Stop worrying about what you are scoring on practice tests.  That's silly and it means nothing.  I took 1 or 2 practices AIMEs maximum last year not counting mock AIMEs.  I never scored above a 5.  My score on the real AIME...11.

I'm not saying to stop taking practice tests.  By all means, doing old AIME problems is the best way to prepare.  I'm just saying that your scores on them aren't [i]that[/i] important.

That sounds like the divine intervention.

Somehow I'm a bit skeptical about that friend of yours.

[quote="yif man12"]Here's the story of my friend.

Last year during Christmas break, he was struggling on practice AMC10s, averaging about a 75. In February, he got a 150 on the AMC12, then went on to get a 14 on the AIME in March and then a 42 on the USAMO.

The moral of this story: stop being so nervous. Christmas is a time for festive cheer, getting fat, rotting away on the sofa watching infomercials, and (hopefully for some) getting some.[/quote]

Hey, don't give away my story! :P

[quote="yif man12"]Here's the story of my friend.

Last year during Christmas break, he was struggling on practice AMC10s, averaging about a 75. In February, he got a 150 on the AMC12, then went on to get a 14 on the AIME in March and then a 42 on the USAMO.

The moral of this story: stop being so nervous. Christmas is a time for festive cheer, getting fat, rotting away on the sofa watching infomercials, and (hopefully for some) getting some.[/quote]

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamo-results.html]*cough* :roll:[/url]

It must have been your imaginary friend, then :D

[quote="1234567890"]All right. I need a reality check in my USAMO quest (and more motivation). Lately, I've been averaging 8-10ish on AIMEs, usually answering 12 or so questions. But I need to see if I'm on the right track. So, all USAMO qualifiers last year, could you post what you were getting in late December of last year, and what you actually got in last year's AIME.

It would really help me. Thanks.[/quote]

AHHHH just keep practicing and work.... knowing what other people get on practices will not help you, seriously. You know that you are aiming for like 11 on AIME, so just keep working to reach your goal. Knowing what I get on practices will not help you in anyway(unless you need confidence boost or something, then I say I was averaging about a  1-2 on practice AIME's. Oh if you are looking for motivation I was averaging 15 on AIME's).

[quote="chess64"][quote="yif man12"]No, my friend is definitely real. He's not in any of the 2005 stats or records because he failed the mandatory MOSP drug test.[/quote]

 :rotfl: I still don't believe you.[/quote]

Isn't it Zabelman, I heard he got a 150 and 14 last year

For some reason, I doubt it was Zach.

[quote="yif man12"]...he failed the mandatory MOSP drug test.[/quote]

[quote="yif man12"]...he was struggling on practice AMC10s, averaging about a 75.[/quote]

:D

Yeah, you're right, that story does seem a little bit too idealistic, to be able to go from a 70 to a 150 in two months. Are you sure he wasn't lying to you about his 70? He must've done math around 20 hours a day to reach that level from where he was :o

Shobhit, they were being facetious  :P Noone aced the USAMO (Brian Lawrence came close with a 41).  Zach did get 150/14, but got like a 28 on the USAMO (still extremely good score...I believe he tied Mildorf and Mildorf went on to get a gold at IMO).  But yeah, no one struggles (if they're taking them seriously) on AMC 10 and then goes on to ace the USAMO within a few months.

[quote="beta"][quote="yif man12"]Here's the story of my friend.

Last year during Christmas break, he was struggling on practice AMC10s, averaging about a 75. In February, he got a 150 on the AMC12, then went on to get a 14 on the AIME in March and then a 42 on the USAMO.
[/quote]

Hahaha that takes super improving skills. BTW Why not 15 on AIME?[/quote]

He tells me his pencil broke and he had to borrow one from his teacher. Sadly, the one he borrowed was a #4 pencil.

In a year Zach did improve from a 3 to a 28 on USAMO, and ThAzN1 went from 4 on AIME to 11 on AIME/ 22ish on USAMO and made MOP.

Guys, obviously yiffy is lying. The "my friend" thing is the most common trick. He's obviously reffering to himself. And he got a 43, not a 42.

But Saraubh, if you can usually get above 8, you should be safe.

[quote="1234567890"]I'll probably need an 11 or 12, though; I am a junior.  :([/quote]

Don't set a cold hard number goal before the test.  Last year I was expecting a 9 or so to get me in.  I went in just trying to get as many as I possibly could ending up with an 11.  Before the test, I would have thought an 11 would get me to the USAMO without a doubt.  After I took the test and was able to judge the difficulty of it.  I started to worry that an 11 wouldn't do.  I made the cutoff by 2 pts (disregard the fact that I was 20 points ahead of anyone else in SC so I could have gotten a lower score on the AIME...not sure what would happen in the case of a tie...perhaps both would go?).

Don't you guys think telling your personal stories of how you went from like 5 to 11 or whatever in a year is just a little more about blowing your own horn than actually helping answer the guy's question? Regardless of your intentions that's the way I'm seeing it and I doubt I'm alone and it happens all over these forums. People who make wonderful improvements over short periods of time have a right to be proud but it seems like every other "how can I improve" post gets littered with a story like this.

Honestly, not many people go from 5 to 11 in the course of a few months. It can happen but overall the goal should be to learn more math and other stuff will follow. Besides, the original poster said he's getting around 8 or 9 so the 5 to 11 stories or whatever are just....

If these forums give the impression that such rapid and drastic improvement is common that's because of a vocal bragging minority. I hope people who visit these forums and don't necessarily post a lot don't get discouraged because they don't show such dramatic improvements quite as rapidly.

[quote="1234567890"]All right. I need a reality check in my USAMO quest (and more motivation). Lately, I've been averaging 8-10ish on AIMEs, usually answering 12 or so questions. But I need to see if I'm on the right track. So, all USAMO qualifiers last year, could you post what you were getting in late December of last year, and what you actually got in last year's AIME.

It would really help me. Thanks.[/quote]

I was answering his question.

My main point was that often times practice test scores are not always indicative of what one will score on the AIME.

[quote="darktreb"]Honestly, not many people go from 5 to 11 in the course of a few months. It can happen but overall the goal should be to learn more math and other stuff will follow. Besides, the original poster said he's getting around 8 or 9 so the 5 to 11 stories or whatever are just....[/quote]

OH, sorry.  I should have cited the time when I was getting 8 or 9's on practice tests and then told him the score I got on the AIME that year...

Goodness...political correctness is taking over everything.  Maybe I should put a disclaimer in my signature *sigh*

I hate when people make a big deal over nothing.  I don't care if it's bragging or not, it's not like people are making up false stories or belittling the person who started this topic.

[quote="h_s_potter2002"]Um, wasn't Yifman's story false?[/quote]

Yes.

The previous few comments regarded what I said earlier.

[quote="yif man12"]Heh, sorry for confusing people with my story. Just thought it would get everyone in the festive mood and all that jazz.

I don't really think it's bragging that much in telling personal stories. I mean, sure, people take pride in that stuff, but it's not really for the purpose of bragging.[/quote]

wait was the guy from florida because I think I know him

Yeah... :blush:.  (Look five posts above this one)

[quote="Calculus"]Just curious Joml88, but how did you get a 11 if you weren't even scoring above a 5 on practice tests, thats a huge gap?[/quote]

I think he meant he never scored above a 5 on mock AIME's.

Well, yeah, pretty much.  Like I said before, in addition to the Mock AIME's, I only took about 2 practice AIMEs under test-like conditions.  Of course, I did a lot of AIME problems nonetheless.  I just didn't have large chunks of time to do a whole test.  The point I'm trying to get across is that it's not so much important to nail every AIME problem you attempt the first time as it is as actually learning something new from practicing old problems.  This is why I don't value practice tests scores as much as other people probably do.

The main thing is to as many practice AIME problems as possible (just make sure they are adequately challenging ones).  Make sure to try each problem for a reasonable amount of time before giving up.  Always look at the solution manual afterwards or the solution on AoPS if that's possible.

Those are just my suggestions though.

Obviously Yifman was joking.

And yes it was joml I was talking about. Basically from my understanding of joml over the last year, it seems that he made some serious improvement, since I seem to remember many of his stories about drastically improving from getting like 2's or whatever. Kudos.

[quote]
I was answering his question.
[/quote]

[quote]
OH, sorry. I should have cited the time when I was getting 8 or 9's on practice tests and then told him the score I got on the AIME that year...

Goodness...political correctness is taking over everything. Maybe I should put a disclaimer in my signature *sigh*
[/quote]

So ... by answering his question you eagerly mention the 5's but nothing else? Doesn't he deserve a full answer instead of one masked in a way that makes it sound like all you were getting were 5's? That's certainly the impression I've gotten all year about your improvement last year, that you were getting not-great scores in practice and then popped a great score in AIME. I can't remember hearing you mention 8/9's in practice last year, though I definitely remember hearing about those 5's you were getting on multiple occassions.

I'm just making a point. Rapid improvement doesn't happen to everybody. And it doesn't mean those who don't make those improvements should feel like that's the norm for someone working hard.

This isn't about political correctness... I really don't frequent these forums that much and yet somehow I can remember multiple joml posts about his wonderful year of improvement. I mean, at a certain point it's like, ok that's five times I've heard this story.

Not everybody that frequents these forums happens to post here. And I do personally know of people that have come here and just been like, "Wow these kids all just improve ridiculously fast and I just feel I have no chance because of some of these stories." Well, not everybody improves ridiculously fast but those who have do seem to be pretty vocal :P

[quote="darktreb"][quote]
OH, sorry. I should have cited the time when I was getting 8 or 9's on practice tests and then told him the score I got on the AIME that year...

Goodness...political correctness is taking over everything. Maybe I should put a disclaimer in my signature *sigh*
[/quote]

So ... by answering his question you eagerly mention the 5's but nothing else? Doesn't he deserve a full answer instead of one masked in a way that makes it sound like all you were getting were 5's? That's certainly the impression I've gotten all year about your improvement last year, that you were getting not-great scores in practice and then popped a great score in AIME. I can't remember hearing you mention 8/9's in practice last year, though I definitely remember hearing about those 5's you were getting on multiple occassions. [/quote]

I was being sarcastic about the getting 9's on practice tests.  He asked about the scores people got on practice AIME's they were taking.  I told him what I got.  But you didn't like hearing my story again (which you were vocal about) and said that not everyone goes from getting < 5's to an 11 that quickly.  I was trying to make the point that if there were other years that I had taken practice AIMEs and gotten 8's or 9's and went on to get a 11, I would have told him...but I there was only 1 applicable year and that year, for the record, I never scored above a 5 or whatever and went on to get an 11..blahblahblah.  OK?  Sorry for the misunderstanding there.

I'll stop posting my AIME scores when other people ask.  Will that make you happy?

Ya know, I don't think it's as uncommon as you think for such drastic improvements such as mine to occur.  Off the top of my head, DPopov improved 11 points on the AIME.  beta didn't even make AIME and went on to get a 9 and qualify for USAMO the next year.  I'm sure there are other examples.  Obviously it takes a TON of work.  But what's wrong with shooting high?  Little kids dream about making it to the MLB, NHL, NFL, NBA, etc.  Obviously they have a slim chance of making it but does that mean they should give up?  Making it to professional sports is harder than making it to USAMO, in my opinion, but the analogy is still there.

yo, let's chill guys.

I went from 3 my freshman year to an 8 my sophmore year without any studying at all, so I guess if you study you'll get an even bigger improvement? I dunno.

Keep in mind that this is all subject to the prevalence of certain subjects on the tests that some people will be better at and some worse, as well as stupid mistakes, bad test-taking conditions, etc. People can have huge changes without studying, or they can have no change despite plenty of studying, all depending on quite a few factors.

Everyone should just relax.  Honestly, you're acting like five year olds.  Whether or not joml88 was bragging or has told his story too many times or whatever, what are you trying to accomplish by admonishing him for it?  Trying to spare yourself the agony of hearing it again or listening to another brag story?  If you're that discomforted by the level of "bragging" or repition in the forums then perhaps you shouldn't visit them.  The bottom line is the goal of these forums were to help math students hone their problem solving skills and talk with peers who enjoy the same levels of mathematical problem solving.  Excursions and arguments like this are useless and unnecessary, and their effect is only a diversion from the benefit and usefulness of these forums.  So, anyway, hopefully the forums can be used more for their original purpose and intent which is math and/or [b]friendly[/b] chat.

Focus on the practice, not how you're doing on it. Everything else should fall into place.

[quote] Ya know, I don't think it's as uncommon as you think for such drastic improvements such as mine to occur.  Off the top of my head, DPopov improved 11 points on the AIME.  beta didn't even make AIME and went on to get a 9 and qualify for USAMO the next year.  I'm sure there are other examples.  Obviously it takes a TON of work.  But what's wrong with shooting high?  Little kids dream about making it to the MLB, NHL, NFL, NBA, etc.  Obviously they have a slim chance of making it but does that mean they should give up?  Making it to professional sports is harder than making it to USAMO, in my opinion, but the analogy is still there.[/quote]

You think most of the people that work hard improve that fast?

I mean, I'm no stranger to the "fast improvement". I daresay I've been in your shoes of improving fast and probably then some. I guess it's not your fault but I wish there'd be a couple more stories about how it's not that easy for everybody to improve quickly. I've seen a lot of people who tried really really hard and just couldn't quite make the same strides as others. It's not their fault, and these forums don't help them realize it's not out of the ordinary either (though there will be exceptions like joml, dpopov, beta, a personal friend of mine thazn1 especially, and more)

i think people are missing the point that the tests don't really matter, regardless, you are learning math and problem solving, that will help you in the future, it is nice to do well in the competitions, but as long as you take something away from your preparation, you are fine, and as for large improvements, i got like a 90 on the amc last year, and i am pretty sure that i will be able to do better this time around, but the only reason that would occur is because i have been working since the end of the previous amc...(that is what motivated me), but the thing is that it doesn't matter if i do well on the competition because i have learned a lot

[quote="blahblahblah"]I think that it can be viewed as a sort of passive bragging, but that's not the vibe I get off of reading this thread.

Anyways, the stories of people improving by 5-7 questions on AIME or 20+ questions on USAMO are usually a matter of someone having a bad day one year and performing closer to their peak ability the next (or also the difficulty of the test itself).  Certainly there's a lot of room for improvement in a year, but soft factors like fatigue (the USAMO is near the end of contest season, after all) and silly mistakes can do a number on someone's score in a hurry.[/quote]

If you're fatigued since the USAMO is at the end of a [b] contes series [/b], then something is wrong.  The USAMO comes about a month after AIME.  Unless you've been doing some heavy duty study/cramming, any sleepiness or fatigue should be from whether you slept well the previous week.

There are other math contests outside of AMC/AIME/USAMO. I think thats what he means.  I definitely know what hes talking about.  :P


Let ABC be an acute triangle, show that
$ \frac {1\minus{}cosA}{1 \plus{} cosA} \plus{} \frac { 1\minus{}cosB}{1 \plus{} cosB} \plus{} \frac {1\minus{}cosC}{1 \plus{} cosC}\leq(\frac {1}{cosA}\minus{}1)(\frac {1}{cosB}\minus{}1)(\frac {1}{cosC}\minus{}1)$

[quote="miwalin"]Let ABC be an acute triangle, show that
$ \frac {1 \minus{} cosA}{1 \plus{} cosA} \plus{} \frac { 1 \minus{} cosB}{1 \plus{} cosB} \plus{} \frac {1 \minus{} cosC}{1 \plus{} cosC}\leq(\frac {1}{cosA} \minus{} 1)(\frac {1}{cosB} \minus{} 1)(\frac {1}{cosC} \minus{} 1)$[/quote]
See here:
http://www.mathlinks.ro/viewtopic.php?t=250853


Prove that y^2 = x^3 + 7 has no integer solutions.

Part of the solution for this problem involves the fact that any integer of the form 4k+3 has a prime factor of the form 4k+3. How can we prove this? IS this result synonymous to saying that the integers of the form 4k+3 are closed under multiplication?

Thanks.

[quote="Fierytycoon"]...any integer of the form 4k+3 has a prime factor of the form 4k+3. How can we prove this? IS this result synonymous to saying that the integers of the form 4k+3 are closed under multiplication?[/quote]

The opposite, in fact -- it involves the fact that integers [i]not[/i] of the form 4k + 3 are closed under multiplication.  (Integers of the form 4k + 3 clearly are not closed: 3 * 3 = 9 = 4k + 1).

Well, examining the equation mod 8 yields the result that x is odd. Then (x+2)(x^2-2x+4)=y^2+1. It is well known that y^2 + 1 does not have any prime factors of the form 4k+3, and x^2-2x+4 is of that form. But since x^2-2x+4 is not necessarily prime, we must prove that it has a prime factor of the form 4k+3. How can we do this?

Use the fact I stated -- things that aren't 4k + 3 are closed.  The 4k + 3's being closed wouldn't help -- that would tell you that such a number [i]might[/i] be divisible by a 4k + 3 prime, but not that it must be.

What does saying that a set is closed under some operation (say, multplication) mean anyways?

It means that if you take any two elements of the set, then their product (or whatever the operation is) is in the set.

[quote]we must prove that it has a prime factor of the form 4k+3. How can we do this?[/quote]

you may want to try a slick proof by contradiction (i.e. what would happen if a number = 3 mod 4 had no prime factors = 3 mod 4)

Oh, can we just say that if n=1 mod 4 then n^2 = 1 mod 4, and since all odd numbers are either 1 or 3 mod 4, x must be divisible by a number that is congruent to 3 mod 4?

x is odd, so x is = 1 or 3 mod 4:

case 1: x = 1 mod 4

x^2-2x+4 = 1^2-2*1+4 = 3 mod 4

case 2: x = 3 mod 4

x^2-2x+4 = 3^2-2*3+4 = 7 = 3 mod 4

so in all cases the RHS = 3 mod 4, and then you can use proof by contradiction to show that it must have a prime factor = 3 mod 4.

Basically -- because numbers not congruent to 3 are closed under multiplication (multiplying them by each other will never get you a number which is congruent to 3), any number congruent to 3 must have a prime factor congruent to 3 as well.


How to calculate this:
$ \sum_{k\equal{}0}^{2^{n}\minus{}1}2^{\minus{}n}$

hello, do you mean
$ \sum_{i\equal{}0}^{2^n\minus{}1}2^{\minus{}i}$?
Sonnhard.

If it's the original version, this is just the sum of $ 2^n$ copies of the same original value $ 2^{\minus{}n}$, which is $ 1$. Repeated addition is multiplication.

The sum is exactly as I wrote $ \sum_{k\equal{}0}^{2^{n}\minus{}1}2^{\minus{}n}$

hello, in this case we get
$ \sum_{k\equal{}0}^{2^n\minus{}1}2^{\minus{}n}\equal{}2^{\minus{}n}\cdot2^{n}\equal{}1$
Sonnhard.

Thank you Dr Sonnhard Graubner
But, could you please explain me why do we take that $ \sum_{k\equal{}0}^{2^{n}\minus{}1}$ equals $ 2^{n}$

The list $ 0,1,2,\dots,15$ has 16 elements.

$ \sum_{i\equal{}a}^bc\equal{}c(b\minus{}a\plus{}1).$

oooo...........
Thank you so much!
it is so obvious, but I'm too tired to see it
Thanks again


geometry
circumcircle
parallelogram
projective geometry
geometry proposed

Let $ ABC$ be a triangle with the centroid $ G$ . Its circumcircle meet again $ AG$ in $ D$ . Denote $ E\in CD\cap AB$ and $ F\in BD\cap AC$ .

Prove that if $ AG$ is perpendicular on the [b]Euler's line[/b] of $ ABC$ , then $ D$ is the centroid of $ ABC$ .  Is it truly and reciprocally ?

Call M the midpoint of AB and N of EF.
Lemma 1: $ EF \parallel BC$ (it's simple to prove it)
Lemma 2: N is the harmonic conjugate of M w.r.t. DA (well know)
$ AG \perp OG \Longleftrightarrow AG \equal{} GD \Longleftrightarrow GM \equal{} MD \Longleftrightarrow \frac{ND}{NA} \equal{} \frac{DM}{DA} \equal{} \frac{1}{3} \Longleftrightarrow$$ \mathrm{D \ is \ the \ centroid \ of \ AEF}$

we have\[ OG \bot AD \Rightarrow AG \equal{} GD \Rightarrow \frac{{DM}}{{MA}} \equal{} \frac{1}{3}\] [$ M$is midpoint of$ BC$]

\[ \Delta AFD,BMC: \frac{{AB}}{{BF}} \times \frac{{FC}}{{CD}} \times \frac{{DM}}{{MA}} \equal{} 1\]

and in other case we have \[ \Delta BFC,AMD: \frac{{BM}}{{MC}} \times \frac{{CD}}{{DF}} \times \frac{{FA}}{{AB}} \equal{} 1 \Rightarrow \frac{{DF}}{{CD}} \equal{} \frac{{FA}}{{AB}} \Rightarrow \frac{{FC}}{{CD}} \equal{} \frac{{FA \plus{} AB}}{{AB}}\]

\[ \to \frac{{AB}}{{BF}} \times \frac{{FA \plus{} AB}}{{AB}} \times \frac{1}{3} \equal{} 1 \Rightarrow \frac{{FA \plus{} AB}}{{BF}} \equal{} 3 \Rightarrow \frac{{BF \plus{} 2AB}}{{BF}} \equal{} 3 \Rightarrow AB \equal{} BF\]

and with this way can prove that $ AC\equal{}CE$ then $ D$ is centeroid of $ AEF$

Very simple! If M is the midpoint of BC, MD=MG, BGCD is a parallelogram, CD||BG, BD||CG, so BA=BE, AC=CF because AG=GD.
Hence, FB meet EC at the centroid D of triangle AEF.


I was wondering if someone could explain to me what they mean by the maximum decreasing subsequence in the dynamic programming text in chapter 2 of the usaco?  Is the maximum decreasing subsequence not just the array sorted in reverse order?

Thats amazing optimism SnowStorm! :P  Wish that problem had been that easy!

Well, you cannot sort the array....
So, the MaxDecreSubSeq of (5 , 8 , 12 , 6 , 4 10 , 1) is 1(12 , 6 , 4 , 1) or (8 , 6 , 4 , 1)

for much help u may consult with coreman's book


Can you use only three 2 to make 20 . ( You can use addition , substraction , fraction ...etc ..)

Example : $2^{2^2}=16$ ( It is 16 here but I want 20  :D )

[size=150][b]Rather, try just using two 2's to make 5[/b][/size]! It would make you think for deeply :lol:

I don't think it can be done without using a $1$ as well...

Bictor717 and JesusFreak197 , both of you are correct . Now , lets use three $n$ where $n$ is integer from 1 to 9 , to make a 20  :lol:

[quote="alan"][hide]$\frac{n+n}{.n}$[/hide][/quote]

 :D  :D

[quote="Geehoon"][quote="Shamik Banerjee"]Rather, try just using two 2's to make 5[/quote]
[hide]$\sqrt{.2^{-2}}$[/hide][/quote]

wow ! kool  :lol:  :D

[quote="minsoens"][quote="shyong"]I also remember a tough one :

Use ten $8$ to make $2004$   :)  Have fun  :P[/quote]
$\frac{8!}{8+8}-8 \cdot 8 \cdot 8-\frac{8 \cdot 8}{8+8}$[/quote]

awesome  :thumbup: a totally different solution than mine  :clap2:

I got a solution, however it uses floor function(hope that it is allowed).

$[\sqrt[8]{(\frac8{.8}+8+8)!\cdot\frac{[\sqrt{8}]}{[\sqrt{8}]+\frac88}}]-8=2004$

Well any form of solution is allowed ..... here is my solution

((8+8+8)x8+8/.8-.8)x8/.8 - 8 = 2004

 :)


inequalities
geometry
circumcircle
inradius
inequalities proposed

$ R,r,s$ are the circumradius, inradius and semiperimeter of a triangle . Prove that at least one of these two inequalities is always true:

$ \frac {9}{4}R^2 \plus{} 4r^2 \plus{} 7Rr \ge s^2$ or

$ 9R^2 \minus{} 5r^2 \minus{} 2Rr \ge s^2$

Of course the second inequality holds
\[9R^2-5r^2-2Rr \ge 4R^2+4Rr+3r^2 \iff (5R+4r)(R-2r) \ge 0\]
which is true.
So it is enough to show $4R^2+4Rr+3r^2 \ge s^2$ which is just Gerretsen.


Right, but i think for any a,b,c,d different from 1, the ineq. is true without =<

You're right. I made a mistake in the spoiler  :blush: . So, < is turned into a =<.

Is the inequality just AM-GM in four variables, or am I missing something?  :?

Yeah, I thought that was pretty much the exact definition of AM-GM, too.
Who knows, maybe there's some subtle twist I didn't catch. :)

[quote="icare"]Hi everyone, im new here so i dont if it's the good place post it.Here s the problem.

For a,b,c,d > 0, prove that:

(abcd)^(1/4) < (1/4)(a+b+c+d)

Hint:

Proving, for x,y >0, (xy)^(1/2) < (1/2)(x+y), could be useful.

Good luck.

Hugo[/quote]

x+y >= 2sqrt(xy) by (x-y)^2 >= 0

so

[(a+b)/2 + (c+d)/2]/2 >= sqrt([(a+b)/2][(c+d)/2])
sqrt([(a+b)/2][(c+d)/2]) >= sqrt(sqrt(ab)*sqrt(cd)) = (abcd)^(1/4)

Hence (a+b+c+d)/4 >= (abcd)^(1/4)

It is worth noting that the standard method of proving AM-GM is to note (as I just showed) that it can be shown to hold easily for n = 2^k, k arbitrarily large, and then shown that if it holds for n, it holds for n-1.


geometry function rectangle rotation calculus derivative integration

Hi.

Suppose an arbitrary function, g(X) is plotted. For any Xi which is X1< Xi <X2 , one normal line to g(X) is drawn so for all Xi which is X1< Xi <X2 Infinite number of normal lines to g will be drawn. These normal lines ( start from g(X) ) have magnitude of constant A>0. The area of the shape created (by these normal lines) is desired in terms of A, X1, X2, g(X) or its derivatives.
( for example if g(X)=straight line, then these normal lines to g(X) will be parallel and the shape created will be rectangle )

Thanks again.

The area, in general, is hard; it behaves like an absolute value, and universal formulas are unlikely. On the other hand, there is a notion of "net area" that we could use; we count a region positively if it is swept out by the ray moving "forward", and negatively if it is swept out by the ray moving "backward". If that region is covered multiple times, we count each of them and add the weights. The part of the ray closest to the curve always sweeps forward, while the part farther away can sometimes sweep backward  if the curve bends tightly enough.

Also, it makes much more sense to go with a general curve, and use the tools of differential geometry.

Parametrize by arclength; let the curve be $ r(s)$ for $ s\in I$ (some interval), with unit tangent $ T(s) \equal{} r'(s)$ and continuous unit normal $ N(s)$, rotated $ 90^\circ$ from $ r'$. The second derivative $ r''$ is parallel to $ N$; define the signed curvature $ \kappa(s)$ as $ r''(s) \equal{} \kappa(s) N(s)$. As $ r$ moves forward $ ds$ along the curve, the opposite end of the ray of length $ A$ in the $ N(s)$ direction moves forward by $ ds$ and backward (for $ k$ positive) by about $ A\cdot |T'|\cdot ds \equal{} A\cdot\kappa(s)\cdot ds$. It sweeps out a net area of about $ \frac12(ds \plus{} ds \minus{} A\cdot\kappa(s)\cdot ds)\cdot A \equal{} (A \minus{} \frac12A^2\kappa(s))\cdot ds$. Integrate that, and the overall net area covered is $ \int_I A \minus{} A^2\kappa(s)\,ds$.

As an example, consider a circle with radius $ R$, and $ N$ pointing outward. The curvature of the circle is the constant $ \minus{} \frac1R$, so the area is $ \int_0^{2\pi R}A \plus{} \frac {A^2}{2R}\,ds \equal{} 2\pi R A \plus{}\pi A^2 \equal{} \pi(A \plus{} R)^2 \minus{} \pi R^2$.

The preceding post used the language of differential geometry, notably including parametrization by arclength. It's easy to parametrize practically any curve, but usually impossible to solve for exact arclength. We need a way to fake it, and calculate anyway even without actually solving for arclength.

In this version, let the curve be $ r(t)$ for $ t\in [a,b]$, with the condition that $ r'(t)$ is never the zero vector in $ (a,b)$ (so as to avoid sharp corners). Assume reasonable smoothness; in this case, that $ r''$ is piecewise continuous. Let $ R$ be the operator that rotates a vector $ 90^{\circ}$ counterclockwise, and define the cross product $ (x_1,y_1)\times (x_2,y_2)=x_1y_2-y_1x_2$. Note that $ u\times R(u)=1$ for any unit vector $ u$.

Now $ T(t)=\frac1{|r'(t)|}r'(t)$ and either $ N(t)=R(T(t))$ everywhere or $ N(t)=-R(T(t))$ everywhere. We have $ r''(t)=\frac{d}{dt}v(t)T(t)=v'(t)T(t)+v(t)T'(t)=v'(t)T(t)+v(t)[\kappa(t)N(t)v(t)]=v'(t)T(t)+(v(t))^2\kappa(t)N(t)$ (where $ v(t)$ is the speed $ |r'(t)|$). We want the curvature, so we need to extract the $ N$ component while ignoring the $ T$ component. That's what the cross product does best. We have $ r'(t)\times r''(t)=v(t)T(t)\times(v'(t)T(t)+v^2(t)\kappa(t)N(t))=v^3(t)\kappa(t) T(t)\times N(t)=\pm v^3(t)\kappa(t)$, with the plus sign if we have the positive sign $ N(t)=R(T(t))$ and the negative sign if we have $ N(t)=-R(T(t))$.

OK, now we know how to calculate $ \kappa$; it's $ \pm\frac{r'\times r''}{|r'|^3}$. In order to integrate $ ds$, note that $ v\,dt=ds$. Our area integral transforms as follows:
$ \int_I A-\frac12A^2\kappa(s)\,ds$
$ =\int_a^b (A-\frac12A^2\kappa(t))v(t)\,dt$
$ =\int_a^b (A- [\pm \frac12A^2\frac{r'(t)\times r''(t)}{|r'(t)|^3}])|r'(t)|\,dt$
In coordinates, let $ r(t)=(x(t),y(t))$. The integral becomes
$ \int_a^b (A- [\pm \frac12A^2\frac{(x'(t),y'(t))\times (x''(t),y''(t))}{[(x'(t))^2+(y'(t))^2]^{\frac32}}])\sqrt{(x'(t))^2+(y'(t))^2}\,dt$
$ =\int_a^b (A- [\pm \frac12A^2\frac{x'(t)y''(t)-y'(t)x''(t)}{[(x'(t))^2+(y'(t))^2]^{\frac32}}])\sqrt{(x'(t))^2+(y'(t))^2}\,dt$
$ =\int_a^b A\sqrt{(x'(t))^2+(y'(t))^2}\pm \frac12A^2\frac{x''(t)y'(t)-x'(t)y''(t)}{(x'(t))^2+(y'(t))^2}\,dt$
(That last version switched order in the cross product for a more convenient $ \pm$ sign)
Note how this splits into two terms; the first is $ A$ times the total arclength, while the curvature term has no square roots and is generally easier to calculate.

Some special cases:
First, a concrete calculation. Consider one arch of a cycloid$ _1$ $ x(t)=t-\sin t$, $ y(t)=1-\cos t$ for $ t\in [0,2\pi]$, and choose $ N$ pointing outward. The area is
$ \int_0^{2\pi} A\sqrt{(1-\cos t)^2+(\sin t)^2}+ \frac12A^2\frac{(\sin t)(\sin t)-(1-\cos t)(\cos t)}{(1-\cos t)^2+(\sin t)^2}\,dt$
$ =\int_0^{2\pi} A\sqrt{2-2\cos t}+ A^2\frac{1-\cos t}{2\cdot(2-2\cos t)}\,dt$
$ =\int_0^{2\pi} 2A\sin\frac{t}{2}+ \frac14 A^2\,dt$
$ =8A+\frac{\pi}{2} A^2$

Second, what if it's a curve described as you initially wrote: $ y=g(x)$, with $ N$ pointing at least partly upward?
In this case, we parametrize as $ x(t)=t$, $ y(t)=g(t)$. "Up" is counterclockwise from the tangent, which has a positive $ x$ component, so we use the $ +$ sign in the $ \pm$. The integral for the net area becomes
$ \int_{x_1}^{x_2} A\sqrt{1+(g'(t))^2}+ \frac12A^2\frac{0\cdot g'(t)-1\cdot g''(t)}{1+(g'(t))^2}\,dt$
$ =\int_{x_1}^{x_2} A\sqrt{1+(g'(x))^2}- \frac12A^2\frac{g''(x)}{1+(g'(x))^2}\,dx$

$ _1$ Why the cycloid? Because it almost always gives a nice answer for any sort of curve integral. I didn't know what this one was going to be until I worked it out.

On second thought, that answer looks too familiar. We can completely avoid integration for the curvature term; it simplifies to $ \frac12A^2$ times the (net) angular change in the normal vector, and we can see this by making $ A$ very large. We get $ \pi$ for the cycloid because the normal goes from $ (-1,0)$ to the antipode $ (1,0)$. Watching the signs is still important.


algebra
polynomial
superior algebra
superior algebra unsolved

Find an irreducible polynomial of $\mathbb{Z}[X]$ which is reducible over $\mathbb{Z}/p \mathbb{Z}$ for every prime number $p$.

$x^4+1, x^4-x^2+1$...
For a proof of reducibility see http://www.mathlinks.ro/Forum/viewtopic.php?t=89188

I see ZetaX proposed the solution found in Milne's course ... When Milne wrote he first was not aware of any elementary solution, what non-elementary solution was he thinking about ?


Let $X$ and $Y$ be subsets of a vector space $V$.
Show that $\langle X \cup Y \rangle = \langle X \rangle+\langle Y \rangle$

If you think in terms of bases, it is trivial.


[quote="Cyberman"]The problem asks me to find the slope-intercept form from this equation.

x = 3/2y - 3

I know the answer is already y= 2/3x + 2

I just can't figure out the exact way to do this problem.

Would it be...

x + 3 = 3/2y
(2/3)x + 3 = (2/3)3/2y
2/3x + 3 = y
[y = 2/3x + 3]
?????????????

I think I understand how to get the problem once seeing the question and answer together. But I wonder if the right way to do this was to multiple 3/2 by its reciprocal. Why should it be done this way? Did I do this correctly?[/quote]

[hide]$x=\frac{3}{2}y-3$

$x+3=\frac{3}{2}y$

$\frac{2}{3}(x+3)=y$

$\frac{2}{3}x+\frac{6}{2}=y$

$y=\frac{2}{3}x+2$[/hide]

Basically, you took the reciprocal of 3/2 and made it into 2/3 and put it on the other side. Wouldn't have using the reciprocal of 2/3 and multiplying it on both sides have been the right thing to do?

Why is it that I take it away from a variable and put it onto another variable. It just doesn't seem right somehow to me.

I guess the reason it has to be turned into a reciprocal is because it solves for y.

Yet when you gave me a reply you didn't multiply it to the y side.

What should I be looking for that hints me in the direction to knowing I should multiply something by a reciprocal?

When Drunken_Math multiplied both sides by $\frac{2}{3}$, the equation went from $x+3=\frac{3}{2}y$ to $\frac{2}{3}(x+3) = y$, because he multiplied both sides by 2/3, so when you multiply the right side ($\frac{3}{2}$) by $\frac{2}{3}$, the numbers cancel out, leaving 1 in front of y.

When you divide by a fraction, you multiply it by its reciprocal. So in an equation like
$\frac{5}{6}x=10$, you multiply both sides by $\frac{6}{5}$ so that on the side with x, $\frac{5}{6}\cdot\frac{6}{5} = 1$, and the otehr side is $10\cdot\frac{6}{5}$ or 12. Thus, x=12.

Hope this helps a little.

In other words, if you can move stuff to the other side of the equation but you see why with some attachment, do whatever you can to turn that attachment into Y and then if you did something to Y on one side of the equation, you have to also do that to the other side.

wow, now i understand the basics of why monkeys rule. amazing.


hey any1 plz help with nsea (national standard examination examination in astronomy)my schools not a member .so how will i apply plz tell me

first off don't mess with the space setting

mods please delete this post

hey dont get so hyped up!
it was just my mistake
dont take it too seriously!
and chill!


AMC
AMC 10
geometry
geometric transformation
reflection

On a certian test, there are 25 questions.  You get 6 points for a correct answer, 1.5 points if you do not answer the question, and 0 points if you get the question wrong.  For each question, you can only get 1 out of those 3 possible point values.  How many different possible final scores are obtainable.

For example, a score of 6 is obtainable if you get 1 correct and 24 wrong or 21 wrong and 4 unanswerd.

However, a score of 6.5 is unattainable.

Your looking for final scores not how many ways to get the score.

[quote="abacadaea"]all i can think of is an organized chart[/quote]
you are thinking way to hard...its very easy..try it starting with a score of 0 and you'll find a pattern (and it will be an arithmetic sequence)..its like a 15/30 on a national sprint..or like a 25/30 on a state sprint so its not too complicated..its just a little trick.

[quote="Nexmus"][hide]

Well, we have a number every 1.5 until 145.5
So, we have $ \frac{145.5}{1.5}+1=98$ (We add the 1 for 0).
Next, we have one more value which is the whole score. Hence, we add 1 to get our answer of $ \boxed{99}$

[/hide]

I might've made a mistake...[/quote]

You are very very close, you made a little mistake as its not every number form1.5 to 145.5

Nice problem, which corresponds to the current scoring of AMC10 and AMC12.
Here are two  variations that it might be interesting to try:

[b]
Variation 1:[/b]  How many different scores were possible under the scoring system used in 2000 and 2001:  6 points for correct answers, 2 points for questions left blank, 0 ponts for incorrect answers?

[b]
Variation 2:[/b]  How many different scores were possible under the scoring system used in 2002 through 2006:   6 points for correct answers, 2.5 points for questions left blank, 0 ponts for incorrect answers?

[hide="Comments"]Variation 1 is similar to the original problem, and perhaps even a bit straightforward, since it doesn't present the trap that Nexmus fell into.  Variation 2 involves somewhat more work than the other two, but with a bit of reflection you can analyze it without setting up an exhaustive table.[/hide]

Um, how can you get to 145.5 using only 1.5 increments if you can only have 25 answers?

[hide]Anyway, I got 25 possible scores using only 1.5 increments and now must count those containing 6's as well.  Using mostly intutition to count these, I got a total of 25+25=50 however, one must remeber to count the score of 0, so there are 51 possible scores. :D [/hide]

[quote="Math Geek"]Um, how can you get to 145.5 using only 1.5 increments if you can only have 25 answers?

[/quote]

145.5=24*6+1.5 or 24 right 1 unanswered 0 incorrect and indianamath got the right answer, just with no solution...

[quote="indianamath"]pretty much nexmus's solution except that 0 is a possible score.[/quote]

Umm that means your answer wouldnve been 100...not 98...Nexmus got 99, if he didnt include 0 (which he even did), and you did include 0 (saying nexmus didnt) then you wouldve gotten 100...

This problem is almost exactly the same as if 1.5 was the only possible score for getting any points at all, seeing as $ 1.5 \equiv 0 \text{ (mod }6)$.
The only difference is that you can achieve higher numbers.


Define a lable graph $G=(V_G,E_G)$ where $V(G)=${$1,...,n$}as follows :{i,j}$\in E(G)$ iff i+j is a prime number.is G Hamiltonian for each n even?

Is it an unsolved problem or an open question? I'm asking because it looks rather freaky :). So, are you sure there's a known solution?

At least until recently, it was an open problem....

http://groups.google.fr/groups?hl=fr&lr=&ie=UTF-8&threadm=9ea9cda9.0206160219.6855fada%40posting.google.com&rnum=1&prev=/groups%3Fq%3Da%2Bcurious%2Bgraph%26hl%3Dfr%26lr%3D%26ie%3DUTF-8%26selm%3D9ea9cda9.0206160219.6855fada%2540posting.google.com%26rnum%3D1

Pierre.


Given that $ a,b,c,d$ are positive reals that satisfy $ ab\equal{}cd$, prove that;

\[ (a\plus{}b)^2(a\plus{}b\plus{}c\plus{}d)^2 \geq (2a\plus{}2c)(b\plus{}c)(a\plus{}d)(2b\plus{}2d)
\]

We need to prove that
$ ( a^2 +b^2+2ab+ac+ad+bc+bd)^2 \ge 4(a^2+ad+ac+ab)(b^2+bc+bd+ab)$
We are done!

we have, $ (2a\plus{}2c)(b\plus{}c)(a\plus{}d)(2b\plus{}2d)\equal{}4(a\plus{}c)(a\plus{}d)(b\plus{}c)(b\plus{}d)$
$ \equal{}4(a^2\plus{}ac\plus{}dc\plus{}ad)(b^2\plus{}bd\plus{}bc\plus{}cd)$
with AM-GM, $ 4(a^2\plus{}ac\plus{}dc\plus{}ad)(b^2\plus{}bd\plus{}bc\plus{}cd)\leq(a^2\plus{}b^2\plus{}2cd\plus{}bd\plus{}bc\plus{}ac\plus{}ad)^2$
$ \equal{}(a^2\plus{}b^2\plus{}2ab\plus{}bd\plus{}bc\plus{}ac\plus{}ad)^2$
then $ (2a\plus{}2c)(b\plus{}c)(a\plus{}d)(2b\plus{}2d)\leq(a^2\plus{}b^2\plus{}2ab\plus{}bd\plus{}bc\plus{}ac\plus{}ad)^2$
$ \equal{}(a\plus{}b)^2(a\plus{}b\plus{}c\plus{}d)^2$


f is a naturel function
f(ab)=f(a)f(b) if a and b are coprime
f(a+b)=f(a)+f(b)  if a and b are prime number.
Show that f(2)=2 , f(3)=3 and f(1999)=1999

[quote="tµtµ"]Again shobber I don't think you can use Cauchy functionnal equation.[/quote]

No, Shobber, since the Cauchy equation only "works" for rationals etc.


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

what is the Sylow 2-subgroup of SL_2(5)?

You can take the subgroup of matrices $ \begin{pmatrix}a&b\\c&d\end{pmatrix}$ satisfying $ b\equal{}c\equal{}0$ or $ a\equal{}d\equal{}0$. owk


integration
logarithms
calculus
real analysis
real analysis solved

Prove that $i^i$ is real, where $i=\sqrt{-1}$

:D  Warning !  You didn't prove that $i^i $ is real, you proved that $i^i$ is not well defined !!  :D

$a,z \in \mathbb C$
$z^a=exp(a.log_{\mathbb C}z)$

Even if they may be correct, these answers are not the ones I was looking for.  There is a very elegant algebraic solution to this one!

$\frac\pi2=\int_0^\infty\frac{dx}{1+x^2}=\frac1{2i}\int_0^\infty\left(\frac1{x-i}-\frac1{x+i}\right)=$ $\frac1{2i}\ln\left(\frac{x-i}{x+i}\right)|_0^\infty=$ $-\frac1{2i}(\ln(-i)-\ln(i))=\frac i2\left(\ln\frac1i-\ln i\right)=-\frac i2\cdot2\ln i=-i\ln i=-\ln i^i$ $\Rightarrow i^i=e^{-\frac\pi2}$

That's calculus, not algebra.

Another Hint: [hide]The solution involves raising the givens to powers.[/hide]

we have
$i^i=(e^{i\frac{\pi}{2}+2k\pi i})^i=e^{-\frac{\pi}{2}-2k\pi}$.
now choose your favourite solution :)

but i think it's not as nice as kuba's solution for example...my solution is just too straightforward to be nice. anyway, i think this problem shouldn't be posted in such an advanced forum.

easy trick,

$e^{i\pi}=-1$
square root both sides,
$e^{i\frac{\pi}{2}} = i$

so $i^i = {(e^{i\frac{\pi}{2}}) ^ i}$

so $i^i = e^{-\frac{\pi}{2}}$

kuba used brute force calc though result is same...
which belongs to real as e is real and exponent is also real -pi/2


Okay...so I can do basic counting.

However there are a few problems that confuse me.

1. Say you have 10 IDENTICAL marbles. There are 3 urns to put them in. How many ways are there to place all of the balls in the urns?

?!?!!?!?!?!!?!?!?!?!?!?!!!?!??!!?

Since you guys are much better than me at math (because I fail) you guys can hopefully help me a little bit....

See the AoPSWiki article on balls and urns.

Okay so the question is basically asking if there are 10 identical marbles how many spaces can u put in them.
For example:
xxx xxx xxxx
xx x xxxxxxx

So because there are
x x x x x x x x x x (9 spaces between the x's 3 groups to put it in)
9!/6!3! = 84

I think this would be right but im not exactly to sure about this.

That is wrong think about it like this:
You have ten balls and 3 urns.
$ xxxxxxxxxx$ Ten balls
$ \parallel{}$ Two dividers to divide the balls into three groups so they can be placed into the urns
$ xxx|xxx|xxxx$ One example

So the formula is $ \binom{\text{Number of Balls}\plus{}\text{Number of Urns} \minus{} 1}{\text{Number of Urns} \minus{} 1}$
So the answer is $ \binom{12}{2} \equal{} \boxed{66}$
@AlphaBetaTheta don't state whether a problem is easy or not because what is easy to you may not be to other people

ok this is stars and bars too... ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=296502[/url])

$ 3\minus{}1\equal{}2$ bars
$ 10$ stars

$ 2\plus{}10$ choose $ 10$

$ \binom{12}{10}\equal{}\binom{12}{2}\equal{}12*11/2\equal{}66$


The questions are attached in images.

[/img]

As a general principle, putting your questions in attachments is a very good way to have them completely ignored. We have $ \text{\LaTeX}$ enabled here for a reason- if you want to see how things are done, quote posts with similar structures to what you want.


i wanna buy a new computer, thinking about a PowerMac
but i only heard it's awesome but know nothing about it.

can anyone give me some advices and overviews on Mac
and anyother stuff that i need to know(i'm windows user)

thanks

Personally, I can't decide what OS is the best. I have heard great things about Macs too, but they are too consumer-oreinted from what I have seen, as opposed to programmer oreinted. If you want a cool computer, look at a Linux. You can get versions off the Internet that you can burn to a CD and boot off of to try it. Or just stick with Windows. I don't know.  :?

Billy

Don't get a Mac unless you're a rich fanboy. PCs cost 2x less for the same specs. As for Windows vs. Linux, almost all games are Windows exclusive/sometimes for Macs too, but otherwise, Linux is OK.

Macs are OK, their operating system is about 20 million times more secure than Windows (and so is Linux).  But it depends on what you're gonna use it for, if its for gaming, go with windows. But due to Macs unpopularity, many softwares aren't capatible with them.

I HATE MACS lol well maybe it's cause i'm use to windows and most sites with stuff require windows or like windows media player like most the sites i go to.  DON'T GET AN APPLE!

elvenchamp777

i really want to play games on my computer, so which is better, windows or mac

I would go for Windows, but consider Linux. I know that games are compatible with Redhat, but I think it costs more.

If you're gonna use the computer for gaming then definitely go with windows, because there aren't that many games for macs. and make sure you get a pretty good graphics card such as GeForce 5200 or higher.

It really depends.  Windows is good for gaming, but bad for anything (everything) else.  Linux is free, but there aren't as many commercial games for Linux.  I'm a Mac persion myself (:D).  It's not true that the Mac isn't a programming OS.  It's based on BSD  and includes everything like GCC, Make, Autoconf, Automake, Xcode (an IDE), Interface Builder (interface designer).  Also, OS X Tiger comes out in 13 days!
http://www.apple.com/macosx/

i heard from a friend that you can buy a program called virtue PC or something like that, and then you can play PC games on Mac. this sounds too good to be true, is it?

Yes you can buy a program called Virtue PC (in fact there is also an emulator for windows that allows you to install Mac OS), but it only supports certain programs and the speed and reliability of these programs decrese dramatically. Im pretty sure that if you do use Virtue PC you won't beable to play those games that require fast computers on your Mac such as Half Life 2 or Doom 3.
So I wouldn't recomment Virutue PC for games, it is generally used for applications
If you want to learn more about Virtue PC go here http://www.microsoft.com/mac/products/virtualpc/virtualpc.aspx?pid=virtualpc
(yes it is made by microsoft. -_-)

I give credit for the Apple folks for being the best years ago, but I must say it's now either Windows or Linux, depending on what you're trying to do.  Windows is better for the average consumer, but Linux is more secure and such, and it doesn't do stupid things like Windows does.  I don't see any reason for a mac unless you want your computer to look cool.

[quote="stupidkid"]Personally, the only reason I think that Microsoft has a monopoly on the OS and software market, is Windows' ease of use.  It is pretty much an OS for dummies and of course the general population isn't that smart (like people here on AoPS), they don't know how to use more complicated OS like Linux or Mac.[/quote]You'ld be amazed at the stats to see how many people on AoPS - ML use Windows. I think very smart people use Windows for good reasons. Ease of use is not a deffect, but actually a quality of a product! ;)

i went to the Dell website, the fastest one is like 3.6 GHz(memory)
but the best Mac is 2.5 GHz, i don't get it
does this mean that Macs are slower and more expensive?

First of all, the hardware of Macs and windows pcs are different, so just because the CPU speed is lower is doesn't mean its slower, because it is used in conjunction with different hardware.  Secondly, the power mac G 5 cpus are 64 bit, and Intels are 32 bit. (i know they came out with 6xx processors, but I dont think they are true 64 bit processors.)

[quote="Valentin Vornicu"]This I think says a lot about Apple: Apple comes with IE 6 preinstalled. Apple has Windows emulators, however I know of no Apple emulators on Windows. I wonder why :D[/quote]

Not IE 6.  IE 5.  Micro$oft is too lazy to support Mac in making IE. (which is a horrible browser anyway.)

EDIT: 1000 posts!

I have a G4 (about1 or 2 years old) ,which is a whole lot faster than any computer I'ved used before.

Someone said that people use Windows for ease of use.  But Windows isn't any easier to use than a Mac!  People just don't want to learn.  (And even learning how to use a Mac doesn't take much longer than 5-10 minutes.)

[quote="juicybooty911"]thanks now i get it.
so generally, which one is faster?[/quote]
Sorry I don't have a Mac so I don't really know how fast they are.

[quote="tarquin"]Down with Intel!
Long live AMD![/quote]

my laptop is AMD. and it's as slow as crap
it's also a windows xp
so that's why i wanna switch to mac

[quote="tetrahedr0n"]I also have a laptop running Windows.

It's a Pentium I MMX, 233 mhz, (Windows 95.)

But ... hm .... I'm not going to switch to mac.[/quote]
LOL...nice 233 mhz.  I use to have a dektop that is celeron 466mhz, but it broke couple months ago.

[quote="xdl615x"]Just buy a Window's. Be like the rest of us.  :D[/quote]

Eh, not necissarily.  I remember a various What OS Do you use? topics, and there were definitely a large number of Linux users. (And then some Mac users like me :) )

anybody going to the Apple Tiger World Premier? i heard they will give out some fun stuff, like ipod, powermac...

[quote="xdl615x"]Just buy a Window's. Be like the rest of us.  :D[/quote]I agree that for the average user, Windows is easier while with Linux you can do a lot more things... that probably you will never really need to do. Nevertheless, for the more expert user, Linux is cheaper and has a lot more freedom than Windows (my personal point of view!)

Note that Linux has WINE to run Windows programs. Reportedly the games work quite well on WINE.

Yeah, Linux has a variety of Windows emulators that work decently.  But for all my purposes, I can just use Windows and make a remote connection to a linux computer if I want to do a little programming or something.  I just wish they didn't make all these stupid features on Windows XP.  Sometimes I prefer Windows 98, but lots of things aren't compatible with that.

I'm a Mac user.  One of the reasons I appreciate OS X is that I have a great GUI and simultaneously have Unix.  No need to reboot... I have essentially two OS all the time.  (Plus, if I can't figure something out using the GUI, I just revert to unix commands.)  And none of my machines ever crash.  Yes, really.  I use three different Macs, and run so many programs simultaneously and have so many windows open for days at a time that people get confused  looking at my screen.  But the OS is VERY VERY stable.

For those who think Windows is easier to use... it really depends on what you're used to.  I always need help maneuvering through the Windows interface.

Tiger release:  Would go (went to the Panther release) but I'll be on the road.

Speed:  About a year ago, the fastest Mac had a lower rating than the fastest PC, but in terms of benchmarks the Mac was much faster.  I think that now they're roughly the same (the 3.5Ghz pc and 2.5Ghz mac).

Games:  Well, it depends.  Many of the major games are made for Mac as well as PC, but there isn't quite the variety of games ported to OS X as there are written for PC.  If you need to play thousands of different games, and you always want the latest hottest game when it comes out, then you need a PC.  But if hundreds of games will do it for you and you don't need the latest and the hottest all the time, a Mac should be fine.

how come Mac games are way more expensive than PC ones? for example, simcity4&rush hour EP cost  $35 at bestbuy,  but  it  cost $80 for Mac?
and also why are they made by different companies?like PC version is made by EA games but Mac version is made by Aspyr media?

will the iMac be automaticly shipped with tiger or panther still? or do you have to customize it?

i brought an iMac G5, it's soooooooo awesome

anybody use iTunes and buy music on iTunes music store?

[quote="juicybooty911"]
anybody use iTunes and buy music on iTunes music store?[/quote]

Yup, of course.  Interestingly, music bought at the iTunes store has some interesting security; it knows what username bought the music, so if you try to share it (as in by file-sharing) then it wants an authorization for the iTunes account.

My major complaint about the iTunes music store is that it doesn't have much of the music I'm looking for, but then again I'm super-picky and often want obscure things.

[quote="juicybooty911"]is there any restictions with the songs brought from iTunes Music store?
ex. can i buy a song, burn it to a cd, then take the cd and upload the songs to another computer?
any restictions on how many times a song can be burned?[/quote]

There are legal restrictions, and some actual restrictions.  That is, I believe you're allowed to put a song on up to three computers which you own, and allowed to burn it to CD 10 times (or a similar number).  I'm not sure whether the security on the AAC files actually restricts you from burning the song more than 10 times, or not.  Certainly you can burn a song to a CD and upload the song elsewhere, as long as you burn it redbook rather than mp3.  That is, burn it as an audio cd, not an mp3 or data CD, and it should be fine.  I'm pretty sure that in the redbook conversion to aiff the aac restrictions go away, but I haven't done extensive testing.

I think rather than upload he can probably rip the songs in MP3 format.  Then there will be no need for an ID.

[quote="juicybooty911"]i was wondering if i buy a song from the itunes music store, then burn it to a cd, give it to a friend, can he can play it on any cd player, right? but if he tries to upload the song onto his computer, he needs my id and password to authorize his computer to play right? so does this mean he can't play the songs i brought on his computer? what if he upload the song into another music software like windows media player rather than itunes?[/quote]stupidkid is totally right. If you want to get rid of such a problem with your songs... just burn them in a regular CD, and then rip it back to MP3.

This is not totally clear for me. As far as I remember sometime I read, when you buy any copyrighted material (books, CDs, DVDs, etc) you have the right to play it and either copy it anywhere as long as it is going to be used only for "close friends and relatives" and you are not going to receive a profit of it. This obviusly exclude the P2P nets (that is, even when I am not that sure the laws specifically say that the download of copyrighted material is illegal, the upload is!), but it shouldn't include the distribution with friends (or at least that is what I think). I think that when you download songs from iTunes, you can  have it in no more than 3 dispositives simultaneously. Well, what if  I have five computers (I do, if that include my wife's computers) plus I want to have it in my MP3 player, then I wouldn't have enough freedom to have my songs on each dispositive I own!

[quote="smbelcas"]My major complaint about the iTunes music store is that it doesn't have much of the music I'm looking for, but then again I'm super-picky and often want obscure things.[/quote]Hehe, what do you mean by obscure things? if you don't mind the question :P
My major complane about iTunes is that it is expensive compared to other companies like Napster, at least if you download songs frequently. (iTunes charge you by each song you download, Napster charge you monthly by the service.)

for me Napster cost too much bc u have to pay every month, it's like 15 bucks which means a whole cd.  i don't need a new cd every month.
and i heard you have to keep your membership or you can't listen to the songs anymore


So has anyone made any progress on showing that the numerator of

1/1 + 1/2 + 1/3 + ... + 1/(p-1)

is divisible by p^2, where p is an odd prime greater than 3?

For example, can you show that the numerator is divisible by p?

I haven't thought too much about this problem, but it seems to me it would involve Wilson's Theorem. Specifically:
\[ \dfrac{1}{1} + \dfrac{1}{2} + ... +\dfrac{1}{p-1} = \dfrac{(p-1)!/1+(p-1)!/2+...+(p-1)!/(p-1)}{1*2*3*...*(p-1)} \]

So the numerator mod p would be something like
\[ -(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{p-1}) \]

Which reminds me of inverses mod primes, but I can't remember how that works.

And a question, are we supposed to reduce the fraction before we show the numerator is divisible by p^2? Because if we are, then that's a problem for my Wilson's theorem thing...


trigonometry
limit
topology
real analysis
real analysis unsolved

In the plane R^2, metric space, consider the subsets C = {(x,y)|x^2 + y^2 = 1} and T = { ((2+t)/(1+t))*(cos(t),sin(t)) | t in [0,00) }.

Is T V C compact? Why?
Is T V C connected? Why?

$T\cup C$ is clearly bounded, so the issue of whether it is compact rests on whether it is closed.

$C$ is simply the unit circle. It is itself closed; any convergent sequence of points in $C$ has a limit in $C.$

The spiral $T$ is not itself closed. So now consider a sequence of points $(x_{n},y_{n})$ in $T$ that converges to a point $(x,y)$ not in $T.$ Define $t_{n}$ so that $x_{n}=\left(\frac{2+t}{1+t}\right)\cos t_{n}$ and $y_{n}=\left(\frac{2+t}{1+t}\right)\sin t_{n}.$

Then we must have $\lim t_{n}=\infty.$ Otherwise, at least infinitely many of the points $(x_{n},y_{n})$ would be confined to the portion of the spiral for which $\sqrt{x^{2}+y^{2}}\ge1+\delta.$ That is a closed set, and the limit point would also have to in that closed set, hence in $T.$ In that case, $\lim\sqrt{x_{n}^{2}+y_{n}^{2}}=1,$ and it follows that $\sqrt{x^{2}+y^{2}}=1.$ But then, $(x,y)\in C$ and thus in $T\cup C.$

This proves that $T\cup C$ is closed and thus compact.

Now, is $T\cup C$ connected? $C$ is itself path-connected, hence connected, and $T$ is itself path-connected, hence connected. The only way for $T\cup C$ to be disconnected would be to have $T$ and $C$ as connected components. But that would require $T$ to be a closed subset of $T\cup C,$ and the argument above shows that it is not that.

That proves that $T\cup C$ is connected.

[But it's not path-connected.]


geometry trigonometry probability topology function \/closed

I've got an idea, create new sections like physics, chemistry and stuff like that.

Each is connected with maths, aren't they?

Science problems are sometimes posted in the Round Table Forum. Also, if you have a suggestion to the administrators, you should post it in the Questions and Suggestions forum. More specifically, one of the wish list topics.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16359
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1418

Those are the wish list topics.

Also, welcome to the AoPS-Mathlinks forums!

that idea has been posted sooo often before...

anyway some pushing never hurts :D

this site is (very) big as is, no need to clutter more forums into it. there are special places for phisics and chemistry forums on the net.

it is possible that at some point in the future, a forum for computer science / combinatorics issues will be created, but that point is still far away in the future! :)

[quote="planck"]thanks :)  So i've posted in the wrong place, sorry

also i want to recommend polish forum http://www.matematyka.pl

I know that most of you can't speak polish but don't worry you can post there in english i'm sure most of polish users can speak english :)[/quote]

About speaking polish, I realize that isn't nessecary, but if we're really going to use it as a forum, shouldn't we need to be able to read polish? (I don't even know what a thread is about)

yeah but you can post a topic in english and everybody would post in english, you know when you start discussion in english there's no problem

but my another aim is to check if polish users are able to discuss in english :)

Polish users would gladly help, most speak English 4 sure.
Here goes the structure:
Teoria chaosu- everything on maths, that does not fit to other threads
--==Matematyka- krolowa nauk:
2+2=?- widely understood algebra
Pitagoras- geometry, incl. trigonometry
Urna z kol. kulkami- probability, statistics, graphs
AnalIza- analysis, "higher" maths
Zbior-ka!- topology, dimensions, function theory etc.
Olimpiada- olympic level problems
--==Matematyka to nie wszystko:
Co kazdy user...- what every user should know
Fizyka- phisics
Chemia- chemistry
Informatyka- informatics
Technika- technical problems, inventions
Hyde Park- everything not related to maths

If I see lot of you, english names would also appear

If english would become the main language, of course you'd see a lot of people :)

(more on the science than on the maths forums probably)

Informatyka - Computer Science
Fizyka - Physics
:)


calculus
derivative
calculus computations

this is a problem from my calculs textbook, and I have trouble finding the answer. If anyone here can show me how to do this problem step by step, I would really appreciate it. Thanks in advance.

f(x)=6/(x-8)
Find f'(x) when the limit as t approaches to x.

Notes: the answer will have x and t in it, and you have to solve it by the defination of the derivative which is (f(x+h)-f(x))/(h) when the limit of h approaches to 0.

its going to the calculus forum :o

just to let you know

Here's what I think this problem wants you to do:
Take the difference quotient $\frac{f(x)-f(t)}{x-t}$
Simplify.
Don't take any limits.

where did you get that forumula from? I don't understand you explaination, can you explain more?

Never mind, I figured it out.  Thank you for all the replies.


[quote="mr. math"]how would u do this problem?

billy tosses one fair six-sided die with faces labeled 1-6. he records the outcome. billy does this 3 more times and the product of his four outcomes is 120. how many possible combinations of 4 rolls could he have rolled.(rolling a 1,1,2,2 is considered the same combination as 1,2,2,1)[/quote]

[hide]120 = 3*5*8.
So the 5 is definitely one of the dice. Now the question is how many ways can we make 3*2*2*2 with 3 dice.
There are 2 ways to use a 6: 6*2*2 and 6*4*1.
Now, without using 6:
3*2*4.

So there are 3 ways (6225, 6415, 3425)[/hide]


$ \frac{c}{ab} \plus{}\frac{f}{de} \plus{}\frac{i}{gh} \equal{}1$

 where a,b,c,d,e,f,g,h,i is the different digits from 1,2,3,4...,9.

How to put these nine digits into the three fractions?

[hide="Hint 1"]If either 5 or 7 is in the denominator, we cannot possibly cancel them when we combine fractions (why?). Therefore, put them in the numerator.[/hide]

[hide="Answer"]$ \frac{5}{8 \cdot 9} \plus{} \frac{7}{2 \cdot 4} \plus{} \frac{1}{3 \cdot 6} \equal{} 1$[/hide]

$ \text{Sorry !}$

$ \text{My problem stated unclear...}$

$ ab ,de, gh \text{ are represented by different double\minus{}digit numbers, not for products.}$


For any positive integer x, let S(x)  be the sum of the digits of x, and let T(x) be abs(S(x+2)-S(x)). How many values of T(x) do not exceed 1999?

im not sure about this assumption, but:

[hide]

wouldn't all numbers except for like 999999999999... work for this? for instance if 29742934927593475934 was used... then S(x) would be a big number, but S(x+2) would still be only 2 more than  S(x)... therefore T(x) is equal to 2. However for 999999999, S(x) is 81, and S(x+2) is 2. Therefore T(x) = 79. If you had enough 9's... about 223, then T(x) > 1999. But that's the only way I can think of.[/hide]

It's not asking how many xs satisfy it. It's asking how many distinct values of T(x) are possible and <= 1999.

well we can narrow it down to be 0-999, since all answers on AIME are integers from 0-999  :wink:

[hide]well, if S(x) ends with 0-7,  then T(x)=2, if S(x) ends with 8, then T(x)=7, unless the second to last number is 9, in which case T(x)=15 if the third to last number is also 9, T(x)=23 etc.. that make 249 values for T(x) if S(x) ends with 8. if S(x) ends with 9, then T(x)=8 and with each 9 added to the end we get another value for T(x) making total possibilits 1+249+248=498

am i right?[/hide]


The integers from $ 1$ to $ 2n$ are partitioned into two groups of $ n$ numbers.
Prove that the pairwise sums of the numbers in each group (the sum of each number with itself included) have the same remainder upon division by $ 2n$ (note that each pair of numbers should be added twice and that each remainder must occur the same number of times in the two groups).

i've moved the problem since it seems to me that this is more of a number theory problem than a combinatorial one ...
besides there were no problems in this forum :)

You can find it either in the 1995-1996 book or in the 1996-1997 book , I can't remember exactly now .

well i belive it's not a question of where you can find it ... since many people do not have access to a lot of math information, but rather how do you solve it ... :)

maybe a clue will best fit for the begining, what do you think?

Looking on some old unsolved problems of the forum....

First note that the additive table of Z/(2nZ) is a latin square, that is each of the numbers 0,1,...,2n-1 occurs exactly once in each row and in each colummn.
Let A,B be a partition of Z/(2nZ) where |A| = |B| = n.
Let C,D be a partition of Z/(2nZ) where |C| = |D| = n.
We will prove that, for each x in Z/(2nZ) the number N(x) of ways that x can be written in the form x = a + c is the same as the number N'(x) of ways that x can be written in the form x = b + d.

To prove that claim, it suffices to construct the additive table of Z/(2nZ), where the n first columns are for the elements of C, and the n last columns for the elements of D. And, the first n rows are for the elements of A, and the last n rows are for the elements of B (it is not easy with a keyboard).
Then, in the cases corresponding to the numbers of the form a + c (the nxn square up-left), the number x occurs exactly N(x) times. Since, its total number of occurences is n in the first n rows (and in the first n colummns), we deduce that x occurs exactly n-N(x) times in the nxn square down-left, and n-N(x) times in the up-right nxn square.
The same reasoning on the n last rows and the n last columns shows that x occurs exactly N(x) times in the nxn down-right nxn square of the table. But, this is exactly the number N'(x) of ways that x can be written as b + d, and we are done.

To solve the problem, it now suffices to choose C = A and D = B.

Note that the conclusion still holds for differences instead of sums, by choosing C = -A and D = -D, and to many other choices...

Pierre.

A word to say that this result for the sums has been discovered empirically by the students of the music composer A. Schoenberg.

Pierre.

[quote=iandrei]ok , I'll post the solution tonight if I have time . cheers ![/quote]

I know that it has only been 18 years, but I would just like to know how when you will post the solution. Sorry for bugging you (I am a very impatient person). I know how busy you have been.


(1)

Suppose we pass over the natural numbers 1,2,3,... and pick out a sequence U by taking: the first odd number (1), the next two even numbers (2,4), the next three odd numbers (5,7,9), the next four even numbers (10,12,14,16), the next five odd numbers (17,19,21,23,25) and so on. So U: 1, 2 ,4, 5, 7, 9, 10, 12, 14, 16, 17, 19, 21, 23, 25,... Find the explicit formula for u_n.

(2)

Consider the complementary sequence V of natural numbers which are not un U. V = 3, 6, 8, 11, 13, 15, 18, 22, 24, 27,... Find the explicit formula for v_n.

Hint: u_n + v_n = 4n

Nice problem  :)

1) Let's notice that u(n) is 'not too far from 2*n'.

Indeed, a(n) = 2*n - u(n) is : 1, 2,2, 3,3,3, 4,4,4,4, .... (by induction)

a(n) is well known to be [1/2+\sqrt(2n)] (i.e the closest integer to \sqrt(2n)).

u(n) = 2*n - [1/2+\sqrt(2n)]

2) I'll use the wonderful result of Moser & Lambek about complementary sequences :

u(n) and v(n) are complementary sequences hence f(n)=u(n)-n and f*(n)=v(n)-n are inverse sequences (nice no  :)  ?)

So if f*(n)=N then f(N) < n <= f(N+1), which solves to N = n + [1/2+\sqrt(2n)].

Hence v(n) = f*(n)+n = 2*n + [1/2+\sqrt(2n)]

And v(n)+u(n) = 4n :D

I know this isn't all that important, but the first problem orl posted here was given at a Romanian TST (can't remember the year).

Well done belenos. I like your solution.

You also may rewrite u_n and v_n.

u_n = 2n - li(1/2*(8n-7)^0.5)
v_n = 2n + li(1/2*(8n-7)^0.5)

li(x) denotes the greatest integer \leq x

The problem is from AMM1960, page 380, problem E1382. So grobber,
did you find out on which Romanian IMO selection paper the problem appeared ?


I'm struggling with this equation, but I simply can't see how to reach the solution. The problem occurs on the left side of the equation, can anyone help me how do come from the first to the second step?

[img]http://img178.imageshack.us/img178/8611/gravitymedervation03engoq0.gif[/img]

Source:
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/GravityME-Dervation.html

The word is "algebraic."  You have two extra "r"s.

It's the distributive property -- there's a $ \frac {1}{2}v_p^2$ factor common to both terms on the left, and they brought it out in front.

Oh, okay. English is not my native language, so I'm just trying my best ;).

I see what you mean, but is it okay to do so?

For example:

2x^2-2(x+(y/z))^2
<=>2x^2(y^2/z^2).

Because my first impresion would be that the result would be:

2x^2-2x^2(y^2/z^2)
<=>y^2/z^2

Neither of your examples looks like what has happened in the original, which is
\[ ab \plus{} ac \equal{} a(b \plus{} c)
\]for $ a \equal{} \frac {1}{2}v_p^2$, $ b \equal{} 1$ and $ c \equal{} \minus{} \left(\frac {1 \minus{} e}{1 \plus{} e}\right)^2$.


inequalities
algebra
polynomial
algebra solved

If everyone post inequalities, why shouldn't I ? So, prove that if a,b,c are the sides of a triangle, then  sum sqrt(abc/(b+c-a))>=a+b+c.

What do you say about this one? Isn't it difficult?

Not a complete soln, actually (don't even know whether it works or not):

We write a=y+z, b=x+z, c=x+y. The inequality becomes                   S1=\sum 1/x^1/2 >= 8^1/2(x+y+z)/[(x+y)(y+z)(z+x)]^1/2=P. But we can see that S1>=S2=9/ \sum x^1/2, so it would suffice to prove S1>=P. I think this works out pretty nicely: we square it etc. I'm in an Icafe and I haven't got any time to do the calculations  :( .

Well the inequality 9/sum x^1/2>=8^1/2 (x+y+z)/ sqrt((x+y)(y+z)(z+x)) is not always true. Just take y=z=1 and x very large.

Surely this very interesting inequality is equivalent to:

for x,y,z > 0 prove that

(a 2 +b 2 )(b 2 +c 2 )(c 2 +a 2 )(ab+bc+ca) 2  \geq a 2 b 2 c 2 (a 2 +b 2 +c 2 ) 2

and this isn't simple to prove.
I have tried also to expand the products.

Sorry, I haxe forgotten to multiply the 2nd member by 8.

Exactly, it is exactly the same inequality (if you put b+c-a=x^2, etc). In fact, I wanted to propse first this form of the inequality, but I thought this would be too difficult. It seems that the problem is much more difficult than expected.

Well, I haven't seen it anywhere by now, so I guess it can be called original.

[quote="manlio"]Surely this very interesting inequality is equivalent to:

for x,y,z > 0 prove that

(a 2 +b 2 )(b 2 +c 2 )(c 2 +a 2 )(ab+bc+ca) 2  \geq 8a 2 b 2 c 2 (a 2 +b 2 +c 2 ) 2

and this isn't simple to prove.
I have tried also to expand the products.[/quote]
==========
Hi all,

I use Manlio's result to solve this inequality. In fact, LSH - RSH =
a^6(b-c)^2(b^2+4bc+c^2) + b^6(a-c)^2(a^2+4ac+c^2) + c^6(b-a)^2(b^2+4ba+a^2) + 2abc(a^2b^2+b^2c^2+c^2a^2)(a^2b+a^c+b^2a+b^2c+c^2a+c^2b-6abc) which is obviously nonegative. Now I try to find the more elegant solution.

Namdung

Wow, can I ask you how did you managed to deduce such an identity? Are there any special techniques?

I tried to start from the initial form of the inequality (I replaced a with y+z etc.) I got this after squaring it:

T(3,2,0)+2T(3,3/2,1/2)+2T(5/2,2,1/2)+2T(5/2,3/2,1)+2T(2,3/2,3/2) >=
3T(3,1,1)+6T(2,2,1), where the T's are those symmetric sums in Murhead's thm. I think this isn't that hard to prove:

for one thing, T(a,b,c)+T(t,u,v)+T(m,n,p) >= 3T((a+t+m)/3,etc.) by AM-GM, which means that 2T(5/2,2,1/2)+2T(5/2,3/2,1)+2T(2,3/2,3/2) >=
6T(7/3,5/3,1) and it's easy to prove that this one is >= 6T(2,2,1), because, for example, x^7/3*y^5/3*z+x^5/3*y^7/3*z >= 2x^2*y^2*z by AM-GM and we add all these up.

On the other hand, T(3,2,0)+2T(3,3/2,1/2) >= 3T(3,5/3,1/3) >= 3T(3,1,1), also by AM-GM, just as above. I think this finishes it, except if I've made some mistake  :D Smb needs to check this all.

[quote="harazi"]Wow, can I ask you how did you managed to deduce such an identity? Are there any techiques?[/quote]
==================
My techniques is based on
1) Hilbert theorem on positive forms
2) The fact that if P(x,y,z) = x^nQ1(y,z) + x^(n-1)Q2(y,z)... >= 0 then Q1(y,z) >=0.
3) Some intuition and computational ability of Maple.

First, I grouped the members with x^6 and, knowing that Q(y,z)>=0, easily to get the factorization. Similarly with y^6, z^6. The rest members are all divide by xyz and I used command factor to get the identity. I think Maple is very usefull for this kind of problems (homogenus, symmetric and polynomial inequality). Almost problems (of this kind!) can be solved by this techniques.

Namdung

Maybe it's time to post my solution. Firstly, it follows easily from Holder's inequality that sum a^3(b+c-a)=sum a^3/ (1/sqrt(b+c-a))^2>= (sum a)^3/ (sum 1/sqrt(b+c-a))^2. So, if is enough to prove that sum a^3(b+c-a)<= abc(a+b+c), which is the same with sum a^2(a-b)(a-c)>=0, which is true (Schur). Of course, this is the kind of "author's solution", but I think that it is quite nice.

Hey, does ne1 have the time to check that nasty thing I posted? :D


geometry
geometric transformation
homothety
geometry proposed

Let $ABC$ be a non-right-angled triangle and let $H$ be its orthocenter. Let $M_1,M_2,M_3$ be the midpoints of the sides $BC$, $CA$, $AB$ respectively. Let $A_1$, $B_1$, $C_1$ be the symmetrical points of $H$ with respect to $M_1$, $M_2$ and $M_3$ respectively, and let $A_2$, $B_2$, $C_2$ be the orthocenters of the triangles $BA_1C$, $CB_1A$ and $AC_1B$ respectively. Prove that:

a) triangles $ABC$ and $A_2B_2C_2$ have the same centroid;

b) the centroids of the triangles $AA_1A_2$, $BB_1B_2$, $CC_1C_2$ form a triangle similar with $ABC$.

(a) It is trivial that $A,B,C$ are the midpoints of $B_{2}C_{2},C_{2}A_{2},A_{2}B_{2}$ respectively. Hence $\triangle ABC$ and $\triangle A_{2}B_{2}C_{2}$ have the same centroid.

(b) Let $A_{3},B_{3},C_{3}$ be the centroids of $\triangle AA_{1}A_{2}, \triangle BB_{1}B_{2}, \triangle C_{1}C_{2}$ respectively. It is trivial that $H,M_{1},A_{3},A_{1}$, $H,M_{2},B_{3},B_{1}$ and $H,M_{3},C_{3},C_{1}$ are collinear, and $\frac{A_{1}H}{A_{3}H}=\frac{B_{1}H}{B_{3}H}=\frac{C_{1}H}{C_{3}H}=\frac{3}{2}$. By homothety about $H$, $\triangle A_{1}B_{1}C_{1} \sim \triangle A_{3}B_{3}C_{3}$. As again, it is obvious that $\triangle A_{1}B_{1}C_{1} \equiv \triangle ABC$, hence $\triangle A_{3}B_{3}C_{3} \sim \triangle ABC$.


Right $ \triangle ABC$ has $ AB \equal{} 3$, $ BC \equal{} 4$, and $ AC \equal{} 5$. Square $ XYZW$ is inscribed in $ \triangle ABC$ with $ X$ and $ Y$ on $ \overline{AC}$, $ W$ on $ \overline{AB}$, and $ Z$ on $ \overline{BC}$. What is the side length of the square?

[asy]size(200);defaultpen(fontsize(10pt)+linewidth(.8pt));
real s = (60/37);
pair A = (0,0);
pair C = (5,0);
pair B = dir(60)*3;
pair W = waypoint(B--A,(1/3));
pair X = foot(W,A,C);
pair Y = (X.x + s, X.y);
pair Z = (W.x + s, W.y);

label("$A$",A,SW);
label("$B$",B,NW);
label("$C$",C,SE);
label("$W$",W,NW);
label("$X$",X,S);
label("$Y$",Y,S);
label("$Z$",Z,NE);

draw(A--B--C--cycle);
draw(X--W--Z--Y);[/asy]

$ \textbf{(A)}\ \frac {3}{2}\qquad \textbf{(B)}\ \frac {60}{37}\qquad \textbf{(C)}\ \frac {12}{7}\qquad \textbf{(D)}\ \frac {23}{13}\qquad \textbf{(E)}\ 2$

[hide]
Let the height of the triangle be $ h$
$ \sin \angle BAC \equal{} \frac{4}{5} \equal{} \frac{h}{3} \Rightarrow h\equal{}\frac{12}{5}$

Notice that $ \bigtriangleup BAC \sim \bigtriangleup BWZ$, so $ \overline{WZ} \equal{} 5k$ and the height of the smaller trangle is $ hk$

$ \Rightarrow h\minus{}hk\equal{}5k \Longrightarrow k \equal{} \frac{h}{h\plus{}5} \equal{} \dfrac{\frac{12}{5}}{\frac{12}{5} \plus{} 5} \equal{} \frac{12}{37}$

$ \Rightarrow 5k \equal{} \frac{60}{37} \Longrightarrow \boxed{B}$[/hide]

[hide="coordinate bashing..."]( ...but not directly on the triangle. That would be too much work. xD )

Consider the line through point $ B$ and parallel to $ AC$. If we choose any point on this line, call it $ B'$, then the square inscribed in $ AB'C$ will have the same dimensions as the one inscribed in $ ABC$.

For convenience, we will choose $ B'$ such that $ \angle B'AC$ is a right angle. Now place $ AB'C$ in a coordinate plane so that $ A$ lies on $ (0,0)$ and $ C$ lies on $ (5,0)$, $ B'$ lies on $ (0,12/5)$.

The equation of line $ BC$ is then $ \frac{x}{5}\plus{}\frac{y}{12/5} \equal{} 1$.

One vertex of the square lies on $ (0,0)$. The opposite vertex lies on $ (s,s)$, where $ s$ is the side length of the sqaure. This point lies on $ BC$, so we have $ \frac{s}{5}\plus{}\frac{s}{12/5} \equal{} 1 \implies s \equal{} \boxed{\frac{60}{37}}$[/hide]

[quote="mathwizarddude"][quote="ocha"][hide]
Let the height of the triangle be $ h$
$ \sin \angle BAC \equal{} \frac {4}{5} \equal{} \frac {h}{3} \Rightarrow h \equal{} \frac {12}{5}$

Notice that $ \bigtriangleup BAC \sim \bigtriangleup BWZ$, so $ \overline{WZ} \equal{} 5k$ and the height of the smaller trangle is $ hk$

$ \Rightarrow h \minus{} hk \equal{} 5k \Longrightarrow k \equal{} \frac {h}{h \plus{} 5} \equal{} \dfrac{\frac {12}{5}}{\frac {12}{5} \plus{} 5} \equal{} \frac {12}{37}$

$ \Rightarrow 5k \equal{} \frac {60}{37} \Longrightarrow \boxed{B}$[/hide][/quote]
How did you get $ \sin \angle BAC \equal{} 4/5$?[/quote]
Because ABC is a right triangle at B! :wink:

Oh yeah, didn't read the word "right" in the problem... guess the picture should have shown the right angle...  :D

I actually just copied the picture nearly exactly as it appeared on the actual AMC. I think the fact that they didn't include a right angle was to make sure people were "awake" when reading the problem and played close attention what was in the problem.

[hide="Solution"]Drop a perpendicular from $ B$ to $ AC$ meeting at $ B'$ and intersecting $ WZ$ at $ Q$. Let $ BB'\equal{}h$. $ \frac{5h}{2}\equal{}\frac{12}{2}\equal{}6\implies h\equal{}\frac{12}{5}$. If $ x$ is the square side length, and $ BQ\equal{}y$ , note that $ \frac{12}{5}\equal{}y\plus{}x$.

Translate $ \triangle WAX$ to the right so that $ WX$ and $ ZY$ become the same line. Call the new location of $ A$ $ A'$. Draw a line passing through $ B$, parallel to $ WZ$, which in turn is parallel to $ AC$. It's obvious now that $ \triangle A'ZC\sim \triangle BWZ\sim \triangle ABC$.

Notice that $ A'C\equal{}5\minus{}x$, and therefore, by similarity, $ \frac{x}{\frac{12}{5}}\equal{}\frac{5x}{12}\equal{}\frac{5\minus{}x}{5}$
Therefore, $ 25x\equal{}60\minus{}12x\implies x\equal{}\boxed{\textbf{(B)}\ \frac{60}{37}}$.[/hide]

[quote="hsm174"][quote="brightzhu"][hide="Faster solution"]
Let $ x$ the side of the square.
Notice that $ \triangle ABC \sim \triangle WBZ \sim \triangle AXW$.
So $ BW \equal{} \frac {3x}{5}$ and $ WA \equal{} \frac {5x}{4}$.
Since $ BA$ is the shorter leg, we have $ \frac {3x}{5} \plus{} \frac {5x}{4}\equal{} 3$.
Solving this we get $ x \equal{} \frac {60}{37} \Rightarrow \boxed {B}$.
[/hide][/quote]
um, how did you get BW=3x/5 and WA=5x/4?[/quote]

It comes from the fact that ABC is imilar to the triangles.  So you cn use similatiy ratios.
Namely:

In BWZ x is the hypotenuse and in ABC 5 is the hypotenuse.  Also BW coressponds to the side length 3 in ABC so we have.

$\frac{x}{5}=\frac{BW}{3}$  which by multiplying gives the desired result.  Similar things should happen with the other triangle.


[quote="Bob"]10 toddlers are seated in a circle and given candy. Ted receives 8 pieces and the number of cadies received by each toddler differs by one from the number received by the adjacent toddler. How many different ways are there to distribute the candy?[/quote]
Is Ted a toddler, or is he distributing candy?

[hide]Starting with Ted and going clockwise, let A mean that the next toddler has 1 more candy than the previous one, and let B mean that the next toddler has 1 less candy than the previous.

Because this progression will come back to Ted, the number of A's must equal the number of B's (i.e. 5 A's and 5  B's).

Each arrangement of 5 A's and 5 B's will result in a unique distribution of candy. There are $ \binom{10}{5}\equal{}\boxed{252}$ ways to arrange 5 A's and B's, and therefore that many ways to distribute candy.
[/hide]


Find a formula for the measure of an angle such that the points a[size=75]j[/size], a[size=75]k[/size], and a[size=75]l[/size] in an n-gon (with vertices a1, a2, a3, ..., an) form the angle (with a[size=59]k[/size] as the vertex of the angle).

Example:  in a square (4-gon) a(n), a(n+1), a(n+2) is always 90, and a(n), a(n+1), a(n+3) is always 45

In order to see the general solution, try [hide]drawing the circumscribed circle.[/hide]


geometry
geometric transformation
reflection
ARML

Usually when people claim that I am arrogant without factual basis, I try to take their claim to the logical extreme with some sarcastic statement. I don't see a much better way to handle the situation; I definitely don't take such statements to be accurate reflections on myself.

Otherwise, I don't think I am particularly arrogant in my every day interactions with people. I usually take a neutral view towards people unless they actively do something to earn my respect or disdain.

I'd probably do the first reply, but that's because I'm pretty arrogant.

[quote="José"][quote="CheesyP18"]I'd laugh and call them retards.[/quote]

That's being arrogant.....

If anyone call you arrogant, just do not waste time and do not answer....[/quote]

Somebody didn't catch the irony in that.   ;)

I'd laugh... simply laugh. Laughing does wonders.

missing option:

apoligize, say you didn't intend to come off that way, and ask what struck them that way?  :roll:

Or, if my mom tells me that, I tell her I learned it from her  :D

[quote="mathman440"]well considering the people talking on this site are usually smarter that the average american public, then a bit of arrogance should be appropriate.  :)[/quote]

Except.... not.

filletwho, how did you replied to the person who said that?  :lol:

I acknowledge that slight arrogance is one of my flaws, and say that I try to fix it.
But people always tell me that I'm at the far extreme opposite: extremely modest.  Hmmm...meh.  I mean, when I'm arrogant, I realize it later, and then try to fix that in future.

Yeah, I'm arrogant sometimes, and I need to fix that... :D

I would just say "I'm sorry you feel that way."  it's the ultimate weasel excuse. i doesn't really prove anything either way, but it does make it seem as if the problem is on their end.  :P

i think for me it depends on who im around, at my school im unbelievably arrogant but at arml or any place where im with lots of smart people i at least try to be less arogant


A flea is on each square of a $9$x$9$ board. In each second, all the fleas move to a diagonally adjacent square. Some squares may contain more than one flea. Find the smallest number of unoccupied squares.

The smallest i've found is [hide] $9$. [/hide] There's a method I use to reduce the square into a square of dimensions 2 less, but i don't know how to prove that this is the smallest.  :)

[quote="boxedexe"]Your answer is correct.
[b]Solution.[/b] Color the columns of the board alternately black and white.  You'll have $36$ of one color and $45$ of the other.  Crawling diagonally changes the color of a flea's square.  As a result, at least $9$ squares must be empty.$\Box$

[b]Remark.[/b] This exercise is widely known in the field of proofs using coloring.[/quote]

Ooh, that makes sense. I'll have to remember to try colorings like this one. Thank you!  :)

([b]EDIT[/b]: I know about the Engel to learn about these types coloring proofs (and can't afford it right now); but do you know of any other (hopefully less expensive) good books, lectures, or websites on these methods? Thanks.  :)  )


Given quadrilateral $ ABCD$ with $ \angle CAD\equal{}45, \angle ACD\equal{}30, \angle BAC\equal{}\angle BCA\equal{}15$. Find $ \angle DBC$.

See the attached diagram.

$ \cos 15^\circ \equal{} \sqrt {\frac {1 \plus{} \cos 30^\circ}{2}} \equal{} \frac {\sqrt {2 \plus{} \sqrt {3}}}{2} \equal{} \frac {1 \plus{} \sqrt {3}}{2\sqrt {2}}$

Therefore $ AC \equal{} 2a\cos 15^\circ \equal{} a\frac {1 \plus{} \sqrt {3}}{\sqrt {2}}$

On the other hand, $ AC \equal{} AH \plus{} HC \equal{} h(1 \plus{} \sqrt {3})$, so

$ h(1 \plus{} \sqrt {3}) \equal{} a\frac {1 \plus{} \sqrt {3}}{\sqrt {2}}\implies a \equal{} h\sqrt {2} \equal{} AD$. Hence $ \triangle ABD$ is equilateral, which gives $ \angle ABD \equal{} 60^\circ\implies\angle DBC \equal{} 90^\circ$

(Almost) obviously $B$ is the circumcenter of $\Delta ACD$, so $m(\angle DBC)=2m(\angle DAC)=90^\circ$!

Best regards,
sunken rock


Median $AD$ of triangle $ABC$ intersect his inscribed circle in points $M,N$.  Find $\measuredangle MON$ ($O$ center of inscribed circle )if $AB+AD=AC$

Hey...I've got this one :) Let $N$ be nearer to $D$ than point  $M$ Let's denote tangent points of our circle to sides $BC,AC,AB$ by $X,Y,Z$ respectively. Then we have $BX=BZ$, $AZ=AY$, $CY=CX$ which together with out condition $AB+AD=AC$ can give us $BX+AD=XC$ But as $BD=DC$ we obtain $XC-BX=2XD$ which gives together \[ AD=2XD \].Now let $P$ be a point symmetric to $X$ with respect to $D$ and $Q$ - point symmetric to $X$  with respect to $O$. Now we  have that points $P,Q,A$ ae collinear ( it'an well known and easy fact...  I can prove it later if you want) Let $R$ be perpendicular projection of $O$ on $AD$ and let $\angle ODR=\alpha$ Then as $XO=OQ$ and $XD=DP$ we get $OD||AP$ and $OD=\frac{QP}{2}$ this means that $\angle DAP= \angle ODR=\alpha$ For tringle $ODR$ we can get \[ OR=OD sin\alpha=\frac{QP}{2} sin\alpha \] But form sine theorem for triangle $ADP$ we may obtain $sin\angle XPQ=2sin \alpha$ so \[ OR=\frac{QP}{2} \frac{sin \angle XPQ}{2}=\frac{QX}{4}=\frac{OX}{2} \] Now let's look on the triangle $ORM$ we get $\frac{OR}{OM}=\frac{OX}{2OX}=\frac{1}{2}$ which means that $\angle ROM=\frac{\pi}{6}$ and finally \[ \angle MON=\frac{\pi}{3} \]

What is the locus of vertex A when B and C are fixed , D is the  midpoint of BC and

                 AB + AD  =   AC    ?

  Thank you.

  M.T.

[b]The proof of the relation[/b] $Q\in AP.$ Denote $L\in BC,\ AL\perp BC.$

$BD=\frac a2,\ BX=p-b,\ PX=2\cdot DX=2|BD-BX|\Longrightarrow PX=|b-c|.$
$BP=p-c,\ BL=c\cos B,\ PL=|BP-BL|\Longrightarrow PL=\frac pa\cdot |b-c|.$
$Q\in AP\Longleftrightarrow \frac{PX}{PL}=\frac{QX}{AL}\Longleftrightarrow \frac ap=\frac{2r}{h_a},$ what is truly $(S=pr=\frac 12 ah_a).$


parameterization algebra polynomial puzzles

Given that the first ten terms of a sequence are $S=\{1,8,9,10,8,9,a,b,c,1\}$, find $(a,b,c)$.

that satisfies this sequence because what YOU have in mind for the pattern of the sequence may be different from what I have in mind which may be different from what OTHERS have in mind.  Unless you narrow the parameters, there is no unique solution.

Okay, so give me at least two possible answers, then tell me what I need to do to make only one answer possible. :huh:

All you have to do is create a polynomial that gives those values for successive x. Therefore a) it has infinite solutions and b) doesn't necessarily have to be expressed by some trick (i.e. it can be purely mathematical, not logical).

Remember that sets are actually unordered.  It makes more sense to say that the first ten terms are $1,8,9,10,8,9,a,b,c,1$.


In the expansion of $ (a\plus{}b)^n$, where $ n$ is a natural number, there are $ n\plus{}1$ dissimilar terms. Find the number of dissimilar terms in the expansion of
$ (a\plus{}b\plus{}c)^n$

I'll expand on JRav's hint:

[hide]If the like terms of the expansion are combined, each term is of the form $ ka^xb^yc^z$, where $ k$ is a numerical coefficient. ($ x,y,z$ could be zero.) What relation can you write between the values $ x,y,z,n$?[/hide]

Once you solve this problem, you should actually be able to determine the answer to your question for any number of terms, and any power.

Actually, it is very simple.
[hide="Solution"]Let a+b=x
$ (a \plus{} b \plus{} c)^n \equal{} (x \plus{} c)^n \equal{} \sum\limits_{i \equal{} 1}^{n}\binom{n}{i}x^ic^{n \minus{} i}$
where i=0 we have 1 term
where i=1 we have 2 dissimilar terms
where i=2 we have 3 dissimilar terms
....
where i=n we have n+1 dissimilar terms
All terms is dissimilar so there are $ \frac {(n \plus{} 1)(n \plus{} 2)}{2}$ dissimilar terms in the expension of $ (a \plus{} b \plus{} c)^n$[/hide]

Also note the distribution solution:
[hide]Each term in the expansion of $ (a\plus{}b\plus{}c)^n$ is going to be in the form $ Ca^xb^yc^z$, where $ C$ is the coefficient and $ x\plus{}y\plus{}z\equal{}n$. Each solution of $ x\plus{}y\plus{}z\equal{}n$ in non-negative numbers corresponds to a different term, so by counting the number of solutions we get the number of terms.

Using balls-and-urns, the answer is $ \binom{n\plus{}3\minus{}1}{3\minus{}1} \equal{} \binom{n\plus{}2}{2} \equal{} \frac{(n\plus{}1)(n\plus{}2)}{2}$.[/hide]

Simple extension:
How many dissimilar terms are in the expansion of $ (a_1\plus{}a_2\plus{}\cdots\plus{}a_n)^r$, where each of the $ a_i$ are independent variables?

I attempted this in the past and I believe sunehra is correct.

[quote="TachyonPulse"][hide="My Solution"]
Using balls and urns we find that it is $ \boxed{\dbinom{n+r-1}{r-1}}$.

[/hide][/quote]

correction
 $ \dbinom{n+r-1}{r} $

That's wrong, it doesn't work for r=2. Also, no need to drag up topics that were already solved.


geometry
geometric transformation
rotation

[quote="Reon"]On a frictionless horizontal surface the body AB is pushed with a force F at some point near the end. Find the point on the body about which it would rotate.[/quote]
You posted this twice.
The force being applied gradually will tend to produce no rotation as will be done by a momentum-imparting-flick wherein the rod would rotate aout the centre of mass unless there is some friction or opposing forces present.


http://www.artofproblemsolving.com/Community/MJImages/AMC8_2004_14.gif

What is the area? I used a very unorthodox method that worked. What others are there?

Pick's Theorem or Shoelace Method would both work nicely.

Pick's Theorem: For any polygon with vertices on lattice points, where I is the number of points on the interior of the figure and B is on the boundary of the figure, the area of this polygon is $I+\frac{B}{2}-1$.


http://www.transience.com.au

At this website you'll find something in the Games lounge called "Pearls before swine (part III)". And I've found a way to always beat that weird arabian guy!

The way the game works is, you have rows of pearls for you to choose from. You can take as many pearls as you want from any one row. Opponent does the same. The loser is the one who is left with the last pearl. In part 3, it's a series of rounds with different amounts and rows of pearls. And so...

(Before you scroll down, play the game and see if you can figure it out first. It is a pretty fun game, and especially becuase of all the funny things the guy does.)

The way i figured out-- okay, i didn't really FIGURE this out, i found out from an old text-based game. But anyway, what you would do, for example, 4 rows with 3,4,5,and 6 pearls.

1. Convert them all to base 2. (11, 100, 101, 110.)
2. Add them like they were base 10. (...=322.)
3. The point is to take away the least possible amount from that number so all the digits are even.

322-100=222. So, we need to remove 4. It works with any of the last 3 rows. That is the best move you can make. He will try another move, but do the same thing you just did and you will have him beat.
Just remember, at the end you might be faced with 1 and 1, and 1+1=2. Oh no! It's even! But take away one. That's it. It's mostly mathematical, but you do need some sense and knowledge of common patterns.

This is actually a very famous game called Nim - http://www.cut-the-knot.org/ctk/May2001.shtml Well, it's actually a variation of the original. In the original, the winner is the one who TAKES the last pearl.

Two other less common variations are: Poker-Nim, where on your turn you can either take as many as you want from one heap, or increasing the size of one (existing) heap using some (or all) of the counters that you gained in previous moves.

And Lasker's Nim, where on your turn you can either take as many as you want from one heap, or split one heap into two smaller non-empty heaps.

Edit: Actually, the strategy that you outlined above is for the original version of the game - where you WANT to take the last pearl. Try it out. A similar idea though will lead you to the winning strategy for the game that you described.