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I am a strong believer that violent video games do not cause kids to commit violent crimes.

Read this article, it's an awesome one:

http://gr.bolt.com/articles/violence/violence.htm

Please post comments...

On the whole, I agree with the article.  But I would like to stress that the games that children should be allowed to played should be seriously monitored.  There's nothing inherently wrong with video games...just be careful what you put in them.

[quote="10000th User"]I think video games are not #1 reason for crimes. If a boy cannot control himself, he is the problem, not video games. But video games should not advertise violence because stupid boys can copy same violence shown in video games and since they are stupid, they cannot control themselves and do stupid violence.[/quote]
This is the point though, all data indicates that this is indeed NOT the case. My opinion is that young people who would otherwise commit a violent crime instead take it out through a video game, and not the other way around. Look at the release of Playstation and the drop in crime thereafter, interesting stuff.

Stupid kids do stupid things regardless of what they play on a game. It appears that games have caused a reduction in crime though.

[quote="Spoon"]This is the point though, all data indicates that this is indeed NOT the case. My opinion is that young people who would otherwise commit a violent crime instead take it out through a video game, and not the other way around. Look at the release of Playstation and the drop in crime thereafter, interesting stuff. [/quote]

Agreed. If I feel like shooting someone, I'd rather do it while playing Doom than in real life.

They are not stupid, they are so young therefore they cannot control theirselves.

I was playing some video games recently (First Person Shooter) and I noticed how crappy the game was with shooting a gun. I mean, I've shot guns before and these games aren't realistic enough. However when I started playing the game I shotout a light from 100 yards away with a pistol. PSI-Ops

With that being said, games relieve stress. They don't create it. They won't teach you how to shoot a gun, they will teach you how to hide from opponents though.

Now about all this, I'll tell you something. They do aid a more violent person with tactical skills.

Back in the days of World Wars people didn't want to shoot one another. Except the Nazis cause they were trained to be violent.
American soldiers didn't want to shoot many people and casualities were created because of that.
Only when they created more realistic targets for them to train with during Boot Camp and Basics in Vietnam did the death count go up.

I think the reasoning to shoot an enemy soldier went up 60%. I read this in a book somewhere.

Now with this video game age of people they are most likely willing to shoot someone in a war scenario.

[quote="Cyberman"]Back in the days of World Wars people didn't want to shoot one another. Except the Nazis cause they were trained to be violent.
American soldiers didn't want to shoot many people and casualities were created because of that.
[/quote]

I don't get it... wouldn't killing or injuring as many hostile soldiers make for better chances of survival? Why wouldn't anyone be as deadly as possible against enemy forces?

A lot of people don't want to have to kill another person. They find it immoral and unjust.

In combat people have to learn to accept, "Kill or be Killed." or else they die. People don't want to die, nor do they want to kill.
I don't know how old you are, but since I'm 18 I've seen some things change. I never went with the norm of people, but I do know how the future generations are going to be made.

Throughout generations, violence in the media and the outside world has altered the minds of the American Youth. There wasn't really any violence back then. I spoke to my dad about Vietnam, he said no body really robbed anybody's house. I was thinking, "My god, I would have robbed a ton of people if I knew they were in wartime and not inside their house." That's how you can see generation change right there. We see crime more and more and get ideas.

Back then, they didn't really have anyone to plant ideas of hate and crime in their head. Just xenophobia and fear of foreign people who were strange.

Nazis on the other hand were bred to kill. Even in todays world they would be scary suckers to mess with for today's youth.

When members were growing up they would be put in a fighting ring at age 12 and fight against someone age 14 or 15. It would be a knock down, drag out fight. All things they learned were to be used in combat situations to kill people. People were emotionally disturbed. They were brainwashed to hate others but somehow accept teamwork.

However video games do something to the mind to somehow enhance the want to kill. I'm not going to say they influence it easily saying it creates a great deal of want, but it raises up the ability and initiative.

I think that when people team up with others they are more sociopaths than regular people. Sociopaths like teaming up with other violent people because they accept that type of behavior. Regular healthy minded people who have lived a life of good don't want to be near sociopaths whatsoever.

I don't believe video games, etc, should be "responsible" for the murders of other people. However, I do believe that video games (and movies) can teach... what is the word... ok, since I can't remember the key word in my argument, I'm just going to describe it for you. Violent media teaches children to [not be as emotionally disturbed by the idea of death and killing]. For example, has anyone here seen [i]Saving Private Ryan[/i]? The opening shots of this movie are some of the most dramatic and gory I have ever seen. We had to watch it in english class last year. Being that I have not watched a lot of violent movies, I had to cover my eyes and ears. However, most of my classmates had very little problem watching this- they were not as disturbed by it, because they had seen so much, they had become desensitized. Now, whether this breeds criminal activity, I am not sure. I would think that it would be easier to kill someone in cold blood, though, if you had seen it happen many times before. It all comes back to the same thing, therefore. There are bad things in this world. As a parent, it is your job to guard your children from these things until they are mature enough to handle it. It is the parent's job to teach their child right from wrong. While I would not buy "Grand Theft Auto" for anyone I know (because I believe it is conflicting to the morals I hold, video game or not), I still believe that it is the First Amendment (? I think that's the right one) right of the video game makers to create such a game, as long as it is marketed to a sufficiently mature age group.

Ok, the word is "desensitized." I just remembered.

[quote="Cyberman"]I don't know how old you are, but since I'm 18 I've seen some things change.[/quote]

I will be 18 in a few months, so I don't think we're so different in age, and perhaps in general experiences?  :)

[quote="Cyberman"]wasn't really any violence back then.[/quote]

You're honestly being naive if you believe this. Crime rates have gone down if anything in the last 30-40 years. Let's keep the various riots throughout the '60s and '70s and the general looting that ensued. Crime has been around forever, and it will always stick around. It's not anything new at all.

[quote="Cyberman"]Back then, they didn't really have anyone to plant ideas of hate and crime in their head. Just xenophobia and fear of foreign people who were strange. [/quote]

This in the era of civil rights movement? No ideas of hate and crime?

Honestly, I think that violence, crime, etc... have been around since the dawn of man. Videogames are hardly a big deal. Sure, 30 years ago there were none, but you could still pick up a James Bond novel and find plenty of action. Heck, there's plenty of "classical" literature out there that is based on violence, rape, murder, crime. Same with music.

I disagree with your view of Nazi soldiers. I'm with you when we talk about a certain group - mostly Nazi officers, who were sick sick people. But the large army was mostly made up of regular men - who went back to their lives after the war. They didn't exactly all disappear.

I didnt read the article yet but my opinion is that video games are a way to relieve stress, but they also influence the [i]way[/i] people relieve stress. Since we have violent video games, people relieve their stress by blowing off some guys head and etc.  If people are stressed and take it out this way, they are prone to be influenced by it, and then teh shooting of the guns becomes kind've relaxing for them in a way, relieving all the time. Which, in some cases, can lead them to believe video games are real and such and they may commit crimes because they are influenced by the games. But this only happens to the extent of what role the video games play in the persons life. As an occasional stress reliever, OCCASIONAL, and only for a short time, one will probably not be very influenced, but when a person depends on the video game for anything, that is to say it becomes their primary way to relieve stress, or of entertainment, the 'game' grows ont hem, and all the shooting desensitizes them. I think that when it gets to a point that someone sees the only way for him/her to get rid of stress is to shoot something i t hink he is too influenced and may be more likely to commit a crime.  If a game is played truly for occasional entertainment I can see no harm. As I said before, I think it all comes down to what role the video game plays in the person's life, thats how much it will influence them.

Well, I do believe that violent video games do cause kids to commit violent crimes. HOwever, I don't think it's the actual video games' fault, bu t rather the parents. Shouldn't they know what games their child is playing, there is an ESRB rating on every game, and mature games are for children older than 17. However, I do believe that children's minds do get corrupted by these games sometimes because they can't tell the difference between reality and a game, they don't know the consequences, etc. Same thing happens with television, remember that 12 year old boy a few years back that tried to imitate a wrestling move on tv on a little girl and ended up killing her. Another occurence was when a little boy pretty much acted out things he had done on Grand Theft Auto and ended up shooting a cop. Therefore, violent video games are only for kids who can handle it and are mature enough to enjoy it without thinking it's reality, and I think that the parents should be more watchful for what their kids are watching and doing.

ToBe: While I can see where your opinion is coming from, I'd also like to know if you have any specific reason, facts or figures, outside of blind guess and assumption, to believe the way you believe? I think before you respond, you should read the article.

And, from personal experience I'd say that assumption is not a very good one, at least not for me. I spent (wasted away) almost 4 years of my life playing counter-stirke everyday. I was completely addicted. I played usually 5-7 hours EVERYDAY (closer to 5 on school nights). For those of you who don't know Counter-Strike (and obviously have been living on another planet since 1999... that was a joke, feel free to laugh), it's a first person shooter where you take real life assult weapons and go around shooting and blowing people's heads off. It's a pretty violent game. I ended up stopping cold turkey, as I one day decided to just uninstall it and never play again. At that point, I didn't feel any more or less inclinded to go out and shoot some because of my experiences with the game. It was a game, nothing more. I don't think any sane, logical person can confuse that. If someone DOES confuse it, then they aren't sane! Which means it's not the games fault in the first place...

[quote="ToBe"]I didnt read the article yet but my opinion is that video games are a way to relieve stress[/quote]

You are absolutely right about this, it's not only entertainment but video games also lessen stress and in a way, this is very safe. If one day they get pissed off and decide to play a video game and as you said, "blow off the other guy's head", then that's much safer than doing something more violent in real life. HOwever, as I said before, parents should make sure what kind of games their kids are playing and if the kid is mature enough to play those kind of games.

Iversonfan,

1) Wrestling is not the same, and you cannot relate the two.

2) You provided an example of 1 incident. One hardly suggests it is a trend. I happen to remember the specific incident you are referring to, and it was the retarded media and victims who immediately pointed the finger at the video game. The boy was not "acting out the video game". He was fooling around with a gun (where did he get the gun? I would point the blame in that direction way before the game) and got scared and ended up shooting a cop. There is definitely not one shred of evidence that ever showed that he was acting out the game and decided to go shoot a cop. Vicitims have to find a reason, humans need reasons, and that was the best they had. It's the same when victims blame music on violent crimes.

You claim kids can't tell the difference between the game and reality. Well clearly, if they cannot tell that difference, then they need serious help, and certianly would not fall under my definition of "sane". And I would like to see a study that supports this view, that kids cannot in fact tell the difference, because I guarantee you cannot find any evidence of it. That is more media manipulation.

I would like to see some kind of statistic or proof for anything claim regarding violent games provoking violent crimes. Show an article, web page, document, anything. In the article I posted above, the author links to several media claims that the games are bad yet they all lack a study, facts, and statistics. It's all meaningless.

The one thing we do know is that since games have come into mainstream America, crime has dropped by 60%. And the recent group of 14-17 year olds (my group), who play the video games the most, have been the least violent of any 14-17 generation in America's entire history. Explain that.

Also, if you are sticking to the parents fault arguement... then it's not the games fault, then is it? We should not blame the game, nor take action against game makers or ban games just because of irresponsible parents, should we? How about I change my arguement: When video games are played by the proper age group (maybe even a few years under the proper age group), they do not cause violence. A 12 year old is clearly too young to be playing a Mature game, but that's not the games fault.

I dont have a solid basis, but I've observed in some cases, for example, when someone loves music, and music is his life. Depending on how the music affects him personally, and I dont think this is something that is easily gauged, he will be influenced. It could happen quickly, it could not happen at all. It all depends on how it affects him. I knew a kid who didnt listen to music much at all, then he started listening to certain types of rap, and his attitude changed, however slowly. The music grew on him, and he became a little more punk-y or whatever. I also know a kid who loves rap metal but it doesnt seem to affect him much at all - real nice guy. Thats the only proof I can point to it depending on personal 'meaning'. So, to you, Counter-Strike (yeah ive played counterstrike  - good stuff, not the point), was just an entertainment source, your hobby. Maybe every now and then you played a little rougher cuz you had a bad day, but it was just that - a game. And the fact that you stopped cold turkey is just further proof that the game didnt affect you much, whatever the reason. Some people ARE affected negatively though, and one person's experience can't say how much another will be affected or influenced.
Another note on stress, consider this scenario: Kid plays violent game to get stress out, but one day he feels so miserable and stressed out - the video game just doesnt cut it. He needs the real stuff. And that could result in a crime being committed. In the article, it proved that violence has gone down. Maybe both sides of the argument are true. Video games have taken the role of stress reliever in MOST would-be crime situations, petty crime. Most crimes are one time things, if I've learned anything from TV and other stuff, but i could be wrong, and dont reflect on the person being a bad person - he just got caught up and had no way to relieve his stress and that was the way he could think of - video games take that place. In that sense video games are good. In another, some people could be influenced by it, but its a lower percentage of people who would commit a crime otherwise. That would account for a decrease in overrall crime, but the existing crime may be supported by video games. I dont know, I dont have the data in front of me - this is all what I see, no hard, statistical data.

P.S. Honestly, I play video games all the time.. PC, Xbox, etc. So really I love video games. Thats not to say I dont see potential harm in (read: some, more easily influenced people (not saying im perfect)) playing them.

I too believe that the video games aren't causing all this youth violence. They're targeting games like "Grand Theft Auto" just because of coincidence. The acts of violence are LOOSELY similar to the story line in GTA.

You didn't see kids jumping on mushrooms and turtles after Mario Bros. came out, did you?

You didn't see kids throwing "Pokemon balls" at animals after Pokemon came out, did you?

You didn't see kids try to curse each other and have magic duels after the Harry Potter games came out, did you?

Exactly.

The article is really good, too.

You dont see kids jumping on mushrooms, throwing pokemon balls, and trying to hex each other in harry potter stuff because its not realistic and reasonable in real life. And, in kids up to like, 11 or 12, they WILL jump on mushrooms, maybe make a makeshift pokemon ball, the works. What you mean is, "you dont see it in the news" - cuz nobody cares about a kid who jumps on mushrooms. It just doesnt make news. A kid who goes to his friends house and brings a twig and tries cursing his friend (harry potter like) will get laughed at, theyll probably even start a game. But when a kid tries to imitate GTA in ways - that stuff draws attention. They haven't arrested Mario in the video games, have they?

Also, another opinion (my sister, shes an S.S. teacher) is that video games themselves dont cause violence themselves, they rather kind of affect kids who [i]are already troubled w/out the video games,[/i] kids who are already disturbed and have problems. The video games may not be good for them, but its certainly not the cause of their actions. This also fits in w/ my opinioni, in that those kids are more easily influenced.

Video games is my hobby.  Been into it for 11 years.

I feel the video game industry has been getting monotonous with these violent games, whether it would be Grand Theft Auto or Halo. My dad and I were shocked because my nephew who's 13 years old plays GTA: SA, and my dad even has to get on him for dressing like a "thug" (i.e. get him to pull his pants up). Don't get me wrong, the only "violent" thing I play would be Street Fighter with my friends, but that's different because in high pressure competitive play it's like a game of chess, you don't know what can happen next so you have to gum up your reflexes and anticipate each other. It's fantasy stuff, no guns grenades and violence made to look real in game form. To me playing SF is another fun way to stimulate my brain other than...math.

That's why not too long ago, I swore only to buy Japanese video games, because there are a lot more genres of games and some of them can also simulate real life. And these same games are more likely to get looked down upon by the group that prefer to play GTA, and throw "nerd" epithets at the people that like what I'm talking about. I can't read them, and have very little knowledge of Japanese language, but there's a very strong appeal.

[quote="ToBe"]Peter VDD, please look at atleast ONE post in the thread before commenting. This.. "mental problem" as you call it - about a fourth of the world id say atleast, has it. Video games aren't little things - and you can't comment on them as being "little" if you  yourself havent been influenced, and havent played them.[/quote]

Who says I haven't played them? I don't play GTA at the moment, as I have more interesting work to do. I used to play videogames all day before I went to university though. (and the age you're talking about is exactly that age, 8-16?)

Besides that, regarding the amount of videogame-players and the amount of criminals, I think your 1/4 estimate is a little off reality. ;) Of course some become criminals due to video games. Some people also become pyromanes by looking at firework. They would most likely have become anyway, maybe some later. I play videogames from the very beginning (commander keen, doom 1, ...) and so do most boys I know here. Looking at who got 'criminal', there is no relation with the amount of gaming.

If you want to prevent kids to turn into criminals, you'd have more effect stopping the lowest educated 5% of the population from reproducing, that would do a lot more than putting your children without their favorite game.

video games are certainly not why kids are causing crime. very few kids murder people or rob banks compared with the number of kids that actually play violent video games.

[quote="ToBe"]This.. "mental problem" as you call it - about a fourth of the world id say atleast, has it. Video games aren't little things - and you can't comment on them as being "little" if you  yourself havent been influenced, and havent played them.[/quote]

WHOA! Where do you come up with that? You just made that up on the spot... There is definitely NO evidence of any form or study that even suggests should a ridiculous statement. Statistics show that crime has dropped by more than half since video games have come about. You cannot claim that 1/4 of the world is being influenced to go commit a crime because of them.

That statement just confirms that everything you are saying is based on your imaginative believe of what "seems logical" to you. Before you develop such opinions and beliefs, you should find some kind of fact or evidence that supports it. You have none. None at all.

[quote="tetrahedr0n"]I disagree. I think the GTA series (or, at least since the 3-D versions were created) is a masterpiece of gaming. It's innovative, it's revolutionary, and like all such creations, it's controversial. Personally, it's up to the parents to enforce their own beliefs about their children's gaming. On the other hand, I do believe that there are plenty of teenagers who are mature enough to play this game without any issues, and even some younger kids. I was playing Doom when I was 9, and when I got to 11, Quake II deathmatches were my thing. I don't think of myself as somehow deeply violent because when I was 11, I fragged five thousand players.[/quote]

Yes, it is in my opinion, the best video game series (other than Final Fantasy) ever made, however, I think that it is unsuitable for children probably under the age of 15, simply because many children under that age can't tell the difference between reality and what is a game.

[quote="Peter VDD"][quote="Iversonfan2005"]I think that it is unsuitable for children probably under the age of 15, simply because many children under that age can't tell the difference between reality and what is a game.[/quote]and who told you that story? ;) do you have any evidence of this 'fact', that many kids can't separate video game vs reality?[/quote]

Obviously there's no "evidence" of this fact, but knowing that a child is still young at that age, he or she is not able to comprehend these kind of things as well as someone who is 17, for which the game is solely intended for.

Well... maybe you should check the research that has been done on this subject, instead of just stubbornly defending that what *looks* obvious to you. (or what the common folk's tales tell you, invented by grannies that are afraid of the video games)

No, video games aren't always that good, and there are more useful things you can do in life, things that are better for your social lift etc. Video games can influence that indeed if you don't watch it. But saying you'll grow criminal from that is way exaggerated. Besides, the disadvantages I stated are not violence-dependent, they're the same for almost all video games.

[quote="ToBe"][quote="Peter VDD"]I would not know what is wrong with playing GTA. I don't play it myself, but if you become a criminal because you played a video game, I think you have a little mental problem.  (since you can become a criminal by such little things)[/quote]

Peter VDD,, please look at atleast ONE post in the thread before commenting. This.. "mental problem" as you call it - about a fourth of the world id say atleast, has it. Video games aren't little things - and you can't comment on them as being "little" if you  yourself havent been influenced, and havent played them.[/quote]

I'm thinking PeterVDD was saying that only people who become criminals by playing video games have mental problems, NOT that everyone who's played a video game has mental problems.

[quote="Iversonfan2005"]but knowing that a child is still young at that age, he or she is not able to comprehend these kind of things as well as someone who is 17, for which the game is solely intended for.[/quote]

Did you consider yourself young when you were 15? I sure didn't consider myself young. In fact, I am quite confident that I was more mature at 15 than MANY 17 year olds. There was never a point in my life that any game, no matter how bloody, gory, violent or whatever that would have ever led me to commit any sort of crime. It's just a ridiculous statement you've made. There is nothing behind it. You didn't state this as even a possibility, you stated it as fact. You must be good friends with ToBe and Hello, because they like to do that too.

I think you need to have some reasons to support your opinions before you go immaturely flaunting them as any sort of fact.

The problem with this discussion is that many of you are at an age in which you really SHOULDN'T be playing M rated video games. They are M rated for a reason. It is (supposed to be) hard to buy them for a reason. No matter how "mature" you say you are, no matter how "mature" you think you are, you shouldn't be playing games like that. Someone said that these video games make you think- well, why can't you play age-appropriate video games that make you think? Why must you play these M rated games, like GTA? They are not meant for kids your age. I hate it when children try and be "grown up" too fast. Relish your childhood for Pete's sake. You shouldn't HAVE to be exposed to such horrible, violent stuff as that. Perhaps I am missing the "entertainment" value of such video games, but I really don't see why young kids (i.e. under the age-appropriate age for M rated games) should play games like that. It desensitizes you to violence and (as I am aware) to other (sexual) behaviors that are really inappropriate, no matter how "mature" you think you are. While you may not be the type to go out and DO the things you do in the games, you probably won't frown upon offenders as heavily as you would have, and therein lies the problem.

1. Bleumoose... You say young children shouldn't play these games just because. Because why? That is not the topic we have been discussing. The topic is not whether children should just "not" play these games, it's if children DO play the games, does it cause them to become more violent. The simple answer: There is absolutely NO evidence that suggests such a thing. NONE. Plain and simple.

2. I will agree that games with sexual activity and nudity should be exclusively for 17+, no matter what. No argument there.

3. However, sometimes the ESRB rating system can be completely bogus. May I remind you of the most notorious and "controversial" "M" rated game to date? Mortal Combat! Remeber those games? Mortal Combat revolutionized the world of gaming. It was an awesome game. It still IS an awesome game. It was also a game that caused more counties around the country to try to BAN it from being sold. Hello? I played Mortal Combat 1, 2, and 3 all when they first came out. I believe I was around 7 when the first one came out. My friends and I used to hold MC tournaments, and we played it all the time. Did it affect any of us? No. Looking back, it is even more absurd when I think about it. Should MC have been rated M? Heck no... Should it have been banned? Of course not... that is ridiculous. It's just a game. And a very fun game at that!

Counter-Strike? I've never played another game that taught teamwork and team strategy as well as Counter-Strike. In fact, as far as strategy goes, I don't believe there any game created to date that is even in the same league as counter-strike. Should CS be rated M? I don't think so. There is no reason for it, it's a very educational game. Especially because there is NO evidence that supports playing a game like counter-strike affects children. Anyone who says so is speculating based on either personal preconception or media manipulation.

May I remind you. The families of the victims HAVE to find an explanation that will justify the crime committer's actions. It is human nature. Therefore they seek out any reason they can possibly find. Nobody wants to believe their loved one was killed because some person was "insane". They need to be able to place blame. They need to be able to point a finger, they need to find a reason, even if that means make one up that "seems" logical. That is exactly what people have been doing with games. Like the article says, in the 50's it was rock-and-roll. Does anyone believe today that rock-and-roll causes people to commit crimes??? Rock-and-Roll was almost BANNED BY CONGRESS. Can you imagine? Games are absolutely no different. This false justification for horrible crimes by grieving families is NOT new, they have just found a new target. And for most people on this forum, games are the only thing they've lived through, so they don't see the resemblance.

Bleumoose, unless you can blurt off the ages of the people who have been responding in this thread (including myself, how old am I?), I think you should not make the implications you have made. I believe you are mistaken.

I didn't say the games should be banned, did I? I just said little kiddies shouldn't be playing them, and that's a parent responsibility thing over which I have no control. You actually proved my point in your own post. The fact that you see "nothing wrong" with the violence and gore in such games shows that you have been desensitized to it. That, in itself, is not going to make a huge huge difference in your life, I wouldn't think- I don't expect you to stand around while someone is shot and not care because of a game. However, it's still kind of disturbing to me that people can play these "games" without batting an eye. I think it is a sign of the... "good time" in this country. You never experience the "real life."

And no, I do not remember mortal combat, or any other game (other than super mario brothers, because my babysitter had a really old nintendo set). My mother did not believe video games and tv were good for me (I was raised in the Waldorf tradition), so I played with legos, and I played outside. I think I learned teamwork and team strategy very well from playing Capture the Flag, and I don't think I missed out. the only video game i ever have any interest in playing is ddr.

Arg! Firstly, I didn't prove your point. I am not desensitized to anything any more than I would have been if I never played "violent" games. Furthermore, you don't know anything about me. I am a pacifist. I think all violence is wrong. I think violence to animals and all living things is wrong. For instance, I do not kill any insect and will not stand for an insect to be killed in my presense, not even ants. I believe they have as much right to live as you do... You call that desensitized? Heh. Violence is wrong, unless it is justified (i.e. the Justice System). I am not "desensitized" in any way shape or form! If anything, I am the opposite. So, as for your analysis of me, which you based soley off my experience with gaming, was... how shall I put it... dead wrong? As you are in the general case.

I see nothing wrong with the violence because it doesn't correlate, relate to in any, or affect real life! At least, it has not been shown otherwise by research or a study. If it IS shown (from a reliable source), I would gladly change my position.

Ok, you don't have experience with games! Well that explains A LOT. It means that you have no idea what you are talking about. If you've never experienced or played these games, then you certainly have absolutely NO BASIS what-so-ever to stake your claims on. You are clearly stating the opinions and views of how you were brought up, and what other people have told you. Your opinion has no merit in this thread... at least not to me, anymore.

Also, just to clear this up. Nowhere did I say or imply that YOU said games should be banned. I referred specifically to COUNTIES wanting to ban specific games. Read before you reply.

[quote="bleumoose"]The problem with this discussion is that many of you are at an age in which you really SHOULDN'T be playing M rated video games. They are M rated for a reason. It is (supposed to be) hard to buy them for a reason. No matter how "mature" you say you are, no matter how "mature" you think you are, you shouldn't be playing games like that. Someone said that these video games make you think- well, why can't you play age-appropriate video games that make you think? Why must you play these M rated games, like GTA? They are not meant for kids your age. I hate it when children try and be "grown up" too fast. Relish your childhood for Pete's sake.[/quote]

I seriously disagree with this. I should mention that I'm "officially" old enough to play GTA (before the rating was idiotically changed of course, because, after all, we should not be exposed to sex at 17 but at 18 anything goes.)

The problem with your statement is that you assume that there exists some sort of "model" behavior that we should all indulge in at a certain age. Why can't we play age-appropriate video games that make us think? First, we play for pleasure, and if thinking is a part of that, so be it, but it's not the objective. Video games aren't scholastic tools, no matter how hard people try to change them to just that. Then there's the issue that there is no such thing as age-appropriate. Not all 7-years olds are the same, not all 15-year olds are the same.

When I was 10, Quake was the game I played. M-rated? You bet. One of the most realistic games to date at that point. The game's objective? To kill as many aliens/monster as possible. The game? It was just awesome.

The idea that I should for some reason not enjoy this game because of someone else's views is just offensive to me. The whole "desensitized" argument is void. What, since when does someone else determine how sensitive I am to anything at any age? The only thing I frown upon is people who think that they must impose their views on all others.

Are you going to provide us with evidence of your 1/4 number? I mean, personally I think it's 1/100,000, but I can't back it up.

[quote="ToBe"]Argh that came out wrong, what I meant to say is that i would see 1/4 as a reasonable estimate of gamers who are AFFECTED in any amount negatively. I would never say that 1/4 are criminals. Maybe 1/30000.... I probably should have reworded that.[/quote]
These are still blind guesses... There is NO evidence to suggest these numbers are correct, as we have stated before. You're just assuming, and I'm not really sure what you are basing your assumption off of. It doesn't matter, because you can't convince anyone with brains by just assuming statistics based on nothing.

*blinks* regardless of topic, the phrase "your opinion is void to me" is rarely, if ever, used to any effect in a discussion.
but you are right, i dont have any experience with games, personally. so what? a lot of my close friends are gamers, i think i can tell something about them and their experiences. are you going to turn into tom cruise on me? "i KNOW, YOU dont"? although that was rather hilarious. *ahem* sorry to get off topic.

Well it's true that you do not know anything about the games or the violence they contain. Thus, you have no basis for an opinion about how "bad" they are. How do you know your friends are a result of "games". You have no way of determining that. It seems to me that you never got to play these games and don't want anyone else to play them either :o.

Okay, the "void" expression was probably not the best one to use. I should have said flawed.


Why is it that when you add salt to boiling water, the temperature will increase (instead of staying constant at 100 degrees Celsius)?

Do you know about boiling point elevation?

Water boils when its vapor pressure equals the ambient pressure.  Adding salt to the water interferes with the liquid water molecules escaping the liquid and turning into vapor molecules, decreasing the vapor pressure of water.  Thus, a higher temperature is necessary to get the water's vapor pressure equal to the external vapor pressure.


this question my homework but i didnt solve  :o
[img]http://img111.imageshack.us/img111/1336/rayyur27jn3zs6.gif[/img]
Thank you...

this question very diffucult :( :(i cant solve :(

hmm...idk either it probably has somehting to do with the fact that that line is the angle bisector of Angle ABC, there might be a formula of something!  But im not sure

[hide]By the angle bisector theorem, we see that $ x/y \equal{} AB/BC$. To find the lengths $ AB$ and $ BC$ we draw an altitude from $ A$ down to a point on $ BC$ we will refer to as $ E$.

$ \triangle ABE$ is $ 30 \minus{} 60 \minus{} 90$ so we see that $ AE \equal{} AB/2$ and $ BE \equal{} AB\sqrt {3}/2$.

We use the properties of $ 30 \minus{} 60 \minus{} 90$ triangles for  $ \triangle AEC$ and find that $ EC \equal{} (x \plus{} y)/2$

and

$ AB/2 \equal{} (x \plus{} y)\sqrt {3}/2 \Rightarrow AB \equal{} \sqrt {3}(x \plus{} y)$.

Now we can substitute $ \sqrt {3}(x \plus{} y)$ for $ AB$ in the equation $ BE \equal{} AB\sqrt {3}/2$.

So $ 3(x \plus{} y)/2 \equal{} BE$. Therefore $ BC \equal{} 3(x \plus{} y)/2 \plus{} (x \plus{} y)/2 \equal{} 2(x \plus{} y)$.

$ AB/BC \equal{} \sqrt {3}(x \plus{} y)/2(x \plus{} y) \equal{} \sqrt {3}/2$.

$ x/y \equal{} \sqrt {3}/2$.[/hide]

After we write our solutions in Latex, do we upload that as an image and include it as an image in our post, or can we write out the latex solution on AOPS itself?

Edit: Fixed

[quote="Illos"]
After we write our solutions in Latex, do we upload that as an image and include it as an image in our post, or can we write out the latex solution on AOPS itself?[/quote]
Just wrap the LaTeX code in dollar signs, and AoPS will render it for you. :D

that it can't be a 30:60:90 triange.Because they give us m(A)>90  :(

Thank you...

Sorry I didn't realize that there was already a point $ D$ in the diagram. Fixed.

$ ABE$ and $ AEC$ are $ 30 \minus{} 60 \minus{} 90$ triangles.


solve this:

$x^2=2^x$

really nice , and not so much easy to solve

have fun :D

my friend gave me this equaltion , last week
he tells that this is his own equaltion :D

i prove in R it only have two results :P

I don't think that it has only two solutions in $\mathbb{R}$:
There are (simple analysis) exactle three:
- $4$
- $2$
- an irrational somewhere between $-1$ and $0$

can you explain more completely ??

i proved it only have {2,4}

The function $f(x):=2^x - x^2$ is continous on $\mathbb{R}$ and fulfills $f(-1)=- \frac{1}{2}$ and $f(0)=1$, so by the intermediate value theorem (hope it's the right term in english) there is one root between $-1$ and $0$.
Now you simply proof that a rational solution would also be an integer, but this number is not, so it's irrational.


algebra
polynomial
linear algebra
matrix
real analysis
real analysis unsolved

Find the general solutions of $ x''\minus{}2x'\minus{}3x\equal{}e^{2t}$

[quote="countdownmath"]Find the general solutions of $ x'' \minus{} 2x' \minus{} 3x \equal{} e^{2t}$[/quote]

$ y(x)\equal{}C_1e^{\minus{}t}\plus{}C_2e^{3t}\minus{}\frac{1}{3}e^{2t}$.

The associated homogeneous equation is $ x'' \minus{} 2x' \minus{} 3x \equal{} 0$, and the corresponding caracteristic polynomial is
$ k^2 \minus{} 2k \minus{} 3 \equal{} 0$, whose solutions are k = 3 and k = -1. Therefore, the general solution $ x_h$ of the homogeneous equation is $ x_h \equal{} C_1e^{3t} \plus{} C_2e^{\minus{}t}$. Let's now use Lagrange's method to find a general solution of the form $ x \equal{} C_1(t)e^{3t} \plus{} C_2(t)e^{\minus{}t}$. To do that we are to solve the following two equations: $ C'_1e^{3t} \plus{} C'_2e^{\minus{}t} \equal{} 0$ and $ 3C'_1e^{3t} \minus{} C'_2e^{\minus{}t} \equal{} e^{2t}$. In matrix notation this is written as

$ \begin{pmatrix}e^{3t}&e^{\minus{}t}\\3e^{3t}&\minus{}e^{\minus{}t}\end{pmatrix} \begin{pmatrix}C'_1\\C'_2 \end{pmatrix} \equal{}\begin{pmatrix}0\\e^{2t}\end{pmatrix}$

We thus have $ \begin{vmatrix}e^{3t}&e^{\minus{}t}\\3e^{3t}&\minus{}e^{\minus{}t}\end{vmatrix} \equal{} e^{3t} \cdot e^{\minus{}t} \begin{vmatrix}1&1\\3&\minus{}1\end{vmatrix} \equal{} \minus{}4e^{2t}$ and so

$ C'_1 \equal{} \frac{\begin{vmatrix}0&e^{\minus{}t}\\e^{2t}&\minus{}e^{\minus{}t}\end{vmatrix}}{\minus{}4e^{2t}} \equal{} \frac{\minus{}e^t}{\minus{}4e^{2t}} \equal{} \frac{1}{4} e^{\minus{}t}$

$ C'_2 \equal{} \frac{\begin{vmatrix}e^{3t}&0\\3e^{3t}&e^{2t}\end{vmatrix}}{\minus{}4e^{2t}} \equal{} \frac{e^{5t}}{\minus{}4e^{2t}} \equal{} \minus{} \frac{1}{4}e^{3t}$.

Integrating we obtain $ C_1(t) \equal{} \minus{} \frac{1}{4}e^{\minus{}t} \plus{} A$ and $ C_2(t) \equal{} \minus{} \frac{1}{12}e^{3t} \plus{} B$. Therefore, the general solution will be

$ x \equal{} (\minus{} \frac{1}{4}e^{\minus{}t} \plus{} A) \cdot e^{3t} \plus{} (\minus{} \frac{1}{12}e^{3t} \plus{} B) \cdot e^{\minus{}t} \equal{} \minus{} \frac{1}{4}e^{2t} \plus{} Ae^{3t} \minus{} \frac{1}{12}e^{2t} \plus{} Be^{\minus{}t}$

$ \equal{}> x \equal{} Ae^{3t} \plus{} Be^{\minus{}t} \minus{} \frac{1}{3}e^{2t}$.


because the inequality looks ugly at first sight, my first proof is ugly :D : just multiply with all denominators, put terms together and use muirhead. but this proof is soo ugly that i'll try to find another one.

Peter

[quote="harazi"]For any positive real numbers $ a,b,c$ we have the inequality
\[\frac{(b+c-a)^2}{3a^2+(b+c)^2}+\frac{(c+a-b)^2}{3b^2+(c+a)^2}+\frac{(a+b-c)^2}{3c^2+(a+b)^2}\geq\frac{9(a^2+b^2+c^2)}{7(a+b+c)^2}.\] Hope someone will solve them[/quote]

Here is the proof of the second one:

WLOG, we may assume $a+b+c=3$. Then the inequality becomes:

\[\frac{(3-2a)^2}{3a^2+(3-a)^2}+\frac{(3-2b)^2}{3b^2+(3-b)^2}+
\frac{(3-2c)^2}{3c^2+(3-c)^2} \geq \frac{a^2+b^2+c^2}{7}.\]

[b]Lemma[/b] For any $0\le x\le 3$, the following inequality holds:

$\displaystyle f(x) = \frac{(3-2x)^2}{3x^2+(3-x)^2} \geq \frac{1}{49}(7x^2-44x+44)$.

The result clearly follows from the lemma since $\displaystyle f(a)+f(b)+f(c) \geq \frac{1}{49}(7(a^2+b^2+c^2) - 44(a+b+c) + 44\cdot3)=\frac{a^2+b^2+c^2}{7}$.

Proof of the Lemma: Let $g(x) = (7x^2 - 44x + 44)(3x^2 +  (3 - x)^2) -  49 (3 - 2x)^2$. Simplifying and factoring, we get $g(x) = (x-1)^2 (28x^2-162x-45)$. It remains to show: $h(x) = 28x^2-162x-45 \leq 0$ when $0\le x\le 3$. But this is almost trivial. Since $h(x)$ is convex, $h(x) \leq \max\{h(0),h(3)\} < 0$.

I would like to see your solution, Harazi.

Thank you very much.

Harazi, pls post your solution now atleast after so many years.

Debanik

[quote="harazi"]Here is an ugly, but strong and interesting inequality. I have a nice solution for it. It says that for any positive real numbers $ a,b,c$ we have
    $\frac{(b+c-a)^2}{17a^2+7(b+c)^2}+\frac{(c+a-b)^2}{17b^2+7(c+a)^2}+\frac{(a+b-c)^2}{17c^2+7(a+b)^2}\geq\frac{a^2+b^2+c^2}{5(a+b+c)^2} $.
     A nicer inequality, based on the same idea is the following:
     For any positive real numbers $ a,b,c$ we have the inequality
     $\frac{(b+c-a)^2}{3a^2+(b+c)^2}+\frac{(c+a-b)^2}{3b^2+(c+a)^2}+\frac{(a+b-c)^2}{3c^2+(a+b)^2}\geq\frac{9(a^2+b^2+c^2)}{7(a+b+c)^2}$. Hope someone will solve them[/quote]
We can prove easily by using cauchy-schwarzt inequality:
$VT= \sum \frac{(b+c)^2(b+c-a)^2}{(b+c)^2(17a^2+7(b+c)^2)} \ge \frac{4(a^2+b^2+c^2)^2}{\sum (b+c)^2(17a^2+7(b+c)^2}$
Finally,we only need to prove that:
$20(a^2+b^2+c^2)(a+b+c)^2 \ge \sum 17a^2(b+c)^2+ \sum 7(b+c)^4$
<=>$a^4+b^4+c^4+abc(a+b+c)-\sum ab(a^2+b^2)+\sum 3ab(a-b)^2 \ge 0$,which is true by schur inequality.
Our proof are completed.Equality occur if and if only a=b=c,or a=b,c=0 or any cyclic permution.
We can prove the similar with the same way :
   [quote]For any positive real numbers $ a,b,c$ we have the inequality
     $\frac{(b+c-a)^2}{3a^2+(b+c)^2}+\frac{(c+a-b)^2}{3b^2+(c+a)^2}+\frac{(a+b-c)^2}{3c^2+(a+b)^2}\geq\frac{9(a^2+b^2+c^2)}{7(a+b+c)^2}$. Hope someone will solve them[/quote]

Can you please post your solution, harazi ?
By the way, here :
[quote="fuzzylogic"]
Assume that $a+b+c=3$. [/quote]
How do you choose the sum you can assume WLOG ? We could pretty much have assumed $a+b+c=1$ or something else...


The equation $ x^{3}+px^{2}+qx+r = 0$ (where p,q,r are non zero) has roots $ \alpha,\beta,\gamma$ such that $ \frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}$ are consecutive arithmetic terms in an sequence. Show that $ beta = \frac{-3r}{q}$

(ii) The eqn $ x^{3}-26x^{2}+216x-576 = 0$ has roots $ \alpha,\beta,\gamma$ such that $ \frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}$ are consecutive arithmetic terms in an sequence. Find the value of $ \alpha,\beta,\gamma$

[hide="solution"]$ \frac{1}{\alpha}+\frac{1}{\gamma}=\frac{2}{\beta}$
Since they are in arithmetical order.

Multiplying it all out we have $ \gamma \beta+\alpha \beta =2 \alpha \gamma$

Now by Viete we have $ \gamma \beta+\alpha \beta+\alpha \gamma = q$

Putting the two together we get $ \alpha \gamma=\frac{q}{3}$

Also by Viete $ \alpha \beta \gamma =-r$

Putting these two together gives the result.

For part 2, subbing in $ q$ and $ r$ in our above formula gives us $ \beta =-8$

From this you can easily get the other 2 roots.[/hide]


Yes, I've looked all over google to no avail. This is really annoying me.

About two days ago, LaTeX gave me an error message whenever I pressed F5. I get this error:

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")] Cannot execute the command.

I really don't want to manually open up the file in acrobat reader. Help would be much appreciated. Thanks!

This happens if you have upgraded Adobe Reader as the path to the program has changed. Go to Build, Define Output Profiles, LaTeX=>PDF, Viewer and browse to find the correct path.

Sigh... if it only was that simple. I have the correct path I believe: C:\Program Files\Adobe\Reader 8.0\Reader\AcroRd32.exe, so that's not the issue.

Here's something that may clarify the situation. You know how you can always compile successfully, even when the pdf is open, and then press F5 to get the new copy? When that happens to me, I get the error that it can't write over my open pdf.

It could be the infamous Adobe Reader version 8.0 problem. There's a long discussion about it [url=http://www.latex-community.org/forum/viewtopic.php?f=31&t=14&start=0]here[/url] so you could try the workarounds there, or upgrade Adobe Reader or use a different (free) pdf reader such as [url=http://www.foxitsoftware.com/pdf/reader/]Foxit[/url] or [url=http://blog.kowalczyk.info/software/sumatrapdf/index.html]Sumatra[/url].

have you written something after putting \begin{document}...?

because if you haven't even written nothing, then that produces the error, you need to write something.

Yes, I have begin and end document. This problem applies to EVERY one of my tex files. And the choice of PDF reader doesn't matter.

[quote="Caelestor"]Ok, there's definitely something wrong with Texniccenter. Sumatra doesn't work either.[/quote]Show us a screen dump of your Viewer tab; there could be an error there.

Take the "%" (percent symbol) out of the file name if you have one. Then try it again.

I'm also having this problem at the moment... I've looked at the possible sources of error given in this thread, and none of them apply to me... hmm

Actually, I stopped asking because it worked again.  :o

Just follow everyone's advice, then wait and hope for the best.


function
integration
trigonometry
complex analysis
complex analysis unsolved

Prove that the function $f(z)=\int\limits_{0}^{1}\frac{\sin tz}{t}dt$ is entire

(Edited) Let $C$ be a closed curve around 0. We have, $\int_{C}\int_{0}^{1}\frac{\sin(tz)}{t}dt dz = \int_{0}^{1}\int_{C}\frac{\sin(tz)}{t}dz dt = \int_{0}^{1}0 dt = 0$ since $\sin(tz)/t$ is entire. The reult now follows by Morera's theorem.

$\frac{\sin tz}t=\sum_{k=0}^{\infty}\frac{(-1)^{k}t^{2k}z^{2k+1}}{(2k+1)!}$

Now integrate that to find that

$f(z)=\sum_{k=0}^{\infty}\frac{(-1)^{k}z^{2k+1}}{(2k+1) (2k+1)!}$

Since this power series has an infinite radius of convergence, it represents an entire function.

--

Ottem - two things: One, it's spelled Morera's theorem. Two, which variable do you want in the denominator of your fraction?

Right you are Kent. I think it works now though.. :)

Thanks, Otten. Note that it doesn't matter whether your curve $C$ encloses zero or not.

Slight generalization: Let $g(t,z)$ be continuous for all $t$ and $z,$ an entire function of $z$ for each fixed $t,$ and an entire function of $t$ for each fixed $z.$ [Note that $g(t,z)=\frac{\sin tx}t$ satisfies this.] Let $\gamma$ be a bounded rectifiable curve (not necessarily closed). [In this case that would be the interval $[0,1].$]

Define $f(z)=\int_{\gamma}g(t,z)\,dt.$ Then $f$ is entire.

Have fun deciding which of these hypotheses are stronger than you really need and how far you can weaken them.


inequalities
LaTeX
inequalities unsolved

For $a,b,c >0$ prove the inequality

$\sqrt{\frac{a}{b+c}} +\sqrt{\frac{b}{c+a}} +\sqrt{\frac{c}{a+b}} +\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}} \geq \frac{9\sqrt 2}{2}$

please, forgive me: i can't use latex (and i'm too lazy to learn)...
so, let's define f(x) = :sqrt: (x/(1-x)) + :sqrt: ((1-x)/x) = 1/ :sqrt: (x-x).
by easy calculation, we have f''(x) > 0 for x belonging to ]0,1[.
so, dividing a,b,c by (a+b+c), we have that a,b,c belong to ]0,1[ and sum to 1.
so we can apply jensen's inequality, to get
f(a') + f(b') + f(c') >= 3f(1/3).
but the left side of inequality is our sum, and the right side is properly 9/ :sqrt: 2

[quote="treegoner"]Why don't we use Cauchy for 9 variables ? :D[/quote]

Dear Treegoner , can you please explain it to me?

Thank you very much.

${6*( \sqrt{ \sqrt{\frac{a}{b+c}} *\sqrt{\frac{b}{c+a}} *\sqrt{\frac{c}{a+b}} *\sqrt{\frac{b+c}{a}}*\sqrt{\frac{c+a}{b}}*\sqrt{\frac{a+b}{c}}}) <=
\sqrt{\frac{a}{b+c}} +\sqrt{\frac{b}{c+a}} +\sqrt{\frac{c}{a+b}} +\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+*\sqrt{\frac{a+b}{c}}}   $

$ ==> $
${ \sqrt{\frac{a}{b+c}} +\sqrt{\frac{b}{c+a}} +\sqrt{\frac{c}{a+b}} +\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+*\sqrt{\frac{a+b}{c}}}    >= 6  $
we know that 6 <= 9/sqrt(2)
so the inequation is true

[quote="various"]we know that 6 <= 9/sqrt(2)
so the inequation is true[/quote]

No.
sign reversed.


modular arithmetic
number theory unsolved
number theory

Let $m,n$ be positive integers. It is given that: $A=\frac{(m+3)^{n}+1}{3m}\in{\mathbb{N}}$. Prove that $A$ is odd.

well some thoughts: first iff $n$ is odd.

iff $m$ is odd, there is nothing to do, Iff m is $8k$, then we have $8|3^n+1$ but this never happens, so in this case we get $m=8k+4$, or $m=8k+2$. iff $m=8k+2$, then the numerator and denominator are both $2$ mod 4, so $A$ is odd.

So we get the final case $m=8k+4$, since $3|m^n+1$, we get $m=-1$ (mod 3), so there is a prime dividing m, different from $2$, such that $p=3k+2$. Now, $p|3^n+1$, i means, since $n$ is odd, that $3$ is a square modulo $p$, iff and only iff $-1$ is also, taking Legendre symbols, and using the law of cuadratic reciprocity we easily get a contradiction...

Now for the secound case $n$ even, we get also to study when $m$ is even, we find $(3+m)^n+1=2$ mod 3, so this is not the case....

Yes Pascual, it's Bul 1998. By the way, i think we can show that in fact $m\equiv2\pmod{12}$ if $A$ is integer. We can prove this result firstly and the the odd property is a simple corollary!

I dont know, if you wan maybe you can do it!

[quote="Pascual2005"]
Now, $p|3^n+1$, i means, since $n$ is odd, that $3$ is a square modulo $p$, iff and only iff $-1$ is also, taking Legendre symbols, and using the law of cuadratic reciprocity we easily get a contradiction...[/quote]
Can you, please, write it in more details?


How do I find the inverse of a function that is defined by two parts.
The function is $ \minus{}x^2$ if $ x\ge 0$ and $ 1\minus{}x^3$ if $ x<0$.

How do I find the inverse of this function?
Do I still express it in terms of x and y and then switch x to y and vice verse  and solve for y?

[hide]if $ x\leq0$, $ f^{\minus{}1}(x)\equal{}\sqrt{\minus{}x}$. If $ x>1$, $ f^{\minus{}1}(x)\equal{}\sqrt[3]{1\minus{}x}$

You find which domain values correspond to each range value, basically.[/hide]

Draw a picture of the function.  Then draw a picture of the inverse function (assuming it really is a function).  Then use what you know about the function to write down the inverse, piecewise.


pagio, αν ξερεις την ανισοτητα αριθμητικου-γεωμετρικου μεσου βγαινει αμεσως ,
ή αλλιως κατευθειαν απο την ταυτοτητα του  Euler οπου το β' μελος ειναι  θετικο η μηδεν. ομως ....
νομιζω οτι ειναι ευκαιρια να πουμε δυο λογια για την ομογενεια μιας συναρτησης ,ας  το κανει καποιος

Παρατήρησε ότι:

$a^{3}+b^{3}+c^{3}-3abc=\frac{1}{2}(a+b+c)[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]\geq 0$

Δηλαδή, αφού έχουμε τετράγωνα, άρα το γίνομενο είναι θετικός επί θετικό, που μας δίνει θετικό ή ίσο με το μηδέν.

[u]Σημείωση[/u]: Αν $a+b+c=0$, τότε $a^{3}+b^{3}+c^{3}=3abc$

[i]Πάνω-κάτω είναι αυτό που είπε ο Κώστας, απλώς στο έδειξα πως παρουσιάζεται ως ανισότητα με σύμβολα.[/i]

[i][color=blue]Α! Τώρα που το είδα, δεν είναι η ανισότητα Cauchy αλλά ΑΜ-ΓΜ![/color][/i]

Ακόμη με ισοδυναμία έχουμε

$a^{3}+b^{3}+c^{3}=3abc$ αν και μόνο αν $a+b+c=0$ ή $a=b=c$

[quote="pagio"]Διαβάζω το βιβλίο του κυρίου Στεργίου και έχω μπερδευτεί με την απόδειξη αυτής της ανισότητας(Cauchy νομίζω...)

a^3+b^3+c^3>=3abc

Θα μπορούσε κάποιος να μου γράψει την απόδειξη??????????[/quote]

[color=red]Pagio ,

 καλώς ήρθες στο forum ! [/color]

 Θα σε συνιστούσα να μη διαβάζεις αποδείξεις ανισοτήτων σε αυτή τη φάση , αλλά μόνο τις εφαρμογές τους !Έτσι κερδίζεις χρόνο. Τις αποδείξεις μπορείς να τις μελετήσεις σε άλλη φάση !

Καλή Επιτυχία !!!

[quote="mostel"] [i][color=blue]Α! Τώρα που το είδα, δεν είναι η ανισότητα Cauchy αλλά ΑΜ-ΓΜ![/color][/i][/quote]

Στο Vietnam Cauchy τη λένε.... :wink:
Και μια παρατήρηση :
Η διαφορά μεταξύ της ΑΜ-ΓΜ για τρεις όρους και της παράπω
$a^{3}+b^{3}+c^{3}\geq 3abc$ είναι ότι η μεν πρώτη ισχύει για μη αρνητικούς αριθμούς ενώ η δεύτερη ισχεύει για όλους τους πραγματικούς με $a+b+c\geq 0$ (η δεύτερη είναι πιο ισχυρή  :wink: )

Ερώτηση Α'
Ποιά είναι η ταυτότητα του Euler?????

Ερώτηση Β'(Προς κ.Στεργίου)
Αν πρέπει να την χρησιμοποίησω π.χ. στον αρχιμίδη δεν Θα πρέπει να την αποδείξω?????????

Ερώτηση Γ'
Σε αυτή την ανισότητα πως φτάνουμε?????

a^3+b^3+c^3 - 3abc >=1/2(α+β+c){(a+b)^2+(b+c)^2+(a+c)^2}

[quote="pagio"]
Άσχετο αλλά πως προέκυψε αυτή η ανισότητα??????
Μπορούμε να φτάσουμε από το α στο β μέλος????[/quote]

Καταρχάς δεν είναι ανισότητα είναι [u]ισότητα[/u] .
Αν κάνεις πράξεις στο δεύτερο μελος θα σου βγει το πρώτο. :)

[quote="pagio"][quote="silouan"]Ταυτότητα  Euler
$a^{3}+b^{3}+c^{3}-3abc=\frac{1}{2}(a+b+c)((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$

Αυτές είναι πολύ γνωστές οπότε δεν χρειάζεται απόδειξη.[/quote]

Άσχετο αλλά πως προέκυψε αυτή η EDIT(ισότητα)??????
Μπορούμε να φτάσουμε από το α στο β μέλος????

PM(Πάνω στην βιασύνη μου Silouan..........)[/quote]

 Pagio !
 Παραξενεύτηκα και εγώ στην αρχή , ψάχνοντας που θα μπορούσε να μπερδευτείς στην απόδειξη. Νόμισα ότι υπήρχε κάποιο τυπογραφικό λάθος ή καμιά ασυναρτησία . Γι αυτό  έλεγξα την απόδειξη και δε βρήκα κανένα κενό .Βλέπω όμως ότι το πρόβλημα ξεκινά από την ταυτότητα Euler που σου διαφεύγει. Δυστυχώς , οι οδηγίες του παιδαγωγικού ινστιτούτου λένε να μην τη διδάσκουν οι καθηγητές στην Α΄Λυκείου στους μαθητές. Αυτό είναι μια βλακώδης οδηγία και δυστυχώς πολλοί καθηγητές την ακολουθούνε. Έτσι , έπεσες και εσύ θύμα αυτής της κατάστασης ισοπέδωσης της μαθηματικής παιδείας προς τα κάτω.
   Τώρα που την έμαθες προσπάθησε και να την αποδείξεις: α) με τον τρόπο που σου είπε ο Σιλουανός(για να σε διευκολύνει) β) πηγαίνοντας από το α΄μέλος στο β΄( για να χρησιμποποιήσεις και μια άλλη ταυτότητα που ίσως σου διαφεύγει . Αν βρεις δυσκολία , εδώ είμαστε για να σου δώσουμε τη λύση !

     [u]Μπάμπης[/u]

 [b]ΣΧΟΛΙΟ[/b]

 [color=blue]Στον ΑΡΧΙΜΗΔΗ δε χρειάζεται να αποδείξεις καμία ταυτότητα ή ανισότητα, από αυτές που στα μαθηματικά θεωρούν Κλασσικές , ακόμα και αν δεν την έχεις μάθει στο σχολείο. ![/color]

[quote="stergiu"]  [color=blue]Στον ΑΡΧΙΜΗΔΗ δε χρειάζεται να αποδείξεις καμία ταυτότητα ή ανισότητα, από αυτές που στα μαθηματικά θεωρούν Κλασσικές , ακόμα και αν δεν την έχεις μάθει στο σχολείο. ![/color][/quote]

Κύριε Μπάμπη αυτό δυστυχώς χρειάζεται μεγάλη συζήτηση και διευκρινησεις .
Να δώσω ένα απλό παράδειγμα ....
Η ανισότητα Ιραν 1996 (πολύ γνωστή αλλά και δύσκολη) (δεν λύνεται από όσο ξέρω τουλάχιστον κλασσικά ) λύνεται με απαλοιφή Muirhead και Schur . (Μια άλλη λύση είναι με SOS λίγο πιο ωραία ) . Το ζήτημα είναι :
Μπορείς να χρησιμοποιήσεις Muirhead ?  :maybe:
(Γιατί αν χρειάζεται απόδειξη αφείστε το  :D )

[quote="pagio"]Ευχαριστω πολύ κύριε Στεργίου για το ενδιαφέρον σας.

Και συγχαριτήρια για το βιβλίο σας.

PM.Την βρήκα την απόδειξη(από α σε β μέλος)....[/quote]

Εύγε !
 α)Οι ταυτότητες είναι πολύ χρήσιμα εργαλεία. Αυτοί που τις ανακάλυψαν , τις βρήκαν από ανάγκη και όχι στην τύχη.Αξίζει να γνωρίζεις μερικές απ'έξω(πχ την de Moivre κλπ)

β) Thanks , but  .... το βιβλίο το έγραψα με το Σκομπρή και τόσο οι ευθύνες όσο και οι έπαινοι πρέπει να μοιράζονται  :) !

  γ) Σιλουανέ , το έχουμε ξαναπεί . Τα επώνυμα θεωρήματα θεωρούνται γνωστά και κανένας δεν κόβει μονάδες , επειδή δεν αποδεικνύονται στο γραπτό.Στην  ΙΜΟ , αν βάλουν άσκηση που στηρίζεται σε τέτοιο θεώρημα ,
 είναι προφανές ότι θα το δεχτούν. Στις εθνικές ολυμπιάδες  κάνει ο καθένας ότι θέλει. Αλλά στις ΙΜΟ επιδιώκουν θέματα που να λύνονται ΚΑΙ με τελείως στοιχειώδη τρόπο. Αλλά η ευθεία Simson για παράδειγμα και το θεώρημα CEVA είναι στοιχειώδη θεωρήματα. Ας πας με το καλό στην ΙΜΟ και λύστα με όποιον τρόπο θέλεις ! Για αυτό υπάρχουν οι συνοδοί καθηγητές . Για να κατοχυρώσουν τις σωστές λύσεις των μαθητών τους , σε περίπτωση που κάποιος εξεταστής κάνει λάθος. Θυμίσου να ρωτήσουμε και τον DULAC για το θέμα αυτό !
 ΜΠΑΜΠΗΣ


Does there exist an infinite topological space $ X$ with only finitely many homeomorphisms into itself?

Take $ \mathbb N$ with the closed subsets $ [0,n]\equal{}\{0,\ldots,n\}$ for every $ n\in\mathbb N$. Then every homeomorphism is forced to leave the finite subsets invariant using a finite counting argument. Because of the order structure of the closed sets this implies that a homeomoprhism is the identity on $ \mathbb N$.

How about an infinite Hausdorff space with finitely many automorphisms then? ;)

i wonder if there's anything less ugly..

consider $ S^1$ as the interval $ [0;1]$ with endpoints identificated. we're building a sequence of discrete sets $ \{A_k\}_{k\ge 2}\subset S^1$ in the following way: $ A_2$ consists of two points $ 0,1/2$. it divides $ S^1$ into two connected components. we're defining $ A_3,A_4$ to be each contained in one of them, consisting of 3 and 4 points respectively, such that they split each component in 4 and 5 components respectively. now we construct $ A_5,\dots,A_13$ in each of the 9 components, made of 5,...,13 points well-spatiated. we finally define $ X\equal{}S^1/\sim$ where $ \sim$ identifies points in $ A_k$ to a single point $ a_k$.

i claim that $ X$ satisfies jmerry's request.
first of all, $ X$ is hausdorff, since every orbit shrinks (is there a general theorem we could quote here? i think i've proved it by hand).
then, the set $ D\equal{}\{a_k\}$ is dense, because the union of the $ A_k$ is. but also the number of connected components of $ X\setminus\{a_k\}$ is $ k$, because further identifications don't mess with the components we've already split. this forces every homeomorphism of $ X$ to fix $ D$, and hence to fix all of $ X$.


When we draw a line from $ A$ to $ B,$ we get a triangle with interior angles $ 150^{\circ}\minus{}90^{\circ}\equal{}60^{\circ}$ and $ 120^{\circ}\minus{}90^{\circ}\equal{}30^{\circ}.$ The last angle must thus be $ 180^{\circ}\minus{}60^{\circ}\minus{}30^{\circ}\equal{}\boxed{90^{\circ}}.$


Consider the sequence defined recursively by $a_1=a$ (any positive numbers) and $a_{n+1} = \frac {-1}{a_n + 1}$, $n=1,2,3..$. For which of the following values of n must $a_n = a$?

A) 14
B) 15
C) 16
D) 17
E) 18

P.S. This question is easier than it looks.

Since no one seems to trying this, I'll give hint:

Say that $a_1 = 1$, then, what is $a_2,a_3,a_4,a_5$?

You should find a pattern there.

Sequence problem often makes it harder than what they actually are.

Using Silverfalcon's hint
[hide]$a_1 = 1 \\
a_2 = \frac{-1}{2} \\
a_3 = -2 \\
a_4 = 1$

The pattern will keep repeating since there's a numver that was already in the pattern.  Looking at the sequence, we can say that if $n \equiv 1\pmod 3$, then $a_n =a$.  The only number in the answer choices that fits this statement is 16, or (c). [/hide]

Can you prove, or at least satisfy yourself, that your solution will work for any starting value of $a$?

[hide]We can just substitute $a$ for $1$, getting this -

$a_1 = a \\
a_2 =  \frac{-1}{a+1} \\
a_3 = \frac{-a-1}{a} \\
a_4 = a$

I tried that at first, but stupid mistakes made me get really complicated fractions.[/hide]

[quote="Silverfalcon"]To be honest, proving is unnecessarily because the question already gave it out.[/quote]

Good point ;-)

I was thinking of a solution like h_s_potter2002's but you are right that it is unnecessary.


By treatment of each one of two isomeric hydrocarbons Y and Z with basic hot potassium permanganate, followed by acidification, we obtain as the only products acetone and carbon dioxide, but in different proportions for the two compounds - 1:1 for Y and 1:5 for Z. What are the structures of Y and Z?

Z is 6-methyl hepta-1,2,3,4,5-pentaene and Y is 2,5-dimethyl hexa-2,3,4-triene

Does that answer look reasonable to you?


calculus
integration
function
geometry
rotation
calculus computations

I've been stuck on the concepts of using integrals to find the volume of certain shapes and such.

Let R be the region given by 0 < x < pi and 0 < y < sin^2x
Find the volume of the solid obtained by revolving R around the x and y axis.
My apologies for not being able to put in the proper symbols.

For the one around the x axis, instead of integrating the function, you want to integrate the discs that the function would create.

Remember that the area of a circle is $ \pi r^2$
The radius of your discs is the height of the function, or f(x)
So you would integrate: $ \int\pi [f(x)]^2 dx$

Around the y axis, instead of integrating the function you integrate the shells that it would make around the y axis. Remember that the surface area of a shell is $ 2\pi r h$
The height of your shell is the height of your function f(x)
The radius of your shell is the distance from the y axis, or x

So this time you would integrate: $ \int2\pi x * f(x) dx$

does this help?

Yeah, that clears things up a bit for me. However... what do I consider the function to be?

I've been working on the problem and I was wondering if someone could help me rearrange the equation y=sin^2 x to x = something?

Instead of rearranging the equation, I suggest you use the shell method.

The shell method allows you to rotate around the y axis while still keeping your equation as a function of x.
Imagine that you are making shells around the y axis and you will see how it works.

If you really must rearrange the equation, you can square root both sides and then take the arcsin


Find all $ f: \mathbb R\to\mathbb R$ such that \[ f(x)f(yf(x)-1)=x^2f(y)-f(x)\] for all $x,y\in\mathbb R$.

[quote="MJ GEO"]ANY BODY????? WHERE ARE YOU??????[/quote]

Huh !!!!

You make an up on a topic after $ 23$ minutes ??????

Wait at least $ 4$ or $ 5$ days before first up.
 :rotfl:

And if you want some mathlinkers find interest in your posts :
1) use Latex
2) thank those who gave you hints / solution / attention

[quote="MJ GEO"]Please help me.I need the solution of this quastion today.its realy important for me.[/quote]

You posted in "Algebra Proposed & own problems" so you are supposed to already have the solution.

If you dont have the solution, post, next time (dont post twice this one), in "unsolved problems"

[quote="MJ GEO"]Find all functions $ f(x)$ from $ \mathbb R\to\mathbb R$ such that $ f(x)f(yf(x)\minus{}1)\equal{}x^2f(y)\minus{}f(x)$ $ \forall x,y$[/quote]

Let $ P(x,y)$ be the assertion $ f(x)f(yf(x)\minus{}1)\equal{}x^2f(y)\minus{}f(x)$

$ P(1,1)$ $ \implies$ $ f(1)f(f(1)\minus{}1)\equal{}0$. Then :
If $ f(1)\equal{}0$ : $ P(1,x)$ $ \implies$ $ 0\equal{}f(x)$ and we got the solution $ f(x)\equal{}0$ $ \forall x$
If $ f(1)\ne 0$ and $ f(1)\ne 1$ then let $ a\equal{}f(1)\minus{}1\ne 0$. We got $ f(a)\equal{}0$ and so $ P(a,x)$ $ \implies$ $ 0\equal{}a^2f(x)$ and we got again the solution $ f(x)\equal{}0$ $ \forall x$

So we'll now consider $ f(1)\equal{}1$
$ P(1,x)$ $ \implies$ $ f(x\minus{}1)\equal{}f(x)\minus{}1$ and so $ f(n)\equal{}n$ $ \forall n\in\mathbb Z$

If $ f(u)\equal{}0$ then $ P(u,1)$ $ \implies$ $ u^2\equal{}0$  and so $ u\equal{}0$

Then $ f(yf(x)\minus{}1)\equal{}f(xf(x))\minus{}1$ and $ P(x,y)$ becomes : $ Q(x,y)$ : $ f(x)f(yf(x))\equal{}x^2f(y)$

$ Q(x,1)$ $ \implies$ $ f(x)f(f(x))\equal{}x^2$ and so $ f(f(x))\equal{}\frac {x^2}{f(x)}$ $ \forall x\ne 0$

Let $ x\ne 0$ then $ Q(x\plus{}1,1)$ $ \implies$ $ f(x\plus{}1)f(f(x\plus{}1))\equal{}(x\plus{}1)^2$ and so $ (f(x)\plus{}1)(f(f(x))\plus{}1)\equal{}(x\plus{}1)^2$ and so $ (f(x)\plus{}1)(\frac {x^2}{f(x)}\plus{}1)\equal{}(x\plus{}1)^2$

$ \implies$ $ (f(x)\plus{}1)(x^2\plus{}f(x))\equal{}f(x)(x^2\plus{}2x\plus{}1)$

$ \implies$ $ (x^2\plus{}1)f(x) \plus{} f(x)^2\plus{}x^2\equal{}f(x)(x^2\plus{}2x\plus{}1)$

$ \implies$ $ f(x)^2\minus{}2xf(x)\plus{}x^2\equal{}0$

$ \implies$ $ (f(x)\minus{}x)^2\equal{}0$ $ \implies$ $ f(x)\equal{}x$, which, indeed, is a solution

Hence the two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$

thank you for this nice and lovely solution :lol:


If $a,b,c$ positive reals such that $abc=1$,
then prove that:
$\frac{(1+ab)^2+(1+bc)^2+(1+ca)^2}{4} \geq \frac{ab+bc+ca+2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{ab+bc+ca+a+b+c-3}$


limit
induction
real analysis
real analysis unsolved

Let $x_n$ be a bounded sequence such that $\lim_{n\to\infty} (x_n+\frac{1}{3}x_{n^2})=1$. Prove that $x_n$ converges.

suppose that $x_{f(n)}$ is a subsequence that tends towards $L$. Then $x_{f(n)^2}$ is a subsequence that tends towards $F(L)$, where $F(x)=3(1-x)$. Induction shows that one can find a subsequence that tends towards $F^{(k)}(L)=F(F(..(L)..))$ for any $k \geq 0$. But a sequence $y_{n+1}=F(y_n)$ is either constant or unbounded. So one can conclude that $x_n$ converges towards the fixed point of $F$.


point set
combinatorics
combinatorial geometry
Coloring
IMO Shortlist

Ten points are marked in the plane so that no three of them lie on a line. Each pair of points is connected with a segment. Each of these segments is painted with one of $k$ colors, in such a way that for any $k$ of the ten points, there are $k$ segments each joining two of them and no two being painted with the same color. Determine all integers $k$, $1\leq k\leq 10$, for which this is possible.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[hide]
First, the cases $k \leq 3$ and $k \geq 6$ are disposed of easily.

Case 1. $k \leq 3$. $k=1, 2$ are obviously impossible. If $k = 3$, note that some color colors at most $\frac{1}{3} \binom{10}{2} = 15$ edges. But in any 10-graph of 15 edges there is an empty triangle, so this case is impossible.

Case 2. $k \geq 6$. For each of $k$ colors, we choose its edges as follows. Split the 10 vertices into $k-1$ groups of 1 or 2. Connect any pair of vertices in the same group. Such a grouping is chosen for each color such that no pair of vertices occurs in the same group for different colors, which is easy. For $k = 6$ a tiny bit extra needs to be done but this is not difficult. $k \geq 6$ is possible.

Case 3. $k = 5$. Let the vertices be numbered $i$, $i = 0, 1, ... ,9$. Consider the following coloring scheme:

Color $1: 0123, 45, 67, 89$
Color $2: 047, 158, 269, 3$
Color $3: 059, 146, 378, 2$
Color $4: 068, 257, 349, 1$
Color $5: 179, 248, 356, 0$

In each string of digits, it means each of these vertices was connected to every other digit of the string with that color, and to no others. $5$ works.

Case 4. $k = 4$. We show that at least 12 edges are needed in a 10-graph to guarantee that no empty quadrilaterals exist for each color, so that at least 48 edges are needed in total, a contradiction since there are only 45.

It is easy to show there is an empty triangle when at most 11 edges are present in a 10-graph. Now, start with this triangle. Every other (out of 7) vertex must be connected to one of its vertices, for 7 edges. Of the 7 other vertices, there must be 5 edges between themselves, or there would be an empty quadrilateral. Therefore 11 edges is impossible, and 12 are required. $k \neq 4$.

Hence the answer is $5 \leq k \leq 10$.

[/hide]

What a problem. Here's my construction for $k=5$ in its full glory.

[asy]
size(8cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6;
pen reddraw,greendraw,bluedraw,purpledraw;
reddraw = RGB(255,85,85);
greendraw = RGB(0,187,0);
bluedraw = RGB(0,102,255);
purpledraw = RGB(170,34,255);
pair A1,A2,A3,A4,A5,A6,A7,A8,A9,A10;
A1 = dir(288);
A2 = dir(324);
A3 = dir(0);
A4 = dir(36);
A5 = dir(72);
A6 = dir(108);
A7 = dir(144);
A8 = dir(180);
A9 = dir(216);
A10 = dir(252);
draw(A2--A3--A4--cycle,reddraw);
draw(A7--A8--A9--cycle,reddraw);
draw(A1--A6,reddraw);
draw(A5--A10,reddraw);
draw(A1--A3--A5--cycle,purpledraw);
draw(A6--A8--A10--cycle,purpledraw);
draw(A2--A9,purpledraw);
draw(A4--A7,purpledraw);
draw(A2--A5--A8--cycle,bluedraw);
draw(A3--A7--A10--cycle,bluedraw);
draw(A1--A9,bluedraw);
draw(A4--A6,bluedraw);
draw(A1--A4--A8--cycle,greendraw);
draw(A3--A6--A9--cycle,greendraw);
draw(A2--A10,greendraw);
draw(A5--A7,greendraw);
draw(A3--A8);
draw(A9--A10--A1--A2);
draw(A4--A5--A6--A7);
draw(A1--A7--A2--A6);
draw(A10--A4--A9--A5);
dot("$A_1$",A1,dir(288));
dot("$A_2$",A2,dir(324));
dot("$A_3$",A3,dir(0));
dot("$A_4$",A4,dir(36));
dot("$A_5$",A5,dir(72));
dot("$A_6$",A6,dir(108));
dot("$A_7$",A7,dir(144));
dot("$A_8$",A8,dir(180));
dot("$A_9$",A9,dir(216));
dot("$A_{10}$",A10,dir(252));
[/asy]

The answer is $k\ge 5$. If $k\le 2$, for every $k$ of the points, there are less than $k$ total segments, so this is not possible. Note that there are a total of $45$ line segments.
\\$~$\\
If $k=3$ then there is a color that appears at most $15$ times, $C$. Suppose for every set of three points, there is at least one edge colored $C$. Now, sum the number of edges $C$ for all sets of three points. Since each edge is counted eight times, we should receive at most $120$. Now, there are a total of $120$ sets of three points, so each set of three points must contain exactly one edge $C$ and there must be exactly $15$ edges $C$. Consider any edge $C$, then neither vertex at the endpoints of this edge can be connected to any other edge $C$. Therefore, only counting this one color, each vertex has degree at most one. However, that means at most $5$ appearances of this color, contradiction. Thus, there is a set of three points that does not contain this color. It is not possible.
\\$~$\\
If $k=4$ then there is a color that appears at most $11$ times, $C$. Suppose for every set of four points, there is at least one edge colored $C$. We already know that there is a set of three points that no edge is colored $C$. Now, for each of the other seven points, the three edges between that point and the three selected points must have one colored $C$. Thus, among the seven, there are at most $4$ edges $C$. We once again pick a triangle such that there is no $C$ edge among them. For each of the four other points, there must be an edge connecting it and one of the points of the triangle that is $C$, leaving no more $C$ edges among the four remaining points, contradiction.
\\$~$\\
If $k \ge 5$, for each color we will create $k-1$ groups of vertices, and we will color all the edges between vertices of the same group. In such a way, when we have $k$ vertices, for each color at least two vertices will be in the same group and thus the edge between them will be colored. For $k=5$ we color
\begin{align*}
C_1 &: 012 369 48 57 \\
C_2 &: 234 581 60 79 \\
C_3 &: 456 703 82 91 \\
C_4 &: 678 925 04 13 \\
C_5 &: 890 147 26 35
\end{align*}
Note that not every edge is colored, but that's okay because we just color them however. For $k\ge 6$ the groups are easy: as long as the edges we color the same color never share an edge, we are fine.


geometry
trigonometry
perimeter
incenter
LaTeX
trig identities
Law of Sines

Let $\triangle ABC$ have right angle at C. $\overline{AB}=5$, and $\overline{AC}=3$. Let $\triangle ACD$ be similar to $\triangle ABC$ with right angle at D. Finally, let the bisector of $\angle ACB$ meet AB at E. Find the area of quadrilateral ADCE.

Are you sure there's enough information? One could use sine, cosine, tangent if an angle measurement was given, you can definately find the perimeter of either of the triangles, but the problem is finding line CE. If you wish to open a bag of worms, click on the hide button if you wish to know the segments and their lengths.

[hide]
Line AD=9/5
Line AB=Given
Line AC=Given
Line BC=radical 34
line DC= (3 radical 34)/5[/hide]

If anyone wishes to help me with math symbols, please e-mail me at zymroz@comcast.net with a topic of Math Symbols. Please and thanks! :D

D is on the outside.

I'd use heron's formula, but we don't know that it works...

Note that the incenter of $\triangle ABC$ lies on $\overline{CE}$. Use that to find $CE$, then that to find the height of $\triangle ACE$.

@kyokun: Click the LaTeX link on the left (below the AoPS symbol and above the Proof-without-words) to find a decent tutorial. It has about everything you need for posting most symbols you might be interested in using.

Let $\theta = m\angle CAE$.  $m\angle ACE = 45 \Rightarrow m\angle AEC = 135-\theta$.  Let $|AE| = x$

By the law of sines, $\frac{x}{\sin45}= \frac{3}{\sin(135-\theta)}\Rightarrow x = \frac{3\sin 45}{\sin135\cos\theta-\cos135\sin\theta}$

$\sin45 = \sin135 = \frac{1}{\sqrt{2}}$

$\cos135 = \frac{-1}{\sqrt{2}}$

By looking at the 3-4-5 triangle, $\sin\theta = \frac{4}{5}$, $\cos\theta = \frac{3}{5}$

After multiplying across by $\sqrt{2}$, we find $x = \frac{3}{\frac{3}{5}+\frac{4}{5}}= \frac{15}{7}$

Now we can find the area of $\triangle CAE$ with the area formula, $K = \frac{1}{2}ab\sin C$, with sides of length $3$ and $15/7$, and angle $\theta$

$K = \frac{3*15*4}{2*7*5}= \frac{18}{7}$

Now for $\triangle ADC$.  It is just the 3-4-5 triangle with sides scaled by a factor $\frac{3}{5}$, resulting in legs of length $\frac{9}{5}$ and $\frac{12}{5}$, giving an area of $\frac{54}{25}$

The sum area is therefore $\boxed{\frac{212}{75}}$


ratio
geometry
trapezoid
rectangle
Pythagorean Theorem

little problem a made up the other day:
find the ratio between the areas of triangle XYZ and trapezoid YZUT where circles z and y are congruent and tangent to circle X, circles t, u, and v are congruent, are tangent to BD, and t and v are tangent to y and z respectively. and AC = 12.
(sorry for bad drawing- t and v are also sposed to be tangent to AB and CD respectively).

it wont let me attach my drawing... any1 know what the problem is?

edit:nvm figured it out
oh i forgot to add the answer and thx JBL for the improved diagram :D
[hide]
since the side of rectangle is 12, we know the radius of big circle is 6 and the radius of the medium circle is 3
thus the height of triangle XYZ by pythagorean theorem is the sqrt of (9^2-3^2) and the base of that triangle is obcviously 6
for the trapezoid:
the small circles' radius are all 2 so we know the base of the trapezoid is 8 and the top is 6.
to find the height u cut two triangles from the trapezoid to get 2 right traingles and a rectangle since the dimensions of the rectangle are 6 and h. the right triangles have hypotenuse 5 and legs h and 1
find h by pythagorean theorem and once u have h use the formula for area of trapezoid.

area of triangle should be 18 sqrt 2 and area of trapezoid should be 14 sqrt 6
thus triangle/ trapezoid = 3( sqrt 3)/7[/hide]

How do you find the area of the trapezoid? I can't figure out how to tell what the height of the trapezoid is. I know the distance from Y to Z and T to V, but I can't figure out the height.

Please help, and thanks in advance.  :)

The area of a trapezoid is $\frac 12 (b_1+b_2)h$ where $b_1$ and $b_2$ are the bases and $h$ is the height.

[hide]Let $r$ be the radii of all the little circles. Then, the radii of the medium and large circles respectively are $1.5r$ and $3r$. By pythagorean theorem we find that the height of the triangle is $\sqrt {(4.5r)^2 - (1.5r)^2} = \sqrt {18r^2} = 3r\sqrt 2$. Then we find that the area of the triangle is $\frac {9r^2\sqrt 2}{2}$. The height of the trapezoid is $\sqrt {(2.5r)^2 - (.5r)^2} = \sqrt {6r^2} = r \sqrt 6$. The area is then $\frac {7r^2\sqrt 6}{2}$. The final answer is $\frac {\frac {9r^2\sqrt 2}{2}}{\frac {7r^2 \sqrt 6}{2}} = \frac {3\sqrt 3}{7}$.[/hide]

tarquin u didnt simplify all the way  :D


Could someone scan in their AMC10 and AMC12 if it is allowed? I missed the AMCs because of weather problems, so I'd appreciate it if I could still take it for practice. That is, if it is allowed. Thanks.

Please do not do this.  This is a violation of the AMC's copyright.


[hide="Solution"]
Let $ m \equal{} \log_23$ and $ n \equal{} \log_25$ so that $ \frac1m \equal{} \log_32$, $ \frac1n \equal{} \log_52$, $ \frac mn \equal{} \log_53$, and $ \frac nm \equal{} \log_35$. From this, we can see that our desired expression is $ \log_{25}{24} \equal{} 3\log_{25}2 \plus{} \log_{25}3 \equal{} \frac3{2\log_25} \plus{} \frac1{2\log35} \equal{} \frac {m \plus{} 3}{2n}$.

Similarly, we can find $ a \equal{} \frac {m \plus{} n}{m \plus{} 1}$ and $ b \equal{} \frac {2m \plus{} 1}{m \plus{} 2}$, and solving the system we get $ m \equal{} \frac {1 \minus{} 2b}{b \minus{} 2}$ and $ n \equal{} \frac {2b \minus{} ab \minus{} a \minus{} 1}{b \minus{} 2}$.

Finally, we can plug in to get $ \log_{25}{24} \equal{} \frac {m \plus{} 3}{2n} \equal{} \frac {\frac {1 \minus{} 2b}{b \minus{} 2} \plus{} 3}{\frac {4b \minus{} 2ab \minus{} 2a \minus{} 2}{b \minus{} 2}} \equal{} \frac {1 \minus{} 2b \plus{} 3b \minus{} 6}{4b \minus{} 2ab \minus{} 2a \minus{} 2} \equal{} \boxed{\frac {b \minus{} 5}{4b \minus{} 2ab \minus{} 2a \minus{} 2}}$.

Does anyone have a simpler solution...??? :maybe: I don't know because the expression looks pretty convoluted.
[/hide]

How did you solve that system? In your explanation, it looks as if you suddenly jumped from one planet to the next in the systems paragraph.

Just clear the denominators and move to one side in both to get a system of equations in $ m$ and $ n$.


In triangle ABC, three segments known as cevians are drawn as shown, one for each vertext to the opposite side. These cevians are concurrent at P. If AE:EC = 2:3 and BP: PE=5:4, find the ratio of AP:PF. D E and F are between AB AC and BC.

I heard MAN+DAD=BMB+CNC

[quote="now a ranger"]
I heard MAN+DAD=BMB+CNC[/quote]

yea that's stewart's theorem.

[quote="now a ranger"]...as shown...[/quote]

Where?

can someone draw me a triangle or link me a website that uses stewart's theorem in a clear and concise way?

Like a good way to learn it.

[hide=" :maybe: .........."]Use Menelaus' Theorem...
Since B,F, and C are collinear,
$\frac{EC}{AC}\frac{PB}{EB}\frac{AF}{PF}=\frac{3}{5}\frac{5}{9}\frac{AF}{PF}=1\implies \frac{AF}{PF}=3$. $\frac{AP}{PF}=2$...[/hide]

@Go Around The Tree, if you have aops volume 2 you should look in there. They have a good explanation of stewart's theorem.


ceiling function
combinatorics unsolved
combinatorics

How many different ways are there to arrange $ n$ white balls and $ m$ black balls on a circle?

$ \lceil \frac {\binom{m \plus{} n}{n}}{m \plus{} n} \rceil$ I believe , it could be wrong haha :P

yes, I'm afraid it is wrong, i thought about $ \frac {\binom{m \plus{} n}{n}}{m \plus{} n}$ too because it's like arranging in line and then every arrangemnt is counted (m+n) times...but as this number is not natural we can't just take the closest natural number as a correct answer  :P

[quote="sunnykim"] and then every arrangemnt is counted (m+n) times...[/quote] I think the problem is that (m+n) is too much..and moreover I believe that the number of times an arrangement repeats depends on the arrangement...maybe another approach would be better :maybe:

This is a standard exercise that can be solved by application of [url=http://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem]Polya's enumeration theorem.[/url]

:maybe: sorry but I can't get the answer from that article...can you post a solution, please?

Neither do I, maybe someone can post an explanation


prove :D

$ 1\minus{}5\left(\frac{1}{2}\right)^{5}\plus{}9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{5}\minus{}13\left(\frac{1\cdot 3\cdot 5}{2\cdot4\cdot 6}\right)^{5}\plus{}17\left(\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot4\cdot 6\cdot 8}\right)^{5}\minus{}\;\;\;\ldots\;\;\;\ldots\;\;\;\plus{}\minus{}\;\;\;\ldots\;\;\equal{}\;\;\boxed{\frac{\Gamma\,^{4}\left(\frac{1}{4}\right)}{2\,\pi^{4}}}$

In Human Language it would be:
\[ \sum_{k\equal{}0}^{\infty }(\minus{}1)^{k}(4 k\plus{}1)\left(\prod_{i\equal{}0}^{k\minus{}1}\frac{2 i\plus{}1}{2 i\plus{}2}\right)^{5}\]
We can make this bizarre formulae

$ (\minus{}1)^{k}(4 k\plus{}1)\left(\prod_{i \equal{} 0}^{k\minus{}1}\frac{2 i\plus{}1}{2 i\plus{}2}\right)^{5}\equal{}\frac{\left(\minus{}2^{\minus{}10}\right)^{k}(4 k\plus{}1) ((2 k)!)^{5}}{(k!)^{10}}$

TaDam! nothing!

 :D


superior algebra
superior algebra solved

This is false: $N_1 \cong N_2 \Leftrightarrow G/N_1 \cong G/N_2$ for $N_1,N_2 \unlhd G$, G finite group.

Does anyone have a counterexample?

let G be the direct product of Z mod 2 (Z2) and Z mod4 (Z4)

let N1 be the group with elements (0,0) and (1,0)
let N2 be the group with elements (0,0) and (0,2)

now G/N1 is isomorphic with Z4
G/N2 is isomorphic with Z2*Z2

interesting enough, this G also proofs that both directions are false

take N1 the group with elements (0,0) (0,1) (0,2) (0,3)
N2 : (0,0) (1,0) (0,2),(1,2)

G/N1 and G/N2 are isomorphic with Z2 (the only group you have of that order)

but N1 and N2 are not isomorphic


calculus
integration
number theory
prime numbers

This is my first post :o, see if you guys can get this one.
This is from me and Integral_e

Prove that for each prime p there are infinitely many positive integers n with the property that if q is a prime divisor of n^2 + p, then there exists a positive integer k less than n^(1/2) such that q divides k^2 + p

Good luck

I don't believe that is the way it is solved, of course, I don't really know what that is. :o

This post has an aweful lot of views for so few posts.... surely someone can come up with a conjecture at least.

someone needs to at least attempt this one.. im not letting it go!

[color=cyan]This isn't an original problem, is it?[/color]

No, this originally came from rrusczyk...

'tis too hard, ViolentBob SpoonPants :P

If you lead us through and give us hints, then maybe there'll be more takers, but threatning them to do it won't help. (unless it's the Chinese torture thingy :) )

I think the biggest problem I had was figuring out what exactly the proof is asking....

 if we pick some prime p, and we must find an n where for each prime divisor q of n^2 + p there exists an int k less than :sqrt:n such that q also divides k^2 + p....

lets do an example,

 choose p = 2, n=5..

 n^2 + p = 25 + 2 or 27.

this one works out nicely because the only q for 27 is 3. so q = 3....

now we must find some k that is either such that 3|k^2+2

we can choose k=1 or k=2

k=1 works because 1+2 = 3 and 3|3....

what we must do is generalize this and then prove that we can choose any prime p, and there will be infinite choices for n that will have a k value to satisfy all of the q's for n^2 + p...

Yes i know it sounds complicated, but it is in fact doable.

This is the advanced forum isnt it?

*sigh* I guess. I'm still chewing on it, but I doubt that I can prove it completely.

heres how I did it:

Test a bunch... see what values out, make a conjecture based on the analysis, and then prove the conjecture...

*still waits for complexzeta to respond with solution*

I got the problem from M&IQ (actually this is a generalized version of the M&IQ problem).

there ya go, now you guys know the origins.

rrusczyk, this problem is quite difficult, and I don't think it's something that can be done at first glance nor even after working for an hour.... at what level on the USAMO/IMO scale would you rate this problem?

This may be stupid to ask, but I don't know what the M&IQ is, so I don't know how difficult it is.

M&IQ stands for Mathematics and Informatics Quarterly. The problems from M&IQ are often quite difficult.

Yeah, I'd call this Olympiad-level difficulty.  The length of time it's gone unanswered on this board is an indication of that.

My main guide when I solved it was to start with a specific p (as the problem was in its original version), stick in values and look for a pattern...

Arne, why don't you give this a try, it seems like something you may find a challenge.

should I post the solution my friend and I came up with??? I mean, does anyone care?

Yes please, I'd like to see it (although I doubt I'd be able to understand it :) )

[quote="Arne"]I'll think of it ... rather tomorrow since it's bedtime now for me.
It's 23 pm in Belgium ![/quote]
..."23 [b]pm[/b]"? :) </jk>

Oh sorry I'm a bit confused ... 23h in Belgium, I think that's "translated" as 11 pm ?

yeah, subtract 12 from your time if you are >12 and you get ours :)

If you are less than 12, ours is same :)...

Ok, ill post the solution I have, maybe it will stem a far less complicated one... and shorter :)

In spoiler:

[hide]

Let M_p denote the set of all n(in the set of Naturals) that satisfy the following property: If q is a prime divisor of n^2 + p then there exists a k(in the set of Naturals) with k < sqrt(n) such that q divides k^2 + p.

Claim: that for all prime numbers p the set M_p is infinite.

Proof:

 First, let p=2, then consider n = 5, we have n^2 + p = 25 + 2 = 27

The prime factors of 27 are 3. if we let k = 1 < sqrt(5) then k^2 + p = 1 + 2 = 3 and so 3 divides k^2 + p.

Hence, 5 is in the set of M_2 and so M_2 is not empty.                                                                                (1)

Now let p be an arbitrary prime with p > 2, then let n = p + 2.

We have n^2 + p = (p+2)^2 + p = p^2 + 4p + 4 + p = p(p+4) + p + 4 = (p + 1)(p + 4)

Now, if q is a prime divisor of n^2 + p, then q divides either p+1 or p+4

Letting k = 1 we see that k^2 + p = p + 1 and hence if q divides p + 1, then q also divides k^2 + p when k = 1.

Letting k = 2 we see that k^2 + p = p + 4 and hence if q divides p + 4, then q also divides k^2 + p when k = 2.

And now since p > 2, sqrt(n) = sqrt(p+2) > sqrt(4) = 2 > 1 and so 1,2 < sqrt(n)

This shows that p + 2 is in the set of M_p for all p > 2                                                                                (2)

By (1) and (2) we have that for all prime p the set M_p is not empty.                                                                         (3)

Now, let p be an arbitrary prime.

By contradiction, suppose M_p is finite,

Since by (3) M_p is not empty

a: = max(M_p) (in the set of naturals) exists

now let n = a^2 + p - a

Since 2 < p + 2 (in the set of M_p) a > 2 and so n = a^2 + p - a > a

We have n^2 + p = ((a^2 + p) - a)^2 + p = (a^2 + p^2) - 2a(a^2 + p) + a^2 + p = (a^2 + p)(a^2 + p - 2a + 1)

Now, if q is a prime divisor of n^2 + p then q divides (a^2 + p) or q divides (a^2 + p - 2a + 1)

If q divides a^2 + p, then since a(is in the set of M_p) there exists a k(in the set of naturals) such that k < sqrt(a) < sqrt(n) and q divides k^2 + p, shown in (2).

If q divides (a^2 + p - 2a + 1) then let k = a - 1 = sqrt((a-1)^2) = sqrt(a^2 -2a + 1) < sqrt( a^2 + p - 2a + 1) = sqrt(n)

And we have k^2 + p = a^2 -2a + 1 + p

And therefore q divides k^2 + p

These two cases are exhaustive and so we have that for each prime divisor q of n^2 + p there exists a k < sqrt(n) such that q divides k^2 + p.

Hence n = a^2 + p - a satisfies the given condition and therefore n(is in the set of M_p). This however, is a contradiction because n = a^2 + p - a > a and "a" was the maximum value in the set M_p.

Thus we have that the set M_p must be infinite.

[/hide]

I hate you all  :twisted:  :evil:  :evil:  :evil:

cause nobody seems to care :(, even though I went through all of that work to paste and clean up the proof for you guys.

If I understood any of this, then I'd comment, but I don't :cry:

Probably all the people interested didn't quite understand and the people who understand don't care since it's easy for them...I share your pain of being ignored :(

lol you are prolly right, it is a bit hard to follow, even when IM THE ONE WHO TYPED IT UP! :)


How many more edges does an octagonal prism have than a cube?

Edges of a n-gonal prism=3n

3(8)-3(4)=12


In the object-oriented programming, what is a "class"?

The literature I've looked through has been too confusing.

Thanks :)

[quote="Wikipedia"]In object-oriented programming, a class is a programming language construct that is used to group related instance variables and methods. Each class expresses structural and behavioral design decisions made by the programmer to indicate what types of objects can exist when the program is executed.[/quote]

You can go [url=http://en.wikipedia.org/wiki/Class_%28computer_science%29]there[/url] for examples, too. In general, the idea of a class is very simple. We want to group a bunch of variables/methods together so they can operate as a single unit.

Oh cool, I always wondered what structs were...

[quote="Hamster1800"]First, you should know about the primitive data types. I'll assume you have that knowledge.

Next, you should look at structures. Structures are a collection of primitive data that has a name. For example:

struct cow {
int height, age;
} a;
Now we have cow a. We can get the height of a by looking at a.height and get the age by looking at a.age.

Now, classes are exactly like structures, except that they can also have methods/functions in them. If we modify our cow to be a class, we could have a method called foo which does something and is called by a.foo();[/quote]

So, for example, we could have a method called eat, which takes an argument called food:

class Cow {
function eat(food) {
return food/2
}}

a.eat(food)

Bad example, I know.