Να βρεθούν όλες οι συναρτήσεις f από το R στο R
ώστε:

$ f(f(x)\plus{}y)\equal{}f(x^2\minus{}y)\plus{}4yf(x)$

για κάθε χ, ψ πραγματικούς.

to post einai ektos askisis!

kosta apithano to moto sou!

ps. prosekste auto pou leei o dimitris.imo 2008. imo 2009? (ba...) bmo 2009 ?(hmmm...) :P

[quote="bilot4"]Βάζω σύντομα τη λύση:
[hide]
1.$ y\rightarrow 0$
2. $ y\rightarrow x^2 \minus{} f(x)$
3.$ f(x) \equal{} 0$ ή $ f(x) \equal{} x^2$ για κάθε $ x$

[b]Εδώ αφήνω την λύση για να προσεχθεί η γνωστή παγίδα![/b][/hide][/quote]
Νομίζω ότι ο τρόπος αυτός οδηγεί στην παγίδα αν εννοείς ότι το γινόμενο δυο συναρτήσεων να είναι μηδέν χωρίς καμιά τους να είναι η μηδενική

[quote="Nick Rapanos"]den exo koitaksei katholou tin askisi, alla fantazomai oti to gnosto kolpo tha einai xrisimo kai edo! dld esto $ x_1,\neq x_2$ tetoia oste $ f(x_1) \equal{} 0$ kai $ f(x_2) \equal{} {x_2}^2$
...[/quote]

Exactly  :wink:  Αυτό ακριβώς πρέπει να γίνει και εδώ... και μετά βγαίνει εύκολα!

Έτσι είναι. Βασικά μόνο αυτό θέλει λίγη δουλειά αλλά είναι όπως το λέει ο Nick και φαντάζομαι όπως το έκανε και ο mim.

Το μότο είναι όντως όλα τα λεφτά. :wink:


AMC
USA(J)MO
USAMO
USAJMO
AIME
AMC 10
AMC 12

This is in response to AMCDirector's call for single-question topics. (In other words, don't put anything about Red MOP here - that's a different subject.)

Should the AMC reduce the number of 11th and 12th graders invited to take the USAMO in order to expand the number of 10th graders and younger invited to take one or the other of the two olympiads?

It appears to me that the 270 USAMO, 230 USAJMO split will have exactly that effect - see in particular post #32 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691&start=20]here.[/url]

No, I feel that the MAA should make it at least a 350/150 split to ensure that the contest has about the same number of under and upperclassmen as before.

The sense of the Committee on the American Mathematics Competitions and the general sense of the USAMO graders was that the "proper" number of participants in the USAMO should be be about 250, as it was in 2003 (250 actual then), 2004 (266 actual then) and 2005 (259 actual then).  In 2006 we increased to 436 actual, and from there to 500+in 2008 and 2009.  The strong sense of the two groups (which overlap) was that many out of the larger numbers of students invited to the USAMO with the increase in size were not getting the good out of the USAMO experience that they should.

Those twin senses led the CAMC to advocate splitting the USAMO into the MO and the JMO, returning to the size of 250 that had seemed to be right.  The size of 270 was chosen to allow for selecting about 250 directly, and then to have representatives from each state not already represented.  The 230 for the JMO then follows from the total resources that we are able to commit to the grading.

Had the 2010 selection procedure been in effect this year in 2009, the invitation index for AMC 12 + 10 * AIME would have been 213.5 for 259 studnets.  We would have found 6 students who took AMC 10 only, then got 11 or better on the AIME.  We would have found 18 states not already represented and invited a the state high index.  This would have given a total of 283 invitees, but we always have a few declines, so we would have been close to our goal of 270.

Under the 2010 rules, we would have have invited:
101 12th graders (instead of 148)
100 11th graders (instead of 143)
45   10th graders (instead of 123)
9     9 th graders (instead of 64)
5   <9th graders  (instead of 43)

For the group of 283 we would have invited under the 2010 rules, the 2009 USAMO Mean score would have been 9.37, and the median 8, with a stdev of 7.87.

Among the 90 11th and 12th graders who were invited this 2009 year, but would not have been invited under the 2010 rules, the 2009 USAMO mean was 4.50 and the median 3 with a stdev of 4.95.

Under the 2010 selection process, about half the students are getting at least one problem right.  The students who would not be selected are getting a few points for starting but not finishing a problem.  That bears out the observations above.

Any split of 500 into two groups will ultimately be somewhat arbitrary, but the size of the split here was guided by the experiences that we had in 2003, 2004, and 2005 compared to 2006 and beyond.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thank you. That is the explicit defense I was looking for. As a matter of policy, you are reversing the expansion of the USAMO, and redirecting the resources used to expand the USAMO into the creation of the USAJMO, which will mean that a larger number of students taking one of your olympiads will be 10th grade or younger.

I have just one quibble:

[quote="AMCDirector"]Had the 2010 selection procedure been in effect this year in 2009, the invitation index for AMC 12 + 10 * AIME would have been 213.5 for 259 students.[/quote]
Had the 2010 selection procedure been in effect in 2009, many of the very top students who took the AMC10 would have elected to take the AMC12 instead. That changes the dynamics, and I would guess the index would have settled in the 215-217 range, and that instead of 201 11th-12th graders you'd have had 180 or so.

Added in edit:

As in the previous linked post, I look at the California list, with the assumption that all high AMC10 scores turn into AMC12 scores that are 20 points lower - in other words, assuming that all of the USAMO-candidate students would elect to concentrate on the AMC12, with the 20 point deduction being an estimate of the difficulty difference between the two.

This time, I'll use 213.5 as the cutoff. At that index, for California we would get 65 USAMO qualifiers, of which:

18 12th graders
19 11th graders
15 10th graders
8 9th graders
5 <9th graders

Note that that's [i]all[/i] of the 9th graders and younger that you were claiming nationally, and fully a third of the 10th graders - just in California alone. The state may be big, but it's not that big.

Using 216.0 as the index eliminates just 6 of those 65:

18 12th graders
16 11th graders
15 10th graders
6 9th graders
4 <9th graders

At 220.0 we would be down to 50 USAMO qaulifiers:

16 12th graders
12 11th graders
12 10th graders
6 9th graders
4 <9th graders

Given the behavior change to taking the AMC12 - and it is going to happen - there are still going be quite a significant number of young USAMO qualifiers.

I am very satisfied with Prof. Dunbar's explanation about 270/230 split. When I initially saw the split, I was thinking about 2005 and couple of years before that when only 250 people selected.

My question now is: will AMC still have A and B on two different days? If so, an underclassman would still be able to do both 10 and 12. How would his/her index get counted? Using AMC12 score and AIME to calculate for USAMO, if not qualify, and then using AMC10 score and AIME to calculate for JMO? Do I understand this correctly?

I understand Prof. Merryfield's point. There are a few 11th and 12th graders with index scores between 213.5 and 215-217, who will be pushed out by a few more underclassmen with high AMC12, but I don't feel a real loss of talent for the purpose of USAMO test. It could be a personal loss for those few students on the borderline, especially 11th graders who never made USAMO in the 9th and 10th grades and ready to apply for college. This 270/230 split would make qualifying for USAMO harder for upperclassmen, but would mean a little bit more for those who made it.

[quote] The strong sense of the two groups (which overlap) was that many out of the larger numbers of students invited to the USAMO with the increase in size were not getting the good out of the USAMO experience that they should. [/quote]
What exactly do you mean by this?

[quote]The strong sense of the two groups (which overlap) was that many out of the larger numbers of students invited to the USAMO with the increase in size were not getting the good out of the USAMO experience that they should.

What exactly do you mean by this?[/quote]

The consensus of the CAMC was that having something like 250 very bright people spending 9 hours over 2 days struggling over the USAMO and then getting a score of 4 or less, not even a full solution on one problem  (the median  score on the  2009 USAMO) is probably not a good way to inspire them or their peers to further study in mathematics, science or engineering.

Having success in fully solving and explaining one, two or even three problems on a six-problem contest is probably a much more satisfying experience, even if by popular standards, it still is "less than half".

Some readers on this forum claim that low scores inspire them to greater energy in studying mathematics, but the CAMC believes that for the majority of students, including those who observe the USAMO participants, it only reinforces the image of mathematics as forbidding, impossible and unwelcoming.  Our task is to create a contest situation that fully challenges the best, while encouraging, or or at least engaging very very good.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote]redirecting the resources used to expand the USAMO into the creation of the USAJMO, which will mean that a larger number of students taking one of your olympiads will be 10th grade or younger. [/quote]

That's true.  It is my observation that most (not all) of the recent IMO team members started young, then developed over several years of Olympiad level competition before making the team.  It is my hope that we may be able to find and develop more of this young talent.  But that was not the impulse behind creating the JMO, the impulse is explained in another one of my posts.  I think this may just be a beneficial side-effect.

[quote]Had the 2010 selection procedure been in effect in 2009, many of the very top students who took the AMC10 would have elected to take the AMC12 instead. That changes the dynamics, [/quote]

I agree, but I don't know how to judge and predict the change in dynamics now.  So I'll measure only what I have now, understanding that is an imperfect measure and can only give a rough guide.

  [quote]Note that that's all of the 9th graders and younger that you were claiming nationally, and fully a third of the 10th graders - just in California alone. The state may be big, but it's not that big. [/quote]

Actually, it might be.  Along with just the population size, the demographics, economics  and culture of the greater Bay areas, parts of the LA area, and the San Diego area might outweigh everything else.  We'll go back and look at the states our 9th graders came form to see if they were all from California, state of origin was not where we concentrating our attention this afternoon.

Steve Dunbar

I have to say that I agree with AMCDirector here. I would have loved for there to have been USAJMO when I was younger. Although I was angry for not qualifying, had I taken USAMO in 9th grade I would only have been discouraging.

[quote]My question now is: will AMC still have A and B on two different days? [/quote]

Yes, having two dates is convenient for school schedules and has more to do with when schools can give the contest than it does with offering it twice to students who wish to improve their chances.

[quote]If so, an underclassman would still be able to do both 10 and 12. How would his/her index get counted? Using AMC12 score and AIME to calculate for USAMO, if not qualify, and then using AMC10 score and AIME to calculate for JMO? Do I understand this correctly?
[/quote]

Yes, that is correct.

[quote]
loss for those few students on the borderline, especially 11th graders who never made USAMO in the 9th and 10th grades and ready to apply for college.
[/quote]

This is not why we offer the AMC contests.

Steve Dunbar
MAA Director, American Mathematics Competitions

I agree with Kent's assertion that basing the projected numbers off of the current set of data is not useful, given the rather reasonable assumption that behavior will change due to the new rules for USAMO/USAJMO qualification.

I am ambivalent about the 270/230 split.  While I think it makes USAMO qualification much more competitive, this might be a good thing if MAA is worried about the low median score on the USAMO.  Diverting some of those students away to the USAJMO should achieve the established goal of making sure that younger students don't hit the wall that is the USAMO and become discouraged.  That being said, I am somewhat worried about the numbers and how they will turn out.  I would be more comfortable if the qualification standard were left vague and flexible, so that the numbers could be adjusted after the fact.

I would also suggest a third option (though this may only make things murkier): 10th graders who take only the AMC12 can be diverted to the USAJMO if they don't make the USAMO cutoff.  I think this alleviates some potential problems with overconfident students who end up making careless mistakes and also the possibility that students who otherwise would have made the USAJMO if they took the AMC10 challenged themselves and took the AMC12 instead getting punished by not qualifying for either the USAJMO or the USAMO.  I don't know if this should be done by using the same AMC12 + 10*AIME index for USAJMO qualification or if there should be some score adjustment.  (And this should only apply to students who were eligible for the AMC10 in the first place.)

[quote="sophia"][quote="Kent Merryfield"][quote="sophia"]I must say that I cringe more than a little bit at calling such students "sharks." [/quote]
Would you have been happier if I'd said "big fish"? The image was meant to evoke something about strength and ability, not anything about behavior. I intend no criticism of those students for behaving rationally.[/quote]

Thanks for the clarification.  I agree that sharks is much more evocative than "big fish."
[/quote]

As a marine biologist (yeah, I know, wrong forum to be on, huh?), I must protest: sharks [i]are[/i] "big fish", and big bony fish (non-sharks) are just as predatory as sharks.  (Though there are examples of both that tend to eat plankton.) :)

[quote="haoye"]I would also suggest a third option (though this may only make things murkier): 10th graders who take only the AMC12 can be diverted to the USAJMO if they don't make the USAMO cutoff.  [/quote]
I like it. I'm not sure how to write the rule, but there should be such a path. What would we get with 1999 data, with AMC10's converted to AMC12's at a 20 point discount, and a USAJMO index cutoff of, say, 195? I don't feel like running this scenario against the California data right now, but I think this would include quite a few of the people you want to have take the USAJMO.

In fact, I wouldn't mind at all if this pathway generated a larger fraction of the USAJMO takers than the AMC10 pathway.

Proof-writing tends to be the generic weakness of the young student, even those young students who know plenty of trigonometry and algebra (and who may even be in a calculus class in the 8th or 9th grade) - the key is the age and the maturity. I can certainly see students who I'd want to challenge themselves with the AMC12 but who would be better served by the USAJMO.

As for the the AMC10 - let it be the sort of body of water in which a smallmouth bass is a big fish. (Just to further torture the ichthyology metaphors.) Let AMC10 takers worry about school honors, and about the chance to maybe find out that the AIME is very difficult.

.[quote]The strong sense of the two groups (which overlap) was that many out of the larger numbers of students invited to the USAMO with the increase in size were not getting the good out of the USAMO experience that they should.[/quote]

Hi. Other than the top students who either qualify for the team, are alternates, and/or end up going to one of the summer MOP programs, how does the "experience" vary whether you are the student just below these cutoffs - lets just say student 101 arbitrarily (as I don't know the exact numbers admitted to each program), or student 502? In other words, with the exception of the students who qualify for further training, I'm not sure how expanding the number of students that qualify (or conversely, reducing that number) has any effect at all on the vast majority of the qualifying students' experience. Could you please elaborate on what you meant? Thank you.

[quote="haoye"]
Diverting some of those students away to the USAJMO should achieve the established goal of making sure that younger students don't hit the wall that is the USAMO and become discouraged.
[/quote]

I am perhaps one of those younger students, I was in 8th grade this year, and had tons of trouble with it this year. This has done the opposite of discouraging me, instead, it offers me a glimpse of the future, one, in which through hard work, I might someday be able to get 7's on the problems. The USAMO has probably been the most encouraging math competition experiences I've had since bombing MATHCOUNTS in 6th grade, both of which push me to work really hard to improve to the level where I [i]don't[/i] fail. Putting some youngsters into USAJMO might steal away their chances of expereincing what I expereinced, and instead, as the analogy was made earlier, make them feel content with being the "big fish in a small pond". I used to be a big fish in my small pond, but had it not been for tests that I had no way of approaching (at the time), I wouldn't have gotten better.

[quote="Kent Merryfield"]
Using 216.0 as the index eliminates just 6 of those 65:

18 12th graders
16 11th graders
15 10th graders
6 9th graders
4 <9th graders[/quote]

That would also eliminate a USAMO honorable mention  :wink: I needed that expansion to make USAMO this year due to an unfortunate AIME score >.>

I still think 230 USAJMO qualifiers is too much. From what I understand, the AMC wants to "soften the blow" of the USAMO to younger students, hoping not to discourage them, but this isn't the case:

[quote="Mewto55555"]
 I was in 8th grade this year, and had tons of trouble with it this year. This has done the opposite of discouraging me, instead, it offers me a glimpse of the future, one, in which through hard work, I might someday be able to get 7's on the problems. [/quote]

THIS is what I believe young students should feel about the USAMO and its high difficulty; those who are good enough to make USAMO definitely have the perseverance to continue to improve to handle USAMO-level problems. (Those who don't probably don't care much anyway [well, not a high priority compared to say, schoolwork] - not many people qualify 2-3 times and get 0s each time.) I feel like this is essentially just trying to make students "feel better about themselves" (let's all be politically correct?)

So I suppose I could say I'm opposed to the existence of the USAJMO because it diverts a segment of the population that could be facing new, steep, challenges into people that are only challenged somewhat. Sure, it may be better for the bottom 25% of "good" young AIME takers (6+ on AIME?) but I believe the middle 50% are hurt by being forced to take an easier contest that doesn't stretch them. However, if the AMC really wants this, I suppose it isn't a bad idea. But the 270/230 split ... I really believe more skilled students should be able to qualify for USAMO as opposed to more younger, but not yet honed, students being able to qualify for USAJMO.

I have an idea.

Why not keep the same selection process for USAMO, then, with the USAMO form-thingy that you must turn in, have all qualifiers (10th grade and under) have an option to take the USAJMO, (and then make them ineligible for MOP?). This solves all of our problems, no? USAJMO becomes an option instead of forced, and we won't have people griping that the change robbed them of the chance to take the USAMO.

No, absolutely not. The selection process as of 2009 is a complete mess, and removing the AMC 10 from the USAMO selection process is an essential part of the plan.

I completely echo mewtwo5555 not_trig and math154's comments, I would be willing to wager that most people who take the USAMO at a young age and score dismally (such as my this yr-9th grade) are more motivated to work harder than discouraged.  I for one definitely am.  I especially like the big fish in a small pond analogy, because that's effectively what I am here in my state, contests like USAMO, ARML, Mathcounts Nationals, etc. reminded me there are other people who are far smarter than me, which definitely motivated me.  Being a big fish in a small pond is good for self-esteem, but definitely not for improvement...

At the same time, we can't take you guys' word for it - I think the people who aren't motivated by failing a 9-hour test are also not the type to frequent AoPS. There's a certain culture here, and we can't forget it.

But it takes a somewhat large amount of work and/or natural talent to even qualify for USAMO. Anyone who makes it that far probably has a good work ethic. I mean come on, is that 4th grader from Texas really going to quit math because he failed the USAMO?

Again, he is not a typical example. We'd need a real survey to actually have valid data for this point.

Yes he clearly isn't a typical example, but then again, [i]none[/i] of the USAMO qualifiers are typical, all of them are exceptional mathematicians.

I'm bumping this old thread mostly to bring up my own post #4 above regarding California, and to check it against the reality of this year. The one thing I didn't foresee was that California would dramatically increase its share of the national olympiad totals. The actual number of USAMO qualifiers for California is 72, partitioned as follows:

22 12th graders
26 11th graders
14 10th graders
7 9th graders
3 <9th graders

I'm going to save any suggestions I might or might not have for a later time.


[quote="Kab^2"]Hint: simultaneous equations

(2x + 5y = 4) x3
(3x + y = 7) x2

6x + 15y = 12
6x + 2y = 14
Subtract the equations,
13y = -2
y = -2/13

3x -2/13 = 7
3x = 7 + 2/13
3x = 93/13
x = 31/13[/quote]

thx a bunch!!

[hide="hint"]2x+5y=4
3x+1y=7

The goal is to be able to add the two equations together and have one variable cancel out. (Do you see why?)

To do this, you have to multiply one equation by a number.

Got it now? If not:

[hide="one solution"]2x+5y=4
-15x-5y=-35 (Note that this is the same equation multiplied by -5)

Adding:
-13x=-31
x=31/13

Now solve for y.[/hide][/hide]

Edit: I posted too slow; two answers have already been posted. Got an error message.

[quote="linzk3"]forgot how to do this problem...

solve for x and y

2x+5y=4
3x+1y=7

pls and thx  :wink:[/quote]

[hide]subtract the 2 equations eliminating x is the easiest way i think.

$6x+15y=12$
$6x+2y=14$
$13y=-2$
$y=-2/13$

sub back in to find x

$3x=7-y$
$3x=93/13$
$x=93/78$
$x=31/13$[/hide]


I've had some trouble understanding how exactly a manometer works. Could someone provide some assistance?  :)

Start by looking [url=http://en.wikipedia.org/wiki/Manometer]here[/url]. What exactly you don't understand?

I specifically have trouble understanding the concept involved with finding the pressure of a gas using the height of the mercury in a manometer and the atmospheric pressure. Basically, why and how this idea works.


search ratio videos Support email algebra function

[quote="13375P34K43V312"]Lips of an Angel by Hinder
Dirty Little Secret by the All American Rejects[/quote]

Interestingly, those are on my Songs I Want To See Parodied The Most list.

Mostly trance music:
(In Windows Media Player, go to "add to library" and then to "add url." Copy and paste the following urls.)

Dreaming (this one is really good):
http://www.fileden.com/files/2006/8/6/157238/Dreaming%20%28Myspace%20Remix%29.mp3

Shining Star:
http://www.fileden.com/files/2006/8/6/157238/Shining%20Star%20%28Mark%20Myspace%20Remix%29.mp3

Elements of Trance:
http://www.fileden.com/files/2006/8/6/157238/elements%20of%20trance%20%28Myspace%20Remix%29.mp3

Kaleidoscope (awesome buildup):
http://www.fileden.com/files/2006/8/6/157238/10%20Kaleidoscope%20%5BExtended%20Mix%5D.wma

PM me if you want to hear more.  :lol:

I like "Call on me" but that was last year or early this year so I don't have a link.

[quote="PenguinIntegral"][quote="13375P34K43V312"]What were your favorite new songs in all of 2006? Mine...

What Hurts the Most by Rascal Flatts
Lips of an Angel by Hinder
Dirty Little Secret by the All American Rejects
Move Along by the All American Rejects
Photograph by Nickelback
Temperature by Sean Paul[/quote]

Hello, MTV/VH1-controlled robot! May I direct you to http://www.pandora.com ?  There you will find reccomendations for songs and artists, based on your current tastes, that [b]don't[/b] suck.

American popluar music is so bad these days. Please watch the documentary [url=http://www.amazon.com/Featuring-Matthews-Costello-Branford-Marsalis/dp/B000KGGP8U]Before the Music Dies.[/url][/quote]

pandora.com seemed pretty cool. i like how it finds a something to listen to that is similar to your search request.

[quote="drunner2007"]
pandora.com seemed pretty cool. i like how it finds a something to listen to that is similar to your search request.[/quote]
Yes. It's pretty much free radio if you get a good login from bugmenot.com .

[quote="Ignite168"]Most of the music I listen to isn't from 2006 but for 2006:

Dani California - Red Hot Chili Peppers
Lean Wit It Rock Wit It - Dem Franchise Boyz
Smack That - Akon
Shake That - Eminem
It's Goin Down - Yung Joc
Gimme That - Chris Brown
How to Save a Life - The Fray[/quote]

See above comments directed at the OP.

Most songs you find on Pandora won't be from 2006 meaning it's offtopic. I could go on and on listing songs that weren't from 2006 as that pretty much excludes almost all of Classic Rock and such.

[quote="Ignite168"]Most songs you find on Pandora won't be from 2006 meaning it's offtopic.[/quote]

I will type slow so you can read this:
I was telling the OP how his/her favorite songs were piles of junk. Lips of Angel uses FOUR CHORDS TOTAL as a background for crappy vocals, except for the "solo", which is a chord progression strummed a few times over. Pandora was invoked in an attempt to show him what better music was. So yes, Ignite, I was on topic. I think it is reasonably excepted that in a Round Table topic, the subject will be debated and discussed, and not devolve into a spam fest like in the G&FF.

If you want to talk about some classic rock, as your quote suggests, then let's talk about Iron Maiden's new album, "A Matter of Life and Death", and Blind Guardian's "A Twist in the Myth."

I cannot express anything less than ultimate fanboy love for those bands.

EDIT: Iced Earth's live DVD, "Alive in Athens", would also be relevant, as it was released this year.

[quote="Elemennop"][quote="PenguinIntegral"]
Hello, MTV/VH1-controlled robot! May I direct you to http://www.pandora.com ?  There you will find reccomendations for songs and artists, based on your current tastes, that [b]don't[/b] suck.
[/quote]

Who the hell are you to pass judgment on other people's musical tastes? You don't even know if that guy hasn't already listened to other types of music and made the decision of listening to what he does now.

And one could make a similar argument against your musical tastes since Iron Maiden and Blind Guardian are some of the most popular bands in heavy metal and power metal, respectively, so it's not as if you are in a position to call people out for listening to popular music. But nobody seems to bother deriding you for that since we're not here to argue about other people's musical tastes. Nothing is more pretentious than having to listen to some person discredit another's musical tastes on the basis of popularity of that music.

Anyways, to stay on topic...I don't even know which albums came out in 2006, but of the ones i'm sure of

Mars Volta - Amputechture
Queensryche - Operation Mindcrime II
Dream Theater - Score
Bela Fleck - The Hidden Land[/quote]

DUDE CHILL , what's with your paranoid posts, and your adamant attitude. lately  :rotfl:  :roll:

Anyways. I haven't listened to that much music this year. I agree with Leetspeak on all but Rascal Flatts. Man, he can be as annoying as the chirping panda in my room which isn't supposed to chirp.

His voice that is.

I also liked Josh Groban's latest CD. I used to like Clay Aiken's songs, but my sister literally hates his songs because it turns out he's gay. HAHA who cares if he's gay or not.

[quote="funcia"]THAT CRAZYYY MAN>>>>>>>GNARLS BARKLEY OR SOMETHING>>>>>>THE CRAZZZZYYY MAN

there that ends our discussion. period.[/quote]

 :rotfl:

indeed a crazy man

oh yeah. did any of you buy or receive CDs for Xmas? sending as a gift, receiving, etc.  :)

At least dragonforce is obviously not taking themselves seriously, with rhapsody i can't tell

and yea that video's cool, they totally just made it as cheesy as possible.. they didn't even bother putting anything on the green screen behind ZP (The singer) during the twin blast duels

[quote="krustyteklown"]mmm are there no mods in games and fun.....
this is hardly a 'debate' anymore (did it start out as one?)[/quote]
Yes, in fact, I [b]do [/b]resent that comment.

[quote="darij grinberg"][quote="PenguinIntegral"]EDIT: [url=http://www.roadrunnerrecords.com/artists/DragonForce/showVideo.aspx?fileID=1728]Click for MegaCheese.[/url][/quote]

 :rotfl: GREAT VIDEO! (I just watched it on [url=http://www.youtube.com/watch?v=Ywxm6zLEjFY]YouTube[/url]...) Still, musically I prefer Hammerfall (their first 4 albums: Glory to the Brave - Crimson Thunder) in this category (and I seem to be one of the few people to enjoy their ballads, which are, by the way, only half as cheesy as an average Rhapsody's song, what still means quite cheesy).

Any of you interested in melodic power metal should also check out Avantasia. I wouldn't call them melodic all through, but there are some very melodic songs: Farewell, The Seven Angels, Neverland, Into The Unknown. Besides, The Final Sacrifice is just cool, even though it is more screamed than sung...

  darij[/quote]
Ok, I'll go check out Hammerfall, some more Rhapsody, and Avantasia...

How goes the Symphony X and Steel attack?

By the way, does anyone know if DragonForce has released any official books of sheet music? I don't want to rely on the internet for complex solos...

[quote="Ignite168"]EVERYTIME WE TOUCH BY CASCADA IS OBV MY FAVORITE NEW SONG OF 2006[/quote]

although that's technically not trance, it's EDM music. check out my links. you might like them.  :)

[quote="SoxFan34"]White and Nerdy-Weird Al Yankovic
How to save a life- the Fray
I'll sue ya -Weird Al Yankovic(this is idiotic, but i like it anyways)
Lazy Sunday-SNL(known to some as the "narnia rap")
Ultimate showdown of ultimate Destiny

And about "Everytime we touch" I made a parody on that:

Cause every time we touch- I hit the ceiling
And every time we kiss- I WISH I COULD DIE!
Don't u understand that I- hate myself
End my stupid life.... :rotfl:

And other songs I like:
Dream On- Aerosmith
Beautiful day-U2
Youre pitiful- Weird Al Yankovic(parody of "You're beautiful" by James Blunt)
Whats really weird is that "You're beautiful" has the "f" word in it. WHAT SLOW SONG HAS THE F WORD IN IT???[/quote]

Wow, I like almost exactly the same songs. :o The only version of "You're beautiful" I've heard was the radio edit :roll:

There needs to be a math version of Hinder's "Lips of an Angel" called "Ellipse of an Angle" or something.

[quote="mathnerd314"][quote="Ignite168"][quote="mathnerd314"]There needs to be a math version of Hinder's "Lips of an Angel" called "Ellipse of an Angle" or something.[/quote]

OH GAWD NO[/quote]

Why not? There's "American Pi" for Don McLean, and "Math Test of Puppets" for Metallica, and...[/quote]

NO..JUST NO

[quote="Alex Chen"]I like Carrie Underwood's songs  :lol:[/quote]

GO DIE

[quote="sapphyre571"][quote="PenguinIntegral"]I really want to get my hands on a dragonforce songbooks, if one exists? I can't seem to find one, though. Does anyone know of some good speed metal bands with lots of complex songs (eg not just OMG powerchords over and over) that has a songbooks out?

I got the Maiden one, but it's kinda dumbed down and repetitive.[/quote]

I saw a tab for "Fury of the Storm"....it was over 100 pages
a whole songbook would be bigger than your head[/quote]
WHERE!?


The four squares pictured share a common vertex. The three interior squares each have a vertex at the center of the next larger square. The largest square has side length 8 cm. How many square centimeters are in the shaded region?

[asy]path p = (0,0)--(1,0)--(1,-1)--(0,-1)--cycle;
fill(scale(4)*p,black); fill(scale(2)*p,white); fill(scale(1)*p,black); draw(scale(8)*p);[/asy]

If the white area in between the black areas, was not there, then the are would be $ 4^2\equal{}16$. Now, the area of the white piece is the area of a complete small white square with side length $ 2$ minus the corner black square which has side length $ 1$. Hence, $ 2^2\minus{}1^2\equal{}3$, and the shaded portion has area $ 16\minus{}3\equal{}\boxed{13}$.


inequalities
function
inequalities proposed

[quote="Shinraehee"]
6. Let $ a,b,c \ge 0 ; a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ 7(ab \plus{} bc \plus{} ca) \le 2 \plus{} 9abc$
[/quote]
[hide]<=>$ 7(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c) \leq 2(a\plus{}b\plus{}c)^3\plus{}9abc$ :)[/hide]

problem3 : use  ineq :
 $ (x\plus{}y\minus{}z)(y\plus{}z\minus{}x)(z\plus{}x\minus{}y) \leq xyz$

[quote="Shinraehee"]1 . Let $ a,b,c \ge 0 ; a \plus{} b \plus{} c \equal{} 3$ . Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 4abc \ge 4$

2. Let $ 0\le x,y,z,t \le 1$ . Prove that:
$ (x^{2}y \plus{} y^{2}z \plus{} z^{2}t \plus{} t^{2}x) \minus{} (y^{2}x \plus{} z^{2}y \plus{} t^{2}z \plus{} x^{2}t) \le \frac {8}{27}$

3. Let $ a,b,c \ge 0 ; a \plus{} b \plus{} c \equal{} 3$ . Prove that:
$ 3(a^2 \plus{} b^2 \plus{} c^2) \plus{} 4abc \ge 12$

4. Let $ x,y,z > 0 ; x \plus{} y \plus{} z \equal{} 1$. Prove that:
$ \frac {1}{1 \minus{} xy} \plus{} \frac {1}{1 \minus{} yz} \plus{} \frac {1}{1 \minus{} zx} \le \frac {27}{8}$

5. Let $ a,b,c \le 0$ . Prove that:
$ 5(a^2 \plus{} b^2 \plus{} c^2) \le 6(a^3 \plus{} b^3 \plus{} c^3) \plus{} 1$

6. Let $ a,b,c \ge 0 ; a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ 7(ab \plus{} bc \plus{} ca) \le 2 \plus{} 9abc$
 :lol:  :D  :ninja:  :arrow:[/quote]
That is easy.All problems we can use Schur Ineq and p,q,r method.Now,I prove problem 5,the condition is $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$ :)
The Ineq can be written $ 5 \minus{} 10q\le 6(1 \minus{} 3q \plus{} 3r) \plus{} 1$
$ \Leftrightarrow 18r \minus{} 8q \plus{} 2\ge 0$
That's true with the Schur Ineq.

[quote="Shinraehee"] 1 . Let $ a,b,c \ge 0 ; a \plus{} b \plus{} c \equal{} 3$ . Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 4abc \ge 4$
6. Let $ a,b,c \ge 0 ; a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ 7(ab \plus{} bc \plus{} ca) \le 2 \plus{} 9abc$[/quote]

Also using $ abc\ge (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)$ :)

[quote="Shinraehee"]5. Let $ a,b,c \le 0$ . Prove that:
$ 5(a^2 \plus{} b^2 \plus{} c^2) \le 6(a^3 \plus{} b^3 \plus{} c^3) \plus{} 1$[/quote]

Wrong!,try $ a\equal{}b\equal{}c\equal{}\minus{}1$  :wink:

[quote="Shinraehee"]2. Let $ 0\le x,y,z,t \le 1$ . Prove that:
$ (x^{2}y \plus{} y^{2}z \plus{} z^{2}t \plus{} t^{2}x) \minus{} (y^{2}x \plus{} z^{2}y \plus{} t^{2}z \plus{} x^{2}t) \le \frac {8}{27}$
[/quote]

Suppose $ x\equal{}max(x,y,z,t)$ and using AM-GM  :)

[quote="Sudoku"][quote="Shinraehee"]5. Let $ a,b,c \le 0$ . Prove that:
$ 5(a^2 \plus{} b^2 \plus{} c^2) \le 6(a^3 \plus{} b^3 \plus{} c^3) \plus{} 1$[/quote]
It's wrong. We have $ 5(a^2 \plus{} b^2 \plus{} c^2) \le 6(a^3 \plus{} b^3 \plus{} c^3) \plus{} 1$ for $ a,b,c\geq 0$[/quote]
I think that $ a\plus{}b\plus{}c\equal{}1$.
We can use Schur inequality:
$ 5(a^2 \plus{} b^2 \plus{} c^2) \le 6(a^3 \plus{} b^3 \plus{} c^3) \plus{} 1$
$ \Leftrightarrow 5(a\plus{}b\plus{}c)(a^2 \plus{} b^2 \plus{} c^2) \le 6(a^3 \plus{} b^3 \plus{} c^3) \plus{}(a\plus{}b\plus{}c)^3$
$ \Leftrightarrow \sum ab(a\plus{}b) \le a^3\plus{}b^3\plus{}c^3\plus{}3abc$
Which is clearly true by Schur inequality.

[quote="Shinraehee"]

4. Let $ x,y,z > 0 ; x \plus{} y \plus{} z \equal{} 1$. Prove that:
$ \frac {1}{1 \minus{} xy} \plus{} \frac {1}{1 \minus{} yz} \plus{} \frac {1}{1 \minus{} zx} \le \frac {27}{8}$
[/quote]

the inequality is equivalent to:

\[ 3\minus{}11(xz\plus{}yz\plus{}xy)\plus{}19xyz(x\plus{}y\plus{}z)\minus{}27x^2y^2z^2 \geq 0\]

\[ 3\minus{}11(xz\plus{}xy\plus{}xz)\plus{}19xyz\minus{}27x^2y^2z^2\geq0\]

by AM-GM :  $ x^2y^2z^2 \leq \frac{1}{27^2}$

so the inequality becomes :
\[ S\equal{}\minus{}11(xy\plus{}xz\plus{}yz)\plus{}19xyz\plus{}\frac{80}{27}\geq 0\]
we put : $ xy\equal{}a$
so :
$ S \equal{} f(a) \equal{} a( \minus{} 11 \plus{} 19z) \minus{} 11z(1 \minus{} z) \plus{} \frac{{80}}{{27}}$ ( f is a linear function)
we have : $ a\ge0$
and :
$ S \equal{} f(0) \equal{}  \minus{} 11z(1 \minus{} z) \plus{} \frac{{80}}{{27}} \ge 0$
and we have : $ a\leq \frac{(x\plus{}y)^2}{4} \equal{} \frac{(1\minus{}z)^2}{4}$
then :
$ S \equal{} f(\frac{{(1 \minus{} z)^2 }}{4}) \equal{} \frac{1}{{108}}(57z \plus{} 23)(3z \minus{} 1)^2  \ge 0$
so \[ f(a) \geq 0\]


ARML has released a new website at [url]http://www.arml.com[/url].

What's so new about it?  It looks exactly the same as before to me.

O you meant [url]http://arml.com/[/url] :D .  Now it looks different.  Thanks for pointing out the new site.  It looks cool.

[quote]O you meant http://arml.com/  . Now it looks different. Thanks for pointing out the new site. It looks cool.[/quote]

Both sites look the same to me. (Well, the new same at least  :) )

joml's DNS cache probably hadn't been updated yet.  He'll see them the same soon!

Ok, they both are the same now.  Sorry about that!


Θα ηθελα να με βοηθησετε στην εξης ασκηση:
$ f(x) \equal{} x^2\sin\frac {1}{x} , x\neq0$
$ f(x) \equal{} 0 , x \equal{} 0$

Δειξτε οτι η f ειναι παραγωγισιμη στο 0 αλλα η παραγωγος της ειναι ασυνεχης στο σημειο αυτο.

Καποια στιγμη βρισκω αυτο
$ f'(x) \equal{} 2x\sin\frac {1}{x} \minus{} \cos\frac {1}{x}$

ομως δεν μπορω να βγαλω το οριο της στο 0

Δεν υπάρχει τό όριο στο μηδέν πάρε πχ δυο υπακολουθίες 1/χ=2νπ και 1/χ=2νπ+π/2

Προφανως το οριο δεν υπαρχει..δεν μας βολευει το δευτερο μερος.
Με τις υπακολουθιες τι κερδιζω δεν καταλαβα?

Edit:Εννοω το οριο μονο το δευτερο μερος δηλαδη μονο το συνημιτονο ενα προς χ και μονο οταν αυτο τεινει στο 0.Για να μην μπερδεψουμε κοσμο

[quote="Anubis"]Προφανως το οριο δεν υπαρχει..δεν μας βολευει το δευτερο μερος.
Με τις υπακολουθιες τι κερδιζω δεν καταλαβα?[/quote]

den boreis na peis oti to orio den uparxei epeidi den einai omorfo to deutero meros. kai gi auto o k. boris sou proteinei mia kataskeui me upakolouthies.

isos na uparxei mia suntaktiki anomalia sto proigoumeno post mou...leo..o k.boris sou proteinei mia kataskeui me upakolouthies gia na apodeiksei oti den uparxei to orio! (sorry an prokalesa sugxush)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that every group of order $ 6$ is isomorphic to either $ \mathbb{Z}_{6}$ or $ S_{3}$.

What are you allowed to use? Sylow's Theorems or only more elementary techniques?

Hi,

Let G be a group of order 6. First, it cannot be true that there is no element of order 3, as then, letting e be the neutral element and g,h elements of order 2, the collection e,g,h,gh would be a subgroup consisting of four elements, which is not possible. Hence there exists an element of order 3, say g. Take k not in A=(e,g,g^2). If k has order 3 too, you see that k^2 cannot be A. The remaining element h has to have order 2 as if not, h^2 has to be in one of the subgroups generated by k and g and hence k would be in the same subgroup, too. We see that G=(e,g,g^2,h,hg,hg^2 ). As A has index 2, its a normal subgroup and thus hgh is in A. If hgh=g, the group is abelian and isomorphic to Z/2Z x Z/3Z. The remaining cases are: hgh=e, g=e, contradiction. hgh=g^2, yielding gh=hg^2. This is sufficient information to write down the table giving the group's multiplication; comparing it to that of S_3 shows that the goups are isomorphic.

props

Carsten


Set $ A_1,A_2,...,A_{27}$ such that:$ |A_i|\equal{}35$ for $ 1\leq i\leq 27$ and $ |A_i\cap A_j\cap A_k|\equal{}1$ for all $ 1\leq i<j<k\leq 27$.
Prove that:$ |A_1\cap A_2\cap....\cap A_{27}|\equal{}1$

[quote="math10"]Set $ A_1,A_2,...,A_{27}$ such that:$ |A_i| \equal{} 35$ for $ 1\leq i\leq 27$ and $ |A_i\cap A_j\cap A_k| \equal{} 1$ for all $ 1\leq i < j < k\leq 27$.
Prove that:$ |A_1\cap A_2\cap....\cap A_{27}| \equal{} 1$[/quote]
I think this is a nice problem, but not too hard to no one answer it.
I want to think of a solution using thePigeonhole Principle but unsuccessful  :(
Can anyone help me?
My solution.

Let $ A \equal{} A_1\cup A_2\cup....\cup A_{27}$ and for each $ x \in A$ let $ n(x)$ denote the number of $ i's$ such that $ x \in M_i$ and $ m \equal{} \max n(x)$. We will prove ve that $ m \geq 8$.
We have:$ \sum n(x) \equal{} 27.35$ and since  $ |A_i\cap A_j\cap A_k| \equal{} 1$ we have $ C_{35}^3 \leq \sum \frac {n(x).(n(x) \minus{} 1)(n(x) \minus{} 2)}{6} \leq \frac {(m \minus{} 1)(m \minus{} 2)}{6}.\sum n(x)$
So $ (m \minus{} 1)(m \minus{} 2)\geq 42$ So $ m\geq 8$
Suppose $ A_1\cap A_2\cap....\cap A_8 \equal{} \{x\}$.We have prove that $ x\in A_i$ for $ i \equal{} 1,2..,27$
If $ x \notin A_i$ , for some $ i \in \{9,10,...,27\}$
Let $ a_{u,v} \equal{} A_u\cap A_v\cap A_i$ for $ 1\leq u < v\leq 8$
If $ a_{u,v} \equal{} a_{z,t}$ then $ a_{u,v} \in A_u,A_v,A_z,A_t$ but $ A_u\cap A_v\cap A_z\cap A_t \equal{} x$,contradiction
So $ a_{u,v} \not \equal{} a_{z,t}$.But have $ C_8^2 \equal{} 28$ pair $ (u,v)$ $ > |A_i| \equal{} 27$,contradiction
So $ x \in A_i$ or $ |A_1\cap A_2\cap....\cap A_{27}| \equal{} 1$
This problem is proved  :D

[quote="math10"]since  $ |A_i\cap A_j\cap A_k| \equal{} 1$ we have $ C_{35}^3 \leq \sum \frac {n(x).(n(x) \minus{} 1)(n(x) \minus{} 2)}{6} \leq \frac {(m \minus{} 1)(m \minus{} 2)}{6}.\sum n(x)$
So $ (m \minus{} 1)(m \minus{} 2)\geq 42$[/quote]

I don't understand this part. :wink:

In another way :
we have $ \frac {26 \times 25}{2}$ to choose $ 1 < i <j$ to form $ A_1 \cap A_i \cap A_j$, which determines an unique element form $ A_1$. Since $ |A_1| \equal{} 35$, the pigeon-hole principle ensures that at least one element in $ A_1$ is obtained at least $ \frac{26 \times 25}{2\times 35} > 9$. Therefore $ m \geq 10$.

Pierre.

[quote="pbornsztein"][quote="math10"]$ C_8^2 \equal{} 28$ pair $ (u,v)$ $ > |A_i| \equal{} 27$[/quote]

Another mistake, since $ |A_i| \equal{} 35$. But, this can be fixed using my previous post.

Pierre.[/quote]
Sorry,I had a wrong calculation  :blush:
We must prove $ m\geq 9$

[quote="pbornsztein"][quote="math10"]since  $ |A_i\cap A_j\cap A_k| \equal{} 1$ we have $ C_{35}^3 \leq \sum \frac {n(x).(n(x) \minus{} 1)(n(x) \minus{} 2)}{6} \leq \frac {(m \minus{} 1)(m \minus{} 2)}{6}.\sum n(x)$
So $ (m \minus{} 1)(m \minus{} 2)\geq 42$[/quote]

I don't understand this part. :wink:
Pierre.[/quote]
Let $ X \equal{} \{(A_i,A_j,A_k,x)|x \in A_i \cap A_j \cap A_k \}$ We have: $ X \equal{} \sum \frac {n(x).(n(x) \minus{} 1)(n(x) \minus{} 2)}{6}$ and
$ X\equal{}C_{35}^3$
but to have $ m\geq 9$, we need better estimates.

I am probably missing something obvious, but can't we just say

induct on $ i$ to find that if $ e \in A_i \cap A_{i \plus{} 1} \cap A_{i \plus{} 2}$, then $ e \in A_{i \plus{} 1} \cap A_{i \plus{} 2} \cap A_{i \plus{} 3}$. Thus $ e \in A_i$ for all $ 1 \le i \le 27$. Consider all $ A_i \cap A_j \cap A_k$ for $ 1 \le i < j < j \le 27$ Since $ e$ is in this set, it is the only element in this set: thus no other element in $ A_i$ is in $ A_j$ nor $ A_k$. We do this for all relevant $ i,j,k$ to find that $ A_1 \cap A_2 \cap A_3 \cap \cdots \cap A_{27} \equal{} {e}$.

I am similarly confused on

[quote="math10"]
 $ A_u\cap A_v\cap A_z\cap A_t \equal{} x$,contradiction
[/quote]


For which nonnegative integers $ n$ and $ k$ is

$ (k\plus{}1)^n \plus{} (k\plus{}2)^n \plus{} (k\plus{}3)^n \plus{} (k\plus{}4)^n \plus{} (k\plus{}5)^n$

divisible by 5?

$ (k \plus{} 1)^n \plus{} (k \plus{} 2)^n \plus{} (k \plus{} 3)^n \plus{} (k \plus{} 4)^n \plus{} (k \plus{} 5)^n \equiv 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n (mod 5)$
$ \equiv (1^n \plus{} ( \minus{} 1)^n) \plus{} (2^n \plus{} ( \minus{} 2)^n) (mod 5)$
$ \therefore$ $ n$ is odd,$ n \equal{} 4k \plus{} 2 ;k \in Z$

[hide="Solution"]$ k\plus{}1$ through $ k\plus{}5$ will be equivalent to some permutation of 0, 1, 2, 3, and 4 mod 5, so all values of $ k$ will hold.
Then we have $ 1^n \plus{} (\minus{}1)^n \plus{} 2^n \plus{} (\minus{}2)^n \equiv 0 \pmod 5$.
This obviously holds for all odd $ n$, so the next step is to try for even $ n$. Set $ n\equal{}2m$, then modify the equivalence:
$ 1^m \plus{} 1^m \plus{} (\minus{}1)^m \plus{} (\minus{}1)^m \equiv 0 \pmod 5$
This is obviously true for all odd $ n$. Applying a similar step, set $ m\equal{}2l$:
$ 1^l \plus{} 1^l \plus{} 1^l \plus{} 1^l \equiv 0 \pmod 5$, which is never
$ k$ is any nonnegative integer, and $ n$ is any nonnegative integer not divisible by 4.
[/hide]

[hide]using mod 5, k does not matter because k+1, k+2, k+3, k+4, and k+5 are a permutation of 0,1,2,3,4 mod 5. The expression reduces to $ 1\plus{}1\plus{}0\plus{}2^n\plus{}2^n$ mod 5 or $ 2\plus{}2^n\plus{}1$ mod 5. $ 2^n$ is periodic with period 4, so n is 2 mod 4.[/hide]

[hide]As others stated above, we can reduce this problem to consider $ 1^n \plus{} 2^n \plus{} 3^n \plus{} 4^n$. k can be any

$ 1^n \equiv 1 (mod 5) \forall n$

$ 2^{1 mod 4} \equiv 2 (mod 5)$
$ 2^{2 mod 4} \equiv 4 (mod 5)$
$ 2^{3 mod 4} \equiv 3 (mod 5)$
$ 2^{0 mod 4} \equiv 1 (mod 5)$

$ 3^{1 mod 4} \equiv 3 (mod 5)$
$ 3^{2 mod 4} \equiv 4 (mod 5)$
$ 3^{3 mod 4} \equiv 2 (mod 5)$
$ 3^{0 mod 4} \equiv 1 (mod 5)$

$ 4^{1 mod 2} \equiv 4 (mod 5)$
$ 4^{0 mod 2} \equiv 1 (mod 5)$

So we see that:
$ 1^{1 mod 4} \plus{} 2^{1 mod 4} \plus{} 3^{1 mod 4} \plus{} 4^{1 mod 4}  \equiv 0 (mod 5)$

$ 1^{2 mod 4} \plus{} 2^{2 mod 4} \plus{} 3^{2 mod 4} \plus{} 4^{2 mod 4}  \equiv 0 (mod 5)$

$ 1^{3 mod 4} \plus{} 2^{3 mod 4} \plus{} 3^{3 mod 4} \plus{} 4^{3 mod 4}  \equiv 0 (mod 5)$

$ 1^{0 mod 4} \plus{} 2^{0 mod 4} \plus{} 3^{0 mod 4} \plus{} 4^{0 mod 4}  \equiv 4 (mod 5)$

Thus n can be any non-negative positive integer not divisible by 4.
[/hide]

Since $\{k+1,k+2,k+3,k+4,k+5\}\equiv \{1,2,3,4,5\}$, the expression reduces to $1^n+2^n+3^n+4^n\pmod 5$. Then observe that $2^1\equiv 2$, $2^2\equiv 4$, $2^3\equiv 3$ and $2^4\equiv 1$, so we can rewrite the expression as $2^n+2^{2n}+2^{3n}+2^{4n}=2^n\left(\frac{2^{4n}-1}{2^n-1}\right)$, and we want this to be $\equiv 0$. Now if $2^n-1\equiv 0\pmod 5\iff n\equiv 0\pmod 4$ we get $2^n+2^{2n}+2^{3n}+2^{4n}\equiv 1+1+1+1\equiv 4\not\equiv 0\pmod 5$, since $2^4\equiv 1\pmod 5$. So clearly $n\not\equiv 0\pmod 4$, and therefore $2^n-1\not\equiv 0\pmod 5$. Hence, going back to the closed form of the geometric series above, we can "multiply through" by $\frac{2^n-1}{2^n}$ (since neither are $0\pmod 5$) to get $2^{4n}-1\equiv 0\pmod 5$, and this is true for all $n$. So the only restriction is what we found, namely $n$ cannot be a multiple of $4$.


MATHCOUNTS
ratio
geometry
probability
3D geometry

Here's a couple problems that are approximately Mathcounts level.

1.  Find the number halfway between $\frac{1}{3}$ and $\frac{1}{5}$

2. Factor completely: $a^2b+4b-4ab$

3. If the ratio of the diameters of two circles is 4:3 and their circumfrences difffer by $4\pi$, then what is the area of the larger circle?

4. A disc of diameter D is thrown along a floor covered with 9 inch square tiles. (9 inch by 9 inch) The disc is a winner if its edges are completely within a square and don't intersect the edges of a square. If the probability of throwing a winner is 1/3, calculate D.

[color=white]

1)
1/3 = 5/15
1/5 = 3/15

Halfway between would be 4/15

2)
a^2b + 4b- 4ab

b(a^2 + 4 - 4a)
(b)(a-2)(a-2)

3)
circumference = dpi
so the radius of the larger one is 8 and the larger one's radius is 6

28pi

4)

Dunno, too hard to type this out while holding an ice cream cone in one hand.

[/color]

You did something funky at the end of #3. Check that one. The others look pretty fine to me.

Good! Next time, just try to put your answers in spoiler, the button under the bold button when you're posting.

sry, but I stilll don't know how to do a spoiler.

Distance between 2 points is: $ \frac{1}{3}\ - \frac{1}{5}\ = \frac{2}{15}\ $

First point + half the distance is: $  \frac{1}{5}\ + \frac{1}{15} = \frac{4}{15} $

1.
[hide]$(1/3+1/5)/2=(8/15)/2$
                   =$4/15$[/hide]
2.
[hide]$a^2b-4ab+4b=b(a^2-4a+4)$
                                           $=b(a-2)(a-2)$[/hide]

3.
[hide]
$4x\pi-3x\pi=4\pi$
$x\pi=4\pi$
$x=4$
diameter of the larger circle=4x=4*4=16
radius of the larger circle is 8

$A=\pi8^2 = 64\pi$[/hide]

4.
[hide]one third of nine is [b]three[/b][/hide]


LaTeX
combinatorics unsolved
combinatorics

All the squares of a board of $(n+1)\times(n-1)$ squares
are painted with [b]three  colors[/b] such that, for any two different
columns and any two different rows, the 4 squares in their
intersections they don't have all the same color. Find the
greatest possible value of $n$.

Can you give a hint? At least an upper bound.

for n=10 we can painted like this(any 11 columns of it satisfied the problem)

for n=11,find all pairs of squares which is in the same column and have the same colour.
In every column,there is at least $C_4^2+C_3^2+C_3^2=12$ pairs.
So there is at least $12\times 12=144$ pairs.
So at least one of the colours has 48 pairs of squares like that.
But their positions in the rows has only $C_{10}^2=45$ different conditions.
So there must be two pairs in the same rows.And they kill the problem.
($C_n^m$ is the binomial coefficient).

Hi everybody. My first post here. My English (and math, I should probably add) is terrible: please forgive my mistakes in advance...

Anyway, with jin's trick you can bound the different boards that can be painted.

More specifically, you can prove the impossibility for 4x19, 5x16, 7x13 and 10x12. It's also quite easy/not too difficult to colour cases 3x<whatever>, 4x18, 6x15, 9x12.

These facts cover all the possible boards, except for 10x10, 10x11 and 11x11. The first turns out to be possible, whereas the other two are not.

Impossibility for 11x11 is a little harder, but 10x11 is definitely the most difficult (supposing I am not mistaken ;-) ). Anyone wants to try?

M.

The example can be generalized for any prime power quantity of colors. Suppose there are $p^k$ colors.

We can build a grid of size $(p^k)^2\times ( p^k\cdot ( p^k + 1) )$ without a rectangle with monochromatic corners.

To do this look at the vector space $\mathbb F_{p^k}^2$ (the space of dimension $2$ over the field with $p^k$ elements).

Each row is going to correspond to one of the vectors in $(\mathbb F_{p^k}^2$.

And therefore every column just corresponds to a coloring of  all the vectors using $\mathbb F_{p^k}$ colors.

Notice that there are $p^k+1$ lines that pass through the origin. Each such line $L$ is going to give us $p^k$ columns as follows:

Look at the lines parallel to $L$, they partition the entire space. We label the $p^{k}$ parallel lines $l_1,l_2,...,l_{p^k}$. We are now ready to define the $p^k$ colorings the correspond to $L$.

The $i'th$ coloring that corresponds to $L$ is going to paint the vectors of line $l_j$ with color $i+j$.

To check that this works we just have to pick two columns and pick a color $c$. We have to check that the vectors of color $c$ in the first column and the second column share at most one vector. Notice that these sets are two lines in a vector space and so they intersect at at most one point. This is unless the two lines are the same, which is clearly never the case.

The case $10\times 12$ is when $p^k=2^1$.


trigonometry
geometry
trig identities
Law of Sines

This is nice and easy:

Let $AP$, $AQ$ and $AR$ be chords in a circle such that $\angle PAQ = \angle QAR = \angle RAS$.

Show that $(AP + AR)\cdot AR = (AQ + AS) \cdot AQ$.

[hide]
All the lengths subtended by the same angles are equal (easily shown by Law of Sines), so

$PQ = QR = RS = x$ and $PR = QS = y$.

Applying Ptolemy's Theorem on $APQR$ and $ASRQ$, we have

$AP \cdot x+AR \cdot x = AQ \cdot y$

$AS \cdot x+AQ \cdot x = AR \cdot y$.

Multiplying the two equations together (LHS of first to RHS of second and vice versa),

$xy \cdot (AP+AR) \cdot AR = xy \cdot (AS+AQ) \cdot AQ$

$(AP+AR) \cdot AR = (AS+AQ) \cdot AQ$

as desired.[/hide]

Indeed. You could also use trigonometry, but this is much nicer.


absolute value
number theory unsolved
number theory

Given a $ 100 \times 100$ array with entries in $ \{ \minus{} 1, 0, 1\}$ and the sum of its entries $ \leq 400$.
prove there exists a $ 25 \times 25$ sub-array with absolute value of the sum of its entries $ \leq 25$.

thanks to mavropnevma.I have re-edit the post to make my ques. more clearly :)

i couldn't understand what you mean by 'form'..
sorry, but could you explain the question more clearly?

Probably it is given a $ 100 \times 100$ array with entries in $ \{ \minus{} 1, 0, 1\}$ and absolute value of the sum of its entries $ \leq 400$.
It is required to prove there exists a $ 25 \times 25$ sub-array with absolute value of the sum of its entries $ \leq 25$.

Can anyone help to solve it?
It seems to be nice and difficult.

it is not so hard if you prove the equivalent contrapositive

If all $ 25 \times 25$ sub-arrays have a sum whose absolute value $ > 25$, then the absolute value of the sum of the full $ 100 \times 100$ array must be $ > 400$

To prove this select one $ 25 \times 25$ sub-array. Its sum $ S$ must either by $ S > 25$ or $ S < \minus{}25$
Now take another $ 25 \times 25$ sub-array which overlaps on $ 24 \times 25$ cells. Its sum $ T$ is $ T > 25$ or $ T < \minus{}25$
But $ |S\minus{}T| \le 50$ because there are only $ 50$ cells that are in one but not the other,
so $ S > 25 \implies T > 25$ and $ S < \minus{}25 \implies T < \minus{}25$, so $ S$ and $ T$ have the same sign.

Since we can go from any $ 25 \times 25$ sub-array to any other by repeatedly shifting position by one so that
consecute sub-arrays in the sequence overlap on $ 24 \times 25$ cells, it follows that the sums of each all have the same signs.

Now just take the $ 16$ $ 25 \times 25$ sub-arrays that cover the whole array without overlapping. Either their sums are all $ < \minus{}25$ or all $ > 25$
It follows that the sum of the full array must be $ < 16 \cdot (\minus{}25) \equal{} \minus{}400$ or $ > 16 \cdot 25 \equal{} 400$ which is the required result.


calculus
integration
Putnam
algebra
function
domain
geometric series

[quote="Jimmy"]Evaluate \[\int_{0}^{1}\int_{0}^{1}\frac{dx \ dy}{1-xy}\] I think this was a past Putnam problem.[/quote] \[\int_{0}^{1}\int_{0}^{1}\frac{dx \ dy}{1-xy}= \int_{0}^{1}\int_{0}^{1}\sum_{k=0}^{\infty}(xy)^{k}\ dx \ dy = \sum_{k=0}^{\infty}\int_{0}^{1}\int_{0}^{1}(xy)^{k}\ dx \ dy =\sum_{k=0}^{\infty}\int_{0}^{1}\frac{y^{k}}{k+1}\ dy\]  \[= \sum_{k=0}^{\infty}\frac{1}{(k+1)^{2}}=\sum_{k=1}^{\infty}\frac{1}{k^{2}}= \zeta (2) = \frac{\pi^{2}}{6}\] where the technique can be made rigorous by replacing the upper bound of 1 in both integrals by $r<1$ so that the convergence of the geometric series is uniform over the domain of integration which justifies the interchage of limiting processes (in particular those of summation and integration and one other [next step]). Now let $r\rightarrow 1^{-}$ and the result follows.

Furthermore, it is notable that for $n=2,3,4,\ldots$ we have \[\zeta (n) = \int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}\frac{dx_{1}\ dx_{2}\ \cdots \ dx_{n}}{1-x_{1}x_{2}\cdots x_{n}}\]

nice solution i never thought of that.
when i tried to do this problem i started with $t=xy$ but quickly ran into problems.

Whats the jacobian transformation for $\frac{1}{1-x^{2}y^{2}}$?

[quote="misan"][quote="Jimmy"]That's the same way I would have done it, but I wasn't sure if on the contest you could assume that $\zeta(2)$ was known.  Anyway, there's another way to do it with a Jacobian transformation.[/quote]

i don't think such a transformation is possible for $\frac{1}{1-xy}$, although i know one has been shown for $\frac{1}{1-x^{2}y^{2}}$ which gives $\frac{\pi^{2}}{8}$

all i am saying is that i am yet to see one for $\frac{1}{1-xy}$!  :maybe:[/quote]

Let $(u,v) = \left(\frac{x+y}{2}, \frac{y-x}{2}\right)$.  Here is the link where I first saw it (page 2):

http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf


$ a,b,c \in R^{\plus{}}$
Prove that $ \sum \frac{a^3}{2a^2\minus{}ab\plus{}2b^2} \ge \frac {1}{3} \sum a$

[hide="Remark"]I only knew an ugly SOS method (not easy at all),can anyone help me to solve it neatly or find a link?[/hide]

Dear hxy09! This inequality is very hard! I solved this ineq by many? but they are not true!!   :mad:
Can anyone solve this problem?

If anyone will solve this problem please post your solution!  :lol:

[quote="hxy09"]$ a,b,c \in R^{ \plus{} }$
Prove that $ \sum \frac {a^3}{2a^2 \minus{} ab \plus{} 2b^2} \ge \frac {1}{3} \sum a$

[hide="Remark"]I only knew an ugly SOS method (not easy at all),can anyone help me to solve it neatly or find a link?[/hide][/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66271


trigonometry
inequalities unsolved
inequalities

Let $ x,y,z$ be positive real numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}1$.

Prove that

$ \frac{x}{1\minus{}x^2}\plus{}\frac{y}{1\minus{}y^2}\plus{}\frac{z}{1\minus{}z^2}\geq\frac{3\sqrt{3}}{2}$

$ \tan ^ 2 A\plus{}\tan ^ 2 B\plus{}\tan ^ 2 C\equal{}1$ with $ 0<A,\ B,\ C<\frac{\pi}{2}\Longrightarrow \tan 2A\plus{}\tan 2B\plus{}\tan 2C\geq 3\sqrt{3}$.

[quote="kunny"] $ \tan 2A \plus{} \tan 2B \plus{} \tan 2C\geq 3\sqrt {3}$.[/quote]

And why is that?

$ x^3\plus{}\frac{1}{3\sqrt{3}}\plus{}\frac{1}{3\sqrt{3}}\geq x\Rightarrow x(1\minus{}x^2)\leq\frac{2}{3\sqrt{3}}$
$ \Rightarrow\frac{x}{1\minus{}x^2}\geq\frac{3\sqrt{3}}{2}x^2$


This problem seemed to be too obvious.. And I think I"m wrong.. :D

Problem:

Consider the sequence $x_0, x_1, x_2,...x_{2000}$ of integers satisfying

\[x_0 = 0, |x_n| = |x_{n-1} + 1|\]

for $n = 1,2,...2000$.

Find the minimum value of the expression $|x_1 + x_2 + ... x_{2000}|$.

[hide="My idea"]

I think I'm wrong but for $x_1$, it's $x_1 = |x_0 + 1| = |0+1| = |1| = 1$ and $x_2 = 2, x_3 = 3,$ and so forth.

So $|x_1+x_2+...x_{2000}| = |1+2+3+4..+2000| = 1+2+3+4..+2000 = \frac{2000 \cdot 2001}{2} = 2001000$. [/hide]

Pretty sure I'm wrong but where did I go wrong?

yes wrong because check this out

$|x_1|=|x_0+1|=|1|$ and this means that $x_1$ can be $x_1=1$ or $x_1=-1$

Ohh..

[hide="Then"]

So $x_1 = -1$ since we want the minimum. Then $x_2 = |x_1+1| = |-1+1| = 0$, right?

so, every odd $x_n$ will be -1 and every even will be 0. Or something close to it..[/hide]

I'm not sure if that makes the minimum though. No wonder it's the last question now..

well when u say every even will be 0 u can say for the odd $|x_n|=1$ therefore u can use 1 and -1 alternating coming to the minimum of 0, because there is sequence is periodic with the period 4, and the first 4 $x_n$ besides $x_0$ are 0 and they repeat themselves 250 times. also 0 is the smallest number possible, since $|x|\ge 0$
done :)

Sweet, an Olympiad problem that I can actually solve! :P

[hide]And the answer should be 1000 for the minimum value; we want to make the value as close as possible to $0$, since we're taking the absolute value, so we have $x_1=-1$, $x_2=0$, $x_3=-1$...[/hide]

Peeta...there's a slight problem...
You can't go from 1 to 0...the thing about this problem is that $|x_n|\le|x_{n+1}|$ when $x_n$ is positive.  In other words, you can't go from 1 to -1.
This one isn't quite as easy as it looks at first...
I'm thinking of one where the minimum is 12...
Basically you can start with 0, then go
-1, 0, 1, -2, -1, 0, 1, 2, -3, ...
going through sets of sizes which are an odd number each, and by the time you get to 22, you'll be at $x_1976$.  24 more numbers.  Then, you go
-23, 22, -23, 22, -23...
But, if there's a way that 2000 - square number=multiple of 23 or 21, then the minimum is 0.
I'm sure there's a solution for that...

Oh, you're right... by a more complicated set, you might be able to get it lower.


Να και μια δικιά μου άσκηση που έφτιαξα πειραματιζόμενος :) .
Μπορεί να σας φανεί εύκολη αλλα τη βάζω γιατί
είμαι περίεργος να δω τη λύση σας:

για $ a , b$ θετικούς ακεραίους ν.δ.ο:

                ${ 2\sqrt[a+b]{ (\frac {a^a + b^b}{2})^2}\geq a+b}$

Νομίζω ότι με jensen στη συνάρτηση $ f(x) \equal{} x^x$, τη σκοτώνουμε την άσκηση  :)

BINGO!!!!! Μπράβο vangelismak :lol: !!!! Ακριβώς αυτό τον τρόπο είχα στο μυαλό μου!!!!! :P
 Θα με ενδιεφερε όμως αν κάποιος αλλος θα μπορούσε να τη λύσει και με διαφορετικό τρόπο!
Απευθύνομαι κυρίως στο Silοuan που απ ότι έχω καταλάβει είναι master στις ανισότητες :wink:


algebra
function
domain
real analysis
real analysis unsolved

Does $ 0$ belong to domain of the following function
\[ f\left( x \right) \equal{} \frac{{4x^2 }}
{{\left( {1 \minus{} \sqrt {1 \plus{} 2x} } \right)^2 }}
\]
 :D

if I understand right your quaestion the this will be answer of your question:
in $ 0$ your function is equivalent to :
$ f(x)$~$ \frac {4}{ (\minus{} 1 \plus{} \frac {x}{2} \plus{} o(x))^{2}}$
by Taylor  :wink:


Find limit : $ lim_{n\to\infty} \sum_{k\equal{}0}^{n} \frac{(\minus{}1)^k}{2^k.k!}$

Well, it's well known that $ \sum_{k\equal{}0}^{\infty} \frac{x^k}{k!}\equal{}e^x$, so putting $ x\equal{}\minus{}\frac{1}{2}$, we obtain:
$ \sum_{k\equal{}0}^{\infty} \frac{(\minus{}1)^k}{2^k.k!}\equal{}e^{\minus{}\frac{1}{2}}$

[quote="Jorge Miranda"] $ \sum_{k \equal{} 0}^{\infty} \frac {x^k}{k!} \equal{} e^x$[/quote]
Why??? Please explain me!!!! :(

[quote="conan_naruto236"][quote="kunny"][url]http://mathworld.wolfram.com/MaclaurinSeries.html[/url] :)[/quote]
In this forum don't explain. Please write your solution for me !!!  :([/quote]
Let, $ f(x)\equal{}e^x$.
$ f^{(n)}(x)\equal{}e^x$. So $ f^{(n)}(0)\equal{}1$
$ \therefore e^x \equal{} \sum_{n\equal{}0}^\infty \frac{f^{(n)}(0)}{n!}x^n \equal{} \sum_{n\equal{}0}^\infty \frac{x^n}{n!}$

I have a similar problem:Find limit
$ lim_{n\to\infty} \sum_{k \equal{} 0}^{n} \frac {( \minus{} 1)^{k \minus{} 1}}{k!}$

the same way
put $ x\equal{}\minus{}1$ and then multiple by $ \minus{}1$
so the sum tends to $ \minus{}e^{\minus{}1}$


LaTeX
algebra
function
domain
complex analysis
complex analysis unsolved

Let $f$,$g$,$h$:$D\subset {\mathbb C}\to {\mathbb C}$ with the following properties:

1) $f$,$g$,$h$ olomorfes on the domain $D$;
2) $\exists z_0\in D$ such that $|f(z)+g(z)+h(z)|\leq |f(z_0)+g(z_0)+h(z_0)|$ for any $z\in {\mathbb C}$;

Show that there exists a $c\in {\mathbb C}$ such that $f\bar {f}+g\bar {g}+h\bar{h}\geq \frac {1}{3}c\bar {c}$,
(where $\bar {x}$ denotes the conjugate of the complex number $x$).

P.S.
.........and...... i've learned $Latex$  :D

The "Solved Problems" forums here are for problems that have been dealt with. If you want someone to look at and perhaps solve it, don't post it in "Solved Problems."

Put them in "Unsolved and Proposed Problems," which is where I'm moving this.

[quote]2) $\exists z_0\in D$ such that $|f(z)+g(z)+h(z)|\leq |f(z_0)+g(z_0)+h(z_0)|$ for any $z\in {\mathbb C}$; [/quote]
By the Maximum Modulus Principle, this says that $f+g+h$ is constant.

Are you sure this is what you meant?

sorry...........
in the future i'll put my work in "unsolved and proposed problems".

yes it is used Maximum Modulus Principle...


geometry
number theory
prime factorization

Derek’s phone number, 336-7624, has the property that the three-digit prefix, 336, equals the product of the last four digits, $ 7 \times 6 \times 2 \times 4$. How many seven-digit phone numbers beginning with 336 have this property?

I think a way to start is by factoring $ 336$, to get $ 2^4 \cdot 3 \cdot 7$.  I don't know where to go from here.

[quote="isabella2296"]Do you have to list all the possible ways? Because that's going to take a while...[/quote]

Not so sure on this one.

Well, we know that exactly ONE of the last four digits must be a multiple of three, and exactly ONE must be a multiple of seven... The digits cannot exceed nine, so we know that one of the digits must be 7. Another one of the digits must be either 3 or 6. If I only had 35 seconds to answer a question like this, I would probably say 72 because it is $ 4\cdot(3 \plus{} 3)\cdot3$. 4 possible positions for the 7, 3 possible positions for the 3 (or 6), and then 3 possible positions for the 2's.

[quote="herefishyfishy1"]Well, we know that exactly ONE of the last four digits must be a multiple of three, and exactly ONE must be a multiple of seven... The digits cannot exceed nine, so we know that one of the digits must be 7. Another one of the digits must be either 3 or 6. If I only had 35 seconds to answer a question like this, I would probably say 75 because it is $ 4\cdot(3 \plus{} 3)\cdot3$. 4 possible positions for the 7, 3 possible positions for the 3 (or 6), and then 3 possible positions for the 2's.[/quote]

Yes, but when you play my games, you only have 12 seconds.

Here are the cases:

splitting up into prime factorization yields $ 2^4*3*7$
since any number multiplied by 7 in this list will force the product to be greater than a one digit number, we can basically assume that all the cases will follow this format:

3 numbers whose product is 48 followed by 7.

A little casework gives

$ 2*8*3*7$ 24 ways
$ 1*8*6*7$ 24 ways
$ 6*4*2*7$ 24 ways
$ 4*4*3*7$ 12 ways

adding it all up yields $ \boxed{84}$

[quote="isabella2296"]Hopefully mewto will turn up with another one of his quirky but useful formulas...[/quote]

 :rotfl: I'm not [b]that[/b] smart. Mewto's formula for this one: Review+mem!!!!

Or use casework! That's how I did it.

[quote="skaldjfhrulez"]whoa! that's my phone number!
I got the answer by using casework[/quote]

Ahem.  Don't reveal that stuff... I might call you.

[quote="ernie"][quote="isabella2296"]You don't have the area code though.[/quote]

Or do I?  dun dun dun........... :ninja:[/quote]
Don't spam *cough* [size=59]ernie![/size] *cough*
casework is like trial and error right?

No. Casework is one when you look at different cases and stuff. It's better explained in Intro To C and P.


Hey Guys,
I'm not even sure if this is in the right forum, but I have a quick question.  Let's say you have 4x^2 = 90.  To get x, obviously you divide by 4 and square root.  But my problem is where do you put the plus minus symbol? Is it before square root 90 (or 45 depending on if you simplify) and the 4 (or the 2 if you simplify)? Or is there just one plus minus symbol before the entire expression?  I'm slightly confused because x could be (unless I'm wrong) plus minus square root 45 divided by plus minus square root 2.

I'm fairly certain it is just one plus or minus symbol before the entire expression. So, in your specific case, $ x \equal{} \pm\frac {3\sqrt {10}}{2}$.


In quadrilateral $ABCD$, angle $B$,$D$ are right angles. Prove that $BD = AC\sin{A}$

[quote="shobber"]Or even easier. Use the definition:
\[l=d\sin{\theta}\][/quote]

What does that mean?

$l$ is the length of the chord, $\theta$ is the angle opposite to $l$. $d$ is the diameter of the circle.

Yeah, just as shobber said, this follows from a certain proof of the law of sines.
Take a triangle $ABC$.  Let $O$ be the circumcenter.  Now let $D$ be a point on the circle such that $BD$ contains $O$.  Then $\measuredangle{BAC}=\measuredangle{BDC}$.  Angle $BCD$ is right, so thus, $\sin{A}=\sin{D}=\frac{BC}{BD}$.  This property can be used to prove the law of sines.  It tells us that $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=R$, and from this property, the proof of this problem follows.  Applied to the problem...
In the triangle $ABD$, $\sin{A}=\frac{BD}{AC}$.  Thus $AC\sin{A}=BD$.

This formula is actually the original definition of $\text{sine}$ which has appeared for nearly 2000 years. The definition we use today was given by Euler.


geometry
3D geometry
LaTeX
algebra
polynomial
number theory
Rational Root Theorem

Question 1: Given that "p" is not a factor of "n" for all primes "p" less than or equal to the cube root of n, show that n is either a prime or a product of two primes.

Use LaTeX!

1. Given that $ p$ does not divide $ n$ for all $ p\le\sqrt[3]{n}$, show that $ n$ is either prime or a product of two primes.

2. Prove that for an integer $ n\ge 2$, $ \sqrt[n]{n}$ is always irrational.

lingomaniac88...I tried downloading latex but it doesn't work...i am unable to compress my files or something like that.

you dont have to download it.
its part of the forums.
just put dollar signs around whatever you want to write.
AKA
[code]$\text{yahh}$ [/code]
is $ \text{yahh}$
theres info on the different codes on the sidebar under LaTeX.
and to see how someone wrote something, just move your mouse over it.

[quote="lingomaniac88"]2. Prove that for an integer $ n\ge 2$, $ \sqrt[n]{n}$ is always irrational.[/quote]

Generalization:  For integers $ a, b$, $ \sqrt[a]{b}$ is either integer or irrational.

Question 3:  If $ p_{n}$ denotes the $ n$th prime number, show that none of the integers $ P_{n}={p_{1}}{p_{2}}\cdots{p_{n}}+1$ is a perfect square

Question 4: Prove that Goldbach's Conjecture that every even integer greater than 2 is the sum of two primes is equivalent to the statement that every integer greater than 5 is the sum of three primes.

Sorry, this isn't a solution, but [b]sludgethrower[/b], is there any particular reason why you're not posting all these problems in one post?

Well I am currently working through some number theory questions justnow, so as I go along, I post the problems that I am having difficulty with to see if anyone else has a better proof or solution.

3: Assume the opposite. Let a = (P_n)^(1/2). Then we have:
p_1*p_2*...*p_n = a^2-1 = (a+1)(a-1), where a+1 and a-1 are integers. Clearly they must have the same parity. However, since 2 is the only even prime, this is impossible. Q.E.D.

p.s. Sorry, I have been too lazy to learn LaTeX.

2 seems easy. the same argument that $ \sqrt{2}$ is irrational works for any prime

if $ n^{2}$ is divided by p then n must be divided by p also since p is prime
assume that a and b are two relativly prime positive integers such that
$ pa^{2}=b^{2}$
then there must be some other integer c such that
cp=b
then you get
$ a^{2}=pc^{2}$
which means there must be some other integer d such that
ap=d

therefore, a and b share the factor p

[quote]Clearly they must have the same parity[/quote]

I like your proof. I did it slightly differently:

Question 3 answer:

$ p_{1}= 2$ and $ p_{2},p_{3},\ldots,p_{n}$ all have form $ 2k+1$ and so $ {p_{2}}{p_{3}}{\ldots}{p_{n}}$ also has the form $ 2k+1$ $ \Longrightarrow$   $ {p_{1}}{p_{2}}{p_{3}}{\ldots}{p_{n}}$ has the form $ 4k+2$ and $ P_{n}= 4k+3$ for certain integers $ k$.

Let $ a$ be any integer.

If $ a$ is even $ a = 2m, m \in Z$
$ \Longrightarrow a^{2}= 4m^{2}= 4k, k \in Z$

If $ a$ is odd $ a = 2m+1, m \in Z$
$ \Longrightarrow a^{2}= 4m^{2}+4m+1 = 4(m^{2}+m)+1 = 4k+1, k \in Z$

No perfect square takes the form $ 4k+3 \Longrightarrow P_{n}$ is never a perfect square.

Question 5

Show that there are infinitely primes of the form $ 6n+5$.

This has been covered [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=154093]here[/url] in a slightly different representation.

illcrowflu...yes, but there is not a proof that there is an infitude of primes of the form $ 6n+5$.

[quote="0714446459923"]2 seems easy. the same argument that $ \sqrt{2}$ is irrational works for any prime[/quote]

You seem to be working with the problem $ \sqrt[2]{p}$, which I addressed in a recent thread.  The question-poser is actually working with the problem $ \sqrt[n]{n}$, which is a little different.  I will work with the generalization I proposed.

[hide="Method 1"] $ \sqrt[a]{b}$ is a real root of the polynomial $ x^{a}-b = 0$.  By the Rational Root Theorem, the only rational roots are the integer factors of $ b$.  So $ \sqrt[a]{b}$ is either irrational or an integer. [/hide]
[hide="Method 2"] Suppose $ \sqrt[a]{b}= \frac{p}{q}$ with $ gcd(p, q) = 1$.  Then

$ q^{a}b = p^{a}$

If $ q = 1$ then we have $ b = p^{a}\implies \sqrt[a]{b}= p$, an integer.

If $ q > 1$ then clearly $ q^{a}| p^{a}$.  But we assumed $ gcd(p, q) = 1$ and we cannot have $ p = 1$.  Contradiction.  Hence $ \sqrt[a]{b}$ is irrational. [/hide]
Edit:  Oh, and

[hide="a proof that the nth root of n is not integer"] [b]Claim:[/b]  $ n \in \mathbb{N}\implies 2^{n}> n$

Proof:  The sequence $ a_{n}= 2^{n}-n$ is monotonically increasing (because $ a_{n+1}-a_{n}= 2^{n}-1 > 0$), and $ a_{1}= 1$.

This implies that $ 2 > \sqrt[n]{n}$.  For $ n > 1$ we clearly cannot have $ \sqrt[n]{n}= 1$.  Hence it is not integer.  QED. [/hide]

[quote="sludgethrower"]Question 2: Prove that for integer "n" greater than or equal to 2, the nth root of n is always irrational.[/quote]

[hide="another surprisingly easy proof of the generalization"]We'll prove that $ \sqrt[m]{n}$ is never an integer, where $ m,n>1$. Assume the contrary: $ \sqrt[m]{n}=\frac{a}{b}$ where $ \gcd(a,b)=1$. then we get $ n=\frac{a^{m}}{b^{m}}$ and thus $ nb=\frac{a^{m}}{b^{m-1}}$.

Now the left hand side is clearly an integer, while the right hand side isn't, since $ a$ and $ b$ have no common factor. This is a contradiction.[/hide]

nayel...you have ingnored the case where $ b = 1$. For instance $ \sqrt{4}$ and $ \sqrt[3]{27}$ are both integers. This generalisation for any $ \sqrt[m]{n}$ is wrong.

Also if you assume the contrary, you would have to assume that $ \sqrt[m]{n}$ is an integer, which is obviously plausible if $ n = a^{m}, a \in Z$

well i thought i'd just skip that part(b=1) cause it's trivial. :P Then my proof is correct!

ps. guess what! I'm in mathlinks as well as on AoPS!

Question 6 (I am truly stuck with this question!)

If $ p$ and $ p^{2}+8$ are both prime numbers, prove that $ p^{3}+4$ is also prime.

[hide="Massive hint"] Only one such $ p$ exists.  ;) [/hide]

[hide="another massive hint"]Try checking $ \mod 3$.[/hide]

Ah...I think I got it!

$ p = 3$  or p takes one of the forms $ 3k+1$ or $ 3k+2$

If $ p = 3$ then $ p^{2}+8 = 3^{2}+8 = 17$ which is prime. $ p^{3}+4 = 3^{3}+4 = 31$ which is prime hence conjecture holds for $ p = 3$.

If $ p = 3k+1$ then $ p^{2}+8 = 9k^{2}+6k+9 = 3(3k^{2}+2k+3)$ which is not prime $ \Longrightarrow p \not= 3k+1$

If $ p = 3k+2$ then $ p^{2}+8 = 9k^{2}+12k+12 = 3(3k^{2}+4k+4)$ which is not prime $ \Longrightarrow p \not= 3k+2$

Hence only one prime, $ p = 3$ allows  $ p^{2}+8$  to also be a prime, and we have shown that for $ p = 3$,  $ p^{3}+4$ is prime.


Let be given a triangle $\Delta ABC$ and an arbitrary point $M$ on the plane. Prove that: $\sum_{cyclic}\frac{MB.MC}{AB.AC}\ge 1$

Lemma 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=136745#136745]http://www.mathlinks.ro/Forum/viewtopic.php?t=21016  post #14[/url], see also http://www.mathlinks.ro/Forum/viewtopic.php?t=6037 and http://www.mathlinks.ro/Forum/viewtopic.php?t=24803 .

  dg


Let $ x$ be a real number such that
\[ \sin{x}\equal{}\sqrt{2}\times \sin{\left(\frac{2\pi}{7}\plus{}x\right)}\]
Find the value of the following:
\[ V\equal{}\frac{\tan{\left(\frac{\pi}{7}\plus{}x\right)}}{\tan{\frac{\pi}{7}}}\]

$ sinx\equal{}\sqrt 2 (sin\frac{2\pi}{7}\plus{}cos\frac{2\pi}{7}sinx)\to b\equal{}tgx\equal{}\frac{\sqrt 2 sin\frac{2\pi}{7}}{1\minus{}\sqrt 2 cos\frac{2\pi}{7}}\equal{}\frac{2\sqrt 2 a}{(1\plus{}\sqrt 2)a^2\plus{}1\minus{}\sqrt 2}$ were $ a\equal{}tg\frac{\pi}{7}$.
Therefore $ V\equal{}\frac{a\plus{}b}{(1\minus{}ab)a}\equal{}\frac{a(Aa^2\plus{}B\plus{}A\minus{}B)}{a(Aa^2\plus{}B\minus{}a^2(A\minus{}B))}\equal{}\frac{A}{\minus{}B}\equal{}3\plus{}2\sqrt 2$ were $ A\equal{}1\plus{}\sqrt 2, B\equal{}1\minus{}\sqrt 2$.


How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits occupy the odd places?

[hide="Not sure if this is right..."][color=red]_[/color] [color=blue]_[/color] [color=red]_[/color] [color=blue]_[/color] [color=red]_[/color] [color=blue]_[/color] [color=red]_[/color]

We need to put the odd digits 1,3,3,1 in the red gaps, or simply just order the digits, which can be done in $ \frac{4!}{2!\cdot 2!}=6$ ways.

Now to put the even digits 2,4,2 in the blue gaps, in the same way we find the number of ways it can be done is $ \frac{3!}{2!\cdot 1!}=3$.

So total number of such numbers $ =6\times 3=18$. [/hide]


welcome all
 :play_ball:
wait the answer about what is wrong with me
thanks

Yes you are right , that is a typo in the enuntiation . This concurrency is very wellknown  :) Also the segment conecting the midpoints of CA and BA passes throug G


For $ 0<u<1$, the sequence $ (u_n)$ is defined by $ u_1\equal{}1\plus{}u$ and $ u_{n\plus{}1}\equal{}\frac{1}{u_n}\plus{}u$ for all $ n \ge 1$. Show that $ u_n>1$ for all $ n$.

By induction on $ n$.  The base case $ n\equal{}1$ is trivially true, and for $ n\equal{}2$, $ u_2\equal{}\frac{1}{1\plus{}u}\plus{}u\equal{}1\plus{}\frac{u^2}{1\plus{}u}>1$.  If the result is true for $ n\minus{}1$ and for $ n$, then $ u_n\equal{}u\plus{}\frac{1}{u_{n\minus{}1}}<u\plus{}1$, or if $ 1\geq u_{n\plus{}1}\equal{}\frac{1}{u_n}\plus{}u$, then $ u_n\geq\frac{1}{1\minus{}u}$, leading to $ (u\plus{}1)(1\minus{}u)>1$, or $ u^2<0$, absurd.  The conclusion follows.


Let $ p$ be a natural number such that $ 1$ is the only perfect square that divides $ p$. Find all natural numbers $ x, y, z$ (in terms of $ p$) that satisfy the relation
$ \dfrac1{\sqrt {x}} \plus{} \dfrac1{\sqrt {y}} \plus{} \dfrac1{\sqrt {z}} \equal{} \dfrac1{\sqrt {p}}$

[quote="aadil"]are x,y,z natural?[/quote]
Yes. Sorry. I edited the question. Now it is corect.

trivial that $ x\equal{}y\equal{}z\equal{}9p$  :P

m trying to find other solutions..

If $ a,b,c \in \mathbb{Z}$ and $ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac {1}{c} \equal{} 1$
then we can take $ x \equal{} a^2p, y\equal{}b^2p, z\equal{}c^2p$

The solutions are $ (a,b,c) \equal{} (3,3,3), (2,4,4), (6,3,2)$ and permutations.

[quote="PhilG"]If $ a,b,c \in \mathbb{Z}$ and $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 1$
then we can take $ x \equal{} a^2p, y \equal{} b^2p, z \equal{} c^2p$

The solutions are $ (a,b,c) \equal{} (3,3,3), (2,4,4), (6,3,2)$ and permutations.[/quote]

Why there don't exist other $ x, y, z$ that satisfy the relation?


$ ABCD$ is a cyclic quadrilateral. If $ \angle B \equal{} \angle D$, $ AC\bigcap BD \equal{} \left\{E\right\}$, $ \angle BCD \equal{} 150{}^\circ$, $ \left|BE\right| \equal{} x$, $ \left|AC\right| \equal{} z$, then find $ \left|ED\right|$ in terms of $ x$ and $ z$.

$\textbf{(A)}\  \frac {z \minus{} x}{\sqrt {3} } \qquad\textbf{(B)}\  \frac {z \minus{} 2x}{3} \qquad\textbf{(C)}\  \frac {z \plus{} x}{\sqrt {3} } \qquad\textbf{(D)}\  \frac {z \minus{} 2x}{2} \qquad\textbf{(E)}\  \frac {2z \minus{} 3x}{2}$

[hide]If $ \angle BAC\equal{}\phi$, then $ \angle BDC\equal{}\phi$, so by Sine Law in $ \triangle BCD$

$ {x\plus{}ED\over{\sin 150^\circ}}\equal{}{BC\over{\sin\phi}}\iff x\plus{}ED\equal{}{BC\over{2\sin\phi}}$

But from $ \triangle ABC$ we get $ BC\equal{}z\sin\phi$, hence $ x\plus{}ED\equal{}{z\over 2}\iff ED\equal{}{z\minus{}2x\over 2}$, so the answer is $ D$.[/hide]


how many complex root theorems are there?can you list them please.

I've got an entire book on roots of the unity, and I am quite sure there are many more around.. so my guess is that there are a lot of theorems may you can google them or even better check them out on planetmath.

i couldnt find a good site tht explains kakayes theorem,kunny can u explain that theorem please.or is it too long for here to be explained?

[color=blue]Kakeya's Theorem[/color]

If $f(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots+a_{1}z+a_{0}$ is a polynomial with real nonnegative coefficients satisfying $a_{n}\geq a_{n-1}\geq \cdots \geq a_{1}\geq a_{0}\geq 0$,

then all roots of $f(z)=0$ lie in $|z|<1$.

Is the opposite true? I mean to say, is:
\[a_{0}\geq a_{1}\geq \ldots \geq a_{n}\geq 0 \implies \text{ roots of }f(z)=0 \text{ have }|z| \geq 1 \]


find the real solution to the equation $x(x+d)(x+2d)(x+3d)=b$

X(X+D)(X+2D)(X+3D)=(sqr(X)+3XD)(sqr(X)+3XD+2Sqr(D))
put sqr(X)+3XD=Y and solve


probability
AMC
AIME
function
AIME I
geometric series

Six fair 6-sided dice are rolled. What is the probability that the sum of the values of the top faces of the dice is divisible by 7?

I didn't know what you meant by "top faces," so I interpreted it as the faces showing (all the faces except the face touching the ground)

I got [hide]0, because the sums would be 20 to 15, and none of them are divisible by 7.[/hide]

Would you mind defining "top faces" for me?

Top faces are just the ones on the very top. Ie you just roll 6 dice, each one shows a number from 1 to 6 and you add these up.

Brute force... I think its definetly possible.

That how I solve most hard questions. I know I won't figure out "that clever way that came up in class a long time ago" :) in 2 minutes, so I use that time and just solve. It won't work on AIME, I guess, because you have to write down the process...
Anyways, it's impossible to solve by brute force since there's 6^6 total nunbers, so that's 46656.
...I got it! (not for this problem, but the other one :) )

[quote]It won't work on AIME, I guess, because you have to write down the process... [/quote]

Huh?

I've never been to AIME, so I'm just guessing, but I don't think AIME's a multiple choice...I thought it was the proof-writing type of test. Correct me if I'm wrong.

You give an answer between 0 and 999 inclusive, so no proof required.

BTW... There are other ways to brute force than counting EACH possibility.  Like multiplying... or like counting the ways of getting a multiple of seven (not that time consuming) and dividing it by possibilities.

I've tried that, but keep in mind there are [b]6[/b] dices, not 2. I mean, it takes a long time figuring out the combination for something that adds up to 21 without any techniques, and currently I have none seeing that I don't have an AoPS book nor have I read it and is in the process of purchasing it.

I was just wondering...could you use generating functions for this problem, and if so, how?

i dont know generating functions too well, but wouldnt it just be the expansion of (x^1 + x^2 + x^3 + x^4 + x^5 + x^6)^6?

yeah, I think that would be the function, but I mean would that approach be simple, or not worth the bother?

well im pretty sure that that would be the  correct function, but how to evaluate that is another question.  As I said, i dont know too much about this stuff... i read the section in AOPS about it, and the book said to evaluate the stuff inside the ( ) first, by using the sum of a geometric series, and then you can raise that to the 6th.  But I dont understand how that would help us with our answer (it might even not), if someone could answer why we would want to ever do that (not just in this question, but in general), I would be grateful.

But if what i mentioned above doesnt help us at all, then i guess the way to do it would to be expand that, which would be extremely difficult and not worth the bother.

Umm...is there a simpler way besides using generating functions? There must be...

A few things:

1) This question wasn't from the AIME. It was from a team contest in which 8 people had an hour to solve 15 questions, and no calculators were allowed.
2) On the actual contest, I tried it with generating functions and made an arithmetic error and got it wrong.
3) There's a MUCH easier method! Find it...

A few quick ideas:

[hide]

If we just consider 5 of the dice, and the last one separate (may have to multiply by 6 at the end, number of ways of choosing this die, not sure..) then if they add to a multiple of 7 the probability is 0, otherwise its 1/6.

So we sorta want the probability 5 add to a multiple of 7 (or its complement).. and then we could keep repeating this..

[/hide]

Keep going. See what you can do with what you have.

Simon: when you said you tried generating functions on the test, did you mean you expanded (x^0+x^1....x^6)^6?  Or is there a different way of doing that out?

Yeah, basically, except it's x+x^2+x^3+x^4+x^5+x^6. There is no x^0 term. You can do it a bit better by writing a die as x(x^6-1)/(x-1), but it's still quite messy.

let P(n) be the probability that the sum of n dice is not a multiple of 7. Then, P(n+1)=5P(n)/6+1-P(n). This is because, if the sum of the n dice is not a multiple of 7 (with probability P(n)), there is exactly 1 possibility that the sixth die will make the sum a multiple of 7, and thus 5/6 probability that it will still not be a multiple of 7.  If the sum of the dice is a multiple of 7 (with probability 1-P(n)), then no matter what happens, the sum of the n+1 dice will not be a multiple of 7. Simplifying our expression for P(n+1), we find that P(n+1)=1-P(n)/6. We know that P(1)=1, P(2)=5/6, etc. it's not that hard to find now.

[quote="Tare"]Oh, whoops. Missed the first "six"
So the answer is...hard to solve by brute force...hmm...there must be an easier way to do this...[/quote]

I'm glad that I found an "intermediate problem" that was easy for me!

Instead of thinking of rolling all the dice at once, consider the situation if each die is rolled individually.  Obviously, if you roll just one die P(7|top face) = 0.  Now think about what happens when a 2nd top face is added to the first.  I know, you're thinking that this is just brute force, BUT things clear up quickly with the right insight, and extending this to 6 dice is not hard at all!  I admit thoough that you might have to calculate out to the third die before seeing the "aha!"

Generating functions should also work, but I'm very rusty in that direction of math-space.

I'll post more hints if desired

Sorry! I read the problem last night and didn't notice Mystic's reply when I came back on today with my solution.  It'll take me awhile to get used to this place!

[quote="ComplexZeta"]Good work, except that you're asked for the probability that the sum IS divisible by 7.[/quote]

um...yeah. so? just take 1-P(6)

Yes, but you should make sure you don't solve the wrong problem.


function
inequalities unsolved
inequalities

Karamata again...

Define a sequence $\left(b_n\right)_{n\in\mathbb{N}}$ by $b_k=\sqrt{k}-\sqrt{k-1}$ for every k. Since

$b_k=\sqrt{k}-\sqrt{k-1}=\frac{\left(\sqrt{k}+\sqrt{k-1}\right)\left(\sqrt{k}-\sqrt{k-1}\right)}{\sqrt{k}+\sqrt{k-1}}$
$=\frac{k-\left(k-1\right)}{\sqrt{k}+\sqrt{k-1}}=\frac{1}{\sqrt{k}+\sqrt{k-1}}$,

it is clear that this sequence $\left(b_n\right)$ is decreasing (since, when k increases, both $\sqrt{k}$ and $\sqrt{k-1}$ increase, so that $\sqrt{k}+\sqrt{k-1}$ increases, and thus $b_k=\frac{1}{\sqrt{k}+\sqrt{k-1}}$ decreases). Hence, for every index k with $1\leq k\leq n$, the numbers $b_1$, $b_2$, ..., $b_k$ are the k greatest numbers among the n numbers $b_1$, $b_2$, ..., $b_n$. Hence, the sum $\sum_{j=1}^k b_j$ is the sum of the k greatest numbers among the n numbers $b_1$, $b_2$, ..., $b_n$. Since

$\sum_{j=1}^k b_j=\sum_{j=1}^k\left(\sqrt{j}-\sqrt{j-1}\right)=\sqrt{k}-\sqrt{0}=\sqrt{k}$,

we thus can state: The sum of the k greatest numbers among the n numbers $b_1$, $b_2$, ..., $b_n$ is $\sqrt{k}$.

For the numbers $a_1$, $a_2$, ..., $a_n$, we don't know how they are arranged, and we can only say that the sum of the k greatest numbers among the n numbers $a_1$, $a_2$, ..., $a_n$ is greater or equal to the sum $\sum_{j=1}^k a_j$ (in fact, this is obvious: the sum of the k greatest numbers is greater or equal to the sum of just some k numbers).

So altogether, for every index k with $1\leq k\leq n$, we have:
- The sum of the k greatest numbers among the n numbers $a_1$, $a_2$, ..., $a_n$ is greater or equal to the sum $\sum_{j=1}^k a_j$.
- By the condition of the problem, $\sum_{j=1}^k a_j \geq \sqrt{k}$.
- The number $\sqrt{k}$ is the sum of the k greatest numbers among the n numbers $b_1$, $b_2$, ..., $b_n$.

Combining this, we see that for every index k with $1\leq k\leq n$, the sum of the k greatest numbers among the n numbers $a_1$, $a_2$, ..., $a_n$ is greater or equal to the sum of the k greatest numbers among the n numbers $b_1$, $b_2$, ..., $b_n$.

Now, we have the following fact:

[color=blue][b]Karamata theorem for "pseudo-majorizing" number arrays:[/b] Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be 2n numbers from an interval $\mathcal{I}$ such that for every index k with $1\leq k\leq n$, the sum of the k greatest numbers among the n numbers $a_1$, $a_2$, ..., $a_n$ is greater or equal to the sum of the k greatest numbers among the n numbers $b_1$, $b_2$, ..., $b_n$ (but we don't require equality for k = n, i. e. we don't necessarily have $a_1+a_2+...+a_n=b_1+b_2+...+b_n$, but we only need $a_1+a_2+...+a_n\geq b_1+b_2+...+b_n$). Also, let f be a function which is [b]convex and monotonically increasing[/b] on the interval $\mathcal{I}$. Then, $\sum_{j=1}^n f\left(a_j\right)\geq\sum_{j=1}^n f\left(b_j\right)$.[/color]

This theorem can be easily deduced from the actual Karamata theorem.

Now, applying this theorem to our numbers $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$, with the interval $\mathcal{I}=\mathbb{R}^+$ and the function $f\left(t\right)=t^2$ which is convex and monotonically increasing on $\mathbb{R}^+$, we get

$\sum_{j=1}^n a_j^2\geq\sum_{j=1}^n b_j^2=\sum_{j=1}^n\left(\frac{1}{\sqrt{j}+\sqrt{j-1}}\right)^2>\sum_{j=1}^n\left(\frac{1}{\sqrt{j}+\sqrt{j}}\right)^2=\sum_{j=1}^n\left(\frac{1}{2\sqrt{j}}\right)^2$
$=\sum_{j=1}^n\frac{1}{4j}=\frac14\sum_{j=1}^n\frac{1}{j}$,

and the problem is solved.

[b]EDIT:[/b] I just saw the same problem in a slightly different version being posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21262 .

  Darij

[quote="cezar lupu"]Darij, it's not Karamata's inequality. Actually this is called Hardy-Littlewood-Polya inequality. ;)[/quote]

Both names are okay. But even if the name is wrong, who cares?  :P

[quote=opptoinfinity]Let us assume that $a_1 \ge a_2 \ge
a_3.....\ge a_n$ .Also let $f_n(x):
\mathbb{R^+}->\mathbb{R^+}$ such that

$f_n(x)=\frac{1}{\sqrt{x_n}}$

Let try to prove $2\sum_{i=1}^{n}a_i^2$

$\ge \sum_{i=1}^{i=n}a_if(x)_i$

Now using Abel's inequality

$ \sum_{i=1}^{i=n}a_i(2a_i - f_i(x))
= $  (a_1 - a_2 )( 2a_1 - f_1(x) ) + $ $(a_2 - a_3)(2a_1 +
2a_2 - $ f_1(x) - f_2(x) ) + .....  ..+ (a_{n-1}$
$ - a_{n} )(2\sum_{i=1}^{n-1}a_i - $ .
 $\sum_{i=1}^{n-1} f_i(x) ) +  x_n(
2\sum_{i=1}^{i=n} a_i -
\sum_{i=1}^{n}f_i(x) )$

Need to prove that $2\sum_{i=1}^{k}a_i
\ge \sum_{i=1}^{k}f_i(x)$

This means we need to show

$\sum_{i=1}^{k}\frac{1}{\sqrt{x}} \le 2\sqrt{k}$

That's very easy to show.[/quote]

[hide = "Smoothing Works"]

Let's fix $n$ and then ignore everything after $a_n$, since they really don't matter for the problem. We claim that the minimal value of $\sum_{j= 1}^{n} a_j^2$ occurs exactly when $a_i = \sqrt{i} - \sqrt{i-1},$ for all $1 \le I \le n$. Upon showing this, we'd be done since $\sqrt{i} - \sqrt{i-1} = \frac{1}{\sqrt{i} + \sqrt{i-1}} > \frac{1}{2 \sqrt{i}}.$ To show this, we will use smoothing. Consider any sequence $a_1, a_2, \cdots, a_n$ which satisfies the necessary conditions and minimizes $\sum_{i = 1}^{n} a_i^2.$ Call a positive integer $i$ $\mathbf{suboptimal}$ if $\sum_{j=1}^{i} a_j > \sqrt{i}$, and $\mathbf{optimal}$ otherwise. We claim that no positive integers can be suboptimal, which proves the claim. Assume the contrary, for contradiction. Then, let $x$ be the smallest suboptimal positive integer. Let $y$ be the smallest optimal number larger than $x$. Then, consider the effect of decreasing $a_x$ and increasing $a_y$ at the same rate until some $t \in [x, y)$ becomes optimal. By assumption, $a_x, a_y$ are replaced by $a_x -c, a_y+c$ respectively for some $c > 0$. But then, since $a_x-c = (a_1 + a_ 2+ \cdots + a_x - c) - \sqrt{x-1} \ge \sqrt{x} - \sqrt{x-1} > \sqrt{y} - \sqrt{y-1} \ge (a_1 + a_2 + \cdots +a_x -c + a_{x+1} + \cdots + a_y + c) - (a_1 + a_2 + \cdots + a_x - c + a_{x+1} + \cdots + a_{y-1}) = a_y+c$, it's clear that $(a_x-c)^2 + (a_y+c)^2 < a_x^2 + a_y^2$, which contradicts the fact that $a_1, a_2, \cdots, a_n$ was a sequence which attained the minimal value of $\sum_{j=1}^{n} a_j^2$ (since $a_1, a_2, \cdots, a_x - c, a_{x+1}, \cdots, a_y + c, a_{y+1}, \cdots, a_n$ attains a smaller value).

Hence, the minimum is attained at $a_j = \sqrt{j} - \sqrt{j-1}$ and we're done.

$\square$

[/hide]

[quote=MithsApprentice]Let $\, a_1, a_2, a_3, \ldots \,$ be a sequence of positive real numbers satisfying $\, \sum_{j=1}^n a_j \geq \sqrt{n} \,$ for all $\, n \geq 1$. Prove that, for all $\, n \geq 1, \,$  \[ \sum_{j=1}^n a_j^2 > \frac{1}{4} \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right).  \][/quote]

Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sum_{j=1}^n a_j^2 \sum_{j=1}^n\frac{1}{4j}
\ge \left(\sum_{j=1}^n \frac{a_j}{2\sqrt{j}}\right)^2.$$
Thus, it suffices to prove that
$$\sum_{j=1}^n \frac{a_j}{2\sqrt{j}}
> \sum_{j=1}^n\frac{1}{4j}.$$

Also, using summation by parts, we have
\begin{align*}
	\sum_{j=1}^n \frac{a_j}{2\sqrt{j}}
	&= \sum_{j=1}^n \left(\frac{1}{2\sqrt{j}} - \frac{1}{2\sqrt{j + 1}}\right)\sum_{i=1}^j a_i + \frac{1}{2\sqrt{n + 1}}\sum_{i=1}^n a_i\\
	&\ge \sum_{j=1}^n \left(\frac{1}{2\sqrt{j}} - \frac{1}{2\sqrt{j + 1}}\right)\sqrt{j} + \frac{1}{2\sqrt{n + 1}}\cdot\sqrt{n}\\
	&= \frac{n}{2} + \frac{\sqrt{n}}{2\sqrt{n + 1}} - \frac12\sum_{j=1}^n \frac{\sqrt{j}}{\sqrt{j + 1}}.
\end{align*}
It suffices to prove that
$$\frac{n}{2} + \frac{\sqrt{n}}{2\sqrt{n + 1}} - \frac12\sum_{j=1}^n \frac{\sqrt{j}}{\sqrt{j + 1}}
> \sum_{j=1}^n\frac{1}{4j}$$
which can be proved by Mathematical Induction.

[u][b]Remark[/b][/u]: There is some similarity between this problem and the following:

[u][b]Problem 1[/b][/u]: Let $a_1, a_2, \cdots > 0$ with $a_1 + a_2 + \cdots + a_n \le n^2$ for each $n \in \mathbb{N}_{>0}$. Prove that, for each $n \in \mathbb{N}_{> 0}$,
$$1/a_1 + 1/a_2 + \cdots + 1/a_n \ge \frac14\log_2 n.$$
[url]https://artofproblemsolving.com/community/c6h2726086p23746406[/url]

My proof for Problem 1 (in the link above):

[hide][b][u]Motivation[/u][/b]: I guess $a_1 = 1, a_2 = 3, a_3 = 5, \cdots, a_k = (2k - 1), \cdots$ which satisfies $a_1 + a_2 + \cdots + a_n = n^2$.

Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\left(\sum_{k=1}^n \frac{1}{a_k} \right)
\left(\sum_{k=1}^n \frac{a_k}{(2k - 1)^2}\right)
\ge \left(\sum_{k=1}^n \frac{1}{2k - 1}\right)^2.$$

Also, using summation by parts, we have
\begin{align*}
	\sum_{k=1}^n \frac{a_k}{(2k - 1)^2}
	&= \sum_{k=1}^n \left(\frac{1}{(2k - 1)^2} - \frac{1}{(2k + 1)^2} \right) \sum_{i=1}^k a_i
	+ \frac{1}{(2n + 1)^2}\sum_{i=1}^n a_i\\
	&\le \sum_{k=1}^n \left(\frac{1}{(2k - 1)^2} - \frac{1}{(2k + 1)^2} \right) k^2 + \frac{1}{(2n + 1)^2}\cdot n^2\\
	&= \sum_{k=1}^n \frac{1}{2k - 1}.
\end{align*}

It suffices to prove that
$$\sum_{k=1}^n \frac{1}{2k - 1}
\ge \frac14\log_2 n
$$
which can be proved by Mathematical Induction.
[/hide]

[quote=Cezar Lupu]Darij, it's not Karamata's inequality. Actually this is called Hardy-Littlewood-Polya inequality. ;)[/quote]

Yes, according to [url]https://artofproblemsolving.com/community/c6h14975[/url],
Karamata published later than Hardy,Littlewood and Polya, and in any case majorization was independently discovered many times before and after all these publications. There is a history in Marshall and Olkin's book on majorization. In the Anglo-Saxon countries it is called "majorization inequality" or "Hardy-Littlewood-Polya" majorization inequality".

The similar thing:

1) Cauchy inequality: before I called "Cauchy-Schwarz", now I call "Cauchy-Bunyakovsky-Schwarz"
[hide]Bunyakovskii published over 150 works on mathematics and mechanics. He is best known for his discovery of the Cauchy-Schwarz inequality, published in a monograph in 1859 on inequalities between integrals. This is twenty-five years before Schwarz's work.

In the monograph Bunyakovskii gave some results on the functional form of the inequality. There are many reasons why certain theorems are not named after their discoverer but after a later rediscoverer. However, in the case of Bunyakovskii there seems no good reason at all why he should not have the credit for his discovery.

One would have to note, however, that the terminology of mathematics is not universal and in some countries his theorem is correctly named, or named after Cauchy, Bunyakovskii and Schwarz.[/hide]

2) Holder inequality: In a book, I saw: 【In 1998, L. Maligranda said it should be called Roger-Holder inequality since Roger discovered it one year earlier than Holder.】


geometry
integration
trigonometry
calculus
calculus computations

a circle of radius 2 at (0,2)
a circle of radius 1 at the origin

from the picture it looks like x is the area in the first quadrant under the top circle inside the bottom circle

$ x^2\plus{}(y\minus{}2)^2\equal{}4 \to y\equal{}2\minus{}\sqrt{4\minus{}x^2}$
$ x^2\plus{}y^2\equal{}1 \to y\equal{}\sqrt{1\minus{}x^2}$

these intersect at $ (\frac{\sqrt {15}}{4},\frac{1}{4})$

so $ A\equal{}\int_0^{\frac{\sqrt{15}}{4}} 2\minus{}\sqrt{4\minus{}x^2}dx\plus{}\int_{\frac{\sqrt{15}}{4}}^1 \sqrt{1\minus{}x^2} dx$

which is just an ugly trig substitution problem if thats what you really intended

[quote="FantasyLover"][geogebra]fcaf432d8cdf9cb968f6a5f32892ca8541458900[/geogebra]

Find the area of $ x$.

Source : Me

How do you solve this problem?[/quote]
I would find the intersection points. (Which are $ (\pm \frac{\sqrt{15}}{4}, \frac{1}{4})$) Call them $ P$ and $ Q$. Call the circles $ C_1,C_2$ and ther midpoints $ M_1,M_2$. Then the area is equal to:
(The area of the halfcircle in $ C_1$ from $ P$ to $ Q$ $ \plus{}$ The area of the halfcircle in $ C_2$ from $ P$ to $ Q$) $ \minus{}$ $ T_{\triangle M_1PQ}$ $ \minus{}$ $ T_{\triangle M_2PQ}$


You have 1000 feet of flexible pool siding and wish to construct a swimming pool.  Experiment with rectangular-shaped pools with perimeters of 1000 feet.

a) How do their areas vary?

b) What is the shape of the rectangle with the largest area?

I don't quite understand a, but for b it would be a square.

way too easy for intermediate.

a) The area of the pool, where x is the length of the pool, is $x(500-x) \text{ feet}^{2}$

Treething,

Thanks for providing me with an answer for a.

a) $\lim_{x\to 50}a=0$ and $\lim_{x\to 0}a=625$
where $x=l-w$ and $l$ and $w$ are dimensions and $a$ is the area of the rectangle.

Part c) Prove that the area of rectangle is maximized when it is a square.

Well the formula is already provided, so

$x(500-x) = 250^{2}-(250-x)^{2}$,

which clearly maximizes at $x = 250$.

I was looking for something more along the lines of
[hide]$p$ is perimeter $l$ and $w$ are dimensions of rectangle
and $c$ is a constant
$p=2c$
$l+w=c$
$a=w(c-w)$
$a'=c-2w=0 \rightarrow w=\frac{c}{2}$
$a=\frac{c}{2}*\frac{c}{2}$
$\therefore$ the max area of the rectangle would be when the dimensions are equal (except the precal way)[/hide]

I love all replies.  This is very informative data.