Make the game fair.

Don't get people to give you ratings.

Act appropriately.

Don't make multiple accounts (most of you have them, get rid of them.)

Play against people of similar ability as you. It's hard to get active new players when everyone just beats up on them.

Welcome new people.

Speak well of one another in chat and in forums.

Think of others before you say or do something.

Follow the Terms of Service (TOS).

Love,
Mom and dad

Bump as it's needed. Too many are acting poorly.

One problem with the Internet is that people tend to say things that they wouldn't say face to face. This is especially true on sites like YouTube...just look at the comments. However, this does not mean you should act the same way. We expect AoPS users to be more mature and friendly than the typical Internet user. We are a small community and feelings can easily get hurt. In addition, the entire math community is quite small. There is more than a trivial chance that you will one day meet up with each other, whether it be through state or national competitions, camps, future schooling, etc. So please, be courteous with one another, not only in FTW, but throughout the site.

If you simply can not get along, please just ignore each other or move on to another site. You can use the /ignore command in FTW. If you use FTW2 this is sticky, if you use FTW1, you will have to do it at each login. Avoid each others blogs. Skip over each others posts.

The admins can not mediate for you.

Hi, I would like to address a few of the more prevalent latent problems on FTW and suggest some possible solutions.

Always be aware of what others might feel about what you say. When you have a, say, 1600 rating, and you say "my rating is pathetic" among a lobby of people all lower than you, that can hurt, even if you didn't mean to say others are more pathetic. This happens a lot in FTW. I made a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=297474]note about this problem[/url] before, and it actually seemed to mitigate the issue a tad bit, but it's going overboard again. If someone says something similar and you feel hurt, try to ignore it and understand that others did not mean any meanness.

Also, don't randomly accuse people of actions unless you have convincing evidence to justify your accusation. Of course, nothing can be 100% proven, but at least, if you have to accuse, make it based on an instance for which it would only be, say 1 out of 20 times, a fluke.

Additionally, the ToS and FAQ have a lot of information that you should read and follow. Although the FAQ was written in a rather jokish and friendly tone, it doesn't mean its context is a joke, and it should definitely be taken seriously.

[i]If you're having problems with another person[/i], try to understand the opposing perspective, and question yourself to see if there's any possible problem on your part. Then, if you're convinced that the other person is being the problem, try to convince them about what is going on. Occasionally, you'll be talking with a jerk that dogmatically annoys people, but most people, at least not here in this high-standard environment, are not like that. There's also always the ignore function, as levans mentioned. If you wish not to ignore, you could just visually ostracize those lines from the chat and tell yourself you're only seeing the lines from other people. Also, asking others for advice is not a bad idea. I have been asked for advice and I have also asked others for advice. Believe it or not, [u]this really helps[/u]. Even if you don't get anything solved by that, simply sharing your situation with someone else usually makes you feel better inside. Do [b]not[/b] report abuse, because that's not what reporting abuse is for (it's for notifying the administration of an infringement of the ToS and/or a possible threat to the FTWian/AoPSian community). Also, don't act like a mirror and fire the same insults and derogatory remarks back; this could get you muted just like the other person, but [i]more importantly[/i], it is worse for your own and the opposing party's mental morale. Take FTW as a calm reception area for fun math with complimentary socializing facilities.

If you going to spend your time on FTW, breaking ToS go start your own chat/game. Please we want to have fun on FTW. There is too much drama. Just ignore the other user and avoid them. Be more mature than the person who is insulting you. There is WAY to much drama on a daily basis. Lets be more mature than most of us are now.
Just because everyone else is breaking ToS doesn't mean you should too.

Just a quick thing before I talk about this topic: I get scared when I see big blocks of text (which is why I almost died after seeing PhireKaLk6781's post), so I'm going to try to keep this nice and simple

[quote="levans"]Bump as it's needed. Too many are acting poorly. [/quote]

I usually don't see any bad behavior when I'm on, but when I see the above quote and when someone says "everyone spams when no mods are on", I get worried (quote from some other thread). The fact of the matter is that mods cannot be on FTW every second of the day. We have lives that include more than sitting at a computer and reading every single thing written on an online math game. And even if we didn't have lives, we still need sleep. The bottom line is that it is simply impossible to have someone watching the chat at any given moment. (of course, there's always language and floodingfilter, but we've learned to not trust those a long time ago  :wink: )

However, the amount of good behavior in the game should not decrease simply because any combination of four people are not on. Not being muted should not be your motivation for being good. Your motivation for being good should be that it helps make the community a better place. We cannot do anything to make your behavior better except for mutes, and that cannot happen if no mods are on at the time of offense. Therefore, you guys need to make the decision to start behaving better.

As Phire said, some things that are said were said without the intention of hurting someone, so please be careful what you say. Any degrading remark, even those about yourself, should be considered carefully to see whether or not it could potentionally hurt someone else.

Anyway, the main idea is that behavior is not going to improve because four people can mute people who behave badly. Behavior can only improve if everyone decides they want to improve behavior. Also, think before you click enter. Think about whether or not what you say could hurt anyone in any way.

[size=75]Hmm... I didn't do a good job keeping it short and simple did I?....
Anyway, I usually like ending long boring speeches with something completely random and hopefully funny like evil brazilian monkeys (yes, I do this even in real life), but I couldn't put that in and make it work here... Oh well.[/size]

[quote="vallon22"]...Anyway, the main idea is that behavior is not going to improve because four people can mute people ...[/quote]

Make that five, with one inactive.

Generally, I think that this is a huge reason why people curse/flood in the lobby:

Being angry or frustrated. It's normal; heck, I've spammed too after some frustration. But keep this in mind: If you flood, you should either make your own passworded room to flood in, whisper to yourself for all your flood (not reliable), open a Word document to spam in (though not as satisfying), or disconnect your keyboard and start pounding on your keyboard (without breaking it).

Cursing is generally used more when one player gets angry at another. So here is a question I have: Is it all right to curse as long as you make sure you don't typo and that you whisper to yourself (incase lang filter doesn't catch it?). I sometimes do this to either kick myself off of FTW so I would go do my HW or I can actually curse without affecting others.

(off topic) Adding to vallon's habit, I generally read the first long post, skim the second, and freak over the third. Vallon was the third. Hope you read my not-so-long post.

[quote="PhireKaLk6781"][quote="dragon96"][quote="vallon22"]...Anyway, the main idea is that behavior is not going to improve because four people can mute people ...[/quote]

Make that five, with one inactive.[/quote]

I was wondering if I should comment on this, but since someone else brought this up, I might as well say that, although no offense is intended (seriously, no offense intended, I could feel some people out there preparing to jump me for saying this), mods that do still come on (getting less and less) have disproportionately long afk periods, and I have seen people swear with mods in the lobby and have nothing happen to them, especially because the voting system seems to have emotional breakdowns.

Nevertheless, you don't have to be a moderator to be able to help. The "Report Abuse" button is always there for you.

There's been another surge of multiple accounts lately. Just...stop. It's definitely not as funny as you think.[/quote]

In our defense, we are getting older and going into high school, which means we will have less time to monitor the chat.
This is why the fifth person who can mute is inactive.
In addition, the times when we can get on are usually times that require us to multitask and do something else as well, so it may be that I will have a tab on FTW, see that chat is not really active at the moment, go off to do my report on the Byzantine Empire or study for my test on protein synthesis, and come back to a really large amount of chat to read through.
While I admit that what you say is true, and that we should spend more time on FTW, it is sometimes simply not possible.

With that said, there are things everyone can do to make the FTW and AoPS community a better place.
The first is be responsible for your own behavior.
Second, you can use the "report abuse" button, like Phire said. You can also directly contact one of the mods and tell us who is misbehaving, when he/she is misbehaving, and what he/she said (i suggest using pm, and please include any and all evidence you can). I personally do not know who receives the "report abuse" reports, so I don't know how often those reports are looked over and acted upon, so I suggest using the second option. BTW, just because we don't go on FTW doesn't mean we don't go on AoPS; there is a good chance any mod will get your pm.
Third, you can inform the person that he/she is misbehaving. Sometime the person does not even realize he/she is hurting someone or doing something that he/she shouldn't be doing. Even if it is obvious that he/she knows it is not allowed, tell them to stop. If they don't, report abuse or pm one of us.

Remember, in order to make the community a better place, each person has to do his/her role. Sometimes it may be behaving well yourself, other times it is informing others to behave well. Levans has shut off chat before, and if chat becomes a problem, he will shut it off again. If that happens, people like me who love chatting in general will go crazy and get on the next spaceship to Mars in the hope that the Martians have a similar math game that allows chat. Unfortunately, I don't particularly want to leave Earth, so let's just make the behavior better so chat isn't shut off in the first place.

Okay, another issue has came up.

Don't join a game and start going "This is Ranking. Make it time." or "Responses fails." or "You forgot to make it timed, not ranking." or something of the such. The reason they are options on that dropdown menu are so they can be used. If you are not fine with a certain scoring method, don't join the game and make your own! You can really be a nuisance bothering the maker of the game like that. Of course, suggesting things like "May we do it timed next time?" is perfectly fine, but once you're getting intrusive, it's peeking out of bounds.

Just leave it alone and [u]don't use it[/u].  It will soon be rendered inactive.

[quote="AndyZ"]how do you get rid of a multi? i made one when i was muted and when i didnt know that they were llegal. now i heard that they are illegal and i dont know what to do with it[/quote]

We also recommend that you tell an authority what the username of your multi was.

Since I am a freshman in high school, I no longer do mathcounts. Being in high school has used up a lot of my

so I can't go on a lot anymore. I've been on a little more recently, but I still don't have time. With tennis

season, it's harder to get on. Even though most of the rule breakers don't read this, I would suggest every

user of FTW to try their best to stop rule-breaking. You can tell cursers to stop cursing, use the report abuse

button, PM A MODERATOR, etc. Just a few ideas

[quote="jjx1"]PM A MODERATOR[/quote]

I would like to kindly point out that you should [u]directly PM a moderator[/u] rather than [u]PM someone to PM a moderator[/u], as that would be quite a redundant thing to do. You could always PM others, though, if you want advice to solve a problem you're having with another user and want some feedback from a different perspective.

Excuse me for asking, but who are the five mods who can directly mute?

[quote="isabella2296"]issabella2296, jjx1, vallon22, and levans (although he's an admin) are the most active, I believe. I don't think iin comes on anymore.[/quote]
You said there were 4 mods with one inactive, just wondering, what was the 5th mod?

[quote="isabella2296"]issabella2296, jjx1, vallon22, and levans (although he's an admin) are the most active, I believe. I don't think [b]iin[/b] comes on anymore.[/quote]

i think that iin is the last one

w00t 400th post

oops, i didnt know that iin was a username

just for reference, like with the new people (like me) here, reading this, where is the report abuse button anyways?
thanks
the new one.

Hi allenjiang,

Welcome to AoPS. Hope you have a great experience with the many features here, including FTW. The "report abuse" button is located on the bar on top of your screen that you play FTW on. It's near the right side, between "Forum" and "My Account."

Hope that helps!

Everyone needs a reminder about some of the rules on FTW.

1) No flooding
2) No spamming
3) No real names

I don't care if there is no mod on or there is an inactive mod. Follow the ToS at all times. Review the ToS periodically. If you break the ToS, we will find you, and we will mute you. You might be able to get away with it when we're busy, but you're not going to be lucky forever.

To those of you trying to be mature and keep chat clean, if you see someone breaking ToS, copy and paste the text and pm it to a mod (include date and time of offense).

For those of you who do not know what the ToS is, it is the Terms of Service and can be found [url=http://www.artofproblemsolving.com/Edutainment/ftwcs3/Pages/tos.php]here[/url].

Well thats sort of like stalking

And what about me and my brother who use the same laptop
??

Cuz xantho and me use the same laptop

[quote="PhireKaLk6781"]Please do not accuse other people of being a multiple account just because you think they are—at least get some sort of evidence that is significantly suspicious. FTW is not Mafia (and even there, I believe, you need reasons).[/quote]
hahahha mafia reference (see game 45)!

[quote="Mrdavid445"]Could we ask mods for I.P adress checks?[/quote]

You could, if you so chose.

Whether you get an informational response, on the other hand, is not that clear-cut.

[quote="Mrdavid445"]Could we ask mods for I.P adress checks?[/quote]

This is from a few years ago and on something else, but

[quote="isabella2296"]Don't accuse based on times. If you think someone did this, ask me to check IP addresses.[/quote]

This means that there's not much harm (if any) in asking a mod if you have legitimate reason to believe that someone is a multi. If you frivolously accuse people of having multis, you are not likely to get an informational response from a mod/admin, as PhireKaLk6781 said

[quote="AFK"]0.o I honestly didnt know that you couldnt make multis, I thought you just couldn't get rating with them.[/quote]

If that were the case, I would have asked that you pretend that that's actually true, so that you won't be tempted to make multis.

Unfortunately, that's not the case. You would have been more convincing if you didn't send me a message about your account on your multi.

[quote="PhireKaLk6781"]Please do not accuse other people of being a multiple account just because you think they are—at least get some sort of evidence that is significantly suspicious. [/quote]

As I have said previously, please don't just openly finger-point in the lobby. If you believe someone is a multiple account, [b]report[/b] them.

[quote="mcqueen"]If the mods are getting older, going into high school, and having less time to monitor the chat, why doesn't AoPS get younger mods, ones with more time to monitor the chat?[/quote]
Just because the mods don't show up in the lobby that much, doesn't mean they can't just look at all the chat logs. I am pretty sure they are pretty active on reviewing the chat logs and looking at the filed reports.
Also, @dy317's sig, ive beaten weegee13!!!

[quote="mcqueen"]If the mods are getting older, going into high school, and having less time to monitor the chat, why doesn't AoPS get younger mods, ones with more time to monitor the chat?[/quote]
And because none of the "younger mods" can't handle it.

(I don't know any mod younger than highschool)

@Mrdavid I think mcqueen meant that FTW should get some new mods that are younger and have more time on their hands.

Hi, I would like to address a few of the more prevalent latent problems on FTW and suggest some possible solutions.

Always be aware of what others might feel about what you say. When you have a, say, 1600 rating, and you say "my rating is pathetic" among a lobby of people all lower than you, that can hurt, even if you didn't mean to say others are more pathetic. This happens a lot in FTW. I made a note about this problem before, and it actually seemed to mitigate the issue a tad bit, but it's going overboard again. If someone says something similar and you feel hurt, try to ignore it and understand that others did not mean any meanness.

Also, don't randomly accuse people of actions unless you have convincing evidence to justify your accusation. Of course, nothing can be 100% proven, but at least, if you have to accuse, make it based on an instance for which it would only be, say 1 out of 20 times, a fluke.

Additionally, the ToS and FAQ have a lot of information that you should read and follow. Although the FAQ was written in a rather jokish and friendly tone, it doesn't mean its context is a joke, and it should definitely be taken seriously.

If you're having problems with another person, try to understand the opposing perspective, and question yourself to see if there's any possible problem on your part. Then, if you're convinced that the other person is being the problem, try to convince them about what is going on. Occasionally, you'll be talking with a jerk that dogmatically annoys people, but most people, at least not here in this high-standard environment, are not like that. There's also always the ignore function, as levans mentioned. If you wish not to ignore, you could just visually ostracize those lines from the chat and tell yourself you're only seeing the lines from other people. Also, asking others for advice is not a bad idea. I have been asked for advice and I have also asked others for advice. Believe it or not, this really helps. Even if you don't get anything solved by that, simply sharing your situation with someone else usually makes you feel better inside. Do not report abuse, because that's not what reporting abuse is for (it's for notifying the administration of an infringement of the ToS and/or a possible threat to the FTWian/AoPSian community). Also, don't act like a mirror and fire the same insults and derogatory remarks back; this could get you muted just like the other person, but more importantly, it is worse for your own and the opposing party's mental morale. Take FTW as a calm reception area for fun math with complimentary socializing facilities.

Hello, please be aware that the above user should learn a lesson in maturity as well.  The above content was copied from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1771036#p1771036]Post #3[/url] of this thread.

So bluecarneal chill maybe x23rocks thought it was a important post and decided to post it again bro

I don't know if it annoys anyone else, but I don't enjoy seeing casual slang in more serious topics.

Bluecarneal does not need to "chill", he is pointing out that x23rocks reposted something one can easily find on the first page.


limit
ceiling function
calculus
integration
real analysis
real analysis unsolved

$\lim_{n \to \infty} \sum_{k=1}^n \left\{ \frac{n}{k}\right\}^2=?$

What does that $\{\}$ symbol mean here? Fractional part?

Assuming it does indeed mean fractional part (if it doesn't, I'm going to be really annoyed):
To save some typing, let $d = \left\lceil \frac{n+1}2 \right\rceil$.

$\sum_{k=1}^n \left\{ \frac{n}{k}\right\}^2 \geq \sum_{k=d}^n \left\{ \frac{n}{k}\right\}^2 = \sum_{k=d}^n \left(\frac{n-k}k\right)^2 = \sum_{k=d}^n 1 + n^2\sum_{k=d}^n \frac1{k^2} - 2n\sum_{k=d}^n \frac1k$

In the limit, the first of these sums is $\frac n2 + O(1)$ while the third is $2n\log2 + O(1)$.  The middle one requires maybe a little work (unless you happen to already know an asymptotic for it), but by bounding $\sum_{i=a}^\infty \frac1{i^2}$ with integrals I was able to get $\frac1{a - 1} - \frac1b > \sum_{i=a}^b \frac1{i^2} > \frac1a - \frac1{b-1}$ and plugging in we get that the middle term is $n + O(1)$.  (The constant part there approaches something between 2 and -3 in the limit.)  Thus, the entire sum is $\left(\frac32 - 2\log2\right)n + O(1)$, which diverges.

[quote="fedja"]It is a little bit more interesting to try to find the limit of the original sum divided by $n$ (assuming that $\{\cdot\}$ stands for the fractional part, of course)[/quote]

Well, I was only able to show it diverged by (essentially) calculating that limit -- can you show it diverges more easily (without calculating the limit of the sum divided by $n$)?

[quote="JBL"]

Well, I was only able to show it diverged by (essentially) calculating that limit [/quote]

Yes, but you calculated the limit only for the easy half of the sum ;).

[quote]
can you show it diverges more easily (without calculating the limit of the sum divided by $n$)?[/quote]

Your way to show the divergence is, probably, the easiest one. The only thing that can be simplified is to reduce summation to the values of $k$ between $\frac 1 2 n$ and $\frac 2 3 n$. For these $k$, we have $\left\{\frac n k\right\}\ge \frac 12$ and therefore the entire sum is at least $\frac 1 4\cdot \frac 1 6 n=\frac n{24}$ for large $n$.

[quote="fedja"][quote="JBL"]Well, I was only able to show it diverged by (essentially) calculating that limit [/quote]
Yes, but you calculated the limit only for the easy half of the sum ;).

[quote] can you show it diverges more easily (without calculating the limit of the sum divided by $n$)?[/quote]

Your way to show the divergence is, probably, the easiest one. The only thing that can be simplified is to reduce summation to the values of $k$ between $\frac 1 2 n$ and $\frac 2 3 n$. For these $k$, we have $\left\{\frac n k\right\}\ge \frac 12$ and therefore the entire sum is at least $\frac 1 4\cdot \frac 1 6 n=\frac n{24}$ for large $n$.[/quote]

Heh, I forgot that I just dropped the first half of the sum  :)

Hi,
The integral you are looking for is equal to: ln(2*Pi)-gamma-1, and the series in question is divergent as it has been proven.
Didilica


geometry
number theory
least common multiple
greatest common divisor
trigonometry
prime factorization
\/closed

In my opinion, the first half of the Intro NT book is easy (if you know the fundamental things about divisibility, prime factorization, LCM/GCD, etc), and the second half is important. So rather than taking the class, you can buy the Intro book, skim through the first part, and focus on the last several chapters.
Also, If you find any of the courses(NT/C&P/geo) hard, you can try the Algebra 2 course as well.

I'm not that good at Number Theory and I hate the prime factorization and those stuff. I despise having to look for a number using clues and stuff like that... so I never really get that good of a grade on it (A or so in reg math, avg is A+ with 100+) I'm really bad at prime factorization of big numbers (1034 for example), and LCM-GCF with big numbers (243 and 526)

Well in that case I think you should take Intro NT.  If you have a good work ethic, buy the book and work through it. :)

[quote="Blazer123"]I think Intro Geometry would be better than Intro NT.  That's just my opinion.  But if don't know much NT, you should take Intro NT.  C & P is a good choice.[/quote]

Number theory does not appear often, but is interesting and worth studying. C&P is a very good choice. It appears on contests A
LOT. However, you really should master basic algebra before Intro to C&P or NT or (especially) Intro to Geometry. Intro to Geometry is the hardest and you should do C&P or NT first. Even I'm using it to prepare for AIME.

[quote="rick8huang"]I'm a 7th grader now. Last year, I had no experience at all with MC questions and I got a really bad score on practice sprint rounds from the last years. I got a really bad 8/30 on my first sprint round (i know, really bad) and there was 3 months in between when I tried to look for problems that will help me. After 1.5 months, we took a practice target and I got 4/8 there, and lastly on my final sprint (in Dec 08) I got 15/30 (somehow my score improved, I don't know why.) I already took a MC basics, and I learned alot through it. Now I registered Intro to Counting and Intro to Number Theory. Is this the right class?[/quote]

I think you are in the right classes, assuming you have some basic algebra under your belt.

I took Honors Pre-Alg last year, and I'm taking Alg 1 this year with some basic trig at school (skipped a grade).

That probably covers AoPS Algebra 1, maybe a bit of Algebra 2.
The AoPS curriculum is not necessarily syncronized with your school's classes.
Read the topics covered in the AoPS classes.

By the way, Mr. Ruscyk, thanks for putting that for all the classes:)

Yeah, and try the pre-tests and post-tests for Algebra 1 and 2 to decide which one(s) you need. A strong basic Algebra experience is required for more math, including C&P and N Theory.

I noticed something odd.  Why is this?

When I took the pre-test for Algebra III, I got about 3/4 of the questions, yet all the questions in the post-test for Algebra II seemed IMPOSSIBLE.  Why is that?

I'm not sure why there's a big difference. However, I'm taking Algebra 3 right now, and I'm sure everything in the Alg 2 post-test is covered in Alg 3.

OH come one....which is easier?
This:  [url]http://www.artofproblemsolving.com/Classes/Algebra2/PostTest.pdf[/url]

or this:  [url]http://www.artofproblemsolving.com/Classes/IntermAlg/PreTest.pdf[/url]

It's common sense.

The pre-test is easier because it covers a review of all the previous lessons in the class.

[quote="AwesomeToad"]Can all of the challenge problems in a chapter be solved using the methods learned earlier?[/quote]

Yes, but it is obviously challenging compared to the review problems. You really need to think how to utilize everything you learned in that chapter and the previous ones, and try to combine those methods creatively.


[quote="Icy"]A boat is crossing a river 260 m wide. In still water, the boat goes 2.2 m/s. When it reaches the other shore, it is 110 m upstream from the starting point. (It's a right triangle, with one leg being 260 m and the other 110 m). To do this, the boat must sail at an angle of 45 degrees away from a straight line perpendicular to the shore it started from. What is the rate of the current?

Well...My biggest problem with this is I don't know what is what. We're technically supposed to solve with vectors...so...please help. I've gotten 3 different answers, and none of them are right. I think the real one is... roughly .9 m/s, but I don't know how to arrive at this number.[/quote]
[hide] Consider the velocity vector of the boat. Its x component (positive is downstream) is $ \minus{}2.2\cos{45^{\circ}}\equal{}\minus{}1.5556$. Its y component (across the river) is $ 1.5556\ \frac{\text{m}}{\text{s}}$. So it takes $ \frac{260}{1.5556}\equal{}167.1343\ \text{s}$ to cross the river.

Let the velocity of the current be $ v_{c}$. The net x component of the boat's velocity will be $ v_{c}\minus{}1.5556$. Thus, it will have traveled $ 167.1343(v_{c}\minus{}1.5556)$ meters in the x direction.

$ 167.1343(v_{c}\minus{}1.5556)\equal{}\minus{}110\iff v_{c}\equal{}0.8974$.[/hide]

Yes, I figured this out after about half of hour at looking at it, and arrived at the same answer, but didn't know how to delete this topic. Thanks anyway though :D


I want to know how to prove this inequality by using Cauchy-Schwars:
$ (1\plus{}\frac{2a}{b})^2\plus{}(1\plus{}\frac{2b}{c})^2\plus{}(1\plus{}\frac{2c}{a})^2\geq\frac{9(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca}$
and a,b,c are positive real numbers

By using Cauchy-Schwars, we have:
$ (1 \plus{} \frac {2a}{b})^{2} \plus{} (1 \plus{} \frac {2b}{c})^{2} \plus{} (1 \plus{} \frac {2c}{a})^{2}$
$ \geq \frac {1}{3}[3 \plus{} 2(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})]^{2} (1)$
Also:
$ (a \plus{} b \plus{} c)^{2} \equal{} (\sqrt {ab}.\frac {a}{\sqrt {ab}} \plus{} \sqrt {bc}.\frac {b}{\sqrt {bc}} \plus{} \sqrt {ca}.\frac {c}{\sqrt {ca}})^{2}$
$ \leq(ab \plus{} bc \plus{} ca)(\frac {a^2}{ab} \plus{} \frac {b^2}{bc} \plus{} \frac {c^2}{ca})$
$ \equal{} (ab \plus{} bc \plus{} ca)(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})$
So: $ \frac {9(a \plus{} b \plus{} c)^{2}}{ab \plus{} bc \plus{} ca}\leq9(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})(2)$
We'll prove that $ \frac {1}{3}[3 \plus{} 2(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})]^{2} \geq9(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})$
[hide]
But up to now, I can't
Does anyone have another proof or continue with that proof?
[/hide]

:lol: I have an idea that we can set $ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \equal{}$ [size=200]A[/size]
So the problem is more easier
[hide]But i still can't solve it :blush: [/hide]

Okay, I'm back
Set $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \equal{} A$
By AM-GM
$ A \geq 3$
We have to proof:
$ \frac{1}{3}(3\plus{}2A)^{2}\geq9A$
$ \Leftrightarrow\frac{1}{3}(9\plus{}12A\plus{}4A^2)\geq9A$
$ \Leftrightarrow3\plus{}4A\plus{}\frac{4}{3}A^2\geq9A$
$ \Leftrightarrow\frac{4}{3}A^2\minus{}5A\plus{}3\geq0$
$ \Leftrightarrow(A\minus{}3)(A\minus{}0.75)\geq0$
Obviously true 'cuz $ A \geq 3$
Equality holds when a=b=c
Hehe, problem's solved

Why doesn't anybody post topic here?
I think my proof is a little bit long and that inequality must have another nicer proof.
I'm really interested in inequalities which has: $ ab,bc,ca,a,b,c,a^2,b^2,c^2$
Especially that inequality


An unusual probability card game is played as follows:

The deck is shuffled into an entirely random arrangement. Six cards are selected from the top of the deck and placed on the table. You and another person are each dealt seven cards, after which you take turns randomly selecting a card from your hand and a card from the six on the table. If the two cards are the same color (red or black), the person gets to keep the cards. If the two cards are the same suit (spades, clubs, hearts, or diamonds), then the person gets to keep the two cards and take another one out of the deck for themselves. If the two cards are the same number or 'letter', then the person gets to keep the two cards and take two more for themselves from the deck. If the two cards are not the same color, suit, or type, the two cards are discarded in a 'trash' pile. At the end of your turn, you replace the card you took from the ones on the table with another from the deck.

If you ever run out of cards, you lose. If one person loses, the other person keeps on playing until they lose or win. The person to draw the last card(s) from the deck wins, whether they won those cards or were using it to replace a card on the table.

a) What is the probability that one of you wins, and

b) What is the probability that both of you lose?


Here is a test I made for second chapter of my General Chemistry class. I don't think anyone will necessarily BENEFIT from this document since most are in AP level but I just want to post this in here so that anyone taking General Chemistry or going to take it in future can see the beginning of it.

Solutions will be posted TOMORROW once I talked with my chemistry teacher whether my answers are correct or not.

Enjoy.  :)

Error. There is no #6. I'm too tired to fix it so oh well.

Why does it contain simple math? Like the 2nd last question? :?  or am I mistaken?

another question: how many significant digits are in "200 rattlesnakes"  :D

I'll post up the PDF form of all the solutions tonight.

Here are #6 which should've been in the PDF. Since I don't want to post new ones, I'll just post questions and solutions for #6 here.

[color=blue]Write in a long-hand form.

A. $1.75 \times 10^6$

B. $27.5 \times 10^{-2}$

C. $4.74 \times 10^3$

D. $0.094 \times 10^{-2}$ [/color]

[hide="Solutions"]

Only thing we have to watch out is for number of significant digits when it's expanded. For A, 1.75 have significant digit of 3. Hence, our expansion must have 3 significant digit as well.

A. $1.75 \times 10^6 = 1750000$ is the answer. Note that there is no decimal point after last zero because adding it will make all these zeros significant, which isn't what we want.

Similar logic applies to rest.

B. $27.5 \times 10^{-2} = 0.275$. This is fine since the leftmost zero is insignificant.

C. $4.74 \times 10^3 = 4740$. Just like A, we shouldn't put decimal point next to 0 since in that case, we'll have significant digits of 4, not 3.

D. $0.094 \times 10^{-2} = 0.00094$. Only significant digit in here is 9 and 4. All zeros are insignificant so our answer is correct. [/hide]

Solution PDF is up.

If anyone has question or think I have wrong answer, please let me know. Thanks.


Give a proof, a combinatorial proof would be great, to:
$ \forall n$
$ (n+2)^{n}=\sum_{k=0}^{n}\left( n \atop k \right) (k+1)^{k}(n-k+1)^{n-k-1}$

You'll have to fix it first.  It doesn't even give integer values (consider the term $ k = n$).  Probably several typos, I guess.

If $ k=n$ it will appear a $ 1^{-1}$ so it's integer, but yes, maybe it's a good idea to change it a little bit before using a combinatorial proof, but even to do a non-combinatorial proof it's quite hard.

As written, the $ k = n$ term gives us a factor of $ (2n+1)^{-1}$ which will not be cancelled by anything else.

try this instead for the combinatorial idea:

$ \sum_{k=1}^{n-1}\left({n \atop k}\right)k^{k}(n-k)^{n-k-1}=(n-1)n^{n-1}$

For a proof see
[url]http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/abel.pdf[/url]

And it's equivalent to [url]http://www.mathlinks.ro/viewtopic.php?t=157460[/url]


advanced fields
advanced fields unsolved

show that $ \alpha$ is a limit ordinal iff $ \alpha \equal{} \omega \cdot \beta$ for some $ \beta \neq 0$.

every ordinal $ \alpha$ may be uniquely written as $ \omega \beta + n$ with $ n < \omega$ (choose $ \beta$ to be the supremum of the $ \lambda$ such that $ \omega \lambda \leq \alpha$). if $ \alpha$ is limit, we must have $ n = 0$. since $ \alpha \neq 0$, we have $ \beta \neq 0$. on the other hand, $ \omega \beta \neq 0$ for $ \beta \neq 0$ and if it has a predecessor $ \omega \beta' + n$, we would have $ \omega \beta = \omega b' + (n + 1)$ which contradicts the uniqueness.


superior algebra
superior algebra solved

I am new here so not sure how this works but I figured this would be the best place to hopefully get my question answered.

I was recently asked the riddle, what is the largest number you can make using only three numbers.

My initial response was 9^99.  I was later told after thinking it through a bit longer and coming up with no other answer that the true answer is 9^9^9.

After thinking it through more, I am still unsure the order multiplication when you have an exponent raised to an exponent.

So is the correct way to solve 9^9^9:

9^387420489

or

387420489^9?

I appreciate any and all advice/help/ridicule if necessary to let me put this problem to bed.

Thank you!

[quote="9^9^9"]So is the correct way to solve 9^9^9:
9^387420489
or
387420489^9?
[/quote]

The former way. Usually the term a^b^c is defined to mean a^(b^c), because you can express (a^b)^c in the form a^(b*c) without the need of a "power tower".

LiMa

Thank you LiMa.  What is a "power tower"?  It has been many years since I have worked on exponents without the use of Excel or a calculator.  In each case, you get a different result.

By "power tower" I simply mean a term of the form a^b^c or with even more powers, written as a "leaning tower" (like the one in Pisa) of powers $a^{b^{c}}$.

Usually, "power tower" refers to terms of the form $x^{x^{x^{...}}}$, where all the occuring numbers are equal. Your number $9^{9^{9}}$ is a power tower in that sense.

LiMa


linear algebra
matrix
superior algebra
superior algebra unsolved

I have another question. It is about finding a matrix P which AP=PJ where J is jordan form of a matrix A.

Given information is

A= 2 1 0
0 2 0
2 3 1
and minimal poly is ((x-2)^2)(x-1)

How can I find P??

thank you

two times, why ??? [url]http://www.mathlinks.ro/viewtopic.php?p=1223632#1223632[/url]


[hide]the chance for Event A to happen is $\frac46$ because there are 4 numbers in the set that aren't 1 or 6. The probability of Event B happening is $\frac36$. So Event A has a greater probability of happening because $\frac46>\frac36$[/hide]

[quote="symmetry"]When a number is chosen at random from the set {1,2,3,4,5,6},
which event has the greatest probability of occurring?  Why?

Event A = NOT choosing either 1 or 6

Event B = choosing a prime number[/quote]

[hide]The probability for Event A to happen is $\frac{4}{6}=\frac{2}{3}$ because there are 4 choices left if you don't pick 1 or 6, out of 6 choices.

The probability for Event A to happen is $\frac{3}{6}=\frac{1}{2}$ because the prime numbers in that set are 2, 3, and 5. Leaving 3 out of six left.

Since $\frac{1}{2}<\frac{2}{3}$, event A has a higher chance of happening.[/hide]

Great explanations.  Good stuff here.

Thanks.


geometry
perimeter
probability
analytic geometry
counting
distinguishability
FTW

I am beginning a tournament for people that would like to join. I am going to post a few problems every week, and I am going to allow the best of all people participating to move on the next round. The rules are simple; you post your answers to all the problems, then you type in your solutions. I will chose from all the people that posted, and I will pick the top 50 people to go to the next round. I will rate each solution out of 10, and if you get a problem wrong, then I will only give you a maximum of 5 points. If you get a full score then you will autimatically go on to the next round. You will post your answers from the days 5/23 to 5/31, and you can only make one reply per question.

Here are the first 5 problems;

1. A square has an area of 25. What is the perimeter of this square?

2. In a school election, there are 14 people running in the election. If they all have an equal chance of winning, then what is the probability that a certain person will win? Express as a decimal to the nearest thousandth.

3. In how many ways can you arrange 7 books on a bookshelf, if they are all seven different books, and they can be arranged in any way possible?

4. What is the value of the following sum?

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + ... + 50

5. What is the sum of the area and the perimeter of a triangle on a coordinate grid with the coordinates (0,0), (6,8), and (6,0)?

Good luck solving these problems, and if you don't do good in this tournament, you may still participate in a later tournament hosted by me again.

PLEASE DO NOT LOOK AT THE BELOW UNLESS YOU ARE

WICKEDESTJR.

Honor code.

[hide="1"]Since the area is $ 25$, the side length must be $ 5$.

So, the perimeter is $ 5 \cdot 4\equal{}20$.[/hide]

[hide="2"]They all have an equal chance of winning, so they

each have probability $ \frac{1}{14}$ of winning.

$ \frac{1}{14}\equal{}.07\overline{142857}$, or at a nearest

thousanth, $ .071$.[/hide]

[hide="3"]There are $ 7$ choices for the first book, $ 6$ for

the second, and so on, for a total of $ 7!\equal{}5040$.[/hide]

[hide="4"]The formula for the sum of an arithmetic series is

$ \frac{(A\plus{}B)n}{2}$, where $ A,B$ are the first and last terms,

and $ n$ is the number of terms.

So, $ \frac{(1\plus{}50)50}{2}\equal{}1275$.

Note that this formula can be generized for $ 1\plus{}2\plus{}3\plus{}\hdots\plus{}n$.

Plugging in, we have $ \frac{n(n\plus{}1)}{2}$ for this formula.

[/hide]

[hide="5"]We have a right triangle, with legs $ 6,8$. So,

the hypotenuse is $ 10$. Therefore, the perimeter is $ 24$.

The area is $ \frac{6 \cdot 8}{2}\equal{}24$.

So, $ 24\plus{}24\equal{}48$[/hide]

If there is more people in the tournament, then I will probably make it get harder at a faster rate because it'll narrow it down to less contestants at a faster rate. What do you guys think about this tournament? Do you think it is going to start getting interesting later when there is more replies?

Good job dynamo729, you made it to the next round.  You also got a full score, and by the way, the last day that you can submit your answers is May 31.

So far here are the four that are going to the next round. This is just so far, if you get a full score, or do really good, then you will make it too;

1. BOGTRO
2. Urc
3. 12markkram34
4. dynamo729

Good job to you four, and there is more to come. If any of you will not be able to participate in the week of June 1st through June 7th inclusive for the next round, then please notify me.

Okay, we have five going on to the next round, but starting now, please private message me the answers.

Please private message me the answers.
Please private message me the answers.
Please private message me the answers .

Here are the five going to the next round so far ;

1. BOGTRO
2. Urc
3. 12markkram34
4. dynamo729
5. isabella2296

Congratulations to the five of you. Everyone else please private message me the answers.
At the next few rounds, I will want you to private message me the answers, so there shouldn't be any more submitting answers, but it is okay if you did, but others will see your answers.

Okay, we have a total of 13 people that have submitted their answers for the competition. I think that everyone, including the two people that got 45's have a big chance of going on to the next round.

1. BOGTRO - 50
2. Urc - 50
3. 12markkram34 - 50
4. dynamo729 - 50
5. isabella2296 - 50
6. Yongyi781 - 50
7. eshen - 50
8. michaelwong256 - 45
9. superllamel - 50
10. kl2836 - 50
11. CountdownKing - 50
12. Canton - 45
13. Ihatepie - 50

Below is information for each player and some tips for upcoming rounds. There is also information for people that are not participating in the next round, but would like to.

 This is information for the following players ; BOGTRO, Urc, 12markkram34, dynamo729, isabella2296, Yongyi781, eshen, superllamel, kl2836, Countdownking, & Ihatepie.
The 11 of you have autimatically made it to the next round. The next round will consist of 5 Algebra related problems. The top 20 contestants in the next round will participate in the following round.

Below is information for the following two players ; michaelwong256, and Canton.

The 2 of you are very close to participating in the next round. If there aren't fifty people that got scores of 50, then you two will participate in the next round. The next round will consist of 5 Algebra related questions. The top 20 in this round will get to participate in the third round.

Below is information for the people that didn't make it to the second round.

 If  you didn't, then don't worry, because if there isn't atleast 20 people that submit their answers next week, then I will post the five problems for the next week on this post. Then I will chose from these people who moves on. Below is the order in which I will chose from the people that participated in the second round.

1. First I will let the top people from the people that actually qualified for the next round, move on.
2. If there isn't twenty people or more that make it to round two that post their answers, then I will post the questions on this post. The people that have the next largest shot at moving on are the people that actually participated in Round 1. I will let people in based on their scores in Rounds 1 & 2.
3. If there is still not twenty people, then I will let people that didn't even participate in
Round 1 move on based on their scores.

If their is something that you do not understand about the tournament, then please private message me the question, or post your question.

Congratulations to the thirteen of you!

Okay, i am no longer excepting answers from new tournament competitors. These are the nineteen people that will be moving on to the next round ;

1. BOGTRO - 50
2. Urc - 50
3. 12markkram34 - 50
4. dynamo729 - 50
5. isabella2296 - 50
6. Yongyi781 - 50
7. eshen - 50
8. michaelwong256 - 45
9. superllamel - 50
10. kl2836 - 50
11. CountdownKing - 50
12. Canton - 45
13. Ihatepie - 50
14. mathcrazed - 50
15. hello.world - 50
16. LarryTheCucumber - 50
17. asethi01 - 45
18. Havingfitz - 30
19. lotsofmath - 50

   These nineteen contestants will be moving on to the next round. I will be private messaging you the questions, and you will be private messaging me the answers. Only the top 18 of these nineteen will be going on to Round 3. Below is a chart of what you have to do to get to each round.

Round 1 - You must be in the top 50.
Round 2 - You must be in the top 18.
Round 3 - You must be in the top 17.
Round 4 - You must be in the top 16.
Round 5 - You must be in the top 15.
Round 6 - You must be in the top 12.
Round 7 - You must be in the top 11.
Round 8 - You must be in the top 10.
Round 9 - You must be in the top 9.
Round 10 - You must be in the top 8.
Round 11 -  You must be in the top 7.
Round 12 - You must be in the top 6.
Round 13 - You must either be in the top 2, or you one of the top 3 in Countdown Round.
Round 14 - You must be in the top 4 in Countdown Round.
Round 15 - You must give your best possible solution to 3 problems that I have made.
Round 16 - You are in the top 2 on a certain set of problems.
Round 17 - In order to be the winner, you must do better than the other person in 2 out of 3 of
               the following;

                1. You must beat the other person in Countdown Round
                2. You must beat the other person in answering a list of hard problems
                3. You must have the highest total score out of everything in the whole tournament.

Congratulations to all nineteen of you, you each have a 5.26% chance of winning.

   If you have any other questions about the tournament, then please private message me your question. You should private message me your answers by June 7th. June 7th is the last day.

  If you would like to know why you got a certain score then please private message me your question.

There is much more to come.

Also, for #5, in order to get the intended answer, eradicate the line where it says

$ x^2 \plus{} \frac 1{x^2} \equal{} 19$

*Eradicate: [i]v[/i]. To complete destroy or remove; erase.

[quote="Wickedestjr"]Urc, there are already people that have gotten the problem right that assumed it was linear.

  BOGTRO, in order to do the countdown round, we'll be scheduling competitions at AoPS for the win, where you'll be competing against others. Don't worry about it now though, because it is a long time from now.

  This is a question to anyone participating in Round 2 and anyone that has a good opinion about this question. If there is a competitor that has autimatically made it to the next round, then should I give them the questions for the following week, or should i just wait til the next week?[/quote]

problems... FTW... Memmers. will. destroy.

Wait til next week to send... otherwise you won't know who's finished and who isn't.

Not to be off topic, but it's like the proverb, "Pride comes before a fall." See you down later.

I forgot to inform you guys that there was a twentieth contestant that had submitted their answers, but I had not noticed, but it was on time, so another contestant that is in the tournament is hurdler.
  So far we only have 6 people out of twenty that have submitted their answers and those people are listed below;

1. Urc
2. isabella2296
3. hello.world
4. Yongyi781
5. BOGTRO
6. mathcrazed

   We can not start the next round unless 12 of the other contestants submit their answers. I have already heard that 12markkram34 and Havingfitz should be submitting his answers, but the other thirteen still need to also. These are the fourteen people that still need to submit their answers. If you know any of these people then please tell them that they need to submit their answers in case they haven't noticed.

People that haven't submitted their answers.

1. 12markkram34
2. dynamo729
3. eshen
4. michaelwong256
5. superllamel
6. kl2836
7. CountdownKing
8. Canton
9. Ihatepie
10. LarryTheCucumber
11. asethi01
12. Havingfitz
13. lotsofmath
14. hurdler

I have seen many of these people posting things on this post, but haven't been submitting their answers, it is okay if you submit three answers one day, and 2 answers another day, or something similar to that, but I will only grade you on what you do submit, so if you submit three answers, then you can only get a score of 30/50.

For number 4 assume that it is linear, and for number five only pay attention to the part that says x + 1/x = 20, that is the only part that is important. It is okay if you don't ignore it, but if you do, then it is important that you get a reasonable answer that deserves a 10 out of 10.

PLEASE SUBMIT YOUR ANSWERS!!!  :thumbup:

When I said I wanted you to submit your answers, I meant that I wanted you to private message me your answers. So, if you don't want to do it because of that, then you were wrong, and I didn't say it clearly enough.

Oh, LOL  :rotfl:   :rotfl:   :rotfl:   :rotfl:   :rotfl:   :rotfl:   :rotfl:   :rotfl:   :rotfl:   :rotfl:

  Okay, so far, of the seven people that have submitted their answers, there are 5 that have scored high enough to autimtically move on to the next round. There are twenty people that are participating in this round, and only the top eighteen people can participate in the next round. There are going to be two contestants that do not make it to Round 3.

  P.S. - Make sure that you practice countdown round, because if you make it to the top 6, then you will be participating in it to make it to the next round. There might also be problems that people have memorized the answers to, so you might want to practice a lot, because you have a couple weeks.

  There is going to be a poll for this tournament that should be up and ready by tonight.

Hmm, so far only Yongyi781 and hello.world are the only people that have solved the problems. I think that CountdownKing quit. CountdownKing, unless you still want to play, you are eliminated. I will wait for you to confirm.

Well, you still did really well, and you probably would have done way better if you had been a little bit more active. Your score was 190, but you only answered the questions for four of the rounds, so a maximum total of 200, so you got a 95%. You did pretty well.

Has anybody seen Canton or asethi01 online at all lately?

I haven't seen canton post in a month.

Asethi01 has only one post...from 1 motnh ago.

Hmm, neither of them post that oftenly anyway. They did confirm though. How long should I give them? I could just start the next round and have them do all ten problems.

Alright, it is now time to start a new round. I will not reveal the scores until the competition is over, but it is obvious that CountdownKing has been eliminated. I will send the next set of questions to the remaining four people.

[quote="Wickedestjr"]I will send the next set of questions to the remaining four people.[/quote]

...Is there any reason why I haven't received the questions yet?

I haven't sent them yet. Sorry about that. I have been really busy. I will send them as soon as I get the chance.

Alright, I have determined the top 2 contestants. They will both recieve one lsat round of questions before the final round.

[hide="Summary"]Twenty people named BOGTRO, Urc, 12markkram34, dynamo729, isabella2296, Yongyi781, eshen, michaelwong256, superllamel, kl2836, CountdownKing, Canton, Ihatepie, mathcrazed, hello.world, LarryTheCucumber, asethi01, Havingfitz, lotsofmath, and hurdler joined a tourney. hurdler had joined a day late, but I was generous enough to let him in.

They answered five questions and none of them got eliminated.

Then round two came. eshen, michaelwong256, and LarryTheCucumber didn't answer the questions, and superllamel only had time for two of the five questions. michaelwong256 and LarryTheCucumber were the only ones who responded to the reminder pm's I sent. So I was forced to get rid of eshen. He would have made it farther if he hadn't been on vacation. He got 20th, and nineteen remained.

[hide]mathcrazed was one of the best contestants from the start, and he had not missed a single question. Unfortunately he quit, which resulted in eighteen more contestants.
[hide]
At round three there were four people that hadn't answered the questions. These four were michaelwong256, Urc, superllamel, and LarryTheCucumber. This was a round in which three people would get eliminated. The three who got eliminated in order were;

michaelwong256 with 45 points. He wasn't one of the better contestants, but he was one of the youngest, so eighteenth was good. Seventeenth place was;

LarryTheCucumber with 50 points. Except for eshen, (s)he was our least active contestant. Sixteenth place was;

superllamel with 70 points. He was only ten points behind Urc. He played a good game apart from his inactivity. The elimination of these three left fifteen contestants who were;

hello.world, Yongyi781, BOGTRO, dynamo729, 12markkram34, CountdownKing, Canton, lotsofmath, hurdler, ihatepie, asethi01, kl2836, HavingFitz, isabella2296, and Urc.

Four of these people had gotten every single question right. They were Yongyi781, 12markkram34, CountdownKing, and hurdler. Only one of these people got all five questions right at round four though, because hurdler quit, leaving us with fourteen. CountdownKing didn't answer the questions at all, and 12markkram34 missed two questions. This was also a round where three people had to get eliminated. So the other two who got eliminated where;

Urc got fourteenth place with a score of 80 points.

kl2836 got thirteenth place with 115 points.

12th place was BOGTRO. He was the first contestant to join, and one of the best, but he was inactive round five, so he lost.

11th place was Ihatepie. He was one of the better contestants also, but he was inactive rounds five and six, so he lost.

That left ten contestants.

Yongyi781 with 300 points. hello.world with 285 points. 12markkram34 with 270 points. Canton with 265 points. asethi01 with 265 points. lotsofmath with 220 points. CountdownKing with 190 points. dynamo729 and Havingfitz with 180 points. And isabella2296 with 176 points.

Yongyi781 and hello.world were the only ones to answer the questions for round 7. Yongyi781 had not missed a single question until this round, and he only got 35 points for that round, while hello.world got 50, putting him in the lead. isabella2296, was one of the best contestants, and also one of the most active, but she got eliminated in round 7.

lotsofmath quit, leaving eight contestants

I sent pm's to the remaining eight contestants asking them if they still wanted to play. One person didn't want to play, and two people didn't respond.

HavingFitz ddin't respond, so he got 8th place.

dynamo729 said he no longer wanted to play so he got 7th place.

Our third best contestant 12markkram34 was the next to lose. He could've gotten third place if he had replied to that pm, but he got 6th.

Then this left us with five contestants.

hello.world - 335 points
Yongyi781 - 330 points
Canton - 265 points
asethi01 - 265 points
CountdownKing - 190 points

CountdownKing realized there was no way he could make it to the top four, because he could get a maximum of 50 points, so he didn't answer the questions at all and he got 5th place. Yongyi781 and hello.world were the only two to answer the questions for this round.

Yongyi781 got 40 points and hello.world got 35 points. Yongyi781 took back the lead.

Canton and asethi01 were both inactive for the rest of the tourney, and they tied for third place.

This left two contestants. hello.world with 375 points, and Yongyi781 with 380 points.

There are two more rounds left.

The second to last round I will have them answer five questions.

The last round will have three bonus points.

One will be awarded to the contestant that has the most points at the beginning of that round.

One will be awarded to the contestant that scores the best on the very last round.

One will be awarded to the winner of a FTW countdown round.

I will send the five questions to them right now. [/hide][/hide][/hide]

It is a courteous thing, what Wickedestjr did.  So when searching for posts, a user doesn't have to go through a big blog of a mess

Sorry, just telling anybody that was curious.  :(

Okay, hello.world is the only one that has answered. Still waiting on Yongyi781.

Alright both contestants have sent there answers for the second to last round, and I am now sending them the questions for the final round.

@Yongyi781 and hello.world - What times can you two be on, because I am wondering when we should do the countdown round.


Can someone explain what reactions take place regarding deuterium and the mechanism/reason behind these? For example, addition of heavy water to $ \ce{CH3MgCl}$ or $ \ce{D2O}$ or $ \ce{D2}$ to $ \ce{C2H6}$.

I think the reactions should be the same as if we use $ \ce{H}$ instead of $ \ce{D}$.

1. $ \ce{CH3MgCl} \plus{} \ce{D2O} \rightarrow \ce{CH3D} \plus{} \ce{Mg(OD)Cl}$

$ \ce{C2H6}$ is saturated so I think you might mean $ \ce{C2H4}$.

2. $ \ce{C2H4} \plus{} \ce{D2} \rightarrow \ce{CH2DCH2D}$

3. $ \ce{C2H4} \plus{} \ce{D2O} \rightarrow \ce{CH2DCH2OD}$

The mechanisms of these reactions can be found in most organic chemistry text books or wikipedia :showoff:.


Divide the euclidian plane into three dense subsets $A$, $B$ and $C$, such that any line meets only two of these subsets.

Hi,

Did you happen to see a reasonable solution ? I have thought about it for quite a while now but I couldn't come up with anything. For 4 denses subsets instead of 3, the problem is of course easy.

Hi, the only construction I know uses the axiom of choice  :lol:

Yeah, I eventually saw this problem in field theory. The solution I read was of the kind I expected, but I couldn't find it myself. Nice result, by the way.


im in for 13. any body who says i wasnt already in, think again. i signed up on page 9,

Post 11

That was after it closed. You will be the first choice for a replacement if someone is inactive.

Please stop creating new threads! This should have been posted in the Mafia Sign-up thread.

This really needs to be deleted, as does the other "13" thread.

kheg_k, keep in mind that the Games and FF forum is not just a mafia forum.  Not everyone who uses this forum does so for the purpose of playing mafia.  So mafia-related stuff should be kept in the signup thread and the threads devoted to each game.


Are u sure on the third equation that $xy=zt$??
Isn't it $sy=zt$?

[hide]

 and  give  and .

From  we have  which used with  gives .  Combined with  this gives

Then we can solve for  in terms of  to get   .  Then we have from  , which allows us to solve for , and hence the other variables.

doing this gives

[/hide]


AMC USA(J)MO USAMO Harvard college MIT geometry

I understand that USAMO is not at all a guarantee for the acceptance to Harvard, MIT and such schools. But I was wondering is there a kind of safe school (please list if you have any suggestion) for all USAMOers? What is the least prestigious university that a USAMOer has ever been to? (no offensive to any university, just curious about this)

Thank you.

You're trying to overgeneralize about students who make the USAMO - the quality of student who does so varies tremendously, and making the USAMO shows little other than that you're good at math.  Many USAMOers go on to extremely prestigious universities (MIT, Caltech, Harvard, etc.), while others don't make it to those schools because they're not very well-rounded, or don't have very good grades in non-math classes, etc.

Making the USAMO is not easy to do, and it certainly looks good on your college application if you make it - however, admissions committees will be more interested to see that it's part of a pattern of dedication to and passion for mathematics.  More importantly, they will evaluate it in the context of the rest of your application, i.e. your grades, test scores, and what else you choose to do with your time.

What zanttrang said. Apply to any reasonable safety school that meets your needs. My definition of a safety school is a school that

1) is pretty much certain to admit my kid, based on its known behavior in acting on admission applications,

2) has a strong program in an area my kid is interested in,

3) is affordable based on its known behavior in acting on financial aid applications,

and

4) is likeable to my kid.

The state university in my state fits those characteristics for my oldest son. It has "holistic" admissions now, which means it can accept or reject any applicant for any reason or for no reason at all, but in actual practice will offer admission to all students who successfully complete the math program there that my son now attends. So that is one school he can get into even if he never qualifies for USAMO.

As for all other schools, once you have chosen a sure-bet safety school, just be sure that they are reasonable choices for YOU, whether or not they are as prestigious as someone else's choices. Apply where you would be happy to attend, and apply to more than one college if you are mainly applying to colleges that are hard to get into.

Good luck. Have you already taken the USAMO before?

you can pick a major and go to usnews for the ranking or any other source and go from there.  They have schools that fit for B students too.

The University of Michigan ended up being my safety school although that wasn't completely intentional.  I meant to apply to USC (South Carolina) and/or Clemson but didn't get around to it.  However, I knew that I wouldn't go in-state so it didn't really matter.

I think UMich is too selective of a school to be a safety. On the other hand, they really like USAMO qualifiers.

It depends...case by case. My safety school this year was pretty much MIT, in the sense that I was fairly confident about it, not that it's somewhere I'd go if I had no other choice. But I definitely also had Duke on my list, in case MIT EA didn't work out.

It seems if the kid is decently well-rounded, a lower-level Ivy or Duke, Michigan, Berkeley, etc. can be the safety.

I suppose the thing about MIT, Caltech being the safety is that you can't REALLY be sure about getting in. But if you EA one or both of those and have a real safety to fall back on for RA, then most USAMOers will be fine. :) Especially ones that qualify many times...

[quote="doctalinh"]:huh: I don't want to get off topic, but....

MithsApprentice, I've read some of your posts recently, and though I understand that you're happy about getting into MIT, you seem pretty cocky about it.  I think doing well on the USAMO's or any math contest is something to be proud of.  But I don't think it can really guarantee admission anywhere.  I think you should also remember that it doesn't mean you're inherently better at math, or smarter, than someone who hasn't done such contests...

 :wink:[/quote]

That's not how I read it, and my experience is that college admissions are a lot less random than most people think for very highly qualified applicants.  I have very rarely been surprised by people who I thought were very qualified being rejected.  I have been surprised in the reverse direction, but that's a different story.

Miths has never really said that USAMO guarantees a spot at College X.  I have read posts where he says that the correlation between a person who makes USAMO and a person who also does well in other relevent fields is generally high.  I'd have to agree with this.

I can see how Miths might come across as cocky.  I would attribute this to the internet effect where it's very easy to misinterpret how someone meant something.

To comment on what Miths said about having an ivy/elite as a back-up, this isn't as crazy an idea as it might first come across as.  In order to do this though you need to be a "shoe-in" for the ivies and send in applications reflecting this.  It's a known fact that top colleges have turned away the cream of the crop seemingly on a whim.  The more top schools a shoe-in applies to, the less likely it is that they'll turn you down.

I should note that I definitely did not consider myself a shoe-in for the elite schools.  I knew I had a good shot at getting in though and just randomly estimated it at 50:50 odds.

When I was applying to colleges, my backup elite was Princeton.  I got rejected.  It's almost like they knew they were my backup :D  Come to think of it, I actually put down an accurate description of myself in my application.  They also knew I was applying to MIT.  Maybe they looked at my app, saw I was a good fit for MIT, and made things easier for me.  Who knows.  On the one hand I was glad I got rejected since that made making a decision easier for me.

By the way, what Miths and I did when we applied may have seem rash since we didn't have any real safeties.  I wouldn't consider them true safeties at least.  However, it's not so much that we (or me, at least) were so arrogant as to think that we MUST get in and we don't need a safety.  It's more that we didn't want to do any more work than necessary.  Had the world collapsed on me and I had been rejected from MIT, UMich, and UVA I'm pretty sure I could still have gotten other schools to take me in even if it was months after the deadline (e.g. UF, UCF, or just any school that will do anything to get Nat'l Merit Finalists).

Well I know that Stanford deferred a friend of mine who happened to be better at computer science than any other undergrad in their CS department, so some schools do do things somewhat whimsically (I am fairly sure he will get in RD, but still...).

Oh and I know he is better because their top ACM guy also went to our school, so I can make a direct comparison with their self-declared "best" person.

Having ACM skills doesn't really make you 'better' at CS than any other set of skills.  Also, Stanford didn't do that well on ACM is previous years so it's possibly not much of a priority ('both' of my schools, SFU and UBC, beat them in the qualifiers for world finals this year).

It's possible to be a complete flake and still do well on math contests; one MOP attendee I remember wound up at a community college due to his bad grades. This is not the norm; most USAMO students also have top-tier grades and test scores. With that profile, USAMO qualification will be noticed by top schools, and certainly helps your chances. You don't need insane breadth if you have depth.
If you don't have the grades and test scores, the USAMO might not help you at all. Lesser schools won't recognize the contest or what it means.

When I applied to colleges, I only sent in three applications. My first choice was Berkeley (in-state), and that was essentially certain. They didn't even need to look at the superlatives; the grades and test scores passed the first cut for automatic admission.
I did get waitlisted by Princeton, but I outright said that Berkeley was my first choice in the interview.

Will USAMO qualification even be considered at all for some not-so-elite schools?

Probably not so much.  It's a bit of a catch-22; the schools who know what it is are harder to get into, which makes things tough on people who have USAMO but nothing else; the schools who don't know what it is tend to be more concerned with the standard measures of high school 'intelligence' - grades, SATs, etc, and may either not know what it is nor weigh it conspicuously more than any other math contest result.

From what I noticed applying this year, MIT and Caltech seem to be the schools most interested in USAMO qualification since there is a spot on the form asking for an AIME score.

[quote="doctalinh"]But I don't think it can really guarantee admission anywhere.  I think you should also remember that it doesn't mean you're inherently better at math, or smarter, than someone who hasn't done such contests...[/quote]

My point is simply that USAMOers (and especially MOPpers) have very high acceptance rates at schools like MIT, Caltech, Pton, Harvard, etc. This doesn't mean that I personally feel like I "deserve" to be admitted at any or all of these places.

USAMO qualifiers in general do demonstrate very high levels of academic and extracurricular achievement, often displaying the "intellectual vitality" and "academic excellence" (to quote Stanford) that the top schools look for. And this certainly neither begins nor ends with me. I'm just proud to be in a group of individuals that have historically (and continue to) demonstrate such remarkable success.

Of course USAMO does not guarantee acceptance anywhere, but it is certainly associated with dedication and achievement that elite schools appreciate. :) And with that in mind, although USAMOers may not [i]expect[/i] to get into every school they apply to, realistically there is no harm in realizing that one's chances are fairly great. Thus there may not be a pragmatic need to apply to "real" safeties like state schools. Like Joe said, it's often a matter of doing what is necessary...not pride or arrogance.


algebra
polynomial
Rational Root Theorem

Prove that if $ p$ and $ q$ are odd integers, the equation $ x^2\plus{}2px\plus{}2q\equal{}0$ cannot have rational roots.

The only way to factor this is $ (x\plus{}a)(x\plus{}b)$, where $ ab\equal{}2q$ and $ a\plus{}b\equal{}2p$.

However, since $ q$ is odd, we will always have one of $ a,b$ odd and the other even, so we can't have an even sum of $ 2p$.

[hide] From the rational root theorem, all rational roots must divide $ 2q$. Clearly, no root can be odd, otherwise we have $ \text{Odd\plus{}Even\plus{}Even\equal{}0}$, which is a contradiction. Thus, if there are rational roots, they must be 2 or -2. Plugging in 2, we have $ 4\plus{}4p\plus{}2q\equal{}0$, so $ 2p\plus{}q\equal{}\minus{}2$. However, 2p is even while q is odd, so their sum must also be odd, which is a contradiction.

Plugging in -2, we have $ 4\minus{}4p\plus{}2q\equal{}0$ so $ q\minus{}2p\equal{}\minus{}2$. q is odd while 2p is even, so their difference must be odd. Another contradiction. Hence, there are no rational roots.[/hide]

[hide="Proof? (Please comment)"][b]Lemma.[/b] The difference between two odd squares is divisible by 4.

[i]Proof.[/i] $ (2n \plus{} 1)^{2} \minus{} (2k \plus{} 1)^{2} \equal{} (2n \minus{} 2k)(2n \plus{} 2k \plus{} 2) \equal{} 4(n \minus{} k)(n \plus{} k \plus{} 1)$, by the "difference of two squares" identity.

The roots of the equation are $ \frac{2p \pm \sqrt{4p^{2} \minus{} 4(2q)}}{2} \equal{} p \pm \sqrt{p^{2} \minus{} 2q}$. Now, for these roots to be rational, $ p^{2} \minus{} 2q \equiv 1 \mod 2$ has to be a perfect square, but that implies $ \frac{p^{2} \minus{} n^{2}}{2} \equal{} q$ is even (by our lemma), which contradicts the problem statement. [/hide]

[hide="A solution"]
\[ x^2 \plus{} 2px \plus{} 2q \equal{} 0 \implies x \equal{} \frac { \minus{} 2p \pm \sqrt {4p^2 \minus{} 4(2q)}}{2} \\
\implies x \equal{} \minus{} p \pm \sqrt {p^2 \minus{} 2q}
\]
If we assume that these are rational roots then

$ \minus{} p \pm \sqrt {p^2 \minus{} 2q} \equal{} \frac {m}{n}$

$ \Rightarrow \sqrt {p^2 \minus{} 2q} \equal{} \frac {r}{s}$, where $ r$ and $ s$ are relatively prime.

$ \Rightarrow s^2\left(p^2 \minus{} 2q\right) \equal{} r^2$

For any odd number $ (2k \plus{} 1)^2 \equal{} 4k^2 \plus{} 4k \plus{} 1 \equiv 1 \bmod{4}$

Likewise for any even number $ (2h)^2 \equal{} 4h^2 \equiv 0\bmod{4}$.

Since $ p$ and $ q$ are both odd, $ p^2 \minus{} 2q \equiv \minus{} 1 \bmod{4}$

Now if for $ (r,s)$ one is even and the other is odd then $ \minus{}s^2 \equiv r^2 \bmod{4} \implies \minus{} 1\equiv 0\bmod{4}\textrm{ or } 0\equiv 1 \bmod{4}$ either of which is a contradiction.

If for $ (r,s)$ both are odd then $ \minus{} s^2 \equiv r^2\bmod{4} \implies \minus{} 1\equiv 1 \bmod{4}$, another contradiction.

If for $ (r,s)$ both are even, then $ r$ and $ s$ are not relatively prime.

Therefore for any $ p$ and $ q$ both odd, $ x^2 \plus{} 2px \plus{} 2q \equal{} 0$ has no rational roots. QED[/hide]

we'll prove that:

i) no root of the equation is an even integer

if x is even, x^2 and 2px are multiples of 4 and 2q is not. this is not possible.

ii) no root of the equation is an odd integer

if x is odd, 2px + 2q is even whereas x^2 is not. this is not possible.

iii) no root of the equation is rational

x^2 + 2px + 2q = 0

=> (x+p)^2 = p^2 - 2q.

from (i) and (ii), x is not an integer. let it be a fraction. but then, the LHS is a fraction while the RHS is not. so, no root of the equation is rational.

therefore, no root of the equation is rational. :)


probability
algebra
polynomial
limit
conditional probability
probability and stats

How to solve this problem: fix a number $ p\in ]0,1[$ and a positive integer $ l\in \{3,4,\ldots,70\}$. A frog starts jumping at the point $ 0$ in the real line and for every integer $ n$, if the frog is at the point $ n$, it has probability $ p$ to go to the point $ n \plus{} 1$ and probability $ 1 \minus{} p$ to go to the point $ n \minus{} 1$. If the frog goes to the point $ \minus{} 1$, it will be eaten by an evil eagle. What is the probability that the frog will reach the point $ l$?

Let $ f(n)$ be the conditional probability of the frog ever reaching point $ \ell$ given that its current position is $ n.$ We have the boundary conditions $ f(\minus{}1)\equal{}0$ and $ f(\ell)\equal{}1.$

In between, $ f$ satisfies the equation $ f(n)\equal{}(1\minus{}p)f(n\minus{}1)\plus{}pf(n\plus{}1).$ This is a second order constant coefficient homogeneous difference equation, with characteristic polynomial $ pr\minus{}r\plus{}(1\minus{}p)\equal{}0.$ The roots of this equation are $ r\equal{}1$ and $ r\equal{}\frac{1\minus{}p}p.$ This leads to the form $ f(n)\equal{}c_1\plus{}c_2\left(\frac{1\minus{}p}p\right)^n.$

Solving this for the boundary conditions gives $ c_1\equal{}\frac1{1\minus{}\left(\frac{1\minus{}p}p\right)^{\ell \plus{}1}}$ and $ c_2\equal{}\frac{\minus{}\frac{1\minus{}p}p}{1\minus{}\left(\frac{1\minus{}p}p\right)^{\ell \plus{}1}}.$

From this the required probability is $ f(0)\equal{}\frac{1\minus{}\frac{1\minus{}p}p}{1\minus{}\left(\frac{1\minus{}p}p\right)^{\ell \plus{}1}}.$

All of this assumed that $ p\ne\frac12.$ If $ p\equal{}\frac12,$ we have a different solution to the difference equation, a linear solution. If $ p\equal{}\frac12,$ we get that $ f(n)\equal{}\frac{n\plus{}1}{\ell\plus{}1},$ meaning that $ f(0)\equal{}\frac{1}{\ell \plus{}1}.$

In the case $ p>\frac12$ (the frog more likely to jump forwards than backwards), we can talk about escaping to arbitrarily far away (and never being eaten). The probability of that would be

$ \lim_{\ell\to\infty}f(0)\equal{}\lim_{\ell\to\infty}\frac{1\minus{}\frac{1\minus{}p}{p}}{1\minus{}\left(\frac{1\minus{}p}p\right)^{\ell \plus{}1}}\equal{}1\minus{}\frac{1\minus{}p}{p}\equal{}2\minus{}\frac1p.$

If $ p\le\frac12,$ the probability of escaping to arbitrarily far away is zero and the probability of being eaten is one.


search
number theory unsolved
number theory

[quote="Beat"]Let $k = 2^{2^n}+ 1$ for some $n\in N$. Show that $k$ is prime if and only if $k$
divides $3^{\frac{k-1}{2}}+1$.[/quote]
It is Pepin test.

Use search button - it works really...

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Taiwan+1997&t=39083


A sequence $x_n,n\ge 0$ ,satisfy $x_{n+1}=x_n+\frac{1}{\sqrt{x_n}}$  .Find $\displaystyle\lim_{n\to\infty}x_n$  if  $x_0> 0$.

$ (x_n) $ is (strictly) increasing, so it admits a limit (not necessarily finite).
then suppose $l \in \mathbb{R}$ to be its limit, plugging it into the recursion we get $\displaystyle l = l + \frac{1}{\sqrt{l}}$, that is $\displaystyle\frac{1}{l} = 0$. contraddiction. then $\displaystyle\lim_{n\to\infty}x_n = +\infty$.

Your solution is similar to the official solution :) .I have showed that $x_n> \sqrt{2n},n\ge 1$.

isn't there a really simple solution for this?
you can define the inifinte series of 1/(sqrt n), which we all know is divergent by P-series, therefore the series of n + 1/(sqrt n), is divergent as well, therefore the limit does not exist and is positive infinity(because you have an increasing series)

THIS IS  THE POWER OF CALCULUS :D

The sequence would still be divergent if $x_{n+1}=x_n+\frac1{x_n^2}$, or indeed $x_{n+1}=x_n+f(x_n)$ for any continuous and strictly positive $f$. Divergence of the series $\sum\frac1{f(n)}$ has nothing to do with it.

blahblahblah,
you are right but i don't think that changes much! you can still aply the same approach in this case.

What you are saying is complete nonsense.  We are talking about a sequence here, not a series.  Hence anything about $p$-series is irrelevant.  Furthermore, the divergence of a sum does not imply that its terms go to infinity - we need only take our favorite harmonic series to see this.


Is there a new upgrade in the rate post part???

Umm... Someone should move/lock/delete this.

Yeah, I just noticed that this should be moved.


inequalities
function
probability
expected value
real analysis
real analysis unsolved

It suffices to prove the case $k = 2$.

Let $X$ be a random variable that is identically distributed to $Y$ but independent of $Y$.  Because $y^\beta$ is an increasing function of $y$, we have
\[(X^{\beta1}-Y^{\beta_{1}})(X^{\beta_{2}}-Y^{\beta_{2}}) \ge 0 \, . \]
Now take the expected value of both sides, and you will get what you want.

This inequality is related to Chebyshev's inequality.


summer program
MathPath
MATHCOUNTS
articles
geometry
3D geometry
pyramid

I thought MathPath was only for high school school students, is it not? Or is it only for middle school students?

Cool article. I know some of the people mentioned in it.

To 236factorial: Mathpath is for middle school students. http://www.mathpath.org

That's right, no wonder [color=darkblue]<xxxxxxxx>[/color] was in there, he's like in 7th grade right?

does anyone know where mathpath will be next year?
i want to try it.

All in all, Mathpath was frankly the best camp I have ever been to!

 :)

The food was pretty good, although breakfast needed a little more variation.
The lectures were awesome; I learned so much! It met if not exceeded my high expectations.
The dorms were relatively awesome, although Larry the Dorm Manager would sort of be like a zombie if you stay out past 10:30. If Larry catches you, you're DEAD. (well not really, but he'd be all like 'what're you doin?'

Overall it was awesome!

Math92: Nobody will know when the next Mathpath is if it's 2 days since this Mathpath has ended. Rumor has it that it's gonna be in California.

Wow, you should definitely have gone to MathPath! I'll post my solutions, I made it (but couldn't go :( ) but my solutions aren't that good. EFuzzy's solutions are amazing!

[quote="10000th User"][quote="G-UNIT"]Since mathpath is over, is it OK if we discuss the solutions to the qualifying quiz, if there is not already a thread in place to do so.  I've only found out about mathpath recently, and solved some of the problems, so it would be great if somecould to post the answers or solutions to the qualifying quiz.[/quote]
I go to High School next year so I cannot go MathPath anymore :(  But I have some answers to Quiz

[hide="Quiz answers"]1.a)1
b)8

2.$x^2+y^2+z^2=\left(\sqrt{\frac{ab}{c}}\right)^2+\left(\sqrt{\frac{ac}{b}}\right)^2+\left(\sqrt{\frac{bc}{a}}\right)^2=\frac{a^2b^2+a^2c^2+b^2c^2}{abc}$

3. I skip this problem. I don't understand it and cannot find age... :blush:

4.Big pyramid $V_B=\frac{1}{3}BH$

Small pyramid $V_S=\frac{1}{3}bh$

Frustum, height $h_f=H-h$, $V_F=V_B-V_S$

$V_B$ and $V_S$ are similar, $\frac{h}{H}=\frac{h}{h+h_f}=\sqrt{\frac{b}{B}}$

$h=\frac{h_f\sqrt{b}}{\sqrt{B}-\sqrt{b}}$

$H=\frac{h_f\sqrt{B}}{\sqrt{B}-\sqrt{b}}$

$V_F=\frac{1}{3}(BH-bh)=\frac{h_f}{3}\left(\frac{B\sqrt{B}-b\sqrt{b}}{\sqrt{B}-\sqrt{b}}\right)$

$V_F=\frac{h_f}{3}(B+\sqrt{Bb}+b)$

5.a)$\frac{1}{3}\times 1\times 1^2=\frac{1}{3}$
b) Put ABCD flat and shortest distance is AC. $AC=2\sqrt{\frac{2}{3}}$

6. $n=11$ because if $n=10$, $|x_1|+|x_2|+\cdots+|x_{10}|<10$

7.a) Let $d$ be digit rotation, $N$ is block of (n-1)-digits never change, $n$ is number of digits in number. $2(10^{n-1}d+N)=10N+d$

$(2\cdot 10^{n-1}-1)d=8N$. $d$ cannot be odd digit because $2\cdot 10^{n-1}-1$ is odd but $8N$ is even.

$d\neq 2$ because $2\cdot 10^{n-1}-1$ is odd but $4N$ is even

$d\neq 4$ because $2\cdot 10^{n-1}-1$ is odd but $2N$ is even

$d\neq 6$ because $3(2\cdot 10^{n-1}-1)$ is odd but $4N$ is even

$d=8$ and $N=2\cdot 10^{n-1}-1$ but impossible because this is n-digit number and $N$ is (n-1)-digit number

Final answer: none

b) $2(10N+d)=10^{n-1}d+N$, $19N=(10^{n-1}-2)d$. But 19 is prime and $d<19$ and $19|10^{n-1}-2$. I work power mods

$10^1\mod19=-9$
$10^2\mod19=(-9)^2=81=5$
$10^3\mod19=-9\cdot 5=-45=-7$
$10^4\mod19=5^2=25=6$
$10^5\mod19=-35=3$
$10^6\mod19=30=-8$
$10^7\mod19=-42=-4$
$10^8\mod19=36=-2$
$10^9\mod19=18=-1$

I can skip because $10^{17}\mod19=10^8\cdot 10^9=(-2)\cdot (-1)=2$, $n-1=17$

But number is cycle, so $n=19k+18$

$N=\left(\frac{10^{19k+17}-2}{19}\right)d$

Because $N$ is 17-digit number, if $d=1$, first digit of $N$ is 0 :)

Final answer: infinite numbers, $\left(10\frac{10^{19k+17}-2}{19}+1\right)d=\left(\frac{10^{19k+18}-1}{19}\right)d$, $d=1,2,3,4,5,6,7,8,9$

8. Too hard, I don't know :blush:[/hide][/quote]

  Thank you for posting your answers.  I'm not sure if im right, but i think this is the answer to[hide="question 3"]

  Lets $a$ detonate the current age of Alice and $b$ the current age of Bob.  We know that Alice is twice as old as Bob was a certain time in the past.  Let the time that has passed from now till then, be represented by $x$.  Thus,

$a=2(b-x)$

  We also know at that time, Alice was 3 times as old as Bob was 3 years earlier than that.  Thus,

$a-x=3(b-x-3)$

  Solving, we see $3b-(2x+9)=a=2(b-x)$ which means,

$3b-(2x+9)=2(b-x)$

$3b=2b+9$

$b=9$

  Solving we see that $b=9$.

  Therefore we know that Bob is 9 years of age, but Alice's age is unknown. [/hide]

The correct answer to 5a is

[hide]
The entire cube has volume $1^2$ and the cut-out pyramid has volume $\frac{1}{3}*1*1^2$. So the solid with the pyramid removed has volume $\frac{2}{3}$.
[/hide]

Sorry, 10000th user. Are you really the 10000th user? :cool:

[hide="Solutions attached"]Be warned: some of them are wrong... :D [/hide]

[quote="236factorial"]I probably shouldn't have went to MathPath, I didn't get the questions at all  :blush:[/quote]

Shouldn't have [b]gone[/b]? :lol:

It's OK. I didn't get the questions either at first, but if you read over them, they start to make sense. I got stuck for several months on #8, and never solved it until someone showed me the answer.

[quote="236factorial"]Hey, don't get smart with my grammar, I only got 98 on my english final!  :mad:[/quote]
 :lol:  :rotfl:  :lol:  :rotfl:  :lol:

Thank you mathnerd314, I forget to subtract pyramid volume  :blush:

[hide="chess64"]  :o  too many advanced topics in your solution, you can solve these problems without advanced topics  :o

Very nice solution problem 1, you find last two digits

Problem 2 try to use symmetry: $xy=a$, $xz=b$, $yz=c$. $x^2=\frac{(xy)(xz)}{(yz)}=\frac{ab}{c}$ and do same with others. Done

Problem 7 there are left rotation numbers. Example: 105263157894736842. Multiply this number by 2 and you get left rotation[/hide]

Example of #6: $x_1=x_3=x_5=x_7=x_9=-1$, $x_2=x_4=x_6=x_8=x_{10}=x_{11}=1$

[quote]Math92: Nobody will know when the next Mathpath is if it's 2 days since this Mathpath has ended. Rumor has it that it's gonna be in California.[/quote]

I heard Bryn Mawr College in Philadelphia was a possibility.

Any way, MathPath is amazing.  I learned more math there than I thought was possible to learn in a month.  I highly recommend it to any one eligible to go next year (a group that sadly does not include me).

[quote="E^(pi*i)=-1"]Any way, MathPath is amazing.  I learned more math there than I thought was possible to learn in a month.  I highly recommend it to any one eligible to go next year (a group that sadly does not include me).[/quote]

You were at Mathpath too? :cool: How come I didn't know? :mad: :D

[quote="rcv"]MathPath ended yesterday, July 24.  Although a general discussion of MathPath is probably off-topic for the MATHCOUNTS Forum, I am designating this specific thread for such discussions.  You can discuss food, dorm life, and general information pertaining to the classes and activities.  This is not a thread to delve into the mathematics of the camp.

New AoPS'ers, recruited by old AoPS'ers are welcome to introduce yourselves here.  Please remember you are never to divulge another student's real name, and you should not divulge your own name without your parents' consent.

Also, there is an online article about MathPath at [url]http://www.gazette.com[/url].  The online article is dated July 25, and will not stick around forever.  So read it soon.[/quote]

wohoo...i was in the picture!

mathpath was fun and i was really sad it ended...

oh and if mathnerd keeps correcting your grammer, dont fell bad...i had to put up with him the whole camp

oh wel...

NOTICE (i editted... a way to stop spam!)
here it my solutions...

[EDIT]: this it the second time i eddited!  *gasp*

oh well...chess64 did you go to camp?

Gifted solve upper-level problems at camp

Amadeus Zhu, 12, responds to a question during a math exercise as part of Math-Path 2005 at Colorado College on Wednesday. Twelve-yearolds Johnny Li, left, and Michael Young are also enrolled in the program.

By BRIAN NEWSOME - THE GAZETTE

In a cavernous hall at Colorado College, some 60 students listen to Professor Tom Drucker lecture on the history of mathematics.

He talks about quaternions and octonions, and the evolution of calculus. He urges his students to use rigor in proving solutions to the complex problems that can take days to solve.

The thing is, school is out of session. And the students, well, they don’t go to CC. They’re in middle school.

It’s called MathPath, a summer camp for the profoundly gifted. The students, who come from 21 states and Canada, descended on CC on June 26. Selected for their unusually high abilities to grasp college-level math concepts, they’ve spent the past month racking their brains seven hours a day, five days a week — and loving it.

For most preteens and teens, MathPath would be the summer camp from hell. Yet for these students it’s a chance to indulge in their enthusiasm for an unpopular topic and meet like-minded friends who share their abilities.

“I’m definitely one of the few who like math at my school,” said Amber Verser, 13, of Loveland. Verser started reading clocks as a toddler. She’s since scored a perfect 800 on the math portion of the SAT. On the American Invitational Mathematics Examination, a test for some of the nation’s brightest math students, she scored a five out of 15, double the average score for a gifted high school student.

George Thomas, a mathematics scholar who started MathPath four years ago, said the program is designed to teach students to “think hard.” The concepts, taught through lectures and hands-on problem solving, are the same that college math majors learn.

“Ideally, the campers who most benefit from this will be those who become mathematicians and who go into mathematics-related fields,” he said.

To find students, MathPath sends applications to about 59,000 middle schools nationwide. About a thousand personal letters are mailed to students who have stood out at various math competitions. Those students make up about 90 percent of the applicants.

To be admitted, a student must successfully solve at least five of eight math problems on the application. Some of the problems, Thomas said, are designed to pose a multiday challenge for a universitylevel math student. Of 400 applications, 68 students were accepted and 66 attended.

“I would say these students are like 1 in 10,000 in the middle schools,” Thomas said.

The camp, which ended Sunday, takes place at different campuses each year. Last year it was in Rhode Island.

This year, five students attended from Colorado, one of them, Allan Gardner, 11, from Colorado Springs. Gardner attends Holy Trinity Catholic School.

Students say it’s hard for them to describe where their passion for frustratingly hard problems comes from. For Anand Natarajan, 12, of Boulder it’s the “visceral pleasure of getting a simple proof to a hard problem.” Charlie Stoddard, 14, of Steamboat Springs said there’s “an internal beauty of math.”

Many of the middle-schoolers are in independent study programs. Stoddard, for example, is taking multivariable differential calculus. He also scored perfect on the math portion of the SAT.

To connect with other math enthusiasts, they rely on the Internet. Their Web site of choice: artofproblemsolving. com.

Though math brings these middle-schoolers together, it’s not their only interest. Stoddard is a ski jumper and selfproclaimed jock. Verser, an avid reader, has read the newest “Harry Potter” book four times, and she wants to study medicine, music or veterinary medicine in college. Natarajan plays the piano.

After Drucker’s lecture Thursday, Natarajan and another student, Daniel Houck, 12, of Jefferson County walked across campus and told a joke that perhaps only the math enthusiast can appreciate:

“What’s the area of a circle?”

“(pi)r².”

“But pies are round.”

CONTACT THE WRITER: 636-0198 or

bnewsome@gazette.com

[quote="jli"]wohoo...i was in the picture!

mathpath was fun and i was really sad it ended...

oh and if mathnerd keeps correcting your grammer, dont fell bad...i had to put up with him the whole camp

oh wel...

NOTICE (i editted... a way to stop spam!)
here it my solutions...

[EDIT]: this it the second time i eddited!  *gasp*

oh well...chess64 did you go to camp?[/quote]
very nice solutions by 10000th User, chess64 and jli. impressive. if compared to when i was in 6th grade, i dont think i could have done them :blush:

a few notes/ideas: read everyone's solutions. many different approaches were presented.
chess64: try to keep the solution more elementary, the more elementary, the closer to 'The Book'-type of solution
jli: try to do problem 7b again and see if you can prove it :)
10000th User: your solution to #6 is wrong. rework it please. $|x_i|<1$

also #8 is a difficult problem and was solved by shobber i think, it's "somewhere" in MathLinks. ill post a link to it if i find it

EFuzzy gave me his solutions, I don't think he'd mind if I shared them with you guys. But I better ask him first. :P

[quote="236factorial"]I can't download jili's solutions because it's a .tex file. what program opens that?[/quote]

YOu don't have $\LaTeX$!??? Download it right now! :D I had to comment out the images because he didn't post the images. Jli, can you either post the images, or, better yet, the PDF file? :)

i do have Latex, it's jsut I get an error message when I open the file saying it can't be found, the address isn't right.

I have received EFuzzy's permission, but he told me to warn you that the pictures aren't there - he drew them in pencil. "However, you may have a hard time reading mine, since I had my pictures hand drawn. Take it as a challenge for you to figure out what I'm talking about. :D"

[quote="jli"]
wohoo...i was in the picture!

mathpath was fun and i was really sad it ended...

oh and if mathnerd keeps correcting your grammer, dont fell bad...i had to put up with him the whole camp

oh wel...

NOTICE (i editted... a way to stop spam!)
here it my solutions...

[EDIT]: this it the second time i eddited!  *gasp*

oh well...chess64 did you go to camp?[/quote]

Dude:
Woohoo has 4 o's, not 3.
Grammar is G-R-A-M-M-A-R not G-R-A-M-M-E-R
The word "don't" has an apostrophe in it.
Feel is F-E-E-L not F-E-L-L
Always capitalize the word I
Well is W-E-L-L not W-E-L
Edited has 1 d and 1 t, and no more.
I think you meant "here is my solutions" instead of "hear it my solutions", and either way, it's "are" not "is"
No, chess64 couldn't go to camp, and don't rub it in.

But seriously folks...

Nice solutions, EFuzzy. Up to about #5, they were exactly what I did. I obfuscated #6 even though my idea was what you had, I did something totally random on #7, and I was stuck on #8, and all I could get was that the areas of the corner triangles added up to that of the center (desired) triangle.

I couldn't open jli's because I don't have $L_AT^EX$

JTLYK, there's a Google group for Mathpath people, called MP2005.

[quote="G-UNIT"]If allowed, would a few mathpathers bother to post some problems they were given in mathpath.  So we can get a feel of what the camp is, if were considering to go next year, and just for the sake of good old problem solving. :D[/quote]
I don't know if MathPath set any formal rules on what students could or could not share after MathPath was over.  In the absence of guidelines from MathPath, I'll offer some general suggestions based on common sense, copyright law, and the rules of this Forum...

In my conversation with camp founder, Dr. Thomas, I am confident he does not intend to reuse questions from the qualifying quiz in future years.  Unless he tells me otherwise, I will not object to solutions from the qualifying quiz being posted in this thread.

I would like to ask that you folks NOT share the Problems of the Day.  These were generally out of the box (brain-teaser) problems.  Those are very likely to be used in future years, so let's not spoil it for future campers.

I would also suggest that you probably shouldn't share complete problem sets without the instructor's permission.  [Sharing a sample problem or two from a long problem set is probably OK.  But make sure the topic does not involve trig or calc or otherwise exceed the level of this Forum.]

I have split a sample problem from this thread to another thread, titled [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=292504#p292504]"A Smooth Game".[/url]


The question talks about perfect squares ending in n identical digits, asking for the smallest perfect square with the maximum N. I got an answer of 538 but I can't find the solutions for this mock aime anywhere. Can someone refer me to the right thread??

And if someone has made a compilation of solutions for the 2005 mock aime could I be directed to it or send a copy? thanks  :lol:

Well, your answer can't be right, [hide="because..."] $538^{2}= 289444$, but if $3$ digits is the maximum then $38^{2}= 1444$ is the actual minimum. [/hide]

oh, good point  :blush: in that case.. i think that answer should be the correct one.. unfortunately i didn't do this problem very elegantly, and also of course i forgot to consider 2 digit numbers.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=15513&start=20
and scroll down to bottom of page

THat is the directory listing, and I think it stays updated

Sorry, this is where you should actually check:

http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME

I can't find the correct AIME 2005 in the list, me@home  :maybe: I don't know which of the mock AIME's it was, and some of them have not been posted at all on the wiki, looks like.

Are you sure its actually in 2005? I checked the entire list and added links to the wiki and can't find it...
It could be either pre-2005 or maybe 2006 (I remember in summer of 06 they already had like three 07 aimes)