Greetings all!
i wanted to alert you to an on-site competition happening in India in December. i attended the last such contest in May of 2002 and plan to go back this year. You can find out some information (but unfortunately not much about the structure of the contest itself) at [url]http://www.cmseducation.org/iymc/[/url]
i am in contact with the organizers to get more information about the structure (a proposal to use tests similar to [url=http://mathleague.org]GPML[/url]/[url=http://www.arml.com]ARML[/url] failed) and hope to have that soon. In the meantime, anyone interested in participating in the contest is invited to contact me and i can help answer any questions and give some information about India.

Tim Sanders
tim@mathleague.org

farmertim, I'm very interested in it. But is there any age restriction for this contest? IWYMIC only allows students of grade 9 or below to join.
In fact, I'm going to seat for IWYMIC on Wednesday this week. Tomorrow I will leave for Macau, the organising city this year.

This is a contest for older students, since the IWYMIC already serves younger students.  It is for students who have not yet completed high school as far as i know.  Good luck at IWYMIC; say hi to Simon Chua for me if he's there, and be sure to let the students you meet know about AoPS and encourage them to join this site..

ha.............
I finally came back from macau.
The horrible combinatorics has driven me to death.............

[quote="linus9891"]Hi guys, I'm also interested with this competition but I am wondering what the problems are like. Do you know where you can get archives of the questions? Also, how long is the trip?[/quote]

There are not so much any resources containing past tests that i know of, and the format is pretty much subject to change at any time if past experience is any indication.  i think the format and rules are intentionally vague at this point because a lot of that will likely be determined on site by the assembled coaches.  The length of the conference is on the website, and the trip itself takes about a full day each way..

Hi ! I'm also interested but I'm far away from India so my questions are:

1.) How to apply ? Is there any test, if yes then what level [ie IMO level ?] and can I take it at home [because it's impossible for me to go to India just to take on an exam].

2.) Are there any "grants" - I mean do they pay for your trip ? or do you have to pay for everything yourself ?

Can you give me more info of as to how I can send school children for this compitition

If anybody knows the link for past paper please share


i disagree
think
in the ctc tourney mewto is  paired up with romaniun gfenuies
and levans could check the ip adress and ban one of them if they had the same ip

RomanianGenius and mewto55555 use computers with different IPs, otherwise mewto could not gain rating from RomanianGenius.  And yes, they [i]do[/i] play each other, or at least have in the past (currently not as often, presumably because of my accusing mewto), which is why superllamel is now ahead of mewto in rating.  Further, mewto has stated many times that RomanianGenius is his brother, thus even if they used the same IP, which they most likely have at one point or another, there is no need to ban.

RomanianGenius is paired with mewto in the CTC team tournament.  However, how does this make sense, considering that mewto beats RomanianGenius every single time he plays him?  mewto doesn't even beat [i]me[/i] every single time.  Thus, it would follow that RomanianGenius is worse than me based upon our recent records against mewto.  It would also follow that RomanianGenius is worse than isabella, who was available as a partner at the time mewto selected RomanianGenius as a partner.  Thus, mewto chose a [i]seemingly [/i]terrible partner instead of a very good one.  This would only make sense if mewto were using RomanianGenius to boost his rating, and is playing in the tournament as both.  Since mewto thinks he's better than isabella, this would make perfect sense.

And this is not the full extent of my evidence, for anyone who's wondering.  I'll give more if provoked.  :P

Note: What I say is not to be taken as fact simply because I'm a moderator on FTW.Think and decide for yourselves based on the evidence presented.

Iin, I am provoking you.  :P I would like to hear all of your evidence. I think it's a little suspicious that mewto always beats RomanianGenius, but I'm not totally convinced that they are the same person.

Okay, super.  The following evidence is based on the [i]habits [/i]of mewto and RomanianGenius.

Back when I played more--about a month ago, perhaps--I CDed quite a lot and basically CDed anyone who wanted to (because that's what I do).  One day, mewto CDed me.  Immediately before the CD, mewto said "I lose," just like he always does before a CD.  I don't remember precisely how the answering went, but when he got a correct answer, he would say precisely how he got the answer and "I lose" or something similar before the next problem started.  He would do this on every problem he got right.  I lost, of course, and immediately afterward, RomanianGenius came on and wanted to CD me.  He proceeded to say "I lose" before the CD started, give the method of solving for the problems he got right, say "I lose" or "you win" in between problems, and beat me, of course.

When confronted, mewto explained this by saying "brothers act similarly," which is an extremely weak argument, given the nature of the similarities.

Anyone who has answered no to this poll, please state a specific reason as to why.  Perhaps I'll convince you, and perhaps you will convince me.  :o

RG is about 200 lower.  Also, mewto told me that he uses modems, which would explain the different IP addresses.  I thikn theyre the same.

EDIT: oh yeah dragon, they have more than 1 compy, and i seriously dont believe that mewto would watch over his shoulder.  hes used that excuse w/ me b4.

http://www.artofproblemsolving.com/Forum/search.php?search_author=RomanianGenius

http://www.artofproblemsolving.com/Forum/search.php?search_author=mewto55555

After looking through RomanianGenius's posts, I have found that mewto55555 and RomanianGenius never post at the same time. I wonder why that is...

I don't think that mewto55555 even has a brother at all.

I'm going through all these arguments 1 by 1.

[quote="bobbymathking123"]Many have come to believe that mewto55555 is also RomanianGenius, not to mention apple_pi and mewto5555. Take the poll and give any evidence you may have have supporting your thoughts on this topic.[/quote]

Mewto5555 has been banned. Don't you think that that's just a little bit obvious of an alternate account?
Who is apple_pi? I seriously have not heard that name before. Maybe cherry_pi has an alternate account?

[quote="iin77"]
1. RomanianGenius and mewto55555 use computers with different IPs, otherwise mewto could not gain rating from RomanianGenius. And yes, they do play each other, or at least have in the past (currently not as often, presumably because of my accusing mewto), which is why superllamel is now ahead of mewto in rating. Further, mewto has stated many times that RomanianGenius is his brother, thus even if they used the same IP, which they most likely have at one point or another, there is no need to ban.

2. RomanianGenius is paired with mewto in the CTC team tournament. However, how does this make sense, considering that mewto beats RomanianGenius every single time he plays him? mewto doesn't even beat me every single time. Thus, it would follow that RomanianGenius is worse than me based upon our recent records against mewto. It would also follow that RomanianGenius is worse than isabella, who was available as a partner at the time mewto selected RomanianGenius as a partner. Thus, mewto chose a seemingly terrible partner instead of a very good one. This would only make sense if mewto were using RomanianGenius to boost his rating, and is playing in the tournament as both. Since mewto thinks he's better than isabella, this would make perfect sense.

And this is not the full extent of my evidence, for anyone who's wondering. I'll give more if provoked.

Note: What I say is not to be taken as fact simply because I'm a moderator on FTW.Think and decide for yourselves based on the evidence presented.[/quote]

I added number so as easily to argue the points.

1. To explain the computer situtation in our house, we have effectively 2 computers which we can use (I say effectvely because the others have no internet). My computer is always open (the computer which I am posting on now) and the other computer is sometimes being used by one of our parents.

2. I don't always pair with th best person available. Iin's logic here (which I admit is sound logic) would point to the fact that I trust RG because he is myself. That's a logical conclusion to draw from that agrument. However, the reason I chose RG is twofold. One, he is my brother and I felt a sense of sibling loyalty, and seeing as if he wa a halfway decent FTW player, I wouldn't have to worry about losing (unless we played Super/Fantasy where we would have lost with izzy or anyone else). Two, being family, RG and I have good team chemistry, whereas without me knowing izzy it would be difficult to comunicate and build a strong team.

[quote="iin77"]Okay, super. The following evidence is based on the habits of mewto and RomanianGenius.

 Back when I played more--about a month ago, perhaps--I CDed quite a lot and basically CDed anyone who wanted to (because that's what I do). One day, mewto CDed me. Immediately before the CD, mewto said "I lose," just like he always does before a CD. I don't remember precisely how the answering went, but when he got a correct answer, he would say precisely how he got the answer and "I lose" or something similar before the next problem started. He would do this on every problem he got right. I lost, of course, and immediately afterward, RomanianGenius came on and wanted to CD me. He proceeded to say "I lose" before the CD started, give the method of solving for the problems he got right, say "I lose" or "you win" in between problems, and beat me, of course.

When confronted, mewto explained this by saying "brothers act similarly," which is an extremely weak argument, given the nature of the similarities.

Anyone who has answered no to this poll, please state a specific reason as to why. Perhaps I'll convince you, and perhaps you will convince me. [/quote]

"Back when I played more--about a month ago, perhaps--I CDed quite a lot and basically CDed anyone who wanted to (because that's what I do)." Do I detect a not-so-subtle attack on my playing habits? :rotfl:

Anyway, this is by far the most convincing evidence given by anyone, and probably the strongest evidence against me. I admit to my part of the situation to be true, and seeing as you have no reason to lie, that my broher also said that. As I said before, brothers do act similarly. However, its not so much because we are brothers. One may notice that RG has only played a few hundred games, whereas I have played in excess of 2000. This is because he joined after me. All of his previous experience with FTW has been from watching me as I play, memorizing answers at the same I did, picking up expressions like gl,gg,and beastificationapwnilized. As such, he realized that my odd habits, like saying i lose, or gg u win, during the games may have had some effect as I pla, possibly psychologically. Maybe they did, maybe they didn't. If you wear a lucky charm and everything goes well, and then you take it off and Earth is struck by a meteor, you would think it had something to do with the charm. Same thing with these words. Quite a few people have told me not to explain solutions to them as it makes them feel inferior, and that affects my game. People whom I beat 4-1 I now beat 4-2. Think what you will, but that is my explanation. You will have to ask RG, who at this time, is asleep.

[quote="dragon96"]Whats RG's rating? If it is really low, then there is a better chance that RG = mewto55555 (because mewto just cds RG and wins all the time). I also remember the time where "RG" was using the computer, and mewto was watching over his shoulder (at least thats what mewto/RG said). If mewto and RG really were different, there would be a higher probablitity their family has 2 or more computers. Why doesn't RG use the other computer? It is possible that RG = mewto55555. I'm not saying that I think so (even though I said yes), but I think I think so.[/quote]

His rating is in the high 1500s. I have addressed the computer argument earlier on.

[quote="herefishyfishy1"]After looking through RomanianGenius's posts, I have found that mewto55555 and RomanianGenius never post at the same time. I wonder why that is...

I don't think that mewto55555 even has a brother at all.[/quote]

One also might notice how few posts RG has. I post in the forums outside of FTWOS when I have nothing better to do, and during that time, RG is doing something productive, like "real math"  :rotfl: And also, I do have a brother. Your real question is whether he plays AOPS and has the account "RomanianGenius".

Anyway sorry for making you read through all of this. I do not think I am RG, mostly based on personal experience :P but also because of all of these arguments, only iin's really make sense, and those I feel to be inaccurate. Logical, but inaccurate.

Surprisingly enough, I expected these [i]exact[/i] arguments.

[quote="mewto55555"]Mewto5555 has been banned. Don't you think that that's just a little bit obvious of an alternate account? [/quote]

This indicates your inherent lack of honesty, particularly regarding the use of multiple accounts.

[quote="mewto55555"]I don't always pair with th best person available. [/quote]

First off, this is a flawed argument, because everyone knows that mewto, of all people, just wants to win.

Also, you think you're the second best player on FTW.  The order in which you called teammates just happened to be:

[quote="mewto55555"]I call superllamel, then rg, then fantasy (in that order,based on availability)[/quote]

RomanianGenius second.  Behind superllamel, who you know is better than you.

[quote="mewto55555"]he is my brother and I felt a sense of sibling loyalty[/quote]

You two always say bad things about each other (presumably to make it more "realistic") and RomanianGenius says bad things about cherry_pi, and now you suddenly feel loyalty towards each other?

[quote="mewto55555"]RG and I have good team chemistry[/quote]

Even if you did have good team chemistry, it doesn't matter.  It's not like you're playing beach volleyball where knowing how your brother plays will help you.  You play your opponents individually in CDs.  It's dubious that you would actually think that "team chemistry" would help.

[quote="mewto55555"]All of his previous experience with FTW has been from watching me as I play, memorizing answers at the same I did, picking up expressions like gl,gg,and beastificationapwnilized. As such, he realized that my odd habits, like saying i lose, or gg u win, during the games may have had some effect as I pla, possibly psychologically.[/quote]

First off, you describe RomanianGenius's thought process as fact.  If you are in fact brothers, you cannot know this.  This fact alone make that whole argument useless.

Second, to preempt your saying "he explained his thought process to me" or something equally absurd, I'll say this:  everyone I know, when playing on the computer, [i]hates[/i] their sibling watching them.  I doubt you would allow him to watch you play hundreds of games.

Third, if he likes the game enough to watch you play it all the time, he would play it himself.  He did eventually, but he probably wouldn't watch you play hundreds of games first.

Fourth, most people remember that when they first started, whenever they knew an answer to a problem, it would still take them a while to answer, simply because either they "freeze up" or they type numbers slowly, usually both.  Anyone who's been gone an extended period of time knows when they come back, they take longer to answer, because their mind just "blanks" despite knowing the answer or they have a slower reaction time (recognizing the problem, then typing the answer takes longer).  RomanianGenius has not had either problem.

[quote="mewto55555"]One also might notice how few posts RG has. I post in the forums outside of FTWOS when I have nothing better to do, and during that time, RG is doing something productive, like "real math"  And also, I do have a brother. Your real question is whether he plays AOPS and has the account "RomanianGenius". [/quote]

One will also notice that the majority of RomanianGenius's posts are in threads in which having two accounts would be beneficial.  The civil war thread, in which each account can only post once an hour, so having two accounts increases your chances of winning. And all the tournament threads, where having two accounts increases your chances of winning.

In regards to mewto having a brother, there's no way to be sure.  However, I'll assume for the sake of argument that mewto has a brother (though he doesn't use the account RomanianGenius).  If he wants to do "real math" then he wouldn't be registered for these CTC tournaments.

Finally, whenever mewto gets muted, he lies and says he meant something else.  A person who lies once with the intention of benefiting themself can lie many times.

If RG is truely your brother, wouldn't he want to join at the same time as you? I checked and he joined a couple months after you.

1) He would have wanted to compete against you, as is the nature of most boys.

2) If he didn't join, he would surely want to start closely after you, if he observed the exciting nature of FTW.  :)

3) After a while, you would have realized the potential benfits of having a multiple account, so you made mewto5555. Having that banned, you realized that you would need a better user name and a better overall story.

Also, I do know a pair of supposed brothers who registered at almost the same time. They have a different personality and while I was working on a challenge set in the FTW room with his older brother, he was quite a pest...

Anyway, just spit out the truth mewto. Thats why they usually give less time in jail if you confess.  :|  :|

@iin: I'm not the best player on FTW, but good evidence. I'm wondering what mewto's excuses will be.

@Dojo: Mewto and RG do compete often. Mewto always wins. I witnessed one game where RG was ahead 3-1, then despite mewto getting problems wrong, RG never got the last point.

@AIME: Good point. Mewto obviously lied there, and people who lie once often lie more.

@mewto: I have a question. Why does RG play you so much if he always loses? I know that my brother would not tolerate playing me so many times if he knew he would lose.

OMG ON FRIDAY I HAD A DREAM ABOUT SOMEONE SETTING UP A FORUM LIKE THIS. I WAS GONNA DO IT BUT DANG YOU GOT IT FIRST . oh well. anyway i have always though rg=mewto.

1. Like i have said before, back when super and mewto were competing for highest rating mewto always played rg like 30 times when super caught up to mewto's rating. and the cds i viewed mewto won like 4-0 and
one time when mewto scored i think he went on the rg's window on accident or something and rg said YES! i got it.

I got more reasons but i can't think right now. i'll post later.

The only real evidence so far seems to come from iin. Everything else is an attack on my character such as this:

[quote="AIME15"]More supporting evidence for iin's typo thing bout mewto:

[quote="mewto55555"]I was wondering about being supreme dictator of CTC and I was like, what the heck. I'll give it a go. Post whether or not you want me as supreme dictator.  Here comes the conspiracy theorists who say this was my idea the whole time...

And this isn't a vote, I just want to know what you think![/quote]

After he got no support:

[quote="mewto55555"]Dang who said anything about supreme dictator? I meant super model....dang typoes seem to confuse people. I'd better reread posts before I post them.[/quote]

From the board member discussion thread.

EDIT: Great evidence iin.  With that, mewto is sure to be banned :DDDD[/quote]

It seems everyone's logic is Mewto lied once. Therefore he is lying about this.

[quote="iin77"]Also, you think you're the second best player on FTW. The order in which you called teammates just happened to be:
[quote="mewto55555"]
I call superllamel, then rg, then fantasy (in that order,based on availability)
[/quote]

RomanianGenius second. Behind superllamel, who you know is better than you. [/quote]

Yes I do think super is better than RG. I don't on the other hand think izzy is, and that thought process was reflected in my choice of team mates.

Also, seeing as the vast majority of people believe I am RG, doesn't that bias the evidence they give? With very few people in support of my position, the evidence that I am not RG goes unnoticed, and when I bring it up, people say "Ha! You just set that up so we would think you were different people!" SO far, iin77 is the only one who has given an unbiased argument. Iin's arguments go as follows: Here is the evidence thus I think Mewto=RG. Everyone elses' arguments look like this: I think mewto is RG, so I will look for all evidnce in favor of that viewpoint, while ignoring all evidence that points to the contrary.

WAIT, seriously?  Do you remember when?  Levans could fish it up and ban both of them!

So... if angie has evidence that mewto told her he was RomanianGenius, I think we can all agree that he has multiple accounts and by the terms of service should be banned, right?

Just what i said.  But levans needs to check the logs 2 make sure angie isnt lying

guys this argument is going nowhere. well what angie said it si. But, lets just let levans or an admin take care of this.

[quote="jjx1"][quote="RomanianGenius"]I voted yes. I mean, wth how obvious can it get. I am mewto, mewto is me. Please ban him, like, now.[/quote]

THERE YOU GO. HE ADMITS IT![/quote]

I was going to post   :rotfl:  :rotfl:  :rotfl: but that would be spam. Oh btw, I'm also Bobby Shen [b]and[/b] Darryl Wu.

the great lord mewto would never make another account

Er...not like THERA and ra5248 and THETRUERA and the 10 other RA's have been banned...

Why is there so much talk about mewto being RG?

I mean, IDK.  I am neutral on this.

But in reality, this is all spam.

I agree with ernie this is all spam.
But just one question. So Mewto is RG? (so maybe I was right for once)

yeah this is really spam. does it really matter whether or not rg is mewto? do we need to devote a whole topic just for this?

Yeah, kl is right.

Do we?

If you wanna make it, make it in the FF. :P

ok guys, it really doesn't matter if mewto is RG
for now, we will let them off
if we find any evidence that mewto is RG, we will continue talking about this
no more posts in here


function
limit
calculus
derivative
real analysis
real analysis solved

Let $f$ a continous function such that $\lim_{h->0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=l_x$ and $l_x >= 0$ .Show that $f$ is convexe.

I think you mean $\lim_{h->0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=l_x$.

This is an aside that doesn't have that much bearing on Killer's problem. If $f$ is twice differentiable, then the limit in question, namely

$\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}$

equals $f''(x).$

An interesting question is this: does the existence of this limit imply the existence of the second derivative? The answer to that is no, as can be seen by the example of $f(x)=x^{\frac13}$ at $x=0.$

That example, however, would not be permitted in Killer's problem.

[b]Case 1.[/b] We will prove statement in the case $l_x\geq\alpha>0$ for all $x\in I$, $I$ is an arbitrary segment.
Suppose contrary. Consider $g(x,y,a)=f(ax+(1-a)y)-(af(x)+(1-a)f(y))$, $(x,y,a)\in I\times I\times[0,1]=K$. It is clear $K$ is a compact and $g$ is continuous on $K$. It follows that $g$ reaches maximum on $K$ at some point $(x_0,y_0,a_0)$. Due to our supposition $g(x_0,y_0,a_0)>0$. It means $x_0\neq y_0$ and $a_0\not\in\{0,1\}$. Let $z_0=ax_0+(1-a)y_0$.
We can find such $h>0$ that $\frac{f(z_0+h)+f(z_0-h)-2f(z_0)}{h^2}\geq \frac{\alpha}{2}$, $z_0+h,z_0-h\in I$. Write $z_0+h$ and $z_0-h$ as $(a+t)x_0+(1-(a+t))y_0$ and $(a-t)x_0+(1-(a-t))y_0$. It is easy to check that $0<f(z_0+h)+f(z_0-h)-2f(z_0)=g(x_0,y_0,a+t)+g(x_0,y_0,a-t)-2g(x_0,y_0,a)$. Therefore $g(x_0,y_0,a+t)>g(x_0,y_0,a)$ or $g(x_0,y_0,a-t)>g(x_0,y_0,a)$ -- contradiction.

[b]Case 2.[/b] General case $l_x\geq 0$. Consider function $f_\alpha(x)+\frac{\alpha x^2}{2}$, $\alpha>0$. We see that $f_\alpha(x)$ satisfies case 1, so $f_\alpha(x)$ is convex, i.e. $af_\alpha(x)+(1-a)f_\alpha(y)\geq f_\alpha(ax+(1-a)y)$ for all $x,y$ and $a\in [0,1]$. Now if $\alpha\to 0$ we immediatelly obtain $af(x)+(1-a)f(y)\geq f(ax+(1-a)y)$.

I think that the continuity of $f$ is a very strong condition. Strong enough that we don't need all of what we were given (i.e. the limit of that expression). To be precise, I think that if $f$ is continuous and we know that for each $x$ there is an $\varepsilon_x>0$ s.t. $f(x+h)+f(x-h)> 2f(x),\ \forall h\in(0,\varepsilon_x)\ (*)$, then $f$ is convex.

Take a chord determined by $A=(a,f(a)),B=(b,f(b))$ on the graph of $f$ and assume there is a point $c\in (a,b)$ s.t. $C=(c,f(c))$ is above the chord $AB$. For $x\in[a,b]$ define $g(x)$ to be the distance from $X=(x,f(x))$ to $AB$ measured "upwards", i.e. if $X$ is below $AB$, then the distance becomes negative. This function has a positive maximum at $x_0$ on [$a,b$] (because it's continuous and it has a positive value at $c$). We see that $(*)$ can't hold for $x_0$, so we get a contradiction.

There are some details to take care of, but that can be done easily. For example, we might have to use a trick like the thing with the $\alpha>0$ that Myth used, but I'm not sure.

Statement is wrong, if $f$ is discontinuous, even if we demand $\lim$ exists.

And, of course, I used only the fact $f(x+h)+f(x-h)>2f(x)$ for some small $h$.

Grobber, did you transit to AoPS forum? :(

Oh, sorry about that, I hadn't read your solution in detail, and I thought you had actually used the limit.

Why would you be upset about me transiting to AoPS? There's nothing wrong with that.. :) Anyway, I can't browse ML anymore, because of nasty network settings (the problem is entirely on my side). I've already mentioned this to some of the Admins, but I don't know if it can be fixed.

First of all, I dislike aops  ;). Secondly, I need to load all images and documents twice: one time from ML and one time from aops (since topic reply notifications points to ML and aops sites alternately) :( It creates BIG unwanted traffic for me. I am not so rich to pay for it...


Hello,

How do I convert a document (specifically a data table) in latex to a picture format (ie. jpeg, gif, etc.)?  I need to import it into Microsoft Publisher without losing the image quality.

Thank you.

1) If it's just a table, compile in the TeXer, right-click and copy the image
or
2) Compile to PDF then take a screen copy (ALT+Print Screen in Windows) then paste into Publisher
or
3) Compile your document at [url=http://sciencesoft.at/latex/?lang=en]ScienceSoft.at[/url] and choose your image format for output.

Thanks for your help.  I already tried to take a screenshot but I lost a lot of the image quality.  The Texer and Science Soft worked fine though.


Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that$ \frac {a}{\sqrt {4a \plus{} 5b^2}} \plus{} \frac {b}{\sqrt {4b \plus{} 5c^2}} \plus{} \frac {c}{\sqrt {4c \plus{} 5a^2}}\leq \frac{3}{\sqrt {17}}$

[quote="Allnames"]Let$ a,b,c\geq 0,a \plus{} b \plus{} c \equal{} 1$
Prove that$ \frac {a}{\sqrt {4a \plus{} 5b^2}} \plus{} \frac {b}{\sqrt {4b \plus{} 5c^2}} \plus{} \frac {c}{\sqrt {4c \plus{} 5a^2}}\leq \frac {3}{\sqrt {17}}$[/quote]
With C-S we have:
$ LHS^2 \le\ \frac {a}{4a \plus{} 5b^2} \plus{} \frac {b}{4b \plus{} 5c^2} \plus{} \frac {c}{4c \plus{} 5a^2}$
Need prove $ \frac {a}{4a \plus{} 5b^2} \plus{} \frac {b}{4b \plus{} 5c^2} \plus{} \frac {c}{4c \plus{} 5a^2} \le\ \frac {9}{17}$
$ \leftrightarrow \frac {b^2}{4a \plus{} 5b^2} \plus{} \frac {c^2}{4b \plus{} 5c^2} \plus{} \frac {a^2}{4c \plus{} 5a^2} \ge\ \frac {3}{17}$
$ \leftrightarrow \frac {b^4}{4ab^2 \plus{} 5b^4} \plus{} \frac {c^4}{4bc^2 \plus{} 5c^4} \plus{} \frac {a^4}{4ca^2 \plus{} 5a^4} \ge\ \frac {3}{17}$
And $ \frac {b^4}{4ab^2 \plus{} 5b^4} \plus{} \frac {c^4}{4bc^2 \plus{} 5c^4} \plus{} \frac {a^4}{4ca^2 \plus{} 5a^4}\ge\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{4\sum ab^2 \plus{} 5\sum b^4}$
Need prove $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{4(\sum ab^2)(a\plus{}b\plus{}c) \plus{} 5\sum b^4} \ge\ \frac {3}{17}$
$ \leftrightarrow \sum_{cyc} (a^2 \minus{} b^2 \minus{} 4ac \plus{} 2ab \plus{} 2bc)^2 \ge\ 0$
We have done!

Very nice ,onlylovemath
But i have just found that Can_hang2007 had a similar problem.Sorry every body :blush:  (although i created it by myself)
And specicaly,he said he has a general for it
Post it,Can_hang2007

[quote="Allnames"]Very nice ,onlylovemath
But i have just found that Can_hang2007 had a similar problem.Sorry every body :blush:  (although i created it by myself)
And specicaly,he said he have a general for it
Post it,Can_hang2007[/quote]
Really? :o Can U post it;please? :wink:


Is there a way to solve $x^{y}=y^{x}$ for $y$ in terms of $x$?

Probably not in terms of elementary functions.  However, consult

http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm

Just to follow up, that website mentions a parametric solution to the equation.  Namely
\begin{align*}x &= t^{1/(t-1)}\\ y &= t^{t / (t-1)}\, , \end{align*}
where $t$ is a positive real number different from $1$.

$x=y \iff x\neq 0$

Thats an obvious solution...

[hide="here's some equations"]$x^{y}=y^{x}$

$\log_{x}y^{x}=y$

$x\log_{x}y=y$

$\log_{x}y=\frac{y}{x}$

$\log_{y}x^{y}=x$

$y\log{y}x=x$

$\log_{y}x=\frac{x}{y}$[/hide]

[quote="bpms"]x=y and 2,4 are the only integer ones I can think of[/quote]
What about +/-?

Those are the only integer ones.

The problem is addressed very nicely in the link that I gave.  I don't think much extra discussion is necessary, unless you are wondering how you would derive the formulas in the link constructively.

For $y=\frac{\ln x}{x}\ (x>0),$ we are to seek $a,\ b$ such that $\frac{\ln a}{a}=\frac{\ln b}{b}.$ Sketch the graph of $y=\frac{\ln x}{x}$ gives you some answer.


calculus
integration
logarithms
trigonometry
geometric series
real analysis
real analysis unsolved

can you :D solve

$ \int_0^1\int_0^1\;\dfrac{1\minus{}y^2}{(1\plus{}(xy)^2)\cdot\ln^2\,(xy) }\;\textbf dx\;\textbf dy$

[quote="misan"]can you :D solve

$ \int_0^1\int_0^1\;\dfrac{1 \minus{} y^2}{(1 \plus{} (xy)^2)\cdot\ln^2\,(xy) }\;\textbf dx\;\textbf dy$[/quote]t\Now i have no time, I think it will be work:
consider :
$ I(a)\equal{}\int\int\frac{(1\minus{}y^{2})(xy)^{a}}{(1\plus{}(xy)^{2})\ln^{2}xy}$
diferentiate it two time by with $ a$
$ I''(a)\equal{}\int\int\frac{(1\minus{}y^{2})(xy)^{a}}{1\plus{}(xy)^{2}}$ and last can be easly find using $ \frac{1}{1\plus{}z}\equal{}\sum (\minus{}1)^{i}z^{i}$ you get answer $ B$ and integrate it two times ..
I'm not sure.

$ \int_0^1 \int_0^1 (1-x)\cdot (xy)^{\alpha-1}\, dx\, dy=\int_0^1 \int_0^1 x^{\alpha-1}\, y^{\alpha-1}\, dx\, dy-\int_0^1 \int_0^1 x^\alpha \, y^{\alpha-1}\, dx\, dy $

$ = \frac{1}{\alpha^2}-\frac{1}{\alpha+1}\cdot\frac{1}{\alpha}=\frac{1}{\alpha^2}-\frac{1}{\alpha}+\frac{1}{\alpha+1} $

Integrate twice with respect to $ \alpha $ :

$ \int_0^1 \int_0^1 (1-x)\, \frac{(xy)^{\alpha-1}}{\log(xy)}\, dx\, dy=-\frac{1}{\alpha}-\log(\alpha)+\log(\alpha+1) $

$ \int_0^1 \int_0^1 (1-x)\, \frac{(xy)^{\alpha-1}}{\log^2(xy)}\, dx\, dy=(\alpha+1)\cdot \log\left(1+\frac{1}{\alpha}\right)-1 $

(Integration constants are zero, since both right hand sides tend to zero for $\alpha\to\infty$)

Substitute $ x\mapsto x^2 \, , \, y\mapsto y^2 $ :

$ \int_0^1 \int_0^1 (1-x^2)\, \frac{(xy)^{2\alpha-1}}{\log^2(xy)}\, dx\, dy=(\alpha+1)\cdot \log\left(1+\frac{1}{\alpha}\right)-1 $

Substitute $ \alpha=\frac{2k+1}{2} $ :

$ \int_0^1 \int_0^1 (1-x^2)\, \frac{(xy)^{2k}}{\log^2(xy)}\, dx\, dy=\frac{2k+3}{2}\cdot \log\left(1+\frac{2}{2k+1}\right)-1 $

Generate a geometric series:

$ \int_0^1 \int_0^1 \frac{1-x^2}{(1+x^2 y^2)\, \log^2(xy)}\, dx\, dy=\sum_{k=0}^\infty (-1)^k \left[\frac{2k+3}{2}\cdot \log\left(1+\frac{2}{2k+1}\right)-1\right] $

$ =\sum_{k=1}^\infty (-1)^{k-1} \left[\frac{2k+1}{2}\cdot \log\left(\frac{2k+1}{2k-1}\right)-1\right] $

$ =\underbrace{\sum_{k=1}^\infty (-1)^{k-1} \left[2k\cdot \frac12 \log\left(\frac{2k+1}{2k-1}\right)-1\right]}_{-\frac12+\frac{2K}{\pi}}+\frac12 \underbrace{\sum_{k=1}^\infty (-1)^{k-1}\,\log\left(\frac{2k+1}{2k-1}\right)}_{-2\log\left(2\cdot\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)} $

It remains to show the underbraced expressions.

$ F(x):=\sum_{k=1}^\infty (-1)^{k-1} \left[2k\cdot \frac12 \log\left(\frac{1+\frac{x}{2k}}{1-\frac{x}{2k}}\right)-x\right] $

$ =\sum_{k=1}^\infty (-1)^{k-1} \sum_{n=1}^\infty \frac{1}{2n+1}\, \frac{x^{2n+1}}{(2k)^{2n}}=\sum_{n=1}^\infty \frac{1}{2n+1}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{(2k)^{2n}}\, x^{2n+1} $ $=\sum_{n=1}^\infty \frac{1}{2n+1}\,\frac{\eta(2n)}{2^{2n}}\, x^{2n+1} $

$ F'(x)=\sum_{n=1}^\infty \eta(2n)\, \left(\frac{x}{2}\right)^{2n}=-\frac12+\frac14\cdot \frac{\pi x}{\sin \frac{\pi x}{2}} $

Since $ F(0)=0 $ we can write $ F(1)=\int_0^1 F'(x)\, dx=-\frac12+\frac14 \int_0^1 \frac{\pi x}{\sin\frac{\pi x}{2}}\, dx=-\frac12+\frac{2K}{\pi} $ .

$ \sum_{k=1}^\infty (-1)^{k-1}\, \log\left(\frac{2k+1}{2k-1}\right)=\sum_{k=0}^\infty (-1)^k \, \log\left(\frac{2k+3}{2k+1}\right) $

$ =\sum_{k=0}^\infty \left(\log\frac{4k+3}{4k+1}-\log\frac{4k+5}{4k+3}\right)=\sum_{k=0}^\infty \log \frac{(4k+3)(4k+3)}{(4k+1)(4k+5)} $

$ =\log\prod_{k=0}^\infty \frac{\left(k+\frac34\right)\left(k+\frac34\right)}{\left(k+\frac14\right)\left(k+\frac54\right)}=\log\frac{\Gamma\left(\frac14\right)\, \Gamma\left(\frac54\right)}{\Gamma\left(\frac34\right)\, \Gamma\left(\frac34\right)}=\log\left(\frac14\cdot \frac{\Gamma^2\left(\frac14\right)}{\Gamma^2\left(\frac34\right)}\right) $

$ =2\log\left(\frac12\cdot \frac{\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}\right)=-2\log\left(2\cdot \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right) $


[url=http://reflections.awesomemath.org/2007_2/3variable.pdf]Is this that you want?[/url]

No,sir.That's not p q r.That's Vo Quoc Ba Can's lemma.

[quote="kunny"]Then what's like $ pqr$ method?[/quote]
That's the method we put $ a\plus{}b\plus{}c\equal{}p,ab\plus{}bc\plus{}ca\equal{}q,abc\equal{}r$

Thanks for your reply! :)

How about this?

[url=http://www.mathlinks.ro/viewtopic.php?t=156796]chapt4Thuan.pdf by pvthuan See P.126[/url]

[quote="Myszon11"]Hey guys, I need a paper about "p,q,r method". I think I've seen it on Mathlinks a few weeks ago, but I just can't find it now  :(  Can anyone help me and give me a link  :?:[/quote]
You can see it at pvthuan's book or see my attach file.

Yes, kunny's second link and Nguyen's are about what I was looking for, still it isn't the exact thing, as I think I have seen a paper about this method in english :) And I am sure it was on mathlinks.

Or instead of a paper, could you provide me with a simple explanation of this method? I know we should put $ p \equal{} a\plus{}b\plus{}c, q \equal{} ab\plus{}bc\plus{}ca, r \equal{} abc$, and what should we do next?

Myszon11, I hope this help you.

[url=http://www.mathlinks.ro/viewtopic.php?t=183200]Take a look BaBak Ghalebi's post[/url]

Yes, this is exactly what I was looking for. Thank you very much kunny  :) I really appreciate your help :)


Let $p$ and $q$ be coprime positive integers such that:
\[ \dfrac{p}{q}=1-\frac12+\frac13-\frac14 \cdots -\dfrac{1}{1334}+\dfrac{1}{1335} \]
Prove $p$ is divisible by 2003.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52497]the solution is hère[/url]. :)

There is a more generalized problem
 If p is grater than 3, q=[2p/3] and let m and n be positive integers such that

m/n = 1- 1/2 +1/3 -1/4 + ... +(-1)exp(q-1) 1/q

then m is divisible by p

You just use Catalan Identity And then group since 1/([(q+1)/2]+k) + 1/(q-k)=p/(..)
(where k is 0,1,2...,(q-1)/2 )
And then p divides m

1335 = [2*2003/3]

[quote=Paragdey12]Old imo lol :rotfl:[/quote]

It is from IMO $1979$


6. A monument made of a certain number of rows consisting of cube-shaped bricks starts with a row of 34 bricks. The row above has 31 bricks, the row above that has 28 bricks, and so on such that each row has three fewer bricks than the row below it. A portion of the monument is shown below. Later a student notices that the total number of bricks used in the monument is just enough to create the brick floor of a rectangular patio that is one layer of bricks, seven times as long as it is wide. (No bricks are broken to make the floor.) How many rows of bricks does the monument have?

[hide]
the solution given suks.

THE ANSWER IS 7[/hide]

[hide]Since this is target and you don't want to make silly mistakes, I would use a calculator. I hope you recognize that the number of bricks in each row decreases by 3 each time. Now, we know that we can arrange the bricks to from a shape with area $ 7x^2$ for some x. I would just keep adding until you get to a point where you can have an integer x.
34+31+28+25+22+19+16=175, which fits. So 7 rows.[/hide]

[quote="Nightmare124"]the answer is 7!!![/quote]
Wow, I can't believe how you got that [/sarcasm]

Le the width of the patio be $ w$. Then the area is $ 7w^2$.  Now obviously, we want $ 34\plus{}31\plus{}\hdots\equal{}7w^2$
If there are $ n$ rows, then the $ n^{th}$ row has $ 34\minus{}3(n\minus{}1)$ bricks in it.  To find the total sum, we take $ \frac{34\plus{}34\minus{}3n\plus{}3}{2}*n\equal{}7w^2$
$ 71n\minus{}3n^2\equal{}14w^2$
$ 3n^2\minus{}71n\plus{}14w^2\equal{}0$
$ n\equal{}\frac{71\pm\sqrt{5041\minus{}168w^2}}{6}$.
err. ya. lol. Let's try for $ w$ instead...

$ w\equal{}\frac{0\pm\sqrt{\minus{}56(3n^2\minus{}71n)}}{6}$.
o.o...
so $ w\equal{}\frac{\pm\sqrt{994n\minus{}42n^2}}{3}$.

A little nicer? I guess =D

So err... $ 994n\minus{}42n^2$ must be a perfect square. And this is really stupid you know. So do it with a calculator. Because it's on a target. xD


If $ a,b,c>0$ such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}1$ prove that $ \frac{a}{b^{2}\plus{}1}\plus{}\frac{b}{c^{2}\plus{}1}\plus{}\frac{c}{a^{2}\plus{}1}\geq \frac{3}{4}(a\sqrt{a}\plus{}b\sqrt{b}\plus{}c\sqrt{c})^{2}$.


algebra
polynomial
calculus
integration
trigonometry
superior algebra
superior algebra solved

$n>0$ integer, $A\in M_{2}(Z)$ such that $A^n=I_2$

Suppose that $A$ has a complex eigenvalue not real

Prove that $A$ satisfy one of the following equation

$A^2=-I_2$ or $A^2+A+I_2=0$ or $A^2-A+I_2=0$

The minimal polynomial of $A$ must be an integral monic polynomial of degree $2$, having as roots two roots of unity, $\cos x\pm i\sin x$. This means that $2\cos x$ is integral, meaning that it must take one of the values $0,\pm 1,\pm 2$. The case $\pm 2$ doesn't work because then the eigenvalues are real, so tat's that.


How many triangles can be formed using the vertices of a regular dodecagon (a 12-sided polygon)?

$ \binom{12}{3}$ because you can choose $ 3$ verticies

[hide=Solution]

Since a triangle has 3 vertices, and a dodecagon is a 12 sided figure, we know that our equation is $\binom{12}{3}.$  Solving gives our answer of $220.$

[hide=Solution]
$\binom{12}{3}=\frac{12\times11\times10}{3!}=\boxed{220}$ triangles.
[/hide]


function
number theory unsolved
number theory

prove that every non negatif rational number can be written on the form
                                        (a^3 + b^3)/(c^3 + d^3)
                        where a,b,c and d are non negatif natural integers

This is ISL 1999. you can find a solution on the kalva homepage.
nice problem by the way.

Find fuction:[tex]f(x)=\sum_{n=1}^ {\infty}F_n.x^n[/tex],
with [tex]F_n[/tex] is n-th Fibônaxi prime

i tink f(x)=1/(x²-x-1)
it called generatrice fonction


[quote="GameBot"]$ ABCDE$ is a five-digit number where $ A,B,C,D,$ and $ E$ are unique digits. Use the clues to determine the number:\\ \\
(1)  None of the digits are prime or zero; \\
(2)  Even and odd digits alternate;\\
(3)  $ C$ is a square number; and\\
(4)  $ A \plus{} D \equal{} B$.[/quote]

By clue 3, $ c\equal{}1,4,9$.

All of the digits are $ 1,4,6,8,9$, which has 2 odds, so A,C,E are all even.

The only way we can have $ A\plus{}D\equal{}B$ is $ 1\plus{}8\equal{}9$. Thus,

$ 8BC1E$.
C is even, so $ 4$.
$ 8B41E$.
B is 9, and E is 6.
So $ 89416$.


Let $ f(x) \equal{} x^2 \minus{} x \plus{} 1$ and $ n \geq 2$ be an integer what is minimal and maximal possible rank of an $ n$x$ n$ matrix  whose $ n^{2}$ entries are precisely the numbers $ f(1),f(2),...,f(n^{2})$ ?

The minimal rank seems to be $ 3$ if $ n\geq{3}$.


inequalities
logarithms
inequalities unsolved

[quote="Orif"]Mongolian inequality:
Let $a_{1}>a_{2}>...>a_{n}>0$ be real numbers.Prove that:

$(a_{1}+a_{2})(a_{2}+a_{3})...(a_{n}+a_{1}) \geq ( a_{1}+a_{2}+a_{3})(a_{2}+a_{3}+a_{4})...(a_{n}+a_{1}+a_{2}){2}^{n}/{3}^{n}$

where $a_{n+1}=a_{1}$ and $a_{n+2}=a_{2}$[/quote]

If I didn't make a mistake, the inequality should be the other way, and it can be generalized to:

$\prod_{i=1}^{n}{\left(\frac{1}{x}\sum_{j=1}^{x}{a_{i+j}}\right)}\geq \prod_{i=1}^{n}{\left(\frac{1}{y}\sum_{j=1}^{y}{a_{i+j}}\right)}$
for $n \geq x \geq y$, $x,y \in \mathbb{Z}^{+}$
with equality at $x=y$ or $\left[\frac{1}{x}\sum_{j=1}^{x}{a_{i+j}}\right] = \left[\frac{1}{y}\sum_{j=1}^{y}{a_{i+j}}\right]$

Proof:
$\prod_{i=1}^{n}{\left(\frac{1}{x}\sum_{j=1}^{x}{a_{i+j}}\right)}\geq \prod_{i=1}^{n}{\left(\frac{1}{y}\sum_{j=1}^{y}{a_{i+j}}\right)}$
$\iff$ $\sum_{i=1}^{n}{\ln\left(\frac{1}{x}\sum_{j=1}^{x}{a_{i+j}}\right)}\geq \sum_{i=1}^{n}{\ln\left(\frac{1}{y}\sum_{j=1}^{y}{a_{i+j}}\right)}$
which is true by Majorization Inequality because $\ln x$ is concave and $\left[\frac{1}{x}\sum_{j=1}^{x}{a_{i+j}}\right] \prec \left[\frac{1}{y}\sum_{j=1}^{y}{a_{i+j}}\right]$

[quote="Orif"]Weiquan you are right!Thanks!But I need for other easier solution!(Whithout majorization inequality)(I have solution from Karamata,Sankt-Peterburg Mathematics Olympiads book,2001, pages 89-102)[/quote]

 :( Haha the first thing I think when I see this kind of problem is Jensen's Inequality then Majorization Inequality if Jensen's doesn't work. So... not very easy for me to see any other simple ways if they exist :P

But of course, if Majorization Inequality works, then smoothing works, with or without induction. You can try (I don't want to because I think it may be messy :lol:) Besides you were given the very helpful condition of strictly decreasing $a_{i}$s which is not required in the Majorization proof but will help in smoothing/induction.


geometry
circumcircle
trigonometry
geometry unsolved

Let $ I$ and $ O$ denote respectively the incircle and the circumcircle of triangle $ ABC$.Given $ \angle AIO\equal{}90^o$,prove that area of triangle $ ABC$ is less than $ \frac{3\sqrt{3}}{4} AI^2$

From hipohtesis we have  \[ AI^2\plus{}OI^2\equal{}R^2\]or\[ AI^2\plus{}R^2\minus{}2Rr\equal{}R^2\]or\[ AI^2\equal{}2Rr\]Thus , it suffices to to show that \[ S\le \frac{3\sqrt{3}}{4}\cdot2Rr\]or\[ 2R^2\sin a\sin B\sin C\le\frac{3\sqrt{3}}{4}2R4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\]or\[ \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\le\frac{3\sqrt{3}}{8}\]wich is clearly true.

$ \left\|\begin{array}{c}
S\equal{}pr\ \ ;\ \ AI^2\equal{}\frac {bc(p\minus{}a)}{p}\\\\
IA\perp IO\Longleftrightarrow a\equal{}2(p\minus{}a)\\\\
abc\equal{}4Rpr\ \ ;\ \ 2p\le 3R\sqrt 3\end{array}\right\|\ \implies\ 2p\cdot a\le 3R\sqrt 3\cdot 2(p\minus{}a)$ $ \Longleftrightarrow$ $ ap\le 3\sqrt 3R(p\minus{}a)$ $ \Longleftrightarrow$

$ bc\cdot ap\le bc\cdot 3\sqrt 3R(p\minus{}a)$ $ \Longleftrightarrow$ $ 4Rp^2r\le 3\sqrt 3Rbc(p\minus{}a)$ $ \Longleftrightarrow$ $ S\le\frac {3\sqrt 3}{4}\cdot\frac {bc(p\minus{}a)}{p}$ $ \Longleftrightarrow$ $ S\le\frac {3\sqrt 3}{4}\cdot AI^2$ .


How many different isoceles triangles have integer side lengths and a perimeter of 23?

\[\text{(A) }2\qquad\text{(B) }4\qquad\text{(C) }6\qquad\text{(D) }9\qquad\text{(E) }11\]

[hide]The base can be 1-11. $\text{C}$[/hide]

[hide] So, we know that the base has to be odd and can't be greater than 11 (if it was, then it'd be greater than the sum of the two congruent sides). So, there's 1, 3, 5, 7, 9, and 11 for a total of 6 - C [/hide]

[hide="answer"]well, sum of two sides has to be greater than third...so, the base cannot be greater than 11..it will also be an odd number,
so the options are: 1,3,5,7,9,11...therefore there 6 different triangles.
ANSWER: $C$  :lol: [/hide]

[hide]2a+b=23 i did a few guess and check and got a>5 a<12,so a is in {6,7,8,9,10,11} there is only one b for each a   :D ,then there are 6 different triangles[/hide]

Yes, the previous solutions are better  :blush:, however this is still correct  :)


If $ f(x)f\left( \minus{} \frac {1}{x}\right) \equal{} \minus{} 1$ for $ x\in \mathbb{R^*}$ and $ f$ is continuous, find $ f(0)$.

I think $ f(0)$ can be any value. For any other real value $ k$, we can pair it up with $ \minus{} \frac {1}{k}$, thus pairing up all the nonzero reals since $ \minus{} \frac {1}{k}$ is the inverse of itself. Then we can make $ f(k)$ and $ f( \minus{} \frac {1}{k})$ whatever nonzero numbers we want as long as they multiply to $ \minus{} 1$.

[quote="dgreenb801"]I think $ f(0)$ can be any value. For any other real value $ k$, we can pair it up with $ \minus{} \frac {1}{k}$, thus pairing up all the reals since $ \minus{} \frac {1}{k}$ is the inverse of itself. Then we can make $ f(k)$ and $ f( \minus{} \frac {1}{k})$ whatever we want as long as they multiply to $ \minus{} 1$.[/quote]

I can't say that i have fully understood your explanation for this problem. Just work with Bolzano's Theorem for a little while to see what's happening...

Okay, suppose I have a function like this:

$ f(0) \equal{} 75894$

$ f(3) \equal{} 7$, $ f( \minus{} \frac {1}{3}) \equal{} \minus{} \frac {1}{7}$
$ f(342) \equal{} 5$, $ f( \minus{} \frac {1}{342}) \equal{} \minus{} \frac {1}{5}$
$ f(11) \equal{} 976$, $ f( \minus{} \frac {1}{11}) \equal{} \minus{} \frac {1}{976}$
.....
This could go on forever randomly and the function would work.

If you want a concrete example of a function, the following works:
$ f(0) \equal{} 3234$, $ f(x) \equal{} 1$ if $ x > 0$, $ \minus{} 1$ if $ x < 0$.

The general solution is:
$ f(0)\equal{}$ any number, $ f(x)\equal{}$ any number for $ x>0$, $ f(x)\equal{}\minus{}\frac{1}{f(\minus{}\frac{1}{x})}$ for $ x<0$.

[quote="dgreenb801"]Okay, suppose I have a function like this:

$ f(0) \equal{} 75894$

$ f(3) \equal{} 7$, $ f( \minus{} \frac {1}{3}) \equal{} \minus{} \frac {1}{7}$
$ f(342) \equal{} 5$, $ f( \minus{} \frac {1}{342}) \equal{} \minus{} \frac {1}{5}$
$ f(11) \equal{} 976$, $ f( \minus{} \frac {1}{11}) \equal{} \minus{} \frac {1}{976}$
.....
This could go on forever randomly and the function would work.

If you want a concrete example of a function, the following works:
$ f(0) \equal{} 3234$, $ f(x) \equal{} 1$ if $ x > 0$, $ \minus{} 1$ if $ x < 0$.

The general solution is:
$ f(0) \equal{}$ any number, $ f(x) \equal{}$ any number for $ x > 0$, $ f(x) \equal{} \minus{} \frac {1}{f( \minus{} \frac {1}{x})}$ for $ x < 0$.[/quote]

Now things are clear. So, you mean that $ f(0)$ isn't dependent from $ f(x)f(\minus{}1/x)\equal{}\minus{}1$ (1).
But let me show you where i get confused.
(1) tells us that there exists a root $ x_0$ that belongs to the space $ (\minus{}1/x,x)$ such that $ f(x_0)\equal{}0$ (2).
If we set $ x\equal{}x_0$ to (1) we get $ 0\equal{}\minus{}1$ thus a contradiction, so it must be $ x_0\equal{}0$. So, from (2) we have $ f(0)\equal{}0$ right :?:

[quote="dgreenb801"]You can only apply the intermediate value theorem (Bolzano's theorem) when the function is continuous. If it is given that the function is continuous then I think your solution is right.[/quote]

Oh my God!!! I forgot to say that $ f$ is continuous. But, even so, what about your solution? I can't find anything wrong on it

[quote="dgreenb801"]If you try to make a continuous function by doing it my way, you will find it will have to end up passing through (0,0) if it is to be continuous from what you said. For example, suppose f(x)=x for all x not equal to 0, then we can't just make f(0) whatever we want or it wouldn't be continuous, it would have to be 0.
The crazy function I made up probably wouldn't have been continuous.[/quote]

Sorry for my idiot mistake :mad: . Anyway did you find any other way to solve it? Because that is the reason for posting this problem :roll:

[quote="Γιώργος"]It's not hard but i'm looking for other ways to prove it rather than this of my teacher(he made the problem).

If $ f(x)f\left( \minus{} \frac {1}{x}\right) \equal{} \minus{} 1$ with $ x\in \mathbb{R^*}$ and $ f$ is continuous, find $ f(0)$.[/quote]
$ f(x)\equal{}x$ is a solution and so $ f(0)\equal{}0$ is possible.

Suppose now $ f(0)\equal{}a\neq 0$. Then $ \lim_{x\to \plus{}\infty}f(x)\equal{}\minus{}\frac 1a$ is of opposite sign than $ f(0)$ and so, since $ f(x)$ is continuous, $ \exists u>0$ such that $ f(u)\equal{}0$, but this is impossible since $ f(u)f(\minus{}\frac 1u)\equal{}\minus{}1$

So $ f(0)\equal{}0$


I know this sounds basic, and I've tried myself but haven't gotten legit figures. How do you calculate your karma earned? Thanks

Oh, I know about that, but thanks anyway. I was talking about how to calculate it [b]now[/b], after I've done all the probs and most of the videos. I've figured out a pretty good formula now, though.

You go to reports then look @ karma points.

Okay, so basically I figured this out now. I take the number you are mentioning, which is the total karma you have, and do some stuff to it to get my karma earned. BTW its ranked 11th, just off leaderboard. :(


quadratics
Gauss
induction
superior algebra
superior algebra unsolved

Let $ a, b \in \mathbb{Z} \minus{} \{0\}$ with $ b\geq 1$ and $ \gcd(a,b) \equal{} 1$. Set $ \zeta_{b}: \equal{} e^{\frac {2\pi i}{b}}$ and $ g_{a,b}: \equal{} \sum_{t \equal{} 0}^{b \minus{} 1}\zeta_{b}^{at^2}$.

For a positive odd prime p not dividing $ a$ and any integer $ r \geq 2$, show that $ g_{a, p^r} \equal{} p.g_{a, p^{r \minus{} 2}}$

Can someone offer a hint on this one?

Sorry for double posting but the section I posted this in is more of a competition section and perhaps the problem is trivial or just uninteresting?

I have tried induction on r but I did not even go pass the base case r = 2.


quadratics
algebra
polynomial
Vieta
number theory proposed
number theory

Given positive integers $ a$ and $ b$ with $ a|(b^2\minus{}1)$ and $ b|(a^2\minus{}1)$, prove that there exists a pair $ (a,b)$ so that $ a\minus{}b\ge m$ for any integer $ m$.

$ b\equal{}1,a\equal{}m\plus{}1$ ?
or nontrivial $ a\equal{}m^2\plus{}2m, b\equal{}m\plus{}1$

[quote="Albanian Eagle"]$ b \equal{} 1,a \equal{} m \plus{} 1$ ?
or nontrivial $ a \equal{} m^2 \plus{} 2m, b \equal{} m \plus{} 1$[/quote]
Whoops; I had not expected it to be that easy  :oops:

I was thinking something like this:
[hide="Solution"]
$ \frac {b^2 \minus{} 1}{a}$ is an integer and $ \frac {a^2 \minus{} 1}{b}$ is an integer.  Clearly, $ a^2 \minus{} 1$ and $ a$ are coprime, meaning that the condition holds if and only if\[ \frac {(b^2 \minus{} 1)(a^2 \minus{} 1)}{ab} \equal{} \frac {a^2b^2 \minus{} a^2 \minus{} b^2 \plus{} 1}{ab} \equal{} ab \minus{} (\frac {a^2 \plus{} b^2 \minus{} 1}{ab})\]is an integer, so we want $ \frac {a^2 \plus{} b^2 \minus{} 1}{ab}$ to be an integer.  Let this value be $ k$.  Then, $ a^2 \minus{} kab \plus{} (b^2 \minus{} 1) \equal{} 0$.  Looking at this in a quadratic in $ a$, we assume that we have a solution, $ (a,b)$ with $ a\ge b$.  Then, $ (b, \frac {b^2 \minus{} 1}{a})$ is a solution.  Clearly, noting that $ a$ and $ b$ are positive, we have that $ \frac {b^2 \minus{} 1}{a} < b$ since $ a\ge b$.  Furthermore, $ b\le a$.  Then, we have a solution $ (x,y)$ with $ x \le a$ and $ y< b$.  We can keep on doing this (without changing $ b$) until $ a$ or $ b$ hits $ 1$.  $ a\ge b$, so let $ b$ hit $ 1$. Then, we get that $ a \equal{} k$.  Hence, the "base" of our Vieta-Jumping procedure starts at $ (k,1)$.  Then, we do the exact of that procedure.  We have $ (1, k)$ is a solution, so $ (k^2 \minus{} 1, k)$ is a solution, so $ (k, k^2 \minus{} 1)$ is a solution.  If $ a \equal{} k^2 \minus{} 1$ and $ b \equal{} k$, then $ a \minus{} b$ is unbounded.  [/hide]

well, that's sort of what got me into the answer above...
But still it's an observation that one could see without doing the Vieta stuff :) (theoretically)


trigonometry quadratics geometry probability AMC AIME trapezoid

In triangle $ ABC$, $ AB \equal{} AC \equal{} 100$, and $ BC \equal{} 56$. Circle $ P$ has radius $ 16$ and is tangent to $ \overline{AC}$ and $ \overline{BC}$. Circle $ Q$ is externally tangent to $ P$ and is tangent to $ \overline{AB}$ and $ \overline{BC}$. No point of circle $ Q$ lies outside of $ \triangle ABC$. The radius of circle $ Q$ can be expressed in the form $ m \minus{} n\sqrt {k}$, where $ m$, $ n$, and $ k$ are positive integers and $ k$ is the product of distinct primes. Find $ m \plus{} nk$.

[hide="It helps to draw a picture as you go along"]
We inscribe the circles as in the hypothesis. From there, we drop the perpendiculars from $ P$ and $ Q$ to $ BC$ to the points $ X$ and $ Y$ respectively.

Let radius of $ Q$ be $ R$. So we know that $ PQ$ = $ R \plus{} 16$.
Now from $ P$ draw segment $ PM \parallel{} BC$ such that $ M$ is on $ QY$. Clearly, $ PM \equal{} XY$ and $ QM \equal{} R \minus{} 16$. Also, we know $ QPM$ is a right triangle.

So let's find $ XY$.

Consider the right triangle $ PBX$. Clearly $ PB$ bisects angle $ ABC$. Let $ \angle ABC \equal{} 2\theta$. So $ \angle PBX \equal{} \theta$. We will apply tangent half-angle identites... dropping altitude from $ A$ to $ BC$, we recognize the ever-popular $ 7 \minus{} 24 \minus{} 25$ right triangle (except it's scaled by 4).

So we get that $ \tan(2\theta) \equal{} 24/7$. From half-angle identity, we can easily see that $ \tan(\theta) \equal{} \frac {3}{4}.$

So $ PX \equal{} \frac {64}{3}$. By similar reasoning in triangle $ QCY$, we see that $ CY \equal{} \frac {4R}{3}$ (recall $ ABC$ is isoceles!)

We conclude that $ XY \equal{} 56 \minus{} \frac {4R \plus{} 64}{3} \equal{} \frac {104 \minus{} 4R}{3}$

So our right triangle has sides:
$ R \plus{} 16$, $ R \minus{} 16$, $ \frac {104 \minus{} 4R}{3}$

By pythagorean and lots of simplification and quadratic formula, we can get
$ R \equal{} 44 \minus{} 6\sqrt {35}$, for a final answer of
$ \fbox{254}$
[/hide]

We had to find m+nk not m+n+k? Major headbash on that one...

man, should've tried this problem; I got it like 5 min. after the test ended.  didn't try it b/c it looks long and I distrust geometry problems...o well

note, this problem is extremly similar to those that are like:
"In a certain triangle, you have a coin of radius one placed at random on the triangle... find the probability that the coin fits completely within it"

the standard approach to those, at least for me, is to do the tangent half angle thing, so when this problem came up, i was liek 'pwned :P'

I liked this more than number 10 on AIME I

[hide]Anyways, draw perpendiculars from P to BC and AC.  Call the one on BC X and the one on AC Y.  Similarily, drop perpendiculars from Q to BC and AB and call them S and R respectively  We want to somehow express R in terms of one of the sides.  Call the point where QS is perpendicular to line P T.  R is the radius of Q

So first of all, XC can be computed using trig on triangle CPX.  Also, SX can be computed becuase it is just PT.  And SB can also be computed.  So we just need all three to add up to $ 56$.

So we find CX first.  Notice that the tan(ACB)=$ \frac {24}{7}$ (easy to tell just drop a height, we have $ 7 - 24 - 25$ triangle.  Inorder to find CX, we need to find $ \tan(\frac {1}{2}ACB$)-because triangles CPY and CPX are congruent.  Using the fact that (since i forgot direct $ \tan(\frac {1}{2}x$) formula) $ \tan(\frac {1}{2}x) = \frac {\sin(\frac {1}{2}x)}{cos(\frac {1}{2}x)}$, we get that $ \tan(\frac {1}{2}ACB) = \frac {3}{4}$ ($ \sin(\frac {1}{2}x) = \sqrt {\frac {1 - cosx}{2}}$, cosine half just adds the cosx on top, note $ \cos(ACB) = \frac {7}{25}$.)  So we have that $ CX = \frac {4}{3}*16 = \frac {64}{3}$.

Now we find SX, clearly from right triangle PQT, SX=$ 8\sqrt {R}$ (use pythagoeran theorm with sides $ R + 16$ and $ R - 16$).

Then we want to find BT.  With the same method to find CX, we know that $ \tan(\frac {1}{2}ABC) = \tan(\frac {1}{2}ACB$)-isoscles-=$ \frac {3}{4}$.  Also traingle BST is congruent to triangle BRT.  So $ BS = \frac {4}{3}*R$

Now $ \frac {4}{3}R + 8\sqrt {R} = 64 - \frac {64}{3}$,
so $ \frac {4}{3}R + 8\sqrt {R} = \frac {104}{3}$
$ 4R + 24\sqrt {R} = 104$ (multiplied both sides by 3)
$ R + 6\sqrt {R} = 26$, $ 6\sqrt {R} = 26 - R$
$ 36R = 676 - 52R + R^2$
$ R^2 - 88R + 676 = 0$
$ R = \frac {88 - \sqrt {88^2 - (4)(676)}}{2}$
we take minus because its still positive and + could make the radius too big (obvious by the answer format and if you draw it you cant fit it)
$ R = \frac {88 - 12\sqrt {35}}{2}$
$ R = 44 - 6\sqrt {35}$
$ 44 + (6)(35) = 254$[/hide]

whoa, i think this is almost the exact same solution as above.
ill try to find a more geometric one.

I just worked through it.
[hide="my method"]
I dropped perpendiculars from $ P$ and $ Q$ to $ BC$

This splits $ BC$ into three segments.

The lengths of these three segments are (from $ B$ to $ C$)

$ \frac{r}{\tan\theta}$, $ \sqrt{(r\plus{}16)^2 \minus{} (r\minus{}16)^2}$, and $ \frac{16}{\tan\theta}$

The first and third are from simple trig where $ \theta$ is one half of the base angle of the original triangle.
The second one comes from the pythagorean theorem on the triangle in the trapezoid formed when the perpendiculars are dropped.

Using the half angle formula, $ \tan\theta \equal{} \frac{3}{4}$.

Setting the sum of those lengths equal to 56 gives

$ 56 \equal{} \frac{r}{\frac{4}{3}} \plus{} \sqrt{(r\plus{}16)^2 \minus{} (r\minus{}16)^2} \plus{} \frac{16}{\frac{4}{3}}$

This simplifies out to

$ 26 \equal{} r \plus{} 6\sqrt{r}$

Solving yields $ r \equal{} 44 \minus{} 6\sqrt{35}$ and we're all set.
I think this way is slightly faster.
[/hide]

not sure if this is the same as the previous ones, but

[hide]
extend $ \overline{PC}$ and $ \overline{BQ}$ so that they meet at a point (the incenter of the triangle, because $ \overline{PC}$ and $ \overline{BQ}$ are angle bisectors (power of a point). drop a perpendicular (the inradius, which can be easily calculated)....

then you get a bunch of nice similar triangles (3-4-5 multiples) so that you can express the base (56) in terms of the radius and solve.
[/hide]

[hide="Outline of solution without trig"]
Set $ r$ as the radius of $ Q$. Since the centers of the circles are on angle bisectors, we draw in the incenter. We can use similar triangles with the incenter to get that the height of the trapezoid created by dropping perpendiculars from the two circles' centers to $ BC$ is $ \frac{104\minus{}4r}{3}$. We now have Pythagorean relationship by dropping a perpendicular from the center of $ P$ to the perpendicular from the center of $ Q$ to $ BC$. This is a quadratic, so we solve for $ r$.
[/hide]

Yeah, I *misread* it on the AIME and spent 30 minutes on a problem that wasn't on the test, didn't have the correct answer form, and requiring solving quadratics with coefficients like $ 79^4$.

[quote="basketball9"]Can someone please nicely explain why PC and BQ are angle bisectors.[/quote]

Points on the angle bisector are equidistant from the two sides that form the angle it bisects. Since circles $P$ and $Q$ are tangent to $\overline{AC}, \overline{BC}$, and $\overline{AB}, \overline{BC}$ respectively, we can see $P$ is on the angle bisector of $\angle{C}$ and $Q$ is on the angle bisector of $\angle{B}$.

Let the center of $P$ be $O_1$ and let the center of $Q$ be $O_2$. Then let the foot perpendicular from $O_1$ to $AB$ be $X$ and define $Y$ similarly. Let the radius of $Q$ be $x$. Then half angle formulas give $\sin \angle YBO_2 = \frac{3}{5}$ and $\cos \angle YBO_2 = \frac{4}{5}$, so $\tan \angle YBO_2 = \frac{3}{4} \implies YB = \frac{4}{3}x$. Similarly, $\tan \angle XAO = \frac{3}{4}$, so $AX = \frac{64}{3}$. We also have $XY = \sqrt{(x+16)^2-(x-16)^2}$, so $\frac{4}{3}x+\frac{64}{3} + \sqrt{(x+16)^2-(x-16)^2} = 56 \implies x = 44 - 6\sqrt{35} \implies \boxed{254}$.

note: this is probably slightly messed up w/points and stuff because I drew the diagram wrong and had to fix stuff in the sol without redrawing, I'll fix when I get home.


geometry
parallelogram
geometry unsolved

This result is true. We can solve this problem, based of the follows lemmas:

[b][size=109]LEMMA  1.[/size][/b] – A quadrilateral $ABCD$ is given and let $E,$ $F$ be, two points ON $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $),$ such that $AE / EB = DF / FC = p.$
We define three points $K,$ $L,$ $M,$ on the segments $AD,$ $EF,$ $BC$ respectively, such that $AK / KD = EL / LF = BM / MC = q.$
Prove that the points $K,$ $L,$ $M,$ are collinear, such that $KL / LM = p.$

[b][size=109]PROOF OF THE LEMMA  1.[/size][/b] – See at: http://www.mathlinks.ro/Forum/viewtopic.php?t=97493

[b][size=109]LEMMA  2.[/size][/b] – A quadrilateral $ABCD$ is given and let $K,$ $L,$ $M,$ $N$ be, the midpoints of $AB,$ $BC,$ $CD,$ $AD$ respectively.
Prove that $(AKN)+(CLM) = (PKN)+(PLM),$ where $P,$ is the intersection point of $KM,$ $LN.$

[b][size=109]PROOF OF THE LEMMA  2.[/size][/b]
Because of the points $K,$ $N,$ are the midpoints of the segments $AB,$ $AD$ respectively, we have that $(AKN) = \frac{(ABD)}{4}$ $,(1)$ and similarly $(CLM) = \frac{(CBD)}{4}$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $(AKN)+(CLM) = \frac{(ABCD)}{4}$ $,(3)$

By the same way we can prove that $(BKL)+(DMN) = \frac{(ABCD)}{4}$ $,(4)$

Hence, we conclude that $(KLMN) = \frac{(ABCD)}{2}$ $,(5)$

From $(5),$ because of the parallelogram $KLMN,$ we have that $(PKL) = (PLM) = (PMN) = (PKN) = \frac{(ABCD)}{8}$ $,(6)$

From $(3),$ $(6)$ $\Longrightarrow$ $(AKN)+(CLM) = (PKN)+(PLM) = \frac{(ABCD)}{4}$ $,(7)$ and the proof of the [b][size=109]Lemma 2,[/size][/b] is completed.

[b][size=109]PROOF OF THE PROPOSED PROBLEM.[/size][/b] - Let us to change the notations as follows:

A quadrilateral $ABCD$ is given and let be pairs of points on its sidesegments as follows:

- The points $E,$ $F$ on $AB,$ such that $AE = EF = FB = AB / 3.$

- The points $G,$ $H$ on $CD,$ such that $DG = GH = HC = CD / 3.$

- The points $K,$ $L$ on $AD,$ such that $AK = KL = LD = AD / 3.$

- The points $M,$ $N$ on $BC,$ such that $BM = MN = NC = BC / 3.$

Also we denote the points $P\equiv KM\cap EG,$ $Q\equiv KM\cap FH,$ $R\equiv LN\cap FH,$ and $S\equiv LN\cap EG,$

We will prove that $(AEPK)+(CHRN) = 2(PQRS).$

$\bullet$ Based on the [b][size=109]Lemma 1,[/size][/b] we conclude that

$KP = PQ = QM$ $,(1)$ $LS = SR = RN$ $,(2)$ $EP = PS = SG$ $,(3)$ and $FQ = QR = RH$ $,(4)$

From the parallelograms $EQSK,$ $QNHS,$ we have that $(PKE) = (PQS)$ $,(5)$ and $(RQS) = (RNH)$ $,(6)$

From $(5),$ $(6)$ $\Longrightarrow$ $(PKE)+(RHN) = (PQRS)$ $,(7)$

$\bullet$ Based on the [b][size=109]Lemma 2,[/size][/b] we conclude that $(AEK)+(RQS) = (PKE)+(PQS)$ $,(8)$

 and $(PQS)+(CHN) = (RQS)+(RHN)$ $,(9)$

From $(8),$ $(9)$ $\Longrightarrow$ $(AEK)+(CHN) = (PKE)+(RHN) = (PQS)+(RQS)$ $\Longrightarrow$ $(AEK)+(CHN) = (PQRS)$ $,(10)$

From $(7),$ $(10)$ $\Longrightarrow$ $(AEPK)+(CHRN) = 2(PQRS)$ $,(11)$ and the proof is completed.

Kostas Vittas.


okay, I can't solve this problem and its driving me nuts:
log[size=67]2[/size][size=100](x-2)[/size]-log[size=67]2[/size][size=100]3[/size]=log[size=67]2[/size][size=100]x[/size]
So what I did was:
log[size=67]2[/size][size=100](x-2)/3[/size]=log[size=67]2[/size][size=100]x[/size]
Then:
x-2/3=x
x-2=3x
x=-1
But this isn't possible, is the answer that there are no x for which this works, or what?

Darn that problem!

please help!
Smiles,
lyra

[quote="surge"][quote="lyra"][quote="sagar_astakala"]Hi...

I think the question is wrong...

If the question has to be

 log(base2) {x-2} + log(base2) 3  = log(base2) x

then the answer  x = 3[/quote]

that was the question, how did you get that answer?

Smiles,
lyra[/quote]

That is not equivalent to your original question.
$\log_2 (x-2)+\log_2 (3) \not=\log_2 (x-2)-\log_2 (3)$

$\log_2 (x-2) +\log_2 (3)=log_2 (x)$
$3(x-2)=x$
$3x-6=x$
$2x=6$
$x=3$[/quote]

Well, the problem says - not +, do you know how to do it?

Thanks,
lyra

$\log_2 3$ is a constant, and for positive numbers, the log of a bigger number is bigger.  so it makes sense that there are no solutions for $\log_2 (x-2)-\log_2 3=\log_2 x$.  since $x$ is greater then $x-2$, $\log_2 x$ is greater than $\log_2 (x-2)$, but in the equation, this is not the case.

Okay, thank you all so much!!!

Smiles,
lyra


it reduces to $1+2^n+3^n+4^n \mod 5$, and $\phi(5)=4$, so the exponents mod 4 only matters, test 0,1,2,3 and 1 and 3 work, so that reduces to $n=2k+1 \, \forall\, n\in\mathbb{Z}$, you are correct

[quote="jelly119"]Determine all positive integers $n$ such that $1981^n+1982^n+1983^n+1984^n+1985^n$ is divisible by $5$.

I thought that $n$ is just the set of all odd natural numbers while the answer in the book is $n=4k+1,k\in\mathbb{n}$. What's wrong with my answer??[/quote]

Modulo 5, the equation reduces to $1^n + 2^n + (-2)^n + (-1)^n \equiv 0 \pmod {5}$. So, $n$ is odd and you are correct.  :)

[quote="chess64"]Modulo 5, the equation reduces to $1^n + 2^n + (-2)^n + (-1)^n \equiv 0 \pmod {5}$. So, $n$ is odd and you are correct.  :)[/quote]

What about 4k+2?
$1^2 + 2^2 + (-2)^2 + (-1)^2 = 10 \equiv 0 \pmod {5}$

I think the wright answer is n=0, and n=4k+p with  1<=p<=3 , with k and p E (0 1 2 3 ...).

Given that this reduces to $1^n+2^n+3^n+4^n$ in mod 5:

Powers of 1 have units digits repeating in the pattern 1, 1, 1, 1
Powers of 2 repeat as 2, 4, 8, 6
Powers of 3 repeat as 3, 9, 7, 1
Powers of 4 repeat as 4, 6, 4, 6

Adding downward, we get 10, 20, 30, 14

So, mod 5, we have 0, 0, 0, 4

This indicates that all integers that are not multiples of 4 satisfy this condition.

Chess 64, remember that in modulus, the left side does not have to EQUAL the right side, it only has to be congruent. So your modulus congruence does not force n to be odd, although any odd n will satisfy it.

Oops, sorry. I didn't think any even $n$ would work.

Yeah, it's 4k + 1,4k + 2,4k + 3.
Look at the numbers in terms of mod 5.Get the total remainder:
(-1) ^ n + 1 ^ n + 2 ^ n+(-2) ^ n + 0
which is 0 when n is odd or n = 4k + 2.

[quote="Mo²"][quote="10000th User"]jimbo is correct. A similar question was asked in an old Hungarian competition (Eotvos???) and the answer I remember was all positive integers except those multiples of 4.[/quote]

n=0 is also correct, and it is also a multiple of 4.[/quote]The question asked for [b]positive integers[/b] $n$, and yes I know $n=0$ is also correct, that's why it's called the [b]trivial[/b] case  :P

[EDIT] Actually, the question asked in that Hungarian competition referred to $1^n+2^n+3^n+4^n$, so $n=0$ would not work :D and I am doubly correct


I have two more questions that are simply insane.  I worked all night but to no avail.

1) Show that log_a (sqrt{x} + sqrt{x - 1}) + log_a(sqrt{x} - sqrt{x - 1}) = 0

2) Find the value of log_2 (4) times log_4 (6) times log_6 (8).

Thank you for your help.

[hide="hint for 1"]
Remember that adding logs of like bases is the same as one log with the product of the arguments.
[/hide]

[hide="hint for 2"]
Use change of base formula for all three logs.
[/hide]

[quote="joml88"]As others have suggested, learning $\LaTeX$ really goes a long way.[/quote]Yep. Our many requests for symmetry to learn $\LaTeX$ (at LEAST just put $\$$ signs around it and see what happens) and to not make many threads of related problems (that seem to have come right out of high school textbooks) have pretty much been ignored.

come on symmetry, learn to use latex, you post many problems, and it is helpful if you use latex, you don't need to download anything, you just need to put dollar signs around your math stuff

[quote="IntrepidMath"][quote="joml88"]As others have suggested, learning $\LaTeX$ really goes a long way.

1) \[\log_a (\sqrt{x} + \sqrt{x - 1}) + \log_a(\sqrt{x} - \sqrt{x - 1}) = 0\]

2) \[(\log_24)(\log_46)(\log_68)\][/quote]
[hide]
1)
$\log_a (\sqrt{x} + \sqrt{x - 1}) + \log_a(\sqrt{x} - \sqrt{x - 1})$
$\log_a((\sqrt{x} + \sqrt{x-1})(\sqrt{x}-\sqrt{x-1}))$
$\log_a(x-\sqrt{x^2-x}+\sqrt{x^2-x}-(x-1))$
$\log_a(x-x+1)$
$\log_a(1)$
If you take the log of 1, regardless of the base, it is zero because $\text{anything}^0=1$, so $\log_a(1)=0$

2)
$\displaystyle (\log_2 4)(\log_4 6)(\log_6 8)=(log_2 4)(\frac{\log_2 6}{\log_2 4})(\frac{\log_2 6}{\log_2 2})$
$\displaystyle (2)(\frac{(log_2 6)^2}{2})\\(\log_2 6)^2$
this is as far as I got.

[/hide][/quote]

Here's a property to consider:

$\log_a{b}*\log_b{c}=\log_a{c}$

[quote="Drunken_Math"]Here's a property to consider:

$\log_a{b}*\log_b{c}=\log_a{c}$[/quote]
[hide="reason"]
$(\log_a b)(\log_b c)$

$=\frac{\log_n b}{\log_n a}\cdot\frac{\log_n c}{\log_n b}$

$=\frac{\log_n c}{\log_n a}$

$=\log_a c$[/hide] :)

Great stuff here.  Thank you all for your tips and help.


Please help me solve this system

x-2x+2x-x=-2
-3x+5x-x-3x=2
2x+3x+3x+5x=-9
4x-x-x-2x=8

Due tomorow, of course. Thanks!

 :!:

Well if you know how to do Gaussian elimination, there is no problem so I'm assuimg you have some trouble with it?  I'm not going to do the problem for you (defeats the purpose) but I'll point you where you want to go.

Put the coefficients into a matrix, in this case,
$ \begin{pmatrix}
1&\minus{}2&2&\minus{}1\\
\minus{}3&5&\minus{}1&\minus{}3\\
2&3&3&5\\
4&\minus{}1&\minus{}1&\minus{}2
\end{pmatrix}$
I think you mean for all those variables to be different, right?  Now you need to create your augmented matrix, and then try to get the matrix into reduced row echelon form.  Don't forget, anything you do on the left MUST be done on the right.  If you aren't familiar with the terms I am using, you really need to go back and re-read your text.

[quote]who's Matrix?[/quote]
What are you talking about?


function
calculus
integration
limit
calculus computations

How to integrate the function on the open interval? the hypothesis of The fundamental theorem of calculus just states for the closed interval. And some kind of improper integral just give an example on interval $(- \infty, \infty)$.
An example
let $ A = \{x: 3 < c < \infty \} $ evaluate $\int_{A} \left( \frac{1}{2} \right )^x dx$

Can we just transform it to $\lim_{b \rightarrow \infty} \int_{3}^b \left( \frac{1}{2} \right )^x dx$ ?

thanks for advance..

Yes; that's even the standard definition. Proper Riemann integrals are only defined for bounded functions on bounded intervals, and any improper Riemann integral has a limit implicit in its definition. If that limit does not exist, we say that the integral diverges.

For integrals that are improper for more than one reason, you need to be careful about convergence. $\int_{-\infty}^\infty\frac{x}{x^2+1}\,dx$ diverges, even though $\int_{-M}^M\frac x{x^2+1}\,dx=0$ for all $M$.


Determine all positive integers k for which there exists a function f:N->Z such that :
 (a) f(1997)=1998
 (b) for all a,b  \in N f(ab)=f(a)+f(b)+kf(gcd(a,b))

take a=b and get f(a 2 )= (k+2)f(a)
and hence f(a^4)=(k+2) 2 f(a)
and on the otherhand
f(a^4)=f(a)+f(a^3)+kf(a)=
(k+1)f(a)+f(a^3)=(2k+2)f(a)+f(a 2 )
=(3k+4)f(a)
compare and get the job done :D

well, as he said:

(k+2)² f(a) = (3k+4) f(a)

scratch f(a) and you get k²+4k+4 = 3k+4
so so k²+k=0 so k = -1 or k = 0

now just check if both are really ok, and you got them

:)
I think k can't equal 0
Because :
  put a=1997,b=0
  we have :f(0)=f(1997)+f(0) +kf(1997)
   so         :  1998(k+1)=0 => k=-1
What do you think about my sol  :D

If we find k with k is integer number then k can't equal 0 :D
is it right?

You should fix the statement of the problem ;)

Let N be the set of the positive integers, let Z be the set of the integers.
Find all integers k for which there exist a function f: N -> Z such that
*f(1997) = 1998
*f(ab) = f(a) + f(b) + kf(gcd(a, b)).

Anyway, even in the Andreescu book for 97-98 it is incorrectly stated:
in the problem they say POSITIVE k, in the answer they say k = 0 or k = -1.
And k = 0 indeed works... An example:
f(p1e1p2e2...pnen) =\sum ekg(pk)

where m is a prime factor of 1997, g(m) = 1998, and g(p) = 0 for all primes p =/= m.

:) ok,your idea is right when N is the set of the positive integers,and k is integer.

but as I known , N is set of the natural numbers(include zero)
                        N* is set of  the positive integers

[quote="Arne"]Can't k be equal to 0? :? As for as I know, 0 is not a positive integer...[/quote]

:? thought 0 was both positive and negative, just not "strictly positive"

Plus, I didn't say that were the eventual solutions, just read what I wrote...

In English positive means our 'strikt positief', Peter.


geometry
geometric transformation
reflection
conics
parabola
ellipse
analytic geometry

A ball moving parallel to the y axis undergoes an elastic collision with a parabolic mirror y^2=2px.Prove that no matter where the prompt of impact lies it will arrive at the mirror's focus.

I believe this can be done in the same way we prove a result for reflection of light in a parabolic mirror.In that equiavalent optics problem,we used Fermat's law of least path.Here it is not available.So,how should we proceed?

The angle of incident is the same angle with which it leaves the surface.

So.... it is basically reflected across the line perpendicular to the surface at the point of impact.

But I cannot see anything using only "law" of reflection.How does it mean that the reflected ball passes through focus(P)?One possibility is that the undeviated ball strikes the directrics and from the definition of parabola,e=1...so,SP=SN.That is (d/dt)SN=(d/dt)SPThis means magnitude of velocity remains unchanged iff the ball reflects along SP...Because,for other points the relation is not true.

I understand the way is hanging around this.But,I cannot develop logically...
by the way,I hope in the same way one can prove for two focii in an ellipse?

kolahalb i think you wanted to say that the ball moves paralell to the x axis.another problem like this is :http://www.mathlinks.ro/Forum/viewtopic.php?p=827996#827996

a parabolic mirror has it's focus situated at p/2 from it's tip if it's ecuation is y^2=2px
you can develop logically like this:proove that anything that comes paralell with axis ox from infinity to the parabola ,no matter how big the height regarding the ox axis is, and reflects, than it passes through the focus

radac_mail,your link is not working.Check it.The information "The topic or post you requested does not exist" is coming.

You are correct.I shouldhave posted parallel to x axis.

Let me get to the problem.A ball parallel to x axis collides the open concave surface of the parabola elastically.We may easily see that that possible trajectory of the ball which passes through the focus on reflection from the parabolic surface is the only one that abides by the principle of least action.This we can prove using the property of a parabola:e=1.
Therefore this is the required path.

Please check if I am correct.

sorry about the bad link.go to page 2 in advanced physics forum and see the topic :optics-mirror with a certain ecuation.this won't help you resolve the problem.it's just another point of view (a harder one)in the same problem.
i don't know what e is but this is an easy geometry problem .proove that for any line of ecuation y=h the reflected line goes through (p/2,0)

Thank you very much.I think I will be able to do it now.

*First you find the co-ordinates of the point of incidence.
*differentiate y^2 and 2px and equate them and find the slope of the normal in terms of co-ordinates of the point of incidence.
*Now you have incident slope 0,normal's slope known,and reflected ray's slope m,say.
*use standard co-ordinate formula that i=r.This gives the value of m.
*The reflected ray's equation can be found.and it is seen that the line passes through (p/2,0)


Hello everyone - I was just wondering where I might be able to find the default Computer Modern font used in LaTeX so that I may use it in Microsoft Office applications (I run a PC, so truetype would be ideal). I found a website that supposedly had them, but there were so many files, and most of the ones I downloaded did not look like the real thing.

Thanks for your help!

Try Aurora ([url]http://elevatorlady.ca[/url]). It has a set of TTFs included.


If two uniform bodies are moved closer to each other along the line which joins them, then the gravitational force exerted between them usually increases. Is this always true?

As far as I am aware in Newtonian gravity this would be true with respect to the inverse square law. I am not sure if relativity has anything more to say on the subject, but I would assume not

gravitational force exerted between 2 bodies always increases when they get closer


logarithms
inequalities
function
inequalities proposed

Since $x\ge y\ge z>0$ and  $xyz=1$, it's easy to see that $x\ge 1$ and $z\le 1$. So $\ln{x} \ge 0$ and $\ln{z}\le 0$.

Case I. If $y\le 1$, then $\ln{y} \le 0$.

$xz\ge xyz=1$ implies that $\ln{x}\geq -\ln{z}$.

Hence both $(\ln{x}, -\ln{z}, \ln{z}, -\ln{x})$ and  $(-2\ln{y}, -\ln{y}, \ln{y}, 2\ln{y})$ are non-increasing sequences.

$xy^2 \geq xyz=1$ implies that $\ln{x} \geq -2\ln{y}$.

$\displaystyle \frac{xy^3}{z} =\frac{y^2}{z^2} \geq 1$ implies  that $\ln{x} + (-\ln{z}) \geq (-2\ln{y}) + (-\ln{y})$.

These should be enough to conclude that:

$(\ln{x}, -\ln{z}, \ln{z}, -\ln{x})$ majorizes $(-2\ln{y}, -\ln{y}, \ln{y}, 2\ln{y})$.

The result follows by applying the Majorization (Karamata) Inequality to the convex function $f(x) = e^x$.

Case II. If $y\geq 1$, do a substitition $u=1/z, v=1/y, w=1/x$, then $v\le 1$, $u\ge v\ge w > 0$ and $uvw=1$. Apply the Case I to $u,v,w$ and we are done.

Come on, or I'm very idiot, or this is trivial. Replace $ z$ with $\frac{1}{xy} $ and consider as a function in x, which is increasing. Done!

Replacing [tex]z[/tex] with [tex]\frac{1}{xy}[/tex] gives:
[tex]x+\frac{1}{xy}+\frac{1}{x}+xy \geq y+y^2+\frac{1}{y}+\frac{1}{y^2}[/tex]

Pulling these into pairs we get:
[tex](x-y)+(xy-y^2) \geq (\frac{1}{y}-\frac{1}{x})+(\frac{1}{y^2}-\frac{1}{xy})[/tex]

Multiplying the fractions on the right and pulling terms on both sides gives:
[tex](x-y)(1+y) \geq \frac{x-y}{xy}+\frac{x-y}{xy^2}=(x-y)(\frac{1}{xy}+\frac{1}{xy^2})[/tex]

Dividing by [tex](x-y)[/tex] and mulitplying by [tex]xy^2[/tex] gives:
[tex]xy^2(y+1) \geq y+1[/tex]

[tex]xy^2 \geq 1[/tex]

[tex]xy\frac{1}{xz} \geq 1[/tex]

[tex]\frac{y}{z} \geq 1[/tex]

And we know this is true since [tex]y \geq z[/tex] QED.

[quote="harazi"]Replace $ z$ with $\frac{1}{xy} $ and consider as a function in x, which is increasing. Done![/quote]

Nice observation!

Tung Lam has been posting exponential problems recently which mislead me to think of using Majorization.

My proof is like "killing a chicken using a knife for cow" as Asian people say.  :)

Oh,yes ! Harazi is right.
Here my solution :
\[ 0 \leq (x-y)(y-z)=xy-xz-y^2+yz= \frac{1}{z}- \frac{1}{y} - y^2 +\frac{1}{x} \]
So we have :
\[ \frac{1}{x}+ \frac{1}{z} \geq  \frac{1}{y} + y^2 \]
\[ \rightarrow \frac{1}{xy} + \frac{1}{yz} \geq \frac{1}{y^2} + y \]
\[ \rightarrow z+x \geq \frac{1}{y^2} +y \]
(q.e.d)

the ineq is equiv to this

xzy+zxy+zy+xy>=zxy^4+zxy^3+zx+zxy

<=>

xy+zy+(z+x-1)y-y^3-zx-zxy>=0

<=>

y^3-(z+x+(-1))y+(-z-x+zx)y-(-zx)<=0

<=>

(y+1)(y-z)(y-x)<=0

which is kindo true !!!!


algebra
polynomial
superior algebra
superior algebra unsolved

How do I compute the Galois group of $ x^4\minus{}5$ over $ \mathbb{Q}(\sqrt{\minus{}5})$ and $ \mathbb{Q}(i)$?

The splitting field of $ x^4\minus{}5$ is $ \mathbb{Q}(\sqrt[4]{5},i)$ and $ [\mathbb{Q}(\sqrt[4]{5},i): \mathbb{Q}(\sqrt{\minus{}5})]\equal{}8, [\mathbb{Q}(\sqrt[4]{5},i): \mathbb{Q}(i)]\equal{}4$?

$ Gal(\mathbb{Q}(\sqrt[4]{5},i)/ \mathbb{Q})$ is the semi-direct product of $ C_4$ and $ C_2$. Hence, the groups you need are $ C_2\times C_2$ and $ C_4$ respectively.


Let $ V$ be a proper subspace of $ \mathbb R^n$. Prove or disprove a)$ V^\circ\equal{}\varnothing$  b)$ \bar{V}\equal{}V$.
PS I've used the Euclidean space because it seemed easier to me when I wanted to write down its answer for my students, one may want to answer it in a more general context.

Let $ v_1,...,v_k$ be a basis of $ V$ and together with $ v_{k\plus{}1},...,v_n$ it's a basis of $ R^n$.

a. True. Suppose the interior is non empty, and that $ a$ is inside. Then $ a\plus{}tv_{k\plus{}1}$ for $ t$ small enough should be in $ V$, but it can't be.

b. True. Suppose furthermore that the basis chosen is orthonormal. Let $ a$ be a point not in $ V$, then the coefficient of at least one of $ v_{k\plus{}1},...,v_n$ is nonzero. For any point within a small ball of $ a$ must also have those coefficients nonzero, thus $ R^n \minus{} V$ is open.

More generally: if $ f\colon \mathbb R^n\to\mathbb R$ is a continuous function, then $ Z: \equal{}f^{\minus{}1}(0)$ is a closed set. If in addition $ f\in C^1$ and $ \nabla f(x)\ne 0$ for all $ x\in Z$, then $ Z$ has empty interior -- in fact, it is a $ (n\minus{}1)$-dimensional manifold.


Two circles are externally tangent to each other and have radii of 1 and 10, respectively. Their common external tangents are drawn to meet at $ A$ and they meet the small circle at $ B,C$, and meet the large circle at $ B'C'$. Find the ratio of the area of $ \Delta ABC$ to the area of $ \Delta AB'C'$.


algebra polynomial calculus integration AMC

I could only find 4 answers to this problem, but my calculus teacher said there were 6.  But then again, I was just guessing and checking, so it'd be a huge help if you could show me how to solve it.

What are all integers K for which X 2 + KX + K + 17 = 0 has integral roots?

This was on the math league today.  I don't believe we should discuss it yet.

[hide=""Shame on you?""]:P[/hide]
Edit: I just saw joml88's post, well if Math League believes I have violated any copyright rights, I sincerely apologize. I just wanted to help.

2nd Edit: yes Phyllis, there are [b]6[/b] values.

If the math league contest was given today though, why would it matter if you posted the answer tonight?  I thought distributors weren't allowed to give the tests out after the specified date.

Well 3X could you or someone else pm it to me?

The Math League ([url]http://www.mathleague.com[/url]) policy is a little confusing.

On their high school dates page, they seem to say that the test may be offered only on time or one week early:
[quote="Math League"]*Alternate contest dates may be scheduled one week before the contest dates, in the event of scheduling conflicts.
[/quote]
But on their rules page, they seem to say that any nearby school day is okay:
[quote="Math League"]In the event of school closings, special administrative functions, testing days, severely inclement weather, or similar disruptions of the normal school schedule, permission is granted to conduct the contest on a proximate school day.
[/quote]
For Contest 2, I think yifman12's school held the contest about 2 days after the scheduled date, because they had a holiday.

So I am not sure when the testing window ends.

Please do not discuss TXML until this week is over.


http://en.wikipedia.org/wiki/Irrational_number#The_square_root_of_2

more easy

I do not know that many arguments of the proof of irrationality of 2 except for the standard one in my intro to logic text. Thanks for the link. Since this argument was not there, would you say my argument is original?

hmm... it's good too

i just think that

[quote]Thus $3b^{2}$ is of the form $4k+3$, but we know this cannot be expressed as a sum of squares[/quote]

you need to proove this. to the proof be complete.

but the easy way is proof that a is even and b is odd and a is odd and b is even, what does not make sense. hm..

[quote="akech"]To prove $\sqrt{2}$ irrational.
Proof: Suppose towards a contradiction that there are $a$ and $b$ with $b$ not equal zero and $(a, b) = 1$ such that $\frac{a}{b}= \sqrt{2}$. Then it follows that $2b^{2}= a^{2}$. Observe that since $a^{2}$ is even, $a$ is even, but then $b$ cannot be even otherwise the $gcd(a,b)$ would not be $1$. Now $2b^{2}+b^{2}= a^{2}+b^{2}$ so that $3b^{2}$ is representable as a sum of two squares. But since b must be odd, call it $2s+1$, then we have $3b^{2}= 3(2s+1)^{2}= 4(3s^{2}+3s)+3$, Thus $3b^{2}$ is of the form $4k+3$, but we know this cannot be expressed as a sum of squares, a contradiction. Thus $\sqrt{2}$ is irrational.
Is this a correct proof?[/quote]

yes that's right u can easily do it using the fact that any sq. is congruent to $0$ or $1$ mod $4$ .

but don't forget the beauty of the original proof ...........

Suppose $\sqrt{2}= \frac{a}{b}$
that is $2b^{2}= a^{2}$.
or, written in binary
$10b_{'}^{2}= a_{'}^{2}$, where the righthandside ends in an even number of zeros,
and the lefthandside in an odd number of zeros.
Impossible.

[quote="petruspan"]
or, written in binary
.[/quote]
What is it?  :huh:

I have heard of that proof about considering base 2.
Thanks,
Akech

[quote="petruspan"]Written in the binary numeral system, or (as you wish) in base-2 number system.
2 (decimal) = 10 (binary)[/quote]
thanks !  :lol:  I know all of them

dont have to be this complicated

Assume $\sqrt{2}$ is rational then $\sqrt{2}= \frac{a}{b}$ and $(a,b)=1$

now $a^{2}=2b^{2}$ hence $a^{2}$ is even, then $a$ is even (because odd number square is odd number)

now, $(2k)^{2}=2b^{2}$
$4k^{2}=2b^{2}$
$2k^{2}=b^{2}$
hence $b^{2}$ is even, therefore $b$ is even

contradiction since (a,b)=1, but a and b are both even

therefore $\sqrt{2}$ is irrational

Following [b]petruspan[/b], I will prove that $\sqrt{9}$ is irrational.

Suppose $\sqrt{9}= \frac{a}{b}$, that is $9b^{2}= a^{2}$, or, written in base $9$, $10b_{'}^{2}= a_{'}^{2}$, where the righthandside ends in an even number of zeros, and the lefthandside in an odd number of zeros. Impossible.


algebra
function
domain
trigonometry
analytic geometry
graphing lines
slope

Why does the graph of tan x repeat every interval of $ \pi$ while the graphs of sin x and cos x repeat every interval of $ 2\pi$?

The domain of the sine and cosine graphs are $ [0,2\pi]$ and for tangent, $ [\minus{}\frac{\pi}{2},\frac{\pi}{2}]$.  The frequency of the sine and cosine graphs is $ 2\pi$ and the frequency of the tangent graph is $ \pi$.  As to why this is so, I don't really know.

If you look at the unit circle definition of $ \sin$ and $ \cos$, it is pretty clear why their cycles are $ 2\pi$. It represents one full wrap around the circle.

Now the tangent function is defined to be $ \tan x \equal{} \frac{\sin x}{\cos x}$
If you add $ \pi$ radians to any angle measurement, then both the sin and cos will become negative. The negative signs will cancel in the fraction and tan will remain unchanged.

Because of this, the period on a $ \tan$ graph is only $ \pi$

[quote="JRav"]The domain of the sine and cosine graphs are $ [0,2\pi]$ and for tangent, $ [ \minus{} \frac {\pi}{2},\frac {\pi}{2}]$. [/quote]

I think you're thinking about the range of $ \arcsin, \arccos, \arctan$.  The domain of $ \sin, \cos$ (as a real function) is $ \mathbb{R}$.

An intuitive explanation based on the unit circle:  $ \tan \theta$ represents the slope of a line from the origin to a point $ (\cos \theta, \sin \theta)$.  When you have rotated $ \pi$ radians (that is, $ 180^{\circ}$), you are on the other side of the origin, but the slope is unchanged.


Two standard six-sided dice are rolled simultaneously for an indefinite number of times, and the sum of the numbers on top is calculated each time. What is the probability of getting 2 sums of 7 before both a sum of 6 and a sum of 8?

I get something uglier.
[hide="Answer, no soluton"]$ \frac {3519}{7744}$?[/hide]

I understand that posting an answer before posting a full solution is much more effective than posting the entire solution right away only to learn that the answer is wrong, but it doesn't work when the answer isn't known in the first place. :D

What could you possibly be implying? :D
Meh. Still not worth it to post.

[hide="Would not it just be"]
$ p(x)$ is the probability of rolling that number.
$ \frac{ p(7) \cdot p(7) \cdot p(6) \cdot p(8)}{3!}$?
[/hide]

no astrophys because if you do that out it is: 1/6^6 which is a little to unlikely xp

The probability of rolling a sum of $ 7$ at any time is greater than both of the others, so the probability should be greater than $ \frac{1}{2}$.

[hide="My Attempt"]We can ignore all rolls except for the first $ 6$, the first $ 8$, and the first two $ 7$s. Thus, we are considering the arrangements of the digits in the four-digit number $ 7768$.

The probability that both $ 7$s will show up before the $ 6$ and the $ 8$ is equivalent to the probability that the last digit is not $ 7$ in that four-digit number.

The probability of that is $ \frac {\frac {1}{6} \plus{} \frac {1}{6}}{\frac {1}{6} \plus{} \frac {1}{6} \plus{} \frac {5}{36} \plus{} \frac {5}{36}} \equal{} \frac {\frac {1}{3}}{\frac {11}{18}} \equal{} \boxed{\frac {6}{11}}$.

I'm pretty sure about everything up to the actual probability calculation; any feedback is appreciated! :)[/hide]

[hide="My attempt."]Well, you can consider only the rolls in which the sum is $ 6$, $ 7$, or $ 8$.  To achieve the desired objective, there are four different arrangements.  The first is the simplest, with $ 2$ sevens in a row.  The second consists of $ 2$ sevens separated by a string of sixes or eights of indeterminate length.  The third consists of such a string followed by $ 2$ sevens.  The fourth is such a string followed by a seven, followed by the opposite string, followed by another seven.  Let $ y$ be the probability of rolling a $ 7$ and $ x$ be the probability of rolling a six (which is equivalent to the probability of rolling an eight).

$ P_1 \equal{} y^2$
$ P_2 \equal{} 2y(x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^n)y$
$ P_3 \equal{} 2(x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^n)y^2$
$ P_4 \equal{} 2(x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^n)y(x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^n)y$

In this case, $ x \equal{} \frac {5}{16}$ and $ y \equal{} \frac {3}{8}$.

$ \sum_{n \equal{} 1}^{\infty}x^n \equal{} \frac {5}{11}$

$ P_1 \plus{} P_2 \plus{} P_3 \plus{} P_4 \equal{} \frac {9}{64} \plus{} \frac {90}{704} \plus{} \frac {90}{704} \plus{} \frac {450}{7744} \equal{} \boxed{\frac {3519}{7744}}$, which is the same answer that someone before obtained.[/hide]
Also, you have a greater chance of rolling a $ 6$ or an $ 8$ than a $ 7$, so the probability is not necessarily greater than $ \frac {1}{2}$.

Yay, I was right! :D
Oh, I programmed it on my TI-84+ to be sure, I got 228/500, which is pretty close to the answer...

[quote="i_like_pie"][quote="Begoner"]Also, you have a greater chance of rolling a $ 6$ or an $ 8$ than a $ 7$, so the probability is not necessarily greater than $ \frac {1}{2}$.[/quote]
There are 6 cases in which the sum is $ 7$, and only 5 for sums of both $ 6$ and $ 8$.

Also, could you expand on your solution? I don't entirely understand it.[/quote]
He means that there is a greater probability of rolling either a $ 6$ [i]or[/i] an $ 8$ than rolling a $ 7$.

Ah, thanks for clearing that up, Brut3Forc3. Now I understand why the probability would be less than $ \frac {1}{2}$. :)

However,
[quote="Begoner"]Let $ y$ be the probability of rolling a $ 7$ and $ x$ be the probability of rolling a six (which is equivalent to the probability of rolling an eight).

...

In this case, $ x \equal{} \frac {5}{16}$ and $ y \equal{} \frac {3}{8}$.[/quote]
Isn't $ x\equal{}\frac{5}{36}$ and $ y\equal{}\frac{6}{36}\equal{}\frac{1}{6}$?

[quote="i_like_pie"]Also, could you expand on your solution? I don't entirely understand it.[/quote]

For example, the following acceptable strings could fall into the four categories.

$ P_1 \equal{} 77$
$ P_2 \equal{} 7888888888887$ or $ 7666667$
$ P_3 \equal{} 888877$ or $ 66666666677$
$ P_4 \equal{} 6666676667$ or $ 887888887$

You can have a string of any number of eights or sixes anywhere in the string, as long as both do not occur simultaneously.  To account for this, an infinite series of sixes or eights is used in three of the categories; together, these represent every single acceptable string.

Edit: well, $ \frac {20}{36}$ of the time, the sum will be neither a seven nor a six or eight, so you can just discount such an occurrence.  The denominator then becomes $ 16$ rather than $ 36$.  To be more accurate, $ y$ should be the probability of rolling a seven [i]before[/i] rolling a six or eight.

This is a really good example of a problem that can be simplified with state diagrams:

[hide]We have positions $ (x, y)$, where $ x$ is the number of 7s we have so far and $ y$ is 0 if we have no 6s or 8s, 1 if we have a 6 but no 8 or vice-versa and 2 if we have a 6 and an 8.  Thus, we start at $ (0, 0)$.  From $ (0, 0)$ we move to $ (1, 0)$ with probability $ \frac {1}{6}$, to $ (0, 1)$ with probability $ \frac {5}{18}$ and to $ (0, 0)$ with probability $ \frac {5}{9}$, and you can calculate these values for each other possibility.  Let $ p_{(m, n)}$ be the probability that we get two 7s before a 6 and an 8 when we start at position $ (m, n)$.  Thus $ p_{(2, 0)} \equal{} p_{(2, 1)} \equal{} 1$, $ p_{(0, 2)} \equal{} p_{(1, 2)} \equal{} 0$ and we want $ p_{(0, 0)}$.

So, we write down a bunch of equations like $ p_{(0, 0)} \equal{} \frac {1}{6} p_{(1, 0)} \plus{} \frac {5}{18} p_{(0, 1)} \plus{} \frac {5}{9} p_{(0, 0)}$, etc.  (Note that "subtracting $ \frac {5}{9}p_{(0,0)}$ from both sides and dividing by $ \frac {4}{9}$" (and the similar calculations you'll make in the other cases) is equivalent to the type of simplification Begoner made by disregarding rolls of any value other than 6, 7 or 8.)  You'll have to be a little careful with what happens to the transition probabilities when you get to a point with second coordinate 1.  (The likelyhood that you stay in the same state increases.)  Then you solve this system, and you do indeed get $ \frac {3519}{7744}$.[/hide]

Here's one reason I like this method: i_like_pie realized the problem was complicated and tried to simplify it, but chose a new problem that wasn't equivalent.  Begoner also recast the problem in a simplified form.  For some reason that seems difficult to explain concisely, Begoner's simplified problem is "the same" while i_like_pie's isn't.  How could we have known that at the beginning?  With state diagrams, it doesn't matter -- we don't need the simplified problem, because the algebra takes care of it for us.  (By the way, if anyone is interested in more uses of this technique, search through the posts of user Ravi B.)

Here are some vaguely related questions, in increasing order of difficulty:

1) Roll dice as in the original question.  What's the probability that you get a 7 before either a 6 or an 8?
2) Again as in the original question.  What's the probability that you get a 7 before both a 6 and an 8?

3) Now, change the way you roll dice: use only one die, and roll it repeatedly.  Then, let the sequence of values you work with be the sum of the first and second rolls, the sum of the second and third rolls, the sum of the third and fourth rolls, etc.  Answer my questions 1 and 2, and the original question, in this context.  (This is probably fairly complicated, but I'm curious about what happens.)

Well, the probability of rolling a seven before a six or an eight with only one die is $ \frac {7}{18}$.  Discounting all the sums that are extraneous to the problem, you can focus solely on obtaining a six, a seven, or an eight.  If your first roll is $ 1$, then there is one way to obtain a six and one way to obtain a seven on the next roll.  If your first roll is $ 2$, $ 3$, $ 4$, or $ 5$, then you can obtain a six, a seven, or an eight on the next roll.  If it's $ 6$, you can only obtain a seven or an eight.  If you don't complete a viable sum on the first turn, you're back where you started, since the last roll has an equal probability of being any number.  That allows you to discard every single other possibility except for those that interest you.  Thus, the answer is $ \frac {1}{2} \cdot \frac {1}{3} \plus{} \frac {1}{3} \cdot \frac {2}{3}$.

As for the second part, the probability is $ 1 \minus{} 2 \cdot \frac {11}{25} \cdot \frac {11}{36}$ $ \equal{}$ $ \frac {329}{450}$.  The chance of obtaining a six, which is equivalent to the chance of obtaining an eight, is $ \frac{1}{2} \cdot \frac{1}{6} \plus{} \frac{1}{3} \cdot \frac{4}{6}$, or $ \frac{11}{36}$.  To be unsuccessful, you can have either a string of indeterminate length of sixes followed by an eight or the opposite.  The probability of such a string is $ \sum_{n\equal{}1}^{\infty}(\frac{11}{36})^n \equal{} \frac{11}{25}$.