[quote="Press alt f4"]anyone else play video games ?[/quote]

A more appropriate question would be:

Does anyone [i]not[/i] play video games?

Very true, but I still know many kids whose parents dont allow them to have Video Games.
Evil parents trying to deprive their kids of one of the many joys of life!   :D

My parents think my grades are "slipping."

I still have straight A's in all gifted classes including a 99 in Geometry.

They think I need to study harder. So they don't let me play video games.

Evil Parents if they denied me of my video games i'd move out.  :lol:

Most games have been in three dimensions for a while now... and most have at least some aspect to it where a good mind is required.

All right, all right, I don't want to get into the video game debate, I get into that enough with my friends. (Adn usually win.) All I'm going to say is, anything good that video games might ever do for you are (a) totally countered by all the bad stuff they do for you and are (b) dwarfed by what other activities can do for you. i.e. [i]human interaction.[/i]

I will say no more on the subject.

You all beat me to while I was preparing my post :blush:

How are video games better for you than TV is?

You're increasing finger strength while at the same time using your mind.

[quote="GoBraves"][quote="09husbri"]I can't stand video games for the life of me. I think they waste your time and kill your mind. And don't pull that hand-eye coordination **** on me...there are a million better ways to practice hand-eye coordination. Guess what? If you pick up a tennis racket, you can practice hand-eye coordination in [i]three[/i] dimensions, and get a breath of fresh air and some exercise! But then you might be up five games to zero in the third set and AAAAAUUUUUGGGGGHHHHHHHH!!!!!!!

 :?

Sorry for being so cynical, I've had a tough day on the courts. Obviously I'm not a fan of Nintendo and Gamecube and all that other stuff, but if you play 'em, all power to you.[/quote]

Wtf?? How are video games a waste of time?? You talk about playing playing tennis and get fresh air. What about in winter?? You arnt going to play tennis then are you?[/quote]What if you live in Texas?  Playing tennis during summer or even late spring for a long time is brutal here, I play tennis mainly in the winter.

All right, I lied, I'm gonna argue this with all I have.

[quote="Press alt f4"]WHAT!!!!!!!!
[quote]I can't stand video games for the life of me. I think they waste your time and kill your mind. And don't pull that hand-eye coordination **** on me...there are a million better ways to practice hand-eye coordination. Guess what? If you pick up a tennis racket, you can practice hand-eye coordination in three dimensions, and get a breath of fresh air and some exercise! But then you might be up five games to zero in the third set and AAAAAUUUUUGGGGGHHHHHHHH!!!!!!!

Sorry for being so cynical, I've had a tough day on the courts. Obviously I'm not a fan of Nintendo and Gamecube and all that other stuff, but if you play 'em, all power to you.[/quote]

Point 1
Wastes Time

Yes we all admit that Video Games waste time yet is that not the point of any game? I do a puzzle, it wastes time, but that exactly the point of a puzzle. No one does a puzzle to actually be productive they just do the puzzle. So why are Video Games any different?

Point 2
Kills Your Mind

This statement is just completely outragous Video Games in no way kill your mind. If Video Games do anything to your mind they help it. Ever played tetris? That actually was proven to stimulate part of your mind! And Games like say the most popular game HALO for instance. Halo may seem on the outside to be a run around and shoot people game yet underneath much strategy and thought is being used. For example people playing wonder to themselves, "should I throw a grenade or burst in with my battle rifle and start shooting everything?" Other games like RPG's (Roleplaying games) Are fantastic for your mind and they include many challenging puzzles and different sidequests which require you to be creative to solve them.

Point 3
You get Exercise playing tennis
You can get exercise playing Video Games as well ever played DDR? That in my opinion takes more coordination than tennis and it gives you almost as much execise in the same amount of time. Try playing DDR on heavy for as long as you play three sets in tennis and tell me your not tired. I'm not dissing tennis I actually like the sport and play with my dad alot so don't take it personally.

My question to you is do you watch TV?
I assume you do, but Video Games which are an alternant choice to TV are much better for you.
I'm not trying to pound on you its just that many of the pionts you made were incorrect.[/quote]

Point 1
When I say wastes time, I don't just mean "lets time run by." Everything takes up time, even sleeping, but where would we be without sleep? The point is, video games take up time with little productivity. A jigsaw puzzle is an activity more stimulating than video games and takes up much less time. I dunno if you only play video games ten minutes at a time, but I have friends who play for hours on end.

Point 2
Yes, I have played Tetris, and I don't doubt that there are a select few games that may stimulate some parts of the brain. How about Myst, which I used to play? Definitely. But for the majority (one could say just about all) video games, one's mind practically turns off when playing. People don't actually even think when playing Halo. I should know, I've seen it played when me and a bunch of kids were at a friend's house. It's not planning ahead for the most part, it's instant reaction and response. And most of the kids knew the levels by heart anyway, reinforcing Point 1 (how much have you played this game again?). Even Tetris, Myst, and the RPGs you claim to be beneficial are nowhere near the average brainpower boost of a productive activity.

Point 3
To bring up DDR and only DDR in trying to disprove my point is downright ridiculous. I like DDR. I have indeed played DDR, and while I prefer conventional exercise outdoors in the fresh air, I don't deny that DDR has a...very high productivity level as video games go. It is indeed great exercise. But out of all the hours kids - and adults! - spend at the controller, what percentage is spent playing a game like DDR? Very, very little. And Lara Croft may be jumping and flipping and I don't know what, but it's she that keeps that great figure, and not you. (That's applicable to both genders.)

To GoBraves: Ever heard of an indoor court? Ever heard of basketball, another sport I play and enjoy?

To Tarquin: Increases finger strength and stimulates your mind? Give me a break! Playing the piano, whether you take it seriously or not, takes ten times the finger strength and ten times the brainpower. Denying that is like denying the Holocaust, without the anti-Semitic forces behind it. It's just ridiculous.

If people play to actually get good or if they are good at the game, then they HAVE to think, there's no way around it.

[quote="Max300"]hey, just wondering. how good are all you DDR players. Like what is your best record on a certain song. i know tarquin owns at DDR. he's awesome![/quote]

I don't think i can remember a best record but I can A A on challenge... (The song is called A). I can't pass Legend Of Max heavy though... I can AA 321 stars sometimes. I can A Dropout Heavy 1.5x.

I don't think anyone here can really say that they play video games for "finger exercise" or "brain stimulation." It's just not what they're meant for. The bottom line is that video games are fun and refreshing. Whenever I get stuck on a really hard problem I'll play them to clear my mind. I don't think anyone on AoPS is one of those addicts who plays all day.

Tennis may be much better exercise, but most people just play that for fun too. It's not the perfect exercise, after all; if you really want to burn calories exercise machines are much better; and tennis can cause injuries. Look at sports like baseball; I can tell from looking at the players that it doesn't require that much physical fitness. Reading a book is a lot of fun for some people, but a lot of books (esp. fiction) don't really do a thing for you.

In response to the original topic, I don't think that a geek is someone who is really smart. I think it's someone who doesn't want to or doesn't know how to have fun. As far as I can see, everyone here has a passion, whether it be tennis or video games or reading books.

P.S. yifman, if we're talking about usefulness, then paintball is hardly any more useful than Halo 2.

[quote="Phelpedo"] Look at sports like baseball; I can tell from looking at the players that it doesn't require that much physical fitness.[/quote]

I hope you're kidding. People in baseball are either incredibly strong or very fast. If you think that anybody could hit a 95 mph fastball, you're definitely kidding yourself.

[quote="Phelpedo"] Look at sports like baseball; I can tell from looking at the players that it doesn't require that much physical fitness.[/quote]

Are you on drugs or something??!!!! Have you ever seen a baseball player??
    You have no    f---ing clue how strong they are!!

I play pc games, but I don't play video games

I don't have any video games systems

my parents dont let be play i have to sneak it

[quote="09husbri"][quote="Phelpedo"]Reading a book is a lot of fun for some people, but a lot of books (esp. fiction) don't really do a thing for you..[/quote]

Are you [i]on[/i] something? First, try reading a Philip Roth novel after a nice long session of video games, I bet you won't have the mental capacity. Second, what [i]don't[/i] books - yes, even fiction - do to benefit your mental health?[/quote]
I will bet you fifty bucks that I can.
Fiction doesn't improve your practical knowledge, but that might not be your mental health.

phelpedo's logic (except the not fit baseball players  :P ) and many of the ppl who have posted is exactly correct (in my opinion). Who cares how much exercise you get playing video games... video games were designed for fun. Tennis is also fun. Reading is fun. Piano can be fun. Differnt people have different interests... Some people find sports more fun than sedentary activities. Some people like videogames books etc. better.
And playing video games is interaction. When i go to friends' houses we play computer games or video games. Just being together is interaction .And u can learn more about each other just like taht. You dont have to have hour long conversations to "INTERACT."

Some people are "extroverts" some people are "introverts". thus, they will like to do different things. will like to interact in differnt ways.

Video games are excellent at two things:  1)  storytelling  and 2)  social interactivity with other people.

Video games are meant to be played in moderation, and all of their positive benefits can only come when moderation is observed.

What is "moderation"?  No one can tell you that - you need to decide that for yourself.  If you're making straight A's in school, or millions of dollars in the professional world, or writing groundbreaking academic papers that will be talked about for centuries, then who am I to tell you not to play video games or watch TV as much as you need to stay productive?

But even if you're not any of those things ... yet ... people need recreation, whether it comes from sports, TV, movies, music, whatever.  If your recreation doesn't involve athletic activity, though, you definitely need to make yourself stay active.  No one's arguing that.

Studying math all day, every day, is every bit as harmful to you as playing video games 24/7.  That's why the very existence of this forum on this board is incredibly important.  Life is not meant to be all about math, believe it or not!

Parents have never understood video games.  But most understand the need for kids to be well-rounded.  I could tell you a thousand stories of what happens to kids of overbearing parents once they go off to college.  I don't think we want to go there.  Most parents want to reward good behavior, fortunately.

And I don't think anyone who's found their way onto this board has a problem with studying or putting the focus on academics, where it belongs.  Thus, this entire argument is ridiculous.

[quote="AHSIMC"]
What is "moderation"?  No one can tell you that - you need to decide that for yourself.  If you're making straight A's in school, or millions of dollars in the professional world, or writing groundbreaking academic papers that will be talked about for centuries, then who am I to tell you not to play video games or watch TV as much as you need to stay productive?[/quote]

Well said. That's the way I see it, too (even though I'm just a kid :lol:). As long as you are successful or achieving your full potential, there is no reason to restrict yourself from some fun. I mean, really, I could theoretically spend ALL of my Spring Break studying for the USAMO and/or my AP tests, but that would be very unhealthy (though I spend quite a bit of time, anyway). Add video games into the equation and voila, it's balanced. Now we are all happy people (or at least I am).

[quote="AHSIMC"]Studying math all day, every day, is every bit as harmful to you as playing video games 24/7.  That's why the very existence of this forum on this board is incredibly important.  Life is not meant to be all about math, believe it or not![/quote]

I agree with you, but nobody ever implied the opposite. A lot of your points make sense, and I salute your sensibility. However, I don't think that video games carry either "story-tellign benefits" (what are those?!) nor interactive benefits (see my earlier posts). Also, well-rounded does not impliy that a child plays video games. I do not like how you argue against the points I have been making by saying that moderation is the key. I have not argued against all recreation. I have argued against video games as a form of recreation, saying it should be replaced by sports, leisure reading, or anything else that doesn't involve a screen, really.
Another point you make that doesn't fit in with my argument is that since video games don't take people's focus off of academics here, the debate is pointless. This is extremely untrue. Academics, while they should be on top of everybody's priority list, are not the only entry in the ideal priority list! Sports, the arts, etc. are all also on that list. What do think is more important, a game of Halo #159 (I'm sure they'll get there eventually) or that a kid take up a musical instrument or a team sport? This is not the extent of my argument, but merely a bit to show you the flaws in your otherwise sensible logic.

[quote="Dark Shadow"]Fiction doesn't improve your practical knowledge, but that might not be your mental health.[/quote]...and video games improve your practical knowledge? They don't improve your "practical knowledge" or your mental health, while reading anything - although preferably a little more than the Sports Illustrated Swimsuit Issue :) - is one of the best things you can do to strengthen your mind.

video games wastes time and are useless, but i don't have the right to say anything bc i play wayyyyy too much PS2

anybody got the new PSP? how is it?should i buy it?

what kinda of VG system do y'll have?
i got PS2 and i love PC games

well now i enjoy most of your points here i am a big videogame fan, atleast 10 hrs a day, now you can say i waste my time, but as posted many times before i am having FUN and you quys are for getting the main reason of videogames NO THINKING Involved, now i have a 3.0 gpa, and when i come home im tired of thinking i want to play a game that dosent use brain power at all, thats what they were designed for hello did u for get that?? now argue all you want, lol mabye im gitting a little personal but i hate people who shoot down video games, expeclay vilonce and games, now they do tend to increass agression in a person, but playing GTA when your mad is better than actually going out on the street and shooting a bunch of people, and they are a distraction, if you have a problem you dont want to deal with at the moment go play a videogame, now you could play tennes or something like that to, i play 10hrs of videogames, but i still am active in sports, im on my football team and i play golf and videogames are a way for me to REST my mind and muscles while still having some amount of fun.


Is it possible to choose $1983$ distinct positive integers, all less than or equal to $10^{5}$, no three of which are consecutive terms of an arithmetic progression?

This is possible, we will construct such a set. Consider the set $S$ of natural numbers $a$ for which the representation of $a$ basis $3$ does not contain the digit $2$.[list][*]Such a list doesn't contiain 3 terms in arithmetic progression: for all distinct $a,b\in S$, take $i$ the last position in which they differ. Then if $(a_i,b_i,c_i)$ is an arithmetic progression, the $i$-th digit must be $2$.[*]The set $S\cap \{1,2,\ldots,10^5\}$ has more than 1983 elements, since $1983=(11110111111)_2$, while $(11110111111)_3=87844<10^5$.[/list]


Determine the number of triples $ (a,b,c)$ of positive integers which satisfy the simultaneous equations
\[ ab \plus{} bc \equal{} 44\]
\[ ac \plus{} bc \equal{} 23\]
$ \textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

EDIT: Problem not needed.

[hide] From the second equation, we have $ c(a\plus{}b)\equal{}23$ This means that either $ c\equal{}1$, abd $ a\plus{}b\equal{}23$, or $ c\equal{}23$, and $ a\plus{}b\equal{}1$ (since $ a$, $ b$ and $ c$ are positive integers0. However, the second scenario is invalid for that very reason, so $ c\equal{}1$ and $ a\plus{}b\equal{}23$. Plugging $ c\equal{}1$ into the first equation, we get $ b(a\plus{}1)\equal{}44$. Substituting $ a\equal{}23\minus{}b$ into the equation, we get that $ (b\minus{}2)(b\minus{}22)\equal{}0$. After verifying that both these cases work, we see that there are $ 2$ solutions, and hence the answer is C.[/hide]


A and B shoot independently and stops as soon as they get their first hit. The probability of each shot for A and B are $ \frac{3}{4} , \frac{5}{7}$ respectively. What is the probability that B requires more shots than A to hit the target??

If you define $ X$ and $ Y$ appropriately then what you require is $ P(Y > X)$. Express this as a double sum and use the fact $ p(x,y) \equal{} p(x) p(y)$. Rest would be an easy exercise. ( ans: $ \frac {3}{13}$ (I hope!))

Thanks I got it.. The answer is correct. The required probability is obtained by

$ P(A' \cap B)\plus{}P(A' \cap A \cap B' \cap B') \plus{} P(A' \cap A' \cap A \cap B' \cap B' \cap B')\plus{}....$

$ \equal{}\frac{3}{4}.\frac{2}{7} \left[ \frac{1}{1\minus{}\frac{2}{28}} \right ]$

$ \equal{}\frac{3}{13}$ :)

What are $ A, A', B, B'$? I meant $ P(Y>X)\equal{} \sum_{x \equal{} 0}^{\infty} \sum_{y \equal{} x \plus{} 1}^{\infty} p(x,y)$

Sorry. I confused myself in presenting the idea.

Assume X and Y are two persons(to avoid ambiguity of the mistake I made in my previous post) and let A be a random variable corresponding to X such that X hits the target, similar analogous statement follows for Y and say the random variable to be B. A' denotes the failure of A and B' denotes the failure of B.

Now, If suppose X  hits in $ m$-shots it means that he has lost $ m \minus{} 1$shots,now to satisfy the condition Y has to hit $ n$-shots such that $ n > m$.

Consider, $ (A'\cap A\cap B'\cap B')$, by this I mean , (failure of A, Success of A, failure of B, failure of B). Here the success of A is attained in second instant, since we need success of Y to be more than that of A , we have two failures for B, thus success for Y occurs only after 2 instants as we have induced that we have first two failures for Y. Similarly , we can make it that X succeeds in his 3-rd instant , thus , now Y can succeed only after 3rd instant thus we have 3 failures for Y  as in $ (A'\cap A'\cap A\cap B'\cap B'\cap B')$. I am so sorry that I didn't know how to represent this idea..


find k is prositive integer so that  min(k) and
       \[ k\cdot 19^{1005}+84^{2004} =m\cdot 17510 \]       :?  :(

Hello!

I just have seen you problem now. It reduces to 4 linear congruences(mod 2,5,17,103) and because of the Chinese Remainder Theorem there is exactly one solution for k mod 17510.

But I think the solution is a result of long computations(especially mod 103) or is there a really nice solution?

Misha


:rotfl:  :rotfl:  :rotfl:

But in a perfect world, there would have been no hijacking.


Let $X$ be a compact  set in $\mathbb{R}^{m}$. Prove that the convex hull of $X$ is compact.

Someone has an idea?

$Tipe$

If I'm not wrong, it follows from Caratheodory's theorem.

I think it's easier to do if you use the fact that a subset of $\mathbb R^{m}$ is compact iff it is closed and bounded. Boundedness is clearly preserved under convexification, and with a little work you can show the same for closedness.

[quote="mlok"]. Boundedness is clearly preserved under convexification, and with a little work you can show the same for closedness.[/quote]
I have problems in prove that "convexification" preserves closedness.

Maybe that's because in general it does not -- only for bounded (hence compact) sets. I can't think of a clean proof right now; maybe you can't avoid [url=http://www.everything2.com/index.pl?node_id=1136581]Caratheodory's theorem[/url] after all.


algebra
polynomial
complex analysis
function
calculus
derivative
complex analysis unsolved

What is the simplest proof of this fact? It is obvious, since they are holomorphic, however, the proofs I know of this fact are somehow very unintuitive for polynomials. Is there a simple way to see this directly, say without resorting to any complex analysis (or at least, complex analysis that isn't exceptionally trivial for polynomials)?

I don't see any way to do it for polynomials that doesn't also work for analytic functions in general.

We can still do that without much complex analysis. Openness is a local property, so we only really need to focus on one point; WLOG, it's zero with $ f(0)\equal{}0$. If the derivative there is nonzero, the Inverse Function Theorem gives us a differentiable local inverse, which is more than enough. If the derivative is zero, we have $ f(z)\equal{}z^mg(z)$ with $ m\ge 2$ and $ g(0)\neq 0$. There is an analytic $ m$th root in a neighborhood of $ g(0)$, so $ h(z)\equal{}z\sqrt[m]{g(z)}$ is an analytic $ m$th root of $ f$ near zero. This has an inverse as before, and taking the $ m$th power keeps it open.

What if we don't assume that the polynomial is analytic? That is, $ f(x \plus{} iy) \equal{} P(x,y) \plus{} iQ(x,y)$ where $ P,Q\in \mathbb R[x,y]$, or, written in another form $ f(z) \equal{} P(z,\bar z)$ where $ P\in \mathbb C[z,w]$. There is an obvious counterexample $ z \plus{} \bar z$. Let's require $ f$ to be surjective -- is it open then?

Added: having typed the above, I realized that the answer is [hide] no, by $ f(x \plus{} iy) \equal{} x^3 \minus{} x \plus{} iy$[/hide]


[quote="GameBot"]If $ 5$ technicians can produce $ 64$ surfboards in $ 4$ days, how many surfboards can $ 8$ technicians produce in $ 10$ days?[/quote]
In 1 day, 5 technicians can produce 64/4=16 surfboards. In 10 days, 5 technicians can produce 160 surfboards. Therefore 8 technicians can produce (8/5)*160= $ \boxed{256}$ surfboards in 10 days.

We see that the number of technicians is directly proportional to the number of surfboards produced and inversely proportional to the number of days.

Hence:
Let $x$=the number of surfboards

$\frac{(4)(5)}{64}=\frac{80}{x}$

$x=\boxed{256}$


I just can't believe what I just heard. There is a new "Wall Street Market" in basically everywhere except USA (who has its own) called the "Hot Air Market" (It has something to do with the Kyoto Treaty). The idea is the following, as far as I understand, somone may be knows more than me about it: there is an international organization that assigns maximum yearly quotas to how much a country can polute. After that, an organization within each country divides that quota withing the local industries.

Now, so far so good. We don't want this guys to pollute, but now they have some maximum quotas, and if they exeed them, they will have big fines. The other part (The countries are chosen randomly, I don't mean to say anything in specific from the country, but from the system itself). If a given industry, say in South Africa, didn't pollute all its quota, then it can sell the overcome of the quota to an industry else where, it can be within the same South Africa, or it can be in Germany or Japan. Therefore, the regions where there are real pollution problems, the companies can [b]buy their right to overpolute[/b].

Now, the fact that the US does not participate on this market is because they have their own, called, I believe, the Chicago market.

What do you think about it?

Sources: What I remember from a CNN report last night!

Why the background of the flash becomes white? It should be black.

[quote="Buffalo"]This is an old news, I'm sure I read it a long time ago.
Before condemning such an action, please look at the famous photo "Hands" first.[/quote]

I must confess that I am deeply ignorant with respect to Asian Languages. I am not able to recognize the language itself (I think is chinease), even less to understand what was written there. Nevertheless, poverty, femine and the other apocalipsis horse riders don't justify to be able to sell the right to polute, at least in my personal opinion. What I can't think is that one company can sell the quota of pollution emitions to other that polutes less. Probably the idea is that in that way the industries will invest more in clean production or so to be able to sell more and more of their own quota, but I think that is a naive idea!

Best regards,

This is a typical language of the emperors. There is a famous idiotic emperor in the history of China called Emperor Hui of Jin, when his minister told him that the people in some place were dying  because they got no rice/grain to eat in the famine, he answered with surprise, "why don't they eat broth instead?"
I don't even think I have the right to blame cannibalism in the famine showed in some of the pictures.


Solve the equation

$ 16^x\plus{}81^x\plus{}(2^x\plus{}3^x)^4 \equal{}3\cdot 6^{2x\plus{}1}$

_________________________________
$ \textbf{Azerbaijan Land of Fire }$ :rotfl:  :roll:

$ 16^x \plus{} 81^x \plus{} (2^x \plus{} 3^x)^4 \equal{} 3\cdot 6^{2x \plus{} 1}$

$ \Longleftrightarrow \left\{\left(\frac 32\right)^{x} \minus{} 1\right\}^2\left\{\left(\frac 32\right)^{2x} \plus{} 4\left(\frac 32\right)^{x} \plus{} 1\right\} \equal{} 0$

$ \Longleftrightarrow (3^x\minus{}2^x)^2(9^x\plus{}4\cdot 6^x\plus{}4^x) \equal{}0$

$ \therefore x \equal{} 0$.


MIT college search AMC AIME USA(J)MO USAMO

MIT has been my dream school for a long time. I see that a lot of people on here have been accepted to it.

Is there any advice you guys can give to me? Im a rising junior (Class of 2009)

[quote="imevil555"]Is there any advice you guys can give to me? Im a rising junior (Class of 2009)[/quote]

1.) Learn to use the search engine.

2.) Learn math: college math via MIT OCW + books, solve math olympiad problems.

common, im not that naive

im a regular poster on college confidential

i solve olympiad problems, made aime this year, took amc 10 and 12

hopefully usamo qualifier next year

mit ocw? opencourseware? how will that be of any help?

i do FIRST robotics, vice-captain of my school's math team, have significant science research experience and have done a research internship

im looking for advice on hooks, essays, etc.

I'm no expert, but I think that with those credentials, you'll stand out in any crowd.  You probably don't need to worry about getting into MIT.

Most likely the best advice on essays is to be yourself.

I think orl meant "learn to search this site", not search in general. There are many discussions of getting into MIT and other similar universities in this particular subforum of this site. Please look for those and read them rather than asking very common questions for what is probably not the second time but the tenth or twelfth at least.

http://www.artofproblemsolving.com/Forum/search.php

The opinion has been expressed here that MIT admission is less predictable than admission to some other colleges. We'll see whether or not that changes under the new management of the admission office.

Well, Caltech is commonly considered "less random" in admissions than MIT, but that's not to say it's any easier to get into, or very predictable in its own right.  Also, Cornell would seem to be more predictable than MIT in admissions, as it is perhaps less competitive to get into than those two Tech schools, but it's still an Ivy.  I won't talk about other schools, because I probably didn't apply to any other relevant ones.

And yes, the questions have probably already been answered elsewhere in this forum.

I can only speak from the perspective of a student who hails from a very competitive San Francisco Bay Area high school:

Here, all east coast top-tier schools are considered "random." Looking down the top 10 or 20 from my high school, one will see the majority of students going to UC Berkeley (where ultimately 50+ students out of 500 end up), with a few Stanfords, MIT's, maybe a Harvard or Yale or Princeton. Looking further down, one will find seeming "random" admits without any significant hook.

Even Stanford has joined the "random" game, admitting much less based on academic rigor, numbers, and awards, but rather on that oh-so-mystical x-factor better known as "intangibles."

i think orl gave the best two advices.
although, you do not have to know how to use the search engine in mathlinks.
but yea, just do math 8 hours a day, every day.

[quote="aznphatso"]... admitting much less based on academic rigor, numbers, and awards, but rather on that oh-so-mystical x-factor better known as "intangibles."[/quote]

A much more plausible explanation than the notion that there is some mysterious unknown factor involved in decisions is that these schools are inundated with applications from those who have essentially equivalent collections of rigorous schoolwork, test scores and other resume items and that as a consequence, final decisions for those in this group of fully qualified individuals [i]actually are[/i] random or pseudorandom.


Two six sided dice are tossed.  What is the probability that the sum of the sides facing upward is:
a. 3
b. 7

[hide="Hint"]Draw a $6 \times 6$ array[/hide]

[hide="A"]There are 36 possible combinations of the two numbers. Of those, only 2 give a sum of 3 (1,2; 2,1).

$\frac1{18}$[/hide]
[hide="B"]Again, there are 36 different possibilities. There are 6 that give a sum of 7 (1,6; 2,5; 3,4; and their inverses).

$\frac16$[/hide]

[quote="i_like_pie"][hide="A"]There are 36 possible combinations of the two numbers. Of those, only 2 give a sum of 3 (1,2; 2,1).

$\frac1{36}$[/hide][/quote]
I think you mean $\frac{1}{18}$.

Oops...

Extension:

What is the probability that the sum of the two numbers facing up is $n$? (1<n<13)

[quote="anyone"][quote="i_like_pie"]
Extension:

What is the probability that the sum of the two numbers facing up is $n$? (1<n<13)[/quote][hide]1, it will always be in those perameters[/hide][/quote]
I meant find a formula or something that expresses the probability of the two numbers having a sum of $n$. I added the condition so there are no questions like "What if $n$ is 14?".

[quote="i_like_pie"]Extension:

What is the probability that the sum of the two numbers facing up is $n$? (1<n<13)[/quote]
[hide]
$P(\text{sum of n})=\frac{6-|n-7|}{36}$
if I'm not mistaken
[/hide]

[quote="i_like_pie"]Extension:

What is the probability that the sum of the two numbers facing up is $n$? (1<n<13)[/quote]

[hide="quick"]There are pairs from $(1,n-1)$ through $(n-1,1)$, so there are $n-1$ pairings. Then the probability is $\boxed{\frac{n-1}{36}}$.[/hide]

Kohjhsd's solution is what I had in mind.

One more extension:

What if there are more than 2 dice being rolled?

[hide="solution"]

After playing Settlers of Catan I have mastered this type of problem :-)

[b]a=1/18[/b] 2 ways to get it, 2/36=1/18
[b]b=1/6[/b]  6 ways to get it, 6/36=1/6
[/hide]


In a general triangle $ ADE$ (as shown) lines $ \overline{EB}$ and $ \overline{EC}$ are drawn. Which of the following angle relations is true?
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);

pair A = origin;
pair B = (1,0);
pair C = (2,0);
pair D = (3,0);
pair E = (1.25,1.75);

draw(A--D--E--cycle);
draw(E--B);
draw(E--C);

label("$A$",A,SW);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,SE);
label("$E$",E,N);
label("$y$",E,3SW + 3S);
label("$w$",E,7S + E);
label("$b$",E,3SE + 4S + E);
label("$x$",A,NE);
label("$z$",B,NW);
label("$m$",B,NE);
label("$n$",C,NW);
label("$c$",C,NE);
label("$a$",D,NW+W);[/asy]
$ \textbf{(A)}\ x \plus{} z \equal{} a \plus{} b\qquad \textbf{(B)}\ y \plus{} z \equal{} a \plus{} b\qquad \textbf{(C)}\ m \plus{} x \equal{} w \plus{} n\qquad \\
\textbf{(D)}\ x \plus{} z \plus{} n \equal{} w \plus{} c \plus{} m\qquad \textbf{(E)}\ x \plus{} y \plus{} n \equal{} a \plus{} b \plus{} m$

Triangle $ ACE$'s angles x, y, w, and n add up to $ 180$ degrees. Likewise, triangle $ BED$'s angles a, b, m, and w add up to $ 180$ degrees. Thus, $ x\plus{}y\plus{}w\plus{}n\equal{}a\plus{}b\plus{}m\plus{}w$, or $ x\plus{}y\plus{}n\equal{}a\plus{}b\plus{}m$, which is $ (E)$.


probability
geometry
3D geometry
complementary counting
expected value

[quote="jeez123"]1.
The good guys have a series of five games to play, with one game each day from Monday through Friday. If it rains on the day of a scheduled game, that game is cancelled. There is a 30% probability of rain each day. How likely is it that the Good Guys will get to play at least two of the five games?

2.
You decided to play a game at the carnival where you can win 50 dollars. It costs 1 dollar to play the game. The game involves picking cubes out of a paper bag. There are 6 cubes in the bag, 4 are green and 2 are blue. If you pick a green cube, you lose. If you pick a blue cube, you get to pick out of a second bag. The second bag has just 4 cubes, 3 are green and 1 is blue. If you pick the blue, then you win the 50 dollars. If you pick a green cube, then you lost the game.

Is it worth it for the carnival to offer this game? In other words, are they making money or losing money in the long run? Use your area or tree diagram to help answer this question.

---------------------------
Thank you very much![/quote]
[hide="1"]
Complementary counting
Play no games: $(\frac{3}{10})^5$
Play one game: $5(\frac{3}{10})^4(\frac{7}{10}$
No games+one game=.03078
Probability they will play at least 2 is 1 minus that.
[/hide]

[hide="2"]Find the expected value: $50(2/6)(1/4)=25/6$ So then it is worth it. [/hide]

This is now officially 11 years ago. Happy New Year everybody!


geometry
perimeter
inradius
geometry proposed

[quote="Cristi Sav"]Let $ABC$ be a triangle with all angles $<90$ degrees, with $G$ the intersection between its medians, and let $M$ be on $[AB]$ and $N$ on $[AC]$ such that $M,N,G$ are on a line. Prove that if $A_{ABC}=2A_{AMN}$ then $P_{ABC}>2P_{AMN}$.($A_{X}$=area of $X$ and $P_{X}$=perimeter of $X$).[/quote]

Denote
$s_1,r_1:$ the semi-perimeter and the inradius respectively of $\triangle ABC$
$s_2,r_2:$ the semi-perimeter and the inradius respectively of $\triangle AMN$

$A_{ABC}=2A_{AMN} \Leftrightarrow$

$s_1r_1=2s_2r_2 \Leftrightarrow$

$\frac{s_1}{2s_2}=\frac{r_2}{r_1} \Leftrightarrow$

$\frac{P_1}{2P_2}=\frac{r_2}{r_1}$

So we have $P_1>2P_2$ if and only if $r_2>r_1$

But the triangle $AMN$ is inside of $ABC$, so $r_2<r_1$

Where did I go wrong?

looks like i've got too far  :) . Please prove that $ABC$ and $ANM$ are similar.


inequalities
absolute value
real analysis
real analysis unsolved

If a_0 and a_1 are positive reals, let a_(n+2)=sqrt(a_(n+1))+sqrt(a_n)

I've shown if E_(n+2)=|a_(n+2)-4| then
E_(n+2)<=(E_(n+1)+E_n)/3

Not sure how this helps me prove the sequence converges?

Any help would be nice,

Aaron

I'd choose a number $\lambda<1$ such that $3\lambda^{2}-\lambda-1>0$ and try to prove that $e_{n+1}<\lambda e_{n}$ for large enough $n$.

This would immediately show that $e_{n}\to0$.

I understand how your method works, but I'm not sure where that characteristic equation comes into play.  Its from my inequality but with the inequality exactly opposite?

Aaron

If I chose $\lambda$ with $3\lambda^{2}-\lambda-1\le0$, it might not be true that $e_{n+1}<\lambda e_{n}$ for large $n$. Consider the extreme case: what happens when $3e_{n+2}=e_{n+1}+e_{n}$?
If I chose $\lambda\ge1$, the inequality wouldn't help.

This doesn't mean that I know how to prove the desired inequality.

We'll assume that we have proved that $e_{n+2}\le \frac{e_{n}+e_{n+1}}3.$

(I haven't checked that myself; for this we're trusting that toadsmack7 can back up his claim.)

Try this: Let $M=\max_{n\le N}e_{n}$ for some nontrivial $N.$

Then $e_{N+1}\le \frac{M+M}3=\frac23M.$

$e_{N+2}\le \frac13\left(M+\frac23M\right)=\frac59M.$

$e_{N+3}\le \frac13\left(\frac23M+\frac59M\right)= \frac{11}{27}M$

Continue this recursively. We wind up with the estimate

$e_{N+n}\le y_{n}M,$ where $y_{n}$ satisfies the linear constant coefficient difference equation

$y_{n+2}=\frac13\left(y_{n}+y_{n+1}\right).$

We can solve this as $y_{n}=c_{1}\lambda_{1}^{n}+c_{2}\lambda_{2}^{n},$ were $\lambda_{1}$ and $\lambda_{2}$ are the two roots of $3\lambda^{2}-\lambda-1=0.$

Since both of these roots have absolute value strictly less than 1, then we have that $y_{n}\to0.$ And that's enough to prove our conclusion.

[quote="Virgil Nicula"][color=darkblue][b]A similar exercise :[/b]
$1.\blacktriangleright$ If the sequence $a_{n}$ is bounded and for any $n\in N^{*}$ , $x_{n+2}\le \frac{1}{2}\cdot\left(x_{n+1}+x_{n}\right)$, then the given sequence is convergent, i.e. $a_{n}\rightarrow l\in R\ .$
$2.\blacktriangleright$ Ascertain the maximum value $K$ (if it exists) for which the following implication is truly :
" The sequence $(a_{n})$ is bounded and for any $n\in N^{*}$ , $a_{n+2}\le K\cdot\left(a_{n+1}+a_{n}\right)\Longrightarrow$ the sequence $(a_{n})$ is convergent ". [/color][/quote]
Do you have a proof for 1?(I have a not very nice proof :) )
And for (2),I think when $K<\frac{1}{2}$ then it is only valid for positive sequence.Obviously,$K=\frac{1}{2}$ is the maximum number(Try the sequences $1,2K,1,2k,\cdot$.)


:D  $f: Z→Z$

Find all possible function f which satisfies

$f(f(m))=m+1$ $m is an integer$

It was in slovenia 199(  )   :D

$f(f(m))=m+1\Longrightarrow f(m+1)=f(m)+1\Longrightarrow f(m)=m+a, \ a=f(0)$, but if f(m)=m+a, then $f(f(m))=m+2a\not =m+1$.

Just to clarify here is how he made first step:

put $f(m)$ for $m$ in given: \[f(m)+1=f(f(f(m)))\] take $f()$ of each side in given: \[f(m+1)=f(f(f(m)))\] And combine these to get this first statement, the rest is easy :)

edit: lol i have exactly 100 posts less than you, Rust!


function
inequalities
calculus
calculus computations

I know that a convex function is a function whose value at the midpoint of every interval is less than or equal to the average value of the endpoints of the interval.  I was just curious if anyone could explain how that is equivalent to

\[ f\left(\lambda x_{1}+\left(1-\lambda\right)x_{2}\right)\leq\lambda f\left(x_{1}\right)+\left(1-\lambda\right)f\left(x_{2}\right)  \]

for $x_{1},x_{2}\in\left[a,b\right]$ and $0<\lambda <1$.  The way I am translating the first sentence is as follows:

\[ f\left(\frac{a+b}{2}\right)\leq\frac{f\left(a\right)+f\left(b\right)}{2}.  \]

Let $\lambda = \frac{1}{2}$.

To get the first from the second, let $a = 2 \lambda x_1, b = 2 (1 - \lambda) x_2$.

Both of them intuitively mean that the secant line between two points on the function is always above the function.

I see how to get it by making the substitution, but what I'm wondering is why that substitution is even made...  What is wrong with the second inequality?

There's nothing wrong with it.  The second one is just a little more general, although not really.


MATHCOUNTS
percent
FTW
AMC
AIME
probability
AMC 10

Which team are you rooting for to win MathCounts Nationals?

[Edited by solafidefarms to make the title look better to his poor eyes]

New Jersey For The Win! We have Alex Clifton and Eric Schneider.

WOOT! GO MISSOURI!

We have Max and Stella Schindler, Runpeng Liu, and Victor Wang.

I picked Maryland, since that's where I'm from. But since I didn't make the team, idk... :no:  :wink:

Go alabama! We have... not had state yet :)

Go Kansas!!  We've got a Indian Darryl Wu with us.

EDIT: That kid is actually 10 years old...he skipped a grade.

My mathcounts coach said that there was this kid in Texas that was in 5th grade last year and was AT LEAST as good as  Bobby Shen. Watch out guys. 6th graders are coming to rule you. I bet it'll be another 6th grade winner.

That would be epic.

I doubt it though.  I'm putting my money on Bobby Shen. (yes real money)

[quote="t0rajir0u"][i]Who.[/i]  "Whom" is an object, not a fancy way of saying "who."[/quote]

In this case, "whom" [b][i]is[/i][/b] the direct object of "for."

Another way of rephrasing this title, perhaps making more clear which words/phrases are objects,

[u]For whom are you rooting in the 2009 MathCounts Nationals?[/u]

@Aime15 Yes it would be epic. I want another 6th grader to win because I want to see that America is leveling its standards to another notch.

Michigan doesn't have 1 super beast person like David Lu, or Allen Yuan, but i think that overall, Michigan is better this year because all 4 people are really good. MI is going to be in the top 5 for sure!

Uh I actually think Michigan this year is [b]weaker[/b] than it was last year.

Then again Michigan made a bunch of mistakes it shouldn't have last year, so if you don't do that, you might do pretty well.

[quote="xpmath"]Uh I actually think Michigan this year is [b]weaker[/b] than it was last year.[/quote]

definitely weaker. try to surprise me, but don't be surprised yourselves when michigan does far worse than you expect.

Well I suppose I should explain what I meant.

I think that the Michigan team is [b]individually[/b] weaker than last year's team (David Lu is pretty hard to beat and Peter Su was pretty good as well). However, Michigan messed up the team round last year which gave them a much lower placing than expected (11th place). As a [b]team[/b], I don't know how good Michigan is this year, which is really what matters for team placing.

Still, all these things are part of why Nationals is fun  :) .

By the way, we should probably all go back to practicing.

[quote="hjoon0125"]i think every state is better than last year[/quote]

Indiana? Texas? I dunno.

It depends on the team's ability to work together... Bet you guys didn't know everyone on Indiana's team is a 7th grader o.o

dude, va ftw :lol:

there is this one 7th grader who got a perfect score, killed EVERYONE in cd, and got 1rst team....so yea. maybe not this year, tough competition this year  :wink: but maybe next.

oh and ttocs45, who is neal wu's(2006 MC nat winner) brother has a good chance once he's in 8th, just 6th rite now and he got 1rst state.

Unless I'm mistaken, ttocs45 is already in 8th.

aime, ttocs45 is in sixth grade.

oh whoops, didn't see xp's comment. anyways, i sort of realized that MI isn't gonna get top five.

go texas!  YEAAA!  Yes, I'm being very, very biased.  I'll bet my shoe on Bobby Shen.

Texas! Because I'm from Texas and I live really close to Bobby Shen and Lilly Shen! I also know them personally!

MASSACHUSETTS FTW! we have 584906328x and ytrewq. pwnage.


1. E.coli cells grown on Nitrogen-15 medium are transferred to N-14 medium and are allowed to grow for two generations (2 rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment and why?

2. About 60 years ago, a drug called DNP was used to help patients with weight loss. DNP uncouples the chemiosmosis machinery by making the lipid bilayer of the inner mitochondrial membrane leaky to hydrogen ions. How does this cause weight loss?

#1. The amount of DNA would quadruple in 2 generations, with the following composition:
-1/2 of the cells would have both helices of DNA containing $^{15}N$.
-The other half would have one helix of $^{14}N$ and one helix of $^{15}N$.

So 3/4 of the DNA would be "heavy", and 1/4 "normal"

#2.  If said membrane becomes leaky, there will be a decreased electrochemical gradient across it; hence, more stored glycogen and fat molecules must be metabolized in order to create the same base amount of ATP, decreasing the patients' weight.  Sounds dangerous though.

[quote="justdudxit"]Sounds dangerous though.[/quote]

I think that method once was used and was banned after it was found to be very dangerous.

i didn't know doctors actually prescribed DNP... it's known by bodybuilders that it can produce pretty extreme fat loss, but it's generally deemed not to be safe and hence i don't think there are many people taking it.  This guy did an experiment and kept a log here: http://www.t-nation.com/readTopic.do?id=1035803

I think that the answer to question #1 is that 1/4 of the DNA is "heavy", N-15 DNA, and 3/4 is "normal", N-14 DNA. This is because originally we have only N-15 DNA. However, as the bacteria replicate on the N-14 medium, all new DNA is N-14 DNA. Becuase the bacteria replicate for 2 generations, the new DNA constitutes 3/4 of the final amount of DNA. Am I wrong?


prove this
24(abc)^3 - 3(ab+bc+ca) 2  + (ab+bc+ca)  \geq  0,given ab+bc+ca \geq 3
best regards

he probably ment a,b,c positive reals, didn't you Galois?

I'm sure that even if they are strictly positive, we can have a triplet (a,b,c) "really close" to the triplet (3, 1, 0) and still get a contradiction. The result is -24, so there's plenty of room for us to make it a bit bigger s.t it still is negative.

yeah - you are right grobber - mislooked the example - thought that he took (-3,1,0) :)

i think i copied the problem wrong.surely i'll check.sorry!!! :D

[quote="galois"]Given nonnegative real numbers $x,y,z$ such that $yz+zx+xy\geq 3$, prove or disprove that
\[24x^{3}y^{3}z^{3}+yz+zx+xy\geq 3(yz+zx+xy)^{2}. \]
[/quote]
If the condition $yz+zx+xy\geq 3$ be replaced by $x^{-\frac{17}{38}}+y^{-\frac{17}{38}}+z^{-\frac{17}{38}}\leq 3$, then the above inequality may be holds.


What is the probability that two tosses of a fair coin result in
exactly one head and one tail? Express your answer as a
common fraction.

Possbilities:

HH
HT
TH
TT

Probablity is $ \frac{2}{4}\equal{}\boxed{\frac{1}{2}}$.

There are 4 possibilities.

2 x 2 = 4.

The possibilities are; Head Head, Tail Tail, Head Tail, and Tail Head.

From this, we can see that two of the combinations have exactly one head and one tail.

$ \frac {2}{4}$ = $ \frac {1}{2}$ . So the answer is $ \frac {1}{2}$


[quote="ashrafmod"]in square ABCD ,AC and BD  meet at E  ,point F is on DC

CAF =FAD ,if  AF meet ED at G
and if  GE =24  find  CF[/quote]
We have that $\frac{DG}{GE}=\frac{AD}{AE}$=>$DG=24\sqrt{2}$.
Also $\triangle{ADG}\sim\triangle{ACF}$=>$\frac{CF}{DG}=\frac{AC}{AD}=\sqrt{2}$=>$CF=48$ :wink:  :lol:

[color=darkblue][b][u]This problem is "wandering" here ![/u][/b]

[b]Remark.[/b] $m(\widehat{DGF})=m(\widehat{DFG})=\frac{3\pi}{8}$ and $\frac{GE}{1}=\frac{GD}{\sqrt 2}=\frac{CF}{2}=\frac{DF}{\sqrt 2}=\frac{1}{2+\sqrt 2}\ .$[/color]


LaTeX
inequalities
algebra
polynomial
function
geometry
geometric transformation

[quote="anhnguyen"]Hehe. I can do it but I gave up on trying to use Latex :)[/quote]Why did you give up? :)

[b]First round(4h):[/b]
1/ $ABCD$ is a parallelogram. $E$ is a point of $BC$ and $F$ a point of $CD$ so that $ABE$ and $BDF$ have the same area. $BD$ crosses $AE $ at the point $M$ and $AF$ at $N$. Prove that $BM, MN, ND$ follows the triangular inequality, and that $E$ and $F$ moves so that $MN$ goes smaller, then the radius of the circle meeting the 3 summits (sorry, don't know how to say in English...) of a triangle which have for lengths $BM, MN, ND$ also goes smaller.

2/ Let the equation $x^2+y^2+z^2+t^2-Nxyzt-N=0 (*)$ ($N\in \mathbb  N,N>0$).
Prove that for an infinity of values of N, (*) have a solution in positive integers, and that for $N=4^k\cdot (8m+7)$ ($k, m$ nonnegative integers), (*) has no solutions in positive integers.

3/ $P$ is a polynomial function of degree 4, with 4 real roots.
Prove that the equation $\frac{(1-4x)P(x)}{x^2}+(1-\frac{(1-4x)}{x^2})P'(x)-P''(x)=0$ has also 4 real roots.
[b]Second round(4h):[/b]
4/ ABC is a triangle with $AB=BC=CA$ (dunno what's this in English...), and $M$ is a point. $D$ is the midpoint of $AD$. $DM$ and $AP$ crosses at point $N$. The images by the symetry of center $M$ of the points $A, B, C$ are $A', B', C'$. Prove that there is an unique point $P$ so that $AA=PB', PB=PC', PC=PA'$, and that when $M$ moves while being different of $D$, the circle meeting $M, N, P$ keeps meeting a point which is not moving.

5/ Find all the functions $f: \mathbb R\to \mathbb R$ so that :
\[f(x\cdot \sqrt {2})+f(x (4+3\cdot \sqrt {2}))=2\cdot f(x (2+\sqrt {2}))\]

6/ Compute $\frac {1}{(n_1!\cdot n_2!\cdot ...\cdot n_{1994}!)(n_2+2n_3+...+1993n_{1994})!}$, with the sum on all the $1994$-uplets $(n_1,...,n_{1994{)}}$ of nonnegative integers so that $n_1+2n_2+...+1994n_{1994}=1994$
It was posted by Antony ;)

Actually I requested a translation of the document in VietNamese which I appended above.  :)

[quote="pigfly"]Can I post VietNam TST 1995,1997,2000 here?I translated it yesterday ;)[/quote]I think it's best you try to post the problems in the [url=http://www.mathlinks.ro/Forum/index.php?f=125]National Olympiads forums[/url], as we all did with the problems like in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21506]this topic[/url].

So you first open up a topic in the national forums, and post the problems, and then, in a new window, you post each problem in a separate topic in the forum where it belongs (algebra, geometry, combinatorics, etc.) and after you have posted a problem in a new topic, click on the topic link and put it in the topic which contains all the problems.

Last attach to that topic (the one which contains all the problems in the national olympiads forums) a .pdf file with all the problems. :) I hope it's clear.
Do this for every year (1995, 1997 and 2000). Don't forget to write the source of each problem when you post them in a separate topic.

If you have further questions on how to do all this, don hesistate to ask.

[quote="pigfly"]Can I post VietNam TST 1995,1997,2000 here?I translated it yesterday ;)[/quote]

I think I have the years 1991 until 2004 already in English (or at least most of them). It is better if you translate those first for which I do not have a translation, e.g. those ones I attached above.  ;)

I will post the years I have as English version in due course.  :)

What a surprise! I even have not these TSTs, thanks [b]orl[/b] so much for sharing them with us (may be your friend lhvbao gave it for you?   ;) ) . May be we need time to translate all of them in English.

By the way, [b]pigfly[/b] will be taking in VNMO 2005, so i think he is ..... strong enough to kill all of them  :D.

[quote="pluricomplex"]What a surprise! I even have not these TSTs, thanks [b]orl[/b] so much for sharing them with us (may be your friend lhvbao gave it for you?   ;) ) . May be we need time to translate all of them in English.
[/quote]

Our sincere thanks goes to [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=2410]lqdschool[/url] who kindly shared the question with us. I also wish her all the best at the VMO 2005.  :)

Hi folks,

my request for the translation is still up-to-date.  :)

@pigfly:i think you can post TST-VMO-04 :D
can you help me to solve problem 6-thanks

@pigfly:
TST-VMO-2004!!!!!
You are so kind and kinder if you post it :D
I don't know it(how boring)

how do vnese people speak english so well!?!? i didnt post any problem yet!! but promise .. i will !! may' nguoi` help me  voi nghen!!

[quote="nguyenkims"]Hi guys,

What do you think about the VietNam TSTs this year? If somebody translats them in English ....? The whole world will be die in laugh about us.[/quote]

  Đúng roài, đừng dịch mà cũng đừng viết tiếng Anh. :D  Các thầy mình ra đề ẩu quá, chán đời :mad:

[quote="orl"]Hi guys,

I have here some VietNam TSTs. Can anybody of you translate them in English and typeset them, possibly in $\LaTeX$ if possible. Thanks.

PS: It is nice. I am the first to post in the mathematical forum of VietNam.  :lol:[/quote]
1991:
1. We observe a set $S$ of $n$ different points in the plan.
1. The distance of two any points of $S$ is less than 1.
(... the next things are missing... Sorry!)


Can you divide all the positive 2-digit integers

into two groups so that the product of the numbers

in each group is the same?

[hide="A fantabulous hint"] $97$ is prime. [/hide]


geometry
perimeter
search
inequalities
triangle inequality

How many isosceles triangles exist which have a perimeter of 10 and integer side lengths?

An isosceles triangle has at least two sides with the same length. Since we can only have integer side lengths in this problem, and 10 isn't divisible by 3, all three side lengths can not be the same in this case.

If one of the congruent sides had length 5, that wouldn't be possible. If one of the congruent sides had length 4, that could work, because the base could be 2. If it had length 3, that could work as well (3-3-4.) 2-2-6, however, wouldn't work, because it doesn't satisfy the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1649450425&t=275736]triangle inequality[/url], so 3-3-4 and 4-4-2 are the only possibilities, giving us an answer of $ \boxed{2}$.


and i think that this statement is true $ [x^{2}]\equal{}[x]^{2}$.

[quote="mayhemkerim"]Why?Can you show that inequality?I can't see it.Thank you.[/quote]
if $ x \equal{} p\plus{}r$(where $ p \equal{} [x]$)
then $ [x^{2}]\equal{}[p^{2}\plus{}2pr\plus{}r^{2}]\equal{}p^{2}\plus{}[2pr\plus{}r^{2}]$
and we have $ p^{2}< p^{2}\plus{}[2pr\plus{}r^{2}] < p^{2}\plus{}2p\plus{}1$ because $ 0\le r < 1$
then $ [x]^{2}< [x^{2}] < ([x]\plus{}1)^{2}$ the rust is evident,

$ arkhammedos$ we have for exemple $ [\sqrt{2}^{2}]\neq [\sqrt{2}]^{2}$


Are there any good online sources for learning more complicated and advanced topics in complex analysis? Thanks!

Depends on what you mean by "advanced topics". For example, B. Ya. Levin's famous book [url=http://www.ams.org/online_bks/mmono150/]Lectures on entire functions[/url] is available for free download now. You may want to see whether you are ready to read it. I think some other fairly advanced books can be found on the internet too though I'm not ready to give links right away. On the other hand, there are important topics like $\bar\partial$-techniques that most textbooks do not cover and for which I do not know any good online sources either.

In general, one still has to learn using a library more than the internet even nowdays. Maybe this will change in the nearest future, I don't know...

Thanks for the resources. That online book is a little out-of-control for me but I will keep looking around and probably buy an introductory book.


Galois Theory
superior algebra
superior algebra unsolved

can anyone help me with some of the galois problems? this is not homework... in fact, that is the problem. there is no hw due in my course and so i try taking some from books i find and they are usually over my head.

(1.) Let $K = \mathbb Q(\sqrt[3]{a}, \sqrt[4]{b})$ where $a,b \in \mathbb Z$ and $a \neq m^3$ and $b \neq \pm n^2$ for some $m,n \in \mathbb Z$. Determine the galois group $\text{Gal} (K/\mathbb Q)$.

(2.) Show that $1 + \zeta_7 + \zeta_7^6$ and $1 + \zeta_7^3 + \zeta_7^4$ are multiplicatively independent units in the cyclotomic field $\mathbb Q(\zeta_7)$.

for #1, the galois group's order is less than or equal to 12, right?

1) I don't think the extension is normal... Do you mean the splitting field of $(x^3-a)(x^4-b)$ ¿

actually, the book I am using uses the notation $\text{Gal} (K/F)$ even when $K: F$ isn't a normal extension (its really old). So if the above extension isnt normal I guess I mean $\text{Aut} (K/\mathbb Q)$.


There is someone on AoPS that keeps rating my posts 1. 1. 1.....

If someone rates you, can a  mod or admin tell who rated him?

Such as this site: (Not the best example... only 2 posts.)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1492300#1492300

I'm not sure if I spammed the second time I posted, but for sure the first time I didn't. And as you can see, someone posted almost all of the posts in a row. That had to be one person. I think. If I can think of some other times people rated my posts 1 for no reason, I'll post it on here.

I think you are underestimating the likelyhood that you simply make a lot of posts that have low information content and that other people find uninteresting.  (You are not the first person to have done this, for sure.)

I'm sure they can tell by looking in the database because I'm sure there's something that saves who rated. Otherwise people could vote more than once.

Rating doesn't really matter. Your reputation will be determined far more by the quality of the proofs and problems you present more than your rating. If people see a post is rated 1 or 6, it is a trivial task to verify the quality of the post.

I rate people 6 to tell them to post similar stuff. I rate people 1 to tell them to stop. If you see that a lot of your posts are rated 1's, you should make sure that your posts contain meaningful content. If you decide that someone's just griefing you, don't sweat it.

Someone rated the OP's post 1. This clearly demonstrates what he is saying.

This demonstrates that I was annoyed by another one who thinks his rating is incorrect. But at least three people voted him very low.

that question should be posted in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=132]MathCounts[/url] forum. It does not pertain to the subject being discussed at the moment in the thread.

[quote="mathlovers1731"]do u guys kno what New Mexico got in Nats? i heard that they were the most improved[/quote]

This is a perfect example of a post that should be rated 1 (In this thread- not necessarily in a MC thread)


I know that we usually have a few mock tests closer to nats and after states, but I was wondering if we could start doing some before states too(that means someone has to make them).

[hide]PLEASE[/hide]

That way people would konw a bit more and have more practice.

I am writing a mock Mathcounts for the local middle school. Hopefully, I'll be able to get some help from others and post it next weekend. In any case, don't expect much.

Yeah, I hope someone makes a Mock MathCounts.

236factorial made a mock MathCounts right here:[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68735]www.artofproblemsolving.com/Forum/viewtopic.php?t=68735[/url]
Yay! My 200th post!

I'm talking about new ones starting now that have not already been done.

I need more practice so a mock mathcounts would really help. :)

does aneone else no ane other old mock mathcounts tests?? i havent done ane of them

Look at posts #90 and #91 in the formulas topic.  I think it's on the 4th or 5th page.

try  previous state competitions  :roll:

i've done "all" of the old states for practice  :?

I've done all the ones that I could find (that is, only the ones from 2001 to 2005) two or three times each. Is that all I can legally get?

If you can find a coach with previous state comps further than that, it's up to them if they wan't to let you have a copy.

Is it possible to get old tests from websites?

Yes it is.  http://goodlymaths.0catch.com

Watch out for the popups!

I think the only way to get those is by a coach or other person.  I've never heard of a site with tests before 2001(and i'm good at finding tests and problems).

Why aren't test before 2001 put on the web?

they're too old and take up space i guess.   :?

They can't take up [i]that[/i] much room, can they?

dont underestimate the size possibility of an adobe file

They really don't take much room, it's just that not many people may have old tests nowadays, and out of that few, only about none would want to give their resources to other people.

I have the 2000 tests right now.  If you buy me a scanner I will send them to you. :D

Hmm... speaking of which... I should do them right now.

In my opinion, a scanner is not worth one year of tests.

Maybe Mathcounts doesn't want the tests before 2000 published.

Don't you mean one year of tests is not worth a scanner? ;)

I've also seen people quote problems from National competitions. Where can I find these?

You have to buy them from the MathCounts site.

Also, I have the 2000 state problems. Also I have scanner at home. Am I allowed to scan them for you guys? Is it legal?

Also, last time I checked, 1998-1999 was available on the site, but that was a few months ago...

yes it is (i have it :) )

goldendomer, it is perfectly legal to openly and freely distribute any MATHCOUNTS material available for purchase, including handbook, competition sets, ...

and NO it is NOT possible to get them anywhere on the internet, except through me :P

[quote="biffanddoc"]
and NO it is NOT possible to get them anywhere on the internet, except through me :P[/quote]
and Arkansas's best coach in Little Rock.

Actually, it's a she, and if you had her as a coach for nats, I bet you would change your mind.

on the game math jam i got zero.
very goot for my 1st math jam. :(
hint: being sarcastic

That's okay.  I even got 0 on my first math jam.  I got 4 this time.

Nice job! How do you answer questions like today's so quickly?Or even instantly think about what to do?

Which ones?  You can look at the transcripts if you need more help.  It's in the math jams tab.

[quote="biffanddoc"]ask robinhe (geez hes like woah)

and xxm0rg I PWNED you 4-0 I didn't think you got any? you weren't on the transcript at all[/quote]

Yeah, I didn't see one thing from xxm0rg07hxx on the transcript.

Actually, my wireless connection was going pretty slow, so I used my dad's account(kchande) because it was connected to the modem.  Kchande(me) got 4 points.

you are kchande? really?

then we tied, but I missed the first question (office hours) so >.<  I win


forall $a,b,c\in[0,1]$ prove that
$(a+b+c)\left(\frac{1}{bc+1}+\frac{1}{ca+1}+\frac{1}{ab+1}\right)\leqslant5$
Don't use too violent method.
[hide="Solution"]$(a+b+c)\left(\frac{1}{bc+1}+\frac{1}{ca+1}+\frac{1}{ab+1}\right)\\ =\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}+\frac{b+c}{bc+1}+\frac{c+a}{ca+1}+\frac{a+b}{ab+1}\\ \leqslant\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}+1+1+1\\ \leqslant\frac{a}{bc+1}+\frac{b}{ca+b}+\frac{c}{ab+c}+3\\ =\frac{a}{bc+1}-\frac{ca}{ca+b}-\frac{ab}{ab+c}+5\\ =a\left(\frac{1}{bc+1}-\frac{c}{ca+b}-\frac{b}{ab+c}\right)+5\\ \leqslant a\left(1-\frac{c}{c+b}-\frac{b}{b+c}\right)+5\\ \leqslant 5$
Then it's done.[/hide]

shalex, nice inequality and nice proof! :lol:

very nice solution shalex!

my solution is using a = 1/(x+1), ... (the case of one of them 0 is separated) and a lot of computations =)

[quote="wxydx00"]It is the SOS you taught me, isn't it?[/quote]

What does it mean? :maybe:

[quote="wxydx00"]I mean :
Did you use SOS in your solution?[/quote]

No I didn't.
You can click on "Solution" to see how it is solved.


inequalities
function
logarithms
triangle inequality

Does Jensen's inequality apply to absolute value functions? Provide examples, if so or if not, please :)

[b]Edit:[/b] [i]I was on drugs or something, ignore[/i]

Thanks! That's what I was thinking, but I wasnt sure and I was definitely desperate to solve an inequality with absolute values. Turns out triangle inequality worked... xD

everything the zuton force wrote is wrong and misleading.  $ f(x) \equal{} |x|$ is a convex function and hence it satisfies the hypotheses (and conclusion) of jensen's inequality.

while differentiability really has nothing to do with convex functions (at least inasmuch as the usual definition is concerned), a convex function is, nonetheless, differentiable at all but countably many points.

[quote="blahblahblah"]everything the zuton force wrote is wrong and misleading.  $ f(x) \equal{} |x|$ is a convex function and hence it satisfies the hypotheses (and conclusion) of jensen's inequality.[/quote]

Can you give an example, please? :)

[quote="blahblahblah"]To show that $ |\cdot|$ is convex, it suffices to show that $ \alpha|x| + (1 - \alpha)|y|\geq |\alpha x + (1 - \alpha)y|$ for all $ \alpha\in(0,1)$ and all real numbers $ x,y$.

But
\begin{eqnarray*} \alpha|x| + (1 - \alpha)|y| & = & |\alpha x| + |(1 - \alpha)y| \\
& \geq & |\alpha x + (1 - \alpha)y| \\
\end{eqnarray*}
by the triangle inequality.

Any convex function automatically satisfies Jensen's inequality, so, for any real numbers $ x_1,\ldots,x_n$ and positive real numbers $ \alpha_1,\ldots,\alpha_n$ summing to $ 1$, we have
\[ \sum_{1\leq j\leq n} \alpha_i|x_i|\geq \left|\sum_{1\leq j\leq n} \alpha_ix_i\right|
\]
[/quote]

So in this case Jensen's inequality IS the triangle inequality (for thsi funciton). That makes a lot of sense, now looking at it (I can't see how I misseed that.. -_-)... Thanks  :D

Or, to be even lazier than blahblahblah, you really only have to show the given statement for one value of $ \alpha$ (I believe).

In most contest problems we use Jensen's Inequality on either exponentials to relate sums and products (for example, AM-GM follows very easily from Jensen's on $ f(x) \equal{} \ln{x}$) or on nasty rational functions that can't easily be broken with Cauchy-Schwarz and that we don't want to expand.  It is always important to homogenize the inequality so that you can get from an inequality of the form $ \sum_{cyc} f\left(x_1,x_2,\cdots, x_n\right) \ge F\left(x_1,x_2,\cdots,x_n\right)$ to an inequality of the form $ \sum_{i\equal{}1}^n g\left(x_i\right) \ge G\left(x_1,x_2,\cdots,x_n\right)$.

[quote="Teki-Teki"]Or, to be even lazier than blahblahblah, you really only have to show the given statement for one value of $ \alpha$ (I believe).[/quote]

as $ |\cdot|$ is continuous, it does suffice to show that it holds for $ \alpha\equal{}1/2$.

No, you don't even have to show it for only that value.  You only have to show it for any value of $ 0 < \alpha < 1$.  So long as the function is continuous in that interval, only one value suffices.  You can pretty much get as close as you want to the desired value.

EDIT: Never mind - misunderstanding.


Solve:

49a + 7b + c = 342, where a,b,c belong to the set of naturals, and
max(a.b.c) is atmost 6.

write 342 in base 7: the three digits of that number will be a, b, c

[hide]$7b+c\leq 48$, so $49a\geq 294$
$a\geq \frac{294}{49}$,  $\boxed{a=6}$
$49(6)+7b+c=342, 7b+c=48$
Obviously, $\boxed{b=6}$ and $\boxed{c=6}$.
Quite a devilish problem (by that i mean the 666, not that it was hard)  :P [/hide]

the base $7$ is a good insight, but this is also obvious through factoring

$7(7a+b)=342-c$, the LHS has a factor of $7$, the the RHS must also, and we have a restricted domain for $c$, we get $c=6$, then it reduces to $7a=48-b$, repeat the logic, $b=6$, and $a=6$

Let me show you an even faster way to it: $342=7^3-1$ so $1000_7-1_7=666_7$ :D

can someone explain to me why when you see a number written in a nice way for a base, you can abandon all algebra standard solving skills and immediately claim you have the solution? aren't there other solutions (at least in some cases) i guess the restricted domain comes into play, is that necessary for all these type of problems?


Notice that the numbers which are in the arithmetic progresson $3n+1$ and which are NOT in the arithmetic progresson $6n+1$ are all even and greater than 2.  Therefore none of them are prime.

[quote="suedenation"][color=olive]Show that if p is a prime in the arithmetic progression 3n + 1, n=1, 2, 3,..., then it is also in the arithmetic progression 6n + 1, n=1, 2, 3....[/color]

Thanks very much for your help!!!:)[/quote]
For the number to be prime in the $3n+1$ sequence, $n$ must be even. Therefore, if a particular $n$ works for the first sequence, $\frac{n}{2}$ works for the second one and creates the same prime number.

Using mods can help.

$p = 3n+1 \equiv 1, 4 \pmod{6}$.

But if $p \equiv 4 \pmod{6}$ then $p$ is even and not prime. So

$p \equiv 1 \pmod{6}\Rightarrow p = 6m+1$.


trigonometry
inequalities
complex numbers
trig identities

Prove that the ecuation  $ \sin x \sin 2x \sin 3x \sin 4x\equal{}\frac{3}{4}$ doesn't have solutions

[hide="Inelegant outline, gotta be something better"]Put into exponential form and distribute out (same as using product-sum). You get a bunch of terms to cancel so you end up with the thing, assuming I made no mistakes, equal to $ \frac{1\plus{}\cos{10x}\plus{}\cos{8x}\plus{}\cos{6x}}{8}$, which is always less than or equal to $ \frac{1}{2}$, so there are no solutions.[/hide]

sorry i don't know how to hide things...
alternatively you can use sum to product identities

We have
[hide]\[ \sin x \sin 2x \sin 3x \sin 4x \&= (\sin 3x \sin x) \cdot (\sin 4x \sin 2x) \\
\&= \displaystyle{\left(\frac {\cos 2x - \cos 4x}{2} \right) \left( \frac{\cos 2x - \cos 6x}{2} \right)} \\
\&= \dfrac {3}{4}
\] Multiplying both sides by 4, we have
\[ (\cos 4x - \cos 2x)(\cos 6x - \cos 2x) = 3
\] By considering a cosine graph, I think we can quickly establish that no values of $ x$ satisfy the equation.[/hide]

[quote="aidan"]We have
[hide]
\[ \sin x \sin 2x \sin 3x \sin 4x \& = (\sin 3x \sin x) \cdot (\sin 4x \sin 2x) \\
\& = \displaystyle{\left(\frac {\cos 2x - \cos 4x}{2} \right) \left( \frac {\cos 2x - \cos 6x}{2} \right)} \\
\& = \dfrac {3}{4}
\]
Multiplying both sides by 4, we have
\[ (\cos 4x - \cos 2x)(\cos 6x - \cos 2x) = 3
\]
By considering a cosine graph, I think we can quickly establish that no values of $ x$ satisfy the equation.[/hide][/quote]That's not really that rigorous, though.

Actually, that's something I've been wanting to ask: how do we define rigorous?

For my solution, the products can be expanded and the product-to-sum rule applied again to get
\[ \frac {\cos 10x \minus{} \cos 8x \minus{} \cos 6x \plus{} 1}{2} \equal{} 3
\] and I guess the obvious follows, since the maximum value of the denominator is 4.

Rigor is mildly subjective; it's how convincing the proof is. The graph makes it look like it is not very convincing. However, use of inequalities and product-sum is rigorous, as we *know* based on the number system we've defined, which has shown itself to be consistent, that these identities are true.

Small thing - I find exponential form far easier than memorizing identities.

[quote="aidan"]I see  :)

Exponential form means the real part of $ e^{2xi}$? Actually i think that after using trig identities so many times, we'd be able to more or less recall them naturally.[/quote]I don't use product-sum often enough to memorize it. Since $ \sin$ and $ \cos$ are odd and even, respectively, we can combine $ e^{ix}$ and its conjugate, $ e^{-ix]}$ in different ways to get rid of either term. $ \frac{e^{ix}-e^{-ix}}{2i}=\sin x$, and $ \frac{e^{ix}+e^{-ix}}{2}=\cos x$. Using these you get a lot of identities completely for free - addition, Pythagorean, product-sum and vice versa, etc. It's faster than using specialized identities in many cases and easier to remember,  too. I've got all the common ones memorized, but not product-sum and sum-product.

That's a nice trick.. Thanks for the tip   :)


function
probability
real analysis
real analysis unsolved

I have a course of theory of measure and it's difficult with me.
In this exercise, I don't know how to resolve, even I have looked in a lot of measure theory textbook.
I try to prove that X is also measurable but I didn't find the way. Someone can help me, please ?

All we need to show that $Y$ is a random variable ([i]i.e.[/i], a measurable function) is to show that the event (set) $\{Y>\alpha\}$ is measurable for all real $\alpha.$

Consider $X^{+}=\max(X,0).$

If $\alpha\ge 0,$ then $\{X^{+}>\alpha\}=\{X>\alpha\}$ which is measurable because $X$ is a (measurable) random variable.

If $\alpha<0,$ then $\{X^{+}>\alpha\}$ is simply the whole probability space, which is measurable.

The argument is somewhat similar for $X^{-}.$

Thanks Kent Merrvfield for your answer of my question. It take me one day to think about your answer and now I understand. Thanks you very very much !  :)


A shell of mass $ 2M$ is fired vertically upwards from the ground with speed $ V.$

At the highest point of its path it breaks into two equal fragments, after an explosion which gives kinetic energy $ mU^2$ to the system, but leaves the momentum unchanged.

Show that the greatest possible distance between the two points where the fragments hit the ground is,

$ \frac{2UV}{g}$ if  U$ \leq$V

$ \frac {U^2 + V^2}_{g}$ if $ U > V$

EDIT.The explosion creates kinetic energy $ MU^2$, NOT $ mu^2$ as written above.My apologies for any confusion.

[quote="Michael Niland"]At the highest point of its path it breaks into two equal fragments, after an explosion which gives kinetic energy $ mU^{2}$ to the system, but leaves the momentum unchanged.[/quote]
What is $ m$?  Do you mean $ M$?

Hey Micheal ! I didn't Get the Cases U>V and U<=V.

However, I was Able to Devise to Possible Cases .
The First case has been Successfully Solved however I wasn't able to Completely Solve the Second Case
See If you can solve the Second One Yourself and If you get the second One , Then Please Inform Me About it too.

All The Best !

 :wink:

The ball attains height $ h \equal{} \frac {V^2}{2g}$.

The particles fly off with velocity $ U$.
Let the smallest angle between one particle and the horizontal direction be $ \theta$.

Then $ v_x \equal{} U \cos{\theta}$ and $ v_y \equal{} U \sin{\theta}$ for both particles.

Let $ A$ be the particle that went up, and $ B$ the one that went down (if they went horizontally, then it doesn't matter which is which).

For A:

$ \frac {gt_A^2}2 \minus{} t_AU\sin{\theta} \equal{} h$
$ t_A \equal{} \frac {U\sin{\theta} \plus{} \sqrt {U^2\sin^2{\theta} \plus{} 2gh}}g$

For B:

$ \frac {gt_B^2}2 \plus{} t_BU\sin{\theta} \equal{} h$
$ t_B \equal{} \frac { \minus{} U\sin{\theta} \plus{} \sqrt {U^2\sin^2{\theta} \plus{} 2gh}}g$

The distance they travel before hitting the ground is $ tv_x$, and the separation is the sum of both distances:

$ S \equal{} t_Av_x \plus{} t_Bv_x$
$ S \equal{} v_x(t_A \plus{} t_B)$
$ S \equal{} v_x(\frac {2\sqrt {U^2\sin^2{\theta} \plus{} 2gh}}g)$

Remembering $ h \equal{} \frac {V^2}{2g}$:
$ S \equal{} v_x(\frac {2\sqrt {U^2\sin^2{\theta} \plus{} V^2}}g)$

Then the distance between them is $ \frac {2U\cos{\theta} \sqrt {U^2 \sin^2{\theta} \plus{} V^2}}g$.

If the separation is maximum, then I believe either $ \theta$ is 0 (when $ U \le V$), or, if $ U > V$, $ \theta$ assumes some special value (which is a function of U and V). But I have no idea how to find that maximum.

@ffao , DOn't You think you are Forgetting Something!!

amangupta2, would you explain what is missing in ffao's post?

Oh ! I am sorry ,   I overread some statement, but later  realised that  i was wrong.
 :oops:

by the way if i am not wrong the extreme cases can be considered by taking 1) "theta"=45 and 2) "theta"=0

Let's define \[ W = U cos \theta \Rightarrow S =\frac {2W \sqrt {U^2  + V^2 - W^2}}{g}\] and we can see that $ S$ is a crescent function of $ W$ while ${ 0 < W < \sqrt {{U^2 +V^2} \over 2}}$
From definition $ W \leq U$ ,than
 for ${ U \leq V,\,\,\,\,\,\,  \sqrt {{U^2 +V^2} \over 2}} \geq U ,\,\,\,$ and maximum $ S$ occurs for $ W = U \rightarrow \theta = 0  \Rightarrow S_{max} = {{2UV} \over g}$

 for  $ U > V\,\, ,$ the extreme happens for ${ W = \sqrt {{U^2 +V^2} \over 2}} \Rightarrow S_{max} = {{U^2+V^2} \over g}$

Sorry; by crescent function I mind increasing function.


function
real analysis
real analysis unsolved

We denote the function $f: [-1,1]\to [-1,1], \ f(x)=ax^{2}+bx+c$ for any real $a,b,c$ with $a\neq 0$. Find the minimum value for a such that $f$ is injective.

$f$ injective iff $f$ monotone iff $f'(x)=2ax+b$ has sign constant on $(-1,1)$, i.e $2ax+b\geq 0\forall x\in (-1,1)$ or $2ax+b\leq 0\forall x\in (-1,1)$.


Let a,b,c be positive reals with sum 1.  Prove:

$(ab)^\frac{5}{4}+(bc)^\frac{5}{4}+(ca)^\frac{5}{4}< \frac{1}{4}$

From Cauchy-Schwartz we have :
$LHS^{2}\leq (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})\leq \left (\sum \frac{a+b}{2}\right)\left( \sum a^{2}b^{2}\right)=\sum a^{2}b^{2}$.
So it's enough to prove that
$\sum a^{2}b^{2}<\frac{1}{16}$ $\Leftrightarrow$ $16\sum a^{2}b^{2}<(a+b+c)^{4}$ ,which can be proved by Schur and rearrangements.


function
geometry
geometric transformation
rotation

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a function with the property $|f(x)-f(y)|=|x-y|$, $\forall x,y \in \mathbb{C}$. Prove that $f$ is an one-to-one function.

or, equivalently,

Prove that any distance preserving transformation of the plane into itself is bijective.

If we choose one point as the origin, And pick two other points. The triangle made by these point is congruent to the triangle under the transformation, Now pick an arbitary point. The distance from this point to the  other three points isconserved, so it must be in the same place relative to those points. Looking at what we started with, the only changes that could have been made by the transformation is translation, rotation or mirrored over. I dont actually know what a bi-jection is, but i think its something along these lines!

bijection = one-to-one = the equation $f(x)=y$ has one and only one soln, $\forall y \in \mathbb{C}$

injective = prove it has $\leq 1$ solutions

surjective = prove it has $\geq 1$ solutions

 and yes, i think it is somewhere along your lines :)

Since i proved its one-to one, is my proof ok?

[quote="blahblahblah"]If we're sending the plane to itself, the transformation is automatically surjective.  Now say $f(x)=f(y)$ for some $x,y\in\mathbb{C}$.  Then $|f(x)-f(y)|=0=|x-y|\implies x=y$ so that $f$ is injective.  The result follows.[/quote]

actually i kind of formulated it incorrectly

we have a function $f$ from plane $P$ to plane $P$. It is not automatically surjective :) you have to prove it, like seamusoboyle did (this is the harder part)

Well, every isometry of $\mathbb{C}$ is of the form $a\overline{z}+b$ or $az+b$ with $|a|=1$.  To see this, you may draw three vertices in $\mathbb{C}$ which form a scalene triangle (man, I haven't used that word in a long time) in $\mathbb{C}$.    Now the image of any other point in $\mathbb{C}$ is uniquely defined by the distances to the images of the three original vertices, so we need only find one transformation that works, and it will be uniquely defined.  This is easily done using transformations of the form $z\to\overline{z}$, $z\to az,|a|=1$, $z\to z+b$


function
algebra
rational function
complex analysis
complex analysis unsolved

Let $ f(z)$ be rational function such that in some neiborhood of $ z\equal{}0$ it is analytic and satisfy conditions $ |f(z)|<1,$ $ \frac{{|f'(z)|}}{{1\minus{}|f(z)|^{2}}}\equal{}\frac{1}{{1\minus{}|z|^{2}}}.$  Prove that $ f$ is linear fractional transformation.

[quote="edoo"]The result is a special consequence of the Schwarz-Pick theorem.

See:  [url]http://en.wikipedia.org/wiki/Schwarz_lemma[/url].[/quote]

Excuse me, can we apply Schwarz-Pick theorem now?
In the problem it is given that $ |f(z)|<1$ in some neiborhood of $ z\equal{}0$ and Schwarz-Pick theorem requires $ |f(z)|<1$ in the whole unit disc...


If $ 12\minus{}\sqrt{x}\equal{}9.3$, what is the value of $ 12\plus{}\sqrt{x}$? Express your answer as a decimal to the nearest tenth.

$ 12\minus{}\sqrt x\equal{}9.3\implies2.7\equal{}\sqrt x\implies12\plus{}\sqrt x\equal{}\boxed{14.7}$.


Gauss
linear algebra
matrix
parameterization

Here is it. I should solve it using Gauss-Jordan

2x1+7x2+3x3+x4=6
3x1+5x2+2x3+2x4=4
9x1+4x2+x3+7x4=2

The main problem for me is that I cant reduce the unknowns.
Whatever I do with all of them I only can get:

11x2+5x3-x4=10
I suppose I should leave two of the unknowns as parameters.
So I get

x4=11x2+5x3-10
and when I put this into first row I get
x1=-9x2-4x3+8

(-9x2-4x3+8, x2, x3, 11x2+5x3-10)
Is this correct?
Every advice will be great appreciated
thanks

hello, do you mean
$ 2x_1\plus{}7x_2\plus{}3x_3\plus{}x_4\equal{}6$
$ 3x_1\plus{}5x_2\plus{}2x_3\plus{}2x_4\equal{}4$
$ 9x_1\plus{}4x_2\plus{}x_3\plus{}7x_4\equal{}2$
Sonnhard.

The $ 3\times 4$ coefficient matrix does indeed have rank $ 2$, as does the $ 3\times 5$ augmented matrix. (In other words, your elimination steps should have given you two rows with pivot elements, and a third row of all zeros.) Hence the solution set should indeed have two free parameters.

You chose to make $ x_2$ and $ x_3$ those parameters; another person solving the same system might instead choose $ x_3$ and $ x_4$ as the parameters. It is possible for the resulting solutions to both be correct but to look different. (Your solution looks different from mine for that reason, and I haven't checked the details of whether yours is correct.)

hello, multiplying the first equation by $ - 3$ and the second by $ 2$ and adding these two equations we got:
Multiplying the first equation by $ - 9$ and the third by $ 2$ and adding we have:
$ 2x_1 + 7x_2 + 3x_3 + x_4 = 6$
     $ - 11x_2 - 5x_3 + x_4 = - 10$
     $ - 55x_2 - 25x_3 + 5x_4 = - 50$
The second and the third equations are the same, so we have only:
$ 2x_1 + 7x_2 + 3x_3 + x_4 = 6$
     $ - 11x_2 - 5x_3 + x_4 = - 10$
From the second equation we have
$ x_4 = - 10 + 5x_3 + 11x_2$ with  $ x_3 = x_3$ and $ x_2 = x_2$ are arbitrary real.
Substituting this in the first equation and solving this with respect to $ x_1$
we have:
$ x_1 = 8 - 9x_2 - 4x_3$
These are the solutions of our system.
Sonnhard.


superior algebra
superior algebra unsolved

Can someone explain a step in a proof (about Krull rings).

(Nagata) Let $K$ be a field, $R_{1},...,R_{n}$ valuation rings of $K$. Set $A = \bigcap R_{i}$. For any $a \in K$, there is an integer $s \geq 2$ such that $(1+a+...+a^{s-1})^{-1}\in A$.

To show that it is in$A$, he fixed some random $R_{i}$ and claimed that if $a \notin R_{i}$, then $a^{-1}\in m_{i}$ and any $s\geq2$ will do...

I don't understand the last step... :(

(Matsumura p87)

[quote="Jacobson"]$a \notin R_{i}$, then $a^{-1}\in m_{i}$[/quote]

well this is clear, isn't it? the other step is probably only understandable if you show us the whole proof ;-).


Prove that there are infinitely many naturals which can't be represented as $ x^3\minus{}y^4$,where $ x$ and $ y$ are naturals.

I think that $ x^3\minus{}y^4\not\equiv6(mod 13)$.


inequalities
search
inequalities proposed

It sounds like [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36391]this problem[/url]  :lol:  ;)

This is exactly Kantorovich's inequality which I mentioned in that thread.

instead $x_i(2 \leq i \leq n-1)$ by $x_1$ or $x_n$

[quote="siuhochung"]how to prove it?[/quote]
I remember that I gave a proof in some thread, try to search "Kantorovich" or something like that.

Here's proof

This ineq is kantorovich Inequality.

$\forall x_i \in [x_1, x_n]$, $(x_i - x_1)(x_i - x_n) \leq 0$

$\Leftrightarrow x_i^2 + x_1 x_n \leq x_i (x_1 + x_n )$

$\Leftrightarrow k_i(x_i + \frac{x_1x_n}{x_i}) \leq k_i(x_1+x_n)$

So, We get $ \sum k_ix_i + \sum \frac{k_i x_1 x_n }{x_i} \leq x_1 + x_n $

Using AD-GD, we get

$x_1+x_n \geq 2 \sqrt {x_1x_n(\sum k_ix_i)(\sum \frac{k_i}{x_i}) }$

$Q.E.D.$


Thank you, nguoivn, for your gift. I proved it by the ugly way :

$ \sum{\frac {\sqrt [3]{a(5a^2 \plus{} bc \plus{} ca \plus{} ab)}}{\sqrt {3(bc \plus{} ca \plus{} ab)}}}$

$ \geq2\sum{\frac {a(2bc \plus{} ca \plus{} ab)}{a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 6abc}} \equal{} 2.$

Can you detail more, Ji Chen ?
Sorry because I really haven't understand your hint yet :oops:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 3$.
Prove that: $ \sqrt [3]{5a^3 \plus{} 3a} \plus{} \sqrt [3]{5b^3 \plus{} 3b} \plus{} \sqrt [3]{5c^3 \plus{} 3c} \geq\ 6$
PS: It exitst an easy proof by Am-Gm, my friend  :)[/quote]
We have
\[ LHS \ge 2\left( a^{3/4}\plus{}b^{3/4}\plus{}c^{3/4}\right) \ge 6\]
;) :)

[quote="nguoivn"][quote="can_hang2007"]
We have

$ LHS \ge 2\left( a^{3/4} \plus{} b^{3/4} \plus{} c^{3/4}\right) \ge 6$

 :)[/quote]
Yes, you saw my way to creat this ineq, my brother.  :oops:[/quote]

Can you explain the last part:
$ 2\left( a^{3/4} \plus{} b^{3/4} \plus{} c^{3/4}\right) \ge 6$
thanks

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 3$.
Prove that: $ \sqrt [3]{5a^3 \plus{} 3a} \plus{} \sqrt [3]{5b^3 \plus{} 3b} \plus{} \sqrt [3]{5c^3 \plus{} 3c} \geq\ 6$
PS: It exitst an easy proof by Am-Gm, my friend  :)[/quote]
We have
\[ LHS \ge 2\left( a^{3/4} \plus{} b^{3/4} \plus{} c^{3/4}\right) \ge 6
\]
;) :)[/quote]
And how you come up with this?

[quote="nguoivn"]You can see the similar problem:
http://www.mathlinks.ro/Forum/viewtopic.php?t=251425
Please think about Am-Gm and Schur ineq, my friend  :)[/quote]

I tried to use Am-Gm and Schur ineq but still nothing....may be you will post detailed solution?


The circumference of a circle is $ 2x^2$. In terms of $ x$, how many square units are in its area?


Let $p_n$ be n'th prime number,and $p_0=1$.

Prove that for every positive integer M we can find $0\le i_1<i_2<...<i_k$ such that

$M=p_{i_1}+p_{i_2}+...+p_{i_k}$

For M=1, 2, 3, 4, 5 we have
1=1;
2=2;
3=3;
4=1+3;
5=5.
If M>5 we'll use method of math. induction. We assume, that for numbers 1, 2, ..., M-1 it's true.
If M is a prime it's evident. Else by the Bertrand's postulate there is a prime p such, that M/2<p<M. If M-p=p[ i[1] ] + ... + p[ i[n] ], then m=p[ i[1] ] + ... + p[ i[n] ] + p, where p[ i[1] ]<...<p[ i[n] ]<M-p<M/2<p, q.e.d.


search
superior algebra
superior algebra unsolved

True or false: if $ I$ is a minimal left ideal in $ R$, then $ M_n(I)$ is a minimal left ideal in $ M_n(R)$?

look at http://www.mathlinks.ro/viewtopic.php?search_id=10331325 . your question comes down to: is $ I \times ... \times I$ a minimal left submodule of $ R^n$ if $ I$ is a minimal left ideal of $ R$? this is clearly false.

Probably because it's a link to a search id  :D
-oo-, you copied the wrong part of the url  :wink:

lazy meets dumb. ;-)

http://www.mathlinks.ro/viewtopic.php?t=193949


vector
calculus
integration
algebra
polynomial
calculus computations

Can anyone explain how to do these problems? I am studying parametrics and vectors and I totally do not understand these problems.

1. At time t=0, a particle is located at the point (1,2). It travels in a straight line to the point (4,1) , has speed 2 at (1,2), and constant acceleration 3[b]i[/b] - [b]j[/b]. Find an equation for the position vector [b]r[/b](t) of the particle at time t. [/b]

2. The position of a kite is given by:

[b]r[/b](t) = (t/8)[b]i[/b] - (3/64)(t)(t-160)[b]j[/b],
where t >= 0 is measured in seconds and distance is measured in meters.

a. How long is the kite above ground? (do I just find arclength?)
b. How high is the kite  (Plug in 40? )
c. At what rate is the kite's altitude increasing at t = 40 seconds?
d. At what time does the kite start to lose altitude?

Any help is vastly appreciated.

I'll give you some hints, and if you still have problems, call out.

1. You know that $\vec{a}=\left<3,-1\right>$. You can find $\vec{v}=\int\vec{a}dt$ and $\vec{r}=\int\vec{v}dt$. You get four integration constants you need to find, and you can use the extra information delivered for that.

2.

a) The height is given by $y=-\frac{3}{64}t(t-160)$. This is a second degree polynomial with $a_2<0$. This means the coefficient before $t^2$ is negative, so it must have a maxima. For which two $t$ is $y=0$? The time interval between these two $t$ must be the interval where the kite is above ground, so the air time is $T=|t_0-t_1|$ if $y(t_0)=y(t_1)=0$.

b) The maximum (or minimum) occurs directly in the middle between $t_0$ and $t_1$. Find the time and you have the height.

c) Simply calculate $\frac{dy}{dt}$ and substitute $t=40$.

d) The kite starts losing altitude when it has reached it's maximum height. (What did you find in b?)

Thanks a lot, TheBB. I think I understand it now.


AMC
USA(J)MO
USAMO
induction
number theory proposed
number theory

Prove that $\forall m\in \mathbb {N^+},\exists n=\overline{a_1a_2...a_k},a_i\neq 0 (\forall i=1,2,...,k)$ such that $5^m|n$

We will prove that for each $m$ there exists $m$-digit number which is divisible by $5^m$ (and has non-zero digits) by induction.

For $m=1$ we have number $5$.

Now suppose we have $m\geq 1$ and a $m$-digit number that is divisible by $5^m$. Let's call this number $x$. We will add 1,2,3,4 or 5 to left side of  the number $x$. They'll change $x$ modulo $5^{m+1}$ by $0,5^m,2*5^m,3*5^m,4*5^m$. But you can see that by choosing the right one we will make number with $m+1$ non-zero digits and divisible by $5^{m+1}$.

Hope it's all right and sorry for not writing more details.  :P
These kinds of problems are often solved by induction... maybe there were many similar problems on this site...

There is also a very similar problem (USAMO 03/1) which has the following problem statement: Show that for each $n \in \mathbb{N}$, we can find an $n$-digit number with all its digits odd which is divisible by $5^n$.

Yes, it can be done the same way: induction

we can replace 5 by another number  n obeys gcd(n,10)=1 ,but besides induction how can you solve it ?

Yes, if you mean to find number fith non-zero digits you can replace it by 3,5,7 or 9.
I don't know of any proof of this not using induction, but is it necessary or helpful?
This induction proof is easy and similar ideas can be applied to similar problems :)
For example you can change condition "having just odd digits" to "having alternating
digits" (odd,even,odd,even,...) ...


limit
logarithms
integration
inequalities
real analysis
real analysis solved

Let $u_n>0$ be a sequence of real number such that:
1. $u_n\to+\infty$,
2. $\lim_{x\to+\infty} \frac{u_{n+1}-u_n}{\ln u_n} = 1$.

Find an equivalent of $u_n$.

[i]Edited by Myth[/i]

of course it is, but what is your proof ?

We will show that $\frac{u_{n+1}}{\ln u_{n+1}}-\frac{u_{n}}{\ln u_{n}}=1$  $(*)$.
Indeed,
\[\left(\frac{u_{n+1}}{\ln u_{n+1}}-\frac{u_{n}}{\ln u_{n}}\right)-\frac{u_{n+1}-u_n}{\ln u_n}=\frac{-u_{n+1}}{\ln u_n\cdot\ln u_{n+1}}\cdot \ln\left({\frac{u_{n+1}-u_n}{u_n}+1}\right).\]
Now we can estimate:
\[\left|\frac{-u_{n+1}}{\ln u_n\cdot\ln u_{n+1}}\cdot \ln\left({\frac{u_{n+1}-u_n}{u_n}+1}\right)\right|\leq
\frac{u_{n+1}}{\ln u_n\cdot\ln u_{n+1}}\cdot \frac{u_{n+1}-u_n}{u_n}\to 0.\]
From $(*)$ we conclude that
\[1=\lim_{n\to\infty} \frac{1}{n}\left(\left(\frac{u_{n+1}}{\ln u_{n+1}}-\frac{u_{n}}{\ln u_{n}}\right)+...+\left(\frac{u_{2}}{\ln u_{2}}-\frac{u_{1}}{\ln u_{1}}\right) \right)=\lim_{n\to\infty} \frac{u_{n}}{n\ln u_n},\]
hence result :D.

[quote="Myth"]
From $(*)$ we conclude that
\[1=\lim_{n\to\infty} \frac{u_{n}}{n\ln u_n},\]
hence result :D.[/quote]

I don't see from now how do you get $u_n \sim n \cdot \ln n$

Why?
It is clear that $\forall a>0$ $\exists N$ s.t. $\forall n>N$ the following inequality holds:
\[(1-a)n\ln n<u_n<(1+a)n\ln n.\]

[quote="alekk"]nice  :)
another approach: $(u_{n+1}-u_n)\frac{1}{ln(u_n)}$ is equivalent to $\int_{u_n}^{u_{n+1}}\frac{dt}{ln(t)}$. Conclusion follows[/quote]

then $\int_{u_n}^{u_{n+1}}\frac{dt}{ln(t)}$ is equivalent to $1$ so
$\int_{7}^{u_{n+1}}\frac{dt}{ln(t)}$ is equivalent to $n$. Hence
$\frac {u_{n+1}} {ln(u_{n+1})}$ is equivalent to $n$. Take $ln$ of both side and you get $ln(u_{n+1})$ is equivalent to $ln(n)$ and you get the result

[quote="Moubinool"][quote="Myth"]
It is clear that $\forall a>0$ $\exists N$ s.t. $\forall n>N$ the following inequality holds:
\[(1-a)n\ln n<u_n<(1+a)n\ln n.\][/quote]

Where is the proof ?

Above you had $\frac{u_n}{\ln u_n} \sim n$ how do you use this

to get $u_n \sim n\cdot \ln n$ ?[/quote]
Are you kidding again?
For example, if $u_n>(1+a)n\ln n$ for infinitely many $n$, then $\frac{u_n}{n\ln u_n}>\frac{(1+a)n\ln n}{n\ln \big((1+a)n\ln n\big)}>(1+a/2)$ (for all large $n$), so $\frac{u_n}{n\ln u_n}\not\to 1$.


geometry
AMC
USA(J)MO
USAMO
3D geometry
tetrahedron

Hi all,

I was just wondering about how to learn 3-D Geometry here in Canada. There's nothing in the curriculum, and I couldn't find anything on Amazon. Are there any training books etc., or texbooks regarding 3-D geometry, to prepare for math contests?

Thanks,

Sunchips :)

The two best sources I know of so far: aops vol 1, and introduction to geometry. Both By Richard Rusczyk ( Sandor Lehocaky was a co-author for aops ). You should seriously consider these books.

it's not stressed in canadian curriculum, but it's not stressed in the contests either.  :D

I suck at 3 dimensional geometry, but it's never seriously been a huge barrier in contests for me either.

So as far as the training for the Canadian math contests goes, don't bother too much about it.. Just try the past problems in contests, especially USAMO, they have extensive problems on tetrahedrons, enough to make you become sick of them for a little while!! (Although nothing beats Fibonacci!  :lol: )


Without quoting the "Erdos squarefree conjecture", prove that the binomial (2^n+1,2^n) is never squarefree for n greater than 2.

by
[quote="LeoYard"]the binomial (2^n+1,2^n)[/quote]
you mean $ \binom{2^{n\plus{}1}}{2^n}$? :maybe:

[quote="Rust"]Obviosly $ \binom{2^{n \plus{} 1}}{2^n}\equiv 2\mod 4$[/quote]
so? :maybe:
the problem asks us to prove that the number isn't squarefree so you must prove that there exists $ p$ s.t $ p^2\mid\binom{2^{n \plus{} 1}}{2^n}$...

BaBaK Ghalebi, yes, it is obvious that we need to prove that. But, perhaps, it would be easier to prove first the following:

Let's assume 'n' is not a power of 2, and prove than (2n,n) is never squarefree for n greater than 1.

The hint that was given is : Show that it is always divisible by 4.

[quote="kevinatcausa"]If $ 4$ divides $ \binom{2n}{n}$ for $ n$ not a power of 2, isn't the original post just a rewording of the (Very hard!) square-free conjecture?[/quote]
It is hard for $ \binom{2n}n$, but I can prove it for $ \binom{2^{n\plus{}1}}{2^n}$, when $ n>N_0$.


Solve the system of equations:

$ 2xy\plus{}(a\plus{}b)x^2\equal{}1$

$ y^2\minus{}abx^2\equal{}\minus{}c$,

where $ a,b,c$ are real numbers such that $ c\leq a<b$

[quote="radio"]Solve the system of equations:

$ 2xy + (a + b)x^2 = 1$

$ y^2 - abx^2 = - c$,

where $ a,b,c$ are real numbers such that $ c\leq a < b$[/quote]

Well,
$ x=\pm \sqrt {\frac {a+b-2c\pm 2\sqrt {(b-c)(a-c)}}{(a-b)^2}}$

All 4 roots are valid so, you have four solutions.
Is that it???
i seriously doubt if i'm going wrong.

Well, there can be more than one solution

Notice that the square root in the numerator can be eliminated (i.e. $ a\plus{}b\minus{}2c\plus{}2\sqrt{(a\minus{}c)(b\minus{}c)}\equal{}(\sqrt{b\minus{}c}\plus{}\sqrt{a\minus{}c})^2$...)

Anyway, you have to pair your answers of $ x$ with answers of $ y$. (So, answer in pairs $ (x,y)$)
And please show your work, so that others understand  :)


A bag contains 15 coins. Each coin is a penny, nickel, dime or quarter. The total value of these coins is
2.16 dollars . There is at least one coin of each denomination, and the number of pennies, nickels, dimes and quarters are all distinct. If there are 5 dimes, how many nickels are in the bag?

[hide] Let the number of pennies be $ x$, the number of nickels $ y$, and the number of quarters $ z$ (we already know the number of dimes). Hence, we have

$ x\plus{}y\plus{}z\equal{}10$

$ x\plus{}5y\plus{}25z\equal{}166$

Subtracting the first equation from the second,

$ 4y\plus{}24z\equal{}156 \implies y\plus{}6z\equal{}39$.

Checking, we see that the only $ y$ and $ z$ which produce distinct solutions are $ y\equal{}3$ and $ z\equal{}6$. Hence, the number of nickeels is $ \boxed{3}$.[/hide]


algebra
function
domain
integration
limit
real analysis
real analysis unsolved

Anyone can show this question for me?  Thanks!

Let $ U$ be a domain of $ \mathbb R^n$ such that $ 0\in U$. Prove that there is no local integrable function $ \delta$ on $ U$ so that
\[ \int_U\delta(x)\phi(x)dx\equal{}\phi(0)\]
for every $ \phi\in C^\infty_0(U)$.

If $ f$ is locally integrable, $ \lim_{\epsilon\to 0}\int_{\minus{}\epsilon}^{\epsilon}|f(x)|\,dx\equal{}0$. Let $ \phi$ be an arbitrary nonnegative test function supported on $ [\minus{}1,1]$, with $ \max_x\phi(x)\equal{}\phi(0)\equal{}1$, and let $ \phi_{\epsilon}(x)\equal{}\phi\left(\frac{x}{\epsilon}\right)$.
We have that
$ \left|\int_{\mathbb{R}}f(x)\phi_{\epsilon}(x)\,dx\right|\equal{}\left|\int_{\minus{}\epsilon}^{\epsilon}f(x)\phi_{\epsilon}(x)\,dx\right|$
$ \le\int_{\minus{}\epsilon}^{\epsilon}|f(x)|\phi_{\epsilon}(x)\,dx\le \int_{\minus{}\epsilon}^{\epsilon}|f(x)|\,dx$,
which goes to zero as $ \epsilon\to 0$. Thus it can't be equal to $ \phi_{\epsilon}(0)$ for all $ \epsilon$, and this $ f$ does not act as $ \delta$.


Find all solutions in natural numbers $x$ and $y$ to the equation
$\frac1x+\frac1y=\frac13$.

Sorry, Splash, I changed it.

[size=75][color=darkred]Please put the question in your post (and not the subject) next time so it is easier to see. Thanks.
-nebula42[/color][/size]

that's impossible as the maximum is 1/1+1/1=2

$\frac{1}{x}+\frac{1}{y}=\frac13$...
[hide="hint"]Simon's favorite factoring trick[/hide]
You might also want to add
x doesn't equal y because then there is the relatively obvious [hide]$\frac16+\frac16$[/hide]

[quote="alex__ib"]6,6  is ONE of the solutions.[/quote]

[hide]the other is $4,12$[/hide]

[quote="alex__ib"]6,6  is ONE of the solutions.[/quote]
It is but it is more obvious. I know a solution but I am used to problems like this.
[hide=" my solution"]$\frac{1}{x}+\frac{1}{y}=\frac13$
$\frac{x+y}{xy}=\frac13$
$3x+3y=xy$
$xy-3x-3y=0$
$xy-3x-3y+9=9$
$(x-3)(y-3)=9$ Complex Zeta's favorite factoring trick
6,6
12,4[/hide]

[hide]
$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{1}{3}$
$3x+3y=xy$
So, with some guess and check, the answer is (6,6), (4,12), and (12, 4).[/hide]

[hide]The three answers are:
6,6
4,12
12,4[/hide]

[hide="another solution"]$3x+3y=xy$ They all have to be integers.
Subtracting and dividing gives $x=\frac{3y}{y-3}$ From here, you know that y>3 and that 3y is divisible by y-3. Although more gues and check is required, it helps if you can't come up with the factoring method.[/hide]