$ \displaystyle{\text{m}\sum\omega\dagger\bigcirc_55^{5_5}5\text{ }\mid\text{s }\int{}e\chi\Upsilon}$

Oh, please, no. You have no idea how many times this has been posted. STOP IT.

[quote="1=2"]$ \displaystyle{\text{m}\sum\omega\dagger\bigcirc_55^{5_5}5\text{ }\mid\text{s }\bigcap\bigcirc\dagger\text{ }\int{}e\chi\Upsilon}$[/quote]

LIES! ALL LIES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! And I ain't never told anybody to post nothing.

[quote="mewto55555"]And I ain't never told anybody to post nothing.[/quote]

Thus you have at least once in your life told somebody to post nothing, or in other words, you have told people to not post. WHO WAS IT?!


Sigh. The world is falling apart.

Iran had an earthquake several days ago. It's estimated 20,000 people were killed (at least 15,000 have already been buried), and another 20,000 were injured. Obviously, the Iranians are in need of some serious humanitarian aid. They've refused to accept any aid from Israel.

Slobodan Milosevic, the former Serbian president, the one responsible for Genocide, crimes against humanity, and war crimes (yeah, I know he's on trial, but he's as guilty as sin), is about to win a seat in the parliament .

In China, a gas well exploded a few days ago, sending poisonous gas over several square kilometers, killing 200 people. It was probably avoidable, but China doesn't require many safety measures. Workers die at their jobs frequently.

In Pakistan, there have been two recent assassination attempts on the Prime Minister. There will likely be more.

Blah.

[quote="MagnusRex"]What made me angry, is how little news coverage the earthquakes got. 20,000. Thats an enormous number of people, 4-5 times as much as 9/11.[/quote]

Granted, 9-11 was a terrorist attack, whereas the earthquakes had a natural cause...but still, you're right, it is a large number.

[quote]
Slobodan Milosevic, the former Serbian president, the one responsible for Genocide, crimes against humanity, and war crimes (yeah, I know he's on trial, but he's as guilty as sin), is about to win a seat in the parliament .
[/quote]

Could you give me a link? I haven't heard about this...

Sure thing. [url=http://www.cnn.com/2003/WORLD/europe/12/28/serbia.election/index.html]CNN article[/url]

Talk about little news coverage!!!  I haven't heard about the earthquake!!!

[url=http://www.cnn.com/2003/US/Southwest/12/29/woman.slain.ap/index.html]This[/url] is completely sick, disgusting, and depraved. The link goes to CNN and there are no pictures, but the story is somewhat graphic, so if you're squeamish, you know what not to click. Regardless... how could anyone do that, and why would they?

That is sick.  I saw a TV show that something like that happened, but what sick mind would do that?  Poor woman, she didn't do anything except get pregnant.  I wonder if her family has anyone to help them?

I just saw on the news that the death toll in Iran has reached 25,000.  I know, the world isn't in good shape at all.

Urgh.  Why would anyone kill another person, let alone someone they [i]knew[/i], just so they could steal their unborn baby and claim it was theirs?

Um I'd say the title's an exaggeration...Asia is not the only thing making up the world you know...maybe AoPS, yes, but not the entire world

Hmm...I did a breif research on abortion but usually human reaches extremeties during/near a human birth so anything is possible, just look through birth-related crimes on the internet or the library...on second thought, that might not be a good idea for your spiritual self.

Tare, what are you talking about? I listed examples from Iran, Yugoslavia, China, Pakistan, and the United States. Of the list, China is in Asia, and Pakistan and Iran could be considered in Asia depending on your definition of Asia. The rest certainly are not in Asia. As for the bit about abortion... it's not exactly coherent, and not related to the topic at hand.

[quote="rippledance"]Horrific. Depraved's the only word to describe it.[/quote]
I agree completely.

[quote="Tare"]usually human reaches extremeties during/near a human birth so anything is possible, just look through birth-related crimes on the internet or the library[/quote]

But that's the whole point: the woman who did the killing was apparently [i]not[/i] about to give birth, right?

[quote]I'll rebute every one of your example:[/quote]

Err...[url=http://www.m-w.com/cgi-bin/dictionary?va=rebut]rebut[/url] or [url=http://www.m-w.com/cgi-bin/dictionary?book=Dictionary&va=refute]refute[/url]?

[quote="Tare"]Not being good at math is your problem not the world's (unless you're the only human in the world which is not true)[/quote]

I hate to use this expression, but it's inevitable: duh!  If I'm not good at math even though I have the resources to become great, it's my fault.  It would be ridculous to try and blame someone else for my own problems, or worse, to blame the world for my problems.

[quote]Again the same, that's your problem.[/quote]

Yes, yes, my problem.  Bad of me to blame the world.  People who blame the world for their problems are [i]whiners[/i].

[quote="Tare"]same[/quote]

Once again, I should have put in some effort to look for my wallet, right?  I shouldn't give up so easily?  Is that what you're saying?

[quote="Tare"]but if you can't help at all then I think you should just ignore them[/quote]

But that's just the point!  You can help, or in the least bit, you can try, and in doing so, you'll attract attention and other people will come to your aid.

All right, then.  Next time you need help with something, don't be surprised if someone who could help you just ignores you.  After all, that's what you just said you would do.  Do unto others as you would have them do unto you.

Do you see what I've done?  I have likened your suggestion to do absolutely nothing, just because the chance that you'll help at all is slim, to that of a person who gives up far too easily, or to that of a person who is unwilling to try.  With the math example, if I work hard, I might make MOP and I might not.  But if I don't work hard, I know I won't make MOP.  Same goes for the running one: if I work hard, I might make a 6 minute mile and I might not.  But if I sit on my butt watching TV and bemoaning the fact that it's pratically impossible for me to accomplish that, then I definitely won't make my goal.  This is the same situation as with trying to stop the flaws of the world: if you try really hard, you might help some and you might not.  But if you turn your back and don't try at all, you will certainly not help.

As for this thing:

[quote]I define "natural disaster" to be an event which you have no control over nor can be stopped. Sure you could lessen the damage but the ground will shake or a cow will die and there's nothing you can do to stop it.[/quote]

Is mad cow disease in an American cow necessarily a natural disaster?  I think it's practically man-made for several reasons:

1. Officials who were responsible for checking the cows to make sure they were healthy failed in their duty.

2. Even if they couldn't tell whether the cow in question had mad cow, humans in general should try harder to find a test for mad cow so that we can protect ourselves from buying cows with mad cow who don't show the symptoms of it yet.

3. Humans have a bad habit of reusing animal parts to save money, even when common sense tells us not to.  How do you think mad cow spread so quickly through Great Britain?  How do you think a bunch of minks displayed BSE-like symptoms, including shakiness, inability to move, and eventually death?  How do you think a disease that was originally only in sheep (Scrapies) made a species jump, became BSE, and infected cows?

We need to have stricter laws and codes regarding importing animals, because they could be harboring anything.  We need to stop feeding "undesirable" animal parts to animals, as this only spreads the disease and causes the disease to appear in animals that didn't originally have it.

[quote]*ahem* Again my point is to argue mad-cow disease is more important than Iranian earthquake[/quote]

You seem to be extremely concerned with not simplifying "mad cow is more important than an earthquake" into "cows are better than humans".  I agree with you on that one; you can't pull parts of an inequality out and say that they satisfy the same inequality.  However, you forget that we're not talking about "an earthquake", we're talking about "33,000 deaths".

aalindsey -- you appear to be equating several things without any really good reason for doing so.  On one hand is military intervention.  On the other hand is humanitarian aid.  (On an entirely seperate third hand is globalization-related goings-on, including aid, loans and other things.  Neither of us is talking about this one, so I'm going to restrict it to this paranthetical and talk about it no more.)

Most of the world was okay with Afghanistan.  And, frankly, the Taliban messed up -- they could have avoided the war entirely if they'd been willing to give up Osama bin Laden, who everyone agrees was responsible for the terrorist attacks of September 11.  But they didn't, so (international opinion held) we were on safe moral ground going in there.  In Iraq, there was an exceptionally different case for many reasons, and as a result the world was not pleased at all, mainly because we went on something of a power trip and invaded a country without managing to get most other nations on our side.  That, combined with our other general approaches to foreign policy which have consistantly been undiplomatic, have alienated much of the world.

Now, in an entirely seperate realm is humanitarian aid.  Everyone tries to pitch in when there's been a disaster.  One thing international organizations like the U.N. have consistantly been able to do without complaint is cleanup.  It's regarded as a good thing, everywhere, that we have sent aid to regions stricken with terrible disasters, such as Iran.  Perhaps the only people who really have issues with it are the guys at the Iranian news agency who have to explain why there are Americans doing good things not befitting a 'satanic nation.'

Now, seperately, a general request -- when you are writing in paragraph form, please do not write as if you are communicating on instant messanger.  It is very hard to read.  Capitalization, spelling, grammar, and punctuation done correctly make it much easier to have a meaningful discussion by eliminating confusion caused by incomprehensible text.  Thank you.  If you'd like to respond to this paragraph, feel free to do so via private message -- don't clutter up this otherwise interesting thread.  Thanks again.

Dear mathfanatic,
            My last post is somehow gone so I'll just say one thing: Read my post again, especially the parts that you quoted. Then reply again.

Sincerely yours,
Tare

[quote="Tare"]Well my last post is somehow gone so I'll just say one thing: Read my post again, especially the parts that you quoted. Then reply again.[/quote]

Who is that directed to?  JBL?  Me?  Confuted?  aalindsey?  You have to be specific, otherwise the whole forum will think you're talking to them.

As for reading your post again, why?  But fine, I read it again.  "Then reply again?"  Why?  I think I said everything I wanted to say, and I said it with a fair degree of clarity.  If you want me to reply again because you don't like what I said, too bad.  I stand by my words.

FYI, those "examples" are demonstrating the point I've been making for a while: it's foolish to give up all hope, surrender, and say "I can never change the flaws in the world".  If you say that, I guarantee it will come true.  But if you try your hardest, you can't help but change the flaws in the world, if only the tiniest bit.

'But those examples are full of fallacies!' you say.  Of course they are! That's the beauty of sarcasm.

And one more thing I didn't address:

[quote]My ultimate point (which is still not really answered): Why make a thread like this? All it does it make us gloomy and sad over things that we can't do anything about and doesn't affect us and are none of our business.[/quote]

First of all, you replied to confuted's initial post.  You didn't have to do that.  In fact, you didn't have to look at the post.  In fact, you didn't even have to be on the AoPS Forum.  I'd say it's basically your choice whether you want to be gloomy and sad."

Second of all, we never established that the events that confuted has been bringing up are things that we can't do anything about, that don't affect us, and that are none of our business.

Also, I disagree that it's none of our business; in my opinion, it's our duty to have respect and empathy for other human beings.  Part of that includes recognizing when someone else gets their foot caught in a trap and helping out in some way.   'Helping out' doesn't have to mean flying to Iran and helping move piles of rubble - rather, it can mean simply spreading awareness of the earthquake, the way confuted has done.  It may mean praying for them (insert equivalent religious ritual) or writing a poem for the poor people who have lost their friends, family, homes, and/or lives.

[quote="Tare"]Are you being sarcastic or serious?[/quote]
I've never been more serious in my entire life.

[quote="mathfanatic"]Also, I disagree that it's none of our business; in my opinion, it's our duty to have respect and empathy for other human beings.  Part of that includes recognizing when someone else gets their foot caught in a trap and helping out in some way.   'Helping out' doesn't have to mean flying to Iran and helping move piles of rubble - rather, it can mean simply spreading awareness of the earthquake, the way confuted has done.  It may mean praying for them (insert equivalent religious ritual) or writing a poem for the poor people who have lost their friends, family, homes, and/or lives.[/quote]

That was very eloquent.

I personally tend to feel that if you're going to do something, it should be something that matters.  Thus, for example, I would prefer those who give blood to those who write poetry in terms of disaster relief.  But doing anything is certainly better, and more human, than doing nothing.

Fine I'll point out the part in which you missed a word or two:
[quote="mathfanatic"]
[quote="Tare"]but if you [b]can't help at all[/b] then I think you should just ignore them[/quote]

But that's just the point!  You can help, or in the least bit, you can try, and in doing so, you'll attract attention and other people will come to your aid.

All right, then.  Next time you need help with something, don't be surprised if someone [b]who could help you [/b]just ignores you.  After all, that's what you just said you would do.  Do unto others as you would have them do unto you.[/quote]
See you brought up a different case then what I said. If you can help then you should but if you can't then why try? (calling for help might count as something you can do so you might want to do that)

Um then your (confuted) comment is ironic, they each improved (or not) the world. However I'm pretty positive Einstein tried to stop the usage of nukes, that's trying (and failing) to improve another flaw in the world.

*sigh* Some day later on in my life I'll come back to this...

[quote]That was very eloquent. [/quote]

Why, thank you.  And I agree that donating blood is preferable to writing poetry - I should have thought of that.

[quote]See you brought up a different case then what I said. If you can help then you should but if you can't then why try? (calling for help might count as something you can do so you might want to do that)[/quote]

I'll accept that I "missed a word" if you can give me an example of where it is absolutely impossible to help someone in any way.

[quote]Um then your (confuted) comment is ironic, they each improved (or not) the world. However I'm pretty positive Einstein tried to stop the usage of nukes, that's trying (and failing) to improve another flaw in the world. [/quote]

Have you noticed that when you ask confuted if he's being sarcastic/serious, he only becomes more sarcastic?  For instance:

mathfanatic: I sure do love ice cream!
confuted: Yeah, me too!  I just looove clogging up my arteries so that I can have a heart attack when I'm 23.  Heart attacks are fun.  So is high cholesterol.
mathfanatic: Are you serious?  Don't you know that heart attacks are bad for you?
confuted: Naw, really?  Gee, I never knew that.  Thanks for telling me!  You've shown me the way, mathfan.  Now I can finally change my ways and become nice and skinny, because I'm oh so fat.
mathfanatic: No problemo!  I'm here to show you the way, any time, any day.  By the way, did I mention that I like ice cream?

Well, I was thoroughly confused, too.  Tone of voice does not carry well into internet forums.

Yeah sarcasm does not do well with words even with italics...

A person dies. Then you can not help because the person you wanted to help is dead. Sure you can help the people around that person like his or her relative but you can not help that specific person anymore. A terminal mentally disabled who doesn't understand anything can not be helped either. There are some thing you just have to let go because it is out of your hands...you aren't God, it is impossible for you to help in every single event and problem in the world and I know you'd rather study for MOP or something then to try and help every single person in the world.

SARS another epidimic that I feel very strongly on. From my previous statements u may have came to the conclusion that I dont give a hoot-n-anie about humanity. Well not true, the fact is I want to be a doctor when I grow up and I also want to be a natrualist and verterinarian. I believe this SARS thing is awful I can see lending aid to people who have suffered from this disease, as I might put it. My cousin was located in singapore during the first outbreak there (hes in the navy) luckily he didnt get it but it was scary he was shipped out for a whole 11 months. I missed him now hes back in California. The only thing I have to say on the whole SARS thing is that it makes me sick to think about the whole disease. I think its awful! More people have died and suffered from that than the amount lost every year from hurricanes and tornados. I mean even dieing in an earthquake would be better than dieing or suffering from this disease at least it happens quick in an earthquake. It turns my stomach to even think about it

[quote="topper"]Headline: China reports second SARS case of season

"The government plans to kill all 10,000 civet cats in the area by Saturday, then move on to the next target: rats. ...The cats will be killed  by drowning or electrocution, their remains boiled into watery mist or burned."  AP Beijing

Couldn't help but think of this thread.

Now what should we do?
Well, what would you do, if your mother asked you?

(Couldn't resist the Cat in the Hat reference.)[/quote]
a.) Get tested (both me and cat)
b.) Get the **** out of China (which I'm never planning to go to anyway)
c.) If I don't know the cat well then I'll just go oh well.

What should we do? Well you could write a letter to your congressman. Still if the people really care they'll just hide the cat just like they can hide babies. After all they can survive a whole lot more stuff than we can.

Whenever I hear statements like, "what is the world coming to?" or "the world is falling apart" - I always have to ask, when was the world ever in a better state?

When (and where) was there a period in history when the world (as a whole) was better than it is today? There have always been diseases. There have always been floods and fires and earthquakes. There have always been wars and famines and genocides and torutures and slavery. These seem to be constancies of the world - not just our time.

Yet the average life expectancy in the world is higher than it has ever been. As many wars as there are, the world is still probably less violent than it was in antiquity. And materially speaking, the stardard of living throughout the world is higher than it has ever been. So if you could choose any place and anytime throughout history in which to live, which would you choose?

[quote]Whenever I hear statements like, "what is the world coming to?" or "the world is falling apart" - I always have to ask, when was the world ever in a better state?
[/quote]

Amen!

People have been complaining about the world falling apart for years.  Guess what?  We're still around.  Infact, as gauss202 pointed out, we're doing better than ever before.  There are a lot of good things going on in the world too-they just don't make headlines.


[quote]The side length of a square is 16 cm long.
The midpoints of each side are joined to form a second square.
The process of joining the midpoints of the sides of the innermost square is repeated.
What is the number of centimeters in the perimeter of the fifth square?[/quote][hide]
The side of each successive square is reduced by a factor of $\frac{1}{\sqrt{2}} $

Hence, the $n^{th}$ square has side: $\frac{16}{(\sqrt{2})^{n-1}} $

For $n = 5$, the side is: $\frac{16}{(\sqrt{2})^4} \;= \;4$ cm.[/hide]

[hide="solution o_o"]
[color=darkred]
Draw a diagram, and you'll find that the length of a side of a square is sqrt(2) times as much as the square formed by joining its midpoints. That means the square's area is twice as large.

First square - 16
Second square - 8
Third square - 4
Fourth square - 2
Fifth square - 1
[/color]
[/hide]


function
trigonometry
induction
algebra unsolved
algebra

Find all continuous functions $f: \mathbb R \to \mathbb R$ such that \[f(x-y)=f(x)f(y)+g(x)g(y)\] for all reals $x$ and $y$.

[quote="eureka-123"]Find all continuous solutions of f(x-y)=f(x)f(y)+g(x)g(y)[/quote]

That's a rather classical equation.

I suppose that $ f(x)$ and $ g(x)$ both are continuous.

Let $ P(x)$ be the assertion $ f(x\minus{}y)\equal{}f(x)f(y)\plus{}g(x)g(y)$

1) Get rid of trivial constant solutions.
If $ f(x)\equal{}c_1$, then $ g(x)g(y)\equal{}c_1\minus{}c_1^2$ and so $ g(x)\equal{}c_2$
Same, If $ g(x)\equal{}c_3$, then $ P(x,x)$ implies $ f(x)^2\equal{}f(0)\minus{}c_3^2$ and so $ f(x)\equal{}c_4$ (since continuous).

So $ f(x)$ constant $ \iff$ $ g(x)$ constant and we have the families of solutions :

$ f(x)\equal{}c$ and $ g(x)\equal{}d$ with $ c,d$ any real numbers such that $ c\equal{}c^2\plus{}d^2$

2) Suppose now neither $ f(x)$, neither $ g(x)$ are constant fonctions
2.1) $ f(0)\equal{}1$, $ g(0)\equal{}0$ and $ f(x)^2\plus{}g(x)^2\equal{}1$ $ \forall x$
$ P(x,0)$ $ \implies$ $ f(x)\equal{}f(x)f(0)\plus{}g(x)g(0)$
If $ g(0)\neq 0$, this implies $ g(x)\equal{}\alpha f(x)$ but then $ P(x,x)$ would imply $ f(0)\equal{}f(x)^2(1\plus{}\alpha^2)$ and so $ f(x)$ constant.
So $ g(0)\equal{}0$
Then $ P(0,0)$ $ \implies$ $ f(0)\equal{}f(0)^2$ and so $ f(0)\equal{}0$ or $ f(0)\equal{}1$. But, if $ f(0)\equal{}0$, $ P(x,x)$ would imply $ f(x)^2\plus{}g(x)^2\equal{}0$ and so $ f(x)$ and $ g(x)$ constant
So $ f(0)\equal{}1$
And then $ P(x,x)$ $ \implies$ $ f(x)^2\plus{}g(x)^2\equal{}1$
Q.E.D.

2.2) $ f(x)$ is an even function and $ g(x)$ is an odd function.
$ P(0,x)$ $ \implies$ $ f(\minus{}x)\equal{}f(x)$ and $ f(x)$ is an even function.
Then, since $ g(x)^2\equal{}1\minus{}f(x)^2$, we get that $ g(x)^2\equal{}g(\minus{}x)^2$
Suppose then that exist a non empty interval $ (u,v)$ such that $ g(x)\equal{}g(\minus{}x)$ $ \forall x\in(u,v)$
$ P(x,\minus{}x)$ $ \implies$ $ f(2x)\equal{}f^2(x)\plus{}g(x)g(\minus{}x)$ and so $ f(2x)\equal{}1$ $ \forall x\in(u,v)$
As a consequence $ g(2x)\equal{}0$ $ \forall x\in(u,v)$

Let then $ y\in(u,v)$. $ P(x,2y)$ $ \implies$ $ f(x\minus{}2y)\equal{}f(x)$ $ \forall x$, $ \forall y\in(u,v)$ and this clearly implies $ f(x)$ constant
So no such interval $ (u,v)$ exist and $ g(\minus{}x)\equal{}g(x)$ $ \forall x$

2.3) $ \exists a>0$ such that $ f(a)\equal{}0$ and $ f(x)>0$ $ \forall x\in[0,a)$
$ P(x,\minus{}x)$ $ \implies$ $ f(2x)\equal{}f(x)^2\minus{}g(x)^2$ $ \equal{}2f(x)^2\minus{}1$
Since $ f(0)\equal{}1$ and $ f(x)$ is a non constant even function, it exists $ u>0$ such that $ f(u)<1$
If $ 1>f(u)>0$, we can build the sequences $ u_n\equal{}2^nu$ and $ v_n\equal{}f(u_n)\equal{}2f(u_{n\minus{}1})^2\minus{}1$ and it's easy to show that it exists some $ v_i\leq 0$
So, since $ f(x)$ is even and continuous, $ \exists x_0>0$ such that $ f(x_0)\equal{}0$

Let now $ A\equal{}\{x>0$ such that $ f(x)\equal{}0\}$ and let $ a\equal{}\inf(A)$; Since $ f(x)$ is continuous, $ f(a)\equal{}0$ and $ f(x)\neq 0$ $ \forall x\in[0,a)$
And, since $ f(0)>0$, $ f(x)>0$ $ \forall x\in[0,a)$
Q.E.D.

2.4) Knowledge of $ f(x)$ for $ x\in[0,a]$ gives exactly full knowledge of two solutions.
First, consider that $ g(x)$ solution implies $ \minus{}g(x)$ solution.
Second, $ f(a)\equal{}0$ implies $ g(a)^2\equal{}1$
So, from there, wlog consider $ g(a)\equal{}1$ (and so $ g(x)>0$ $ \forall x\in(0,a]$)

We have $ f(a)\equal{}0$ and $ g(a)\equal{}1$ and so $ f(\minus{}a)\equal{}0$ and $ g(\minus{}a)\equal{}\minus{}1$
Then, $ P(x,\minus{}a)$ $ \implies$ $ f(x\plus{}a)\equal{}\minus{}g(x)$
So $ f(2a)\equal{}\minus{}g(a)\equal{}\minus{}1$ and $ g(2a)\equal{}0$ and $ f(\minus{}2a)\equal{}\minus{}1$ and $ g(\minus{}2a)\equal{}0$
Then, $ P(x,\minus{}2a)$ $ \implies$ $ f(x\plus{}2a)\equal{}\minus{}f(x)$

So, if $ f(x)$ is defined over $ [0,a]$ :
Then $ f(x)^2\plus{}g(x)^2\equal{}1$ and $ g(x)\geq 0$ over $ [0,a]$ $ \implies$ $ g(x)$ defined over $ [0,a]$
Then $ f(x\plus{}a)\equal{}\minus{}g(x)$ $ \implies$ $ f(x)$ defined  over $ [0,2a]$
Then $ f(x\plus{}2a)\equal{}\minus{}f(x)$ $ \implies$ $ f(x)$ defined  over $ [0,\plus{}\infty)$
Then $ f(\minus{}x)\equal{}f(x)$ $ \implies$ $ f(x)$ defined  over $ \minus{}\infty,\plus{}\infty)$
Then $ g(x)\equal{}\minus{}f(x\plus{}a)$ $ \implies$ $ g(x)$ defined  over $ \minus{}\infty,\plus{}\infty)$
Q.E.D

2.5) $ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ $ \forall x$
We obviously have $ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ for $ x\equal{}a$
Using then $ f(2x)\equal{}2f(x)^2\minus{}1$ and $ f(x)>0$ $ \forall x\in[0,a)$, it's easy to show :
$ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ and $ g(x)\equal{}\sin(\frac{\pi}{2a}x)$ for $ x\equal{}\frac{a}{2^k}$, $ \forall k\in\mathbb N\cup\{0\}$ (remember we said $ g(a)\equal{}1$)

It's then easy with induction to show that :
$ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ and $ g(x)\equal{}\sin(\frac{\pi}{2a}x)$ for $ x\equal{}\frac{pa}{2^k}$, $ \forall k\in\mathbb N\cup\{0\}$ $ \forall$ integer $ p\in[0,2^k]$

And continuity gives :
$ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ and $ g(x)\equal{}\sin(\frac{\pi}{2a}x)$ $ \forall x\in[0,a]$

And 2.4 above gives the final result :
$ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ and $ g(x)\equal{}\sin(\frac{\pi}{2a}x)$ $ \forall x$

3) Synthesis of solutions :
$ f(x)\equal{}c$ and $ g(x)\equal{}d$ with $ c,d$ any real numbers such that $ c\equal{}c^2\plus{}d^2$

$ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ and $ g(x)\equal{}\sin(\frac{\pi}{2a}x)$ $ \forall x$ and any real $ a$

$ f(x)\equal{}\cos(\frac{\pi}{2a}x)$ and $ g(x)\equal{}\minus{}\sin(\frac{\pi}{2a}x)$ $ \forall x$ and any real $ a$

[quote="eureka-123"]yeah its a classsical eqn..with a trivial answwer...............but can anyone give  a  shorter proof to it????[/quote]

You're welcome

[quote="eureka-123"]yeah its a classsical eqn..with a trivial answwer...............but can anyone give  a  shorter proof to it????[/quote]

Here is a shorter solution. I hope you'll enjoy it :

Constant solutions are $ f(x) \equal{} c$ and $ g(x) \equal{} d$ with $ c,d$ any real numbers such that $ c \equal{} c^2 \plus{} d^2$. Let us look now for non constant continuous solutions :

$ y \equal{} 0$ gives $ f(x)(1 \minus{} f(0)) \equal{} g(x)g(0)$ and so, for non constant solutions, $ f(0) \equal{} 1$ and $ g(0) \equal{} 0$. As a consequence $ f(x)^2 \plus{} g(x)^2 \equal{} 1$

$ x \equal{} 0$ $ \implies$ $ f( \minus{} y) \equal{} f(y)$ and so $ f(x)$ is an even function. Then, since $ g(x)^2 \equal{} 1 \minus{} f(x)^2$, $ g( \minus{} x)^2 \equal{} g(x)^2$ $ \forall x$. But if, $ g(x) \equal{} g( \minus{} x)$ over a non empty interval, $ f(2x) \equal{} f(x)^2 \plus{} g(x)^2 \equal{} 1$ over a non empty interval and so $ g(x) \equal{} 0$ over a non empty interval, from which it is easy to show that $ f$ and $ g$ are constant functions. So $ g( \minus{} x) \equal{} \minus{} g(x)$ $ \forall x$

From $ f(2x) \equal{} 2f(x)^2 \minus{} 1$, $ f(x)$ non constant and $ f(0) \equal{} 1$, we get $ f(x) \equal{} 0$ for some $ x$ and, since $ f(x)$ is continuous, $ \exists a > 0$ such that $ f(\frac {\pi}{2a}) \equal{} 0$ and $ f(x) > 0$ $ \forall x\in[0,\frac {\pi}{2a})$

From $ f(0) \equal{} \cos(a\times 0)$, and $ f(\frac {\pi}{2a}) \equal{} \cos(a\frac {\pi}{2a})$ and $ f(2x) \equal{} 2f(x)^2 \minus{} 1$, it's immediate to conclude $ f(x) \equal{} \cos(ax)$ $ \forall x \equal{} \frac {\pi}{2^na}$, then  $ f(x) \equal{} \cos(ax)$ $ \forall x \equal{} \frac {k\pi}{2^na}$, then, using continuity, $ f(x) \equal{} \cos(ax)$ $ \forall x$

Hence the non constant solutions : $ (f,g) \equal{} (\cos(ax),\sin(ax))$ and $ (\cos(ax), \minus{} \sin(ax))$


geometry
trigonometry
perpendicular bisector
Pythagorean Theorem

A piece of wooden board in the shape of an isosceles right triangle, with sides, 1, 1, and √2 is to be sawn into two pieces. Find the length and location of the shortest straight cut which divides the board into 2 parts with equal area.

Let the triangle be $ \triangle ABC$ with right angle $ C$. We have two cases.

Case 1:

We draw a perpendicular bisector from $ C$ to $ AB$, and intersects at $ M$. Then we find that $ CM \equal{} \frac{\sqrt{2}}{2}$.

Case 2:

We draw a cevian from $ A$ to $ BC$, and intersects at $ N$. Since $ \triangle ACN$ and $ \triangle ANB$ share an altitude, they have equal areas if their bases are equal, or $ CN \equal{} NB \equal{} \frac{1}{2}$. By Pythagorean theorem, we have

$ AN \equal{} \sqrt{AC^2 \plus{} CN^2} \equal{} \sqrt{(1)^2 \plus{} (\frac{1}{2})^2} \equal{} \frac{\sqrt{5}}{2}$.

Clearly, $ CM$ is shorter so the answer is $ \boxed{\frac{\sqrt{2}}{2}}$.

Building a line parallel to a leg and perpendicular to the other, you can get $ \frac{\sqrt{6}}{3}$.

Building a line parallel to the hypotenuse you get $ \frac{2\sqrt{2}}{3}$. I can't see any lines shorter than what he's shown.

isn't the answer mainly just the altitude from the hypotenuse. Knowing that the area of the triangle is 1/2 then we get that hsqrt2=1 so we get that the answer is sqrt2/2

You have to prove that the other possibilities aren't shorter.

[quote="AIME15"]None of the endpoints necessarily have to be placed on vertices :wink:[/quote]

Thanks, I didn't realize that.  :blush:

[hide="Full solution"]Let the triangle be right angled at $ B$. We have two cases to consider:

Case 1:

A cut with one endpoint on $ AB$ and one endpoint on $ AC$. This case is equivalent to the case of one endpoint on $ BC$ and one endpoint on $ AC$. Let this cut be $ XY$ and let $ AX=x$ and $ AY=y$ and $ XY=l$ Since the area of triangle $ AXY$ is known to be $ \frac{1}{4}$ then we can write:

$ \frac{1}{2} xy \cdot \sin 45 = \frac{1}{4}$

$ \frac{xy}{\sqrt{2}}=\frac{1}{2}$

$ xy=\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

Now by the cosine rule in triangle $ AXY$:

$ l^2=x^2+y^2-\frac{2xy}{\sqrt{2}}$

$ l^2=x^2+y^2-1$

$ l^2=(x-y)^2+2xy-1$

$ l^2=(x-y)^2+(\sqrt{2}-1)$

So it is clear that $ l$ achieves its minimum when ${ x=y=\sqrt{\frac{1}{\sqrt2}}}$

Using our formula for $ l$ this gives:

$ l^2=\sqrt{2}-1$

$ l=\sqrt{\sqrt{2}-1}$

Case 2:

A cut with one endpoint on $ AB$ and the other on $ BC$. The same method will work here. Use the same terminology as before.

We have that:

$ \frac{1}{2} xy \cdot \sin 90 = \frac{1}{4}$

$ xy=\frac{1}{2}$

So by the cosine rule (it's actually Pythagoras):

$ l^2=x^2+y^2$

$ l^2=(x-y)^2+2xy$

$ l^2=(x-y)^2+1$

And it is now clear that the minimum of $ l$ is $ 1$. Evidently this is longer than in Case 1.

So the shortest cut is located with the endpoints distance $ \sqrt{\frac{1}{\sqrt{2}}}$ from $ A$ along each of $ AB$ and $ AC$. The length of this cut is $ \sqrt{\sqrt{2}-1}$[/hide]


trigonometry
inequalities unsolved
inequalities

$\sin A\sin B+\sin B\sin C+\sin C\sin A \geq (1+\sqrt{2\cos A\cos B\cos C})^{2}$

[quote="pvthuan"]The Mathscope Solution Booklet is coming soon.[/quote]

How soon, may I know?


function
real analysis
real analysis unsolved

Does there [i]exist [/i] a function $ f: \emptyset \to Y$? What are the conventions of what a function [i]is[/i]?

Yes.  There exists precisely one such function, and it does nothing.

A function $ f : A \to B$ is nothing more and nothing less than a rule that assigns to every element of $ A$ an element of $ B$.  In the case that $ A$ is the empty set, there is only one such rule, and it is the empty rule.

This is the best convention in part because it is the one that makes the following statement true: if $ A, B$ are finite sets, then the number of functions $ A \to B$ is $ |B|^{|A|}$.  (Incidentally, this convention implies that $ 0^0 \equal{} 1$.)


trigonometry
geometry
cyclic quadrilateral
geometry proposed

Let $ABCD$ be a cyclic quadrilateral with: $CD<AB, BC<AD$ and let$P$ and$Q$ are point choisen on segments $AB ,AD$ respectively such that:$AP =CD$and $AQ= BC$   Let $M$ be the intersection of $PQ$ and $AC$.
Prove that:$PQ=QM$

Firstly, there is a typo. It should be PM=MQ rather than PQ=QM (it's pretty obvious :) )
$\frac{MQ}{MP}=\frac{AQ\sin\angle DAC}{AP\sin\angle BAC}=\frac{BC\cdot DC}{DC\cdot BC}=1$, as desired.


Let $ a,b,c>0$.Prove that:
$ \frac{4a^2\minus{}b^2\minus{}c^2}{a(b\plus{}c)}\plus{}\frac{4b^2\minus{}a^2\minus{}c^2}{b(c\plus{}a)}\plus{}\frac{4c^2\minus{}a^2\minus{}b^2}{c(a\plus{}b)}\leq 3$

[quote="Inequalities Master"]Let $ a,b,c > 0$.Prove that:
$ \frac {4a^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \plus{} \frac {4b^2 \minus{} a^2 \minus{} c^2}{b(c \plus{} a)} \plus{} \frac {4c^2 \minus{} a^2 \minus{} b^2}{c(a \plus{} b)}\leq 3$[/quote]
It can be solved by SOS

[quote="Inequalities Master"]Let $ a,b,c > 0$.Prove that:
$ \frac {4a^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \plus{} \frac {4b^2 \minus{} a^2 \minus{} c^2}{b(c \plus{} a)} \plus{} \frac {4c^2 \minus{} a^2 \minus{} b^2}{c(a \plus{} b)}\leq 3$[/quote]
It's equivalent to $ \sum_{sym}(a^4b^2 \minus{} a^4bc \plus{} a^3b^2c \minus{} a^2b^2c^2)\geq0,$ which obvious.
The maximal $ k$ for which the following inequality
\[ \frac {ka^2 \minus{} b^2 \minus{} c^2}{a(b \plus{} c)} \plus{} \frac {kb^2 \minus{} a^2 \minus{} c^2}{b(c \plus{} a)} \plus{} \frac {kc^2 \minus{} a^2 \minus{} b^2}{c(a \plus{} b)}\leq \frac{3(k\minus{}2)}{2}\]
holds for all positive $ a,$ $ b$ and $ c$ is $ k\equal{}4.$ :wink:

The following inequality is also true!
Let $ a,b,c>0$.Prove that:
$ \frac{4a^2\minus{}b^2\minus{}c^2}{a(b\plus{}c)}\plus{}\frac{4b^2\minus{}a^2\minus{}c^2}{b(c\plus{}a)}\plus{}\frac{4c^2\minus{}a^2\minus{}b^2}{c(a\plus{}b)}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2abc(a+b)(b+c)(c+a)}\leq 3$


Let $A_{1}, A_{2}, \cdots$ be an infinite sequence of infinite sets. Then there is an infinite sequence $B_{1}, B_{2}, \cdots$ of mutually disjoint sets such that $A_{i}\cap B_{j}\neq \emptyset$ for all $i, j$.

Every infinite set contains some countable subset.  So, for each $i$ let a countable subset of $A_{i}$ be $a_{i1}, a_{i2}, \ldots$.  Then let $B_{j}= \{a_{ij}\mid i \in{\bf N}\}$.

It doesn't follow that your $B_{i}$'s are mutually disjoint.

Oh, bother, you're right.  I thought that seemed a little easy  :)

I don't think the statement is very clear :?. Where are these sets $B_{i}$ supposed to "be"? What is their relation to the $A_{i}$'s?

Edit:

Assuming you meant for the $B$'s to simply be subsets of the union $\bigcup_{i}A_{i}$, we could do this:

We construct a set $\{x_{ij}\ |\ i,j\ge 1\}$ with $x_{ij}\in A_{i},\ \forall i$ (and all $x_{ij}$ distinct). This is done recursively, by progressively enlarging the set $S_{n}=\{x_{ij}\ |\ i,j\in\overline{1,n}\}$ as $n$ increases. Given the fact that all $A_{i}$ are infinite, this is clearly possible. Then set $B_{j}=\{x_{ij}\ |\ i\ge 1\}$.

Sets don't require a habitat to be contained in.

By the way, good proof.

Indeed this problem is inspired by grobber's proof of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=116827]existence of continuum many disjoint subsets of positive outer measure[/url], where an intermediate step is to get a continuum many disjoint subsets $\{B_\mu\}$ of the real line with $A_\lambda \cap B_\mu \neq \emptyset$ for all $\lambda$,$\mu$ from a given continuum many continuum-cardinality subsets  $\{A_\lambda\}$ of of the real line.


Prove that in a company of 17 persons, in which each person is acquainted with exactly 4 other persons, there exist two, who are not acquainted and have no mutual acquaintance.

Let's consider the simple undirected graph $ G$ whose vertices are the persons, two being connected by an edge if and only if they are acquainted with each other.
We know that each vertex has degree $ 4$, and it is asked to prove that $ G$ has diameter at least $ 3$.

It would be pertinent to draw a figure to make the following easier to understand...

Assume, for a contradiction, that $ G$ has diameter $ 2$ (clearly, it is not a complete graph).
Let $ A$ be one of the vertices, and $ B,C,D,E$ be its neighbours. It left $ 12$ vertices, say $ A_1,\cdots,A_{12}$ which are not neighbours of $ A$. Then each of the $ A_i$'s must be connected to at least one of $ B,C,D,E$. But each of these last four vertices can share at most $ 3$ edges with the $ A_i$'s (they have degree $ 4$ and they are connected wih $ A$). It follows that, wlog, we may assume that $ B$ is connected with $ A_1,A_2,A_3$, that $ C$ is connected with $ A_4,A_5,A_6$, that $ D$ is connected with $ A_7,A_8,A_9$ and that $ E$ is connected with $ A_{10},A_{11}, A_{12}.$

In particular, the four neighbours of $ B$ are $ A,A_1,A_2,A_3$, and $ B$ cannot be connected with any of the following $ 12$ vertices $ C,D,E,A_4,\cdots,A_{12}.$
As above the neighbours of $ B$ have to be connected to exactly three non-neighbours of $ B$. We already know the neighbours of $ A$ (and those of $ B,C,D,E$), so each of $ A_1,A_2,A_3$ must have its neighbours (except $ B$) among the $ 9$ vertices $ A_i$'s for $ i \geq 4$, and each of these $ A_i$'s has to be adjacent to one of $ A_1,A_2,A_3$.
Thus, $ A_1$ and $ A_2$ are not adjacent, and they are not adjacent to a common $ A_i$. It follows that $ A_1$ and $ A_2$ are not adjacent and do not have a common neighbour. A contradiction.

Pierre.


I am now writing a Mock AMC F. All the problems will be verified by me as well as beta. THis time, I WILL write ORIGINAL problems. As before, I will have a secure way of transmitting answers that is both easy for everyone.

I will attach it as soon as I am done.

Can you fix the mistakes in your old mock AMC before making a new one?

That's officially over. Also, if you saw, I had typed fromi my paper wrong to the screen for two of the probllems :(

i'll do it
but it better not be as bad as the AMC 10. :lol:

[quote="anirudh"]I'm trying to restrict it to Geometry Honors level.[/quote]

Then it would an AMC 8

No it wouldn't AMC 10 doesn't have alg II

Yes it does. AMC 8 is up to geometry, AMC 10 is up to algebra two/trig and AMC 12 is up to pre-calc.

Up to. Not Up through.

http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC10.php

[quote][b]The AMC 10 is designed for students who have not yet studied curriculum from Algebra II and higher.[/b] Topics tested include concepts from Algebra I and Geometry I along with some number theory, basic logic, counting methods, and probability. As the test progresses, students are challenged to find answers to problems that require a significant depth of understanding of the curriculum as well as a creative eye toward mathematical techniques. This means that the AMC 10 is designed to test problem solving skills as much as curriculum.[/quote]

Therefore, no algebra II

well i guess ur mock amc 10 was definitely an AMC 12.

Whatever; whoever wants to sign up, just sign up here. You can argue about standards ini the other thread that I made like 'AmC curriculum'

UPDATE: Instead of doing the Mock, I am going to run the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=717290]Competition de Battle[/url], which will be about the same level.

Sorry.

[quote="now a ranger"]don't be sorry

 :P[/quote]

okay :maybe:  :maybe:  :maybe:

NO MORE REPLIES!!!


calculus
integration
function
logarithms
calculus computations

i've encountered the following integral:

integral of [Exp^(-x) * x^(-1)]dx

i tried playing around with it using integration by parts, nasty substitutions, but im still not getting close to anything logical. I've tried "The Integrator", and it gave me this answer:
Exp^(Integral of[Exp*i*[-x]])

no idea how it got this answer.

does anybody have any suggestions?

many thnx for ur time,
X

I do not think this integral can be expressed by elementary functions, but you can expand $e^x$ into a power series and integrate term by term:

$\int{\frac{e^x}{x} dx} = \int {\frac{dx}{x} + \int{ \sum_{n=1}^{\infty}{\frac{x^{n-1}}{n!} dx} = \int {\frac{dx}{x} +  \sum_{n=1}^{\infty}{\int{\frac{x^{n-1}}{n!} dx}} =}}}$

$= \ln{|x|} + \sum_{n=1}^{\infty}{\frac{x^n}{n!\ n}} + C =  \ln{|x|} + \sum_{n=0}^{\infty}{\frac{x^n}{n!\ n}} + C'$

thnx for the suggestion.
i'll try to play around with the series and see where it lead. however so far it looks like a dead end :)

many thnx for ur time.

Cheerz :)


If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:

$ \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$

The original triangle is similar to the first by SAS similarity, so all of its dimensions are doubled, so it is multiplied by a factor of $ 4 (C)$

Alternatively, the area of the original triangle is given by $ A \equal{} \frac{ab\sin{C}}{2}$ and the area of the new triangle is given by $ B \equal{} \frac{2a\cdot2b\sin{C}}{2} \equal{} 4A$. Therefore, the answer is $ \boxed{\text{(C)}}$.


The ratio of the length to width of a rectangle is $ \frac{3}{2}$. If the width is 60cm, what is the number of centimeters in the perimeter of the rectangle?
[asy]size(75); import olympiad; defaultpen(linewidth(0.8));
draw((0,0)--(3,0)--(3,2)--(0,2)--cycle);
draw((1,0)--(1,2)^^(2,0)--(2,2),dashed);
draw((0,1)--(3,1),dashed);
label(rotate(90)*"60 cm", origin--(0,2),W);[/asy]

hello, let $ a$ be the length and $ b$ be the width, the we have $ \frac{a}{b}\equal{}\frac{3}{2}$, if $ b\equal{}60$ then follows $ a\equal{}90$ and the perimeter is $ 2(a\plus{}b)\equal{}2\cdot 150cm\equal{}300cm$.
Sonnhard.


Find all functions $ f: N* \minus{} \minus{} > N*$ :
i)   $ f(f(n)) \equal{} f(n) \plus{} n$
ii)  $ f(f(n \plus{} 1) > f(n)$ , for every $ n \in N*$
iii)  $ f(1) \equal{} 2$

See this....

$ f[f(n)]\equal{}f(n)\plus{}n \\ \implies f(f(f(f(n))))\equal{}f(f(f(n)\plus{}n))\equal{}f(f(f(n)\plus{}n))\equal{}f(f(f(n)))\plus{}f(f(n))$

Which resembles exactly the Cauchy Eqn.


geometry
projective geometry
geometry solved

Searching for a solution with projective geometry please.

Let M be the midpoint of chord XY of a circle. AB and CD are chords passing through M.
E = AD ^ XY and F = BC ^ XY. Prove EM = MF.

If O is the center of your circle, then the point M, being the midpoint of the chord XY, is the orthogonal projection of the point O on this chord XY (in fact, in any circle, the midpoint of a chord is the orthogonal projection of the center of the circle on that chord). Hence, the line XY is the line passing through the point M and perpendicular to the line OM. And the whole problem follows from the problem discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18781 or http://www.mathlinks.ro/Forum/viewtopic.php?t=13061.

[b]EDIT:[/b] This is now Theorem 2 in [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i].

  Darij


IN THE NAME OF GOD

Find the population density for each country in the table. Round to the

nearest tenth of a person per square mile.repectively

Country                       Population                          ,                    Area (in hundred Population density (in
                                   (in millions)                                                         thousand square miles

Germany                           84.1                              ,                                        1.38

India                                 967.6                             ,                                       12.22

Japan                                125.7                             ,                                        1.46

Turkey                               63.5                               ,                                        3.01

United States                      268.0                            ,                                      36.75

THANKS
SALAH

Hi salahcool.

Are your figures correct? They don't seem to agree with those on Wikipedia. See:

http://en.wikipedia.org/wiki/Germany

http://en.wikipedia.org/wiki/India

http://en.wikipedia.org/wiki/Japan

http://en.wikipedia.org/wiki/Turkey

http://en.wikipedia.org/wiki/United_States

(My CTRL key is working!)

Don't even understand the unit u r talking about
which is which

if you had my math teacher this is a sure 0[/hide]

are you saying he doesn't have one  :P
and read! Area (in hundred Population density (in
(in millions) thousand square miles

Hello
 In each state the population estimated in the millions, and the area  in thousands of square meters
What is required is to calculate the population density of each State
 what is The wrong , my brothers??

May GOD reward you.

SALAH

Now I see what it means: the first number is the population in millions, the second is the area in units of a hundred thousand square miles. So, using salahcool's figures, the population density of Germany would be $ 84.1$ divided by $ 0.138$ which is $ 609.4$ to the nearest tenth of a person and similarly for India, Japan, Tukey and the USA. (By the way salahcool , to add to the list of polygons given in your geometry post, we also have the alternative archaaic forms derived from the Greek: trigon for triangle and tetragon for quadrilateral.)

Hi salahcool. Our posts crossed.


Let $f$ be a function from the set $Q$ of the rational numbers onto itself such that $f(x+y)=f(x)+f(y)+2547$ for all rational numbers $x,y$.
Moreover $f(2004) = 2547$.
Determine $f(2547).$

Pierre.

There is a really easy solution Kunny  :D

Pierre.

yes, the knowledge of cauchy's makes quite a few problems quite obvious  :D

it's been a while since the last time !:)
ok?
Can we talk about a derivation int the Q set of rationals????

[quote="Peter VDD"][quote="pbornsztein"]Note that $g(x)=f(x)+2547$ does satisfy the Cauchy equation.

Pierre.[/quote]what is the cauchy-equation? Is this pre-olympiad? :P[/quote]
f(x)+f(y)=f(x+y) only has f(x)=cx as a solution where c is a constant for f a continuous function on R or a function on Q, Z, N.

as a side note, anything which stops the function being dense in the whole plane (like boundedness in an interval) would rule out the mad functions in R.


calculus
integration
trigonometry
calculus computations

hello, i have got
$ \int\frac{\cos(2x)^2}{\sin(x)^4\cos(x)^2}\,dx\equal{}\frac{7}{3}\cot(x)\minus{}\frac{1}{3}\cot(x)\csc(x)^2\plus{}\tan(x)\plus{}C$.
Sonnhard.

Both answers are equivalent, since $ \csc^2 x \equal{} 1 \plus{} \cot^2 x$.

$ \int\frac{\cos^{2}2x}{\sin^{4}x\cos^{2}x}\ dx \equal{} \minus{}2[\minus{}2\csc^2(2x)] \plus{} 3(\minus{} \csc^2x) \minus{} (\minus{} \csc^2x) \cot^2x$


calculus
derivative
function
induction
real analysis
real analysis unsolved

$ f: \mathbb{R}\rightarrow \mathbb{R}$ is infinitely many times differentiable,  $ |f(x)| < 1$ for all $ x \in \mathbb{R}$. Prove  $ f^{(2002)}(x) = 0$ for some $ x \in \mathbb{R}$.

We can assume there is $ y \in R$ such that $ f^{2001}(y) \neq 0$(otherwise it is trivial)
also wlog $ y=0$,and $ f^{2001}(0) > 0$
use the Taylor series $ f(x)=f(0)+f'(0)x+f^{2}(0)\frac{x^{2}}{2!}+\cdots+f^{2001}(0) \frac{x^{2001}}{2001!}+f^{2002}(\epsilon_{x})\frac{x^{2002}}{2002!}$
Consider when $ x$ positive large enough and negative large enough to prove there is $ \epsilon_{x_{1}},\epsilon_{x_{2}}$,such that $ f^{2002}(\epsilon_{x_{1}})>0,f^{2002}(\epsilon_{x_{2}})<0$
then use Darboux's Theorem.

Unfortunately zhaobin's proof will not work since it does not make use of the fact that $ |f(x)|<1$ for all $ x \in \mathbb{R}$.

Consider for example the function:

$ f(x)=\frac{2}{2004!}(x+1)^{2004}+\frac{1}{2003!}(x+1)^{2003}$.

Then we have, $ f^{(2001)}(0) > 0$ and $ f^{(2002)}(x)=(x+1)^{2}+1 > 0$ for all $ x \in \mathbb{R}$.  So there is no way to get a negative value in order to use Darboux.

[quote="edoo"]Unfortunately zhaobin's proof will not work since it does not make use of the fact that $ |f(x)|<1$ for all $ x \in \mathbb{R}$.
.[/quote]
In fact I have used that,but you didn't see :wink:
You can do something yourself.

Okay, I see what you meant.  $ f^{(2002)}(\epsilon_{x})\frac{x^{2002}}{2002!}$ will be the dominant term for very large $ x$, but if $ |f(x)|<1$ for all $ x \in \mathbb{R}$, then the coefficient of this term must go to zero or be negative at some point.

Realizing that this is what you were doing was not clear to me when I first read your solution.

[quote="jmerry"]How about this stronger claim:

Show that $ f^{(2007)}$ has at least $ 2006$ distinct zeros.

(I decided to update the year on this year problem; it's the same argument either way)[/quote]
I think we can prove that:
Assume $ x_{1}<x_{2}<\cdots<x_{k}$ are all roots of $ f^{(n)}(x)=0$
then we may assume $ f^{(n)}(x)>0,x>x_{k}$
use Rolle's Theorem to prove there exist $ y_{1}>x_{k}$ such that $ f^{(n+1)}(y_{1})>0$,and use the discussion above to prove there exist $ y_{2}>x_{k}$ such tat  $ f^{(n+1)}(y_{2})>0$
so there is a roots of $ f^{(n+1)}(x)=0$ in the interval $ (x_{k},+\infty)$
Similar for the interval $ (-\infty,x_{1})$
and use induction. :)

That's pretty close; the argument is certainly about Rolle's theorem and its variants.
The specific behavior we need at $ \infty$ is that for any $ M$ and any $ \epsilon>0$, there exists $ x>M$ (and $ y<-M$) such that $ |f^{(n)}(x)|<\epsilon$ (and $ |f^{(n)}(y)|<\epsilon$. That's a Mean Value Theorem argument, using induction. For the first derivative, apply the MVT to the intervals $ [N,2N]$; we get points with derivative less than $ \frac2N$, and these points tend to $ \infty$. For each successive derivative, use intervals of increasing length with lower endpoints going to $ \infty$ drawn from the preceding sequence; the MVT produces a sequence tending to $ \infty$ with $ (n+1)$st derivatives tending to zero.

Now, let $ x_{1},x_{2},\cdots,x_{n-1}$ be distinct zeros of $ f^{(n)}$. By Rolle's theorem, we have at least $ n-2$ distinct zeros of $ f^{(n+1)}$, one in each interval $ (x_{k},x_{k+1})$. Now, choose $ x>x_{n-1}$ randomly. There are two cases: either $ f^{(n)}(x)=0$, or $ f^{(n)}(x)\neq 0$. In the first case, we have another zero of $ f^{(n+1)}$ by Rolle's theorem. In the second, we can find $ y>x$ with $ |f^{(n)}(y)|<|f^{(n)}(x)|$, and $ |f^{(n)}|$ has a positive local maximum in $ (x_{n-1},y)$. This local maximum is a zero of $ f^{(n+1)}$, so we get a zero in $ (x_{n-1},\infty)$ either way. Similarly, there is a zero of $ f^{(n+1)}$ in $ (-\infty,x_{1})$, and we have at least $ n$ total zeros of $ f^{(n+1)}$, completing the induction step.


Find all integers x,y,z such that (x+2y)(x+4z)(y+2z)+8xyz=1-(x+2y+4z)³.
[i](Paolo Leonetti)[/i]

$ (x \plus{} 2y)(x \plus{} 4z)(y \plus{} 2z) \plus{} 8xyz \plus{} (x \plus{} 2y \plus{} 4z)^3 \equal{} 1$
$ (x \plus{} 2y \plus{} 8z)(y \plus{} 2z \plus{} x)(4z \plus{} x \plus{} 4y) \equal{} 1$

There are four factorizations to consider $ (x \plus{} 2y \plus{} 8z, y \plus{} 2z \plus{} x, 4z \plus{} x \plus{} 4y) \equal{} (a,b,c) \equal{} (1,1,1),(\minus{}1,\minus{}1,1),(\minus{}1,1,\minus{}1)$ or $ (1,\minus{}1,\minus{}1)$

Liniear elimination gives $ 16z \equal{} 3a \minus{} 2b \minus{} c$ which does not lead to an integer solution in any of the four cases


inequalities
number theory unsolved
number theory

Well i read the following result deu to Polya on Bull.Amer.Math.Soc sounds like as:

[b]Problem[/b]: Let be given $a,b,d\in{\mathbb Z}_+$ and $\gcd(a,b)=1$ and $a,b>1$. Prove or disprove that the relation $|a^x-b^y|\leq d$ has finitely solutions in postive integer $(x,y)$.

Does any one know the full solution to this? PLZ show it   ;)

It's been discussed before (even if in a very slightly modified form) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4724&postorder=asc]here[/url], and it was mentioned several times on ML. As far as I know, no elementary solution has ever been posted though.


Hey!
I was wondering where I could find some puzzles for my brilliant 8 year old neighbour who is extremely bored with what they teach in school. Any good site I could get challenging problems for much younger kids?(8 or 9 year olds preferrably)
Thanx!

I like the Towers of Hanoi, that is pretty well known.

For what the Towers of Hanoi is:

http://en.wikipedia.org/wiki/Tower_of_Hanoi

Yeah I bought him one of that but do you know any good websites he could get puzzles from? It's weird but his mental age is twice of some teachers in his school, with his physical age being about one fourth of theirs!  :D
Any good links...?

http://www.cut-the-knot.org/index.shtml

For people interested in math, the site has something for everybody.

when he gets older: [url=http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml][wu:riddles][/url]

I think some chess puzzles such as the "Eight queens puzzle" and the "Knight's tour" are good fun. :lol:

Thanx!! I've given the links to him...
p.s.-Google really was useless in this case...needed you to go through a looong list and too much irrelevant spam...

[quote="abacadaea"][quote="speedcuber"]THE RUBIKS CUBE![/quote]

oh yeah, but thats for like 4th grade and up. hence the 8+ on the box[/quote]
Note that the person in question is 8 years old...


Let $ a_1, a_2, ..., a_n$ are non negative reals such that $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}1$. Prove that
$ a_1^3\plus{}a_2^3\plus{}...\plus{}a_n^3\plus{}n^2 \le (n\plus{}1)(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)$

[quote="puujee"]Let $ a_1, a_2, ..., a_n$ are non negative reals such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 1$. Prove that
$ a_1^3 \plus{} a_2^3 \plus{} ... \plus{} a_n^3 \plus{} n^2 \le (n \plus{} 1)(a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2)$[/quote]

Set $ a_1\equal{}a_2\equal{}\frac{1}{2}>0$. There is obviously $ a_1\plus{}a_2\equal{}1$ but the inequality does not hold beause we deduce the contradiction
$ \frac{1}{8}\plus{}\frac{1}{8}\plus{}4\le 3\cdot\left (\frac{1}{4}\plus{}\frac{1}{4}\right ).$

Actually the inverse inequality is true, because $ a_1^2 \plus{} \cdots \plus{} a_n^2 \leq (a_1 \plus{} \cdots \plus{} a_n)^2 \equal{} 1$ so $ (n\plus{}1)(a_1^2 \plus{} \cdots \plus{} a_n^2) \leq n \plus{} 1 < n^2$ for $ n \geq 2$. For $ n \equal{} 1$ direct substitution yields equality.


In case you haven't received an e-mail message or aren't on thte USAMTS mailing list, the problems for the first round of Year 17 of the USA Mathematical Talent Search are now available.

i believe there's a global annoucement on the forum


What does $ [n]$ indicate in the combinatorial sense? Is this $ [n] \equal{} \{1,2,...,n\}$? If this is the case then we have for example $ [a\plus{}b]\equal{}\{1,2,...,a\plus{}b\}$,no?

Thanks in advance,

Los

Yes.  (It's one of those standard notations that nonetheless always needs to be defined :).)


A gardner plants three maple trees, four oak trees and five birch trees in a row.  He plants them in random order, each arrangement being equally likely.  Let $\frac{m}{n}$ in lowest temrs be the probability that no two birch trees are next to one another.  Find $m+n$.

Again, this problem has been discussed many times in AoPS, please use SEARCH or RESOURCES page. THANK YOU.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66513]http://www.mathlinks.ro/Forum/viewtopic.php?t=66513[/url]


ratio
greatest common divisor
arithmetic sequence
AMC

Quote:Let S_n and T_n be the respective sums of the first n terms of 2 arithmetic series. if S_n/T_n=(7n+1)/(4n+27) for all n, the ratio of the 11th term of the 1st series to the eleventh term of the 2nd series is...?

Spoiler:

[hide]Let us assume that GCF(S_n, T_n) = x.  Then

S_n/T_n = (7n+1)x/(4n+27)x.

Therefore, S_n = (7n+1)x and T_n = (4n+27)x

S_10 + (term 11 of seq. S) = S_11,

so (term 11 of seq. S) = S_11 - S_10

similarly,

(term 11 of seq. T) = T_11 - T_10

Since S_11 = (7*11+1)x = 78x and S_10 = (7*10+1)x = 71x,

(term 11...S) = 7x.

Since T_11 = (4*11+27)x = 71x and T_10 = (4*10+27)x = 67x,

(term 11...T) = 4x.

Therefore, (term 11...S)/(term 11...T)  = 7x/4x = 7/4[/hide]

I'm not sure that this much work was needed, but I didn't want to take any chances...

Hmm, that's weird. THe answer key I got says th answer is 4/3, but I'm pretty sure your work is correct...

I like 4/3. For one thing, notice that if the first sequence is 14n-6 and the second is 8n+23, then we get 4/3. Now the question is how we get there.

What I did was to consider general arithmetic sequences an+b. When we sum that, we get a(a+1)/2*n+bn. But what we have is a ratio of linear terms. So why don't we multiply each by n? Then we can actually find the arithmetic sequences. Then you will get 14n-6 and 8n+23. When you put in n=11, you get 148/111=4/3.

Well, i used a different method to get 4/3 as well, but now i need to find the error in mathfanatic's work...

Hey I got 7/4 also like Mathfanatic.  Does anyone know which answer is correct!?!  Mathfanatic's explanation makes sense, but it seems too easy so I am confused.

Mathfanatic, your solution is incorrect.  For one thing, you seem to be assuming that the value of x is independent of n, but it's not.

Right -- the gcf of two terms is not a constant.  x will be different for different n.

I believe I saw something similar to this on a contest, and it threw me for a loop when i saw it, since i had never seen something like this in practice. (Was fun showing my math teacher tho!)

[hide]

 :Sigma: S = n/2*(2*S_1+(n-1)*d_1)

 :Sigma: T = n/2*(2*T_1+(n-1)*d_1)

Taking S/T... we notice that for the same n values,

S/T=

        (S_1+(n-1)*d_1/2)

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

        (T_1+(n-1)*d_2/2)

Noticing now that it is the ratio of terms in the two sequences, by using n=21, this will break down to the ratio of the 11th terms. So S_21/T_21 = the ratio of the 11th terms.

(7(21)+1)/(4(21)+27)= 148/111 = 4/3

[/hide]

I'm not quite sure if this isn't the same as what Geomar and Complex Zeta did, but heres what I did and it came out to 4/3:

Let S_n=a_1+a_2+...+a_n.
So S_n=(n/2)(a_1+a_n)

Similarly, Let T_n=b_1+b_2+...+b_n
So T_n=(n/2)(a_1+a_n)

This yields S_n/T_n=(a_1+a_n)/(b_1+b_n), and hence:
a_n+a_1=7n+1
b_n+b_1=4n+27

Although these could be actually multiplied by some constant, solving for a_n, and b_n, as I will do now, and then determining a_11/b_11 would yield the same value with or without such a multiplication.

Let a_n=7n+k.
Then, 7n+k+a_1=7n+1, and since a_1=7(1)+k, this equation conveniently reduces to k+7+k=1.
so k=-3.
a_n=7n-3

Similarly, b_n=4n+k+b_1=4n+27, k+b_1=27, but b_1=4+k, so:
2k+4=27, k+23/2, and:
b_n=4n+23/2.

a_11/b_11=(7(11)-3)/(4(11)+23/2)=74/(11/2)=4/3[tab]

I got 4/3.

[hide]Explanation:

Let the first sequence have a initial term a_1 and common difference of p.  Have the second sequence have initial term b_1 and common difference q.  The n stands for the nth term in the sequence.

a_1, a_1 + p, a_1 + 2p, a_1 + 3p, ... a_1 + (n-1)p

b_1, b_1 + q, b_1 + 2q, b_1 + 3q, ... b_1 + (n-1)q

Since we're finding a ratio between the sums of the series for corresponding n values, we can just set S_n/T_n = (7n+1)/(4n+27) = 2a_1 + (n-1)p / 2b_1 + (n-1)q.

If we're finding the ratio between the 11th terms of the sequences, we can just set a_1 + 10p, and b_1 + 10q as the terms.  Looking at the ratio of S_n/T_n, we can see that (a_1 + 10p)/(b_1 + 10q) is equal to S_21/T_21 because it's 2a_1 + (21-1)p / 2b_1 + (21-1)q.  So S_21/T_21 = (7*21+1)/(4*21+27)=4/3.[/hide]  Whew I'm done and I see ComplexZeta and Geomar already posted before me.  Oh well.

It looks like several of you (ComplexZeta, GeoMar, lotrfan, WarpedKlown) found the correct solution.  Nice going.  I think there was a simpler way to see it though.

Don't forget that in a (finite) arithmetic sequence, the median is equal to the average.  For example, the 11th term in the first sequence equals S21 / 21.  Similarly, the 11th term in the second sequence equals T21 / 21.  Now divide.

Sorry misread previous entry. Yeah good thinking.


This refers to the limit of a function as its variable tends to some number.  $\lim_{x\to\infty}\frac{1}{x}=0$ as the denominator becomes so large the fraction itself approaches $0$

its the limit.

$\lim_{x\to a}f(x)$ is the limit as x approaches a.

A limit is the function value a function approaches as x approaches a certain value.

$\lim$ isn't a function.

As has already been pointed out, $\lim_{x \to a}f(x)$ denotes the limit of f(x) as x approaches a.

In plain (straightforward, simple, non-rigorous) English:

If f(x) is undefined at a, then obviously you can't find a value for f(a). However, what you can do is have x get closer and closer to a. In that case, if f(x) gets closer and closer to a particular number, that would be $\lim_{x \to a}f(x)$.


[DOM]
Hmm.. I first thought of DFS. But I don’t think it will work so nicely. The maximum possible size is 100*100.

Well, we can place some obvious domino at those square which has already three sides. But when stalks of those squares will finish then what will you do?

You can do maximum flow algorithm, or something similar. Just think it over a bit;)

well i also thought so but can it be within time limit. our algo will be N^4.. that means 10^8. :(

Remember that E = O(V) and there is no time limit, the problem has an open data set.

DFS with pruning gets 16 of the 20 tests within 1 sec. :D

So kubus do u really think that it is possible to get All Correct by maxflow?

Whats ur opinion weeblie?

and what's ur pruning idea?

Sheet i forgot that the tests are given! :blush:

That means i can run my code as long as i wish.

But did u code it? Did it produce all the answers? What was the longest time?

And weeblie for u what was the longest time?

[quote="domino"][quote="Weeblie"]I don't think that maximum flow is correct approach to this task. You have to print out how you are putting the dominoes (print out which edges/vertices you use), which (as far as I know) isn't possible with MF...

Pruning wasn't the correct word. Optimisation with "place dominoe if you have to" is probably a better one. I.e. if one of the squares only can be occupied by one sort of dominoe (like, "up" dominoe), do it! :D

I did a quick solution for the task that makes the tests 1 to 14 and 18 and 19 (or something like that). Theother tests will probably run for a very long time... (solution can probably be changed to make them too, but I'm too lazy to do it right now :P).[/quote]

This is a pretty obvious bipartite matching problem, if you color the board like a chessboard. You have white squares on the left, black squares on the right -> bipartite graph ; you add edges according to the placement of the lines. Every vertex has at most 4 neighbours, so E = 4*V^2. Bipartite matching is O(V*E) = O(V^4) in this case. I implemented it and it worked pretty fast (most of them under 1second)... you can always do bipartite matching with maximum flow, i don't see how you can't print out the result...[/quote]

Aha! Bipartite matching... Didn't think of that one.  :blush:

It's pretty obvious now, after you've mentioned it. :D (btw, IOI usually doesn't have MF tasks... looks like they are going to start with them now).

[quote="mahbub"]well my idea was same with a bit more constrain. I considered every sqr has only 2 neighbours one up and one right. Because if A and B are neighbour then B and A are as well. Though i have not completed my code and did not think over the problems (i am not getting my time) but i am sure it is MF (huh nice simple name) and i am not sure though whether my simplification idea is correct. I will think on that later. if u want then u can also think :D[/quote]

Nice idea, but I it's much harder to create the graph that way... (if it's possible at all) :)

[quote="kubus"]Maximum flow and bipartite matching are almost the same, of course, this case looks more like BM but the solution itself does usually work like maximum flow (unless you implement Hopcroft-Karp, which I somehow doubt) ...
With this, all the input tasks ran almost instantly (less than 0.2 s) on 1,4 GHz 256MB RAM. The fact that you are guarranteed there is a result makes things easier.[/quote]

Well, it doesn't really matter for BM if the result is guarranteed or not... Otherwise, it will return the "best placement" which the task probably would ask for, or, if not, just check if the returned number of edges is "totSize/2". :D

Btw, what maximum flow algorithm are you using to get 0.2 seconds?

[quote="Weeblie"]
Btw, what maximum flow algorithm are you using to get 0.2 seconds?
[/quote]

Well, I said its not really a classic MF algorithm, just using the augmenting path idea. In fact all I did was that MN/2 times I added a domino. I found a free square and did a kind of DFS until I found another free square after a sequence of dominoes. And then just switched the dominoes so that in the end I could add another one. I can give you my code (it's really fluffy, I didn't have time to make it pretty and simple and it worked fine for the first time) if you want.

hmm.. it seems a bit interesting! well, u can send me ur code at shanto86@yahoo.com I am still not getting time for implementing in MF/BPM


S27

There are 42 points $ P_{1}, P_{2}, P_{3}, \ldots, P_{42}$, placed in order on a straight line so that each distance from $ P_{i}$ to $ P_{i+1}$ is $ \frac{1}{i}$ where $ 1 \leq i \leq 41$. What is the sum of the distances between every pair of points?

Sum of distance between every 2 points=41+40+39+38+...+1=861

That's correct but how did you get $ 1+2+3+\ldots+41$ as the answer

Try it with two points, then with three, then with four, and with five. See the pattern


Aratati ca lim(n→∞)a(n)  este un nr irational,unde a(n)=1/2^1! +1/2^2! +….+1/2^n!


I thought that a way to stop spam would be to make your rating public and delete the post count and ranking depending on the post, cause most people who spam do it for the post count, and if the rating is pubilc, then people will want to be more careful.  Of course, it won't stop delibrate spammers, but it'll stop unintentional (public rating) spammers, and post count spammers. :maybe:

P.S. I know that there's already a topic for this, but it's a totally different idea.
If you're not allowed to do this, please notify me, Thanks! :wink:

This would be a good idea, but post rating is notoriously unreliable. The post count usually isn't an accurate measure of a non-spammer's post quality, especially if only a few (or worse, one) people have rated that poster, and may cause good posters to be judged unfairly due to a low (but inaccurate) post rating.

[quote="yunhua98"]delete the post count and ranking depending on the post[/quote]
Could you clarify what you mean by this? :maybe:

Oops, messed up, I meant Just don't show the post count at all, or just make it private, so only you can see it.

In fact, post count is much more reliable than post rating.

Just think about it:
No matter how hard someone spams, it's quite hard to get any number of posts quickly. Looking at the Memberlist, I can tell you that I know of 8 of the top 10 posters as very reliable people that contribute greatly to the AoPS forums, and only 3 on the first page (first 50; I obviously won't mention (user)names) are major spammers. As we progress in declining posts amount, there is a greater density of spammers.
No matter how ineffective their rating contribution is, they can still purposefully rate others 1 and bring their rating down. If anything, I think the post rating system should be removed, as so many people are abusing it and very few people care about it.

Yes, I agree with r15s11z55y89w21 that post rating should be removed. I believe that it should be replaced with a "mark as spam" button. For reasons, see my posts [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1524371#1524371]here[/url].

[quote="hsiljak"]@OP: Let me ask a silly question: what kind of prestige does a high post count bring? I rarely look at people's post count: but, interesting enough, I do look at people's rating. Making ratings public would probably increase the number of multis (multiple-account users), don't you think so (for obvious reasons, boosting their rating since 'it counts now')?[/quote]

You can't see other people's rating.

Being a [b][color=green]moderator[/color][/b], I can (no need to quote entire preceding post, though  :wink:  )

Sorry, didn't know, anyways, how DO you not quote the entire preceding post? :huh:

You don't have to quote it at all, common sense implies you're replying to the preceding post. And you can always quote just a part of post by deleting the unnecessary parts of the text between the quote tags.

Back to the topic:

I accept the fact that I might not feel the spamming problems like you guys, since I spend most of my time in Olympiad, College and National subfora, where spam is expressed in other manners and ways. Still, I don't see the need for no-post-count revolution. And the rating is just there to motivate you to post productively.

[quote="yunhua98"]Now that you mention it, I quite agree with you on the rating, but the post count will do more harm than good, if we delete it, then the honest contributers would keep on doing it if they're not doing it for the post count, but the spammers won't. :!:[/quote]

Post count does signify a good contributor, though. As I have mentioned in my previous post, some of the best contributors have the highest post counts. There are very few spammers that get bored of spamming after, say, 500 posts.

Additionally, and most importantly, most spammers, the consistent spammers at least, [b]spam not for a higher post count but rather to deface forums of the website[/b]. In most cases, it is probably because they are jealous that the forums in this website are so nice, clean, and pristine. Eliminating post count would obviously destroy more benefits in judgment than it brings.


inequalities
number theory proposed
number theory

Find all pairs of positive integers (m,n) such that $ 7(m^5 \minus{}n^5)\equal{}41m^2 n^2 \plus{}1$.
(V.Yasinskyy). :o

__________________________________
$ \textbf{Azerbaijan Land of Fire }$

[hide="hint"]
put $ m\equal{}n\plus{}t$ and then use inequality and try to prove that $ t\equal{}1$.the rest will be trivial.[/hide]


calculus
integration
function
analytic geometry
calculus computations

is there any way to integrate an implicit function of y when it is not possible to solve for y=f(x)?

Green's theorem, or something similar, should work well (or a change of coordinates via the jacobian).

oh hmm...unfortunately i am still unfamiliar with such topics


In any month with five Mondays, a school declares the fifth Monday a Marvelous Monday.  School begins on Monday, September 4.  What is the date of the first Marvelous Monday after school starts?

[hide="Solution"]October 30th. The difference between the first and last monday in a month with a marvelous monday is 28. 4+28=32, so September has four mondays and October 2nd is a monday. 2+28 is 30, which is less than 31, so October 30th is the first marvelous monday.[/hide]


I was in the Mastering MATHCOUNTS class this fall, which has now ended.  My forum for that is gone, and there is no archive forum either.  Help!!!

hi, I have the exact problem. I can get to the forum because of a bookmark, but I can't find the archive.

The class archive should be here - http://www.artofproblemsolving.com/Forum/viewforum.php?f=152


factorial
calculus
integration
function
calculus computations

integrate this

$\int^{8}_{0}\frac{8!}{n!(8-n)!}dn$

xplain how you do it plz

it makes sense alot to me,i think its 1. anyone knows how to prove?

[quote="MathTeX"]:rotfl:

kunny, don't be so hard

Explain 'im why it doesn't.[/quote]

I can't understand what you say. :P

The factorial function is a discrete function; it does not have a natural definition on arbitrary real numbers.  The nearest way to try to make what you've written sensible is to take $n! : = \Gamma(n+1)$, which leads to the results sos stated.


2007 IMO problem 2. A good geometry problem.


The method used above is known as a difference of squares. So for your problem, you see that $x^2$ is a perfect square and $100$ is a perfect square. So, $x^2-100=(x-10)(x+10)$.

x^2-100 factors into (x+10)(x-10), using the (a+b)(a-b)=a^2-b^2 relationship.

$x^2-100=x^2-10^2=(x+10)(x-10)$

There is probably a section in your book that tells you about this rule.

I am taking an online class and the book don't show you how to work these problems out.

This is DOTS.  That's an acronym for Difference Of Two Squares. I learned that in 7th grade, and i'm in 7th grade right now.


Which pairs (a, b) or positive integers satisfy a2 - b7 = 7?

then maybe you could post me the solution by private message, Arne ?

I do not know if Arne sent you a private message, but I will try to give some thoughts:
   If b is even, we find that 4|a 2 +1, false. Suppose now that b is 1 mod 4. Then we can find a prime factor p of the form 4k+3, which divides b+2. Since b^7+2^7=a^2+11^2, it follows that p|a 2 +11 2  and thus p|a and p|11. So p=11. But also (b^7+2^7)/(b+2) is 3 modulo 4, so we can find a prime of the form 4k+3, let it be q, such that q|(b^7+2^7)/(b+2). So q|a^2+11^2 and q=11. It follows that 11|b+2 and 11| (b^7+2^7)/(b+2), false. So, b must be -1 modulo 4. In this case (b^7+2^7)/(b+2) is -1 modulo 4, so we can find a prime p which is 3 modulo 4 and such that p|a^2+11^2. Thus, p=11. So, 11^2| b^7+2^7. But also x^110 is 1 modulo 11^2, so from b^7=-2^7 it follows that b^112=2^112 and so b^2=2^2 in Z_11. Also, b^7=-2^7, so b=-2 in Z-11, so 11|b+2, false.

just a quick remark, this is not a Romanian IMO problem. It's (maybe) a problem used in the training of the team back then. :)


1.Let a, b, c > such that $ ab \plus{}ac\plus{}bc \leq abc$. Prove that :
$ \frac{b\plus{}c}{a^2}\plus{}\frac{c\plus{}a}{b^2}\plus{}\frac{a\plus{}b}{c^2}\plus{}2 \geq \frac8{a\plus{}b}\plus{} \frac8{b\plus{}c}\plus{} \frac8{c\plus{}a}$

2.Let  $ 0\leq a<b<c\leq2$.Find Min:
$ P\equal{}\frac1{(a\minus{}b)^2}\plus{}\frac1{(b\minus{}c)^2}\plus{}\frac1{(c\minus{}a)^2}$

[quote="talaaiha"]1.Let a, b, c > such that $ ab \plus{} ac \plus{} bc \leq abc$. Prove that :
$ \frac {b \plus{} c}{a^2} \plus{} \frac {c \plus{} a}{b^2} \plus{} \frac {a \plus{} b}{c^2} \plus{} 2 \geq \frac8{a \plus{} b} \plus{} \frac8{b \plus{} c} \plus{} \frac8{c \plus{} a}$
[/quote]
$ 2\geq2\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\geq\sum_{cyc}\left(\frac8{b \plus{} c} \minus{} \frac {b \plus{} c}{a^2}\right),$
The last inequality is obviously true because
$ 2\sum_{cyc}\frac {1}{a}\geq\sum_{cyc}\frac {4}{b \plus{} c}$ and $ \sum_{cyc}\frac {b \plus{} c}{a^2}\geq\sum_{cyc}\frac {4}{b \plus{} c}.$

2:
\[ \frac{1}{(b\minus{}a)^2}\plus{}\frac{1}{(c\minus{}b)^2}\geq \frac{1}{2}\left( \frac{1}{b\minus{}a}\plus{}\frac{1}{c\minus{}b}\right)^2\]
\[ \geq \frac{8}{((b\minus{}a)\plus{}(c\minus{}b))^2}\equal{}\frac{8}{(c\minus{}a)^2}\]
\[ \Rightarrow \frac{1}{(b\minus{}a)^2}\plus{}\frac{1}{(c\minus{}b)^2}\plus{}\frac{1}{(c\minus{}a)^2}\geq \frac{9}{(c\minus{}a)^2}\geq \frac{9}{4}\]
Equality holds for $ a\equal{}0,b\equal{}1,c\equal{}2$.


algebra
polynomial
quadratics
algebra unsolved

The real polynomials $p(x), q(x), r(x)$have degree $2, 3, 3$ respectively and satisfy $p(x)^2 + q(x)^2 = r(x)^2$. Show that either $q(x)$ or $r(x)$ has all its roots real

We can assume that the polynomials have positive leading coefficients. Without loss of generality, assume that $p(x)$ has degree 2 and $q(x)$ and $r(x)$ have degree 3. We get that $p(x)^2 = \left(r(x) - q(x)\right)\left(q(x) + r(x)\right)$. Since $p(x)^2$ is a polynomial of degree 4, and since $r(x) + q(x)$ is a polynomial of degree 3, $r(x) - q(x)$ must be a linear polynomial. Define $s(x) = r(x) - q(x)$, then $p(x)^2 + q(x)^2 = (q(x) + s(x))^2$ and hence $2q(x)s(x) = (p(x) - s(x))(p(x) + s(x))$. Since $s(x)$ is linear, it has a real root $a$, so $0 = 2q(a)s(a) = (p(a) - s(a))(p(a) + s(a))$, and $p(a) = 0$. Hence $a$ is a root of $p(x)$ as well, and hence $a$ is a root of $p(x) - s(x)$ and $p(x) + s(x)$. Since $p(x) + s(x)$ is a quadratic polynomial with one real root, both its roots are real. Similarly for $p(x) - s(x)$, so $(p(x) - s(x))(p(x) + s(x))$ has 4 real roots. Hence $2q(x)s(x)$ has 4 real roots, and this implies that $q(x)$ has three real roots.


Well even though I didnt make NATS, im still wondering how things are like there in terms of competition and difficulty to make top 10, also can someone tell me how much effort some of these kids put into winning the national mathcounts?

I believe top 12 makes national countdown, and difficulty to make the countdown varies each year

They put in a heck of a lot of effort and are really good to begin with. It is also very, very difficult to make the top 12.

effort to them is like fun...so it makes it a heck of a lot more enjoyable.

yeah and the countdown is in brackets where the first plays the last palce (in top 12), etc.

Actually, the top 4 written get byes for the first round.

In my opinion, there are three factors that determine how well you do at math competitions.

1)Practice
2)Natural Math Ability
3)Experience

[quote="alkjash"]Actually, the MOST IMPORTANT factor is unquestionably luck.[/quote]

...then why are you people trying to predict outcomes?  Furthermore, why did Kevin, who was predicted by many to win, actually win?

Of course you can't deny that there is some luck, but the truly good people are actually quite consistent.

or, the most important factor is luck, and those that don't need it :P

[quote="cognos599"][quote="uldivad9"]However, luck is very important. If you get lucky and do well, you may surprise everybody and make countdown or whatever. The opposite is also true though.[/quote]

while that is true, it takes a decent amount of skill to get lucky enough for it to count. for example, being unintelligent as i am (almost everyone in IN will back me up on that especially mathcrazed), if i were to take nats will all of you smart folk and say i was really really really lucky and i randomly guessed 6 sprints right. that would still make my total score <10. however if someone who would normally get 15th or 16th were to guess to few questions right, they would make top 12 due to luck, but it takes a certain amount of skill to be 15th or 16th without luck. see what i am saying?[/quote]
no... guessing a few sprint questions is not the luck that most people are talking about

I think they're more talking about, say, my brain is reacting faster today and is working more well today.
Or, my brain is making as many careless mistakes today.
Or, the problems are just my topic.

Or, it could be that rush of adrenaline.

could you guys say all of the different things that people get

(including awards for top students, awards that evry1 gets, awards that only countdown gets,plaques,trophies,etc.)

I have always thought that the most important thing is your preparation for nationals.  You not only have to work a lot of problems but you have to spend a lot of time going over the problems that you miss to make sure you really understand how to do them and why you missed them.  You also have to be working on the right kind of problems....usually old state and national tests.  You have to be solving Mathcounts problems.

You have to also really get to know your teammates and their strengths and weaknesses to be able to do well on the team portion.

then on the day of the competition you have to have a good day and hope that the problems on the tests match up with your abilities.

The real secret is work, work and more work.

Good luck to everyone who is going.......

[quote="JBoyd"]
The real secret is work, work and more work.
[/quote]

And this works on more than just MATHCOUNTS....

More important than genius.  More important than skill.

Maybe even more important than luck.  But you can't control luck...


A machine gun is mounted in a car resting on a frictionless horizonta surface.The mass of the car+gun system is M at a particular instant.At the same instant the gun is firing bullets of mass m,n per unit time.u and v are the velocity of bullet and car w.r.t. the floor.Find the accleleration of the car.

Relevant equations

M(dv/dt)=F-w(dm/dt) where w is the relative velocity

I got the following answer:Please check if I am correct
(dv/dt)=-mn(u-v)/M

regarding the use of the formula I will say one thing.The formula is commonly encountered in texts in rocket equation section.
But,the same formula can be regorously used in different problems...

Please check if I am correct.


[hide]The equation results $x^2 - x - m^2 - m = 0$.
The roots are equal if the discriminant is $0$.
So it is $4m^2 + 4m + 1 = 0$ hence $m = -1/2$[/hide]


If $ a$ is $ 50\%$ larger than $ c$, and $ b$ is $ 25\%$ larger than $ c$,then $ a$ is what percent larger than $ b$?

$ \textbf{(A)}\ 20\%\qquad \textbf{(B)}\ 25\%\qquad \textbf{(C)}\ 50\%\qquad \textbf{(D)}\ 100\%\qquad \textbf{(E)}\ 200\%$

$ a\equal{}1.5c$, $ b\equal{}1.25c$ then:

$ 1.25c \cdot x \equal{} 1.5c$
$ x\equal{}1.2$

Therefore, it is A, 20% larger.

[hide="Somewhat Simpler Solution"]
Makes it easier to compute if you start out with fractions:

$ a \equal{} \frac{3}{2}c\\
b \equal{} \frac{5}{4}c \implies c \equal{} \frac{4}{5}b\\
\\
a \equal{} \frac{3}{2}\cdot \frac{4}{5}b\\
a \equal{} \frac{6}{5}b$

$ 6/5$ = $ 120\%$ so the answer is $ \boxed{20\%}$.
[/hide]

[b]I.[/b]

[b] [u]3a[/u] > c[/b]
[b]2[/b]

[b] [u]5b[/u] > c[/b]
[b]4[/b]

[b]II.[/b]

[b] [u]3a[/u] > [u]5b[/u] [/b]
[b]2[/b][color=white].......[/color][b]4[/b]

Then...

[b] [u]12a[/u] > b[/b]
[b]10[/b]

Therefore;
it is [b]20%[/b] larger than [b]b[/b]

[color=red]****harrrr.... using codes is so annoying how do you guys make it  faster??[/color] :blush:

Or, simply, $ A/B\minus{}1\equal{}\frac{3c}{2}\cdot\frac{4}{5c}\minus{}1\equal{}\frac{1}{5}\equal{}\boxed{\textbf{(A)}\ 20\%}$.

[quote="maximos"]****harrrr.... using codes is so annoying how do you guys make it faster??[/quote]
What do you mean by codes?

Also, in your solution, $ \frac{3a}{2}>c$ and $ \frac{5b}{4}>c$ does not imply that $ \frac{3a}{2}>\frac{5b}{4}$. (Counterexample: $ 5>4$, $ 6>4$, $ 5>6?$). Actually, you want to be using equality, not inequality.

[quote="grn_trtle"]Or, simply, $ A/B \minus{} 1 \equal{} \frac {3c}{2}\cdot\frac {4}{5c} \minus{} 1 \equal{} \frac {1}{5} \equal{} \boxed{\textbf{(A)}\ 20\%}$.

[quote="maximos"]****harrrr.... using codes is so annoying how do you guys make it faster??[/quote]
What do you mean by codes?

Also, in your solution, $ \frac {3a}{2} > c$ and $ \frac {5b}{4} > c$ does not imply that $ \frac {3a}{2} > \frac {5b}{4}$. (Counterexample: $ 5 > 4$, $ 6 > 4$, $ 5 > 6?$). Actually, you want to be using equality, not inequality.[/quote]

... oops sorry for that,.. I mean tags,, is there any list of tags so that I can be guided??

thank you for correcting me anyway and I appreciate it so much......the question was "[b]a[/b] is how much larger than [b]b[/b]" and I took it as a guarantee that [b]a[/b] is already larger than [b]b[/b] but how many percent larger??? so I interpreted it this way:

$ \frac {3a}{2} > \frac {5b}{4}$.

But then it all turned out that my interpretation was wrong... Well, thank you anyway....

 :lol:

Ah, if you want to learn more about the tags, you should look up BBCode. However, it'll be much more rewarding to you if you take the time to learn $ \text{\LaTeX}$.

Anyway, $ a$ is definitely larger than $ b$ as per the problem statement. Indeed, $ \frac{3}{2}a>\frac{5}{4}b$ for any $ a>b>0$. However, $ \frac{12}{10}a>b$ does not imply that $ a$ is $ 20\%$ greater than $ b$ (even though it worked out in the end). It instead means that $ 120\%$ of $ a$ is greater than $ b$, which isn't very helpful, even though it's perfectly true. What you want to do is to say that $ b=c+.25c$, and $ a=c+.5c$, and then use the relations to find a way to [b]equate[/b] $ a$ and $ b$. (For example, factor and rewrite the above to get $ b/c=1.25$ and $ a/c=1.5$, and then divide the second by the first in order to find $ a/b$).

[quote="grn_trtle"]Ah, if you want to learn more about the tags, you should look up BBCode. However, it'll be much more rewarding to you if you take the time to learn $ \text{\LaTeX}$.

Anyway, $ a$ is definitely larger than $ b$ as per the problem statement. Indeed, $ \frac {3}{2}a > \frac {5}{4}b$ for any $ a > b > 0$. However, $ \frac {12}{10}a > b$ does not imply that $ a$ is $ 20\%$ greater than $ b$ (even though it worked out in the end). It instead means that $ 120\%$ of $ a$ is greater than $ b$, which isn't very helpful, even though it's perfectly true. What you want to do is to say that $ b = c + .25c$, and $ a = c + .5c$, and then use the relations to find a way to [b]equate[/b] $ a$ and $ b$. (For example, factor and rewrite the above to get $ b/c = 1.25$ and $ a/c = 1.5$, and then divide the second by the first in order to find $ a/b$).[/quote]

Wow!! Thank You so much it helped a lot :lol:


Factorise x^2005 - 2005x + 2004 into irreducible polynomials in Z(x)

Firstly, why did you post the same question twice?

So far I have $ (x\minus{}1)^{2}(x^{2003} \plus{} 2x^{2002} \plus{} 3x^{2001} \plus{} \cdots \plus{} 2004)$. The last one seems irreducible, but appearances can be deceiving...


Let $ x, y$ be numbers and $ (x,y) \equal{} 1$. Prove that for infinitely many primes $ p$ the exponent of $ p$ in
$ x^{p \minus{} 1} \minus{} y^{p \minus{} 1}$ is odd.

Two nice solutions posted by Tomci0 and bilarev can be found in http://www.mathlinks.ro/viewtopic.php?t=196188  :)


geometry
circumcircle
ratio
power of a point
radical axis
perpendicular bisector
geometry proposed

6. The triangle $ABC$  circuscribed in the circle and in the $ABC$ iscribed a circle, which tangents the sides $AB,BC$ and $AC$ at $A_1,B_1$ and $C_1$ respectively .$P\in{B_1C_1}\cap{BC}$,and $M$ is the midelpoint of  $PA_1$.
Prove that the touch lines segments from $M$ to the circumscribed and inscribed circles are equal.$\clubsuit$

Let a = BC, b = CA, a = BC and WLOG, assume b < c. The circle (M) centered at M and with radius $MA_1$ is perpendicular to the triangle incircle (I). We have to show that it is also perpendicular to the triangle circumcircle (O), because then the tangents from M to (I) and (O) would be equal. To do this, it is sufficient to show that the radical axis of the circle (M) and the vertex C is the perpendicular bisector of the side BC. Then the circle (O) belongs to a pencil of circles intersecting at B, C and the circle (M) belongs to a pencil perpendicular to the first one. By Menelaus theorem for the triangle $\triangle ABC$ cut by the line $B_1C_1,$

$\frac{AC_1}{BC_1} \cdot \frac{BP}{CP} \cdot \frac{CB_1}{AB_1} = 1$

$\frac{BP}{CP} = \frac{BC_1}{AC_1} \cdot \frac{AB_1}{CB_1} = \frac{s - b}{s - a} \cdot \frac{s - a}{s - c} = \frac{s - b}{s - c} = \frac{BA_1}{CA_1}$

Thus $A_1$ cuts the segment a = BC in the ratio $\frac{s - b}{s - c}$ internally, while P cuts this segment in the same ratio externally. Hence

$A_1P = A_1C + CP = \frac{a(s - c)}{(s - b) + (s - c)} + \frac{a(s - c)}{(s - b) - (s - c)} = \frac{2(s - b)(s - c)}{c - b}$

Let A' be the midpoint of BC. Then

$A'A_1 = (s - b) - \frac a 2 = \frac{c - b}{2}$

$A'P = A'A_1 + A_1P = \frac{c - b}{2} + \frac{2(s - b)(s - c)}{c - b} =$

$= \frac{(c - b)^2 + a^2 - (c - b)^2}{2(c - b)} = \frac{a^2}{2(c - b)}$

$A'A_1 \cdot A'P = \frac{c - b}{2} \cdot \frac{a^2}{2(c - b)} = \frac{a^2}{4} = A'C^2$

Thus the powers of the midpoint A' to the circle (M) and to the vertex C are equal, which means that A' lies on the radical axis of te circle (M) and the vertex C. Of course, their radical axis is perpendicular to their center line $MC \equiv BC$, i.e., identical with the perpendicular bisector of BC, which is what we wanted to prove.


Find all primes $ p,q$ such that for the sequence \[ a_{n\plus{}2}\equal{}pa_{n\plus{}1}\minus{}qa_n\] with $ a_0\equal{}0,a_1\equal{}1$ there is a positive integer $ k$ such that $ a_{3k}\equal{}\minus{}3.$

$ a_n=\frac{x_1^n-x_2^n}{\sqrt{p^2-4q}},x_1_,_2=\frac{p\pm \sqrt{p^2-4q}}{2}.$
Therefore $ a_3|a_3k=-3$ and $ a_2=p,a_3=p^2-q$.
If $ p$ and $ q$ odd, then $ a_3$ even is not divisor $ -3$.
1. p=2 give variants $ q=3,5,7$ work only $ p=2,q=7$.
2. q=2 give not variants, suth that $ a_3=\pm 1, \ \pm 3$.

[quote="Rust"]
1. p=2 give variants $ q \equal{} 3,5,7$ work only $ p \equal{} 2,q \equal{} 7$.
[/quote]
Prove that only $ (2,7)$ works...for $ (2,3)$ it is obvious but what about $ (2,5)?$

Consider $ a_p|a_n$ for $ p|n$. We get $ a_{2k}$ is even - not work, for odd prime p $ a_p\equal{}\sum_{k\equal{}0}^{[(p\minus{}1)/2]}C_p^{2k\plus{}1}(\minus{}1)^k4^k\equal{}(\minus{}1)^k\mod p$. If $ a_{3k}\equal{}\minus{}3$ for one odd prime divisor 3k must be $ a_p\equal{}\pm 3 \not \equal{}\pm 1\mod p$.


Form a $ 2000*2002$ screen with unit screens. Some screens are on and some screens are off. In any $ 2*2$ screen as soon as there are $ 3$ unit screens off, the $ 4th$ screen turns automatically off. Find the largest number of screens that can be on initially, so that finally the whole screen is off.

OK, maybe it's too hard so I'll give a hint. Prove the following Lemma.

[b][u]Lemma:[/u][/b]

[i]There can't be more than $ 1999*2001$ screens initially on.[/i]

I have been able to prove the above Lemma but I can't work the problem any further... any ideas?

P.S. The above Lemma was an exercise at the "102 Combinatorial Problems" of Titu Andreescu and Zuming Feng.


Id like to post my proof to the following problem. Please critique without regards to feelings.

problem: 2n+1 persons are placed in the plane so that their mutual distances are different. Everybody then shoots his nearest neighbor. Prove that a. At least one person survives

b. nobody is hit by more than 5 bullets

c. the paths of the bullets dont cross

d. the set of segments formed by the bullet paths doesnot contain a closed polygon

Proof:

[hide]b. Lets say G is hit by 5 bullets, where the distances E_i -- E_j are all different and larger than any E_i--G. Adding in one more E_i and keeping the same constraints, we then draw all of the connections of neighboring E and then all of the E to G. For 6 to hold up, angle E_i--G--E_i+1 > both E_i--E_i+1--G and E_i+1 -- E_i -- G for all i. Lets assume that this holds, then the G_i (all the angles that meet at G) >= 60 degrees, but the sum of the G_i = 360, so if any G_i is strict, there isnt equality there. Thus G_i <= 60 and a contradiction is met.

c. Assume that the paths are the shortest possible. Now lets say that CD and AB cross paths, then one of CD would not be shortest, because they would be closer to either A or B.[/hide]

Ill put up my solutions for the other 2 parts later (its in my notepad thingy already)

For b, you don't quite have a contradiction yet because [hide]x <= 60 and x >= 60 are not contradictory -- you still need to deal with the case x = 60 for all x.[/hide]

For the second, you are not correct.  CD could be the shortest.  However, in that case, AB wouldn't be the shortest.  It needs much more work and rigor.

Ok Thanx JBL... Ill work on those

d.

[hide]Consider in the group of E_n  k elements that would form a k-gon. Now consider the element E_b in the k elements with the max distance from its neighbor. In the polygon, every vertice must be connected 2 others. E_b is going to connect with its other neighbor in the k group, and the neighbor that is on E_b's other side isnt going to use the max distance either, choosing instead to battle along the shorter path. Thus, this "edge" is unused and the polygon is open.[/hide]

d looks good to me.

a is a really great problem.  Have you looked for a counter-example to (a) for the even case?

Hehe this is a great math problem and funny too.

Anyways heres my attempt at a)

[hide]

A bit lengthy:

Select the person who is farthest from his closest neighbour. The distance to his closest neighbour is d1.  Then the distance from the neighbour to the neighbour's closest neighbour is d2, and so on.

The sequence {d} must be monotonic decreasing.  Otherwise assume

d_i < d_(i+1).  Then some person shot his neighbour at a distance of d_i, while his neighbour shot someone else at a distance d_(i+1).  But this cannot be since each person must shoot the closest neighbour.  So starting from the top, each next person was shot at a closer distance.

Since each of the mutual distances were different, and the d_i describes the shortest distance from a certain person, distance d_(2n+1) is the smallest of all distances amongst any 2 persons.  As such they must shoot each other. One of these 2 people also had another neighbour Bob who stood at a distance of d_2n.  This is the next smallest distance after d_(2n+) and Bob has no choice but to shoot one of those 2 people who already have been shot.  Now a bullet has been wasted, and thus someone must survive.

[/hide]

a) was on Oakland University's Math Summer Insitute Qualifying Quiz. I think that the application period is over though, so posting a solution should not be a problem.

Let my try a solution: [hide]

Assume that there is not a person that isn't hit. That means that each person get hit by 1 bullet, and each person shoots a different person. We can form pairs of people that shoot each other and remove them from the 2n+1 people completely, leaving us with 2k+1 people. Then name them A_1, A_2,...,A_2k+1 such that A_1 shoots A_2, A_2 shoots A_3, ... A_2k+1 shoots A_1. Call the distance between A_i and A_i+1 D_i. We get that D_1 > D_2 > D_3 > ... > D_2k+1. But this implies that the distance between A_2k+1 and A_1 is less than the distance between A_1 and A_2, so A_1 would shoot A_2k+1. Contradiction.[/hide]

EDIT - the person above beat me to it.

radagast -- where did you use the fact that you're dealing with an odd number of people?  If you didn't use it, you aren't correct.  You may have used it in a subtle way that I missed, though.

Tetrahedr0n -- I would only suggest that you've left out the possiblity that some even number of people, greater than 2, all wipe each other out.  Not that it really affects your argument.  It basically just breaks down into "by part d," actually.

[quote]Tetrahedr0n -- I would only suggest that you've left out the possiblity that some even number of people, greater than 2, all wipe each other out. Not that it really affects your argument. It basically just breaks down into "by part d," actually.
[/quote]

JBL, I'm a bit confused by your statement. What do you mean by an even number of people? The problem already requires an odd number of people. Can you elaborate? Thanks.

Tetrahedr0n wrote:We can form pairs of people that shoot each other and remove them from the 2n+1 people completely...

Without knowing the result of part d yet, we might also have to cast out quadruples of people, rather than just pairs.  This is a very minor point.  Probably wasn't worth mentioning.

By the way, here is my version of this proof:

[hide]Consider the two people with the smallest distance between them.  They shoot each other.  Now consider the next shortest distance among any two people.  If it is between one of them and a third person, one of them gets shot twice and we are not left with enough bullets to kill everyone else.  Thus, assume that the next shortest distance is between two un-shot people.  This process continues until we are left with only 1 person, who must shoot someone else, and thus goes unkilled.[/hide]

I finished them last night, so ill post it all here now (mom wouldnt get off comp last night)

a.

[hide]Assume No one can survive. Then in the group of 2n+1 people, the ones connected with the shortest edge are going to hit each other. If another different pair is connected by the second shortest edge, then we are at the same place we were after the first edge connection. This cant go on indefinitely and use all the people because there is an odd amount and 1 would live. So we have to mix it up and form connections to those already involved in battle. But everyone only gets 1 bullet, so having E_1 and E_2 attack E_3 leaves some E_i alone that will survive. Thus the contra is met.[/hide]

(Continuation) b.

[hide]If the G_i are all 60 deg, then one of 2 things could happen.

Case 1: One of the other angles is larger. In this case, the line opposite this angle vanishes because the guy using that line to attack would choose a shorter line instead.

Case 2: All of the angles are equal, and thus it is an equilateral traingle, making the lines equal in lenght, a contradiction[/hide]

(Continuation) c.

[hide]Draw AB and CD with F their intersection point (F is between A and B). A and B are the shortest path to each other, and so is C and D. Thus CD < CA and BD and AB < CA and BD. By tri ineq, CF + FA > CA and BF + FD > BD. Adding these 2 ineqs, we that AB + CD > CA + BD, but we already found that AB, CD<CA, BD, so a contradiction is found, and we see that the paths cannot cross.[/hide]

Here is another one:

In the plane, n lines are given (n>=3), no 2 parallel. Through every intersection of 2 lines, there passes at least an additional line. Prove that all lines pass through one point.

My solution (there is a more elegant one if someone wants to try getting it)

[hide]Lines L_1 and L_2 are drawn and intersect at I. L_3 is then added to intersect at I.

Now lets try to add a new line L_4 that doesnt intersect at I, but meets L_1 at I_1, L_2 at I_2, and L_3 at I_3. Now we have to take care of these new intersection points (i.e. add another line to them), so L_5 is added to satiate I_1.

This new line adds 2 more I_k, and in general, when a new line is added, because it can only take care of 1 I_k, it adds R-2 more I_k, where R is how many lines there were before the new addition.

The line cant meet more than 1 I_k because then it would be a line that has already been added.

So we have a function of line adding that increases without bound. The only way to stop it and thus keep the number of lines finite is to have all of the lines intersect at 1 point.[/hide]