THE GAME IS OVER.

[b][u]Alive Players[/u][/b]
06. kheg_k - inspector
11. Ars [color=white]akalra1[/color] - roleblocker
25. b-flat - rock collector

[b][u]Dead Players[/u][/b]
13. maky - Citizen, killed by the moderator Night 0
16. basurp - Citizen, killed by the moderator Night 0
12. ZAF007 - Citizen, killed by the moderator Night 0
00. Ignite 168 - MODERATOR, killed by the bad guys Night 0
27. Treething - Citizen, stoned by the town Day 1
05. Captain Sugar - Mason, shot to death Night 1
[color=red]14. Should I tell you? - Werewolf Minion, stoned by the town Day 2[/color]
[color=green]23. Peter VDD - Cultist/Deputy, bitten to death, Night 2[/color]
04. rem - Fallen Angel, shot to death, Night 2
[color=red]03. nutz_for2.718281828 - Werewolf King/Mason, choked to death, Night 2
07. dts - Werewolf Roleblocker, stoned by the town Day 3[/color]
[color=violet]10. daine - ???, choked to death, Night 3[/color]
21. jli - Miller Witch, killed by the moderator Day 4
08. ragnarok23 - Forced Vigilante, stoned by the town Day 4
[color=violet]10. daine - ???, revived, Night 4[/color]
[color=green]02. Sly Si - Cult Recruiter, bitten to death, Night 4
15. mad_skillz_aops - Cultist/Fairy, shot to death AND throat slit, Night 4
01. bubka - Cultist/Cult Finder, stoned by the town Day 5[/color]
[color=red]09. perfect628 - Werewolf Minion, stoned by the town Day 6 Morning
22. me@home - Werewolf Minion, stoned by the town Day 6 Afternoon[/color]
10. daine - Supernatural Being, stoned by town Day 7
26. BoesFX - Doctor, shot to death, Night 7
24. Jashper - Citizen, stoned by the town Day 8
[color=blue]20. laughinghead505 - Insomniac Serial Killer, stoned by the town Day 10[/color]
18. junggi - New Role Detector, stoned by the town Day 11
19. pieterminate- Tracker, stoned by the town Day 12
[color=red]17. Klebian - Werewolf Minion, stoned by the town Day 13[/color]

[b][u]Current Time[/u][/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=610305#610305]Day 1[/url] - Page 1
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=615985#615985]Day 2[/url] - Page 6
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=621026#621026]Day 3[/url] - Page 11
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=624414#624414]Day 4[/url] - Page 15
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=629427#629427]Day 5[/url] - Page 25
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=636197#636197]Day 6 Morning[/url] - Page 35
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=636531#636531]Day 6 Afternoon[/url] - Page 36
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=637719#637719]Day 7[/url] - Page 37
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=641651#641651]Day 8[/url] - Page 40
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=643595#643595]Day 9[/url] - Page 41
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=653725#653725]Day 10[/url] - Page 46
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=657264#657264]Day 11[/url] - Page 47
It is now: Game Over

Rules :

1. Do not talk outside of this thread, unless your role PM explicitly says you may do so, and even then, you are restricted to night-time for all out-of-thread conversation.
2. Please do not delete your posts, you may edit posts as much as you want before someone else posts but not after that. This is to avoid confusion that has happened in past games.
3. Do not quote your role PMs. This also applies to night conversations. You may, however, paraphrase your PM at any time but it must be more than a couple of words changed.
4. You may request replacements, clarifications, and votecounts from me directly in the thread.
5. Once you are dead, then you are dead. Do not post any relevant information to the game whatsoever. You are, however, allowed a typical “Bah!” post of some sort.
6. Please direct all questions about your roles to PM, do not post them in thread as it may unbalance the game and hint toward other players your role, if I think it should be made public, then I will make it public.

Game play
1. Each day the town must decide on a lynch. Once a majority has been reached on a player, that player will be lynched, regardless of any subsequent unvoting.
2. To lynch somebody, first they must be accused, seconded, and then you must wait at least 48 hours for them to give a defense. Once a defense is given or 48 hours have passed, you may begin voting aye or nay.
3. Vote retractions are not allowed, however, you may change your vote at anytime.
4. If the game lags, I will place a deadline. This deadline will be flexible depending if activity picks up or not.
5. When a deadline has been placed and no one has been lynched by that deadline, there will be no lynch and the game will automatically go into night.
6. You may vote no-lynch if you wish. A majority voting for no-lynch will end the day, and the game will progress directly into night.
7. The first night will last 5 days or until all submissions are in. Nights after that will last 3 days or until all submissions are in. If you have not sent me a night-choice by that time (should you have one), for the first night of the game, I will randomize your choice. For all nights afterwards, I will consider you as choosing that you do not wish to use your action unless your role says otherwise.
8. If you feel that you may not be able to post in-thread for a period of 72 hours or longer, please say so in-thread or send me a PM as warning. If I find it necessary, I will have you replaced, but I would much rather keep all players who have signed up.

[size=200]PLEASE READ THIS POST, IT IS VERY IMOPRTANT[/size]

In this game, people can change alignments, your confirmed innocent investigator could suddenly be working against the town the next day and nobody knows. More might be revealed on this later.

In this game, bad guys will NOT be allowed to communicate to each other on the first day OR night 0. They will only be allowed to communicate during nights after that.

Also, in this game, no deleting of posts is permitted and editting is only permitted if someone else has not posted first.

In this game, retractions of ayes/nays are not permitted, however, you may change your vote as much as you want. You may retract an accusation unless it has been seconded.

For this game, there are MANY new roles. However, nobody knows the list of roles in this game. Not the bad guys, not the town.

[i]In the peaceful town of Aopsville everybody lived a great life. However.. one day, an evil came to the town. No one saw it coming. All they knew was after one peaceful night's rest, they woke up and they found to their shock and horror, 4 dead bodies.. These were maky, an ordinary citizen, basurp, an ordinary citizen, and ZAF007, an ordinary citizen. However, one more person was dead who was not an ordinary citizen. He was Ignite168. His spirit parted his body but did not leave the town. He now knew everything. He knew what special abilities some towns members have been keeping secret as well as the killers. The one thing he did not know however was how this would all play out. The town was shocked that this could happen to their own peaceful town and they decided they must lynch someone everyday! They knew that was the only way ordinary citizens could fight back against this evil, to lynch this evil and perish it from the town.
However, while the mob of the townspeople was talking, one person found a note on a tree. It said as following:

Dear fellow townsmates,
BoesFX has always been a good citizen. He has never once acted suspicious and has always been on good terms with everybody. We should reward him and everybody should promise not to lynch him today.
-Anonymus

After the note was passed round and round, everybody agreed that BoesFX indeed, was an outstanding citizen. They all decided not to lynch him today. A majority of 12 ayes will be needed to lynch someone.[/i]

In summary,
13. maky - Citizen, killed by the moderator Night 0
16. basurp - Citizen, killed by the moderator Night 0
12. ZAF007 - Citizen, killed by the moderator Night 0
0. Ignite 168, MODERATOR, killed by the bad guys Night 0
BoesFX may NOT be lynched today.
It takes a majority of 12 to lynch.

It is now Day 1. Wake up town!

hmmm I must say that the intro story is a bit weird  :huh:

This game is going to be a bit different than the others (all the roles that are unknown and the side-switching bit).

To answer some questions..

Mafia, Werewolves (if there are any), etc. will not be able to change sides. Only townspeople will.

All roles were done through random.org. I did NOT pick who gets what role, I did it randomly.

Only 1 person had a night action for Night 0 so no, unless I specifically PMed you saying send in something for night 0, you did not have a night action.

I will not tell you about failed kills at night. I will also not say who killed who but the killing methods should remain the same.

There will be one votecount on each page as well as the first post. I will edit the one on the page we're on and the 1st post one only. I will try and make the votecount so it's the first post on the page.

[quote="Ignite168"]To answer some questions..

Mafia, Werewolves (if there are any), etc. will not be able to change sides. Only townspeople will.

All roles were done through random.org. I did NOT pick who gets what role, I did it randomly.

Only 1 person had a night action for Night 0 so no, unless I specifically PMed you saying send in something for night 0, you did not have a night action.

I will not tell you about failed kills at night. I will also not say who killed who but the killing methods should remain the same.[/quote]

Will we know the role of whoever died?

Yes, I will reveal their Role PM and any other PMs they were sent with any important parts like Mason partners or Mafia partners or investigation results cut out.

*snore....* HEY! T'is day! I'm sleepy! *yawn*

Oh well. Sorry, ignite said it was night before.

nutz stop talking

i propose we lynch nutz just to shut him and his annoying comments up, anybody want to second this? (i'm accusing nutz)

How about Treething, in the grand tradition of lynching the first to accuse? :wink:

In case that isn't fully obvious, the above is a joke.

I'm not sure who to accuse yet, but I'm not really tempted to go after nutz.  Hey, at least he's active, which is more than can be said about some participants.  I think I'll wait until it becomes obvious that there are some inactives before I accuse (it's bound to happen in such a large game).

yeah I know and also one thing I figured Klebian was scum after your lynch but dismissed the idea at first at it came back as why is he not dead yet/ cultist and also if not you but someone else was lynched you guys would have won or I would have killed you off if it was no lynch

Bah. I assumed there was a mafia when I called for a death of dts. Smart modding by ignite. Ars was on to me for a while; I would've killed/lynched him but he seemed to be the only person who I could call guilty... And I nearly won, until b-flat whimsically voted me. I HAD that game won...

Maybe I shouldn't bus so much next time I'm guilty.

Well, sly si cleverly put the pushing for a roleclaim in his 2 or so posts that he had to. Ha.

ANALYSIS AND THOUGHTS FROM THE MODERATOR

My goal in this game, was to bring lots of new flavors to AoPS, and let them "try it out" so to say. I was trying to get people away from assuming things, and to get a lot more thinking involved. Moves like to have only werewolves instead of mafia, were subtle, but led people to not assume things.

I was disappointed with the Treething lynch Day 1, because I felt people were not thinking of what mafia would be more likely to do, versus what town would be more likely to do, but rather just voting on baseless reasons. I was surprised with how open the werewolves made their alliance through voting.

Day 2 I felt was rather early for bussing in a large game such as this and the implications of that showed to Day 3. I would've definitely have been suspicious of Klebian for not mentioning that he's a miller to the town, as that's definitely something you want to get out in the open immediately regardless of what side you were on and was disappointed that the town kept letting minor things go throughout the game such as not asking Klebian for his results and actions anymore.

Day 4 I was wondering why ragnarok23 thought his role was so bad. I believe that a Vigilante is the second MOST powerful role next to an Inspector, because it gives the town the power of an extra kill and one of my goals in this game was to show how beneficial a vigilante could be and get people more confident of the kills. Sadly, that did not occur.

Day 5 was just silly and people need to realize it is a game.

Day 6 was very sloppy play on the part of the werewolves, and I'm surprised that they weren't lynched sooner and I'm glad that the town finally got their act together.

Day 7 was just way too silly. People needed to think and realize that in the hands of a guilty, that role would be way too powerful.

Day 8 Jashper didn't really put up a defense and should've.

Day 9 I didn't have that much thoughts on.

Day 10 I felt bad for laughinghead505 cause even with all the breaks, the Inspector did him in in the end. I was surpised he didn't off kheg_k earlier though.

Day 11 was more silliness, as junggi was practically confirmed from his investigations, and people were saying things like "A role like that is useless." junggi's roles was one of the most powerful in this game as any Werewolf Minion claiming to be things like a Citizen or such would get caught and definitely a mistake on the side of the town.

Day 12 was also extremely silly considering pieterminate had virtually confirmed himself and was a powerful role yet the town still lynched him. I'm still wondering how Klebian wasn't lynched Day 11.

Day 13 town finally did what they needed to do.

Day 14 wasn't as funny as I expected it to be. Oh well.

Overall, I felt that town didn't really deserve this win and their night actions saved them. They weren't really thinking that much and should've thought out each action more carefully and constant complaints of that the day was going too long or that this game is too long annoyed me as in all other mafia games on other sites, days go MUCH longer and are much more pages because discussion is beneficial to the town. People should've slowed down and been more careful and thought things out more, if that happened, town would've probably won much earlier with all the luck they received.

Sly Si and laughinghead505 played very well, the latter especially as shown by Sly Si wanting to induct him. The cult played TOO good which led to their downfall in the end as every role wanted to kill them but I'd definitely award the MVP to the non-werewolf/town roles, the cult and the SK.

So if nutz_for had lived, would he have been required to quote Shakespeare in a foreign language in every post for the rest of the game?

Also, until Klebian told me on AIM that he was a werewolf, I was convinced that day 13 was a sham like day 14, and possible earlier days too.

Hey you must not blaim the entire town I had figured it out but could not convince anybody until it became deadly obvious because of all the bad lynches

[quote="Ars"]I even supplied you with the pieterminate lynch (werewolf tracker) so that I would get a wise town[/quote]
I would've attempted to lynch pieterminate regardless of your actions. I used you as an excuse to make the town go with it.

Yeah you really needed me dead I just gave him as he was acting stupid by doubting b flat and I would not be able to convince him to vote you

YAY!!!

So what was the point of those last 2 nights/days??? :huh:
oh well.

 :D  :)  :lol:

[b][i][/i][/b][b][i][/i][/b][b][i][/i][/b][b][i][/i][/b][b][i][/i][/b]

Yay we won.  :lol:   Klebian seemed [i]slightly[/i] suspicious to me when he didn't fully roleclaim at first, but dismissed it until Klebian became bent on lynching me.

Yay!! Go me!! :lol:

There is a slight error, during Night 5 I chose to kill pieterminate.

I should have killed Ars instead of BoesFX, but I wanted to lynch BoesFX because he was annoying me by saving the person I wanted to kill two night in a row. :P

Actually I expect that we are losing, especially when cultist start bringing townies with them to the graves. :D [Note: mad_skill_aopz bring ragnarok while bubka let himself be lynched]

Well, about no mafia in this game is really interesting, with that assumption I think that we are losing when bubka is lynched, I just hope that there are as less mafia as possible. About klebian maybe I did suspect him for being mafia (or SK) but werewolf. Wow. Quite shocking. Nice job for klebian and ignite168 [the latter for breaking the assumption I told above and for the last day]. I should be more careful next time. :P

[quote="Ignite168"]
Day 7 was just way too silly. People needed to think and realize that in the hands of a guilty, that role would be way too powerful.
[/quote]

I don't know why I miss writing the possibility of laughinghead being SK :roll:

[quote="Ignite168"]
Day 11 was more silliness, as junggi was practically confirmed from his investigations, and people were saying things like "A role like that is useless." junggi's roles was one of the most powerful in this game as any Werewolf Minion claiming to be things like a Citizen or such would get caught and definitely a mistake on the side of the town.
[/quote]

Oops... ehm... maybe the werewolves coincidentally do not say that they are citizen. The roles are mostly made so I don't know. :roll: [I feel a bit guilty that I also said that the role like that is useless when I was alive.]

Anyway this is the first online mafia game and our team won it. Not bad. :D

I can't believe that I had to post in a foreign language for the rest of the game! I had a feeling that the same person did those two things, but I never knew it was jli!
Also, for nights 1 and 2, it was as if someone knew what I was :huh: !


[quote="akimatsu"]hi, i am desperately in need for someone to explain to me the procedure for factoring radicals within radicals. I learnt it but i forgot  :P

for example: write $\sqrt{7-\sqrt{48}}$ in its simplest form[/quote]
you want to find the square root of whatever's under the radical.  $(\sqrt{a}+\sqrt{b})^{2}=a+b+2\sqrt{ab}$

so in the example with $\sqrt{7-\sqrt{48}}$, $a+b=7$ and $ab=12$ (because $\sqrt{48}=2\sqrt{12}$).  so just solve for a and b, and then use a - sign instead of +, and there's your answer.


[quote="bryantacademy"]The sum of the numbers in an arithmetic sequence (a list of numbers) can be derived with the following formula.

n(a+z)/2

where n is the number of terms in the sequence, a is is the lowest term, and z is the highest term.

I figured out the following formula that works just as well, but beginners [b]do not use this! It has not been rigorously tested!!![/b]

n(n+1)/2

where n the highest number in the sequence.[/quote]

ummm... isnt that gauss's formula? (I'm not positive it is)
i havent even heard of the 1st one

don't you prove this by induction?

[hide]
$\displaystyle\frac{n(n+1)}{2}$ for 1 is true

prove that k implies k+1

$1+2+3...k+(k+1)$

$=\frac{k(k+1)}{2}+(k+1)$

$=\frac{k(k+1)+2(k+1)}{2}$

$=\frac{(k+1)(k+2)}{2}$
[/hide]

I've never seen the other one though. :)

[quote="10000th User"][quote="sirorange"]While I feel bad deleting excellent, thoughtful posts, those formulas are far too different for the MOEMS forum, and if repeat offenses do occur, you may be banned. So please, consider your posts carefully.[/quote]Alright, I did not know the simple proof was not even acceptable in middle school  :oops: . Last year when I was in 8th grade, that's the way my teacher taught us in Algebra I, and my school follows the American system...  :?
And I spent 1 hour thinking and typing all of that in the post :(  :stretcher:[/quote]

I'm sorry, but by absolutely no means was that a MOEMS level approach to solving an answer. Perhaps, you could have posted this somewhere in Getting Started, and you could have linked to it. I'm extremely sorry.

[quote="236factorial"][quote="nonie"]i havent even heard of the 1st one[/quote]

You havent? You should know it....  :)[/quote]

Actually, I think I've seen it, but in a different form.

The first one is simple.  N is the last number and $(a+z)/2$ is the number of terms so if you had 2 4 6 8 10 you would have $10(1+5)/2$ which is 30 and is true.  If you are wondering about $(1+5)/2$ you take the entire set and divide it by 2.  I came up with this formula when I was working on a generalization for a homework problem for MathCounts.

[quote="solafidefarms"]That only works on sequences that start at 1 and end at n, whose terms differ by 1...e.g. 1,2,3,4,5,6. It does work though. Here's a demonstration of why it works: http://www.artofproblemsolving.com/Images/Side2.swf[/quote]

[b]Solafide: I believe you forgot the last step that is important. (Part 2 is what I added in that I think Solafide forgot about)[/b]

[b][hide="Part 1"][/b][b]Take the highest number [/b]($n$)
[b]Do [/b]$\frac{n\times(n+1)}2$[/hide]
[b][hide="Part 2"][/b][b]Take the number 1 less than your lowest number in the sequence[/b]($o$)[b]. Take[/b] $o$ [b]and do [/b]$o+1$. Take $o$ and do $o\div2$. [b]Take the two numbers that you got for the last two steps and multiply them[/b] $(o+1)\times(o\div2)$. [b]Minus that answer from your answer to Part 1[/b][/hide]

[b][hide="Click for Example that Puts Both Parts Together"][/b]
[b]My numbers are from 23 to 47.[/b]
$\frac{47\times(47+1)}2=1,128$
[b]My lowest number is 23. So one lower than that is 22.[/b] $(22+1)\times(22\div2)=253$
$1,128-253=875$
[b]875 is the answer[/b][/hide]
[b]
P.S. If the sequence starts at 1 just go to Step 2 and forget 1.-Hope this helps-It took a long time to code this and to think this all out :) [/b]


If x,y,z are positive real numbers with $ x^2\plus{}y^2\plus{}z^2\equal{}1$ , then

\[ \frac{x^4}{y^2}\plus{}\frac{y^4}{z^2}\plus{}\frac{z^4}{x^2}\ge 3(x^3\plus{}y^3\plus{}z^3)^2\]

First multiply the left side by $ 1^{2} \equal{} (\sum_{cyc}x^{2})^{2}$

First let's adnote that one of the following inequalities must hold:
1)$ \sum_{cyc}x^{2}y^{4}\geq \sum_{cyc}x^{3}y^{3}$
2)$ \sum_{cyc}x^{4}y^{2} \geq \sum_{cyc}x^{3}y^{3}$
(an obvious fact, assuming the opposite and summing them up we get a contradiction)

1)
From rearrangement $ \sum_{cyc}\frac {x^{8}}{y^{2}}\geq \sum_{cyc}x^{6}$
From Am-Gm $ 2\sum_{cyc}(\frac {x^{2}y^{6}}{z^{2}} \plus{} z^{2}x^{4})\geq 4\sum_{cyc}x^{3}y^{3}$
From Am-Gm and our assumption we get $ \sum_{cyc}(\frac {x^{4}y^{4}}{z^{2}} \plus{} z^{2}y^{4})\geq 2\sum_{cyc}x^{2}y^{4}\geq 2\sum_{cyc}x^{3}y^{3}$
$ 2\sum_{cyc}x^{6} \equal{} 2\sum_{cyc}x^{6}$
Adding those four up we get the desired result

2)
From rearrangement $ \sum_{cyc}\frac {x^{8}}{y^{2}}\geq \sum_{cyc}x^{6}$
From our assumption $ 2\sum_{cyc}x^{4}y^{2}\geq 2\sum_{cyc}x^{3}y^{3}$
$ 2\sum_{cyc}x^{6} \equal{} 2\sum_{cyc}x^{6}$
From Am-Gm $ \sum_{cyc}(x^{2}y^{4} \plus{} \frac {y^{4}z^{4}}{x^{2}} \plus{} \frac {x^{6}z^{2}}{y^{2}} \plus{} \frac {x^{6}z^{2}}{y^{2}})\geq 4\sum_{cyc}x^{3}z^{3}$

Adding those 4 up gives us the desired result.
Since in both cases the inequality holds, than our inequality is also true Q.E.D. :wink:

From Cauchy

$ (x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)\left(\frac {x^4}{y^2} \plus{} \frac {y^4}{z^2} \plus{} \frac {z^4}{x^2}\right)\ge (x^3 \plus{} y^3 \plus{} z^3)^2$

So we need to proove
$ (x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \le \frac {1}{3}$

Fom our restiction

$ (x^2 \plus{} y^2 \plus{} z^2)^2 \equal{} 1 \equal{} x^4 \plus{} y^4 \plus{} z^4 \plus{} 2(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)$

Assuming wlog $ x^2>y^2>z^2$, rearrangement shows

$ x^4 \plus{} y^4 \plus{} z^4 \ge x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2$

$ \Rightarrow x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \le \frac {1}{3}$

And we are done  :lol:

EDIT: beaten  :(

[quote="ocha"]
EDIT: beaten  :([/quote]

Well, but your proof is much easier, so we should consider it a draw :D (although nobody mentioned we should treat it as a contest)


let $ x$ , $ y$ , $ z$ $ \in [0, \plus{} \infty[$ , prove that :

$ (xy)^3 \plus{} (yz)^3 \plus{} (zx)^3 \plus{} 3(xyz)^2 \geq x^3yz(y \plus{} z) \plus{} y^3xz(x \plus{} z) \plus{} z^3yx(y \plus{} x)$

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ , $ z$ $ \in [0, \plus{} \infty]$ , prove that :

$ (xy)^3 \plus{} (yz)^3 \plus{} (zx)^3 \plus{} 3(xyz)^2 \geq x^3yz(y \plus{} z) \plus{} y^3xz(x \plus{} z) \plus{} z^3yx(y \plus{} x)$[/quote]
With Schur$ a^3\plus{}b^3\plus{}c^3\plus{}3xyz \ge\ xy(x\plus{}y)\plus{}yz(y\plus{}z)\plus{}zx(z\plus{}x)$
And we change $ a\equal{}xy;b\equal{}yz;c\equal{}zx$. We have
$ (xy)^3 \plus{} (yz)^3 \plus{} (zx)^3 \plus{} 3(xyz)^2 \geq x^3yz(y \plus{} z) \plus{} y^3xz(x \plus{} z) \plus{} z^3yx(y \plus{} x)$


[quote="Leon"]Let $ f$ be a function defined on the set of non-negative integers and taking values in the same
set. Given that

(a) $ \displaystyle x \minus{} f(x) \equal{} 19\left[\frac {x}{19}\right] \minus{} 90\left[\frac {f(x)}{90}\right]$ for all non-negative integers $ x$;

(b) $ 1900 < f(1990) < 2000$,

find the possible values that $ f(1990)$ can take.
(Notation : here $ [z]$ refers to largest integer that is $ \leq z$, e.g. $ [3.1415] \equal{} 3$).[/quote]

$ x\minus{}f(x)\equal{}19[\frac x{19}]\minus{}90[\frac{f(x)}{90}]$ $ \equal{}19(\frac x{19}\minus{}\{\frac x{19}\})$ $ \minus{}90(\frac{f(x)}{90}\minus{}\{\frac{f(x)}{90}\})$ $ \iff$ $ 19\{\frac x{19}\}\equal{}90\{\frac{f(x)}{90}\})$ $ \iff$ $ \{\frac{f(x)}{90}\}\equal{}\frac{19}{90}\{\frac x{19}\}$

$ \iff$ $ \frac{f(x)}{90}\equal{}h(x)\plus{}\frac{19}{90}\{\frac x{19}\}$ where $ h(x)\equal{}[\frac{f(x)}{90}]\in\mathbb N_0$ and so $ f(x)\equal{}90h(x)\plus{}19\{\frac x{19}\}$

so $ f(1990)\equal{}90n\plus{}19\{\frac {1990}{19}\}\equal{}90n\plus{}14$ and, since $ 1900<f(1990)<2000$, $ f(1990)\in\{1904,1994\}$

[quote=INMO 1990 P3]Let $ f$ be a function defined on the set of non-negative integers and taking values in the same
set. Given that

(a) $ \displaystyle x \minus{} f(x) \equal{} 19\left[\frac{x}{19}\right] \minus{} 90\left[\frac{f(x)}{90}\right]$ for all non-negative integers $ x$;

(b) $ 1900 < f(1990) < 2000$,

find the possible values that $ f(1990)$ can take.
(Notation : here $ [z]$ refers to largest integer that is $ \leq z$, e.g. $ [3.1415] \equal{} 3$).[/quote]
[b][color=#000]Solution:[/color][/b] $$x-f(x)=19 \left \lfloor \frac{x}{19} \right \rfloor - 90 \left \lfloor \frac{f(x)}{90} \right \rfloor $$
Plug in $1990$,
$$\implies 1990-f(1990)=19 \left \lfloor \frac{1990}{19} \right \rfloor - 90 \left \lfloor \frac{f(1990)}{90} \right \rfloor
 \implies \left \lfloor \frac{f(1990)}{90} \right \rfloor=\frac{f(1990)}{90}-\frac{14}{90} \implies \left \{ \frac{f(1990)}{90} \right \} = \frac{14}{90}$$
Now, $$\frac{1990}{90} < \frac{f(1990)}{90} < \frac{2000}{90} \implies \left \lfloor \frac{f(1990)}{90} \right \rfloor =21,22$$
Hence, $\frac{f(1990)}{90}=\frac{14}{90}+21 , \frac{14}{90} +22 \implies \boxed{f(1990)=1904, 1994 } \qquad \blacksquare$


abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ be a ring that has the ACC on prime ideals. Is it true that $ Spec(A)$ is a Noetherian space?

No. Check $ A\equal{}\mathbb F_2^\mathbb N$.


inequalities
LaTeX
trigonometry
geometry proposed
geometry

[quote="warut_suk"]In a triangle, let $a,b,c$ denote the side lengths and $h_a, h_b, h_c$ the altitudes to the corresponding side. Prove that $(\frac{a}{h_a})^2+(\frac{b}{h_b})^2+(\frac{c}{h_c})^2 \geq 4$[/quote]

Well, the inequality is eqivalent with $ a^4+b^4+c^4\geq 16A^2 $ which is trivial and it has been disscused at
 least two times. ;)

Here's another solution: By Cauchy-Schwarz inequality we have
$3((\frac{a}{h_a})^2+(\frac{b}{h_b})^2+(\frac{c}{h_c})^2)\geq(\frac{a}{h_a}+\frac{b}{h_b}+\frac{c}{h_c})^2$

and it remains to show that $ \frac{a}{h_a}+\frac{b}{h_b}+\frac{c}{h_c}\geq2\sqrt3 $. Well we do know that $ah_a=2A$, and the inequality reduces to Weitzenbock's  $a^2+b^2+c^2 \geq 4A\sqrt3$ ;).

Here is a better inequality: $a^2b^2+b^2c^2+c^2a^2\geq 16A^2 $.

Here's another interresting problem based on the same idea:

 Prove that in any triangle $ABC$ the following inequality holds:

         $(\frac{a}{r_a})^2+(\frac{b}{r_b})^2+(\frac{c}{r_c})^2\geq4$  Good luck!  :lol:  ;)

How to do this?
Substitute $r_a = \frac{2A}{b+c-a}$?

@ Cezar Lupu: Some LaTeX tips: The product of two numbers a and b is best written as $a\cdot b$ (code: [color=red]a\cdot b[/color]), not $a.b$. The sign "doesn't equal" is $\neq$ (code: [color=red]\neq[/color]). (This also goes to Levi.) The square root of a number x is $\sqrt{x}$ (code: [color=red]\sqrt{x}[/color], not just [color=red]sqrt(x)[/color]). It also took me some time to learn all of these things, but it was worth doing.

And a note to [b]many[/b] members: [b]Please do define your notations![/b] So, don't just write "in any triangle ABC, the inequality $(\frac{a}{r_a})^2+(\frac{b}{r_b})^2+(\frac{c}{r_c})^2\geq4$ holds", but write "in any triangle ABC with exradii $r_a$, $r_b$, $r_c$, the inequality $(\frac{a}{r_a})^2+(\frac{b}{r_b})^2+(\frac{c}{r_c})^2\geq4$ holds" (you can omit defining the sides and the angles, but the rest should be defined). This is not just because different people use different notations, but also since it should be possible to find your post when searching for all posts containing the string "exradi*". You can look at http://www.mathlinks.ro/Forum/viewtopic.php?t=21455 for details.

___________________________________________________________________________________________________

Now to the problem.

[quote="cezar lupu"]Here is a better inequality: $a^2b^2+b^2c^2+c^2a^2\geq 16A^2 $.[/quote]

Indeed, there's a stronger inequality: $bc+ca+ab\geq 4\sqrt3A$. And, even stronger: $\left(a^2+b^2+c^2\right)-Q\geq 4\sqrt3A$, where $Q=\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2$. See http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 for proofs of this.

[quote="cezar lupu"]Prove that in any triangle $ABC$ the following inequality holds:

         $(\frac{a}{r_a})^2+(\frac{b}{r_b})^2+(\frac{c}{r_c})^2\geq4$[/quote]

[color=blue][b]Problem.[/b] Prove that in any triangle ABC with exradii $r_a$, $r_b$, $r_c$, the inequality $\left(\frac{a}{r_a}\right)^2+\left(\frac{b}{r_b}\right)^2+\left(\frac{c}{r_c}\right)^2\geq 4$ holds.[/color]

[i]Solution.[/i] According to http://www.mathlinks.ro/Forum/viewtopic.php?t=42538 post #4, we have $a=r_a\cdot\left(\tan\frac{B}{2}+\tan\frac{C}{2}\right)$, so that $\frac{a}{r_a}=\tan\frac{B}{2}+\tan\frac{C}{2}$. Denoting $\tan\frac{A}{2}=x$, $\tan\frac{B}{2}=y$, $\tan\frac{C}{2}=z$, this becomes $\frac{a}{r_a}=y+z$. Similarly, $\frac{b}{r_b}=z+x$ and $\frac{c}{r_c}=x+y$. On the other hand, the equation $\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}+\tan\frac{A}{2}\tan\frac{B}{2}=1$ proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=29067 becomes yz + zx + xy = 1. By a well-known inequality, $x^2+y^2+z^2\geq yz+zx+xy$, so that $x^2+y^2+z^2\geq 1$. Thus,

$\left(\frac{a}{r_a}\right)^2+\left(\frac{b}{r_b}\right)^2+\left(\frac{c}{r_c}\right)^2=\left(y+z\right)^2+\left(z+x\right)^2+\left(x+y\right)^2$
$=2\left(x^2+y^2+z^2\right)+2\left(yz+zx+xy\right)\geq 2\cdot 1+2\cdot 1=4$,

and the problem is solved.

  Darij


Let $ ABC$ be an isoceles triangle with $ AC\equal{}BC$. A point $ P$ lies inside $ ABC$ such that \[ \angle PAB \equal{} \angle PBC, \angle PAC \equal{} \angle PCB.\] Let $ M$ be the midpoint of $ AB$ and $ K$ be the intersection of $ BP$ and $ AC$. Prove that $ AP$ and $ PK$ trisect $ \angle MPC$.

Hints:

Let $ \angle PAB\equal{}\alpha,\angle PAC\equal{}\beta$. Do some angle chasing. $ \angle CPK\equal{}\angle APK$ is straighforward. We need to prove that $ \angle MPA\equal{}\alpha\plus{}\beta$.

Extend $ AP$ until it cuts $ BC$ at $ X$. $ BPX$ is similar to $ APB$, and $ \angle BPX\equal{}\alpha\plus{}\beta$. So we want to prove that $ PX$ is a median. Compare the area of CPX and PXB, and you'll get the desired result.

[quote="Raja Oktovin"]Let $ ABC$ be an isoceles triangle with $ AC \equal{} BC$. A point $ P$ lies inside $ ABC$ such that
\[ \angle PAB \equal{} \angle PBC, \angle PAC \equal{} \angle PCB.
\]
Let $ M$ be the midpoint of $ AB$ and $ K$ be the intersection of $ BP$ and $ AC$. Prove that $ AP$ and $ PK$ trisect $ \angle MPC$.[/quote]
Let $ (O)$ is the circumcircle of $ \Delta APB$ then $ CA,CB$ are two tangents of $ (O)$. $ CP\cap (O)\equal{}\{J\}$
It's easy to see $ \angle CPK\equal{}\angle APK$
We need to prove that $ \angle CPK\equal{}\angle JPB\equal{}\angle APM$ or $ \angle APJ\equal{}\angle BPM$
Applying Ptoleme's theorem:$ AP.JB\plus{}AJ.PB\equal{}PJ.AB$ and $ AP.JB\equal{}PB.AJ$ we get
$ AJ.PB\equal{}PJ.MB$ then $ \frac{AJ}{PJ}\equal{}\frac{MB}{PB}$
So $ \Delta AJP$~$ \Delta MBP$
Therefore $ \angle APJ\equal{}\angle MPB$

Let $ \angle PCB=\alpha ;\angle PBC=\beta $
$PA$ meets  $BC$ at $N$
We have:
$ \angle NCP =\angle NAC \rightarrow NC^2=NP.NA $
$ \angle NBP =\angle NBC \rightarrow NB^2=NP.NA $

Therefore ,$NC=NB$ otherwise , $N$ is the midpoint of $BC$ (*)
It's easy to show that $ \triangle PAB \sim \triangle PBC$ and in addition , we have :$N$ is midpoint of AC
So :$\angle PNB=angle MPA \rightarrow \angle PAB=\angle NPM=180-\alpha-\beta$
$= \angle CPK=\angle KPA$
Q.E.D

1) Easily, $\angle APK=\angle CPK \ (\ 1 \ )$
2) $CP$ is symmedian of $\triangle ABP$, done!
I thing it has been posted before, or, at least, similar!

Best regards,
sunken rock


$ f_n\equal{}34f_{n\minus{}1}\minus{}f_{n\minus{}2}\plus{}2$

$ f_0\equal{}0$
$ f_1\equal{}1$

proof that $ 8f_{n}\plus{}1$ is square number when $ n$ is a natural number.

[quote="Lomisedu"]$ f_n \equal{} 34f_{n \minus{} 1} \minus{} f_{n \minus{} 2} \plus{} 2$

$ f_0 \equal{} 0$
$ f_1 \equal{} 1$

proof that $ 8f_{n} \plus{} 1$ is square number when $ n$ is a natural number.[/quote]

Direct resolution gives $ f_n \equal{} \frac {(17 \plus{} 12\sqrt 2)^n \plus{} (17 \minus{} 12\sqrt 2)^n \minus{} 2}{32}$

So $ 8f_n \plus{} 1 \equal{} \frac {(17 \plus{} 12\sqrt 2)^n \plus{} (17 \minus{} 12\sqrt 2)^n \plus{} 2}{4}$ $ \equal{} \left(\frac {(3 \plus{} 2\sqrt 2)^n \plus{} (3 \minus{} 2\sqrt 2)^n}2\right)^2$ $ \equal{}\left(\sum_{k\equal{}0}^{[\frac n 2]}\binom n{2k}3^{n\minus{}2k}2^{3k}\right)^2$

Hence the result


inequalities
function
inequalities unsolved

This is [url=http://www.mathlinks.ro/viewtopic.php?t=36873]Vasc's inequality[/url].

[quote="neworder"]Let $f(x)=\frac{1}{1-x}$. Of course f(x) is concave, so we may use Jensen, for $x_{1}=ab, x_{2}=bc, x_{3}=ca$ and $\alpha_{1}=\alpha_{2}=\alpha_{3}=\frac{1}{3}$, so: ${\frac{1}{3}(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}) = \frac{1}{3}(f(ab)+f(bc)+f(ca)) \leq f(\frac{1}{3}(ab+bc+ca)} = \frac{1}{1-\frac{ab+bc+ca}{3}}$.
Now we have to prove that $\frac{1}{1-\frac{ab+bc+ca}{3}} \leq \frac{3}{2} \Longleftrightarrow \frac{2-3(1-\frac{ab+bc+ca}{3})}{2(1-\frac{ab+bc+ca}{3})} \leq 0$. Of course $1=a^{2}+b^{2}+c^{2} \geq ab+bc+ca$, so the numerator is negative, and the denominator is positive, so the inequality is true.[/quote]

it's convex.

Right, I forgot that in this case $x \leq 1$

[quote="neworder"]Let $f(x)=\frac{1}{1-x}$. Of course f(x) is concave, so we may use Jensen, for $x_{1}=ab, x_{2}=bc, x_{3}=ca$ and $\alpha_{1}=\alpha_{2}=\alpha_{3}=\frac{1}{3}$, so: ${\frac{1}{3}(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}) = \frac{1}{3}(f(ab)+f(bc)+f(ca)) \leq f(\frac{1}{3}(ab+bc+ca)} = \frac{1}{1-\frac{ab+bc+ca}{3}}$.
Now we have to prove that $\frac{1}{1-\frac{ab+bc+ca}{3}} \leq \frac{3}{2} \Longleftrightarrow \frac{2-3(1-\frac{ab+bc+ca}{3})}{2(1-\frac{ab+bc+ca}{3})} \leq 0$. Of course $1=a^{2}+b^{2}+c^{2} \geq ab+bc+ca$, so the numerator is negative, and the denominator is positive, so the inequality is true.[/quote]

convex or concave?????

convex.

And just a stupid proof. Multiply by $(1-ab)(1-bc)(1-ca)$, homogenize with $a^2+b^2+c^2=1$. Then we get something of the form $f(a,b,c)\geq 0$. Assume that $a\geq b\geq c$ and set $b=c+x$, $a=c+x+y$. Then $f(c+x+y,c+x,c)$ is the sum of terms $c^kx^ly^m$ with positive coefficients, qed.

I guess Vasc's EVT can also helpful(Leave $a^2+b^2+c^2=1$ and $abc$ fixed. We have to take the function $f(x)=\frac{1}{1-\frac 1x}$ which most likely fulfils the conditions to apply the EVT).

That's a bit of brute force  ;) What's EVT?

Has the quite simple proof method?    Thanks!

[quote="scpajmb"][quote="arqady"]
.....
$=a^4+b^4+c^4-a^3b-a^3c-b^3a-b^3c-c^3a-c^3b+abc(a+b+c)+$
$+(a-b)^4+(a-c)^4+(b-c)^4\geq0.$ :)[/quote]
why?[/quote]
Since
$a^4+b^4+c^4-a^3b-a^3c-b^3a-b^3c-c^3a-c^3b+abc(a+b+c)\geq0\Leftrightarrow$
$\Leftrightarrow a^2(a-b)(a-c)+b^2(b-a)(b-c)+c^2(c-a)(c-b)\geq0.$ This is Schur.


abstract algebra
group theory
advanced fields
advanced fields unsolved

Let $ G$ be a group, $ Set_G$ a category of $ G$-sets with canonical topology. Let $ S$ be a sheaf of abelian groups on $ Set_G$. Describe $ S(*)$ in terms of $ G$-module $ S(G)$. Similarly describe $ S(G/H)$ ($ H$ is a normal subgroup of $ G$).[It should be proved that $ S(*) \equal{} S(G)^{G}$]

I assume you are talking about Grothendieck topologies. The problem becomes clear when you have established the equivalence of categories {left $ G$-sets} = {sheaves on $ Set_G$}, $ \mathcal{F} \mapsto \mathcal{F}(G)$, $ X \mapsto \mathrm{Hom}_G(\minus{},X)$ (every sheaf is representable).


Let $ a, b, c$ non-negative real numbers such that $ abc \equal{} 1$. Prove or disprove that

$ \frac{(ab\plus{}1)(a^2\plus{}b^2)}{c} \plus{} \frac{(bc\plus{}1)(b^2\plus{}c^2)}{a} \plus{} \frac{(ca\plus{}1)(c^2\plus{}a^2)}{b} \ge 12$

____
 :maybe:

[quote="Dimitris X"]${ \frac {(ab + 1)(a^2 + b^2)}{c} \ge 4 \frac {ab \sqrt {ab}} {c}}$

And $ a^2 b^2 \sqrt {ab} + b^2c^2 \sqrt {bc} + c^2 a^2 \sqrt {ca} \ge3$[/quote]


calculus
integration
trigonometry
calculus computations

hello, let $ t=1-x$, then we have $ -\int\frac{1}{t\sqrt{1-(1-t)^2}}\,dt$, now we setting $ u=\frac{1}{t}$
and we have $ dt=-\frac{1}{u^2}du$ and our integral is $ \int \frac{1}{\sqrt{2u-1}}\,du$, with $ z=\sqrt{2u-1}$ we get finally $ \int1\,dz=z+C$
Sonnhard.

Let $ x \equal{} \sin t$

$ \int {\frac {{dx}} {{(1 \minus{} x)\sqrt {1 \minus{} x^2 } }}}$
$ \equal{} \int\frac {1}{(1 \minus{} \sin t)\cos t}\cos tdt$
$ \equal{} \int\frac {1 \plus{} \sin t}{\cos^2 t}dt$
$ \equal{} \int \sec^2 tdt \minus{} \int\frac {1}{\cos^2 t}d\cos t$
$ \equal{} \tan t \plus{} \frac {1}{\cos t} \plus{} C$


[quote="arcsin"]find $ y$
$ y \plus{} \sqrt {b^{2} \plus{} y^{2}} \equal{} \frac {5b^{2}}{ b^{2} \plus{} y^{2}}$[/quote]
What is the condition for $ x,b$  :?:

condition for $ x$ ? i can't understand you clear.
$ y$ is here the unknown , so we have to find $ y$

the solution should be for example $ y\equal{}2b$
sorry for my English please.

[quote="arcsin"]condition for $ x$ ? i can't understand you clear.
$ y$ is here the unknown , so we have to find $ y$

the solution should be for example $ y \equal{} 2b$
sorry for my English please.[/quote]
I was meaning $ y$ no $ x$... :)
So $ y,b \in R$ :?:

Are these problems you're posting from your national olympiad ? (Gjilan  :lol: )

hello, an idea, rewriting your equation in the form
$ y\plus{}\sqrt{b^2\plus{}y^2}\equal{}\frac{5b^2}{(\sqrt{b^2\plus{}y^2})^2}$ and setting $ t\equal{}\sqrt{b^2\plus{}y^2}$ we will get
$ \pm\sqrt{t^2\minus{}b^2}\plus{}t\equal{}\frac{5b^2}{t^2}$.
Sonnhard.


Show that in a number with $16$ digits, there's always a group of consecutive digits such that the product of those digits is a perfect square.

Not necessarily true. Consider the number:
$1212121212121212$

The product of the consecutive digits is $2$, which is clearly not a perfect square.

I think you maybe missing a condition.

Length of group isn't neccesarily 2.
In your case, the group $1212$ has a product of 4, a perfect square.


Since the polynomial has all real coefficients, then if $x_1 = 2+i$ is a root, then its conjugate $x_2 = 2-i$ is also a root.  So we know a factor of the polynomial is  $(x-(2+i))(x-(2-i)) = x^2-4x+5$.  Also, the other factor is $3x+n$ for some n.  However, the coefficient of x^2 of $(x^2-4x+5)(3x+n)$ must be $-13$, so n = -1.  Our polynomial is thus $3x^3 - 13x^2 + 19x - 5$.
$19 + 5 + 2\sqrt{2^2 + 1} + 1/3 = 73/3 + 2\sqrt{5}$

[quote="Slizzel"]I got the polynomial same form.  I've just replaced $x_{1}$ and I got $a,b$.  :)[/quote]

Oh heh  :P

If we first put the values of $a$ and $b$ in terms of roots(Vieta's formulae) and then simplify, is it easier??

Seems it will be more difficult!!!


If the answers I got from those at TJ are right, then I got an 8.

Ineedmathhelp, who's also in VA got a 3.


parameterization
algebra
system of equations

Hi, can someone give me a hint to solve this scary looking system of equations:
$ ax\plus{}by\plus{}cz \equal{} 0$
$ bcx\plus{}cay\plus{}abz \equal{} 0$
$ xyz\plus{}abc(a^3 x \plus{} b^3 y \plus{} c^3 z) \equal{} 0$

Thanks

Are we solving for $ (x,y,z)$, with $ a,b,c$ as parameters?

Where did you get this problem from? Apart from the trivial solution $ (0,0,0)$, the general solution to the system is pretty gross.


trigonometry
geometry
geometric transformation
reflection
real analysis
real analysis unsolved

Compute

$ \frac {\Gamma(1/14)\Gamma(9/14)\Gamma(11/14)}{\Gamma(3/14)\Gamma(5/14)\Gamma(13/14)}$

Source: Glasser

Used the duplication formula six times and got $ 2$.

So now we can prove
\[ \cos \left( \pi/7  \right) \equal{} {\frac { \minus{}\left( \Gamma  \left( {\frac {11}{14}} \right)
 \right) ^{2} \left( \Gamma  \left( {\frac {9}{14}} \right)  \right) ^
{2}}{ 8\left( \Gamma  \left( {\frac {13}{14}} \right)  \right) ^{2}\pi
}}\plus{}1\]

Can you also show that $ \frac {\Gamma\left(\frac {1}{24}\right)\,\Gamma\left(\frac {11}{24}\right)}{\Gamma\left(\frac {5}{24}\right)\,\Gamma\left(\frac {7}{24}\right)} \equal{} \sqrt {3}\cdot\sqrt {2 \plus{} \sqrt {3}}$ ?

[quote="Doe John"]Can you also show that ...?[/quote]
Yes, I can :)

I'm too lazy to compute the exact values, so I'll be denoting by $ C$ any easily computable constant.

Abbreviate $ g(x) \equal{} \Gamma(x/24)$.
\[ \frac {g(1)g(11)}{g(5)g(7)} \equal{} \frac {g(1)g(23)\cdot g(11)g(13)}{g(5)g(13)g(7)g(23)}\]
\[ \equal{} C\frac {g(15) g(21) }{g(5)g(13)g(21)\cdot g(7)g(15)g(23)} \equal{} C\frac {g(15)g(21)}{g(15)g(21)} \equal{} C.\]

Yes, this does the trick. Good job!

Let me repeat your computation in slow motion:

First you expand the fraction

$ A: = \frac {\Gamma\left(\frac {1}{24}\right)\, \Gamma\left(\frac {11}{24}\right)}{\Gamma\left(\frac {5}{24}\right)\, \Gamma\left(\frac {7}{24}\right)}$ with $ \Gamma\left(\frac {13}{24}\right) \Gamma\left(\frac {23}{24}\right)$ :

$ A = \frac {\Gamma\left(\frac {1}{24}\right)\, \Gamma\left(\frac {23}{24}\right) \;\cdot \; \Gamma\left(\frac {11}{24}\right) \, \Gamma\left(\frac {13}{24}\right)}{\Gamma\left(\frac {5}{24}\right)\, \Gamma\left(\frac {13}{24}\right) \, \cdot \, \Gamma\left(\frac {7}{24}\right)\, \Gamma\left(\frac {23}{24}\right)}$

After Euler's reflection formula

$ \Gamma\left(\frac {1}{24}\right) \Gamma\left(\frac {23}{24}\right) = \frac {\pi}{\sin\left(\frac {\pi}{24}\right)}$ and $ \Gamma\left(\frac {11}{24}\right) \Gamma\left(\frac {13}{24}\right) = \frac {\pi}{\cos\left(\frac {\pi}{24}\right)}$ .

So $ A = \frac {\pi^2}{\sin\left(\frac {\pi}{24}\right)\cos\left(\frac {\pi}{24}\right)}\cdot\frac {1}{\Gamma\left(\frac {5}{24}\right)\, \Gamma\left(\frac {13}{24}\right) \, \cdot \, \Gamma\left(\frac {7}{24}\right)\, \Gamma\left(\frac {23}{24}\right)}$ .

Now you expand the fraction with $ \Gamma\left(\frac {15}{24}\right) \Gamma\left(\frac {21}{24}\right)$ :

$ A = \frac {2\pi^2}{\sin\left(\frac {\pi}{12}\right)}\cdot\frac {\Gamma\left(\frac {15}{24}\right) \cdot\Gamma\left(\frac {21}{24}\right)}{\Gamma\left(\frac {5}{24}\right)\, \Gamma\left(\frac {13}{24}\right)\, \Gamma\left(\frac {21}{24}\right) \, \cdot \, \Gamma\left(\frac {7}{24}\right)\,\Gamma\left(\frac {15}{24}\right)\, \Gamma\left(\frac {23}{24}\right)}$

Then you use Gauss's multiplication formula

$ \Gamma(z)\;\, \Gamma\left(z + \frac13\right)\, \Gamma\left(z + \frac23\right) = \frac {2\pi\sqrt {3}}{3^{3z}}\,\, \Gamma(3z)$

to write $ \Gamma\left(\frac {5}{24}\right)\Gamma\left(\frac {13}{24}\right) \Gamma\left(\frac {21}{24}\right)$

$ = \Gamma\left(\frac {5}{24}\right) \Gamma\left(\frac {5}{24} + \frac13\right)\Gamma\left(\frac {5}{24} + \frac23\right)$ as $ \frac {2\pi\sqrt {3}}{3^{\frac {15}{24}}}\;\,\Gamma\left(\frac {15}{24}\right)$

and $ \Gamma\left(\frac {7}{24}\right)\Gamma\left(\frac {15}{24}\right) \Gamma\left(\frac {23}{24}\right)$

$ = \Gamma\left(\frac {7}{24}\right) \Gamma\left(\frac {7}{24} + \frac13\right)\Gamma\left(\frac {7}{24} + \frac23\right)$ as $ \frac {2\pi\sqrt {3}}{3^{\frac {21}{24}}}\;\,\Gamma\left(\frac {21}{24}\right)$ .

Hence ${ A = \frac {2\pi^2}{\sin\left(\frac {\pi}{12}\right)}\cdot \frac {3^{\frac {15}{24}}\cdot 3^{\frac {21}{24}}}{2\pi\sqrt {3}\cdot 2\pi\sqrt {3}}}$ and this is ${ \frac {\sqrt {3}}{2\cdot\sin\left(\frac {\pi}{12}\right)}}$ .

Since $ \sin\left(\frac {\pi}{12}\right) = \sqrt {\frac {1 - \cos\left(\frac {\pi}{6}\right)}{2}} = \sqrt {\frac {1 - \frac {\sqrt {3}}{2}}{2}}$

$ 2\cdot \sin\left(\frac {\pi}{12}\right) = \sqrt {2 - \sqrt {3}}$ and $ \frac {1}{2\cdot \sin\left(\frac {\pi}{12}\right)} = \sqrt {2 + \sqrt {3}}$ .

And therefore $ A = \sqrt {3}\cdot \sqrt {2 + \sqrt {3}}$ .


inequalities
logarithms
real analysis
real analysis unsolved

How do you prove that ln(1+x)<=x when -1<x where ln is the natural log? I can do the case x => 0  but I need help with -1<x<0.

And then how do you use

$ \frac {x}{x \plus{} 1} \leq \log (1 \plus{} x) \leq x$

to prove that $ \log(1 \plus{} x) \equal{} x \plus{} O(x^2)$ as $ x \to 0$ (which is of course obvious from Taylors Theorem)?

[quote="FibonacciFan"]How do you prove that ln(1+x)<=x when -1<x where ln is the natural log? I can do the case x => 0  but I need help with -1<x<0.[/quote]

concavity.  what's the tangent line to $ \ln(1\plus{}x)$ at $ x\equal{}0$?


If $ a,b,c>0$, then

$ \frac{a^2\plus{}bc}{2a\plus{}b\plus{}c} \plus{} \frac{b^2\plus{}ca}{a\plus{}2b\plus{}c} \plus{} \frac{c^2\plus{}ab}{a\plus{}b\plus{}2c}\leq \frac{1}{2}(\frac{a^2}{b\plus{}c} \plus{} \frac{b^2}{c\plus{}a} \plus{} \frac{c^2}{a\plus{}b} \plus{} \frac{a\plus{}b\plus{}c}{2})$

[quote="lipako"]If $ a,b,c > 0$, then

$ \frac {a^2 \plus{} bc}{2a \plus{} b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{a \plus{} 2b \plus{} c} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b \plus{} 2c}\leq \frac {1}{2}(\frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b} \plus{} \frac {a \plus{} b \plus{} c}{2})$[/quote]
The inequality is not sharp, because by Cauchy Schwarz, we have
$ \sum \frac{4(a^2\plus{}bc)}{2a\plus{}b\plus{}c} \le \sum \left( \frac{a^2\plus{}bc}{a\plus{}b}\plus{}\frac{a^2\plus{}bc}{a\plus{}c}\right)\equal{} \sum \frac{a^2\plus{}b^2}{a\plus{}b} \plus{}a\plus{}b\plus{}c\equal{}2\sum \frac{a^2}{a\plus{}b} \plus{}a\plus{}b\plus{}c$
It remains to show that
$ \sum \frac{a^2}{b\plus{}c} \ge \sum \frac{a^2}{a\plus{}b},$
which follows from the Rearrangement Inequality. ;) :)

[quote="lipako"][quote="can_hang2007"][quote="lipako"]If $ a,b,c > 0$, then

$ \frac {a^2 \plus{} bc}{2a \plus{} b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{a \plus{} 2b \plus{} c} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b \plus{} 2c}\leq \frac {1}{2}(\frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b} \plus{} \frac {a \plus{} b \plus{} c}{2})$[/quote]
The inequality is not sharp, because by Cauchy Schwarz, we have
$ \sum \frac {4(a^2 \plus{} bc)}{2a \plus{} b \plus{} c} \le \sum \left( \frac {a^2 \plus{} bc}{a \plus{} b} \plus{} \frac {a^2 \plus{} bc}{a \plus{} c}\right) \equal{} \sum \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} a \plus{} b \plus{} c \equal{} 2\sum \frac {a^2}{a \plus{} b} \plus{} a \plus{} b \plus{} c$
It remains to show that
$ \sum \frac {a^2}{b \plus{} c} \ge \sum \frac {a^2}{a \plus{} b},$
which follows from the Rearrangement Inequality. ;) :)[/quote]

Thank you my friend.
Do you work with software?
I work with paper and pen... :blush:[/quote]
Sorry, but I dont understand you? What software? I haven't used one!  :mad:
Let me explain more my solution
$ \sum \left( \frac {a^2 \plus{} bc}{a \plus{} b} \plus{} \frac {a^2 \plus{} bc}{a \plus{} c}\right) \equal{}\sum \frac{a^2\plus{}bc}{a\plus{}b}\plus{}\sum \frac{a^2\plus{}bc}{c\plus{}a} \equal{}\sum \frac{a^2\plus{}bc}{a\plus{}b} \plus{}\sum \frac{b^2\plus{}ca}{a\plus{}b} \equal{}\sum \frac{a^2\plus{}b^2\plus{}c(a\plus{}b)}{a\plus{}b} \equal{}\sum \frac{a^2\plus{}b^2}{a\plus{}b} \plus{}a\plus{}b\plus{}c,$
and
$ \sum \frac{a^2}{a\plus{}b} \minus{}\sum \frac{b^2}{a\plus{}b} \equal{}\sum \frac{a^2\minus{}b^2}{a\plus{}b} \equal{}\sum (a\minus{}b) \equal{}0.$
So
$ \sum \frac{a^2\plus{}b^2}{a\plus{}b} \equal{}2\sum \frac{a^2}{a\plus{}b}.$
Is it all ok now?


x, y and z are the numbers 2, 3 and 5 in some order. What is the mean of all the possiible values of (2^x)(3^y)(5^z)?

The geometric mean is $ \sqrt[6]{(2^2\cdot 3^2\cdot 5^2)^{x\plus{}y\plus{}z}}\equal{}30^{10/3}$.

I'm guessing he wants the arithmetic mean.  Fortunately, there are only six such values, so the calculation can just be done explicitly.


What is the probably that 5 randomly drawn cards will be a full house (3 cards of a rank and 2 cards of another rank)?

An incorrect solution said, assuming that:

The first card can be any card- 52 ways to pick this card.
The second card can be any card of different rank - 48 ways to pick this card.
The third card matches the first card in rank - 3 ways to pick this card.
The fourth card matches the first card in rank - 2 ways to pick this card.
The fifth card matches the second card in rank - 3 ways to pick this card.

the answer is 52*48*3*2*3 divided by the number of possible hands, C(52,5). That is around 1.7%.

The correct approach was to count to the number of triples of any rank times the number of pairs of the remaining ranks divided by the number of possible hands, 13*C(4,3)  *  12*C(4,2)   /  C(52,5). That is around 0.14%.

The 2nd number seemed much more reasonable and I looked it up to confirm that it is the correct answer. But can anyone tell me what is wrong about the method the first solution is using?

The first approach is logically sound (the thing about choosing).

The problem lies in that a wrong number has been chosen as the denominator.

$ 52\times51\times50\times49\times48\times\frac{1}{10}$ should be used.

Your method is wrong because it overcounts. In this problem, it doesn't matter in what order you choose the cards, but, your method is counting two different orders of cards as two different ways to pick them.

Well, they are different cause then you have twice the probability of choosing it. Like when you roll two dice, you count 7 ways to toss a sum of 7, since you count {1,6} and {6,1} as different.

[quote="Guni"]Why do you divide the denominator by 10?[/quote]

ABAAB isn't the only sequence of events that could amount to a full house.

AAABB, for example, is another.

There are altogether $ 10$ such events.


Determine all positive integers which cannot be represented as $\frac{a}{b}+\frac{a+1}{b+1}$ with $a,b$ being positive integers.

n=a/b+(a+1)/(b+1)=(2ab+b+a)/b(b+1) ==> b|a, a=kb, k \in N  ==> (b+1)|(2kb+k+1) ==> (b+1)|(k-1) ==> k=m(b+1)+1, m \in N+{0} ==> n=(2b+1)m+2  ==> if n>2 then n-2 contains odd multiplier ==> n-2 is not power of 2, and for all numbers n<>2k+2 we will find appropriate values for b and m.

Answer: n<>2k+2.

Is that correct?


calculus
derivative
algebra unsolved
algebra

The derivative of x^2, with respect to x, is 2x.  However, suppose we write x^2 as the sum of x x's, and then take the derivative:

Let f(x) = x + x + ... + x  (x times)
Then f'(x)	= d/dx[x + x + ... + x]  (x times)
 	= d/dx[x] + d/dx[x] + ... + d/dx[x]  (x times)
 	= 1 + 1 + ... + 1  (x times)
 	= x

This argument appears to show that the derivative of x^2, with respect to x, is actually x.  Where is the fallacy?

Tried to write $\sqrt{2}$ summands of $\sqrt{2}$ ;)

You're holding one of the $x$'s in $x^{2}$ constant when you write it as $x+x+\ldots+x$.


geometry
cyclic quadrilateral
projective geometry
geometry unsolved

I don't know if this has been posted before. Someone proposed this problem to me and said that he hasn't been able to solve it. So I tried and I haven't been able to solve it either. It goes like this:

Given a cyclic quadrilateral $ ABCD$ with extensions of sides $ AB$ and $ DC$ meeting at $ P$, extensions of sides $ DA$ and $ CB$ meeting at $ Q$, and the diagonals $ AC$ and $ BD$ meeting at $ M$, prove that the line $ QM$ meets the circle (around $ ABCD$) at the points of tangency from $ P$ to the circle.

(This would mean that you can construct a tangent by drawing two lines through $ P$ intersecting the circle to form the quadrilateral and the tangent as in the theorem).

PS: I have come across a similar construction on another website (again without proof) where one of the lines $ PBA$ or $ PCD$ goes through the centre of the circle - I don't know if this is a requirement, but the more general version seems plausible.

i want to change the words.$ ABCD$ is cyclic and $ AC,BD$ intersect at $ K$ ,$ AD,BC$ at $ T$ and $ AB,CD$ at $ P$.let $ Q_1,Q_2$ be the intersection of $ TK$ with $ AB,CD$. its well known that $ (ABQ_ 1P)\equal{}\minus{}1$ and $ (DCQ_2P)\equal{}\minus{}1$.but if $ L_1,L_2$ be the tangency of tangents from $ P$ we have this again and so $ T,Q_1,Q_2,L_1,L_2$ and the pionts of tnagency lie on a line.

Thanks
I also found the problem of this construction in the topic Examples of Using Harmonic Division and Polarity at about post 85-90 (?). BTW that is a very nice topic to visit.


Prove this:
$[a^{m+n},a^mb^n,b^{m+n}]=[a,b]^{m+n}$ $a,b \in Z^*$ and $m,n \in N$

[hide]I presume $[a,b]=lcm(a,b)$ ?

Writing $a = \prod p_i^a_i$ and $b = \prod p_i^b_i$ with $a_i$ or $b_i$ null eventually

Equality comes directly from $[a,b]= \prod p_i^{max(a_i,b_i)}$[/hide]


geometry
analytic geometry
trigonometry
symmetry

Let $A=(0,0)$ and $B=(b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB=120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m+n.$

No one has posted a solution to this problem yet, so I will.

[hide="Yay I solved it"]

Just by geometric intuition, the y-coordinates going around counterclockwise (assume $b$ is positive) are 0,2,6,10,8,4.

Notice that 8=2*4, so if this geometric intuition is correct, one of our points (E) is $(0,8)$. We know about the point $(b,2)$, so our equilateral hexagon has side length $\sqrt{b^{2}+4}$, so point F is at $(-\sqrt{b^{2}-12},4)$. This means C lies at $(b+\sqrt{b^{2}-12},6)$ and D at $(b,10)$.

We now have our 6 points and, by surveyor's method (or adding parts of areas of the hexagon), our area is $8b+8\sqrt{b^{2}-12}$, depending on $b$, and the only thing left in the problem is making $\angle{FAB}=120$.

Letting X be a point on the positive x-axis and Y a point on the negative x-axis, we know that $\angle{BAX}+\angle{FAY}=60^\circ$

[hide="Using angle subtraction and some rearrangement"]
Let $\angle{BAX}=\theta$.
Notice that $\sin\theta=\frac{2}{\sqrt{b^{2}+4}}$ and $\sin(60-\theta)=\frac{4}{\sqrt{b^{2}+4}}$ so
$\sin(60-\theta)=2\sin\theta$
$\frac{\sqrt3}{2}\cos\theta-\frac{1}{2}\sin\theta=2\sin\theta$
$\frac{\sqrt3\cos\theta}{2}=\frac{5\sin\theta}{2}$
$\cot\theta=\frac{5}{\sqrt3}$[/hide] we get $\cot\angle{BAX}=\frac{5}{\sqrt3}=\frac{b}{2}$, so $b=\frac{10}{\sqrt3}$

This means our area is $8(\frac{10}{\sqrt3})+8\sqrt{\frac{64}{3}}=\frac{144}{\sqrt3}=48\sqrt3$

Our answer is $48+3=\boxed{051}$

Kalva decided to agree with me on this one.
[/hide]

You can use complex coordinates to get b. We have
$ Im((b + 2i)(cis120)) = 4$
$ \frac {b\sqrt {3}}{2} - 1 = 4$
$ b = \frac {10\sqrt {3}}{3}$
Using the distance formula, we have
$ C=(6\sqrt{3},6)$
$ F=(\8\sqrt{3}/3,4)$
Using symmetry, we have
$ D=(10\sqrt{3}/3,10)$
$ E=(0,8)$
Using shoelace, the area is $ 48\sqrt{3}$.

Sorry for the big bump (I did the 2003 AIME II as a practice test), but can someone please explain how sapphyre571 got $E=(0,8)$?

[quote=Skupp3]@awin

That's kinda mean. He just wanted information. It's fine if you bump like that.[/quote]

Ummm, I wasn't intending to be mean, I was just saying that 10 years is a long time!

[hide=sketch]Let the coordinates of $B$ be $(b,2)$ and the coordinates of $A$ be $(0,0)$, and the coordinates of $F$ be $(a,4)$.
First of all, if the angle $AB$ makes with the $x-$axis is $\theta$, then the angle $FA$ makes with the $x-$ axis is  $60 - \theta$. We know that therefore, $$\tan \theta = \frac{2}{b}$$ $$\frac{4}{a} = \tan (60 - \theta) = \frac{\tan 60 - \tan \theta}{1+\tan 60 \tan \theta} = \frac{\sqrt{3} - \frac{2}{b}}{1+\frac{2\sqrt{3}}{b}} = \frac{\sqrt{3}b-2}{b + 2\sqrt{3}}$$
Therefore,
$$a = \frac{4b+8\sqrt{3}}{\sqrt{3}b-2}$$
However, we also know that $$FA^2 = AB^2 \implies b^2 + 4 = a^2 + 16 \implies b^2 -12 = \left(\frac{4b+8\sqrt{3}}{\sqrt{3}b-2} \right)^2 \implies (4b+8\sqrt{3})^2+12(b\sqrt{3}-2)^2=b^2(b\sqrt{3}-2)^2$$
I’ll spare the computation, but in the end we end up with a 4th degree polynomial in terms of $b$. With some inspection, we speculate that $b = x\sqrt{3}$ where $x$ is rational, so we write the polynomial in terms of $x$, and we can find $x$ from here.   Therefore, we can find $a$ and $b$ and the area of the hexagon.
[/hide]

Hmm. this is not that bash stuffs I guess :)


abstract algebra
vector
linear algebra
matrix
superior algebra
superior algebra theorems

G=(Z+Z+Z)/N where Z denote the integers and + is direct sum and
N = <(7,8,9), (4,5,6), (1,2,3)> or the smallest submodule of Z+Z+Z containing these 3 vectors.

How would you describe G?

One could use a presentation to describe G which in this case is G = <a,b,c |7a+8b+9c=0, 4a+5b+6c=0, a+2b+3c=0> where a,b,c are the standard basis vectors for R^3.

However I do not know what G looks like in this form either.

One way to look at G is G=X/(q:X->X)

where X is R^3. q is the linear transformation given by the transformation matrix
A=
7 4 1
8 5 2
9 6 3

But what does this form of G look like?

Since $(4,5,6)-(1,2,3)=(3,3,3)$ and $(7,8,9)-(1,2,3)=(6,6,6)$, the submodule $N$ is actually generated by $(3,3,3)$ and $(2,1,0)$. Therefore, $N=\{(a,b,c)\colon c\equiv 0\mod 3, a-c=2(b-c)\}$, or  $N=\{(a,b,2b-a)\colon 2b\equiv a\mod 3\}$.
Now pick a convenient set of generators of $\mathbb Z\oplus \mathbb Z\oplus \mathbb Z$, such as $s_{1}=(0,0,1)$, $s_{2}=(2,1,0)$, and $s_{3}=(1,1,1)$. The image of $s_{1}$ in $G/N$ has infinite order, the image of $s_{2}$ is the zero element, and the image of $s_{3}$ has order $3$. Therefore, $G/N\cong \mathbb Z\oplus \mathbb Z_{3}$.


Counting \u0026 Probability
Counting and Probability
complementary counting

Liz has 11 marbles of different colors. In how many ways can she choose 3 of the marbles, if she refuses to choose the green one and the purple one at the same time?

There are $ 11C3\equal{}165$ total ways to choose 3 marbles. 9 of these have the purple and green marbles together, so the answer is $ 165\minus{}9\equal{}\boxed{156}$

I did this the same way as tinytim -- I think it's more elegant than the case-based method given in the official solution.

I agree- C(11,3) - 1*1*9 is much simpler. To cut an L, why cut and then rejoin the pieces if you can just slice out the part that shouldn't be there?

I used complimentary counting.

[hide="Easiest method"]

There are $\dbinom{11}{3}$ to choose any combination of 3 marbles from 11. Then, if Liz chooses both the green and the purple one, there are $9$ choices, because there are $9$ other choices. Therefore, the answer is $\dbinom{11}{3}-9=165-9=\boxed{156}$ ways.

Oh I got the answer wrong because i did 11C3-9*3 (As there are 3 ways to arrange the green and purple marble in the set of 3). Silly mistake done by me. I got 165-27=138 instead.

I did Complementary Counting As well. AoPs's Solution is much harder

[quote=davidaops][quote=NotMyRealNameII]How does "at the same time" affect the answer?[/quote]

I don't think it does. Correct me if I'm wrong.[/quote]

It does. At the same time means green and purple won't work.

Use complementary counting. 165 ways to choose 3 marbles from 11, subtract the 9c1 = 9 cases in which both the green and purple ones have been selected at the same time. 165 - 9 = [b]156.[/b]

[quote=V_V][i]imagine using casework....[/i] LOLL[/quote]

i hate casework

everyone does, but it's needed sometimes  :wacko:

but casework would be... not very good for this

yeah i used complementary counting, but case work could work

¯\_(ツ)_/¯

[quote=Ryan2010][quote=hansenhe]welp complementary counting is good[/quote]

is that postfarming?[/quote]

back at you  :coool:


If N differs from two successive squares between which it lies by x and y respectively, prove that N - xy is a square.

Let n2 & (n+1)2 be the consequtive squares then we have
x= N-n 2 , y= (n+1) 2 -N
N-xy= N 2 -2Nn(n+1)+n 2 (n+1) 2
So we have the result


Suppose you have a clay sphere with radius 1. You can divide it into two equal amounts of clay and then sculpt spheres out of those, which will result in the same volume but a greater surface area. Likewise if you divide the one into 3 spheres.

Let S be the total surface area after dividing one into N spheres. As N approaches infinity, does S also approach infinity? If not, what value of N makes S the largest?

[hide]
Let $n$ be the number of spheres.

$\frac{4}{3}\pi = n\left(\frac{4}{3}\pi r^3\right) \Rightarrow r = \frac{1}{\sqrt[3]{n}}$.

So

$S = n(4\pi r^2) = \frac{4\pi n}{n^{\frac{2}{3}}} = 4\pi \sqrt[3]{n}$.

Thus $\lim_{n \rightarrow \infty} S = \infty$.[/hide]

[hide]Since the total volume will always equal $\frac{4}{3}\pi (1)^3=\frac{4\pi}{3}$, we know that $\frac{4\pi}{3}=n(\frac{4}{3}\pi r^2$, where $n$ is the number of spheres and $r$ is their radii. Therefore $r=\frac{1}{\sqrt[3]n}$.
$S=n(4\pi r^2) = n(4\pi \frac{1}{(\sqrt[3]n)^2}) = 4\pi \sqrt[3]{n}$, so as $n\rightarrow \infty$, $S\rightarrow \infty$[/hide]


function
complex analysis
complex analysis theorems

We know the uniqueness theorem for power series :

Suppose Sum(from 0 to infinity) Cn z^n  is zero at all points of a nonzero sequence {zk} which converges to zero. Then the power series is identically zero.

How can we get use of this theorem in the following problem ?

Prove that there is no power series f(z) = Sum(from 0 to infinity) Cn z^n
such that f(1/k) = 1 for k = 2,3,4,..., and f'(0) > 0.

By the continuity of f at the origin the constant term is found to be 1. How can it be proved that Cn is zero for n different from 0?

The function $g(z)=1$ evaluates to 1 at each $\frac1k$. By the uniqueness theorem, $f(z)-g(z)$ is identically zero and $f(z)=1$.

Thanks for the reply, but can you explain a little more? For instance, why uniqueness thm. tells us that f x - g x = 0 ? Why did we choose g(x) = 1 ?


Consider a $n$-sided polygon inscribed in a circle ($n \geq 4$). Partition the polygon into $n-2$ triangles using [b]non-intersecting[/b] diagnols. Prove that, irrespective of the triangulation, the sum of the in-radii of the triangles is a constant.

hint: http://mathworld.wolfram.com/CarnotsTheorem.html

solution: http://mathworld.wolfram.com/JapaneseTheorem.html

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=1094 and http://agutie.homestead.com/files/Geoproblem_B.htm#Sangaku_Old . I proved the particular case for n = 4 in http://www.mathlinks.ro/Forum/viewtopic.php?t=45622 post #3; the proof for generic n is similar.

  darij

I proved it by induction.

Quite east that way too.

Although u have to prove for a quadrilateral (use r=4*R*sin(A/2)*sin(B/2)*sin(c/2))


function
symmetry
geometry
geometric transformation
reflection
algebra
domain

I found it to be $ f(x) \equal{} \frac{1}{x\plus{}1} \minus{}1$...although don't ask me for my working!

Jus tell me atleast the idea , please  :(

I have never studied functional analysis...so the reasoning is going to be quite elementary...

$ f(x + f(y) + xf(y)) = y + f(x) + yf(x)$

Note this holds for all $ x$ and $ y$. So we just swap $ x$ and $ y$ to obtain: $ f(y + f(x) + yf(x)) = x + f(y) + xf(y)$
We have implicitly assumed here that $ y + f(x) + yf(x)  \neq -1$

But we can substitute for the argument on the left to get: $ f^2(x + f(y) + xf(y)) = x + f(y) + xf(y)$.

So $ f^2$ acts identically on the set: $ I = \{x(1 + f(y)) + f(y) \in \Re\backslash\{-1\}: (x,y \in \Re\backslash\{-1\})\wedge(y(1 + f(x))+ f(x) \neq -1)\}$

So $ f(x) = f^{-1}(x)$ for all $ x \in I$. So this means that if one draws a graph $ Y = f(x)$, we would see that at all points where $ f$ doesn't evaluate to $ -1$, we would have reflectional symmetry in the line $ Y=x$.

We first look at the case where there exists $ x_0 \neq -1$ such that $ f(x_0) = -1$. Then substituting into the formula at the top, we have for any $ y \neq -1$, $ f(x_0 + f(y) + x_0f(y)) = -1$. The condition that the argument $ \neq -1$ is the same as $ f(y) \neq -1$. Suppose that there exists $ y_0 > -1$ such that $ f(y_0) \neq -1$.

Then we take a neighbourhood around $ y_0$, in which the function evaluates to $ \neq -1$ - this is possible due to the differentiability, hence continuity, of $ f$. So we extend this neighbourhood to $ (y_1, y_2)$ in which $ f(y)\neq-1$ and either $ y_1 = sup\{y \in (-\infty,y_0): f(y)=-1\}$ or $ y_2 = inf\{y \in (y_0,\infty): f(y)=-1\}$. Without loss of generality, we look at the infimum case. As $ y \to y_1$ we have $ x_0+ (1+x_0)f(y) \to -1$. So there must exist $ \epsilon > 0$ such that every element of $ (-1,-1+\epsilon)$ or $ (-1-\epsilon,-1)$ can be written as $ x_0 + f(y) + x_0f(y)$. Now any element in this open interval therefore must evaluate to $ -1$ under $ f$.

Cases:

Suppose that $ f(y_0) > -1$.  Then the region (see above) is $ (-1,-1+\epsilon)$. The "right branch" of the function would then appear to emanate from a point $ (-1+\eta,-1)$ where $ -1 + \eta = inf\{x \in (-1,\infty): f(x) \neq -1\}$ and increase towards $ f(y_0)$ continously (there can only be one discontinuity in $ f$). But consideration of the reflection would mean $ f$ is double valued on some non-empty subset of the interval $ (-1,-1+\eta)$. So this case cannot occur.

Suppose that $ f(y_0) < -1$. Then the region is $ (-1-\epsilon,-1)$. The "right branch" of the function would then appear to emanate from the point $ (-1-\eta,-1)$ (for $ \eta$ similarly defined as above) to reach $ (y_0,f(y_0))$. It must actualy have negative gradient in order for $ f$ not to be multi-valued when the reflection is accounted for. So me must in fact have $ x_0 < -1$, since we cannot have such a value on the right branch. Also, by considering the reflection of the right branch, we must have $ f$ unbounded above. $ x_0+ (1+x_0)f(y)$ then would allowed to be unlimitingly large and negative. And so one should see that on the far left of the "left branch" we would have $ f(x) = -1$. This cannot happen, as the left branch cannot doubleback on itself - it would mean having a multivalued right branch.

I believe the argument for $ y_0 < -1$ would be identical, but you should really check. So this completes the proof that $ f$ cannot simultaneously have to such points $ x_0$ and $ y_1$. So $ f$ is either identically $ -1$, which can be trivially checked to be true, or it acts on the domain, and so one can write $ f(x) = f^{-1}(x)$ where we can now see $ f$ as a function $ \Re\backslash\{-1\} \to \Re\backslash\{-1\}$.

I need a break now...but I will return with the next portion of a proof. Even I don't know what it will actually be about, but most probably I will be analysing the existance and properties fixed points of the function. There is probably a really simple way to do all of the above...but...it probably invloves some theorems that I don't know.

Haha...rereading I have found a flaw (probably of many) in my argument.

It is in the nature of $ y_1$ or $ y_2$. I think everything is OK if $ x_0$ is greater -1. But if $ x_0 < \minus{}1$ we cannot be sure that f is continuous on $ (y_1,y_2)$ - it may include -1, over which the function is not continuous...

[quote="vijaymenon"]If  $ f: R \minus{} ( \minus{} 1 ) \rightarrow R$ and $ f$ is  differentiable function satisfies :

$ f ( x \plus{} f(y) \plus{} xf(y) ) \equal{} y \plus{} f(x) \plus{} yf(x)$

$ x, y \epsilon R \minus{} { ( \minus{} 1) }$ then find $ f(x)$[/quote]
taking y as constant differentiate wrt $ x$
$ f'(x \plus{} f(y) \plus{} xf(y)) (1 \plus{} f(y)) \equal{} f'(x) (1 \plus{} y )$
Similarly take x constant differentiate wrt y
 $ f'(x \plus{} f(y) \plus{} xf(y))( f'(y)(1 \plus{} x)) \equal{} (1 \plus{} f(x))$
Diving both
$ \frac { 1 \plus{} f(y)}{f'(y) (1 \plus{} y )} \equal{} \frac {(1 \plus{} x)f'(x)}{ 1 \plus{} (x)} \equal{} k$
$ f'(y) \equal{} \frac {1 \plus{} f(y)}{k(1 \plus{} y)}$
$ f'(x) \equal{} k \frac {1 \plus{} f(x)}{ 1 \plus{} x}$
$ k \equal{} \frac {1}{k}$
$ k \equal{} \plus{} 1$or$ \minus{} 1$
$ \frac {f'(x)}{1 \plus{} f(x)} \equal{} [\plus{}1 or (\minus{}1)]\frac {1}{1 \plus{} x}$
$ f(x) \equal{} \alpha (1 \plus{} x)^{ \plus{}1 or (\minus{}1) } \minus{} 1$
$ f (f (0)) \equal{} f(0)$
$ \alpha^{ 1 \plus{} or (\minus{}1) } \equal{}\alpha$
+1 will give $ \alpha \equal{} 0 or 1$
f(x)=-1 or x
but both take -1 in x or y .
-1 will give $ \alpha \equal{} 1$

$ f(x) \equal{} \frac { \minus{} x}{1 \plus{} x}$
that's d answer .  :)

Yeah solution looks good, but can't help thinking that it lacks rigour

That differential equation $ \frac{f'(x)}{1\plus{}f(x)}\equal{}\pm\frac1{1\plus{}x}$ is easily solved with separation of variables; integrating, we get $ \ln(1\plus{}f(x))\equal{}\pm\ln(1\plus{}x)\plus{}C$, or $ f(x)\equal{}A(1\plus{}x)^{\pm 1}\minus{}1$.

Testing these: if it's $ A(1\plus{}x)\minus{}1$, the defining equation becomes
$ A(1\plus{}x\plus{}f(y)\plus{}xf(y))\minus{}1\equal{}y\plus{}f(x)\plus{}yf(x)$
$ A(x\plus{}1)(f(y)\plus{}1)\equal{}(y\plus{}1)(f(x)\plus{}1)$
$ A^2(x\plus{}1)(y\plus{}1)\equal{}A(y\plus{}1)(x\plus{}1)$ and $ A^2\equal{}A$, for either $ A\equal{}0$ and $ f(x)\equal{}\minus{}1$ or $ A\equal{}1$ and $ f(x)\equal{}x$.

If it's $ \frac{A}{1\plus{}x}\minus{}1$, the defining equation becomes
$ \frac{A}{1\plus{}x\plus{}f(y)\plus{}xf(y)}\minus{}1\equal{}y\plus{}f(x)\plus{}yf(x)$
$ \frac{A}{(x\plus{}1)(f(y)\plus{}1)}\equal{}(y\plus{}1)(f(x)\plus{}1)$
$ \frac{A}{x\plus{}1}\cdot\frac{1\plus{}y}{A}\equal{}(y\plus{}1)\cdot\frac{A}{x\plus{}1}$ (WLOG, $ A\neq 0$; that case is already covered)
$ A\equal{}1$, for $ f(x)\equal{}\frac{1}{x\plus{}1}\minus{}1$.

Side note to kabi: \pm is a useful code.

These three solutions work, and they are everything- at least when restricted to one side of $ \minus{}1$. Can we mix and match?

For convenience, write $ g(x)\equal{}f(x\minus{}1)\plus{}1$; the equation becomes $ g((x\plus{}1)(g(y\plus{}1))\equal{}(y\plus{}1)(g(x\plus{}1))$, or shifting the variables, $ g(x\cdot g(y))\equal{}y\cdot g(x)$. The pieces we build from are $ g(x)\equal{}x$, $ g(x)\equal{}0$, and $ g(x)\equal{}\frac1x$.
If $ g(x)\equal{}x$ for $ x>0$, we have $ g(x\cdot g(y))\equal{}xy$ for $ x>0,y<0$. The three possibilities for the left-hand side are $ xy$, $ 0$, and $ \frac{x}{y}$; only the first works, and $ g(x)\equal{}x$ globally.
If $ g(x)\equal{}0$ for $ x>0$, we have $ g(x\cdot g(y))\equal{}0$ for $ x>0,y<0$. The left-hand side is either $ xy$, $ 0$, or $ \frac{x}{y}$, so again $ g(x)\equal{}0$ globally. This one stretches the original functional equation; we need to extend to zero for the expressions to make sense.
If $ g(x)\equal{}\frac1x$ for $ x>0$, we have $ g(x\cdot g(y))\equal{}\frac{x}{y}$ for $ x>0,y<0$. As before, this forces $ g(x)\equal{}\frac1x$ globally.
There is no mixing, and these three (or two, if we don't allow the extension) solutions are everything.


geometry
analytic geometry
graphing lines
slope

The triangle with vertices at $ (2,4), (2,6) and (2\plus{}\sqrt{3},5)$ is

[b](a) right angled.
(b) right angled and isosceles
(c) equilaterlal
(d) obtused[/b]

Its answer is (a). But there is a problem i.e. if I solve it using two different ways it gives different answers.Plz help.

[b]Solution 1[/b] : Answer: [b](a)[/b]

Find out the slope of the line joining [b](2,4)[/b] and [b](2,6)[/b] is

$ m\equal{} (6\minus{}4)/(2\minus{}2) \equal{}$[b]infinity ,so m=tan(theta)=90 degree (say theta is the angle which this line makes with the x-axis)[/b]

so the tringle becomes right angled.

[b]Solution 2[/b] : Answer: (c)

[b]To find out the distance of each line[/b]

a)line joining $ (2,4) and (2,6)$
$ \sqrt{(6\minus{}4)^2 \plus{} (2\minus{}2)^2} \equal{} \sqrt{4\plus{}0}\equal{}2$

b)line joining $ (2,4) and (2\plus{} \sqrt{3} , 5)$
$ \sqrt{(5\minus{}4)^2 \plus{} ((2\plus{} \sqrt{3})\minus{}2)^2} \equal{} \sqrt{1\plus{}3}\equal{}2$

c) line joining $ (2\plus{} \sqrt{3} , 5) and (2,6)$
$ \sqrt{(6\minus{}5)^2 \plus{} (2\minus{}(2\plus{}\sqrt{3})^2} \equal{} \sqrt{1\plus{}3}\equal{}2$

So the three lines are equal.So triangle should be equilateral. Please tell me where I am wrong. Thank you so much.

hello, let $ A(2,4)$, $ B(2,6)$ and $ C(2\plus{}\sqrt{3},5)$ so we have
$ \overline{AB}\equal{}\sqrt{2^2}\equal{}2$
$ \overline{BC}\equal{}\sqrt{3\plus{}1}\equal{}2$
$ \overline{CA}\equal{}\sqrt{3\plus{}1}\equal{}2$
so your triangle ist equilateral.
Sonnhard.

Because the slope between $ (2,4)$ and $ (2,6)$ is infinite doesn't mean that there is a right angle.  The point $ (2 \plus{} \sqrt {3},5)$ is above the point $ (2,4)$, and thereofre would make the angle $ A$ less than $ 90^\circ$.  It follows then because of everyone else's work that the answer is $ c$.

Thanks to both of you for your help. Thanks to "rd5493" who made me clear how it is not a right-angled triangle.I got that.Thank you so much.


factorial
limit
logarithms
algebra
polynomial
inequalities
real analysis

Stirling's Approximation for the factorial is a well-known and useful tool. Using it, we know that

$\lim_{n\to\infty}\binom{2n}{n}\cdot \frac{\sqrt{\pi n}}{4^{n}}= 1$

But in the interests of asymptotic analysis we might wish to know something about the rate of convergence. In other words, what can we say about the difference

$d(n) = \binom{2n}{n}\frac{\sqrt{\pi n}}{4^{n}}-1$

when $n$ is large? Computations suggest that $d(n) = O(n^{-3/2})$ and that
$\lim_{n\to\infty}d(n)n^{3/2}= \frac{-1}{8\sqrt{\pi}}$

I have not yet had any ideas about how to approach proving this. Anyone interested? :)

The full asymptotic series version of Stirling's formula:
For each $N$,
$\lim_{z\to\infty}z^{2N}\left(\log \Gamma(z)-\left((z-\frac12)\log z-z+\frac12\log (2\pi)+\sum_{m=1}^{N}\frac{B_{2m}}{2m(2m-1)z^{2m-1}}\right)\right)$
$=0$, where $B_{2m}$ are the Bernoulli numbers. Since the Bernoulli numbers grow faster than any exponential, the series diverges as $N\to\infty$ for all $x$.
The Bernoulli numbers alternate between being positive and negative; we can estimate the error in any partial sum by either the last term added or the next term after that. This error is minimized when $N$ is approximately $\pi z$, with a value of about $e^{-2\pi z}$ (ignoring some polynomial factors).

The best way to find Bernoulli numbers with large indices is, ironically, Stirling's formula: $B_{2n}=2\zeta(2n)\cdot\frac{(2n)!}{(2\pi)^{2n}}$. We can estimate $\zeta(2n)$ very accurately, and there are ways to bound the denominator of $B_{n}$ so that the asymptotic series determines $B_{n}$ uniquely.

There's a derivation of the above series in Ahlfors, although he never states it in that form.
This is serious overkill for your problem, of course. RIP gave the estimate up to the first term of the sum, which is sufficient.

[quote="RIP"]
\[d(n)=-\frac1{8n}+O(\frac1{n^{2}}) \]
Use
\[n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{\frac1{12n}+O(\frac1{n^{2}})}\]
if I'm right.[/quote]
Okay just for posterity and my own edification I will work it out here.

$d(n) = \binom{2n}{n}\frac{\sqrt{\pi n}}{4^{n}}-1$

$= \frac{\sqrt{4\pi n}(2n/e)^{2n}e^{1/(24n)+O(1/n^{2})}}{2\pi n (n/e)^{2n}e^{1/6n+O(1/n^{2})}}\cdot \frac{\sqrt{\pi n}}{4^{n}}-1$

$= e^{-1/(8n)+O(1/n^{2})}-1$
$= 1-\frac{1}{8n}+O(\frac{1}{n^{2}})-1$
$=-\frac{1}{8n}+O(\frac{1}{n^{2}})$

Thanks very much guys. :)
Similarly by taking more of the series in the exponential we can make a whole asymptotic expansion for $d(n)$, if we really wanted to.

I have heard that the following inequality is valid for all $n\in \mathbb{N}$:
\[\sqrt{2 \pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12n+1}}<n!<\sqrt{2 \pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12n}}.\]


Find the best constant $ c$ such that for any positive real numbers $ a_1, a_2,\ldots , a_n$ and any $ n \geq 3$ the following inequality holds:
\[ \frac {a_1}{a_2 \plus{} a_3} \plus{} \frac {a_2}{a_3 \plus{} a_4} \plus{} \cdots \plus{} \frac {a_n}{a_1 \plus{} a_2}\geq cn.\]
.

I think it's very hard. But I know c is bigger than $ 2^{1/2}\minus{}1$

yes, the inequality holds for $ c\equal{}\sqrt{2}\minus{}1$ but that is not the optimal value..what I know is that only one man has ever solved this inequality--Vladimir Drinfield..but I think harazi, a mathlinker, also knows the solution...


Are there an infinite number of primes of the form $n^{2}+1$?
I found this problem in a book (with no answer).  I think that the answer is yes, but how do prove it?

One of the most posted open problems here.
(especially there is no proof known)

Oh...thanks.  I feel much better now that I could not prove it!


$Question 1$
Two identical rectangles having integer length and width are placed on top of each other, with their centres coinciding and at right angles.  The outline of the resulting figure is a fat cross.  Suppose that the area of the cross is $703687 cm^{2}$.  Find its perimeter.

$Question 2$
$23, 235, 2357, 235711, 23571113 \cdots d$ etc.
In the above sequence $23$ and $2357$ are primes.  Now continue this series. When do you encounter the next prime?

[hide="HINT Question 1"]
$l = \text{length}\ , \ w = \text{width}$
$2 \left( l-w \right) \left( w \right)+w^{2}= 703687$
[/hide]

Solution to #1:

[hide]Note that 703687 has only two factors: 751 and 937.

a = length, b = width and a > b, we have

2ab-b^2 = 703687 or
2a = (703687)/b + b = 751 + 937
a = 844

Perimeter = 4a = 3376 cm[/hide]

Question #2 is more difficult.

[quote="xxxyyyy"][quote="HTA"][quote="stuck"][hide="Hint Question 2"]
The sequence is formed by appending the next prime to the resulting number. So the next number in the sequence is $2357111317$, or $23571113$ with $17$ appended to it.
[/hide][/quote]
i think you understood wrong the problem ,the question is to find out the next prime in this sequence not only continue this sequence  :wink:[/quote]

The fourth in the series is a concatenation of $128$ primes ending in $179$, It has some special name too which I don't remember. This number contains $355$ digits and is:

[hide]
$23571113171923293137414347535961677173798389971011$
$03107109113127131137139149151157163167173179181191$
$19319719921122322722923323924125125726326927127728$
$12832933073113133173313373473493533593673733793833$
$89397401409419421431433439443449457461463467479487$
$49149950350952152354154755756356957157758759359960$
$16076136176196316416436476536596616736776836917017$
$09719$[/hide][/quote]
Hmm, this is an interesting problem.  Are there infinitely many primes of this form?

[quote="miyomiyo"]Are there infinitely many primes of this form?[/quote]

Now THAT'S interesting!!!

I honestly don't know.


geometry AMC AIME quadratics algebra polynomial logarithms \/closed

On Thursday, October 14th I will be hosting a Math Jam discussing problems similar to those that will be used in the upcoming AMC-12 Problem Series class.

Student who took last years combined AMC-12/AIME problem series will see many new problems as we now have a problem series for each of these exams.  We have also expanded the number of practice problems that will be distributed on the Message Board.

We recommend this course most highly to students who believe they are likely to score between 100 and 135 on the AMC-12.  Students who believe they cannot yet achieve a 100 on the AMC-12 are probably better off enrolling in our new AMC-10 Problem Series.

The structure of the courses have also changed dramatically.  This year the problems will be discussed by topic area so that students have a chance to master each concept one idea at a time.  Topic areas are chosen according to what we consider to be the most important or broadly covered areas of mathematical problem solving tested by the AMC and AIME exams.

Last year a dramatic majority of the students enrolled in the AMC/AIME problem series class took the AIME and many took the USAMO.  A few attended the Mathematical Olympiad Summer Program.

The following is the course outline for the AMC-12 Problem Series:

Lesson I: Arithmetic, Quadratics, Absolute Values
Lesson II: Algebraic Techniques
Lesson III: Number Theory
Lesson IV: Polynomials, Logarithms, and other Functions
Lesson V: Binomial Expansion, Inequalities, Optimization, Systems of Equations
Lesson VI: Counting and Probability
Lesson VII: Statistics, Sequences, and Series
Lesson VIII: Complex Numbers, Trigonometry, Law of Cosines
Lesson IX: Right Triangles, Quadrilaterals, Pick's Theorem
Lesson X: Circles and Spheres
Leson XI: Triangles, Polygons, Polyhedra
Lesson XII: Coordinates, Graphs, and Geometric Inequalities

[quote="MCrawford"]We recommend this course most highly to students who believe they are likely to score between 100 and 135 on the AMC-12.[/quote]

What my family learned last year from JBL is that the best way to get an AIME-qualifying score on an AMC test is to dare not to answer all the questions. A test-taker who answers fifteen questions CORRECTLY on the AMC 12 and leaves the other ten questions blank ought to be able to break 100, by the scoring rules, so I think even the seventh-graders on AoPS (who are a pretty special group of seventh graders, on average   ;)  ) ought to try to take the AMC 12, which suggests that they ought to take the online class.

I would only recommend this class to the strongest seventh graders.  The class is not simply AMC-12 problems, but focuses on the harder problems in particular.

[quote="tokenadult"]...the best way to get an AIME-qualifying score on an AMC test is to dare not to answer all the questions....A test-taker who answers fifteen questions CORRECTLY on the AMC 12 and leaves the other ten questions blank ought to be able to break 100, by the scoring rules,...[/quote]

I'm not sure I'd recommend that strategy myself.  I have a friend who uses (or, at least, used to use; not sure if he will this year) that strategy, answering only as many problems as he need to to make AIME, but he always missed the qualifying score because he made one mistake and got a problem wrong.  Better to have a bit of a buffer for instances like that (but you should still not answer questions you don't know how to do).  Besides, for all you 11th and 12th graders out there, if your goal is USAMO, you want, in addition to a high AIME score, a high AMC score (since it's the sum of the two that counts).

The strategy must fit the student.  Those not expecting to qualify for the USAMO might tend to answer fewer questions to be on the safe side.

I know that some people would disagree with me here, but I wanted to mention it anyway.  When I took the the AHSME (the exam that was basically the AMC-12 in the past) and the AIME, I never once thought about strategy.  I simply worked problems until I solved them.  I still believe this is the best strategy, though I know that people are looking for all kinds of ideas to fit certain goals so I do talk strategy a bit.

I m going to agree with Mr. Crawford on a few things. First, there are not very many 7th graders for whom this class would be accessible to.  But that doesn't mean there aren't 7th graders who are well prepared for the class.  As always it depends on the person.  The problems in the class are mostly the harder AMC questions.  I  would say that you should be able to answer(or at least make decent progress) a good deal of the #10-#20 level questions on the AMC 12 if you really want to get the full benefit of the class.

Next, I agree with him on the strategy for taking the AMC.  Try and answer as many questions as you can.  Don't just settle for a 100 when you can get a 105, or a 110 when you could get a 115, etc...  What I did, however, last year is I knew how many I needed to make AIME.  From when the test started I pretty much forgot about that and worked through as many problems as I could do or looked like I could do.  I ended up getting around a 106.5.  So I guess basically have somewhat of a strategy but nothing can make up for good old practice.

[quote="mna851"]If a student can already get in between a 100 and 135 on the AMC 12 what purposes would this online class serve them.[/quote]Because there is a big difference between (say) a 100 and a 140 with respect to USAMO qualifying, or even between (say) a 125 and a 135.

[quote="mna851"]If a student can already get in between a 100 and 135 on the AMC 12 what purposes would this online class serve them.[/quote]

Another thing I would note is that the AMC 12 Problem Series focuses on building up the skills that help students tackle the harder problems on the exam.  These skills certainly also transfer to the AIME.  Essentially, the AMC 12 Problem Series is already designed for students who make between a 105 and a 135 on the AMC 12.

Students who hover around the 80-105 range on the AMC 12 are currently better off taking the AMC 10 Problem Series.


how many solution, in $\mathbb{N},$ does the equation:
$n!+3=3^{n-1}$
have?
calculators (or computers :) ) are not allowed! :D

[hide="solution"]$n!+3 = 3^{n-1}$

if $n \ge 4$, then $n! > 3^{n-1}$. so n could be 1,2, or 3.
since $n! = 3^{n-1} - 3$, $n!$ has to be divisible by 3.
so the only solution is 3.[/hide]


Sa se determine functia $f: [0,\infty)\rightarrow\mathbb{R}$, astfel incat
\[f(x)+\sqrt{f^{2}([x])+f^{2}(\{x\})}=x,\]
oricare ar fi $x\in [0,\infty).$

Dându-i valori lui x ,  am gasit :

$f(x) = x-\frac{1}{2}\sqrt{x^{2}-2[x]\{x\}}$ . Sper ca nu am gresit undeva ..

Nu cred ca e buna solutia. Puteti detalia?

Pai , eu m-am gandit asa :

Primadata pentru $x \gets 0$ avem :

$f(0) = 0$ (1) ;

apoi pentru $x \gets [x]$ obtinem :

$f([x]) = \frac{[x]}{2}$ (2).

pentru $x \gets \{x\}$ obtinem :

$f(\{x\})= \frac{\{x\}}{2}$ (3) .
Inlocuim $(2)$ si $(3)$ in ipoteza si obtinem :

$f(x)+\sqrt{f^{2}([x])+f^{2}(\{x\})}= f(x)+\sqrt{\frac{[x]^{2}+\{x\}^{2}}{4}}= x$

De aici $f(x) = x-\frac{1}{2}\sqrt{[x]^{2}+\{x\}^{2}}= x-\frac{1}{2}\sqrt{x^{2}-2[x]\{x\}}$ . Pentru numerele carora le-am facut eu verificare a mers ... acuma nu stiu ..poate mi-a scapat ceva ..

[quote="Marius Damian"][quote="cckek"][quote="Marius Damian"]Sa se determine functia $f: [0,\infty)\rightarrow\mathbb{R}$, astfel incat
\[f(x)+\sqrt{f^{2}([x])+f^{2}(\{x\})}=x, \]
oricare ar fi $x\in [0,\infty).$[/quote]

$f(x)=x-\frac{k}{2}$ când $x \in [k,k+1), k\in \mathbb{N}$ ???[/quote]
Puteti detalia?[/quote]

Evident $f(0)=0, f(k)=\frac{k}{2}$. Cand $x\in [0,1)$ atunci $x=\{x\},\quad [x]=0$. Deci cand $x\in [0,1)$ atunci $f(x)+\sqrt{f^{2}([x])+f^{2}(\{x\})}=f(x)+\sqrt{f^{2}(x)},\implies f(x)=\frac{x}{2}$
Cand $x\in[k,k+1)$ atunci $[x]=k,\{x\}=x-[x]\in [0,1)$ deci
$f(x)+\sqrt{\frac{k^{2}}{4}+(\frac{x-k}{2})^{2}}=x\implies f(x)=x-\sqrt{\frac{x^{2}-2kx+2k^{2}}{4}}$
Dar $k=[x]\implies f(x)=x-\sqrt{\frac{x^{2}-2[x]x+2[x]^{2}}{4}}$

Ceace este echivalent cu rezultatul lui turcas_c:)


algorithm
factorial
number theory
least common multiple

An interesting problem which I believe can be solved by a logical approach. But the 2nd part is more interesting.

Given a positive integer $ n$, construct $ n$ consecutive composite numbers.

For Example: If n=5
One of the answers: 24,25,26,27,28.

Give the ans in terms of n. Also is your answer the smallest possible n consecutive numbers. (ie: In above case ans may also be, 62,63,64,65,66). If not try deriving an algorithm for smallest possible case. :P

~ Shreyas

[quote="hell_ever"]first part is very famous.
Consider (n+1)!+2, (n+1)!+3,.....,(n+1)!+(n+1)[/quote]

Right you are, but something smaller. Why factorial?

Ans for first part
[hide] lcm(1 to n+1)+2, lcm+3,.....[/hide]


Anyone try Google's new web accelerator? I did, and it makes loading aOPS A LOT faster...you should too

I just downloaded it minutes ago. I don't see much difference yet.

Did you set your preferences? When I changed the setting from default DSL option to my connection speed I really saw a difference.

Go to google.com, click on "more," scroll down to labs, click on labs, then the first one you see is google web accelerator...make sure you set your preferences if you download it...(connection speed on preferences is important)

It seems to me that it doesn't help my windows 98.

very fast..... yay... i have DSL from MSN, and it feels like Comcast now.

hmm...I'm not too sure about this whole giving Google the right to look through everyplace you go thing...

Do you have something to hide, MithsApprentice?  Are you a spy??? :D

it is done being offered. It says they are doing furthur  stuff to increase the capacity.

Does anyone know any other places to download it? Because I really want to try it. Or can some one send me the install file?

Yeah I would enjoy downloading it also.. Is it connected to Google? Because then even if you send it to us it wont work... It's worth a try though.

Yes, it is connected to Google, so even if you get it unofficially, it won't work for you. Sorry! I was really disappointed myself when I saw nobody else was being accepted.

Do you mean they closed off the download?

question- i have the fastest DSL available... will i benefit from this? anyone with DSL?

I have Comcast. It claims that it's saved 1.3 hours, but I don't notice too much except maybe a second off on every AoPS page because they are sloooooooooooow ;). Oh well, it can't hurt to try I guess.

Google Accelerator runs on a pre-fetch system; meaning, not that great of an idea!

Links:
[url=http://www.37signals.com/svn/archives2/google_web_accelerator_hey_not_so_fast_an_alert_for_web_app_designers.php]Signal vs. Noise[/url]
[url=http://www.boingboing.net/2005/05/06/google_accelerator_i.html]BoingBoing[/url]
[url=http://www.eweek.com/article2/0,1759,1813761,00.asp?kc=EWRSS03119TX1K0000594]EWeek[/url]
[url=http://radar.oreilly.com/archives/2005/05/google_web_acce_1.html]O'Reilly[/url]

or just do a Google News search
[url=http://news.google.com/news?q=google%20accelerator&hl=en&lr=&client=firefox-a&rls=org.mozilla:en-US:official&sa=N&tab=wn]GNews: Google Accelerator[/url]

Yea somehow I don't think it's completely accurate. Shoot like the first 100 websites I went to the lil ticker jumped up about .4 seconds each time. Now it goes .4 minutes...

How do you download this after all??
I went to[url]http://webaccelerator.google.com[/url] but there was nowhere for me to download.......

They took it down after like a week because of their user capacity. The tool is only beta. However, I think that the real issue is the bugs that have been found.

[quote="joml88"]grrr...it keeps screwing up the way my blog showed up for me so I'm turning it off.[/quote]

maybe you can give it to me?

Well anyone who's downloaded it can give it to you. You can still install it if you have the file.

In the meantime, mine now shows 1.5 hours. Funny how it increases so fast now. It used to be a fraction of a second saved. I guess now it's a fraction of an hour.

i would appreciate if someone give it to me. PM or email. (Well I guess PM sorta doesnt work for file sending :lol: ....so email would be great :) )


A data set for a class of 25 sixth graders has their ages listed as the integer values of either 10 or 11 years. The median age in the data set is 0.36 years greater than the mean. How many 10-year-olds are in the class?

the median must either be 10 or 11, so the mean must be either 10-0.36=9.64 or 11-0.36=10.64

9.64*25=241, but the minimum age is 10, so teh minimum total age is 250. so 241 isnt possible.

10.64*25=266; so theres either 6 11's or 16 11's

6*11+19*10=256, so thats not right
so theres 16 11's and 9 10's

9

kool
i'll just latex

the median must either be $ 10$ or $ 11$, so the mean must be either $ 10\minus{}0.36\equal{}9.64$ or $ 11\minus{}0.36\equal{}10.64$

$ 9.64 \times 25\equal{}241$, but the minimum age is $ 10$, so teh minimum total age is $ 250$. so $ 241$ isnt possible.

$ 10.64 \times 25\equal{}266$; so theres either $ 6 11$'s or $ 16 11$'s

$ 6 \times 11\plus{}19 \times 10\equal{}256$, so thats not right
so theres $ 16 11$'s and $ 9 10$'s

$ \boxed {9}$


function
algebra
domain
calculus
calculus computations

I have $ f(B)=\{n+1; n \in B \})$

The questions were

1. Is this a surjective function?
2. prove this

my answer is yes, but how to prove this?

thanks guys

Checking surjectivity requires knowing the domain and the range, which you didn't specify. Anyway, the general idea is to take an arbitrary $ C$ in the range and find $ B$ in the domain such that $ f(B)\equal{}C$. Here you do not have much choice: the only possible $ B$ is $ \{n\minus{}1: n\in C\}$.


1-2sin^2(5pie/12)

sorrie i dont know how to put the pie symbol on but its five pie over 12 thanks.

just use the identity $1-2\sin^{2}\theta = \cos 2\theta$

how do you simplify though using 5pie/12?  like can anyone show me the full step to the end after its simplified thanks..

Well, you can rewrite that equation as
$\cos\frac{10\pi}{12}$
or
$\cos\frac{5\pi}{6}$

Which is fairly straightforward.

[quote="Treething"]If $\theta = \frac{5\pi}{6}$, then $2\theta = \frac{5\pi}{6}$...[/quote]
nice....

@krnboy32: it's spelled pi, not pie :wink:

If $\theta = \frac{5\pi}{12}$.

Holy crap I'm making a lot of really stupid mistakes...


What is the largest integer $ N$ for which we can say that $ n^{5} \minus{} 5n^{3} \plus{} 4n$ is divisible by $ N$ for every integer $ n$?

[hide="Solution"]\begin{align*}n^5-5n^3+4n&=n(n^4-5n^2+4)\\
&=n(n^2-4)(n^2-1)\\
&=n(n-2)(n+2)(n-1)(n+1)\\
&=(n+2)(n+1)(n)(n-1)(n-2)\end{align*}
In every $ 5$ consecutive integers, there is one number that is divisible by $ 2$, one that is divisible by $ 3$, one that is divisible by $ 4$ (not the same as the one that is divisible by $ 2$), and one that is divisible by $ 5$.  Therefore, $ N=\boxed{120}$[/hide]

[quote="jxl28"]
\begin{align*}n^5 - 5n^3 + 4n & = n(n^4 - 5n^2 + 4) \\
& = n(n^2 - 4)(n^2 - 1) \\
& = n(n - 2)(n + 2)(n - 1)(n + 1) \\
& = (n + 2)(n + 1)(n)(n - 1)(n - 2)\end{align*}
In every $ 5$ consecutive integers, there is one number that is divisible by $ 2$, one that is divisible by $ 3$, one that is divisible by $ 4$ (not the same as the one that is divisible by $ 2$), and one that is divisible by $ 5$.  Therefore, $ N = \boxed{120}$[/quote]

To make your last statement more rigorous, you could note that $ (n + 2)(n + 1)(n)(n - 1)(n - 2) = \binom{n + 2}{5}$ which means that $ \frac {(n + 2)(n + 1)(n)(n - 1)(n - 2)}{5!}$ is an integer for all $ n$, and hence 5! will be the largest divisor of $ (n + 2)(n + 1)(n)(n - 1)(n - 2)$ for all integers $ n$ so our answer is $ N = 5! = \boxed{120}$

as n^5-5n^3+4n is the product of 5 consecutive numbers,it is divisible by 5!=120


It's not true for $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$...


geometry
geometric transformation
Support

There are said to have been only eleven times in history where the "F" word has been considered acceptable for use.

They are as follows:

11.) "What the **** do you mean we are sinking?"
*** Captain E.J. Smith of RMS Titanic, 1912

10.) "What the **** was that?"
Mayor of Hiroshima, 1945

9.) "Where did all those ****ing Indians come from?"
Custer, 1877

8.) "Any ****ing idiot would understand that."
Einstein, 1938

7.) "It does so ****ing look like her."
Picasso, 1926

6.) "How the **** did you work that out?"
Pythagoras, 126BC

5.) "You want WHAT on the ****ing ceiling?'
Michelangelo, 1566

4.) "Where the **** are we?"
Amelia Earhart, 1937

3.) "Scattered ****ing showers, my ass!"
Noah, 4314BC

2.) "Aw c'mon. Who the **** is going to find out?"
Bill Clinton, 1999

..........and a drum roll for ...................................

1.) "Geez, I didn't think they'd get this **** mad."
Sadam Hussein, 2003
 :D

lol

I like the last one and number 10 was good and same with number 2. Are these really true though?

I highly doubt these are true. but funny nevertheless.
 :D

[quote="juicybooty911"]hahaha, awesome, those are ****ing funny :D
Where did Clinton say that sentence?[/quote]I suppose that in precence of his former assistant!  :D but that should have been a lot before 1999

Ha ha. Yeah, those are pretty good.  :)  :)  :lol:  :P

[quote="j3lincl"]Actually all of them are fake, a friend of mine showed them to me from some random site.[/quote]Of course they are!! C'mon, it is only some humor. You need to be very fool to not notice it. First, some of those people never speak english (in some cases, english was not even created yet), and in the others, well.... c'mon....

[quote="aznness"]i dont know if the f-word was around at 126bc[/quote]Well, probably some translation to the languages spoken back then, but for sure not that word. According to a website, the first time that it was recorded such a word was in a few erotic poems written during the first decade of the sixteenth century (well, at the begining it was not that dirty at it is considered today, or at least I don't think that a poet of this days would seriusly place such a word in erotic poems) but any way, for sure, it was not around by 126BC!

[quote="aznness"]i think this is getting kinda off topic.[/quote]So what? it is G&F forum. As long as it is kinda related and has a respectful (but also informal is acceptable) tone, I suppose there shouldn't be a problem!  :P

Those are great - I think the Custer one came from a Far Side cartoon or something...

I would say the clinton and the custer ones are the best.

General CUster, mid 1800's I think, famous Indian killer, but he got misinformed and surrounded by said Indians who then killed him and all his men. That battle was the Battle of Little Big Horn. Little Big Horn is just inside the border of Montana, and Big Horn is in Wyoming.  :?

and battle of little bighorn- typical referred to as "custer's last stand." he was ambushed pretty bad, they killed all his men, and thus he was heroified. pity, really, but thats americanism. we love our heroes.

of course, he was scalped by the indian tradition of sign of victory. That must have been pretty horrifying.

Hahaha...that's great. You don't mind if I copy-paste onto my Xanga do you?

#1 seems misplaced but oh well. :-)

[quote="stupidkid"]Haha, now i acutally understand number 9.  I don't think the first one is really that funny.[/quote]Neither do I, but well... say that it is somehow a psicological support for the troops there!