logarithms
trigonometry
calculus
calculus computations

Answer the following question.

[1] $\frac{d}{dx} \frac{x}{\sqrt{x^2+x+1}}$

[2] $\frac{d}{dx} \ln (x+\sqrt{x^2+1})$

[3] $\frac{d}{dx} \sin (1+\ln x)$

[4] Let $x=\frac{e^{3t}}{1+t^2},y=\frac{t}{1+t^2}$, find $\frac{dy}{dx}$

[5] Sinsyu University Faculty of Textile Sience and technology

(1) Let $x^2-y^2=a^2$, find $\frac{d^2y}{dx^2}$

(2) Let $y=\ln (1+\cos x)^2$, find the value of $\frac{d^2y}{dx^2}+2e^{-\frac{y}{2}}$

hello, to the first we have by the chaine  and quotient rule
$ \frac{d}{dx}\frac{x}{\sqrt{x^2\plus{}x\plus{}1}}\equal{}\\
\frac{\sqrt{x^2\plus{}x\plus{}1}\minus{}x\frac{1}{2}(x^2\plus{}x\plus{}1)^{\frac{1}{2}}(2x\plus{}1)}{x^2\plus{}x\plus{}1}\equal{}\\
\frac{x^2\plus{}x\plus{}1\minus{}x\frac{1}{2}(2x\plus{}1)}{(x^2\plus{}x\plus{}1)\sqrt{x^2\plus{}x\plus{}1}}\equal{}\\
\frac{x\plus{}2}{2(x^2\plus{}x\plus{}1)^{\frac{3}{2}}}$
Sonnhard.

[quote="kunny"]Answer the following question.

[1] $ \frac {d}{dx} \frac {x}{\sqrt {x^2 \plus{} x \plus{} 1}}$

[2] $ \frac {d}{dx} \ln (x \plus{} \sqrt {x^2 \plus{} 1})$

[3] $ \frac {d}{dx} \sin (1 \plus{} \ln x)$

[4] Let $ x \equal{} \frac {e^{3t}}{1 \plus{} t^2},y \equal{} \frac {t}{1 \plus{} t^2}$, find $ \frac {dy}{dx}$

[5] Sinsyu University Faculty of Textile Sience and technology

(1) Let $ x^2 \minus{} y^2 \equal{} a^2$, find $ \frac {d^2y}{dx^2}$

(2) Let $ y \equal{} \ln (1 \plus{} \cos x)^2$, find the value of $ \frac {d^2y}{dx^2} \plus{} 2e^{ \minus{} \frac {y}{2}}$[/quote]

[2] $ \frac {d}{dx} \ln (x \plus{} \sqrt {x^2 \plus{} 1})\equal{} \frac{1}{x\plus{}\sqrt{x^2\plus{}1}} .(1\plus{}\frac{x}{\sqrt{x^2\plus{}1}})$
3/  $ cos(1\plus{}lnx) .\frac{1}{x}$

[quote="kunny"][2] You can express your answer in simple form.[/quote]
yes,  $ (2)\equal{} \frac{1}{\sqrt{x^2\plus{}1}}$

can I do like this::?
[4] $ \frac{dy}{dx}\equal{}\frac{dy}{dt}\cdot\frac{dt}{dx}$

where $ \frac{dy}{dt}\equal{}\frac{(1\plus{}t^2)\minus{}2t^2}{(1\plus{}t^2)^2}$
and $ \frac{dx}{dt}\equal{}\frac{3e^{3t}(1\plus{}t^2)\minus{}2e^{3t}t}{(1\plus{}t^2)^2}\rightarrow\frac{dt}{dx}\equal{}\frac{(1\plus{}t^2)^2}{3e^{3t}(1\plus{}t^2)\minus{}2e^{3t}t}$
so $ \frac{dy}{dx}\equal{}\frac{dy}{dt}\cdot\frac{dt}{dx}\equal{}\frac{(1\plus{}t^2)\minus{}2t^2}{(1\plus{}t^2)^2}\cdot\frac{(1\plus{}t^2)^2}{3e^{3t}(1\plus{}t^2)\minus{}2e^{3t}t}\equal{}\frac{1\minus{}t^2}{e^{3t}(3t^2\minus{}2t\plus{}3)}$

can I also make like this?
[5]
(1) $ x^2 \minus{} y^2 \equal{} a^2$

$ 2x \minus{} 2y\frac {dy}{dx} \equal{} 2a\frac {da}{dx}\rightarrow\frac {dy}{dx} \equal{} \frac {x}{y} \minus{} \frac {a}{y}\frac {da}{dx}$

$ 2 \minus{} 2(\frac {dy}{dx})^2 \minus{} 2y\frac {d^2y}{dx^2} \equal{} 2\frac {da}{dx} \plus{} 2a\frac {d^2a}{dx^2}$

$ 1 \minus{} (\frac {x^2}{y^2} \plus{} \frac {a^2}{y^2}(\frac {da}{dx})^2 \minus{} \frac {2xa}{y^2}\frac {da}{dx}) \minus{} y\frac {d^2y}{dx^2} \equal{} \frac {da}{dx} \plus{} a\frac {d^2a}{dx^2}$

$ y\frac {d^2y}{dx^2} \equal{} 1 \minus{} \frac {x^2}{y^2} \minus{} \frac {a^2}{y^2}(\frac {da}{dx})^2 \plus{} \frac {2xa}{y^2}\frac {da}{dx} \minus{} \frac {da}{dx} \minus{} a\frac {d^2a}{dx^2}$

$ \frac {d^2y}{dx^2} \equal{} \frac {1}{y} \minus{} \frac {x^2}{y^3} \minus{} \frac {a^2}{y^3}(\frac {da}{dx})^2 \plus{} \frac {2xa}{y^3}\frac {da}{dx} \minus{} \frac {1}{y}\frac {da}{dx} \minus{} \frac {a}{y}\frac {d^2a}{dx^2}$

(2)$ y \equal{} \ln(1 \plus{} \cos x)^2 \equal{} 2\ln(1 \plus{} \cos x)$
$ y' \equal{} \minus{} 2(\frac {\sin x}{1 \plus{} \cos x}$
$ y'' \equal{} \minus{} 2(\frac {\cos x(1 \plus{} \cos x) \plus{} \sin^2x}{(1 \plus{} \cos x)^2} \equal{} \minus{} 2(\frac {1}{1 \plus{} \cos x})$

$ 2e^{ \minus{} \frac {y}{2}} \equal{} 2\frac {1}{e^{\frac {y}{2}}} \equal{} 2(\frac {1}{1 \plus{} \cos x})$
so $ y'' \plus{} 2e^{ \minus{} \frac {y}{2}} \equal{} 0$

sorry if I made a mistake..
thx

hello, to [5] 1), here i have $ 2x - 2yy^' = 0 \Leftrightarrow y^' = \frac {x}{y}$.
By differentiating again we get $ y^{''} = \frac {y - xy^'}{y^2} = \frac {y^2 - x^2}{y^3}$. By $ y^2 - x^2 = - a^2$ we get $ y^{''} = -\frac {a^2}{y^3}$.
Sonnhard.

are you sure $ a$ doesn't contain $ x$
what if $ a(x)$?? :maybe:

hello, i have assumed that $ a$ is a constant.
Sonnhard.


I think Iran 1996 might be useful... Have you got a proof of your inequality?

I found it on polish mathematical forum (probably You know limes123 matematyka.pl) I can't prove this.

[quote="matex (L)(L)(L)"]$ a,b,c > 0$

$ \frac {1}{6a^2 \plus{} bc} \plus{} \frac {1}{6b^2 \plus{} ac} \plus{} \frac {1}{6c^2 \plus{} ab}\ge\frac {9}{7}\frac {1}{ab \plus{} bc \plus{} ca}$[/quote]
It is a privet case of the following Vasc's inequality.
Let $ a,$ $ b$ and $ c$ are positive numbers and $ r\geq3\plus{}\sqrt7.$ Prove that
\[ \frac{1}{ra^2\plus{}bc}\plus{}\frac{1}{rb^2\plus{}ac}\plus{}\frac{1}{rc^2\plus{}ab}\geq\frac{9}{(r\plus{}1)(ab\plus{}ac\plus{}bc)}\]

[quote="matex (L)(L)(L)"]Where I can find prove of this generalization?[/quote]
Here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100399


geometry
3D geometry
modular arithmetic
floor function
ceiling function

If a and b are natural numbers and $ 5a^3\equal{}6b^4$, what is the smallest possible value of $ a \plus{} b$?

[hide="My Solution"]
Rearranging the equation in terms of $ a$, we have $ a^3 \equal{} \frac{6b^4}{5}$.

We have $ 6 \equal{} 2\cdot 3$ and the numerator is being divided by $ 5$ so we can conclude that $ b \equal{} 2^h\cdot 3^i\cdot 5^j$. Substituting that, we have

$ a^3 \equal{} \frac{6(2^h\cdot 3^i\cdot 5^j)^4}{5} \equal{} 2^{4h\plus{}1}\cdot 3^{4i \plus{} 1}\cdot 5^{4j \minus{} 1}$.

In order to minimize $ a$, we must minimize $ a^3$ and doing that will minimize $ b^4$. Because $ a$ is a perfect cube, we must have the following:

$ 4h \plus{} 1 \equiv 0 \pmod{3} \implies h \equiv 2 \pmod{3}\\
4i \plus{} 1 \equiv 0 \pmod{3} \implies i \equiv 2 \pmod{3}\\
4j \minus{} 1 \equiv 0 \pmod{3} \implies j \equiv 1 \pmod{3}$

In order to minimize $ a$, we must minimize the powers of $ 2$, $ 3$, and $ 5$. In order to do that, we must have $ h \equal{} i \equal{} 2$ and $ j \equal{} 1$. So we have:

$ 5a^3 \equal{} 6(2^2\cdot 3^2\cdot 5)^4 \implies a \equal{} 2^3\cdot 3^3\cdot 5 \equal{} 1080$

It follows that $ b \equal{}  180$. Therefore, $ a \plus{} b \equal{} 1080 \plus{} 180 \equal{} \boxed{1260}$.
[/hide]

alternatively, proceed using divisibility, through repeated application of the lemma:

if $ p|a^n$, where $ p$ is prime, then $ p|a$.

[hide="Proof"] Start from euclid's lemma ($ p|ab$, where $ p$ is prime, implies $ p|a$ or $ p|b$).  The statement results from repeated application of this lemma.  Let $ b\equal{}a^{n\minus{}1}$.  Either $ p|a$ or $ p|a^{n\minus{}1}$.  But $ p|a^{n\minus{}1}$ implies $ p|a$ or $ p|a^{n\minus{}2}$, and continuing in this way we find that all cases lead to $ p|a$.
[/hide]

Note that $ p^m|a^n$ only if $ p|a^n$.  In fact the more general result that $ p^m | a^n$ implies $ p^{\lfloor m/n \rfloor} | a$.
[hide="Proof"]This can be shown by noting $ p^m|a^n \implies p|a^n \implies p|a$.  Now let $ a\equal{}bp$; we have $ p^m|b^np^n$, hence $ p^{m\minus{}n}|b^n$ if $ m \ge n$.  Hence we can apply this result up to as many times as $ \lceil m/n \rceil$ since this is the number of times you have to subtract $ n$ from $ m$ to get a nonpositive which will result in the $ m < n$ case.
[/hide]

Last comment: for the general composite numbers, we can put together the result for each prime factors to get $ p_1^{e_1}p_2^{e_2} \cdots | a^n$ implies $ p_1^{\lceil e_1/n \rceil} p_2^{\lceil e_2/n \rceil} \cdots | a$.

[hide="Solution"]
$ 5|6b^4$ implies $ 5|b$ so let $ b\equal{}5b'$
$ 6|5a^3$ implies $ 6|a$ (since $ 6$ is squarefree) so let $ a\equal{}6a'$.

Plugging back in
$ 5(6a')^3 \equal{} 6(5b')^4$
$ 6^2 a'^3 \equal{} 5^3 b'^4$

Now $ 5^3|36a'^3$ implies $ 5|a'$.  Let $ a'\equal{}5a''$
$ 6^2|125b'^4$ implies $ 6|b'$.  Let b'=$ 6b''$.

Plugging in again
$ 6^2 5^3 a''^3 \equal{} 5^3 6^4 b''^4$
$ a''^3 \equal{} 6^2 b''^4$

Now $ 6^2|a''^3$ implies $ 6|a''$.  Let $ a''\equal{}6a'''$
Substituting,
$ 6^3 a'''^3 \equal{} 6^2 b''^4$
$ 6a'''^3 \equal{} b''^4$

$ 6|b''^4$ implies $ 6|b''$ so let $ b''\equal{}6b'''$.  Substituting again
$ 6a'''^3 \equal{} 6^4 b'''^4$
$ a'''^3 \equal{} 6^3 b'''^4$

$ 6^3|a'''^3$ implies $ 6|a'''$.  Now letting $ a'''\equal{}6a''''$ we have
$ a''''^3 \equal{} b'''^4$

The smallest positive integer that is both a cube and a 4th power is $ 1$ (and the sequence of cube 4th powers are given by $ a_n\equal{}n^{12}$), so our minimum is achieved at $ a''''\equal{}b'''\equal{}1$.  Working backwards:
$ a\equal{}6a'\equal{}6(5a'')\equal{}30(6a''')\equal{}180(6a'''') \equal{} 1080$
$ b\equal{}5b'\equal{}5(6b'') \equal{} 30(6b''') \equal{} 180$

Thus the minimum $ a\plus{}b$ is $ 180\plus{}1080 \equal{} 1260$.[/hide]


Hey Everyone,

I used to see Alcumus, but I didn't pay much attention to it. But now I want to do good so as to get in all-time high scores (inspiration by nikeballa96). Everyone who's really smart already have the position. I'm terrible in FTW so I atleast want to get the high scores in this. Currently, my goal is to get in Daily Top Ten everyday.

Is this enough or should I set like a certain number of points and karma a day? Can someone in the list give me some advice? I'm a newbie so I need a lot of help.

It really depends on how much time you are willing to spend. I used to do a 100 points a day, but after you complete all the problems, you get the ones you got wrong the first time again for only one point. I f you want to be on Top 10 daily, you should aim for about 45 points and 400 Karma points.  Also, remember to rate problems!

That shouldn't be the point. Who really cares how high in the rankings you are? Do it becuase you have fun, want to get better, and love problems! :)

Haha, yeah...
I've only gotten into daily top 10 like 2-3 times and that was when I actually did alcumus for a while, not like the 30-45 minutes I usually do.


have you been there before?

how many advance?

i suppose (i can't think of his name) but the guy who led the team from ny to nationals could help you :)

-jorian

my chapter in 14
my state in 42

[quote="kyyuanmathcount"]Just a reminder to all, once your chapter competition is done, DO NOT discuss the competition.  Many chapters don't finish until March.

I guess as advice, be careful and don't be nervous.  If the competition is hard, it is hard for everyone.  If it's easy, just be really careful, because then, the person w/ least dumb mistakes wins.[/quote]

The person with no mistakes also wins. :roll:

I hate school because that is where I make the mistakes, chapter where I make mistakes.

bah, smartness is just a delusion, one that people are under when they make it to national mathcounts.

Yeah, chapter cd is fun. Luckilly, my chapter isn't great, b/c it's where I make all of my mistakes. I got 43 last year... :blush: I get really nervous there, though. at states, I'm almost TOO confident. I'd rather not discuss my composure at nats...

[quote="bpms"][quote="1=2"][quote="egghead91"]Oh my goodness, my region's competition is on Feb. 3rd. Exactly two weeks. :( It'll be my first, can anyone from the Rochester, NY area tell me what it's like?[/quote]interesting... My chapter competition is on the same day...

@egghead: Where are you taking it? I'm taking it at Niskayuna...

Well, my first year was last year, so I only know what an easy competition is like.

You get there, I get a little shaky, maybe I'm a sugar low guy...

The sprint round is a little boring, the target round is a little better, but the team round is REALLY fun! Also, if you get into the countdown round, it should be really fun.[/quote]
Really, I always thought the opposite. THe sprint round is a otn of fun (the last ten problems are actually good ones), the target is boring (the county target problems have always been much easier than they should, at least in my opinion), and the team round is boring because the questions are too easy considering that you have four people to work on it, the problems are also way too varied in difficulty (four of the first five are usually easy enough to be in the first 10 of the sprint, with one of them being hard enough for 11-20. Then 3 of the last five could be easy enough to be 1-15 and two of them could be 16-25 difficulty wise).[/quote]

Same here, I always thought that sprint was the most fun because of the time pressure (although there isn't much at chapters.) Chapter target is boring because you finish really early, because usually the problems are pretty easy (chapter targets could be more fun if a calculator was prohibited). Team is okay..... but this is my opinion

[quote="laughinghead505"][quote="jhredsox"]haha, 7 teams from my chapter advance, 12 individuals :)

-jorian

and i'm 2nd on my team (behind guess who)

oh, i would like to bring up that dan's actually a lot smarter than i thought[/quote]

DUDE, are you serious?!?!? Two teams from our chapter advance and I don't remember how many individuals, but definitely less than ten. Maybe three or four.

I went the last two years and did really bad. (I got in the top 30 and got scores of around 30-ish :oops: ) I plan on doing a lot better this year (hopefully).

I personally think team rounds are pretty fun, but there's really no one that smart at my school so they just leave all the problems that look hard to me. :roll: This year though, we might actually have a chance because last year we got third (errrr!!! So close!!) and this year we have a new kid whose pretty smart. :lol:[/quote]
Last year a 32 would've gotten you in countdown in my county.

Whoa.  our cutoff, in redneck country, was a 38 or 39.

and on the number of people sent to state, we send 2 teams and 4 individuals.  but it really turns out to be like 2 teams and 1 individual because normally at least 3 of the top 4 individuals were on one of the two top teams.

[quote="laughinghead505"][quote="jhredsox"]haha, 7 teams from my chapter advance, 12 individuals :)

-jorian

and i'm 2nd on my team (behind guess who)

oh, i would like to bring up that dan's actually a lot smarter than i thought[/quote]

DUDE, are you serious?!?!? Two teams from our chapter advance and I don't remember how many individuals, but definitely less than ten. Maybe three or four.

I went the last two years and did really bad. (I got in the top 30 and got scores of around 30-ish :oops: ) I plan on doing a lot better this year (hopefully).

I personally think team rounds are pretty fun, but there's really no one that smart at my school so they just leave all the problems that look hard to me. :roll: This year though, we might actually have a chance because last year we got third (errrr!!! So close!!) and this year we have a new kid whose pretty smart. :lol:[/quote]

Yeah 7 teams is a lot.... it might be because he lives in a small state.

Well, my opinion of the sprint being boring was from the 2006 chapter.(that's the only one I've been to)

I believe you are allowed to say things about rankings, team or individual, just not anything about scores, problems, solutions, difficulties, etc.

[quote="perfect628"]u got second, and think ur an idiot...ur NOT!!![/quote]

Can a person who got seventh think he is an idiot? :fool:

yeah. in m ine, though, yes. i can't wait 2 c who that is...

[quote="star99"]second place, ubemaya.  oh my god, what a tragedy.[/quote]
It might be a tragedy for ubemaya.

not according to his goals 4 this year(in his sig)

Okay I changed my goals. Anyway, it's not the place that sucks, it's the score that I got.

[quote="jhredsox"]uhh, just so a mod/admin doesn't see that, ur not supposed to say that ur score (sucked/was great/was average/etc.)

-jorian[/quote]
But who knows what ubemaya considers to suck?
I HAVEN'T DONE CHAPTERS YET SO THE BELOW IS ALL SPECULATION
Let's say the chapter was hard. I got a 45. I am content. I got a 44. I am content. I got a 43, I am unhappy. I got a 42, I am sad.

Let's say it was easy. I got a 46. I am happy. I got a 45 I am sad.

Big difference.

Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132111]my announcement[/url] on what is and is not allowed to be discussed.

In a nutshell: you can say your ranking and your school's ranking.  Nothing else.  [b]No scores.[/b]  This includes whether you thought you did better or worse than your were expecting.

We'll tell you when you can talk freely, but it won't be until after all the chapters are completed in late February.

T-T. Saturday: I took a Tylenol for my flu and went off to MATHCOUNTS. I'm always sick before/after math competitions / piano recitals. EH, I had to leave at like 1 something. Did not do well at ALL. And I'm still sick, missing a field trip tomorrow... *stabs the flu* :mad:  :mad:  :mad:  :mad:

Also, there are like 30 schools in the SCV chapter so if we sent like 7 teams... :/

i wish they sent seven....they only sent 5...

I thought SCV was like the biggest chapter or something. I could just be making things up, but... SCV is so competitive!


I don't know if this has been asked before (I didn't find it).

The MiKTex download site does not include Ubuntu as one of the compatible systems. Is there another software (comparable to MiKTeX and TeXnic center) for Ubuntu?

[code]sudo apt-get install tetex-base tetex-bin tetex-extra[/code]

also http://ubuntuforums.org/showpost.php?p=26528&postcount=3

teTeX is now obsolete and has been superceded by TeX Live; also Kile uses KDE which may not be suitable if sticking to Gnome.
I would look in the package manager to see if Tex Live and Texmaker (or Kile if you want to use it) are there but, if not, then there are instructions at http://ubuntuforums.org/showthread.php?t=131507.

TeX Live can be found [url=http://www.tug.org/texlive/]here[/url] with Ubuntu files [url=http://packages.ubuntu.com/search?keywords=texlive]here[/url].

Wait wha?

My server still uses tetex. What's wrong with it/

There's nothing wrong with it and you can continue to use it for as long as you wish. But it's no longer maintained so will never be updated. See the De-support notice at [url=http://www.tug.org/tetex/]The teTeX Homepage[/url], the recommendation to use TeX Live and CTAN's announcement at http://www.mail-archive.com/ctan-ann@dante.de/msg01091.html

I just stumbled across this older post while reading through the forum. I've been using LaTeX on Ubuntu Linux since some years, perhaps I could give some advice on it, if you still have questions.
I'm using Kile on GNOME, without any problems, it just installed some needed KDE parts, and I'm very satisfied with it. Further I'm using TeX Live mentioned above.

Concerning the first question regarding MiKTeX: the MiKTeX package manager has been ported to Unix and can be used on Ubuntu Linux too. I've used it on Ubuntu 7.04, 7.10 and the current 8.04, the mpm has been very useful for me. [url=http://texblog.net/latex-archive/linux/mpm-miktex-package-manager/]Here[/url] I've described the installation of the MiKTeX package manager on Ubuntu 8.04 with the needed packages and commands.

Just to mention, TeX Live 2008 has been [url=http://texblog.net/latex-archive/latex-general/texlive-2008/]released[/url], and it comes together with it's own package manager [url=http://tug.org/texlive/doc/texlive-en/texlive-en.html#x1-410006]tlmgr[/url].

Stefan


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? \\ \\
$ \text{(A)}\ 5724 \qquad \text{(B)}\ 7245 \qquad \text{(C)}\ 7254 \qquad \text{(D)}\ 7425 \qquad \text{(E)}\ 7542$ \\ \\
Select the most appropriate letter.

Consider the multiples of some smaller numbers that are made up of $ 2$, $ 4$, $ 5$ and $ 7$.

$ 2457$: $ 4914$, $ 7371$, ...

$ 2475$: $ 4950$, $ 7425$, ...

We notice that $ 7425$ is also made up of the digits $ 2$, $ 4$, $ 5$ and $ 7$, so the answer is $ \boxed{D}$


trigonometry
limit
function
ratio
calculus
derivative
calculus computations

Find:

$ \lim_{x\to\infty}\frac{x \minus{} \cos{x}}{x}$

intuitively i think since,

$ \frac{x \minus{} \cos{x}}{x} \equal{} 1 \minus{} \frac{\cos{x}}{x}$

and as x goes to infinity, $ \cos{x}$ will keep oscillating between $ \minus{}1$ and $ 1$, but, since $ x$ gets much bigger the second term goes to zero and limit reaches $ 1$. This [u]is[/u] the answer at the back of the book. Is there any other better way of solving this perhaps using l'Hospital or simple limit techniques? Also, when $ x\to\infty$ what would one do in general with a trig limit since trig functions don't have any limit at infinity?

First of all, your intuitive sense is fine. You don't need any more techniques than that.

The issue I hammer on repeatedly with my class is that you have to understand size. You have to know what grows faster than what else. Looking at this, $ x\to \infty,$ and in fact $ x$ is a pretty simple size all by itself. Now, how big is $ \cos x?$ I usually state that as "one, only wigglier." That is, the observation that $ |\cos x|\le 1$ is the only thing you need to know. When you add (or subtract) two things of different size, the larger one dominates and the smaller one becomes inconsequential, so in $ x\minus{}\cos x,$ it's the $ x$ that really matters (unless you do something that in effect subtracts that $ x$ back off).

What about L'Hôpital? This problem breaks L'Hôpital. To be more specific, the statement of L'Hôpital is this: if the original limit is a $ \frac00$ or $ \frac{\infty}{\infty}$ indeterminate case, and [b]if the ratio of the derivatives has a limit[/b], then the original limit is the limit of the ratio of the derivatives. But what if the ratio of the derivatives doesn't exist? In this case it doesn't - but the original limit exists anyway.

I have no doubt that the textbook author intentionally planted this problem to demonstrate that particular shortcoming of L'Hôpital.

I also spend a lot of effort in my class trying both to talk students out of using L'Hôpital for everything and also talking them out of trying to make everything a simple equality (which is what I think you're uncomfortable with, since you can't write $ \lim_{x\to\infty}\cos x$ as anything meaningful). The understanding of size is a big part of that.

[quote]which is what I think you're uncomfortable with, since you can't write $ \lim_{x\to\infty}\cos x$ as anything meaningful[/quote]

Yes this was the main reason I asked the question. Thanks a lot for the detailed explanation.

hello, it is
$ \lim_{x \to \infty}\frac{x\minus{}\cos(x)}{x}\equal{}1$
Sonnhard.


The roots of $ (x^2\minus{}3x\plus{}2)(x)(x\minus{}4)\equal{}0$ are:

$\textbf{(A)}\ 4\qquad
\textbf{(B)}\ 0\text{ and }4 \qquad
\textbf{(C)}\ 1\text{ and }2 \qquad
\textbf{(D)}\ 0,1,2\text{ and }4\qquad
\textbf{(E)}\ 1,2\text{ and }4$

[quote="Smartguy"]The roots of $ (x^2 \minus{} 3x \plus{} 2)(x)(x \minus{} 4) \equal{} 0$ are:
(A) 4
(B) 0 and 4
(C) 1 and 2
(D) 0, 1, 2, and 4
(E) 1, 2, and 4[/quote]

Factor $ x^2\minus{}3x\plus{}2$ to get $ (x\minus{}1)(x\minus{}2)$ so its D

You can also use process of elimination. Since x is one of the terms, then one of the roots is 0 and so that eliminates all answers except B and D. Clearly, substituting one into the quadratic results in another 0. Therefore, D.


Let $x, y , z$ three nonnegative real numbers. Prove that \[1+\frac{xyz}{(x+y)(y+z)(z+x)}\geq\frac{27xyz(x+y+z)}{8(xy+yz+zx)^{2}}.\]

$1+\frac{xyz}{(x+y)(y+z)(z+x}=\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$
So our inequality rewrites as:
$8(xy+yz+zx)^{3}\geq 27xyz(x+y)(y+z)(z+x)=27(zx+yz)(xy+zx)(yz+xy)$
Let $xy=a, yz=b, zx=c$. We have to prove that:
$8(a+b+c)^{3}\geq 27(a+b)(b+c)(c+a)$
But this is just AM-GM for the numbers $a+b, b+c, c+a$ after cubing.


superior algebra
superior algebra unsolved

Let $ R$ be a ring, $ l,n\in\mathbb{N}$ and $ \omega\in R$ is a primitive nth root of unity.
Prove that for any $ 1<l<n$, $ \omega^{l}\minus{}1$ is not a zero divisor in $ R$.

 :P note:
$ \omega$ is an nth root of unity if $ \omega^{n}\equal{}1$ and $ \omega$ is primitive if $ \omega^{n/t}\minus{}1$ is not a zero divisor for any prime divisor $ t$ of $ n$.

Hmm, interesting problem~~

I will take careful thinking.

PS: Could you tell me which book it is and the author?

I've come up with a proof, but I'm not sure whether my solution is right.
Consider $ (l,n) \equal{} d$, assume $ l_{0}\equal{}\cfrac{l}{d}$, so exists $ \alpha,\beta\in\mathbb{Z}$ such that $ \alpha l_{0}\plus{}\beta\cfrac{n}{d}\equal{} 1$
if there exists an $ l$ that $ \omega^{l}\minus{}1$ is a a zero divisor in $ R$, we can assume $ \omega^{l}\minus{}1$ is a left zero divisor.

So there exists $ r\in R(r\neq 0)$ satisfies: $ (\omega^{l}\minus{}1)r \equal{} 0$.
So $ (\omega^{(\alpha\minus{}1)l}\plus{}\omega^{(\alpha\minus{}2)l}\plus{}\cdots\plus{}\omega^{l}\plus{}1)((\omega^{l}\minus{}1))r \equal{} 0$
$ \Longleftrightarrow (\omega^{\alpha l}\minus{}1)r \equal{} 0$
$ \Longleftrightarrow (\omega^{\alpha l_{0}d}\minus{}1)r \equal{} 0$
$ \Longleftrightarrow (\omega^{d}(\omega^{n\beta})^{\minus{}1}\minus{}1)r \equal{} 0$
$ \Longleftrightarrow (\omega^{d}\minus{}1)r \equal{} 0$
So $ \omega^{d}\minus{}1$ is a left zero divisor.

But for $ \forall d < n,d|n$, there must exist a prime $ t_{0}$ such that $ d|\cfrac{n}{t_{0}}$.
So we have
$ (\omega^{(\frac{n}{t_{0}d}\minus{}1)d}\plus{}\omega^{(\frac{n}{t_{0}d}\minus{}2)d}\plus{}\cdots\plus{}\omega^{d}\plus{}1)(\omega^{d}\minus{}1)r \equal{} 0$
$ \Longleftrightarrow (\omega^{\frac{n}{t_{0}}}\minus{}1)r \equal{} 0$
So $ \omega^{\frac{n}{t_{0}}}\minus{}1$ is a left zero divisor, contraction!

For $ \omega^{l}\minus{}1$ is a right zero divisor, the proof is similar.

Forgive my boring Chinese English.
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

I should thank Zhaobin for pointing out a serious mistake in my post, now I've corrected my proof.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I'm hoping someone can help me out with the following question.

In this problem you may use without proof the fact that the alternating groups $A_n$ are simple for $n \ge 5$.

(a) Use Sylow's theorems to show that the symmetric group $S_5$ can be made to act transitively on a set of size six.

(b) Show that $S_6$ contains at least two conjugacy classes of subgroups that are isomorphic to $S_5$.

(c) If $X$ and $Y$ are sets, show that a bijection between them induces an isomorphism of the symmetric group of $X$ with the symmetric group of $Y$.

(d) Show that there is a non-trivial outer automorphism of $S_6$. [Recall that an outer automorphism is an automorphism that is not the conjugation map of an element.]

Just for reference, I know how to do parts (a) and (c), but do not quite see how to do (b) and (d). Part (c) is a very standard result, so feel free to skip its solution.

Thanks!

Part (b) is a simple application of part (a). One conjugacy class is the obvious set of subgroups with a fixed element. We want to construct another subgroup isomorphic  to $S_5$, which fixes no elements. To do this, take the group and action from part (a), considered as permutations of the set they act on. Since it is transitive, it does not belong to the conjugacy class we already know about.

We need to show that no element of $S_5$ acts trivially; this should be included in part (a).

Part (d) is less obvious.

Here's where it's been discussed before (jmerry solved part (d) there):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphism&t=16615[/url]

Thanks guys for sharing those details. I was actually aware of the ideas discussed in the post by jmerry et al, wherein one maps transpositions to disjoint triples of transpositions. In fact that method is outlined in a couple of Dummit and Foote exercises.

I wonder, however, if for the specific setup in my problem, there is a solution for (d) that doesn't require this idea but rather builds upon parts (a) - (c), as is what one might intuitively expect.

I'd appreciate if anyone makes further progress on this. Thanks!


trigonometry
inequalities unsolved
inequalities

[quote="Harry Potter"]$ m,n,p$ are 3 positive real numbers such that : $ mn \plus{} np \plus{} mp \plus{} mnp \equal{} 1$ . Show that : $ \sqrt {mn} \plus{} \sqrt {np} \plus{} \sqrt {mp}\leq \frac {17}{12} \plus{} mnp$[/quote]
See similar here.
http://www.mathlinks.ro/viewtopic.php?t=236977

[quote="Harry Potter"]My Problem is Very Harder Than The Problem in your Links :D
Thank You[/quote]
Maybe because I haven't tried it yet.  :P

Are you sure it is correct?  :wink:  In fact, if we call
\[ f(a,b,c)\equal{}17/12\plus{}abc\minus{}\sqrt{ab}\minus{}\sqrt{bc}\minus{}\sqrt{ca}\]
Then I will show that $ f(a,b,c)<0$  :D
We try:
\[ a\equal{}b\equal{}\frac{\sqrt{3}\minus{}1}{2}\equal{}x,c\equal{}1\]
Notice that it such that: $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}1$
\[ f(x,x,1)\equal{}\minus{}0.025...<0\]

Now, I think we don't have anything to do  :P

I would make a trigonometric substitution...$ \cos{\alpha} \equal{} \sqrt {mn}$ or something else...

Positive reals $ m,n,p$ satisfy $ mn \plus{} np \plus{} pm \plus{} 2mnp \equal{} 1$.
Prove that $ \sqrt {mn} \plus{} \sqrt {np} \plus{} \sqrt {pm} \leq \frac {17}{12} \plus{} mnp$

[hide="Hint"]
From the given condition, we can deduce that there exist positive reals $ x,y,z$ s.t. $ m \equal{} \frac {x}{y \plus{} z}, n \equal{} \frac {y}{z \plus{} x}, p \equal{} \frac {z}{x \plus{} y}$
[hide="Why?"]
The given condition is equivalent to $ \frac {1}{m \plus{} 1} \plus{} \frac {1}{n \plus{} 1} \plus{} \frac {1}{p \plus{} 1} \equal{} 2$.

Let $ \frac {1}{m \plus{} 1} \equal{} a, \frac {1}{n \plus{} 1} \equal{} b, \frac {1}{p \plus{} 1} \equal{} c$.
We have $ m \equal{} \frac {1 \minus{} a}{a} \equal{} \frac {b \plus{} c \minus{} a}{2a}$ and analogously for $ n,p$.
Setting $ b \plus{} c \minus{} a \equal{} x, c \plus{} a \minus{} b \equal{} y, a \plus{} b \minus{} c \equal{} z$, we have the desired result. ($ x,y,z$ are positive because $ m,n,p$ are positive)
[/hide]
[/hide]

[quote="Harry Potter"]$ m,n,p$ are 3 positive real numbers such that : $ mn \plus{} np \plus{} mp \plus{} 2mnp \equal{} 1$ . Show that : $ \sqrt {mn} \plus{} \sqrt {np} \plus{} \sqrt {mp}\leq \frac {17}{12} \plus{} mnp$

I edited Problem >.<[/quote]

$ k\equal{}\frac {17}{12}$ is the best constat?! it such that:
\[ f(m,n,p)\equal{}\sqrt {mn} \plus{} \sqrt {np} \plus{} \sqrt {pm} \minus{}mnp \leq k ?!\]

Do you know when does equality hold?

[quote="zaizai-hoang"][quote="Harry Potter"]$ m,n,p$ are 3 positive real numbers such that : $ mn \plus{} np \plus{} mp \plus{} 2mnp \equal{} 1$ . Show that : $ \sqrt {mn} \plus{} \sqrt {np} \plus{} \sqrt {mp}\leq \frac {17}{12} \plus{} mnp$

I edited Problem >.<[/quote]

$ k \equal{} \frac {17}{12}$ is the best constat?! it such that:
\[ f(m,n,p) \equal{} \sqrt {mn} \plus{} \sqrt {np} \plus{} \sqrt {pm} \minus{} mnp \leq k ?!
\]
Do you know when does equality hold?[/quote]

When m=n=p :)
What Held ?  :D

To Zaizai : Tớ dùng đạo hàm đế CM bài này  :lol:  Nhưng fải đạo hàm một cách thông minh  :blush:  Nếu cậu có giải thì post lên nhé . Cám ơn nhiều ^^

When $ m\equal{}n\equal{}p\equal{}1/2 ?!$ Did you means like that?!
But when $ m\equal{}n\equal{}p\equal{}1/2$ :
\[ LHS\equal{}3/2 \neq RHS\equal{}37/24\]

I am also misunderstand  :maybe:

[quote="Harry Potter"]Oh SOrry ^^
I don't know about equality >.< I created it from Other Problem ^^
It ís true 100% :D[/quote]

You don't know this? Really? But you said this one was solved by you by "special Derivative's method", totally what do you mean?

Oh My GOD  :blush:   :blush:

[hide="To Zaizai"]Tớ đã rất lịch sự khi dùng Tiếng Việt Zaizai ạ , thiếp nghĩ việc tớ viết mấy câu trên cũng không tớ mức khiến một học sinh suất sắc về bất đẳng thức như cậu phải tò mò xăm xoi như vậy chứ  :( hay đó chỉ là cái mã của cậu thôi  :wink:  . Nếu như cậu có lời giải hay những ý tưởng cho bài toán thì post lên còn nếu không thì không nên post spam những thứ không liên quan , tớ là chủ topic và tớ không thích người ta spam trong Topic của mình !  [/hide]


inequalities
function
inequalities unsolved
n-variable inequality

let $ x_1$ , $ x_2$ ,.....,$ x_n$ positive reals numbers , $ n > 1$ , and $ \frac{1}{x_1 \plus{}1}\plus{}\frac{1}{x_2\plus{}1}\plus{}.....\plus{}\frac{1}{x_n  \plus{}1}\equal{}1$

prove that :

$ \sqrt[n]{x_1 x_2 ...x_n} \geq n\minus{}1$

[quote="y-a-s-s-i-n-e"]let $ x_1$ , $ x_2$ ,.....,$ x_n$ positive reals numbers , $ n > 1$ , and $ \frac {1}{x_1 \plus{} 1} \plus{} \frac {1}{x_2 \plus{} 1} \plus{} ..... \plus{} \frac {1}{x_n \plus{} 1} \equal{} 1$

prove that :

$ \sqrt [n]{x_1 x_2 ...x_n} \geq n \minus{} 1$[/quote]
We have:$ \frac{x_1}{x_1\plus{}1}\equal{}\frac{1}{x_2\plus{}1}\plus{}....\plus{}\frac{1}{x_n\plus{}1} \ge (n\minus{}1) \sqrt[n\minus{}1]{\frac{1}{(x_2\plus{}1)...(x_n\plus{}1)}}$
Similar, we have Q.E.D :)

Since the function $ f(x) \equal{} \frac {1}{1 \plus{} e^x}$ is convex, we have from Jensen's Inequality

$ \frac {1}{n} \left( \frac {1}{1 \plus{} e^{y_1}} \plus{} \frac {1}{1 \plus{} e^{y_2}} \plus{} ... \plus{} \frac {1}{1 \plus{} e^{y_n}} \right) \ge \frac {1}{1 \plus{} e^{\frac {y_1 \plus{} y_2 \plus{} ... \plus{} y_n}{n}}}$ where $ e^{y_i} \equal{} x_i$

Hence $ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_2} \plus{} ... \plus{} \frac {1}{1 \plus{} x_n} \equal{} 1 \ge \frac {n}{1 \plus{} \sqrt [n] {x_1x_2x_3...x_n}}$

From which the inequality follows

[quote="hsbhatt"]Since the function $ f(x) \equal{} \frac {1}{1 \plus{} e^x}$ is convex,...[/quote]
Are you sure?

[quote="quykhtn-qa1"]We have:$ \frac{x_1}{x_1\plus{}1}\equal{}\frac{1}{x_2\plus{}1}\plus{}....\plus{}\frac{1}{x_n\plus{}1} \ge (n\minus{}1) \sqrt[n\minus{}1]{\frac{1}{(x_2\plus{}1)...(x_n\plus{}1)}}$
Similar, we have Q.E.D :)[/quote]

Don't need to revive.. But, i still didn't understand. What should we do after get

$ \frac{x_1}{x_1\plus{}1} \ge (n\minus{}1) \sqrt[n\minus{}1]{\frac{1}{(x_2\plus{}1)...(x_n\plus{}1)}}$

[quote=y-a-s-s-i-n-e]let $ x_1$ , $ x_2$ ,.....,$ x_n$ positive reals numbers , $ n > 1$ , and $ \frac{1}{x_1 \plus{}1}\plus{}\frac{1}{x_2\plus{}1}\plus{}.....\plus{}\frac{1}{x_n  \plus{}1}\equal{}1$ prove that $$ \sqrt[n]{x_1 x_2 ...x_n} \geq n\minus{}1$$[/quote]
[url=https://artofproblemsolving.com/community/c6h2330399p18696367]2001 Kazakhstan[/url]


I do not consider it ridiculous just because the answer is monstrous, but I find it totally spectacular that we can express the result in a closed form. This is magnificient to this problem. And I did not post it to receive a solution (though I would be more than impressed to see a solution!), but just like a curiosity: here is the answer to an apparently innocent question. What about 4 variables? Probably this case is not so ridiculous.

i don't think it's unreasonable to expect a closed form.  the expression to be maximized has nice properties.

There are so many nice expressions which are waiting for their closed forms...

so you say the source is Hilbert... the expression to be maximized looks like a piece of a determinant; and the expression that gives the maximum reminds me of the expression for the determinant of the Hilbert matrix...

just some ideas; maybe there's a connection.

do you know when the equality case is reached?


floor function
number theory unsolved
number theory

Let $ n$ be a positive  integer and let $ p$ be prime number.Prove that if $ p^p|n!$ then $ p^{p\plus{}1}|n!$

[quote="math10"]Let $ n$ be a positive  integer and let $ p$ be prime number.Prove that if $ p^p|n!$ then $ p^{p \plus{} 1}|n!$[/quote]

$ n<p^2 \Rightarrow \lfloor \frac{n}{p} \rfloor \plus{} \lfloor \frac{n}{p^2} \rfloor \plus{} .... \le p\minus{}1$
$ \Rightarrow p^p \nmid n!$, contradiction.

So $ n \ge p^2$.

$ \lfloor \frac{p^2}{p} \rfloor \plus{} \lfloor \frac{p^2}{p^2} \rfloor \plus{} .... \equal{}p\plus{}1$
$ \Rightarrow p^{p\plus{}1}|(p^2)!|n!$, done.


I read about this last night and was horrified.  It may be of interest to some of you.

http://www.nytimes.com/2004/05/13/national/13custody.html

Basically, there's a case going on in which a foster family, which was temporarily taking care of the daughter of an immigrant family from China, claims that they have adopted the girl.  The biological parents did sign the appropriate papers, but they were (deliberately?) misled into thinking that they could get their daughter back at any time.

There's more to this story, though.  The biological parents originally put their daughter up for temporary care because they were in legal, financial, and physical trouble: the father was accused of sexual assault, causing him to lose his position at the University of Memphis.  As a result, he lost his stipend and his student visa.  (Later, he was acquitted, but it took four years!)  Meanwhile, the mother was weak from birth and an assault in a supermarket.  So they put their daughter in a foster home for 90 days.  After that time expired, the foster family offered to keep the girl for longer, but said that the parents could get her back at any time.  They gave them papers to sign, saying that the papers would allow them to put the daughter on their insurance plan.  So the parents signed, not knowing the true nature of the papers...

Some time passed, and when the girl was two, her biological parents wanted to get a photo done with her.  But the foster parents wouldn't let them.  Fighting ensued, and the foster parents called the police, who ordered the biological parents to stay away from the house.  So they did.  Then, the foster parents claimed that the biological parents had abandoned the daughter since they hadn't been visiting.  (However, they couldn't - they were ordered to stay away!)  This abandonment, they say, shows that the biological parents are not fit to be parents, and the foster parents should get custody of the girl.

The parents have tried several times to get their daughter back, but each time, they have been denied by a certain judge.  (The judge was later dismissed on the grounds of bias, and another one has taken his place.)  In the case today, the parents' parental rights were terminated on the grounds of their "fail[ing] to establish a meaningful relationship", the fact that their home is unstable ("the decision calls Ms. He impetuous and calculating. Citing Mr. He's answers on loan and visa applications, it says his conduct is "marked by deceitfulness and dishonesty.""), etc.

There are several issues at stake, though.  First, there's the cultural ignorance factor.  The foster parents say that they can give the girl a better life in their home, and that if she goes back to her immigrant parents, she'll probably be sent back to China (note that this is speculation).  But who is to say what is a 'better' life?  And what is so bad about going back to China?  Furthermore, the foster parents' lawyer has tried to get the biological parents, along with their other children, deported - if they were so against sending people back to China, then why are they trying to send the parents back?  The foster family also says that if the girl goes back to her immigrant parents, she'll have to live in a crowded house with a bunch of relatives (more speculation).  Again, who says there's anything bad about that?  It's traditional in China to live with relatives, and (almost) everyone lives in crowded houses.  You don't need to have 3,000 square feet of living space for one small family!

Then, there's the stuff-that-doesn't-stand-up-in-court factor.  The foster parents claim that there was a verbal agreement between the two families to raise the girl up to adulthood.  Verbal agreement my foot!  Considering that at the time, the mother spoke no English, and the father very little...how could a verbal agreement take place?  Furthermore, how could the immigrants be expected to uncover the nuances in a legal contract if they barely understand it?  And two of the deciding factors in the case were character evidence (a person's character can be portrayed in practically any way with a skillful lawyer) and speculation (see previous paragraph).  That doesn't hold up in court too well.

And there's the unfair accusation factor.  The biological parents are accused of abandoning their child and failing to establish a meaningful relationship.  However, they visited as much as they possibly could; they were ordered to stay away, and the foster parents tried to really restrict their visits.  As for 'establishing a meaningful relationship', well, the foster mother helped prevent a 'meaningful relationship' between the parents and the daughter.

I admit that the foster parents have put in a lot of time and money into caring for this girl.  And yes, they think they're doing the right thing by preventing the girl from [i]a cursed fate in China[/i] (sarcasm).  But that doesn't instantly mean that they should get the child.  A child should stay with her parents unless there is some legitimate reason why she shouldn't.  In this case, I don't believe there is a solid reason (that is, one that does not draw upon cultural ignorance, speculation, character evidence, or unpreventable situations).

Your thoughts?

by the way, i think you've gotta be a new york times member to see it.

judging by your description, i'd think the foster parents are just trying to kinda cheat the biological parents out of their kid.

It sounds like a typical legal case in that both parties look sleazy if you can believe what you read in the newspapers. I studied family law when I was in law school, because of my interest in parental rights in education, and several aspects of this story on both sides don't quite pass the smell test.

[quote="mathfanatic"]In the case today, the parents' parental rights were terminated on the grounds of their "fail[ing] to establish a meaningful relationship", the fact that their home is unstable ("the decision calls Ms. He impetuous and calculating. Citing Mr. He's answers on loan and visa applications, it says his conduct is "marked by deceitfulness and dishonesty.""), etc. [/quote]

It is a VERY BAD idea to play fast and loose with the truth if you are a party to a legal proceeding. Judges don't like that kind of thing. Many factual disputes in legal cases are resolved by the fact-finder deciding which of two contradictory witnesses to believe. Someone with a track record of speaking untruthfully can have his statements on other issues disregarded by the fact-finder.

[quote="mathfanatic"]The foster parents claim that there was a verbal agreement between the two families to raise the girl up to adulthood.  Verbal agreement my foot!  Considering that at the time, the mother spoke no English, and the father very little...how could a verbal agreement take place?[/quote]

The "verbal agreement" claim looks far-fetched, but so does the claim of weak language proficiency. A person holding a student visa has to have a minimal degree of English proficiency as a requirement for obtaining the visa--that's what the TOEFL test is all about.

[quote="mathfanatic"]A child should stay with her parents unless there is some legitimate reason why she shouldn't.[/quote]

That would be my gut reaction too, but the legal standard is actually different. The standard that a judge deciding such a case must follow is a determination of the best interest of the child. A child born in the United States, brought up in an English-speaking home, whose biological parents previously surrendered care of that child, presents a rather strong case for a decision in favor of keeping the child in the familiar environment that the child already knows in the foster home. I'd have to see the bulky case file--NOT just a newspaper report--to know what my gut reaction would be in this particular case, but it's not just blowing smoke to say that the foster family has a strong claim. Determinations that one situation or another is in the best interest of a child are notoriously "fact bound," and thus very difficult to reverse on appeal. The biological parents had better do their utmost to be truthful and prudent in their behavior before the trial judge, or they are unlikely to get a second chance.

As to the general issue of whether life is better in China or in the United States, can there really be any doubt about most people's opinion on that issue when you look at immigration flows? I happen to like living in FREE Chinese-speaking countries, but even at that it is much more convenient to provide my children with a good education here in the States than it is in most parts of the Chinese-speaking world.

[quote]As to the general issue of whether life is better in China or in the United States, can there really be any doubt about most people's opinion on that issue when you look at immigration flows? I happen to like living in FREE Chinese-speaking countries, but even at that it is much more convenient to provide my children with a good education here in the States than it is in most parts of the Chinese-speaking world.[/quote]

But how does one define 'better'?  If you like overpopulation [replace this with any characteristic of China], then life in China may well be considered 'better'.  I don't think it's appropriate to include 'where life is better' in the court case, since it's such a subjective word.

And yes, I know what the general opinion is - in general, people believe that life is better in the U.S.  I agree with them.  But should a child's destiny be determined by what the general opinion is?  There are plenty of things that the general public agrees on which are not good for a child.  For instance, in general, people think spending hours playing slave to a funny box that shows pictures is fun.  But watching TV for hours isn't good for any child.

Of course, it's a moot point to discuss whether life in China or the U.S. is 'better', because we don't know for a fact that the biological parents will be going back to China.  They could end up staying in the U.S.  (Granted, I think it's more likely that they'll return to China, but it's not certain.)

[quote]That would be my gut reaction too, but the legal standard is actually different. The standard that a judge deciding such a case must follow is a determination of the best interest of the child.[/quote]

So the interests of the parents (or foster parents) play no role in the case?  That surprises me - suppose I were kidnapped by a multimillionaire family and allowed to lead a lavish life.   You could argue that it would be in my own best interest to stay with the multimillionaires, since I would go to the best schools, eat the best foods, get the best medical care, and so on.  This is a silly example, but it raises the question: do parents' rights change anything?

[quote]I'd have to see the bulky case file--NOT just a newspaper report--to know what my gut reaction would be in this particular case, but it's not just blowing smoke to say that the foster family has a strong claim. Determinations that one situation or another is in the best interest of a child are notoriously "fact bound," and thus very difficult to reverse on appeal. The biological parents had better do their utmost to be truthful and prudent in their behavior before the trial judge, or they are unlikely to get a second chance.[/quote]

Alright, I agree that reading a news report will not tell all of the facts.  I'm not sure how much better reading ten news reports is, but I have done that.

[quote="mathfanatic"][quote]
But how does one define 'better'?  If you like overpopulation [replace this with any characteristic of China], then life in China may well be considered 'better'.  I don't think it's appropriate to include 'where life is better' in the court case, since it's such a subjective word.
[/quote][/quote]

I'm not bashing China or anything but who would favor overpopulated?  And why?  I can understand liking to be around a lot of people but overpopulated usually has a bad connotation and means that there is a huge excess of people living in one area.  I have never been to China so I can't say which place is better based on my experiences.

[quote="tokenadult"]As to the general issue of whether life is better in China or in the United States, can there really be any doubt about most people's opinion on that issue when you look at immigration flows? I happen to like living in FREE Chinese-speaking countries, but even at that it is much more convenient to provide my children with a good education here in the States than it is in most parts of the Chinese-speaking world.[/quote]

Wait a minute! The argument that more people are moving from China to the US than from the US to China does not mean that most people think life is better is the US than in China at all! I don't know whether the people who aren't immigrating  aren't immigrating because they can't or because they think life in China is much better. There are still way more people in China than there are in the US, so there is no way you can possibly draw accurate conclusions about which one people prefer simply from immigration flow.

[quote="joml88"][quote="mathfanatic"][quote]
But how does one define 'better'?  If you like overpopulation [replace this with any characteristic of China], then life in China may well be considered 'better'.  I don't think it's appropriate to include 'where life is better' in the court case, since it's such a subjective word.
[/quote]

I'm not bashing China or anything but who would favor overpopulated?  And why?  I can understand liking to be around a lot of people but overpopulated usually has a bad connotation and means that there is a huge excess of people living in one area.  I have never been to China so I can't say which place is better based on my experiences.[/quote][/quote]

There are people who can afford to move i n china.  The parents must have enough money.  If life was worse in china, why would they move bacK? and why dont the people who have enough money to move in china not move?

[quote="Scrambled"]
There are people who can afford to move in china.  The parents must have enough money.  If life was worse in china, why would they move bacK? and why dont the people who have enough money to move in china not move?[/quote]

I don't really see how what you said relates to what I said.  I never said that life was worse in China.  I was asking why would someone want to be overpoplulated?  I couldn't judge which place is better because I ve only lived in the US.

[quote="Scrambled"]If life was worse in china, why would they move bacK? and why dont the people who have enough money to move in china not move?[/quote]

Without taking any larger position on anything related to this discussion, I just wanted to point out possible answers to your questions:

Quite possibly because of immigration rules.  The U.S. might not allow them to stay here.  Also, because there families are there, they grew up there, etc., etc.  People have ties to places, even if those places don't have the best standards of living or the best whatever else.  None of those things necessarily carries over to a young child.

[quote="mathfanatic"]... Your thoughts?[/quote]

From the position and facts that you gave us, I'm inclined to agree with you. However, from the way you present those facts, I really have to say that (what I perceive as) your bias prevents the rest of us from looking at it accurately. Either we have a gut instinct to go against you, or we are drawn in by your argument and agree. I have no idea what the law says about this type of case, but I have a feeling that the biological parents were not quite capable enough in the English language to fully grasp the terms of the agreement they signed. I personally don't believe that the foster parents would be a great influence for that little girl. On the other hand, I'm not sure that the daughter should live with her biological parents, either. If they're prone to being assaulted and accused of felonies, they're not exactly your average couple here. The best option IMO may be to let the girl live with a DIFFERENT set of foster parents; as tokenadult says (though in a rather different context):

[quote]Several aspects of this story on both sides don't quite pass the smell test.
[/quote]

Not really a happily-ever-after type story, eh?

[quote="JBL"] Quite possibly because of immigration rules.  The U.S. might not allow them to stay here.[/quote]

Right. The US tends to be pretty hostile to immigrants as well (speaking from my family's experience). Recently they've started limiting the number of visas that they give out / renew. People often have no choice but to go back, regardless of their wishes.

[quote="Scrambled"] why dont the people who have enough money to move in china not move?[/quote]
I know people in Singapore who would like to leave but either 1) can't get a job here/ a visa to come in and 2) are afraid of USA's safety record (or rather, lack thereof) - they think of 9/11 and Columbia and the 2nd Amendment and decide it's safer to stay put. 3) Fact: America's elementary-to-high-school education isn't great. Some immigrants with kids would rather have their children grow up under another education system and build a strong foundation before coming to the universities here.
I don't know how often these cases hold true, but I hope they provide further insight into your question.


A constant thrust rocket is just able to rise vertically from a launching pad when its total mass is M kg.
As fuel  burns ,mass is lost and the rocket accelerates upwards.
When the total mass is reduced to M/2 what is the upward acceleration of the rocket in metres per second?

Do the planet's gravity also play a part, as this will make a very big difference!

I would recommend using Newton 2, and if you want gravity also included, combine it with the Gravitational law.

Acceleration is measured in m.s^(-2) (last line! :?: )

However the question becomes much more interesting if you consider the velocity instead of acceleration  :D


[quote="connaissance"]I am, and I know many others, are going to AMSP while also taking AoPS classes. It has been said that you can only have internet if you have an online class, but I'm wondering where you would have your online class? Is it at the library, or some busy hallway? Also, is there enough time for you to take 2 AoPS online classes, doing the problems for the classes, working through various AoPS books, and doing all the homework for AMSP?

I also know that many people are going to be practicing instruments at camp. My violin can reach 100 decibels when I'm playing fortissimo (Yes, it is VERY loud), so I'm wondering if that is too loud to play in the dorms. I'm assuming that there is a practice area, but I'm wondering how many ppl usually are there at once.

Final Q: Will I be bored out of my head when camp is over, and have to continue going to school?[/quote]

You need to have a laptop or borrow one.  I think some counselors will lend you theirs, but I wouldn't be completely dependent on that.  You take the classes in your apartment, or wherever you get the laptop.  There was no open lab (at least that we were allowed to use) last year at UTD.  No reason to assume there will be this year.

There aren't too many people practicing constantly, if that's what you're thinking, but yeah, that's probably too loud for the dorms.  Would you play that in the middle of an apartment complex in the suburbs?  Probably not.  There are also actual college people who take college summer classes as well, though I'm not sure whether dorms will be shared with them.

There isn't someone telling you "go play from this time to that time. he plays then to then" etc.  You have to work it out with whoever is there.  That's part of the individuality and independence the came allows.  There's nobody watching over your shoulder.

The last question I say depends on the level of your motivation and how much sleep you get during camp.  I recommend you get a proper amount.  From experience.

Hey you play the violin!
Cool, I used to play (until my body was to big for it) so now I play the viola.
I bet I could learn something from you (considering that I'm not all that good, well I'm not a noob either, just decent).
Maybe we could make a practice group?
Are we allowed to bring computers even if we don't take a class?
I like to stay in contact with people via email and skype.

You [i]could[/i] bring it, but e-mail or skype?  Nope.  The wifi has a password.  And it's not anything too easy to predict, or so I heard.  It's not worth the hassle.

It should actually be nice to get away from all that for a little while.  After a while, you won't really want to use it as much.  Anyways, it's a camp, so part of reason to go is to also do things with people as well.  Don't worry.  There's always something to do.

Thanks for telling me all this. So, you can only use the internet if you have an online class? oh...

hmm they say that internet's only available to people taking online classes..
but what happened last year among the girls was that few people got password&username and passed that around.
and i think same happened for the guys?

you will need a laptop to take online courses. so bring one if you have one to take your classes.
and you take them anywhere you want to. i just did my classes in my dorm. [but ended up goofing off with my roomies half of the time]

and violin practice-i wouldnt recommend practicing in your dorms unless you have a metal practice mute. the practice rooms at UTD are WAYY TOO FAR to walk in that ridiculous weather...plus there's only 2..like container box looking thingies and NO AC. gAWD it was SOO HOT in there. and there were like 5 people in each boxes-_-  =inefficient practicing. so after the first week, i just gave up practicing[which resulted me in practicing my arse off when i came back to get ready for my competitions] but the point is, just bring a metal practice mute and do it in your bedroom. not in the ocmmon area of the dorms if youre at UTD. i have no idea how santa cruz will be like.

Okay so I was at CTY in Santa Cruz last year, and I am quite sure that there are no practice rooms of any sort anywhere. You probably can practice in your room. The particular area that we were in had an upright piano with about 10 broken keys, one with the hammer completely snapped off, completely dead action and completely out of tune, one caster missing, and lots of wood missing from the case. Oh and pedals that don't work. The place where they had the talent show and dances had a nice piano though.

Yes, you're right sooozy, yet my point was that you won't want to bring it assuming you'll be able to get the password.  And you want to sleep at night, not go online.  Remember what happened that one night with facebook?

actually, the internet use is all up to the counselours...if your counselour is nice and ok with you using their laptop, you can use it all you want...but my counselour was never there in our dorm...and if he was he came back at 2 in the morning...but there are computers in the library...or whatever it was in the middle of nowhere in the campus...OR you could hog the apple computers that they have at the tech store across from where we eat(the bookstore)

about instruments...i suggest you dont bring them... AT ALL...unless you play piano, you can bring ur music...and play on the bad pianos they have or if the grand piano isnt being used by a college student, you can use that as well. and i do remeber the facebook nite

its not a bad idea to bring violin
its just that there's really no adequate place to practice without getting bothered or bothering others.

alexhmun: errrrr what fb incident?
i know all the ocunselors were like WTF IS SUZY DOING cuz i was on fb with my phone 24/7
referring to when meena got a call at like 11:30?

So could we use our own internet then? (aka bring our personal wireless modem)

I guess this means I shouldn't bring my drum set. (darn)

Hopefully a practice pad would be quiet enough for in-dorm use?


geometry
geometric transformation
rotation
reflection

is it just[hide](7C4)/7=5

hmm...that doesn't seem quite right for some odd reason[/hide]

jorian

[quote="Fanatic"]Uhh... my answer key says jhredsox's answer was correct.[/quote]

My brain tells me Jorian's method is correct.

-_- I solved it and got it right a week ago, and had it checked by Mr. Fischer's answer key -_-.

[quote="vishalarul"]Seven points are equally distributed on a circle. How many non-congruent quadrilaterals can be drawn w/ vertices chosen from among the seven points?[/quote]
jorian is correct
[hide="answer"]
first point has 6 to choose from, then decreases
(6*5*4)/(3*2) =20
5*4*3)/(3*2)=10
4*3*2)/(3*2)=4
6/6=1
20+10+4+1=35
each shape is repeated 7 times.
35/7 equals 5.
[/hide]

[quote="Fanatic"]Uhh... my answer key says jhredsox's answer was correct.[/quote]
[hide="Please show your work."]
If the answer is 5, as several of you have claimed, please prove you are correct, by listing five non-congruent quadrilaterals?  E.g., Label the seven points, A, B, C, D, E, F, and G, in order around the circle.  The quadrilaterals are ABCD, ABCE, etc.  Answer keys have been known to be incorrect.[/hide]

It sounds like many are treating this like a key chain problem, but since some possible quadrilaterals are the same as others that are not only rotations but reflections, while others are not repeated as reflections, I am not sure you can treat it that way.  I tried actually making them, and I can't get 5 unique ones no matter how hard I try and it seems clear there aren't.

Hmm....

i am normal and i do not am not and will not be good at this kind of quetion because i have not just broken my scooter today!!!!!!!
                 :arrow:  :ninja:  :idea:

oh i'm sorry

perhapas 1,2,4,7

i know there's another pattern

the point is, you divide by 7 because each can be repeated 7 times

-jorian

Why would you do that-I like ALtheman's way and he got 4, the same answer as mine

there are 7 arc lengths and each side of the quadrilateral can contain the arc lengths that add up to 7

they can be 2 2 2 1
3 1 1 1
2 1 1 3
2 1 3 1

apart so 4 (for example 2 3 1 1 would not work because you can rotate that-imagine a circle, into 2 1 1 3)

and fanatic-are you sure that your teacher's answer key correct?

Not positive, I guess it COULD be wrong... but I don't think so.

I did this problem under time constraints, and I know I'm bad at visual stuff. Thus, I just did Jorian's method 7 choose four divided by 7 and it worked... not quite sure why, but intuitively it seemed correct.

You're probably missing a quadrilateral somewhere...

[hide="think i got it if you want"]

The $_{7}C_{3}$ divided by 7 idea is a really good one, but there is one problem.  You are dividing by 7 because we assume there are 7 of each of the quadrilaterals.  However since the Quad created by three points in a row and then skipping 1 can be created 14 ways since it is not symmetric and can be created by taking 3 in a row and then skipping 2.

In the picture you can see the 5 you get by doing $_{7}C_{3}$ divided by 7, but the top 2 are the same just flipped, so I believe there are only 4.

Oh, and I checked the answer key  [hide]and 4 is the answer[/hide]

[/hide]

don't yu mean 7C4?

how would you figure this out on the test?

jorian

[quote="Altheman"][hide]the number of arcs that each side subtends uniquely defines a quadrilateral; suppose we have the arcs subtended as a,b,c,d, we have that a,b,c,d>=1 and a+b+c+d=7, and no circular permutations of a solution, this is the same as w+x+y+z=3, with w,x,y,z>=0

3,0,0,0 [1 way]
2,1,0,0 [2 ways]
1,1,1,0 [1 way]

so 4[/hide][/quote]

Could you elaborate on this a bit more? It seems interesting but I don't get the concepts behind it.

Is there any other way besides listing and this? I'm very bad at listing >__>


How to prove that the group $ G$ defined by the presentation $ <a,b,ab^{2}=b^{3}a,ba^{2}=a^{3}b>$ is reduced to the neutral element ?
I proved that $ a^{2}=b^{-1}a^{3}b$, $ a^{4}=b^{-1}a^{6}b$, etc... but if somebody can give me an idea, thanks for advance! Bob.


For the USAMO, if you get an answer but you can't prove how you got it, should you bother writing it down? How much is the actual answer to a problem worth, if anything?

99% of the time, a USAMO problem starts with "Prove that..." so there is no actual answer; it is given to you, and you need to prove it. However, there are some problems where it might say "find, with proof...", and I think the answer is worth 0 points there. (The USAMO is very strict, remember).

But always write down WHATEVER you have, it doesn't hurt (you can't get a negative score ;))

Depends on the problem. On many problems, such as most functional equations, the answer will be obvious. Obvious answers get you no points, but if the answer is not obvious, you might get 1 point. If you can come up with an intuitive argument why it is the answer, you might get 1 or 2 point.

Just to throw out an extreme example, the answer for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1186813#1186813]this IMO shortlist problem[/url] is in Zuming's words "worth a lot of points", because half of the problem is solved when you realize what it is. When it was given in the TST 2008 no one got the answer right during the test, and when told what the right answer actually was afterwards several people came up with solutions very quickly.

But generally, pythag's description is accurate, though I would caution that the "intuitive argument" better have some ideas used in the actual solution to merit more points.


Hello. I stoped to receive topics reply notification messages. I remember I didn't visit two topics of which notifications I received. As I understand, no more notifications will be sent until I visit these topics. The trouble is that I don't remember what these topic were since all messages in my mail box are marked as "read". How do I continue receiving reply notifications?

You can just click "view your posts" (you don't have that many) and visit all of them.

I found email notifications somewhat distracting and turned them off - the "view your posts" list serves the same purpose.

Yes. But it it also the case that I subscribe to a topic I am interested in, which I didn't create and for which I haven't posted answer yet. Such topics are not in "view your posts" section. Without notifications I must visit the forum very often so that be informed about answers for all messages of that sort.

And these two topics I didn't visit was among those I didn't create and gave no answer for.

You could add the topic to your favorites, and each time you wished to access it, you could click the favorites button in your userbar.

(Too add the topic to your favorites: There is a star under the page numbers for the topic. Clicking it will add it. To view your favorites, there's a button saying "Favorites" next to "profile". At least in MathLinks theme. :maybe:)


How many unique sets of three prime numbers exist for which the sum of the members of the set is 44?

$ 44$ is even number.
So, even+odd+odd  i.e.  $ 2\plus{}\text{odd}\plus{}\text{odd}\equal{}44$.
So, sum of the two prime number is $ 42$.

$ 44\minus{}2\minus{}3\equal{}39 \not\in \text{prime}$
$ 44\minus{}2\minus{}5\equal{}37 \in \text{prime}$
$ 44\minus{}2\minus{}7\equal{}35 \not\in \text{prime}$
$ 44\minus{}2\minus{}11\equal{}31 \in \text{prime}$
$ 44\minus{}2\minus{}13\equal{}29 \in \text{prime}$
$ 44\minus{}2\minus{}17\equal{}25 \not\in \text{prime}$
$ 44\minus{}2\minus{}19\equal{}23 \in \text{prime}$

Therefore, $ (2, \ 5, \ 37)$, $ (2, \ 11, \ 31)$, $ (2, \ 13, \ 29)$, $ (2, \ 19, \ 23)$.
Thus $ \boxed{4}$.

just curious, but could you divide 44 by 11, because 11 is it's biggest prime, and get 4 that way?


Given 16 distinct positive integers not exceeding 100, prove that it is possible to choose 4 distinct integers $ a$, $ b$, $ c$ and $ d$ such that $ a\plus{}c \equal{} b \plus{} d$.

Is this statement still true for 15 distinct positive integers?

[hide]
Suppose we have $ a,b,c,d$ distinct integers with $ a<b$ and $ c<d$ and $ a\plus{}b \equal{} c\plus{}d$
This implies that $ a<c<d<b$. Therefore we can write the expression as $ b\minus{}c \equal{} d\minus{}a$
Where $ b\minus{}c$ and $ d\minus{}a >0$

So the question becomes: given $ 16$ positive integers less than $ 100$, we can find two pairs who have the same difference.

Now we have $ 1\le |a_i \minus{} a_j| \le 99$. However there are $ \binom{16}{2} \equal{} 120$ pairs $ (a_i,a_j)$
 So by the box principle we have two pairs such that; $ a_i \minus{} a_j \equal{} a_k \minus{} a_l$

Or; $ a_i \plus{} a_l \equal{} a_k \plus{} a_j$[/hide]

[quote="ocha"][hide]
Suppose we have $ a,b,c,d$ distinct integers with $ a < b$ and $ c < d$ and $ a \plus{} b \equal{} c \plus{} d$
This implies that $ a < c < d < b$. Therefore we can write the expression as $ b \minus{} c \equal{} d \minus{} a$
Where $ b \minus{} c$ and $ d \minus{} a > 0$

So the question becomes: given $ 16$ positive integers less than $ 100$, we can find two pairs who have the same difference.

Now we have $ 1\le |a_i \minus{} a_j| \le 99$. However there are $ \binom{16}{2} \equal{} 120$ pairs $ (a_i,a_j)$
 So by the box principle we have two pairs such that; $ a_i \minus{} a_j \equal{} a_k \minus{} a_l$

Or; $ a_i \plus{} a_l \equal{} a_k \plus{} a_j$[/hide][/quote]

how do u know that $ a_i, a_j, a_k, a_l$ are distinct.


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group of order $ p^2$. Prove that $ G$ is abelian.

nontrivial $ p$-groups have nontrivial center. so in our case $ G/Z(G)$ has order $ 1$ or $ p$. so $ G/Z(G)$ is cyclic, thus $ G$ is abelian.


algebra
polynomial
trigonometry
ratio
conics
parabola
complex numbers

Just a reminder that we will have All-State Math Team practices on Saturday, May 9th, 16th, and 25th. If you haven't sent Mr. R your information sheet and your answers to the problems you were asked to work on, do so as soon as possible.

By the way, there are some really good problems floating around in the Intermediate and Pre-Olympiad section lately.

Good job at the practices today for those who came. For those in my room I hope you got something out of our talk about Complex Numbers. I know it went by pretty fast. Here are some practice problems to reinforce what we talked about:

1.) Find all complex numbers $ z$ such that $ z^2 \equal{} i$

2.) Find all complex numbers $ z$ such that $ z^3 \equal{} \minus{} 8$

3.) Find all complex numbers $ z$ such that $ z^6 \plus{} z^3 \plus{} 1 \equal{} 0$

4.) $ P(x)$ is a 6th degree polynomial with integer coefficients such that $ 4$, $ \sqrt {3}$, and $ 2 \minus{} i$ are roots. If $ P(2) \equal{} 17$, find $ P(0)$.

-----------------------------------------------------------------------------
Here are the solutions to Set 1 of the practice problems given when you were accepted to the team:

1.) If the number $ A3640548981270644B$ is divisible by 99, compute the ordered pair of digits $ (A,B)$

[hide="Solution:"] A number is divisible by 9 if and only if the sum of its digits is divisible by 9. So $ A \plus{} B \equiv 1 \bmod 9$

A number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. So $ A \minus{} B \equiv 8 \bmod 11$

$ A \equal{} 9$, $ B \equal{} 1$ satisfies these conditions, so the answer is $ \boxed{(9,1)}$[/hide]

2.) If $ n$ is the smallest positive integer for which positive $ a$ and $ b$ exist that make $ (a \plus{} bi)^n \equal{} (a \minus{} bi)^n$, compute the numerical value of $ \frac {b}{a}$

[hide="Solution:"] Rewrite $ a \plus{} bi$ in polar form as $ re^{i\theta}$. Then $ a \minus{} bi$ will be $ re^{ \minus{} i\theta}$.

So $ (a \plus{} bi)^n \equal{} (a \minus{} bi)^n$ means
$ (re^{i\theta})^n \equal{} (re^{ \minus{} i\theta})^n$
$ r^ne^{in\theta} \equal{} r^ne^{ \minus{} in\theta}$

This implies either $ r \equal{} 0$ or $ e^{in\theta} \equal{} e^{ \minus{} in\theta}$
Since we know that $ a$ and $ b$ are positive, the former cannot be true (because then both $ a$ and $ b$ would be zero).

So the latter must be true, and also $ \theta$ must be between $ 0$ and $ \frac {\pi}{2}$ (exclusive).

$ e^{in\theta} \equal{} e^{ \minus{} in\theta}$ means $ e^{2in\theta} \equal{} 1$
So $ 2n\theta \equal{} 2\pi k$ for some integer $ k \implies \theta \equal{} \frac {2\pi k}{n}$

The smallest positive integer value of $ n$ that will make $ 0 < \theta < \frac {\pi}{2}$ is $ n \equal{} 3$ (with $ k \equal{} 1$).

So $ \theta \equal{} \frac {\pi}{3}$ and $ \tan{\theta} \equal{} \frac {b}{a}$. So $ \boxed{\frac {b}{a} \equal{} \sqrt {3}}$[/hide]

3.) A mixture of wine and water is made in the ratio of wine:total = k:m. Adding $ x$ units of water or removing $ x$ units of wine $ (x \neq 0)$ each produces the same new ratio of wine:total. Compute the numerical value of this new ratio.

[hide="Solution:"] Let $ a$ be the amount of wine and $ b$ the amount of water.
Then $ \frac {a}{a \plus{} b \plus{} x} \equal{} \frac {k}{m} \implies am \equal{} ka \plus{} kb \plus{} kx$,
and $ \frac {a \minus{} x}{a \plus{} b \minus{} x} \equal{} \frac {k}{m} \implies am \minus{} mx \equal{} ka \plus{} kb \minus{} kx$

Subtracting the second equation from the first gives $ mx \minus{} kx$
So $ m \equal{} k$ and $ \boxed{\frac {k}{m} \equal{} 1}$[/hide]

4.) The length of each side of regular hexagon $ ABCDEF$ is $ 2$. $ A$, $ B$, and $ C$ are the centers of three mutually extermally tangent circles. Compute the radius of the smallest of these three circles.

[hide="Solution:"] First, draw a picture (not shown here). Let $ r_1$ be the radius of the circle centered at $ B$ (which will be the smallest circle), and let $ r_2$ be the radius of the circles centered at $ A$ and $ C$ (notice these circles will have the same radius).

Since $ AB \equal{} 2$ we have $ r_1 \plus{} r_2 \equal{} 2$,
and because $ AC \equal{} 2\sqrt {3}$ (by 30-60-90 triangles, or by law of cosines), we have $ 2r_2 \equal{} 2\sqrt {3}$

Solving, we find the $ \boxed{r_1 \equal{} 2 \minus{} \sqrt {3}}$[/hide]

5.) The parabola $ y \equal{} ax^2 \plus{} bx \plus{} c$ crosses the x-axis at $ (p,0)$ and $ (q,0)$, both to the right of the origin. A circles passes through these two points. Find the length of a tangent from the origin to the circle, in terms of one or more of the coefficients $ a$, $ b$, and $ c$.

[hide="Solution:"] I assumed that the circle was centered on the x-axis (hey, they didn't say that it couldn't be!!). So now draw a picture (not shown).

The center of the circle will be midway between the roots, at $ \left(\frac {p \plus{} q}{2},0\right)$, and the radius of the circle will be $ r \equal{} \frac {p \minus{} q}{2}$

The tangent line from the origin to the circle will make a right angle with the radius of the circle, so by Pythagoras' Theorem the length of the tangent line will be
$ L \equal{} \sqrt {\left(\frac {p \plus{} q}{2}\right)^2 \minus{} \left(\frac {p \minus{} q}{2}\right)^2} \equal{} \boxed{\sqrt {pq}}$ (using difference of squares to simplify)[/hide]

6.) In triangle $ ABC$, $ D$ is the midpoint of $ BC$. If $ \angle BAD \equal{} 15^o$ and $ \angle CAD \equal{} 30^o$, compute the tangent of angle $ C$.

[hide="Solution:"]Draw a picture!! Let $ BD \equal{} CD \equal{} x$ and let $ AD \equal{} d$. Let angle $ C \equal{} \theta$, making angle $ B \equal{} 135^o \minus{} \theta$.

By the law of sines on triangle $ ABD$,
$ \frac {x}{\sin{15}} \equal{} \frac {d}{\sin{(135 \minus{} \theta)}}$
and by the law of sines on triangle $ ACD$,
$ \frac {x}{\sin{30}} \equal{} \frac {d}{\sin{\theta}}$

Combining these equations gives
$ \frac {\sin{15}}{\sin{(135 \minus{} \theta)}} \equal{} \frac {\sin{30}}{\sin{\theta}}$

$ \implies \sin{15} \cdot \sin{\theta} \equal{} \sin{30} \cdot \sin{(135 \minus{} \theta)}$

Note: $ \sin{15} \equal{} \sin{(45 \minus{} 30)} \equal{} \frac {\sqrt {2}}{2} \cdot \frac {\sqrt {3}}{2} \minus{} \frac {\sqrt {2}}{2} \cdot \frac {1}{2} \equal{} \frac {\sqrt {6} \minus{} \sqrt {2}}{4}$

So the equation becomes
$ \frac {\sqrt {6} \minus{} \sqrt {2}}{4} \sin{\theta} \equal{} \frac {1}{2} \left[\frac {\sqrt {2}}{2} \cdot \cos{\theta} \plus{} \frac {\sqrt {2}}{2} \cdot \sin{\theta} \right]$

$ \frac {\sqrt {6} \minus{} 2\sqrt {2}}{4} \sin{\theta} \equal{} \frac {\sqrt {2}}{4} \cos{\theta}$
$ \tan{\theta} \equal{} \frac {\sqrt {2}}{\sqrt {6} \minus{} 2\sqrt {2}} \equal{} \frac {1}{\sqrt {3} \minus{} 2} \equal{} \boxed{ \minus{} 2 \minus{} \sqrt {3}}$[/hide]

[quote="gauss202"]
3.) A mixture of wine and water is made in the ratio of wine:total = k:m. Adding $ x$ units of water or removing $ x$ units of wine $ (x \neq 0)$ each produces the same new ratio of wine:total. Compute the numerical value of this new ratio.

Let $ a$ be the amount of wine and $ b$ the amount of water.
Then $ \frac {a}{a \plus{} b \plus{} x} \equal{} \frac {k}{m} \implies am \equal{} ka \plus{} kb \plus{} kx$,
and $ \frac {a \minus{} x}{a \plus{} b \minus{} x} \equal{} \frac {k}{m} \implies am \minus{} mx \equal{} ka \plus{} kb \minus{} kx$

Subtracting the second equation from the first gives $ mx \minus{} kx$
So $ m \equal{} k$ and $ \boxed{\frac {k}{m} \equal{} 1}$[/quote]

Subtracting the second from the first gives $ \minus{} ( \minus{} mx) \equal{} kx \minus{} ( \minus{} kx) \equal{} 2kx$
So $ m \equal{} 2k$ and $ \boxed{\frac {k}{m} \equal{} \frac {1}{2}}$

After all, you are adding $ x$ units of water in order to obtain the ratio one way, and as $ x$ cannot be $ 0$ the final ratio cannot indicate that all that is left is wine.

5.) (Continued)
The product of the roots of any polynomial is given by the constant term (possibly negated, but not in the case of a polynomial of even degree) divided by the coefficient of the term of greatest degree.  This is expressed in this problem as $ \frac {c}{a}$.

Thus, $ \displaystyle\sqrt {pq} \equal{} \sqrt {\frac {c}{a}}$.

[hide="1.)"] $ \displaystyle z^2\equal{}cis(\frac{\pi}{2})$
$ \displaystyle z\equal{}cis(\frac{\pi}{4}\plus{}\frac{2\pi k}{2})$ where $ k\equal{}0,1$
or
$ \displaystyle z\equal{}\pm\frac{\sqrt{2}}{2}(1\plus{}i)$[/hide]

[hide="2.)"] $ \displaystyle z^3 \equal{} 8cis(\pi)$
$ \displaystyle z\equal{}2cis(\frac{\pi}{3}\plus{}\frac{2\pi k}{3})$ where $ k\equal{}0,1,2$
or
$ \displaystyle z\equal{}\minus{}2, 1\pm\sqrt{3}i$[/hide]

[hide="3.)"] $ (z^3\minus{}1)(z^6\plus{}z^3\plus{}1) \equal{} 0(z^3\minus{}1)$
$ z^9 \equal{} 1$
$ \displaystyle z\equal{}cis(\frac{2\pi k}{9})$ where $ k\equal{}0,1,2,3,4,5,6,7,8$

However, we must take away the roots supplied by the extra $ z^3\minus{}1$, so $ k\equal{}1,2,4,5,7,8$[/hide]

One question on number 4:  If a polynomial has integer coefficients and an even root (like 4), can $ P(2)$ ever be odd?

Correct. I should have said $ P(2) \equal{} 8$


inequalities
rearrangement inequality
inequalities unsolved

Let $ a,b,c\geq0$.Prove that:
$ 2(x^5\plus{}y^5\plus{}z^5)\plus{}x^4y\plus{}y^4z\plus{}z^4x\geq3(xy^4\plus{}yz^4\plus{}zx^4)$

by rearrangement inequality :
$ 2(x^5 \plus{} y^5 \plus{} z^5) \equal{} 2(x^4x \plus{} y^4y \plus{} z^4z)\geq2(zx^4 \plus{} xy^4 \plus{} yz^4)$
and :
$ x^4y \plus{} y^4z \plus{} z^4x \geq x^5 \plus{} y^5 \plus{} z^5$

(not mistake ??)

Yes, I meant $ x,y,z\geq0$.By the way ,mehdi cherif, $ x^4y\plus{}y^4z\plus{}z^4x \geq x^5\plus{}y^5\plus{}z^5$ is not a correct inequality.

[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ 2(x^5 \plus{} y^5 \plus{} z^5) \plus{} x^4y \plus{} y^4z \plus{} z^4x\geq3(xy^4 \plus{} yz^4 \plus{} zx^4)$[/quote]
Try the following reasoning:
let $ x\equal{}\min\{x,y,z\},$ $ y\equal{}x\plus{}u$ and $ z\equal{}x\plus{}v.$ :wink:

The most dirty solution I think,,

By Muirhead inequality,
   (5,0,0)>(1,4,0)and (4,1,0)>(1,4,0) so,,
 x^5+y^5+z^5>=xy^4+yz^4+zx^4
 x^5+y^5+z^5>=xy^4+yz^4+zx^4
 x^4y+y^4z+z^4x>=xy^4+yz^4+zx^4
    If we sum all of these inequalities,,
    We can get  2(x^5+y^5+z^5)+(x^4y+y^4z+z^4x)>=3(xy^4+yz^4+zx^4)
    I hate Muirhead inequalities,, but I use it often,,,

[quote="GoKorea"]
 x^4y+y^4z+z^4x>=xy^4+yz^4+zx^4
    [/quote]
It's wrong. Try $ x\equal{}1,$ $ y\equal{}2$ and $ z\equal{}0.$  :wink:

I just notice that this problem is pretty hard,,,

Sorry,arqady,I still cannot get your method to the end.Can someone help me? :wallbash:  :?:

For you, Inequalities Master! :lol:
[quote="Inequalities Master"]Let $ x,y,z\geq0$.Prove that:
$ 2(x^5 \plus{} y^5 \plus{} z^5) \plus{} x^4y \plus{} y^4z \plus{} z^4x\geq3(xy^4 \plus{} yz^4 \plus{} zx^4)$[/quote]
Let $ x \equal{} \min\{x,y,z\}$ and $ y \equal{} x \plus{} u,$ $ z \equal{} x \plus{} v.$ Then I have obtained:
$ \sum_{cyc}(2x^5 \plus{} x^4y \minus{} 3xy^4) \equal{} 8(u^2 \minus{} uv \plus{} v^2)x^3 \plus{} 6(2u^3 \plus{} u^2v \minus{} 3uv^2 \plus{} 2v^3)x^2 \plus{}$
$ \plus{} 4(2u^4 \plus{} u^3v \minus{} 3uv^3 \plus{} 2v^4)x \plus{} 2u^5 \plus{} u^4v \minus{} 3uv^4 \plus{} 2v^5\geq0.$ :wink:

For the collection:
For all reals $ a,$ $ b$ and $ c$ prove that:
\[ 2(a^4+b^4+c^4)+abc(a+b+c)\geq3(a^3b+b^3c+c^3a)\]

Actually,by Schur we know that $ a^4\plus{}b^4\plus{}c^4\plus{}abc(a\plus{}b\plus{}c)\geq 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$.
And the inequality $ (a^2\plus{}b^2\plus{}c^2)^2\geq 3(a^3b\plus{}b^3c\plus{}c^3a)$ is true,but very hard to prove.
How to prove easier your inequality, arqady? :?:

[quote="Inequalities Master"]Actually,by Schur we know that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c)\geq 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$
[/quote]
Which wrong.
Try $ a\equal{}\minus{}1$ and $ b\equal{}c\equal{}1.$ :wink:

But I meant is true for positive reals.Is your inequality true for all reals? :o

[quote="Inequalities Master"]Is your inequality true for all reals? :o[/quote]
Yes! :lol:

[quote="Inequalities Master"]OK, arqady, how to prove it? :?:[/quote]
Try the following way:
Let $ a\equal{}\min\{a,b,c\},$ $ b\equal{}a\plus{}x$ and $ c\equal{}a\plus{}y...$ :wink:


Why are there two websites for AoPS? Which is the main one? All my posts on this site are on mathlinks. I don't get this, why are there two sites for the same thing?

One additional fact: the merge was in July 2004.

[url=http://www.mathlinks.ro/]Going to the Mathlinks site[/url] and seeing the portal on the homepage might help.  :)

Additionally, [url=http://www.lmgtfy.com/?q=define%3Aportal]Google can also be a great friend.[/url]


[quote="Interval"]On the starship Enterprise, Captain Picard continually asks the food replicator to brew him a cup of Earl Grey tea.  The proper brewing temperature for Earl Grey tea is 210º F plus or minus 5 degrees.  Write and solve an absolute value equation representing the maximum and minimum brewing temperatures for Earl Grey tea.[/quote]

[hide]ummm...

$|210\pm 5|=205---215$.

I dont know what an absolute value equation is.[/hide]


geometry
rectangle
combinatorics solved
combinatorics

More than three quarters of the circumference of a circle is colored black. Prove that there exists a rectangle such that all of its vertices are black.

Actually the result holds if "three quarters" is replaced by "one half"...

Actually, I hope something like this is ok.
So, consider point $A$ and it's simetric with respect to the center of the circle $A'$. By continuously moving $A$ on the black region, you get a position when $A'$ is also black. Now remove point $A$ and $A'$. The area is still bigger than one half. In the same way, find $B$ and $B'$. Ready :)


This is so stupid question but I keep wondering, what's TST?

 :D

Thank you and Happy Holiday!

Hardest test in an national olympiad. They select the team that participates at the International Mathematical Olympiad representing the country. In USA TSTs are only taken by the first 12 USAMOers *with exceptions sometimes* ;)

[quote="h_s_potter2002"]I never even knew they had a TST in the U.S.  Does it take place in MOP?[/quote]US always had a TST since they began participating in IMO (I think). I'm not sure exactly when I takes place, but I think it's at the beginning of MOP. IMO Team members from the site can tell us more (or Steve Dunbar for that matter).

At the USAMO official page, on the download part, where you find the USAMOs and the APMOs you will also find the TSTs from the past 4-5 years.

The US has only been using a specialized TST to select the team from among the top 12 on the USAMO for the past five years (the first one was in 2000).  That is, the TSTs on the AMC website are the only ones that have ever been given (at the top they say "43rd IMO Team Selection Test" and such, but don't let that fool you: it's the the test to pick the team for the 43rd IMO, not the 43rd test).  It's generally been held on the first Friday and Saturday of MOP.  It isn't necessarily true that the USA TST is harder than the corresponding USAMO -- I think that the last couple years the top 12 as a whole have done roughly as well or better on the TST thanoon the USAMO.  There have been some quite hard TST's though:  I believe that the first TST, in 2000, didn't do as good a job of picking a team as it did of showing how much better Reid Barton was than anyone else. ;-)  But they seem to have mostly worked the kinks out since then.

[quote="fleeting_guest"][quote]The US has only been using a specialized TST to select the team from among the top 12 on the USAMO for the past five years (the first one was in 2000).  That is, the TSTs on the AMC website are the only ones that have ever been given.[/quote]

The names of the US IMO team and USAMO top 6 are
published for the years before 2000, and the two lists
differ substantially.  TSTs were used, in other words.[/quote]Both Richard and Mathew were involved in the USAMO / TST tesing back then, (Richard was #4 if I remember correctly on USAMO - but he didn't go to the IMO - so tests were issued somehow).

[quote="orl"][quote="Alison"] I believe that the first TST, in 2000, didn't do as good a job of picking a team as it did of showing how much better Reid Barton was than anyone else. ;-)  But they seem to have mostly worked the kinks out since then.[/quote]

Where do they publish the individual score of contestants or at least score distributions per problem ?  :)[/quote]

Actually I referred to TST scores and not those of the USAMO...  ;)

Team selection for the US has varied quite a bit over the years.  Some years it was just the top 6 (or 8 if you go back far enough) on the USAMO.  When Dave & I were at MOP (late 80s), the team was determined by the USAMO plus 3 or 4 more tests in the first week or two of MOP (plus a little discretion of the IMO deputies).   I think the process is more highly defined now, and therefore better.

[quote="Valentin Vornicu"]Both Richard and Mathew were involved in the USAMO / TST tesing back then, (Richard was #4 if I remember correctly on USAMO - but he didn't go to the IMO - so tests were issued somehow).[/quote]

Me too -- I was tied for #6 on the USAMO in 1988, but did really badly on the TSTs (there were 3 my year), so I didn't go to the IMO.  (Not that I would have necessarily went anyway; they would have had to break the tie for #6 somehow.)

In the early days (late 1970s/early 1980s) and also for a while in the 1990s I believe, the US did away with the TSTs and just took the top 6 scorers on the USAMO for the IMO team.

TST scores are not made public, usually even the team members don't know their exact scores.  But little bits of information get leaked around MOP ;-)...

Hmmm... fascinating.  I only knew about the team selection process from when I was at MOP and what I'd heard from veterans and staff members who were there in the late 90's, although I think I was vaguely aware that at one point people's performance on MOP tests fed into the team selection process.    (Somebody needs to write a history of MOP.  :-)

The TSTs were done away with after 1993.  In 1994 and 1995 the team was selected solely from USAMO scores.  I don't know how long that lasted.  They probably thought it was a good idea at first because 1994 was the year the US team had the only team ever to have 6 perfect 42's at the IMO.


This one has been locked.
http://www.mathlinks.ro/Forum/viewtopic.php?t=227462

Can you show me how to write in latex. I am so sorry

can you show your solution so that we can discuss together, please

[quote="shfdfzhjj"][quote="Cai Ni"]I can only tell you this is really an easy problem.I got the solution.[/quote]
Sorry,I don't think it is an easy problem.
Can you send me your proof?[/quote]
I think it quite nice and not too easy .

My hint is use AM-GM then use a known Vasc 's lemma  :wink:

To Cai Ni : Sorry but I don't want you post solution because this problem haven't limited on MYM in Vietnam

[quote="onlylove_math"]

My hint is use AM-GM then use a known Vasc 's lemma  :wink:[/quote]

I think we get the same proof. :)

[quote="onlylove_math"]

To Cai Ni : Sorry but I don't want you post solution because this problem haven't limited on MYM in Vietnam[/quote]

Sorry,I'm not good at English.Actually,I don't understand what's the meaning.It seems that in some problem we can't show our answer?
 :?:
OK.I use Holder,then AM-GM.
We may easily remember the similar problem.(The inequality from 42th IMO)
So I try it and luckily I managed to get the solution.
Oh,yes.I think it's not a very easy problem,just I solved it quickly.

[quote="Cai Ni"][quote="onlylove_math"]

To Cai Ni : Sorry but I don't want you post solution because this problem haven't limited on MYM in Vietnam[/quote]

Sorry,I'm not good at English.Actually,I don't understand what's the meaning.It seems that in some problem we can't show our answer?
 :?:
OK.I use Holder,then AM-GM.
We may easily remember the similar problem.(The inequality from 42th IMO)
[/quote]
Please check your proof,I think the step you use AM-GM is wrong.

Dont discuss more,please
[quote ="Cai Ni"
]Sorry,I'm not good at English.Actually,I don't understand what's the meaning.It seems that in some problem we can't show our answer? [/quote]
It means here is the new problem inthis issue of MYM(in our country)which is running
besides
[quote="Cai Ni"]Please check your proof,I think the step you use AM-GM is wrong[/quote]
Sure????,I have a genaral one for it (and it is similar to an ineq that I posted it) and I think i wasn't wrong (I used AMGM):lol:

[quote="Allnames"]Please check your proof,I think the step you use AM-GM is wrong[/quote]
Sure????,I have a genaral one for it (and it is similar to an ineq that I posted it) and I think i was wrong (I used AMGM):lol:[/quote]

I am not sure,I haven't seen the proof yet.Would you like to send your proof or genaral one to me? :maybe:
Thank you very much.

Mod please lock this topic. :mad:  This is Vo Quoc Ba Can 's problem in this month's issue of Mathematics and Youth Magazine!

@Lamminh: còn là học sinh LHP thì đừng bao giờ làm cái trò này nhé.Thật xấu hổ!

[quote="shfdfzhjj"][quote="lamminh_LHP_TC"]I think this is a pretty hard inequality, shall we discuss its solution[/quote]
I hope deadline has past,here is my proof :)  :[/quote]
Yes shfdfzhjj,here is also the proof that I said to you above,of course,the general one is also not hard,You can kill it by  the same way
$ \sum \frac{a}{\sqrt [k] {8c^3\plus{}1}}\ge 3^{\frac{k\minus{}2}{k}}$
Can_hang problem is the case $ k\equal{}2$ :)b

[quote="lamminh_LHP_TC"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1,$ then we have this inequality:

$ \frac {a}{\sqrt {8c^3 \plus{} 1}} \plus{} \frac {b}{\sqrt {8a^3 \plus{} 1}} \plus{} \frac {c}{\sqrt {8b^3 \plus{} 1}}\geq1.$[/quote]From $ abc \equal{} 1$, we may substitute $ a \equal{} \frac {x}{y},b \equal{} \frac {y}{z},c \equal{} \frac {z}{x},$ and the switching inequality is

$ 1\leq\sum_{cyc}{\frac {a}{\sqrt {8c^3 \plus{} 1}}} \equal{} \sum_{cyc}{\sqrt {\frac {x^5}{y^2(8z^3 \plus{} x^3)}}}.$

Using the Hölder inequality,

$ \left(\sum_{cyc}{\sqrt {\frac {x^5}{y^2(8z^3 \plus{} x^3)}}}\right)^2\sum_{cyc}{xy^2(8z^3 \plus{} x^3)}\geq\left(\sum{x^2}\right)^3,$

we have to prove $ \left(\sum{x^2}\right)^3\geq\sum_{cyc}{xy^2(8z^3 \plus{} x^3)}.$

But $ \left(\sum{x^2}\right)^3 \minus{} 8xyz\sum_{cyc}{yz^2} \minus{} \sum_{cyc}{y^4z^2}$

$ \equal{} \sum_{cyc}{(y \minus{} z)^2(2x^3y \plus{} 2x^2y^2 \plus{} 4xyz^2 \plus{} 2xz^3 \plus{} y^4 \plus{} y^2z^2)}\geq 0.$
[quote="Cai Ni"]OK.I use Holder,then AM-GM.

We may easily remember the similar problem.(The inequality from 42th IMO).[/quote]See : http://www.mathlinks.ro/viewtopic.php?t=15294
[quote="shfdfzhjj"]$ \sum_{cyc}{\frac {a}{\sqrt {8c^3 \plus{} 1}}}\geq\sum_{cyc}{\frac {a}{2c^2 \plus{} 1}} \equal{} \sum_{cyc}{\frac {x^3}{y(2z^2 \plus{} x^2)}}\geq\frac {(\sum{x^2})^2}{\sum{xy(2z^2 \plus{} x^2)}}\geq 1$[/quote]$ \Longleftarrow \left(\sum{x^2}\right)^2 \minus{} 2xyz\sum{x} \minus{} \sum_{cyc}{x^3y} \equal{} \sum_{cyc}{(y \minus{} z)^2\left(\frac {x^2}{2} \plus{} xz \plus{} y^2\right)}\geq0.$


Let X(i), i=1,…,n, be an exchangeable sequence of random variables with P[X(i)>0]=1 for each i. For 1 < k < n fixed, find
E[sum{X(i)(i=1 to k)}/sum{X(i)(i=1 to n)}]
Hint: first prove that, conditioned on sum{X(i)} (i=1 to n), the sequence X(i), i=1,…,n is still exchangeable.

Honestly I have no idea where to start here.