Mafia
AMC
AIME
percent
big brother
geometry

The game of [b]Big Brother[/b] is starting! These are the basic rules:

1. We well keep track of "days" (not the real days), and the game will be divided into "weeks."
2. On day 0 of a week (the last day of the week, excluding the first week), the HOH competition will be held. The winner is the HOH.
3. On day 1, there may be a luxury/reward competition on some weeks, but not all.
4. On day 2, the HOH will nominate 2 people for eviction.
5. On day 3, 6 people, including the HOH and 2 nominees, compete for the veto, and the power to veto one of the nominations. He does not have to use this power. 6. If used, the HOH must nominate someone else, other than the vetoholder, on day 4.
7. On day 7, houseguests, other than the HOH and nominees, will vote who to evict. The evicted houseguest will leave immediately.
8. Immediately afterward, all remaining houseguests other than the outgoing HOH will compete to become the new HOH.

This is how it goes in normal weeks; that is, weeks other than the first (the days will just be a little scattered, that's all), and when there are too few people for this to work. Then, the rules change a little; we'll get into it.

Some ground rules:

1. You are encouraged and expected to communicate with people about the game outside this thread.
2. You may [b]not[/b] communicate with anyone about the game in or out of the thread when voting or a challenge is in progress.
3. No double accounts, or any of that nonsense.
4. At no time may you communicate with others outside of the game about the game.
5. At no time may you communicate with others previously evicted from the house about the game, unless you are both on the jury, or both have no more role in the game.
6. You are always on the honor system. It's not like I'll ever know, but just follow the rules.

Now, houseguests: you may enter the house.

Houseguest 1: [color=red]erabel[/color]
Houseguest 2: [color=blue]xpmath[/color]
Houseguest 3: [color=blue]perfect628[/color]
Houseguest 4: [color=green]Klebian[/color]
Houseguest 5: [color=blue]crazychinesemaniac[/color]
Houseguest 6: [color=red]1=2[/color]
Houseguest 7: [color=blue]mathling235[/color]
Houseguest 8: [color=blue]andersonw[/color]
Houseguest 9: [color=blue]distracted523[/color]
Houseguest 10: [color=blue]Cyrazeno[/color]
Houseguest 11: [color=green]AIMEman[/color]

Welcome to day 1! The HOH competition will begin shortly (tomorrow afternoon, in which it will still be day 1). Use the time now to mingle and aquaint yourselves with each other.

Good luck!

Note: if you wanted to be in the house but didn't make it (or came late), you can still have a role! pm me for details.

Hm, this is a pretty interesting mix. 4 Mole players (and Catalyst is a mole player as well), 5+ Mafia players, and a couple of newbies. I would fit in the last category.

Hey, mathling uses the same numbers as me in his/her username... :|

Um, the air. A question: What are days 5 and 6 for?

[quote="Klebian"][quote="crazychinesemaniac"]crazychinesemaniac IS IN DA HOUSE!!!!!!!! lol so whats up people?[/quote]he's out first guys[/quote]

QFT

[quote="Klebian"][quote="crazychinesemaniac"]crazychinesemaniac IS IN DA HOUSE!!!!!!!! lol so whats up people?[/quote]he's out first guys[/quote]

What? Why?

yo all... whats up? Odds are, I will lose very soon... but hi anyways!

[size=200]HOH Competition: Week 1[/size]

Prior to entering the house, each of you gave a statement outlining your personality and how you plan to play, and ultimately win, this game. Below are ten quotes, one for each player in the game. Your task is to match up the quotes with the people.

There is a catch: after your submission, I will immediately PM you with the number of correct answers you had. Then, you will send me a new set of answers (which may or may not be the same as before). [b]If you assign a person's name to two different quotes, or vice versa, your score will be zero for whichever submission you are working on.[/b] Keep in mind: your score can go down. The houseguest with the most correct on their second submission is our first Head of Household.

[b]No discussion in our out of the thread about the competition or the game as a whole until the HOH is announced.[/b]

Quotes (random order):

1. "Make alliances with more people than I plan to take with me to the final few. "
2. "I am a very smart person (but probably not as smart as most of everyone else on aops)..."
3. "Most of the time I tend to sit back and watch others do their thing and just jump the bandwagon..."
4. "Sometimes I make a random quote to [i]try[/i] to eek out a laugh."
5. "Then I'll win. See it's easy."
6. "...I'm the kind of person that likes to work independently and deceive everyone else."
7. "I also want to play enfluentially so I can get people to vote the same way I do."
8. "I'm usually quiet and reserved, but will speak out if I feel I should."
9. "If I don't try to be friends with everybody...it could go really badly, so that's what I'll focus on most, especially being nice to the HoH."
10. "I'll have a better idea once we start."

Houseguest choices (in random order)

a. 1=2
b. andersonw
c. crazychinesemaniac
d. Cyrazeno
e. distracted523
f. erabel
g. Klebian
h. mathling235
i. perfect628
j. xpmath

umm.. Catalyst, you pmed me and said I was a houseguest...

You didn't respond to the first task. If you still wish to take a role in the game, pm me.

[quote="CatalystOfNostalgia"][b]If you assign a person's name to two different quotes, or vice versa, your score will be zero for whichever submission you are working on.[/b][/quote]

Is this for two separate PMs, or just one PM? Like can you submit these:

1. a
2. b
3. c

Then

1. c
2. a
3. b

?

Yeah, you can do that. What I don't want you to be doing is something like

1. a
2. a
3. a

on the same submission.

Oh man, this is probably the worst week possible. I have AIME tomorrow and States on Saturday.

I'm frustrated that i haven't had enough time to really respond to these but I know I will by tomorrow. I'm really sorry.

[quote="xpmath"]
[b]Klebian[/b]: Well, reading perfect's question and your reply helped answer some of my questions too. If you could have chosen anyone else to be in the Final 2 with you, why would you choose them and who would it be (if this is more than one question, don't answer the who.)[/quote]
I think this is again the misconception of Final 2 that is making my evicting perfect look worse than it should. Because I am viewing it as so, I will answer the question for final 3, and then final 2. I would have chosen you and perfect to be with me in the final 3 because we were an alliance from the start and I think up till the final 3 alliances should be honored as much as possible. It hurt to lose you early but we didn't have enough influence then to sway the vote. Now if we understand that you had been evicted, I would have chosen the weakest person in the final 3 to go with me to the final 2. Anyone else is a difficult question but obviously it would have been somebody who could have managed to get in the final 3 and then I would have picked them to be in the final 2 with me. This would probably have been chinesemaniac. From what I could see I would've stood a good shot against him in the final 2 even if I had evicted an ally of mine. I think ideally for me I would've made it to the final 2 without being the final HOH, as I stated earlier because in evicting someone I would lose at least 1 vote I otherwise would have surely had. I will reiterate myself in stating that I believe that in the final 2 you want the person you have a shot of beating. In the end you have to realize that the point of winning HOH in the final 3 would be to increase your chances of winning and you would try to take with you the person who was the smaller threat in terms of jury influence. I would say I did so. I mean if I had taken perfect, I would have near-surely lost (giving him at the very least AIMEman's vote and his 2 allies' vote). I felt that I had a slightly better chance of winning if I took AIMEman. Again, the jury should consider this aspect of my eviction.
Hopefully I answered your question; I went a bit off topic because I felt it was a bit too narrow for me to answer as stated.

[quote="distracted"][b]Klebian[/b]: You voted me. How do you justify that in order to win my vote?[/quote]
As such. Recall that it was you or AIMEman to vote for, and perfect and chinese were the other 2 remaining. If I had voted off AIMEman, it would have been you 3 left with me. Even though perfect and I were allied, I had pretty much no shot of winning if both you and chinese were left in the game. So essentially voting you was the only way to ensure my survival and you must realize that you would have done the same thing in my situation. So I ask you to consider more than that. AIMEman and I were temp allied by that time and so if our positions had been switched (and I had his reward) the same outcome would have occurred that week. So you should not see your eviction as my doing but rather our doing, and so you should not base your vote on that week. Consider other weeks and consider how little AIMEman really ever did.

[quote="crazychinesemaniac"]Klebian: Who did you see as the biggest threat againist you and why.[/quote]
Really you and distracted both. You guys had been consistently winning HOHs and vetos and I was surprised no others really viewed you guys as a threat enough to band up. Even with perfect on my side I knew that with the power you guys always had you could overcome his say at one point and I felt the same even when I had allied with Aimeman. Once distracted was out I was lucky because you couldn't win the next HOH and so you lost your power right at the end where it mattered the most. Thus the threat against me had, really, been diminished and I took advantage of this to try my hardest to win the final HOH, which as I have said I shouldn't have done.

[quote="mathling235"]Was there a time in this game where you felt that you had no way of winning? (Please explain)[/quote]
Somewhat. Week 6 (when cyrazeno was evicted) I wasn't sure if I could survive out of there. I felt that if I could get through that week I had at least a shot at winning. I had stupidly not submitted a Half the average submission and I wasn't sure if I could team up with AIMEman to get his vote. I knew that at that point perfect was on my side but I didn't know if distracted would vote out cyrazeno as well.
I actually think that after I won the final HOH I felt like I had a smaller shot of winning than really at any point earlier. This mistake was pretty crucial to what I think will end up being my loss, as both of my final 3 members were my allies, in a sense. I knew that I would lose 'less' votes by evicting AIMEman but I felt that perfect had a stronger influence on the jury members, partly just because of his character in general. It was sort of a catch 22 for me at that point because I knew as I evicted perfect that I was losing at least 1 vote and quite possibly more but I felt that the fact that AIMEman had done next to nothing and coasted his way into the final 3 (and then final 2) would help my cause somewhat. Still, I felt that either way my decision would result in a loss for me. (Slightly ironic, isn't it?)

I'll answer the remaining 2 questions ASAP

hmmmm, thanks for the detailed responses Klebian, someone please prod AIMEman (I don't feel like doing it).

perfect628 is still in charge btw

I already prodded him. Three times actually.

In that case, if nothing happens in 24 hours, start the voting. It should remain on the minds of the jury that he did not provide satisfactory answers.

So sorry.  Was at MSSB.  Will get to more questions later today.

[quote="andersonw"]Klebian:
Are you playing this game for: 1)Winning or 2)Having fun? Pick the 1 that describes you most and explain.[/quote]
I think it's somewhat clear that we were all playing in this game to win. I do think I had fun and that was part of the reason I played but really I believe that all decisions made had winning the game as the top consideration. Especially since this is an online game and there are no 'real' interactions, I think you lose a lot of the fun that could be had in a big brother type game.
I don't know if I'm going to answer this to the extent that you want but that's pretty much all I can say- Playing to win describes me best but I think it describes everyone best. Hopefully this will also explain my rationale for my late game decisions, and then my answers to other question asked of me.

[quote="Cyrazeno"][b]Klebian[/b]- What did you like most about this game? [/quote]
I appreciated the organization that the mod had. Everything went smoothly on his part and prevented this game from being abandoned like it had the potential to do. Because of this I'm sure that the game was a better experience for everyone and in the end it doesn't feel like the effort put in the game went to waste. The challenges were creative and of decent difficulty and participation was pretty reasonable. I hope I answered your question.

sorry.  had no computer availability Thursday - Sunday.

[quote="xpmath"][b]AIMEman[/b]: Well, I don't know much about what you did, but I would just like to ask you what your game plan and strategies were.[/quote]

Hmm, game strategies.  Well, as most people figured out, I laid low for a while, which was kind of a strategy.  Also, in the beginning, I analyzed every move I made and thought about how it would affect my game later.  I soon figured out that those strategies were not working.  After winning HOH, I tried to keep some modest control over my status and how close I was to being evicted.  And then, after my alliance with maniac sort of faded, Klebian asked me about that impromptu alliance.  I saw that as my only way to keep my head above water.  And here we are.

[quote="distracted523"][b]AIMEman[/b]: Be honest: how much did chance play a factor, the week of my eviction, in allowing you to survive that week?
[/quote]

Well, the factor that chance really played was when Catalyst put me and Klebian next to each other on the game board that week.  If Klebian stuck to his alliance once he won and did not vote for me when I was up for eviction with you, then I would nullify perfect's vote with my reward and Klebian's vote would evict whoever else maniac put up.

I want to make the point that I was not plotting to evict you, I was just trying to save myself.

Will get to more questions later tonight.

[quote="crazychinesemaniac"]AIMEman: Why do you think you made it this far. Also, who do you think will win and why?[/quote]

I think I made it this far because of a few things.

1) [b]Lying low for 1st half or so of the game.[/b]  This did not mean inactivity, just not taking any big risks or anything like that.
2) [b]My reward.[/b]  It certainly saved me from getting eliminated that pivotal week.
3) [b]My alliance with klebian.[/b]  Definitely the reason we are both here.  (And same reason as #2)

I think it will be fairly close in this final round.  I also think that it will come down to what the jury makes of the houseguests' answers.  (I don't read klebian's answers because I don't think its fair that he posted answers before me and wouldn't have been able to read mine, so I don't know what the jury made of his answers.)  So I honestly have no idea.

[quote="mathling235"]Was there a time in this game where you felt that you had no way of winning? (Please explain)[/quote]

In short, at least the whole first half.  But I will go into detail:

The first four weeks were unbearable.  Like many other houseguests, I am sure, I was very discouraged when klebian and distracted each won HOH for two of the first four weeks of the game.  I had no alliance with either of them, and I felt like I was next on the chopping block, since I had not won any contests.  The only alliance I had was with maniac.  After klebian's anti-reward expired, he and distracted were still there and I felt even more hopeless.  But then, when klebian and i teamed up, things started looking up.

Get to others in an hour or so.

[quote="andersonw"]AIMEman:
Do you believe that you made the right decisions in this game and why?
[/quote]
I think I made mostly the right decisions.  One of the hardest decisions I had to make was when klebian and cyrazeno were up for eviction (week 6).  Klebian offered me to start an alliance, and if I hadn't accepted, we both probably would not be here now.  So it saved both of us.  Even though I am competing against klebian, we needed each other to get here.

[quote="Cyrazeno"]Eh, okay...
[b]AIMEman[/b]- Do you plan to play again? Why or why not?[/quote]
I don't see any reason not to play again.  I thought the challenges were great and very creative, especially using the ones on AIM.  I think the game functioned well and went exactly how it was supposed to.  It was a very big success.

Ok, so now I am done.  Take it away, mod.

Voting begins NOW. No more talking. The seven jury members will cast a vote in perfect628's PM box (NOT MINE). perfect628, you must vote first, since you are also on the jury. You may not open any votes until you have cast your official vote.

I'll give you more information about making the final game post. perfect628 is still in charge, but I'm taking the liberty of prodding you all.

(edit) I mentioned this in all your PM's, but when submitting your vote you will give a final statement to everyone that will be read before all of the votes are read. Your statement should not give away your vote.

64 Days, 59 Cameras, 112 Microphones, Countless Betrayals, 9 Houseguests, and 1 Moderator later, and here the Final Two stand!

[b]----------------[/b]

I'm your host, [b]perfect628[/b], replacing [size=59]CatalystOfNostalgia[/size] because of a recent accident with some sort of cleaning implement. (I didn't really ask.) Within the hour, we will know the Big Brother winner. [b]BUT FIRST[/b], [i](reference to Big Brother on TV)[/i]

We'll invite the first two evicted Houseguests back on to the premises...(we almost didn't, but the producers forced us.) Please welcome back the only ones without an influnce tonight,

[b]1=2[/b]

[b]erabel[/b]

Thank you for returning. But now, on to more important business. [i](boos from their guests)[/i]

Here are our seven Jury Members, in the order they were evicted:

[b]mathling235

xpmath

andersonw

...Cyrazeno seems to be missing...

distracted523

crazychinesemaniac

perfect628[/b]

We'll begin the live vote for the winner of Big Brother now. Jury Members, one at a time, you will come up to the box, make one final statement to our Final Two, and cast your vote [b]for the winner[/b], not for whom to evict. We'll begin with

[hide="xpmath"] It was fun playing this with all of you. It was hard to decide who I would vote for, but I've made my decision.

*inserts key*

Next up, [hide="mathling235"]May the best man win.

*inserts key*

You're up, [hide="crazychinesemaniac"]well, its down to the two of you. Let's start with AIMEman, to tell you the truth I'm suprised you lasted so long. At first you weren't really winning and it didn't even seem like you were trying. However, you had a strategy and that strategy got you here. Now for Klebian, I realize it was a mistake not to make allies with you. That might have gotten me a little further. However, I probably still would have lost. You were a formidable opponent and for that I respect you. At this time I shall throw one of the keys into the box....

*inserts key*

You're next Thanks to everyone for a great game. I had a lot of fun, and I'm sure everyone else did, too.

*inserts key*

I'll go next, so As I said, I was pretty shocked by my eviction. I thought I was totally in control of every aspect of this game, and saw my victory inevitable once I reached Final 4 when I didn't have to take any part in distracted's eviction. Just ask Catalyst, I was sure I was omniscient and omnipotent.

Needless to say I was wrong.

Probably the cornerstone of my entire game was trusting no one except for one person whom I needed to trust absolutely. This was my flaw, which I now see. (20-20 hindsight I know.)

So, now that you know that, I cast my vote keeping only one thing in mind: respect. Honestly, I only remotely respected one of your games, and I totally respected that person. I was sure about my vote before I read statements, and I was not swayed.

Good luck to you both.

*inserts key*

Please go now, Oh right she's not here...

*doesn't insert key*

And last but not least, I don't doubt my vote, and the questioning didn't change my mind.

*inserts key*

And that's it, all the votes have been cast, and the decision is final. The winner has been decided...Let's not stretch this out. I shall now reveal the votes in order:

[b]Klebian[/b]

It's 1-0 [b]Klebian[/b].

[b]Klebian[/b]

That makes it 2-0 [b]Klebian[/b].

[b]AIMEman[/b]

Now it's 2-1 [b]Klebian[/b].

[b]Klebian[/b]

It's 3-1, so [b]Klebian[/b] only needs one more vote to win.

Congratulations, [b]Klebian[/b], you are the winner of Big Brother!

The final vote was 5-1, so [b]distracted523[/b] voted for [b]Klebian[/b] too. [/hide][/hide][/hide]

Congratulations again to our big winner, and we hope to see you all next season.

Here is the [url=http://www.quicktopic.com/41/H/qqKW8FgBYcr]Jury Quicktopic[/url] (some shockers in there, at least to me):

PLUG ALERT: I will be running BB2 starting with signups in mid to late August. Sign up.

Thanks to perfect628 for taking over moderation, and for successfully coordinating the endgame (although it did take long) while a cleaning implement was in my way (but an extremely worthwhile 'setback'). Also, thanks to all the players for helping this run smoothly and making it a great game (especially to the final five).

And of course, congratulations to Klebian for winning and to AIMEman for finishing as runner-up.

I have attached a complete table of voting history. It isn't the most exciting thing, as there were a total of six votes for the opposition over the entire game, with one being nullified. But it's nice to have a summary of the game. I took the table format from the Wikipedia pages for the US Big Brother seasons.

thanks for the game catalyst
good game all

Very well organized.  Good game to everyone, especially Klebian.


OK. A person has to ask a question for a value, and the next person answers and posts another one. Make the questions ahrder as they go. Lets see how much we can add up.

Values:
100
200
300
500
1K
2K
4K
8K
16K
32K
64K
128K
256K
500K
1M
2M
3M
5M
10M
20M
40M
80M
160M
250M
500M
1B

for 100$, what is the capital of the usa :rotfl:

Washington D.C.

Do I have to get it right to post another one?

[quote="Ars"]Washington D.C
Who is the President of Iran?[/quote]

Just to let you know, I doubt that'd be a 200 question.

We just had middle east current events so for me it is also google it

You just have to solve all of them, and then you can make them up.

The capital of the USA is Washington DC,
the President of Iran is Mahmoud Ahmadinejad.
for $\$300$, What is my username spelled backwards?

capital of USA is Washington DC

President of Iran is Mahmoud Ahmadinejad

Tnatsetorp     i think, unless if you are asking for your real name

for $500, what is 1+1???

For 500$answer, 2! So now I collect: [hide: ed604fc185] 100+200+300+500=900$. I shall continue[/hide]

(Please include a sum of your scores, and indicate whether you will take your cash, or venture forward. If you get a question wrong, you lose all of your cash. Beware! :D )

For $\$1000,$ what is the biggest country, by land area, in the world?

Russia of course so that makes $\$1900$ and I will continue lets see next question
What lake has the most fresh water in the world?

For the $1000 question, I answer Russia, and that is correct!  :clap2:  :clap:
I now have:
[hide]$1900[/hide]

For the 2K question, Lake Baikal.

For the next question, which city has the second greatest population in Wisconsin :D

Umm that's not how who wants to be a millionaire works, the money values are culmulative.

I'm just going to guess the capital for this question, is it Madison?

[quote="Ignite168"]I'm just going to guess the capital for this question, is it Madison?[/quote]
That is correct.

For $\$4000,$ how many permanent seats are there in the United Nations?

$\$4000$

How many permanent members of the UN Security Council are there?

[quote="Ignite168"]Five. :arrow:  :arrow:[/quote]
That is correct.

$\$8000$

What two countries were involved in the Cold War (1947-1991)?

[quote="ProtestanT"][quote="D.P.L"]$\text{Russia VS USA}$ :ninja: [hide=" :ninja: "]:ninja: [/hide][/quote]

No, it's the Soviet Union (USSR) vs the United States (USA).

[quote="nutz_for2.718281828"]For 16 grand,

In the TV show "The Simpsons", what is the Simpson's religious neighbor?

a. Moe the Bartender
b. Reverand Lovejoy
c. Krusty the Klown
d. Ned Flanders[/quote]

I'm guessing B) Reverand Lovejoy

for 32K,
How many centimeters are in an inch?
(rounded DOWN)
A) 2.4
B) 2.5
C) 2.6
D) 2.7[/quote]
I believe it's B, the exact number is 2.54

The National Geography Bee includes religion, so...

64K
What country has the largest Muslim population?

[quote="ProtestanT"][quote="nutz_for2.718281828"]For 16 grand,

In the TV show "The Simpsons", what is the Simpson's religious neighbor?

a. Moe the Bartender
b. Reverand Lovejoy
c. Krusty the Klown
d. Ned Flanders[/quote]

I'm guessing B) Reverand Lovejoy[/quote]

WRONG!!!! The correct answer[hide="was"]D) Ned Flanders[/hide]

[quote="anirudh"]For the religion one, teh naswer is Indonesia, and I will now take home 64K :trampoline: . I shall continue. [b]DPL is now officially out.[/b][/quote]
You usually don't get the money when someone already posted the answer.

And nutz, did you not notice that in the first post it says who wants to be a millionare, GEOGRAPHY EDITION?

$\$128000$

How many continents have only one country?

A) 1
B) 2
C) 3
D) 5

But i still get partial credit because USSR=[b]Russia[/b]+Some other countries

You do not get partial credit in millionare and also Russia is no where close to what the Soviet Union was politically so unless you want to say that every Russian is a socialist I recomend not arguing your point.

For 128000, I would go with A.

@i_like_pie: actually, australia is part of oceania, which is what the NGB uses, so a choice should actually be 0. and one more thing-NGB doesn't have MCQ


function
algebra
polynomial
algorithm
algebra unsolved

Find all functions $ f: \mathbb{R} \to \mathbb{R}$ such that for all reals $ x$ and $ y$, the equation
\[ \left(f\left(x\right)\right)^{2} \minus{} \left(f\left(y\right)\right)^{2}\equal{}f\left(x \minus{} y\right)f\left(x \plus{} y\right)
\]
holds.

http://www.mathlinks.ro/Forum/viewtopic.php?t=114493 for all continuous $ f$.

Its not solved there.

There was a typo. Now i believe i have solved the problem... finally.. after 4 edits...lol

For those interested in my approach here it is;

[hide]

The first idea is to prove that f(2x) =2f(x). to prove this it is sufficent to prove that f(2)=2f(1) which i am not going to do here but i can give an argument for it if one needs it.... so let me go on from this right now:

f(x+2y) ^2 - f(x)^2 = f(2y) f(2x+2y) = 4 f(y)f(x+y)

Now f(x-2y)^2 - f(x)^2 = -4f(y)f(x-y)

so f(x+2y)^2 - f(x-2y)^2 = 4f(y) ( f(x+y) + f(x-y))

so f(4y) f(2x) = 4f(y) (f(x+y) +f(x-y))

and therefore f(x+y) + f(x-y) = f(2x)
this leads to the following: f(a) + f(b) =f(a+b) so long as f(a-b) is not zero.

[/hide]

So i have a given a prove that if f(x)= 0 only for x=0 then f(x) is cauchy.
i.e: f(x+y) = f(x) +f(y)
Now what happens in the more general case where there may be non-trivial solutions to f(x)=0??
[hide]
in this case i'm going to show that again f(a+b) = f(a) + f(b) even if f(a-b)=0. (lets assume that f(a) is not zero becasue in this case clearly the statement is valid!)

for if f(a-b)=0 then f(a+b)^2 - 0 = 4f(a) f(b)
Also f(a)^2 - f(b)^2 = f(a-b) f(a+b) = 0
so f(a) ^2 = f(b)^2 but since f(a) f(b) >0 therefore
f(a+b)^2 = 4 f(a)^2 and therefore f(a+b) = +/- 2f(a) = f(a) +f(b)
one only needs to consider the case f(a+b) = -f(a) - f(b)
in this case f(a+b) +f(a) = f(b) and therefore f(2a+b) =f(b)
but then this implies that either f(a) is zero or f(a+b)=0
in any case contradiction can be reached.
[/hide]

To prove that f(2) =2f(1)

note that one set of equations is

f(2)^2 - 1 = f(3) f(1)
f(3)^2 - 1 = f(2) f(4)
f(4)^2 - 1 = f(3)f(5)

and also

f(3)^2 - f(2)^2 = f(1) f(5)

this will give us two ways of calculating f(5) in terms of f(1) and f(2)

In fact i got a polynomial of order 8 in x = f(2) /f(1) to which of course x=2
was a solution but i don't know if this was the only solution.... i sure hope so:)
but even if there are other roots. one can develope computer programs to go about the same algorithm for larger numbers and eventually i believe it shouldn't be that hard ( at least [b]impossible[/b]) for us to prove that f(2)=2f(1)

please post any other solutions that you have in mind.... [/b]

You presumed that f(1)=1???
If f(x)=sinx then $f(2) \neq 2f(1)$
$f(2x)\neq 2f(x)$

ok i didn't assume that f(1) =1
i just said f(2) = 2 f(1)

also if f(x) = sinx

then sin^2x  - sin^2 y = sin (x-y) sin(x+y) is not true!!!!

ummmm , i thought i edited my self.... :blush: :blush:  :lol:

yeah what i said was absurd and i realized it right after i posted it.
indeed your solution is one that i can n't explain and it proves me wrong;

that is the criteria f(2x) = 2f(x) is not nessacarily true as i presumed...

ummm, this forces me to give this more thought now...
please keep your ideas coming.

Ali


I'm a homeschool parent and have just discovered this website, forum, and curriculum today.  My son is loves math. He's managed to teach himself a couple of years of math just from paying attention to his older sister's lessons and spends his free time practicing at our white board.  He's progressing through our curriculum at such a rate that I've been trying to figure out what to do with him in the future.  I think I may have found the answer here.

We've been using which does a good job of teaching just the basics.  It lacks variety of problems and problem solving, though.  He's 7 (officially in 2nd grade) and has covered their Alpha, Beta, Gamma, and Delta levels (approximately 1st through 4th grade levels).   He has a strong mastery of addition, subtraction, multiplication, division, basic area, finding average, things like that.  These include concepts like multiple-digit multiplication and long division.  Of the little basic algebra (ie. y=2x+3, if x is 42, what is y) I've shown him, he's understood with no problem.  His assessment test scores are equal to the median for 6th grade but he hasn't covered topics like order of operations, fractions, decimals, exponents, prealgebra, and algebra yet.

What would be the suggested sequence of texts to cover, starting next year, considering that he wants to complete calculus before graduation?  I realize that he'll need something before starting the Intro to Algebra text but what should that be?  Sticking with Math-U-See with some sort of problem solving supplement?  Or is there something that will cover everything he needs?

I appreciate any suggestions.  I have no idea what to do to best meet his needs but am looking forward to learning math all over again with him.[/url]

The place to start with AoPS is our Intro to Algebra text (along with the Algebra 1 class if he's taking classes).  Prior to AoPS, homeschoolers seem to have a great deal of success with the Singapore elementary series.  Whichever route you choose prior to AoPS, spicing up the curriculum with some problems from [url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_MOEMS.php]MOEMS[/url] would give him some exposure to problem solving.

He sounds like our kind of student, and should have no trouble reaching calculus before graduating high school.

I was thinking of sticking with our current program through algebra with some supplementing from something like Singapore's Challenging Word Problems or MOEMS or contest samples from The Math League and then moving into the Intro to Algebra.  Would that work well?  That will put him in the Intro to Algebra at age 9/10 and 5th grade unless prealgebra or algebra slows him down.  I would be surprised if they slow him down though since he already understands some of the basics.

Should be be taking a basic algebra course before doing Intro to Algebra?

I'm having a hard time considering Singapore as our math curriculum. I tried using it with my daughter but it was such a different method from what I was taught that I had a hard time figuring out even the lower levels well enough to teach it.

Your plan should be fine, particularly if he remains so strongly attracted to math.

For students who are that young, it's not a bad idea to have them go through a less rigorous basic algebra class before our Intro to Algebra class.  (Sometimes 4th-5th graders who use our Intro Algebra book will make two passes through it, with the second pass spent focusing on the more challenging material.)  Supplementing the Math-U-See with tougher problems from Singapore and MOEMS is a good idea, and will help ease him into the greater challenge our curriculum will offer.  At some point, he'll escape what you can do for/with him.  Then your job goes from being a teacher to being an enabler.  Sounds like you're already thinking well ahead to that day, which is very good.


geometry
rectangle
analytic geometry
graphing lines
slope

[hide]Split the rectangle into 24 1-by-3 rectangles. This gives us a 24-by-24 square. Obviously, the diagonal will pass through 24 of these squares. Since the squares consist of 1-by-3 triangles, there will be 3 small squares that the diagonal passes through for each of the larger 'squares.' Thus the diagonal passes through $3*24=72$ squares.[/hide]

[quote="bpms"]A 24ft by 72 ft rectangular floor is covered by 1x1 tiles, and then the opposite corners are connected by a diagonal. Through how many interiors of the tiles does this diagonal go through?
Although there is a formula (PIE), I know another way for this problem (although that way only works in this case). Any other solutions?[/quote]

[hide="NON-PIE solution"]
divide the 24*72 rectangle into 24 of 1*3 rectangles, which would result in a 24*24 rectangle/square.. 3*24=72 because there are 3 of these 24*24 squares[/hide]

[quote="bpms"]Is anyone going for the non-PIE solution? PIE is better for this problem but I have another solution if you don't know PIE. It's been 5 days and no one has gone for it. Here is my non-PIE solution
[hide]A 24x72 rectangle is similar to a 1x3 rectangle. In the 1x3 the diaganol goes through all three tiles. Multiplying by 24 gives you 72.[/hide][/quote]
Well, that's not exactly a "non-PIE" solution.  :wink:  That is, in fact, the method from which the PIE formula is derived. Here's an explanation:

[quote="I once"]There's a simple formula for these kinds of problems. If the line goes through $(0,0)$ and $(a,b)$, the answer is \[a+b-\gcd{(a,b)}\] squares. This is because if $m$ and $n$ are relatively prime, then the line that goes through $(0,0)$ and $(m,n)$ passes through $m+n-1$ points.
This is because it passes through $m-1$ horizontal line segments and $n-1$ vertical line segments. Each time it passes through one of these, it enters a new square. Also, it cannot pass through a horizontal line and a vertical line at the same time because $m$ and $n$ are relatively prime. This means that the number of squares is $(m-1)+(n-1)+1=m+n-1$ (the extra 1 is because of the first square).
(We can also extend this to three dimensions in a similar way.)
If we multiply $m$ and $n$ by some number $k$, then we are multiplying the number of squares the line passes through by $k$ (since we "copy" it $k$ times). This means that the line that goes through $(0,0)$ and $(mk,nk)$ passes through $mk+nk-k=mk+nk-\gcd{(mk,nk)}$ points. Setting $a=mk$ and $b=nk$ gives us our desired result.[/quote]

See attachment for general method and applications.

[quote="bpms"]A 24ft by 72 ft rectangular floor is covered by 1x1 tiles, and then the opposite corners are connected by a diagonal. Through how many interiors of the tiles does this diagonal go through?
Although there is a formula (PIE), I know another way for this problem (although that way only works in this case). Any other solutions?[/quote]
[hide]
it has a slope or something with like 3 then you multiply by 24 because it repeats 24 tiems or something[/hide]


Prove that in a circle the measure of the arc that an angle subtends is equal to double the angle.

Clarification: an inscribed angle

[hide="outline of proof"]

Case 1: One of the sides of the angle passes through the center of the circle

Draw a line between the center of the circle and the intersection of the side with the circle.
Use the angles in the triangle to show the 2:1 relationship between the central angle and the inscribed angle

Case 2: The center of the circle is between the sides of the angle
Case 3: The center of the circle is outside the sides of the angle

Draw a line from the vertex of the angle through the center of the circle. Apply the result of Case 1.
[/hide]


function
calculus
derivative
real analysis
real analysis unsolved

Suppose that  $ f ,f^',f^{''},f^{'''}$ are positive, and we have $ f>f^{'''}$. Prove that $ 2f>f^{'}$.

[quote="JRav"]Well the fact that both $ f'$ and $ f''$ are positive means that the function is concave up and increasing.  So in such a scenario, $ 2f$ would have to be greater.[/quote]
Not necessarily. For example, if $ \mathrm{f}(x)\equal{}\mathrm{e}^{3x}$, then all the derivatives of f are positive but $ 2\mathrm{f}\not>\mathrm{f}'$. You still need the condition $ \mathrm{f}>\mathrm{f}'''$.

edit:Oh,Sorry,I didn't get your meaning a minite ago.

f  is a solution of the differential equation y''' - y = h < 0

you can solve it with variations of constant and express f with the complex roots of X^3-1

and h  and some constant.

Use the condition f ,f' , f'' , f''' are positive will give condition on the constant

Finally study the sign 2f - f'


calculus integration trigonometry function

this is an easy one...

Put $ {x}{\equal{}}{acos^2y\plus{}bsin^2y}$... and then the integral is split according  to the powers...

i think multiplication and division by $ (\cos^2 x)^{p \plus{} q}$ would be a possible next step, though i doubt it would be the easiest way to solve it  :|

no man not wallis method...(though u can also aplly that ...very good idea)...

I am talking abt the integratiuon by substitution.... that kk sir taught us in class.

[quote="Cyber-Shreyas"][quote="valeriummaximum"]depending upon thewhether p is odd or even /q is odd or even we decide the substitution and if both are even then we do it by complex no method...[/quote]

Wallis method? :D[/quote]
umm.. how is wallis method applicable here?
it isn't even of that form, in case you guys've not noticed.
 :huh:

[quote="elastiboysai"]Answer:
[code]
$\frac{$(p+1)!(q+1)!$(b-a)^{p+q+1}$}{(p+q+2)!}$
[/code]
oops my 1st latex code .flop :o
the output was fine bt how do i paste it[/quote]

Wallis 4mula is 1 method

Another method wud be to use properties to bring it to the form of beta function
n den use d direct formula.
Is dat wat u were lookin 4??

dei semma kattan polairrukku, 3:40 morning varil ne awake ka?  :o  Apparom eppo thungi eppo eyundhuruppa?  :rotfl:

FIITJEE enna sollirukkan paru, 10:30 pm varil than awake ka irrukkanam  :rotfl:

[quote="elastiboysai"]dey
3.50 - 9 30 is not inhuman n im not a nerd :blush:
u r d baby nerd  :rotfl:  :rotfl:[/quote]
you wake up at 3:50 ah??? surely u are a kattan!!!
but madness never goes to sleep thats a diff case to consider!!! :rotfl:

Hey srini stop all this da otw you shall surely be doomed  :D   :rotfl:   :P

Since I haven't specified anything about p & q, okay. But both complex no and walli's method are right. :D


Can some one is brave enough to factor this or using Muirhead to verifiy that this is true:

Denote Symmetric Sum = Sym in the Muirhead sense. [b]Show that:

Sym(a^4.b) + 3.Sym(a^3.b^2) + 5.Sym(a^2.b^2.c) >= 9.Sym(a^3.b.c)

where a, b, c > 0.[/b]
Thanks in advance.

Note: I knew that the first two Syms are bigger than Sym(a^3.b.c) but overall the last Sym of LHS is smaller than Sym(a^3.b.c).

Minh Can, the present for you:
Let $a,$ $b$ and $c$ are non-negative real numbers. Prove that
$\sum_{sym}(a^{6}-6a^{5}b+3a^{4}b^{2}+2a^{3}b^{3}+12a^{4}bc-24a^{3}b^{2}c+12a^{2}b^{2}c^{2})\geq0.$

Arqady: thank you for the solution. I was going to ask how you factor the expression out so quickly ? I am sure other Mathlinkers have asked similar questions on how to group the terms so that you can split it into sum of squares ? or just use your skillful problem solving ? or the last resort the computer software factor package ? btw, I will attempt the "present" you gave. cheers  :lol:  :P

[quote="Minh Can"]... how to group the terms so that you can split it into sum of squares ? [/quote]
If $a=b$ then
$\sum_{sym}(a^{4}b+3a^{3}b^{2}+5a^{2}b^{2}c-9a^{3}bc)=2b(b-c)^{2}(2b-c)^{2}.$
From here it is already possible to guess, that
$\sum_{sym}(a^{4}b+3a^{3}b^{2}+5a^{2}b^{2}c-9a^{3}bc)=\sum_{cyc}a(b-c)^{2}(b+c-3a)^{2}$
and to check up it.
But it is possible and to solve system from equality:
$\sum_{sym}(a^{4}b+3a^{3}b^{2}+5a^{2}b^{2}c-9a^{3}bc)=\sum_{cyc}a(pb^{2}+qc^{2}+\gamma ab+\beta ac+\alpha bc)^{2}.$
First, certainly to substitute $c=0.$


trigonometry
calculus
calculus computations

How would you find the Fourier Series of $\sin^{3}(x)$


geometry
perimeter
area of a triangle
Heron\u0027s formula

Find all ordered triple $(a,b,c)$ with $a \leq b \leq c$ and $a,b,c$ are integer which are also the sides of the right triangle, so that $Perimeter = Area$

(5, 12, 13) works, as well as (6, 8, 10)...but that's simply because we have $4a+5=a^2$ and $c^2-b^2=a^2=2c+2b$...but as the sides get larger, it doesn't hold really...for example, just look at 7-24-25.  The area is much too large!  As you get smaller, it doesn't hold either...
Let us have such a triangle abc, with a as the shorter leg and c the hypotenuse.  Let $c-b=n$.  Then $c^2-b^2=a^2=n(c+b)$.
When n=1, we have $b+c=a^2 \implies a^2+a=\frac{ab}{2}$.  Therefore, $a+1=\frac{b}{2}$, which gives (5, 12, 13)
Let n=2.  Then $2(b+c)=a^2 \implies 2(a+b+c)=a(2+a) \implies b=a+2$.  This yields (6, 8, 10).
Let $n\ge3$.  Then we have $n(a+b+c)=a^2+an \implies \frac{bn}{2}=a+n$.  There are no more integer solutions...the LHS gets too big...

your answer is correct, but you must prove that there are no other ordered triple satisfy that.
[hide]
I denote $A = Area$ and $P = perimeter$
$(a+b)^2=a^2+b^2+2ab=c^2+4A=c^2+4P$
$(a+b)^2-c^2=4(a+b+c)$ and thus we have
$(a+b+c)(a+b-c)=4(a+b+c)$ or $a+b-c=4 .........(1)$
And By Heron's formula we have
$A= \sqrt{s(s-a)(s-b)(s-c)}$ with $s = \frac{a+b+c}{2}$
and $s= \frac{P}{2} = \frac{A}{2} \rightarrow A=2s$
and we have $4s^2= s(s-a)(s-b)(s-c)$
$4s=(s-a)(s-b) \frac{(a+b-c)}{2} = (s-a)(s-b) \frac{4}{2}$
$2s= (s-a)(s-b)$ , since $A= \frac{1}{2} ab$ so we have $s= \frac{1}{4} ab$  thus we have
$\frac{1}{2} ab= \frac{(ab-4a)(ab-4b)}{16}$ this equivalent to

$8ab=ab(a-4)(b-4)$ or $8=(a-4)(b-4)$
since $8$ has the prime factor $8=(2)(2)(2)$ so the only posible thing is$8=(4)(2)$ and $8=(8)(1)$
thus we have $b-4=8$ and $a-4=1$ or
$b-4=4$ and $a-4=2$
so $a=5$ and  $b=12$ or $a=6$ and $b=8$.
{(5,12,13),(6,8,10)}
[/hide]

I posted the same problem in  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48782]this post[/url] about two days ago ;)


search
geometry
geometric transformation
reflection
function

Hi

i don't know if you are familiar with sql but I have a little question.

Are there(exists)  2 distinct sets of numbers (positive, >0 , different numbers) that have the same MIN,MAX,SUM,AVG,STDEV(standard deviation) and COUNT.

sure, assumng that there are  enough  (>5) numbers in each set.

can you give me an example or prove to me that.

The 2 sets i'm comparing will have a maximum of 10 numbers. The sets will have the same number count. (so let say 10 in one set and 10 in another).  The numbers in the sets will not be greater then 400, so (1 - 400).
Can you give me an example of 2 sets that have the same  MIN,MAX,SUM,AVG,STDEV.

MIN(Set1) = MIN (Set2)
MAX(Set1) = MAX(Set2)
SUM(Set1) = MAX (Set2)
AVG(Set1) = AVG(Set2)
STDEV(Set1) = STDEV (Set2)

Thank you

Why don't you just try constructing one?  It's actually not so hard to do.  As Gyan said, 5 points are all you need, and you know quite a lot about them already -- for instance, the minimum and maximum of each set have to be the same, and it should be clear that the actual values you choose for them are irrelevant.  So start off by setting both sets to have maximum 99 and minimum 1 (or any other values you happen to like -- I chose those values because it gives my interval a nice median value).  Since the sum and average conditions are equivalent (same number of elements), you just have to figure out a way to add three points to each set to get the same standard deviation and same sum.  Standard deviation is "harder," so you probably want to do something that will ensure equal standard deviations.

The smallest example in positive integers involves 5 (non-distinct) values selected from the set {1, 2, 3, 4, 5}.

Yes, not really that hard, you can start with say $(0,0,2,2,3))$  and  get  $(0, \frac{3-\sqrt 5} 2 , 1 , \frac{3+\sqrt 5} 2 , 3 )$

[hide](Basically you select  three numbers  (almost arbitrary)  say (0,2,2)  and try to get ,  say (1,x,y)  such that  ($x+y=3$) and ($x^2+y^2= 7$) and solve for x and y.

 (We want to keep the  sum  as same as before (0+2+2=4=1+x+y) and the same goes  for  the  sum of the squares .($1^2+x^2+y^2=0^2+2^2+2^2$)
[hide]
Note: If sum and sum of squares are same for the two sets. the STD DEV would be equal too.[/hide]
(Just pad them on the both sides so that MIN and MAX are same)[/hide]

ok, my mistake

i shoud have mention that the number are N (natural numbers) , (1,2,3,4,5......... 400)
and that there will be no duplication in the sets like (0 , 0,  2, 2 , 3) , ther will be only distinct numbers.  (0,1,2,3,4,5)

Sorry again, hope you don't get upset.

I'll try to think again for myself and find a solution in N. But if you have it, please show it.

edit:

so as I think of it:

set 1 = ( a, b , c , d , e)
set 2 = ( a, b , x, y ,z)

a > b , c , d , e , x, y, z
b < a , c , d , e , x, y, z

c+d+e = x+y+x

c^2 + d^2 + e^2 = x^2 + y^2 + z^2

and a,b,c,d,e,x,y,z belong to |N

Those restrictions strike me as rather pointless.  It's not too hard to find two sets of five distinct integer values selected from {1, 2, 3, 4, 5, 6, 7, 8, 9} with the same "specs."

If it is not that hard , can you give an example. I must be stupid.

I could, but I'd rather not.  I can tell you what thought process I used to generate my example, however: suppose we have five numbers, $a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5$.  You want those $\leq$'s to be <'s, but it's the same process either way.  I want to preserve the mean and the standard deviation of this list.  So, the "structure" has to stay the same.  What sort of simple operations preserve the standard deviation?  Translating my entire list, for instance, wouldn't change the variance, but it would change the mean.

Here was my example with repeated elements -- maybe it will help you figure out what I'm getting at.  The sets 1, 1, 4, 4, 5 and 1, 2, 2, 5, 5 have the same minimum, maximum, standard deviation and mean.  So, how did I get from one to the other, and why does it work?

In fact, since you have your two equations and I gave you a range to search in, you could simply check the (reasonably small) set of possibilities -- it would only take 70 simple calculations to list out every single possibility, but of course finding a single solution doesn't require that much work.

JBL thank you very much, you really helped me.
And thank you for not giving me the solution.

So, as a very lazzy person I just wrote a small c# program to do that calculation for me.
(I don't know why i did not do that from the start :( , i guess i thought the answer is no)

so i came with this numbers for example:
1 2 6 7 9    and 1 3 4 8 9

That was my small example as well.  The "trick" I used to generate it was to find a set whose mean was equal to the center of the interval, but which wasn't arrayed symmetrically around the center.  (You need 5 points in order for this to be possible.)  Then, if you reflect the entire interval through its center, you get another set with the same variance (because we haven't changed the shape at all) and mean (because the mean was just the center of the interval).

My problem was that I had 1 set of numberes (in sql) and wanted to find another set of numbers identical with the first one.
The limitation was that I need to use only aggregate functions like above.

Do you see a solution for this problem?

I'm not sure I understand -- you had data, and you wanted to collect enough information about it to guarantee uniqueness?  Presumably, you wanted a fixed number of conditions which would encode sets of arbitrarily large size?


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ R_1\equal{}\mathbb{Z}_p[X]/<X^2\minus{}2>$ and $ R_2\equal{}\mathbb{Z}_p[X]/<X^2\minus{}3>$.Consider when $ R_1$ isomorhic to $ R_2$ in the ways : $ p\equal{}2,p\equal{}5$ and $ p\equal{}11$

What do you mean by $ \mathbb{Z}_p$?
$ \mathbb{F}_p\equal{}\mathbb{Z}/p\mathbb{Z}$ or really $ \mathbb{Z}_p$ the $ p$-adic integers?
And do you mean isomorphic as rings?

I have a feeling that he means the former.

This question mostly boils down to the irreducibility of $ x^2\minus{}2$ and $ x^2\minus{}3$ in $ \mathbb{Z}_p$.  If one is irreducible and the other is not, these rings cannot be isomorphic since one has zero divisors and the other does not.

There's a little more work for $ p\equal{}2$, since both polynomials are reducible.  But you can show that an isomorphism from $ R_1$ to $ R_2$ must send $ a\plus{}bX$ to $ a\plus{}b(X\plus{}1)$.  Now just decide if that's a valid isomorphism.


HMMT
geometry
3D geometry
tetrahedron
AMC
AIME
geometric transformation

This is from the HMMT 2004 problem on the geometry test. It is also a 3D version of a problem on the AIME 2 in 2003.

Given a regular tetrahedron of side length 1, reflect it about its center. What is the area of the intersection of the original tetrahedron and the reflected tetrahedron?

I am having trouble visualizing the intersection. Any ideas?

Aww, come on, you can do it

Hints:[hide]Try and figure out what each side will look like.  Then, count them.[/hide]

[quote="Scrambled"]i think you can imagine two tetrahedron with a side length 1/2 placed back to back.[/quote]

What do you mean by "back to back"? If they were back to back, they wouldn't be intersecting. :/

Oh. Maybe I'm just being stupid.

*grumble* You all have such high IQ's. Do you see the picture easily in your heads?

By the way, does anyone know of any 3D math software that can help me see something like this?

oops.  i just realized i had misunderstood the problem.
This is how id visualize the intersection:

imagine a tetrahedron
imagine a tetrahedron with 1/3 the side length.
subtract 4 1/3 length tetrahedrons from each corner.  That is your answer.

im only in 6th grade so you might not want to take my word for it.

[quote="Fierytycoon"]What is the area of the intersection of the original tetrahedron and the reflected tetrahedron?[/quote]

you do mean volume right?

I think the vertices of the two intersecting tetrahedra will form the vertices of a cube with side 1/ :rt2: , and the intersection will be an octahedron with vertices at the centre of the cubes faces.

Here's how I did it --

a tetrahedron is really just 4 triangles, right?  And it's got all sorts of nice symmetry.  So just reflect one of the triangles.  Use that to "cut" the original tetrahedron.  Then preform the same action on the 4 other sides.

I did not see this immediately, I've tried to figure it out on my own before.

There really isn't good 3D graphing software that I know about, although Mathematica does have some limited abilities in that respect, and you can probably use 2-D graphing software creatively to accomplish the same thing.


Tori's mathematic test had 75 problems: 10 arithmetic, 30 algebra, and 35 geometry problems. Although she answered $ 70\%$ of the arithmetic, $ 40\%$ of the algebra, and $ 60\%$ of the geometry problems correctly, she did not pass the test because she got less than $ 60\%$ of the problems right. How many more questions would she have needed to answer correctly to earn a $ 60\%$ passing grade?

70% of 10 = 7 problems
40% of 30 = 12 problems
60% of 35 = 21 problems

7+12+21=40 problems total
60% of 75 is 45 problems

Tori needs 45-40=5 more problems to barely pass.
(she really needs improvement in math)

EDIT: did my math wrong the first time so forgot to change from 19 to 21. Thanks!

[quote="dragon96"]70% of 10 = 7 problems
40% of 30 = 12 problems
60% of 35 = [b]21[/b] problems

7+12+[b]19[/b]=40 problems total
60% of 75 is 45 problems

Tori needs 45-40=5 more problems to barely pass.
(she really needs improvement in math)[/quote]

a little contradiction, which i have highlighted in bold
could you change?

then again, 7+12+19 is not 40, so the answer is still correct


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite cyclic group of order $ n$. Show that for each positive integer $ d$ dividing $ n$, there exists a subgroup of order $ d$.

Suppose $ n\equal{}k \cdot d$. The order of $ <x^k>$ is $ d$.


What do you people think? Theres the youtube
Do not laugh in high school
24 hour tag
Do not laugh in police station
Spring inn
Memorisation game for dancing
Quiet library game

And some more of Downtown's "game shows".

Those are hilarious!!!  :rotfl:

I like high school
especially the english video, the part after PE, and the science class
I also like the part when Yamazaki is given leniency then Endo laughs but gets caught anyway  :rotfl:  :rotfl:  :rotfl:

I saw the first two minutes from this one: http://www.youtube.com/watch?v=cz2VDdf-X6Q and I think that the video is really funny  :D  :D  :D


Κάναμε στα  μαθ.κατ στο σχολείο για την έλλειψη και μου γεννήθηκε η εξής απορία...Ρώτησα και τον καθηγητή μου αλλά δεν ήξερε...

Αν μας δίνεται τυχαία έλλειψη υπάρχει τρόπος να βρούμε τις εστίες της με κανόνα και διαβήτη;

Φέρνεις το ΑΑ' (όπου Α,Α' οι κορυφές του μεγάλου άξονα) και απο το μέσο Ο γράφεις κύκλο (Ο,ΟΑ). Στη συνέχεια, φέρνεις μια τυχαια εφαπτομένη της έλλειψης και προβάλλεις στην ΑΑ' τα σημεία τομης της εφαπτομένης με τον κύκλο.Αυτές είναι οι εστιες!

1) Πώς θα βρω με κανόνα και διαβήτη τον μεγάλο άξονα(μας δίνεται απλά η έλλειψη χωρίς τους άξονες)
2)Πώς θα φέρω εφαπτομένη με κανόνα και διαβήτη;

Αν θέτεις έτσι το πρόβλημα δε ξέρω αν μπορούμε να κάνουμε τέτοια κατασκευη.Λίγο μπακαλίστικα που το σκέφτηκα τώρα ειναι να κάνουμε το εξής. Αφου ξέρουμε οτι το ΑΑ' είναι το μεγιστο ευθυγραμμο τμήμα που ενωνει 2 σημεία της έλλειψης, μπορούμε να το προσδιορίσουμε.
Όσον αφορα την εφαπτομενη έλλειψης για να την κατασκευάσεις θα πρεπει γενικα να γνωρίζεις τις εστίεσ της. Μπορείς όμως να βρεις τον μικρο αξονα ο οποιος ειναι καθετος στην ΑΑ' στο σημειο Ο. Έπειτα εχοντας βρει τα Β,Β' φερνεις παραλληλη στον μεγαλο αξονα κ εχεις την εφαπτομενη στο Β (ή Β' αντιστοιχα).

(Θα προσπαθήσω να σκεφτω αν υπαρχει πιο μαθηματικοσ τροποσ να βρουμε το ΑΑ')

Φέρνουμε δύο ζεύγη  παράλληλων ευθειών που τέμνουν την έλλειψη
(να μην είναι όλες παράλληλες ).

Στα παράλληλα τμήματα που ορίζονται από τα σημεία τομής βρίσκουμε τα μέσα
και φέρνουμε τις ευθείες που περνουν από αυτά .

Οι δύο ευθείες τέμνονται στο κέντρο της έλλειψης
( έχω μια απόδειξη για αυτό αλλά δεν είμαι σίγουρος ) .

Έχοντας βρεί το κέντρο φτίαχνουμε κύκλο με κέντρο το κέντρο της έλλειψης
και ακτίνα τέτοια ώστε ο κύκλος τα τμήσει την έλλειψη .

Αν τα παραπάνω είναι σωστά το υπόλοιπο είναι εύκολο .  :D  :)

[quote="Ramanujan"]Οι δύο ευθείες τέμνονται στο κέντρο της έλλειψης
( έχω μια απόδειξη για αυτό αλλά δεν είμαι σίγουρος )[/quote]

Αποδεικνύουμε πρώτα το ακόλουθο λήμμα:
[color=blue]
[size=117]Λήμμα[/size]
Έστω $\mathcal{C}: \ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ η δοθείσα έλλειψη.

Έστω τυχαία χορδή $AB$ της έλλειψης και $M$ το μέσον του $AB$.
Η ημιευθεία $OM$ τέμνει την έλλειψη στο $P$.
Τότε η εφαπτομένη από το $P$ είναι παράλληλη στην $AB$
[/color]

[size=117][hide="[color=red"]Απόδειξη][/color][/size]
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \Rightarrow$

$b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}\ \ (1)$

$AB: \ y=mx+k$

Τότε $y^{2}= m^{2}x^{2}+2kmx+k^{2}$. Αντικαθιστούμε  στην $(1)\Rightarrow$

$(b^{2}+a^{2}m^{2})x^{2}+2kma^{2}x+k^{2}-a^{2}b^{2}=0$

Οι ρίζες του τριωνύμου μας δίνουν τις τετμημένες των $A,B$.
Επομένως η τετμημένη του $M$ θα βρεθεί από το γνωστό $-\frac{b}{2a}$
Δηλαδή:
$x_{M}=-\frac{kma^{2}}{b^{2}+a^{2}m^{2}}$

Το [b]Μ[/b] ανήκει στην [b]ΑΒ[/b], επομένως $y_{M}= m\cdot x_{M}+k$

$\frac{y_{M}}{x_{M}}= m+\frac{k}{x_{M}}= m-k\frac{b^{2}+a^{2}m^{2}}{kma^{2}}= m-\frac{b^{2}+a^{2}m^{2}}{ma^{2}}=-\frac{b^{2}}{ma^{2}}$

$\frac{y_{P}}{x_{P}}= \frac{y_{M}}{x_{M}}\Rightarrow$

$\frac{y_{P}}{x_{P}}=-\frac{b^{2}}{ma^{2}}\Rightarrow m =-\frac{b^{2}x_{P}}{a^{2}y_{P}}$

Από την $(1)$ παίρνουμε ότι η πολική του $P$ είναι η $(b^{2}x_{P})\cdot x+(a^{2}y_{P})\cdot y=a^{2}b^{2}$

Δηλαδή ο συντελεστής της διεύθυνσής της είναι $-\frac{b^{2}x_{P}}{a^{2}y_{P}}=m$. Επομένως είναι παράλληλη με την $AB$.
$\hline$
[/hide]
Έστω τώρα χορδή $CD\parallel AB$ και $N$ το μέσον της.
Η ημιευθεία $ON$ ορίζει σημείο $P'$ πάνω στην έλλειψη και αυτό λόγω του λήμματος οφείλει να είτε να ταυτίζεται με το $P$ ή να είναι το συμμετρικό του ως προς το $O$. Σε κάθε περίπτωση βρίσκουμε ότι το $O$ θα ανήκει στην ευθεία $MN$.

Λουκάς

Πολύ ωραία....Σας ευχαριστώ και τους δύο παιδιά.. :D  :D  :D

[quote="pontios"][quote="dimos"]Να κατασκευασθεί (με κανόνα και διαβήτη) έλλειψη της οποίας δίνονται μια εστία και 3 σημειά της....[/quote]
Αν μπορούσαμε να κατασκευάσουμε υπερβολή με κανόνα και διαβήτη, θα ήταν εύκολο...  :D[/quote]

Κ αν μιλούσαμε για παραβολή θα ήταν ακόμη πιο εύκολα(αν κ θα αρκούσαν 2σημεία). :lol:
Είναι όντως κάπως δύσκολη η κατασκευή της παραπάνω έλλειψης. :wink:

Όχι...
εννοούσα ότι μπορούμε να βρούμε τη δεύτερη εστία στην τομή δύο υπερβολών

Έστω $E,F$ οι δύο εστίες και $A,B,C$ τα σημεία. Τότε:

$EA+FA = EB+FB \Rightarrow  EA-EB = FB-FA$.

Επομένως κατασκευάζουμε την υπερβολή που έχει εστίες τα $A,B$ και περνάει από το $E$.
Τότε το $F$ θα ανήκει στον άλλο κλάδο της ίδιας υπερβολής.

Ομοίως κατασκευάζουμε μία ακόμη υπερβολή με εστίες τα $A,C$ που περνάει από το $E$

Και το $F$ βρίσκεται στην τομή των υπερβολών (στον κλάδο πάντα που δεν περιέχει το $E$)


I wish they made it to be more of a competition rather than a marathon.

I did. I didn't get that many points though. Some people had over 100,000!!! :o I wonder how they got so much time. I was also kind of disappointed that it was more of who spent the most time on Mathletics than actual competition.

I also did it, but didn't enjoy it very much.  I only had aound 3000 points.  Seriously, I thought it was going to be like FTW instead of really basic stuff.  And I also didn't like that it was more of a marathon than a countdown-style competition.


logarithms
linear algebra
matrix
vector
limit
real analysis
real analysis unsolved

Let $a>0$ and $\left( x_n \right)_{n \geq 1}$ be a sequence defined by $x_1=a$ and \[ x_{n+1} = \frac{x_1}{n+1} + \frac{x_2}{n+2} + \ldots + \frac{x_n}{n+n}, \, n \geqslant 1 . \,  \]

Prove that $\left( x_n \right)_{n \geq 1}$ is convergent and that its limit is $0$.

[i]Problem 98 of the Romanian MO Shortlist 2005, 11th grade[/i]

Denote by $a_n=\max\{x_1,x_2,\dots,x_n\}$. We have that $a_{n+1}=\max(a_n,x_{n+1})$. From the given relation, we get that $x_{n+1}\le \frac{a_n} {n+1}+...+\frac{a_n}{2n}\le (n+1)\frac{a_n}{n+1}<a_n$. Hence $a_{n+1}=a_n$. Hence, $x_n$ is bounded. Let $M>|x_n|,\forall n$.

If $\epsilon>0$ we have that $x_{n+1}\le \frac{M}{n+1}+\frac{M}{n+2}+...+\frac{M}{2n}\le M \ln 2+\epsilon$,for all $n\ge 1$. Doing the same thing we get that $x_n\le M(\ln 2)^m+\epsilon$ for all $n\ge N_m$ and $\epsilon$,  hence $x_n\to 0$.

[quote="xirti"]$x_{n+1}\le (\frac{M}{n+1}+\frac{M}{n+2}+\frac{M}{2n}\le M \ln 2$,for all $n\ge 1$. Doing the same thing we get that $x_n\le M(\ln 2)^m$ for all $n\ge m$, hence $x_n\to 0$.[/quote]

Could you do this explicitely? I'm trying to do "the same thing", but I'm having problems with some of the terms... :blush:

The claim is actually as i edited it now. And to prove it we have:

\[ x_{n+1}\le M (\frac 1{n+1}+..+\frac 1{n+N_m})+(M (\ln 2)^m+\epsilon) (\frac 1 {n+N_m+1}+...+\frac 1 2m )\le \]
\[ \frac{MN_m}{n}+ \epsilon \ln 2+ M(\ln 2)^{m+1}\le M (\ln 2)^{m+1}+\epsilon \] for $n>N_{m+1}$.

Sorry but.... I don`t quite understand  :blush:. Could you pretty please explain it again?

I finally understood : $x_{n+1}<M\ln2+\epsilon$ implies $x_{n+1}<(M\ln2+\epsilon)\ln2<M(\ln2)^2+\epsilon$ and so forth. Sorry for my nagging.

One can prove that $x_n > x_{n+1}$ by writing the recurrence relation for $n,n+1$. Next, let $\ell$ be the limit.

Suppose it's $\neq 0$. Take $\epsilon$ such that $\ell > ( \ell + \epsilon) \ln 2$. Then for big enough $k$ we have $x_k < \ell + \epsilon$. (say $k \geq k_0$)

As xirti, we divide the sum which appears in the recurrence formula in two: one before $k_0$ and another after $k_0$.

Take the limit of this. In the LHS we get $\ell$.

The first part of the RHS $\to 0$, while the second part is smaller than smth that tends $( \ell + \epsilon ) \ln 2$. Hence, the contradiction we wanted.

I hope it is OK. xirti's tricks seem "unfindable" to me and I was trying to come up with something which looks more reasonable to me. (based on his solution)

[quote="perfect_radio"]One can prove that $x_n > x_{n+1}$ by writing the recurrence relation for $n,n+1$. Next, let $\ell$ be the limit.
[/quote]
 :idea:
If you can prove this then the sequence clearly has limit so if you apply Toeplitz you get $\ell=\ell \ln2$ from where $\ell=0$.

I didin't noticed that the sequence in decreasing. :D that's why I introduced anothr sequence $a_n$ to prove it's boundness.

\[ \displaystyle x_{n+1} > x_{n+2}  \]
\[ \displaystyle \Leftrightarrow x_{n+1} > \sum_{k=1}^n \frac{x_k}{n+k+1} + \frac{x_{n+1}}{2n+2}  \]
\[ \displaystyle \Leftrightarrow \frac{2n+1}{2n+2} \cdot \sum_{k=1}^n \frac{x_k}{n+k} > \sum_{k=1}^n \frac{x_k}{n+k+1}  \]

But $\displaystyle \frac{2n+1}{2n+2} \cdot \frac{1}{n+k} > \frac{1}{n+k+1}$ $\, \Longleftrightarrow \, (2n+1)(n+k+1) > (2n+2)(n+k)$ $\, \Longleftrightarrow \, 2n^2 + 2(k+1) n + n + k + 1 > 2n^2 + 2nk + 2n +2k$ $\, \Longleftrightarrow \, n + 1 > k$, so $x_{n+1} > x_{n+2}$ holds for all $n \geq 1$. (also, we trivially have $x_1 > x_2$)

How do you use Toeplitz theorem to conclude what you said?

Is this theorem considered as "well-known" in the Romanian Math Olympiad?

PS: My proof works, or not?

Suppose you have sequence $x_n$ and an lower triangular matrix $A$ of infinite dimension,we will use the notation $V_n$ for a column vector which contains the first $n$ terms of the sequence $x_n$:
If :
1)${\sum_{k=1}^n a_ {nk}=1}$
2)${\lim_{n\to\infty} a_{nk}=0, \forall k\in N}$
3)${\exists \lim_{n\to\infty} x_n=\ell}$
Toeplitz theroem states that $y_n$ where $y_n$ is the sequence of terms in column vector $V_nA$ converges to $\ell$.

In other words if $\left(\begin{array}{c} x_1\\x_2\\{...}\\x_n \end{array} \right ) \left (\begin{array}{ccccc} a_{11}&0&0&0&{...}\\a_{21}&a_{22}&0&0&{...}\\{...}&{...}&{...}&{...}&{...}\\a_{n1}&{...}&a_{nn}&0&{...}\\{...}&{...}&{...}&{...}&{...} \end{array}\right )=\left(\begin{array}{c} y_1\\y_2\\{...}\\y_n \end{array} \right )$ and $x_n$ converges to $\ell$ then $y_n$ converges to $\ell$.

The first condition can be improved : if $\lim_{n\to\infty}{\sum_{k=1}^n a_ {nk}}=\alpha \leq 1$ then $y_n\to \alpha\cdot \ell$.

In our context $a_{nk}=\frac{1}{n+k}$ .So $y_n$ converges to $\ell \ln 2$ but $y_n=x_{n+1}$ so $\ell=\ell \ln 2$ so $\ell=0$ q.e.d.


What is the maximum number of 3 x 4 rectangles that will fit, without overlap,
within a 20 x 20 square?

For some reason, I keep getting 32, but the answer is 33.  :wallbash_red:

Thank you!

The area of the square is $ 20(20)\equal{}400$. The area of each rectangle is $ 3(4)\equal{}12$. Hence, $ \frac{400}{12}\equal{}33\frac13$.

33 is the answer, but for future problems like this, do you just divide the area of the big shape by the area of the little shape?

For figures like squares and rectangles, yes. (This may generalize, but I'm not completely sure).

NO.  The solution is incorrect, though the answer is correct.

For example, how many $ 30 \times 1$ rectangles will fit in a $ 20 \times 20$ square?  It's not $ \left \lfloor \frac{400}{30} \right \rfloor$...:wink:

You can fit four 12x8 rectangles in 20x20 and left a 4x4 rectangle in the center.
This gives you 4*8+1 = 33  3x4 rectangles

So do you just guess and check until you find one that might work?  :maybe:


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that the following inequality holds
${ \frac{\sqrt{bc}}{a^2+bc}}+\frac{\sqrt{ca}}{b^2+ca}+\frac{\sqrt{ab}}{c^2+ab} \ge \frac{9}{2}\cdot \frac{a^3+b^3+c^3+10abc}{(a+b+c)^4}.$
When do we have equality? :)[/quote]

Using the AM-GM inequality,we have
\[ 6ab(a^2-ab+b^2)^2 \leq \left[\dfrac{2(a^2-ab+b^2)+6ab}{3}\right]^3=\dfrac{8(a+b)^6}{27}, \]
or
\[ \sqrt{ab}(a^2-ab+b^2) \leq \dfrac{2(a+b)^3}{9}. \]
We obtain
\[\begin{aligned} \dfrac{\sqrt{ab}}{ab+c^2}=\dfrac{\sqrt{ab}(a+b)^3}{(a+b)^3(ab+c^2)} &\geq \dfrac{9}{2}\cdot \dfrac{\sqrt{ab}.\sqrt{ab}(a^2-ab+b^2)}{(a+b)^3(ab+c^2)} \\ &=\dfrac{9}{2}\cdot \dfrac{ab(a^3+b^3)}{(a+b)^4(ab+c^2)} \end{aligned}.\]
Therefore,it suffices to show that
\[ \sum \dfrac{ab(a^3+b^3)}{(a+b)^4(ab+c^2)} \geq \dfrac{a^3+b^3+c^3+10abc}{(a+b+c)^4}, \]
or
\[ \sum \dfrac{ab(a^3+b^3)(a+b+c)^4}{(a+b)^4(ab+c^2)} \geq a^3+b^3+c^3+10abc. \]
Since
\[\begin{aligned} (a+b+c)^4 &=(a+b)^4+4(a+b)^3c+6(a+b)^2c^2+4(a+b)c^3+c^4\\ &\geq (a+b)^4+2c(a+b)\left[2(a+b)^2+3c(a+b)+2c^2\right]\end{aligned}\] ,we have
\[\begin{aligned} \dfrac{ab(a^3+b^3)(a+b+c)^4}{(a+b)^4(ab+c^2)}  &\geq \dfrac{ab(a^3+b^3)}{ab+c^2}+\dfrac{2abc(a^3+b^3)\left[2(a+b)^2+3c(a+b)+2c^2\right]}{(a+b)^3(ab+c^2)} \\ & \geq\dfrac{ab(a^3+b^3)}{ab+c^2}+\dfrac{abc\left[2c^2+3c(a+b)+2(a+b)^2 \right]}{2(ab+c^2)}\\ & =\dfrac{ab(a^3+b^3)}{ab+c^2}+abc\left[1+\dfrac{3c(a+b)}{2(ab+c^2)}+\dfrac{a^2+ab+b^2}{ab+c^2}\right].\end{aligned}\]
The proof completed if we have
$ 1) \ \ \ \dfrac{ab(a^3+b^3)}{ab+c^2}+\dfrac{bc(b^3+c^3)}{bc+a^2}+\dfrac{ca(c^3+a^3)}{ca+b^2} \geq a^3+b^3+c^3. $
$ 2) \ \ \ \dfrac{c(a+b)}{ab+c^2}+\dfrac{a(b+c)}{bc+a^2}+\dfrac{b(c+a)}{ca+b^2} \geq 2. $
$ 3) \ \ \ \dfrac{a^2+ab+b^2}{ab+c^2}+\dfrac{b^2+bc+c^2}{bc+a^2}+\dfrac{c^2+ca+a^2}{ca+b^2} \geq \dfrac{9}{2}>4. $

Which are the well know inequalities.
Equality holds if and only if $ (a,b,c)=\left(\dfrac{7+3\sqrt{5}}{2}t,t,0\right) $ and any cyclic permutations.


The triangle vertices $(0,0), (0,1), (2,0)$ is repeatedly reflected in the three lines $AB, BC, CA$ where $A=(0,0), B=(3,0), C =(0,3)$. Show that one of the images has vertices $(24,36), (24,37),(26,36)$.


Given integers m and n such that 1<=m<n, in terms of m and n, determine the number of solutions of -1-x^m+x^n=0.

There are $ n$ solutions to any polynomial of degree $ n$, no matter what the other coefficients are. So, if that was the problem given, the answer would simply be $ m$.

That aside, I'd like to think that the problem you posted was not the actual problem. There's a reasonable chance you omitted one or both of the following two restrictions:

i. The number of [i]distinct[/i] roots
ii. The number of [i]real[/i] roots

[hide="Approaching restriction 1"]
This will take a bit of calculus.

Consider $ P(x)=(x-a)^{k}\prod_{j=1}^{n-k}(x-r_{j}), \; k > 1$

Then, $ P'(x) = k(x-a)^{k-1}\prod_{j=1}^{n-k}(x-r_{j})+(x-a)^{k}\frac{d}{dx}\left( \prod_{j=1}^{n-k}(x-r_{j}) \right)$
$ \implies P'(a) = 0+0=0$

Let a repeated root be $ a$. Then,
$ a^{m}=a^{n}+1$
$ ma^{m}=na^{n}\implies a^{m}= \frac{n}{m}a^{n}$
$ a^{n}+1 = \frac{n}{m}a^{n}$
$ a = \left( \frac{m}{n-m}\right)^{\frac{1}{n}}$

We can plug this back into our original equation. Let $ m=n+k$.

$ \frac{m}{-k}\left( \left( \frac{m}{-k}\right)^{\frac{1}{k}}-1\right)= 1$
$ \frac{m}{-k}=\frac{n^{k}}{m^{k}}$
$ m^{k+1}+k n^{k}=0$

This has no solution for positive $ m,n,k$, so the roots must be distinct.

[b]Edit:[/b] Actually it's a bit more complicated than that, the expression for $ a$ should have a roots of unity term in it. But it comes out to the same final result.
[/hide]

[u]Approach for Restriction 2:[/u]

If the roots must be real, then it's trivial by inspection. You need to consider only four cases: both parities of $ m$ combined with both parities of $ n$, paying particular attention to which terms are increasing and decreasing when.

Oh shoot, I forgot to indicate that the solutions must be [b]positive reals[/b]...lol though not neccessarily distinct


[quote="Nerevar"]given that $ab + bc + ca \geq3\sqrt{abc}, a,b,c >0$

prove $2(ab+bc+ca)+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq9.$[/quote]
Let $\sqrt{\frac{ab}{c}}=z,$ $\sqrt{\frac{ac}{b}}=y$ and $\sqrt{\frac{bc}{a}}=x.$
Hence, $c=xy,$ $b=xz,$ $a=yz$ and  $2(ab+bc+ca)+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq9\Leftrightarrow$
$\Leftrightarrow2xyz(x+y+z)+\frac{1}{x^2y^2}+\frac{1}{x^2z^2}+\frac{1}{y^2z^2}\geq9\Leftrightarrow$
$\Leftrightarrow x^2+y^2+z^2+2x^3y^3z^3(x+y+z)\geq9x^2y^2z^2,$ where $x+y+z\geq3.$
But $x^2+2x^4y^3z^3\geq3\sqrt[3]{x^{10}y^6z^6}=3x^2y^2z^2\cdot x^{\frac{4}{3}}.$
$f(x)=x^{\frac{4}{3}}$ is convex function.
Hence, $\sum_{cyc}(x^2+2x^4y^3z^3)\geq3x^2y^2z^2\sum_{cyc}x^{\frac{4}{3}}\geq$
$\geq9x^2y^2z^2\left(\frac{x+y+z}{3}\right)^{\frac{4}{3}}\geq9x^2y^2z^2.$ :)


calculus derivative function algebra domain limit integration

Find [color=red]all functions  [/color]$f$  that have domain the interval $(0 , +\infty)$  , have derivative for all $x$ in this interval  and
     $f(x) = mf(mx) , \forall x > 0$ for some positive real  $m$ (clearly $m$ is different from $1$ .

 [u]Babis[/u]

[quote="DiaDi"]take $f(x)=\frac{1}{mx}.$
$mf(mx)=m*\frac{1}{{m^2}x}=\frac{1}{mx}=f(x), \forall x \in (0, \infty).$
$f(x)$ has a derivative on $(0, \infty)$ equal to $f'(x)=-\frac{1}{mx^2}.$

PS:this is a simple observation of the result :)[/quote]

In fact all $f = \frac{1}{cx}$ with $c$ not $0$ work.

uhm...yes...he said for a given positive real m so whichever you choose m real and positive to be, it works. :)

For $m = 1$ any function which is derivable on that interval works. So I guess we should assume that $m$ is not $1$

Sorry if I was not very specific. I wrote << find the function...>> meaning << [color=red]find all functions [/color]....>>.I did not write << find[color=red] a [/color]function...>> , so I thought that I would not be misunderstood.

Now , I think we must try much more.( [i]I have not the solution, too.[/i])

   [u]Babis[/u]

In English, both [i]a[/i] and [i]the[/i] solution indicate you mean only one solution. If you said [i]the solutions[/i] then it would be clear you meant all of them.

Meanwhile, do you mean for particular $m$ or all $m$? As RaduSorici pointed out, there can be more or less functions depending on your choice of $m$.

Well, really thanks for  clearing to me this notation.
 The problems asks :

 << For a given $m > 0$ , find all functions $f$ with the given property >>. I have observed that any function of the form $f(x) = \frac{k}{x}$ , where k is a non zero constant , works.

 [u]Babis[/u]

[quote="DiaDi"]sorry...I'm tired and I'm making stupid mistakes :blush: . the solution goes like this :

we can observe[as you said] that $f(x)=\frac{k}{x}$ is a solution to the problem, $\forall k \in \Re$.
let us now prove that this is the unique solution. suppose $f(x)=\frac{k}{x}*\varphi(x)$ then, from the condition $f(x)=mf(mx)$ we get $\varphi(x)=\varphi(mx),$ for k not 0, which conducts us to $\varphi(x)=constant$ hence to the only solution $f(x)=\frac{k}{x}$. :)[/quote]

Can you explain how you made the conclusion that the second function is constant. I have been thinking about it for a while now and it does not seem that clear to me.

Yeah, I think $\varphi(x)$ should be periodic with period $m$.

Xevarion,$\varphi$ would have been periodic if $\varphi(x)=\varphi(x+m)$. try to see that $\varphi(x)=\varphi(xm^n)$ for n integer. then calculate the limit for $n \rightarrow \infty$ and $n \rightarrow -\infty$. :)

Dear friends , good work! But ,  can we define the function $\phi$  at $x_0 = 0$ and then prove that it is continuous at $x_0 = 0$  ?If yes  , then the problem becomes easier and $\phi$ is equal to $\phi (0)$ for all $x \geq 0$.(using the limit of a suitable sequence)

  Babis

since we've established that the only solution to the problem is $f(x)=\frac{k}{x}$ where k is a real constant different of 0 and $\lim_{x \searrow 0} f(x)=\infty$, (or $-\infty$ if k is negative) then for f to be continuous it would be necessary to have $f(0)=\infty$, (or again $-\infty$) impossible.

but then again you could take $x_0=1$ and two restrictions of f on $(0,1)$ and $[1,\infty)$ and work on these two restrictions with appropriate sequences. :)

Mmm, this  seem to me better,  but how could I work for $x_0 = 1$ ? This point seems to be better.

[color=red]The initial problem:[/color]

 << Find all functions f with derivative on $(0 , +\infty )$  and

           $\int_{ax}^{bx} f(t)dt = \int_a^b f(t)dt$

where $a , b$  are not equal  possitive real numbers >>.

 Babis

For the original problem, I think there are many functions besides $f(x) = \frac{k}{x}$ that work.  For example, when $m = e^{2\pi}$, the following function works:
\[ f(x) = \frac{\cos(\ln x)}{x} \, .  \]

[quote="DiaDi"]$\varphi$ would have been periodic if $\varphi(x)=\varphi(x+m)$. try to see that $\varphi(x)=\varphi(xm^n)$ for n integer. then calculate the limit for $n \rightarrow \infty$ and $n \rightarrow -\infty$. :)[/quote]

got my mistake :blush: : we don't know that $\lim_{n \to \infty}{\varphi(xm^n)}$ exists. that's why I skippied other solutios... :oops:

[quote="DiaDi"] we get $\varphi(x)=\varphi(mx),$ for k not 0, which conducts us to $\varphi(x)=constant$[/quote]
Take any function $g$ with period $\ln m$ and let $\varphi (x)=g(\ln x)$. Then $\varphi (mx)=g(\ln(mx))=g(\ln x +\ln m)=g(\ln x)=\varphi(x)$.

[b]TO RAVI[/b]

  Have you  made the calculations ? I'll try again but the first time I tried ,  the funtion you gave does not satisfy the integral equation(the integral is easy to find). Of course I'll see it again after , two min.

 [b]TO BOGDAN[/b]

  I know that in Romany you like this kind of problems. Any help is welcome!

  babis

My example was for your original question in Post #1, not for your integral question.

Thank you Ravi B.
 Yes , it really works and it is a very good example! This makes the problem  much complicated  but more interesting. I'm afraid , we ''ll have trouble with it !!!
    But some friend told me he saw the problem proposed in a greek newspaper(in the preparation section for our panhellenic school exams for the university).I'll be waiting to see their solution , if of course they give one!

[u]Babis[/u]

I think the problem is already solved :)

Yes, now I understood your hint.

But is  any function   $\varphi$ with $\varphi(x) = \varphi(mx)$ of the form   $\varphi(x) = g(lnx)$  , where  $g$ has period $\ln m$ ?

Yeah, it has to be. You can always write $\varphi(x) = g(\ln(x))$, and you need $g(\ln(mx)) = g(\ln(m) + \ln(x)) = g(\ln(x))$, so $g$ having period $\ln(m)$ is necessary and sufficient.

Very well .

   I think we have nothing more to do now and the problem is solved.Thanks to you and  all of you , who

 helped !

  [u] Babis[/u]


find the sum of the greatest odd positive divisor of all natural numbers from $ 1$ to $ 2^n$

Similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288498]this[/url].

[hide="solution"] Define $ d(n)$ as the largest odd divisor of $ n$.

Then, we claim $ d(n) \plus{} d(n \plus{} 1) \plus{} ... \plus{} d(2n \minus{} 1) \equal{} n^2$.

[hide="proof"] We proceed by induction. For $ n \equal{} 1$, we have $ d(1) \equal{} 1 \equal{} 1^2$, as desired.

Assume it is true for $ n$. For $ n \plus{} 1$, we have $ d(n \plus{} 1) \plus{} ... \plus{} d(2n \plus{} 1) \equal{} n^2 \minus{} d(n) \plus{} d(2n) \plus{} d(2n \plus{} 1)$.
Clearly $ d(n) \equal{} d(2n)$. Then, as $ 2n \plus{} 1$ is odd, $ d(2n \plus{} 1) \equal{} 2n \plus{} 1$, so $ d(n \plus{} 1) \plus{} d(n \plus{} 2) \plus{} ... \plus{} d(2n \plus{} 1) \equal{} n^2 \plus{} 2n \plus{} 1 \equal{} (n \plus{} 1)^2$, as desired.[/hide]

Now, $ d(1) \plus{} d(2) \plus{} ... \plus{} d(2^n)$

$ \equal{} (d(1)) \plus{} (d(2) \plus{} d(3)) \plus{} (d(4) \plus{} ...d(7)) \plus{} ... \plus{} (d(2^{n \minus{} 1}) \plus{} ...d(2^{n} \minus{} 1)) \plus{} d(2^{n})$

$ \equal{} 1^2 \plus{} 2^2 \plus{} 4^2 \plus{} 8^2 \plus{} ... \plus{} (2^{n \minus{} 1})^2 \plus{} 1$

$ \equal{} 2^0 \plus{} 2^2 \plus{} 2^4 \plus{} 2^6 \plus{} ... \plus{} 2^{2n \minus{} 2} \plus{} 1$

$ \frac {2^{2n} \minus{} 1}{3} \plus{} 1$

If you wanted, you could verify that by induction as well.[/hide]

Hi [b]xpmath[/b], can you please tell me how you know $ d(n)\plus{}d(n\plus{}1)\plus{}...\plus{}d(2n\minus{}1) \equal{} n^{2}$ when you were working on this problem?

Do you mean how I thought of it? If so, I suppose it was just experimenting with small values of $ n$ and noticing that the sum was $ n^2$. This problem involved powers of 2, so it was reasonable to experiment with powers of 2s. In the problem I linked to, the number 503 seemed like a strange number to start with, and the only interesting thing about it that I could think of was that it was approximately a fourth of 2009, so I just tried random things.

I guess I don't really have a good reason  :oops:

Y'a, know...

First take the odd numbers.
\[ 1 \plus{} 3 \plus{} 5 \plus{} \dots \plus{} (2^n \minus{} 1) \equal{} 2^{2(n \minus{} 1)} \equal{} 4^{n \minus{} 1}.\]
Then, for the remaining numbers, divide by 2. We have the same list as before but only half the size, with the largest term being $ 2^{n \minus{} 1}$. Take the odd numbers again:
\[ 1 \plus{} 3 \plus{} 5 \plus{} \dots \plus{} (2^{n \minus{} 1} \minus{} 1) \equal{} 2^{2(n \minus{} 2)} \equal{} 4^{n \minus{} 2}.\]
Then we repeat until we have $ 4^{n \minus{} 3},4^{n \minus{} 4},\dots,4^1,4^0$. However, this $ 4^0$ was the result of taking the odd numbers from $ \{1,2\}$, so when we do this step one more time, we have an additional $ 1$ that we must add to the sum.

All but the last term is a geometric series which has common ratio $ 4$, and so the sum is $ 1 \plus{} \sum_{i \equal{} 0}^{n \minus{} 1} 4^i \equal{} \frac {4^n \plus{} 2}3$, which, of course, is the same answer that xpmath had.

[hide="Another perspective on the lemma xpmath had"]The set $ \{d(n), d(n \plus{} 1), \ldots d(2n \minus{} 1) \}$ is equal to the set $ \{1,3,5,\ldots 2n \minus{} 1\}$ and the lemma follows directly.

Here is some intuition behind this that should make the proof of the above claim fairly obvious. We're looking at the interval of integers $ I \equal{} [n,2n \minus{} 1]$. The reason that $ I$ is special is that for $ k \leq 2n \minus{} 1$, the sequence
$ k, 2k, 4k, 8k \ldots$
has exactly one integer in $ I$. The interval is just wide enough so doubling a number can't jump past it, but just small enough so you can't stay inside when you double.

So it follows there's one power of 2, one power of 2 times 3...[/hide]


Well, this is it. Probably the last time I post on here before leaving for Detroit tomorrow morning. In two short days, we will know the winners. We will know the rankings.
I wish all of you the best of luck, and I will see you in Detroit!

Yay!
I'm going to see the Penguin! House
and set the penguins free.

You guys all miss school for one day, right?

I miss two... I had the impression that everyone missed at least two (the farther the more). Unless you're flying to Detroit on the morning of competition, you're probably missing two as well.

Nats is always Thursday, Friday, Saturday, Sunday I think.  Good luck to everyone (who has already departed :P)!  Illinois team better win or else :D

[quote="math92"][quote="alc"][quote="Phelpedo"]I miss two... I had the impression that everyone missed at least two (the farther the more). [/quote]

Hmm... How about people living in Detroit?[/quote]never thought about that.  I guess they don't miss school.  stinks for them.[/quote]

Don't they still miss Thursday and Friday? Or is there something I don't see?

Not necessarily. Because the testing starts so early, they would probably stay in the hotel overnight, but they don't have to get there until 6PM Thursday so they don't have to miss school that day.

But if they had a choice, they would probably miss it, and they would probably have a choice.

Not if their parents would have any say in it.

Were there any kids from Detroit this year anywasys?


$f(x)=\sum_{k=0}^{\infty} a_k cos(kx) + b_k sin(kx)$ a trigonometric series.
$E=f^{-1}(0) \cap [0;2 \pi]$ is the set where $f$ vanishes.

1/ if $E=[0;2 \pi]$, does it follow that $a_k=b_k=0$ for all $k$
2/ if $E=[0;2 \pi]-F$, where $F$ is a finite set, does it follow that $a_k=b_k=0$ for all $k$

Salut Alek,

Jette un coup d'oeil a:

Kahane/Salem, Ensembles parfaits et series trigonometriques, Hermann (193, Chap. 5)

@+

justement, c est a partir de ce bouquin que mon ami a tire ce pb ...


geometry
parallelogram
circumcircle
geometric transformation
reflection
angle bisector
perpendicular bisector

As Shown in the picture Below:

In parallelgoram $ABCD$ , $BE=BC$ and $AF$ is angle bisector of $\angle DAB$.
Prove that: $FC\bot BD$

It is confusedly. I didn't understand this enunciation. See again !

Let $ABCD$ be a parallelogram and pick a point $E$ on the ray $AB$ such that $BE = BC$ (where $B$ lies in between $A$ and $E$). Take a point $F$ on the angle bisector of $\angle{DAB}$ such that $EF$ is perpendicular to $AB$. Show that $FC$ is perpendicular to $BD$.

[i] Cute problem for beginning![/i]

If $AF$ meets $CD$ at point  $Z$ , then $DZ = DA = BC = BE$, so $DZEB$is parallelogram, that is $ZE // BD$.But $EC \perp AF$ , since $\angle AEC + \angle A / 2= \angle ABC / 2 + \angle A /2 = 90$. Obviously , $ZC \perp EF$ , so $C$ is orthocenter in $\triangle FZE$. This means that  $FC \perp EZ$ , or  $FC \perp DB$

        [u]Babis[/u]

Excellent Stergiu Because that was my solution. :D
Another Sol that I know is using Pythagorean

easy one .
let $AB = DC=a, AD = BC=b$. we choose the point $G$ on line $AD$ such that $AG=AE$. with Pythagorean if we prove that :
$BC^2-DC^ = BF^2-DF^2$, problem will solve . now we have :
$BC^2-DC^2=b^2-a^2$ and$BF^2=BE^2+EF^2=b^2+EF^2$ and $DF^2=DG^2+FG^2=a^2+FG^2$ (because $AGF~AEF$ is a right triangle).and we have $FE=FG$ so $BF^2-DF^2= b^2-a^2=BC^2-DC^2$ and problem solved.

nice solution Behzad.you solved it nice and clear.(i think Karno's theorem is one of the best.) :D

A solution I am just posting because of the triangle geometry connection which is not that clear from the other proofs:

First we rewrite the problem:

[color=blue][b]Problem'.[/b] Let ABCD be a parallelogram, and E a point on the extension of the segment AB beyound B such that BE = BC. Then, the perpendicular to BD through C, the perpendicular to AB through E, and the angle bisector of the angle DAB concur at one point.[/color]

Now we rename points en masse:

[color=blue][b]Problem''.[/b] Let AZXY be a parallelogram, and E a point on the extension of the segment AZ beyound Z such that ZE = ZX. Then, the perpendicular to YZ through X, the perpendicular to AZ through E, and the angle bisector of the angle YAZ concur at one point.[/color]

Now, let's solve this problem: Consider everything from the viewpoint of triangle XYZ. Since AZXY is a parallelogram, the point A is the point of intersection of the parallel to ZX through Y with the parallel to XY through Z. Let B be the point of intersection of the parallel to XY through Z with the parallel to YZ through X, and let C be the point of intersection of the parallel to YZ through X with the parallel to ZX through Y.

Sice BX || YZ and BZ || XY, the quadrilateral BXYZ is a parallelogram, so that BX = ZY. Similarly, the quadrilateral CYZX is a parallelogram, yielding XC = ZY. Hence, BX = XC, and thus the point X is the midpoint of the segment BC.

Let the angle bisector of the angle CAB meet the circumcircle of triangle ABC at a point R (apart from A). Then, the point R lies on the angle bisector of the angle YAZ (in fact, we know that R lies on the angle bisector of the angle CAB, but this is the same as the angle bisector of the angle YAZ). Also, since < BAR = < CAR, the chords BR and CR in the circumcircle of triangle ABC have equal chordal angles; therefore, these chords must be equal in length, so we have BR = CR. Thus, the point R lies on the perpendicular bisector of the segment BC. In other words, $RX\perp BC$ (since X is the midpoint of the segment BC). Since BC || YZ, this becomes $RX\perp YZ$, so the point R lies on the perpendicular to YZ through X.

In the following, we WLOG assume that the points A, Z, B, E lie on the line AB in this order.

Since $RX\perp BC$, we have < RXB = 90°. Thus, in the right-angled triangle RXB, we have < BRX = 90° - < RBX. Since the point R lies on the circumcircle of triangle ABC, we have < RBC = < RAC. Since R lies on the angle bisector of the angle CAB, we have $\measuredangle RAC=\frac{\measuredangle BAC}{2}$, and CA || ZX yields < BAC = < EZX. But since ZE = ZX, the triangle EZX is isosceles, so that its base angle < ZEX equals $\measuredangle ZEX=90^{\circ}-\frac{\measuredangle EZX}{2}$. Thus,

$\measuredangle BRX=90^{\circ}-\measuredangle RBX=90^{\circ}-\measuredangle RBC=90^{\circ}-\measuredangle RAC=90^{\circ}-\frac{\measuredangle BAC}{2}$
$=90^{\circ}-\frac{\measuredangle EZX}{2}=\measuredangle ZEX=\measuredangle BEX$.

Hence, the points B, X, R, E lie on one circle. Since < RXB = 90°, the segment RB is a diameter of this circle, what yields < REB = 90°. Thus, the point R lies on the perpendicular to AZ through E.

Combining the above, we see that the perpendicular to YZ through X, the perpendicular to AZ through E, and the angle bisector of the angle YAZ concur at one point - namely, at the point R. Problem solved.

  darij

[quote="darij grinberg"]...
$=90^{\circ}-\frac{\measuredangle EZX}{2}=\measuredangle ZEX=\measuredangle BEX$.
...[/quote]

nice solution darij,But I think there is a typo.
$\measuredangle ZEX=\measuredangle BEX$ should be $\measuredangle ZEX=180^{\circ}-\measuredangle BEX$

[quote="Amir.S"]nice solution darij,But I think there is a typo.
$\measuredangle ZEX=\measuredangle BEX$ should be $\measuredangle ZEX=180^{\circ}\measuredangle BEX$[/quote]

Not on my sketch...

[quote="darij grinberg"]In the following, we WLOG assume that the points A, Z, B, E lie on the line AB in this order.[/quote]

Or did I miss another necessary "WLOG" assertion about the arrangement of the points?

  darij

oh
in your sketch: $XY < XZ$ then $\angle BEX=\angle ZEX$

and if  $XY> XZ$ then $\angle BEX=180^{\circ}=\angle ZEX$

[color=darkblue][b]Proof 1 (metric).[/b] Denote : $AB=CD=a$, $AD=BC=BE=b$ ; the point $H\in BD$ for wich $CH\perp BD$ ; the point $F\in CH$ for which $FE\perp AB$ ; the point $I\in AD$ for which $FI\perp AD$ ; $ID=x\ .$ From $CH\perp BD$ results $HB^{2}-HD^{2}=b^{2}-a^{2}\ .$ Apply the Carnot's lemma in the triangle $ABD$ for the points $E\in AB$, $H\in BD$, $I\in DA\ :$ $(EA^{2}-EB^{2})+(HB^{2}-HD^{2})+(ID^{2}-IA^{2})=0$ $\Longleftrightarrow$ $\left[(a+b)^{2}-b^{2}\right]+\left(b^{2}-a^{2}\right)+\left[x^{2}-(x+b)^{2}\right]=0$ $\Longleftrightarrow$
 $x=a$ $\Longleftrightarrow$ $AI=AE$, i.e. $FI=FE$. Therefore, the point $F$ belongs to the bisector of the angle $\widehat{BAD}\ .$

[b]Remark.[/b] So far, in the my opinion, [u]the Stergiu's proof is the best ![/u] For its sake, I"ll reproduce it here :

[b]Proof 2 (synthetic).[/b] Denote : the point $F$ which belongs to the bisector of the angle $\widehat{BAD}$ and for which $FC\perp BD$ ; the intersection $Z\in DC\cap AF$. From the relation $m(\widehat{ZAE})+m(\widehat{CEA})=\frac{1}{2}\cdot \left[m(\widehat{DAB})+m(\widehat{ABC})\right]=90^{\circ}$ obtain $EC\perp ZF$. From $DZ=AD=BC=BE$ $\Longrightarrow$ $DZ=BE$ $\Longrightarrow$ the quadrilateral $BDZE$ is a parallelogram $\Longrightarrow$ $ZE\parallel BD$. From $FC\perp BD$ obtain $FC\perp ZE$, i.e. the point $C$ is the orthocenter of the triangle $EFZ$ $\Longrightarrow$ $ZC\perp FE$. But $ZC\parallel AB$. In conclusion, $FE\perp AB$.
[/color]

Reflect $ E$ over the line $ AF$ to form point $ X$. Then $ \triangle AEF$ and $ \triangle AXF$ are congruent. Thus, since $ AD \equal{} BC \equal{} BE$, $ DX \equal{} AB \equal{} DC$. Let $ BC$ meet $ FX$ at $ Y$ and $ DC$ meet $ EF$ at $ Z$. Let $ K$ be the perpendicular from $ F$ to $ BD$. Then since $ \angle BEF \equal{} \angle FHB \equal{} \angle FYB \equal{} 90$, $ FYKBE$ is cyclic, similarly $ FZKDX$ is cyclic. Also, it is easy to see that quadrilaterals $ XDCY$ and $ BCZE$ are similar, so $ \triangle DCY\sim \triangle BCZ$, so $ BDYZ$ is cyclic. So by the radical axis theorem, the radical axes of $ BDYZ$, $ FYKBE$, and $ FZKDX$, which are $ DZ$, $ BY$, and $ FK$ concur, so $ FK$ passes through $ C$, so $ K\equal{}H$.


A and B represent non-zero digits.  If the base-ten numeral AB divides the base-ten numberal A0B(without remainder, the second number is a three digit number with the middle digit as 0), find the sum of all possible values of AB.

[hide]
This is equivalent to
$ \frac{100A\plus{}B}{10A\plus{}B}\equal{}n$, where n is a positive integer.

$ 100A\plus{}B\equal{}10An\plus{}Bn$

$ 100A\minus{}10An\equal{}Bn\minus{}B$

$ 10A(10\minus{}n)\equal{}B(n\minus{}1)$
We can just plug in values of n from here, with 1<n<10.
We find that if n<6, then the LHS is always larger than the RHS.
When n=6,
$ 40A\equal{}5B\rightarrow$
$ A\equal{}1$ and $ B\equal{}8$.
When n=7,
$ 30A\equal{}6B\rightarrow$
$ A\equal{}1$ and $ B\equal{}5$
No solutions for n=8
When n=9,
$ 10A\equal{}8B\rightarrow$
$ A\equal{}4$ and $ B\equal{}5$
So, the answer is
$ 18\plus{}15\plus{}45\equal{}\boxed{78}$
[/hide]
I'm pretty sure there's a better way, but I haven't thought of one yet.


I uninstalled miktex and put it on another drive, so how I can I get TeXnic Center to know to switch locations?

Build, Define Output Profiles, Wizard (button at bottm left) then tell TexnicCenter where latex.exe is (in /your_pathmiktex/bin).


Σήμερα έχασα μία ώρα για να brute forceάρω την παρακάτω ανισότητα αλλά βρήκα και μία πιο έξυπνη λύση:

$ a,b,c$ θετικοί

$ \sum\frac {a^2 \plus{} bc}{(b \plus{} c)^2}\geq \frac {3}{2}$

[color=red]edit Γραψε ενα \sum ρε Κώστα [/color]

This is just one line  :P
$ \sum\frac {a^2 \plus{} bc}{(b \plus{} c)^2}\geq \frac {3}{2}$ equivalent to

$ \sum\frac{(a\minus{}b)^2(a\plus{}b)(a\plus{}b\plus{}2c)}{(b\plus{}c)^2(c\plus{}a)^2}\geq 0$   qed

Σιλουανέ, είναι μια γραμμή αν κάποιος είναι τόσο καλός όσο εσύ και το δει αυτό. Και στο βιβλίο του κύριου Στεργίου λύνεται με πρόσθεση δύο ανισοτήτων σχετικά γνωστών αλλά δεν είναι εμφανή αυτά στο μέσο μάτι. :D

Εγώ έθεσα $ 2x\equal{}a\plus{}b$ κτλ. και έκανα πράξεις και βγαίνει πάλι σχεδόν αμέσως. :wink:

Αλλά στην αρχή είπα να πειραματιστώ με brute force και την έπαθα... Βγήκε πάντως!!! :o

Η απαλοιφή παρονομαστών, το άνοιγμα των παρενθέσεων και η κατάλληλη σε αναλογία χρήση των ανισοτήτων Schur και Muirhead για τα γιγαντιαία συμμετρικά αθροίσματα που προκύπτουν. :o

Δεν τη συνιστώ διότι είναι χρονοβόρα αλλά σου λύνει σίγουρα το θέμα δεν πα να γίνει σεισμός. :D

Κάτι ελαφρώς άσχετο, αλλά τόσον καιρό είχα την απορία γιατί καμία άσκηση δε την έλυνα με Murhead, ενώ μου είχε αρέσει και τη θυμόμουν καλά. Το έψαξα και βρήκα στο ελληνικό φόρουμ ελλαχιστότατες λύσεις (2 αν θυμάμαι καλά).

Και μόνο χθες ανακάλυψα ότι η Murhead είναι απλά μια εφαρμογή της αναδιάταξης και ότι ως φανατικός χρήστης της τελευταίας, απεδείκνυα κάθε φορά υποσεινήδητα τη Murhead χωρίς να δίνω σημασία στι γεγονός. :o

Τελικά, αυτή είναι η μαγεία των μαθηματικών: ακόμα και αν αγνοείς τη θεωρία, για να λύσεις μια άσκηση που χρειάζεσαι αυτή τη θεωρία, αργά ή γρήγορα αναγκάζεσαι να την αποδείξεις (εκτός και αν βρεις τελικά άλλο τρόπο ή τεχνική). :lol:

Κώστα δεν νομίζω ότι όλες οι ανισότητες λύνοντι με brute force .
Όσο για αυτό που λες maniopa ότι η Muirhead είναι ειδική περίπτωση της αναδιάταξης πιστεύω ότι είναι λάθος διότι θα το είχαν δει τόσοι και τόσοι μέχρι τώρα και δεν θα αποδείκνυαν την Murihead σε καμια μιση-μία σελίδα με χρήση της θεωρίας του Majorazation . Θα την αποδείκνυαν σε τρεις γραμμές που αποδείκνυεται η αναδιάταξη . Εκτός αν έχουν όρεξη για πράξεις .... :P

Έχεις δίκιο. Εκ παραδρομής το είπα. Μερικές είναι σφικτές και η muirhead έχει τη σφικτότητα του AM-GM στο περίπου (όσο πιο μεγάλη είναι οι εκθέτες τόσο πιο σφικτή, όπως και η Schur). :D


So, do you use Limewire or not, personally, I don't because it's unlawful and you could get fined if anyone finds out, like the guy who repairs your computer when it crashes.

Limewire doesn't increase the threat to your computer security.
Unless you count other people able to download anything off the designated folder that you use for the shareware.

Music file that requires a "codec" that is actually a virus is [i]not[/i] a secure thing.  Or the file could be a virus in itself.  I just listen off of Seeqpod.  I don't carry my music absolutely everywhere so uhhhh, yeah.

um i definetely dont use limewire although a lot of my friends do...it just stupid first off i dont have an ipod or mp3 and second you could go to youtube and listen to the songs for free without downloading them or being illegal...now i suspect if you record the songs off of youtube and put them on your ipod or mp3 that would be illegal but just listening to the songs from there is completely safe and thank gosh LEGAL!!!!

Dude Limewire Sucks. It just sucks in music.

Talk about real p2p.

Its free, just make sure your share folder isn't set to default, or anything in your documents folder can and will be downloaded by others.
Like secret information
Like Social security numbers and bank accounts.
But whatever.
I still don't use it because it's ILLEGAL!

i used to..
[quote="MustItAlwaysBeMe]Where's the "1% of us don't want viruses on our computers" option?[/quote]
exactly my point.. :D

[quote="undefined117"]Any BitTorrent users out there?

Azureus.rules.[/quote]

I use Opera for BitTorrent downloads...

...now there's some [i]real[/i] P2P.

Torrents are technically illegal too...
Well it's not illegal to dl torrents, but the media on them is usually illegal...

Who said I downloaded illeagal media.  Like I said, I use Seeqpod(and Youtube) for music and video; I don't download it.

for listening, (w/o computer) i use the radio
for listening with computer, i use music.aol.com hehe

and i hardly view videos.. sometimes i go on YouTube.. :D

Azeurus is using bloated java stuff.

Use utorrent dude.

yeah I used to use utorrent on my windows machine, but I'm on a mac now.. the best lightweight mac torrent client is transmission, but it's too basic..

I have linux on a crappy old laptop, but linux on a macbook? nah :D

Vista is essentially the same as XP, but with a good graphics framework, and worse GUI :D .

If anyone was wondering: I use Opera because,
[list][*]My computer is terrible and I can't run more than 3 apps at once.
[*]It is very customizable and you can put bookmarklets nearly anywhere.
[*]You can use either your mouse or keyboard, but you don't have to switch back and forth continuously(mouse gestures plus MANY, MANY keyboard shortcuts, which can be customized in any way you can think of)
[*]No need for extensions, though I really wish there was firebug for Opera :) .[/list]

So there.

I love limewire, though my sister got a hefty fine (1000 buckaroos) for downloading 500 songs in one day. My MP3 is about 99% limewire downloads and i have over 400 songs. The only reason I have any legal songs is because my friend gave me an itunes gift card and I said I minus well use it.

I don't use limewire but I do burn CDs I get from my library onto my iPod/MP3  :P

Its those using limewire who get caught.

[quote="undefined117"]Any BitTorrent users out there?

Azureus.rules.[/quote]

Depends on what kind of torrents to be illegal right? I only do Dattebyo (anime fan subber)
I use microtorrent (utorrent)

They don't care if you downloading a few songs. If this were the case all the kids in my school would have payed a couple hundred dollars in fines each. Also the police would need a special unit called, "Catching people who use Limewire unit" and it would need about 100 people per every city with more than 30000 people...

yeah the network itself isn't illegal. It's just illegal when the content is copyrighted.

CNN said there was like this group of "cyperpolice" in China that track these networks and locate p2p users who download illegally..

maybe they should be charged for searching without warrant XD

[quote="#H34N1"]Macs suck.[/quote]

Why is everyone insulting me?!

i love macs!!! i want macs!!!

i have windows...

MY MACBOOK IS BEAUTIFUL

but macs have zero viruses (literally)

mine hasn't crashed yet, and it's already been several months (:

My windows hasn't crashed yet, and I've had it for 6 years.
So yeah.
And Macs cant play any games or anything.
:|

Well people, iMacs use laptop parts. Don't deny that.

I built my desktop.

I have not had a virus for like half a decade.

[quote="ZhangPeijin"]Dude, macs have like, a million bugs.
People only get macs because they look 'pretty'.[/quote]

aha, i'm not alone

[quote="ZhangPeijin"]And Macs cant play any games or anything.
:|[/quote]

This is becoming not true.

http://store.apple.com/1-800-MY-APPLE/WebObjects/AppleStore.woa/wa/RSLID?nnmm=browse&mco=2E630EB&node=home/shop_mac/software/action_games

Guitar Hero 3
Call of Duty 2
Indiana Jones and the Emperor's Tomb
The Sims 2 University
X-Plane 8 Deluxe

and growing.

[quote="math92"][quote="ZhangPeijin"]Windows sucks.
>.>
Especially Vista.[/quote]

True.  Macs > Linux > Windoze  :D

Yes, I use Limewire.  Why pay for it when you can download it for free?[/quote]

Linux is greater than Macs.

Both Mac OSX and Windows(all of them) are based off Linux.  And the fact that you use Limewire says something about you knowledge of this stuff.

And I personally like my Windows 2000 computer from so many years ago better than I would a MacBook Air.  (I just want Quicksilver)

Personally, I've never used Linux before, and my experience of mac is horrible, I mean doing something as simple as typing a document on AppleWorks can get the computer frozen.

[quote="1=2"]Call of Duty 2[/quote]

Whoa, ancient.

And I thought this topic was called "Limewire", not "Macs".