's say that there are four planes that cuts a cube. We can make sure that no plane is horizental by slightly tilting it(this is possible since there are only finite planes). Now place a bead on each pieces. They fall down because of gravity. If we look at this four planes from two dimemsion that is horizontal, we would see four lines. So $4+1+_4C_2=11$ of them will drop eternally, and each intersections of triple of planes will have a bead. Therefore, the maximum number of pieces in three dimemsion is $11+_4C_3=15$.[/quote]

That is in deed a very nice solution lightrhee!

My only comment is that while some of your assertions (like the one below) are intuitively clear, you should probably prove them in order to shore up your argument.

[quote="lightrhee"]"4+1 of them will drop eternally"[/quote]

Sorry. I was going to post a detailed solution yesterday, but I had no time to do that.
I have a bad English(I lived in Canada for only 3 years, I am a Korean). I'm not good at explaining things that are complicated.
Can you help me to explain this proof, gauss202?


function
number theory unsolved
number theory

For [tex]f(n)=n+[\sqrt{n}][/tex] with [tex]n\in Z+[/tex].Prove that at leat there exist [tex]m[/tex] so that:[tex]f_m(n)[/tex] is perfect square

It'm sure it's supposed to be $f$ composed with itself $m$ times.

Suppose $n-k^2=t\in(0,2k+1)$. We have $f(n)=n+k=k^2+k+t$. If $t>k+1$, then $f(n)=(k+1)^2+t-(k+1),\ t-(k+1)>0$, if $t=k+1$ then $f(n)=(k+1)^2$, and if $t<k+1$ then $f(f(n))=k^2+2k+t=(k+1)^2+t-1$ with $t-1\in(0,2(k+1)+1)$.

In all the cases treated above, given a distance $t$ between $n$ and the largest perfect square not exceeding $n$, we can always reach a smaller such distance by iterating $f$. We repeat this procedure until the distance between $f^{(m)}(n)$ and the largest perfect square not exceeding $f^{(m)}(n)$ becomes $0$.

Ah, I see. Now the question makes sense.

Nice!


$F_{n}=2^{2^n}+1$,
where n is a nonnegative integer.
Prove that $641|F_5$

[quote="http://en.wikipedia.org/wiki/Fermat_prime"]... Leonhard Euler in 1732 [...] showed that

   \[ F_5 = 2^{2^5} + 1 = 2^{32} + 1 = 4294967297 = 641 \cdot 6700417 \];

It is interesting to note how Euler found this factorization. Euler had proved that every factor of $F_n$ must have the form $k2^{n+1} + 1$. For $n = 5$, this means that the only possible factors are of the form $64k + 1$. It did not take Euler very long to find the factor $641 = 10 \times 64 + 1$.[/quote]

Legend has it that Euler noticed:

$641 = 64\cdot10+1 = 2^7\cdot 5 + 1$ (which means $2^7 \cdot 5 \equiv -1 \bmod 641)$
and also $641 = 625+16 = 5^4+2^4$   (which means $5^4 \equiv -2^4 \bmod 641$)
So if we take all mod 641 we see

$2^7\cdot 5 \equiv -1$
so $(2^7 \cdot 5)^4 \equiv 1 \equiv 2^{28} \cdot 5^4 \equiv 2^{28} \cdot (-2^4) \equiv -2^{32}$
so $2^{32}+1 \equiv 0$
[size=150]QED.[/size]

Also, mathmanman:
 one can further improve:  that  any prime factor of $2^{2^n} +1$ has to be of the form $k\cdot 2^{n+2}$   so Euler need to test  (had he known this) only 5 values before he hit the jack_pot (not to mention,  he need not try for k=1, or 3, or 4 ..:)   (eg k=1 ===>  129 is divisible by 3 so could be ignored)   What a luck !!! Next value after 2 he really needed to check was 5 .... and  he hits the jack_pot!!!!


inequalities
calculus
Euler
inequalities unsolved

n is a positive integer. Prove that
$0<\frac{4n^3+3n}{2}-3\sum_{k=1}^{n^2}\sqrt{k}<1$

Is it so hard?
It's something from Moldova TST...
I think a Moldova friend may help me...  :P

This is a very nice inequality, I had no idea you could make such a sharp estimate of this sum.  I've re-phrased it as follows:  Prove that
\[\frac{4n^3 + 3n - 2}{6} < \sum_{k = 1}^{n^2} \sqrt{k} < \frac{4n^3 + 3n}{6}.\]

The upper bound is not too bad, but I found the lower bound to be tricky.  Do you have a nice solution?

I get it from Moldova TST 2005, See there:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=190516#190516]www.mathlinks.ro/Forum/viewtopic.php?p=190516#190516[/url]
Someone gave a solution with calculus, while I'm wondering a elementary way.

I think it can be proven using the Euler-Maclaurin summation formula (but that's not elementary nor nice). Anyway, I'm wondering if the identity:
<center>[tex] \sum^{n^2}_{j = 1} \sqrt{j} = n^3 - \sum_{j = 1}^{n^2 - 1} j ( \sqrt{j + 1}
   - \sqrt{j} )[/tex]</center>
is of any use... (I strugled with this prob for quite a long time... :( )


There exists $ m$ different color sets of beads. Set i $ (i \equal{} 1,2,...,m)$ consists of $ a_i$ beads. In how many ways can one make a necklace using all of the beads?
thank for your nice solution  :)

There is an elementary solution, but you can readily solve this using the [url=http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem]PET[/url].


inequalities
vector
inequalities proposed

Let $x_1, x_2, \cdots, x_n >0$ be real numbers.

Prove that:

$\sqrt{x_1 + x_2 + \cdots + x_n} + \sqrt{x_2 + x_3 + \cdots + x_n} + \cdots + \sqrt{ x_n} \geq \sqrt{x_1 + 4x_2 +\cdots+ n^2x_n}$

Fabiola Bravo      :pilot:
Lima -PERU

A classic

[hide]Let $v_1 = (\sqrt{x_1}, \sqrt{x_2}, \dots, \sqrt{x_n})$, $v_2 = (0, \sqrt{x_2}, \dots, \sqrt{x_n})$, ..., $v_n = (0, \dots, 0, \sqrt{x_n})$.  Then we apply a certain inequality to the vectors $v_i$.[/hide]


I registered for the AMC 10 Problem Series late- and can't find the message board for homework for this class. How do I access it?


quadratics trigonometry geometry probability function perimeter calculus

Hi Guys this is my first post here and i wanted to make it a long one. I was wondering whether any of you guys could help me out with these qquestions because i am completely stuck. If it helps I have been studying Quadratic Graphs, Trigonometry, Geometry, Measurement, and Geometry (with triangles)

X = multiply		x=x

Boxes
A Sheet of cardboard measures 20cm X 30cm. A square measuring xcm X xcm is cut from each corner so that an open box can be formed.
a)	Express the length and width of the base of the box in terms of x
b)	If the area of the base of the box is 264 cm form an equation (Factorised) involving the area and x
c)	Solve the equation to find x
d)	What are the dimensions of the box when the area of the base is 264cm²

The Captain and the Lighthouse
A ship is sailing on a bearing of 200º T. At 1200 hours the captain of the ship observes a lighthouse on a bearing of 225º T, and after travelling 30km the captain observes the same lighthouse on a bearing of 250º T.
a)	Draw a diagram using A for the position o the ship at 1200 hours, L for the position of the lighthouse and B for the final position of the ship. Clearly label all bearings and true north directions
b)	At the time of the second sighting find the distance that the ship is from the lighthouse
c)	If the ship travels at 20km an hour how long does the journey take

Red and Black Marbles
a)	A bag contains 4 red marbles and 6 black marbles. 3 marbles are drawn from the bag each marble is replaced after it is drawn and the bag is shaken. Find the probability of selecting
i)	3 Black Marbles
ii)	Exactly 2 Red Marbles
iii)	At Least 1 Red Marble
b)	From the same bag of marbles as in part A 2 marbles are selected without replacing the first marble. Find the probability that the selection contains
i)	2 Red Marbles
ii)	A Black and a Red Marble

The Fence and the Farmer
A farmer wishes to build a rectangular paddock using 200 metres of fencing, he wishes to make the paddock as large as possible (That is have the largest possible are)
a)	Let the width of his paddock be x metres. What is the length of the paddock in terms of x?
b)	Show that the area of the paddock A m² is given by equals 100x-x²
c)	Plot the graph of A as a function of x for {x:0 <_ x <_ 100}
d)	Hence find the maximum area of the paddock and its dimensions
e)	If 20<_x<_70 use the graph from part C to find the smallest possible area for the paddock and its dimensions

[hide="marbles"]a) i) Since we are replacing marbles, the probability for drawing a black marbles is the same each time, $\frac{3}{5}$. Three consecutive draws calls for $(\frac{3}{5})^3=\frac{27}{125}$.

ii) We want two red ones and a black one, so the probability is $\frac{2}{5}\cdot\frac{2}{5}\cdot\frac{3}{5}=\frac{12}{125}$. However, we could also draw a red one, a black one, and then a red one, and still end up with the same result. Also, drawing a black first and then two reds can also work. There are three ways to draw exactly two marbles, so the actual probability is $\frac{12}{125}\cdot3=\frac{36}{125}$.

iii) We basically want to find the probability of getting none as red and then subtract from 1. Getting no reds is the same as getting three blacks, which we found above to be $\frac{27}{125}$. This from 1 gives us $\frac{98}{125}$.

b) i) after taking one red away, which is a proability of $\frac{2}{5}$, we are left with three red marbles and 9 total marbles. Selecting another one is a $\frac{1}{3}$ chance, and the probability of choosing two marbles that are red without replacement is $\frac{2}{5}\cdot\frac{1}{3}=\frac{2}{15}$.

ii) The probability of choosing a red after a black is chosen is $\frac{4}{9}$. The probability is then $\frac{3}{5}\cdot\frac{4}{9}=\frac{4}{15}$. THis can also be reversed, so the actual probabiltiy is $\frac{4}{15}\cdot2=\frac{8}{15}$. [/hide]

your "boxes question has been answered:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57791

[hide="Farmer"]
a)Total Perimeter of the field:
x+x+nx+nx=200
2x+2nx=200
x+nx=100
x(1+n)=100
1+n=100/x
n=(100/x)-1

Therefore, length of the field:
[(100/x)-1)]x
100-x
-----------------------
b)Area of the paddock:
Since:
Width=x
Length=100-x
Area=Length x Width

Area=x(100-x)
Area=100x^2
-----------------------
c)
[img]http://i34.photobucket.com/albums/d104/aommaster/Graph.jpg[/img]
-----------------------
d)The maximum area of the graph can be found by differentiation:
Differentiate the area:
100x-x^2 will become ---> 100-2x
Since the maximum is a stationary point:
100-2x=0
2x=100
x=50 --->You now know that the x-co-ordinates=50
y=100(50)-(50)^2
y=5000-2500
y=2500 --->that is the maximum area of the paddock

The dimensions of the paddock,therefore, is:
Width:x = 50
Length:100-x = 100-50 = 50
--------------------------------------------------------
e)When the graph is sketched, the smallest possible area is when x=20
Therefore, the area of the paddock is 100(20)-(20)^2 = 1600
The dimensions:
Width: x = 20
Length:100-x = 80

[/hide]

Hope that helps!


modular arithmetic
geometry
3D geometry
Euler
number theory

Should that be $2002^{2003}$?

This looks a little difficult for HSB.

[hide]since $n^{5}$ is congruent to n mod 10 the expression is equivalent to $1^{3}+2^{3}+....+2002^{3}+2003^{3}$ which is equivalent to $(1^{2}+2^{2}+....2002^{2}+2003^{2})^{2}$
Using the formula for sum of square we get $\frac{2003*2004*4007}{6}$ which ends in a units digit of $4$. So $4^{2}$ ends in a units digit of [b]6[/b][/hide]

Would this be a good place to use synthetic division?

On what?

[hide="Heres an idea"] $1^{2003}+1999^{2003}= (1+1999)( \text{Stuff})$

And so forth.  So we only need to calculate

$2000^{2003}+...+2003^{2003}\bmod 10$

The first term is obviously $0$, so simplifying with some modular properties and Totient Theorem, we just want

$1^{3}+2^{3}+3^{3}\equiv \boxed{6}\bmod 10$. [/hide]

[quote="t0rajir0u"]This looks a little difficult for HSB.[/quote]

Not too difficult...

[hide]$n^{5}\equiv n\pmod{10}\Rightarrow1^{2003}+2^{2003}+\cdots\equiv1^{3}+2^{3}+\cdots\pmod{10}$.

Using the sum of consecutive cubes formula, this becomes

$(1^{2}+2^{2}+3^{2}+\cdots+2003^{2})^{2}\pmod{10}$

We know that the sum of the first $n$ squares is $\frac{n(n+1)(2n+1)}6$, so

$\left(\frac{2003(2004)(4007)}6\right)^{2}\pmod{10}$

$\left[3(4)(7)\right]^{2}\pmod{10}$

Thus, $1^{2003}+2^{2003}+\cdots+2003^{2003}\equiv\boxed{6}$.[/hide]

[quote="13375P34K43V312"]Yeah, Farenhajt is right about the formula. I like t0rajir0u's solution, it's cool.[/quote]
That's because he is a beast. I like number theory, it is the most interesting branch of math...

This might be considered similar to t0rajir0u's solution. Here it is:
[hide="Solution"]
Looking at everying modulo 10, we get many terms like $1^{2003}+2^{2003}+3^{2003}+4^{2003}+5^{2003}-4^{2003}-3^{2003}-2^{2003}-1^{2003}+0^{2003}$ in the sum. Because there are an even number of terms divisible by 5, the fives cancel out, leaving $2001^{2003}+2002^{2003}+2003^{2003}$. We can simplify this using the Euler-Fermat Theorem to $1+8+27=36$. When divided by 10, 36 leaves a remainder of 6, so the answer is $\boxed{6}$.[/hide]

[quote="Farenhajt"]I'm sorry, [b]Max300[/b] and [b]b-flat[/b], but regarding that sum of the cubes: As far as I know, $1^{3}+\dots+n^{3}=(1+\dots+n)^{2}$, and not $(1^{2}+\dots+n^{2})^{2}$.

Here it happens that the incorrect formula gives the correct result, but that's just a chance.

(I mean, just stop and compare $1^{3}+2^{3}, (1+2)^{2}$ and $(1^{2}+2^{2})^{2}$)[/quote]

ack you're right i don't know what i was thinking  :oops:
i must not be getting enough sleep   :D

lol, I can't believe I made the same mistake...

I will edit my post now  :roll:


modular arithmetic
algorithm
Diophantine Equations

Solve in positive integers the equation $10^{a}+2^{b}-3^{c}=1997$.

This equation does not have solutions modulo $ 80$.

We have
$ (10^a\bmod 80)\in\{ 10, 20, 40, 0 \}$
$ (2^b\bmod 80)\in\{ 2, 4, 8, 16, 32, 64, 48 \}$
$ (3^c \plus{} 1997\bmod 80)\in\{ 0, 6, 24, 78 \}$
[size=59]but not two elements from the first two sets sum up to an element of the third.[/size] oops! See correction below.

Happily, this still suffices as $ 10^a \equiv 20 \mod 80$ for exactly one $ a$ (namely $ 2$).
Similar, $ 2^b \equiv 4 \mod 80$ has only one solution $ b\equal{}2$, thus this combination does not give a solution.

No, that's still wrong. $ 64 \plus{} 40 \equiv 24 \pmod{80}$.

Ok, so from the mod $ 80$ we see that we must have $ a\equal{}3$ and $ b \geq 6$. Thus, we are left with the equation $ 2^b \minus{} 997 \equal{} 3^c$. This equation DOES have the solution in case you haven't noticed that already, namely $ b\equal{}10$ and $ c\equal{}3$. It's also easy to check that for $ c\equal{}1$ and $ c\equal{}2$ there are no solutions, so let's assume that $ c \geq 4$.

Order of $ 2$ mod $ 81$ is $ 54$ so the equation $ 2^x \minus{} 997 \equiv 2^x \minus{} 25 \equiv 0 \pmod{81}$ has the unique solution for $ x \pmod{54}$, namely $ x \geq 46 \pmod{54}$, thus $ b \equiv 46 \pmod{54}$, in particular we have $ b \geq 10 \pmod{18}$ and $ 2^b \equiv 2^{10} \equiv 17 \pmod{19}$. It follows that $ 3^c \equiv 8 \pmod{19}$ and since order of $ 3$ mod $ 19$ is $ 18$ this implies that $ c \equiv 3 \pmod{18}$.

Now we do modulo $ 433 \equal{} 8 \cdot 27 \plus{} 1$ which is a prime number. It's easy to check that order of $ 2$ mod $ 433$ is $ 72$ and as we remember that $ b \equiv 10 \pmod{18}$ we have $ 2^b\minus{}997 \in \{2^{10} \minus{} 997, 2^{28}\minus{}997, 2^{46}\minus{}997\} \equal{} \{27, 6, 144\} \pmod{433}$. The order of $ 3$ mod $ 433$ is $ 27$ and $ 3^c \in \{27, 6, 144\} \pmod{433}$ implies that $ c \equiv 3 \pmod{27}$.

Now, we do modulo $ 262657 \equal{} \frac{2^{27} \minus{} 1}{2^{9}\minus{}1}$. This number is prime (but unfortunately we have to verify this with computer). Since the order of $ 2$ is $ 27$ and we know that $ b \equiv 19 \mod{27}$ we have that $ 2^b \minus{} 997 \equiv 2^{19} \minus{} 997 \equiv 260634 \pmod{262657}$. And now luckily it happens that order of $ 3$ is $ 54$ and since we know that $ c \equiv 3 \pmod{27}$ we have only to check $ 3^3$ and $ 3^{30}$ mod $ 262657$. But this values are $ 27$ and $ 258554$ respectively, a contradiction.

I guess that a contradiction could be reached also with some prime number in the interval $ [600, 2000]$. You can try to find some better number than $ 262657$ but I'm pretty sure that it can't be less than $ 600$.

Sorry for confusion. Modulo $ 80$ works but explanation is a bit more involved:
$ 10^a\bmod 80: 10, 20, 40, [0]$ (in brackets is the period)
$ 2^b\bmod 80: 2, 4, 8, [16, 32, 64, 48]$
$ 3^c \plus{} 1997\bmod 80: [0, 6, 24, 78]$

It is clear that there is no solution in the [b]periods[/b] of these functions. While the remaining cases is easy to check manually (with the help of computers).

You are still wrong! Modulo $ 80$ gives us only that $ a\equal{}3$ and it's not easy to check this manually. In fact, the whole difficulty of this problem is the case $ a\equal{}3$.

Well, it may not be easy to test manually (i.e., without computer) but anyway there is an easy algorithm for such problems. See http://www.mathlinks.ro/viewtopic.php?t=48431

For $ a \equal{} 3$ the problem is to solve $ 2^b \equal{} 3^c \plus{} 997$. My software easily produces the number $ 42143976$ (minimality was not aimed) modulo which there is no periodic solution.

[quote="TomciO"]I guess that a contradiction could be reached also with some prime number in the interval $ [600, 2000]$.[/quote]
Just for curiosity I've tested all moduli below $ 9000$ - no luck.

$ 2^b\equal{}3^c\plus{}997\mod p$ had not solution for $ p\equal{}73$ and for $ p\equal{}601$.

maxal your number is pretty good since it is only $ 8 \cdot  243 \cdot 19 \cdot 163$ so it doesn't contain any big prime factor. It is probably an advantage of doing modulo $ 243$ instead of $ 81$.

Rust, this congruences have solution: $ b \equal{} 10, c \equal{} 3$. There is no possibility to obtain a contradiction without using that $ b \equiv 46 \pmod{54}$ or $ c \equiv 259 \pmod{512}$, i.e. we have to choose numbers with $ \varphi$ divisible by at least $ 27$ or a big power of two. So $ 73$ and $ 601$ aren't best choices.

Yes, $ b\equal{}10,c\equal{}3$ unique solution.
Consider by mod $ 73$ we get $ b\equal{}10\mod 9,c\equal{}3\mod 12$.
Consider by mod $ 271$ we get $ b\equal{}10\mod 135,c\equal{}3\mod 30$ or $ b\equal{}19\mod 135 , c\equal{}27\mod 30$.
Consider by mod $ 30$ we get $ b\equal{}0\mod 5,c\equal{}3\mod 30$ or $ b\equal{}2\mod 5,c\equal{}15\mod 30$.
Therefore $ c\equal{}3\mod 60$ and $ b\equal{}10\mod 135$. If $ b\equal{}10\mod 135$, then $ 81\not |2^b\minus{}997$. Therefore $ c\le 3$

[quote="TomciO"]Rust I think that you've meant considering mod $ 31$ instead of mod $ 30$. And you are right, mod $ 271$ works! I've thought that I had checked it but I probably have made mistake in calculations. Your solution can be also little simplified, because it is enough to consider mod $ 19$ instead of mod $ 73$.[/quote]
Yes $ mod 31$. I choose $ p$, suth that minimal period 2 $ 27|T_p(2)$. minimal of them $ 271$, suth that $ 3^3|T_p(2)\equal{}135.$


Έστω $ a_{1},a_{2},...,a_{n}$ μη αρνητικοί πραγματικοί αριθμοί με $ a_{1}<a_{2}<...<a_{n}$. Αποδείξτε ότι η εξίσωση $ \sum^{n}_{i\equal{}1}\frac{a_{i}}{a_{i}\minus{}x}\equal{}n$ ισχύει για $ n$ πραγματικές τιμές του $ x$ εάν όλα τα $ a_{i}$ έχουν το ίδιο πρόσημο. Τι συμβαίνει εάν $ a_{1}<0<a_{n}$ :?:

Πέρασαν τόσες μέρες, κανένας ρε παιδιά??

Ως "παιδι" και εγω ας απαντησω ( με γνωσεις Γ Λυκειου)
Η συνάρτηση που προκύπτει αν φερουμε το $ n$ στο 1ο μελος εχει οριο στο πλην απειρο τον $ \minus{}n$ και πλευρικά στα α1,α2,....,αn αντίστοιχα +απειρο ,-απειρο,.....-απειρο .Σε $ n$ διαστηματα ανηκει το $ 0$ και αφου η ισοδυναμη της ανάγεται σε πολ/μο $ n$ βαθμου ειναι $ n$ ακριβως οι ριζες .
Αν κ οι ανητικοι $ ai$ τοτε......... εχει $ n\minus{}k$ ριζες


Find the greatest positive integer $n$ which is divisible by all positive integers $k < n^{1/3}$.

[quote="Treething"]I'd go for 420.[/quote]

Yup, that's right.

I'll guess 420, because that's the highest I can come up with.

[hide]
It needs to be divisible by 2 and 3, so it has a factor of 6.

It needs to be divisible by 4, so it has an extra factor of 2.

It needs to be divisible by 5.

It needs to be divisible by 7.

$6\cdot 2\cdot 5\cdot 7=420$.

Notice that $\sqrt[3]{420}$ is between 7 and 8.

If you try to make $n$ divisible by 8 by adding another factor of 2, it doesn't work because $\sqrt[3]{840}$ is between 9 and 10 (so you would have to add another factor of 3 to make it divisible by 9, etc.)[/hide]


vector
analytic geometry
pigeonhole principle
Galois Theory
6th edition

In problem 3, isn't there an error?
Shouldn't z be non zero?

I join to the question (I have also noticed this)

:blush: Yes, of course it should be not null (that's why I made the null vector definition there :P).
The statement has been corrected.

Toumaf, did you solve problem 3? I didn't manage to do it.

The only I got was 2, but I gave a very ugly proof...

Valentin, I have a question: does $r>n-1$  mean $\forall r \in \mathbb{R} : r>n-1$ or $\forall r \in \mathbb{N} : r>n-1$?

[quote="harazi"]It is a real number.[/quote]

is this your problem?

[quote="harazi"]No, it's just a small part of a writtten exam given at ENS Paris. But I found the problem extremely nice (and difficult ;) and this is the reason for which it appears in Mathlinks Contest. I have to confess that I have no complete solution for this problem.[/quote]

For heavens sake ,this tells how hard was this exam ,please, i was so excited about the beginnig of the 6 edition ,but when i saw the problems ,Well i completly gave up

there were so hard ,even the problem 1 ,was so hard ,i mean SO HARD .

Are the rest of the rounds are going to be like this too ?(please say no)

ENS exams (especially those in 6 hours) are really math exams, that is they give reaserch problems to students. Of course, the subject is very long and there are many many indications and steps, but from time to time they give very hard questions such as this one without many indications. Really, I respect the one who manages to finish a 6 hours exam at ENS. I think there weren't more than 8 people in the last 20 years who finished the exam.
   Anyway, it is clear that this round was difficult, probably too difficult, but you must admit the problems were very nice.  Even problem 1 gave me headaches for quite a long time before finding Peter's solution. Anyway, good problems for this round. I hope the other rounds will be equally nice, if not so difficult.

[quote="lomos_lupin"]Are the rest of the rounds are going to be like this too ?(please say no)[/quote]I cannot say, but certainly there will be easy problems in the competition, as always (but of course not very many :D).

In this round problem 2 was not that hard and there were many ways to approach it. Of course it was also not an easy problem :)

So, anyone having a complete solution to this problem, which I think was the (real) mathematical problem of this round?

Well, isn't Round 1 Problem 3 quite a straightforward generalization of the well-known 2-dimensional case? At least, it looked like that to my mind.

First, assume that $r<n-1$.

Fix some (large) natural number $N$ which we will choose later on. For all vectors $z=(z_1,...,z_n)\in \mathbb{Z}^n$ with $0\leq z_i<N$ for all $i=1,...,n$, consider $x\cdot z$. Let $M$ be the maximal absolute values of the achieved values of this expression.
By the box principle, there are $z^{(1)}$ and $z^{(2)}$ such that $|x\cdot z^{(1)} - x\cdot z^{(2)}| \leq \frac{2M}{N^n}$. But it is very easy to see that $M\leq nN$. Therefore, we have, taking $z^{(0)} = z^{(1)} - z^{(2)}$, that

$|x\cdot z^{(0)}|\leq \frac{2n}{N^{n-1}}$.

On the other hand, all coordinates of $z^{(0)}$ lie between $-N$ and $N$. Therefore $||z^{(0)}|| \leq \sqrt{n}N$. It follows that

$|x\cdot z^{(0)}| \leq \frac{2n\sqrt{n}^r}{||z^{(0)}||^rN^{n-1-r}}$.

Now choose $N$ large enough so that $N^{n-1-r}\geq 2nk\sqrt{n}^r$. So we're done with the first part.

Now, assume that $r>n-1$. We will construct an explicit point lying in $A(k,r)$ for some $k$.

Take some galois number field K of degree $n$ over $\mathbb{Q}$ which can be embedded into the real numbers – from now on, consider it to be embedded into $\mathbb{R}$. The existence of such a field should not be too hard, yet I didn't think much about it. Choose linearly independent algebraic integers $x_1,...,x_n$ in the ring of algebraic integers of $K$. Let $x^{\prime}= (x_1,...,x_n)$. The galois automorphisms of $K$ are $\sigma_1 = id, \sigma_2, ..., \sigma_n$.

Take any vector $z=(z_1,...,z_n)\in \mathbb{Z}^n$. Then $|\sigma_1(x^{\prime}\cdot z)\cdots \sigma_n(x^{\prime}\cdot z)|\geq 1$. On the other hand, $x^{\prime}\cdot z=\sigma_1(x^{\prime}\cdot z)$ and $\sigma_k(x^{\prime}\cdot z)\leq C||z||$ for some constant $C$ since $\sigma_k(x^{\prime}\cdot z)$ is linear in the coordinates of $z$ and we can take $C$ to be $n$ times the maximal coefficient of some coordinate of $z$. Scaling $x^{\prime}$ to have unit length and doing some manipulation this immediately gives the result.

Peter.

For the existence of such a field: Take some prime $p$ such that $p\equiv 1\mod 2n$. Look at the maximal real subfield $K^{\prime}$ of $\mathbb{Q}(\zeta_p)$ where $\zeta_p$ is a $p$-th root of unity. Then this has degree $\frac{p-1}2$ over $\mathbb{Q}$. Since $n$ divides $\frac{p-1}2$ and the Galois group of $K^{\prime}$ is cyclic, there is a subfield of order $n$ which can, of course, be embedded into the real numbers.

Peter, for those who know Galois theory it may seem straightforward... for the other poor people this is not so quick...

As I am one of these poor people who does not know Galois Theory, can anybody present more elementary proof ?

Or at least give any good references for a beginner to start with it?  :)  (preferred books with  interesting problems). Thanks in advance.

[quote="perfect_radio"]I can't help it wonder... when does one learn Galois Theory?[/quote]2nd year of college usually. But obviously you can learn it quicker :D

Hmm, maybe I should start to send my solutions to the MLC contest to Valentin - I've always been to lazy to do so, but as nobody gave a solution for this problem and only harazi seems to have found one for problem 1, it looks like I would rule the MLC ;).

[quote="Peter Scholze"]Hmm, maybe I should start to send my solutions to the MLC contest to Valentin - it looks like I would rule the MLC ;).[/quote]Maybe you should ;) :D

I think there is supposed to be a 0  :huh:


Putnam
function
geometry
perimeter
college contests

Determine all real numbers $a>0$ for which there exists a nonnegative continuous function $f(x)$ defined on $[0,a]$ with the property that the region
$R=\{(x,y): 0\le x\le a, 0\le y\le f(x)\}$
has perimeter $k$ units and area $k$ square units for some real number $k$.

This is possible for all $a>2$ and impossible for all $a\le2$.

If $a>2,$ we can let $f(x)=b,$ a constant. The area is $ab,$ the perimeter is $2a+2b,$ and the two are equal when $b=2+\frac2{a-2}.$

Now suppose $a\le2.$ The function $f$, being a continuous function on a closed bounded interval, achieves its maximum. Call that maximum value $M.$ Then the area is less than or equal to $aM,$ which is in turn less than or equal to $2M.$ But the perimeter is strictly greater than twice the length of a vertical segment of height $M.$ That is, the perimeter is strictly greater than $2M.$ Thus the perimeter is greater than the area and the two cannot be equal.

Just saying it has to be $(a-2)(b-2)=4$ and hence $b=\frac{4}{a-2}+2$.


Given that $ g(x) \equal{} \frac2x$, what is the value of $ g(\frac14) \minus{} g(\minus{}\frac14)$?

The answer is 2/(1/4) - -2/(1/4) = 8 + 8 = 16

You just plug 1/4 and -1/4 in for x in the equation.


Actually, it will be odd for all primes since $n=2p+1$ is not divisible by $2$.  Also, consider $p=3\implies n=7$.

Do you have a solution?  I think there exist infinite such primes.

[quote="Karth"][quote="amcavoy"]For which primes $p$ is $n=2p+1$ also prime?[/quote]

[hide]The answer is $p=2$ only, because all other primes are odd. Substituting an odd $p$ into the equation would yield an even number which is not $2$ (apart from $\frac{1}{2}$), which isn't prime. So, the only $p$ that is possible is $2$, which yields $n=5$.[/hide][/quote]

Counterexample: p=3

OK then, sorry about that.  I was thinking about it yesterday and figured there might be a solution, but I guess not...  Thanks for the help though. :)


$\boxed{1}$ Simplify $\frac{1-2i}{3-i}-\frac{5+i}{5i}.$

$\boxed{2}$ Find the real values of $a,\ b$ such that $\frac{(1-2i)(a+bi)}{3-2i}=3+4i.$

$\boxed{3}$ Let $\alpha ,\ \beta$ be the roots of the quadratic equation $x^{2}+x+1=0.$ Find the values of $\alpha^{2}+\alpha \beta+\beta^{2},\ \alpha^{3}+\beta^{3}$ and $\frac{1}{\alpha+2}+\frac{1}{\beta+2}.$

$\boxed{4}$ Let $p$ be real number. Distinguish the roots of the quadratic equation of $x^{2}-4(p+1)x+8=0.$

[quote="kunny"][quote="cincodemayo5590"]
Distinguish? Do you mean find when the quadratic has roots? [/quote]

I meant that what kind of roots the given quadratic equation has.[/quote]
[quote="cincodemayo5590"][hide="quoted solution, #4"]
\[b^{2}-4ac = 16(p+1)^{2}-4(1)(8) = 16p^{2}+32p-16 = 16(p^{2}+2p-1) \]

\[b^{2}-4ac \geq 0 \implies (p+1)^{2}\geq 2 \implies p \in (-\infty,-1-\sqrt{2}] \cup [-1+\sqrt{2}, \infty) \]
[/hide][/quote]

[hide="Alright, finishing this..."]
There are two real roots when $p \in \eft(-\infty,-1-\sqrt{2}\right) \cup \left(-1+\sqrt{2}, \infty \right)$; a double root when $p =-1 \pm \sqrt{2}$; and two roots with nonzero imaginary part when $p \in \left(-1-\sqrt{2},-1+\sqrt{2}\right)$. [/hide]


function
floor function
ceiling function
inequalities

I'll use [ ] for floor and < > for ceiling.

Exercise 21-16 in Volume 1 says:

[quote]If f(x)=[x] and g(x)=<x>, express g(x) in terms of f(x) and x.[/quote]

Looking at the solutions, [x] - <x> = 0 if x is an integer and the <x> -  [x] = 1 if x is anything else.

To find the expression a(x) which does this...  If a(x)=1-[b(x)], b(x) being 1 when x is an integer and between 0 and 1 otherwise.

Why does the last phrase "between 0 and 1 otherwise" apply?  If x wasn't an integer, then wouldn't b(x) strictly be zero, causing 1-0=0, and a(x)=1?

Think like this: The floor (usually denoted $\lfloor\,\rfloor$) is the largest integer not exceeding the given number, and the ceiling (usually denoted $\lceil\,\rceil$) is the smallest integer not underruning the given number. Hence

$\lfloor x\rfloor\leq x<\lfloor x\rfloor+1$
$\lceil x\rceil-1<x\leq\lceil x\rceil$

Note that the "equal" possibility is placed differently in the two inequalities.

Now try negating the second inequality:

$-\lceil x\rceil\leq -x<-\lceil x\rceil+1$

We actually get the first type inequality, of the floor kind. Hence we can write

$\lfloor -x\rfloor=-\lceil x\rceil$ or $\lceil x\rceil =-\lfloor -x\rfloor$

With the notation given in the problem, we get $g(x)=-f(-x).$


Given $ a,b,c\geq0$ satisfying  $ a\plus{}b\plus{}c\equal{}5$ Prove that :

$ 16(a^3b\plus{}b^3c\plus{}c^3a)\plus{}640\geq11(ab^3\plus{}bc^3\plus{}ca^3)$

[quote="M.KELAMLEDBA"]Given $ a,b,c\geq0$ satisfying  $ a \plus{} b \plus{} c \equal{} 5$ Prove that :

$ 16(a^3b \plus{} b^3c \plus{} c^3a) \plus{} 640\geq11(ab^3 \plus{} bc^3 \plus{} ca^3)$[/quote]
I think IT's not ineq! :rotfl:  :P

:wink:  :wink:
sorry Mr.2424 if you can't slove this inequality dont post anything. Or disprove it if you say "IT's not ineq"  :wink:

It's an old problem of Nguyen Anh Cuong :)
We can solve it by can_hang's lenma.


function
real analysis
real analysis unsolved

How would I solve the following 3 problems:

1. Find $\parallel F \parallel_{D}$ if $D=${${(x,y) \in \mathbb{R}^{2}: x^{2}+y^{2}\leq 1}$} and  $F: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ is defined by
$F(x,y) = (x+1,y+1)$

2. If K is a compact set, show that a series
$\sum_{k=1}^\infty F_{k}(x)$
of functions from $K \subset \mathbb{R}^{q}$ to $\mathbb{R}^{q}$ converges unifomly on K if the series of numbers
$\sum_{k=1}^\infty \parallel F_{k}(x) \parallel_{K}$
converges.

3. Does the series $\sum_{k=0}^\infty x^{k}y^{k}$ converge uniformly on the disc
{$(x,y) \in \mathbb{R}^{2}:-1 < x < 1,-1 < y < 1$}?
Justify your answer.

#1: Just a calculation.
#2: Find a candidate limit (the pointwise sum). How big can the errors be?
#3: Calculate the sum-it's unbounded. Can a sequence of bounded functions converge uniformly to an unbounded function?

On #3, the disc I meant to specify is:
{$(x, y) \in \mathbb{R}^{2}: x^{2}+y^{2}\leq 1$}
Now, I would prove this converges uniformly by the Weierstrass M-test, but what is the series bounded by? What would be the $\sum_{k=0}^\infty M_{k}$ such that $\parallel F_{k}(x,y) \parallel \leq M_{k}$ for all k and all $(x,y) \in D$. I know that $\sum_{k=0}^\infty x^{k}y^{k}= \frac{1}{1-xy}$ if $|xy|<1$ and since $x^{2}+y^{2}\leq 1$ then that must be true.

$|xy|\le\frac{x^{2}+y^{2}}{2}$. Raise this to the $n$th power, and you've got your bound for the terms.


linear algebra
matrix
linear algebra unsolved

Prove that any matrix is product of matrices of trace 0. Can we choose at most $N$ of them with $N$ ansolute constant?

harazi,

would you mind posting a hint/solution?

Thanks


geometry
geometric transformation
rotation

Can anyone help with this question?

Optically active hydrocarbon $ A$ reacts with 2 molar equivalents of hydrogen to produce diastereomers $ B$ and $ C$, both of which are optically inactive. Compound $ B$ has a smaller heat of combustion than $ C$. Ozonolysis and dimethyl sulfide reduction of $ A$ affords pyruvaldehyde $ D$ ($ C_{3}H_{4}O_{2}$ and $ S \minus{} isopropylsuccindialdehyde (E)$. What are the structures A-E? What are the sign and value of the optical rotation of A?

From googling, I know what D and E look like, but I am confused about piecing them together to form A. How can I have to diastereometers B and C that are both optically inactive? I can't find a example of this where this is indeed possible. Usually, with two stereogenic centers, if we have optically inactive SR, then the diastereomer would be active SS or RR, b/c RS is an enantiomer.

Can anyone help me find what A, B, and C are?

From the ozonolysis products, the only reasonable structure for A that I see is 5-isopropyl-3-methylcyclohexa-1,3-diene.

[quote="Bythecliff"]How can I have to diastereometers B and C that are both optically inactive?[/quote]

Are the cis and trans isomers of an alkene optically active?

never mind, I got the same compound for A as you did. But what about B and C?

What happens to the double bonds with hydrogenation?

they become saturated single bonds, but I think one of the cis/trans isomers is optically active. How can they both be optically inactive?

This seems pretty straightforward to me.  :huh:
A is 2 methyl, 5 isopropyl,1,3cyclohexadiene. Notice that C5 is  chiral carbon. B is a disubstituted cyclohexane in which 1,4 substituents are trans, whereas in C they are cis. They are both optically inactive.


by Cauchy-Schwarz; \[ L.H.S. \geq \frac{(\sqrt{a^3}\plus{}\sqrt{b^3}\plus{}\sqrt{c^3})^2}{a^2\plus{}b^2\plus{}c^2\plus{}3}\]
\[ \geq  \frac{9abc}{a^2\plus{}b^2\plus{}c^2\plus{}3}\] by am-gm

yes i see. but for what $ a,b,c$ do we have equality ?

$ a\equal{}b\equal{}c\equal{}1$
 :rotfl: