integration
vector
analytic geometry
trigonometry
calculus
geometry
calculus computations

Given any nonvertical plane $ P$ paralle to the x-axis, let $ C$ be the curve of intersecttion of $ P$ with the cylinder $ x^2\plus{}y^2\equal{}a^2$.  Show that $ \int_C[(yz\minus{}y)dx\plus{}(xz\plus{}x)dy]\equal{}2\pi a^2$.

assume the curve is oriented in the counter-clockwise direction when viewed from high above the x-y plane.

The snag I'm hitting is that I can't get the normal vector.  When you apply Stoke's Theorem, you take the curl of the vector field, and I'm pretty sure the vector field here is given by $ \textbf{F}\equal{}(yz\minus{}y,xz\plus{}x,0)$, so $ \nabla\times\textbf{F}\equal{}(x,y,z)$ I think.  Once you find the normal vector, it's straightforward, just multiply the curl by the normal vector, switch to polar coordinates to make it easier, and evaluate.

i get $ \nabla\times\textbf{F} \equal{} (\minus{}x,y,2)$ :maybe:

Actually, if that is indeed the curl, then I think the solution works.  Here is my solution (check it):
$ \nabla\times\textbf{F}\equal{}(\minus{}x,y,2)$
The normal vector I think is given by $ (\minus{}2x,\minus{}2y,1)$ so by Stokes' Theorem,
$ \iint_{S}(\minus{}x,y,2)(\minus{}2x,\minus{}2y,1)\mbox{ d}\textbf{S}\equal{}\iint_{S}2x^{2}\minus{}2y^{2}\plus{}2\mbox{ d}\textbf{S}$
Now move into polar coordinates to make it easier:
$ \int_{0}^{2\pi}\int_{0}^{a}r\left(2(r^{2}\cos^{2}\theta\minus{}r^{2}\sin^{2}\theta)\right)\plus{}2r\mbox{ d}r\mbox{ d}\theta\equal{}\int_{0}^{2\pi}\int_{0}^{a}2r^{3}\cos{2\theta}\plus{}2r\mbox{ d}r\mbox{ d}\theta\equal{}\int_{0}^{2\pi}\frac{a^{4}}{2}\cos{2\theta}\plus{}a^{2}\mbox{ d}\theta\equal{}\frac{a^{4}}{2}\frac{\sin{2\theta}}{2}\plus{}a^{2}\theta\large|^{2\pi}_{0}\equal{}2\pi a^{2}$

$ \langle -2x,-2y,1\rangle$ is not the normal vector to a plane. The normal vector to a plan is constant. Furthermore, it's not normalized. But then, you didn't really have $ d\mathbf{S}$ either.

You don't always have to worry about normalization, however. If you have a surface $ z=g(x,y)$ and we choose to use $ (x,y)$ to parameterize the surface so that we will compute an integral $ dx\,dy,$ then:

$ \mathbf{n}=\frac{\left\langle-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1\right\rangle}{\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}}$

But the scalar surface area element is:

$ dS={\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}\,dx\,dy}$

which makes the oriented vector surface area element

$ d\mathbf{S}=\mathbf{n}\,dS=\left\langle-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1\right\rangle\,dx\,dy$

A nonvertical plane parallel to the $ x$ axis can be given by $ z=my+b.$

Then, $ d\mathbf{S}=\langle 0,-m,1\rangle\,dx\,dy$

That makes $ (\nabla\times\mathbf{F})\cdot d\mathbf{S}=(my+2)\,dx\,dy.$

You should be able to finish this from here. In particular, note that $ my$ integrates to zero over a disk centered at the origin.


The average of five numbers is $ 10.6$. Four of the numbers are 10, 4, 5 and 20. What is the value of the fifth number?

The sum of th five numbers is 53.

Let $ x$ be the fifth number.

Then we have $ 10\plus{}4\plus{}5\plus{}20\plus{}x\equal{}53$
$ 39\plus{}x\equal{}53$
$ x\equal{}\boxed{14}$


Despite the creation of this sticky, I found 3 new topics asked and answered by the same person. Waiting 5 minutes doesn't change IP addresses, by the way.

ER what's a red dot?

EDIT:To ernie:

3. If someone posts spam,

-- Rate the post 1 cube.

-- Private message the person detailing why it is spam, in case they did not know, and ask them to kindly abstain from spamming in the future.
[b]
-- Do not make a post saying, "Hey! That's spamming! Don't do it, __________!" [/b]

Stop spamming.  And a red dot is a thread you have posted (or in your case, spammed) in.

[quote="ernie"]I did this once because I wanted a better explanation of how to do it.[/quote]

That is okay. What's bad is when you want to raise your post count or feel like seeing more red dots, so you ask a question you know how to solve and then answer it as if you didn't know how. If you want a better explanation, or are unsure if your solution was correct, that's fine.  :)

[quote="gauss1181"]Well izzy for some reason mewto55555 was just repeating people's answers/solutions when they were already posted. Therefore I rated his posts 1 cube.[/quote]

Planning to PM an admin tonight.  Obviously, mewto is trying to get more posts.

[quote="gauss1181"]Well izzy for some reason mewto55555 was just repeating people's answers/solutions when they were already posted. Therefore I rated his posts 1 cube.[/quote]

this might be due to both people entering at about the same time
say i click to answer a question
while i'm typing the answer, mewto clicks to answer the question
after i click enter, he's not going to see my post until he clicks enter

one thing to check: look at the times before you accuse people
if they're within a minute or two, it's probably due to this

Most of the times for mew's posts were days in between...or hours...

We're getting a little off topic.

Don't accuse based on times. If you think someone did this, ask me to check IP addresses.

i am sorry if these was an announcement describing what the "Ask on Forum" button did, but i did not see it, and i accidentally created the most recent thread
...
the creators of the thread should be able to delete it

Just one point:
FTW is also a speed game, so why people post solutions which take them more than minute to figure out and solve?
Like hand counting of some numbers instead of using a formula: http://www.mathlinks.ro/viewtopic.php?t=237155
I'm sure that one of them would never do like that in FTW, just wants a higher post count.

Always comes back to mewto doesnt it? Not everything in life revolves around FTW. That's great in a cd, but if you have extra time (because your opponent missed or something), its good to check your answer with a different method.

I am continuing to find people asking and answering questions...I wonder how pointless it would be to create a sticky that said "Read the Stickies!"...

Bump. Someone and his multi-accounts apparently did not read this sticky, in addition to a few other people.

It appears some of us need another reminder.

Two points I would like to make. Please read through everything before you post a response.

1) I know this thread is old, but there has recently been a report about the same user asking and answering, which is why I've decided to revive this thread rather than start a new one. Please don't do this. As the first post in this thread says, it's really nothing more than spam. If you don't know how to solve a problem, post here. If you do, then posting the question is really quite pointless.
2) I don't see why Isabella's posts have been rated as spam. To me, that says people don't care about keeping the forums clean and avoiding unnecessary posts. True, these posts were two years ago, maybe (hopefully) the attitude has changed.

And of course, there are those of you who didn't read through that because it seemed long and boring, so here is a simplified version of everything up there:
[color=red]Don't post a question and then answer it yourself! It's pointless and clutters up the forum.[/color]

You may now start attacking me for reviving a two-year old thread/whatever else you may come up with, if you wish.

Post ratings were abused a lot more in the past then they are now.

To anyone considering asking questions and answering them: READ THIS THREAD!!! It contains many good reasons why you SHOULD NOT do so! Threads in which the same person asked and answered the question WILL BE DELETED, so there is NO POINT in asking and answering a question!

I have a question about this topic. One that I'm semi-concerned with.

Is it wrong to answer a question that the Gamebot posted?

Sometimes, well most of the time, I go on to the FTW Problems Discussion forum, and when a new question is posted by Gamebot, I answer them. This is due to the fact that many other people on the forum do the same.

Is that wrong?

I may be wrong here but I really do hope for a clarification. To my knowledge, asking a question and answering them is basically posting a new topic with a question, then going on to post again to answer it again right? Then why am I being accused of SPAMMING by Mrdavid445?

Here is his PM:

[quote="Mrdavid445"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2424464#p2424464]Subject: #29941 Billy is hiking in Colorado. He walks eastward four m[/url]

[quote="SilentBunny"][quote="GameBot"]Billy is hiking in Colorado. He walks eastward four miles, then turns $60$ degrees northward and walks six miles. How far is he from his starting point? Express your answer in simplest radical form.[/quote]

This problem forms a triangle and we have to solve for the missing piece. Two sides of the triangle are $4$ and $6$.

We also know that by turning northward 60 degrees, the obtuse angle of this triangle is 120 degrees.

Using the Law of Cosine $c^2=a^2+b^2-2ab\cos(gamma)$

We get $c^2 = 76$

Simplify that down to $2\sqrt(19)$

$ Answer = 2\sqrt(19)$[/quote]

Hello,

Please do not ask and answer questions. It's considered spam.

Thanks[/quote]

Now that I do not understand. He accused me of spamming when I was just trying to legitimately solve a question posted by Gamebot. You can verify with my other posts on FTW Discussion Forum. I NEVER ask questions and then answer them, I just answer questions posted on there by Gamebot. And in addition to that, I never post Spam answers. All my answers have justification and have good cause.

When I tried to tell him this he responded.

[quote="Mrdavid445"]Sigh...

When you're on FTW and you click "discuss a problem on forum", you only do it because you don't understand. You don't answer the problem you asked. It was quite obvious that you did it- There was 1 reply (yours) and only 1 view (must've been yours)[/quote]

Honestly, I have never went on to FTW to click discuss a problem on forum.  I certainly do not at all care about my post number for that is just a meaningless number. I come on this website to practice math, solve math and propose solutions to unanswered questions about math, not to gain stupid post count.

Now all I want is some clarification. If what I am doing is wrong. By contributing to the site by answering questions. Then I shall stop and not post any longer. If Mrdavid445 is right and I am some how breaking some forum rule that I was not previously aware of, then I am sorry.

Thank you,

SilentBunny

The thing is, some people think it is wrong to make solutions on the posts that you clicked "Discuss on Forum". I do not think so, neither is it considered spam. So you should be fine. Just ignore him.

Also, if you have a really long post like the one on the bottom, please use
[code][hide=Hidden Awesome Texts!]Text here![/hide][/code]
which yields

[hide="Hidden Awesome Texts!"]Text here![/hide]

[quote="TrueshotBarrage"]The thing is, some people think it is wrong to make solutions on the posts that you clicked "Discuss on Forum". I do not think so, neither is it considered spam. So you should be fine. Just ignore him.[/quote]

It is wrong to click 'Ask on Forum' and then answer the question that you asked. The point of this forum is to help each other solve problems. For example, if user A doesn't understand this particular problem x, then he can Ask on Forum. If user B sees this problem and knows how to solve it, he can write a solution. It doesn't make sense for user A to Ask on Forum and then answer the question that he asked, as it doesn't help.

SilentBunny (though he doesn't even come on AoPS I think) is saying that he never answered the problem that he asked.

[quote="bestwillcui1"]It doesn't make sense for user A to Ask on Forum and then answer the question that he asked, as it doesn't help.[/quote]
So you're saying that if I click "Ask on Forum" and answer it in advance for the people in the future who might click "Ask on Forum" for that same question, that is spamming and wrong? What if someone else clicks Ask on Forum later but sees no answer posted? Then that means that, that person will not know the answer to it and will be disappointed.

[quote="hshiems"]Sorry for the revive, but I have two questions:

1. I know that you'll get it trouble for using multiple accounts to ask questions with Gamebot and answer them. But what if you and a friend are under the SAME IP address (i.e. at school or at the same house)? Then you'll get in trouble for asking a quesiton using Gamebot and having your friend answer that question.

2. Also, suppose you ask a question on FTW using Gamebot. What if the Gamebot post is a few years old and nobody has answered it ever since? If you're desperate for help on that problem, wouldn't you have to revive the Gamebot post to attract peoples' attention and get help?[/quote]

1. You don't get in trouble for asking a question with GameBot and answering the same question.

2. You can do that any time, without waiting a few years :P

So... we can hit Ask on Forum and answer our own questions??

[quote="isabella2296"]Recently, I have noticed that quite a few people are asking questions and then answering them just to get more red dots or posts. This is spam and needlessly clogs up the forum. May I remind you that the purpose of this forum is to ask questions [b][i]you don't know how to solve[/i][/b] so that [b][i]someone else[/i][/b] can answer them. Topics that have been asked and answered by the same person have been deleted.[/quote]

Agreed.  I don't do it, but other people...

Is this forum moderated at all?

What are cubes for? Is it like stars on Amazon?

It's called post rating, yes, it's like stars on Amazon.


Euler trigonometry function algebra polynomial search

well i understan the proof that Euler provided for $\displaystyle\sum_{n=1}^\infty{\frac{1}{n^2}}$ and have done it myself and was able to get the result of $\frac{\pi^2}{6}$.  However i read that he was able to use the same method and find the value of $\displaystyle\sum_{n=1}^\infty{\frac{1}{n^4}}$ i have tried to expand on what i did with the other proof but am getting nowhere.

can someone please explain to me what i should do ore if i am even attacking the problem in the right manner.  :blush:
Thanks ahead of time.

Maokid7 - If you remember the sine product ( [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53316]URL)[/url] ), problem you  posted,   one of the solution's "by-product" is  a nice proof of this theorem....(One can use $((sin(nx))/sin(x))$ type formula etc)

Anyway ..
Euler's idea was something like this. You start with $\sin (x)= x - \frac{x^3} {3!} + \frac{x^5} {5!} \dots$

Assume the function $f(x^2) = \frac{\sin x} {x} = 1 - \frac{x^2} 6 + \frac{x^4}{5!} \dots$

The polynomial  has roots when $x^2 = \pi^2 , 2^2 \pi^2, 3^2 \pi^2 \dots$

So sum of reciprocals of roots = $(1/6)$
And sum of squares of (reciprocals of the roots) = $((1/6)^2 - (2/120)) = (1/90)$

So, $\displaystyle\sum_{n=1}^\infty{\frac{1}{n^2} = (\pi^2 / 6)}$

and  $\displaystyle\sum_{n=1}^\infty{\frac{1}{n^4} = (\pi^4 / 90)}$

(This can be made fairly rigorous  - You can see  more details by doing a search here..blah^3 has posted a link with many proofs)

Euler's actual method:

$\frac{\sin \pi x}{\pi x}=\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right)$. Don't worry about justifying this; Euler didn't really.

Now equate the coefficients to the known power series and use Viete's relations, again not worrying about justification:
$\sum_i \frac1{i^2}=\frac{\pi^2}6$
$\sum_{i<j}\frac1{i^2j^2}=\frac{\pi^4}{120}$
$\sum_{i<j<k}\frac1{i^2j^2k^2}=\frac{\pi^6}{5040}$
and so on.

It takes more algebra to find $sum_n\frac1{n^{2t}}$, but we can derive it from the values above. For example,
$\sum_n\frac1{n^4}=\left(\sum_i \frac1{i^2}\right)^2-2\cdot\sum_{i<j}\frac1{i^2j^2} =\frac{\pi^4}{36}-\frac{\pi^4}{60}=\frac{\pi^4}{90}$

Euler ran this up to $\sum_n\frac1{n^{26}}$. That's a prodigious feat of calculation.


I am trying to solve $ \sqrt{1\plus{}\sqrt{1\plus{}\frac12\sqrt{1\plus{}\frac13\sqrt{1\plus{}\dots}}}}$.

I am able to show that it does in fact converge to a real number by comparison to $ \sqrt{1\plus{}\sqrt{1\plus{}\sqrt{1\plus{}\dots}}}$. Then I tried to prove that $ \sqrt{1\plus{}\sqrt{1\plus{}\sqrt{1\plus{}\dots}}}$ actually did converge. I started by showing that it was bounded but then when I tried to show that it was monotonic I ran into some problems. Could someone please help me with this. thanks.

$ t_n \equal{} \sqrt {1 \plus{} \sqrt {1 \plus{} \ldots \plus{} \sqrt {1}}}$
$ t_{n \plus{} 1} \equal{} \sqrt {1 \plus{} \sqrt {1 \plus{} \ldots \plus{} \sqrt {1 \plus{} \sqrt {1}}}}$
So, $ t_{n \plus{} 1} > t_n$.

sorry but i don't quite follow what you are trying to say.

I know that but I don't see how you are able to say that $ t_{n\plus{}1}>t_n$.

Actually you have just shown that both $ t_n$ and $ t_{n\plus{}1}$ are bounded by $ \frac{1\plus{}\sqrt{5}}{2}$.

I don't understand. You doesn't agree with solution?  :mad:  :D You read not carefully  :wink:

I've shown that if $ t_n < \frac {\sqrt {5} \plus{} 1}{2}$ then $ t_{n \plus{} 1} < \frac {\sqrt {5} \plus{} 1}{2}$. So, all $ t_n < \frac {\sqrt {5} \plus{} 1}{2}$.

But unequality $ t_{n \plus{} 1} > t_n$ is true $ \left(t_{n}^2 \minus{} t_n \minus{} 1 < 0\right)$ if $ t_n < \frac {\sqrt {5} \plus{} 1}{2}$. So it is obviously true and sequence $ \left\{t_n\right\}$ is monothonously increesing and bounded by $ \frac {\sqrt {5} \plus{} 1}{2}$ :wink: Obviously it has limit $ L$, which satisfies the equation $ L \equal{} \sqrt {1 \plus{} L}$. This gives $ L \equal{} \frac {\sqrt {5} \plus{} 1}{2}.$

Now about your fist sequence:

Let $ a_n \equal{} \frac {1}{n}$. Then $ a_{n \minus{} 1} \equal{} \frac {1}{n \minus{} 1}\sqrt {1 \plus{} a_n}$. So, $ a_1 \equal{} \sqrt {1 \plus{} \frac12\sqrt {1 \plus{} \frac13\sqrt {1 \plus{} \dots}}}$
Let's find such $ \alpha \equal{} const$ that $ a_n < \frac {\alpha}{n}$ for all $ n$.
Suppose $ a_n < \frac {\alpha}{n}$. Then $ a_{n \minus{} 1} \equal{} \frac {1}{n \minus{} 1}\sqrt {1 \plus{} a_n} < \frac {1}{n \minus{} 1}\sqrt {1 \plus{} \frac {\alpha}{n}} < \frac {\alpha}{n \minus{} 1}$
This gives $ n > \frac {\alpha}{\alpha ^2 \minus{} 1}$. It should be for all $ n\geq 1$. So $ \frac {\alpha}{\alpha ^2 \minus{} 1} < 1$ this gives $ \alpha > \frac {\sqrt {5} \plus{} 1}{2}.$

Now $ a_1 < \frac {\alpha}{1} < \frac {\sqrt {5} \plus{} 1}{2}$.

To prove that $ a_{n \minus{} 1} > a_n$ use same method than above.

Thanks, I now see how you got the $ t_n<t_{n\plus{}1}$. ( i wasn't looking at the fact that $ x^2\minus{}x\minus{}1<0$ when $ x<\frac{1\plus{}\sqrt{5}}{2}$).


Hello Every one.

   Prove or disprove that there exists a map from positive integers to positive  integers such that for any positive integer we have :
   a )       f(n) < n^8 ,
   b )     For any  a_1 ,....... a_k distinct positive integers f(n) is not equal to f(a_1) + f(a_2) ,.... + f(a_k) .

a) requires f(1) < 1^8 = 1

Therefore no such mapping exists since 1 cannot be mapped to a positive integer less than 1.

However, if you want [tex]f(n) \leq n^8[/tex] then I'll need to ask for some clarification on b.

Is it saying
[tex]f(100)\not=f(1)[/tex]
[tex]f(100)\not=f(1)+f(2)[/tex]
.
.
.
[tex]f(100)\not=f(1)+f(2)+...+f(99)[/tex]

or can we choose any k elements adding restrictions like [tex]f(100)\not=f(1)+f(4)+f(25)+f(97)[/tex] ??

Thanks.

-Tom

I have already broken my head on this problem (and not only me :) )
I don't know why I still can't get a solution.

With the condition b, we could prove that an/ n tends to infinity. So i think the answer for this question is yes, there exist.

Such a function, indeed, does exist. Slightly restated, the problem asks for a subset $ E$ of the set of positive integers such that no element of $ E$ is a sum of distinct smaller elements of $ E$ and $ \#(E\cap[1,n^8])\ge n$ for all $ n\ge 1$. Let me do it with $ 7$ instead of $ 8$. Suppose that we already created a finite subset of $ [1,M^7]$ containing $ M^3$ elements with the desired property and $ M > 1$. Now add to this set the numbers $ M^{20} \plus{} M^{11}, M^{20} \plus{} 2M^{11},\dots, M^{20} \plus{} (M^9 \minus{} M^3)M^{11}$. Note that the new numbers are all strictly between $ M^{20}$ and $ 2M^{20}\le M^{21}$, so if we want to represent one of them as a sum of smaller ones, we can use only one new number, after which we'll still have to cover the distance $ M^{11}$ with numbers $ \le M^7$. But we have only $ M^3$ of those!. Thus, we obtain $ (M^3)^3$ numbers in the interval $ (M^3)^7$ and can do the same trick with $ M^3$ instead of $ M$. If $ n\in[M,M^3]$, we have
\[ \#(E\cap[1,n^7])\ge \#(E\cap[1,M^7]) \equal{} M^3\ge n\,.
\]
So, to run this inductive construction, it is enough to find $ 8$ numbers from $ 1$ to $ 128$ with the desired property. $ 1,100,102,104,106,108,110,112$ would be one way of doing it.

This construction can give $ f(n)\le Cn^6$ but breaks down below that. On the other hand, one can show that $ f(n)\ge cn^{2/(\sqrt 5 \minus{} 1)}$ for infinitely many $ n$. I wonder whether this gap can be reduced from either side :).

Thank you very much, Fedor!
Can you explain last estimate from below?

Certainly. Sorry for the delay. Let $ \alpha\equal{}\frac{\sqrt 5\minus{}1}2$ be the positive root of the equation $ \alpha\plus{}\alpha^2\equal{}1$. Take $ n>1$. Let $ x_1<x_2<\dots<x_m$ be the elements of $ E$ not exceeding $ n^\alpha$. Let $ F\equal{}E\cap[1,n]$. Then the sets $ F,F\plus{}x_1, ,F\plus{}x_1\plus{}x_2,\dots, F\plus{}x_1\plus{}\dots\plus{}x_m$ are pairwise disjoint and contained in $ [0,mn^\alpha\plus{}n]$. If $ m\le n^{\alpha^2}$, we conclude that there are only $ (n^\alpha)^\alpha$ elements of $ E$ in $ [1,n^\alpha]$. If $ m>n^{\alpha^2}$, then $ E$ contains not more than $ n^\alpha\plus{}\frac nm\le 2n^\alpha$ elements in $ [1,n]$.

Sorry,why exist in [1,M^7],M^3 elements ?How many are in [1,M^8]?

Please,clearly solution for this problem!


In $ \triangle ABC$, Prove that
$ \cot \frac{B}{2}\cot \frac{C}{2}\sec^2\frac{A}{2}\plus{}\cot \frac{C}{2}\cot \frac{A}{2}\sec^2\frac{B}{2}\plus{}\cot \frac{A}{2}\cot \frac{B}{2}\sec^2\frac{C}{2}\geq \frac{1}{3}\left(\sec^2\frac{A}{2}\plus{}\sec^2\frac{B}{2}\plus{}\sec^2\frac{C}{2}\right)$

[quote="ityaa"]In $ \triangle ABC$, Prove that
$ \cot \frac {B}{2}\cot \frac {C}{2}\sec^2\frac {A}{2} \plus{} \cot \frac {C}{2}\cot \frac {A}{2}\sec^2\frac {B}{2} \plus{} \cot \frac {A}{2}\cot \frac {B}{2}\sec^2\frac {C}{2}\geq \frac {1}{3}\left(\sec^2\frac {A}{2} \plus{} \sec^2\frac {B}{2} \plus{} \sec^2\frac {C}{2}\right)$[/quote]

the follwoing form is right
$ \frac {1}{3}\left(\cot \frac {B}{2}\cot \frac {C}{2}\sec^2\frac {A}{2} \plus{} \cot \frac {C}{2}\cot \frac {A}{2}\sec^2\frac {B}{2} \plus{} \cot \frac {A}{2}\cot \frac {B}{2}\sec^2\frac {C}{2}\right)\geq \left(\sec^2\frac {A}{2} \plus{} \sec^2\frac {B}{2} \plus{} \sec^2\frac {C}{2}\right)$

Your inequality need not be true always.
Itis true only for trinagles with semiperimeter greater than or equal to $ \sqrt {3}$.
By using the well known relations for half angles, we get that the inequality is equivalent to
$ (\sum_{cyclic} (\frac{(z + x)(x + y)}{x}) ( {\frac {(x + y + z)^2 - 3}{3}) \ge 0}$ where  $ a = y + z$, b= z+x$ , c = x + y$.
This necessitaes $ s^2 \ge 3$ where $ s$ is the semi perimeter.

[quote="ityaa"]
the follwoing form is right
$ \frac {1}{3}\left(\cot \frac {B}{2}\cot \frac {C}{2}\sec^2\frac {A}{2} \plus{} \cot \frac {C}{2}\cot \frac {A}{2}\sec^2\frac {B}{2} \plus{} \cot \frac {A}{2}\cot \frac {B}{2}\sec^2\frac {C}{2}\right)\geq \left(\sec^2\frac {A}{2} \plus{} \sec^2\frac {B}{2} \plus{} \sec^2\frac {C}{2}\right)$[/quote]

By Chebyshev : $ LHS \ge \frac{1}{9}(\cot\frac{A}{2}\cot\frac{B}{2}\plus{}\cot\frac{B}{2}\cot\frac{C}{2}\plus{}\cot\frac{C}{2}\cot\frac{A}{2})$ $ (\sec^2\frac{A}{2}\plus{}\sec^2\frac{B}{2}\plus{}\sec^2\frac{C}{2}) \ge \sec^2\frac{A}{2}\plus{}\sec^2\frac{B}{2}\plus{}\sec^2\frac{C}{2}$

$ (\cot\frac{A}{2}\cot\frac{B}{2}\plus{}\cot\frac{B}{2}\cot\frac{C}{2}$ $ \plus{}\cot\frac{C}{2}\cot\frac{A}{2}\equal{}1\plus{} 4\frac{R}{r} \ge 9)$


let polynomial : $ P(x) \equal{} a_{0}x^n \plus{} a_{1}x^{n \minus{} 1} \plus{} ... \plus{} a_n$
      $ a_0,a_1,...,a_n \in R, a_0 \neq 0, n\in N$
set m= min{$ a_0,a_0 \plus{} a_1,...,a_0 \plus{} a_1 \plus{} ... \plus{} a_n$}
prove that: $ P(x) \geq mx^n$ ,    $ \forall x \geq 1.$

[quote="darkdieuguerre"]Forgive me for not understanding, but what does it mean for $ m$ to be equal to a set; do you mean that there exists an element in the set for which the claim is true?[/quote]
sr.^^.i edited one.


100 people from 25 countries attended in a meeting ( each country has 4 ). They sit around a round table.
Prove that we can divide them into 4 groups such that : in each group there is exactly 1 person of each country, and there is no 2 people that sit next to the other in a same group.
Sorry for my bad English, I cannot write more clearly  :blush:


Kevin's order of two hamburgers and three sodas costs $ \$4.13$ at the Burger Sport. Alana orders four cheeseburgers and six sodas. Cheeseburgers cost 15 cents more than hamburgers. How many dollars does Alana's order cost? Express your answer as a decimal to the nearest cent.

Let $ a$ = hamburgers
Let $ b$ = cheeseburgers
Let $ c$ = sodas

$ 2a \plus{} 3c \equal{} 4.13$

$ b \equal{} 0.15 \plus{} a$

Let's rewrite our second equation so that it looks something like our first.

$ b \minus{} a \equal{} 0.15$

What happens if we add our equations?

$ a \plus{} b \plus{} 3c \equal{} 4.28$

We want our equations to look something like this: $ 4b \plus{} 6c \equal{} \text{whatever}$

This is as far as I got. Does anybody have any new insight?

None of that is need.

I will be using the same variables as isabella.

The second equation can be rewritten as $ 2(2b\plus{}3c)$. Now $ 2b\equal{}2(a\plus{}15)$, so by substitution, you get $ 2(4.43)\equal{}\boxed{8.86}$

Oh. Nice.

I feel kind of stupid for all that wasted effort... :blush:


In most of the textbooks it is given that to find the eigen values of a matrix $A$ solve \[ det(A-\lambda I)=0. \] Assume $A$ is $n\times n$
why is it so?
I mean why is the matrix $A-\lambda I$ should be singular?

thank you

sri hari

I think that in order a linear system $Bx=0$ has non-trivial solutions(other than $x=0$) you need $det(B)=0$.

In this case $(A-\lambda I)x=0$ where $x$ are your eigenvectors.

can you please explain in more detail why the system to have nontrivial solution should satisfy the condition $det(B)=0$?

If $Bx=0$ then

$\begin{displaymath} \mathbf{Bx} = \left( \begin{array}{cc} b_{11}}& b_{12} \\ b_{21} & b_{22} \end{array} \right) \end{displaymath}\left( \begin{array}{c} x_1}& x_2 \end{array} \right)=0$

$b_{11}x_1+b_{12}x_2=0$
$b_{21}x_1+b_{22}x_2=0$

if you solve for $x_1$ and $x_2$

$x_1=-\frac{b_{12}}{b_{11}}x_2$

$x_2(\frac{b_{11}b_{22}-b_{12}b_{21}}{b_{11}})=0$

$x_2(\frac{det(B)}{b_{11}})=0$

So, inorder $x \neq 0$ then $det(B)=0$

I guess you can extend this to $n \times n$ matrices :)

simply, if $B$ is invertible (thus det$B\neq0$) you just multiply the equation $Bx=0$ at the left with $B^{-1}$ and you get $x=0$. So in order to have a non trivial solution $B$ must be singular, i.e. det$B=0$.


A complete cycle of a traffic light takes 60 seconds. During each
cycle the light is green for 25 seconds, yellow for 5 seconds, and
red for 30 seconds. At a randomly chosen time, what is the
probability that the light will NOT be green?  Express your answer as a common fraction.

it is green for 5/12 of the time so 1-5/12 =  $ \boxed{\frac{7}{12}}$


Any NYC math team alums who want to go to ARML, email me at jcocoros@stuy.edu

Remember the final practice is the tuesday after Memorial Day at Stuy at 3:30 (immediately following the NYCIML awards)

Jim Cocoros

Congrats to jimmyc on winning a Sliffe Award for Distinguished Teaching!  See
[url]http://www.unl.edu/amc/f-miscellaneous/f1-archive/2006-07-fa/07hssliffe.shtml#NY[/url].

mr.cocoros could you help me. for some reason i didn't receive a registration email from arml. what can i do?

well, spiral191, the first thing I need is who you are, once I get that, I can inform ARML :)

Thanks Ravi, but it really is an award for the nyc math team and the stuyvesant math team--I am nothing without those wonderful and talented students!!!

We were supposed to receive registration emails??

as of now, I have 92 students registered for NYC and housing for the lot.  I will fire off an email to the arml crew to see about your not getting an email.

Mr. C :)

We're going to fit 92 students plus coaches on two buses?  Wow.

Also (sort of unrelated/out of the blue question), is the Queens bus actually leaving from Forest Hills like it says on the permission slip?

Well, Mr. C informed me that the buses were larger this year--59 slots per bus, if my memory serves me correctly.  So 92 students isn't too much of a stretch, and would still leave plenty of room for the adult coaches and the lowly peon coaches like me.

~KC

So I double-posted.  Sue me.

Anyway, the trip is finalized.  Check nycmathteam.com for the details, as well as those students who are taking the make-up SAT exam.

The weather at Penn State will be ~85 each day, with about a 2/5 chance of rain.  Bring an umbrella, just in case--the weather has a habit of messing with everyone, especially at ARML.

[quote="JBL"]I've only seen the partial results that were posted in the ARML forum, but it sounds like NYC A deserves some congratulations for a fabulous 3rd place finish.  Also, it seems that we can put to rest the story of how NYC does poorly on the relays.  Nice work, everyone, and I hope you had a good time![/quote]

Also don't forget that we had the ARML WINNER, if you weren't told already...

[quote="JBL"]No, I'm allergic to threads where everyone complains about their scores, so I hadn't seen.  So, congratulations to Chen, as well.  I gather there were a lot of 7s but that number 8 was a killer?[/quote]

There are around 6 people got #8, surprisingly, nobody got a 8.. Brian Lawrence, USAMO top scorer 3 years in a row, got #8, but he missed #4, 5 , 6, according to Arnav Tripathy.

According to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152302

This year's scores are really close  :ninja:
It's kinda scary to see 3rd-7th differ by 1 point.

BTW, great work guys  :D

Indeed.  Working in the scoring room was kinda nail-biting, as I was pretty much focused on calculating the scores as they were being posted.  I focused primarily on the Division A scores, and I noticed that our A team had a small deficit going into the relay round (in the scoring room, mind you, we get Power scores LAST.)  The relay scores were [i]quite[/i] encouraging, and I believe tied for the top on the Penn State site.  We calculated scores alphabetically, and by the time New York came up, a score of 36 was sufficient to take first place.  We got 30, which wasn't too bad.

To quote Jack, when he saw my pseudo-Excel spreadsheet for the scores, "Damn, Kevin.  I don't keep track of my stocks like that!"

There was also the matter about the penalty levied against the A team.  Luckily, the penalty levied was only one point, and did not affect our ranking.  If the penalty nullified the entire pair, we would have fallen a rank.

~KC

It's true.  I've seen hedge funds that do less analysis on numbers than what Kevin did in the scoring room.


[quote="StefanS"]Find all functions $ f : R \to R$ such that for all $ x,y \in R$ :
$ f(f(x) \minus{} f(y)) \equal{} (x \minus{} y)^2f(x \plus{} y)$.[/quote] i found  f(x)=f(-x) , f(f(-x))=f(f(x)) , f(0)=0 , and f(2f(x))=0 , so f(x)=0  :oops:

[quote="StefanS"]Find all functions $ f : R \to R$ such that for all $ x,y \in R$ :
$ f(f(x) \minus{} f(y)) \equal{} (x \minus{} y)^2f(x \plus{} y)$.[/quote]
Assume that $ f(x)$ isn't the function $ f(x) \equal{} 0 \forall x \in \mathbb{R}$.
So there exist an $ t$ such that $ f(t) \neq 0$.
Let $ P(x,y)$ be the assertion $ f(f(x)\minus{}f(y)) \equal{} (x\minus{}y)^2f(x\plus{}y)$, and $ Q(a,b) \equal{} P \left ( \frac{a\plus{}b}{2}, \frac{a\minus{}b}{2} \right ) \iff f \left ( f \left ( \frac{a\plus{}b}{2} \right ) \minus{} f \left ( \frac{a\minus{}b}{2} \right ) \right ) \equal{} b^2f(a)$.
Looking at $ Q(x,t)$ it is easily seen that $ f$ is surjective on either $ [0;\plus{}\infty)$ or $ (\minus{}\infty;0]$.
$ P(x,0) \iff f(f(x)) \equal{} x^2f(x)$
$ P(0,x) \iff f(\minus{}f(x)) \equal{} x^2f(x)$
So $ f(f(x)) \equal{} f(\minus{}f(x))$ and since $ f$ is surjective on either $ [0;\plus{}\infty)$ or $ (\minus{}\infty;0]$ it is apparent that $ f(x) \equal{} f(\minus{}x) \forall x \in \mathbb{R}$.
$ Q(a,b) \iff f \left ( f \left ( \frac{a\plus{}b}{2} \right ) \minus{} f \left ( \frac{a\minus{}b}{2} \right ) \right ) \equal{} b^2f(a)$
$ Q(b,a) \iff f \left ( f \left ( \frac{a\plus{}b}{2} \right ) \minus{} f \left ( \minus{}\frac{a\minus{}b}{2} \right ) \right ) \equal{} a^2f(b)$
Since $ f \left ( \frac{a\minus{}b}{2} \right ) \equal{} f \left ( \minus{}\frac{a\minus{}b}{2} \right )$ we see:
$ b^2f(a) \equal{} a^2f(b)$.
Letting $ b\equal{}1$:
$ f(a) \equal{} a^2f(1)$
So $ f(x) \equal{} cx^2, c \in \mathbb{R}$
Inserting gives us $ c^3\minus{}c \equal{} 0 \iff c\equal{}1$ or $ c\equal{}\minus{}1$ (Since it was not the function $ f(x) \equal{} 0 \forall x \in \mathbb{R}$). There are both seen to fit the equation.
So we have three solutions:
$ f(x) \equal{} x^2 \forall x \in \mathbb{R}$
$ f(x) \equal{} \minus{}x^2 \forall x \in \mathbb{R}$
$ f(x) \equal{} 0 \forall x \in \mathbb{R}$

Let $ P(x,y)$ the assertion $ f(f(x) \minus{} f(y)) \equal{} (x \minus{} y)^2f(x \plus{} y)$

$ P(x,x) \iff f(0)\equal{}0$

$ P(x,0) \iff f(f(x))\equal{}x^2f(x)$

Let $ x_1 \ne x_2 \in \mathbb{R}$ and let suppose $ f(x_1)\equal{}f(x_2)$.
We have $ f(f(x_1))\equal{}f(f(x_2))$ and so $ x_1^2f(x_1)\equal{}x_2^2f(x_2) \qquad (*)$.

Then we have the following cases:

case $ a)$
The $ (*)$ is a contradiction and so $ f(x_1) \ne f(x_2)$ and so $ f(x)$ is injective.
$ P(x,\minus{}x) \iff f(f(x) \minus{} f(\minus{}x)) \equal{} 0 \iff f(x)\minus{}f(\minus{}x)\equal{}0 \iff$
$ \iff f(x)\equal{}f(\minus{}x) \; x\in \mathbb{R}$ and this is again a contradiction since an even function cannot be injective.

case $ b)$
$ f(x_1)\equal{}f(x_2)\equal{}0$ for every $ x_1 \ne x_2$ and so $ f(x)\equal{}0, x\in \mathbb{R}$

case $ c)$
$ f(x_1)\equal{}f(x_2) \ne 0$ and so $ x_1^2\equal{}x_2^2$. This implies $ x_1\equal{}\minus{}x_2$ since $ x_1\equal{}x_2$ is a contradiction. So the $ f(x)$ is an even function.
From now we can proceed as before as posted by mathias introducing the assertion $ Q(x,y)$ and concluding that the solutions are three:

$ f(x) \equal{} x^2 \forall x \in \mathbb{R}$
$ f(x) \equal{} \minus{} x^2 \forall x \in \mathbb{R}$
$ f(x) \equal{} 0 \forall x \in \mathbb{R}$

@Prester:
Your proof doesn't make much sense. You assume that there exists $ x_1 \neq x_2$ such that $ f(x_1) \equal{} f(x_2)$, but you do not consider the case where there doesn't exist such $ x_1,x_2$. (This should be similar to case $ a$)
Case b) Just because $ x_1 \neq x_2$ and $ f(x_1) \equal{} f(x_2) \equal{} 0$ we don't necessarily know that $ f(x) \equal{} 0 \forall x \in \mathbb{R}$. (Consider for example $ g(x) \equal{} x^2\minus{}x$ where $ 1 \neq 0$ and $ g(1)\equal{}g(0)\equal{}0$.)
Case c) Just because $ f(x_1) \equal{} f(x_2)$ and thus $ x_1\plus{}x_2\equal{}0$ for one pair of $ x_1,x_2$ it isn't necessarily true that $ f$ is an even function. (e.g. $ g(x) \equal{} x^3\minus{}x\plus{}1$, $ 1 \equal{} \minus{}(\minus{}1)$ and $ g(1) \equal{} 1 \equal{} g(\minus{}1)$)

@Mathias_DK
Yes, I see what you say. I meant for any $ x_1$ and $ x_2 \ne x_1$ but now I agree that it does not make sense...
I have still many things to learn...thank you :(

As you proved $ f(x)\equal{}f(\minus{}x).$ Now let $ u\equal{}x\plus{}y$ and $ v\equal{}x\minus{}y.$ So we have
$ v^2f(u)\equal{}f(f(x)\minus{}f(y))\equal{}f(f(x)\minus{}f(\minus{}y))\equal{}u^2f(v)$ implying that
$ \frac{f(x)}{x^2}\equal{}c$ for some constant $ c$ and the original equation restricts this $ c$ to be -1,0,1.
So all solutions are $ f(x)\equal{}\pm x^2$ and $ f(x)\equal{}0.$

[quote="Mathias_DK"] $ Q(a,b) \equal{} P \left ( \frac {a \plus{} b}{2}, \frac {a \minus{} b}{2} \right ) \iff f \left ( f \left ( \frac {a \plus{} b}{2} \right ) \minus{} f \left ( \frac {a \minus{} b}{2} \right ) \right ) \equal{} b^2f(a)$.
Looking at $ Q(x,t)$ it is easily seen that $ f$ is surjective on either $ [0;\plus{}\infty]$ or $ ( \minus{} \infty;0]$.
[/quote]how do you know that f can't be a surjective on $ ( \minus{} \infty;\plus{} \infty)$  :?:

[quote="Dumel"][quote="Mathias_DK"] $ Q(a,b) \equal{} P \left ( \frac {a \plus{} b}{2}, \frac {a \minus{} b}{2} \right ) \iff f \left ( f \left ( \frac {a \plus{} b}{2} \right ) \minus{} f \left ( \frac {a \minus{} b}{2} \right ) \right ) \equal{} b^2f(a)$.
Looking at $ Q(x,t)$ it is easily seen that $ f$ is surjective on either $ [0; \plus{} \infty]$ or $ ( \minus{} \infty;0]$.
[/quote]how do you know that f can't be a surjective on $ ( \minus{} \infty; \plus{} \infty)$  :?:[/quote]
I can't. If it is surjective on $ ( \minus{} \infty; \plus{} \infty)$ then it is also surjective on $ [0; \plus{} \infty )$ and $ ( \minus{} \infty;0]$. ;)

[quote="StefanS"]Let $ P(x,y)$ be the assertion $ f(f(x) \minus{} f(y)) \equal{} (x \minus{} y)^2f(x \plus{} y)$, and $ Q(a,b) \equal{} P \left ( \frac {a \plus{} b}{2}, \frac {a \minus{} b}{2} \right ) \iff f \left ( f \left ( \frac {a \plus{} b}{2} \right ) \minus{} f \left ( \frac {a \minus{} b}{2} \right ) \right ) \equal{} b^2f(a)$.
Looking at $ Q(x,t)$ it is easily seen that $ f$ is surjective on either $ [0; \plus{} \infty)$ or $ ( \minus{} \infty;0]$.

Do you mean looking at $ Q(t,x)$ in order not to have f(a)=0 which will unable the function to be surjective?[/quote]
I mean: If $ b \equal{} \sqrt{\frac{t}{f(a)}}$, where $ t$ has same sign as $ f(a) \neq 0$, we get:
$ Q(a,b) \iff f f \left ( f \left ( \frac {a \plus{} b}{2} \right ) \minus{} f \left ( \frac {a \minus{} b}{2} \right ) \right ) \equal{} x$. And since we can vary $ x$ as we like the conclusion follows.

This is the first post.
--------------------------------------------------------------------------
[quote="StefanS"]Find all functions $ f : R \to R$ such that for all $ x,y \in R$ :

$ f(f(x) - f(y)) = (x - y)^2f(x + y)$.[/quote]
Let $P(x,y)$ be the assertion.
First,see that if $f$ is constant then $f(x)=0\forall x\in \mathbb{R}$ is the only solution.
we will now consider those cases when $f$ is non-constant.
$P(x,x)\implies f(0)=0$
$P(x,0)\implies f(f(x))=x^2f(x)$
Let $f(x)=f(x')$ ($x,x'\in\mathbb{R}$)
Assume $f(x)\not =0$
$\implies f(f(x'))=(x')^2f(x')\implies $ if $f(x)=f(x') $ then $x^2=(x')^2\cdots\cdots \star$.
$P(x,-x)\implies f(f(x)-f(-x))=0=f(0)\implies (f(x)-f(-x))^2=0$(from $\star$) hence $f(x)=f(-x)$.
$P(x,y)\implies f(f(x) - f(y)) = (x - y)^2f(x + y)\cdots\cdots\star\star$
$P(x,-y)\implies f(f(x)-f(-y))=(x+y)^2f(x-y)$
$\implies f(f(x)-f(y))=(x+y)^2f(x-y)\cdots\cdots\star\star\star$
Comparing $\star\star\text{ and }\star\star\star$, we get,
$(x+y)^2f(x-y)=(x-y)^2f(x+y)$
Let $x-y=a,x+y=b$
Hence $b^2f(a)=a^2f(b)$
Taking $b=1$,
we get $f(a)=f(1)a^2$
Let $f(1)=c$
Then $f(a)=ca^2$
Plugging this in the original equation,we get,
$c=1$ or $c=-1$
Hence all solutions are $f(x)=0$ or $f(x)=x^2$ or $f(x)=-x^2\forall x\in\mathbb{R}\blacksquare$
Can somebody check this,I make a lot of mistakes in functional equations.
--------------------------------------------------------------------------------------------------------
EDIT:
ThirdTimeLucky just told me that my proof of $f(x)=f(-x)$ is wrong,but I think that the rest is fine.
$f(x)=f(-x)$ is ture, but my proof is wrong, please see the previous poster's proof using surjectivity.

[quote=Mathias_DK]So $ f(f(x)) = f(-f(x))$ and since $ f$ is surjective on either $ [0;+\infty)$ or $ (-\infty;0]$ it is apparent that $ f(x) = f(-x) \forall x \in \mathbb{R}$.[/quote]
sorry, do I understand correctly that f is surjective on either $ [0;+\infty)$ or $ (-\infty;0]$ means all function which are answers to the question satisfy surjectiveness on either $ [0;+\infty)$ or $ (-\infty;0]$  only. So, for example, consider a function that  is surjective on only $ [0;+\infty)$ . then we can't conclude that   $ f(f(x)) = f(-f(x))$ imply $ f(x) = f(-x) \forall x \in \mathbb{R}$. I think that it imply $ f(x) = f(-x) \forall x \in \mathbb{R+}$ only.
so please tell me the point that i misunderstand.
thanks in advance, srry for my bad english.

We claim that the the only solutions are $ f(x) = 0\ \forall x \in \mathbb{R}$, $ f(x) = x^2 \ \forall x \in \mathbb{R}$ and $ f(x) = -x^2 \forall x \in \mathbb{R}$. Let $P(x,y)$ be the given assertion. $P(0,0)$ gives that $f(0)=0$. Clearly, we have that all three of the above possibilities are indeed solutions. We now show that they are the only solutions. Assume that there exists some $k$ such that $f(k)\not =0$. If $f(k)$ is positive, then we see that because of the $(x-y)^2$ term, $f$ is surjective onto $\mathbb{R}_0$. If $k$ is negative, then we see that $f$ is surjective onto the set of non-positive real numbers.
$P(x,0)$ yields that $f(f(x))=x^2f(x)$, and $P(0,x)$ yields that $f(-f(x))=x^2f(x)$. From our previous findings, we have that for any $x$, there exists some $t$ for which either $f(t)=x$ or $-f(t)=x$, hence we find that $f$ is even. From $P(x,0)$, we find that $f(f(x))=x^2f(x)$. Tripling the involution, $f(f(f(x)))=f(x)^2f(f(x))=x^2f(x)^3$. Let $f(1)=k$. Plugging in $x=1$ yields: $f(k)=k$, and $f(f(f(1)))=k=k^3$. Hence, we see that $k=f(1)=-1$, $0$ or $1$.
Going back to the original assertion, we have that since $f$ is even, $(x-y)^2f(x+y)=f(f(x)-f(y))=f(f(x)-f(-y))=(x+y)^2f(x-y) \implies (x-y)^2f(x+y)=(x+y)^2f(x-y)$. Letting $a=x+y$, and $b=x-y$, we find that $a^2f(b)=b^2f(a) \ \forall a,b \in \mathbb{R}$. Let $Q(a,b)$ be the assertion $a^2f(b)=b^2f(a)$. $Q(a,1)$ gives that $a^2f(1)=f(a)$. If $f(1)=0$, we see that $ \boxed{f(x) = 0\ \forall x \in \mathbb{R}}$. If $f(1)=-1$, we see that $\boxed{ f(x) = -x^2 \forall x \in \mathbb{R}}$. Finally, if $f(1)=1$ we find that $\boxed{ f(x) = x^2 \ \forall x \in \mathbb{R}}$. It follows that these are the only solutions, so we are done. $\square$

[quote=Delray]$P(x,0)$ yields that $f(f(x))=x^2f(x)$, and $P(0,x)$ yields that $f(-f(x))=x^2f(x)$. From our previous findings, we have that for any $x$, there exists some $t$ for which either $f(t)=x$ or $-f(t)=x$, hence we find that $f$ is even.[/quote]

Please, could anyone explain why the second sentence is necessary? Can we not just say $$f(f(x)) = x^2f(x) = f(-f(x))$$ thus the function is even? Why do we need to use surjective here?

you need surjectivity otherwise you can't conclude that f(y)=f(-y) for all y since f(x) doesn't necessarily have a range of all real numbers


hi everyone ok let's get started
x,y,z in R* and x+y+z=1

prove that
0=<xy+yz+zx-2xyz=<7/27

good luck

[url=http://www.mathlinks.ro/viewtopic.php?p=366604#366604]IMO 1984, Problem 1[/url]


Evaluate the sum:
$\sum_{i=0}^{n}\frac{1}{(i+1)(i+2)(i+3)}$

Also, can someone help me on figuring the generalization where there are $k$ consecutive numbers in the denominator?

[hide="A transformation"] Here I make repeated use of the identity $\frac{1}{n}-\frac{1}{n+k}= \frac{k}{n(n+k)}$.

$\frac{1}{(i+1)(i+2)(i+3)}= \left( \frac{1}{i+1}-\frac{1}{i+2}\right) \cdot \frac{1}{i+3}= \frac{1}{2}\left( \frac{1}{i+1}-\frac{1}{i+3}-\frac{2}{i+2}+\frac{2}{i+3}\right) = \frac{1}{2}\left( \frac{1}{i+1}-\frac{2}{i+2}+\frac{1}{i+3}\right)$[/hide] [hide="Hint to the generalization"] The generating functions of the generalizations are given by the repeated integration of $1+x+x^{2}+...+x^{n}= \frac{1-x^{n+1}}{1-x}$. [/hide]

well, using the separable fraction trick twice...
[hide]
$\frac{1}{(i)(i+1)(i+2)}=\frac{1}{2}[\frac{1}{i}+\frac{1}{i+2}-\frac{2}{i+1}]$
[/hide]

which still telescopes

edit: heh, t0ra beat me to it


What is the sum of the seven smallest distinct positive integer multiples of 9?

We have $ 9\cdot1\plus{}9\cdot2\plus{}9\cdot3\plus{}9\cdot4\plus{}9\cdot5\plus{}9\cdot6\plus{}9\cdot7$.  We can factor this to get $ (1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7)(9)\equal{}(28)(9)\equal{}\boxed{252}$.

Or, because this forms an arithmetic sequence, we take the middle term, or the median and multiply it by the amount of terms. 36*7=252


test game

if the above person's signature has a emoticon, reply to the signature.

Otherwise, reply to the above post.

LOL  :)

[hide]i need to make this post longer. read the next hide. [/hide]

[hide]i need to make this post longer. read the next hide, it's different. [/hide]

[hide]i need to make this post longer. read the next hide, it's different-- seriously. [/hide]

[hide]how do you hide things with a link other than 'click to reveal hidden content?'[/hide]

[code][hide=your text goes here if u feel like it, but it doesnt hafta]yr text might go here[/hide][/code]

[hide=" hopefully this works"] thanks![/hide]

[hide=" click here to find true happiness"]

Haha, all I did was expand your page

[/hide]

Yep!!!!!

[hide="no...this is true happiness"]

 :wallbash_red:
hehe

[/hide]


In the Fancyhdr section of the LaTeX guide on this site, there is a typo:

It says

"
Once you have made fancyhdr accessible to your $\LaTeX$, you include it in your source file in the usual way:

\usepackage{facnyhdr}
"

It should be "\usepackage{fancyhdr}.


AwesomeMath
summer program
MathPath
Mathcamp
HCSSiM

We are in the process of lining up representatives from various summer programs to conduct Math Jams.  These Math Jams are an opportunity for you to learn more about the various programs and to ask questions about the application process.

The list of currently scheduled Math Jams (all are at 7:30 p.m. Eastern / 4:30 p.m. Pacific):

Mon, Jan 29: [b]AwesomeMath[/b] (Zuming Feng)

Thu, Mar 1: [b]MathPath[/b] (George Thomas)

Tue, Mar 20: [b]Mathcamp[/b] (Dan Zaharopol)

Thu, Mar 29: [b]Hampshire College Summer Studies in Mathematics[/b] (sarah-marie belcastro)

Tue, May 1: [b]MathPath[/b] (George Thomas)

Watch this space -- we hope to add more of these Math Jams in the near future!

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here to view to complete Math Jam schedule.[/url]

Remember, if you miss a Math Jam, you can always [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]read the transcript afterwards[/url].


Ring Theory
superior algebra
superior algebra unsolved

Prove that every infinite local field with finite characteristic is isomorphic to Laurent series field F_q((X)).

Let $K$ be our field, which we assume to be complete with respect to a discrete valuation $\nu$ such that the residue field $k$ is the finite field $\mathbb F_{q}$ (the field with $q$ elements, where $q$ is a prime power $p^{n}$). Also let $A$ be the valuation ring of $\nu$, i.e. $A=\{x\in K\ | \ \nu(x)\ge 0\}$, and $\frak p$ the unique maximal ideal of $A$, i.e. $\frak p=\{x\in K\ |\ \nu(x)>0\}$.

If $\pi\in K$ is such that $\nu(\pi)=1$, and $S\subset A$ is a system of distinct representatives modulo $\frak p$ which contains $0$, then it's easy to show that every element of $K$ can be written uniquely in the form $\sum_{n\in\mathbb Z}s_{n}\pi^{n}$ with $s_{n}\in S$ being $0$ for sufficiently small integers $n$. If we manage to prove that $\mathbb F_{q}$ can be embedded in $K$, then we can take $S=\mathbb F_{q}$, and this says precisely that $K=\mathbb F_{q}((X))$, as desired. Finally, the fact that $\mathbb F_{q}$ can be embedded in $K$ follows from [url=http://mathworld.wolfram.com/HenselsLemma.html]Hensel's lemma[/url] together with the fact that $K$ has characteristic $p$: the equation $f(x)=x^{q}-x=0$ has $q$ solutions in $k$, which by Hensel's lemma we can lift to $q$ solutions in $A$.


search
number theory unsolved
number theory

Problem 3:
Let $ A$ be a set of real numbers such that $ A$ has at least four elements.Suppose $ A$ has the property that $ a^{2} \plus{} bc$ is a rational number for all distinct numbers $ a, b, c$ in $ A$. Prove that there exists a positive integer $ M$ such that $ a\sqrt{M}$ is a rational number for every $ a$ in $ A$.

The following statements are for each a,b,c,d $ \in$ A, all distinct:

1. if $ 0 \in A$, $ 0^2 \plus{} ab \equal{} ab \in A$

2. let's suppose now that 0 is not in A
$ a^2 \plus{} bc \in Q$, $ d^2 \plus{} bc \in Q$, from here by substraction: $ d^2 \minus{} a^2 \in Q$ [1]
$ a^2 \plus{} bc \in Q$, $ b^2 \plus{} ac \in Q$, again by substraction: $ a^2 \minus{} b^2 \plus{} c(b \minus{} a) \in Q$ [2]
From [1], putting d->b, $ b^2 \minus{} a^2 \in Q$ and adding to [2] yields into: $ c(b \minus{} a) \in Q$ [3]
From [3], putting c->d, $ d(b \minus{} a) \in Q$ and dividing by [3]: $ \frac{d}{c} \in Q$ because c is not 0.
So d->b, $ \frac{b}{c} \in Q$ and multiply by [3]: $ b(b \minus{} a) \in Q$, $ b^2 \minus{} ba \in Q$. Add this to $ c^2 \plus{} ab \in Q$ goes into: $ b^2 \plus{} c^2 \in Q$, $ c^2(\frac{b^2}{c^2} \plus{} 1) \in Q$, but $ \frac{b}{c} \in Q$ so $ c^2 \in Q$.We conclude $ c^2 * \frac{b}{c} \equal{} bc \in Q$
Finally from initial condition we found $ ab \in Q$ for each $ a,b \in A$ [distinct or not]. The reverse is also true

From $ a^2 \in Q$ for each $ a \in A$ we get $ a \equal{} e\sqrt{\frac{x}{y}}, e \equal{} 1$ or $ e \equal{} \minus{}1, x,y \in N$, y not 0, $ (x,y) \equal{} 1$
Let's consider a non zero element $ a_0 \equal{} e\sqrt{\frac{x_0}{y_0}}$ of A such that $ x_0 \geq 1,y_0 \geq 1$, and $ (x_0,y_0) \equal{} 1$
If $ a \in A$ is another element, then $ a a_0 \equal{} q \in Q$, so $ a \equal{} \frac{q}{a_0}$, therefore $ A \equal{} \{a_0\} U \{\frac{q}{a_0} | q \in Q\}$
We may consider $ M \equal{} x_0 y_0$ and check easily that $ a \sqrt{M} \in Q$ for each $ a \in A$


Let $a,b,c,d>0$ such that $a+b+c+d=12$.
Prove that $27+ab+cd+ac+bd+ad+bc\geq abcd$.


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

Let $ P\in\mathbb{C}^{m\times m}$ be a nonzero projector. Show that $ \parallel{}P\parallel{}_2\ge 1$ with equality iff $ P$ is orthogonal.

(I have the first part but can't prove the equality case. Also what is the right syntax for $ \parallel{}P\parallel{}$?)

Hi Peter, some questions have come up to my mind, what is the definition of projector that you've mentioned ? and what does $ \parallel{}.\parallel{}_2$ mean ?

The syntax you're looking for is \|P\|, which gives $ \|P\|.$

We have that $ P$ is unitarily similar to $ \begin{pmatrix}I&B\\0&0\end{pmatrix}.$

(This follows from Schur's lemma that any square matrix is unitarily similar to an upper triangular matrix, together with the fact that the minimal polynomial for $ P$ divides $ x^2\minus{}x.$)

We're trying to show that $ \|P\|_2\equal{}1$ if and only if $ B\equal{}0.$

Suppose $ B\ne 0.$ Note that $ BB^*$ is Hermitian and positive semidefinite. There exists $ u\ne 0$ such that $ BB^*u\equal{}\lambda u$ for some $ \lambda >0.$

Now let $ x\equal{}\begin{pmatrix}u\\ B^*u\end{pmatrix}$

$ \|x\|^2\equal{}u^*u\plus{}u^*BB^*u\equal{}(1\plus{}\lambda)\|u\|^2.$

$ Px\equal{}\begin{pmatrix}I&B\\0&0\end{pmatrix} \begin{pmatrix}u\\ B^*u\end{pmatrix}\equal{}\begin{pmatrix}u\plus{}BB^*u\\0\end{pmatrix}\equal{} \begin{pmatrix}(1\plus{}\lambda)u\\0\end{pmatrix}$

Hence $ \|Px\|^2\equal{}(1\plus{}\lambda)^2\|u\|^2.$ Since $ \lambda>0,$ we have that $ (1\plus{}\lambda)^2>1\plus{}\lambda$ and $ \|Px\|^2>\|x\|^2.$

This means that $ \|P\|$ is strictly greater than $ 1.$ The only escape from this example is to have $ B\equal{}0,$ which means that $ P$ is an orthogonal projection.

@ Ehsan: a projector is a square matrix $ P$ with $ P^2 \equal{} P$, and it is orthogonal iff $ P^* \equal{} P$ (i.e. $ P$ is hermitian).
The two-norm of a vector $ x$ is $ \|x\|_2 \equal{} \sqrt {\sum |x_i|^2}$ and the two-norm of a matrix $ A$ is $ \max_{\|x\|_2 \equal{} 1}\|Ax\|_2$.

[quote="Kent Merryfield"]The syntax you're looking for is \|P\|, which gives $ \|P\|.$[/quote]Thanks!

[quote="Kent Merryfield"]We have that $ P$ is unitarily similar to $ \begin{pmatrix}I & B \\
0 & 0\end{pmatrix}.$

(This follows from Schur's lemma that any square matrix is unitarily similar to an upper triangular matrix)
[hide="rest of the solution"]
 together with the fact that the minimal polynomial for $ P$ divides $ x^2 \minus{} x.$

We're trying to show that $ \|P\|_2 \equal{} 1$ if and only if $ B \equal{} 0.$

Suppose $ B\ne 0.$ Note that $ BB^*$ is Hermitian and positive semidefinite. There exists $ u\ne 0$ such that $ BB^*u \equal{} \lambda u$ for some $ \lambda > 0.$

Now let $ x \equal{} \begin{pmatrix}u \\
B^*u\end{pmatrix}$

$ \|x\|^2 \equal{} u^*u \plus{} u^*BB^*u \equal{} (1 \plus{} \lambda)\|u\|^2.$

$ Px \equal{} \begin{pmatrix}I & B \\
0 & 0\end{pmatrix} \begin{pmatrix}u \\
B^*u\end{pmatrix} \equal{} \begin{pmatrix}u \plus{} BB^*u \\
0\end{pmatrix} \equal{} \begin{pmatrix}(1 \plus{} \lambda)u \\
0\end{pmatrix}$

Hence $ \|Px\|^2 \equal{} (1 \plus{} \lambda)^2\|u\|^2.$ Since $ \lambda > 0,$ we have that $ (1 \plus{} \lambda)^2 > 1 \plus{} \lambda$ and $ \|Px\|^2 > \|x\|^2.$

This means that $ \|P\|$ is strictly greater than $ 1.$ The only escape from this example is to have $ B \equal{} 0,$ which means that $ P$ is an orthogonal projection.[/hide][/quote]Yes, that way it works indeed. Thanks!

Strange though, at that point in the course we have not seen Schur's lemma yet (only trivial properties of unitary matrices, norms, singular value decomposition and projections). Does anyone happen to know a solution using these tools only?

[quote="Peter"]a projector is a square matrix $ P$ with $ P^2 \equal{} P$, [/quote]
I would probably have used the word "idempotent" for this. But as you can see, I knew exactly what you meant.

$ P^2\equal{}P$ then $ \parallel{}P\parallel{}\leq{\parallel{}P\parallel{}^2}$ and $ \parallel{}P\parallel{}\geq{1}$ because $ P$ is non zero.
Now assume $ P$ non orthogonal that is to say $ ker(P)$ and $ Im(P)$ are not orthogonal. There exists $ z$ in the orthogonal of $ ker(P)$ and not in $ Im(P)$. $ z\equal{}x\plus{}y$ where $ x\in{Im(P)}$ and $ y$ a non zero vector of $ ker(P)$. $ \parallel{}z\parallel{}^2\equal{}\parallel{}x\parallel{}^2\minus{}\parallel{}y\parallel{}^2<\parallel{}x\parallel{}^2$ and $ P(z)\equal{}x$ then $ \parallel{}P\parallel{}>1$.


calculus
integration
number theory proposed
number theory

hallo,
find all integral solutions of :
$19x^{3}-84y^{2}=1984$.  :)

This equation had not solutions, because $19(x^{3}-100)=84(y^{2}+1)$, give $19|(y^{2}+1)$. It is contaditon, because $(\frac{-1}{19})=-1$.


Multiply.

(x-2)4 (the 4 is the the 4th, I don't know how to make the number smaller)

Please show me every step to get the answer!!! Thank you!

Code for $ (x\minus{}2)^4$: (x-2)^4 - under dollar signs
Do you know [url=http://en.wikipedia.org/wiki/Binomial_theorem]Binomial Theorem[/url]?
Other method:
$ (x\minus{}2)^4\equal{}((x\minus{}2)^2)^2\equal{}(x^2\minus{}4x\plus{}4)^2\equal{}x^4\plus{}16x^2\plus{}16\minus{}8x^3\plus{}8x^2\minus{}32x \equal{}
x^4\minus{}8x^3\plus{}24x^2\minus{}32x\plus{}16$
, if you know how to square binomials.

The Binomial Theorem states that $ (x \plus{} y)^n \equal{} \sum_{k \equal{} 0}^{n}\binom{n}{k}x^{n \minus{} k}y^k$, where $ \binom{n}{k} \equal{} \frac {n!}{(n \minus{} k)!k!}$.
This looks confusing but it actually is pretty simple.

For your problem you would use it like this:

$ \sum_{k \equal{} 0}^{4}\binom{4}{k}x^{4 \minus{} k}( \minus{} 2)^k$

$ \equal{} \binom{4}{0}x^4( \minus{} 2)^0 \plus{} \binom{4}{1}x^3( \minus{} 2)^1 \plus{} \binom{4}{2}x^2( \minus{} 2)^2 \plus{} \binom{4}{3}x^1( \minus{} 2)^3 \plus{} \binom{4}{4}x^0( \minus{} 2)^4$

$ \equal{} x^4 \minus{} 8x^3 \plus{} 24x^2 \minus{} 32x \plus{} 16$

If you play through Alcumus long enough, you can get to the AoPS Binomial Theorem videos, which are rather helpful.

Ah, thanks a bunch! That really helped me out~


Find all positive integers n for which the quadratic equation
$ a_{n\plus{}1}x^2\minus{}2x\sqrt{\sum_{k\equal{}1}^{n\plus{}1}a_k^2}\plus{}\sum_{k\equal{}1}^{n}a_k\equal{}0$
has real roots for every choice of real numbers $ a_1,a_2,...,a_{n\plus{}1}$.

[quote="tamal banerjee"]Find all positive integers n for which the quadratic equation
$ a_{n \plus{} 1}x^2 \minus{} 2x\sqrt {\sum_{k \equal{} 1}^{n \plus{} 1}a_k^2} \plus{} \sum_{k \equal{} 1}^{n}a_k \equal{} 0$
has real roots for every choice of real numbers $ a_1,a_2,...,a_{n \plus{} 1}$.[/quote]
Let $ P(n,p)$, $ n,p\in\mathbb N$ be the assertion : $ \exists$ $ n$ real numbers $ \{a_k\}_{k\equal{}1}^n$ such that $ (\sum_{k\equal{}1}^na_k)^2>p\sum_{k\equal{}1}^{n}a_k^2$

[u][b]Lemma [/b][/u]: For $ n>1,p>0$ $ P(n,p)$ $ \iff$ $ P(n\minus{}1,p\minus{}1)$
$ P(n,p)$ $ \iff$ $ \exists$ $ n\minus{}1$ real numbers $ \{a_k\}_{k\equal{}1}^{n\minus{}1}$ such that the inequation $ (x\plus{}\sum_{k\equal{}1}^{n\minus{}1}a_k)^2>p(x^2\plus{}\sum_{k\equal{}1}^{n\minus{}1}a_k^2)$ has a solution
$ (x\plus{}\sum_{k\equal{}1}^{n\minus{}1}a_k)^2>p(x^2\plus{}\sum_{k\equal{}1}^{n\minus{}1}a_k^2)$ $ \iff$ $ (p\minus{}1)x^2\minus{}2x\sum_{k\equal{}1}^{n\minus{}1}a_k\plus{}p\sum_{k\equal{}1}^{n\minus{}1}a_k^2\minus{}(\sum_{k\equal{}1}^{n\minus{}1}a_k)^2<0$
And so $ (x\plus{}\sum_{k\equal{}1}^{n\minus{}1}a_k)^2>p(x^2\plus{}\sum_{k\equal{}1}^{n\minus{}1}a_k^2)$ has a solution $ \iff$ $ (\sum_{k\equal{}1}^{n\minus{}1}a_k)^2>(p\minus{}1)(p\sum_{k\equal{}1}^{n\minus{}1}a_k^2\minus{}(\sum_{k\equal{}1}^{n\minus{}1}a_k)^2)$ $ \iff$ $ (\sum_{k\equal{}1}^{n\minus{}1}a_k)^2>(p\minus{}1)\sum_{k\equal{}1}^{n\minus{}1}a_k^2$ (if $ p>0$)
Hence the lemma's proof.

[u][b]Main proof [/b][/u]: the required problem is equivalent to :
Find all positive integers n for which $ \sum_{k\equal{}1}^{n\plus{}1}a_k^2\ge a_{n\plus{}1}\sum_{k\equal{}1}^{n}a_k$ for every choice of real numbers $ \{a_k\}_{k\equal{}1}^{n\plus{}1}$
$ \iff$ $ a_{n\plus{}1}^2\minus{}a_{n\plus{}1}\sum_{k\equal{}1}^{n}a_k\plus{}\sum_{k\equal{}1}^{n}a_k^2\ge 0$ for every choice of real numbers $ \{a_k\}_{k\equal{}1}^{n\plus{}1}$
$ \iff$ $ (\sum_{k\equal{}1}^{n}a_k)^2\minus{}4\sum_{k\equal{}1}^{n}a_k^2<0$ for every choice of real numbers $ \{a_k\}_{k\equal{}1}^{n}$

So the required problem is "Find all positive integers n for which $ P(n,4)$ is false"

For $ n\equal{}1$, $ P(1,4)$ $ \iff$ $ \exists a\in\mathbb R$ such that $ a^2>4a^2$, false
For $ n\equal{}2$, $ P(2,4)$ $ \iff$ $ P(1,3)$ $ \iff$ $ \exists a\in\mathbb R$ such that $ a^2>3a^2$, false
For $ n\equal{}3$, $ P(3,4)$ $ \iff$ $ P(1,2)$ $ \iff$ $ \exists a\in\mathbb R$ such that $ a^2>2a^2$, false
For $ n\equal{}4$, $ P(4,4)$ $ \iff$ $ P(1,1)$ $ \iff$ $ \exists a\in\mathbb R$ such that $ a^2>a^2$, false
For $ n>4$, $ P(n,4)$ $ \iff$ $ P(n\minus{}4,0)$, obviously true

Hence the result : $ \boxed{n\in\{1,2,3,4\}}$


The sides of a right triangle are $ a$ and $ b$ and the hypotenuse is $ c$. A perpendicular from the vertex divides $ c$ into segments $ r$ and $ s$, adjacent respectively to $ a$ and $ b$. If $ a : b \equal{} 1 : 3$, then the ratio of $ r$ to $ s$ is:

$ \textbf{(A)}\ 1 : 3\qquad
\textbf{(B)}\ 1 : 9\qquad
\textbf{(C)}\ 1 : 10\qquad
\textbf{(D)}\ 3 : 10\qquad
\textbf{(E)}\ 1 : \sqrt{10}$

[quote="sunehra"]... A perpendicular from the vertex [which vertex?] divides $ c$ into...[/quote]

Which vertex is the perpindicular drawn from?

he means vertex C

$ b^2 \equal{} s.a \ , \ a^2 \equal{} r.a$ , therefore $ r/s \equal{} a^2/b^2 \equal{} 1/9$ $ (B)$


geometry
perpendicular bisector
angle bisector
\/closed

Well, I've had Intro to Geometry for a long time, but I've never actually gone through more than the first four chapters. I was thinking of going through it now, but I'm worried that I'll already know so much of the material that the time will be somewhat of a waste. I took the post test online, and I can solve all of them besides #10, since I don't know how to construct anything more advanced than a perpendicular bisector, angle bisector, etc :P . Also, I basically solved every question on sight, or knew exactly what to do, besides questions #6 and #8, which took some time.

Should I just attempt all the challenge problems at the end of chapters, and on those chapters where I can't do all but a couple problems, should I go over that chapter?

Thank you.

[quote="xpmath"]
Should I just attempt all the challenge problems at the end of chapters, and on those chapters where I can't do all but a couple problems, should I go over that chapter?
[/quote]

This is exactly what you should do.  And read/work through the problems in each of the Construction sections.


Yeah, there was an error in the Advanced Topics test at PUMaC, and the summation was somewhat harder, because Vandermonde trivialized the intended one. But I'm pretty sure of my answer on this one.

$\sum_{n=0,6\le r\le12,12-r\ge6-n}^{r}{r\choose6}{12-r\choose 6-n}$.

[hide="the obvious hint"]factor out ${r\choose6}$.[/hide]

I don't understand the notation at all. Could you explain? I understand simpler notation, but not this.

well it's pretty self explanatory?

just the regular $\sum_{k=0}^{r}$, then with the other restrictions that $6\le r\le12$, etc.?

\substack is a useful command for this:

$\sum_{\substack{0 \leq n \leq r \\ 6 \leq r \leq 12 \\ 6-n \leq 12-r}}\binom{r}{6}\binom{12-r}{6-n}$

Usually, this means you sum over all choices of the variables which meet the conditions.  In this case, it might be better to write it like this:

$f(r) = \sum_{\substack{0 \leq n \leq r \\ 6-n \leq 12-r}}\binom{r}{6}\binom{12-r}{6-n}$ for $6 \leq r \leq 12$.  Compute/simplify the expression for $f(r)$.


Let $a;b;c>0$.Prove that:
$\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}\geq a\sqrt{ab}+b\sqrt{bc}+c\sqrt{ca}$

Use AM-GM, we obtain:
$\frac{a^{3}}{b}+ab \ge 2a^{2}$
and
$\frac{a^{3}}{b}+b^{2}\ge 2a \sqrt{ab}$
so that
$2 \sum \frac{a^{3}}{b}+\sum ab+\sum a^{2}\ge 2 \sum a^{2}+2\sum a \sqrt{ab}$
On the other hand, we have:
$ab+bc+ca \le a^{2}+b^{2}+c^{2}$
Hence
$\sum \frac{a^{3}}{b}\ge \sum a \sqrt{ab}$

It can be solved with the same method as [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=137346[/url].


Is there a word $x=x_1x_2...x_{2005}$ ($x_i\in\{a,b\}$) s.t. it is impossible to partition $x$ to at most
a) 811 palindromes?
b) 669 palindromes?
c) 668 palindromes?

Well, $811$ can clearly always be made. We can see by inspection that we can always split a word of length 5 into 2 palindromes; thus by partitioning into groups of 5 we can do it with $2(401)=802$ palindromes.

Strange that they used an old [b]Kvant[/b] problem for the tournament. It's about [b]baabab[/b]s, isn't it?

It is not strange. Kvant contains thousands problems.

[quote="Myth"]Kvant contains thousands problems.[/quote] That's true. But not too many of them were devoted two special articles to. Besides, these articles were really hard to forget, so funny was the story. First somebody published an article praising the power of computer and titled "Um horosho, a pjat' luchshe" (a brain is good but five brains is better) in which he considered three problems, the palindrom one being one of them. He advertised using a computer (with penti[b]um[/b] processor) for them because they seemed to him impossible to solve by ordinary methods. Next (probably in the very next issue, but I am not sure about this now) somebody else under the pen name [b]Baabab[/b]ov published an article titled something like "Pentium-horosho, a um luchshe" (Pentium is good but a brain is better) in which he gave an elementary solution to each of the three problems mentioned in the first article. Anyway, that's beside the point. Let's see whether somebody will find a short solution for b)   ;)


A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 19.0 m/s when it reaches the end of the ramp, which has length 128 m.

What is the acceleration of the car?

The traffic on the freeway is moving at a constant speed of 19.0m/s. What distance does the traffic travel while the car is moving the length of the ramp?

For the first part, we use $ v_{f}^{2}=v_{i}^{2}+2ad$ so by plugging in, we get
$ 19^{2}=0^{2}+2a(128)\Rightarrow a\approx 1.41\mbox{m}/{\mbox{s^{2}}$.

For the second one, $ v_{f}=v_{i}+at$ and $ \bar{v}=\frac{d}{t}$.  I'll leave the rest to you.


Given $ a,b,c \ge 0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ \frac{a}{ab\plus{}3}\plus{}\frac{b}{bc\plus{}3}\plus{}\frac{c}{ca\plus{}3} \le \frac{3}{4}$
It posted in this forum but the solution is wrong,so I hope the author can edited it :)

Can_hang2007,you can edited your solution?It 's wrong

[quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} \le \frac {3}{4}$
It posted in this forum but the solution is wrong,so I hope the author can edited it :)[/quote]
What a pity that I got a silly mistake in calculating! Very sorry and thank you very much for pointing out my mistake, quykhtn. I have changed the solution by the following, it is ugly so I wait for a nicer proof

[u]Solution:[/u]
[u]Lemma:[/u] Let $ a;b;c$ be positive numbers. We have:
$ 1)\sum\limits_{cyc} {ab^3 }  \le \frac{1}{3}(a^2  \plus{} b^2  \plus{} c^2 )^2  \le 3$
$ 2)\sum\limits_{cyc} {ab^2 }  \ge \frac{{9abc(a^2  \plus{} b^2  \plus{} c^2 )}}{{(a \plus{} b \plus{} c)^2 }} \ge \frac{{27abc}}{{(a \plus{} b \plus{} c)^2 }}$
Applying the AM-GM inequality,we obtain:
$ ab \ge \frac{{4a^2 b^2 }}{{(a \plus{} b)^2 }}$
So,our job is to prove that
$ \sum {\frac{{a(a \plus{} b)^2 }}{{4a^2 b^2  \plus{} 3(a \plus{} b)^2 }}}  \le \frac{3}{4}$
We may rewrite inequality as
$ \sum {\left( {\frac{1}{3}a \minus{} \frac{{a(a \plus{} b)^2 }}{{4a^2 b^2  \plus{} 3(a \plus{} b)^2 }}} \right)}  \ge \frac{1}{3}(a \plus{} b \plus{} c) \minus{} \frac{3}{4}$
$ \Leftrightarrow \frac{4}{3}\sum {\frac{{a^3 b^2 }}{{4a^2 b^2  \plus{} 3(a \plus{} b)^2 }}}  \ge \frac{1}{3}(a \plus{} b \plus{} c) \minus{} \frac{3}{4}$
By the Cauchy-Schwarz inequaliy,we have:
$ \sum {\frac{{a^3 b^2 }}{{4a^2 b^2  \plus{} 3(a \plus{} b)^2 }}}  \equal{} \sum {\frac{{a^2 b^2 }}{{4ab^2  \plus{} 3\frac{{(a \plus{} b)^2 }}{a}}}}  \ge \frac{{(ab \plus{} bc \plus{} ca)^2 }}{{4\sum\limits_{cyc} {ab^2 }  \plus{} 3\sum\limits_{cyc} {\frac{{(a \plus{} b)^2 }}{a}} }}$
We need to prove that:
$ (ab \plus{} bc \plus{} ca)^2  \ge \left[ {4\sum\limits_{cyc} {ab^2 }  \plus{} 3\sum\limits_{cyc} {\frac{{(a \plus{} b)^2 }}{a}} } \right].\left[ {\frac{1}{3}(a \plus{} b \plus{} c) \minus{} \frac{3}{4}} \right]$
 $ \Leftrightarrow 3\sum\limits_{cyc} {ab^2 }  \plus{} \frac{9}{4}\left[ {3(a \plus{} b \plus{} c) \plus{} \sum\limits_{cyc} {\frac{{b^2 }}{a}} } \right] \ge \frac{{10}}{3}(ab\plus{}bc\plus{}ca)\plus{}9\plus{}\frac{4}{3}abc(a \plus{}b \plus{}c)\plus{}\frac{4}{3}\sum\limits_{cyc} {ab^3 }  \plus{} (a \plus{} b \plus{} c)\sum\limits_{cyc} {\frac{{b^2 }}{a}}$
Applying our lemma,it suffices to prove that:
$ \left( {\frac{9}{4} \minus{} a \minus{} b \minus{} c} \right)\sum\limits_{cyc} {\frac{{b^2 }}{a}}  \plus{} \frac{{81abc}}{{(a \plus{} b \plus{} c)^2 }} \ge \frac{{10}}{3}(ab \plus{} bc \plus{} ca) \plus{} \frac{4}{3}abc(a \plus{} b \plus{} c) \plus{} 13$
$ \Leftrightarrow \left( {\frac{9}{4} \minus{} a \minus{} b \minus{} c} \right)\left[ {\sum\limits_{cyc} {\left( {\frac{{b^2 }}{a} \minus{} 2b \plus{} a} \right)} } \right] \plus{} \frac{{81abc}}{{(a \plus{} b \plus{} c)^2 }} \minus{} 3(a \plus{} b \plus{} c) \plus{} \frac{{16}}{3}(\sum {a^2 }  \minus{} \sum {ab} ) \plus{} \frac{4}{3}(\sum {a^2 b^2 }  \minus{} abc\sum a ) \plus{} \frac{4}{9}\left[ {(\sum {a^2 } )^2  \minus{} 3\sum {a^2 b^2 } } \right] \plus{} 12.\frac{{3\sum {a^2 }  \minus{} (\sum a )^2 }}{{a \plus{} b \plus{} c \plus{} 3}} \ge 0$
$ \Leftrightarrow \sum {(a \minus{} b)^2 \left[ {\frac{9}{4}a \plus{} \frac{{3(a \plus{} b \plus{} 7c)}}{{2(a \plus{} b \plus{} c)^2 }} \plus{} \frac{8}{3} \plus{} \frac{2}{3}c^2  \plus{} \frac{2}{9}(a \plus{} b)^2  \minus{} \frac{{12}}{{3 \plus{} a \plus{} b \plus{} c}}} \right]}  \ge 0$
Morever, $ (3 \plus{} a \plus{} b \plus{} c)S_c  > \frac{{27}}{{4a}} \plus{} \frac{{10}}{3}a \plus{} \frac{9}{4} \plus{} \frac{9}{{2(a \plus{} b \plus{} c)^2 }} \plus{} \frac{3}{{2(a \plus{} b \plus{} c)}} \plus{} 8 \minus{} 12 > 0$.
That yields $ S_c>0$. Similarly,$ S_a>0;S_b>0.$
Equality occurs if and only if $ a\equal{}b\equal{}c\equal{}1$.  :lol:

Sorry, it's wrong. It's my mistake in calculation.  :D  :blush:

I'll try it again without helping of computer.
Firstly,we have a equality:
$ 2\sum\limits_{cyc} {\frac {a}{{ab \plus{} 3}}} \equal{} \sum {\frac {{a \plus{} b}}{{ab \plus{} 3}}} \plus{} \frac {{3(a \minus{} b)(b \minus{} c)(c \minus{} a)}}{{(ab \plus{} 3)(bc \plus{} 3)(ca \plus{} 3)}}.$
So,our job is to prove that:
$ \frac {{3\prod {(a \minus{} b)} }}{{\prod {(ab \plus{} 3)} }} \le \sum {\left( {\frac {1}{2} \minus{} \frac {{a \plus{} b}}{{ab \plus{} 3}}} \right)} \equal{} \frac {1}{4}\sum {\frac {{(a \plus{} b \minus{} 2)^2 }}{{ab \plus{} 3}}} \plus{} \frac {1}{4}\sum {\frac {{c^2 \minus{} 1}}{{ab \plus{} 3}}}$
Now,we'll prove the inequality:
$ \frac {1}{4}\sum {\frac {{c^2 \minus{} 1}}{{ab \plus{} 3}}} \ge \frac {{3\prod {(a \minus{} b)} }}{{\prod {(ab \plus{} 3)} }}$
$ \Leftrightarrow \frac {1}{{12}}\sum {\frac {{(a \minus{} b)^2 (a \plus{} b)c}}{{(ac \plus{} 3)(bc \plus{} 3)}}} \ge \frac {{3\prod {(a \minus{} b)} }}{{\prod {(ab \plus{} 3)} }}$
$ \Leftrightarrow \sum {(a \minus{} b)^2 (ab \plus{} 3)c(a \plus{} b) \ge } 36\prod {(a \minus{} b)}$
Setting $ f(a;b;c) \equal{} \sum {(a \minus{} b)^2 (ab \plus{} 3)c(a \plus{} b) \minus{} } 36\prod {(a \minus{} b)}$
Very easy to verify that $ f(a;b;c) \ge f(a \minus{} \alpha ;b \minus{} \alpha ;c \minus{} \alpha )$ with $ \alpha \equal{} \min (a;b;c)$
Thus,we only prove our inequality with the case $ a \equal{} 0$ or $ b \equal{} 0$ or $ c \equal{} 0$. But it's obvious.
Equality occurs if and only if $ a \equal{} b \equal{} c \equal{} 1$. :lol:

[quote=arqady][quote=tronghieu]
$ \Leftrightarrow \sum {(a \minus{} b)^2 (ab \plus{} 3)c(a \plus{} b) \ge } 36\prod {(a \minus{} b)}$
[/quote]
It's wrong. Try $c=0$.[/quote]

I think it is right if c=0. The RHS is always $\leq$ 0.

[quote=tobylong]
I think it is right if c=0. The RHS is always $\leq$ 0.[/quote]
Are you sure? For $c=0$ we have $$(a-b)(b-c)(c-a)=(b-a)ab.$$


vector
induction
blogs
linear algebra
linear algebra unsolved

Hi there
Let a_1,...,a_n be n vectors,where their magnitude in not greater than one .let C=+-a_1+-a_2+-...+-a_n.
Show that ther exists a combination of + - signs such that the magnitude of Cis not greater than 2^(1/2).

Thanks

Back when I was in $9$'th grade, this was used in the Romanian National Olympiad. I remember being very happy about solving it during the contest :).

If $n=2$, it's easy, so we assume from now on that $n\ge 3$. We can use induction as follows: since $n\ge 3$, choose three vectors, $a_1,a_2,a_3$. Among the $6$ vectors $\pm a_1,\pm a_2,\pm a_3$ there must be two with an angle of at least $\frac {2\pi}3$ between them (and at most $\pi$). Assume WLOG that these two vectors are $a_1,a_2$. Then it's easy to see that $\|a_1+a_2\|\le 1$, and we can use the induction hypothesis on the system of vectors $a_1+a_2,a_3,\ldots,a_n$.

Here is a somehow similar problem [url=http://www.mathlinks.ro/weblog_entry.php?p=1264379#1264379]on my blog[/url], which you may find interesting.


Translate:

Six rounds, all of radius 1 cm, are arranged on the plane as shown in pictures on the right:

a) Calculate the area of triangle ABC

b) Calcule a área do paralelogramo MNPQ e compare-a com a área do triângulo ABC

[img]http://img179.imageshack.us/img179/565/deletar.jpg[/img]

By symmetry, the triangle is equilateral and so it has sides of length $ \sqrt {3} \plus{} 2 \plus{} 2 \plus{} \sqrt {3}$ and thus an area of $ \sqrt {3}(2\plus{}\sqrt{3})^{2}\equal{} 12 \plus{} 7\sqrt{3}$.

Part (b) doesn't really need translating but it is

(b) Calculate the area of the parallelogram $ MNPQ$ and compare it to the area of triangle $ ABC$.

By the way, luiseduardo, the English for circulo is circle and not round, so the question should start Six circles. But I'm sure everyone will understand.

Edited after seeing Mewto55555's correction.

Sorry Andrew, I use google translate too.
But Thanks for your help

a. The side is $ 4\plus{}2\sqrt3$ so the area is $ \frac{\sqrt3(4\plus{}2\sqrt3)^2}{4}\equal{}\frac{\sqrt3(28\plus{}16\sqrt3)}{4}\equal{}7\sqrt3\plus{}12$


$ xy \plus{}yz \plus{}zx \equal{} xyz (x,y,z\geq1)$ Prove that !!

$ : \sqrt{x\minus{}1} \plus{}\sqrt{y\minus{}1} \plus{}\sqrt{z\minus{}1} \leq\sqrt{2(x\plus{}y\plus{}z)}$

[quote="bigbang195"]$ xy \plus{} yz \plus{} zx \equal{} xyz (x,y,z\geq1)$ Prove that !!

$ : \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1} \leq\sqrt {2(x \plus{} y \plus{} z)}$[/quote]

$ xy \plus{} yz \plus{} zx \equal{} xyz \iff \frac{{x \minus{} 1}}{x} \plus{} \frac{{y \minus{} 1}}{y} \plus{} \frac{{z \minus{} 1}}{z} \equal{} 2,$

by CS

$ \sqrt {2(x \plus{} y \plus{} z)} \equal{} \sqrt {\left(\frac{{x \minus{} 1}}{x} \plus{} \frac{{y \minus{} 1}}{y} \plus{} \frac{{z \minus{} 1}}{z}\right)(x \plus{} y \plus{} z)}$

$ \ge \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}.$

done.

[quote="kuing"][quote="bigbang195"]$ xy \plus{} yz \plus{} zx \equal{} xyz (x,y,z\geq1)$ Prove that !!

$ : \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1} \leq\sqrt {2(x \plus{} y \plus{} z)}$[/quote]

$ xy \plus{} yz \plus{} zx \equal{} xyz \iff \frac {{x \minus{} 1}}{x} \plus{} \frac {{y \minus{} 1}}{y} \plus{} \frac {{z \minus{} 1}}{z} \equal{} 2,$

by CS

$ \sqrt {2(x \plus{} y \plus{} z)} \equal{} \sqrt {\left(\frac {{x \minus{} 1}}{x} \plus{} \frac {{y \minus{} 1}}{y} \plus{} \frac {{z \minus{} 1}}{z}\right)(x \plus{} y \plus{} z)}$

$ \ge \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}.$

done.[/quote]

thanks you !!!!! It's Interesting


linear algebra
matrix
algebra unsolved
algebra

Let A be an mxn matrix, show that the following are equivalent:
(a) A has a left inverse, that is, there exists an nxm matrix B such that BA=I.
(b) The system Ax=b has at most one solution x for each b in Rm.
(c) The columns of A are linearly independent.
Show that any one of (a),(b), or (c) implies that m >= n.

THanks...

Does this belong to the Olympiad Section? It's just an easy 0815 problem of linear algebra :-)


I'd like something with lots of problems for sequences and continuity; I already own Rudin's book and the book by Bartle and Sherbert.

G. Pólya and G. Szegö, Problems and theorems in analysis.


$ k \left(\sqrt{1-x}+\sqrt{1+x}\right) = mx+n \sqrt{1-x^{2}}$

for some positive numbers $ k,m,$ and $ n$.

If $ \frac{4k^{2}+m^{2}+n^{2}}{2k \sqrt{m^{2}+n^{2}}}= 2$ and $ x = \frac{r}{s},$ find $ r+s$.

[quote="sylow_theory"][hide="A start"]
Let $ x=\cos \theta$ then $ \sqrt{1\pm x}= \sqrt{1\pm \cos \theta}= \sqrt{2}\cdot \sqrt{\frac{1\pm \cos \theta}{2}}$

So the equation becomes,
$ k\sqrt{2}\left( \cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right) = m\cos \theta+n \sin \theta$

$ k \sqrt{2}\cdot \sqrt{2}\sin \left( \frac{\theta}{2}+\frac{\pi}{4}\right) = \sqrt{n^{2}+m^{2}}\sin (\theta+\phi)$ where $ \phi = \tan^{-1}(m/n)$

[/hide][/quote]

You almost had it, I think, because

[hide]$ \frac{4k^{2}+m^{2}+n^{2}}{2k\sqrt{m^{2}+n^{2}}}\geq 2$ with equality iff $ 4k^{2}= m^{2}+n^{2}$.  :wink:

So if we let $ x = \cos 2\alpha$ with $ 2\alpha\in[0,\pi]$, we obtain $ \sin\left(\frac{\pi}{4}+\alpha\right) = \sin\left(\chi+2\alpha\right)$, where $ \sin\chi = \frac{m}{2k}$.

As $ \frac{\pi}{4}+\alpha \in \left[\frac{\pi}{4},\frac{3\pi}{4}\right]$ and $ \chi+2\alpha\in[0,2\pi]$, this leads to
$ 2\alpha = \frac{\pi}{2}-2\chi$, so $ x = \cos(2\alpha) = \sin(2\chi) = \boxed{\frac{mn}{2k^{2}}}$
OR
$ 2\alpha = \frac{\pi}{2}-\frac{2}{3}\chi$, so $ x = \cos(2\alpha) = \sin\left(\frac{2}{3}\chi\right)$, so $ x$ is a solution to $ 4x^{3}-3x+\frac{mn}{2k^{2}}= 0$. So, silverfalcon: are you sure about the problem statement? Or did I just make a mistake?  :roll:
[/hide]

[quote="Silverfalcon"]Hmm... Here's what I did

[hide="Let see"]

It's very similar to what you guys did and here is my next to last step.

$ k \sqrt{2}(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}) = m \cos \theta+n \sin \theta$

$ 2k \sin (\frac{\pi}{4}+\frac{\theta}{2}) = \sqrt{m^{2}+n^{2}}\sin (\alpha+\theta) \ \blacksquare$

Note that on RHS, we have $ \alpha$ because you can't find angle since coefficients were $ m$ and $ n$.

Now, here comes really tricky part.

$ \frac{4k^{2}+m^{2}+n^{2}}{2k \sqrt{m^{2}+n^{2}}}$
$ = \frac{4k^{2}}{2k \sqrt{m^{2}+n^{2}}}+\frac{m^{2}+n^{2}}{2k \sqrt{m^{2}+n^{2}}}$
${ = \frac{2k}{\sqrt{m^{2}+n^{2}}}+\frac{\sqrt{m^{2}+n^{2}}}{2k}}$

Moreover, since these two are reciprocals of each other, and so let one be $ z$.

$ z+\frac{1}{z}= 2 \rightarrow z = 1$

Now back to equation $ \blacksquare$

We now can simply the equation to:

$ \sin (\frac{\pi}{4}+\frac{\theta}{2}) = \sin (\alpha+\theta)$

From here, it's pretty straightforward. You just solve for $ \theta$ in terms of $ \theta$ and then use $ x = \cos \theta$ and substitute. It should come out nicely and that...[/hide]

[hide="Answer that I got"]

$ x = \frac{24}{25}$ and final answer is $ 24+25 = 049$. [/hide][/quote]

You say "you just solve for $ \theta$", but your $ \alpha$ has dissappeared, as well as $ k,m$ and $ n$. Do not forget that $ \alpha$ depends on $ m$ and $ n$.

It should've said:

Solve for $ \theta$ in terms of $ \alpha$.

[hide="Ok let me finish then.."]

Set the inner angles to each other.

$ \frac{\pi}{4}+\frac{\theta}{2}= \alpha+\theta$
$ \frac{\theta}{2}= \frac{\pi}{4}-\alpha$
$ \theta = \frac{\pi}{2}-2 \alpha$

Now, remember that $ x = \cos \theta$

$ x = \cos \left(\frac{\pi}{2}-2 \alpha\right)$

$ = \cos \frac{\pi}{2}\cdot \cos 2 \alpha+\sin \frac{\pi}{2}\cdot \sin 2 \alpha$

The first part is zero. The last part is then

$ \sin 2 \alpha = 2 \sin \alpha \cos \alpha$ [/hide]

Oh man... I see where this went wrong. I had to give more information... Well... it was supposed to be that $ \sin \alpha = 3/5$ and $ \cos \alpha = 4/5$. But I guess it slipped.  :oops:


[quote="andre.l"]Let $x_1, x_2, ..., x_{2004}$ be a sequence of integer numbers such that $x_{k+3}=x_{k+2}+x_{k}x_{k+1}$, $\forall  1 \le k \le 2001$. Is it possible that more than half of the elements are negative?[/quote]

The sequence start with -e, -e, -e is a negative sequence if e<1/2002, which causes (-e)*(-e)<e/2002.

[quote="Darkseer"]The sequence start with -e, -e, -e is a negative sequence if e<1/2002, which causes (-e)*(-e)<e/2002.[/quote]

Hmmm... but the sequence only contains integer numbers...

[quote="cyshine"]Hmmm... but the sequence only contains integer numbers...[/quote]
Sorry for the stupid mistake...
However, I think I've found a real solution now.
Consider the sequence starting with $-n^2$, $-n^3$, $-n^4$, where n is a very big interger. Then we see by induction that
(1) $x_n>>x_{n-1}$.
(2) $x_n<<x_{n-1}x_{n-2}$.
(3) sign($x_n$)=sign($x_{n-2}$)sign($x_{n-3}$)
And the sign of the sequence is:
 - - - + + - + - - - + + - + ...
Now we see that $\frac{4}{7}$ of the sequence is negative.


Which of these four numbers $ \sqrt{\pi^2},\,\sqrt[3]{.8},\,\sqrt[4]{.00016},\,\sqrt[3]{\minus{}1}\cdot \sqrt{(.09)^{\minus{}1}}$, is (are) rational:

$ \textbf{(A)}\ \text{none}\qquad
\textbf{(B)}\ \text{all}\qquad
\textbf{(C)}\ \text{the first and fourth}\qquad
\textbf{(D)}\ \text{only the fourth}\qquad
\textbf{(E)}\ \text{only the first}$

[hide="Solution"]
Check each one for rationality:

$ \sqrt{\pi^2}\equal{}\pi$ and since $ \pi$ is irrational, option $ \mathbf{(C)}$ and $ \mathbf{(E)}$ are out.

$ \sqrt[3]{.8} \equal{} \sqrt[3]{\dfrac{4}{5}}$ neither $ 5$ nor $ 4$ is a perfect cube, so  this is not rational.

$ \sqrt[4]{.00016} \equal{} \sqrt[4]{\dfrac{16}{100000}} \equal{} \sqrt[4]{\dfrac{2^4}{10^5}} \equal{} 20\cdot \sqrt[4]{\dfrac{1}{10}}$ so this is also not rational.

$ \sqrt[3]{\minus{}1} \cdot \sqrt{(.09)^{\minus{}1}} \equal{} \minus{}1 \cdot \dfrac{10}{3} \equal{} \minus{}\dfrac{10}{3}$ which is the only rational number of the four so the answer is $ \mathbf{(D)}$.
[/hide]


Perhaps this has already been discussed, but maybe everyone's opinions have changed. Which authors and books do you like best? My latest favorite author is Ayn Rand, and I'm reading The Fountainhead. Do you like a specific genre of books? For example, I'm into philosophy, but I read mostly fiction of random varieties.

Mystery: Agatha Christie
Sci-Fi/Suspense: Michael Crichton

Absolute favorite: Harry Potter.
Second favorites: A Series of Unfortunate Events, Phantom Tollbooth, Holes, Artemis Fowl, anything by Edgar Allan Poe, a couple other random short ghost stories I can't remember either the title or author of, and a lot of other good books and authors I can't think of off the top of my head.
As you can see I mostly read fantasy and ghost stories, though sci-fi and contemporary are good too...speaking of sci-fi, I just finished Fahrenheit 451 a couple days ago. I've also been trying to read as many classics as possible, so I won't feel ignorant when others start talking about them...

Ah, yes, of course.  How could I forget the author of my signature...Poe's short stories.  Incredible stuff.

Yes. It doesn feel bad when people are talking about books, and you have no clue what the heck they are talking about. That's why I look at book lists, and read straight down the list.

i love non fictions.. political books are good..

although i am more liberal, i love bill oreilly
and i havent read jon steward's america(the book) yet but im sure its gonna be good

I like R. A. Salvatore for fantasy. And I read almost any science fiction book.

Dan Brown is an awesome writer. I love Angels and Demons.

Who's john brown? I also like anything by dan brown.

Books are wonderful.  I am only to about page 20 of [i]Inkheart,[/i] but it promises to be good.  Also, I just finished [i]The Grapes of Wrath[/i], and [i]Countdown[/i] is marvelous.

I'll read any book by Billy Amend. Foxtrot En Masse, Camp Foxtrot, Take Us To Your Mall, Death by Field Trip, Yo Mama Thinks Square Roots are Vegetables, etc.  :D

I'll pretty much read any book if I find it interesting.

Shogun by James Clavell is awesome, though it can take a while to get through.

I really liked Shogun when I read it a couple years back. I also enjoyed Ayn Rand, most of all, her epic "Atlas Shrugged." I liked it far more than "the Fountainhead." Right now, I'm reading Goethe's Faust.

I like all science fiction and some history. Artemis Fowl ROX!! Lemony Snicket is good too.
Has anyone read The Outsiders? Is it good?

[quote="yif man12"]I'll read any book by Billy Amend. Foxtrot En Masse, Camp Foxtrot, Take Us To Your Mall, Death by Field Trip, Yo Mama Thinks Square Roots are Vegetables, etc.  :D
[/quote]

Woo-hoo!  Foxtrot is the best!

Ahem...sorry, I couldn't help myself...

The Outsiders is a good book. Very realistic. I was just finishing up Tarzan of the Apes today, and I found it to be a most interesting comparison of humanity, apes, and african tribes during the age when Africa was the Dark Continent. Most contrasting to what we believe today.

Hmm...I'll read anything. Favorites?

The Count of Monte Cristo by Alexandre Dumas
The "HIs Dark Materials" trilogy by Philip Pullman
Daniel Pinkwater is hilarious!
1984 by George Orwell
Harry Potter would be awesome if it weren't ruined by all this marketing crud
And I used to be into Redwall until the 15th came out and it was, like, exactly identical to the other 14  :maybe:
And the Lemony Snicket books are easy, but they're pretty funny and I love the guy's style - plus, he's a genius putting in all those literary allusions (Isadora Duncan, all those crows being reminiscient of Alfred Hitchcock's The Birds, etc) except the new one stunk
Clive Barker's abarat is the only worthwhile "fantasy" book I'd recommend that nobody's discovered yet
Newton's Pricipia Mathematica (just kidding)

And my all time favorite:

Harper Lee's To Kill A Mockingbird

*Phew*

i dont really like ayn rand. she's clever, but her philosophy doesnt agree with me. Tobias Wolff wrote a book called  [u]Old School[/u], which is fiction, in which he critisizes ayn rand, and it influenced me.

I have a blog with book reviews, although I've been getting lax lately; go to http://www.bibliofilesindex.blogspot.com and comment on a book you've read... i would be most grateful.

[/u]


Prove that $\frac{5^{125}-1}{5^{25}-1}$ is a composite number.

[quote="Peter"]Prove that $ \frac {5^{125} \minus{} 1}{5^{25} \minus{} 1}$ is a composite number.[/quote]

An epsilon remark. This was proposed by Korea  :)

With the similar method,one can prove that when p=4k+1 is prime,then $ \frac {p^{p^{3}} \minus{} 1}{p^{p^{2}} \minus{} 1}$is a composite number.

Yes! What are the motivations of these factorizations ?

[quote=lightstar_01]Can I just ask, how does one come up with these factorisations? The solutions described above are very beautiful, I just cannot see how to come up with them![/quote]

Had the same thought. Someone told me that it's a whole class of factorisation known as aurifeuillean factorization :D


How many perfect cubes are there between 100 and 1100?

List the cubes.
4^3=64
5^3=125
etc
10^3=1000
11^3=1331

5 to 10 cubed works, so 5,6,7,8,9,and 10 is 6 numbers.

Note: Also, to calculate the numbers of x to y inclusive, do y-x+1.


What is the size of the set of transcendental numbers in relation to the algebraic reals?  In particular,

1)  Does there exist a surjection from the algebraic real numbers to the transcendental numbers?  If so,

2)  Does there exist a bijection from the algebraic real numbers to the transcendental numbers?

No and no. Since the algebraic numbers are countable and the real numbers are not, the transcendentals have the same cardinality as $\mathbb{R}$.


trigonometry
geometry
rectangle
vector
circumcircle
3D geometry
tetrahedron

All-Russian Mathematical Olympiad 2006 took place in Pskov from 21st until 26th of April. You can find the problems [url=http://olympiads.mccme.ru/vmo/32/sheets.pdf]here.[/url] Can some people, as Darij for instance, who are more fluent in Russian than me, please translate the problems into German or English. Otherwise it would take me quite some time. Many thanks!  :)

[b]11.1.[/b] Prove that $\sin\sqrt{x}<\sqrt{\sin x}$ for every real $x$ such that $0<x<\frac{\pi}{2}$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86557

[b]11.2.[/b] The sum and the product of two purely periodic decimal fractions $a$ and $b$ are purely periodic decimal fractions of period length $T$. Show that the lengths of the periods of the fractions $a$ and $b$ are not greater than $T$.
[i]Note.[/i] A [i]purely periodic decimal fraction[/i] is a periodic decimal fraction without a non-periodic starting part.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86559

[b]11.3.[/b] On a $49\times 69$ rectangle formed by a grid of lattice squares, all $50\cdot 70$ lattice points are colored blue. Two persons play the following game: In each step, a player colors two blue points red, and draws a segment between these two points. (Different segments can intersect in their interior.) Segments are drawn this way until all formerly blue points are colored red. At this moment, the first player directs all segments drawn - i. e., he takes every segment AB, and replaces it either by the vector $\overrightarrow{AB}$, or by the vector $\overrightarrow{BA}$. If the first player succeeds to direct all the segments drawn in such a way that the sum of the resulting vectors is $\overrightarrow{0}$, then he wins; else, the second player wins.
Which player has a winning strategy?
http://www.mathlinks.ro/Forum/viewtopic.php?t=86560

[b]11.4.[/b] The angle bisectors of the angles $ABC$ and $BCA$ of a triangle $ABC$ intersect the sides $CA$ and $AB$ at the points $B_1$ and $C_1$ and intersect each other at the point $I$. The line $B_1C_1$ intersects the circumcircle of triangle $ABC$ at the points $M$ and $N$. Prove that the circumradius of triangle $MIN$ is twice as long as the circumradius of triangle $ABC$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86561

[b]11.5.[/b] Two sequences of positive reals, $\left(x_n\right)$ and $\left(y_n\right)$, satisfy the relations $x_{n+2}=x_n+x_{n+1}^2$ and $y_{n+2}=y_n^2+y_{n+1}$ for all natural $n$. Prove that, if all numbers $x_1$, $x_2$, $y_1$, $y_2$ are all $>1$, then there exists a natural $k$ such that $x_k>y_k$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86562

[b]11.6.[/b] Consider a tetrahedron $SABC$. The incircle of the triangle $ABC$ has the center $I$ and touches its sides $BC$, $CA$, $AB$ at the points $E$, $F$, $D$, respectively. Let $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be the points on the segments $SA$, $SB$, $SC$ such that $AA^{\prime}=AD$, $BB^{\prime}=BE$, $CC^{\prime}=CF$, and let $S^{\prime}$ be the point diametrically opposite to the point $S$ on the circumsphere of the tetrahedron $SABC$. Assume that the line $SI$ is an altitude of the tetrahedron $SABC$. Show that $S^{\prime}A^{\prime}=S^{\prime}B^{\prime}=S^{\prime}C^{\prime}$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86563

[b]11.7.[/b] Assume that the polynomial $\left(x+1\right)^n-1$ is divisible by some polynomial $P\left(x\right)=x^k+c_{k-1}x^{k-1}+c_{k-2}x^{k-2}+...+c_1x+c_0$, whose degree $k$ is even and whose coefficients $c_{k-1}$, $c_{k-2}$, ..., $c_1$, $c_0$ all are odd integers. Show that $k+1\mid n$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86564

[b]11.8.[/b] At a tourist camp, each person has at least $50$ and at most $100$ friends among the other persons at the camp. Show that one can hand out a t-shirt to every person such that the t-shirts have (at most) $1331$ different colors, and any person has $20$ friends whose t-shirts all have pairwisely different colors.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86565

[b]10.1. = 9.1.[/b] Given a $15\times 15$ chessboard. We draw a closed broken line without self-intersections such that every edge of the broken line is a segment joining the centers of two adjacent cells of the chessboard. If this broken line is symmetric with respect to a diagonal of the chessboard, then show that the length of the broken line is $\leq 200$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86763

[b]10.2.[/b] If an integer $a > 1$ is given such that $\left(a-1\right)^3+a^3+\left(a+1\right)^3$ is the cube of an integer, then show that $4\mid a$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86764

[b]10.3. = 9.3.[/b] Given a circle and $2006$ points lying on this circle. Albatross colors these $2006$ points in $17$ colors. After that, Frankinfueter joins some of the points by chords such that the endpoints of each chord have the same color and two different chords have no common points (not even a common endpoint). Hereby, Frankinfueter intends to draw as many chords as possible, while Albatross is trying to hinder him as much as he can. What is the maximal number of chords Frankinfueter will always be able to draw?
http://www.mathlinks.ro/Forum/viewtopic.php?t=86765

[b]10.4.[/b] Consider an isosceles triangle $ABC$ with $AB=AC$, and a circle $\omega$ which is tangent to the sides $AB$ and $AC$ of this triangle and intersects the side $BC$ at the points $K$ and $L$. The segment $AK$ intersects the circle $\omega$ at a point $M$ (apart from $K$). Let $P$ and $Q$ be the reflections of the point $K$ in the points $B$ and $C$, respectively. Show that the circumcircle of triangle $PMQ$ is tangent to the circle $\omega$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86767

[b]10.5. = 9.5.[/b] Let $a_1$, $a_2$, ..., $a_{10}$ be positive integers such that $a_1<a_2<...<a_{10}$. For every $k$, denote by $b_k$ the greatest divisor of $a_k$ such that $b_k<a_k$. Assume that $b_1>b_2>...>b_{10}$. Show that $a_{10}>500$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86768

[b]10.6.[/b] Let $K$ and $L$ be two points on the arcs $AB$ and $BC$ of the circumcircle of a triangle $ABC$, respectively, such that $KL\parallel AC$. Show that the incenters of triangles $ABK$ and $CBL$ are equidistant from the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86769

[b]10.7. = 9.8.[/b] Given a quadratic trinomial $f\left(x\right)=x^2+ax+b$. Assume that the equation $f\left(f\left(x\right)\right)=0$ has four different real solutions, and that the sum of two of these solutions is $-1$. Prove that $b\leq -\frac14$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86770

[b]10.8.[/b] A $3000\times 3000$ square is tiled by dominoes (i. e. $1\times 2$ rectangles) in an arbitrary way. Show that one can color the dominoes in three colors such that the number of the dominoes of each color is the same, and each dominoe $d$ has at most two neighbours of the same color as $d$. (Two dominoes are said to be [i]neighbours[/i] if a cell of one domino has a common edge with a cell of the other one.)
http://www.mathlinks.ro/Forum/viewtopic.php?t=86771

[b]9.2.[/b] Show that there exist four integers $a$, $b$, $c$, $d$ whose absolute values are all $>1000000$ and which satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{abcd}$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86779

[b]9.4.[/b] Given a triangle $ABC$. Let a circle $\omega$ touch the circumcircle of triangle $ABC$ at the point $A$, intersect the side $AB$ at a point $K$, and intersect the side $BC$. Let $CL$ be a tangent to the circle $\omega$, where the point $L$ lies on $\omega$ and the segment $KL$ intersects the side $BC$ at a point $T$. Show that the segment $BT$ has the same length as the tangent from the point $B$ to the circle $\omega$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86780

[b]9.6.[/b] Let $P$, $Q$, $R$ be points on the sides $AB$, $BC$, $CA$ of a triangle $ABC$ such that $AP=CQ$ and the quadrilateral $RPBQ$ is cyclic. The tangents to the circumcircle of triangle $ABC$ at the points $C$ and $A$ intersect the lines $RQ$ and $RP$ at the points $X$ and $Y$, respectively. Prove that $RX=RY$.
http://www.mathlinks.ro/Forum/viewtopic.php?t=86781

[b]9.7.[/b] A $100\times 100$ chessboard is cut into dominoes ($1\times 2$ rectangles). Two persons play the following game: At each turn, a player glues together two adjacent cells (which were formerly separated by a cut-edge). A player loses if, after his turn, the $100\times 100$ chessboard becomes connected, i. e. between any two cells there exists a way which doesn't intersect any cut-edge. Which player has a winning strategy - the starting player or his opponent?
http://www.mathlinks.ro/Forum/viewtopic.php?t=86782

[quote="Tiks"][quote="orl"][quote="nguyentrang"]Is there the fourth round problem like last year?[/quote]

This is the fourth round my friend.  :P[/quote]
I don't think so orl,this is the 5th and final round of All RMO ;)
Here is the translations of all grade problems ;) .

[/quote]

Anyway I meant it is the final round. I just forgot that the fifth round in the final round.  :D

[quote="Aizana"][quote="nguyentrang"]So, no fourth round?[/quote]
Do you know that final round of Kazakh national olympiad and 4th round of Russian olympiad have the same problems?!  :o[/quote]

Can you post the problems?

[/quote]

Can you post the problems?[/quote]
I can post only 10th grade in Russian!


[quote="Silverfalcon"]A positive number $x$ satisfies the inequality $\sqrt{x} < 2x$ if and only if

A. $x > \frac{1}{4}$
B. $x > 2$
C. $x > 4$
D. $x < \frac{1}{4}$
E. $x < 4$[/quote]

[hide]
$\sqrt{x} < 2x$
(x & $\sqrt{x}$ are always positive so) $x < 4x^2$
(x is always positive so) $1 < 4x$
$\frac{1}{4} < x$

A.
[/hide]

You looked at it the wrong way. Your inequality is correct, but your answer choice is wrong. ;)

[quote="Silverfalcon"]A positive number $x$ satisfies the inequality $\sqrt{x} < 2x$ if and only if

A. $x > \frac{1}{4}$
B. $x > 2$
C. $x > 4$
D. $x < \frac{1}{4}$
E. $x < 4$[/quote]
[hide]
Squaring, we have:
x<4x^2, 1<4x, x>1/4.
A, is our answer.
[/hide]

[hide]
To get rid of the sparue root sign, you square it:
$x>4(x)^2$ Since $x$ is positive, you get $1<4x$
So you get $x>\frac{1}{4}$ $A$[/hide]


The layout of a Little League playing field is a square, 60 feet on a side.
Overlay a rectangular coordinate system on a Little League baseball diamond so that the orgin is at home plate, the positve x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base.

(a) What are the coordinates of first base, second base and third base? (Use feet as unit of measurement.)

(b) If the right fielder is located at point (180, 20), how far is it from there to second base?

(c) If the center fielder is located at point (220, 220), how far is it from there to third base?

[quote="Interval"]Good job but how did you get part a?[/quote]
If you draw a square (in the first quadrant) with side lengths of 60 where one of the points is on the origin, you will get those points.


Prove that from 11 infinite decimals, we can select 2 numbers a,b so that their decimal representation have the digits at infinitely many corresponding places.

At each place, by the Dirichlet's Principle there must be a match in at least one pair of numbers, you have $ \left( 11 \atop 2\right)$ pairs , supose all pairs only match a finite number of times $ <M$ than there would be only, at most, $ \left( 11 \atop 2\right)M$ places, but the number of places is infinite, contradiction.


Does there exist a quadrilateral with the vertices in order $ (\minus{}1,4),(5,6),(2,9),(x,x^{2})$ and area $ \frac{15}{2}$ sq. units

:arrow: [b]TIP[/b]
[img]http://img144.imageshack.us/img144/443/ryszy4.gif[/img]

$ A( \minus{} 1,4), B(5,6), C(2,9), D(x,x^{2})$

The area of triangle $ \Delta ABC$ equals 12, so only concave quadrilateral might exist.

$ S_{ABCD} \equal{} S_{\Delta ABD} \plus{} S_{\Delta BDC}$

indeed it turns out to be concave solving for $ x$ we get $ (x,x^{2})$.But the formulae i applied the area formula holds only for convex figures :rotfl: