On the side $BC$ of a triangle , points $B_1$ and
$C_1$ are marked so that $CC_1 = BB_1$ and $<BAB_1 = <CAC_1$.
prove that $AB=AC$

Just apply the sine rule a few times and the solutions follows trivially I think.

Call $F$ the midpoint of $BC$. Sine rule: $\frac{BB_1}{\sin{BAB_1}}=\frac{AB}{\sin{AB_1B}}=\frac{AB.AB_1}{AF.\sin{AFB_1}}$
Analoguously, we prove that $\frac{CC_1}{\sin{CAC_1}} = \frac{AC.AC_1}{AF.\sin{AFC_1}}$
So combining these two leads us to the conclusion that $\frac{AC}{AB}=\frac{AB_1}{AC_1}$, and together with the fact they have equal heights, we conclude that $\Delta ABC_1 \cong \Delta ACB_1$

$\frac{sin\angle BAB_1}{sin\angle B_1AC_1}=\frac{BB_1.AC} {B_1C_1.AB}$

$\frac{sin\angle CAC_1}{sin\angle B_1AC_1}=\frac{CC_1.AB_1} {B_1C_1.AC}$

then:

$\frac{CC_1.AB_1}{AC}=\frac{BB_1.AC_1}{AB}$

and:

$AB_1.AB=AC.AC_1$

and it means that triangles $ABB_1,ACC_1$, are simmilar to each other.

and because of $BB_1=CC_1$,

they are equal to each other and, $AB=AC$ ;)


Can anyone explain the difference between maximum and maximal? Thank you

Well, I don't know if this will really help you, but, anyway, check this out :
http://mathworld.wolfram.com/Maximum.html
http://mathworld.wolfram.com/MaximalElement.html
http://mathworld.wolfram.com/MaximalOrder.html


For any positive interger $j$, denote by $k =k(j)$ the smallest integer such that $\sum \frac{1}{i} \geq j$ ($i$ runs from $1$ to $k$). Prove that as $j$ approaches to infinity, $\frac{k(j+1)}{k(j)} $ approaches to $e$.

Hint : Use the fact that $\sum_{i=1}^{n} \frac{1}{i} - ln(n) \rightarrow \gamma$, as $n\rightarrow \infty$ and
that $k(j) \rightarrow \infty$, $j \rightarrow \infty$. So, does $j-ln(k(j))$ converge?  ;)

[quote="grobber"]This has been discussed before. I think harazi even proposed a stronger version.[/quote]
Yes, here is the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17746]link[/url].

Actualy, there's another thread, because I distinctly remember posting what I believed to be a proof of that extension of harazi's :) (nobody bothered to say whether it was correct or not :)).


ratio
geometry
trigonometry
trig identities
Law of Cosines

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively.  There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$  The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime.  What is $a^2+b^2+c^2$?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=438[/img]

[quote="joml88"]The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c}$...[/quote]
I don't understand :oops: ... how can you say the ratio of the area of one triangle to the area of another triangle is a fixed number? This would be the same as saying that the ratio of $x$ to $y$ is, for instance, $4$, instead of $1: 4$ or $4: 1$.

It's the same as saying divide [ABC] by [PQR].  So, for example, say they are 10 cm^2 and 5 cm^2, respectively.  The ratio of [ABC] to [PQR] is 10/5=2.  You could also say 2:1.  They're basically interchangeable.

[hide]Let (CD)=1 and EQ intersect AC at G

Triangle CDQ is similar to Triangle EQG because of vertical angles and a 60 degree angle in both triangles.

Let (QD)=x

then (DP)=x^2

x+x^2=1

solve for x and use law of cosines to figure out (QR)^2 and you get 83 or 7+3sqrt(5)[/hide]


MATHCOUNTS Yale college Support probability articles search

[quote]when Bush "won" the 2004 election[/quote]

Umm... is there some question about it?

[quote="nat mc"]THAT'S AWESOME!!! ( I hate Bush)[/quote]

me too

Are you saying that you actually like having a monkey for president

[quote="JesusFreak197"][quote="bubala"]http://whatreallyhappened.com/2004votefraud.html[/quote]

It's sad when people believe such obviously false claims just because they can't accept what really happened.[/quote]

What really happened is 60,000,000 voted for an idiot.
I know it's just a stupid conspiracy theory. :)  But it helps to lie to yourself.  It's very hard for me to believe that my country actually elected such a silly-goose-head.

However, I will have you note that the Republicans DO do their fair share of cheating.  I don't know if it would have swayed the election results this year because Bush won Ohio by over 100,000 votes, but if you honestly think that Bush won the 2000 election...no comment.

Anyhow, it is peculiar how white districts always seem to have the shortest lines, and lines were 4 hours long in predominantly black districts, and how the polling places seem to be harder to commute to in predominantly black districts...and those e-voting "glitches"

Not that the Dems don't every cheat, either.  I don't know any examples offhand, but I know back in the 60s and 70s and such, sometimes they'd do things--particularly in Chicago.  But today, the Republicans are by far the worst offenders.

He's brainwashed everybody (in the words of David Benjamin about Bill Gates, but now I'm saying it about the monkey)

please, nat mc, Bush is many things, but he's certainly not a monkey.

This conversation just shows the difference between Democrats and Republicans. If Kerry was elected, would I be complaining that the votes were rigged, or that the people who voted for him were total idiots, or that he looks like a horse?

But not every anti-Bush person on this thread is doing those things...and how do we know you wouldn't? Or if not you, other Bush supporters? This demonstrates nothing. Don't generalize.

except that

1) The votes WERE rigged--but I admit not enough to sway the election.  Again, that link I posted is a stupid conspiracy theory--it was just supposed to be funny.  It amazes me how people refuse to believe the fact that honest voters were disenfranchised, but hey, IT HAPPENED (and it happened in 2000, too, btw).
2) I didn't say that everyone who voted for him was a total idiot--I said that he was a total idiot.  However, many people voted for him were probably total idiots.  Frankly, a lot of people are idiots :)  Of course, many of the people who voted for Kerry were also total idiots.
3) I didn't say he looked like a horse--that was someone else.  I think he looks like a common man who everyone can relate to--unlike John Kerry--because hey, it's not like Dubya wasn't born into privilege or anything.
4) I don't know what you'd be doing if Kerry had won.  Don't ask me, ask yourself.

Wow. You really hate Bush, Babala. Whatever you post, you always relate it with Bush. :rotfl:

[quote="nat mc"]looks like one smells like one talks like one acts like one
maybe I exaggeratd a litttle[/quote]

how does he smell like a monkey

[quote="xtrememath8"][quote="JesusFreak197"][quote="bubala"]http://whatreallyhappened.com/2004votefraud.html[/quote]

It's sad when people believe such obviously false claims just because they can't accept what really happened.[/quote]

That is a conspiracy theory. Looking back on it it may seem unlikely Bush would be reelected, but Kerry made some seriously damaging comments in the presedential debates. His shot at Bush's daughter turned the media against him for a while, and that was very close to voting day. And I think name-calling is too trivial for politics. Bush may very well be a monkey, but calling him that in a political debate adds nothing. And though you may disagree with his actions with Iraq, Making a strong statement about it is pointless. Bush had to make a decision. He had a clear understanding of the politics involved. He had all the information and statistics. We don't. It's easy to accuse the president of making a bad decision, but as bush doesn't read off all his top secret Iraq info to America, most accusations are unfounded.[/quote]

Yes, we've already discussed that it's a stupid conspiracy theory.

Kerry won all the Presidential debates!  He was a much better candidate.  Kerry's mistakes were really at the Democratic National Convention.  He didn't attack Bush at all; he just said good things about himself.

On the other hand, at his Convention, Bush tried to scare the heck out of the American people and give them the idea that voting for Kerry would mean we would get hit again.  BS.  "Terrorism," "Danger," and "Kerry" were said hundreds and hundreds of times during the convetion.  They even held their convention in our backyard--New York--to show the idea that he was the "terrorism" President.  The fact is, most of America needs not worry about terrorism.  They're not going to attack Omaha, Nebraska.  If they attack again, they'd attack Manhattan most likely.  And Manhattans chose who was best to protect them: Kerry 87% to 13%.

Second, what you seem to be advocating is blind patriotism boarding on nationalism.  "Most accusations are unfounded."  Are you kidding?  I really hope so.  One of the founding principles of our great nation is freedom of speech and stuff like that.  The people are supposed to question the government.  Unfortunately, through "terrorism" and other mumbo-jumbo, Bush has convinced most of the country to blindly follow everything he says and not question anything.  This is dangerous!!  Remember what happened when everyone in Germany let Hitler do whatever he wanted to?  (Just for the record, that's just an example.  I know Bush is nothing close to an evil man like Hitler and he's not going to start WWIII, but he has started an unnecessary war already, and I wouldn't be surprised if he starts the "Francian (as he calls it)" War in 2007.

Saying that he had a clear understanding of the politics involved is true.  There were no Weapons of Mass Destruction.  There was no link to Al Qaeda (This was verified by the 9/11 commission led by a REPUBLICAN who Bush himself appointed!).  Bush wanted to go to war, and he did.  It was that simple.  However, he understood that the country would follow him because he was "protecting" America by going into a country, getting Arabs mad at us, fueling terrorist causes while ignoring Al Qaeda and the search for Bin Laden ( he didn't say this, of couse).   He got support by lying to the American people.  So yeah, he had all the top-secret documents, and then did exactly what he wanted to, regardless!

Seriously, though, questioning the government is supposed to be encouraged.  That's what our country is *supposed to be* all about--the power of the people.  Blind loyalty is not the answer.

No, blind loyalty is not the answer, but I think our opinions about large issues like bush declaring war are hindered by our lack of understanding about the situation. Yes, Kerry won the debates, but while Kerry was cold and concise, bush was more of a "real" person. America was able to connect with him more then Kerry, and that is a factor. And Kerry's shot at bush's daughter did turn media against him for a while. I'm not saying we can't have an opinion, but attacking bush for his larger decisions is unfounded without proper understanding and perspective. And the war was not started unneccesarily. Hitler started war for power, bush started was for protection. We were attacked first, during 9/11.

I personally think Bush is an idiot. That's why I didn't vote for him. He may have a speech impairment, but that doesn't make up for his lack of knowledge in foreign affairs and of different languages which create diplomatic resolve.

I voted for Kerry, the lesser of two evils. What's interesting is how the majority of peopel I come across actually hate Bush. Perhaps there needs to be another vietnam so we can push through the stupid american citizens heads what war really is. Maybe that will make them hate Bush. It was sickening to find out that U.S. Military troops liked him because he would give them more jobs. No duh, when you start more war, you need more jobs.

Sadly only 30% of America has a four year degree.
One out of five people drop out from highschool.

That leaves a population of 80% of people who could actually make an intelligent vote.
Since not everyone votes and most people don't care about the president there is a lack of voting for the more positive side.

Anyways, anyone who truly believes in Jesus wouldn't support war or want someone in office who did.

I refuse to accept anyone's opinion here unless they came from Africa, South Korea, or Iraq.

All other countries have a crappy grasp American government.

I also agree that such things as the money and soldiers could have done better things and the education in America could increase. Seeing as how this country is usually run by military expedentures(?sp) and income through taking over foreign lands, I see this war as purely idiotic. It's far away from America there's no reason to be there.

From what I've been told from ex-military is that it's a very huge drug scene. There's a lot of underground drug trafficking and that's why we are there. Not just because Bush wants to take over the country, but because he wants to get his hands on anything they have of value and seize it for himself. By doing these things, he makes a country go broke.

If this were a real war, I would expect more casualities. From the numbers I know of, this seems like terrorist work, and terrorists usually do work with either drugs or weapons.

[quote="MithsApprentice"][quote="xtrememath8"]No, blind loyalty is not the answer, but I think our opinions about large issues like bush declaring war are hindered by our lack of understanding about the situation.[/quote]

Yes, thank you. That's exactly what many people don't understand. Just because the President is not always able to back up his decisions by publicizing every bit of intelligence that they were working with, doesn't mean he has acted irrationally. And I remind you that the largest decision-making factor that Americans cited was "moral values".

Oh and uh...bubala, if you're in 8th grade and you haven't started puberty...uh...okay yeah no comment. By the way, under no circumstances can any department in the US lock you away at Guantanamo without a trial given that you are a legal permanent resident. Now if you're not aware of your own legal rights, that's your problem. Oh and btw, no idiot would have any sort of government agent "check out" a person for making a post on an online forum referring to a hypothetical event concerning a candidate that was never elected. By the way, even if such an event did ocur it wouldn't be the Secret Service taking care of it.

Do question the government. Do so all you want. Don't insinuate that others are "blind" because they trust that the President they elected wouldn't purposefully send the nation to war. Kerry had the same information from Bush did and from the same source. He supported the war.[/quote]

I said I haven't "gone through" puberty.  IN other words, I haven't [i] finished [/i].  Of course I've started a while ago.  LOL  However, female gymnasts who basically go through child abuse in order to win their gold medals and whatnot rarely start puberty until after they retire--in their late teens or early twenties.

Also, many Democrats say "Oh, well, we trusted the President and he went to war the wrong way."  I agree with you that that's bs.  Their votes were purely political.  I'm ashamed of them. :mad:  Kerry made a huge mistake by voting for the war.  The mistakes that President Bush made, however, are far worse (like going to war without letting the inspector's finish their job and without a plan--that was really smart).

How do you konw the President would never send the nation into war when he didn't have to?  Name one reason why the war was necessary.  WMD? Nope.  Connection to 9/11?  Nope.  Defending national security?  Nope.  In fact, Al Qaeda attacked us, not Saddam.  By going after Saddam, we took the focus off the real war on terror.  Mistake.  (And by the way, what makes you trust the President?  Have you ever met him to verify he's a good person?  Look at the things he's done in office.  What has he accomplished for America?  "No Child Left Behind"  "Clear Skies"  "Healthy Forests" (cough) ? )

On the topic of why there aren't many people voting, some people just don't want to follow the election so closely. Maybe they're just not interested in the election. If that's their fault, so be it. Go phone them and yell at them for being an embarrassment to America.

[quote="bubala"]However, female gymnasts who basically go through child abuse in order to win their gold medals and whatnot rarely start puberty until after they retire--in their late teens or early twenties.  [/quote]

Yup, today's early favorite for the "Fact that we really didn't have to know" prize. But, the day's only 27 minutes old, so who knows what kind of comments are waiting out there.

Personally, I don't like most politicians. Even though I don't like Bush at all, it's not like I favor Kerry a lot either. Kerry has lied to the public, so has Bush. Kerry has done some things that would shock most of America if they knew....and so has Bush. What America needs most is politicians that care more about the country than their reputation.

[quote="Cyberman"]Anyways, anyone who truly believes in Jesus wouldn't support war or want someone in office who did.

I refuse to accept anyone's opinion here unless they came from Africa, South Korea, or Iraq.

All other countries have a crappy grasp American government.

I also agree that such things as the money and soldiers could have done better things and the education in America could increase. Seeing as how this country is usually run by military expedentures(?sp) and income through taking over foreign lands, I see this war as purely idiotic. It's far away from America there's no reason to be there.

From what I've been told from ex-military is that it's a very huge drug scene. There's a lot of underground drug trafficking and that's why we are there. Not just because Bush wants to take over the country, but because he wants to get his hands on anything they have of value and seize it for himself. By doing these things, he makes a country go broke.

If this were a real war, I would expect more casualities. From the numbers I know of, this seems like terrorist work, and terrorists usually do work with either drugs or weapons.[/quote]

What about Charlemagne? What about the Crusades? Admittedly, these are dated references, but that first comment is a sweeping generalization.

It is my opinion that it is a bad thing to categorically disregard the opinions of everyone except individuals from Africa, South Korea, or Iraq.  Plenty of people from elsewhere have good heads on their shoulders. Whether you agree with them is totally your decision. Whenever possible, I even try to expose myself to views differing from my own, at least to get a more complete picture if not alter my thoughts.

As for America being run by military expenditures, I assure you that this at best an incomplete representation of the federal budget. The overwhelming majority of money spent by the government is uncontrollable spending on entitlements (Social Security, Medicare, Medicaid, etc.) As for discretionary spending, yes, the military is the largest line item. Without the war, the hundreds of billions of dollars in emergency appropriations would have been saved. As for income through foreign lands, America is running a trade deficit, and the numbers in that department pale in comparison to the $2,600,000,000,000 annual budget.

As for wanting to make countries go broke, has the Bush administration not been visibly involved in international negotiations to cover billions of dollars of debts owed by various nations? Moreover, the nations of the G8 summit recently promised increased aid to Africa.

Finally, the military activity in Iraq hardly constitutes a real war. It is more of an occupation - one that I don't see ending soon I might add. Despite their fanaticism, the insurgency does not have the capacity of mounting conventional assaults on the U.S. Army. Thus, casualties are low when compared to "real war." What have been accumulating, however, are casualties deriving from the widespread use of crude homemade bombs.

...

That said, Bush has endorsed some policies that I do not particularly like.

[quote]Sadly only 30% of America has a four year degree.
One out of five people drop out from highschool.
That leaves a population of 80% of people who could actually make an intelligent vote.
[/quote]

I think this is a bleak, snobbish way to look at our population.  This attitude makes people hate the "liberal elite."  Not everyone will be blessed with the education that you have, and in a capitalist society, not everyone should.  Just because all your peers will graduate high school and most likely college, doesn't mean that everyone else is an idiot and should be looked down on.

People voted for Bush out of fear.  Preachers in the conservative Christian community(the Christocrats) are getting more and more aggressive.  In some communities where, for example, believing in evolution is a sin, the church community has more control over people's lives than you can even imagine.  When God comes before all else in your life, a preacher's word can become the law, because they are the closest connection to the afterlife to many people.

In addition to this, the media has been scaring the crap out of us since 911.  In the two or three years immediately after Iraq, how many obnoxious, false, vauge alerts did we experience?  Bush has done an excellent job of presenting himself has the resolute, confident choice.  As bull-headed and damaging as his policies are today and especially for future generations, he is extremely consistent, and I think Americans are searching for this in these "unsure times", which is phrase that the neoconservatives have been stressing.

Calling people who voted for Bush stupid is not going to solve anything.   It results in bitterness, hatred, and division.  The worst part of this last four years is the drastic polarization of our country.  In my 30 mile commute between my house and Alpine, I start in a flag-waving navy community, pass through one of the most liberal areas of San Diego, and end in a town reminiscent of rural Texas.  These separate peoples don't meld their ideas together and don't understand each other, and only suspicion and mistrust results.  What will keep this from happening again (i.e. keep Bill Frist out of the White House) is the free spread of ideas and flow of information, so people can make decisions through logic and reason, free of other mind-clouding influences.  This won't happen overnight, and as long as we have the cycle of power/money between the corporations, media, and the politicians, it will be squelched.

He [i]probably[/i] won fair in 2004 (defiently not in 2000). He won because of several reasons. He makes it sound like he is extreemly religious, giving him an excuse to start a war just to get oil. He provides taxcuts mainly only to the weathyest 1%.  He lies fluently that everything is going perfectly in Iraq, and that the economy is doing great. And, he has  :furious: Carol Rove :furious:  makeing the other candidate sound like a devil. In this case, he made a war hero sound like a draft dodger and a draft dodger sound like a war hero. There is also a ton of other stuff I'm not going to get into. (Oh, did you hear about :furious: Rove :furious: leaking that CIA name?)

[quote="xtrememath8"][quote="wdragyn"]In addition to this, the media has been scaring the crap out of us since 911. In the two or three years immediately after Iraq, how many obnoxious, false, vauge alerts did we experience? [/quote]

America needs to be cautious, even bordering on paranoid. Mabey if we got more "obnoxious, vaugue" alerts three years ago, 9/11 could have been avoided. Yes, we've been careful since 9/11, but terrorism is a real threat. It isn't much to ask to watch news about possible terrorist actions, when the goverment's actions decrease the chances of another 9/11.[/quote]

No...vague warnings would not have stopped 911.  What would have stopped 911 would be investment in domestic security (especially airports) and recognizing and acknowledging the threat of Al Qaeda before it became worthwhile for Bush (i.e. political advantage).  Telling the airport security about the August 6 brief Bush didn't read about bin Laden being determined to attack in the US would have helped.  The people who died on the planes didn't let the hijackers through security.
Saying that there might be a terrorist attack in America at some time in the near future doesn't prevent terorrist attacks from happening.  It creates fear, which makes people more willing to give power to the government and to not question them (Patriot Act).  It also gets people to watch more Fox News, so they continue to produce factless crap.  Paranoia is [b]never[/b] positive.  We stop thinking and we lose control of logic and reason, which is the foundation of our democracy.

I wasn't referring to the media, but to the caution of the government. And a border on security paranoia (for the government, if not the citizens) does make a terrorist attack less likely.

its all bush :diablo:'s fault  :cursing:  :bomb:  :furious:


I want to do a science fair project that's related to math... but I'm not sure what's a good topic to look into.

I did a project on fractals and chaos theory once, and it was pretty fun.

I want a topic that is interesting and applicable... can you guys help me out?

Thanks in advance!

Hm, i want something more "math-like" than physics.

Thanks for the suggestion though.

An interesting/non-trivial extension to a USAMTS problem would be a good  scifairproj. Or you can check out the lowest article on http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php, which was my 9th grade science fair project. It took me to ISEF.


superior algebra
superior algebra unsolved

Prove that for finite field with $ q$ element, the monomial $ f(x) \equal{} x^d$ is one to one iff $ \gcd(d,q\minus{}1)\equal{}1$.

More general (and posted before):
Let $ G$ be a finite group. Then $ x \mapsto x^k$ is a bijection iff $ \gcd(|G|,k) \equal{} 1$.

Can you be more specific?Give a complete solution :D

If some prime $ p|\gcd(k,|G|)$, then by [url=http://en.wikipedia.org/wiki/Cauchy%27s_theorem_%28group_theory%29]Cauchy[/url] we find some $ 1\not\equal{}y\in G$ of order $ p$.
Then $ y^k\equal{}1\equal{}1^k$ gives the map $ G\to G,x\mapsto x^k$ not injective.

If $ \gcd(k,|G|)\equal{}1$, then $ uk\plus{}v|G|\equal{}1$ for some $ u,v\in \mathbb{Z}$
and by [url=http://en.wikipedia.org/wiki/Lagrange%27s_theorem_%28group_theory%29]Lagrange[/url] $ x^{uk}\equal{}x^{1\minus{}v|G|}\equal{}x$ for all $ x\in G$,
so $ G\to G,x\mapsto x^u$ is the inverse map to $ G\to G,x\mapsto x^k$.


Newton Middle School has a hall with 500 lockers, all of which are closed. Five hundred students start down the hall. The first student opens every locker. The second student closes all the lockers that are multiples of 2. The third student changes (closes an open locker or opens a closed one) all multiples of 3. The fourth student changes all multiples of 4. And so on. After all students have finished with the lockers, how many lockers will be open?

basically a variation on 1000 lockers.  see my explanation for 1000 lockers

1^2 through 22^2
22 lockers opened

[quote="DPopov"]Every locker that is a perfect square will remain open. This is a question that is often asked on many contests.[/quote]

It would be great if you can post an explanation of why this is so.

I think this was also on the Putnam competition once, too.

[hide]
The way I did it was to consider 1 - 20 instead of 1 - 100.
0 - closed
1 - opened
Locker 1:0,[b]1[/b]

Locker 2:0,1,0

Locker 3:0,1,0

Locker 4:0,1,0,[b]1[/b]

Locker 5:0,1,0

Locker 6:0,1,0,1,0

Locker 7:0,1,0

Locker 8:0,1,0,1,0

Locker 9:0,1,0,[b]1[/b]

Locker 10:0,1,0,1,0

Locker 11:0,1,0

Locker 12:0,1,0,1,0,1,0

Locker 13:0,1,0

Locker 14:0,1,0,1,0

Locker 15:0,1,0,1,0

Locker 16:0,1,0,1,0,[b]1[/b]

Locker 17:0,1,0

Locker 18:0,1,0,1,0,1,0

Locker 19:0,1,0

Locker 20:0,1,0,1,0,1,0

As you can see, all the ones that are open at the end are perfect squares.
[/hide]

Now I'll explain why.

[hide]
After the first student, every locker is open. In order to end up open, the locker must be opened/closed an odd number of times. This means it has to have an odd number of factors.

$\paragraph{Lemma 1.}$ If a number $n$ has an odd number of factors, then $n$ is a perfect square.

Proof: Factors come in pairs. If a number has an odd number of factors, one factor must be unpaired. The only way this can happen is if that factor is multiplied by itself. In other words, the number is a perfect square. For example, $16=(1,16),(2,8),(4,4)$. 16's factors are $1,2,4,8,16$. It has an odd number of factors because 4 is unpaired-it is multiplied by itself. $\blacksquare$ QED.

Tell me if that's confusing. BTW, yes, I did write that solution myself.

[/hide]

Cool, I thought of another explanation.

$n=p_1^{q_1} \cdot p_2^{q_1} ...$.
$f=(q_1+1)(q_2+1)...$

Where $f$ is the number of factors. If $f$ is odd, or not even, $q_1+1$, etc. must be odd. So $q_1$, etc. is even. If the exponents are even, $n$ is a perfect square. $\blacksquare$ QED.

Is this one confusing?

Chess, I think that's a great explanation!  Nicely done.  (I was referring to the first one -- the second one is cool too, but definitely more confusing.)


Let $ A,B,C,B,E,F \in M_n{(\mathbb{C})}$
Prove that Inequality rank of matrix :
$ 1$)
$ rank(A)rank(B) \plus{} rank(C)rank(D) \leq rank(AC)rank(BD) \plus{} n^2$
$ 2$)
$ rank(AC)rank(BD) \plus{} rank(CE)rank(DF) \leq rank(C)rank(D) \plus{} rank(ACE)rank(BDF)$
 :w00tb:  :w00tb:  :w00tb:

[b]Lemma 1)[/b]  [color=blue]Loup blanc[/color] was proof I want give a different solution.
 $ X,Y \in M_n{(\mathbb{C})}$ then $ rank(X\otimes{Y}) \equal{} rank(X)rank(Y)$
[b]lemma 2)[/b] Let $ X,Y,Z \in M_m{(\mathbb{C})}$
then  $ rank(X) \plus{} rank(Y) \leq rank(XY) \plus{} m$ and $ rank(XY) \plus{} rank(YZ) \leq rank(Y) \plus{} rank(XYZ)$
 [b]Proof lemma 1)[/b] Let $ I_k$ is $ k$x$ k$ identity matrix ,$ k \in \mathbb{N^{*}}$
Put $ rank(X) \equal{} r,rank(Y) \equal{} s$ then the exist $ (P,Q) \in GL_n{(\mathbb{C})}xGL_n{(\mathbb{C})}$,$ (R,S) \in GL_n{(\mathbb{C})}xGL_n{(\mathbb{C})}$
 Such that : $ A \equal{} P\begin{pmatrix}I_r & O \\
O & O\end{pmatrix}Q$,$ B \equal{} R\begin{pmatrix}I_s & O \\
O & O\end{pmatrix}S$
 We have $ A\otimes{B} \equal{} (P \otimes R)( \begin{pmatrix}I_r & O \\
O & O\end{pmatrix}\otimes{\begin{pmatrix}I_s & O \\
O & O\end{pmatrix}})(Q \otimes{S} )$
By $ (P \otimes R) \in GL_{n^2}{(\mathbb{C})},(Q \otimes{S} ) \in GL_{n^2}{(\mathbb{C})}$
Then $ rank(A\otimes{B}) \equal{} rank(( \begin{pmatrix}I_r & O \\
O & O\end{pmatrix}\otimes{\begin{pmatrix}I_s & O \\
O & O\end{pmatrix}})) \equal{} rs$
 [b]Proof lemma 2)[/b]
 Use $ Ker(X|_{Im{Y}}) \subset Ker(X)$,$ Ker(X|_{Im{(YZ)}}) \subset Ker(X|_{Im{Y}})$ , by $ Im{Y} \subset M_{n1}(\mathbb{C}),Im(YZ) \subset Im{Y}$
Hence $ \dim(Ker(X|_{Im{Y}})) \leq \dim(Ker(X))$ or $ rank(Y) \minus{} rank(XY) \leq m \minus{} rank(X)$
 and     $ \dim(Ker(X|_{Im{(YZ)}})) \leq \dim(Ker(X|_{Im{Y}}))$ or $ rank(YZ) \minus{} rank(XYZ) \leq rank(Y) \minus{} rank(XY)$
 The end.
 We have :
1) $ rank(A \otimes{B}) \plus{} rank(C \otimes{D}) \leq rank((A \otimes{B})(C \otimes{D})) \plus{} n^2 \equal{} rank((AC) \otimes{(BD)}) \plus{} n^2$
 Or $ rank(A)rank(B) \plus{} rank(C)rank(D) \leq rank(AC)rank(BD) \plus{} n^2$
2) We use lemma 2)
 $ rank((A \otimes{B})(C \otimes{D}) \plus{} rank((C \otimes{D})(E \otimes{F}) \leq rank((C \otimes{D})) \plus{} rank((A \otimes{B})(C \otimes{D})(E \otimes{F}))$
  $ rank((AC)\otimes{(BD)}) \plus{} rank((CE)\otimes{(DF)}) \leq rank((ACE)\otimes{(BDF)})$
  $ rank(AC)rank(BD) \plus{} rank(CE)rank(DF)\leq rank(C)rank(D) \plus{} rank(ACE)rank(BDF)$
Easy from 1) If $ B$ , $ D$ $ \in GL_n{\mathbb(C)}$ we can find lemma 1) and
        from 2) If $ BDF \in GL_n{\mathbb(C)}$ we can find lemma 2)


inequalities
pigeonhole principle
inequalities proposed

Let a,b,c be real numbers such that $a+b+c=0$.
Prove that $\ (|a|+|b|+|c|)^2\geqq 2(a^2+b^2+c^2).$
When does equality holds?

Because of $a+b+c=0$ you can rewrite the inequality as follows:

$(|a|+|b|+|a+b|)^2\geq 2(a^2+b^2+(a+b)^2)$ which implies:

$|a||b|+|a+b|(|a|+|b|)\geq a^2+b^2+ab$

Now I consider 2 cases: $ab\geq0$ and $ab<0$:

First case: the left side is greater or equals to $|a||b|+(a+b)^2\geq a^2+b^2+ab$. The second inequality follows from $2ab\geq 0$. Equality holds for $ab=0$ so for $a=0,b=c$ and for every cyclic premutation of this statement.

Second case: we can assume $a<0$ and $b\geq 0$ because the inequality is symmetric in $a$ and $b$. The inequality becomes:

$-ab+|a+b|(b-a)\geq a^2+b^2+ab$ which is the same as

$|a+b|(b-a)\geq(a+b)^2$ and $b-a\geq |a+b|$ (because of $(a+b)^2=|a+b|^2$ and $|a+b|\geq0$). In this case it is the same as $|a|+|b|\geq|a+b|$. Th equality cann't hold in this case because of $a<0$.

Misha

You are right ,Misha123.

I will post my solution later.

kunny

I suppose you solution is shorter than mine.

Misha

Another solution...

By the pigeonhole principle, (at least) two of the numbers a, b, c have equal sign. WLOG assume that two such numbers are b and c. Then | b | + | c | = | b + c |. Since a + b + c = 0, we have | b + c | = | - a | = | a |. Hence, | b | + | c | = | a |. Denoting | b | = x and | c | = y (of course, x and y are positive), we get | a | = x + y. But the inequality to be proven,

$\left( \left| a\right| +\left| b\right| +\left| c\right| \right) ^{2}\geq 2\left( a^{2}+b^{2}+c^{2}\right) $,

can be rewritten as

$\left( \left| a\right| +\left| b\right| +\left| c\right| \right) ^{2}\geq 2\left( \left| a\right| ^{2}+\left| b\right| ^{2}+\left| c\right| ^{2}\right) $,

i. e. as

$\left( \left( x+y\right) +x+y\right) ^{2}\geq 2\left( \left( x+y\right) ^{2}+x^{2}+y^{2}\right) $,

what is clear since

$\left( \left( x+y\right) +x+y\right) ^{2}-2\left( \left( x+y\right) ^{2}+x^{2}+y^{2}\right) =4xy\geq 0$.

The equality case is left to the reader (I'm lazy).

  Darij

Thank you for your reply,Darij.

Your solution is one of the solution.
I will post my solution.

kunny

Kunny, please post you solution. I would like to see it.

Misha

Hello!      Misha 123,darij

I am sorry to have kept you waiting for you.

I will show you  my solution as follows.

Since $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)$ and for real number $A$,$|A|+A\geqq0$,equality holds when  $A\leqq0$,

we have $(|a|+|b|+|c|)^2-2(a^2+b^2+c^2)=2|ab|+2|bc|+2|ca|-(a^2+b^2+c^2)$

$=2(|ab|+ab)+2(|bc|+bc)+2(|ca|+ca)\geqq0$ Q.E.D

The equality holds when $ab\leqq0$and$bc\leqq0$and$,ca\leqq0$,then $ab\cdot bc\cdot ca=(abc)^2\leqq0$,as $(abc)^2\geqq0$ yielding $abc=0$

kunny


function
topology
real analysis
real analysis unsolved

Let $ E$ be the set of all real functions on $ I\equal{}[0,1]$. Prove that one cannot define a topology on $ E$ in which $ f_n\rightarrow f$ holds if and only if $ f_n$ converges to $ f$ almost everywhere.

Let
\[ I_{2^m\plus{}k}\equal{}\left[ \frac{k\minus{}1}{2^m}, \frac{k}{2^m} \right] \ \ (m\equal{}0, 1, 2, \ldots; k\equal{}1, 2, \ldots, 2^m),\]
and denote by $ f_n$ the characteristic function of $ I_n$. Then $ \{ f_n \}$ will nowhere converge to 0;but if $ \{ f_{n_k} \}$ is an arbitrary subsequence and $ x_{n_k} \in I_{nk},$ where $ \{ x_{n_{kl}} \}$ conberges to a number $ x \in I$, then obviously $ \{ f_{n_{kl}} \} \to 0$ everywhere but at $ x$, so in particular, almost everywhere. So if $ T$ is a topological space and $ x_n \not \to x$, $ x_n \in T$, $ x \in T$, then there exist, a neighborhood of $ x$ such that for some subsequence $ x_{n_k}, x_{n_k} \not \in V$, and hence for any subsequence $ x_{n_{kl}}, x_{n_{kl}} \not \to x$.


Hi.

Could anyone point me out some demography models webpages?

Or, could someone explain to me what it is?

Thanks.


[quote="mathemonster"]prove that the largest number you can make by repeatedly summing two number m and n,
is given by n x m - (n + m)

for example: if m = 2 and n = 5 then you can make 9 by 2 + 2 +5, and the largest number you can't make is 3.

Have fun![/quote]The wording is slightly off but to be precise, this is the famous AoPS theorem: the Chicken McNugget Theorem. This has been proved plenty of times on AoPS.

Then can u show me where i can find a proof

[url]http://www.mathlinks.ro/Wiki/index.php/Chicken_McNugget_Theorem[/url]

AoPSWiki is our friend... :lol:

The following is also really good: [url]http://mathforum.org/library/drmath/view/62084.html[/url].


geometry
blogs
number theory
prime factorization

Develop a complete theory for
\[x^{2}+2y^{2}= z^{2}\]
By complete theory, something like the $(r^{2}-s^{2}, 2rs, r^{2}+s^{2})$ for pythagorean triples.

What is $\mathbb{Z}[\sqrt{-2}]$ called?  I've seen this notation before, but I don't know what to research without a name...

Edit:  posted the same thing!

Looks like the field of $\sqrt{-2}$??? Could you post a more pedestrian solution ?  :wink:

elementary solution (yo yo wtf is z(rt-2) HUH)

[hide]

We'll generate prime triplets, so assume that there is no common divisor for (x,y,z).

z must be odd. To show that this is true, suppose that it is even; then the LHS must be even, but this is only true if both x and y are even, and then 2 is a common divisor for our triplet, and this contradicts our original statement. So z is odd.

x is odd. To show this, suppose it is even. Then the left hand side is even, and z^2 is odd, so x cannot be even.

y is even. To show this, suppose it is odd; then the LHS is 3 mod 4, but the RHS must be 1 mod 4, so by contradiction y must be even.

Let . Then we have . Since z and x are odd, z+x and z-x are even, so we can rewrite this as . Let  and . We'll show that  and  are coprime.

Assume that , . Then  and , but if both  and  had some common divisor than  must also be divided by , and then we do not have a primitive triplet, so , and .

Since  and , then we can infer from the statement  that  and  or  and . Both of these yield the form , although one has the two variables switched; this can be ignored. Consequently, , and it follows that . Since these generate the primitive triples (by our beginning assumption), we need only multiply by some factor  to get all nonprimitive triples. Thus, .

[/hide]

$\mathbb{Z}[\sqrt{-2}]$ is the ring $\{a+b\sqrt{-2}: \ a,b\in\mathbb{Z}\}$. boxedexe seems to have forgotten that this is the HIGH SCHOOL forum. Granted, while there are high school programs that might delve into this area, there are also more elementary solutions to this problem.

yea i bet like half the people have no clue what a ring is

or maybe that's just me

[quote="boxedexe"]Use $\mathbb{Z}[\sqrt{-2}]$ to get $(x,y,z)=(u^{2}-2v^{2},2uv,u^{2}+2v^{2})\in\mathbb{Z}^{3}$.[/quote]
could you please explain it? and obviously the second solution is $(x,y,z)=(2r^{2}-s^{2},2rs,2r^{2}+s^{2})$

[quote="paladin8"][quote="chess64"]so err this $\mathbb{Z}[\sqrt{-2}]$ thing... does it generalize?[/quote]

For the most part, though in some rings you lose unique prime factorization, so the proof would fail I think. One that works, for example, is $\mathbb{Z}[\sqrt{-1}] = \mathbb{Z}[i] = \{a+bi | a, b \in \mathbb{Z}\}$, our very own Gaussian integers. You can use this to solve the original equation $a^{2}+b^{2}= c^{2}$.[/quote]

For example, unique prime factorization fails in $\mathbb{Z}[\sqrt{-6}]$ because $(2+\sqrt{-6})(2-\sqrt{-6}) = 2*5$.

I've seen this crop up reasonably often enough that I wrote a blog post about it.  Link in sig :)


Prove that there exists a triangle which can be cut into 2005 congruent triangles.

[quote="mecrazywong"]yptsoi has an ingenious design for this problem, so I'll leave this for him :D[/quote]
 :blush: It is not ingenious at all... but anyway here is my solution:
$2005=41^2+18^2$, so a right-angled triangle with adjacent sides lengths $1681$ and $738$ can be cut into $41^2$ congruent right-angled triangle with adjacent sides lengths $41$ and $18$. Similarly a right-angled triangle with adjacent sides lengths $738$ and $324$ can be cut into $18^2$ congruent right-angled triangle with adjacent sides lengths $41$ and $18$. Combining these two triangles gives another big right-angled triangle which can be cut into $2005$ congruent triangles. Therefore such triangle exists.

If I only realized that 2005=41 2 +18 2 .....

Nice solution, nevertheless, though I believe this is about the only way to solve the problem...

[quote="Severius"]Also $2005=39^2+22^2$.

To find these you can use Gaussian integers (although it is something I realized the next day...):

$2005=5 \times 401=(2+i)(2-i)(20+i)(20-i)$

Now multiplying out in this fashion: 1st-3rd, 2nd-4th you get the pair $39, 22$, while doing 1st-4th, 2nd-3rd gives $41, 18$.[/quote]

Severius: this is a nice method to decompose an integer as the sum of two squares, but it works only for very special cases: for example you cannot apply it to prove that $73=8^2+3^2$ and $97=9^2+4^2$. Do you know a general algorithm which allows us to decompose an integer as the sum of two squares.

[quote="Darkseer"]A number can be expressed as the sum of two squares if and only if it does not contain prime factor of the form 4k+3

Still it is clearly possible for 3 or 6. I don't know whether that is possible for 7 or not.[/quote]

Not quite. A number can be expressed as the sum of two squares iff every prime factor of the form 4k+3 is raised to an even power.

yes, that's right... and to al.M.V., i don't think it was necesary to realize that $2005=41^2+18^2$.. Just make the construction in general (i.e., for $n=a^2+b^2$) and then since $2005=4k+1$ it can be written in the form $a^2+b^2$ for some $a,b$, so the triangle exists (it didn't say you had to put an example of such triangle.. just show its existance) ...

That's right, billzhao. Thanks for pointing out my mistake.

By the way, 21=4*5+1 can not be expressed as the sum of two squares.

yes sorry... i wanted to say that the prime divisors were of that form (2005 is not a prime!!! ahh!!  :D ), $2005=5.401$ (and hope 401 is prime!) ...

[quote=David_Kim_0202][quote=john0512]2005 is 5*401, and both 5 and 401 are 1 mod 4, so 2005 is a sum of two squares.

Consider an acute triangle $\triangle ABC$. Let $D$ be the foot from $A$ to $BC$. Split right triangles $\triangle ABD$ and $\triangle ACD$ into $b^2$ and $c^2$ smaller similar triangles respectively such that $b^2+c^2=2005.$[/quote]

how do you know those triangles are congruent?[/quote]

oh nvm I misread the problem


What is the positive difference between the two greatest prime factors of 2730?

2730 is:
$ 2*1365$
$ 2*5*273$
$ 2*3*5*91$
$ 2*3*5*7*13$
the 2 greatest prime factors are 7 and 13, which have a difference of $ \boxed{6}$


calculus
integration
calculus computations

Just give a minor hint and dont solve it completely

Famous problem - I'm sure it's been posted, and I've used it for my problem-solving class. Hints, you say?

(a) Integrate by parts to set up a recursion.

(b) Long division.

Yes, You had posted certainly the very problem before. I have it.

The first two parts are done

Can anyone help prove the last part?

[quote="debanik2"]could u cite the link?[/quote]

Sure,  :)

Here is: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=125923#125923[/url]


geometry
analytic geometry
graphing lines
slope
geometric transformation
reflection
rhombus

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $ 50$ miles per hour along the highways and at $ 14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $ m/n$ square miles, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n.$

What would the picture look like?
I get a circle and then there are 2 rhombuses that intersect the circle at 4 points...I think thats wrong since the answer doesnt have pi in it.

There is no circle. The shape is a star, with edges at slope $ \pm\frac{7}{24}$ or reciprocals. The shortest path to a point always includes some of the highway.
The corners are at $ (5,0)$, $ (\frac{35}{31},\frac{35}{31})$, and reflections of these.

Alright, I thought of that (since there was no pi)
but
I get the enpoints to be (5,0) and (7/5,0) of the two rhombus-when i draw in that line of a side of a rhombus (and the negatives of them) so the lines have slopes +-7/25 and recipricals...

If he stays on the road he can get up to 5 miles.
If he goes on the prairie initially and never comes back to the road, doesnt he go 14/60*6=14/10=7/5 miles (it can never be that but it gets really close to it when you travel 0 mintes on the road and 6 minutes on the prairie)

So I got the intersection points as 35/32 (not 35/31)...

So when I did the area I was doing that triangle with base 5 and height 35/32 muiltiply that by 8 (for 8 reflections) then times1/2 since they are all tirangles,

I get 175/8.
THat answer is very close to 700/31

(the first one is 21.875 and the real answer is 22.58ish)

What did I do wrong here (theres sometrhing wrong with the (7/5,0) but I dont know what)-im pretty sure there is a circle (its when you start in the prarie in a given direction, it will have a circle  of r=7/5), I just dont know how to find the intersection between that line and the circle, that should give the equation of the line
Thanks.

Your slopes of $ \frac{7}{25}$ come from only traveling exactly north/south/east/west on the prairie. That isn't the fastest way; we can get to the same point faster by leaving the road a little earlier.

ok thanks

but how would you calculate the slopes exactly?
sorry i dont want to waste your time doing all this, but is there a way to figure out the fastest way.

The easiest way is probably calculus.

Alternately, the tangent to the "ball" should be perpendicular to the path used to get there. Draw in the right triangle, and use the ratio of the sides we know.

k thanks for the help

ill try and work on it

i know some basic calculus so i think i can do that (although i always thoguht AIME makes would make the "calculus" solution a lot more tedious)


Do grad schools prefer one degree over the other for pure math? The advantage of a BA for me is that it would allow me to take more grad-level math classes in my 3rd or 4th year. (With a BSc I would be required to take more non-math science classes.)

And which degree are the math majors here pursuing?

At my uni (Princeton) math is BA only.

It probably depends on what you want to do with your degree. If you want to go on to grad school in math and just do pure math, probably you should do the BA. If you think you might want to move into a more applied field or work in industry, you might want to go the other route.

Ah, thanks. That cleared things up greatly.

Can you explain whats the difference between Ba and BSc? Please keep in mind that I don't know what either of it means.

[quote="ktm"]The advantage of a BA for me is that it would allow me to take more grad-level math classes in my 3rd or 4th year. (With a BSc I would be required to take more non-math science classes.)[/quote]

BA = Bachelor of Arts, BSc = Bachelor of Science.  I agree with Xev.

Ahh... so pure math is considered an Art while applied math a Science??

The usual distinction between a B.Sc. and a B.A. is in what kind of elective courses you will take while you study for your degree.


abstract algebra
vector
search
linear algebra
linear algebra unsolved

Let $ V_1$ and $ V_2$ be vector spaces and $ V_1^*$ and $ V_2^*$ their respective dual spaces.
Let $ A: V_1\rightarrow V_2$ be a linear map. For  $ l\in V_2^*$ we define by $ A^*l$ the map $ l\circ A$. The tranpose map $ A^*: V_2^* \rightarrow V_1^*$ is defined by taking $ l$ to $ A^*l$.
Prove that $ Im(A^*) \equal{} Ann(Ker(A))$
where $ Ann(Ker(A))$ denotes the annihilator of $ Ker(A)$. (search wikipedia if you do not know the definition of annihilator.

[hide="Hint"]Reduce to the case $ ker(A)\equal{}0$.[/hide]

Yes, this case is easy but why can we reduce it to this case. what if A is not injective?

The standard reduction technique uses one of the homomorphism theorems such that $ A$ can be composed by a surjection followed by an injection: $ V_1\to V_1/ker(A)\hookrightarrow V_2$.


[quote="Adam4444"][quote="ckck"][quote="Adam4444"]Sometimes I see you use weird signs which I don't know the meaning of. So I just wonder if there exists any good homepage wear one can read about those signs?[/quote]
It would be a good idea to tell us what it looks like because we use many different signs.[/quote]Here is just a few examples:
$\not< \theta \phi \bigcup$
Basically I don't understand any of those greek signs and I still wonder if there exists any webpage whish list all of them with a litle explanation.[/quote]
This site has some. But I've never heard about $\bigcup$.
http://mathworld.wolfram.com/letters/
Sorry if it takes you a long time searching for the letters, its just that I don't know anywhere else.

[quote="ckck"]http://mathworld.wolfram.com/letters/[/quote]Bookmark on that one. :wink:

[quote="Adam4444"]Sometimes I see you use weird signs which I don't know the meaning of. So I just wonder if there exists any good homepage wear one can read about those signs?[/quote]
Usually, you can hold your mouse over a latex formula and it will give you the code and then you know what it's called and then you can search it on [url]http://mathworld.wolfram.com/[/url] and then you'll know what it means. Also, if it's on this site, you can post a question saying what's that mean? and most of the time you'll get an answer.

The U could mean union.  Like if set A = {1, 3, 4, 5} and set B = {1, 2, 4, 7} then A U B = {1, 2, 3, 4, 5, 7}

But usually the U for union is smaller...

According to [url]http://www-sop.inria.fr/apics/tralics/doc-b.html[/url], $\bigcup$ is a symbol for a sum of some type

 :rotfl: sum of some type.....

$\lceil{7.54^{3}}\rceil^{th}$ post

[quote="buzzer11"]Maybe it works the same as $\sum$!

$\bigcup_{i=1}^{3}{3i}$

Interesting...[/quote]
That looks very unusual.

Maybe that's not what it's for... :huh:

the $\not <$ just means not less than....
the $\theta$ usually is a variable that represents an angle measure
the $\phi$ usually does the same as $\theta$ or is used in the phi function which is not classroom level

http://www.scholarpedia.org/article/Help:Math#Special

Go to      "3. Subscripts, Superscripts, Itegrals"
/bigcup/ is in the last row of the chart.

[quote="buzzer11"]well, holding my mouse over the $\bigcup$ yields "bigcup".  :)
hmm.[/quote]

$\bigcup$ is a generalized union in the same way that $\sum$ is a generalized sum.

For example, given numbers $a_{1}, a_{2}, ... a_{n}$ then

$\sum_{k=1}^{n}a_{k}= a_{1}+a_{2}+...+a_{n}$

Well, given [b]sets[/b] $A_{1}, A_{2}, ... A_{n}$ then

$\bigcup_{k=1}^{n}A_{k}= A_{1}\cup A_{2}\cup ... \cup A_{n}$

Edit:  For those of you who are less familiar with set ideas, a set is just a collection of (different) things.  The union of two sets is just the collection of things that belong to either set.


function
complex analysis
complex analysis unsolved

Let $ G \equal{} \{z: |\text{Im } z| < \frac {\pi}{2}\}$, and suppose $ f: G\to\mathbb{C}$ is analytic and $ \limsup_{z\to w} |f(z)|\le M$ for $ w\in\partial G$. Also, suppose $ A < \infty$ and $ a < 1$ can be found such that
\[ |f(z)| < \exp[A\exp(a|\Re z|)]
\]
for all $ z\in G$.

Examine $ \exp(\exp z)$ to see that this is the best possible growth condition, can we take $ a \equal{} 1$ above?

Source Conway: Functions of One Complex Variable

Sorry, I forgot to add the statement: Show $ |f(z)|\le M$ for all $ z\in G$.


Question: If I register late for the USATMS and do only say round 3 problems and later, is that allowed? I do not necessarily wish to win or anything, I just want to do proofs and get feedback. I just found out about the program recently, and already registered online, so can I just start with the most recent problems? Thanks!

This is discussed in number 5 of the official rules which can be found [url=http://usamts.org/Rules/U_RulesComplete.php]here[/url].


Let $ A,B,C$ be a point on circle $ O$ radius 1 that $ AB \equal{} BC \equal{} CA$ , and $ P$ is any point on the plane , Suppose that $ PA \equal{} a$ , $ PB \equal{} b$ , $ PC \equal{} c$.
Prove that the triangle that have sides $ a,b,c$. The area of the triangle depends on $ PO$ only.

is simple to get that this triangle is similar to the pedal triangle of P with ratio $ \frac{2}{\sqrt{3}}$ and his area is $ \frac{4}{3} A'$ where A' is the area of pedal triangle of P that depend by PO (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=214577]here[/url])


calculus
integration
function
LaTeX
trigonometry
logarithms
calculus computations

Bad notation won't get you answers. It seems somewhat unlikely that you'd just randomly add $ 1$ to a function whose antiderivative you're asking for, but I don't really know for sure what you're asking for. Please write this in an intelligible way.

Sorry, my mistake, what's the integral (antiderivative) of xcotx?

hello, why don't you use $ \text{\LaTeX}$? do you mean $ \int x\cot(x)\,dx$?
Sonnhard.

You wanted to evaluate $ \int x \cot x dx$, right? Such integral does not exist. However, if you were asking for evaluation of $ \int_{0}^{\frac{\pi}{2}} x \cot x dx$, then the answer is as follows:
Setting $ u \equal{} x$ and $ v'\equal{}\cot x$, then $ u' \equal{}1, v \equal{} \int \cot x dx \equal{} \int \frac{(\sin x)'}{\sin x}dx \equal{} \log \sin x$.

We get: $ I \equal{} \int_{0}^{\frac{\pi}{2}} x \cot x dx \equal{} \minus{} \int_{0}^{\frac{\pi}{2}} \log \sin x dx \equal{} \frac{\pi}{2} \log 2$.

[quote="akech"]Such integral does not exist.[/quote]
Don't say that! Of course it exists, at least on any interval on which the integrand is continuous.

I guess I'm showing that I'm an analyst at heart, and I take "the integral exists" to mean what it means in analysis. Any continuous function is (Riemann) integrable, and hence by the Fundamental Theorem of Calculus has an antiderivative.

Of course there's the question, which is at heart algebra and not analysis, of whether a particular elementary function has an elementary antiderivative. That isn't asking whether the integral exists - that's asking whether you know its name.

Here's a parallel question, not about integrals. Suppose I asked you, "solve $ x\equal{}\cos x$." Would you say that the answer doesn't exist? Again, by analysis (the Intermediate Value Theorem in this case), there does exist a real number that satisfies this equation. I can approximate it closely and efficiently, but I don't know its name. Or, to put it another way, I don't know any simpler or shorter name for it other than "the real number such that $ x\equal{}\cos x.$"

hello, i have got
$ \int x\cot(x)\,dx\equal{}x\ln(\sin(x))\minus{}\int\ln(\sin(x))\,dx$
Sonnhard.

To sum up what Kent is saying, existence and succinct/closed/elementary expression are two different things.

. . . Mathworld . . . . cotangent . . . scroll down, scroll down, scroll down:

\[ \cot(x)\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n 2^{2n}B_{2n}}{(2n)!} x^{2n\minus{}1}\]

therefore:

\[ \int x\cot(x)\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^{n}2^{2n}B_{2n} x^{2n\plus{}1}}{(2n\plus{}1)!}\]

(where ever such converges)


trigonometry
function
analytic geometry
geometry
geometric transformation
rotation
linear algebra

:D

Um... can somebody explain sines,cosine,and tangents for me?
and I already no SOHCAHTOA so give me like problems with explnations and stuff

and what's TI-89 and TI-93??;;;

allyujusehyo

baesm:babo

Follow this link for lessons on this topic.

[url]http://www.regentsprep.org/Regents/math/ALGEBRA/AT2/indexAT2.htm[/url]

is it just me, or do they only have 3 of the six basic trigonometric functions?

anyways, there's another way where you take the unit circle and plot angle $ \theta$ against the $ x$-axis.  Then, where the ray hits the circle has coordinates $ (\cos \theta , \sin \theta)$. This allows you to take sine and cosine of angles that don't exist in right triangles.

Some people prefer to digest one lesson before going for more.

[cosθ -sinθ]     [x coordinate]    [new X]
[sinθ   cosθ] *  [y coordinate] = [new Y]

Rotaion Matrix!! :)


let $ t,z$ be real numbers, such that $ t < z$, prove that: $ t^3 \minus{} 3t \minus{}2 \le z^3 \minus{}3z \plus{} 2$

$ z \equal{} t \plus{} h,(h > 0)$
Then
$ \leftrightarrow t^3 \minus{} 3t \minus{} 2\leq (t \plus{} h)^3 \minus{} 3(t \plus{} h) \plus{} 2$
$ \leftrightarrow 3{h}t^2 \plus{} 3{h}^{2}t \plus{} h^{3} \minus{} 3h \plus{} 4$
Consider $ \Delta < 0$

[quote="TTsphn"]$ z \equal{} t \plus{} h,(h > 0)$
Then
$ \leftrightarrow t^3 \minus{} 3t \minus{} 2\geq (t \plus{} h)^3 \minus{} 3(t \plus{} h) \plus{} 2$
$ \leftrightarrow 3{h}t^2 \plus{} 3{h}^{2}t \plus{} h^{3} \minus{} 3h \plus{} 4$
Consider $ \Delta < 0$[/quote]

note that the inequality is $ t^3 \minus{} 3t \minus{} 2\le (t \plus{} h)^3 \minus{} 3(t \plus{} h) \plus{} 2$ and not $ t^3 \minus{} 3t \minus{} 2\geq (t \plus{} h)^3 \minus{} 3(t \plus{} h) \plus{} 2$

and what do you mean by: "Consider $ \Delta < 0$"

I want to consider $ \Delta$.Easy to check it negative.

$ \Delta\equal{}9h^4\minus{}12h(h^3\minus{}4h\plus{}4)\equal{}\minus{}3h(h^3\minus{}12h\plus{}16)\equal{}\minus{}3h(h\minus{}2)^2(h\plus{}4)\leq 0$
Do you understand? :wink:

[quote="waelalghamdi"]let $ t,z$ be real numbers, such that $ t < z$, prove that: $ t^3 \minus{} 3t \minus{} 2 \le z^3 \minus{} 3z \plus{} 2$[/quote]
$ (z^3 \minus{} 3z \plus{} 2)\minus{}(t^3 \minus{} 3t \minus{} 2) \equal{} (z\minus{}t)(z^2\plus{}zt\plus{}t^2) \minus{}3(z\minus{}t)\plus{}4 \ge \frac{(z\minus{}t)^3}{4} \minus{}3(z\minus{}t)\plus{}4\equal{}\frac{1}{4}(z\minus{}t\minus{}2)^2(z\minus{}t\plus{}4) \ge 0$

If you read my solution it is one when you let $ h\equal{}z\minus{}t$


[img]http://i40.tinypic.com/2q0o7kx.jpg[/img]

how to do part (ii)

At some point, you've proved or stated a theorem of the form, "Every cumulative distribution function has the following properties."  Then (perhaps as part of the same theorem), you've proved or stated the converse: "Every function with the following properties is a cumulative distribution function."  Go open your notes or textbook, find these results, and then check the properties one by one.  If some part of this is giving you difficulty, come back here, write down what's problematic and what you've done, and I'm sure someone will help you.


floor function
geometry
geometric transformation
reflection
pigeonhole principle
ceiling function
graph theory

Given a regular 2007- gon , find the smallest positive integer k such
that in any set of k vertices of the polygon, there are four vertices
which form a quadrilateral with three edges of the 2007-gon.

This is what I tried.
In a 2007-gon, call the vertexes A(1), A(2), ..., A(2007).
Then, divide the set of vertexes into

A(4k+1), A(4k+2), A(4k+3), A(4k+4), with k=1(1)500

Now select the first three A(i)'s from each set, and also choose A(2005) and A(2006).

This gives us a total of 1500+2 = 1502 vertexes, which follows the following property:

There are NO FOUR vertexes in that set, which when joined, gives a quadrilateral with three edges of the 2007-gon.

But select ANY OTHER vertex, and this set of 1503 vertexes will NOT follow the aforementioned property.

So is the answer 1503?

Also, can one please tell me if a better (= more beautiful/more rigorous) solution exists, if this is indeed the solution?

THanK YoU.

Ain't the answer $ 1505$?

Let $ P : \equal{} P_{1}P_{2}\dots{P_{2007}}$ be the polygon.  Define $ S$ to be the set containing all $ P_{i}$'s such that $ i\minus{}1$ is not divisible by $ 4$.  Therefore, $ \left\lfloor\frac{2007\minus{}1}{4}\right\rfloor\plus{}1 \equal{} 502$ vertices are not in $ S$.  Thus,
\[ |S|\equal{}2007\minus{}502\equal{}1505\,.\]
A quadrilateral occupies three edges in $ P$ if and only if it contains four consecutive vertices in $ P$.  We call a quadrilateral with this property to be a "cool" quadrilateral.  Therefore, $ S$ does not contain any cool quadrilateral and hence, $ S$ satisfies the condition.

To show that $ S$ is maximal, we let $ T$ be any set with no cool quadrilaterals.  Let $ G$ be a bipartite graph consisting of two sets of vertices $ V_{1}\equal{} T$ and $ V_{2}\equal{}\{1,2,\dots,2007\}$.  All vertices in $ V_{1}$ reflects elements of $ T$.  The element $ k$ in $ V_{2}$ represents the set $ X_{k}: \equal{}\{P_{k},P_{k\plus{}1},P_{k\plus{}2},P_{k\plus{}3}\}$ in $ P$ (the indices are considered modulo $ 2007$).  A vertex $ v$ in $ V_{1}$ is connected to a vertex $ k$ in $ V_{2}$ if and only if $ v\in X_{k}$.  Since $ T$ does not contain cool quadrilaterals, $ \deg(k)\leq 3$ for all $ k\in V_{2}$.  Moreover, $ \deg(v) \equal{} 4$ for all $ v\in V_{1}$.  Note that
\[ 4|T|\equal{}4|V_{1}|\equal{}\sum_{v\in{V_{1}}}\deg(v)\equal{}\sum_{k\in V_{2}}\deg(k)\leq 3|V_{2}|\equal{}3(2007)\,.\]
Therefore, $ |T|\leq\frac{3}{4}(2007) < 1506$ and we are done.

You mean the answer is $ 1505\plus{}1\equal{}1506$

Yes it is 1506. A simpler method is as follows:

If $ k < 1506$, the construction to derive a contradiction is easy.
If $ k\geq 1506$,
Let $ a_{i}$ be 1 if vertex $ i$ is chosen and 0 otherwise. For simplicity define $ a_{i+2007}= a_{i}$.
Then $ \sum_{i = 1}^{2007}{\sum_{j = 0}^{3}{a_{i+j}}}= 4\sum_{i = 1}^{2007}{a_{i}}= 4\times 1506$
By pigeonhole $ \exists i|\sum_{j = 0}^{3}{a_{i+j}}\geq\lceil\frac{4\times 1506}{2007}\rceil = 4$
which gives the desired result.


Numerele reale pozitive a,b verifica relatia a+b+ab=3.
Demonstrati ca $a+b \ge \sqrt{a}+\sqrt{b}$.
*-macar asta e mai grea

asta a fost misto. care a fost cel mai mare punctaj la voi?

[quote="pohoatza"]$\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$, deci ne ramane $2\leq a+b$.

acum ne ducem in $a+b+ab=3$, deci $(a+b)+\frac{(a+b)^{2}}{4}\geq 3 \Longleftrightarrow s^{2}+4s-12 \geq 0$, cu $s=a+b$, acum rezolvand ecuatia asta pentru cand e $=0$, o sa obtinem $s = 2$ (celalalt $s$ da $-6$ , da nu se poate ca $a,b$ pozitive) , dar cum functia e crescatoare, rezulta concluzia.[/quote]

super solutie, dar nu de locala.
Aplicam un Cauchy si ajungem la:
$\sqrt{a}+\sqrt{b}\leq 2$, adica la $a+b+2\sqrt{ab}\leq 1+a+b+ab$. Scadem $a+b$ si ramane inegalitatea mediilor.

[quote="Virgil Nicula"][u][b]O usoara generalizare :[/b][/u]

$\{a,b,c\}\subset (0,\infty )\ ,\ a+b+ab=c(c+2)\Longrightarrow \frac{a+b}{\sqrt a+\sqrt b}\ge \sqrt c\ .$[/quote]

:) metoda lui pohoatza ... ramane de demonstrat ca

$a+b \ge c$ , dar avem ca $(a+b)+\frac{(a+b)^{2}}{4}\ge c (c+2)$ si  .. din semnul functiei de gradul II ( tinand cont ca $a+b > 0$ ) avem ca

$a+b \ge 2\sqrt{1+c(c+2)}-2 = 2c$  .. ( cctd )


quadratics
floor function
blogs
number theory proposed
number theory

Let $ n$ be a  positive integer and greater than 1  .Suppose $ \{x_1,...,x_n\}\equal{} \{1,..,n\}$ such that $ 1^{x_1},...,n^{x_n}$ is also a complete residue mod n . Prove that n must have form $ p$ or $ 2p$ where $ p$ is a prime number .

The trick here is to count quadratic residues

In the permutation $ \{x_1,...,x_n\}$ there are $ \lfloor\frac{n}{2}\rfloor$ even numbers and when $ x_i$ is even $ i^{x_i}$ is a quadratic residue.
Therefore the set $ \{ 1^{x_1},...,n^{x_n}\}$ has to contain at least $ \lfloor\frac{n}{2}\rfloor$ quadratic residues.

But how many quadratic residues can there be $ \bmod{n}$?

Subdivide residues into two sets $ A \equal{} \{a : (a,n) \equal{} 1\}$ and $ B \equal{} \{a : (a,n) \ne 1\}$

$ a \in B \implies \minus{}a \in B$ and $ a^2 \in B$. Since $ a^2 \equal{} (\minus{}a)^2$ and $ a \equiv \minus{}a$ has at most two solutions $ \bmod n$,
the number of quadratic residues in $ b$ is at most $ \lfloor \frac{|B|}{2} \rfloor \plus{} 1$

$ a \in A \implies \minus{}a \in A$ and $ a^2 \in A$ suppose now that $ n \equal{} rs, r > 2, s > 2, (r,s) \equal{} 1$

For $ a,b \in A$, there is a unique solution $ \bmod n$ to $ a \equiv b \bmod r$ and $ a \equiv \minus{}b \bmod s \implies a^2 \equiv b^2 \bmod n, b \ne a, b \ne \minus{}a$
there is also a unique solution $ \bmod n$ to $ a \equiv b \bmod s$ and $ a \equiv \minus{}b \bmod r \implies a^2 \equiv b^2 \bmod n, b \ne a, b \ne \minus{}a$

This means that the  residues in $ A$ can be divided into quadruples giving the same quadratic residue and the number of quadratic residues in $ A$ is at most $ \frac{|A|}{4}$

So the number of quadratic residues is at most $ R \equal{} \frac{|A|}{4} \plus{} \lfloor \frac{|B|}{2} \rfloor \plus{} 1$ but $ |A| \ge 8 \implies R < \lfloor\frac{n}{2}\rfloor$

This elimnates all cases except when $ n$ cannot be factorised into two coprime factors greater than $ 2$ ie. $ n \equal{} 2^k p$ for some prime $ p$

To complete the problem we need to eliminate $ n$ when $ 4 \mid n$. This can be done easily by counting the number of residues which are multiples of $ 4$ in
$ \{ 1^{x_1},...,n^{x_n}\}$. If $ i$ is even and $ x_i > 1$ then $ i^{x_i}$ is a multiple of $ 4$. This gives too many residues that are multiples of $ 4$ if $ n > 4$.

This solves the problem but begs the question, for which $ n \equal{} p$ or $ n \equal{} 2p$ does such a permutation exists?

I also found that if p satisfies this condition then $ p\minus{}1$ must be a square free . But I have not had a complete solution .

I don't have a solution either but it should be possible to make some progress.

It helps if you use an equivalence relation between residues defines by $ a \sim b$ iff $ \exists x,y : a^x \equiv b$ and $ b^y \equiv a \bmod n$
This equivalence relation partitions the residues into equivalence classes with same power residues. $ i^{x_i}$ defines a permutation and
its cycles have to contain elements that are in the same equivalence class. You can work out which exponents work for each class.

For example lets take $ n \equal{} 31$ you get equivalence classes corresponding to each factor of $ \phi(n) \equal{} 30$. You find that $ 5 \sim 25$ because
these are tenth power residues and these are the only residues in their class. The powers to use with these numbers must be $ 10$ and $ 20$
because they always map to these classes. You find $ 5^{10} \equiv 5, 5^{20} \equiv 25, 25^{10} \equiv 25, 25^{20} \equiv 5 \bmod 31$. From this we see
that the residues $ 5$ and $ 25$ cannot form cycles using both exponents, so there is no solution $ \bmod 31$

Edit: This example is not coplete, $ 10$ and $ 20$ could also be exponents for $ 1$ or $ 31$, however, $ 30$ must also be an exponent of $ 1$ or $ 31$
and from the six power residues we find that one of $ 6,12,18,24$ must also be an exponent for $ 1$ or $ 31$. That makes three exponents for two
numbers so the contradiction follows.

To solve the problem you have to generalise this method.

Together with TTsphn and PhilG, I have proved more general results in [url=http://bboyjordan.wordpress.com/2009/10/26/exponential-complete-residue-system/]this page[/url] of my blog:

[i]1) A odd number $ n>1$ works if and only if $ \frac{n\minus{}1}{2}$ is a Sophie-Germain prime.

2) A even number $ n>1$ works only if $ \displaystyle \frac{n}{2}$ is prime and $ \displaystyle \frac{n\minus{}2}{2}$ is squarefree.[/i]

Although this, only a necessary condition is reached in point 2. Can someone do better?


probability
combinatorics unsolved
combinatorics

what is the probabilty of distributing r non identical balls into n cells so that at least
one cell will be left empty .

I think the solution is $P(E) = \frac{(n-1)^{r}}{n^{r-1}}$, where $E$ is the event "$r$ nonidentical balls are placed into $n$ cells and one cell is left empty".

First note that the sample space is $n^{r}$, since for each of our $r$ balls, we have a choice of $n$ cells to place them in.  Therefore we need only to find $N(E)$.

There are $n$ different cells that may be left empty.  After we choose one, there are $n-1$ cells left to ditribute the balls into.  The total number of ways we can do this is $(n-1)^{r}$.  Therefore $N(E) = n(n-1)^{r}$.

So, $P(E) = \frac{N(E)}{n^{r}}= \frac{n(n-1)^{r}}{n^{r}}= \frac{(n-1)^{r}}{n^{r-1}}$.  This holds as long as $r \geq n$.  If $r < n$, then $P(E) = 1$.


Treething has seven tortilla chips, and they are all either corn tortilla chips or guacamole tortilla chips. Four of them are definitely guacamole tortilla chips, however, Treething dropped three of them in the mud, making it impossible to tell what flavor they are. (The mud masks the flavor a lot.) What is the probability that exactly six of the tortilla chips are guacamole-flavored, given that Treething knows that he has at least five guacamole-flavored chips? Round to the nearest percent.

[quote="Treething"]Treething has seven tortilla chips, and they are all either corn tortilla chips or guacamole tortilla chips. [b]Four[/b] of them are definitely guacamole tortilla chips, however, Treething dropped three of them in the mud, making it impossible to tell what flavor they are. (The mud masks the flavor a lot.) What is the probability that exactly six of the tortilla chips are guacamole-flavored, given that Treething knows that he has at least [b]five[/b] guacamole-flavored chips? Round to the nearest percent.[/quote]

I'm confused

That's because three tortilla chips are of undetermined flavor.

[quote="lotrgreengrapes7926"]So there's an equal probablility that it's corn or guacamole? Is this a trick question or something? Do you have to do something with finding the probability of the 4 chips not dropped in mud to be all guacamole?
[hide="Doubt this is right"] $\frac{_{7}C_{6}}{_{7}C_{5}+_{7}C_{6}+_{7}C_{7}}=\frac{7}{21+7+1}=\frac{7}{29}$ [/hide]
[hide="Possibly"] $\frac{3}{7}$ ? :? Listing the possible outcomes for the 3 "mystery" chips, we get gcc, cgc, ccg, ggc, gcg, cgg, ggg. Only 3 of those work. Of course, the probability of getting some of those may be higher thatn getting other combinations. [/hide][/quote]

Second one is correct.

This is kind of like that son/daughter problem.
[hide]
The ccc possibility is eliminated, and 3 of the last 7 all have 2 g's.

Yay, 3/7
AKA 43%[/hide]

Edit: whoops, had misspelled daughter XD


conics
geometry
3D geometry
parabola
ellipse
hyperbola
symmetry

How would you prove that taking a cross section of a cone such that the plane of intersection is "less tilted" than a generating line of the cone makes a curve that is always the set of points such that there is constant sum of the distances from each of those points to 2 foci. Same question for parabolas and hyperbolas. (i.e. that the 2 definitions of an ellipse always match)

I can see pretty intuitively and clearly why this is true for a circle, and I can also see that the the set of points defined as that the sum of their distances to 2 foci is constant would be VERY SIMILAR as what you would get if you cut a cone such that it is less tilted than a generating line of the cone, but I don't have a firm intuitive grasp of being able to visualize this.

Basically, I am having a hard time believing that no matter how you cut the cone, you will always get a locus that fits one of the 3 definitions of the 3 conic sections.

I am sorry if this should go into a more advanced forum, but I thought a definition might be a rather easy question, and that maybe I am missing something obvious.

I'm not sure how one might go about proving them, but [url=http://mathworld.wolfram.com/ConicSection.html]these[/url] [url=http://www.artofproblemsolving.com/Wiki/index.php/Conic_section#Nondegenerate_conic_sections]few[/url] [url=http://en.wikipedia.org/wiki/Conic_section]links[/url] might give you an intuitive understanding. There are several degenerate cases of these conic sections, which occur when the plane of intersection passes the apex of the two cones.

Hello
Sorry as don't have the graphics right now.So kindly try to draw them my statements only.
Consider a (3-D) double right circular cone with a semi-vertical angle $ \theta$.
Let $ \alpha$ be the angle an intersecting plane makes with the height of the  double cone.
For parabola- take any intersecting plane parallel to the slant of the cone.$ \alpha\equal{}\theta$
For ellipse - take an intersecting plane at an angle $ \alpha>\theta$
NOTE-$ \alpha\equal{}\frac{\pi}{2}$ is case of circle.
For hyperbola-take an intersecting plane at an angle $ \alpha<\theta$:in this case the plane actually cut both the top and bottom halves of double cone.
Degenerate cases-
point-A point is formed if the plane passes through only the vertex.
straight line-If a plane is tangent to the double -cone's surface.
Pair of straight lines-It is formed if the plane is passer through the vertex and intersects both the top and bottom halves of the cone.
Thank u.

Thanks guys, but my question is still not really answered. Let me try again to explain my difficulties in understanding conics. I can solve most basic problems that involve conics, but I can't really see intuitively, for instance, why taking a cross section of the cone as described in my earlier post (about the ellipse) would necessarily always (or even EVER give, except in the case of the circle) a figure symmetrical about its minor axis (which I also cannot intuitively picture in my head would necessarily exist since I can't picture the symmetry), as is obvious in the case where you draw an ellipse using its foci definition.

Thanks again.


Let $ ABC$ is  a triangle, let $ BH \perp AC, CK \perp AB$ with $ H \in AC$ and $ K\in AB$. Let $ E, F$ are midpoints of $ AC$ and $ AB$  respectively. Let $ EF$ meet $ HK$ at point $ I$. Prove that $ AI$ perpendicular with the Euler’s line of  $ ABC$ triangle.

[quote="thanhnam2902"]Let $ ABC$ is  a triangle, let $ BH \perp AC, CK \perp AB$ with $ H \in AC$ and $ K\in AB$. Let $ E, F$ are midpoints of $ AC$ and $ AB$  respectively. Let $ EF$ meet $ HK$ at point $ K$. Prove that $ AK$ perpendicular with the Euler’s line of  $ ABC$ triangle.[/quote]

See problem $ 3$ and the solution of this problem: http://www.mathlinks.ro/viewtopic.php?t=301127

Can you post solution of this problem using Pole and Polar? Thank you very much.

[quote="thanhnam2902"]Let $ EF$ meet $ HK$ at point $ K$.[/quote]
Something is wrong here.


AAAAAAAAAAANNND ladies and gentlemen, in this corner, its the boys of AoPS, with 1 point on their scoresheet, ready to go on to round III! And on this corner, the girls of AoPS, also with 1 point, ready to fight it out in the most grueling game in the Fun Factory! That's right, folks, the game is tied. Who knows where this will go next. PLACE YOUR BETS!!!

20

@powerofpi: no.. add the number of stars (post #) you have

12

[b]19[/b]

Was that supposed to be a 13?

[b]12[/b]

@Powerofpi: hmm i think girls will win

14

you know, if i stop posting, than izzy comes and she and i posts, then we could win, cuz technically if i keep posting i'm helping the boys.. by adding one for them each time so this is my last post until some girl comes !! XD bb

NOOOOOOO

A new reply has been posted or the last post has been deleted while you were replying. Check the "Topic review" at bottom of this page, and resubmit your post if appropriate.

A new reply has been posted or the last post has been deleted while you were replying. Check the "Topic review" at bottom of this page, and resubmit your post if appropriate.

A new reply has been posted or the last post has been deleted while you were replying. Check the "Topic review" at bottom of this page, and resubmit your post if appropriate.

A new reply has been posted or the last post has been deleted while you were replying. Check the "Topic review" at bottom of this page, and resubmit your post if appropriate.
 A new reply has been posted or the last post has been deleted while you were replying. Check the "Topic review" at bottom of this page, and resubmit your post if appropriate.

Uh, guys, read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=139&t=330236]this thread[/url].  and mods lock

Locking, the problem with the thread being your revival.


function
algebra
functional equation
algebra solved

This one is from Poland 02 or so, I think third or fourth round.
I know for sure it's a Poland problem. I'll look it up.

The other problem (F.E. 1) comes from a Singaporean TST.

"Functional equation 2" comes from the Poland Math Olympiad 2000, 2nd round.
(There are three rounds in the Poland MO, not four :))

I really don't know if this works, but here's the fction I thought of (tell me if it's wrong):

f(0)=0;
f(2^a*t)=2^a*f(t) where a is natural and t is odd;
f(4k+1)=4k+3 for k natural;
f(4k+3)=2(4k+1).

Oh I see. Yes, this might work. I didn't spend much time on the problem not yet.
But anyway, I'm bad at "constructing" functions.

Arne, out of curiousity, how do you get Singapore problems? (I only know [i]one[/i] website to get problems from, the rest are all training handouts!)  And I know this is quite embarrassing, but what does TST mean?

eg Singapore TST

he means the selection tests for the IMO team

cheers! :D :D

Er, no, that doesn't come from the selection test for the IMO team; it's from the selection test for the national training team (of 15 or so people every year).

[quote="Arne"]Does there exist a function f : N -> N such that f(f(n)) = 2n ?[/quote]
===============
We can construct solutions of this equation by following way

First, we divide the set of odd natural number to pairs: (a1, b1), (a2, b2), ... and we construct function f as:
          f(2n) = 2f(n)
          f(ai) = bi, f(bi) = 2ai

I think, that are all solutions.

Interesting related question: Is there an increasing function with mentioning condition?


Μια ερώτηση που αφορά ανάπτυγμα σε σειρά:
Πως αναπτύσσεται (με ποιο τύπο και μέθοδο) η $ \frac{1}{\sqrt{1\minus{}sinx}}$ γύρω από το ανώμαλο σημείο $ \frac{\pi }{2}$ ;

Είδα το ανάπτυγμα της που έχει για πρώτο όρο το $ \frac{\sqrt{2}}{x\minus{}\frac{\pi }{2}}$ αλλά δεν ξέρω από ποιο τύπο βγήκε και πως ακριβώς προέκυψε η δύναμη -1 του $ x\minus{}\frac{\pi }{2}$

Το πλαίσιο στο οποίο πρέπει να εργαστεί κανείς είναι εκείνο των μιγαδικών αριθμών.   Θα χρησιμοποιήσουμε σειρές Taylor και

Laurent τις οποίες θα υπενθυμίσουμε. Για λόγους ευκολίας θα εργαστούμε με κέντρο το 0. Θεωρούμε δεδομένη την έννοια της

παραγώγισης συναρτήσεων μιγαδικής μεταβλητής. Συναρτήσεις οι οποίες είναι παραγωγίσιμες σε ένα σύνολο που είναι ένωση ανοικτών

δίσκων ονομάζονται αναλυτικές. Εξετάζουμε τρεις περιπτώσεις:

- Αρχικά ξεκινάμε με την περίπτωση που η συνάρτηση $ f$ είναι παραγωγίσιμη σε ένα δίσκο της μορφής $ \{z: |z|<r\}$ όπου $ r>0$. Μια

τέτοια συνάρτηση έχει ανάπτυγμα της μορφής:

\[ f(z)=\displaystyle\sum_{n=0}^{+\infty}a_nz^n\]

Οι συντελεστές είναι μοναδικοί και δεν εξαρτώνται από την επιλογή της ακτίνας του δίσκου. Επιπλέον το ανάπτυγμα αυτό ισχύει σε

{\it κάθε} δίσκο με κέντρο 0 στον οποίο η συνάρτηση είναι παραγωγίσιμη. Οι ολοκληρωτικοί τύποι του Cauchy δίνουν τους συντελεστές

οι οποίοι βέβαια χαρακτηρίζονται και μέσω των ανωτέρων παραγώγων της συναρτησης στο 0.Δε θα χρειαστούμε καμμία από τις δυο μορφές

παρακάτω.

-Έπειτα, μια συνάρτηση $ f$ μπορεί να οριστεί ως αναλυτική στο άπειρο. Αυτό συμβαίνει αν είναι παραγωγίσιμη σε ένα σύνολο

τηςμορφής $ \{z: |z|>R\}$ και υπάρχει το όριο $ \displaystyle\lim_{z\to\infty}f(z)$ ως μιγαδικός αριθμός. Σε αυτή την περίπτωση η
συνάρτηση $ f(1/z)$ είναι αναλυτική στο  0. Χρησιμοποιώντας την προηγούμενη περίπτωση, συνάγουμε ότι η  $ f$ has έχει ανάπτυγμα της
μορφής

\[ f(z)=\displaystyle\sum_{n=-\infty}^{0}a_nz^n\]

- Στην τρίτη περίπτωση θεωρούμε μια συνάρτηση $ f$ που είνα παραγωγίσιμη σε ένα δακτύλιο της μορφής $ \{z: r<|z|<R\}$. Εδώ ισχύει

$ r<R$
ενώ το  $ r$ μπορεί να είναι και 0. Αυτό σημαίνει ότι η περίπτωση αυτή περιλαμβάνει την περίπτωση μεμονωμένης ανωμαλίας. Μια

τέτοια συνάρτηση είναι άθροισμα δύο συναρτήσεων, μιας από την πρώτη περίπτωση που αναφέραμε και μιας από τη δεύτερη. Οι δύο αυτές

συναρτήσεις είναι μοναδικές ως προς προσθετική σταθερά και μας δίνουν το ανάπτυγμα  Laurent:

\[ f(z)=\displaystyle\sum_{n=-\infty}^{+\infty}a_nz^n\]

Τα $ a_n$ είναι μοναδικά  και το ανάπτυγμα ισχύει σε οποιοδήποτε δακτύλιο έχει μη κενή τομή με τον αρχικό μας δακτύλιο, και στον

οποίο βέβαια η συνάρτηση
$ f$ είναι παραγωγίσιμη. Διαισθητικά μπορούμε να θεωρήσουμε το δακτύλιο ως την τομή ενός δίσκου με κέντρο το 0 και ενός που έχει

κέντρο το άπειρο. Η συνάρτησή μας είναι άθροισμα δύο συναρτήσεων που ορίζονται στους δίσκους αυτούς. Είναι λοιπόν φυσιολογικό το

άθροισμά τους να ορίζεται στην τομή.
Ας έρθουμε επιτέλους (!) στο αρχικό σου ερώτημα. Όταν ξέρουμε ότι το ανάπτυγμα υπάρχει, τότε χρησιμοποιούμε κάποιες αλγεβρικές

μεθόδους για να προσδιορίσουμε τη μορφή του. Τυπικά, μεταφέρουμε το κέντρο στο 0 όπου τα πάντα είναι πιο απλά. Μιας και

δουλεύουμε με μιγαδικους, θα χρησιμοποιήσω το συμβολισμό   $ 1/\sqrt{1-\sin z}$. Το πρώτο πράγμα που πρέπει να παρατηρήσει κανείς

είναι ότι υπάρχει ένα πρόβλημα με το συμβολισμό. Το σύμβολο "τετραγωνική ρίζα" δεν έχει νόημα στην περίπτωσή μας γιατί δεν

υπάρχει μόνο μία αλλά δύο συναρτήσεις των οποίων το τετράγωνο είναι $ 1-\sin z$ και το γεγονός αυτό χρειάζεται απόδειξη. Για να

παρακάμψουμε τις δυσκολίες αυτές θα εργαστούμε με τη συνάρτηση $ 1/\(1-\sin z)$ και τελικά θα βγάλουμε συμπεράσματα για την

"τετραγωνική ρίζα" της.Τώρα θα μεταφέρουμε το κέντρο στο 0. Αυτο σημαίνει ότι θα αντικαταστήσουμε το
 $ z$ με $ w+\frac{\pi}{2}$. Συνεπώς έχουμε τη συνάρτηση $ 1/(1-\cos w)$. Αν τώρα θεωρήσουμε την $ w^2/(1-\cos w)$, βλέπουμε ότι

είναι αναλυτική στο 0 και μάλιστα με μη μηδενική τιμή. Αυτό σημαίνει ότι η $ 1/(1-\cos w)$ έχει πόλο τάξης 2
στο  0. Συνεπώς το ανάπτυγμά της είναι της μορφής:

\[ \dfrac{1}{1-\cos w}=\sum_{n=-2}^{+\infty}a_nw^n\]

Ακριβώς επειδή η συνάρτηση αυτή έχει πόλο τάξης 2, αυτό σημαίνει ότι έχει "τετραγωνική ρίζα" η οποία είναι αναλυτική στο 0.

Άρα κάθε συνάρτηση της οποίας το τετράγωνο είναι  $ 1/(1-\cos w)$ έχει πόλο τάξης  1 στο  0.Ας συμβολίσουμε με  $ h(w)$ μια από τις

δύο αναλυτικές συναρτήσεις με την ιδιότητα αυτή, τότε το πρόβλημα ανάγεται στον προσδιορισμό του  $ b_{-1}$ where:

\[ h(w)=\sum_{n=-1}^{+\infty}b_nw^n\]

Για το λόγο αυτό παρατηρούμε (τετραγωνίζοντας το ανάπτυγμα της $ h(w)$) ότι $ a_{-2}=b_{-1}^2$. Από το όριο:

\[ \lim_{w\to 0}w^2/(1-\cos w)=2\]

προκύπτει ότι $ a_{-2}=2$ που σημαίνει ότι $ b_{-1}=\pm\sqrt{2}$. Και οι δυο τιμές είναι δυνατές αλλά σύμφωνα με τα όσα γράφεις

έχει επιλεγεί το θετικό πρόσημο. Αυτό σημαίνει ότι

\[ h(w)=\dfrac{\sqrt{2}}{w}+\sum_{n=0}^{+\infty}b_nw^n\]

Που σύμφωνα με το συμβολισμό σου, συνεπάγεται ότι:

\[ \dfrac{1}{\sqrt{1-\sin x}}=\dfrac{\sqrt{2}}{x-\frac{\pi}{2}}+\sum_{n=0}^{+\infty}b_n(x-\frac{\pi}{2})^n\]

Σε ευχαριστώ πολύ.

Η ερώτησης αυτή προέκυψε στο εξής πρόβλημα φυσικής:

Να βρεθεί ο χρόνος που χρειάζεται ένα υλικό σωματίδιο να ανέβει στην κορυφή ενός ημικυκλίου αν η ενέργεια που το δόθηκε είναι ακριβώς όση χρειάζεται για να φτάσει στην κορυφή (με μηδενική ταχύτητα). Με άλλα λόγια, σε πόσο χρόνο προσεγγίζεται το σημείο ασταθούς ισορροπίας; Πεπερασμένο ή άπειρο;

Όποιος θέλει μπορεί να το δοκιμάσει.

Το τελικό αποτέλεσμα βγαίνει και με άλλο τρόπο, αλλά η ερώτηση που έκανα μπορεί κατά τη γνώμη μου να οδηγήσει σε μια γενικότερη μεθοδολογία.


Given finitely many points in a plane, it is known that the area of the triangle formed by any three points of the set is less than 1. Show that all points of the set lie inside or on boundary of a triangle with area less than 4.

Consider the triangle of largest area, ABC. It's area is < 1. Draw a line through A parallel to BC. Any point D must lie on this line or on the same side as BC, otherwise BCD would have area > 1. Likewise draw lines through B and C parallel to AC and AB, respectively. All points must lie on or inside the triangle made up of the three lines we have drawn. This triangle has 4 times the area of ABC, so its area is less than 4.

I solved it in exactly the same way.
Anyway, what I'm interested in is that is there any solution to the problem using convex hull?

Oddly enough, this is one of the examples in Engel.

That problem was also in AACOPS I think.

I think that the constant 4 can be improved. Can anyone prove or disprove this claim?


search
geometry
algorithm
function
Euler
number theory
relatively prime

For a positive integer $ n$ and an integer $ a$ relatively prime to $ n$, there exists a prime congruent to $ a \bmod n$.

hello,

I believe another area of research is an algorithmic in nature. I have been working on binary factoring algorithm and I have an algorithm that calculates remainder (modular) of a factor to a given n. I used a special form of phi function to calculate the periodicity of factor. It may not be a proof but possibly the use of euler phi function might give lights to other solutions.

cheers,

jaime

Does $ a$ necessarily need to be relatively prime to $ n$ for this to be true?

If $ a$ is neither prime nor relatively prime to $ n$, numbers congruent to $ a \bmod n$ have a nontrivial divisor, so no primes of that form exist.  If $ a$ is prime and divides $ n$, exactly one prime congruent to $ a \bmod n$ exists (which is $ a$).

jaimeM:  I'm not sure what you're trying to say.  Binary exponentiation is already a well-known algorithm, and might the "modified phi function" you're using be the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url]?  In any case, I don't see a direct connection between the problem of determining orders $ \bmod n$ and the problem of primes in a particular residue class.

Ah, of course. I was thinking of the proof in terms of Dirichlet's Theorem which I had forgotten has the requirement that $ \gcd(a, n) \equal{} 1$, which led to my question.

It has not been proven yet in either direction. For some a and n you can prove it without Dirichlet, but the conjecture is that not for all of them. If you assume Hypothesis H (of Schinzel) then it would follow that you can't prove it without Dirichlet's theorem unless $ a^2 \equal{} 1 \mod n$. The paper with this proof can be found here: http://www.math.dartmouth.edu/~ppollack/euc4.pdf


I guess I will have to say "other".  I am [b]not[/b] a genetic mutation. (no offense)

WHAT is with the choices????? is 1=2 red haired? altho blue probably wouldnt hav been up there anyway... :(

why here are no other choices????    as blond, or black or brown or i do not know.

[color=orange]I have blond hair ;)[/color]

Ummmmm.............
I think there should be more options than:
Red
Other
Most people don't have red hair.

Only red and other? You should expand other to blue, green, red, maybe blonde or brown, probably not black, and ... oh-don't forget him $ \rightarrow$  :D (bald)

l[color=brown]ol red... umm no i have darkdarkdarkbrownhair that some ppls like to call black [/color] :D


Prove that $ (p\minus{}1)!\plus{}1$ is divisible by $ p$ for all prime numbers, $ p$.

[hide="My proof using concepts recently learned..."]

Let $ p$ be a prime number.
If $ x$ is an integer that satisfies $ x^2 \equiv 1 \bmod p$, then $ p$ divides $ (x\minus{}1)(x\plus{}1)$, which means it divides either of the two. Therefore, if $ 1\le x \le p\minus{}1$ and $ x^2 \equiv 1 \bmod p$, then $ x$ is either $ 1$ or $ p\minus{}1$. For each integer $ x$ that satisfies $ 1 \le x \le p\minus{}1$, there is exactly one integer $ y$ satisfying $ 1 \le y \le p\minus{}1$ and that $ xy \equiv 1 \bmod p$.  In addition, $ y \ne x$ when $ 2 \le x \le p\minus{}2$. It also follows that $ (p \minus{} 2)!$ is a product of numbers of the form $ xy$, where $ x$ and $ y$ are distinct integers from $ 2$ and $ p\minus{}2$, and $ xy \equiv 1 \bmod p$. Therefore, $ (p\minus{}1)! \equiv 1 \bmod p$ and $ (p\minus{}1)! \plus{}1 \equiv 0 \bmod p$. QED
[/hide][/hide]


use modular arithmetic
5^2 = 1 (mod 8)
5^136 = 1 (mod 8)
5^137 = 5 (mod 8)
so our answer is 5
check the aops wiki for more details

[quote="isabella2296"]How do you solve this problem?

What is the remainder of 5^137 divided by 8?[/quote]

We use Modular Arithmetic, which is basically says that

$ n \cong n\plus{}ap (mod p)$, for any $ n$ and $ a$, and $ p \ge 1$.

So, we see that $ 25 \cong 1 (mod 8)$, because in this,

$ n\equal{}25$, $ p\equal{}8$, and $ a\equal{}\minus{}3$. If you don't understand this, the Wiki

is a great place.

So, we can take off $ 5^2$ numerous times, to get it down to

"What is the remainder of 5 divided by 8?", which is obviously $ 5$.

NOTE: THE $ \cong$ SHOULD LOOK LIKE AN EQUALS SIGN, WITH

3 LINES, I JUST DON'T KNOW THE LATEX FOR IT.

Some people write modular arithmetic over the computer as ==

Note that would probably not be a typo

use $ \equiv$ for congruence in modular arithmetic :wink:


geometry
3D geometry
tetrahedron
trigonometry
Euler
linear algebra
matrix

Give one tertrahedron ABCD. Have six lengths of edges AB, AC, AD, BC, BD, CD, compute volume of tetrahedron folow them.

well there definitely is a way. It's just a horribly horrible way.

Volume of tetradedron is 1/3 B*h.

B = area of any one of the triangles, you can use Heron's Formula for that.

To get the height, um, well, you can use Law of Cosines 3 times, with three variables, that's solvable...

There is a formula due to Euler, which I think is true in any number of dimensions.  I don't remember all details of the formula, but the essential idea is that,  up to a constant factor (of 288),  $V^{2}$ is the determinant of a matrix whose entries are the squared distances between vertices,  plus an extra row and column consisting of 1's.

MathWorld calls the formula the Cayley-Menger determinant.  See
[url]http://mathworld.wolfram.com/Cayley-MengerDeterminant.html[/url].

If $U, V, W, u, v, w$ are lengths of edges of the tetrahedron (first 3 form a triangle; u opposite to U and so on), then

$Volume=\sqrt{\frac{(-a+b+c+d)\*(a-b+c+d)\*(a+b-c+d)\*(a+b+c-d)}{192\*u\*v\*w}}$

where
$a=\sqrt{x\*Y\*Z}$

$b=\sqrt{y\*Z\*X}$

$c=\sqrt{z\*X\*Y}$

$d=\sqrt{x\*y\*z}$

$X = (w-U+v)\*(U+v+w)$
$x = (U-v+w)\*(v-w+U)$
$Y = (u-V+w)\*(V+w+u)$
$y = (V-w+u)\*(w-u+V)$
$Z = (v-W+u)\*(W+u+v)$
$z = (W-u+v)\*(u-v+W)$


number theory
greatest common divisor
number theory proposed

m$\in$N.Let gcd$(m,6)=1$. $S_{m}$ is the set of numbers which is smaller than m and coprime to m.And $\sum_{{n}\in{S}_{m}}\frac{1}{i}=\frac{A}{B}$ , ${A,B}\in{N}$. Prove that $m^{2}$ divide $A$

Cool problem.

I will establish if $\sum_{n\in S_m}\frac1n = A/B$ with $(A,B)=1$ then $m^2\mid A$, provided $(m,6)=1$. In what follows below, I do all arithmetic in modulo $m$. Note that,
$$
\sum_{n\in S_m}\frac1n = \sum_{n\in S_m:n\leqslant \frac{m-1}{2}}\left(\frac1n+\frac{1}{m-n}\right)= \sum_{n\in S_m:n\leqslant \frac{m-1}{2}}\frac{m}{n(m-n)}.
$$
Therefore, it suffices to establish
$$
m\mid \sum_{n\in S_m:n\leqslant \frac{m-1}{2}}\frac{1}{n(m-n)}.
$$
Note that $n(m-n)\equiv -n^2\equiv -(m-n)^2\pmod{m}$. Hence,
$$
\sum_{n\in S_m:n\leqslant \frac{m-1}{2}}n^{-2}\equiv \frac12\sum_{n\in S_m}n^{-2}\pmod{m},
$$
thus it suffices to establish
$$
m\mid \sum_{n\in S_m}n^{-2}\iff m\mid \sum_{n\in S_m} n^2.
$$
Now, note that in modulo $m$, the sets $S_m$ and $\{2a:a\in S_m\}$ coincide (this uses the fact that $2\nmid m$). Hence,
$$
\sum_{n\in S_m}n^2\equiv 4\sum_{n\in S_m}n^2 \pmod{m}\Rightarrow m\mid 3\sum_{n\in S_m}n^2.
$$
Since  $(m,3)=1$, the conclusion follows.

@above. Can you explain $\sum_{n\in S_m} n^2\equiv 4\sum_{n\in S_m} n^2\mod m $

Yes. As I said, the sets $S_m$, and $2S_m=\{2a:a\in S_m\}$ coincide modulo $m$: note that for any two entries $2a$ and $2a'$ in $2S_m$, $2a\equiv  2a'\pmod{m}\iff m\mid 2(a-a')$. Since $(m,2)=1$, this holds iff  $a=a'$. Thus, the sets indeed coincide modulo  $m$. From here, $\sum_{x\in S_m}x^2\equiv \sum_{y\in 2S_m}y^2\pmod{m}$, which is precisely what I wrote above.


I think we should have a meetup at mc nats.  Any1 have any ideas on places to meet?  AoPS will probably have a table and Mr R will be there.

Post here if ur going to nats : )

[quote="mewto55555"]I don't know whether or not there will be a table, but we should all wear signs saying we are AOPS or carry a sign. Maybe this:
[img]http://www.tribuneindia.com/2007/20070629/spr6.jpg[/img][/quote]
Hmmmmm...propaganda?

hahahahaha  :rotfl:

I'm going, maybe I could come.

It would be pretty interesting if people actually get this together.  I'm predicting that means at least 50 people attempting to converge in the same place?

ahahaha... i was just thinking of starting a topic like this
yeah i'm going... cya all there! =D

ok, I think we should all meet in a single elevator and see if that works :P

nah, jk

cya there ppl :)

Hmm...we could try.

Anyway, I'll see you guys too! :)

Dude like, obviously we can't meet in normal times... we are with our state teams....

so i suggest that we meet during lunch  :P .

Maybe... copy the stuff from the AoPS-Tshirt for the sign?

Hmm...maybe print out the logo and tape it to your shirt :lol:

or, tape the aops sign to one of the tables, we could all meet there during lunch.

Put a nametag with AoPS logo and your AoPS sn.
I didn't even make nats...

How about we wear AoPS shirts? I will make a sign there saying AoPS, it's one sheet of paper. And I'll just non-chalantly carry it around, and people will be staring at me like "What is the matter with you?". :)

Actually, I will get there Wednesday night. I will get up on Thursday, and people will be coming in. I will just wait until people do, and then I will hold up a sign saying "AoPS".

I don't have an AoPS shirt and I don't really want to buy one.

You could use my approach which consists of trying to collect all 228 folks' contact info, consisting of email address and username. What, you don't have a username? :o (Plus, if you do this, I'll fuse it with my list from three years ago.)

(They give everyone a yearbook with everyone's name and picture.)

If I was homeschooled I would totally go Wednesday...but I mean at my school missing 2 days is already kind of a lot.  Can't wait to meet everyone  :D !

[quote="Ravi B"][/quote][quote="frost13"][quote="1=2"][quote="Quickster94"]So, does 6-8 pm the day we get there work?  That day the coaches are eating separately, so they won't be with us[/quote]

I get there Wednesday. Is it on thursday?[/quote]

So NY is arriving Wednesday too?  Starting to be a popular choice :lol:[/quote][quote="Ravi B"]1=2 is arriving Wednesday, but the rest of the NY team (including my daughter) is arriving Thursday.  In retrospect, arriving early would have been a good idea.  Gives you some time to relax and get used to the time difference.[/quote]

[quote="CatalystOfNostalgia"]I think you guys should chill a little more on Thursday afternoon. Get all the work done beforehand. Yeah, do a team round to keep your brains warm. But I think for me, it added some stress to do all that work on Thursday.

Then again, in 2006 I did nothing but play cards and make fun of people's hobbies on Thursday and came in 112th...[/quote]

OMG!  Same here!  Except I didn't bring cards and I didn't do as bad, though my rank was only a bit better than 112th.  The thing that stressed me out was my lack of sleep on Thursday, cuz there were fireworks going on outside for no apparent reason, and my teammate in my room was snoring like crazy.  lol, but the hobbies were so funny, no offense to those who have them and are on AoPS.  "Researching game theory" was my favorite from last year.

I think i get to the airport around noon.

write your aops username on your nametag.

You get a badge that you must wear at all timesin order to get into contest, meals, buses, etc.  I'll show you my old ones at our next practice.

We, the MI team, are arriving around 10:00 or so Denver time on Thursday. We're also going to win, as a sidenote. (Well, maybe not.)
Heh, I really don't need to put my sn on my tag... It's probably already there, except without a nine and backwards.

AH-HA! Now I know your name is one of the following:

Davidlu
D Avidlu
Da Vidlu
Dav Idlu
Davi Dlu
David Lu
Davidl U

Eh, it's David Lu, you could have asked and I would have told you.

The MO crew is arriving a short while after noon

Every1 arrives by thurs afternoon right?

i thought the earliest you could arrive is wednesday.

[quote="JBoyd"]ra5249, you are going too be busy with [u]TEAM [/u]activities up to Saturday evening.  Your first responsibility is to you teammates and TEAM.[/quote][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=495557#495557]:lol:[/url] lol