HMMT ARML AMC USA(J)MO USAMO inequalities MATHCOUNTS

what is the best thing to do on the day before a math competition?

Preparation, in terms of sleep, begins at least a few days before. If you are like me (average midnight sleeper, or even later), then you probably want to work in the few days before going to sleep a little earlier. The day before, try and go to sleep as early as you can (for me, this would probably be 9, but it depends on your average sleep). However, if you do happen to get less sleep than you wanted, don't stress over it. It probably doesn't even make that much of a deal (just keep telling yourself that).

Let's see...I wouldn't do any math the day before. Maybe just some puzzles, to keep your brain sharp and relax. I would also highly recommend listening to some good music.

Definitely have fun, and get a lot of sleep. To tell a quick story, the night before 2009 HMMT, I stayed up until about 1 AM and got practically no sleep. Did not do too well, even though I had tons of fun. In stark comparison, 2009 ARML was also fun, and I managed to get about seven hours of sleep. I almost qualified for tiebreakers. I'm not sure if there's causation there, but there's definitely a correlation.

You definitely shouldn't do too much math. Maybe review a couple of formulas or something, but nothing significant. You don't want to stress yourself out before the competition.

I think getting sleep is a must, in addition to good breakfast. Eat something that perks you up.

Also don't do too much math the same week.

Just stay calm about the competition. Listen to some good music or something that calms you down, as said previously. Get a good night sleep, and also eat a healthy breakfast. Eat like things that give you long hours of energy, like starchy food, eg. potatoes? Also, if the competition is like more of a time bound competition, try practicing same/similar level problems in a lesser time limit than they are supposed to be done in. Finally, just enjoy... :)

Play basketball...even if you hate basketball. That's what I did the morning of USAMO this year, because I was told I didn't need to go to school in the morning. It really clears your mind.

Also, the night before, listen to some music that makes you feel angry. This has a similar effect.

If you really want, you can also write a haiku about olympiad inequalities so that you can practice using creativity while thinking about inequalities while not thinking too hard, seeing as you'll think hard enough the next day.

ok, thanks for the advice everyone :D

lol i wonder what the last line of my haiku should be.....

why dont you treat everyday as the day before the competition??

because then, the day of the competition would be the most miserable day of your life  :P

Also note that, if you treat every day as the day before the competition, whatever you do before the competition no longer becomes special because it is a quotidian activity.

[quote="pythag011"][quote="Hamster1800"]My method of preparing for contests is to work up until two days before the competition. As soon as it is the day before, do absolutely no math. Relax and be well rested. Then the morning of maybe do a bit of cramming if you want, but this is unlikely to help. It might make you feel more confident though.

Same thing for afternoon of USAMO day 1. I would recommend doing no math other than perhaps going over what happened on day 1.[/quote]

Do you recommend going over day 1 stuff on afternoon?[/quote]

I'm pretty sure everyone does whether it's good or not. It's hard not to think about the problems that you just worked on for 4.5 hours... but don't dwell on it I guess?

Man why are you asking about this? Don't you know about this already?

That's not an easy question to answer and it might depend on the person how helpful/detrimental it is. (I don't see why anyone should definitely know the answer to that... well maybe someone like Zuming.)

In my opinion it depends on how confident you are in your preparation. If you aren't confident, hearing how to solve the problems you missed might help you figure out why you didn't think of certain steps and reorient your thinking for the next day. If you've prepared well, anything you hear is probably going to be much more distracting than instructive.

Personally, I never discussed the day 1 problems with anyone else (except for people at my school right after the test), although I'd often turn them over in my head over the course of the day - as said above, it's hard not to.

Maybe try some easyi-ish problems the day before the test to warm up your brain.  Such as, say, Mathcounts School Rounds, or Math League competitions....

Nothing significant should be done. Getting sleep is good. Remember, if you want to prep for a competition do it over a long period of time. In other words, don't cram in the last week. Also, relax and enjoy. Most importantly learn.

[quote="PowerOfPi"]If you do practice tests, you should do easy ones. If you do easy ones, you will probably do good and become confident. If you do hard ones, you will get frustrated if you don't do good. You usually do as good as you think you will do.[/quote]

You do need to make sure you don't do too easy ones. I did and got overconfident :oops:

On the question of USAMO day 1, I think hearing how to do problems can help you get them out of your head quickly and allow you to stop dwelling on the ones you missed. I think this probably helped me on #2 last year, since I knew it was easy, but after not getting it after 4 hours I just had to hear a solution.

[quote="pianoforte"]If you find it hard to get to sleep, do a physical activity before sleeping.[/quote]

Also, don't drink anything but water after 8:00.

hahah and if its MATHCOUNTS, dont drink water at all!!! lol.
(on the day of)

its true though

[quote="PowerOfPi"]iYOA: Why not?[/quote]

because it causes you to have to go really badly.  it might be nerves too though, but i doubt it.

Of course, AFTER you've finished your tests, have fun, eat pizza, and drink sugary soda pop and enjoy your free lunch at Purdue University....and then find out you only got 23/46 and didn't make state cd :((yes it was sad for me)


inequalities
logarithms
rearrangement inequality
inequalities proposed

For the first inequality,I prove it use only AM-GM's inequality
Solution:
Use the well-known inequality:$(x+y+z)^2 \geq 3(xy+yz+xz) (x,y,z \in R)$
So we have $(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b})^2 \geq 3(\sum \frac{ab^2c}{ac})$
we need to prove:$\sum \frac{ab^2c}{ac} \geq 3$(*)
(*) equavilent to $\sum ab^3c \geq 3abc$(Obviously true from the condition $a^2+b^2+c^2 =3$)
So the problem a) is proved  :)  :)  (Is your solution shotter than my solution,Hungkhtn?)
For the second inequality,I will prove this inequality with  require conditions:$a^3b^3+b^3c^3+c^3a^3 \geq 3abc$
We have $\sum a^3b^3+abc(\sum_{sym}a^2b) \geq 3abc+abc(\sum_{sym}a^2b)$ (It's true because we have the condition $\sum a^3b^3 \geq 3abc$) (**)
Apply AM_GM's inequality we have $a^2b+b^2a+1 \geq 3ab$
                                                   $b^2c+c^2b+1 \geq 3bc$
                                                   $c^2a+a^2c+1 \geq 3ac$
Sum up these inequalities we have (**) $\geq 3abc(ab+ca+bc)$
So that $(\sum a^2b^2)(\sum ab) \geq 3abc(ab+bc+ca)$
Hence $\sum a^2b^2 \geq 3abc$ and it equavilent to $\sum \frac{ab}{c} \geq 3$(the problem is proved) :)  :)

[quote="hungkhtn"]1. Let $a,b,c$ are positive reals sthat $a^2+b^2+c^2=3$, prove that
\[ \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \ge 3  \]

2. Let $a,b,c$ are positive reals sthat $a^3+b^3+c^3=3$, consider the inequality
\[ \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \ge 3  \]
$(i)$ Prove that the above inequality can't be true for all $a,b,c$.

$(ii)$ Prove that this one will be true if $a,b,c$ require the conditional
\[ a^3b^3+b^3c^3+c^3a^3 \ge 3abc  \]

3. Find the best constant $k$ such that the following inequality is true for all $a,b,c \ge 0$
\[ \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \ge 3\sqrt[k]{\frac{a^k+b^k+c^k}{3}}  \][/quote]

The first one, as you probably know, is NOT new.
Following is my new solution.

If we put $\frac{bc}a=x,$ $\frac{ca}b=y$, and $\frac{ab}c=z$, the given condition becomes $xy+yz+zx=3$. And the desired inequality becomes $x+y+z\geq3$. But this follows from the well-known ineq
\[ (x+y+z)^2\geq 3(xy+yz+zx). \]

Ok, the  first one 's easy but the second one is not. Of course, nhat's solution is very nice.

for the general case we can put $a=c \rightarrow 0$ and we find that $k\le \frac{\ln 3}{\ln 3 -\ln 2}$
but im still working on it...

First two inequalities are easy (and true for all positive numbers),

Because inequality is cyclic, we can take $a\geq b\geq c$

We want to proove that

\[ \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a} \geq 3  \]

$a\geq b\geq c$ forces $ab\geq ac \geq bc$

Now using rearrangement inequality we have:

\[ \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq \frac{ab}{b}+\frac{ac}{a}+\frac{bc}{c}=3  \]

OK. Nhat's solution is very suprising, because I created and solve it by SOS.

Any Idea for the last?

The last can be solved by n-1 Equal Variable Theorem if we write it as

$abc(\frac 1 {a^2}+\frac 1 {b^2}+\frac 1 {c^2} \geq 3\sqrt[k]{\frac{a^k+b^k+c^k}{3}}$.

For $abc=constant$ and $\frac 1 {a^2}+\frac 1 {b^2}+\frac 1 {c^2} =constant$, the expression $a^k+b^k+c^k$ is maximal when $0<a=b \leq c$. Thus, we may consider $a=b=1$ and $c \geq 1$.

Actually, the inequality is equivalent to the known inequality

$(\frac{x^{2/k}+y^{2/k}+z^{2/k}}{3})^k \geq \frac{xy+yz+zx}{3}$.

Yes.  :lol:  This approach's use many time  and really nice.

[quote="hungkhtn"]Oh, my solution's very nice and short.[/quote]Can you post this very nice and short solution ? :!:

Ok, suppose that $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}=3$, let
\[ x=\frac{ab}{c},y=\frac{bc}{a},z=\frac{ca}{b} \]

The problem equals to

If $x+y+z=3$, find $r \max (r=k/2)$ for which the following inequality is true :

\[ (xy)^r+(yz)^r+(zx)^r \le 3 \]

And It's really a well-known problem? :)

Of course. I published and solved it in Crux. Math., in 2004. ;)

But my improvement with 4 numbers hasn't any solution in this forum.

Yes, Of course. It's really hard problem. It will appear in my book.


A couple has two children. If we are given that female and male births are NOT equally likely and that identical twins are necessarily the same sex, what is

a) the probability that both are girls given that the oldest is a girl
b) the probability that both are girls given that one of them is a girl

Obtain a general expression for answers in terms of the probability of a female birth and the probability of identical twins (assume that identical twins are just as likely to be female as the overall female probability).

a)$ 1/2$
b)apparently $ 1/3$
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox


What is the least number of children in a family in which each child has at least one brother and two sisters?

Each brother has at least one other brother, so two brothers at least
Each sister has at least two other sisters, so three sisters at least

5.


$ (\forall)n\in \mathbb{N}$ prove that :

$ \sin \frac{\pi}{2n}\cdot \sin \frac{2\pi}{2n}\cdot ...\cdot \sin \frac{(n\minus{}2)\pi}{2n}\cdot \sin \frac{(n\minus{}1)\pi}{2n}\equal{}\cos \frac{\pi}{2n}\cdot \cos \frac{2\pi}{2n}\cdot ...\cdot \cos \frac{(n\minus{}2)\pi}{2n}\cdot \cos \frac{(n\minus{}1)\pi}{2n}$

Isn't this just $ \sin \frac{k\pi}{2n} \equal{} \cos \left(\frac{\pi}{2} \minus{} \frac{k\pi}{2n}\right) \equal{} \cos \frac{(n\minus{}k)\pi}{2n}$?

[quote="azjps"]Isn't this just $ \sin \frac {k\pi}{2n} \equal{} \cos \left(\frac {\pi}{2} \minus{} \frac {k\pi}{2n}\right) \equal{} \cos \frac {(n \minus{} k)\pi}{2n}$?[/quote]

Yes  :)


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[url=http://nbcsports.msnbc.com/id/23978408/]Protesters attacked the Olympic Torch[/url]
[url=http://www.msnbc.msn.com/id/23998157/]Protests in Tibet[/url]
[url=http://www.msnbc.msn.com/id/23964840/]More Protests[/url]
[url=http://www.msnbc.msn.com/id/23629811/]MoreProtests,Again[/url]
So, what do you guys think?
I actually don't get how their human rights are being harmed or anything, but maybe I just don't read these things carefully enough.

there are worse problems in china than tibet.....

many many problems....very very bad problems...

actually...theres soemthing wrong with the whole society.

What? Whole society? Can you ellaborate?

yeah...the whole society is pretty much driven by the want of money..

and anyone with any sort of power abuses it

everywhere u turn, theres bribery and corruption...

 :mad:

but i guess its common for a developing country...

Consider that most countries don't exactly have the best track record either, like the USA and gitmo. Consider also that Chinas has a very large population that other countries simply don't have.

Just putting it out there. I don't know enough to point fingers.

[quote="serialk11r"]In a place without money, money=happiness for 99% of all values of happiness.
In China corruption is bad, but it does exist in the US too. As far as I know, if you're caught with some kind of scandal in China, they shoot you.

How is the whole society bad? The want to rise up, to be wealthier, is what makes China powerful economically. Students in China work much harder than us lazy American bums, because they have VALUES.[/quote]

the fact that they shoot you for scandals makes that a good thing?  :huh:

it exists in US, but just not as much

and hey! i have values!

and we are not lazy...they are just hard working  :D

and in my opinion, the whole world has problems...lots of problems..

i think you can agree with that right? and china is part of the world, so that means china has problems...i would use fancy math symbols to show it, but i don't know how.  :P

and i agree that u.s.'s history is not something that should be too proud of...

but back to tibet....i really liek that design with the handcuffs...very creative.  :lol:

but who wouldn't want to have the chance to extinguish the olympic chance and humiliate one of the biggest governemtn in the world?

First off GO FRANCE
2nd, george bush sucks to much to boycott
3rd, we should
4th, free tibet, and end the war in iraq! :D

Who wouldn't want to? Hmm...maybe because police would arrest you and you would spend a lot of time in jail. You'd also get tons of Chinese people (like me) to hate you.
Also I believe in capital punishment so yes, I think it IS A GOOD THING.

So how exactly do you plan to free Tibet, when not even everyone in America has a negative opinion towards China? Most people have a negative opinion of staying in Iraq. Pulling out will occur sometime or later. Having China give up Tibet is another thing.

Also, have you heard that the Dalai Lama supports the Beijing games? :) An attempted boycott will be useless. My family goes to China every year, and this year ALL the tickets were booked, so we won't be able to go. Try to stop those thousands of people from going to attend the Olympics, who already bought their expensive 1500 dollar tickets.

On a side note, I don't know why this was moved to G&FF, but I will argue seriously about this, so I suggest you people to use slightly more proper English, or you'll lose credibility if you want to argue. May I humbly suggest that this be moved back to the old forum, as I think it is a serious issue that should not be flooded with posts typical to that of G&FF?

[quote="serialk11r"]Then go read them carefully. I don't think human rights are being violated, and I did read them carefully.
The protesters in Paris have no right to go attack other people. They deserve to be put in prison.
And also, China is not the world's biggest prison. The US is.
The Olympics is about international cooperation and peace. People who protest are against the peace.[/quote]
If you want a serious discussion without raising flame wars, you can have one. I'll be glad to participate. However, [b]note that[/b] at the first sign of flaming towards other users, people responsible for that will be banned. The warning has been given, so make sure you respect it.

On the topic now. I'll take each of the 4 sentences above in order.

1) Human rights are being violated for the Tibetans, as they are not let to have their own faith, and/or are prosecuted because of that. There is no denying that China is definitely not one of the top countries in the world in terms of human rights.

2) I agree with this point. People think that by violence they can stop violence which is not really a good course of action.

3) This is the most laughable statement ever. I will not even try to comment it.

4) It is so. The people who protest are not against peace, but they feel that this is the only chance of the Tibetans to make their cause heard to the world (read: otherwise nobody would care that much).

[quote="serialk11r"]Also I believe in capital punishment so yes, I think it IS A GOOD THING.[/quote]As I said ANY off-topic comments and/or starting of a flame will lead to the lock of this topic and/or banning the user who does it. So make sure you are posting on topic.

[b]There will be no other warnings![/b]

1. "United we stand, divided we fall."
   first at all, without judging the human rights and all the stuff... I think we all agree that the media actually tried to control and influence people's thought...and the US definitely don't want china to be united as it will lead china became more powerful. that is why the US is back on both the tibet's problem and taiwan's as well.

2. those protests are not the same side with Dalai Lama.
   those are different groups... those protestants are againt dalai, so the early words about human rights dalai had spoke is not even a reason for their protest.. what they want is something else...(continue reading)

3. what the protestant want:
   return to the "Traditional" life of tibet...which is [b]slavery![/b], religional slavery, where the maximum government is religion and there is slaves who are selled  between masters and slaves "voluntarily like to be slave" because it is a "religious belief" of the slaves. a little similar to ancient Rome empire.

4. get into the statement: "china have a lot problems"
   due to the inexistence of utopia, implies that every country have problems. this is not the great deal... the real problem is why all focused problems of china not problems of other country, in which some are bigger than china's... please think rethink what media wanna us to think, and how we are conditioned(read Brave New World it is a great book). think want the media said about Russia during the cold war...

5. the so called protest are in fact violental acts.
   some American, Canadian, and European reporters in tibet reveals those protestant are really violent, what they did is killing Han-ethinity person, assault and rob their goods etc. chinese police and army have orders to not deffend... that is racial and painful.

[quote="serialk11r"][quote="Valentin Vornicu"]
3) This is the most laughable statement ever. I will not even try to comment it.
[/quote]
Laughable?  I take offense to that...
http://www.commondreams.org/headlines06/1209-01.htm
Many sources indicate that the US has THE most criminals, as well as the most inmates.[/quote]You shouldn't take offense. You should maybe revise your grammar and/or vocabulary skills. When one says "China is the world's biggest prison" it doesn't refer to the fact that China has the largest number of people in jail (or otherwise known as inmates). If you do not understand that then I think your role in this discussion has just ended.

[quote="JBL"]Now, I ask: what is preventing such a process from happening in Tibet?[/quote]
Other than the obvious, how about the fear that throwing out the current government and attempting to install a new government may result in long-term instability or an equally oppressive new government rather than a peaceful and open one? Or that the Tibetan people are so culturally unprepared for democracy that they're unable to run it themselves and someone else has to do it for them for a significant period of time (cf American escapades in the Middle East -- but if the "someone else" who stepped in to help was China then the whole ordeal might even turn out to solve nothing).

[quote="JBL"][quote="beta"]What exactly is the will of the Tibetan people?[/quote]
This is a really good question.  Where I'm from, the way one would figure this out is by allowing the people of Tibet to hold free elections, unencumbered by restrictions on the press, censorship of the internet or police actions against those who support one or another electoral outcome.  Now, I ask: what is preventing such a process from happening in Tibet?[/quote]

The problem is, China is currently ruled by "communism"(eh are these people really communists)... here is the problem: are Tibetans bitter toward the fact that they have to be Chinese or are they simply bitter toward the current government? Based on history we know that Tibetans actually had a pretty good relationship with the Qing Dynasty before the British invaded, so it is probably the latter.

So if Tibetans are only people in China now that do not like the government, then maybe they do have to right to be independent. But if they aren't the only group of people, then we can't just divide China up into 23498249 countries based on regions(most of them were happy to live with one another before the current government) based on regions just because all of them do not like the current government. instead maybe it's better to work together with other to improve the current government.

[quote="beta"]So if Tibetans are only people in China now that do not like the government, then maybe they do have to right to be independent. But if they aren't the only group of people, then we can't just divide China up into 23498249 countries based on regions(most of them were happy to live with one another before the current government) based on regions just because all of them do not like the current government. instead maybe it's better to work together with other to improve the current government.[/quote]
Yes, exactly! But now this presents another very difficult problem! Do you have any ideas? I don't even speak Chinese, so I feel pretty impotent with respect to having an impact on Chinese politics myself or helping people in China to secure more freedom from the government...

I've noticed a common sentiment among my friends that although they still consider themselves Chinese and feel a very strong emotional connection to China, they don't want to live there again. They like the freedom in America. And since they know they won't have to go back, they don't seem very motivated to try to solve anything back home.  :|

That's very interesting Xevarion.

I know that most of the chinese people I know do not feel that way.  Even though I was born here in America, I still consider myself Chinese and I feel a strong connection to China

[quote="Quickster94"]I know that most of the chinese people I know do not feel that way.  Even though I was born here in America, I still consider myself Chinese and I feel a strong connection to China[/quote]
Your friends don't feel which way? I didn't say my friends don't consider themselves Chinese or that they don't feel strong connection to China. In fact they do! (So much that they have influenced me a fair amount in that regard, as you might deduce from this thread...) Anyway, the problem isn't that they don't care about China. It's that they don't feel motivated to try to effect political change in China. They might agree that it would be good if Chinese state-run media wouldn't make false claims all the time, and that there should be more religious freedom and less censorship and a more open and malleable government and so on, but they don't care enough about these issues to want take real action, because they are not really affected.

The reason why some American Chinese don't care about China is because they're living the good life in America, where everyone can drive fat cars, eat out everyday, etc.

serialk11r:  Like I said before, I don't know what to say in response.  I don't think I can convince you of the absurdity of your argument (about the relatives thing) if you can't already see it for yourself.  I accept much of the blame for widening the discussion beyond Tibetan independence, because I did so intentionally.  (I contemplated starting a new thread on China but decided that my post fit this one well enough.)

Enough with the comparisons between China and the United States.  They're completely inapposite, and they don't seem to illustrate any point.

Regarding the Olympics:  Earlier, I discussed some of the reasoning behind boycotting the Olympics (and also why I'm not crazy about the boycott).  But I'd like to stop to rebut extreme statements like, "sport and politics have nothing to do with each other, so there should be no boycott."  A hypothetical:  Would you go to the IMO if it were being sponsored by a state that was conducting genocide?  I sure hope not.  Now I'm NOT saying that China is that awful (although calls for a boycott originally grew out of concerns about Darfur).  What I am saying is that a moral person should and would boycott the Olympics because of politics, if you thought the situation warranted it.  The question is whether your personal line has been crossed.  So in other words, you can go ahead and say that the boycotters are wrong, but I disagree that a political boycott of a sporting event is prima facie unjustifiable.  One more bit about the boycott:  During its bid for the Olympics, China itself tried to argue that one of the collateral benefits of having the Olympics in Beijing would be an improvement in the political climate in China, e.g. human rights.

JBL makes an excellent point that ANY legitimate argument about why Tibet should be a part of a China must contemplate the will of the Tibetan people.  This comment was followed by some discussion about what the Tibetans might or might not want, but I'd like to point out that most of the One China posters on this forum don't seem to care whether the Tibetans want to be a part of China.  After all, there's big important stuff like History and Culture going on here, which are far more important than trivial things such as the desires of the living human beings who are affected.

In general, I'm seeing a lot of bizzarro world reasoning on this thread, so I don't know how much longer I can keep responding.

Finally, I *really* don't want to get into the business of defending CNN, but... characterizing the cropping of a photograph as an act of "photoshopping" is disingenuous.  (Yes, I can understand why the cropping might be considered evidence of bias, but in this case I do not believe that it rises to the level of an actual "lie.")

[Sigh]
What I hope you can see is, by the relatives thing, I mean that since Taiwan has so many mainlanders, that makes it a reason, because if your Taiwan does not = Chinese argument was valid, then the "Chinese" (non-Taiwanese if Taiwanese were not Chinese) in Taiwan would be reason enough for unification. There is a good reason why a slight majority of people in Taiwan want unification.
Canada and the US don't really have anything definite culturally, because there is no real American race. Natives make up a tiny percentage of people, and have little influence. Taiwan as I said, was no less similar to any other province in China than the provinces of China are similar to each other.

[quote="Xevarion"]Other than the obvious, how about the fear that throwing out the current government and attempting to install a new government may result in long-term instability or an equally oppressive new government rather than a peaceful and open one? Or that the Tibetan people are so culturally unprepared for democracy that they're unable to run it themselves and someone else has to do it for them for a significant period of time (cf American escapades in the Middle East -- but if the "someone else" who stepped in to help was China then the whole ordeal might even turn out to solve nothing).[/quote]  These are not answers to my question.  These are reasons that a democratic Tibet might not work very well, or that Tibetans might want to vote to continue their current governmental system.  You're once again muddling up the idea that "Tibetans should have the right to decide whether or not to be independent" with the idea that "Tibet should be independent."  It's quite possible that, given the chance to vote, Tibetans [i]would[/i] choose to remain part of China (though I wouldn't bet on it).  The only reason such an arrangement hasn't been made is because the Chinese government is authoritarian and antidemocratic.  (By the way, American escapades in the Middle East don't provide any evidence of anything like a cultural inability to have democracy -- if anyone wants to hear about what they [i]do[/i] provide evidence for, we can have a new thread.)

[quote="yenlee"]JBL makes an excellent point that ANY legitimate argument about why Tibet should be a part of a China must contemplate the will of the Tibetan people.  This comment was followed by some discussion about what the Tibetans might or might not want, but I'd like to point out that most of the One China posters on this forum don't seem to care whether the Tibetans want to be a part of China.  After all, there's big important stuff like History and Culture going on here, which are far more important than trivial things such as the desires of the living human beings who are affected.[/quote]  Thank you, and I think you put this extremely well.

[quote="beta"]So if Tibetans are only people in China now that do not like the government, then maybe they do have to right to be independent. But if they aren't the only group of people, then we can't just divide China up into 23498249 countries based on regions(most of them were happy to live with one another before the current government) based on regions just because all of them do not like the current government. instead maybe it's better to work together with other to improve the current government.[/quote] I see an assertion here that doesn't make any sense to me.  Why, if everyone is miserable and oppressed under Chinese rule, would that be an argument [i]against[/i] dissolving the Chinese state?  It seems to me that if the Chinese are so incompetent that nearly every group of people in the country wants to leave, this would argue in favor of something new.  "Working together to improve the current government," of course, is something that the Chinese government seems to discourage (cf 1989, Tianamen).

[quote="yenlee"]

JBL makes an excellent point that ANY legitimate argument about why Tibet should be a part of a China must contemplate the will of the Tibetan people.  This comment was followed by some discussion about what the Tibetans might or might not want, but I'd like to point out that most of the One China posters on this forum don't seem to care whether the Tibetans want to be a part of China.  After all, there's big important stuff like History and Culture going on here, which are far more important than trivial things such as the desires of the living human beings who are affected.

In general, I'm seeing a lot of bizzarro world reasoning on this thread, so I don't know how much longer I can keep responding. [/quote]

How many Tibetans that are living in Tibet right now actually want independence?

Last time I checked even the Dalai Lama himself do not want independence. Also all these human rights issues with Tibet are probably happening in the same way in other provinces of China as well. So do people also want Manchuria independence, Shanghai Independence,
Henan Independence, Shangdong Independence as well because you can probably find some human rights abuses in those provinces as well?

[quote="JBL"]These are not answers to my question.  These are reasons that a democratic Tibet might not work very well[/quote]
Isn't the possibility that taking this course of action would lead to an even worse situation in Tibet a reason against trying it? Clearly you don't think it's a sufficiently strong reason, but denying that it's a reason at all seems a bit weird.

I also don't understand why "the US has repeatedly failed miserably at creating democratic governments in other places" is exclusively a statement about our inability, and says nothing about the inappropriateness of the action.

Sure, in theory, a free election would be great! But can you really expect that even if China would let go of Tibet, that a fair election would be held? Let alone quickly! And that the outcome is so much more likely to be better than the current situation, rather than worse, that it's worth trying anyway?

I suppose the main thrust of your argument against my position is that I'm way too optimistic about the ability to get the Chinese government to change. I just think you're too optimistic about how much better it will be if someone wrests control of Tibet from China's grasp and attempts to give it to the Tibetan people.

[quote="Xevarion"]Isn't the possibility that taking this course of action would lead to an even worse situation in Tibet a reason against trying it? Clearly you don't think it's a sufficiently strong reason, but denying that it's a reason at all seems a bit weird.[/quote]  I don't see why you think you or I should be the ones making the risk analysis here, rather than the people (i.e., Tibetans) who would actually be affected by the possible outcomes.  And again, "remaining a fully integrated part of China" is certainly a possible outcome of such an election (it was the outcome in Quebec, for example).

[quote="Xevarion"]But can you really expect that even if China would let go of Tibet, that a fair election would be held? Let alone quickly! And that the outcome is so much more likely to be better than the current situation, rather than worse, that it's worth trying anyway?[/quote]  I don't see why not -- international bodies have stepped in to help many regions run free and fair votes on questions of independence over the years, and there is no theoretical reason that such a thing couldn't happen in Tibet.  It's totally unclear to me why you think this is a bad idea [i]in principle[/i].  (Obviously, the bad behavior of the Chinese government and the fact that I don't think anyone should start a war over Tibet are [i]practical[/i] reasons that what I'm suggesting isn't going to happen.)  Also, this quoted section seems to be full of rhetorical deck-stacking -- why is holding an election "quickly" a precondition for holding an election?  Why does an outcome you prefer need to be "so much more likely" than an unfavorable outcome?  (There was a bit of this in the preceding paragraph, too -- does the possibility of a bad outcome usually prevent you from making decisions?  If so, I imagine you spend a lot of time indoors, huddled under your blankets :P :))

[quote="beta"]Also all these human rights issues with Tibet are probably happening in the same way in other provinces of China as well. So do people also want Manchuria independence, Shanghai Independence,
Henan Independence, Shangdong Independence as well because you can probably find some human rights abuses in those provinces as well?[/quote]
Please see my preceding post.  (I imagine it went up while you were typing.)  I'd love to see a good explanation of how "The Chinese government is bad for all of its citizens" serves to support the legitimacy of the current government rather than to undermine it.  This is, as yenlee said, bizzaro-world thinking that you're using.

[quote="beta"]How many Tibetans that are living in Tibet right now actually want independence?[/quote] And again -- how do you propose we answer this question, if not through free elections?

[quote="JBL"]

Please see my preceding post.  (I imagine it went up while you were typing.)  I'd love to see a good explanation of how "The Chinese government is bad for all of its citizens" serves to support the legitimacy of the current government rather than to undermine it.  This is, as yenlee said, bizzaro-world thinking that you're using.
[/quote]

  Splitting the country up is a legitimate way to "undermine" the government---> this goes back to the idea of unification is a lot better than separation (and this idea is not uniquely Chinese, I see slogans like "United We Stand, Divided We Fall" in America all the time). And to be honest, even if you replace the current government in China, you will just get another oppressive government in Western standards (Chinese culture is different and Chinese people have lived under Emperors for too long)

[quote="JBL"]I don't see why you think you or I should be the ones making the risk analysis here, rather than the people (i.e., Tibetans) who would actually be affected by the possible outcomes. [/quote]
This is a catch-22. We can't ask them whether they think it's a good idea to ask them without first asking them! I think that experts in international politics and Tibetan leaders (e.g. the Dalai Lama himself) should be making the risk analysis.

[quote="JBL"]I don't see why not -- international bodies have stepped in to help many regions run free and fair votes on questions of independence over the years, and there is no theoretical reason that such a thing couldn't happen in Tibet.  It's totally unclear to me why you think this is a bad idea [i]in principle[/i].  (Obviously, the bad behavior of the Chinese government and the fact that I don't think anyone should start a war over Tibet are [i]practical[/i] reasons that what I'm suggesting isn't going to happen.)  Also, this quoted section seems to be full of rhetorical deck-stacking -- why is holding an election "quickly" a precondition for holding an election?  Why does an outcome you prefer need to be "so much more likely" than an unfavorable outcome?  (There was a bit of this in the preceding paragraph, too -- does the possibility of a bad outcome usually prevent you from making decisions?  If so, I imagine you spend a lot of time indoors, huddled under your blankets :P :))[/quote]
I don't think it's a bad idea in principle. But I don't think that just because something is right in principle, we are justified in putting that into practice -- other factors in the situation may not warrant it.

Regarding speed & fairness: it's quite possible that I vastly underestimate the necessary resources to hold a fair election in an undeveloped nation. But it seems to me that there's not sufficient infrastructure in place and a lot of money and effort would have to be spent to hold the election. Where is this coming from? And how to prevent China from trying to interfere with the election?

And about gambling on uncertain outcomes: you're a mathematician! What are you thinking? Of course I'm not so risk averse that I won't make any decision that has a potential bad outcome. But I'm not going to take some action with uncertain outcome if I expect the outcome to be bad, and I think that the potential good outcomes are not that great, or that very good outcomes are highly improbable.

To summarize: I don't care if it's "right" that the Tibetans should be able to decide their own government. What I care about is what's actually going to happen to the people there. And I'm still not convinced that the best process would lead to the best outcome here. But as I said, this is in part because I'm optimistic about Chinese government, and this may be misguided.

Regarding splitting up China further and so on: I agree with beta. There's no reason to expect that the new governments are going to be so much better. They might improve on some small area -- if each small country fits well with the boundary of some ethnic group, that ethnic group will have (much) more freedom than it currently does in China. But I really don't think that the solution to the world's sociopolitical problems is to divide each country into such small chunks that everyone inside is the same. It won't work (i.e. there's no clear stopping point -- smaller and smaller groups will just complain about their lack of freedoms; do you really want NAMBLA to carve its own little city-state out of some unwanted land in the USA?), it's economically inefficient in a variety of ways, and there's no reason to expect that in general the new governments will be better than the old ones. I don't mean to say that this approach can never be right, but I get the impression that many people believe that it's [i]usually[/i] right, and this bothers me.

[quote="Xevarion"][quote="fredbel6"]I think there is a huge difference between "nationalism" and "patriottism".  Usually, when nationalism leads to conflict, it is because one nation claims land occupied by the other nation.  A quick look at the map suffices to see the difference with the war in Iraq and Afghanistan.[/quote]
So you think that if the motivation for war is retaliation (or preemption of aggression) against the country, support of the war is patriotism, while if the motivation for war is connected to actual ownership of territory, it's nationalism? Honestly I think this is really a technical rather than ideological difference. One could say, "patriotism is a name for nationalism once the nation has conclusively asserted control over all of the land it believes to be its own."
[/quote]
I still think that there is a difference with conflicts in Northern ireland, wars in the Balkan, ....  In all these conflicts, people feel that something "close" to them (of high cultural value) for instance is being threatened or simply in the wrong hands.
There has never been an "American minority" in Iraq or vice versa, there are no Iraqi monuments or monasteries in the USA,.....
This is a geopolitical conflict, in my opinion portrayed as a conflict of civilizations (Bush literally said "we wage a war to save civilization itself")


geometry
analytic geometry
geometric transformation
reflection
Pythagorean Theorem

Find the side length of the internal square and the radii of the internal circles, in terms of $a$.

[quote="computer_nuke"]Well I guess you can use coordinate geometry to figue everything out, but you don't have to.[/quote]

Thats right, When I started with co-ordinate geometry, it became messy... may be i goofed up somewhere... so I am looking for an elegant answer.

[quote="lotrgreengrapes7926"][hide] One corner of the large square is at the origin. The equations of the quarter circles are $x^{2}+y^{2}=a^{2}$ and $(x-a)^{2}+y^{2}=a^{2}$. Let's call the side length of the small square $s$. The equation of the right side of the small square is $x=\frac{a+s}{2}$. Substituting in, we get

\begin{eqnarray*}(\frac{a+s}{2})^{2}+y^{2}&=&a^{2}\\ \frac{a^{2}+2as+s^{2}}{4}+y^{2}&=&a^{2}\\ y^{2}&=&\frac{3a^{2}-2as-s^{2}}{4}\\ y&=&\frac{\sqrt{3a^{2}-2as-s^{2}}}{2}\end{eqnarray*}

This is equal to $s$.

\begin{eqnarray*}s&=&\frac{\sqrt{3a^{2}-2as-s^{2}}}{2}\\ s^{2}&=&\frac{3a^{2}-2as-s^{2}}{4}\\ 3a^{2}-2as-5s^{2}&=&0\\ (3a-5s)(a+s)&=&0\\ s&=&\boxed{\frac{3a}{5}}\end{eqnarray*}. [/hide][/quote]

You had to also find the radii of the other two circles. The one on top of the square and the one on that of that one LOL. Have fun  :D

[quote="JBL"][quote="computer_nuke"]You had to also find the radii of the other two circles. The one on top of the square and the one on that of that one LOL. Have fun  :D[/quote]

After you know the side of the square, it's not so hard -- in fact, there's no need for coordinate geometry here at all, just adding a lot of radii and right triangles.  The side-length of the square could have been done this way as well, you just would have called on the Pythagorean Theorem instead of the formula for a circle.[/quote]

Could you give a solution of it without coordinate Geometry? I kept on drawing the picture and triangles like you said but just can't see anything. Thanks

[quote="lotrgreengrapes7926"][quote="computer_nuke"]Could you give a solution of it without coordinate Geometry? I kept on drawing the picture and triangles like you said but just can't see anything. Thanks[/quote][hide] Draw a segment from the lower left corner of the large square to the upper right of the small square. Then we have a right triangle.

$\begin{eqnarray*}(\frac{a+s}{2})^{2}+s^{2}&=&a^{2}\\ \frac{a^{2}+2as+s^{2}}{4}+s^{2}&=&a^{2}\\ a^{2}+2as+s^{2}+4s^{2}&=&4a^{2}\\3a^{2}-2as-5s^{2}&=&0\\(3a-5s)(a+s)&=&0\\3a&=&5s\\s&=&\frac{3a}{5}$.

Which is easier than the coordinate geometry way.[/hide][/quote]

Nice Solution.

[quote="lotrgreengrapes7926"][quote="computer_nuke"]Could you give a solution of it without coordinate Geometry? I kept on drawing the picture and triangles like you said but just can't see anything. Thanks[/quote][hide] Draw a segment from the lower left corner of the large square to the upper right of the small square. Then we have a right triangle.

$\begin{eqnarray*}(\frac{a+s}{2})^{2}+s^{2}&=&a^{2}\\ \frac{a^{2}+2as+s^{2}}{4}+s^{2}&=&a^{2}\\ a^{2}+2as+s^{2}+4s^{2}&=&4a^{2}\\3a^{2}-2as-5s^{2}&=&0\\(3a-5s)(a+s)&=&0\\3a&=&5s\\s&=&\frac{3a}{5}$.

Which is easier than the coordinate geometry way.[/hide][/quote]

How about the radii of two inner circles? No one has attempted that yet...

[color=darkblue]Denote : the square $ABCD$, where $AB=1$ ; the circle $w=C(B,1)$ ; the reflection $E$ of the point $A$ with respect to the point $B$ ($\{C,E\}\subset w$) ; the small square $MNPQ$, where $Q\in w$, $M\in (AB)$, $N\in (MB)$ ; $MN=2x$. Then $QM=2x$, $AM=\frac{1}{2}-x$, $ME=\frac{3}{2}+x$. Thus, $QA\perp QE$ $\Longleftrightarrow$ $QM^{2}=MA\cdot ME$ $\Longleftrightarrow$ $(2x)^{2}=\left(\frac{1}{2}-x\right)\left(\frac{3}{2}+x\right)$ $\Longleftrightarrow$ $20x^{2}+4x-3=0$ $\Longleftrightarrow$  $x=\frac{3}{10}$, i.e. the lenght of the small square is $\boxed{\ 2x=\frac{3}{5}\ }$ and in this case $AM=NB=\frac{1}{5}$. Prove easily that the length-radius of the inner circle is $\frac{39}{320}$ and the length-radius of the outer circle is $\frac{1}{16}\ .$[/color]

[quote="xxxyyyy"][quote="computer_nuke"]And similartly to what lotrgreengrapes7926 did, I found that the radius of the outer circle (the one on top of the circle on top of the square) to be $a/16$. Any confirmations?[/quote]

I think its correct... how did you arrive at this?[/quote]

You have $(a+r)^{2}=(a-r)^{2}+(a/2)^{2}$

Then solve to get $r=a/16$


[img]http://img84.imageshack.us/img84/2313/matrix1.jpg[/img]

Here is the answer

[img]http://img84.imageshack.us/img84/4820/matrix2c.jpg[/img]

It seems that we cannot sub s=1,t=0,q=0 in this case.Why? A unit vector cannot be orthogonal to the vecors?

No, you can make that choice. There is a three-dimensional space of vectors orthogonal to the three given vectors, and the question only asked for one element of the space. Don't mistake an example answer for the only answer.


Let $ ABC$ be an acute-angled triangle with $ AB>AC$ and $ \angle BAC\equal{}60^{\circ}$. Denote the circumcenter by $ O$ and the orthocenter by $ H$ and let $ OH$ meet $ AB$ at $ P$ and $ AC$ at $ Q$. Prove that $ PO\equal{}HQ$.

Too easy :wink: [hide="For Nayel bhai"][eimatro ghum theke uthlam.uthei solve koresi.So,vaab marlam :P ]
[/hide]
We know that, $ AH \equal{} 2R \cos A \equal{} 2R \cos 60^{\circ} \equal{} R$ and $ AO \equal{} R$
So, $ AO \equal{} AH$
And also, $ \angle AHP \equal{} \angle AOQ$ and $ \angle HAP \equal{} \frac {\pi}{2} \minus{} \angle B$
From $ \triangle CAO$ we get, $ \angle CAO \equal{} \frac {\pi \minus{} \angle B}{2} \equal{} \frac {\pi}{2} \minus{} \angle B$

So, by ASA we get, $ \triangle HAP \cong \triangle OAQ$
Which implies, $ QO \equal{} HP$ and $ HQ \equal{} OP$

tora sob danob hoye, jaccis.................

toooo ez.CD and BE are altitudes
let M be the midpoint of AB.
$ <OAM\equal{}<HAE\equal{}90\minus{}C$  for isogonal conjugate
AE= ABsin30=AM

so AOM and AHE are congruent.
now $ <QHE \equal{}180\minus{}<EHA\minus{}<AHO\equal{}180\minus{}<AOM\minus{}<AOH\equal{}<MOP$
so QHE and POM are congruent and we are done :rotfl:


abstract algebra
superior algebra
superior algebra solved

For the group Sn of all permutations of n distict symbols, what is the number of elements on Sn? Show that Sn is not abelian for n>2.

Uh, $|S_n|=n!$ (how many ways can you permute the numbers $1,2,\ldots,n$?)

Not only is $S_n$ not abelian for $n>2$, it has trivial center.  Let $\sigma\in S_n$ be a permutation such that $\sigma(a)=b$ with $a\not=b$.  Pick $\tau\in S_n$ such that $\tau(a)=c$ and $\tau(b)=b$, with $a\not=c$

Then $\tau\circ \sigma(a)=b$, whereas $\sigma\circ \tau(a)=\sigma(c)$.  But $\sigma(c)\not=b$, since $\sigma(a)=b$.


Is the function

$f(x,y)= 0$ when $y \neq 0$, $x^{2}$ when $y=0$

differentiable at the origin?

$f(x,y)=\begin{cases}0~\text{when}~y\neq 0\\ x^{2}~\text{when}~y=0\\ \end{cases}$

Should have continued in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129599]original thread[/url].

Your function is differentiable. Try something more drastic: a function that is equal to $0$ unless $0<y<x^{2}$, but equal to $1$ when $y=x^{2}/2$ and $x>0$. This one surely is not differentiable at $0$ - it's not even continuous there.


How many interior diagonals does an icosahedron have?  (An $ \emph{icosahedron}$ is a 3-dimensional figure with 20 triangular faces and 12 vertices, with 5 faces meeting at each vertex.  An $ \emph{interior}$ diagonal is a segment connecting two vertices which do not lie on a common face.)

[quote=brainomega][hide="sol 1"]12 vertices * 6 "non-adjacent" /2 for overcounting =36[/hide]
[hide="sol 2"]$ \binom{12}2\minus{}20*3/2\equal{}36$
20*3/2 for edges[/hide][/quote]

I have a question: how did you find $-20*\frac 3 2?$ I am confused...

@above, not really a nice comment; no need for 1 year bump
@2above, They found that beacuse of the edges.

wow the first post of this was almost 12 yrs ago and people are still posting.

[hide=Solution]
There are $12$ vertices in an icosahedron. So there are $\binom{12}{2}=66$ total diagonals.

However, this counts the edges of the icosahedron. There are  $3$ diagonals per face, but that counts each one $2$ times (each edge is part of $2$ faces) so you have to divide by $2$.

There are $20$ faces, so the number of edges is $\frac{3\times20}{2}=30$.

We subtract $30$ from the total number of diagonals, which is $66$, to get $66-30=\boxed{36}$ interior diagonals in an icosahedron.
[/hide]


function
limit
real analysis
real analysis theorems

f:R^2 -> R

f(x,y) = $ e^{ \minus{} x^{2}/y^{2}}$ if y is not 0, and 0 if y is 0

a) At (1, 1), is f continuous?
b) At (1, 0), is f continuous?

For a) I think it is continuous at (1, 1). My reasoning is that $ e^{ \minus{} x^{2}/y^{2}}$ is continuous on $ R^{2} \minus{} \left\{(x, 0)\right\}$, and $ (1,1) \in R^{2} \minus{} \left\{(x, 0)\right\}$.

For b) I don't think this is true... but not sure how to show it. I am trying to work with the definition: "The function f is continuous at the point c if for every sequence (x_n) in X with limit lim x_n= c, we have lim f(x_n) = f(c)."

So if f is continuous at (1, 0), according to the definition above, that means for any lim (x_n, y_n) = (1, 0), we must have lim f(x_n, y_n) = f(1,0) = 0.

But if I choose (x_n, y_n) = (1, 1/n), the limit is (1,0). But lim f(x_n, y_n) = lim $ e^{ \minus{} n^{2}}$, which does not equal 0. Does this show that f is not continuous at point (1,0)?

How does that sound? Thank you for any help in advance.

$ \lim_{n\to\infty}e^{\minus{}n^2}$ [i]does[/i] equal zero.

This function is continuous - in fact, smooth - at $ (1,0).$ The only point at which it's not continuous is $ (0,0).$

[quote="Kent Merryfield"]$ \lim_{n\to\infty}e^{ \minus{} n^2}$ [i]does[/i] equal zero.

This function is continuous - in fact, smooth - at $ (1,0).$ The only point at which it's not continuous is $ (0,0).$[/quote]

Thank you, I didn't catch that. Does my reasoning for part a) make sense, though?

Sure. A composition of continuous functions is continuous. We only need to spend any real energy here investigating what happens as $ y\to 0.$


This problem 's posted before . Try searching with the key word : " good ineq " (author : Suweijie)  ;)


blogs
function
geometry
geometric transformation
reflection
quadratics
analytic geometry

Well at the first look at it, $(1,1)$ is a solution to the system.

[hide]$\ 2-x^3=y$
$2-y^3=x$

$2-x^3=(-x+2)^{\frac{1}{3}}$
$(2-x^3)^3=(-x+2)$
$-x^9+6x^6-12x^3+x+6=0$

Use a little synthetic division, only one that works is $x=1$. Subsitute back into the equation...

Unless my calculations are flawed (really tired)...
$\boxed{x=1;y=1}$[/hide]

Also take a look at the graph.  You'll see that more solutions can't exist.

$2-x^3=y$
$2-y^3=x$

$2=x+y^3$

$(x+y^3)-x^3=y$

$y^3-y=x^3-x$

Now in that last step, i don't know how to prove $x=y$ even though it is evident, so I just kept on going:

$2-x^3=x$
$x^3+x-2$
$(x-1)(x^2+x+2)$

Since both x and y are real numbers we omit the $x^2+x+2$ factor and we $x=1$ and the solution set id $(1,1)$ after all..........

[quote="Jhonny Ruiz"]$y^3-y=x^3-x$
Now in that last step, i don't know how to prove $x=y$ even though it is evident, [/quote]
Unfortunately, by the time you get to this point you've thrown away too much information: the function $f(x) = x^3 - x$ has many instances such that $f(x_0) = f(x_1)$ but $x_0 \neq x_1$.  For instance, $f(-\frac 1{\sqrt{3}}) = f(\frac 2{\sqrt 3})$.  That is, the function is not injective (or one-to-one).

rearrange these, they are inverses, thus the intersection occurs at x=y

[quote="mna851"]This problem appeared on chess64's blog but it is currently unsolved:

$\ 2-x^3=y$
$\ 2-y^3=x$

$\ x$ and $\ y$ are reals.[/quote]

Well going on probability1.01's hint: (which was also posted in chess64's blog)

[hide]
Let $y=f(x)=2-x^3$. Then $2-[f(x)]^3=x\implies \sqrt[3]{2-x}=f(x)$.

So $2-x^3=\sqrt[3]{2-x}$.

But $2-x^3$ and $\sqrt[3]{2-x}$ are inverses. The inverse of a function is its reflection over $y=x$, so if they are ever equal, the point must lie on $y=x$.

Thus we have $2-x^3=x$ or $x^3+x-2=0$. Checking possible roots by RRT, we find that $x=1$ works and it factors as $(x-1)(x^2+x+2)=0$. The discrimant of the quadratic factor is $-7<0$ so it doesn't have any real roots.

$\therefore (x,y)=\boxed {(1,1)}$.
[/hide]

[quote="chess64"]
But $2-x^3$ and $\sqrt[3]{2-x}$ are inverses. The inverse of a function is its reflection over $y=x$, so if they are ever equal, the point must lie on $y=x$.
[/quote]

That's not true.  Consider the functions $y = 2 - x^2$ and $y = \sqrt{2 - x}$.  The point
\[ \left( \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} \right) \]
lies on both graphs, but the coordinates are not equal.

[quote="kokosai"][quote="surge"][hide]$\ 2-x^3=y$
$2-y^3=x$

$2-x^3=(-x+2)^{\frac{1}{3}}$
$(2-x^3)^3=(-x+2)$
$-x^9+4x^6-4x^3+2x^6+x+6=0$[/hide][/quote]I think you maybe meant $6+x-12 x^3+6x^6-x^9=0$.  I like surge's approach because it leads to the complex solutions as well (although the problem isn't asking for them).[/quote]

yeah, that's what I mean. :) didn't combine like terms and messed up somewhere with the x^3 term.

[quote="nsato"]...Consider the functions $y = 2 - x^2$...[/quote]

This function doesn't have an inverse...

Well, let's make it simpler.  Consider the function $y = -x$.  The inverse function is also $y = -x$.  The point $(1, -1)$ lies on both graphs, but does not have equal coordinates.

[quote="chess64"]This function doesn't have an inverse...[/quote]

That's true, but it's not really the point.  The point is if I take a graph and reflect it over the line $y = x$, the intersection points do not have to lie on the line $y = x$.

If you need another example, take $y = -x^3$.  The inverse is $y = -x^{1/3}$, and these two graphs intersect at the points (-1,1), (0,0), and (1,-1).  The points (-1,1) and (1,-1) do not lie on the line $y = x$.

it only matters the fixed points $f(f(x))=x$, which don't neccesarily lie on $y=x$ unless $f(x)$ is always increasing or decreasing.

[hide="Isnt it as simple as..."]
$x^3 + y= y^3 +x \implies x^3 - y^3 = x-y \implies (x-y)(x^2+xy+y^2)=x-y$
$\implies (x-y)(x^2+xy+y^2 - 1)=0$. Therefore (by zero-product rule) either $x=y$ or $x^2+xy+y^2=1$.

$x=y \implies 0=x^3+x-2$ with roots $x=1,  x=\frac{-1 \pm i\sqrt{7}}{2}$. This gives $(1,1)$.

$x^2+xy+y^2 = 1 \implies x=\frac{ -xy \pm \sqrt{-3x^2 y^2}}{2x^2}$. For $x$ to be $\in \mathbb{R}$ we must have $-3x^2 y^2 \geq 0$ which is only possible for either $x=0,  y = 0$. There is no value such that $2=x^3=x$ so this case is ruled out and the only solution is $\boxed{(1,1)}$.[/hide]

[quote="chess64"][quote="cincodemayo5590"]$x^2+xy+y^2 = 1 \implies x=\frac{ -xy \pm \sqrt{-3x^2 y^2}}{2x^2}$[/quote]

Umm, no. Go look up the quadratic formula again.[/quote]

oh wow i really messed that up... ugh ok whatever i meant $x^2+xy+y^2 = 1 \implies x=\frac{ -xy \pm \sqrt{(xy)^2-4x^2 (y^2-1)}}{2x^2}$

[quote="cincodemayo5590"]
oh wow i really messed that up... ugh ok whatever i meant $x^2+xy+y^2 = 1 \implies x=\frac{ -xy \pm \sqrt{(xy)^2-4x^2 (y^2-1)}}{2x^2}$[/quote]

that is not correct either, you are considering the polynomial in x, therefore the coefficient of the first degree term is y, not xy,

i think you are going for

$x^2+(y)x+(y^2-1)=0$ where the coefficients are in parenthesis

so... $x=\frac{ -y \pm \sqrt{y^2-4(1)(y^2-1) } }{2}=\frac{ -y \pm \sqrt{4-3y^2 } }{2}$

this again does not really get us anywhere, when we substitute it back in, we get a 6th degree polynomial (after squaring to get rid of the square root)

i will think a bit as to another solution, i think i have seen a similar problem before...is this from aops btw?

ugh whatever i'm tired and i can't do math after gaming... im gonna go back to gaming :P

oh, ok i got it now

let $f(x)=2-x^3$,

our system becomes

$f(x)=y$
$f(y)=x$

assume for sake of contradiction that $x$ and $y$ are different, WLOG $x>y$
either by graphing, or a little calculus, $f(x)$ is strictly decreasing, so

$x>y \implies f(x)<f(y)$, substituting the equalities into $x>y$

gives: $f(y)>f(x)$

contradiction

the only possibility left is: $x=y$

$x^3+x-2=0=(x-1)(x^2+x+2)$

clearly the other factor produces imaginary numbers, so $\boxed{x=y=1}$ is the only [i]real[/i] root

----

also on the topic of inverses, i agree with nsato (of course  :) ), but there is still some information that we can use with the fact that they are inverses, however, it might not be useful here, more specifically, suppose that $(x_0,y_0)$ is a solution to the system, then $(y_0,x_0)$ is also a solution

[quote="Ravi B"][quote="Altheman"]let $f(x)=2-x^3$,

our system becomes

$f(x)=y$
$f(y)=x$

assume for sake of contradiction that $x$ and $y$ are different, WLOG $x>y$
either by graphing, or a little calculus, $f(x)$ is strictly decreasing, so

$x>y \implies f(x)<f(y)$, substituting the equalities into $x>y$

gives: $f(y)>f(x)$

contradiction[/quote]Where is the contradiction?[/quote]

woops...lol i guess that only works for [i]increasing[/i] functions, this method should work, i cannot seem to resolve this now...the only thing that i have found (as i said before) that the intersections are symmetric about y=x


what happened to Alcumus?
It is not letting me on it!!!

I believe Alcumus is upgrading.  See the following topic:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294119

this is the third time someone mentioned this  :(
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293979[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294119[/url]

well sorry, i wz jst wndring...dsnt meen U have to rate my post low!


A class of 40 children is 60$ \%$ boys. How many boys would have
to be replaced by girls to make a majority of the class girls?

If 60% of a class of 40 students are boys, that means that 24 are boys and 16 are girls. We want there to be more than 50% of the class to be girls, so let's try some values.

We have 24 boys and 16 girls. If four boys were replaced by girls, we would have exactly 50% girls and 50% boys. But we want a majority to be girls. Hence, our answer is $ \boxed{5}$.

Please tell me if this explanation was unclear, or if you have any other questions.

btw im the one who clicked "ask on forum"
is there some kind of formula you use, though?

Ok use of inequalities:

Let $ x$ be the number. The inequality that follows is
\[ (40\cdot.6) - x < (40\cdot.4)+x
\]
Solving:
\begin{align*} (40\cdot.6) - x & < (40\cdot.4) + x \\
24 - x < 16 + x \\
8 < 2x \\
4 < x \end{align*}
Since there is no such thing as a fractional student, the smallest amount is $ \boxed{5}$.

There is no formula.


function
limit
calculus
absolute value
calculus computations

Hi,

I had a question about limits that no one else could answer.

In this book I have, it talks about using the sandwich theorem to prove that lim x->0 (x sin(1/x)) = 0.

The example goes on to show that lim x->0 abs(x sin(1/x)) = 0,

and then says, lim x->0 (x sin (1/x)) = 0, by the definition of a limit.

What I don't get it, how do you go from proving that the limit of the absolute value of a function is zero to saying that the limit of the function itself is zero?

Maybe i'm stupid, but i don't get it.

Please help,

Vinay

Theorem:

If $\lim_{x\to a}|f(x)|=0$ then $\lim_{x\to a}f(x)=0.$

Proof:

Let $\epsilon>0.$ By the definition of limit, there exists $\delta>0$ such that if $0<|x-a|<\delta$ then $|\,|f(x)|-0|<\epsilon.$

But $|\,|f(x)|-0|=|f(x)|=|f(x)-0|.$ Hence we have shown that for $x$ in the region specified, $|f(x)-0|<\epsilon$ and hence we have proved that $|f(x)|\to0.$

Now assume $L>0.$ Are the following proposed theorems true or false? Why or why not?

Proposed theorem 1:

If $\lim_{x\to a}f(x)=L,$ then $\lim_{x\to a}|f(x)|=L.$

Proposed theorem 2:

If $\lim_{x\to a}|f(x)|=L,$ then $\lim_{x\to a}f(x)=L.$

Wow thanks,

I think both of your proposed theorems are false.  I can't provide a proof, but I can try to give a counter example...

Let f(x) = -x^2 and a = 2

So, lim x->2 (f(x)) = -4, but...

lim x->2 |f(x)| = 4...

so, both the theorems aren't true

right?

Vinay

[quote="Kent Merryfield"]I did specify that $L>0$, so that's not a counterexample.[/quote]

Proposed Theorem 1 - True

$
| {f( x ) - L} | < \varepsilon \,\,\, \Rightarrow \,\,\, L - \varepsilon \,\, < \,\,f( x ) < \,\,L + \varepsilon
\, $  therefore taking $0 < \varepsilon  < L \,$ ...

Proposed Theorem 2 - False

Take $ a = 0$  and $f( x ) = \{ \begin{array}{l}
 \,\,\,1\,\,\,if\,\,\,x \ge 0 \\
  - 1\,\,\,otherwise \\
 \end{array}  \, \, \, \, \, $

Then $ L=1 \, >0 \, \, $ but...

[quote="Kent Merryfield"]That's correct, fskilnik.[/quote]
 ;)

Those "proposed theorems" should appear in every Calculus book, Mr. Merryfield.


Given $ a, b, c, d \ge 0$ satisfy $ a\plus{}b\plus{}c\plus{}d\equal{}4$.
Prove that: $ a^2bc\plus{}b^2cd\plus{}c^2da\plus{}d^2ab \le 4$  :maybe:

[quote="nguoivn"]Given $ a, b, c, d \ge 0$ satisfy $ a \plus{} b \plus{} c \plus{} d \equal{} 4$.
Prove that: $ a^2bc \plus{} b^2cd \plus{} c^2da \plus{} d^2ab \le 4$  :maybe:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960


If you have ever played pinball (the original version), post your list of high scores here.

Please do not post PhotoShopped (or in any other way edited) pictures as I want to see how far people can actually get in this game.

Here are my records:

[img]http://i3.tinypic.com/67z398k.png[/img]

I had racked up over 7 million points at one point, but that was before my hard drive died. Now I don't have time to play.  :(

i think ive gotten 30 million+ or 3million, but 3million is easy, so maybe 30 million, idk

maybe it was 17 million

During my highest scoring game (which was before the second highest), I would've gone on indefinitely, but I got distracted by something for a mere fraction of a second, and that was all it took. But still, 11 million is pretty good for someone who rarely plays.

5 million. I've gotten it three times in my lifetime: when I was like 5, and there wasn't that much time for pinball, and twice this year.

I like getting to Ensign.

psh, i can get way more than you guys. i cheat though :P  (any of you know the "hidden test" cheat?)

[quote="miyomiyo"]I have a mac, but I know I've gotten at least 10 million on other peoples computers.[/quote]

Welcome to the club! Do you have parallels?

[quote="i_like_pie"]@ Nerd_of_the_Ages - With cheats, I can get [whatever the maximum score is], because there's a cheat that lets you edit the high scores list.[/quote]

do you mean a cheat that was originally built in to the game? or using a program or something. cuz my cheat is actually in the game (kind of like an easter egg or something)

When I was like 8 I got 9M but I'm not sure what it is now.

wow....

i suck worse then you do...person that posted above me...

3,700,000 ish....

I could have gone on and on...

[img]http://img225.imageshack.us/img225/5035/pinballaw2.png[/img]

@ Anirudh - Did you not read the first post at all? I requested real pictures, not PSed! For God's sake, stop trying to show off with your editing "skills".

And don't even [i]think[/i] about telling me that it's possible to get 19M on level 2. Just don't.

[quote="ashpotter"]technally, i would have gotten so high of a score that the computer would wreck (this also happens to the old 9Xs when my dad played Tetris.
But if that happens, you get so many free balls that you can't lose.
This is how you achieve the score:
At the beginning of the game, type "hidden test" and then you would be able to drag around the ball. what you can do then is drag it into hyperspace and keep it there untill you break the computer. Have fun!![/quote]

Where do I type it???

ok... Sorry for double posting but I just typed h, and the scores are now

5- 2 million and something,

4 and 3- 3 mil and something

2- 1 billion???????

1- 7 million and something....

[hide="extreme cheats"]when you are at the beginning, player 1, ball 1, score 0, type "hidden test". Then you can type the following letters:

e(or was it r?): instant promotion
h: sets second place in the high score list to 1 billion.

you can also click and drag the ball around.[/hide]

[quote="i_like_pie"][quote="#H34N1"]Where is hyperspace?[/quote]
So you're saying you got 19M without even knowing what hyperspace is? :roll:[/quote]

I finally figured it out. Its that little white flapper thing :D

Skill:

[img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/skill.gif[/img]

[quote="#H34N1"][quote="i_like_pie"][quote="#H34N1"]Where is hyperspace?[/quote]
So you're saying you got 19M without even knowing what hyperspace is? :roll:[/quote]

I finally figured it out. Its that little white flapper thing :D [/quote]

...

Nice try, but it's impossible to get down from the red wormhole all the way to where your ball is while the "Hit Red Wormhole" thing is still written there. Also, you over-did the "pro-looking" field with all 4th level bumpers and stuff. If you're going to do that, then at least don't forget the 10x field multiplier. :wink:

Other than that, however, nice PS skills. :P

Dude the 4th level bumpers are easy to to. Just keep hitting it up there.

Ok. I videotaped the screen getting 10x and the reds and w/e. So yeah. Then I got sleepy and crashed, but yeah. It shows I can get all reds/10x. So I'll post that on youtube and put it on sometime next week, after I get uSAMtS done.

I never thought I of all people would be saying this to someone, but you guys have no life.  Seriously, do homework.  Read.  Hang out with friends.  Go to http://www.familyguy.com.  But virtual pinball....?

[quote="i_like_pie"]4th level bumpers + 10x field multipliers isn't hard to get, and I never said it was. I said that combined with Fleet Admiral level + 23M score is quite impressive.

I taped a game of mine today: 8.2M first try. :D It'll be up on YouTube soon...

EDIT: YouTube is being stupid and not letting me upload a 249 Mb file. So, no vid. :(

[hide="A brief description"]Ball 1: 5M; an opportunity for an extra ball, but completely messing up on that; all red for approximately 1 second (as soon as I got the attack bumpers red, the others went back to yellow); got to 2nd rank, but completed my first mission only after 500K points (couldn't get it in the ramp). To complete a mission I purposely let the ball sink (I had a replay as insurance). Unfortunately, it took only a couple minutes after that for the ball to sink unintentionally.

Ball 2: 200K. Took less than a minute; stupid mistake.

Ball 3: 3M. Some close calls at first, but then things cooled down. Again a chance for an extra ball, and again a failure to claim it. Almost got gravity field, but missed hyperspace so many times. After completing one mission, I started another one, but very quickly died due to an unavoidable situation (ball goes in black hole and gets shot out, bounces off two walls, and flies straight between the flippers).[/hide]

I'll tape each ball separately tomorrow, and upload it in three (or maybe more) parts.[/quote]

Ok. I'll then try not to upgrade myself again :D. I'll make 23M. Also, change the fps to 4 in Hypercam.

what is your average score. geez i must suck best: 9 mil.  avg:1 mil


APMO result has been posted...I got HM

http://www.kms.or.kr/competitions/apmo/

Congratz to all winners!

Congratulations to all winners and ppl in the HM!

how can you participate in the APMO? are there requirements?

If they haven't changed it from previous years, APMO participants are selected based on their performance on the COMC (plus people who have attended past Winter Camps, and random high level things like that).  I think less than 50 people in Canada write it, though, so if you screw up on COMC, but want some hard olympiad-writing experience, check with a local math teacher to see if you can write it unofficially with official participants in your area.  Or, if you don't particularly care about writing it in a "real time" contest atmosphere, just download the problems once they're posted online and work on them.


induction
linear algebra
matrix
superior algebra
superior algebra unsolved

Wrong for $ n\equal{}2$ if $ A_1\supset A_1\cap A_2\subset A_2$ have all the same measure. E.g. if $ A_1\equal{}A_2$.

[quote="hamidreza"]Let $ A_i$  for $ i \equal{} 1,2,\dots,n$ be the sets such that $ 0 < \mu_L(A_i) < \infty$ where $ i \equal{} 1,2,\dots ,n$.
Let $ A \equal{} (a_{i,j})_{n \times n}$ where $ a_{i,j} \equal{} \mu_L(A_i \cap A_j)$. Show that det$ (A) > 0$.[/quote]

please replace det$ (A) > 0$ with det$ (A)\geq 0$  :oops:

Imam Ali (AS):

There is no greater wealth than wisdom; no greater poverty than ignorance; no greater heritage than culture;

and no greater friend and helpmate than consultation.

Proof by induction under the weakened premises $ \mu(A_i)<\infty$. Assume $ \mu(A_1)=1$ wlog and apply the induction hypothesis to $ B_i=A_i\setminus A_1$, $ i=2,\ldots,n$, to get the existence of the Cholesky decomposition $ A=L'L$ with $ L=\left(\begin{array}{cc}{1}& x'\\0 &\tilde{B}\end{array}\right)$.

ps: alf shukriin!

for every $ x_1, \ldots x_n \in \mathbb{R}$:
\[ 0 \leq \| \sum_{i\equal{}1}^n x_i 1_{A_i}\|_2^2 \equal{} \sum_{i,j} x_i x_j \mu(A_i \cap A_j) \]
so the conclusion since the matrix $ A$ is thus symmetric positive.

[quote="alekk"]for every $ x_1, \ldots x_n \in \mathbb{R}$:
\[ 0 \leq \| \sum_{i \equal{} 1}^n x_i 1_{A_i}\|_2^2 \equal{} \sum_{i,j} x_i x_j \mu(A_i \cap A_j)
\]
so the conclusion since the matrix $ A$ is thus symmetric positive.[/quote]

What's the definition of $ 1_{A_i}$?


modular arithmetic
algebra
system of equations
algebra proposed

Find the total number of solutions to the following system of equations: \[ \begin{cases} a^2\plus{}bc\equiv a\pmod {37}\\ b(a\plus{}d)\equiv b\pmod {37}\\ c(a\plus{}d)\equiv c\pmod{37}\\ bc\plus{}d^2\equiv d\pmod{37}\\ ad\minus{}bc\equiv 1\pmod{37}\end{cases}\]

$ a,d$ roots of $ x^2\minus{}x\plus{}bc\equal{}0$ in field $ F_{37}$, therefore $ a,d\equal{}\frac{1\pm\sqrt{1\minus{}4bc}}{2}$.
If $ a\equal{}d$, then $ a^2\equal{}bc\plus{}1\to a\equal{}d\equal{}2bc\plus{}1\to bc(4bc\plus{}3)\equal{}0$.
If $ 4bc\plus{}3\equal{}0$, then $ a\plus{}d\equal{}4bc\plus{}2\equal{}\minus{}1$, therefore $ b\equal{}c\equal{}0$ - contradition.
If $ bc\equal{}0$, then $ ad\equal{}1\to a\equal{}d\equal{}1\to b\equal{}c\equal{}0$. We found one solution $ a\equal{}d\equal{}1,b\equal{}c\equal{}0$.
If $ a\not \equal{}d$, then from Viet theorem $ a\plus{}d\equal{}1,ad\equal{}bc$ - contradition with $ ad\equal{}bc\plus{}1$.
Therefore unique solution $ a\equal{}d\equal{}1,b\equal{}c\equal{}0$.

Let
\[ M \equal{} \left[
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right].\]
Then from the given equations, $ M^2 \equiv M \pmod{37}$, and $ M^{\minus{}1} \pmod{37}$ exists.  Multiplying both sides by $ M^{\minus{}1}$, we get $ M \equiv I \pmod{37}$.  Hence, $ a \equiv d \equiv 1$ and $ b \equiv c \equiv 0 \pmod{37}$.

Very nice solution! Anyway with case analysis we get easely the result starting with b divisible or not by 37


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ with $ f(xf(x)\plus{}f(y)) \equal{} [f(x)]^2\plus{}y$ for all real numbers $ x,y$.

[hide="Solution"]$ f(x)\equal{}x$ or $ f(x)\equal{}\minus{}x$[/hide]

[hide="Hint"]Show that $ [f(x)]^2\equal{}x^2$.[/hide]

Noone?

[hide="The first steps"]Setting $ x\equal{}0$ and $ a\equal{}f(0)$, we have $ f(f(y)) \equal{} a^2\plus{}y$. Thus $ f$ has to be bijective; We can then choose a real number $ x_0$ with $ f(x_0)\equal{}0$. Setting $ x\equal{}x_0$, we get $ f(f(y)) \equal{} y\implies a\equal{}0$.[/hide]

[quote="nayel"]Letting $ x \equal{} 0,y \equal{} t$ we get $ f(f(t)) \equal{} t\,(*)$ for all $ t\in\mathbb R$. Therefore $ f$ is surjective.

Let $ f(a) \equal{} f(b)$ for some $ a,b\in\mathbb R$. Then $ f(xf(x) \plus{} f(a)) \equal{} f(xf(x) \plus{} f(b))$ implies $ (f(x))^2 \plus{} a \equal{} (f(x))^2 \plus{} b\implies a \equal{} b$. Thus $ f$ is injective, which combined with the surjectivity implies $ f$ is bijective.

Now verify that $ x\mapsto f(x)$ with $ (*)$ implies $ f(xf(x) \plus{} f(y)) \equal{} x^2 \plus{} y$ for all $ x,y\in\mathbb R$, since $ f$ is bijective. Thus $ (f(x))^2 \equal{} x^2$ which implies $ f(x) \equal{} x$ or $ f(x) \equal{} \minus{} x$ which are the only solutions.

PS. I didn't see any of the hints or solutions. :P[/quote]

2 flaws:

The first equation should be $ f(f(y)) \equal{} [f(0)]^2\plus{}y$... this already implies bijectivity, anyway.

Also, you only showed that for each $ x,\; f(x) \equal{} x \lor f(x) \equal{} \minus{}x$ - you still have to prove that the same sign apllies for all reals ;)

[quote="bertram"]Noone?

[hide=The first steps]...we have $ f(f(y)) \equal{} a^2 \plus{} y$. Thus $ f$ has to be bijective[/quote]

Would you please explain why that equation gives us both injection and surjection?  :blush:

[quote="polskimisiek"][quote="bertram"]Noone?

[hide="The first steps"]...we have $ f(f(y)) \equal{} a^2 \plus{} y$. Thus $ f$ has to be bijective[/hide][/quote]

Would you please explain why that equation gives us both injection and surjection?  :blush:[/quote]

Of course :D

[b]Theorem: [/b]Suppose $ f(f(x)) \equal{} h(x)$ for a bijective function $ h$. Then $ f$ is bijective, too.
[b]Proof: [/b]First, let's prove injectivity:
$ f(a) \equal{} f(b) \implies f(f(a)) \equal{} f(f(b)) \implies h(a) \equal{} h(b) \implies a \equal{} b$ since $ h$ is injective.
Now surjectivity:
Since $ h$ is surjective, for every real numer $ a$ there is a real number $ b$ with $ h(b) \equal{} a \implies f(f(b)) \equal{} a \implies f$ is surjective, since for each $ a$ the numer $ b \equal{} f(h^{\minus{}1}(a))$ satisfies $ f(b) \equal{} a$.

As a corollary, if $ f(f(x)) \equal{} ax\plus{}b$ and $ a \neq 0$, then $ f$ is bijective.

[quote="bertram"][quote="polskimisiek"][quote="bertram"]Noone?

[hide="The first steps"]...we have $ f(f(y)) \equal{} a^2 \plus{} y$. Thus $ f$ has to be bijective[/hide][/quote]

Would you please explain why that equation gives us both injection and surjection?  :blush:[/quote]

Of course :D

[b]Theorem: [/b]Suppose $ f(f(x)) \equal{} h(x)$ for a bijective function $ h$. Then $ f$ is bijective, too.
[b]Proof: [/b]First, let's prove injectivity:
$ f(a) \equal{} f(b) \implies f(f(a)) \equal{} f(f(b)) \implies h(a) \equal{} h(b) \implies a \equal{} b$ since $ h$ is injective.
Now surjectivity:
Since $ h$ is surjective, for every real numer $ a$ there is a real number $ b$ with $ h(b) \equal{} a \implies f(f(b)) \equal{} a \implies f$ is surjective, since for each $ a$ the numer $ b \equal{} f(h^{ \minus{} 1}(a))$ satisfies $ f(b) \equal{} a$.

As a corollary, if $ f(f(x)) \equal{} ax \plus{} b$ and $ a \neq 0$, then $ f$ is bijective.[/quote]

Thank you  :D  :D  :P


algebra
polynomial
inequalities
absolute value
complex analysis
complex analysis unsolved

What is wrong with this.

I want to show this: Let a_1,...,a_n be positive real numbers. Prove that the polynomial
$ P(x) \equal{} x^n\minus{}a_1 x^{n\minus{}1}\minus{}...\minus{}a_n$ has a unique positive zero.

$ Q(x) \equal{} x^n\plus{}a_1 x^{n\minus{}1} \plus{} ...\plus{} a_n$ has n complex nonzero zeroes. For each of them, we have that

$ 0 \equal{} |x^n\plus{}a_1 x^{n\minus{}1} \plus{} ...\plus{} a_n | \geq \parallel{}x|^n\minus{}a_1 |x|^{n\minus{}1} \minus{} ...\minus{} a_n |$

which implies that |x| is a zero $ y^n\minus{}a_1 y^{n\minus{}1} \minus{} ...\minus{} a_n$. But that implies that there could be more than one unless it is somehow true that all of the zeros of Q(x) lie on a circle in the complex plane!

You appear to be abusing the triangle inequality.  Your inequality is false: try $ a_k \equal{} 1, x \equal{} \minus{} 1$.

This is a special case of [url=http://www.artofproblemsolving.com/Wiki/index.php/Polynomials#Descartes.27_Law_of_Signs]Descartes' Law of Signs[/url].

Edit:  It seems like you're trying to use the inequality $ |x \plus{} y| \ge |x| \minus{} |y|$.  But it's not at all true that $ |x \plus{} y| \ge \parallel{}x| \minus{} |y\parallel{}$; the above inequality holds because there are many cases where $ |y| > |x|$.

[quote="t0rajir0u"] But it's not at all true that $ |x \plus{} y| \ge \parallel{}x| \minus{} |y\parallel{}$;[/quote]

Yes it is. Give me a counterexample.

The inequality you're using is OK, but only if you remove the absolute value signs.  What t0rajir0u posted isn't right, you need at least 3 variables for a counterexample--$ |a\plus{}b\plus{}c| \geq \parallel{}a|\minus{}|b|\minus{}|c\parallel{}$ is clearly false, for example, as when you let $ a\equal{}0$ you get the opposite of the triangle inequality.

So, taking away the absolute value signs, all you've proved is that $ P(|x|)\leq 0$, not that $ P(|x|) \equal{} 0$.  Which seems to imply only that all the complex zeros are on or inside the circle around the origin that passes through the positive real zero.

[quote="LydianRain"]The inequality you're using is OK, but only if you remove the absolute value signs.  What t0rajir0u posted isn't right, you need at least 3 variables for a counterexample--$ |a \plus{} b \plus{} c| \geq \parallel{}a| \minus{} |b| \minus{} |c\parallel{}$ is clearly false, for example, as when you let $ a \equal{} 0$ you get the opposite of the triangle inequality.[/quote]

Thanks for clarifying!


geometry
circumcircle
geometric transformation
reflection
Euler
projective geometry
perpendicular bisector

Consider a triangle ABC,let (O) be its circumcircle and let the BC-perpendicular bisector meet CA and AB at $ A_2,A_1$,define $ B_1,B_2,C_1,C_2$ cyclically.$ A_2B_1 \cap AB \equal{} {M},B_2C_1 \cap BC \equal{} {J},C_2A_1 \cap CA \equal{} {K}$.Prove that:
a/Five points $ A_2,B_1,A,B,O$ are concyclic.
b/$ AJ,BK,CM$ concurrent at one point lie on circle (O)
[img]http://i588.photobucket.com/albums/ss328/pmkhoa_photo/gift.jpg[/img]

PS:Happy birhday to my classmates,Bao Anh and Trung. :P

[quote="No Reason"]Consider a triangle ABC,let (O) be its circumcircle and let the BC-perpendicular bisector meet CA and AB at $ A_2,A_1$,define $ B_1,B_2,C_1,C_2$ cyclically.$ A_2B_1 \cap AB \equal{} {M},B_2C_1 \cap BC \equal{} {J},C_2A_1 \cap CA \equal{} {K}$.Prove that:
a/Five points $ A_2,B_1,A,B,O$ are concyclic.
b/$ AJ,BK,CM$ concurrent at one point lie on circle (O)
[/quote]
a)
Let $ C'$is middle point of $ AB$, difine $ A',B'$ similary
1)$ A,B,A_2,B_1$are concyclic :
$ AB$is parallel to $ A'B'$which is antiparallel to $ A_2B_1$ respect to $ C$, so we have $ AB$ is antiparallel to $ A_2B_1$ respect to $ \angle ACB$, $ \Rightarrow A,B,A_2,B_1$are concycic.
2)$ A,B,A_2,O$are concyclic :
$ AA_2$is parallel to $ C'A'$ which is antiparallel to $ BO$ respect to $ A_1$, so we have $ AA_2$is antiparallel to $ BO$ respect to $ \angle BA_1O$, $ \Rightarrow A,B,O,A_2$are concyclic.

[quote="No Reason"]Consider a triangle ABC,let (O) be its circumcircle and let the BC-perpendicular bisector meet CA and AB at $ A_2,A_1$,define $ B_1,B_2,C_1,C_2$ cyclically.$ A{}_2B{}_1 \cap AB = {M},B{}_2C{}_1 \cap BC = {J},C{}_2A{}_1 \cap CA = {K}$.Prove that:
a/Five points $ A_2,B_1,A,B,O$ are concyclic.
b/$ AJ,BK,CM$ concurrent at one point lie on circle (O)
[/quote]

I have just show that $ AJ,BK,CM$ concurrent , but I still have trouble with the rest , so I 'll try harder
To prove $ AJ,BK,CM$ concurrent , it is equivalent with :
$ \frac{MB}{MA}.\frac{KA}{KC}.\frac{JC}{JB}=1$ (Ceva's Theorem)
Easy to show that : $ \triangle ABC \sim \triangle AC{}_1B{}_2$ and $ \frac{CC_1}{AC_1}.\frac{AB_2}{BB_2}.\frac{BJ}{CJ}=1 \Rightarrow \frac{BJ}{CJ}=\frac{AB^2}{AC^2}.\frac{JB_2}{JC_1}$
$ \Rightarrow \frac{MB}{MA}.\frac{KA}{KC}.\frac{JC}{JB}=\frac{CB^2}{CA^2}.\frac{MB_1}{MA_2}.\frac{BA^2}{BC^2}.\frac{KA_1}{KC_2}.\frac{AC^2}{AB^2}.\frac{JC_1}{JB_2}=\frac{MB_1}{MA_2}.\frac{KA_1}{KC_2}.\frac{JC_1}{JB_2}$
Apply Menelaus 's Theorem again, we also have :
$ \frac{KC_2}{KA_1}.\frac{BC}{AB}.\frac{AA_1}{CC_2} \Rightarrow \frac{KA_1}{KC_2}=\frac{BC}{AB}.\frac{AA_1}{CC_2}$
$ \Rightarrow \frac{MB_1}{MA_2}.\frac{KA_1}{KC_2}.\frac{JC_1}{JB_2}=\frac{AA_1}{CC_2}.\frac{BB_1}{AA_2}.\frac{CC_1}{BB_2}=\frac{A{}_1A{}_2}{C{}_1{}_2}.\frac{sin2C}{sin2A}.\frac{B{}_1B{}_2}{A{}_1A{}_2}.\frac{sin2A}{sin2B}.\frac{C{}_1C{}_2}{B{}_1B{}_2}.\frac{sin2B}{sin2C}=1$
That means $ \frac{MB}{MA}.\frac{KA}{KC}.\frac{JC}{JB}=1$.
Q.E.D

(a) Trivial. $ AB_2 \equal{} CB_2, AC_1 \equal{} BC_1$ $ \Longrightarrow$ $ \angle BB_2C \equal{} \angle BC_1C \equal{} \angle BOC \equal{} 2 \angle A \mod 180^\circ$ $ \Longrightarrow$ $ B_2BOCC_1$ is cyclic.

(b) No calculation, just easy collection of well-known theorems. Suppose $ BCC_1B_2$ is arbitrary cyclic quadrilateral (convex or reflex, does not matter) with a circumcircle $ (O_a),$ not necessary $ O \in (O_a),$ with opposite sides intersecting at $ A \equiv BB_2 \cap CC_1, J \equiv BC \cap B_2C_1,$ its Steiner line connecting orthocenters $ H, H', H_b, H_c$ of $ \triangle ABC, \triangle AB_2C_1, \triangle JBB_2, \triangle JCC_1$ goes through its diagonal intersection $ D \equiv BC_1 \cap CB_2.$ Reflection of this line in any 3 sidelines of $ BCC_1B_2$ concur at the pole of this Steiner line lying on the circumcircle of the corresponding triangle, therefore these 4 reflections and 4 circumcircles $ (O) \equiv \odot (ABC), \odot(AB_2C_1), \odot(JBB_2), \odot(JCC_1)$ conur at a point $ P.$ $ BC, B_2C_1$ are pairwise radical axes of $ (O), (O_a)$ and $ (O_a), \odot (AB_2C_1),$ $ J$ is their radical center $ \Longrightarrow$ radical axis $ AJ$ of $ (O), \odot (AB_2C_1)$ goes through $ P.$ Assume now that $ O \in (O_a).$ Diagonal intersection $ D$ of $ BCC_1B_2$ is then on Euler line $ OH$ of $ \triangle ABC$ (easy by Pappus theorem), so that $ P \in (O)$ is now Euler reflection point (pole of Steiner line identical with the Euler line $ OH$). This makes $ P$ a unique point of $ \triangle ABC$ $ \Longrightarrow$ $ BK, CM$ also go through $ P \in (O).$


vector
analytic geometry
algorithm
probability
logarithms
probability and stats

Let $ S_n$ be the set of vectors of length $ 2n$, each entry being $ 0$ or $ 1$,  and such that there are exactly $ n$ entries with value $ 1$.
How would you do to pick a random element of $ S_n$ (uniform distribution over $ S_n$)

A naive way would be to start with the vector $ (0,0, \ldots, 0)$, and then pick a random integer $ i$ between $ 1$ and $ 2n$ and then switch the $ i$-th coordinate to $ 1$. One can do that $ n$ times, taking care that sometimes the coordinate that one randomly chooses can already be set to $ 1$: in this case, one simply do it again until one finds a new coordinate. This algorithm needs to generate $ Cn\ln(n)$ random number on average (and the worst case is... infinite time :-)

Another way would be to start with $ (1,1, \ldots,1,0,0, \ldots, 0) \in S_n$ and then to perform random transpositions. This is an approximate algorithm: how many random transposition does one have to perform in order to get a good approximation ?

ideas .. ?

Here's what I'd do: generate $ 2n$ random numbers uniformly on $ [0,1]$; most computers have this built in. Find the $ n$ largest and replace them with $ 1$. Find the $ n$ smallest and replace them with $ 0$.

This generalizes easily to choosing something other than half of the numbers, and the uniform distribution isn't important. Any distribution with a low probability of generating the same number twice works. It costs $ 2n$ random numbers and about $ n\log n$ comparisons.

that's nice! and it even works in $ O(n)$ since one can find the median of a set in linear time.

[quote="jmerry"]Can you outline an algorithm for finding the median faster than a full sort? Even better, the $ k$th largest of $ n$ elements? I'm curious.[/quote]

[url=http://www-math.mit.edu/~djk/18.310/Lecture-Notes/median_finding_2007.html]see here[/url] (i did not know this either)

OK. Incidentally, with the uniform distribution I started with, I can find the $ k$th largest element rapidly: less than $ 2n$ comparisons on average. Start by counting the number of elements greater than $ 1 \minus{} \frac {k}{n}$; it's $ k \plus{} O(\sqrt {n})$. Run a [url=http://en.wikipedia.org/wiki/Tournament_sort]tournament sort[/url] on the side which has more elements than it should to find the element of the appropriate rank. Since that sort is memory-intensive, we might want to run a second cut against a fixed element first if we were off by a relatively large amount.


induction
function
number theory
totient function
relatively prime

Show that, for any fixed integer $\,n \geq 1,\,$ the sequence
\[2, \; 2^{2}, \; 2^{2^{2}}, \; 2^{2^{2^{2}}}, \ldots (\mbox{mod}\; n) \]
is eventually constant.

[The tower of exponents is defined by $a_{1}= 2, \; a_{i+1}= 2^{a_{i}}$. Also $a_{i}\; (\mbox{mod}\; n)$ means the remainder which results from dividing $\,a_{i}\,$ by $\,n$.]

Uh, why are you posting old USAMO's in here? This one is 1991/3, if I recall correctly. ;)

You mean Active.

Two different verbs man.

[hide]
Proof by induction

Base Case: n=1
Everything is 0 mod 1.

Indutive Step: Prove it's true for n given its true for all positive integers less than n.

Let $n=2^{k}m$ where $2^{k}$ is the greatest power of 2 dividing n and m is odd. I need to prove the proposition is true $\mod 2^{k}$ and $\mod m$ and then they can be put together by the Chinese Remainder Theorem.

The $\mod 2^{k}$ part is easy, since the sequence will eventually be $0\mod 2^{k}$. For the $\mod m$ part:

Let $\phi(m)$ denote the totient function, the number of positive integers relatively prime to m and less than m. Since x is of course not relatively prime to x, $\phi(m)<m$. Thus the proposition must me true $\mod \phi(n)$ by the inductive hypothesis. So there is such integer r and s and a sequence $b_{0},b_{1},...$ such that:

$a_{s}=b_{0}\phi(m)+r$
$a_{s+1}=b_{1}\phi(m)+r$
$a_{s+2}=b_{2}\phi(m)+r$
...

Then since $a_{i+1}=2^{a_{i}}$

$a_{s+1}=2^{b_{0}\phi(m)+r}$
$a_{s+2}=2^{b_{1}\phi(m)+r}$
...

Euler's totient theorem says that $2^{\phi(m)}\equiv 1\mod m$

$a_{s+1}\equiv 2^{r}\mod m$
$a_{s+2}\equiv 2^{r}\mod m$
...

Thus the sequene remains constant.

[/hide]

[quote="#H34N1"]I wanna make this forum popular.[/quote]
and "incidentally" increase your own post count.  Posting old AMC series problems will not increase the quality of this forum, it will only create more junk to sift through.

I just understood your screenname.

I read the The Road Not Taken by Robert Frost.  Good Poem.

And yeah, I guess these are good problems to people who haven't seen it before, but yeah.

I think this problem is wrong because if you give n=3, then 2,2^2,2^2^2,... has not the same remainder(it has 2 remainders:1 and-1!) :(

For n=3 the sequence is

$ 2,1,1,1,1,...$

While the first term may be different from the other terms, all terms starting with the second are the same. The problem stated that the sequence become [i]eventually[/i] constant.

The problem is indeed correct.

Oh I can't understand why n=3, the sequence is 2,1,1,1...??The sequence is the constant:2,2^2,2^2^2,...!And the sequence become constant with mod m mean the reaminder of each number in sequence deivide m has the same remainder?

Yes it means the remainders of the terms of the sequence when divided by n are constant. For n=3, the terms are 2, 4, 16, 65536,...

$ 2\equiv 2\mod 3$
$ 4\equiv 1\mod 3$
$ 16\equiv 1\mod 3$
$ 65536\equiv 1\mod 3$

So you see the sequence of remainders is eventually just 1's.


prove first line!!!
[img]http://sweb.cz/Felix.Phan/combin.JPG[/img]

thanks

Suppose you have $n$ men and $m$ women, and you want to take: $m$ people to be first-class citizens, some of the remaining women (possibly all or none) to be second-class citizens, and the rest as slaves.

To do this, if in your choice of first-class citizens you take $k$ men, then you have $\binom{n}{k}\binom{m}{k}2^k$ choices: you pick the men in $\binom{n}{k}$ ways, you pick the women in $\binom{m}{m-k}=\binom{m}{k}$ ways, and you take some women ($2^k$ choices). Since you can take any number of men between $0$ and $m$, this is =RHS.

On the other hand, you can first pick the second-class women. If you choose $m-r$, for some $0 \leq r \leq m$, you have $\binom{m}{m-r}=\binom{m}{r}$ ways of doing this, and there are $n+r$ people left among which you can choose your $m$ first-class citizens. This gives the LHS.


if all six angles of ABCDEF are 120 degrees, AB=BC=CD, AF=DE
area FCE=60
find area of ABCDEF

say |AB| = y
and |AF| = x

let's first suppose x = y (so a regular hexagon)

then the surface would be 3 * 60 = 180

but er... if you move EF backwards/forwards... you will have no fixed surface number any more...

are we supposed to express it related to some other surface (not the 60 of course :))  or is there something missing here :?

(or am I wrong somewhere?)

I got the same thing: we have an equation and 2 unknown quantities: y and x (Peter's notation).


Please if anyone can help for this , do as soon as you can . Thanks in advance

I will attach the file from the original source.
It is the 3rd question from the list .

A satellite of mass m orbits a planet with gravitational force $\frac{\mu m}{r^2}$ directed toward the center of the planet O. At time t = 0 the satellite is at point S, a distance a from O and its velocity has magnitude U directed perpendicularly to OS, so that at t = 0, $\dot r = 0$ and $a \dot \theta = U$, where $\theta$ is angle measured from OS.

In curvilinear coordinates, Newton's equations are best derived from the principle of the least action. In polar coordinates, kinetic energy of the satellite is $T = \frac m 2\ (\dot r^2 + r^2 \dot \theta^2)$ and its potential energy is $V = -\frac{\mu m}{r}$.

$L(r, \dot r, \theta, \dot \theta) = T - V = \frac m 2\ (\dot r^2 + r^2 \dot \theta^2) + \frac{\mu m}{r}$

is the lagrangian, $S = \int_{t_1}^{t_2} L\ \text dt$ is action from time $t_1$ to $t_2$ and the satellite moves so that action between any 2 times is minimum. This happens when

$\frac{\text d}{\text dt}\left(\frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} = 0$

$\frac{\text d}{\text dt}\left(\frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} = 0$

which leads to Newton's equations of the satellite in polar coordinates:

$m\left(\ddot r - r \dot \theta^2\right) + \frac{\mu m}{r^2} = 0$

$m\left(r^2 \ddot \theta + 2r \dot r \dot \theta\right) = 0$

If a coordinate does not appear in the lagrangian ($\theta$ in this case), it is called cyclic and it always leads to an integral of motion (something conserved). Obviously, because then $\frac{\partial L}{\partial \theta} = 0$, so that $\frac{\text d}{\text dt}\left(\frac{\partial L}{\partial \dot \theta} \right) = 0$ and $\frac{\partial L}{\partial \dot \theta} = \text{const}$. In this case, it is the orbital momentum, i.e., nothing new:

$\frac{\text d}{\text dt}\left(\frac{\partial L}{\partial \dot \theta} \right) = \frac{\text d}{\text dt}\left(mr^2 \dot \theta\right) = 0$

$mr^2 \dot \theta = \text{const},\ r^2 \dot \theta = \frac{\text{const}}{m} = \text{const} = h$

Multiplying the 1st Newton's equation by $\dot r$ and substituting $\dot \theta = \frac{h}{r^2}$,

$m\left(\ddot r \dot r - r \dot r \dot \theta^2\right) + \frac{\mu m \dot r}{r^2} = m\left(\ddot r \dot r - \frac{h^2 \dot r}{r^3}\right) + \frac{\mu m \dot r}{r^2} = 0$

which is easily integrated with respect to time using $\frac{\text d}{\text dt} \left(\dot r^2\right) = 2 \dot r \ddot r$, $\frac{\text d}{\text dt} \left(\frac{1}{r^2}\right) = -\frac{2\dot r}{r^3}$, and $\frac{\text d}{\text dt} \left(\frac{1}{r}\right) = -\frac{\dot r}{r^2}$ :

$m \frac{\text d}{\text dt} \left(\frac{\dot r^2}{2} + \frac{h^2}{2r^2} - \frac{\mu}{r}\right) = 0$

$\frac{\dot r^2}{2} + \frac{h^2}{2r^2} - \frac{\mu}{r} = \text{const} = E$

Using the initial condition $r(0)= a,\ a \dot \theta(0) = U$ leads to $a^2 \dot \theta(0) = aU = h$ (h is conserved). Substituting this to the above equation for E and using the other initial condition $r(0) = a,\ \dot r(0) = 0$,

$E = \frac{\dot r^2(0)}{2} + \frac{h^2}{2a^2} - \frac{\mu}{a} = \frac{U^2}{2} - \frac{\mu}{a}$


The problem has a mistake.When n=o,1.It's wrong.
When $ n\ge 2,\frac {n \plus{} 1}{2}\ge1,So (1 \plus{} na)^{(n \plus{} 1)/2}\ge1 \plus{} \frac {na(n \plus{} 1)}{2}\ge1 \plus{} (n \plus{} 1)a$

n belongs to natural numbers and 0 is not a natural number its a whole number or an integer... :)

I'm so tired.
But it seems that n=1.It's wrong?