Prove that $p^{2}\leq 4R^{2}+4Rr+3r^{2}$ in a triangle.

[quote][color=darkred]Prove that in any triangle $ABC,$ we have:[/color]
\[16Rr-5r^{2}\le\boxed{p^{2}\le 4R^{2}+4Rr+3r^{2}}\]
[/quote]
[color=indigo]It's a famous inequality in triangle and to prove it I use a strong result (not use trigonometric):
\[\begin{cases}2R^{2}+10Rr-r^{2}+2(R-2r)\sqrt{R(R-2r)}\ge p^{2}\\ p^{2}\ge 2R^{2}+10Rr-r^{2}-2(R-2r)\sqrt{R(R-2r)}\end{cases}\]
Or to prove $p^{2}\ge 16Rr-5r^{2},$ you can use the relation: $9\cdot IG^{2}=p^{2}+5r^{2}-16Rr$[/color]

[quote="April"][quote][color=darkred]Prove that in any triangle $ABC,$ we have:[/color]
\[16Rr-5r^{2}\le\boxed{p^{2}\le 4R^{2}+4Rr+3r^{2}}\]
[/quote]
[color=indigo]It's a famous inequality in triangle and to prove it I use a strong result (not use trigonometric):
\[\begin{cases}[b: 0090216da9]2R^{2}+10Rr+r^{2}+2(R-2r)\sqrt{R(R-2r)}\ge p^{2}[/b: 0090216da9]\\ p^{2}\ge 2R^{2}+10Rr-r^{2}-2(R-2r)\sqrt{R(R-2r)}\end{cases}\]
Or to prove $p^{2}\ge 16Rr-5r^{2},$ you can use the relation: $9\cdot IG^{2}=p^{2}+5r^{2}-16Rr$[/color][/quote]

Thanks, April. But I think it should be
$2R^{2}+10Rr-r^{2}+2(R-2r)\sqrt{R(R-2r)}\ge p^{2}$

Moreover, how to prove those?

I come up with this inequality when I did this problem.
$(1-cosA)(1-cosB)(1-cosC)\geq cosAcosBcosC$

I try expanding, then use some identities, and that leads to that inequality and I couldn't deal with it.

[quote="bibobeo"]... I come up with this inequality when I did this problem.
\[(1-\cos A)(1-\cos B)(1-\cos C)\geq \cos A\cos B\cos C\]
I try expanding, then use some identities, and that leads to that inequality and I couldn't deal with it.[/quote]
[color=indigo]Thank you, bibobeo! I'll try it soon. And you can prove $p^{2}\le 4R^{2}+4Rr+3r^{2}$ by use the relation: [/color]\[\boxed{IH^{2}=4R^{2}+4Rr+3r^{2}-p^{2}}\]  :wink:

Oh April, could you please give me some hints to prove those identities and ineualities you gave?
I've tried, but stuck ...

Let $t_{a},t_{b},t_{c}$ be $\tan\frac{A}{2},\tan\frac{B}{2},\tan\frac{C}{2}$
So we have the given inequality is
$\Longleftrightarrow 8t_{a}^{2}t_{b}^{2}t_{c}^{2}\geq (1-t_{a}^{2})(1-t_{b}^{2})(1-t_{c}^{2})$
$\Longleftrightarrow tanAtanBtanC\geq \cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}$
which is easy.
 :D

Btw, April, I still cannot prove your relations and the two inequalities  :oops:

I'm sorry, I forgot it  :blush:
[hide="Hint."]$\begin{cases}a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=0\\ aIA^{2}+bIB^{2}+cIC^{2}=abc\\ aHA^{2}+bHB^{2}+cHC^{2}=4R^{2}(a+b+c)-\left(a^{3}+b^{3}+c^{3}\right)\end{cases}\implies \boxed{IH^{2}=4R^{2}-\frac{a^{3}+b^{3}+c^{3}+abc}{a+b+c}}$[/hide]

Two inequalities are very hard and my proof is very ugly. And I think it's not very important, so you you needn't to know it.


geometry
geometric transformation
reflection

that's why every kid I know thinks that the school districts are stupid

[quote="Ignite168"]This 1st grader was suspended from school for 10 days for brining a 2 inch squirt gun into school.

This seems absolute ridicolous to me and I believe the school board should be more focused on actual important things such as the bullying that goes on in school that often starts at a young age.

http://www.newsnet5.com/education/9891863/detail.html[/quote]
Well, it's a gun... Lol just kidding, that is [i]really[/i] ridiculous. Who cares about squirt guns? (besides the district :roll:)

They make little things big, but they don't care about the things that are already big. :censored:

Over a 2-inch squirt gun?  Some people keep water guns at school and shoot each other at track practice (but those are high school seniors...)

When I was in third grade, there were kids who threw rocks at me whenever I played handball.  I would say that's much worse.  But it didn't matter beacuse I still beat them all... :D

Here in the UK, I believe it is a criminal offence to carry any kind of replica gun (even a 2-inch water pistol) in any public place, without a licence.

How crazy. 10 days suspension: what is that intended to achieve?

Gentlemen, you seem to be under the illusion that a 10-day holiday for a 1st grader is some sort of a [i]punishment[/i]. Now I have known that word for many years now, and howsoever may I agree to bend its meaning, it will possibly take something along the lines of a black hole to warp it enough so as to include the above stated phenomenon.

[quote="bubka"]Gentlemen, you seem to be under the illusion that a 10-day holiday for a 1st grader is some sort of a [i]punishment[/i].[/quote]

I don't know about anyone else here, but I'm definitely not.  What would give you that impression?

The punishment lies not in the holiday, but in the fact that it is recorded as a suspension on his permanent record. (which is, of course, something that a 1st grader would probaby struggle to realise). I think the idea is that you go home and reflect on your actions for long enough that you really regret them. It has always seemed a little odd to me...

If you're smart enough, you regret your actions right away. In that case, the suspension becomes a vacation. Unless you did something really bad. :P

Uhm, when we were punished in school, we either had to rewrite the same sentence ("Ik mag niet...") over and over again (sometimes for like three pages!!) or we had to sit out a couple of hours of detention.

EDIT : Bubka please, your picture is so huge it stretches the width of this topic.  I sent you a smaller copy : USE IT.  I'm surprised moderators haven't commented on it yet... :|

[quote="fredbel6"]Uhm, when we were punished in school, we either had to rewrite the same sentence ("Ik mag niet...") over and over again (sometimes for like three pages!!) or we had to sit out a couple of hours of detention.[/quote]

That's for minor incidents..

[quote="fredbel6"]Ignite I'm sorry but you haven't even finished high school yet...
If we did something more serious, we would have been kicked out of school.  Either permanently or for a few weeks.  When you are sixteen or something you start to realize that the burden is on YOURSELF because you will get behind on all sorts of things, fail to take tests and thus flunk etc...
You can choose not to care about that, but that will results in losing a year (at least one year)

I fail to see the educational purpose of doing it to a firstgrader... unless you are "attacking" the kid indirectly via its parents.  They will have to either stay at home to watch him while he can't go to school or pay someone else to do it.  And that won't make them happy.[/quote]

And what's that supopsed to mean? I know the consequences for punishment, thank you very much.

I also fail to see what you posted about knowing who the burden is on at 16 is in anyway related to this topic.

[quote="fredbel6"]Uhm, when we were punished in school, we either had to rewrite the same sentence ("Ik mag niet...") over and over again (sometimes for like three pages!!) or we had to sit out a couple of hours of detention.
[/quote]
just three pages. here they would give you to write it at least 100 times (for that age there were such loooooong sentences), stand on your feet the whole hour, or stand in the corner. then your mother would be supposed to come to school, and you would go to the headmaster and school psychologist (this was for serious problems)

[quote="fredbel6"]Ignite I'm sorry but you haven't even finished high school yet...
If we did something more serious, we would have been kicked out of school.  Either permanently or for a few weeks.  When you are sixteen or something you start to realize that the burden is on YOURSELF because you will get behind on all sorts of things, fail to take tests and thus flunk etc...
You can choose not to care about that, but that will results in losing a year (at least one year)

I fail to see the educational purpose of doing it to a firstgrader... unless you are "attacking" the kid indirectly via its parents.  They will have to either stay at home to watch him while he can't go to school or pay someone else to do it.  And that won't make them happy.[/quote]

if a student have 30 or more unattended lessons (without a proper reason) he is automatickly kicked out of school

what is that kid going to learn from all this?

woooooooooooow thats lame.
this 10th grader got suspended 2 days for bringing a real knife and selling it. thats way worse

hmm.

This one kid, he hacked into the school system and changed his grades. He got caught, he got suspended, and while he was waiting at home for his dad to come home and further punish him, he hanged himself.

He's an idiot - I could do the same without getting caught (remote access, networking services, the whole nine yards), but I wouldn't want to.

[quote="kyyuanmathcount"]I remember that some kid in a school somewhere in michigan got expelled from carrying a cafeteria plastic knife to class.

The parents only got him back in by sueing the school for arming their child.   :rotfl:[/quote]

XD

yeah, schools do a lot of things that may seem pointless, but it's a start, i mean, if somebody in your school walked in with a carving knife and guillotined you and your classmates to death would you like it?

Sporks to the rescue! Part spoon, part fork, and turn it around, and you can spread cream cheese on your bagle with the handle! 100 for 5 dollars at your nearest supermarket!

Okay mustafa, now that's spam.

All of us on this community agree that the American education system is useless in more areas than one - the prominent one being teaching math. I think it does a decent job of disciplining kids though. I mean, sure there are cases where the oddball nutjob will screw things up for everyone by suspending a kid for having a squirt gun. But in the few years that I was in the system, I found that most, if not all, of my teachers were very fair and understanding. That, of course, can be explained partially by the fact that my district happens to be one of the best-funded around but also by the care that most school systems try to exercise in selecting their employees.

Okay so, despite that, they still can't teach math. But generally you couldn't say that they're too mean to a kid. In fact I think there's not enough disciplinary measures in the public schools and the few rare examples to the contrary don't prove anything.

[quote="MithsApprentice"] I found that most, if not all, of my teachers were very fair and understanding. [/quote]
Obviously you didn't spend enough years in the system. The problem is almost never the teachers (in this kind of situation). It's usually the administrators who cause problems like these. The worst a teacher can do is teach poorly, and in that case you can still even blame the administrators by saying that they shouldn't have hired such a lame teacher (or should try to get rid of him/her).

You know, at my school, teh administration deters teachers from failling kids by making it a huge hassle for them to let a student fail (and sometimes even by explicitly telling them not to let people fail).

[quote="i_like_pie"][quote="kyyuanmathcount"]I remember that some kid in a school somewhere in michigan got expelled from carrying a cafeteria plastic knife to class.

The parents only got him back in by sueing the school for arming their child.   :rotfl:[/quote]
What can you do with a plastic knife that would hurt somebody? :huh:[/quote]
I saw someone get stabbed with a spork yesterday...

plastic? how does that hurt? unless it was really hard...

[quote="Xevarion"]You know, at my school, teh administration deters teachers from failling kids by making it a huge hassle for them to let a student fail (and sometimes even by explicitly telling them not to let people fail).[/quote]

Okay yeah...I didn't spend enough years in the system. But I do know that this not letting kids fail thing happens. Our principal basically banned teachers from failing kids (and we were a Bluebonnet School b/c of it! :-D).

I was kind of referring to the disciplinary problems though. -.-

[quote="MithsApprentice"]
Okay yeah...I didn't spend enough years in the system. But I do know that this not letting kids fail thing happens. Our principal basically banned teachers from failing kids (and we were a Bluebonnet School b/c of it! :-D).
I was kind of referring to the disciplinary problems though. -.-[/quote]
Uhm what system?  In my country it's six years of elementary, six years of secondary.  How can you get out of the system?

I had a teacher who had a few very arrogant kids once(who kept hanging around instead of going to his class when our schoolbell rang) and he... let them pass so he would never see them again. :o   That's a convenient punishment.

[quote="fredbel6"]Uhm what system?  In my country it's six years of elementary, six years of secondary.  How can you get out of the system?[/quote]
I think in USA we have to be in school until we are 16 years of age. But you can avoid the public school system by going to a private school or by being home-schooled (as is MithsApprentice). [NB: A private school does not necessarily protect you from the problems we cite about public schools.]

[quote="MithsApprentice"]I was kind of referring to the disciplinary problems though.[/quote]
Well, in my opinion, the fault is still most often with the administrators. For stupid legal reasons, teachers aren't allowed to dole out effective punishments. When a student causes problems serious enough to come to the attention of the main office, the main office usually deals with it through measures like in-school and out of school suspension or detention. These don't do anything.

[quote="Xevarion"]When a student causes problems serious enough to come to the attention of the main office, the main office usually deals with it through measures like in-school and out of school suspension or detention. These don't do anything.[/quote]

Yeah agreed. And I don't know if you guys have this, but some districts around here have an "alternative learning center" for "troubled" kids apparently. So I suppose if you started going on killing sprees, decidedly to bring illegal drugs on campus, or destroy district property, they'd send you to one of those. Guess how effective they are. No really, guess.

[quote="MithsApprentice"]
Yeah agreed. And I don't know if you guys have this, but some districts around here have an "alternative learning center" for "troubled" kids apparently. So I suppose if you started going on killing sprees, decidedly to bring illegal drugs on campus, or destroy district property, they'd send you to one of those. Guess how effective they are. No really, guess.[/quote]
In Belgium you CANNOT send minors (17 or younger) to a real prison, that would be considered harmful.
The consequence : a 16-year old beat up his teacher, but there was NO ROOM for him in any juvenile detention center...so he was gonna be released immediately.  Eventually, due to outrage, he was given TWO WEEKS in a detention center.
About a month after his crime, he was caught robbing a man in Leuven off his MP3-player.  Again he was gonna be released immediately due to no room at all.  Eventually a few days later they did find room for him.
Maybe our system doesn't really scare criminals, when police catch a youth doing something illegal, first thing they do is check if there is any room. :o

juvenile hall is what we have in US, but really idn'y that a bit harsh? 2 inch squirt gun...

I think the entire problem is that it's not harsh enough. Sure, bringing a 2-inch squirt gun to school may not merit time in a detention center or, perhaps, even suspension. But that's why it made the news in the first place. It was outrageous. If it's what normally happens, newspapers wouldn't have covered it.

But the thing is, I'm not so sure examples like that are a bad thing. After all, more punishment = less crime.

[quote="MithsApprentice"]After all, more punishment = less crime.[/quote]
From one point of view, that is very true. The more punishment one gets, the less crime he will commit.

However, I sometimes look at it from a different point of view. The more punishment one gets, the more likely he will be to commit another crime because he is mad at the people who punished him. I'm pretty sure this happens less often than the former, but there is a small chance that more punishment will lead to more crime.

This thing just shows one thing: a lot of idiots are following the orders from above without thinking.
(Not that this was ever different...)

[quote="MithsApprentice"]After all, more punishment = less crime.[/quote]
I can't agree on this generally. For example here, I see no way how it lowers crimes (in fact I can even think of stupid people now bringing (phake¿) guns just to test if they are caught; this could result in their usage and...).

By the way: you didn't say anything about the punishment being justifiable ;)

Also, our criminal justice system has the huge problem of basically throwing criminals together and letting them hang out all day. I know that we have different prisons for different levels of crime, but in any given prison there will still be "more heinous" and "less heinous" criminals. And there's a very good chance that the "not so bad" guys will learn the evil ways of the "horrible" guys.

Well, my statement about more punishment = less crime concerns - as far as I dare take it - only to criminals with no previous record. I believe that greater punishment will deter people who are trying to decide whether or not they'd want to risk the judicial penalties at all. But yes, I can see why someone who's already been incarcerated, fined, etc. may be more eager to cause more havoc if their sentences were increased.

Anyway, I think prisons in the US in general are too lenient. First, like Xevarion said, we need separation. And secondly, we need more conduct regulations across the board. To many people (especially underprivledged individuals), having a room (with a bed, mattress, comforter, and toilet), good food, and recreational facilities may not seem like such a terrible way of life. *shrugs* Problem is what they had in the way of a lifestyle wasn't so much better than what they faced. And it's the government's job to create that relative difference - not by welfare, but by education.

Education...now that's something that the US needs to work on. Which brings us back to the topic of why math isn't taught in public schools, since that would [i]clearly[/i] be a good way to fix a lot of social problems.

[quote="MithsApprentice"] To many people (especially underprivledged individuals), having a room (with a bed, mattress, comforter, and toilet), good food, and recreational facilities may not seem like such a terrible way of life. [/quote]
Apparently [url=http://claudiarosett.pajamasmedia.com/2006/09/28/guantanamo_weigh.php]the government has decided[/url] that this is a way to keep terrorists off the streets...

[quote="Xevarion"]
Apparently [url=http://claudiarosett.pajamasmedia.com/2006/09/28/guantanamo_weigh.php]the government has decided[/url] that this is a way to keep terrorists off the streets...[/quote]

Well with the new torture regulations, this may seem like a pretty effective option to the military. You just overfeed them until they're too fat and lazy to be effective. Next they should considering spiking the drinks there and get the prisoners drunk. Maybe we'll finally get some useful information out of them.

4200 calories a day. Ugh. I don't even get that many.

4200 a day...that's actually ridiculous...I don't get much more than half that...even during track season...

[quote="mustafa"]I think I had about 4200 calories for dinner tonight :rotfl:[/quote]

Maybe if you were eating lard by the tub...


If $ \frac {x}{y} \equal{} \frac {3}{4}$, then the incorrect expression in the following is:

$ \textbf{(A)}\ \frac {x \plus{} y}{y} \equal{} \frac {7}{4} \qquad\textbf{(B)}\ \frac {y}{y \minus{} x} \equal{} \frac {4}{1} \qquad\textbf{(C)}\ \frac {x \plus{} 2y}{x} \equal{} \frac {11}{3}$
$ \textbf{(D)}\ \frac {x}{2y} \equal{} \frac {3}{8} \qquad\textbf{(E)}\ \frac {x \minus{} y}{y} \equal{} \frac {1}{4}$

[hide="Click for solution"]
$ x\equal{}\frac{3}{4}y \implies \frac{x\minus{}y}{y}\equal{}\minus{}\frac{1}{4}$, so $ \boxed{\textbf{(E)}}$.
[/hide]

[hide="hint"]Just plug in the ratio values in each question and solve. Find which equation is not equal. [/hide]

[hide="answer"]$ \frac{x\minus{}y}{y}\equal{}\frac{1}{4}$. Plug in $ x\equal{}3$ and $ y\equal{}4$ to get $ \frac{3\minus{}4}{4}\equal{}\frac{1}{4}$. Simplify and you get that $ \frac{\minus{}1}{4}\neq\frac{1}{4}$.

So the answer is:  $ E$[/hide]


im getting kindof nervous because i only answered 7 and 2 other partial solution

so i wanted to know if i really should be nervous

Psh. You should be fine. I've heard tell of people that only really solved FOUR PROBLEMS and got in. I only did six anyway, with some (kind of failish) attempts at the other problems.

much better than me. 5 and one partial solution, which is totally wrong. turns out i factored completely wrong (for number 6).

At least you were able to kind of DO number 6...I repeatedly rearranged/combined equations and failed it, basically.

well
ive truly mastered the art of making careless mistakes

for number 6 i forgot negative factors existed so i said it was not possible but i got the factoring portion.

did eight. not bad, but i did number 4 in a completely stupid-dumb-way, but got everything else pretty quickly.

im not sure how well i explain stuff at 2 in the morning

my explanations wer  1 page each

I was just too fail at TeX to do it in LaTeX.  :maybe:

[quote="AIME15"]Me too.  I even made an effort to write neatly :D

(too lazy to TeX up)[/quote]

My neat handwriting looked like a 5-year-old wrote it.  :rotfl: A 5-year-old with horrible handwriting.

[quote="sjyoo93"]Hmm.. Some of you guys who voted you solved all ten.. .. ..

I was able to first solve 6 with ease, in order, 1, 10, 5, 4, 8, 3... Then I thought about 2, and got down that.

Then I solved 6.. not really. I just kind of got around the factoring and all that.. So. 7 solve with 1 partial. Hm. Good enough, I hope..[/quote]

That's definitely good enough. Cause if I got 5 and 1 partial and got in...I'm sure you def would.

Got all of them...

The poll should have been 'If you got accepted to AMSP, how many problems did you answer?'

Problems in order I solved them: 1, 2, 3, 4, 7, 8, 10, 6, 9, 5.
Problems' difficulties (My thought):
1, 2, 3, 4, 6, 7, 8, 10 - trivial
5, 9 - pretty good problems, but not hard enough

did the first 7, but 6 was a fail.
My explanation for number 2 sucked.
I need to learn how to prove things better...

Why does everyone think #5 was a pretty difficult problem and #6 was easier... #6 was kind of harder..

#5 was like so easy, but maybe because I think differently and have different fields of math that I'm good at...

But the key for me to solving #5 was Gauss's way of adding however many consecutive integers from 1 to x...

I'm probably going to be the dumbest person at UC Santa Cruz...I feel really stupid now.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14747[/img]

 :(

Let M the midpoint of OA and N of OB. I think theat you missed the condition that P can't be on the segment MN, if it's this D it's the intersection of the circle with a circle with center P and radius PO cause $ \angle ODP \equal{} \frac{\angle COA \minus{} \angle BOD}{2}\equal{}\angle BOD$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14753[/img]

 :thumbup:


geometry
3D geometry
arithmetic sequence
number theory proposed
number theory

We already know that we can't find four squares $a^2, b^2, c^2 ,d^2$ where $a, b, c,d\in N$

such that $a^2, b^2, c^2, d^2$ are arithmetic sequence.

Then, What is the limit number of cubes such that cubes make arithmetic sequence?

And What is the limit number of $p$th-powers such that they make arithmetic sequence? ($p$ is a prime number)

It has been proven on this forum that no three positive cubes can form an arithmetic progression. Of course, you have the progressions of the type $-k^3,0,k^3$ when you allow the cubes to be negative.


$ f$ is a function defined on all reals in the interval $ [0, 1]$ and satisfies $ f(0) \equal{} 0, f(\frac{x}{3}) \equal{} \frac{f(x)}{2}, f(1 \minus{} x) \equal{} 1 \minus{} f(x)$. Find $ f(\frac{18}{1991})$.


integration
trigonometry
calculus
inequalities
limit
real analysis
real analysis solved

Let $I_n=\int^\frac{2}{n}_\frac{1}{n} \frac{\cos t}{\sqrt{t}} dt (n=1,2,\cdot\cdot\cdot)$.

Find the range of the real number $\alpha$ such that sequence ${n^{\alpha}I_n}$ converges and then the value of limit.

$I_n=\int_{\frac1n}^{\frac2n}\frac{\cos x}{\sqrt x}dx=2\int_{\frac1n}^{\frac2n}(\sqrt x)'\cos xdx=\frac2{\sqrt n}\left(\sqrt 2\cos\frac2n-\cos\frac1n\right)-2J_n$
where $J_n=\int_{\frac1n}^{\frac2n}\sqrt x\sin x$.
Now $\int_{\frac1n}^{\frac2n}\sqrt x>J_n>\int_{\frac1n}^{\frac2n}x\sin x$ (*) (because ($x,\sin x\in(0,1)$.
But $\sqrt n\int_{\frac1n}^{\frac2n}\sqrt x=\frac23\frac1{n}\left(2\sqrt 2-1\right)\to 0$ when $n\to \infty$ (1).
And now from integration by parts the right member of the inequality (*) becomes:
$\sqrt n\int_{\frac1n}^{\frac2n}x\sin xdx=-\frac2{\sqrt n}\cos{\frac2n}+\frac1{\sqrt n}\cos{\frac1n}+\sqrt n\sin{\frac2n}-\sqrt n\sin{\frac1n}\to 0$ (2).

So from (1) and (2) we get that $\sqrt nJ_n\to0$.
But $\sqrt nI_n=2\left(\sqrt 2\cos{\frac2n}-\cos{\frac1n}\right)-2\sqrt nJ_n\to 2(\sqrt 2-1)$.
So $\alpha=\frac12$ and $\lim_{n\to\infty}\sqrt nI_n=2(\sqrt 2-1)$

Thank you for you reply, Kuba.

Your answer is right partly.You need to find another the range of $\alpha$.
Please try again!

kunny

The integration by parts, and the exact computation of the integral, are not needed. We know that as $t\to0, \cos t=1+O(t^2)$. Hence, $\frac{\cos t}{\sqrt{t}} = t^{-\frac12}+O(t^{\frac32}).$ Integrate that and evaluate at $\frac2n$ and $\frac1n$ to get:

$I_n=2(\sqrt{2}-1)n^{-\frac12}+O(n^{-\frac52}).$

$n^{\alpha}I_n=2(\sqrt{2}-1)n^{\alpha-\frac12}+O(n^{\alpha-\frac52}).$

This converges for $\alpha\le\frac12$ but there are two cases.

If $\alpha=\frac12$, then $\lim_{n\to\infty}n^{\alpha}I_n=2(\sqrt{2}-1)$.

If $\alpha<\frac12$, then $\lim_{n\to\infty}n^{\alpha}I_n=0$.

Kuba was right except that he missed the case $\alpha<\frac12.$

Thank you for your reply, Kent Merryfield.

You are right.I will post my solution.Please wait for a while.

kunny

[quote]Thank you for you reply, Kuba.

Your answer is right partly.You need to find another the range of .
Please try again! [/quote]
actually I was really confused when I saw this... but... after that, when I was about to asleep I remembered that I forgot the case when $\alpha<\frac12$  :blush:

oh.. and Kent.. nice proof.. :)

In Japan,senior high school students learn Calculus at the age of 17 or 18.
Some school deal with Calculus in junior high school such as private school.
They don't study  $\epsilon-\delta$ ,L'Hopital's theorem,Taylor expansion,Landau sign etc. in this stage.Tha'ts why I make problems so that they can solve ellementarily as possible.
As you know,the problems which I have been posted except my oiginal are all Entrance Exam for Entering University.

Here is my solution.

As $1-\frac{t^2}{2}\leqq\cos t\leqq1 (t\geqq0)$  ,for   $t>0$,we have $\frac{1}{\sqrt{t}}-\frac{1}{2}t^{\frac{3}{2}}<\frac{\cos t}{\sqrt{t}}<\frac{1}{\sqrt{t}}$.Integrating  at $\left[\frac{1}{n},\frac{2}{n}\right]$  the formula of each side,

we obtain $2(\sqrt{2}-1)n^{\alpha-\frac{1}{2}}-\frac{4\sqrt{2}-1}{5}n^{\alpha-\frac{5}{2}}<n^\alpha I_n<2(\sqrt{2}-1)n^{\alpha-\frac{1}{2}}  $,which follows Kent Merryfield's solution.

Thank you.

kunny


Avem $n \geq 2$ prizonieri azteci.

Ei sunt informaţi că a doua zi vor fi tunşi ca la închisoare şi că unora li se vor vopsi căpăţânile în roşu, altora în albastru. Apoi, o să fie aliniaţi în curte în şir indian, i.e. sistem $>>>>>>>>>\ldots>>>$.

Vor fi întrebaţi pe rând, începând cu cel din spate, ce culoare are ţeasta personală. Cine nu răspunde corect, o fură.

Numărul de culori folosite nu se ştie. Pot să fie toţi roşii, pot să fie toţi albaştrii. Daca vreunul foloseşte un semnal de avertizare, ton vocal diferit, gesticulare de mână etc. vor fi omorâţi toţi.

Care este numărul maxim de prizonieri ce pot scăpa (indiferent de felul cum cei $n$ prizonieri sunt coloraţi în cap) şi ce strategie pot să dezvolte într-o noapte de gândire?

PS: Presupun că e foarte cunoscută problema, dar de-abia acuma am aflat-o şi eu. Pentru cei pentru care nu este la "first sight", rog să-i lase pe ceilalţi.

Mi-a luat ceva timp sa ma prind :).

Primul prizonier poate doar sa ghiceasca, asa ca pot fi salvati cu siguranta maxim $n-1$. Partea buna e ca chiar pot fi salvati $n-1$.

Primul spune 'rosu' sau 'albastru' dupa cum vede in fata lui numar par sau, respectiv, impar de capete rosii. Acum fiecare din ceilalti (dupa ce a auzit ce spun cei intrebati inaintea lui, desigur) are toate informatiile necesare ca sa determine raspunsul corect.


function
rotation
geometry
rectangle
calculus
calculus computations

Find the range of $ k$ such that there exists a square with centered the origin and 4 vertices on $ y\equal{}x^3\minus{}kx.$

No, I don't have it.  I just like making plots:

[img]http://img139.imageshack.us/img139/2296/squareincubejp6.jpg[/img]

If anybody would like to experiment with this, here's the Mathematica code I used to find it:

[code]Manipulate[Show[{Plot[x^3 - k*x,
         {x, -5, 5}], Graphics[
         Rotate[{EdgeForm[{Thick, Blue}],
             FaceForm[Opacity[0]], Rectangle[
               {-size, -size}, {size, size}]},
           t]]}, PlotRange -> {{-10, 10},
         {-10, 10}}, AspectRatio -> 1],
   {t, 0, 2*Pi}, {size, 0.1, 5}, {k, -15, 15}][/code]

Well, I got $ \left|k\right|\geq2\sqrt {2}$, but I'm not sure it's correct because I did a horrible calculation..


let $a,b \in \mathbb{R^+}$ such that $b^2+b^6\leq a^2-a^6$.

Show that $a\leq 1$ and $b<\frac{2}{3}$

[hide="A_part"]Assume that $a>0$.  Then $a^6>a^2$ and $a^2-a^6<0$.  And since $b^2+b^6>0$, by the laws of transitivity we have $b^2+b^6>a^2-a^6$, so $a\leq1$.[/hide]

[quote="jrshoch"] Assume that $a>0$.  Then $a^6>a^2$ and ... etc .. [/quote]
A typo there,  it should read  [quote]Assume that $a> 1$[/quote]

[hide="my part"]To show $b < 2/3$, we must find the minimum value for $a^2 - a^6$, which (I'm not sure if we're allowed to use calculus or not), if we use the first derivative test, has to be at the zeros of

$2a - 6a^5$

$2a (1 -3a^4)$

$2a(1+\sqrt{3}a^2)(1-\sqrt{3}a^2)$

$2a(1+\sqrt{3}a^2)(1+3^\frac{1}4a)(1-3^\frac{1}4a)$

Since we want positive zeros less than one, the only one that works comes from the last factor, and $a=3^\frac{-1}4$.

That makes the minimum value of the rhs ${3^\frac{-1}4}^2 - {3^\frac{-1}4}^6 = 3^\frac{-1}2 - 3^\frac{-3}2$

Let $\sqrt{3} =x$.

Minimum is $\frac{1}x - \frac{1}{x^3}=\frac{x^2-1}{x^3}=\frac{2}{x^3}$

So we need to find $b$ such that $b^2 + b^6 \leq \frac{2}{\sqrt{3}^3}$

Is that right so far?..... I'm kinda stuck now though....:([/hide]

p.s. how do i make my fractional exponents prettier?

[quote="lasagnaman"][hide=my part]To show $b < 2/3$, we must find the minimum value for $a^2 - a^6$, which (I'm not sure if we're allowed to use calculus or not), if we use the first derivative test, has to be at the zeros of
[/quote]

umm $3^{\frac{-1}{4}}$ actually turns out to be a max, if you test some more points.

[quote]Is that right so far?..... I'm kinda stuck now though.[/quote]
[hide]Well you do know (if   $0 \ge a \ge 1$ )

 $a^2-a^6 \le \frac 2 {3^{3/2}}$      ................................ $\ and \ b \ge 2/3$   will give a  contradiction.)

[hide="because"]as someone pointed out,  $a=(1/3)^{1/4}$  you have a  [i]maxima [/i]so the value of the function is less than this at other places...
[hide="You can also use AM-GM"]
$(2/3)^3 = (\frac{(2a^2) + (1+a^2) + (1-a^2) } 3 )^3 \ge (2a^2)(1+a^2)(1-a^2) = 2(a^2-a^6)$ [/hide]
[/hide]
and if  $b \ge 2/3$  $\implies$ , $b^2+b^6 \ge b^2 \ge 4/9 > \frac 2 {3^{3/2}}$  it will give a  contradiction...etc. [/hide]


Sorry if this is the wrong area and sorry for my dumbness but i really need help working out a percentage...percentages were always my sticking point!! I had my final exam today and it went really well but then i got home and my coursework score had been emailed to me and was rubbish to say the least!! For this module there is a 60/40 split in the marking with the coursework contributing 60% of the overall mark and the exam contributing 40% of the overall mark. Well i only got 34.5% for the coursework and giving that i require 40% overall from the 2 marks to pass the module can anyone tell me what percentage i need to get in the exam to get 40% overall, given the 60/40 split 60%coursework 40%exam?? If it were 50/50 then i know 45.5% in the exam would give me the 40% overall target but this 60/40 business has me a bit stumped...thanks in advance for any feedback on this  :)

This is not the right place for this -- it would have been better in one of the math fora.  Rather than move it, though, I'll just answer your question for you:

Your final grade is computed as a weighted average of your coursework and exam grades.  In particular, if your coursework grade is $C$ and your exam grade is $E$, the instructor will compute your final grade as $F = C\cdot 0.6 + E \cdot 0.4$ (because 60% = 0.6 and 40% = 0.4).  Now, you know $C$ and your desired value of $F$, so we can substitute these values to find your desired $E$ (again changing percents to decimals first): $0.4 = 0.6\cdot 0.345 + 0.4\cdot E$.  $0.6 \cdot 0.345 = 0.207$ so we subtract this and get $0.193 = 0.4\cdot E$.  We just divide by 0.4 to find that the score you need is $\frac{0.193}{0.4} = 0.4825$ which is 48.25%.


geometry
circumcircle
perpendicular bisector

[quote="Poincare"]I am having trouble with this(I know its probably very simple):
In right triangle ABC (with BC as the hypotenuse), draw a median to the hypotenuse, AM. Prove that AM=BM.[/quote]

Note that the [url]http://en.wikipedia.org/wiki/Circumcenter[/url] is on the midpoint of the hypotenuse.

Proof: Consider AB. The perpendicular bisector of this creates a triangle C'A'B. This is similar to ABC, so

$ \frac{BA'}{BA}\equal{}\frac{C'A'}{CA}\equal{}\frac{1}{2}$ (since we know that A' is the midpoint of AB).

So C' is the midpoint of the hypotenuse $ (\frac{BC'}{BC}\equal{}\frac{1}{2})$. We do the same process for BC.

Thus the perpendicular bisectors meet at the midpoint of the hypotenuse.

---

By the definition of the circumcenter, all points are the same distance from the circumcenter (they are all radii). So letting the circumcenter be O, $ OA\equal{}OB\equal{}OC$. (Or M, meaning that $ AM\equal{}BM$).

We are finished.

[quote="xpmath"][hide="alternatively"] Consider the circumcircle of a right triangle. Since a right triangle has a right angle by definition, the hypotenuse must be a diameter (so the right angle lies on a semicircle and is thus one-half of 180 degrees). The midpoint of the hypotenuse must be the center of the circle then.

Now, if you draw the circumcircle, you notice that AM and BM are both radii, and thus are equal.[/hide][/quote]

Which is what I said...

First of all, you didn't hide your solution.

Second of all, your proof for stating that the center of the circumcenter is the midpoint of the hypotenuse was pretty different from mine.


[quote="yechanhong"]Thank you all  :D

Wanna Beta Test My Game?
http://www.canissoft.com  :lol:[/quote]

very nice game! well it's not  bad but the graphics can be better and it was a little laggy (i think it was my comp) other than that your original webiste needs to be faster (the Macromedia takes too long)...

Rating: 8.5 out of 10

BTW:oh and what is the next mission? after the first one the thing loops back and forth

wow very nice game. i just don't get how you die. I walked through the whole mob and din't die on the last part of the first mission

-edit-
yeah i do see how there are 3 lives.

way to steal sprites from megaman

0/10 for copyright content


Prove that if
c=a^2+b^2
Then
c*(2^m)=x^2+y^2

Obviously a,b,c,m,x, and y are non-negative integers.

I know of both an elegant and an inelegant solution to this.  Odds are no one will find the inelegant one, but I think that's the one that's much more fun.

[hide]c = a2 + b2

c2 = (a2 + b2)2

= a4 + b4 + 2a2b2

= (a2 - b2)2 + (2ab)2

we can repeat the steps for (c2)2 and ((c2)2)2 and so on, since c2 and(c2)2 can also be written as the sum of 2 squares[/hide]

Wrong question.  You aren't raising c to some power of two, you're multiplying by a power of two.

So, for example:
74=5^2+7^2
2^2*74=296=10^2+14^2.

If m is even, say m=2n, then

[hide](2^m)c = ((2^n)^2)c = ((2^n)a)^2 + ((2^n)b)^2

i.e. just multiply a and b by 2^n[/hide]

If m is odd, say m=2n+1 then

[hide](2^m)c = 2c' where c'=(2^(m-1))c

m-1 is even so c'=(a')^2 + (b')^2 where a'=(2^n)a and b'=(2^n)b as above. Then

2c' = 2(a')^2 + 2(b')^2 = (a'+b')^2 + (a'-b')^2[/hide]

That last line is all you really need.  That was the "elegant" solution I mentioned.

[hide]

(a+b)^2+(a-b)^2=2*(a^2+b^2)

[/hide]

Then all you have to do is repeat it to get the 2^m, meaning, you say that this new number, 2*c is the sum of two squares, so 2*2*c will also be, so on and so forth.


You should be very careful when multiplying infinite sums by a variable...

For example, if you take the generating function for the Fibonacci numbers, which gives the sum of a geometric series with Fibonacci coefficients, although the difference of two series is always rational, the series do not necessarily converge, i.e. f(1) = sum of all Fibonaccis = -1 is clearly a false statement.

However, in this case, it is correct to say that if the latter sequence converges, so does your first sequence.

In your case, try a ratio test...your general term is nx^(n-1)/3^n. Using the ratio test, you get lim x ->  :inf: (n+1)(x^n)(3^n)/((3^(n+1))(x^(n-1))n) < 1. Cancelling, we see that ((n+1)/n)(x/3)<1, or that x<3. This is also the bound for your new sequence, so in this case, yes: If the latter sequence converges, so does your first sequence.

Anyway, I would suggest using the ratio test...and be very careful when multiplying by variables.

It is not correct to say that it will converge for all x.  It will converge for all x such that |x| < 3, however.  As will the original sum.  Generally, if you have something that is acting like a geometric series, doing things like multiplying the first term by 1, the second term by 2, and so on, will not change convergence.  Exponentials are better than polynomials.

Usually, though, it is okay to play along with summations as if they converged without checking at every step to see if they still do.  The entire theory of generating functions is based around that fact.  Things don't usually self-destruct when you preform simple operations on them.

Yep, pretty much.  If you wanted to be rigorous, you could toss in something about the ratio test -- I wouldn't have, though.

What's the purpose of multiplying a series by something?  It looks like the general term of the series is  1/x+ ((nx^n-1 ) / 3^n)
where n goes from n=2 to infinity.   I think the correct answer would be -3 less or equal to x less than or equal to 3

[quote="cats..."]The question i was given was S= 1/x +2x/(3^2) + 3x^2/(3^3) +4x^3/(3^4)+....
Find where the series converges.

I multiplied S by x/3, then did S-Sx/3 to get 1/3 + x/3^2 + x^2/3^3+...
Is it correct to say that S will converge for all x that make 1/3 + x/3^2 + x^2/3^3+... converge?[/quote]

What are you talking about with holes?

[quote="cats..."]So there would be nothing wrong with doing what I did on that series?  If I were to do that on the AP exam, there would be no holes ?[/quote]

He means holes in his proof.  He was asking if he was being fully rigorous.


Let $P$ be a convex $2006$-gon. The $1003$ diagonals connecting opposite vertices and the $1003$ lines connecting the midpoints of opposite sides are concurrent, that is, all $2006$ lines have a common point. Prove that the opposite sides of $P$ are parallel and congruent.

First prove opposite sides are parallel (by noticing a property of a trapezoid).  Then based on the fact the 1003 diagonals are concurrent, prove that the point of concurrency is the midpoint of all diagonals (by using ratio).

I don't see anything wrong with my solution (probably the way I write it was quite obscure).

Okay, let $A_{1}A_{2}...A_{2006}$ be that polygon such that $A_{1}A_{1004}, A_{2}A_{1005}, .... , A_{1003}A_{2006}$ and $M_{1}M_{1004}, M_{2}M_{1005}, ..., M_{1003}M_{2006}$ are concurrent at $O$, where $M_{1}, M_{2},....$ are the midpoints of $A_{1}A_{2},A_{2}A_{3},...$

Since $A_{1}A_{1}004, A_{2}A_{1005}, M_{1}M_{1004}$ are concurrent, we obtain $A_{1}A_{2}//A_{1004}A_{1005}$. Similarly, we obtain any opposite sides are parallel.

Now we have by applying the ratio theorem repetitively, we obtain

$\frac{OA_{1}}{O_{1004}}= \frac{O{}_{A}{}_{2}}{OA_{1005}}= .... = \frac{OA_{1003}}{OA_{2006}}= \frac{OA_{1004}}{OA_{1}}= \frac{OA_{1005}}{OA_{2}}= .....=\frac{OA_{2006}}{OA_{1003}}$.

This implies that O is the midpoint of every main diagonals. Therefore, all opposite are congruent and parallel.

Another one:
Proving for quads is just a property of trapezoids.
Now take any 3 consecutive vertices in the 2006-gon and their corresponding opposite points. Note that by Brianchon's Theorem and convexity, we have and ellipse tangent to the hexagon.

Also the opposite sides of the hexagon are parallel. This implies the lines joining opposite lines of the hexagon are tangent to antipodal points on the ellipse. But this gives us rotational symmetry implying the opposite sides are equal.

Hence, we prove all sides of the 2006-gon are congruent with opposite sides parallel to each other. $\blacksquare$

Enumerate the points of $P$ as $P_1, \ldots, P_{2006}$ in clockwise order. Let the common point be $O$. Since $P$ is convex, $O$ lies inside segment $P_i P_{i+1003}$ for all $i$. Let the midpoint of segment $P_i P_{i+1}$ be $M_i$. For all $i$, taking indices modulo 2006 we have by Ratio Lemma
\[ \frac{OP_i \sin \angle P_i O M_i}{OP_{i+1} \sin \angle P_{i+1}OM_i} = 1 = \frac{OP_{i+1003} \sin \angle P_{i+1003} O M_{i+1003}}{OP_{i+1004} \sin \angle P_{i+1004} O M_{i+1003}} \implies \frac{OP_i}{OP_{i+1}} = \frac{OP_{i+1003}}{OP_{i+1004}} \]
upon cancelling equal angles. $\angle P_i O P_{i+1} = \angle P_{i+1003} O P_{i+1004}$, thus triangles $P_i O P_{i+1}, P_{i+1003} O P_{i+1004}$ are similar. So $P_i P_{i+1}, P_{i+1003} P_{i+1004}$ are parallel.

Now let $OP_i = x_i$ for each $i$. Triangles $P_i O P_{i+1}, P_{i+1003} O P_{i+1004}$ are similar for all $1 \le i \le 1002$, so chaining equalities for these similarities
\[ r = \frac{x_{1004}}{x_1} = \frac{x_{1005}}{x_2} = \frac{x_{1006}}{x_3} = \cdots = \frac{x_{2006}}{x_{1003}} > 0 \]
However similar triangles $P_{1003} O P_{1004}, P_{2006} O P_1$ gives
\[ \frac{x_{2006}}{x_{1003}} = \frac{x_1}{x_{1004}} \implies r=\frac{1}{r} \implies r=1 \]
This yields the conclusion.

I think the condition that the lines connecting the midpoints of opposite sides can be loosened to the lines connecting points $Q_i, Q_{i+1003}$ in the interior of opposite sides $P_i P_{i+1}, P_{i+1003}P_{i+1004}$ such that $\frac{P_i Q_i}{Q_i P_{i+1}} = \frac{P_{i+1003} Q_{i+1003}}{Q_{i+1003}P_{i+1004}}$; my solution works the same.

I did something quite diferent in my solution:

Hint:
[hide]Prove that $ABCD$ are four points in the polygon, with $A,C$ opposite from the center, and $B,D$ also opposite, then $ABCD$ is a parallelogram.[/hide]

To prove our hint I'll use projective geometry.

On the hint, let $P$ be the concurrent point. Because of the colinearity, we may project through $P$ to $AB$ and $DC$, and noiticing that the infinity point of the lines are the same ($M$ and $N$ are the midpoints of $AB$ and $CD$, and $R,Q$ the infinity points of $AB,CD$. Because $M,N,P$ are collinear, note that $P(A,B;M,R)=P(D,C;N,Q)$. Thus, $AB || CD$. Now, similarly, $BC || AD$. Hence the polygon is a parallelogram.
Now, note that every principal diagonal meet each other at their midpoint.

$Q.E.D$


geometry
perpendicular bisector
area of a triangle
Heron\u0027s formula

In triangle $\bigtriangleup ABC$, side $\bar {AC}$ and the perpendicular bisector of $\bar {BC}$ meet in point $D$, and $\bar {BD}$ bisects $\angle ABC$. IF $AD = 9$ and $DC = 7$, what is the area of triangle $ABD$?

Solve this without trigonometry.

Call $E$ the midpoint of $BC$. The perpendicular bisector

$DE$ of triangle $DBC$ gives $DB=DC=7$ and $\angle
ACB= \angle DBC = \angle DBA$. Since $\triangle ADB$ and $\triangle ABC$

also share $\angle BAC$, $\triangle ADB \sim \triangle ABC$. So $\frac{9}{AB}= \frac{AB}{16}$ and $AB=12$. $\triangle ADB$ has sides $7,9,12$ and we

can use Heron's formula to find its area = $\sqrt{14(2)(5)(7)} = 14 \sqrt{5}$

You probably want to use \overline{AB} rather than \bar{AB}

\overline{AB} yields $\overline{AB}$ whereas \bar{AB} yields $\bar{AB}$

I get the feel that u've created the problem urself and there is some confusion as to whether point D is on which side of the triangle. a picture will definitly help


Prove if a,b,c,x,y,z>0 then
$ \sum x\sqrt{\frac{2a}{b\plus{}c}}>\equal{} \sum\frac{xy}{z}$

[quote="tranquoc"]Prove if a,b,c,x,y,z>0 then
$ \sum x\sqrt {\frac {2a}{b \plus{} c}} > \equal{} \sum\frac {xy}{z}$[/quote]
Try $ a\equal{}b\equal{}c\equal{}1,x\equal{}y\equal{}1,z\equal{}1/100$ ;)


inequalities
function
triangle inequality
absolute value
complex analysis

Suppose $ f$ analytic on the unit disk, $ |f(z)|\leq 1$ on $ |z|<1$, and $ f(0)\equal{}1/2$. Prove that $ |f(1/3)|\geq 1/5$.

Now I have done the standard thing and applied the Schwarz-Pick or whatever of the several names you want to call it.

$ \left|\frac{f(0)\minus{}f(1/3)}{1\minus{}\overline{f(0)}f(1/3)}\right|\leq \left|\frac{0\minus{}1/3}{1\minus{}0}\right|\equal{}1/3$

Thus $ \frac{|\frac{1}{2}\minus{}f(1/3)|}{|1\minus{}\frac{1}{2}f(1/3)|}\leq 1/3\Rightarrow \
|1/2\minus{}f(1/3)|\leq\frac{1}{3}|1\minus{}\frac{1}{2}f(1/3)|$. So triangle inequality both ways (reverse the one way) and get $ \frac{1}{2}\minus{}|f(1/3)|\leq \frac{1}{3}\plus{}\frac{1}{6}|f(1/3)|\Rightarrow |f(1/3)|\geq 1/7$.

Possibility one: There is a typo and it meant $ 1/7$ not $ 1/5$.
Possibility two: Some of those inequalities can be tightened?
Possibility three: I made a mistake somewhere.
Possibility four: There are other approaches.
Possibility five: You decide...

Any takers?

Try to build an extreme function. A function $ f$ which is bijective on the disk has the form $ f(z)\equal{}a\frac{z\minus{}b}{1\minus{}\overline{b}z}$ with $ |a|\equal{}1$. In this case, we need $ \minus{}ab\equal{}\frac12$, and we get $ f(\tfrac13)\equal{}a\frac{1\minus{}3b}{3\minus{}\overline{b}}\equal{}ab\frac{1\minus{}3b}{3b\minus{}|b|^2}\equal{}\frac12\frac{3b\minus{}1}{3b\minus{}\frac14}\equal{}\frac{2(3b\minus{}1)}{12b\minus{}1}$
We get $ \frac15$ for an absolute value of $ \frac15$ by setting $ b\equal{}\frac12$. This is a minimum for this form.

The tricky part here is that both terms inside that fraction have the same sign. Your method would have worked well for a point beyond $ \frac12$, but there's cancellation closer in.

Keeping that in mind, we can do better in the estimate. We know $ \left|\frac{\frac12\minus{}f(\frac13)}{1\minus{}\frac12f(\frac13)}\right|\le \frac13$, so set $ frac{\frac12\minus{}a}{1\minus{}\frac12a}\equal{}\frac13b$ and solve for $ a$:
$ 1\minus{}2a\equal{}\tfrac13(2\minus{}a)b\equal{}\tfrac23 b\minus{}\tfrac13 b$
$ 3\minus{}2b\equal{}6a\minus{}ab$
$ a\equal{}\frac{3\minus{}2b}{6\minus{}b}$, with $ |b|\le 1$.
This is minimized when $ b\equal{}1$, in which case $ f(\tfrac13)\equal{}a\equal{}\frac15$. The minimizing function is the one I constructed above.


Given 7 hours after school (Mon-Fri) and this situation what would you do?

School is going to start in about a week (in Provo, Utah). You are going to be taking AP Chem, AP Stats, and AP U.S. History. You will also be doing at least two clubs. You also want to prepare for both the PSAT and AMC/AIME. How would you spend your time, such that you do all these things?

...This is kind of stated like a math problem. :P

Set up a schedule for yourself.  7 hours is a nice big chunk of time, so you can fit everything in easily.  Get all your homework out of the way quickly at the beginning of each day, and then divide the rest equitably between your clubs, PSAT studying, and math contest stuff.  Sticking to a schedule is hard at first, but over time you get into a sort of rhythm, and it's nice to know that your time ahead if you is planned, rather than potentially wasted.

There is no need to study for AP's until about 1.5-2 weeks before the exam.

Same goes for PSAT; it won't really help you that much to study for months before.

Clubs probably won't cut into your time much, if at all.

The biggest time waster is AMC/AIME.

[quote="1/(ln x)"]

...This is kind of stated like a math problem. :P[/quote]

then the answer is $3$

anyways, I agree with elemmnop about APs, I mean, there's no need to start preparing for the actual test months before the test since your AP class is supposed to help you do that anyways. Also, for PSAT, you might want to go over a few practice tests before you actually take the real thing.

For AMC/AIME...uh...do a lot of math

I never made up a schedule in high school for after school activities.  There wasn't really a point.  Cross country was always at the same time.  Math team was always at the same time.  The clubs I was in only met during school.  So after I got home from cross country, I would start doing my homework and math studying in whatever order I felt.  As long as you get everything done that you want to, this should work out.

[quote="krustyteklown"]I'm not too sure about how seriously Elemennop is taking some of the other tests. Balancing time greatly weighs upon how much you weight potential results on the tests yourself.[/quote]

serious enough to get good scores on them. AP's aren't all that hard if you just cram the week before the test. The same can be said in a sense to MANY high school tests (AMC's and up excluded).

[quote="SplashD"]you can memorize all the vocabulary for the PSAT in 1.5-2 weeks?!? :o[/quote]
I never studied vocab for the SAT explicitly - if you read a lot and look up words you don't know (and try to remember and use them when you can), then that's a lot better than memorizing words for a test.  Our school also gives us weekly vocab tests with 30 words a week, which probably helped as well; in any case, studying that many words took about half an hour, so you don't really need to plan a lot of vocab studying - just make sure you learn new words occasionally.

You know, if you could find a vocabulary list for the PSAT, I'd appreciate it alot.

Since there isn't one, you can see that it's pretty useless picking random words and studying them hoping for the small, tiny chance that they might appear on the PSAT.

well not only cramming for the exam

some schools' classes are actually quite work-heavy, and the OP might be interested in excelling in his class as well as a 5 on the exam

Schools usually don't have very heavy workloads, even in the AP classes. Most of the homework can be done during a study period or just working in other classes.

Unless he's in a school like Exeter or TJ; but then he has other things to worry about besides how he should do his homework.

I do have a list of PSAT vocab words. It's from a book. There's like 40 pages and 50 words on each page. I started memorizing at the beginning of summer and still isn't half away through all of it.

*moved to Round Table*

not much in this thread about AMC, other than an initial mention as a task...

I'm sorry, but "a list" of PSAT words is not neccesarily legit. There is no gaurantee that all or any of the words on the PSAT will be on that list, so I think it's a waste of time to study vocab for an exam with a possible pool of words in the upper 300,000. There's better things you can do with your time.

Seva, you already have a good vocabulary.  It's not fair for you to tell people (who might not have as large a vocabulary) that they shouldn't study.  There are words that very commonly show up on the PSAT/SAT's or that are frequently used in intelligent conversation or books that one should know.  Using root words and context clues should be emphasized more in the process of studying for the SAT's, in my opinion.

bah.

get warcraft 3, buy a lot of mountain dew, play wc3 until 2 AM, then do hw and study for tests on the bus

Can anyone give me an idea of how much homework an average AP class gives? I have only taken one AP Class (Calculus BC) so far. I would like to do well in my AP classes and on the AP tests.

Would anyone suggest doing some of my homework during lunch?

dude chill out, you shouldn't be studying 24 hours a day...

ap homework daily should take about 45 minutes to an hour.

don't freak out about stuff

study a week before major tests, a day before minor tests

study three wekks before uber major tests

by major, i mean spend bout 20 min two times a day

The AP that I think drains your time the most is AP Art. For summer, we had to make two random drawings, and it took me about two weeks to complete both of them (about an hour a day).


function
inequalities proposed
inequalities

For $ a,b,c\in\mathbb{R}^\plus{}$ . Prove that -
$ \frac{a^2}{b^2} \plus{} \frac{b^2}{c^2} \plus{} \frac{c^2}{a^2} \plus{} 9 \ge 4(\frac{a}{b} \plus{}\frac{b}{c} \plus{}\frac{c}{a})$

[quote="geniusbliss"]For $ a,b,c\in\mathbb{R}^ \plus{}$ . Prove that -
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \plus{} 9 \ge 4(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})$[/quote]

try $ a\equal{}15,b\equal{}12,c\equal{}10$... :wink:


A natural number  is called a lucky number if the sum of its digits is 7. If we arrange all lucky numbers in increasing order, we get the sequence A1,A2,A3.....if An=2005 If , then what is A5n ?PLEASE NOTE THAT A1 ,A2 ,A3. A5n  ALL ARE SYMBOLS.

use the "bars and shorts" method: the number of solutions in non-negative integers of $ x_1\plus{}x_2\plus{}\cdots \plus{}x_n\equal{}m$ is $ \binom{m\plus{}n\minus{}1}{n\minus{}1}$.

please explain  "bars and shorts" method


There are $2^n$ words of length $n$ over the alphabet $\{0,1\}.$ Prove that the following algorithm generates the sequence $w_1, \ldots, w_{2^n-1}$ of all these words such that any two consecutive words differ in exactly one digit.

(i) $w_0 = 00\ldots0$ ($n$ zeros).
(ii) Suppose $w_{n-1} = a_1a_2 \ldots a_n,$ $a_i \in \{0,1\}$. Let $e(m)$ be the exponent of 2 in the representation of $n$ as a product of primes, and let $j = 1 + e(m)$. Replace the digit $a_j$ in the word $w_{m-1}$ by $1-a_j.$ The obtained word is $w_m.$

Simple induction on $n$ seems to work fine: consider the first $2^{n-1}$ (all of these have the last digit equal to $0$) sequences and the next $2^{n-1}$ (with the last digit equal to $1$). For each group apply the induction hypothesis disregarding the last digit (work only with the first $n-1$ digits).


I'm doing okay. (like I'm getting, 80% of the problems right? Maybe 70%, depending on the concept.)

It doesn't matter if you're the oldest in the class (besides, the assistant and instructor are older than 13 :P) ... you can always lie about your age :P

If you want to take the class, then you should.  Even if it's a little bit too easy, the hardest problems in the problem sets should still challenge you and it will be a good review.

I took that class when I was 13, and it was my first AoPS test.  I'm still 13 right now!

[quote="myyellowducky82"]Hi, I recently got the Intro. to C&P book and have worked through some of it. I am wondering, at 13 years old, if I'm too old to take the class? It's not the easiest class, but I'd still like to take it (with the risk that I'm the oldest). Any suggestions?[/quote]

You won't be the oldest.  You'll probably be right in the middle.  Besides, don't worry about your age...  (Maybe I should follow that advice!)

thanks for the advice everyone. Seeing that I'm one of the youngest in my grade, I just don't like the whole "age" thing. :P


Let $ n > 1$ be an integer.
Given an [i]ordered[/i] set of $ k$ positive integers $ (a_i)$ whose sum is $ n$, namely $ \sum_{i \equal{} 1}^{k} a_i \equal{} n$, compute the product $ k! \prod_{i \equal{} 1}^k( \minus{} a_i)$.
Prove that the sum, over all the possible ways to express $ n$ as the [i]ordered[/i] sum of some positive integers (so, with $ k$ ranging from $ 1$ to $ n$), of the reciprocals of the corresponding products, is $ 0$.

Here's an example with $ n \equal{} 3$, just to make the statement clearer.

$ (3) \rightarrow \minus{} 3$
$ (2,1) \rightarrow 2\cdot 2$
$ (1,2) \rightarrow 2\cdot 2$
$ (1,1,1) \rightarrow \minus{} 3!$

Therefore, the sum of reciprocals is $ \minus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{4} \minus{} \frac {1}{6} \equal{} 0$.

A couple of variations on this problem:

1) Same stuff, but replace $ k! \prod_{i \equal{} 1}^k( \minus{} a_i)$ with $ k! \prod_{i \equal{} 1}^k a_i$. Prove that now the sum of reciprocals is $ 1$.

2) Same stuff, but replace $ k! \prod_{i \equal{} 1}^k( \minus{} a_i)$ with $ k \prod_{i \equal{} 1}^k( \minus{} a_i !)$. Prove that the sum of reciprocals is again $ 0$.


calculus
integration
geometry
3D geometry
sphere
vector
parameterization

Hi all,

would anyone be able to give me a little bit of help with this question.

[img]http://img222.imageshack.us/img222/4540/picture1qy8.png[/img]

for part (a) i did this....

our sphere is given by $x^{2}+y^{2}+z^{2}= 25$

putting in z = 3 to find out part of the sphere gives:
$x^{2}+y^{2}= 16$

so we change to the parametric representation of the part of the sphere we are interested in sphere which is

$\vec{r}(\theta,\psi) = (4sin\psi cos\theta) \vec{i}+(4sin\psi sin\theta) \vec{j}+(4cos\psi) \vec{k}$

then

$\vec{r_\theta}(\theta,\psi) = (-4sin\psi sin\theta) \vec{i}+(4sin\psi cos\theta) \vec{j}$

$\vec{r_\psi}(\theta,\psi) = (4cos\psi cos\theta) \vec{i}+(4cos\psi sin\theta) \vec{j}+(-4sin\psi) \vec{k}$

so we get that

$\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi) =-16sin^{2}\psi cos\theta)\vec{i}-16sin^{2}\psi sin\theta \vec{j}-16sin\psi cos\psi \vec{k}$

our unit normal vector is given by:

$\vec{\hat{n}}=-\frac{\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)}{|\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)|}= \frac{16sin^{2}\psi cos\theta)\vec{i}+16sin^{2}\psi sin\theta \vec{j}+16sin\psi cos\psi \vec{k}}{|\vec{r_\theta}(\theta,\psi) \times \vec{r_\psi}(\theta,\psi)|}$

does this look like i am heading in the right direction for part (a)?

cheers! :)

-Sarah

No. It is great that you have combined the equations of the sphere and plane into one - but what you get this way describes the [b]intersection[/b] of these surfaces (circle of radius $4$ at height $3$ over the $xy$-plane). This is important for determining the parameter domain, but it has no effect on parametrization. The sphere still has radius $5$.

Also, I would recommend avoiding spherical coordinates whenever possible. It is possible here, since you only have a piece of the sphere in the upper half-space. Take $x$ and $y$ as your parameters; then $z=\sqrt{25-x^{2}-y^{2}}$ and here is the parametrization: $\vec r(x,y)=\langle x,y,\sqrt{25-x^{2}-y^{2}}\rangle$ and the parameter domain: $D=\{(x,y)\colon x^{2}+y^{2}\le 16\}$.

Refer to what I wrote [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107956]here[/url].

As of now, I have one seat available in my Vector Calculus class that begins this Monday (current enrollment 84/85)...   :wink:

ok, i am quite new to this stuff, and so a little confused.

i read that surface integral is $\int_{S}\vec{F}\cdot d\vec{S}$

but i can't find anywhere how to evaluate that, i can only find how to evaluate $\int \int_{S}\vec{F}\cdot d\vec{S}$

are these the same thing?

84/85!  :o , that sounds like it is going to be pretty intense! good luck! :)

[quote="mlok"]Some people are too lazy or too sophisticated to write two integral signs in a surface integral. I am neither. By the way, LaTeX has a special command for this: \iint (or \iiint for triple).[/quote]

ah ok, i see, so they are the same.  that is a relief!  :P

i'll just try the question again, hold on...

so we want to calulate this for part (a)? \[\iint_{S}\mathbf{F}\cdot d\mathbf{S}=\iint_{D}\mathbf{F}(\mathbf{r}(x,y))\cdot \left(\frac{\partial \vec{r}}{\partial x}\times\frac{\partial \vec{r}}{\partial y}\right)\,dx\,dy\] where $\mathbf r(x,y)=\langle x,y,\sqrt{25-x^{2}-y^{2}}\rangle$

and so $\mathbf{F}(\mathbf{r}(x,y)) = \langle x\sqrt{25-x^{2}-y^{2}}, y\sqrt{25-x^{2}-y^{2}}, 1 \rangle$

is that ok so far?

hmm i get different answers for parts (a) and (b), thats not good now is it... :S

for part (a) i get S1 = $144\pi$ , S2 = $16\pi$

for part (b) i get $128\pi$

for part (b) i evalutated the following integral to get my answer: [img]http://img246.imageshack.us/img246/1948/picture3cb8.png[/img]

[quote="mlok"]I presume you used the upward normal for both $S_{1}$ and $S_{2}$. This is perfectly fine, but since closed surfaces are usually assumed oriented outward, $\iint_{S}\vec F\cdot d\vec S=\iint_{S_{1}}\vec F\cdot d\vec S-\iint_{S_{2}}\vec F\cdot d\vec S=128\pi$. Which is what I get from the divergence theorem.  :P
Indeed, ${\rm div}F=2z$, so the triple integral of divergence is (after conversion to polar coords) \[\int_{0}^{4}\int_{0}^{2\pi}\int_{3}^{\sqrt{25-r^{2}}}2zr\,dz\,d\theta\,dr=2\pi\int_{0}^{4}(16r-r^{3})dr=128\pi\] You forgot the Jacobian $r$, and probably made one more mistake somewhere in computations.[/quote]

yep yep i forgot the r, i was just trying ot edit my post before, but you had already replyed!

oh ok so i have to minus s1 and s2, that makes things work out! yay! :D  how come the minus didnt come in automatically when when we caluclated s2 ?

thanks for the help once again! :)


Let be $ ABC$ an acute-angle triangle . Prove that :

$ \frac{a}{c\sin B}\plus{}\frac{b}{a\sin C}\plus{}\frac{c}{b\sin A}\le \frac{2p}{3r}\Longleftrightarrow \frac{a}{c\cos B}\plus{}\frac{b}{a\cos C}\plus{}\frac{c}{b\cos A}\le 6$

hello, your inequality $ \frac {a}{c\cos(\beta)} \plus{} \frac {b}{a\cos(\gamma)} \plus{} \frac {c}{b\cos(\alpha)}\le 6$ is
equivalent with
$ a^6 \plus{} b^6 \plus{} c^6 \plus{} 3a^2b^2c^2\geq 2a^4b^2 \plus{} 2b^4c^2 \plus{} 2c^4a^2$
but after a hint from Darij holds: if  $ a,b,c$ are the sidelength of an acute triangle $ \Leftrightarrow$ then
are $ a^2,b^2,c^2$ the sidelength of an traingle and we need to prove that
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\geq 2a^2b \plus{} 2b^2c \plus{} 2c^2a$ and here you can find the proof from Darij:
http://www.mathlinks.ro/Forum/viewtopic.php?t=35295
Sonnhard.

hello, to your  first inequality, setting $ \sin(\alpha)\equal{}\frac{2A}{bc}$, $ \sin(\beta)\equal{}\frac{2A}{ac}$, $ \sin(\gamma)\equal{}\frac{2A}{ab}$ and $ r\equal{}\frac{A}{p}$ we get after a simplification
$ \frac{a^2}{2}\plus{}\frac{b^2}{2}\plus{}\frac{c^2}{2}\le \frac{2p^2}{3}$
and this is equivalent to
$ a^2\plus{}b^2\plus{}c^2\le ab\plus{}ac\plus{}bc$
but this is only possible for $ a\equal{}b\equal{}c$.
Sonnhard.


The "AMC Prep" is a series of questions to help you prepare for the AMC10/12 this spring.

Source: MMPC 1966, part 1

N is an integer such that N divided by 10 leaves the remainder 9, N divided by 9 leaves the remainder 8, N divided by 8 leaves the remainder 7, ...... , N divided by 2 leaves the remainder 1.  In other words, N divided by n leaves the remainder n-1 for every integer n in the interval (1,10].  Which of the following conclusions about N is false?

(A) N+1 is divisible by 8, 9, and 10
(B) N+1 is divisible by 12, 20, and 28
(C) N+1 is divisible by 6, 15, and 21
(D) N+1 is divisible by 4, 12, and 20
(E) N+1 is divisible by 9, 18, and 27

E is false.  Since N leaves a remainder of n-1 for all positive integers n from 2 to 10 inclusively, N+1 is divisible by all positive integers from 2 to 10.

N+1 [b]l[/b] 2,3,4,5,6,7,8,9,10

N+1 has to be divisible by 9 and 18 though because it will have at least 2:^3: (8) and a 3:^2: (9) in its prime factorization.  N+1 doesn't need to be divisible by 27 because it can have a factorization with only 3:^2:.

Uhh...right!

(there are no answers,  :-x )


limit
integration
calculus
real analysis
real analysis unsolved

Find \[ \lim_{n\to\infty}n\left(\int_{0}^{\infty} e^{-x^{n}}dx-1\right). \]

If we make the change of variables $w=x^{n}$ the integral is

$\\ \frac{1}{n} \int_{0}^{\infty} w^{\frac{1}{n} -1} e^{-w}\, dw =\frac{1}{n} \Gamma{\left( \frac{1}{n} \right) }$

and the limit is $\lim_{n\to\infty}\left(\Gamma{\left( \frac{1}{n} \right) }-n\right) = - \gamma$

 :)


Find all positive integers $k$ for which the following assertion holds:
If $F(x)$ is polynomial with integer coefficients ehich satisfies $F(c) \leq k$ for all $c \in \{0,1, \cdots,k+1 \}$, then \[F(0)= F(1) = \cdots =F(k+1).\]

I think the right statement is $0 \leq F(c) \leq k$.
In that case it is from ISL 1997.

Pierre.

I'm also confuse about that, anyone know the original problem?
i solve it by assuming the  pbornsztein's restrcition.
so the problem  is need to be corrected? :lol:

I have checked and Pierre is right $0\leq F(c)\leq k$

Find all positive integers $k$ for which the following assertion holds:
If $F(x)$ is polynomial with integer coefficients ehich satisfies $0  \leq F(c) \leq k$ for all $c \in \{0,1, \cdots,k+1 \}$, then \[F(0)= F(1) = \cdots =F(k+1).\]


calculus
integration
trigonometry
calculus computations

i am having problem to solve $ \int \frac{dx}{x^2 \sqrt{(x\minus{}1)}}$ and $ \int \frac{dx}{(x^2\plus{}1)^2}$ can someone help

For the second, let $ x \equal{} \tan \theta$.

when it doubt use trigonometry.

$ \int \frac{dx}{x^2\sqrt {x\minus{}1} }\;\qquad \overbrace{\qquad\equal{}\qquad}^{x\equal{}\sec^2\;\theta }\;\qquad \int \frac{2\sec^2\;\theta \; \tan\; \theta \;d\theta}{\sec^4\;\theta \;\tan\;\theta}$

$ \equal{}\int 2\cos^2\;\theta\;d\theta \equal{}\int 1\plus{}\cos\;2\theta\;d\theta\equal{}\theta \plus{} \sin\;\theta\;\cos\;\theta  \plus{} C$

$ \equal{} \sec^{\minus{}1}\;\sqrt x \plus{} \frac{\sqrt {x\minus{}1}}{\sqrt x} \cdot\frac{1}{\sqrt x} \plus{} C$

We have $ \int\frac {1}{x^2 \sqrt {x \minus{} 1}}\,dx \equal{} k \frac {\sqrt {x \minus{} 1}}{x^n} \plus{} f(x) \plus{} C$, and differentiating both sides we get

$ \frac {1}{x^2 \sqrt {x \minus{} 1}} \equal{} k \frac {(1 \minus{} 2n)x \plus{} 2n}{2x^{n \plus{} 1} \sqrt {x \minus{} 1}} \plus{} f'(x)$

From this its clear that we must have n = 1 and k = 1, and so we get

$ \frac {1}{x^2 \sqrt {x \minus{} 1}} \equal{} \frac {2 \minus{} x}{2x^2 \sqrt {x \minus{} 1}} \plus{} f'(x) \equal{} \frac {1}{x^2 \sqrt {x \minus{} 1}} \minus{} \frac {1}{2x \sqrt {x \minus{} 1}} \plus{} f'(x)$

and so f(x) must verify $ f'(x) \equal{} \frac {1}{2x \sqrt {x \minus{} 1}} \equal{} \frac {\frac {1}{2\sqrt {x}}}{\sqrt {x} \sqrt {(\sqrt {x})^2 \minus{} 1}} \equal{} > f(x) \equal{} \sec^{\minus{}1}(\sqrt {x})$.

Finaly, $ \int\frac {1}{x^2 \sqrt {x \minus{} 1}}\,dx \equal{}  \frac {\sqrt {x \minus{} 1}}{x} \plus{} \sec^{\minus{}1}(\sqrt {x}) \plus{} C$

It's also clear that $ \int\frac {1}{(x^2 \plus{} 1)^2}\,dx \equal{} \frac {Ax \plus{} B}{x^2 \plus{} 1} \plus{} \int\frac {Cx \plus{} D}{x^2 \plus{} 1}\,dx$. Now differentiate both sides:

$ \frac {1}{(x^2 \plus{} 1)^2} \equal{} \frac { \minus{} Ax^2 \minus{} 2Bx \plus{} A}{(x^2 \plus{} 1)^2} \plus{} \frac {Cx \plus{} D}{x^2 \plus{} 1} \equal{}$

$ \equal{} \frac {Cx^3 \plus{} (D \minus{} A)x^2 \plus{} (C \minus{} 2B)x \plus{} (A \plus{} D)}{(x^2 \plus{} 1)^2}$, from which we get $ A \equal{} D \equal{} \frac {1}{2}$, $ B \equal{} C \equal{} 0$, and so

$ \int\frac {1}{(x^2 \plus{} 1)^2}\,dx \equal{} \frac {x}{2(x^2 \plus{} 1)} \plus{} \frac {1}{2}\int\frac {1}{x^2 \plus{} 1}\,dx \equal{} \frac {x}{2(x^2 \plus{} 1)} \plus{} \frac {1}{2} \arctan x \plus{} C$

The two integrals are actually related. Observing a subsititution of $ u\equal{}\sqrt{x\minus{}1}$ we differentiate to find $ 2u\text{ d}u\equal{}\text{ d}x$. We also rewrite the first integral as \[ \int \frac{1}{(u^2\plus{}1)^2u}\text{ d}x\equal{}\int \frac{2}{(u^2\plus{}1)^2}\text{ d}u\] whose solution in $ u$ follows from above. Someone may want to check me on that but I'm pretty sure that's a valid substitution.

Good observation Igiul, that substitution looks perfectly ok.


Each of two boxes contains both black and white marbles, and the sum of all marbles in the two boxes is $25$. One marble is taken out of each box randomly. The probabillity that both marbles are black is \frac{27}{50}. What is the probabillity that both marbles are white? :)

I didn't understand if there are
a) a total of $25$ marbles in each box, or
b) a total of $25$ marbles combining the two boxes.[hide="Solution of case b"]
If there are $c$ marbles in the first box ($a$ of them black) and $25-c$ marbles in the second box ($b$ of them black), the probability of extracting two black marbles is
$\frac{a}{c}\times\frac{b}{25-c}=\frac{27d}{50d}$.

In positive integers, $c(25-c)=50d$ must be
(i) $c=5$, $25-c=20$, $d=2$,
(ii) $c=10$, $25-c=15$, $d=3$
(or symmetrical $c\to 25-c$, that doesn't add solutions).

Then, we must have $a\leq c$, $b\leq 25-c$, $ab=27d$.
(i) $a\leq 5$, $b\leq 20$, $ab=54$ $\to$ $a=3$, $b=18$.
(ii) $a\leq 10$, $b\leq 15$, $ab=81$ $\to$ $a=9$, $b=9$.

Anyway, the probability of extracting two white marbles is
$\frac{c-a}{c}\times\frac{(25-c)-b}{25-c}=\frac{1}{25}$[/hide]


I tried to send my solutions in at about 12:00 AM EST (which is okay, because the deadline is 12 AM PST, right?) but Outlook was unable to send the email because my pdf file was 16 MB. I tried several different ways of arranging my solutions - making each solution a separate file, keeping my scanned images as jpgs instead of pdfs, but nothing worked.

I ended up using a web service where you upload a file and it sends a link to it in an email. Is this going to be okay? It was in before the 12 AM PST deadline, but I dont know if the fact that the actual file wasn't in the email is going to be a problem.

Email us the details (at solutions@usamts.org) and we'll take it under advisement.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[size=150][color=red][b]Problem.[/b][/color][/size]

- Is there an infinite group with a finite number of proper subgroups?
- Give an example of an infinite group with no infinite proper subgroup.
- Give an example of a nonabelian group with abelian automorphism group.

Happy solving to you all! I look forward to reading your replies a-s-a-p.  :wink:

MATH-ly,

[b]José.[/b]

P.S. Last month's proposal can be found [url=http://www.mathlinks.ro/viewtopic.php?t=270550]here[/url].

1. Assume $ G$ is such a group. Now there are only elements of finite order in $ G$, because $ \mathbb{Z}$ has an infinite number of subgroups. But there are only a finite number of subgroups of order $ n$ so there can only be a finite number of elemnts of order $ n$ too. Since $ G$ is infinite this means that there are infinitely many orders of elements of $ G$. So we can choose elements $ g_1, g_2, ...\in G$ with distinct orders. So the subgroups $ \langle g_1\rangle, \langle g_2 \rangle, ...$ are distinct. This is a contradiction, no such $ G$ can exist.

2. $ \mathbb{Z}_{p^\infty}$

[b]1.[/b]

Your argument is superb, esteemed [b][color=orange]Gockel[/color][/b].  :thumbup:

[b]2.[/b]

You still need to provide a justification for this entry, [b][color=orange]Sir[/color][/b].

[b]3.[/b]

I look forward to reading your reply for this one.  :wink:

[quote="Gockel"]2. $ \mathbb{Z}_{p^\infty}$[/quote]

It's a torsion group. So an infinite subgroups of $ \mathbb{Z}_{p^\infty}$ must be infinitely generated.
Let's $ S$ be an infinitely generated subgroup of $ \mathbb{Z}_{p^\infty}$ and $ \{g_i\}_{i\in\mathbb{N}}$ the set of generators of $ S$.
Let's consider a monotonically increasing infinite integer sequence $ \{a_i\}_{i\in\mathbb{N}}$... such that $ p^{a_i}$ is the order of the generator $ g_i$. If $ a_i\equal{}i$ for every $ i\in\mathbb{N}$ then $ S\cong\mathbb{Z}_{p^\infty}$.
Every $ \langle g_i \rangle$ contains a subgroups for every divisor of $ p^{a_i}$ and so for every integer lower or equal to $ a_i$. Since the sequence $ \{a_i\}_{i\in\mathbb{N}}$ hasn't an upper bound we can add to the generator set of $ S$ an elements for every integer not in the sequence $ \{a_i\}_{i\in\mathbb{N}}$. But this means that $ S\cong\mathbb{Z}_{p^\infty}$

3. I'm thinking about it...

P.S: I'm tired and so I hope my English is understandable...  :maybe:

The smallest non-abelian group with abelian automorphism group has order 64. Other examples are constructed in:

D. Jonah, M. Konvisser,[i] Some non-abelian $ p$-groups with Abelian automorphism groups[/i],  Arch. Math., [b]26[/b] (1975), pp.131-133.

When $ p$ is an odd prime the smallest  non-abelian $ p$-group with abelian automorphism group has order $ p^7.$

[quote="vict85"][quote="Gockel"]2. $ \mathbb{Z}_{p^\infty}$[/quote]

Since the sequence $ \{a_i\}_{i\in\mathbb{N}}$ hasn't an upper bound we can add to the generator set of $ S$ an elements for every integer not in the sequence $ \{a_i\}_{i\in\mathbb{N}}$.

[/quote]

Hi there, [b][color=blue]vict85[/color][/b],

Could you please rewrite this part of your argument?

[quote="theba"]The smallest non-abelian group with abelian automorphism group has order 64.[/quote]

Hi there, [b][color=blue]theba[/color][/b],

Could you state an explicit presentation of this minimal counterexample that you've just mentioned?
--
Regards,

José.

P.S. Thank you both for taking the time to follow these monthly proposals of [i]mine[/i].


125 cL is the same as all of these except

1. 1.25 liter

2. 1250 mililiter

3. 12.5 deciliter

4. 125 cubic liter

5. 125 centiliter

i choose # 4 is that correct.

I guess you are correct since i don't  know if cubic liter really exists.  :P


[quote="kunny"]Let $a,b,c,s,t,u$ be positive numbers such that $a+b+c=1.$ Prove the following inequality.

\[ \frac{sa^2+tb^2+uc^2}{sa+tb+uc}+\frac{sb^2+tc^2+ua^2}{sb+tc+ua}+\frac{sc^2+ta^2+ub^2}{sc+ta+ub}\geq1.  \][/quote]

  Nice, Kunny! Use your technique at http://www.mathlinks.ro/Forum/topic-85613.html can solve it. :D

Thank you, Gibbenergy.  :) Yes, this problem was made by conceive's problem today.  :P

Let
$\sum s = s + t +u$
$\sum sa^2 = sa^2 + tb^2 +uc^2$
From Cauchy-Schwarz :--
$(\sum s)(\sum sa^2) \geq (\sum as)^2$
which implies:--
$\frac{(\sum sa^2)}{(\sum as)} \geq \frac{(\sum as)}{\sum s}$
Adding similar expressions we get:--

Given Expression $\geq a + b + c = 1$
QED


Any three noncollinear points determine a unique...

1. circle
2. arc
3. plane
4. triangle

Anything else?

[quote="DiscreetFourierTransform"]Parabolas work as well.[/quote]
I meant geometric shapes.

If the determine ellipses, don't they also determine hyperbolas?

Also, i_like_pie, what do you mean by "geometric shape?"
Isn't a parabola a geometric shape, or do you mean figure of finite length (in which case a hyperbola would not work)?

I think hyperbola should be one of them.  :wink: but duno depends what pie means by that


[color=darkblue][b]Find $ n \in N^*$ such as $ x_1,x_2,...,x_n \in [0,\infty)$ are existing, with the property that \[ (1\plus{}x_1)(1\plus{}x_2)...(1\plus{}x_n)(x_1\plus{}x_2\plus{}...\plus{}x_n)\equal{}2n^2x_1x_2...x_n\][/b][/color]

[quote="lingomaniac88"]I know that $ N$ represents the set of natural numbers, but what exactly is $ N^*$?[/quote]

In a context where it is unclear whether $ \mathbb{N}$ includes $ 0$ or not, $ \mathbb{N}^{*}$ refers to the positive integers (although I would prefer $ \mathbb{Z}^{*}$ or simply $ \mathbb{Z}_{> 0}$ myself :)).

[color=darkblue][b]Nice problem[/b][/color] [hide="solution"]  Will prove that $ (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_n)(x_1 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_n)\geq 2n^2x_1x_2x_3...x_n$ and will decide the equality case . Aplying Huygens $ (1 \plus{} x_1)(1 \plus{} x_2)...(1 \plus{} x_n)\geq(1 \plus{} \sqrt [n]{x_1x_2...x_n})$ and AM-GM $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n\geq n\sqrt [n]{x_1x_2...x_n}$ and the inequality becomes $ \boxed{(1 \plus{} \sqrt [n]{x_1x_2...x_n}) \cdot n\sqrt [n]{x_1x_2...x_n} \geq 2n^2x_1x_2...x_n}$
Dividing by $ x_1x_2...x_n$ and note $ x \equal{} \displaystyle\frac {1}{\sqrt [n]{x_1...x_n}}$ and remains to prove that $ (1 \plus{} x)^n\geq 2nx \implies x \plus{} 1 \geq \sqrt [n]{2nx}$.But with AM-GM we have that $ x \plus{} \frac {1}{n \minus{} 1} \plus{} \frac {1}{n \minus{} 1} \plus{} ... \plus{} \frac {1}{n \minus{} 1} \geq n\sqrt [n]{x \frac {1}{(n \minus{} 1)^{n \minus{} 1}}} \geq \sqrt [n]{2nx}$ $ \implies$ $ \boxed{(1 \plus{} \frac {1}{n \minus{} 1})^{n \minus{} 1} \geq 2}$      wich is Bernoulli inequality.The last inequality transforms into equality only if $ \boxed{n \equal{} 2}$ and then $ \boxed{x_1 \equal{} x_2 \equal{} 1}$   [/hide]


Let $k$ be a positive integer. Let $log(cosx)_k = log(cos(log(cos(...(log(cosx)))))$ with $k$ times of $log(cos($
Find all possible $x$ when $2pi/3 < absolute(x) < 9pi/2$ such that sum(k=1 to n) of $k(log(cosx)_k)$ is maximal

Let $k$ be a positive integer. Let $\lg{(\cos{x})_k}=\lg{(\cos{(\lg{(\cos{\cdots})})}}$ with $k$ times.
Find all $x$ when $\frac{2\pi}{3}<|x|<\frac{9\pi}{2}$ such that ... ...

I editted your post so that it is more comfortable for people to read.


Prove that for any real numbers $ a_1, ..., a_N$,

 $ \displaystyle\frac{1}{\sqrt{N}} \sum_{i\equal{}1}^N a_n\leq\sqrt{\sum_{i\equal{}1}^N {a_n}^{2}}$

hint: can use Cauchy-Schwartz inequality

$ \left( {\sum\limits_{i \equal{} 1}^N {a_i } } \right)^2  \le N.\sum\limits_{i \equal{} 1}^N {a_i^2 }  \Leftrightarrow \frac{1}{{\sqrt N }}.\left| {\sum\limits_{i \equal{} 1}^N {a_i } } \right| \le \sqrt {\sum\limits_{i \equal{} 1}^N {a_i^2 } }$

and $ \sum\limits_{i \equal{} 1}^N {a_i }  \le \left| {\sum\limits_{i \equal{} 1}^N {a_i } } \right|$


There's a tip that use the following formula:

$ \tan 3\theta \equal{} \tan \theta (\frac {3 \minus{} \tan ^2 \theta}{1 \minus{} 3\tan ^2 \theta})$

And let
$ A_k \equal{} \tan [\frac {\pi}{3} (1 \plus{} \frac {3^k}{3^n \minus{} 1})]$

$ B_k \equal{} \tan [\frac {\pi}{3} (1 \minus{} \frac {3^k}{3^n \minus{} 1})]$

So $ \prod_{k \equal{} 1}^{n}{A_k B_k} \equal{} 1$

But how to prove that

 $ \prod_{k \equal{} 1}^{n}{A_k B_k} \equal{} 1$ ?


I am currently using TexnicCenter (1 beta 7.01) with MiKTeX (2.5 according to the help manual) on Windows XP. I attempted to compile a file with the line:

[code]\usepackage{apalike}[/code]

at which point I got the error message: 'File "apalike.sty" not found'.

I attempted to install the file with MiKTeX file manager, which appeared to be successful, as the package was shown as installed in the according to the 'Packages' tab of the 'MiKTeX Options' dialogue box.

I then pressed the 'Refresh FNDB' and 'Update Formats' buttons on the 'General' tab.

Unfortunately, when I try to compile the file, I still get the error message about the missing file. What else do I need to do?

The file apalike.sty is currently located at "C:\tex\texmf\source\apalike-src.cab" on my hard drive. Does this make any difference?

Any help gratefully received as I've spent ages searching for a solution!

[quote="OHradicals"]The file apalike.sty is currently located at "C:\tex\texmf\source\apalike-src.cab" on my hard drive. Does this make any difference? [/quote]Yes, that is just a compressed file containing apalike.sty and is why LaTeX can't find it. I suspect the reason that MiKTeX has done this is that it is an ancient package meant for use wth previous versions of LaTeX or even Plain TeX. I don't think you will have much joy trying to use it.

There should be other apalike files in C:\tex\texmf\bibtex\bst\apalike, C:\tex\texmf\tex\plain\apalike and C:\tex\texmf\doc\bibtex\apalike but if not, then you can install them as below.

But if you want to try (it won't harm anything) then do it manually. Go to http://tug.ctan.org/tex-archive/biblio/bibtex/contrib/apalike/ and download apalike.zip. Unzip the file and put
apalike.bst, apalike2.bst in C:\tex\texmf\bibtex\bst\apalike
apalike.doc in C:\tex\texmf\doc\bibtex\apalike
apalike.tex in C:\tex\texmf\tex\plain\apalike
apalike.sty in C:\tex\texmf\tex\latex\apalike
Then, very importantly, Refresh FNDB. LaTeX will now find the package but I suspect you will have problems using it, though I haven't tried it.


It will end in 28 days, 6 hours, 42 minutes and 12 seconds.


inequalities
trigonometry
inequalities unsolved

Let $ x\in \mathbb{R}$ and $ n\in \mathbb{N}^*$,
prove that:
$ |\sum_{k = 1}^{n}{\frac {\sin kx}{k}| \leq 2\sqrt {\pi}}$


How many common fractions with a value between 0 and 1 have 6 as their denominator?

There is $ \frac{1}{6}$, $ \frac{2}{6}$, $ \frac{3}{6}$, $ \frac{4}{6}$, and $ \frac{5}{6}$. But we can simplify $ 3$ of these, leaving $ \boxed{2}$.

The first 5 natural numbers can be the numerators. However, the even numerators can be simplified, leaving us with 2 different fractions.


pigeonhole principle
geometry
combinatorics unsolved
combinatorics

This may sound very familiar, but I've been trying to find the answer like hell. If we have 9 points in a unit square, a simple pigeonhole principle shows us that there is at least one triangle with area 1/8. Now, does the conclusion follow if we have only 8 points? or 7 points? I haven't been able to prove nor disprove it.

Thanx

This is Heilbronn numbers problem. Have a look here :
http://mathworld.wolfram.com/HeilbronnTriangleProblem.html

Pierre.

Wow, nice link. I hadn't heard about this before.

Neither had I. Thank you very much.

By the way, does anyone have a prove of the fact that it's always possible to find a triangle whose area is smaller than or equal to one eigth?


In a round-robin tournament with n teams, the number of games that must be played is $ (n^2\minus{}n)/2$. How many teams are in a round-robin tournament in which 55 games are played?

For this problem, you need to use algebra.

The proper way:

$ \frac {n^2 \minus{} n}{2} \equal{} 55$

$ n^2 \minus{} n \equal{} 110$

$ n^2 \minus{} n \minus{} 110 \equal{} 0$

$ (n \minus{} 11)(n \plus{} 10) \equal{} 0$

$ n \equal{} 11 \text{ or } n \equal{} \minus{} 10$

Since negative teams cannot exist, the answer is $ \boxed{11}$.


[d] is the mid pt. of side [bc] of triangle [abc] &[e] is the mid pt. of [ad]. [be] produced meets [ac] at pt. [m]. prove that [be]+3*[em].   can anyone suggest a short proof fro it ...... :?:
also , can anyone tell me what reinmann conjecture, or other subtitles given in the forum mean?

Riemann Conjecture/the other millenium problems are subtitles corresponding to the number of posts you have - take a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1056489#1056489]this thread[/url].

[hide="Proof"]
Concept of area ratios: If two triangles share the same height then their areas are in the ratio of their bases, or vice versa. This is obvious from the formula $ A \equal{} \frac {1}{2}bh$.

Let $ [\ldots]$ denote area. Then $ \triangle BDE, CDE$ share the same height, have equal bases, so $ [BDE] \equal{} [CDE]$; similiarily $ [BDE] \equal{} [BAE]$ and $ [CDE] \equal{} [CAE]$. Therefore $ [BDE] \equal{} [BAE] \equal{} [CDE] \equal{} [CAE] \equal{} \frac {1}{4}[ABC]$. Also, $ \triangle ABC, AEC$ share the same base, so $ \frac {[AEC]}{[ABC]} \equal{} \frac {EM}{BM} \equal{} \frac {1}{4}$, and $ BE \equal{} 3EM$.
[/hide]

Umm you do know that [abc] stands for the area of triangle abc?
So yeah, I don't think you should get used to using brackets as separators.


Is it possible to mathematically model a universe that has the property of time (changes of position and changes of state) without elevating time to the status of a dimension? For instance, if 3-space were a reconfiguring 3-manifold which was moving through a 4-space, couldn't we describe changes of position and state in relation to the 4-space without requiring a dimension of time? I'm trying to determine if such a model might be able to describe a dynamic and evolving (vice a static and deterministic) universe.

What is time supposed to be if it's not a dimension?


How can you show that if you modify finitely many elements in a Cauchy sequence, the resulting sequence is equivalent to the original one?

thanks!

What do you mean by equivalent ??

Do you mean that $|x_{n}-y_{n}| \to 0$ as $n \to \infty$ ?

[quote="benorin"]Converges to the same value.[/quote]

It only matters what is happening for big enough $n$.

If $a_{n}$ is the sequence and you modified $a_{n_{1}},...,a_{n_{k}}$

such that $n_{k}> n_{k-1}>...>n_{1}$ then since $n \to \infty$ you can choose $n > n_{k}$ .

Sorry if my explanation is messy...  :D

[quote]How can you show that if you modify finitely many elements in a Cauchy sequence, the resulting sequence is equivalent to the original one?
[/quote]

I don't understand what do you mean?


1. Is there a function $ f: \mathbb{R}\to \mathbb{R}$ satisfying $ f(f(x)) \equal{} \minus{} x$?
2. Is there a function $ f: \mathbb{Z}\to \mathbb{Z}$ satisfying $ f(f(x)) \equal{} \minus{} x$?

[quote="MeKnowsNothing"]1. Is there a function $ f: \mathbb{R}\to \mathbb{R}$ satisfying $ f(f(x)) \equal{} \minus{} x$?
1. Is there a function $ f: \mathbb{Z}\to \mathbb{Z}$ satisfying $ f(f(x)) \equal{} \minus{} x$?[/quote]

Yes and yes, infinitely many in both cases :

For 1.
Split $ \mathbb R^ \plus{}$ in two equinumerous sets $ A$ and $ B$ and let $ h(x)$ a bijection from $ A\to B$. Then :

$ f(0) \equal{} 0$
$ \forall x\in A$ : $ f(x) \equal{} h(x)$ and $ f( \minus{} x) \equal{} \minus{} h(x)$
$ \forall x\in B$ : $ f(x) \equal{} \minus{} h^{[ \minus{} 1]}(x)$ and $ f( \minus{} x) \equal{} h^{[ \minus{} 1]}(x)$

For 2.
Split $ \mathbb N$ in two equinumerous sets $ A$ and $ B$ and let $ h(x)$ a bijection from $ A\to B$. Then :

$ f(0) \equal{} 0$
$ \forall x\in A$ : $ f(x) \equal{} h(x)$ and $ f( \minus{} x) \equal{} \minus{} h(x)$
$ \forall x\in B$ : $ f(x) \equal{} \minus{} h^{[ \minus{} 1]}(x)$ and $ f( \minus{} x) \equal{} h^{[ \minus{} 1]}(x)$

Example : $ A \equal{} \{2p,p\in\mathbb N\}$, $ B \equal{} \{2p \minus{} 1,p\in\mathbb N\}$ and $ h(x) \equal{} x \minus{} 1$. So :
$ f(0) \equal{} 0$
$ f(2n) \equal{} 2n \minus{} 1$ $ \forall n > 0$
$ f(2n \minus{} 1) \equal{} \minus{} 2n$ $ \forall n > 0$
$ f( \minus{} 2n) \equal{} \minus{} 2n \plus{} 1$ $ \forall n > 0$
$ f( \minus{} 2n \plus{} 1) \equal{} 2n$ $ \forall n > 0$


Compute \[\frac{1}{m^2} \sum_{k=1}^{m} \sum_{l=1}^{m} \sin\left(\frac{(2k-1)r}{2m}\pi \right) \sin\left(\frac{(2l-1)r}{2m}\pi \right) \cos\left(\frac{(k-l)r}{m}\pi \right),\]
with $r \in \mathbb{N}$ and $r <m$.

Come on Tibo! As a french man, you should have read 'Supermath' exercise Al.54.

The answer is $\frac 1 4$, except for $r=0$ for which it is $0$.

Pierre.

[quote="harazi"]What is "Supermath"? As a quarter of a french man, I am interested to know what is this Bible of french mathematics.  ;) . By the way: I have never seen such an ugly sum. I guess there must be some nice idea behind it.[/quote]

It is one of my books  :D  but I think it would be presumptuous to call it Bible  ;)
It might be some interseting idea but the computations are straightforward (expand, use geometric sums to see that some terms are $0$, that's all).  :(

Pierre.

Upppsss, ok, I understood.  :blush: . I will try to find it... the sooner... the better.  :)

Oh... I have borrowed this book to a friend so...
Can you please give me just a hint Pierre ?

In fact there are several Bibles (French mathematics are a kind of polytheistic religion !) : GERLL & GIRARD (1st one ?), SOULAMI, Pierre's books (Supermath, Hypermath, Mégamath) and AASILA's books.
But "the" Bible is Soulami's book (="the" blue book).

Tss....I think I already gave some big hints above. I don't think I'm wanting to do more for such an heretic.... :P

Pierre.


For this problem, I guessed and checked, and got the right answer.  But how can I do it faster without guess and check?

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5, and 7 exactly once.  Only one of these four-digit numbers is a multiple of another one.  Which of the following is it?

A) 5724              B)7245             C) 7254               D) 7425              E) 7542

[hide]Answer is D[/hide]

[hide="HINT"]
What number must any number using the digits 2,4,5, and 7 have to be divisible by?
[/hide]

Not was I was looking for...

What OTHER number?

Well in some cases its 11, and who keeps rating my posts spam?

The answer I'm looking for is $ 3$...

Units digits of the numbers are 2, 4, 5 and 7, so their doubles will end in 4,
8, 0 and 4, respectively. Choice (A) 5724 ends in 4 but 5724=2 = 2862, not
one of the 24 numbers. Likewise, choice (C) 7254 produces 7254=2 = 3627,
also not one of the numbers. When the units digits are tripled the resulting
units digits are 6, 2, 5 and 1 and choices (B) 7245, (D) 7425 and (E) 7542
are possibilities. Division by 3 yields 2415, 2475 and 2514 respectively. Only
the second of these numbers is one of the 24 given numbers. Choice (D) is
correct.


calculus
integration
trigonometry
limit
calculus computations

Let $ F(t)\equal{}\frac{1}{t}\int_0^{\frac{\pi}{2}t}|\cos 2x|\ dx$ for $ 0<t\leq 1.$

(1) Find $ \lim_{t\rightarrow 0} F(t).$

(2) Find the range of $ t$ such that $ F(t)\geq 1.$

I think $ \int \cos{2x} dx \equal{} \frac12 \sin{2x} \plus{} C$

so $ f(t) \equal{} \frac {1}{2t} \sin{\pi*t}$

the $ \lim_{t\to0} f(t) \equal{} \frac {\pi}{2}$

in interval $ 0<t\leq1$ we know that $ \sin{\pi t}\leq1$
so we must have the denominator less or same as $ 1$
$ 2t\leq1\longrightarrow t\leq\frac12$

the solution of $ t$ may be $ 0<t\leq \frac12$


inequalities
function
inequalities proposed

Positive numbers $ a$, $ b$, $ c$ are such that $ \left(a \plus{} b\right)\left(b \plus{} c\right)\left(c \plus{} a\right) \equal{} 8$. Prove the inequality
\[ \frac {a \plus{} b \plus{} c}{3}\ge\sqrt [27]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]

If we put $ x\equal{}a\plus{}b\plus{}c, y\equal{}ab\plus{}bc\plus{}ac, z\equal{}abc$, we have: $ (1) xy\minus{}z\equal{}8$, and we need to prove that:
$ \frac{x}{3} \ge (\frac{x(x^{2}\minus{}3y)\plus{}3z}{3})^{\frac{1}{27}}$
If we plug (1), we need to prove that: $ \frac{x}{3} \ge (\frac{x^{3}\minus{}24}{3})^{\frac{1}{27}}$, or:
$ f(x)\equal{}x^{27}\minus{}3^{26}x^{3}\plus{}24*3^{26} \ge 0$. We have $ f(3)\equal{}0$, and: $ f'\equal{}27x^{26}\minus{}3^{27}x^{2}\equal{}27x^{2}(x^{24}\minus{}3^{24})$, so f is increasing on $ [3, \infty)$, hence non-negative for $ x \ge 3$. But $ x \ge 3$:
$ 8\equal{}\prod (a\plus{}b) \le (\frac{a\plus{}b\plus{}b\plus{}c\plus{}c\plus{}a}{3})^{3}\equal{}(\frac{2x}{3})^3 \implies x \ge 3$, and we're done. Equality occurs when $ x\equal{}3$, or: $ a\equal{}b\equal{}c\equal{}1$

[quote="April"]Positive numbers $ a$, $ b$, $ c$ are such that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} 8$. Prove the inequality
\[ \frac {a \plus{} b \plus{} c}{3}\ge\sqrt [27]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}
\]
[/quote]
It's my and Virgil Nicula's inequality! :D
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106225
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106260

$ \frac {a \plus{} b \plus{} c}{3}\geq \sqrt [27]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}\Longleftrightarrow (a \plus{} b \plus{} c)^{27}\geq 3^{26}(a^3 \plus{} b^3 \plus{} c^3)\ \cdots [*]$.
Let $ a \plus{} b \equal{} 2x,\ b \plus{} c \equal{} 2y,\ c \plus{} a \equal{} 2z$, we have that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} 8\Longleftrightarrow xyz \equal{} 1$ and $ 2(a \plus{} b \plus{} c) \equal{} 2(x \plus{} y \plus{} z)\Longleftrightarrow a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$.$ (a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a)\Longleftrightarrow a^3 \plus{} b^3 \plus{} c^3 \equal{} (x \plus{} y \plus{} z)^3 \minus{} 24$. $ \therefore [*]\Longleftrightarrow (x \plus{} y \plus{} z)^{27}\geq 3^{26}\{(x \plus{} y \plus{} z)^3 \minus{} 24\}$.
Let $ t \equal{} (x \plus{} y \plus{} z)^3$, by AM-GM, we have that $ x \plus{} y \plus{} z\geq 3\sqrt [3]{xyz}\Longleftrightarrow x \plus{} y \plus{} z\geq 3$, yielding $ t\geq 27$.
Since $ y \equal{} t^9$ is an increasing and concave up function for $ t > 0$, the tangent line of $ y \equal{} t^9$ at $ t \equal{} 3$ is $ y \equal{} 3^{26}(t \minus{} 27) \plus{} 3^{27}$.We can obtain $ t^9\geq 3^{26}(t \minus{} 27) \plus{} 3^{27}$, yielding $ t^9\geq 3^{26}(t \minus{} 24)$, which completes the proof.