Just putting this out there:

Who else on AoPs likes playing ultimate and is able to play on a team?

I think you must rephrase the question: who here [b]doesn't like[/b] frisbee and why?! (I imagine most reasons will be "cuz I don't know what is that ..." :) ).

I have better obssessions than ultimate frisbee.... :D

I LOVE TABLE TENNIS!!!

Ultimite Frisbee would be pretty fun if one didn't have incompetent teammates who can't throw a frisbee or catch one for that matter.

I'm with Shinwoo: TABLE TENNIS RULES!!! :D

I find Ultimate Frisbee boring, team handball is better, but most of all I would prefer playing basketball, or football.

what????????

Ultimate Frisbee boring???

No Way, there is constant movement and you always have to work to get the disc. Also, the momentum can swing in an instant, and to score you can have a fast counter attack or a succession of tosses.
How can it be boring? argh.

I am not obsessed with Frisbee, I just think it is really fun. I also like to play table tennis, because that is also fun. WHat makes these sports fun is that they are not really completely mainstream sports, so there is more of a community between the players. I used to play soccer (still do, a bit) but that became really unfun after it became all about winning and there was loads of pressure as you got older and it became more professionalized.

Ultimate Frisbee is the best game.
It even beats out speedball and handball.
Unfortunately, almost no one in my area agrees with me...
 :D

I think the most weird poll would be: given a small soccer field, a frisbee and a soccer ball, which game would you play with your other 13 friends: soccer of frisbee (supposedly you'ld all vote for that ...)

Using my super-admin powers :P I made this topic into a poll with that question, so feel free to vote. I admit I would first vote soccer, and then frisbee :D

DOnt get me wrong, I love playing soccer, but ultimate is awesome. Tough decision Valentin  ;) There should be a both option. :D

I like playing recreational soccer, like pickup soccer.

I've added the coin toss option just for undetermined persons :D :D ... what's pickup soccer? (I think I know what it is, but it is called different in Romania) ... :?

Soccer is SO much more fun than Ultimite Frisbee. Well, i've only played ultimate frisbee in P.E., and we had to pass it to each teammate before we could score, so it got kindof boring...

oK, now I understand. Yes, that kind of frisbee is boring, because it was 'mandatory fun' so to speak.

In my school, there is an ultimate team, and we play teams around us. (I live in Central New Jersey, so there are a lot of schools) We host tournaments and play and it is very fun.

SOCCER SOCCER SOCCER SOCCER SOCCER!!!!!

I absolutely LOVE soccer!!! I really enjoy ultimate as well adn used to play every monday with people from my neighbourhood but then the guy that organised it moved to california so we don't play anymore. it's a cool game tho...but soccer beats it any day!

It would be a hard decision but ultimate frisbee rules
then again I'm with shinwoo and plokoon51, PING-PONG IS AWESOME! - the only problem is if you have a bunch of friends with you who all want to play at the same time, soccer or ultimate sounds a lot better

[quote="mcalderbank"]It is just when you get together with friends and set up goals with old clothes/trees/ anything and get a ball and just play with no explicit rules

Competitive soccer is playing with a registered team and having a coach and sponsors and all that jazz.[/quote]In Romania, we get together, rent a field (or find a free one :) ) for 1-2 hours, and play (with the rules) 6-6, 7-7 (depending on the length of the rented field) ..

That sounds like fun. Here in America, you can play soccer at Communtiy Park, which includes grass fields and sometimes goals.

With the rules, the games I play are often very physical and competitive and people tend to ignore some of the rules since there is no referee. :)  Anyone who says soccer is not a contact sport has never played.

As for frisbee, there are no referees, and in competitive play (tournaments) there is no referee and the fouls are called by other players.

What I like about both sports is that they are very simple, in terms of equipment
Soccer: Non-flat ball
Frisbee: Unbent disc

Goals can be fabricated and other equipment (shoes) help but are not required. Still, In the grand scheme of things, soccer is more elegant than frisbee, because incredibly skilled soccer players can do wonders with a ball. With frisbee, not so much.

I see nobody voted for coin toss yet, Valentin.
Somebody just did, aaaaaah, why? :P

I'm undecided. Frisbee is so frustrating when nobody can make the right passes or there's wind. Soccer, wow.... It's just running around with a ball and waiting to score at the last moment. Personally, I like football because there's action the WHOLE game. But if I was forced to vote.... I'd have to give it to Ultimate because you're more involved. Soccer is really dominated by a few players unlike Ultimate which is a TEAM sport.

I like Ultimate better because there are sudden turnarounds, that force you to run from one side of the field to the other. There's also some variety in skills in that you need some ability to catch, throw and somehow block. Final reason for my preference of ultimate over soccer: no one I know plays soccer.

[quote="Jhyon8149"]Soccer is really dominated by a few players unlike Ultimate which is a TEAM sport.[/quote]You are VERY VERY wrong here. For example, Steaua Bucharest, local Romanian team (which hasn't been playing more than a couple of games in the europeans cups each year since the beginning of the 90's), with a team of players all worth less (summed up) than each player (taken separate) of Valencia CF (en-titre UEFA Cup Holders, Champions of Spain - the most powerfull soccer championship in the world, en-titre holders of the SuperEuropean Cup of Soccer), with a 0-2 match disadvantage and still won and managed to qualify further in the UEFA Cup just a few days ago. And it was all a TEAM's work :)

I have to totally agree with Valentin on this one. Soccer is DEFINATELY a team sport. While there may be the several better players on the team, they couldn't accomplish what they do without the rest of the team there. Everyone has to work together. Passing, and plays definately require teamwork. When a team plays individually, they lose. There's no question about that. That's what happened to our varsity team this past season. They wouldn't work together, and they lost. By the end of the season, they got their act together and started to work together as a team, and they began to win. So yeah...I completely disagree with the statement that soccer is not a team sport.

[quote="MathFiend"]So yeah...I completely disagree with the statement that soccer is not a team sport.[/quote]

I never said that soccer is not a team sport. What I said was:

[quote="Jhyon8149"]Soccer is really dominated by a few players unlike Ultimate which is a TEAM sport. [/quote]

I can understand the part about how teammates are required for soccer. But what I was trying to say is that there is USUALLY a star player in soccer, one who will take the shot with 10 seconds to go. I mean, where would Brazil be without players like Ronaldo, Pele, Rivaldo, etc. Or what about France in 1998 without Zinedine Zidane? Argentina without Maradona? England without David Beckham? So sorry if people thought I was trying to say that soccer is not a team sport-trust me, I think it's a team sport. I was just trying to make the point that there is always that one special player in soccer while in Ultimate, there isn't.

[quote="Jhyon8149"]I mean, where would [...] England be without David Beckham?

 So sorry if people thought I was trying to say that soccer is not a team sport-trust me, I think it's a team sport. I was just trying to make the point that there is always that one special player in soccer while in Ultimate, there isn't.[/quote]
England without Beckham would have probably played in the Final against Greece at the Eurocup 2004, so you didn't get this quite right (Beckhan missed 2 penalties in that EuroCup being the worst player for England).

I belive it is the same situation in Ultimate (at a pro level, it's just that we haven't been exposed to so much publicity as we have been to soccer).

Yeah, I would have to agree with Valentin on this, at least a bit. In Ultimate, there is usually one or two people from whom the offense revolves around: the handlers.


$ x^{2}\plus{}y^{2}\plus{}xy\plus{}y\equal{}0$
$ x^{2}\plus{}x\plus{}y\plus{}2yz\equal{}0$
$ 3x^{2}\plus{}8y^{2}\plus{}8xy\plus{}8yz\minus{}2x\minus{}4z\equal{}2$


All of u in particular those selected please post ur marks.
Here are mine
[b]NSEP[/b]:
$ A_{1}$ : [b]-12[/b] :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o
$ A_{2}$:[b] 12[/b] :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o

[b]$ \boxed{TOTAL: 0}$ [/b]:mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:
[color=red]
[size=200][b]I WAS EXPECTING 132 in objective[/b][/size][/color]

[b]
NSEC[/b]:148 :rotfl:  :rotfl:
[b]
A MUST READ[/b]
A friend of mine who was [b] Absent[/b] got 50 .Can u beleive it a PERSON WHO IS ABSENT GOT HIS MARKS.A BIG **** ON THEM AND THEIR CHECKING

Did you really get 0?  :o
Well i got 95 + 19 = 114  :D . Yay i beat you  :P

yes i got 0 :rotfl:  :rotfl:  :rotfl:  :huh:  :huh:  :huh:  :huh:  :huh:  :huh:  :huh:  :huh:  :huh:  :huh:  :huh:
 whats ur NSEC mark

[quote="madness"]Did you really get 0?
Well i got 95 + 19 = 114   . Yay i beat you  [/quote]
Poi sollara mannan da :rotfl:  :rotfl:  :rotfl:

[quote="chemrock"]
Poi sollara mannan da :rotfl:  :rotfl:  :rotfl:[/quote]
anyways he would 'beat me' THAT Sh*** SAYS MY MARKS ARE 0 :huh:  :huh:  :huh:  :huh:  :starwars:  :starwars:  :starwars:  :starwars:  :starwars:  :starwars:  :starwars:  :bruce:  :bruce:  :bruce:

@pardesi : u believed that duh!!!  :P  I was lying all the way even though im sure i will get more than you  :P  And we still dont know our marks da  :D

it doesn't matter da if u get more than me or less.....what matters is I GOT 0 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
i am disheartened  :(

[quote="chemrock"]I know mine bcs chitra told me da.
And sundari miss was giving away the NSO certificates.
We have exam on 24th[/quote]
dei loose i am talking of the Physics,Chemistry and Astronomy olympiads da  :rotfl:

It happens that a boy appeared NSEP in Mumbai and got through...

But of course, since he had initially registered in Bhubaneswar, he was marked absent in that particular centre...

But even by remaining absent he managed to do some black magic....

Roll No     Signature                A     B     Grand Total
002         Absent                   32    18     50

Mystery of the not-missing marks....

yes actually i didn't tell it was u as i didn't want them to get confused :)

[quote="chemrock"]No no no I was saying that chitra told us our NSEP marks da.
I got 102 that's all :(  :(  :([/quote]
ok so what about others Shre,Chappli and all

Lemme ask tomorrow  :D  Im sure i would have got less than 50  :rotfl:

Of course you can guess who that boy was....

It was Soumyashant Nayak...

He gave NSEP in Mumbai and got through....
He did not give NSEP in Bhubaneswar..yet he gets 50 marks when he was clearly absent...

I got a pathetic 80 in physics(expected atleast 20 more)
got 150 in chem
still got through the 10% for both :rotfl:

by the way Nitish u got 78 in phy .....,

I am more concerned about getting my NSEP nation top 1% certificate than my marks...
But for any of them, I have to contact the Mumbai centre coordinator...
Damn!!

I have no idea how to get it... :(

i got 148 marks in total and i was expecting 235+
A friend of mine was expecting 250 and he got just 152 marks

those marks are your part A marks...

So you might as well be getting 235+ after adding part B...

taht means he got 152 in PART A :oops:
ok why aren't they mailing our Part B marks there is a seperate column for it in the marks sheet

[quote="anitsahu"]i dont understand a thing
my school is yet to recieve our marks[/quote]

Was your school your centre as well??

again same question was ur school a offical NSE centre??

sorry but had to make my post longer or wudnt have been able to submit it


Let $ 0<a,b,c<1$. Prove the inequality:

$ \frac{a}{1\minus{}a}\plus{}\frac{b}{1\minus{}b}\plus{}\frac{c}{1\minus{}c} \ge \frac {3 \sqrt[3]{abc}}{1\minus{} \sqrt[3]{abc}}.$

Determine the cases of equality.

[quote="moldovan"]Let $ 0 < a,b,c < 1$. Prove the inequality:

$ \frac {a}{1 \minus{} a} \plus{} \frac {b}{1 \minus{} b} \plus{} \frac {c}{1 \minus{} c} \ge \frac {3 \sqrt [3]{abc}}{1 \minus{} \sqrt [3]{abc}}.$

Determine the cases of equality.[/quote]

$ f(x)\equal{}\frac{x}{1\minus{}x}$ - increase and convex $ (0,1)$

$ LHS \ge 3 f(\frac{a\plus{}b\plus{}c}{3}) \ge 3f(\sqrt [3]{abc})$

[i]NOTE: Same proof as above but with a bit more detail about being increasing and convex.[/i]

    Let $\displaystyle f(x) = \frac{x}{1 - x}$. First we prove that $f$ is convex for $0 < x < 1$. Differentiating twice, we have
    $$
    f'(x) = \frac{1}{(1 - x)^2}, \quad \quad \text{and} \quad \quad f''(x) = \frac{2}{(1 - x)^3}.
    $$
    For $0 < x < 1$, $f''(x) > 0$ and thus $f$ is convex. Then, applying Jensen's inequality,
    $$
    \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c} = f(a) + f(b) + f(c) \geq 3f\left(\frac{a + b + c}{3}\right).
    $$

    Now we prove that $f$ is monotonic increasing. For $0 < a,b < 1$,
    \begin{align*}
        f(a) &\geq f(b) \\
    \iff \frac{a}{1 - b} &\geq \frac{b}{1 - b}\\
    \iff a(1 - b) & \geq b(1 - a) \\
    \iff a - ab &\geq b - ab \\
    \iff a & \geq b,
    \end{align*}
    so this is true.

    Using the monotonicity of $f$, we can apply AM-GM to get
    $$
    3f\left(\frac{a + b + c}{3}\right) \geq 3f(\sqrt[3]{abc}) = \frac{3 \sqrt[3]{abc}}{1 - \sqrt[3]{abc}},
    $$
    so we have
    $$
    \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c} \geq  \frac{3 \sqrt[3]{abc}}{1 - \sqrt[3]{abc}},
    $$
    as required.

    The use of AM-GM implies a necessary condition for equality is $a = b = c$, and we find that is sufficient as
    $$
    \frac{a}{1 - a} + \frac{a}{1 - a} + \frac{a}{1 - a} = \frac{3a}{1 - a} = \frac{3 \sqrt[3]{a^3}}{1 - \sqrt[3]{a^3}}.
    $$

Let $f(x)=\frac{x}{1-x}$ furthermore $f'(x)=\frac{1}{(1-x)^2}\text{ and }f''(x)=\frac{2}{(1-x)^3}>0, \forall x\in(0,1)$ thus $f$ is convex for $x\in(0,1)$
Now $\sum_{cyc}\frac{a}{1-a}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f\left(\frac{\sum_{cyc}a}{3}\right)=\frac{3\sum_{cyc}a}{3-\sum_{cyc}a}$
Therefore the inequity is equal to $\frac{3\sum_{cyc}a}{3-\sum_{cyc}a}\ge\frac{3\sqrt[3]{abc}}{1-\sqrt[3]{abc}}\Longleftrightarrow\sum_{cyc}a-\sqrt[3]{abc}\sum_{cyc}a\ge3\sqrt[3]{abc}-\sqrt[3]{abc}\sum_{cyc}a$
This transforms into $\sum_{cyc}a\ge3\sqrt[3]{abc}$ which is just $\text{AM-GM}$ $\blacksquare$.


$ \begin{cases} \sqrt{x^2\plus{}21}\equal{}\sqrt{y\minus{}1}\plus{}y^2\\ \sqrt{y^2\plus{}21}\equal{}\sqrt{x\minus{}1}\plus{}x^2 \end{cases}$

Assuming $ x,y\in\mathbb{R}$:

[hide="Calculus bash"]Add the two to get $ \sqrt {y - 1} + y^2 + \sqrt {y^2 + 21} = \sqrt {x - 1} + x^2 + \sqrt {x^2 + 21}$. Define
\[ f(u) = \sqrt {u - 1} + u^2 + \sqrt {u^2 + 21}\]$ f(x) = f(y)$. Obviously, $ x,y\ge1$ because of the $ \sqrt {x - 1}$ term and the fact that the LHS in each original equation is real. If we try WLOG $ x = 1$, we have $ y^2 = - 20$, so $ x,y > 1$.
\[ \frac {df}{du} = \frac {1}{2\sqrt {u - 1}} + 2u + \frac {u}{\sqrt {u^2 + 21}}\]This is trivially positive for $ u > 1$. So $ f(u)$ is one to one on $ (1,\infty)$, and since $ f(x) = f(y)$, $ x = y$. Obviously, (2,2) is a solution. We claim that this is unique. Consider the function
\[ g(x)=\sqrt {x - 1} + x^2 - \sqrt {x^2 + 21}\]The set of roots of $ g$ is equal to the set of solutions $ (x,x)$ to the system. Taking the derivative,
\begin{eqnarray*}
\frac{dg}{dx}&=&\frac {1}{2\sqrt {x - 1}} + 2x - \frac {x}{\sqrt {x^2 + 21}}\\
&=&\frac {1}{2\sqrt {x - 1}} + x\left(2 - \frac {1}{\sqrt {x^2 + 21}}\right)
\end{eqnarray*}$ x^2>1$, so we have:
\begin{eqnarray*}
x^2+21&>&22\\
&>&16\\
\sqrt{x^2+21}&>&4\\
\frac{1}{\sqrt{x^2+21}}&<&\frac14\\
2-\frac{1}{\sqrt{x^2+21}}&>&\frac74
\end{eqnarray*}Since $ x$ is positive, we have that the function is monotonically increasing for all $ x>1$, and thus has only one root for $ x>1$. Since all solutions $ (x,y)$ have $ x>1$, there is only one solution to the system, $ \boxed{(2,2)}$.
[/hide]

[quote="phuong"]$ \begin{cases} \sqrt {x^2 \plus{} 21} \equal{} \sqrt {y \minus{} 1} \plus{} y^2 \\
\sqrt {y^2 \plus{} 21} \equal{} \sqrt {x \minus{} 1} \plus{} x^2 \end{cases}$[/quote]
My method:
Subtracting sides by sides 1st equation to 2nd equation, we get: (x-y)(x+y)/[sqrt{x^2+21}+sqrt{y^2+21}]=(y-x)(y+x)+(y-x)/[sqrt{y-1)+sqrt{x-1}]
Now we have:(x-y)((x+y)/(sqrt{x^2+21}+sqrt{y^2+21})+(y+x)+1/[sqrt{y-1)+sqrt{x-1}])=0
We see that the sum in the parentheses >0 since y,x>=1
So the system above has the unique solution:x=y.
Replacing x=y into either 1st or 2nd equation, we have:sqrt{x^2+21}=sqrt{x-1}+x^2
This equation is hard to solve without calculus. I saw the 1st solution is 2 but not sure whether it is a unique solution or not. Can anybody give elementary solution to this equation? I'd like to see a nice solution. :wink:

I'm pretty sure it's a unique solution, and taking a really ugly derivative of $ f(x)\equal{}x^2\plus{}\sqrt{x\minus{}1}\minus{}\sqrt{x^2\plus{}21}$ to show that it is monotonically increasing would verify this.


This is from the 1999 Canadian Open. I tried solving it but I'm getting a wierd answer, and I've no idea why. Here's the question.

Determine all x which satisfy: $2\sin^3{x}+6\sin^2{x}-\sin{x}-3=0 , 0<x<2\pi$

Here's what I tried. First, for the sake of easier notation, let $b=\sin{x}$.

So we can write $2b^3+6b^2-b-3=0$
Luckily, that factors to $(b+3)(2b^2-1)=0$
Now substitute back in to get $(\sin{x}+3)(2\sin^2{x}-1)=0$
So either the first or second term equals zero. There's two options.

$\sin{x}+3=0$
$\sin{x}=-3$ ... this is clearly not in the domain. So no solution for the first part. On to the second part.

$2\sin^2{x}-1=0$
$\sin^2{x}-1=0$
$\sin^2{x}=1$
$\sin{x}=1$
$x=\sin^{-1}{1}=0$
$x=0$

But $x=0$ is outside of the domain of the original question, $0<x<2\pi$, so clearly I did something very wrong :( Can anyone point this out?

[hide="hint"]
doesnt that line(before you dropped the 2) kind of look like a double angle formula...
[/hide]

[quote="Altheman"][quote="Hikaru79"]

$2\sin^2{x}-1=0$
$\sin^2{x}-1=0$
$\sin^2{x}=1$
$\sin{x}=1$
$x=\sin^{-1}{1}=0$
$x=0$

[/quote]

you just dropped the 2 in that first line there...[/quote]

Oh, wow >.< I feel so stupid. I divided both sides by two and didn't notice there were two terms on the left side. Thanks for this :) I hate it when I make silly little arithmetic mistakes like so.

So, taking this mistake into account, I continue from above

\[ 2\sin^2{x}-1=0 \]
\[ \sin^2{x}=\frac12 \]
\[ \sin{x}=\frac{1}{\sqrt{2}} \]
\[ x=45 \]
Is this the only solution? Am I missing anything? :)

[quote="daveed"]when you took the square root, you forgot to include $-\frac{1}{\sqrt{2}}$[/quote]
 :blush: Wow, lots of mistakes today.

\[ \sin^2{x}=\frac12  \]
\[ \sin{x}=\pm\frac{1}{\sqrt{2}}  \]
$x=45$ or  $x=315$

Better?  :blush:  :blush:  :blush:

yes, that is correct, though probably easier to use $2\sin^2 x-1=-\cos 2x$ so the equation simply becomes $\cos 2x=0$.

[quote="goran_milicic"]You forgot 2k(PI)       k is in domain of Z numbers
+-45 + 360k
+-315 + 360k[/quote]

Can you explain? I'm not sure what "Z" numbers are...

goran_millicic, the question only asks for solutions $0 < x < 2 \pi$, so those solutions are invalid.

He means that k is $\in \mathbb{Z}$ - that is, k is an integer (positive or negative).  So $45 + 360, 45 - 360, 45 + 10 \times 360$ are all valid solutions... but not in $0 < x < 2 \pi$.

what about
$\frac{3\pi}{4}, \frac{5\pi}{4}$

yes, they are $\frac{(2n+1)\pi}{4}$, $n=0,1,2,3$ because they are all between 0 and $2\pi$ and satisfy $\cos 2x=0$.


group theory
abstract algebra
superior algebra
superior algebra solved

Hi - I am trying to solve a problem from section 2.7 of Herstein's Topics in Algebra.

If N,M are normal subgroups of G, prove that:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/398e52397cba25c6a5dcff7c2a97aa73.gif[/img]

Thanks.

Just find an onto homomorphism from $NM$ to $N/(N \cap M)$ with kernel $M$

Take the obvious epimorphism $N \to NM / M$, which has obviously kernel $N \cap M$. ;-)


inequalities
n-variable inequality
httpbit.ly3vjUcdg

This one easy  :lol:

By Caushy Swarz Inequality, Am-Gm,... algebraic substitution all of them kill this problem.
 :rotfl:  Applying C-S : (LHS)(Sum a_n+Sum b_n)>= (Sum a_n)^2 so our problem is killed if you want i can do that by Am-Gm too

Davron

$\sum \frac{a_i^2}{a_i+b_i} \ge \frac{\left(\sum a_i\right)^2}{\sum a_i+ \sum b_i}$

by Cauchy, but since $\sum a_i+\sum b_i = 2 \sum a_i$, we have

$\frac{\left(\sum a_i\right)^2}{\sum a_i+ \sum b_i} = \frac{\left(\sum a_i\right)^2}{2\sum a_i} = \frac{\sum a_i}{2}$.

My proof is based on only AM-GM.
\[ \frac{a^2_{i}}{a_{i}+b_{i}} + \frac{a_{i}+b_{i}}{4} \ge a_{i}\]

Therefore, $\sum \frac{a^2_{i}}{a_{i}+b_{i}} + \sum \frac{a_{i}}{2} \ge \sum a_{i}$. so we get our result.

This is just Titu's Lemma applied once :P

[quote=shobber]Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$.  Show that
\[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \][/quote]

Let $f(x) = \frac{1}{1+x}$. f is convex and we have:
\[\frac{1}{\sum a_{i}}\sum \frac{a_{i}^2}{a_{i}+b_{i}}=\sum \frac{a_{i}}{\sum a_{i}}f(\frac{b_{i}}{a_{i}}) \geq f(\frac{\sum b_{i}}{\sum a_{i}})=f(1)=\frac{1}{2}\]

Yet another proof:
In what follows $\sum $ denotes $\sum_{i=1}^{n}$.

Let $A=\sum \frac{a_i^2}{a_i+b_i}$ and $B=\sum \frac{b_i^2}{a_i+b_i}$. Note that $$A-B=\sum\frac{a_i^2-b_i^2}{a_i+b_i}=\sum a_i-b_i=\sum a_i-\sum b_i=0\implies A=B$$
Also note that for all $i$, $\frac{a_i^2+b_i^2}{a_i+b_i}\ge \frac{a_i+b_i}{2}\iff \left(a_i-b_i\right)^2\ge 0$, which is true.
Thus $$A=\frac 12 (A+B)\ge \frac 12 \sum \frac{a_i^2+b_i^2}{a_i+b_i}\ge \frac 12 \sum  \frac{a_i+b_i}{2}=\frac 12 \left(\sum a_i\right)$$
QED.

[quote=shobber]Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$.  Show that
\[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \][/quote]

[hide=hint]Titu lemma :D

[b]Solution.[/b]
By Titu's Lemma we have that $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{(a_1+a_2+\cdots +a_n)^2}{a_1+a_2+\cdots +a_n+b_1+b_2+\cdots +b_n}$,

$a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n \implies$ $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}.\blacksquare$

this problem can be easily solved by  using
Titu's Lemma
By titu's lemma we can say that,
$\sum \limits_{i=1}^{n}$$\frac{a_i^2}{a_i+b_i}\geq \frac{(a_1+a_2+a_3+ \cdots +a_n)^2}{a_1+b_1+a_2+b_2+ \cdots +a_n+b_n}$

$As,a_1+a_2+a_3+ \cdots +a_n=b_1+b_2+b_3+ \cdots +b_n$
(stated in the problem)

So,$a_1+b_1+a_2+b_2+ \cdots +a_n+b_n=2(a_1+a_2+ \cdots +a_n)$
Now,we can easily say that,

$\sum \limits_{i=1}^{n}$$\frac{a_i^2}{a_i+b_i}\geq \frac{(a_1+a_2+\cdots +a_n)^2}{2(a_1+b_1+a_2+b_2+ \cdots +a_n+ b_n}$

$\Rightarrow \sum \limits_{i=1}^{n}\frac{a_i^2}{a_i+b_i}  \geq \frac {a_1+a_2+ \cdots a_n}{2}$

This problem is a direct application of titu lemma

[quote=shobber]Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$.  Show that
\[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \][/quote]

Direct titu

[hide = :P]
T2 $$\sum \frac{a_i ^2}{a_i + b_i} \ge \frac{(\sum a_i)^2}{\sum a_i + \sum b_i} = \frac{(\sum a_i)^2}{2 \sum a_i} = \frac{\sum a_i}{2}$$[/hide]

Let $A=\sum_{k=1}^na_k$ and $B=\sum_{k=1}^nb_k$. We have $A=B$.
$$\sum_{k=1}^n\frac{a_k^2}{a_k+b_k}\overset{\text{T2}}\ge\frac{A^2}{A+B}=\frac A2.\square$$

This does not even require Cauchy-Schwarz or Engel form. Just $$\dfrac{a+b}{2}\geq \dfrac{2ab}{a+b}$$ is enough to prove this. It's easy to forget that Engel form was not as well known in 1991.
See the following
                                             \begin{align*}
							\dfrac{a_{1}^{2}}{a_{1}+b_{1}}+\ldots+\dfrac{a_{n}^{2}}{a_{n}+b_{n}}
								& \geq\dfrac{a_{1}+\ldots+a_{n}}{2}\\
							\iff \dfrac{a_{1}^{2}}{a_{1}+b_{1}}+\ldots+\dfrac{a_{n}^{2}}{a_{n}+b_{n}}+\dfrac{a_{1}+\ldots+a_{n}}{2}
								& \geq a_{1}+\ldots+a_{n}\\
							\iff \dfrac{a_{1}+\ldots+a_{n}}{2}
								& \geq \left(a_{1}-\dfrac{a_{1}^{2}}{a_{1}+b_{1}}\right)+\ldots+\left(a_{n}-\dfrac{a_{n}^{2}}{a_{n}+b_{n}}\right)\\
								& = \dfrac{a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{a_{n}b_{n}}{a_{n}+b_{n}}\\
							\iff a_{1}+\ldots+a_{n}
								& \geq \dfrac{2a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{2a_{n}b_{n}}{a_{n}+b_{n}}\\
							\iff \dfrac{1}{2}(a_{1}+b_{1}+\ldots+a_{n}+b_{n})
								& \geq \dfrac{2a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{2a_{n}b_{n}}{a_{n}+b_{n}}\\
							\iff \dfrac{a_{1}+b_{1}}{2}+\ldots+\dfrac{a_{n}+b_{n}}{2}
								& \geq \dfrac{2a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{2a_{n}b_{n}}{a_{n}+b_{n}}
						\end{align*}
The last inequality is evidently true.

Hello, I am new to Olympiad math . In very likely case that this is wrong, I would appreciate some feedback on what my mistake is.

$a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$ wouldn't this mean that  $a_n= b_n$?

Therefore, \[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \]

Can be also written as \[ \frac{a_1^2}{2(a_1)} + \frac{a_2^2}{2(a_2)} + \cdots + \frac{a_n^2}{2(a_n)} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \]

When both sides are multiplied by 2, we get that
$a_1 + a_2 + \cdots + a_n \geq $$a_1 + a_2 + \cdots + a_n$
I think this is a viable proof.
Please correct me if there are mistakes

@above,

$$a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$$
doesn't mean that $a_n= b_n$, for example
$$1 + 2 + 3 = 1 + 1 + 4$$

I presumed that there would be some kind of formula behind the a an b values. Are you saying that the a and b values can be from sets with no pattern?

@above
Yeah the only condition ${a_i}$ and ${b_i}$ must satisfy is that they are positive real numbers and,
$$a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$$

other than that they need not have any pattern. You can refer to previous posts for the correct solution

1990 USSR or 1990 All-Russian
[url]https://artofproblemsolving.com/community/u229790h1895500p27134878[/url]

[hide=My solution][quote=shobber]Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$.  Show that
\[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \][/quote]
Notice that by Cauchy-Schwartz we have that: $\sum_{i=1}^n\frac{a_i^2}{a_i+b_i}\ge\frac{(\sum_{i=1}^na_i)^2}{\sum_{i=1}^na_i+\sum_{i=1}^nb_i} $

Futher more recall that $\sum_{i=1}^na_i=\sum_{i=1}^nb_i$ $\Longrightarrow$  $\sum_{i=1}^n\frac{a_i^2}{a_i+b_i}\ge\frac{(\sum_{i=1}^na_i)^2}{2\sum_{i=1}^na_i}$ which implies the desired result.
And we are done! :starwars: [/hide]

Another:
$$\frac{a^2}{b} \ge 2a-b, \text{ because } (a-b)^2 \ge 0$$. Now inequality is obviously.

$\color{red} \boxed{\textbf{SOLUTION}}$

From the $\textbf{Engel Form of Cauchy-Schwarz Inequality,}$
$$\sum \frac{a_i^2}{a_i+b_i} \ge \frac{\left(\sum a_i\right)^2}{\sum a_i+ \sum b_i} = \frac{\left(\sum a_i\right)^2}{2\sum a_i} = \frac{\sum a_i}{2}\blacksquare$$

WHOA!! DO WHATEVER YOU WANT ...TITU,C-S, WHATEVER

Observe that from Titu's lemma, $$\sum_{i = 1}^n \frac{a_i^2}{a_i+b_i} \ge \frac{\left( \sum_{i = 1}^n a_i\right)^2}{\sum_{i=1}^n a_i + b_i} = \frac{\left( \sum_{i = 1}^n a_i\right)^2}{2\sum_{i=1}^n a_i} = \frac{\sum_{i=1}^n a_i}{2},$$ qed.$\square$


A woman was $ \frac{3}{8}$ of the way across a bridge when she heard the Wabash Cannonball Express behind her, approaching the bridge at 60 miles per hour. She quickly calculated that she could just save herself by running to either end of the bridges at top speed. What is her top speed in miles per hour?

If she goes $ \frac{3}{8}$ of the bridge's length either way, the train will be at the end of the bridge the woman came from.
If she went $ \frac{3}{8}$ towards the end away from the train and then goes $ \frac{2}{8}\equal{}\frac{1}{4}$ of the bridge length, the train will have traveled the entire bridge length.  Since those take the same time, the speed of the woman is $ 60\cdot \frac{1}{4}\equal{}\boxed{15}$

does anyone care to make that clearer? :P


Solve the equation $ a^{3} \equal{} 6(a \plus{} 1)$

hello, we have $ a_1\equal{}{2}^{2/3}\plus{}\sqrt [3]{2}$, the other solutions you can get by polynomial long division.
Sonnhard.

[quote="darkdieuguerre"][quote="Dr Sonnhard Graubner"]hello, we have $ a_1 \equal{} {2}^{2/3} \plus{} \sqrt [3]{2}$, the other solutions you can get by polynomial long division.
Sonnhard.[/quote]

What is the procedure for solving cubics with irrational zeroes?[/quote]This is a reduced cubic (notice that the $ a^2$ term is missing) so I bet that Dr Sonnhard used the cubic formula and somehow (out of luck) it turns out that the roots are "nice".


HAPPY NEW YEAR!!! TO INFINITY AND BEYOND!  :lol:

As the New Year has started (at least for me), this can be moved into the Solved Problems section. ;)

  Darij

wow, you must live in europe or china or something. Well, In Taiwan they don't celebrate until Chinese New Year in a month or two.  :lol:

Happy Day After New Year's!

AKA 1st day of the year 2007

Happy new year!
To celebrate the new year, I've deleted the NumberTheoryThread concerning this ;)

Happy new year to all.
We had a general announcement last year that was pretty suitable.

I am not sure why valentin didnt put one this year.


search
number theory unsolved
number theory

let a, b be integers greater than 1. Sequence $ x_n$ is defined as $ x_0\equal{}0;x_1\equal{}1$ and
$ x_{2n}\equal{}ax_{2n\minus{}1}\minus{}x_{2n\minus{}2}$ ; $ x_{2n\plus{}1}\equal{}bx_{2n} \minus{} x_{2n\minus{}1}$
for every positive integer n.
Prove that for every pair of positive integers m,n the product $ x_{n\plus{}m}x_{n\plus{}m\minus{}1}...x_{n\plus{}1}$ is divisible by the product $ x_{m}x_{m\minus{}1}...x_1$

[quote="kihe_freety5"]l is divisible by the product $ x_{m}x_{m \minus{} 1}...x_1$[/quote]

Sorry but isnt this .... $ x_{m}x_{m\minus{}1}$ ....2001 St Petersburg City Mathematical Olympiad-
http://www.mathlinks.ro/viewtopic.php?search_id=742871651&t=75956


find all ordered n-tuples $ (a_1,a_2,\cdots a_n )$ such that if $ s\equal{}a_1\plus{}a_2\plus{}\cdots\plus{} a_n$, then $ a_i | s$ for all $ i$

[hide]Wouldn't it just be the factors of perfect numbers?[/hide]

Are they positive integers, or any integers?

For i_like_pie: Not necessarily (even if they are just positive integers).  The numbers all divide their sum, but that doesn't mean that they together form the set of divisors of any number.  Consider, for example, $ (2, 4, 6)$.

[quote="K81o7"]Consider, for example, $ (2, 4, 6)$.[/quote]

This is exactly double the set $ (1, 2, 3)$, which is the set of proper divisors of $ 6$, so this isn't actually a great example.  :)

[hide="Discussion"] This is equivalent to

$ 1 \equal{} \frac{1}{b_1} \plus{} \frac{1}{b_2} \plus{} ... \plus{} \frac{1}{b_n}$

where $ b_k \equal{} \frac{s}{a_k}$.  See http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Graham_conjecture .[/hide]


geometry
3D geometry
number theory
least common multiple

Let the integers be A, B, C.

We want $ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{a\plus{}b\plus{}c}$
This is $ a^{2}\minus{}ab\minus{}ac\plus{}b^{2}\plus{}2bc\plus{}c^{2}\minus{}\frac{3bc(b\plus{}c)}{a\plus{}b\plus{}c}$

From here, I'm stuck :/

[quote="xxxyyyy"]Take three positive integers, either all different or at least one different from the other two. Divide the sum of the cubes of the three integers by the sum of the three integers.

Can the division result in a value equal to the product of any of the three integers without any remainder ?[/quote]

It does in some cases : $ \frac{1^{3}\plus{}2^{3}\plus{}3^{3}}{1\plus{}2\plus{}3}\equal{}6$ and $ 1,2$ and $ 3$ divide 6.
It does not in some cases : $ \frac{1^{3}\plus{}2^{3}\plus{}4^{3}}{1\plus{}2\plus{}4}\equal{}\frac{73}{7}$ which is not an integer.

So, could you precise your question : do you want all the cases $ (a,b,c)$ for which $ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{a\plus{}b\plus{}c}$ is an integer which can be divided by $ a$, by$ b$ and by $ c$ ?
Something else ?

Thanks for any precision.

[quote="pco"][quote="xxxyyyy"]Take three positive integers, either all different or at least one different from the other two. Divide the sum of the cubes of the three integers by the sum of the three integers.

Can the division result in a value equal to the product of any of the three integers without any remainder ?[/quote]

It does in some cases : $ \frac{1^{3}\plus{}2^{3}\plus{}3^{3}}{1\plus{}2\plus{}3}\equal{} 6$ and $ 1,2$ and $ 3$ divide 6.
It does not in some cases : $ \frac{1^{3}\plus{}2^{3}\plus{}4^{3}}{1\plus{}2\plus{}4}\equal{}\frac{73}{7}$ which is not an integer.

So, could you precise your question : do you want all the cases $ (a,b,c)$ for which $ \frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{a\plus{}b\plus{}c}$ is an integer which can be divided by $ a$, by$ b$ and by $ c$ ?
Something else ?

Thanks for any precision.[/quote]

yes, all the cases. But note that two of these numbers can be same.

[quote="t0rajir0u"][quote="xxxyyyy"]Can the division result in a value equal to the [i]product of any of the three integers[/i] without any remainder ?[/quote]

Can you clarify what you mean?  Do you mean a number of the form $ kabc$ or $ k\cdot\text{lcm}(a, b, c)$ or $ ka, kb, kc$?[/quote]

The numbers can be $ r$, $ s$ and $ t$ which have no relation to $ a$, $ b$, $ c$.

[quote="Kyron"][quote="   yyyy"]Can the division result in a value equal to the product of any of the three integers without any remainder ?[/quote]

Could you clarify this part? In specific the "without any remainder part." Does this mean that if the fraction is $ \frac{55}{3}$ that we treat it as 18?

Thanks in advance for the info.[/quote]

It means when you do the division, there is no remaider. Like $ 42$ is completely divisible by $ 3$ but $ 55$ is not.


If $x,y,z>0$ prove that $x^3(y+z)+y^3(x+z)+z^3(x+y)\geq 6xyz\sqrt[3]{xyz}$.

Yes, he means   xyz and not  abc.

  [u]Babis[/u]

Mmm, setting $x=y=z=a=b=c=1/2$  :D reverses the inequality (which is then false without extra conditions).

I fixe the ineq. Sorry for the mistake before.

Isn't this a completely trivial problem - just apply AM GM once?


Is it just me, or can't x + y + z go to negative infinity, since x and z can be as negative as they want, and y can be very close to zero to 'pull' the LHS down again?

assume x,y and z to be positive. now try.

[quote="numbergeek"]If
$x^{2}y^{3}z^{4} = 120$

find the minimum value of $x + y + z$ ?

x,y and z are real nos.

Ok Java man, I correct the condition as
x,y and z being positive nos.

Now solve. I saw this question somewhere. I shall recheck again but i remember, we must take x,y,z to be positive nos.[/quote]

[hide]Consider the AM-GM Inequality.

Let the numbers be $\frac{x}{2}$, $\frac{x}{2}$, $\frac{y}{3}$, $\frac{y}{3}$, $\frac{y}{3}$, $\frac{z}{4}$, $\frac{z}{4}$, $\frac{z}{4}$, and $\frac{z}{4}$.

So, by AM-GM, I will simplify all at once at the same time

$\frac{x+y+z}{9}$>=$((\frac{x^{2}}{4})(\frac{y^{3}}{27})(\frac{z^{4}}{256}))^{\frac{1}{9}}$

So $(x+y+z)$>=$9(\frac{5}{1152})^{\frac{1}{9}}$

I know I didn't simplify but the answer is

$9(\frac{5}{1152})^{\frac{1}{9}}$[/hide]

Tell me about it.

If this was Mathcounts, though it shouldn't be, what would our correct have to be in order to comply with "Mathcounts Rules".

[hide="hmm..."][quote="mathgeniuse^ln(x)"]Consider the AM-GM Inequality.

Let the numbers be $\frac{x}{2}$, $\frac{x}{2}$, $\frac{y}{3}$, $\frac{y}{3}$, $\frac{y}{3}$, $\frac{z}{4}$, $\frac{z}{4}$, $\frac{z}{4}$, and $\frac{z}{4}$.

So, by AM-GM, I will simplify all at once at the same time

$\frac{x+y+z}{3}$>=$((\frac{x^{2}}{4})(\frac{y^{3}}{27})(\frac{z^{4}}{256}))^{\frac{1}{9}}$[/quote]
Shouldn't the LHS of the inequality be $\frac{x+y+z}{9}$ instead of $\frac{x+y+z}{3}$?[/hide]

Yes it should, I was being stupid.  I will edit that.

Thanks for catch.

[quote="#20002 pwns"]yes, it is correct. That wasn't a very fun problem![/quote]

Actually, I got that question wrong. :(

FRT was aware of my mistake.


by the way, i love chemistry  :lol: and now i feel better about bio :maybe:

Seriously, was there any reason to post that?

Ah i read it & now i'm thinking i sholdn't have read it

Guys I hope my post is the last reply to this silly topic.  :(  Now I hope someone will lock it.

hey man---will this topic ever end!!!!!!!

I don't like Star Wars and I don't like this thread but the emoticon is cool. :starwars:

:spam:  :spam:

[hide]   :rotfl: [/hide]


Given a sequence $a_1,a_2,\ldots $ of non-negative real numbers satisfying the conditions:

1. $a_n + a_{2n} \geq 3n$;
2. $a_{n+1}+n \leq 2\sqrt{a_n \left(n+1\right)}$

for all $n\in\mathbb N$ (where $\mathbb N=\left\{1,2,3,...\right\}$).

(1) Prove that the inequality $a_n \geq n$ holds for every $n \in \mathbb N$.
(2) Give an example of such a sequence.

[b](a)[/b] We have

$a_{n+1} + n \leq 2\sqrt{a_n\cdot\left(n+1\right)}$

by our problem hypothesis. On the other hand, the AM-GM inequality yields

$2\sqrt{a_n\cdot\left(n+1\right)} \leq a_n + \left(n+1\right)$.

Thus, $a_{n+1} + n \leq a_n + \left(n+1\right)$, so that $a_{n+1} - a_n \leq 1$. Similarly, $a_{n+2} - a_{n+1} \leq 1$, $a_{n+3} - a_{n+2} \leq 1$, ..., $a_{2n-1} - a_{2n-2} \leq 1$, $a_{2n} - a_{2n-1} \leq 1$. Adding these inequalities together, we get

$a_{2n} - a_n$
$= \left(a_{2n} - a_{2n-1}\right) + \left(a_{2n-1} - a_{2n-2}\right) + ... + \left(a_{n+3} - a_{n+2}\right) + \left(a_{n+2} - a_{n+1}\right) + \left(a_{n+1} - a_n\right)$
$\leq 1 + 1 + ... + 1 + 1 + 1 = n$,

so that

$a_n - a_{2n} \geq -n$.

Adding this to $a_n + a_{2n} \geq 3n$, we get $2a_n \geq 2n$, so that $a_n \geq n$, what proves [b](a)[/b].

[b](b)[/b] Just take $a_n = n+1$ for every n. The condition $a_n + a_{2n} \geq 3n$ is clearly satisfied ($\left(n+1\right)+\left(2n+1\right) = 3n+2 \geq 3n$), and the condition $a_{n+1} + n \leq 2\sqrt{a_n\cdot\left(n+1\right)}$ is also satisfied (even the $\leq$ sign can be replaced by the = sign, since

$a_{n+1} + n = \left(n+2\right) + n = 2\left(n+1\right) = 2\sqrt{\left(n+1\right)^2}$
$= 2\sqrt{\left(n+1\right)\cdot\left(n+1\right)} = 2\sqrt{a_n\cdot\left(n+1\right)}$

).

  Darij

I have an alternate proof of (a).

Assume that, for the sake of contradiction, that for some $n$, $a_n<n$.
Thus, $a_{n+1}+n<2\sqrt{a_n(n+1)}<2\sqrt{n^2+n}<2\sqrt{n^2+n+\frac{1}{4}}<2n+1$.
Thus, $a_{n+1}<n+1$.
Repeating this $n-1$ times, $a_{2n}<2n$.  Thus, $a_{2n}+a_n<3n$, contradiction.  And so $a_n\ge n$ for all positive integers $n$.

[quote="mathkhtn"][quote="TT-P"][quote="GoldenFrog1618"]Assume that, for the sake of contradiction, that for some $n$, $a_n<n$.
Thus, $a_{n+1}+n<2\sqrt{a_n(n+1)}<2\sqrt{n^2+n}<2\sqrt{n^2+n+\frac{1}{4}}<2n+1$.
Thus, $a_{n+1}<n+1$.
Repeating this $n-1$ times, $a_{2n}<2n$.  Thus, $a_{2n}+a_n<3n$, contradiction.  And so $a_n\ge n$ for all positive integers $n$.[/quote]
This proof is wrong. Because you have $a_n<n$ for some n, and $a_{n+1}<n+1$ for some n only, too. So we cannot "repeat" them for other n.[/quote]

No I don't think so.........Because we can assume n + 1 = A ....and it is right with A+1 so on[/quote]
No, your proof is not true, sorry.


Identify the major product  :maybe:  :maybe:

[hide]Ethylene, together with some 3-methylcyclohexene.[/hide]


Let an be the nth positive integer k such that the greatest integer not exceeding sqrt(k) divides k, so the first few terms of {a_n} are {1,2,3,4,6,8,9,12, . . .}. Find a_10000

[hide="Solution"]Suppose $ \lfloor\sqrt{k}\rfloor =m$.  Then $ m|k$, and $ m^{2}\le k<m^{2}+2m+1$.  The possible values of $ k$ are $ m^{2},m^{2}+m,m^{2}+2m$.  Therefore, $ a_{1000}=334^{2}$ and $ a_{10000}=3334^{2}$.[/hide]

$ 3334^{2}=111155556$, which is totally pointless yet pretty cool.

[quote="Lazarus"]$ 3334^{2}=111155556$, which is totally pointless[/quote]

It actually has a pretty interesting interpretation in terms of $ 10$-adic numbers.  And even if you don't want to get that advanced, it's a concrete realization of the fact that $ 0.\overline{3}^{2}= 0.\overline{1}$.


induction
inequalities
combinatorial geometry
triangle inequality

There are 2n balls in the plane such that no three balls are on the same line and such that no two balls touch each other. n balls are red and the other n balls are green. Show that there is at least one way to draw n line segments
by connecting each ball to a unique different colored ball so that no two line segments intersect.

[hide]Induction!

base case: n = 1,  I hope you see why this is true.

assume for 2n balls, this can be done.

Look at 2n+2 balls, since we have a finite number of balls in the plane, if we take their convex hull, it has a finite number of vertices on the boundry.   Remove any two adjacent points on this boundry, no three collinear means there are non in between them, and by def of a convex hull it follows that no other line can intersect this.  Once these are removed we use the inductive hypothesis, and we have a pairing for those 2n points, and we simply add back in the two we removed, which again we have guaranteed not to violate the condition.[/hide]

[hide="another way"]

Choose the way of picking the segments such that the total sum of the lengths of the segments is minimal. Assume that two segments intersect. Then we have two red balls A and B and two green balls C and D such that segment AC intersects BD at some point, call it E.

By the triangle inequality:
$ AE+DE>AD$
$ BE+CE>BC$
$ AE+BE+CE+DE>AD+BC$
$ AC+BD>AD+BC$

So choosing AD and BC instead of AC and BD would give a lower total sum of all segments, a contradiction; so none of the segments intersect.
[/hide]

The first looks like it would be a good solution, but I don't think it generalizes to the actual problem given, where you have $ n$ red balls and $ n$ green balls and you need to connect different colors.

EDIT: Correction.

tjhance's solution is completely correct.  What do you see wrong with it?

nateharman's solution, on the other hand, assumes that he can find two adjacent balls on the convex hull of different color, which is not true.


hi-
use sum or diffenrece identiteis to solve the equation exactly

cos 3x cos x= sin 3x sin x

how do i start/do this problem. i have the list of identites in front of me but i dont know what do to.
can you help me?
thanks

i was told to use the identity cos(a+b)=cos a cos b -sin a sin b
with a= 3x and b=x
but i dont get how to apply it

[quote="IluvChocolate"]
$ cos 3x cos x\equal{} sin 3x sin x$
[/quote]

$ \cos 3x \cos x \minus{} \sin 3x \sin x \equal{} \cos (3x \plus{} x) \equal{} \cos 4x \equal{} 0$

thank you for the help but i am still a little confused....how did you go from cos 3x cosx - sin3x sinx to just cos(3x+x) what happened to the sin? then how did you go from cos 4x to 0??
thanks

[hide]$ \cos(a \plus{} b) \equal{} \cos{a}\cos{b} \minus{} \sin{a}\sin{b}$, so you just recognize terms.[/hide]


calculus
analytic geometry
graphing lines
slope

[quote="g1sk8terz"]How many points are there on the curve $f(x)=4x^{5}-3x^{4}+15x^{2}+6$ whose tangent line pass through the origin?[/quote]
Won't this require calculus??
[hide] The tangent line has slope $f'(x)$ and passes through point $(x,f(x))$. So, $f(x)=f'(x)\cdot x+b\Longrightarrow 4x^{5}-3x^{4}+15x^{2}+6=x(20x^{4}-12x^{3}+30x)+b\Longrightarrow b=-16x^{5}+9x^{4}-15x^{2}+6$. So, we need to find the roots of $-16x^{5}+9x^{4}-15x^{2}+6$. [/hide]

[quote="g1sk8terz"]How many points are there on the curve $f(x)=4x^{5}-3x^{4}+15x^{2}+6$ whose tangent line pass through the origin?[/quote]

i suppose i could do this through calculus

$f(x)=4x^{5}-3x^{4}+15x^{2}+6$
$f'(x)=20x^{4}-12x^{3}+30x$

$x(20x^{4}-12x^{3}+30x) = 4x^{5}-3x^{4}+15x^{2}+6$
$20x^{5}-12x^{4}+30x^{2}= 4x^{5}-3x^{4}+15x^{2}+6$
$16x^{5}-9x^{4}+15x^{2}-6 = 0$

go do it from there or something... might have made a mistake.

edit: yeah, he beat me to it.

good job guys..

..im sorry i posted this problem in this forum... i just realized there was a thread.." do not post calculus problems in this forum!"...sorry!

but good job so far guys..


Does there exist a countable subset of $ \mathbb{R}$ whose set of limit points is also countable?

$ \bf Z$ is its own boundary in $ \bf R$.

What is the boundary of a convergent sequence?  :wink:

Sorry; that's not the question I wanted to ask.  Edited.

Clearly I can construct a sequence such as $ 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, ...$ whose set of limit points is a countable set, but I have not yet been able to construct an intermediate example between the set of limit points of a finite "union" of convergent sequences and $ \mathbb{Q}$.

Edit:  Oy.  The countable union of disjoint convergent sequences works just fine, doesn't it?  (I'm wondering if this still works on $ [0, 1]$.)

It works as long as the sequences are well separated so that they don't interact with each other and create new limit points ($ %Error. "mathh" is a bad command.
{Q}$ is the countable union of the constant sequences, and it has a lot of limit points). I don't see why it doesn't work on $ [0,1]$. Choose the nth sequence to be contained in $ (1 - 2^{ - n}, 1 - 2^{ - n - 1})$ and converge to the middle, which only creates the additional limit point $ 1$ (unless I am wrong  :P )


Sa se arate ca pentru orice numar natural $n\geq 2$, exista n puncte conciclice astfel incat distanta oricaror 2 puncte este un numar intreg.

[quote="harazi"]Cezar, din fericire matematica nu prea are tendinte din acestea de stadion, asa ca te-as ruga ca in locul unor expresii din acestea ''gigesti'' si ''becalesti'' sa prezinti altfel satisfactia ta pe forum. Nu de altceva, dar vazand ultimele ''tendinte'' ale forumului romanesc (care este, citatie din clasici in viata, din pacate), nu m-ar mira ca in curand sa facem dedicatii in manele si seminte langa problemele propuse. Parerea mea... Cat despre problema, a fost dicutata de multe ori pe forum, use search button. S-a aratat chiar ca exista o multime densa in cercul unitate cu distantele intre punctele sale rationale.[/quote]
Subscriu la cele scrise mai sus.

[quote="harazi"]Cezar, din fericire matematica nu prea are tendinte din acestea de stadion, asa ca te-as ruga ca in locul unor expresii din acestea ''gigesti'' si ''becalesti'' sa prezinti altfel satisfactia ta pe forum.[/quote]

Corect, ai dreptate aici. Am vrut sa fiu in ton cu moda cum se zice. Auzi, sunt curios de ce spui de stadion? Sa stii ca la noi in Parlament se fac lucruri cu mult mai de joasa speta decat pe stadion. Te asigur. :wink:
Cat despre Becali, no comment. Sa nu-l mai bagam atata in seama pe "voievod". Daca pana si aici pe forum aduci vorba de el, deja incep s a ma sperii. Uite chiar azi a fost decernat cu locul 2 la politicianul anului la
Gala "10 pentru Romania", dar in Romanica nostra nu tre' sa stii prea multe ca sa fii mare si tare.

P.S. Ca tot zici tu de Becali, vezi ca e posibil ca "voievodul" nostru sa iasa si presedinte.  :rotfl:

dupa ce ca forumul asta e plin de mortaciuni, se mai ia lumea si de ala care incearca sa mai destinda si el atmosfera asta de RIP. multumim domnule Cezar!  :P

Gabi, imi dai si mie un link, ceva, te rog? :)

Citeste postul lui Pierre. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=distances+integer&t=24838


a uniform chain of mass $ m$ and length $ l$ overhangs a table with two third part on the table.
find the work to be done by a person to put the hanging part back on the table

Is there any friction?
[hide="Hint:"]Mechanical energy is conserved[/hide]

Why do we need mechanical energy here? And why would it be conserved? We are injecting work into the system...

We just need to convert the gravitational potential energy of the hanging part, i.e. our work will be the amount of gravitational potential energy the hanging part has:

Let $ x$ be the vertical coordinate from the edge of the table. Then assuming uniform mass distribution throughout the chain the mass of a small piece will be: $ dm\equal{}\frac{m}{l}dx$ So the total potential energy will be:

\[ \int_{0}^{\frac{l}{3}}\frac{m}{l}gx\,dx\equal{}\frac{mgl}{9}\]

yup yeahnigga even i am getting the same answer as u got
but at the back it is given $ \frac {mgl}{18}$ .

well sorry if this is dumb
cant i do without calculus
like this $ (\frac{m}{3})(\frac{l}{3})(g)$

yeahnigga, why do u have to assume the chain to have uniform mass isnt that the meaning of  " uniform"

To find the PE,just assume that the Centre of mass of the hanging part lies at half the height( $ \frac{l}{3*2}$).
Mass of the hanging part = $ \frac{m}{3}$
Therefore PE =$ \frac{l}{3*2} * \frac{m}{3} * g$
              =$ \frac{mgl}{18}$

the mass hanging is$ \frac{m}{3}$ and you have to raise it to a height of $ \frac{l}{3}$
as the initial and final velocities of the  that chain are $ 0$ .by work energy theorem we have $ dK\equal{}0$
so work done by gravity is negative of work done by a person raising it.
as the chain is raised to a height of $ \frac{l}{3}$  ,the work done by the gravity as it is being raised is $ \minus{}(\frac{m}{3})(\frac{l}{3})(g)$

[quote="srinath.r"]the mass hanging is$ \frac {m}{3}$ and you have to raise it to a height of $ \frac {l}{3}$
as the initial and final velocities of the  that chain are $ 0$ .by work energy theorem we have $ dK \equal{} 0$
so work done by gravity is negative of work done by a person raising it.
as the chain is raised to a height of $ \frac {l}{3}$  ,the work done by the gravity as it is being raised is $ \minus{} (\frac {m}{3})(\frac {l}{3})(g)$[/quote]

No,
We need not raise the entire part by $ \frac {l}{3}$.
We raise the Centre of mass by $ \frac {l}{3*2}$
See it like this:
The part near the table will be raised only by a small distance,while the bottommost point has to be raised by $ \frac {l}{3}$.
So, we just see the COM,as it is uniform wire.

my ideology and approach and setup of the integral was correct of course, i just calculated the integral wrong. So the integral:

\[ \int_0^{\frac{l}{3}}\frac{m}{l}gx\,dx\equal{}\frac{mg}{l}\left[\frac{x^2}{2}\right]_0^{\frac{l}{3}}\equal{}\frac{mg}{2l}\frac{l^2}{9}\equal{}\frac{mgl}{18}\]

You guys could have noticed... :D

ya thanks arjunarul .i realize my stupidity. :blush:
wow how stupid i was

[quote="yeahnigga"]Why do we need mechanical energy here? And why would it be conserved? We are injecting work into the system...
[/quote]

In the sense $ W_{done} \equal{} \Delta M.E$

Which is what Arjun did.

well sorry if i am dumb
arjunarul has [b]assumed[/b] centre of mass of the  lower part of chain to be at the middle  of $ \frac{l}{3}$
but why do we have to assume so .the centre of mass is at  the middle for $ this$ continuous body.
i am just asking if there is any real significance of the word [b]asuumed[/b]

The centre of mass is at the geometric centre for a body with uniformly distributed mass. Perhaps Arjun did not notice "A [b]uniform[/b] chain of...." so he [b]assumed[/b] it is uniform.

But that is not "mechanical energy is conserved"..

It doesnt make any sense talking about conservation here, since there isnt any dynamics involved, nor there is a change in the system...

You are mixing it up with the "energy stored in the system", there is no conservation here.. that is a different concept

Energy is injected if you consider only the chain as your system. If you (who's pulling the chain), alongwith the chain are the system, the work you do is utilised to increase the PE of the chain. So you see, Energy is conserved!


$ R,r,s$ are the circumradio,inradio an semiperimeter of $ \Delta ABC$..Show

$ 2s^2 \ge 27Rr$

$ s^2 \ge 12Rr \plus{} 3r^2$ ( this one is easy  :P )

I think it's rather easy...
By Gerretsen ：s^2≥16Rr-5r^2....and the two inequalities becomes R≥2r,which is obvious.

The first :  $ a \plus{} b \plus{} c\ge 3\sqrt [3]{abc}$ and you know that $ a \plus{} b \plus{} c \equal{} 2s$ , $ abc \equal{} 4Rsr$ a.s.o.

The second : $ s^2\equal{}\left[(s\minus{}a)\plus{}(s\minus{}b)\plus{}(s\minus{}c)\right]^2\ge 3\sum[(s\minus{}b)(s\minus{}c)] \equal{}3\cdot r(4R\plus{}r)$ a.s.o.

You can use the relations $ ab\plus{}bc\plus{}ca\equal{}s^2\plus{}r^2\plus{}4Rr$ or $ r_a\plus{}r_b\plus{}r_c\equal{}4R\plus{}r$

and $ S\equal{}\sqrt {s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\equal{}sr\equal{}(s\minus{}a)r_a\equal{}(s\minus{}b)r_b\equal{}(s\minus{}c)r_c$ .


[quote="swanny1080"]\combi _{ 1 }{ a },...\combi _{ n }{ a }are\quad not\quad zero(?)integer.\\ \\ such\quad that\combi _{ 1 }{ a }+\frac { 1 }{ \combi _{ 2 }{ a }+\frac { 1 }{ \combi _{ 3 }{ a }+...\frac { 1 }{ \combi _{ 1 }{ a }+\frac { 1 }{ x } } } }=x\\ prove\quad that\quad n\quad is\quad even\quad number

[/quote][/quote]
What do you mean? :wink:


number theory
relatively prime
number theory unsolved

[quote="PTynan89"]Then, multiplying by the product of all $a_{i}$, we have that
$(a_{1})^{2}*a_{3}*a_{4}*...*a_{n}+(a_{2})^{2}*a_{4}*a_{5}*...*a_{n}*a_{1}+...$ is divisible by $a_{i}$, $\forall i \in \mathbb{N}_{n}$.

Well, it's also obvious here that for any given $a_{i}$, every term is divisible by it, other than the term containing the factor $(a_{i-1})^{2}$.  Therefore, $a_{i}$ must divide this other product as well, in order to divide the sum.  And since all of the terms in that product are relatively prime to $a_{i}$, and $a_{i}$ divides the product, then $a_{i}= 1$.  Since this was for an arbitrary $i$, it holds $\forall a_{i}$.

Now, $a_{i}= 1$, $\forall i \in \mathbb{N}_{n}$ works, with the result that the sum is equal to $n$, and is therefore the only solution.[/quote]
Condition $GCD(a_{1},a_{2},...,a_{n})=1$ isn't equivalent to condition, that $a_{1}, a_{2}, ..., a_{n}$ are relatively prime.

Let n is even and $a_{i}=a, i-odd, \ a_{i}=b, i-even$, were $a|n,b|n, (a,b)=1$. Then $GCD(a_{i})=1$, and $\sum_{i}\frac{a_{i}}{a_{i+1}}=\frac{n}{2}\frac{a^{2}+b^{2}}{ab}$ is integer.


Check the sign of this expression
$\huge P=\log_{a}b-\log_{c}d$

[quote="Svejk"]it's very general .you must have some conditions for a,b,c,d[/quote]
help me ??? :roll:

As [b]Svejk[/b] said, you have to put some more condition for $a,b,c,d$ then the problem can be solve?

Supporting the previous answers:

The question is nearly the same as to ask: check the sign of the expression $Q=ab-cd$

Indeed

$P=\frac{\ln b \cdot \ln c-\ln a \cdot \ln d}{\ln a \cdot lnc}$

So P>0 if and only if (for $0<a,c\ne 1$)

$(a-1)(c-1)>0$ and $\ln b \cdot \ln c-\ln a \cdot \ln d >0$ or

$(a-1)(c-1)<0$ and $\ln b \cdot \ln c-\ln a \cdot \ln d <0$

P.S. If you don't know what does mean $\ln x$ stand instand it $\log_{e}x$ for any $e>1$.


Consider a circle $(O)$ and a point $P$ lie outside $(O)$ . The two tangents from $P$ to this circle touch $(O)$ at $A ,B$.
Any line passing through $P$ cuts $(O)$ at $C$ and $D$. Prove that $AB$ is the $A$-symmedian of triangle $ADC$.

I have:AC*BD=AD*BC
But accord ptoleme'S theorem.wehave:AC*BD+AD*BC=AB*CD,therefore
AC*BD=AD*BC=AB*CD/2.
Let M be the midpoint of CD then AC*BD=AB*CM
We have:triangle ACM is similar to triangle ABD.Therefore $angleMAC =angleDAB$->AB is the -symmedian of triangle ADC.
To cuopbientcd:this problem is nice but not hard

$AB$ is the polar of $B$, so $(BCSD)$ is harmonically, where $S$ is the intersection of $AB$ and $CD$, but $PA$ is tangent to $ACD$, so the conclusion follows.


The ratio of the radii of the unshaded semicircles is 1:2, and the unshaded semicircles are tangent to each other. What is the number of square inches in the area of the shaded region? Express your answer in terms of $ \pi$.

[asy]path a,b,c;
a = Arc((0,0),6,-180,0)--cycle;
b = Arc((2,0),4,-180,0)--cycle;
c = Arc((-4,0),2,-180,0)--cycle;
fill(a,grey); fill(b,white); fill(c,white);
draw(a); draw(b); draw(c);
draw((-6,.5)--(6,.5),Arrows);
label("12 in",(-6,.5)--(6,.5),N);[/asy]

$ \frac {1}{2} \cdot 6^2\pi \minus{} \left(\frac {1}{2} \cdot 2^2\pi \plus{} \frac {1}{2} \cdot 4^2\pi\right) \equal{} 18\pi \minus{} (2\pi \plus{} 8\pi) \equal{} \boxed{8\pi}$.


geometry
geometric transformation
reflection

two points $A,B$ are fixed .$A$ is reflected at all straight lines through $B$. find the locus of all

images..

We call $C_0$ the circle with midpoint $B$ and $r=AB$.
We take just any straight line $l$ through point $B$.
We know that $A \in C_0$, we call $A'$ the reflection of $A$ at line $l$, and call $O$ the intersection of $AA'$ and $l$. Now, since $AA' \bot l$, $A \in C_0$ and $AO=OA'$, we can conclude that $AA4$ is a chord of $C_0$, and therefor that $A' \in C_0$.

So it is proven that circle $C_0$ is the locus we are searching.


Just substitute $ a=x/y, b=y/z, c=z/x$, then it's a trivial case of Cauchy...


AMC AIME USAMTS MIT college USA(J)MO USAMO

Just wondering how ya'll pronounce AIME; is there a standard/correct pronunciation, or is it up to the speaker to determine it? What pronunciation is most popular?

I put Amy, but for some reason I pronounce everything phonetically (I don't know any better).

Anyone like the idea of pronoucing USAMTS "you-sam-mites"?

i have just gotten used to AIME as amy- i guess cause that's what my teacher called it, but i was very surprised when i was talking to someone who called it the A I M E, it just didn't sound right :(

i say aim for the AIME and u-s-a-m-t-s for USAMTS

I have other two pronounciation problems:
1: Cauchy
2: L'hospital

[quote="bookaholic"]A. I. M. E. and you-samts. No, I don't know why I spell the letters out for one and pronounce the other phonetically. I just do.[/quote]

Same with me.

So I keep thinking you're LyenelleYe (sp?) cause of your lightbulb avatar.

I say Amy and U-S-A-M-T-S.  Also Yoo-sa-mo for USAMO, although I used to spell that one out.

Cauchy is Coe-shee, L'Hospital (or is it L'Hopital?) I think is Lope-it-al.  The H is silent.

I'm pretty sure Fourier is Fore-ee-ay.

I say AIME as Amy and A I M E.  Depends, because a lot of people wouldn't know what I meant if I said Amy.
USAMTS...I generally say U S A M T S, but sometimes you-sam-tis.

It's obviously "amy" and "you-samts." The idea is to make it as easy as possible to say :).

i say amy or aim.  what does aime stand for?

AIME = American Invitational Math Examination. Comes after the AMC 10/12 or USAMTS.

Amy, and Us-am-ts(like tssss snake sound)

USAMO=U-SA-MOH, Rite?

The reason I think USAMTS should be pronounced "USAMites" is because that way it sounds like little USAMOs, or something. Which is sort of what it is. :lol:

[quote="zeb"]Aren't they "coashy" and "la-hope-it-all"?

And what about Fourier? (don't laugh at me).[/quote]
Fourier=Four-E-Ay(in a french tone, 4EA)

Wow I pronounced AIME different from anyone else, I say "Eye May"  :rotfl:

[quote="beta"]IMO =I.M.O or I-Moh?[/quote]
I.M.O in my opinion.

"Amy", as immortalized in the phrase "You know you're an AoPSer if you tell your friends that your girlfriend's name is AIME." (Somewhere else on the site.)

[quote="Hellstar"]How do you pronounce USAMO? I'm not sure if its U.S.A.M.O or just  you-say-moh[/quote]

I say you-sah-moh.

I have another pronounciation problem: $\LaTeX$.  :)

I used to say LAY-teks (like you'd think it's pronounced), but I was then told it was LAH-tek, so that's what I say.  I've also heard other people say la-TEK, but I don't think it's as easy to say.

Maybe... because the $\chi$ in the Greek word that $\LaTeX$ was derived from has a hard C sound? I asked a similar question somewhere in the LaTeX forum. At least I think I did...

But [b]X[/b] is pronounced [b]ecks[/b] in English... so unless you're saying that $\LaTeX$ is actually a Greek word in this case, that wouldn't be true. :roll: Also, it wasn't quite a hard [b]C[/b] in Greek, it was more like the [b]ch[/b] in "church."

In the 2000-2001 ( i think) school handbook, there was a problem suggesting that AIME was pronounced Amy, where they had somebodies name as Aimee.

http://www.tex.ac.uk/cgi-bin/texfaq2html?label=latexpronounce
http://www.math.upenn.edu/tex_docs/help/faq/uktug-faq/FAQ2.html

It's supposed to be pronounced either "Lay TeX" or "Lah TeX" where "TeX" is pronounced the same way as tech.

But I say it "Lay Teks"...perhaps I should start changing that.

I'm a Lay-teks pronouncer too, and I'm not changing :P

I say AIME as AIME. I say USAMTS as U.S.A.M.T.S and USAMO as U.S.A.M.O. I say LaTeX like lay tecks.


logarithms
calculus
calculus computations

Show the following series converges:

$ \sum_{n \equal{} 1}^{\infty}(\frac {n}{n \plus{} 1})^{n^2}$

Let $ a_n \equal{} \left(\frac {n}{n \plus{} 1}\right)^{n^2} \equal{} \left(1 \minus{} \frac1{n \plus{} 1}\right)^{n^2}$

Then $ \ln a_n \equal{} n^2\ln\left(1 \minus{} \frac {1}{n \plus{} 1}\right)$

$ \equal{} n^2\left( \minus{} \frac {1}{n \plus{} 1} \plus{} O(n^{ \minus{} 2})\right) \equal{} \minus{} n \plus{} O(1).$

Hence $ a_n \equal{} e^{ \minus{} n \plus{} O(1)}.$ This gives us that $ \frac {a_n}{e^{ \minus{} n}}$ is bounded above and below.

Hence, $ \sum_na_n$ converges by comparison to $ \sum_ne^{ \minus{} n}.$

[Of course, the root test should work fairly simply.]

can you explain the big O notation.  I have seen it before, but I am rather confused how to work with it.

Read Kent's excellent sticky in this subforum :)


Could someone tell me if he/she had seen the following problem somewhere? Find the largest $M$ such that there exists a real number $x$ such that for all natural numbers $n$ we have $\{nx\}+\{x^n\}>M$ ($\{\}$ is the fractional part; 0 is not considered natural number). I like it very much, though it is very simple.

[i]Edited by Myth[/i]

I know, harazi is a super mind. I was thinking for 25 minutes! It seems it is a time to prepare my grave.
Answer is $M=1$.

No, Myth, really you should not prepare your grave. Just think about the huge quantity of problems that would remain unsolved in case you did so.  :) .


I know the answer can't be an integer. But I am assuming it is 1 or 0 because those are two values these numbers can equal.

[hide="Hint"]No, when $ \sin x$ is 0, $ \cos x$ is 1, and vice versa.

Think about the original meaning of sine and cosine, and try to draw a triangle where they're equal. What do you know about this kind of triangle?[/hide]

[hide]Let $ \sin{x} \equal{} a.$ Then $ \cos{x} \equal{} \sqrt {1 \minus{} a^2}$. Setting these equal, squaring and rearranging, we have $ 2a^2 \equal{} 1$, so $ a \equal{} \sin{x} \equal{} \pm \frac {\sqrt {2}}{2}$, and thus $ x \equal{} \frac {\pi}{4} \plus{} k\pi$.[/hide]

This should definitely be in HSB.

$ \sin x \equal{} \cos x$
$ \sin x \minus{} \cos x \equal{}0$
$ \frac{1}{\sqrt{2}}\sin x \minus{} \frac{1}{\sqrt{2}} \cos x\equal{}0$

Now just remember that $ \sin45\equal{}\cos45\equal{}\frac{1}{\sqrt{2}}$, then

$ \cos45\sin x\minus{}\sin45\cos x\equal{}0$

Appliying the rule for the sin of a diference:

$ \sin(x\minus{}45)\equal{}0$

Wich implies $ x\equal{}45\plus{}180n$, $ n$ an integer (so $ x$ is an integer in degrees)

 :)


Never Mind: This is the real problem. Marty has 150 000 grapes that are all different colors, but comprise of shades of purple, green, and blue (yes blue) in a 4:5:6 ratio. If he picks 330 grapes what is the probability that he has equal numbers of all the grapes

Is there any restriction on the number of words she can form with those letters?

Like is there any restriction for the number of spaces a rearrangement can have?

Do each of these words have to be the same length as the original, since you are rearranging?

Um i don't know
i think you have to divide al of the numbers of grapes
no wait add up the ratio
then divide the total number of grapes (800)
by the sum of the ratio and go from there

Wait...I don't understand the problem. So all the 150,000 grapes are different colors, but they comprise of shades of purple:green:blue in a 4:5:6 ratio? How is that possible, since they are [i]all[/i] different colors?

[quote="darkdieuguerre"][quote="limac"]Wait...I don't understand the problem. So all the 150,000 grapes are different colors, but they comprise of [b]shades[/b] of purple:green:blue in a 4:5:6 ratio? How is that possible, since they are [i]all[/i] different colors?[/quote]

He probably meant shades throughout, but didn't write it explicitly.[/quote]

Yes that' what he meant. its an interesting problem......

Hmmmm...........The problem isn't worded specifically enough in order to solve it.......... confuzing......... *_____*!


could anyone give the proof that each real number is a limit of a rationnal sequence.
many thanks.

The real numbers are frequently constructed as the completion of the rationals (i.e. as the set of numbers that are a limit of some sequence of rationals).  What's your definition of the reals?

Step 1: between any two real numbers lies at least one rational number.

Proof: Suppose $ a<b.$ Find $ n$ such that $ n>\frac1{b\minus{}a}.$ (Note that I am explicitly invoking the Archimedean property of the real numbers.) Let $ k$ be the largest integer such that $ k<nb.$ Then, $ nb\minus{}k\le 1$ while $ nb\minus{}na\equal{}n(b\minus{}a)>1,$ so $ nb\minus{}k<nb\minus{}na$ or $ na<k.$ We now have that $ na<k<nb$ or $ a<\frac{k}{n}<b.$ We have found a rational number between $ a$ and $ b.$

Finishing the result: Let $ a$ be an arbitrary real number. From the lemma in step 1, we can find a rational number $ x_n$ such that $ a<x_n<a\plus{}\frac1n$. It follows from the squeeze theorem that $ \lim x_n\equal{}a.$

---

An alternate approach: if you accept that every real number has a decimal expansion, then the truncations of that decimal expansion are a sequence of rational numbers converging to the number in question.

For instance, the sequence $ 2,2.7, 2.71, 2.718,2.7182,2.71828,\dots$ converges to $ e.$

The real number $ x$ is the limit of the sequence of rationals $ a_n \equal{} \frac{[10^n x]}{10^n}$ where $ [y]$ stands for the floor of $ y$.

A last fun approach, although something of a tautology:  consider the sequence of convergents of the continued fradction of a given irrational number.


algebra
binomial theorem
geometric series

That idea only leads to the geometric-series solution.

[hide="Discussion"] Let $ \phi \equal{} \frac {1 \plus{} \sqrt {5}}{2}, \varphi \equal{} \frac {1 \minus{} \sqrt {5}}{2}$ be the two roots of $ x^2 \equal{} x \plus{} 1$.  Note that $ \phi \varphi \equal{} \minus{} 1$, so that our summation can be rewritten

$ \varphi^{10} \minus{} \varphi^8 \plus{} \varphi^6 \minus{} \varphi^4 \plus{} \varphi^2 \minus{} 1 \plus{} \phi^2 \minus{} \phi^4 \plus{} \phi^6 \minus{} \phi^8 \plus{} \phi^{10}$.

The numbers $ L_n \equal{} \phi^n \plus{} \varphi^n$ are the [url=http://en.wikipedia.org/wiki/Lucas_number]Lucas numbers[/url], satisfying

$ L_0 \equal{} 2$
$ L_1 \equal{} 1$
$ L_n \equal{} L_{n \minus{} 1} \plus{} L_{n \minus{} 2}$.

We can therefore quickly calculate the values of $ L_2, L_4, L_6, L_8, L_{10}$ to find our desired sum.  :)

Edit:  A minor adjustment.  Our desired sum is

$ L_{10} \minus{} L_{8} \plus{} L_6 \minus{} L_4 \plus{} L_2 \minus{} 1$

and we can use the recursion to rewrite this as

$ L_9 \plus{} L_5 \plus{} 2$ [/hide]

[quote="mathgeniuse^ln(x)"][quote="fibointepi"]Actually, there's an easier way to do this.[/quote]

Wait, there is

It looks reminiscent of the Binomial Theorem, but there are no binomial coefficients.[/quote]
Well it is a geometric series, so if you multiply by $ \phi\minus{}\overline{\phi}$, a lot of it cancels.


function
limit
integration
induction
algebra proposed
algebra

the  problem it was a oral problem for polyt in france .
find all function such $\lim_{x\to\{1}}{\frac{f\left(x\right)}{x}=1$
and $f\left(x+y\right)\le{f\left(x\right)+f\left(y\right)}$.

 :roll:

no  one interst. the function is definied at ${R}$ and ${\lim_{x\to\{0}}{\frac{f\left(x\right)}{x}}=1$ and now you have all propretes about $f$. please find all functions :mad:

Your question is not correct. For example, if $g(t)$ is any decrease function with $g(0)=1$, then \[f(x)=\int_{0}^{x}g(t)dt\] satisfyed your condition.

$f: \mathbb{R}\to\mathbb{R}$ with
(1) $\lim_{x\to 0}{f(x)\over x}=1$
(2) $f(x+y)\leq f(x)+f(y)$ for all $x,y\in\mathbb{R}$

Then $f(0)=f(0+0)\leq f(0)+f(0)$ and $0\leq f(0)$
and for all $x\in\mathbb{R},n\in\mathbb{N}$ we get $f(nx)\leq nf(x)$ by induction using (2).
So $f(x)\leq nf({x\over n})=x\cdot{f({x\over n})\over{x\over n}}\mathop{\longrightarrow}_{n\to\infty}x$ for all $x\in\mathbb{R}\setminus\{0\}$ by (1)
and so for all $x\in\mathbb{R}\setminus\{0\}$ we get
$0\leq f(0)=f(x+(-x))\leq f(x)+f(-x)\leq x+(-x)=0$ again using (2).
This equality gives $f(0)=0$ and $f(x)=x$ for all $x\in\mathbb{R}\setminus\{0\}$.

So $f(x)=x$ for all $x\in\mathbb{R}$.

yes mr olorin. my solution is :
we have $f\left(x\right)-\frac{1}{n}f\left(-\frac{y}{n}\right)\le{f\left(x+y\right)}\le{f\left(x\right)+\frac{1}{n}f\left(\frac{y}{n}\right)}$.
we use 1 $f\left(x+y\right)=f\left(x\right)+y$ and we are done.


Let $ f: \mathbb{N} \rightarrow \mathbb{N}$ be an increasing function such that $ f(f(n))\equal{}3n$ for all positive integers $ n$. Determine $ f(1992)$.

If im not mistaken, this is from bulgaria balkan tst.
I found that $ f(1)\equal{}2,f(2)\equal{}3,f(3)\equal{}6$, $f(4)\equal{}7,f(5)\equal{}8,f(6)\equal{}9$, $f(7)\equal{}12,f(8)\equal{}15,\ldots$ and so on, but should I do this until $ 1992$?  :oops:

i think balkan tests  are only of romania?  am i right?

Very nice problem, let's solve it

Lemma: $ f(3^{k}) \equal{} 2 \times 3^{k}$ and $ f(2 \times 3^{k}) \equal{} 3^{k \plus{} 1}$.

Proof: In fact, it can easily be checked $ f(3) \equal{} 6$ and $ f(6) \equal{} 9$. So assume that is true for

$ j \leq k$. Then $ f(3^{k \plus{} 1}) \equal{} f(f(2 \times 3^{k})) \equal{} 3 \times 2 \times 3^{k} \equal{} 2 \times 3^{k \plus{} 1}$.

By the other hand, $ f(2 \times 3^{k \plus{} 1}) \equal{} f(f(3^{k \plus{} 1})) \equal{} 3^{k \plus{} 2}$.

But now, between $ 3^{k}$ and $ 2 \times 3^{k}$ there are $ 3^{k} \plus{} 1$ numbers, and between $ 2 \times 3^{k}$

and $ 3^{k \plus{} 1}$ there are $ 3^{k} \plus{} 1$ numbers, too.

Since the function is increasing it immediatly implies $ f(3^{k} \plus{} i) \equal{} 2 \times 3^{k} \plus{} i,$ for all $ 0 \leq i \leq 3^{k}$.

$ 1992 \equal{} 3 \times 664 \Rightarrow f(1992) \equal{} f(f(f(664))) \equal{} 3 \times f(664)$.

By the other hand $ 664 \equal{} 2 \times 3^{5} \plus{} 178$ and then $ 664 \equal{} f(3^{5}\plus{}178) \equal{} f(421) \Rightarrow f(664) \equal{} f(f(421)) \equal{} 1263$,

and finally $ f(1992) \equal{} 3 \times 1263 \equal{} 3789$.

[quote="Lousin Garckz"]Very nice problem, let's solve it

Lemma: $ f(3^{k}) \equal{} 2 \times 3^{k}$ and $ f(2 \times 3^{k}) \equal{} 3^{k \plus{} 1}$.

Proof: In fact, it can easily be checked $ f(3) \equal{} 6$ and $ f(6) \equal{} 9$. So assume that is true for

$ j \leq k$. Then $ f(3^{k \plus{} 1}) \equal{} f(f(2 \times 3^{k})) \equal{} 3 \times 2 \times 3^{k} \equal{} 2 \times 3^{k \plus{} 1}$.

By the other hand, $ f(2 \times 3^{k \plus{} 1}) \equal{} f(f(3^{k \plus{} 1})) \equal{} 3^{k \plus{} 2}$.

But now, between $ 3^{k}$ and $ 2 \times 3^{k}$ there are $ 3^{k} \plus{} 1$ numbers, and between $ 2 \times 3^{k}$

and $ 3^{k \plus{} 1}$ there are $ 3^{k} \plus{} 1$ numbers, too.

Since the function is increasing it immediatly implies $ f(3^{k} \plus{} i) \equal{} 2 \times 3^{k} \plus{} i,$ for all $ 0 \leq i \leq 3^{k}$.

$ 1992 \equal{} 3 \times 664 \Rightarrow f(1992) \equal{} f(f(f(664))) \equal{} 3 \times f(664)$.

By the other hand $ 664 \equal{} 2 \times 3^{5} \plus{} 178$ and then $ 664 \equal{} f(3^{5} \plus{} 178) \equal{} f(421) \Rightarrow f(664) \equal{} f(f(421)) \equal{} 1263$,

and finally $ f(1992) \equal{} 3 \times 1263 \equal{} 3789$.[/quote]

Thank you very much!
I did not notice  $ f(2 \times 3^k)$ more.
In fact I see something in period $ 18$, LOL.  :blush:

you are welcome  :)  ... by the way, what is the source of this problem? is this from a recently olympiad? because I think I solved this problem a long time ago....but i'm not sure


trigonometry
inequalities proposed
inequalities

If $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$,x,y,z>0,prove that:
$ 2(xy\plus{}yz\plus{}zx)\plus{}9 \geq 7(x\plus{}y\plus{}z)$

You can set $ x\equal{}\cos A,\ \cos B,\ \cos C\ \left(0<A,\ B,\ C<\frac{\pi}{2}\right)$.

[quote="kunny"]You can set $ x \equal{} \cos A,\ \cos B,\ \cos C\ \left(0 < A,\ B,\ C < \frac {\pi}{2}\right)$.[/quote]
so????What is your solution>?

[quote="tuandokim"]If $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$,x,y,z>0,prove that:
$ 2(xy \plus{} yz \plus{} zx) \plus{} 9 \geq 7(x \plus{} y \plus{} z)$[/quote]

http://www.mathlinks.ro/viewtopic.php?p=1331805#1331805

[quote="ye109"][quote="manjil"][quote="tuandokim"]If $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$,x,y,z>0,prove that:
$ 2(xy \plus{} yz \plus{} zx) \plus{} 9 \geq 7(x \plus{} y \plus{} z)$[/quote]

I think this can be sharpened. Can anyone give me some sharper forms?[/quote]

a    sharpened   one.....
If $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$,x,y,z>0,prove that:
$ 4(xy \plus{} yz \plus{} zx) \plus{} 15 \geq 12(x \plus{} y \plus{} z)$[/quote]

What do you think about:
If $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$ and $ x,y,z > 0,max\{x,y,z\} \equal{} x$. Prove that:
\[ 4(xy \plus{} yz \plus{} zx) \plus{} 27 \geq 12(2x \plus{} y \plus{} z) \plus{} (4x^2 \plus{} 6)(2x \minus{} 1)^2
\]
which is sharper than your  :P

[quote="zaizai-hoang"][/quote][quote="ye109"][quote="manjil"][quote="tuandokim"]If $ x^2 + y^2 + z^2 + 2xyz = 1$,x,y,z>0,prove that:
$ 2(xy + yz + zx) + 9 \geq 7(x + y + z)$[/quote]

I think this can be sharpened. Can anyone give me some sharper forms?[/quote]

a    sharpened   one.....
If $ x^2 + y^2 + z^2 + 2xyz = 1$,x,y,z>0,prove that:
$ 4(xy + yz + zx) + 15 \geq 12(x + y + z)$[/quote][quote="zaizai-hoang"]

What do you think about:
If $ x^2 + y^2 + z^2 + 2xyz = 1$ and $ x,y,z > 0,max\{x,y,z\} = x$. Prove that:
\[ 4(xy + yz + zx) + 27 \geq 12(2x + y + z) + (4x^2 + 6)(2x - 1)^2
\]
which is sharper than your  :P[/quote]

$ x->1,y->0,z->0$
$ LHS=27,RHS=24+10=34$
so i think it's wrong....

ohh sorry it should be:
If $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$ and $ x,y,z > 0$ and $ max\{x,y,z\} \equal{} x$ Prove that:
\[ 4(xy \plus{} yz \plus{} zx) \plus{}15 \geq 12(x \plus{} y \plus{} z) \plus{} (4x^2 \plus{} 6)(2x \minus{} 1)^2
\]

[quote="ye109"][quote="zaizai-hoang"]ohh sorry it should be:
If $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$ and $ x,y,z > 0$ and $ max\{x,y,z\} \equal{} x$ Prove that:
\[ 4(xy \plus{} yz \plus{} zx) \plus{} 15 \geq 12(x \plus{} y \plus{} z) \plus{} (4x^2 \plus{} 6)(2x \minus{} 1)^2
\]
[/quote]

wrong agian...
$ x \minus{} > 1,y \minus{} > 0,z \minus{} > 0$
$ LHS \equal{} 15,RHS \equal{} 12 \plus{} 10 \equal{} 22$
$ LHS < \equal{} RHS$ :wink:[/quote]

Sorry, your are right. The min value of the  $ f(x,1\minus{}2x^2,1\minus{}2x^2)\equal{}0$. I have made some mistakes in calculating  :wink:


calculus
integration
trigonometry
logarithms
calculus computations

Compute integral of $\int^{\dfrac \pi 4}_0 \ln(sin x +\cos x )\ dx$.

split up and integrate! (I think)
integral of sin x = - cos x + C
integral of cos x = sin x + C
This should get you started.

First note that $\ sin x + \ cos x = \sqrt {2} \ sin (x+ \frac {\pi} {4}) $. Then if you know that:
\[
\displaystyle \int^{\frac {\pi} {2}}_0 \ ln(sinx) \,dx=\frac {\pi} {2} \ln \frac  {1} {2}
\]
all is done.

[quote="seshadri"]i don't see why this is posted in this forum.
but anyway,
$ \int^{\pi/4}_0 \ln(\sin x +\cos x )\ dx=\int^{\pi/4}_0 \ln(\sqrt 2)\frac{\sin x+\cos x}{\sqrt 2}\ dx=$

$\int^{\pi/4}_0 (\ln(\sqrt 2)\ dx + \int^{\pi/4}_0\cos(\dfrac \pi 4-x)\ dx=\dfrac \pi 4\ln(\sqrt 2) + \int^{\pi/4}_0 \ln(\cos x)\ dx.$
the second integral is a standard calculus problem..[/quote]
By "this forum," seshadri is referring to another place where passangd posted this same problem.

In fairness, to sky9093, his post made more sense before passangd edited his problem.

passangd: please do not post the same problem in multiple locations. That doesn't get any more people to see your problem, and mostly just annoys people who read the same problem from you multiple times. This (Calculus Computations and Tutorials) was the right place for this problem all along, but if you get that wrong, the moderators may move it for you.

:?  To be completely honest, I do not see how knowing $\int_0^{\pi/2}\ln\sin x\,dx$ can help to evaluate the integral $\int_0^{\pi/4}\ln\cos x\,dx$. Neither have I ever seen it computed in any textbook. Am I missing something simple here? :?

I think you can get help from
http://www-2.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm
and
http://mathworld.wolfram.com/CatalansConstant.html


abstract algebra
superior algebra
superior algebra unsolved

How many homomorphisms are there from $Z_n$ to $Z_n$? and what are they?

How about from $Z_{20}$ to $Z_{30}$?

A good idea would be considering the Ideals of $Z_n$ since these homomorfisms are i thing characterized by their kernel, or at least it is easier to count those which leave some kernel.


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

#9.

[hide]Divide both sides by x, we get the answer x-1/x=4.[/hide]

#6

[hide] Solve for m = 15n/2n-15.  2n-15 divides 15^2 or 3^2*5^2.
I tried 3 and it works. m=15, n=9, so the sum is 24.[/hide]

Well, since we're in the middle of MathCounts, these problems are appropriate.  And it doesn't matter how old they are, they're unsolved!


so i was looking at an old AMC 12 and saw this question;
$\displaystyle \sqrt{1984}=\sqrt{x}+\sqrt{y}$ find the number of integral $(x,y)$ solutions such that $0<x<y$ .

i got an answer of three, and was wondering if this was right.
also could you please provide a proof with your answer.

First factor out $1984$ to get $2^6 \cdot 31$, now $y=(\sqrt{1984}-\sqrt{x})^2 \Longrightarrow y=1984-\sqrt{1984x}+x^2$. Thus $1984x$ must be a perfect square, and also notice that $x$ must be in form $31a^2$  for $1984x$ to be perfect square and $x$ must be less than $1984$ but I don't know where to go from there :(

[quote="frt"][hide]
$\sqrt{1984}=8\sqrt{31}$

Letting $\sqrt{x}=a\sqrt{31}$ and $\sqrt{y}=b\sqrt{31}$ (where $a$ and $b$ are positive integers), we want to find the number of ordered pairs $(a,b)$ such that $a+b=8$ and $b>a$.
That gives three solutions:
$(a,b)=(1,7),(2,6),(3,5)\\ \Longleftrightarrow (x,y)=(31,1519),(124,1116),(279,775)$[/hide][/quote]

That was a MUCH better solution than mine


For each $ r>0$, there exists infinitely many positive integers $ n$ for which $ n^2\plus{}1$ has a prime divisor $ p > 2n\plus{}\sqrt{2rn}$.

(Just curious.  It'd be great if someone comes up with a solution.)

Also, please take a look at this one:
http://www.mathlinks.ro/viewtopic.php?t=214331.


trigonometry
integration
calculus
calculus computations

Here's the problem. I have been solving this for some time but i can't sem to figure out how to do it.

Given $r(t) = <t^{2}, \cos t+t\sin t, \sin t-t\cos t>$ with $t\geq0$how do you find the value of $t$ when the pt is moved $\frac{\sqrt{5}}{8}\pi^{2}$ units from  starting pt. $(0,0,1)$.

Find the speed $\|r'(t)\|$. It's rigged to come out nicely.
Once you find the speed, you can look at the arclength $\int_{0}^{t}\|r'(s)\|\,ds$, and find $t$ that makes this the desired length.

Incidentally, you have the starting point wrong.

I see sir. What do you mean by having the starting pt. wrong? In the problem It said that the initial pt of the particle is $(0,0,1)$. Do you mean there is no t that satisfies the initial pt. when evaluated at $r(t)$. That was also my problem.

I mean that $(0,0,1)$ is not on the curve- but $(0,1,0)$ is.


Let $ a,b,c \geq\ 0$ .Prove that or disprove that
 $ \sqrt{\frac{a^2\plus{}2bc}{b\plus{}c}}\plus{}\sqrt{\frac{b^2\plus{}2ca}{c\plus{}a}}\plus{}\sqrt{\frac{c^2\plus{}2ab}{a\plus{}b}}\geq\ 3\sqrt{\frac{a\plus{}b\plus{}c}{2}}$

[quote="nguoivn"]@arqady: Do U think ineq above is trues ?  :maybe:[/quote]
It seems true. Nice inequality.

Thank you. And I think it is really hard. It's stronger than:
$ \sum\ \frac{a^2\plus{}2bc}{b\plus{}c} \geq\ \frac{3(a\plus{}b\plus{}c)}{2}$

[quote="PHONGVAN"]Let $ a,b,c \geq\ 0$ .Prove that or disprove that
 $ \sqrt {\frac {a^2 \plus{} 2bc}{b \plus{} c}} \plus{} \sqrt {\frac {b^2 \plus{} 2ca}{c \plus{} a}} \plus{} \sqrt {\frac {c^2 \plus{} 2ab}{a \plus{} b}}\geq\ 3\sqrt {\frac {a \plus{} b \plus{} c}{2}}$[/quote]

Nice, my friend ! And I think that the following is true too:

$ \sqrt {\frac {a^3 \plus{} abc}{b \plus{} c}} \plus{} \sqrt {\frac {b^3 \plus{} abc}{c \plus{} a}} \plus{} \sqrt {\frac {c^3 \plus{} abc}{a \plus{} b}}\geq a \plus{} b \plus{} c$

[quote="dduclam"][quote="PHONGVAN"]Let $ a,b,c \geq\ 0$ .Prove that or disprove that
 $ \sqrt {\frac {a^2 + 2bc}{b + c}} + \sqrt {\frac {b^2 + 2ca}{c + a}} + \sqrt {\frac {c^2 + 2ab}{a + b}}\geq\ 3\sqrt {\frac {a + b + c}{2}}$[/quote]

Nice, my friend ! And I think that the following is true too:

$ \sqrt {\frac {a^3 + abc}{b + c}} + \sqrt {\frac {b^3 + abc}{c + a}} + \sqrt {\frac {c^3 + abc}{a + b}}\geq a + b + c$[/quote]
Yes,it is true and here is my proof:

By Holder we have:

$ (\sum{\sqrt {\frac {a^3 + abc}{b + c}}})(\sum{\sqrt {{\frac {a^3(b + c)}{a^2 + bc}}})(\sum{a}) \geq (a + b + c)^3,}$

Hence It suffices to prove that:

${ \sum \sqrt {\frac {a^3(b + c)}{a^2 + bc}}} \leq a + b + c,$

By C.S we have:

$ LHS \leq \sqrt {(a + b + c)(\frac {a^2(b + c)}{a^2 + bc} + \frac {b^2(a + c)}{b^2 + ac} + \frac {c^2(b + c)}{c^2 + ab})},$

It suffices now to prove that:

$ \frac {a^2(b + c)}{a^2 + bc} + \frac {b^2(a + c)}{b^2 + ac} + \frac {c^2(a + b)}{c^2 + ab} \leq a + b + c,$

$ \Leftrightarrow \sum \frac {a(a - b)(a - c)}{a^2 + bc} \geq 0,$

WLOG we can assume that $ a \geq b \geq c,$

we have:

$ \frac {c(c - a)(c - b)}{c^2 + ab} \geq 0,$

It suffices to prove that:

$ \frac {a(a - b)(a - c)}{a^2 + bc} + \frac {b(b - a)(b - c)}{b^2 + ac} \geq 0,$

$ \Leftrightarrow (a - b)(a + b - c) \geq 0$ which is true.

You're right, rachid. Infact, we only use Cauchy-Schwarz Inequality: $ (\sum{\sqrt {\frac {a^3 + abc}{b + c}}})(\sum{\sqrt {{\frac {a^3(b + c)}{a^2 + bc}}}) \geq (a + b + c)^2}$

Besides, we can solve only with Vornicu-Schur. :)

We still have some nother results as follows

$ \sum a\sqrt{\frac{b+c}{a^2+bc}}\le \sum\sqrt a$

and   $ \sum \sqrt{\frac{a^3+abc}{b^2+c^2}}\ge \sum\sqrt a$

[quote="dduclam"]
$ \sum \sqrt {\frac {a^3 \plus{} abc}{b^2 \plus{} c^2}}\ge \sum\sqrt a$[/quote]
We' ll prove: $ \sum\ \frac{a^3\plus{}abc}{b^2\plus{}c^2} \geq\ a\plus{}b\plus{}c$
With Voni Schur: $ LHS \geq\ \sum\ \frac{a^2(b\plus{}c)}{b^2\plus{}c^2} \geq\ a\plus{}b\plus{}c$
$ \Leftrightarrow \sum\ \frac{ab(a\plus{}b)(a\minus{}b)^2}{(b^2\plus{}c^2)(c^2\plus{}a^2)} \geq\ 0$ (always trues)
PS: It is also my old problem  :)

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that :
$$\sqrt[4]{\frac{a^2+2bc}{b+c}}+\sqrt[4]{\frac{b^2+2ca}{c+a}}+\sqrt[4]{\frac{c^2+2ab}{a+b}}\geq\ 3\sqrt[4]{\frac{a+b+c}{2}}$$

[quote=can_hang2007][quote="PHONGVAN"]Let $ a,b,c \geq\ 0$ .Prove that or disprove that
 $ \sqrt {\frac {a^2 \plus{} 2bc}{b \plus{} c}} \plus{} \sqrt {\frac {b^2 \plus{} 2ca}{c \plus{} a}} \plus{} \sqrt {\frac {c^2 \plus{} 2ab}{a \plus{} b}}\geq\ 3\sqrt {\frac {a \plus{} b \plus{} c}{2}}$[/quote]
It is true and nice, thach. It is stronger than the following inequality of VasC
\[ \sum \frac{\sqrt{a^2\plus{}2bc}}{b\plus{}c} \ge \frac{3\sqrt{3}}{2}\]
By the way, we can prove all these inequalities with Cauchy Schwarz. ;) :)[/quote]

But who can do this:

\[{\it \sum} \left( {\frac {\sqrt {{x}^{2}+2\,yz}}{y+z}} \right) \geq \frac{1}{2}\,{
\it \sum} \left( \sqrt {{\frac {{x}^{2}+2\,yz}{ \left( y+z \right)
 \left( x+y+z \right) }}} \right) \sqrt {6}\]

[quote=dduclam]Let $ a,b,c \geq\ 0$ .Prove that
$$ \sqrt{\frac{a^3 +abc}{b +c}}+\sqrt{\frac{b^3 +abc}{c +a}}+\sqrt{\frac{c^3 +abc}{a +b}}\geq a+b+c$$[/quote]
[url]https://artofproblemsolving.com/community/c6h260824p5938592[/url]
[url]https://artofproblemsolving.com/community/c6h2027096p14272252[/url]


vector
rotation
geometry
geometric transformation
trigonometry
linear algebra
matrix

if we rotate the vector $ \vec{a}$ around the vector $ \vec{b}$ by $ \theta$ clockwise , what is the new components for $ \vec{a}$.

 :(

what is rotation?
turn into complex
$ (x,y) \to z\equal{}x\plus{}iy\equal{}re^{i\theta}$
so rotation by $ \alpha$ is $ R_\alpha (z)\equal{}R_\alpha (re^{i\theta}) \equal{} re^{i(\theta\plus{}\alpha)}\equal{}ze^{i\alpha}$

so $ R_\alpha$ is a linear map that takes

$ (x,y) \equal{} x\plus{}iy \equal{} z \to ze^{i\alpha} \equal{}(x\plus{}iy)(\cos\alpha\plus{}i\sin\alpha)\equal{}$

$ (x\cos\alpha\minus{}y\sin\alpha)\plus{}i(y\cos\alpha \plus{}x\sin\alpha)$

i cant draw matrices =(, but the matrix representation of rotation by $ \theta$ is then
$ A_\theta\equal{} [[\cos\theta, \minus{}\sin\theta]^T, [\sin\theta, \cos\theta]^T]$

so then your transform on vectors is just $ \vec b \plus{} A_\theta \vec a$

[quote="kenn4000"]

i cant draw matrices =(, but the matrix representation of rotation by $ \theta$ is then
$ A_\theta \equal{} [[\cos\theta, \minus{} \sin\theta]^T, [\sin\theta, \cos\theta]^T]$

so then your transform on vectors is just $ \vec b \plus{} A_\theta \vec a$[/quote]

I still misunderstand

This is linear algebra or vector geometry: how do you build the rotation matrix over $ \mathbb{R}^3$ for rotation by $ \theta$ around a fixed direction $ b$?

kenn4000: I read this as a three-dimensional problem; rotation around s random vector doesn't make sense in the plane. Also, note the "clockwise".

Assume WLOG that $ b$ is a unit vector. Complete it to an orthonormal basis; find $ v$ and $ w$ both unit vectors so that $ b,v,w$ are mutually perpendicular, and let $ B$ be the matrix $ \begin{bmatrix}b&v&w\end{bmatrix}$
With respect to this new basis $ B$, the desired rotation has matrix $ R\equal{}\begin{bmatrix}1&0&0\\0&\cos\theta&\sin\theta\\0&\minus{}\sin\theta&\cos\theta\end{bmatrix}$ - or maybe its matrix is $ R^T$, interchanging $ \theta$ for $ \minus{}\theta$. "Clockwise" has no independent meaning, and you have to pick a three-dimensional orientation for it to make sense. Under the usual right-hand rule convention, my $ A$ is what you want when $ B$ has determinant $ 1$.

So far, we have the transformed matrix. To get back to the original, use similarity and let $ R'\equal{}BRB^{\minus{}1}$. This $ R'$ is the matrix of the original rotation, and $ a$ goes to $ R'a$.

[quote="jmerry"]This is linear algebra or vector geometry: how do you build the rotation matrix over $ \mathbb{R}^3$ for rotation by $ \theta$ around a fixed direction $ b$?

kenn4000: I read this as a three-dimensional problem; rotation around s random vector doesn't make sense in the plane. Also, note the "clockwise".

Assume WLOG that $ b$ is a unit vector. Complete it to an orthonormal basis; find $ v$ and $ w$ both unit vectors so that $ b,v,w$ are mutually perpendicular, and let $ B$ be the matrix $ \begin{bmatrix}b & v & w\end{bmatrix}$
With respect to this new basis $ B$, the desired rotation has matrix $ R \equal{} \begin{bmatrix}1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & \minus{} \sin\theta & \cos\theta\end{bmatrix}$ - or maybe its matrix is $ R^T$, interchanging $ \theta$ for $ \minus{} \theta$. "Clockwise" has no independent meaning, and you have to pick a three-dimensional orientation for it to make sense. Under the usual right-hand rule convention, my $ A$ is what you want when $ B$ has determinant $ 1$.

So far, we have the transformed matrix. To get back to the original, use similarity and let $ R' \equal{} BRB^{ \minus{} 1}$. This $ R'$ is the matrix of the original rotation, and $ a$ goes to $ R'a$.[/quote]

How you make $ R$...please

[quote="kenn4000"]yeah three makes more sense  :blush:[/quote]

me too dude  :blush:  :blush:


This is ugly solution.

$\iff abc(2\sum_{sym}a^{3}-3\sum_{sym}a^{2}b+\sum_{sym}abc)\ge0$
This is trivial by Muirhead & Schur.

Equality holds iff $a=0$ or $b=0$ or $c=0$ or $a=b=c$  :!:


geometry
probability
analytic geometry
inequalities
AMC
AIME

Hello,
Can anyone help me to solve this problem, please? I know that we have to use geometry. However, I can't figure out how to draw the line thingy. Here's the problem:
Consider the square having as vertices the following points (-1,1),
> (1,1), (1,-1), and (-1,-1) . Find the probabilty that a randomly selected point inside the square will have the sum of its coordinates greater than 1/11 .Write your answer as a common fraction in lowest term. *
 please help me to solve this problem!!!
> Thank you!!! :D

If you can graph two-variable inequalities, then this shouldn't be that hard. First plot the points of the square on the coordinate plane. Then, you have that $ x \plus{} y > \frac{1}{11}$ from the problem. Graph that inequality and the area common to the inequality and the interior of the square should be put over the area of the square to find your answer.

hmm....this is like my first time doing this so it might be a pretty bad solution  :P

so you just graph the square and the line x+y > 1/11 and the square with those vertices and find the area of the region inside the square that satisfies the inequality (the darker region in the diagram) y=-x+1/11 and y=1 intersect at (-10/11 , 1) so FB=21/11=BG because BGF is a 45 45 90 triangle.  So the area of BGF is ((21/11)^2)/2=(441/121)/2=441/242.  Then the area of pentagon ADCGF=2^2 - 441/242=4 - 441/242 = 527/242.[/code]

Note that you do not want the area of the 'good' region, but the area of the 'good' region over the total area of the region. Since the area of the big square is 4, the probability is bowser's answer divided by 4.

aaaa thanks I forgot  :blush: heh.
thats what made me fail aime this year......4 careless mistakes...... >.<


Yup,that's right, $\frac{-1}{\sqrt{3}}$ :)


[quote="stancioiu sorin"]
$x^{4}+y^{4}+z^{4}\leq\ 1$
[/quote]
Inegalitatea asta nu este adevarata; daca facem $x\longrightarrow 0$ si $y\longrightarrow 0$, atunci, din conditia $x^{2}+y^{2}+z^{2}=\sqrt{2}$, obtinem $z^{4}\longrightarrow 2$, deci $x^{4}+y^{4}+z^{4}\longrightarrow 2.$

Da, asa e .Atunci cred ca ar merge o reducere la inegalitatea lui Nesbitt.O sa ma mai gandesc maine.

O sa demonstrez ca $\sqrt{x^{4}+y^{4}+(x+y)^{4}}\leq \frac3{\sqrt 2}\left( x^{2}+y^{2}\right)$.
[hide="Demonstraţia inegalităţii de mai sus."]
$\, \Longleftrightarrow \, \, x^{4}+y^{4}+(x+y)^{4}\leq \frac92 \left( x^{2}+y^{2}\right)^{2}$
$\, \Longleftrightarrow \, \, 2x^{4}+2y^{4}+2(x+y)^{4}\leq 9 \left( x^{2}+y^{2}\right)^{2}$
$\, \Longleftrightarrow \, \, 4x^{4}+4y^{4}+8 x^{3}y+8 x y^{3}+12 x^{2}y^{2}\leq 9 x^{4}+18 x^{2}y^{2}y^{2}+9 y^{4}$
$\, \Longleftrightarrow \, \, 8 x^{3}y+8 x y^{3}\leq 5 x^{4}+6 x^{2}y^{2}+5 y^{4}$
$\, \Longleftrightarrow \, \, 0 \leq 5 \left( x^{4}+y^{4}-x^{3}y-x y^{3}\right)+3 \left( x^{2}y^{2}+x^{2}y^{2}-x^{3}y-x y^{3}\right)$
$\, \Longleftrightarrow \, \, 0 \leq 5 \left( x^{3}(x-y)+y^{3}(y-x) \right)+3 \left( x^{2}y (y-x)+x y^{2}(x-y) \right)$
$\, \Longleftrightarrow \, \, 0 \leq 5 (x-y) \left( x^{3}-y^{3}\right)+3 (x-y) \left(-x^{2}y+x y^{2}\right)$
$\, \Longleftrightarrow \, \, 0 \leq 5 (x-y)^{2}\left( x^{2}+xy+y^{2}\right)-3 (x-y)^{2}xy$
$\, \Longleftrightarrow \, \, 0 \leq (x-y)^{2}\left( 5 x^{2}+2xy+5y^{2}\right)$,
care e clar adevarata$. \ \ \ \ \ \ \ \ \ \ \ \ \square$
[/hide]

Asadar, \[\sum_\textrm{cyc}\frac1{\sqrt{x^{4}+y^{4}+(x+y)^{4}}}\geq \sum_\textrm{cyc}\frac{\sqrt 2}{3(x^{2}+y^{2})},\] deci e suficient sa demonstram ca \[\sum_\textrm{cyc}\frac1{x^{2}+y^{2}}\geq \frac9{2 \sqrt 2}.\] Dar din Cauchy avem \[\sum_\textrm{cyc}\frac1{x^{2}+y^{2}}\cdot \sum_\textrm{cyc}\left( x^{2}+y^{2}\right) \geq 9 ,\] i.e. \[\sum_\textrm{cyc}\frac1{x^{2}+y^{2}}\geq \frac9{2 \sqrt 2}.\] Sper ca n-am gresit nicaieri.

Nu ai gresit; solutia ta este foarte eleganta.
Cam pe aceeasi idee am mers si eu; am folosit identitatea \[2\bigg( x^{4}+y^{4}+(x+y)^{4}\bigg)=\bigg( x^{2}+y^{2}+(x+y)^{2}\bigg)^{2}.\]


calculus
derivative
function
real analysis
real analysis unsolved

let $f$ be a $C^n$ function such that $lim_{x-> +\infty}f(x)= lim_{x-> +\infty}f^{(n)}(x)=0$.
What can be said about $lim_{x-> +\infty}f^{(k)}(x)$ ?


probability
combinatorics proposed
combinatorics

Given a group of N balls, which consists of C colors (repersent as color 1,2,3........,C).

Number of Balls in each color are represented as n1, n2, n3,......,nC.

Therefore, n1+n2+n3+.......+nC = N

If we randomly pick A balls from this group of balls (A<=N). What is the probabilities that A balls which have been picked consist of

a1,a2,a3,.....aC

where, ai = number of balls in color i that have been picked

(a1+a2+a3+....+aC = A)

There are $C_N^A$ possibilities to choose $A$ balls.
Further, thre are $C_{n_i}^{a_i}$ possibilities to choose $a_i$ balls from $n_i$ balls. Thus we have $C_{n_1}^{a_1}\cdot C_{n_2}^{a_2}\cdot ...\cdot C_{n_C}^{a_C}$ possibilities to make a right choice. Therefore, required probability is
\[\frac{C_{n_1}^{a_1}\cdot C_{n_2}^{a_2}\cdot ...\cdot C_{n_C}^{a_C}}{C_N^A}.\]

Yes, but I couldn't pass problem with such title. If you want I can tell you in PM what it partially mean in russian.  :P  :P  :P  :lol:  :lol:  :lol:

Many Thanks. Myth,  :D  I post this problem because I"m not sure in my solution. Now I " m sure. !

Your pseudo should be 'Truth', not Myth. Even if Myth is myth  ;)

Pierre.

I will think about your offer.
 :lol:  :lol:  :lol: