Ok so it seems there is an ongoing debate right now about lowering the drinking age in US from 21 to 18. For those of you not residing in US, this legal age is very, [b]very[/b] strictly enforced in US, the only exception being college (frat) parties, where people usually get a hold of ridiculous amounts of booze and ... drink it.

So what's your opinion on this?

Even if I am not from US, I can tell that 21 year old limit is a very good move in US, and it shouldn't be lowered. There are many people who destroyed their life with alcohol and there should be laws that totally forbid alcohol.

I like the way you begin with how it's very very strictly enforced before pointing out that it's not. I've never been to a college frat party (wait no I may have been to one briefly) but have still had fairly easy access to alcohol. In high school I would have had to go a bit further, but still not very far, but already considering that the majority of 18+ year olds in the United States go to college where alcohol access is basically trivial, the law is essentially unenforced. I don't really care either way, but there really doesn't seem to be any point in having the drinking age be as high as it is.

By "enforced" I think he is quite correct. People go to jail for serving minors. Minors are arrested for having alcohol in their system. There are zero-tolerance laws. It isn't some slap on the wrist offense. Law enforcement takes this very seriously. Easy access and punishment for being caught are considerably different. I am friends with a German exchange student, and this concept is completely unfathomable to her. She drinks underage, but is sort of reckless about it. We try to explain that she can't be caught, but it is a silly idea in her mind.

That being said, I really think this is a good idea. I don't have any studies to back me up, but it is often said that alcohol related accidents are precisely because of the high age. Kids are forced to sneak around instead of being in a safe controlled environment. Other kids who don't drink until the accepted age or close to it tend to not know how to handle themselves, and drink too much since they don't have the experience. People who grow up having a glass of wine at dinner or whatever don't find such a novelty in drinking, so don't feel the need to go overboard or be irresponsible with it.

Maybe I'll look for some of these studies. I know I've seen them before.

Many kids go into college with little to know idea of how much alcohol does what to their system. They might hear someone go "YO I CAN DRINK 12 NO PROBLEMS" and then be like, if that guy can drink 12, I can at least drink 8, and then get mad drunk. Kids are like, geez, noone can pressure me into drinking, I'll just say no. But if they don't know their limits, they can't understand the pressure being put on them. Just the other day, I saw a pretty innocent freshman girl get pressured into drinking a bit, and half an hour later she was acting like a total slut.

By very very enforced I meant outside the college area. Like there is no way you can get in bars, clubs, or get alcohol by yourself at the supermarket. Not to mention the punishments for being caught driving under influence (night in jail, etc.)

I'm trying to compare with European Laws which basically have 18 as a legal drinking age (with the mention that pretty much everyone drinks starting at an early age anyway, because they are not enforced). And yet there is a lower rate of people dying because of drinking and driving than in US. And also the average college student in Europe consumes much less alcohol than the average college student in US ...

IMO, it won't really help unless you [b]remove age limits on drinking completely[/b], which is what the US should do. This way there's no sudden urge to try it out on the first few days of the year that one gets old enough for the beverage. In addition, it also shows less concern for alcoholic beverages, which brings less attention to it and results in fewer people taking it excessively. (Look at China)

hmm. people from other countries with less strict drinking laws should feel free to correct me, but I don't think this urge to start drinking as soon as one is legally able to exists even though those countries do still have drinking ages (just, lower than ours).

The laws against underage drinking effect drinking and driving.  If it wasn't so taboo among authority figures, then kids wouldn't be forced to hide it... and less accidents would result from it.  The logic seems funny, but I've seen it happen way too many times.  And these silly laws can ruin a kid's life, just because he happened to be in the same car as a kid who was drinking a beer... The whole situation is ridiculous.  If I am "responsible" enough to decide whether or not I want to join the army or buy a gun at age 18, I think I should be able to have a few beers.

[quote="HilbertThm90"]That being said, I really think this is a good idea. I don't have any studies to back me up, but it is often said that alcohol related accidents are precisely because of the high age. Kids are forced to sneak around instead of being in a safe controlled environment. Other kids who don't drink until the accepted age or close to it tend to not know how to handle themselves, and drink too much since they don't have the experience. People who grow up having a glass of wine at dinner or whatever don't find such a novelty in drinking, so don't feel the need to go overboard or be irresponsible with it.

[/quote]

i think kids start drinking at a pretty early age in russia.. and people get drunk a lot more than they do in the u.s...

and why dont we let them try smoking at a safe, controlled enviroment too? heck, throw in crack, cocaine, pot, and all the other drugs, and let the kids try them so that they won't use it excessively later in life

[quote="(^_^)"]and why dont we let them try smoking at a safe, controlled enviroment too? heck, throw in crack, cocaine, pot, and all the other drugs, and let the kids try them so that they won't use it excessively later in life[/quote]

I guess that was sarcasm? Anyway, in my opinion, its just really sad that we would even need to find the legal drinking age/other conditions which would psychologically minimize people's desire to drink excessively or irresponsibly. I find it stupid that some people are so anxious to drink a ton, when everybody knows or should know that irresponsible drinking is obviously bad for you and causes deaths, etc. Apparently the main cause is peer pressure; I just still have a hard time understanding how this pressure occurs. Was it just some stupid people originally who decided it was fun to get so wasted that they couldn't stand up and then they pressured people slightly younger than them, who then grew up and pressured people slightly younger than them, etc.? Now, the idea does make sense about how kids are forced to sneak around and it would be better if they had a safe controlled environment, but I just can't wrap my mind around the fact that people actually want to go get drunk in an unsafe uncontrolled environment when the other option is closed because of the law which they should realize is probably there for a reason.

Alcohol poisoning is another big one.

I was at a Carolina football game with a friend when we encountered one of her good friends sitting on a ramp throwing up all over the place.  My immediate (and extremely ignorant) reaction was to get him out of the stadium, which was overflowing with cops, induce vomiting, and get as much water into his system as possible (that's the "normal" thing to do).  We were lucky enough to run into a cop who forced us to take the kid to the medical tent, and promised not to take any kind of police action.  The kid ended up getting his stomach pumped, and apparently he had alcohol poisoning.

The point is, we got lucky.  What kind of teen is gonna call the cops on his friend and tell them he's been drinking?  Especially with the prevalent "invincible" mentality among most kids.  Just let them throw it up, make sure they pass out on their sides rather than on their backs so they don't choke on their vomit, and they'll be fine in the morning.

If underage drinking wasn't illegal, teens with alcohol poisoning would be treated in hospitals with proper equipment, instead of being dragged into the bathroom and instructed to vomit into the toilet.

Okay, I have my doubts about this whole "safe environment for people to drink" thing. The only part I go for is mysmartmouth's example of people who get completely sick from alcohol poisoning and then won't be [i]as[/i] afraid to get medical attention.

See, if the drinking age was lowered/removed, I think most people would be like "yeah, freedom, now we don't have to hide it" and then they might drink even more and lose any little sense of "this is too much alcohol" that they may have already had. Would people choose this supposed "safe environment"? I think people who like to get drunk probably like to get drunk with other people in social situations such as parties where they can "show off" their drunkedness.

Even then, mysmartmouth's argument might not apply. If the person gets alcohol poisoning, they have probably reached the point where they lose enough judgment that they don't realize they need medical attention, nor would they be surrounded by the type of people who would have this judgment.

So, lowering the age may make create some situations, but I think it may also make things worse is some aspects.  :wink:

And I guess i am still a little sickened by the logic that making something stupid legal would help those who decide to do that stupid thing.

eh, in general I think personal responsibility is much more effective than government-enforced responsibility, almost by definition. If you remove the latter, people will be forced to learn the former and hence be able to better control themselves. so in a way, having the government around forbidding you to do certain things almost makes you less responsible as a result.

[quote="tjhance"] Apparently the main cause is peer pressure; I just still have a hard time understanding how this pressure occurs. Was it just some stupid people originally who decided it was fun to get so wasted that they couldn't stand up and then they pressured people slightly younger than them, who then grew up and pressured people slightly younger than them, etc.? Now, the idea does make sense about how kids are forced to sneak around and it would be better if they had a safe controlled environment, but I just can't wrap my mind around the fact that people actually want to go get drunk in an unsafe uncontrolled environment when the other option is closed because of the law which they should realize is probably there for a reason.[/quote]

Apparently, you've never experienced peer pressure or ever drank anything alcoholic.

Many people like to drink into that 'tipsy' stage...where things are lighter, more entertaining, and things are in a much different perspective.  There is nothing wrong with this stage.  The problem is when people over-estimate their boundaries, and drink way past that point until they are throwing up.

So clearly, if a bunch of guys are having fun, and they happen to be drinking, why not have a couple drinks with them and join in the fun?  That's what peer pressure is.  In very few cases are there actually people who push you into drinking and getting wasted.  I don't know about you, but most people don't like drinking to the point of being completely wasted, it ruins your evening.  In many cases, drinking is a social thing, you go to a bar, chat with a beer, people want to fit in, and so they drink.

I think the drinking laws are both stupid, and smart.  The fact that there is a law is to keep people safe, and prevent alcohol induced accidents, which is terrific.  On the other hand, because there is this law, people go around trying to get alcohol under-age, wanting to try it out, and without knowing the facts they get wasted too quickly.  It's like the whole "cookie in a jar" thing for children.  Because they can't have cookies whenever they want, some will try and sneak around and steal a cookie from the jar when you're not looking.

People need to be educated about the facts of alcohol, but the fact that there's this societal taboo surrounding it, is what makes is such a big deal when you're 21, and so you're more likely for a few years to go past your boundaries instead of having just a drink or two.

mysmartmouth and hilbertthm are pretty much right on.  denying 19/20 year olds (who, by most reasonable standards, are adults) the ability to freely obtain alcohol results in them indulging excessively when they do get access to it and results in most of the underage drinking horror stories that you hear about in the news.  lowering the drinking age would [i]mitigate[/i] this.  i personally think people are much less likely to binge drink or to drink excessively once they are of age, because there isn't really any incentive to once you have done it before and felt the aftereffects, and you can also buy more booze any time you want.

I basically agree with blahblahblah.  One important point that doesn't seem to have been discussed, however, is drunk driving.  The single best reason to have a high drinking age is to limit drunk driving (and this explains, for example, why MADD flips out when anyone prominently suggests lowering the drinking age).  This is less a problem in Europe, for example, in large part because Europe tends to be far less car-dependent than the US.  A [url=http://www.eschatonblog.com/2007_04_15_archive.html#117699170983806473]prominent blogger[/url] I read has suggested that people have a choice of a drinking license or a driving license but not both until age 21.

Well, in China there isn't a legal age for drinking.
It doesn't really matter if they change or not
If people really desired to get alcohol, they would get it in some way and going to jail won't destroy their desire.
People should look after themselves...

This is a hot topic. Yes, removing the age limit might make things better in a "perfect" world, but probably not so in the real world. If all parents taught their kids reponsibility for themselves and their actions, removing the age limit would probably fix some problems. However, that is not quite so in the real world. One of the reasons why we have an age limit in the first place is because some parents are not responsible for their children. They let them drink and drink all they wanted, living forever as street winos/drunks (lives wasted).

MAIN IDEA: You should be allowed to drink when you want IF AND ONLY IF you are responsible for yourself and your actions.

so basically you don't understand the arguments presented and added a useless nonsensical argument.  thanks for your valuable contributions to this thread!

btw 18 or 19 year olds are not children; 18 is the age of majority in 47 states, 19 is the age of majority in two others, and only one state has the age of majority as 21.

Forget it. If you have nothing nice to say about other people's opinions, don't say it.

[b]@12markkram34:[/b]

$ \boxed{1}$ He was talking to you. In less impassioned terms, he meant that allowing people to drink at will iff. they are responsible is not a legal standard that people can be held to. And if legal guidelines as to what constitutes a responsible record were put into place, their enforcement would just clutter the legal system.

$ \boxed{2}$ A summary of your argument is as follows: "we do not know what will happen if the drinking age is changed, so we should be just be risk-averse and keep things the way they are". Risk-averseness is more a type of personal bias (resistance to change) than it is a concrete reason.

If this is not what you meant, you may want to detail why the risks outweigh the potential benifits of such a change; either because the risks are simply worse, or because there is a small risk of something absolutely terrible happening, or whatever.

If the risks do not outweigh the potential benefits, then you may want to explain (if you have a reason) why you think an even balance of risks and benefits means that we should sit on our bums rather than try to do something.  :wink:

[quote="tjhance"]See, if the drinking age was lowered/removed, I think most people would be like "yeah, freedom, now we don't have to hide it" and then they might drink even more and lose any little sense of "this is too much alcohol" that they may have already had. Would people choose this supposed "safe environment"? I think people who like to get drunk probably like to get drunk with other people in social situations such as parties where they can "show off" their drunkedness.
[/quote]You see this is where you are wrong. I come from Romania, where there is no legal age basically, people drink at 14 or even lower. Parties with beer start in 8th grade usually.

But I must tell you that I have never ever seen such huge amounts of alcohol being consumed at a single party as I have seen at some US colleges.

The fact that it is available in Romania makes teens [b]not[/b] drink it so much overall, because it's not a novelty. You can have 2 beers and go home, it's fine.

In the end US is the [b]only[/b] country in the world with drinking age 21. Everywhere else you have 18, 19 or no drinking age at all. Saying that "we don't know what would happen if we lower the drinking age" is basically like saying "we are the only people on the planet, or all the countries in the world have drinking age at least 21".

The logic seems to be "if you allow something currently prohibited, it will be less attractive and, thereby, people will do it less frequently or to lesser extent" seems very weird to me. Following the same logic, we should replace the death penalty for murder with $ \$$15 fine and hope that it'll improve the public safety :P. In my opinion, the main question is not how to change the law but how to enforce it, whether the cutoff is 18 or 80 (or even the current 21).

As to the change itself, I believe that the age should be left as it is not because of 18-21 years olds but because of 15 year olds. It is the same story as with the speed limit: under normal road conditions, you can safely drive above 80mph on most USA highways without any problem. Still, the speed limit is kept at 65mph, the "gray zone" 65-75mph is not really enforced (except by some overzealous policemen at the end of the month) and driving over 80mph will almost surely result in a heavy fine. The result is a more or less steady flow at 70-75mph, which seems to be OK with everybody: the drivers get their "excitement" from violating the law and the police is sure that the speed is within the safe limit (which is really around 80-85mph on most Midwest roads) and can make a clear case (speeding by 15mph or more) against anybody whose speed is really excessive. I'm in favor of the same scheme with drinking: legal age at 21 (or even higher) with the penalty gradually diminishing as you approach it.

I think the difference between drinking laws and murdering laws is that the former is a crime to oneself (except in cases involving drunk driving, where I do support a far harsher penalty) while the latter is a crime to others. Certainly, government should exist to eliminate or at least reduce as far as possible one's tendency to inflict harm upon others and this is the reason for jail sentences and occasionally the death penalty for murder, but with regard to decisions made by oneself and having effects upon oneself, it seems far more logical to relegate this to the arena of personal responsibility.

and I don't really have a problem with 15-year olds drinking. I'm not saying the legal age needs to be reduced down to 15 right now as I don't think America is ready for such a drastic change, but starting to move in that direction doesn't seem inherently flawed.

[quote="fedja"]The logic seems to be "if you allow something currently prohibited, it will be less attractive and, thereby, people will do it less frequently or to lesser extent" seems very weird to me. Following the same logic, we should replace the death penalty for murder with $ \$$15 fine and hope that it'll improve the public safety :P. In my opinion, the main question is not how to change the law but how to enforce it, whether the cutoff is 18 or 80 (or even the current 21).[/quote]As Mystic pointed out there is a difference between having a beer and killing somebody. Such a comparison is really out of the context.

As I told you, I've witnessed way (WAY!) more drinking (illegally) in US than I've seen in Romania where it is legal. I don't think Romania has a drinking problem on their teens, although there is no legal age for drinking (you can drink at 12 y.o. if you really want.) What does that show? We can extend this to all of Europe if you wish (there is no enforced legal drinking age in most European countries.)

I'm not saying that killing and drinking beer is the same, I'm just questioning the logical template that, stripped to its bare essense is "if we want to prevent some undesireable behavior $ X$, it is better to allow it and rely upon people's ability to develop individual responsibility than to outlaw it and try to enforce the law". Everybody seems to cheer in favor of it when $ X$ is "underage drinking" but when I just change the value of $ X$ to "murder", the number of supporters suddenly goes down. Of course, this just means that the template is not universal. But, since it is not accepted as universal any more, I have serious grounds for questioning it with the original value of $ X$.

As to Romania and Europe, the only thing it shows is that not having the law is somehow correlated with not having a problem (like swinging of the trees is somehow correlated with the presence of the wind). Now does the wind blow because the trees swing or is it the other way around, is not as clear as Valentin tries to present. If you want to make a convincing argument against the law, you'll have to trace it to the moment it was introduced and compare the statistics of underage drinking a couple of years before and a couple of years after its introduction. That will at least show something and, maybe, you'll be right and I'll be wrong. But comparing to other countries the way it is done and trying to draw the conclusion that the law should be relaxed or abandoned is like saying that St-Petersburg, Russia, doesn't have tornado shelters and nobody has seen a tornado there, so, probably, people in Manhattan, Kansas, should destroy theirs too and the problem will be solved.

I lived in a country with no drinking age; was served beer in a cafe at age 10 once, but did not drink beer again until past 21. There were no college freshmen dying of alcohol poisoning there. In fact, I saw few drunks. I say make drinking legal at any age.

The killing as a counterexample is interesting, because I believe that the US is one of the only developed countries to still have the death penalty, yet the US has one of the highest (maybe the highest) homicide rate out of developed countries. So there is some empirical data that suggests allowing people to do things will cause them to lose interest (even in hard to imagine cases). Even though it is somewhat counterintuitive, I think that lots of psychological theories give good reason as to why this occurs such as cognitive dissonance.

[quote="fedja"]The logic seems to be "if you allow something currently prohibited, it will be less attractive and, thereby, people will do it less frequently or to lesser extent" seems very weird to me. [/quote]

i don't think that's quite the argument here.  i fully expect that lowering the drinking age would result in a higher total consumption of alcohol by 18 and 19 year olds.  i do, however, expect that it would curtail binge drinking.  i don't know what i expect the net effect on drunk driving would be.


show the composition of two contraction mappings is a contraction mapping:

i.e.

If $ \exists f_1, f_2$ such that $ f_1,f_2: X \to X$, $ X$ a metric space.

Also:

$ \| f_1 (y) \minus{} f_1 (x) \| \le \kappa_1 \| y \minus{} x \|$ with $ 0 \le \kappa_1 < 1$
AND
$ \| f_2 (y) \minus{} f_2 (x) \| \le \kappa_2 \| y \minus{} x \|$ with $ 0 \le \kappa_2 < 1$

Show that $ \| (f_1 \circ f_2)(y) \minus{} (f_1 \circ f_2)(x) \| \le \kappa_3 \| y \minus{} x \|$ with $ 0 \le \kappa_3 < 1$.

Also, show that $ \kappa_3 \equal{} \kappa_1 \cdot \kappa_2$


I went to report and my subject progress spot was empty and I didn't have an alcumus score but it said I had done problems. After that, I went to lessons and it said you currently have no lessons available and on the daily top 10 an alcumus score all of them say 0.00. I hope all my info still exists :(

and when i did a  problem and clicked the let alcumus choose button it gave me a blank screen. help :(


Prove that $ [5x]\plus{}[5y]\geq [3x\plus{}y]\plus{}[3y\plus{}x]\plus{}[x]\plus{}[y]$


inequalities
trigonometry
algebra
polynomial
geometry
circumcircle
inequalities unsolved

[quote="songoku2006"]This is the execise that my teacher gave me but i can solve it.Pleasehelp me
$\sum(sin^2A+sin^2B)cos(C/2)\leq 3\sqrt{3}/2(cosA+cosB+cosC)$[/quote]
do you think about use law of sine and cosin to sovle it
i think it is usefull ;)

Yes,my teacher solved it and it very long,and i want a more beautiful solution.Its solution is use law of sine and cosin .
Thank you eagerkill

[quote="songoku2006"] $\sum(\sin^{2}B+\sin^{2}C)\cos\frac{A}{2}\leq \frac{3\sqrt{3}}{2}\sum \cos A$.[/quote]
This is not ture, for a counterexample, $A=\frac{\pi}{5},B=C=\frac{2\pi}{5}$:

$(\sin^{2}B+\sin^{2}C)\cos\frac{A}{2}+(\sin^{2}C+\sin^{2}A)\cos\frac{B}{2}+(\sin^{2}A+\sin^{2}B)\cos\frac{C}{2}$

$=\frac{5(1+\sqrt{5})}{8}+\frac{\sqrt{25+10\sqrt{5}}}{4}=3.7430198865263354830147\cdots$,

which is a root of the irreducible polynomial $256s^{4}-640s^{3}-1200s^{2}+125$;

$\frac{3\sqrt{3}}{2}(\cos A+\cos B+\cos C)$

$=\frac{3\sqrt{3}(3\sqrt{5}-1)}{8}=3.7075872116450150107538\cdots$,

which is a root of the irreducible polynomial $256t^{4}-9936t^{2}+88209$.
[hide="Exercises"]
If $A,B,C$ are the the angles of a  triangle, $m_{a},m_{b},m_{c}$ are the medians, $R$ is the radius of circumcircle and $r$ is the radius of incircle, then

$\sum(\sin^{2}B+\sin^{2}C)\cos\frac{A}{2}<(2+\sqrt{2})\sum \cos A$;

$\sum(\sin^{2}B+\sin^{2}C)\cos\frac{A}{2}\leq\frac{\sqrt{3}}{2R}\sum m_{a}$;

$\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}}\leq\frac{2}{3r}(\cos A+\cos B+\cos C)$.
[/hide]


algebra
polynomial
number theory unsolved
number theory

$ n^2.(n\plus{}1)\minus{}1$

Find the different integer values of n which makes the above expression prime...

[quote="epsilon07"]$ n^2.(n \plus{} 1) \minus{} 1$

Find the different integer values of n which makes the above expression prime...[/quote]

Where is this problem coming from (class, book, contest, ... )?
Are you sure there is a solution?

Just testing the very first values gives a lot of values for $ n$ : $ 2,4,5,6,9,$ $ 11,12,14,19,$ $ 22,25,26,27,29,32,34,36,...$

And it seems (I have no proof) that there are infinitely many such numbers with no obvious general formula. Btw, any general formula would be a prime generator formula and it would be quite surprising that such a problem is proposed in an olympiad preparation.

Could you confirm that you are sure there is a regular solution, please ?

[quote="epsilon07"]QUESTIN IS FROM A TEXT BOOK WHICH CONTAIN OLYMPIAD PROBLEMS FROM TURKEY ( MERAKLISINA İLK ADIM MATEMATİK - WRITTEN BY RECEP YUCESAN )  NOW I SEND YOU ORIGINAL FORM OF THE QUESTION;

$ 3.(1^2 \plus{} 2^2 \plus{} 3^2 \plus{} ... \plus{} n^2) \minus{} (1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} n) \minus{} 1$

how many integer n values makes the above expression a prime number?

A) 0       b) 1        c) 3        d)  4         e) infinitely many[/quote]

So neither a), neither b) neither c), nor d)

So e)

$ 3(1^2 \plus{} \dots \plus{} n^2) \minus{} (1\plus{} \dots \plus{} n) \minus{} 1 \equal{} \frac{n(n\plus{}1)(2n\plus{}1)}{2} \minus{} \frac{n(n\plus{}1)}{2} \minus{} 1 \equal{} n^2(n\plus{}1) \minus{} 1,$ which is presumably how he got the problem.

To the original poster, the multiple choice nature of the problem means you only have to verify the first finite choices are incorrect. Regardless, showing there are infinitely many values of $ n$ which make this prime would be a better question than finding the values of $ n$ which make the expression prime. Also, please refrain from using all-caps in posts; it looks like shouting.

For future reference, given a polynomial in one variable, with integer coeficients and degree greater than $ 1$, it is not known up to now if it takes infinitely many prime values on integer argument. In some sense, Dirichlet's theorem is the best result on this line.

In general "we" conjecture it to be truth.


Putnam
Harvard
college
Stanford
college contests

Manolescu particpated in 1999; he was a member of the Harvard team but did not place in the top 50 or so.

A perfect score would be 120 points. Here are the actual highest scores since 1999, along with the names of the Putnam Fellows from each year. The person with the highest score must be one of the Putnam Fellows, but which one is not a matter of public record. The order in which the Putnam Fellows' names are listed is always alphabetical.

1999: 74. Cautis, Kisman, Kumar, Maulik, Mihelich, Percival
2000: 96. Carroll, Kumar, Manolescu, Pylyavskyy, Schwartz
2001: 101. Barton, Carroll, Lacker, Lee, Siwanowitch
2002: 116. Barton, Carroll, Cebikins, Schwartz, Wood
2003: 110. Barton, Caraiani, Carroll, Furmanick, Kane
2004: 101. Barton, Barzov, Caraiani, Kane, Pixton

The story, at least according to Valentin, is that he had a headache that year.  Given that he was a Putnam fellow the other 3 times around, I'm inclined to give it some credence.

[quote="Cezar Lupu"]I'm sorry to tell you this, but in my opinion [color=blue] Ciprian Manolescu[/color] is the best ever to participate
in olympiads and math contests. :)  ;)  He is a genious. ;)[/quote]

why sorry? this word has a heavy pessimism impact...by the way, is Manolescu working as a reseacher now? is Ioan Manolescu a member of his family? this guy got the french nationality last year to study in ENS (a great french school of science research with an extremely hard entrance exam)!!
Romanians brains are really amazing , is the secret in the eduational systems? or they are born... ;) !?

I don't know about Ioan but Ciprian's homepage can be retrieved here: http://www.math.columbia.edu/~cm/

[quote="maskman"]why sorry? this word has a heavy pessimism impact...by the way, is Manolescu working as a reseacher now? is Ioan Manolescu a member of his family? this guy got the french nationality last year to study in ENS (a great french school of science research with an extremely hard entrance exam)!!
Romanians brains are really amazing , is the secret in the eduational systems? or they are born... ;) !?[/quote]Ioan Manolescu is not a relative of Cirpian Manolescu.

[quote="alcachofo"]Did Ciprian Manolescu got perfect scores in Putnam like he did for 3 times in IMO? if no, have anyone gotten a perfect score? Other question: did he participated in 1999? if no, why?[/quote]

His homepage includes a CV which states "first place on Putnam competition" for 1997,1998, and 2000, and that he held the Putnam fellowship (offered to one of the top 5) for 3 years at Harvard.   I don't know if the fellowship is always awarded to the highest scorer within the top 5.

[quote="Kent Merryfield"]2001: 101. Barton, Carroll, Lacker, Lee, Siwanowitch[/quote]
It's Siwanowicz

A question related: What is Ciprian Manolescu doing now? Is he researching??

Google loves you.

http://www.math.columbia.edu/~cm/

[quote="blahblahblah"]Google loves you.[/quote]

Jejejjeej!!!! Thank you!!!

my best friend saw him a week ago in walking around in stanford, cal...

Oh I see.. and what about Reid Barton? what is he doing now??? I couldn't find anything about him...

He's doing grad school at Harvard.

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

Even now, after Alex Song has broken the record and many stand above him on the leader board, I still think that it counts for something to have never dropped a point at IMO.


Let $f$ and $g$ are two functions. So that $f: \mathbf{R^{n}}\longrightarrow \mathbf{R}$ and  $g: \mathbf{R^{n}}\longrightarrow \mathbf{R}$
$(a)$ Can we define $fog=f(g)$ ?  How?
$(b)$ Can we define inverse function $f^{-1}$ ?  How?
Thanks for your helps.

(a) you cannot define f.g because $f: \mathbf{R^{n}}\longrightarrow \mathbf{R}$ and $g: \mathbf{R^{n}}\longrightarrow \mathbf{R}$
you can define f.g if and only if $f: B\longrightarrow C$ and $g: A\longrightarrow B$ and in that case f.g(x) is just f(g(x))
(b) you can define inverse only if f is bijective and if it is then the inverse is such that $f.f^{-1}(x)=f^{-1}.f(x)=x$ for all x.

Thanks Baronov. I ask my question again.If $f: \mathbf{R^{n}}\longrightarrow \mathbf{A}$ and  ${g: \mathbf{B}}\longrightarrow \mathbf{R^{n}}$ Now, can we define $fog=f(g)$ ?  How?

Definition:
For each $x\in B$ then $g(x)=\left(h_{1}(x),h_{2}(x),...,h_{n}(x)\right) \in\mathbb{R}^{n}$ where $h_{i}(x)$ is a function from $\mathbb{R}\to\mathbb{R}$,  $i=\overline{1,n}$
Therefrom:
$f(g(x))=f\left(h_{1}(x),h_{2}(x),...,h_{n}(x)\right)$


You missed two solutions:

[hide]$u=5\pm 2\sqrt{6}\implies x^{2}-3=\pm 1\implies x^{2}\in\{4,2\}\implies x=\pm 2\lor x=\pm\sqrt{2}$[/hide]

Sorry,sorry. I was crazy  :blush: [hide]
I though $5-2\sqrt6$ is negative [/hide]:wallbash:

Wait. how did you see so quickly that 5-2root6 is 5+2root6 ^-1?

I just ran into a very similar problem yesterday which required that knowledge.  Whenever you see a problem like this you should look for that.


find all real solutions to
$3x^{3}-\lfloor{x}\rfloor=3$

Let $\lfloor x\rfloor=n$. Then $x=\sqrt[3]{1+{n\over 3}}$. From $0\leqslant x-\lfloor x\rfloor<1$ we get two inequalities:

1. $\sqrt[3]{1+{n\over 3}}-n\geqslant 0\iff 3n^{3}-n-3\leqslant 0\iff (n-1)(3n^{2}+3n+2)-1\leqslant 0$

The expression $3n^{2}+3n+2$ is always positive and its minimum in $\mathbb{R}$ is ${5\over 4}$. Assume $n-1\geqslant 1$. Then $(n-1)(3n^{2}+3n+2)-1\geqslant(3n^{2}+3n+2)-1>{5\over 4}-1>0$, hence there's no solution. It's obvious that for $n-1\leqslant 0$ the inequation is satisified. Therefore, we have $n\leqslant 1$.

2. $\sqrt[3]{1+{n\over 3}}-n<1\iff n\left(n^{2}+3n+{26\over 3}\right)>0$

Since $n^{2}+3n+{26\over 3}$ is always positive, that's equivalent to $n>0$

Hence $0<n\leqslant 1\implies n=1$, and the only solution to the initial equation is $x=\sqrt[3]{{4\over 3}}$

When the floor of $x$ is the integer $k$, we can always find $x$ so that $k+3 = 3x^{3}$ can take any of the values $\in [3k^{3}, 3(k+1)^{3}) \cap \mathbb{N}$.  So we have the necessary condition $3k^{3}\leq k+3 < 3(k+1)^{3}$.  The first inequality holds if $k < 2$.  The second inequality holds if k > 0.  So $k = 1$ is a necessary condition.

So $3x^{3}= 4$ is a necessary condition.  Checking, $x = \sqrt[3]{\frac{4}{3}}$ works.


for Individual Examples,Given Actual Expressions,Easy to see ,or Research.

BQ


Find all positive integer of equation:
X^2=2Y^2+Z^3
where gcd(X,Y)=1.

[quote="Rust"]For all prime divisors of z we have $(\frac{2}{p})=1\Longrightarrow p=\pm 1(mod 8)$ It give $a_{p},b_{p}\ : \ a_{p}^{2}-2b_{p}^{2}=p$.
Let $z=\prod_{p}p^{k_{p}}$, were $p=\pm 1 (mod 8)$.
All solutions given by: $(x+y\sqrt 2)=(3+2\sqrt 2 )^{n}\prod_{p|z}(a_{p}+b_{p}\sqrt 2 )^{3k_{p}}$.[/quote]
I don't know  :wallbash_red:  :wallbash_red:


Prove that for all positive reals $ a; b; c$ such that $ a \plus{} b \plus{} c \equal{} 3$,
$ \sum_{cyc} \frac{a}{ab\plus{}1} \ge \frac{3}{2}$

maybe it have been posted before, but i can`t find it.

[quote="xuheiedgar"]Prove that for all positive reals $ a; b; c$ such that $ a \plus{} b \plus{} c \equal{} 3$,
$ \sum_{cyc} \frac {a}{ab \plus{} 1} \ge \frac {3}{2}$

maybe it have been posted before, but i can`t find it.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689

Uff, that was a long one.
First multiply both sides by $ (ab \plus{} 1)(bc \plus{} 1)(ca \plus{} 1)$ and substitute 3 on the right side with a+b+c.
Then after multiple transforms it suffices to prove:
$ (a^{2}bc \plus{} ab^{2}c \plus{} c^{2}ab) \plus{} (a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} (\frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a})\leq 3 \plus{} (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} (\frac {1}{bc} \plus{} \frac {1}{ac} \plus{} \frac {1}{ab})$
Now, we know that $ abc\leq (\frac {a \plus{} b \plus{} c}{3})^{3} \equal{} 1$
Now, transforming again:
First:
$ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} (a \plus{} b \plus{} c)^{2} \minus{} 2(ab \plus{} bc \plus{} ca) \equal{} 9 \minus{} 2(ab \plus{} bc \plus{} ca)$ thus
$ L\leq (a^{2}bc \plus{} ab^{2}c \plus{} c^{2}ab) \plus{} 9 \minus{} 2(ab \plus{} bc \plus{} ca) \plus{} (\frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a})$
Now, we know that $ \sum_{cyc}bc\leq \sum_{cyc}\frac {1}{a}$ from $ abc\leq 1$
And from inequality beetween harmonical and arithmetical mean we have $ \sum_{cyc}\frac {1}{a}\geq \frac {9}{a \plus{} b \plus{} c} \equal{} 3$ therefore:
$ L\leq (a^{2}bc \plus{} ab^{2}c \plus{} c^{2}ab) \plus{} 9 \minus{} 2*3 \plus{} (\frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a})$
Next we have to prove:
$ \sum_{cyc}\frac {1}{bc}\geq (a^{2}bc \plus{} ab^{2}c \plus{} c^{2}ab)$
It's obvious that:
$ \sum_{cyc}\frac {1}{bc}\geq a \plus{} b \plus{} c\geq (a \plus{} b \plus{} c)abc \equal{} a^{2}bc \plus{} ab^{2}c \plus{} c^{2}ab$
Therefore we only need to prove
$ \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a}\leq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ which I think can be proved by Muirhead :wink:

EDIT: Hehe, I was too late. :wink:

[quote="polskimisiek"]
Therefore we only need to prove
$ \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a}\leq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ which I think can be proved by Muirhead :wink:
[/quote]
But it's obviously wrong because $ \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a}\leq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\Leftrightarrow(a\minus{}b)(b\minus{}c)(c\minus{}a)\geq0.$ :wink:

They are equivalent, both arqady and you made a typo only.

[quote="polskimisiek"]Ehrm $ \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b}\leq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ sorry
and this I think is not equivalent to this :wink:[/quote]
$ \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b}\leq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\Leftrightarrow(a \minus{} b)(b \minus{} c)(c \minus{} a)\geq0.$
Here is equivalent! :D
Thank you, Soarer! :)

Aaah, yes, you're right :wink: sorry :blush:


Given positive odd numbers a,b,c,d with $ a<b<d$ and $ ad=bc, a+d=2^{k}, b+c=2^{m}$ where k and m are integers, then a=?

DELONG!

Honestly, the easiest way to do this question is to bash it out. You can tell that the numbers should be fairly small, and you can find out that $ a=1$ and $ d=15$.

umm isn't there multiple answers?
eg 2,6,10,30

[hide="general solution"]We get
\[ a=2^{2m-2-k}, d=2^{k}-2^{2m-2-k}\]
\[ b=2^{m-1}-2^{2m-2-k}, c=2^{m-1}+2^{2m-2-k}\]

Why: Assume $ a<b \le c<d$.  Cleary $ m<k$.  We have $ a \equiv-d \mod{2^{m-1}}$, $ b \equiv-c \mod{2^{m-1}}$.
But
\[ a+d=a+\frac{bc}{a}=a+\frac{b(2^{m}-b)}{a}=2^{k}\]
\[ \implies a^{2}-b^{2}=2^{k}a-2^{m}b\]
\[ \implies (a+b)(a-b) \equiv 0\mod{2^{m}}\]
So then $ 2^{m-1}$ divides exactly one of $ a-b, a+b$.  So $ a \equiv \pm b \mod{2^{m-1}}$.  But $ a< b \le 2^{m-1}$.  Therefore, $ a \equiv-b \mod{2^{m-1}}$.

Let $ b=2^{m-1}-x, c=2^{m-1}+x$ for some nonnegative integer $ x$.  Then $ a=x, d=2^{k}-x$.  Plugging this into $ bc=ad$,

\[ 2^{2(m-1)}-x^{2}=2^{k}x-x^{2}\]
\[ x=2^{2m-2-k}\]

Indeed, this quadruple is a general solution in integers as long as $ 2m-2-k \ge 0$.[/hide]

[quote="MellowMelon"]You missed a crucial word in the first post.

[quote]positive [b]odd[/b] numbers[/quote][/quote]

Good point.  In that case, we take out all even solutions, leaving us with

[hide]$ a=1, d=2^{2m-2}-1, b=2^{m-1}-1, c=2^{m-1}+1$, so indeed $ a=1$.[/hide]

how do you guys get $ a=1$ at the start. i cant think of any way to do it except for brute force


floor function
induction
number theory unsolved
number theory

Prove that $k^3!$ is divisible by $k!^{k^2+k+1}$.

[i]Edited by Myth[/i]

I have a solution based on well known formula $p^m\,|\,n!$ iff $m\leq \sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor$.
But I guess that such problem could have combinatorical core.

Could someone please post the solution ?

More generally, we will prove that $k^n!$ is divisible by $(k!)^{k^{n-1} + \cdots + k + 1}$ for all positive integers $n$.
For $n=1$ this trivially comes to $k!$ being divisible by $k!$. Let us proceed by induction.

We have $\prod_{j=0}^{k^{n-1}-1} \frac {(k^n - jk)!} {k!(k^n - (j+1)k)!} = \frac {k^n!} {(k!)^{k^{n-1}}}$ to be an integer, as being a product of binomial coefficients.
But $\frac {(k^n - jk)!} {k!(k^n - (j+1)k)!}$ is divisible by $k^{n-1} -j$, since $\frac {\gcd (a,b)} {a+b} \cdot \frac {(a+b)!} {a!b!} = \frac {ua+vb} {a+b} \cdot \frac {(a+b)!} {a!b!} = u\frac {(a+b-1)!} {(a-1)!b!} + v\frac {(a+b-1)!} {a!(b-1)!}$ is an integer, while $k^n-jk=k(k^{n-1}-j)$ and $\gcd (k, k^n - (j+1)k) = k$.

But the numbers $k^{n-1} -j$, $0 \leq j \leq k^{n-1}-1$, are the numbers $1,2,\ldots, k^{n-1}$, therefore our number is divisible by $k^{n-1}!$, hence $k^n!$ is divisible by $(k!)^{k^{n-1}}k^{n-1}!$. But, by the induction step, $k^{n-1}!$ is divisible by $(k!)^{k^{n-2} + \cdots + k + 1}$.


What is the sum of all of the four-digit positive integers that
can be written with the digits 1, 2, 3 and 4 if each digit must be
used exactly once in each four-digit positive integer?

Put a 1 into the last digit, and note that the other 3 digits can permute 6 ways. 1 will be in the units digit 6 times in total. The same principle applies for 2,3, and 4, and for every single digit. 1,2,3,4 will each appear 6 times in every digit, so the sum of all numbers is $ 6(1\plus{}2\plus{}3\plus{}4)(1000\plus{}100\plus{}10\plus{}1)\equal{}60(1111)\equal{}66660$.


Hi all,

I'm new to vectors and am having a bit of trouble with the problem of finding the distance from a point to a line. I'm using AoPS Vol. 2 (p. 167), and it presents the problem like this:

There is a line of the form $Ax+By+C=0$, and a point $(x_0,y_0)$. The point is called Z and the line l. Then, a point W on l is placed arbitrarily, with coordinates $(x_1,y_1)$. Then, it says to draw line through Z parallel to line l (obviously, the distance between these two parallel lines is the same as the distance between the point and the line, this I understand). Then, it says that the normal vector $\vec{n}$ to line l through W intersects the line through Z at Y. (Now, this following part I don't understand.) It states that the desired distance is the length of the projection of $\vec{WZ}$ onto $\vec{n}$.

Why is that? Isn't that projection equal to the distance of ZY, and not the distance WY? Perhaps, then, I don't understand the concept of the dot product, but I can't seem to grasp this.

Any help is greatly appreciated. Thanks! :)

The dot product, $\vec A \cdot \vec B$, is $AB\cos \theta$, where $\theta$ is the angle between the vectors.
The vector projection of $\vec B$ onto $\vec A$ is $proj_{\vec A}(\vec B)=\frac{\vec A \cdot \vec B}{|\vec A|^2} \vec A$.
Basically, the vector projection of $\vec B$ onto $\vec A$ is the part of $\vec B$ that points in the same direction as $\vec A$.

In the distance from a point to a line, the vector projection of $\vec {WZ}$ onto $\vec n$ is $\vec {WY}$, not $\vec {ZY}$. The magnitude of $proj_{\vec n}(\vec {WY})$ is the distance from $W$ to $l$.

FYI, $|proj_{\vec A}(\vec B)|$ is also known as the scalar projection of $\vec B$ onto $\vec A$, or $comp_{\vec A}(\vec B)=\frac{\vec A \cdot \vec B}{|\vec A|}$.


A fair 6-sided die is rolled.  What is the probability that the number rolled is a divisor of 6?

To begin, it says to find the probability of rolling a divisor of six on a fair 6-sided dice. Divisors are basically the numbers able to divide into the given number to produce a whole number. 6's divisors are 1,2,3, and 6. Look back at the problem. This gives you a probability of 4/6, or [u][b]2/3[/b][/u]

The numbers on the dice are 1,2,3,4,5 and 6. Of them, 4 are divisors of 6. 4/6=2/3.

[hide=Solution #3]There are 4 factors of 6, and 6 faces. The answer is $\boxed{\frac{4}{6}=\boxed{\frac{2}{3}}}.$[/hide]

[hide]There are 4 divisors of 6, namely $1,2,3,6$. So the answer is $\dfrac46=\boxed{\dfrac23}$.  :pilot:[/hide]


[quote="sxgfly"]$ let\,\,x,y,z > 0.\,\,such\,\,that\,\,x \plus{} y \plus{} z \equal{} 1 \\
prove\,\, that: \,\,\frac {1}{1 \minus{} xy} \plus{} \frac {1}{1 \minus{} yz} \plus{} \frac {1}{1 \minus{} zx}\le \frac {27}{8}$[/quote]

It's old problem
"p;q;r" method can slove it  :P

[quote="Dr Sonnhard Graubner"]hello, then give us your solution if you know it.
Sonnhard.[/quote]

Ineq
With:$ q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc;p\equal{}a\plus{}b\plus{}c\equal{}1$
$ <\equal{}>8(3\minus{}2q\plus{}r) \le 27(1\minus{}q\plus{}r\minus{}r^2)$
$ <\equal{}>3\minus{}11q\plus{}18r\plus{}r(1\minus{}27r)$
We have:
$ 1 \ge 27r$
And:$ 18r \ge 2(4q\minus{}1)$-shur
Done!  :P


Suppose that we win $ \$2$ if we flip a heads on a coin toss, but lose $ \$1$ if we flip tails.  What is the expected value, in dollars, of our winnings after one flip?

[color=black]I have put 1/2 as the answer, and the machine accepted that as correct. However, amounts of money are normally reported with decimals, and so would the machine accept 0.5 as correct?[/color]

I have had this problem before, and I put in .5, and it accepts it.
So to answer your question, yes  :D

Nobody has posted a solution...just an answer. :(
[hide=Solution] The probability we get heads is $1/2$.
$$1/2*2=1$$
The probability we get tails is also $1/2$.
$$1/2*-1=-1/2$$
Now we add this up:
$$1-\frac12=\boxed{0.50}$$


Can someone explain to me the importance of $ M$ and $ L$ series in the proof?
Also, how was the Frey Elliptical Curve Formulated and proved to be non-modular?

WIth regards to the epsilon conjecture: http://en.wikipedia.org/wiki/Frey_elliptic_curve


To those of you who have this book, to what score range on the AMC 12 would you recommend it? For example, I'm expecting myself to score somewhere between 100 and 110. Do you think this book could help me bump it up to more like 130-140? (I'm sure it wouldn't hurt to get it, but I'm curious as to whether or not you think it's worth it).

i would suggest buying the intro books and the problem books instead. Because that would in a lot more depth  :)


modular arithmetic
group theory
number theory

In one recent math competition, there was the problem:

What is the remainder of $\frac{100!}{1}+\frac{100!}{2}+\frac{100!}{3}+...+\frac{100!}{100}$ when divided by $101$?

Well, I figured out how to solve it by noticing that mod 101, the terms in the sum turn out to be a rearrangement of $1+2+3+...+100$ (a cyclic group) since 101 is prime (if it weren't prime they would all give sums of 0).

Well, now I've been trying to figure out exactly why it forms a rearrangement if $\frac{n!}{1}+\frac{n!}{2}+\frac{n!}{3}+...+\frac{n!}{n}\pmod{p}$, and I can't seem to form a proof, and it's been bugging me.  I think it might have something to do with Wilson's Theorem, but I'm not sure.

Thanks.

The second general expression you gave should really have $n = p-1$.

This is simply the fact that every nonzero element of $\mathbb{Z}_{101}$ (the integers $\bmod 101$) has a unique multiplicative inverse $\bmod 101$, so if you invert the set $\{ 1, 2, 3, ... 100 \}$ (by "invert" I mean "for $a$ find $b$ such that $ab \equiv 1 \bmod 101$) then you get the same set back.

This set is called $U_{101}$, or the set of units in $\mathbb{Z}_{101}$.  It includes every nonzero element of $\mathbb{Z}_{101}$ because $101$ is prime.

Edit:  More concisely,

[b]Definition:[/b]  $\frac{a}{b}\equiv c \bmod m \Leftrightarrow a \equiv bc \bmod m$

[b]Claim:[/b]  Given a prime $p$,

$\{ 1, 2, 3, ... p-1 \}= \{ \frac{1}{1}, \frac{1}{2}, ... \frac{1}{p-1}\}$

(Where inversion is taken $\bmod p$.)

For some reason I think it holds for all p less than n!

Less than $n$ or less than $n!$?  Both are false for $\frac{n}{2}< p < n$.

hmm p=5, n=3 works
other small cases work

Are we debating the same statement?

[b]My Claim:[/b]  $\frac{n!}{1}+\frac{n!}{2}+...+\frac{n!}{n}\equiv 0 \bmod p$ when $p$ is an odd prime and $n = p-1$.

(Outline of proof:  $\sum_{k=1}^{p-1}\frac{1}{k}\equiv \sum_{k=1}^{p-1}k = \frac{p(p-1)}{2}\equiv 0 \bmod p$)

[b]Your Claim:[/b] (as I interpret it) The above holds for $n$ such that $p < n!$.

But this is obviously false.  Using your example,

$\frac{3!}{1}+\frac{3!}{2}+\frac{3!}{3}= 11$

Which is clearly not divisible by $5$.

I think he is saying it works when we have

$n!\left(1+\frac{1}{2}+\cdots+\frac{1}{p-1}\right)$ (even though this is an incorrect interpretation of the problem).

But this is obvious because if $1+\frac{1}{2}+\cdots+\frac{1}{p-1}\equiv 0 \pmod{p}$, then multiplying won't change anything.

Nevermind what I was saying, I was just looking for a generalization because the proof seemed too easy and I though I saw something nice about symmetric sum this represents


Find all [tex]n\in N[/tex] such that exists [tex]x,y,p\in N[/tex], [tex]p[/tex] is odd prime, sastify: [tex]x^n+y^n=p[/tex]

I think such $n$ are powers of 2.

First of all, $n$ must be even (excluding trivial case $n=1$) otherwise $x^n + y^n$ has a non-trivial factor $x + y$.
On the other hand, it is clear that if $n=2k$ is not a power of two then it is divisible by some odd prime $q$ implying that $x^n + y^n$ has a non-trivial factor $x^{n/q} + y^{n/q}$.

So $n$ must be a power of 2. But it is not clear if every $n=2^m$ admits some prime of the form $p=x^n + y^n$.


Fie ABC triunghi echilateral si D un pct pe BC;construim DE un segment egal cu BD ; Stiind ca m(DEC)=30(grade) aflati masurile unghiurilor trg BDE;

Sunt patru pozitii (cel mult) posibile pentru punctul $E$. Despre care dintre ele este vorba ? Sau in toate cazurile ?

Pai nu cred ca exista decat un caz posibil pt E aaaaaaaahhhhhhh am uitat sa precizez ca BD>DC  :blush: Cred ca asta reduce cele 4 cazuri la unu singur .

Poti sa-mi spui ce rost are atunci triunghiul echilateral, mai exact varful $A$ ?
Citeste cu atentie enuntul problemei tale.
Dupa parerea mea, cred ca punctul $E\in AB\cup AC$.
Te rog, sterge copiile ultimului tau reply (dupa ce il voi sterge pe al meu).

Acum miam dat seama m-am grabit cand am scris-o ; Deci ABC echilateral si D un pct pe BC a.i. BD>CD Pe AD luam un punct E a.i. DE=BD ; Stiind ca masura unghiului DEC este de 30 de grade Aflati masurile unghiurilor trg BDE :oops:

Iar ai gresit. Te rog mai multa grija in redactarea textelor. Incepe sa devina simptomatica chestia asta. Pai daca $D\in BC$, ce-i acela triunghiul $BDC$ ? Apoi cred ca unghiul $DEC$ si nu $BEC$ este de 30 grade !


SMO2007, Open #11

Determine the largest value of $a$ such that $a$ satisfies the equations $a^{2}-bc-8a+7 = 0$ and $b^{2}+c^{2}+bc-6a+6 = 0$ for some real numbers $b$ and $c$.

[hide]$a^{2}-8a+7=bc=(a-7)(a-1)$
Substitute that into equation 2.
One gets $b^{2}+c^{2}+(a-13)(a-1)=0$
Now, add 2bc to both sides, substitute bc in terms of a on the right side only and combine like terms.
$(b+c)^{2}=(a-1)(2a-14+13-a)$ or $(b+c)^{2}=(a-1)^{2}$
a doesn't have a largest value.[/hide]
Did I do this right?

But you didn't check that your solutions satisfy both original equations.

Edit:
[hide="Solution"] We have the equations

$bc = a^{2}-8a+7 = (a-1)(a-7)$
$b^{2}+bc+c^{2}= 6a-6 = 6(a-1) \Leftrightarrow$
$b^{2}+c^{2}=-a^{2}+14a-13 = (a-1)(13-a)$

So we see that $a \le 13$.  But there's more!  By AM-GM,

$b^{2}+c^{2}\ge 2bc \Leftrightarrow$
$(a-1)(13-a) \ge 2(a-1)(a-7) \implies$ (for $a > 1$)
$13-a \ge 2a-14 \Leftrightarrow$
$27 \ge 3a \Leftrightarrow$
$\boxed{ a \le 9}$

Indeed, for $a = 9$ we have $b = c = 4$. [/hide]

Err.. $a$ has to have a largest value because there's an integer answer to all SMO problems.. hmm.

Wow.. the AM-GM was nice, didn't see that one.  :)  That should be correct.


[i]Let $S$ be a set and let $*$ be a binary operation on $S$ satisfying the two laws
\[\forall x\in S,\ x*x=x\]
and
\[\forall (x,y,x)\in S^{3},\ (x*y)*z=(y*z)*x.\]
Prove that $*$ is commutative.[/i]

Masoud Zargar

$xy = (xx)(yy) = (x(yy))x = ((yy)x)x = ((yx)y)x = (yx)(yx) = yx$.

edit:ah, someone else posted in the meantime :)


this year there were 3 boys and 3 girls in the spanish team, i know this is rather unusual; but what's completely astonishing, girls outscored boys : 42-15
i'll try to find out any sort of explanation to this. i dont know what the boys were doing during the contest! maybe thinking about the girls  ;)
does anybody remember any similar results in the imo??

well last year there were about 4-5 gold medalists girls, so that can count, right? :)
for example Ana Caraiani ranked 7th with 38p, and was way ahead the next Romanian Team member, andreis with 25p.

three girls also succeeded in getting a gold medal this year also, and the strange fact is that they are from some of the top countries in the imo (alison in usa, the russian and uckranian girls).

There were two girls in the Dutch team this year, and we were the best of our team. 15+10 for the girls and 10+9+8+1 for the boys.

There were also 3 girls from macedonia, I think, but I don't now their scores. Indeed, I know that Macedonia1 made 41 pts and the others all < 10 !! Unbelievable...

this is only tangentially relevant, but...
 some countries have a requirement from their dept of education or whoever oversees their olympiad representation that each olympiad team has to have at least 1 (or whatever number) girl. Several years ago there was a proposal at the IOI that every country should be required to have a spot for a girl, which they would forfeit if they didn't have a girl to fill it.  The US delegation supported the proposal, even though there has never been a female on the US IOI team. Interestingly, the Iranians have done pretty well with girls being at the IOI (not winning, just being there). The proposal didn't pass. Also interesting is that one of the of the most vocal opponents was a female delegate from Argentina.

Some countries also limit how many olympiads one student can participate in, following a "spread the wealth" philosophy. The Netherlands is one of the countries that do this (I got it from the Dutch coach at the IOI), but I think there are others. The physics and chemistry olympiads in the US will not allow students to participate in one of those int'l olympiads (or training camps) and any others in the same year.

[quote="texas137"]Some countries also limit how many olympiads one student can participate in, following a "spread the wealth" philosophy. The Netherlands is one of the countries that do this (I got it from the Dutch coach at the IOI), but I think there are others. The physics and chemistry olympiads in the US will not allow students to participate in one of those int'l olympiads (or training camps) and any others in the same year.[/quote] happens also in Romania :(

and the limitation happens in China, too.

[quote="luishhh"]this year there were 3 boys and 3 girls in the spanish team, i know this is rather unusual; but what's completely astonishing, girls outscored boys : 42-15
i'll try to find out any sort of explanation to this.[/quote]
Maybe they were just better??

[quote="texas137"]Several years ago there was a proposal at the IOI that every country should be required to have a spot for a girl, which they would forfeit if they didn't have a girl to fill it. [...] The proposal didn't pass. Also interesting is that one of the of the most vocal opponents was a female delegate from Argentina.[/quote]
I can totally understand that. I wouldn't want to be at the IOI (what is IOI, by the way? Informatics?) just because of such a rule. It gives you a feeling of being not as qualified and good and much worth as the boys who go to the IOI. I would want to qualify just the way the boys do. Otherwise nobody would recognize my abilities.

Well, Canada too has had exceptional female IMOers. Olena B. last year won Gold Medal with 31 points. I think she becomes the first female gold medalist in North America. (This I am not quite sure, but she's definitely the first Canadian female gold medalist.) Talk about female wit!

I completely agree with Annika. If girls would be chosen because they're girls it's like girls are by defenition inferior at mathematics and need such rules to qualify. That's not good for your self-esteem...

And yes, in the Netherlands you can only participate in one olympiad per year. You can't even join two training-and-selection camps, because most of them are in the same week :(

[quote="Iriz"]I completely agree with Annika. If girls would be chosen because they're girls it's like girls are by defenition inferior at mathematics and need such rules to qualify. That's not good for your self-esteem...

And yes, in the Netherlands you can only participate in one olympiad per year.([/quote]

Are you from the Netherlands? What about the rule the Dutch IOI coach told me that Dutch teams have to include a girl? Does that include math too, or just IOI ("Int'l Olympiad in Informatics", btw. It's one of the 5 int'l olympiads, and involves algorithmic computer programming)? I think there are more strong female mathematicians than there are strong female computer programmers, although the type of programming involved in the IOI is very mathematical. There has been a lot of cross-representation btwn the IMO and IOI (like Reid Barton, Anders Kaseorg, Tiankai Liu). Maybe some of the talented female mathematicians on this forum should try it out. http://www.usaco.org

[quote="texas137"]What about the rule the Dutch IOI coach told me that Dutch teams have to include a girl? [/quote]

China has a similar policy, for olympiad in informatics only. Because they think that there are too few girls attending olympiad in informatics, so this policy is made to encourage more girls to participate in it.

The Dutch IMO team does not have to include a girl. And as I said I'm glad about that. I don't know about the IOI team though...

[quote="Annika"][quote="luishhh"]this year there were 3 boys and 3 girls in the spanish team, i know this is rather unusual; but what's completely astonishing, girls outscored boys : 42-15
i'll try to find out any sort of explanation to this.[/quote]
Maybe they were just better??[/quote]

This year i was one of the teachers in the brief training of the spanish imo team, and of course i noticed the girls were more likely to do better in greece, but i didn't think there would be such difference between them. In fact the girls seemed to have the boys under their control  ;)
I do hope there will be much more female presence for the future olympiads!!

In Spain you are not allowed to take part in more than one between imo, ipho & icho; but it's possible to participate in the ioi & one of the others

[quote="Tasar"]The Dutch IOI team indeed has to to include a girl (at least they had to a few years ago), and at least in 1998 the girl participated in the same selection test as the boys and scored much lower than most of the boys.

BTW I think it is still possible in the Netherlands to participate both in the IOI and the IMO (it was in 1999), since they are always pretty far apart in time. I can understand, however, people saying you can't participate in for example the IMO and the IPhO if this means flying straight from the closing ceremony of the one to the opening ceremony of the other.  In fact this year the IPhO opening ceremony was last friday, whereas the IMO closing ceremony was last saturday, so the IPhO and IMO overlapped.

It is unfortunate though that it has become impossible to attend both selection camps, since first of all you have to choose what to do and run the risk of not getting selected anywhere and secondly even if you do not go to the international olympiad you always learn a lot. But in the Netherlands they are all organised at the same time, because that time is the only time when you know all students are free from school (at least students in their last year).[/quote]
I think that the best thing to do is to choose one olympiad and train yourself upto at least the level of a medal before you start concentrating on a second olympiad.

Furthermore I think the IOI rule to choose 2 boys and 2 girls in each team is extraordinarily sexistic, which is itself a reason to boycot it completely (at least I boycotted it when I was a high-school student, also because I found the problems in the Dutch olympiad too easy).

Have U ever heard that we had girl in our team?

The results are now posted on the IMO-website, and the results are split between boys and girls. Apparently girl power is not magnificent since the girls scored an average 14.1 versus the boys 16.4. However I don't know whether the well performing countries brought along as many girls as the less performing ones, so this statistic might not be relevant. More interesting however is the fact that girls scored more on problems 1 and 5, and were beaten on the other 4 problems. Does this mean that girls are better at geometry, or is this another coincidence?


MATHCOUNTS
probability
algebra
polynomial
geometry
modular arithmetic
AMC

This is a progressive difficulty test I made up by myself.  I think I know the answers, so lets see if you can come up with them(I might be wrong)!  If you can get all ten right, you are definitely prepared for the Mathcounts chapter and even Mathcounts state competition.
1) What is the value of $ 3^5 \minus{} 3^x$ if x equals 3?
2) If 2x+5=29, what is the value of $ x(x \minus{} 1)$?
3) If 3/5 of the students at a certain middle school are taking gym class,  75% are taking Pre-Algebra, and 10% are taking both.  If there are 1,000 students in the middle school, how many are taking neither?
4) In an arithmetic sequence, the first term is 3 and the eighth term is 45, and the ninth and tenth terms are defined as a and b, what is the value of $ b! \minus{} a!$?
5) Sherman is taking a multiple choice test.  There are 4 choices for each problem, and there are 20 problems.  If he randomly guesses on each problem, what is the probability that he will get at least a 50%?  Express your answer as a common fraction.
6) Expand the polynomial $ x(2x^2 \plus{} 3x^5 \minus{} 2x)$
7) What is the least common multiple of 5!, 4!, 3!, and 2!?
8) What is the units digit of 3 to the ninety ninth power?
9) A man has a certain age.  If you reverse the two digits of his age, multiply the result by two, and subtract 4, the result is his age.  What is the man's age?
10) What is the area of a figure enclosed by the equations y=x+1, y=5x+3, and y=x-2?
Good luck. :lol:

[hide="Solutions"]
1. $ 3^5\minus{}3^3\equal{}243\minus{}27\equal{}\boxed{216}$.
2. $ 2x\plus{}5\equal{}29 \implies 2x\equal{}24 \implies x\equal{}12 \implies x(x\minus{}1)\equal{}\boxed{132}$.
3. Er...3/5+3/4-1/10 is greater than 1 :wink:
4. $ 3,10,17,24,31,38,45,52,59$...$ 59!\minus{}52!$ is quite too large to calculate.
5. Again, too large to calculate.
6. $ 2x^3\plus{}3x^6\minus{}2x^2$
7. $ 5!$, obviously...
8. $ 3^{99} \equiv 3^{3} \equiv \boxed{7} \pmod{10}$
9. $ 2(10y\plus{}x)\minus{}4\equal{}10x\plus{}y \implies 2x\plus{}20y\minus{}4 \equal{} 10x\plus{}y \implies 8x\plus{}4\equal{}19y \implies x\equal{}9, y\equal{}4$, and the number is $ \boxed{94}$.
10. Find all the boundary points, shoelace it.
[/hide]

Did you actually try to solve the problems?  Also, MATHCOUNTS has many more word problems and problems that aren't total brute force :wink:

Mathcounts.

On the same page, do you mean thread? (it's the same for everyone)

[quote="varunrocks"]Is this mathcounts or MOEM because my computer puts mathcounts,moems, and AMC8 on the same page![/quote]

They merged the middle school forums into the contests and classroom. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280659]here[/url].

[quote="varunrocks"]Is this mathcounts or MOEM because my computer puts mathcounts,moems, and AMC8 on the same page![/quote]
It's mathcounts.  This is less of real problems and more of a "skill check".  Also, this assumes the use of a calculator.  Also, disregard Problem 4.  I got it from a dud math book(no, it was not an AoPs book, of course!)


number theory
relatively prime
number theory unsolved

1) Solve the equation: $ 3^x\plus{}4^y\equal{}5^z$ in integer numbers x,y,z
2) Solve the equation: $ 7^x\plus{}12^y\equal{}13^z$ in integer numbers x,y,z

1) checking by modulo $ 3$ and $ 4$ we can conclude that $ x$ and $ z$ are even. Therefore $ 3^x \equal{} 3^{2a}$ and $ 5^z \equal{} 5^{2b}$ are perfect squares, $ 4^y \equal{} 2^{2y}$ is by itself a perfect square. therefore $ 3^a, 2^y,5^b$ is a primitive pythagorean triple. meaning $ 2^y \equal{} 2mn$ for some relatively prime $ m,n$. leading to the only posibility of $ (x,y,z) \equal{} (2,2,2)$

2) checking my modulo $ 7$ and $ 13$ we conclude that $ x$ and $ y$ are divisible by 3. therefore $ 7^{3a} \plus{} 12^{3b} \equal{} (7^a \plus{} 12^b)(7^{2a} \plus{} 7^a\cdot 12^b \plus{} 12^{2b}) \equal{} 13^z$ implying $ 7^a \plus{} 12^b \equal{} 13^c$ for some $ c\leq z$. this way we can construct ever smaller solutions. infinte descend kills it... no solutions

3) checking by modulo $ 4$ we get $ x \equal{} z (mod2)$, by modulo $ 7$ we conclude that $ x,z$ are odd. now by modulo $ 8$ we get the unique solution of $ (x,y,z) \equal{} (1,1,1)$

[quote="Jure the frEEEk"]
3) checking by modulo $ 4$ we get $ x \equal{} z (mod2)$, by modulo $ 7$ we conclude that $ x,z$ are odd. now by modulo $ 8$ we get the unique solution of $ (x,y,z) \equal{} (1,1,1)$[/quote]
I don't think so. I'm sure that by the above modulo, we only get $ y\equal{}1$. How do you show that $ x\equal{}z\equal{}1$?

Let's solve the equation: $ 3^x\plus{}4\equal{}7^z$


y^2=1+x+x^2+x^3

Hi, I found this D.E. in a book with the hint:"use that x^4+y^4=x^2 have no positive integers solutions" i don't know how to use it, however I use that y^2=(1+x)(1+x^2),...

are we supposed to prove this possible or impossible?

22=1+1+12+13

I cannot seem to find any other simple examples.

he means y=20 .... 400 = 1+7+49+343

I take back my previous statement of no other simple solutions and replace it now with no solutions with x<30 other than x=1,7.

Also, quick calculations give x=0,1,or7 (mod8)

hmm.. nice!

are there more such solutions? (can be of any form 1+x+x^2+...+x^n?)

Let's consider the equation \[1+x+ \cdots +x^{n-1} = y^m\]
where $x,y,n,m$ are integers with $n \geq 3,m\geq 2, |x| > 1$.

It has been proved that :

a) if $n=0$ mod[4] then the unique solution is $n=4,m=2,x=7,y= \pm 20$ (see [1]).

Note that it is our given problem...even if it may be solved more easily (to be continued...).

b) If $m=2$ the solutions are those above and $n=5,x=3,y = \pm 11$ (see [2] and [3])

c) If $n=0$ mod[3] the solutions are $m=n=3,y=7,x \in \{-19;18 \}$ (see [2])

d) If $m=3$ and $n \neq -1$ mod[6] the solutions are those of c) (see [2]).

Moreover, if $x,y$ are given, with $x > 1$ and $y$ primes, then (see [4]) the equation above has at most one solution $(m,n)$. And in the case we have a solution, $n$ is also a prime and $n$ is the order of $x$ modulo $y$, and $m$ is divisible by the order of $y$ modulo $x$.

References :
[1] T. Nagell, 'Note sur l'quation indetermine $\frac {x^n - 1} {x-1} = y^q$', Norsk. Mat. Tidsskr. (1920), p.75-78.
[2] T. Nagell, 'Sur l'quation indetermine $\frac {x^n - 1} {x-1} = y^2$', Norsk. Mat Forenings Skrifter I, n3 (1921), p.17.
[3] W . Ljunggren, 'Noen setninger om ubestente likninger av formen $\frac {x^n - 1} {x-1} = y^q$', Norsk. Mat. Todsskr. (1943), p.17-20.
[4] H.M. Edgar, 'The exponential equation $1+a+ \cdots + a^{x-1} = p^y$', A.M.M., 1974, p.758-759.

Pierre.


Let $ x,y,z \ge 0$ and $ x\plus{}y\plus{}z\equal{}1$. Prove that $ x^3\plus{}y^3\plus{}z^3\plus{}6xyz \ge \frac{1}{4}$

[hide]Hint:
Homogenizing the RHS and applying 3-degree Schur's inequality[/hide]

Please give me a full solution. Thank you very much.

[quote="trantienhung"]Let $ x,y,z \ge 0$ and $ x \plus{} y \plus{} z \equal{} 1$. Prove that $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \frac {1}{4}$[/quote]
we can prove the inequality than stronger: $ x^3\plus{}y^3\plus{}z^3\plus{}\frac{15xyz}{4}  \ge \frac{1}{4}$
It is equivalent to:
$ 4(x^3\plus{}y^3\plus{}z^3)\plus{}15xyz \ge (x\plus{}y\plus{}z)^3$.
$ <\equal{}> x^3\plus{}y^3\plus{}z^3\plus{}3xyz \ge x^2(y\plus{}z)\plus{}y^2(z\plus{}x)\plus{}z^2(x\plus{}y)$
It is Schur 'inequality.We have Q.E.D

[quote="trantienhung"]Let $ x,y,z \ge 0$ and $ x \plus{} y \plus{} z \equal{} 1$. Prove that $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \frac {1}{4}$[/quote]
Here is my solution
Suppose $ x\geq y\geq z$
We have $ x^3\plus{}y^3\plus{}z^3\plus{}6xyz\geq x^3\plus{}(y\plus{}z)^3\geq\frac{1}{4}(x\plus{}y\plus{}z)^3\equal{}\frac{1}{4}$
We have done

[quote="DCTPKTCSPHN"][quote="trantienhung"]Let $ x,y,z \ge 0$ and $ x \plus{} y \plus{} z \equal{} 1$. Prove that $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \frac {1}{4}$[/quote]
Here is my solution
Suppose $ x\geq y\geq z$
We have $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz\geq x^3 \plus{} (y \plus{} z)^3\geq\frac {1}{4}(x \plus{} y \plus{} z)^3 \equal{} \frac {1}{4}$
We have done[/quote]
Your proof can't solve : $ x^3\plus{}y^3\plus{}z^3\plus{}\frac{15xyz}{4} \ge \frac{1}{4}$


Problem 2a, Round 1 asks
"Find, with proof, all ordered pairs..."

Problem 5 Round 1 asks
"Find, with proof, an increasing sequence..."

The first one obviously needs all answers.
The second one can take any answer.

So my question is: If it doesn't ask for all values and if it doesn't ask for only one, do I need to find all values, and do I need to prove that I have found them all / that there is only one?

For 5a, Round 1, you could write any sequence, but there is only one solution.

No, you do not need to find all solutions (I think I know which problem you are referring to), only one EDIT: solution.

BOGTRO, you accidentally referred to #3 on round 1.
Yes, for #2, you must find all ordered pairs.

@BOGOTRO
I think I know what problem you think I am referring to, and if I'm right about that, your wrong, and if I'm wrong your probably right. :huh:

anyway...

I should give an example

$ 27 x^3 \minus{} 135 x^2 \plus{} 225 x \minus{} 125 \equal{} 0$
Find x.

Must I prove that 5/3 is the unique value?

Or...

$ 27 x^3 \minus{} 126 x^2 \plus{} 195 x \minus{} 100 \equal{} 0$
Find x.

Must I find both 4/3 and 5/3 and prove that they are the only values?

EDIT: Or can I assume that I only need to prove that 5/3 works.

If your just guessing, tell me because I don't want to lose a point on this.

I can always prove it - it's just a pain.

I would just prove it, especially if it doesn't take too much work. Better to be safe.
But if the proof of uniqueness is ridiculous, maybe you can just skip over it.

Ahh...

So when it says "find all answers" you need to find every answer and prove that (1) they are actually answers and (2) they are the only answers.

When it says "find an answer" you need to find one answer and prove that it works.


At some point they publish everything you could need/want to know - like how many in each grade scored what, and how many perfects there were, etc -- but everytime I check, the site sends me somewhere and then says they redid the site or something - might anyone know how soon after the tests are done that they compile these stats? Are these stats here already and I'm just not finding them? Anyone else besides me care?

the statistics aren't up yet, I don't think. When they are, you won't be redirected to the homepage.

Patience, young man. You can see from the other threads that we will all know when those statistics are posted by visiting the home page of the AMC Web site.

http://www.unl.edu/amc/index.shtml

Meanwhile, PLEASE don't spam up this forum with repeated instances of the same question that has been asked in a bunch of other threads.

Good luck in your competition activities.

I seem to remember state stats coming out by the end of April in the past, but it's not much before that.  They come out quite a while after national stats.

According to the datestamps on the AMC website, the 2006 state stats came out 3/13/2006 (the date the page was last updated).  I might be reading this wrongly but that's all the information I've got.  So... another week or two is my guess.


topology
inequalities
triangle inequality
advanced fields
advanced fields unsolved

I'm going off of [url]http://zaki.web.ugm.ac.id/web/mod.php?mod=download&op=visit&lid=122[/url], problem 2.12 on page 26.

I started with proving the cases where x and y lie in H, and then without loss of generality, we may state that d(x, H) > d(y, H).

So the problem reduces to:

$ d(x, H)-d(y, H) \leq d(x, y)$

Which is of course:

$ d(x, H) \leq d(x, y)+d(y, H)$

Which looks exactly like the triangle inequality.  However, d(x, H) is defined as the inf of {d(x, h) : h is in H}.  Can I still just apply the triangle inequality?  What more has to be done to do so?

I was too lazy to download the PDF, but since you were too lazy to copy the problem I hope that's o.k.  :maybe: If $ A(h)\leq B(h)$ for all $ h\in H$, then $ \inf_{h\in H}A(h)\leq\inf_{h\in H}B(h)$. owk

Sorry, I should have indicated that you didn't actually need to go to the problem to understand, I put all the relevant information in my post.

But thanks for the response.  Not sure why I didn't think of that...


A bottling plant completely fills 0.5-liter bottles from a 56-liter tank of soda. A flat consists of 24 bottles. When all of the bottles of one flat are completely filled, how many total liters of soda are in the flat bottles?

there's no need of 56.

the answer is just 24 * 0.5 = 12.

answer : 12


[hide]B.)945

just 90*10+45.
45 is all of the units digits added up together.[/hide]

so easy you just take the average which is 94.5 and multiply it by the number of numbers which is 10 so the answer si 945

Do you realize that this thread was started in 2009 and that the last post was in 2016?

Imagine bumping a thread that's probably older than some of the people on the site

[quote=Sirfrankie_II]why the bump

this is 11 years old[/quote]

more like 12 years old :P

[quote=OronSH][quote=megarnie]The answer is trivially $\boxed{\textbf{(B)}\ 945}$, by QM-AM-GM-HM.[/quote]

arnev bad

[hide=Solution]Let $f(x)=x+(x+1)+(x+2)+(x+3)+...+(x+9).$ We can simplify this to $10x+45.$ Now we can test all the answer choices as follows: let $A$ be an answer choice. We want to find the value of A such that $g(x)=10x+45-A$ has root $x=90.$ Make a guess for $x,$ suppose $0.$ Then we have $g(0)=45-A.$ We also have $g'(0)=10.$ Therefore we have that the root is approximately $\frac{45-A}{-10}$ so setting this equal to $90,$ we can trivially solve to get that $A$ is approximately equal to $945$ which is answer choice $\boxed{B}.$

Note: While this may not be rigorous, the full iteration of Newton's Method is not necessary in a timed competition like the AMC 8.[/hide][/quote]

sobad imagien bad solution

The sum is the number of terms times the average, so $94.5\cdot 10=\boxed{945}$.

[hide=sol]the average of the terms is 94.5, and there are 10 terms in the sequence, so the sum is 94.5 * 10=945[/hide]

[quote=AIME15]$ 90\plus{}91\plus{}92\plus{}93\plus{}94\plus{}95\plus{}96\plus{}97\plus{}98\plus{}99\equal{}$

\[ \textbf{(A)}\ 845 \qquad
\textbf{(B)}\ 945 \qquad
\textbf{(C)}\ 1005 \qquad
\textbf{(D)}\ 1025 \qquad
\textbf{(E)}\ 1045
\][/quote]

Me be like okay I just add up all the terms cuz I forgot all the formulas for sequences. I bash and get 945

[hide=Solution]The solution is B, 945[/hide]
Very simple problem, just find the pattern

[quote=huashiliao2020]y is this in msm, isnt there an aops esm (elementary)? Like there should exist one, considering aops wants to encourage elementary kids[/quote]

The distribution of stuff in esm is probs a rlly tiny subset of msm so its fine if it ocassionally comes here.

I am the first person to post here in 2023

Some posts on this thread have suggested multiplying the average by the number of terms, but how would you find the average without finding the sum?

[rule]
@above just why?

I am the third person to post here in 2023

[quote=Facejo]Some posts on this thread have suggested multiplying the average by the number of terms, but how would you find the average without finding the sum?[/quote]

Group $90$ and $99$, $91$ and $98$, and so on. All of these pairs have the same average. So the average of all of these numbers must also be the average of any of the pairs.

I am the 4th person to post here in 2023

[quote=Facejo]Some posts on this thread have suggested multiplying the average by the number of terms, but how would you find the average without finding the sum?[/quote]

Since they are consecutive, the average is just the median, or $94.5$. With $10$ terms, the sum is going to be $10\cdot94.5=\boxed{945}$, or $\boxed{\text{(B)}}$


I answered a question right on the 1st try, then my alcumus score dropped 0.02%. Then, I answered another correct, this time on my 2nd try, and my alcumus score dropped another 0.80%.

Sounds bad, right? But get this...

I solved a question that was much easier than the 1st 2-it was worth only 1 point-and my alcumus score [i]increased[/i] 0.05%.

I don't like how alcumus score is so confusing. Can this be fixed?  :(

Other people were using Alcumus at the same time, probably getting problems that you had already seen right, which would make the impact of those questions (if you got them correct) go down, thus lowering your Alcumus score.


Prove that $ n!>\left(\dfrac{n\plus{}1}{2}\right)^n$ for $ n\equal{}2,3,4,\dots$.

[quote]$ n!\equal{}1*2*...*n\leq(\frac{1\plus{}2\plus{}...\plus{}n}{n})^{n}\equal{}(\frac{\frac{n(n\plus{}1)}{2}}{n})^{2}\equal{}(\frac{n\plus{}1}{2})^{n}$[/quote]with IAG you can prove the opposite of this inequality, I think that she is wrong

[quote="Mathnerdmo"][quote="peine"][quote]$ n! \equal{} 1*2*...*n\leq(\frac {1 \plus{} 2 \plus{} ... \plus{} n}{n})^{n} \equal{} (\frac {\frac {n(n \plus{} 1)}{2}}{n})^{2} \equal{} (\frac {n \plus{} 1}{2})^{n}$[/quote]with IAG you can prove the opposite of this inequality, I think that she is wrong[/quote]I don't know what IAG is, but the only thing wrong about zaya_yc's post is the typo with the squared term instead of the nth power...[/quote]

IAG = inequality of arithmetic and geometric means (or AM-GM)

[quote="#H34N1"]Prove that $ n! > \left(\dfrac{n \plus{} 1}{2}\right)^n$ for $ n \equal{} 2,3,4,\dots$.[/quote]

$ 3!\equal{}6\newline \left(\frac{3\plus{}1}{2}\right)^3\equal{}8\newline \implies 3!<\left(\frac{3\plus{}1}{2}\right)^3$

[quote="jvenezuela716"][quote="#H34N1"]Prove that $ n! > \left(\dfrac{n \plus{} 1}{2}\right)^n$ for $ n \equal{} 2,3,4,\dots$.[/quote]

$ 3! \equal{} 6\newline \left(\frac {3 \plus{} 1}{2}\right)^3 \equal{} 8\newline \implies 3! < \left(\frac {3 \plus{} 1}{2}\right)^3$[/quote]

Yeah, sorry, that was a misunderstanding on my part; I saw (was given  :P ) the wrong inequality and told #H34N1 that the direction sign should be reversed.

[quote="#H34N1"]Prove that $ n! < \left(\dfrac{n \plus{} 1}{2}\right)^n$ for $ n \equal{} 2,3,4,\dots$.[/quote]
[hide]Square: $ (n\cdot 1)((n \minus{} 1)(2))((n \minus{} 2)(3))\cdot (1\cdot n) < (((n \plus{} 1)/2)^2)^n$.
Is true if $ (n \minus{} k)(k \plus{} 1) < ((n \plus{} 1)/2)^2$, for all k=0,...,n-1.

Which is true by AM-GM.[/hide]

[quote="#H34N1"]Prove that $ n! < \left(\dfrac{n \plus{} 1}{2}\right)^n$ for $ n \equal{} 2,3,4,\dots$.[/quote]

supposing $ n!<(\frac{n\plus{}1} {2})^{n}$ and proving  $ (n\plus{}1)!<(\frac{n\plus{}2} {2}){n\plus{}1}$  wich is too easy to prove it :roll:

[quote="mathemonster"]Prove that $ n! < \left(\dfrac{n \plus{} 1}{2}\right)^n$ for $ n \equal{} 2,3,4,\dots$.
 Induction: let $ n \equal{} 2, 2! \equal{} 2 < (\frac {2 \plus{} 1}{2})^2 \equal{} 9/4.$
 Assume that $ n! < \left(\dfrac{n \plus{} 1}{2}\right)^n$ for $ n \equal{} 2,3,4,\dots$. Multiplying both sides by $ n \plus{} 1$
we have $ (n \plus{} 1)! < \left(\dfrac{n \plus{} 1}{2}\right)^n * \frac {2(n \plus{} 1)}{2} \implies \left(\dfrac{n \plus{} 2}{2}\right)^{n \plus{} 1}$
Q.E.D.

zaya_yc's prove by AM-GM was definately the best tho  :D[/quote]

i think the best solution has to be without theorems that what i know  :maybe:

Since $ \ln x$ is concave, by Jensen's we have

$ \frac{\ln 1 \plus{} \ln 2 \plus{}...\plus{}\ln n}{n} < \ln {\frac{1\plus{}2\plus{}3\plus{}...\plus{}n}{n}}$

$ \Rightarrow \ln n! < n \ln \left(\frac{n\plus{}1}{2} \right)$

$ \Rightarrow n! < \left(\frac{n\plus{}1}{2}\right)^n$


[quote="jmerry"]Suppose X, A, B, C are points in a plane such that XA = 6, XB = 11, and XC = 14. Find the maximum possible area of triangle ABC[/quote]

75 sq units?
well this is how I thought of it:

c_______x______b
                |
                a

that gives me (14+11)*6/2

is this right?

Too small. By the same reasoning, an area of at least 11*14/2 + 6*14/2 = 119 is possible, and the maximum is larger than that.

It is also easy to establish an upper bound- the area is less than 11*14/2 + 6*14/2 + 6*11/2 = 152

If we keep angle AXC fixed for a bit and just vary point B, in order to maximize the area of triangle ABC we must put B as far away from line AC as possible. This happens when XB :perp: AC. Likewise, XA :perp: BC and XC :perp: AB is necessary for the maximum area to be reached. Thus, X is the orthocenter of triangle ABC.

I'm sure there are better ways to find the area, but this works:

Label XF=a. From similar triangles XFC and XEB, we get EX=11a/14. Then, from similar triangles BFA and CEA, we get

[tex]\displaystyle\frac{11+a}{\sqrt{36-a^2}}=\frac{14+\frac{11a}{14}}{\sqrt{36-\frac{121}{196}a^2}}[/tex],

whose positive solution is a=4. Thus, [tex]\displaystyle AC = AF+FC=\sqrt{36-16}+\sqrt{196-16}=8\sqrt{5}[/tex], and so the area is

[tex]\displaystyle \frac{1}{2}(8\sqrt{5})(15)=60\sqrt{5}[/tex].

That's a nice solution. Seeing that X is the orthocenter is easy, finding the right area is hard. Your equation simplifies to a cubic
2*11*a :^3:  + (6 :^2:  + 11 :^2:  + 14 :^2: )*a :^2:  - 6 :^2: *14 :^2:  = 0, which happens to have the root at 4, as well as two irrelevant negative real roots. The other two corresponding versions (with XD or XE as variable) would give you 22/3 and 22/7 as roots.

My solution, leading to a different cubic:
Let R be the circumradius of ABC. Then XA = 2Rcos :ang: A, XB = 2Rcos :ang: B,
XC = 2Rcos :ang: C = -2Rcos( :ang: A +  :ang: B)
2R*XC = -XA*XB +  :sqrt: (4R :^2: - XA :^2:) * :sqrt: (4R :^2: - XB :^2:)
(2R*XC + XA*XB) :^2:  = (4R :^2: - XA :^2:)*(4R :^2: - XB :^2:)
4R :^3:  -  (XA :^2:  + XB :^2:  + XC :^2: )*R - XA*XB*XC = 0
4R :^3:  -  353R - 924 = 0
This has positive root 21/2 and two irrelevant negative roots. The sides of the triangle are then AB =  :sqrt: (4R :^2:  - XC :^2: ) = 7 :rt5: , BC = :sqrt: (4R :^2:  - XA :^2: ) = 9 :rt5: , and AC = :sqrt: (4R :^2:  - XB :^2: ) = 8 :rt5: .
The area is XA*XB*XC/(4R) = 60 :rt5: .

There's no way around solving at least a cubic- I built the problem by working backwards from a relatively nice triangle, which guaranteed a rational root.


geometry
circumcircle
inradius
trigonometry
perimeter
inequalities
geometry proposed

[u][i] Problem [/i] [/u][i]  Let $ S , R , r$ be respectively the area , the circumradius and the inradius of triangle

       $ ABC$ . Let $ BC \equal{} a , AC \equal{} b , AB \equal{} c$ Prove that

  $ \frac{2S}{R} \leq (b\plus{}c\minus{}a)sin\frac{A}{2} \plus{}(c\plus{}a\minus{}b)sin\frac{B}{2}\plus{}(a\plus{}b\minus{}c)sin\frac{C}{2} \leq \frac{S}{r}$

     When does one equality occur?[/i]

[i]
   Love An forever[/i]
   [i]Virgil Nicula , Erken ...can you help me?[/i] :blush:

[color=darkblue]Observe that, if $ A',B',C'$ is the points where the incircle of the triangle $ ABC$ touches the sides,

then $ (b \plus{} c \minus{} a)\sin\frac {A}{2} \plus{} (c \plus{} a \minus{} b)\sin\frac {B}{2} \plus{} (a \plus{} b \minus{} c)\sin\frac {C}{2}$ is the perimeter $ P'$

of $ \triangle A'B'C'$ . Now you can apply the well-known chain $ 6\sqrt 3r'\le P'\le 3\sqrt 3R'$ in $ \triangle A'B'C'$ .

Thus, the your inequality is equivalently with the classical chain $ 6\sqrt 3r\le P\le 3\sqrt 3R$ in $ \triangle ABC$ .[/color]

Very beautifull indeed... :blush: ....both the problem and the proof


group theory
abstract algebra
topology
superior algebra
superior algebra unsolved

A group $ G$ is said to be profinite if it is a compact Hausdorff topological group whose open subgroups form a base for the neighbourhoods of the identity.
Let $ G$ be a profinite group. Prove that every open subgroup contains an open normal subgroup of $ G$.

Sorry, it is so easy. Let $ H$ be an open subgroup of $ G$. $ H$ has finite index, say $ n$,  because $ G$ is compact. The action of $ G$ in the set of left coclases of $ H$ induces an homomorfism $ G\rightarrow S_n$. Then the kernel $ K$ of this homomorfism is a normal subgroup contained in $ H$ of finte index. It is easy to see that $ K\equal{}\cap_{g\in G} gNg^{\minus{}1}$ so it is closed and since it is a subgroup of finite index, then it is open.


For each integer $ n$ of the form $ n\equal{}p_1 p_2 p_3 p_4$, where $ p_1,p_2,p_3,p_4$ are distinct primes, let $ 1\equal{}d_1<d_2<...<d_{15}<d_{16}\equal{}n$ be the divisors of $ n$. Prove that if $ n<1995$, then $ d_9\minus{}d_8 \not\equal{} 22$.


group theory
abstract algebra
function
induction
number theory
totient function
superior algebra

I'm wondering there exist a direct proof for the following? :

[b]Problem.[/b] If $ k$ is a field and $ G$ is a finite subgroup of the multiplicative group $ k^\times\equal{}k\setminus\{0\}$, then $ G$ is cyclic.

PS: I got a proof, but it uses some facts on polynomials. I don't like it.  :)

For finite field you can use the following lemma:

Let $ G$ be a finite group with $ n$ elements such that for every $ m\in\mathbb{Z}_\plus{}$ the equation $ x^m\equal{}1$ has at most $ m$ roots. Then $ G$ is cyclic.

A standard proof of this lemma--you can weaken the condition by assuming $ m | n$, by the way--involves the [url=http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula]Möbius inversion formula[/url], which is a tool to construct the value of a function at some integer based on its behavior on divisors of that integer.

Essentially, you prove that the number of elements of order $ d$ is exactly $ \phi(d)$, where $ \phi$ is [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]Euler's function[/url].

LR, you have to be careful about characteristic issues (though once you note that the order of the group cannot be divisible by the field characteristic, these disappear).

But that is the idea. It suffices to show that the $ |G|$th roots of unity are cyclic, since obviously $ G$ embeds in those and any subgroup of a cyclic group is itself cyclic. So, what you have to show is that for all $ n$, there exists a primitive $ n$th root of unity. And you can see this with a simple counting argument and induction, as LR indicated.

Yes they are all really interesting, thank you.  :)


search
floor function
limit
number theory unsolved
number theory

There are given $ p,q \in \mathbf{N}, (p,q)\equal{}1$. Find (if they exist) all $ c,d \in \mathbf{R}$ such that sets $ A\equal{}\{ [\frac{np}{q}] : n \in \mathbf{N}\}$ and $ B\equal{}\{ [cn\plus{}d] :  n \in \mathbf{N} \}$ satisfy  that  union $ A$ and $ B$ is set $ \mathbf{N}$ and  also $ A$ and $ B$ has no common elements.

Maybe I am wrong but here is an approach . You know that there is a sequence of irrational numbers that have limit a rational number and vice versa . After this use this well-known fact
http://www.mathlinks.ro/viewtopic.php?search_id=912218370&t=3499

which stays in fact that if $ a,\ b$ are irrational numbers such that $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} 1$ then the sets
$ A \equal{} \left\{\left\lfloor na\right\rfloor \ |n \equal{} 1,2,3...\right\}$  and  $ B \equal{} \left\{\left\lfloor nb\right\rfloor \ |n \equal{} 1,2,3...\right\}$

are disjoint and their union is $ N^{*}$

Notice that $ \frac {p}{q}$ isn't irrational...

If $ p<q$, then $ A\equal{}\{0,1,2,3,...\}$, contradiction. If $ p\equal{}q\equal{}1$ then $ A\equal{}\mathbf{N}$, contradiction.
If $ p>q$, then $ \lim_{n\to\infty}\frac{card(A\cap\{1,2,...,n\})}{n}\equal{}\frac{q}{p}$ and $ \lim_{n\to\infty}\frac{card(B\cap\{1,2,...,n\})}{n}\equal{}\frac{1}{c}$ so $ c\equal{}\frac{p}{p\minus{}q}$.
This proof with irrational numbers is here almost correct. But if $ d\equal{}\minus{}\frac{1}{2(p\minus{}q)}$, we have no problems with the numbers divisible by $ p$. ($ kp\equal{}\frac{(kq)p}{q}\equal{}\frac{(k(p\minus{}q))p}{p\minus{}q}$).

[quote="mazur89"]Notice that $ \frac {p}{q}$ isn't irrational...
[/quote]

I know that  :)  . But as I wrote since there is a sequence of irrational numbers that have limit this rational number we can choose and irational number $ g$ such that the sets
$ A \equal{} \{ [\frac {np}{q}] : n \in \mathbf{N}\}$ and $ G \equal{} \{ [gn] : n \in \mathbf{N} \}$  to be the same  :wink:


Find all $a,b,c,d$ of integers which satisfy:-

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=2$

$2(a+b+c+d)=ab+cd+(a+b)(c+d)+1$

[hide]If one factors the right side on the second equation, ab+cd+ac+ad+bc+bd+1.  In other words, every possible multiplication between any two numbers in the solution set {a,b,c,d} is represented there.  Therefore, {a,b,c,d} will create the same answer as {a,c,b,d}.
Now, examine the second equation in binary.
0=ab+ac+ad+bc+bd+cd+1.  Only if there's 2 or 3 odd numbers in the solution set would this work.
The first equation makes it impossible for 2 or more numbers in the solution set to be negative.
Now, let's show that {-n,n,1,1} is impossible.
$3=-n^2-2n+2n+1$ or $2=-n^2$ which is clearly false.
Also, {-n,1,c,d} is impossible if c and d are greater than 1 because the greatest they can each be is 1/2.  2-1/n never equals 2.
But, if one of them is equals one, then 2-1/n+1/d=2 or n=d, which we already proved was impossible.

Now, let's restrict {a,b,c,d} to positive cases.
Obviously, if one of the variables exceeds 2, then at least one of the other variables is 1.
$\frac{a-1}{a}=1/b+1/c$ where a is greater than 2.  Now, we can easily show since {1,3,3,3} and {1,2,4,4} which fails the parity test, if a exceeds 3, b or c must be 2.  Now, $c=\frac{2a}{a-2}$.
A simple check shows that {1,2,3,6} yields 24=2+3+6+6+12+18+1=48 which of course, is false.
The graph of $\frac{2a}{a-2}$ is that of a hyperbola.  As a grows to very large values, c tends to 2.  Therefore, since both c and a are integers, there's only a fixed amount of possible values.  In other words, the positive values that satisfy equation 1 are {2,2,2,2} {1,2,3,6} {1,3,3,3} {1,2,4,4}, none of which that satisfy the second equation.
THere's no solution.[/hide]


read my answer math92. You've got the steps right, just the assumption wrong.

oh,  :wallbash:
i knew i was wrong.
nice job 236!
 :omighty:


Triangle $ ABC$ is a right triangle. If the measure of angle $ PAB$ is $ x$ degrees and the measure of angle $ ACB$ is expressed in the form $ Mx\plus{}N$ with $ M\equal{}1$, what is the value of $ M\plus{}N$?

[asy]draw((-10,0)--(20,0),linewidth(1),Arrows);
draw((0,0)--(10,10/sqrt(3))--(10+10/3,0),linewidth(1));

draw((10,10/sqrt(3))+dir(-150)--(10,10/sqrt(3))+dir(-150)+dir(-60)--(10,10/sqrt(3))+dir(-60),linewidth(1));

dot((-3,0));

draw(dir(180)..dir(105)..dir(30),linewidth(1));

label("P",(-3,0),NW);
label("A",(0,0),S);
label("$x$",(-1,1),N);
label("B",(10,10/sqrt(3)),N);
label("C",(10+10/3,0),NE);[/asy]

x can be expressed as sum of angle B and angle C.

angle B is 90 degrees.

so x = 90+C.

so, C can be expressed as x-90.

it's 1x-90.

so M=1, N=-90.

so M+N is -89.

answer : -89


algebra
system of equations
number theory
prime factorization

[b]1.[/b] How many different ways are there to write 100 as the sum of consecutive integers?

[b]2.[/b] If (x,y) is a solution for the system of equations below, find the maximum value of x:^2:+y:^2:,

2x:^2:+5xy+3y:^2: = 2
6x:^2:+8xy+4y:^2: = 3

Did you mean for the second one to end up quite so ugly?

1. [hide]we classify all such sequences according to the number of elements (n) and the middle element (m) of that sequence. n will be positive. Either n is odd, and m is a whole number, or n is even, making m 1/2 more than a whole number (such as .5, 1.5, 2.5, 3.5, etc.). Alrighty, this means that in general, if we are trying to find how many sequences of consecutive integers x can be written as, we are trying to factorize x in the form m*n, where either m is whole and n is odd, or m is a half and n is even. Note that in the second case, we can multiply m by 2 and divide n by 2 to get that m is odd and n is (any) integer. This will obviously yield the same number of sequences as case 1, since it's the exact same conditions. So we concentrate on case 1, and later we will multiply the number of solutions by 2. x=m*n, m any integer, n odd. Clearly, if x=2^r*a, where a is odd, m=2^r*b (n must be odd). So we have 2^r*a=2^r*b*n. Cancelling the 2^r, we have a=b*n. So we must factor a as any two integers. Since n must be positive, we don't need to worry about signs (the sign of b is automatically the sign of a.) Ok now, the number of ways to factor a=b*n will simply be the number of factors of a. So, our final answer is 2d(a), where d(z) signifies the number of factors of z. A simple formula for d(z) is as follows: If z, written in it's prime power factorization, is :pi: (i=1 to :inf:) pi^ei (the :pi: represents the product), then d(z) is given by :pi: (i=1 to :inf:) (ei+1). Thus, if the prime factorization of x (our original number) is :pi: (i=1 to :inf:) pi^ei, then 2d(a) will be 2:pi: (i=2 to :inf:) pi^ei. We finally solve our problem. 100=2^2*5^2. It's prime power factorization is e1 is 2, e2 is 2, and all other ei=0. Thus, using our formula, we find that the number of ways it can be written as a sequence is 2*(2+1)=6[/hide]

Thats the same answer I get.. although I used a slightly different (shorter?) method.. seeing someones already posted the answer I don't feel guilty about posting mine in this forum

[hide]

Summing k terms in a row starting with a is (k/2)(2a + k - 1) using the sum of an A.P. So we get k(2a+k-1) = 200. Now all we do is notice that those two brackets have opposite parities, and thus all we need to do is count how many ways 200 can be factored as odd x even. 1x200, 5x40, 25x8 and vice versa, giving 6.

[/hide]

not really shorter. i generalized, and i tried to explain every step very clearly (which as you can see i'm not too good at doing concisely ;)) but same method pretty much.

Yeah that's pretty much what I did for the first one when I did it.

[quote="TheSouthpaw"]For number 2, I'm pretty sure we use Simon's favorite factoring trick.[/quote]

I don't know about that. We have an x2 term alone and a y2 term alone, but the combination is unfortunately only an xy term.


I'm trying to figure out how for E1 (p.2) he comes to the conclusion that lim x(n) =lim y(n)=x, with x^2=AB.

Thanks for your consideration.

You'd get a much better response if you posted the entire problem -- otherwise, only people who (1) own Engel and (2) are willing to trudge over to their bookshelf, open it up and read enough to figure out what you're talking about will be able to help you.

Unrelatedly, many problems from Engel have appeared on the forums before, so you might try a search.

Thanks for the suggestion. Mind you I haven't been doing this for 5 years, so this may seem simple enough. I'm trying to understand the logic behind this question (or whatever it is).

[url=http://img505.imageshack.us/my.php?image=e1xg7.jpg][img]http://img505.imageshack.us/img505/4159/e1xg7.th.jpg[/img][/url]

[quote="t0rajir0u"]From his previous argument he is able to conclude that $ 0 < x_{n+1}-y_{n+1}< \frac{x_{n}-y_{n}}{2}$.  In other words, the sequence of differences $ d_{n}= x_{n}-y_{n}$ is

- Positive, and
- Decreases by at least half when $ n$ increases by $ 1$

Which means that it approaches $ 0$.[/quote]

Thanks for the help.

How does he conclude that the lim = x, with x^2=ab?

[quote="JBL"]We know $ x_{n}y_{n}= ab$, and we know $ \lim x_{n}= \lim y_{n}= x$.  Thus $ \lim x_{n}y_{n}= x^{2}$.  But $ \lim x_{n}y_{n}= \lim ab = ab$.[/quote]

You guys are too fast  :)

I've got this part, but how does he conclude that the limit Xn/Yn = X?

You probably want to reconsider that notation you used  :wink:

He showed they have the same limit.  He gave this limit the name $ x$.  So, it's by definition.