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For a long time I have been interested in the Myers-Briggs personality typing and its efficacy.  I am curious as to the personality make-up of the AoPS community.  For those who are interested, taking [url=http://www.humanmetrics.com/cgi-win/JTypes1.htm]this quiz[/url] and report your results.

I can't seem to take those tests, since a lot of the questions are to me a bit ambiguous in what they're after, and I can't help but start to guess what they are getting at, which screws the whole thing up.  (Heck, I had a hard time with political compass, and that's a lot less trouble, because the propositions you are agreeing/disagreeing with are not about yourself.)  It seems to me that that's a really big weakness of this sort of tests -- it's easy to influence the outcome of the test to reflect how you perceive you are, rather than how you really are; or (if you realize the aforementioned and actively try to prevent it) influence the outcome against how you perceive you are.  I wonder if there is a way to write such tests in order to avoid that kind of problem.

But everyone who knows me and has seen fit to guess has told me that I'm INTP, so I'll put that.

setofallsets, part of the idea behind the questions is that they gauge your perception of the questions themselves.  Just go with that.  Though, having worked with you for a summer, I feel like INTP is a very safe bet for you.

I'm an ESTP:

Extroverted = 89%
Sensing = 1%
Thinking = 12%
Perceiving = 22%

Hmmm....can someone tell me what that means?

INTJ

Strength of the preferences %
Introverted 11%
Intuitive 50%
Thinking 1% (Great, heh)
Judging 33%

ESTP...

I dunno, I don't really agree with it.

[quote="sirorange"]I'm an other, or an ENFJ. Hmm. . . . I don't really buy into these polls too much though, as I don't feel that 72 questions could accurately determine how a person will behave. It's still interesting stuff though.[/quote]

You don't think it's pretty amazing that 3% of people are INTx's, yet 65ish percent of the people who find their way to this survey through AoPS are INTx's?  Applying Bayes Theorem, that suggests that INTx's are around 60-70 times more likely to visit AoPS.  To me that seems to be a remarkably accurate predictor of behavior given the nature of the questions.

It's not meant to make complete predictions of the ways in which people behave.  However, I recommend doing a bit of reading about the subject to anyone interested.  These particular brands of personality tests are remarkably effective in predicting behavior within groups.  For instance, despite the fact that INTP is the most rare personality type and INTJ is also rare, nearly all Phd mathematicians fall into one of those two personality types.  Judging from a survey in the 2 + 2 poker forum, most professional poker players fall into one of those two personality types. Intelligence agencies around the world suggest that they use similar personality typing to profile criminals and other people of interest to them.  There is even a famous consultant for NFL teams who has been spot-on predicting which drafted quarterbacks would flop and which would fail solely on the basis of their personality tests.

On the note of asking a few questions to determine a lot about a person...when artificial intelligence machines play games against each other, outcomes are clear after only around 30 moves (such as in the rock/scissors/paper tournaments played between AI programs).  It may be that broad patterns of behavior are fairly predictable after a very small number of observations (or questions).

INTP (44,75,25,11)

And I feel like the results are pretty darn accurate.

ENTJ

Extroverted 1( :huh: )

Intuitive     38

Thinking       12

Judging         22

ENTJ {1, 50, 25, 44}

Fictional ENTJs:
Geordi LaForge (STNG)

That's not bad :)

humanmetrics rated me as being a EIFJ. I find this ironic, as everyone else on this website is apparently everyone on this board is the exact opposite of me. However, having read the profile, I'm surprised I am so rare on the forum.

Hmm... ESFJ. What does that mean? :?:  :huh:

I'm INFP as well. I've taken the test several times and get the same report every time.

INTJ...Interesting how many people here are the same as me.

Henceforward, any post on this thread that merely states a personal result without also making a substantive comment or responding to an earlier comment will be deleted.

[quote="Marble-101"]INFP...
Don't even know what it is though...[/quote]
INFP means

Introverted
iNtuitive
Feeling
Perceiving

This just means that based on the results of the questionnaire you appear to better fit these categories as opposed to their respective complements which are:

Extroverted
Sensing
Thinking
Judging

See [url=http://en.wikipedia.org/wiki/Myers-Briggs_Type_Indicator]here[/url] for more info. Personally, after taking multiple tests I appear to fluctuate (so far) between moderate ENTJ, ENFP, and ENTP. I have yet to get ENFJ. :) Due to the very nature of AoPS INTJ is clearly going to be the most popular result.

Why do you think INTJ the most prevalent (aside from peoples' responses?) As pretty much a laymen to the subject it seems like ENSP or something because they are supposedly the "innovative" ones (I don't know how accurate that is either).

It seems as if INTJ is the idealistic 'mad scientist' personality where the person utterly detaches himself from reality in the relentless and obsessive pursuit of knowledge...i.e. 'the absentminded professor.'

I scored INTJ, but in a very peculiar way. I scored 100% for Introverted, 62% for iNtuitive, 62% for Thinking, and 1% for Judgement. I could have just as easily been an INTP, depending on my mood; I share characteristics of both. I originally thought that I was just moderately introverted, but now I realize that I am introverted almost to the point where it is a problem, and maybe past that point.

I - and I am guessing many other users in this forum - have an innate romantic attraction (not romantic in the 'love' sense) to the personality mentioned in the first paragraph. This is why one of the people whom I admire is Albert Einstein, because he just seems to have radiated this personality when he was still alive. I have gone almost as far as comparing our early childhoods together, even though I know that sounds obsessive. :blush:

I even figured out that his political orientation was similar to that of mine, which raises an interesting question: are people who share similar scores likely to have similar political views?

[quote="Fraqtive42"]

I even figured out that his political orientation was similar to that of mine, which raises an interesting question: are people who share similar scores likely to have similar political views?[/quote]

Probably.. If you have similar personalities, you will probably have a higher change of thinking down similar ideologies.. However, there is also the influence of education and experience. (Though both those influence your personality as well).

[quote="NuncChaos"]However, there is also the influence of education and experience. (Though both those influence your personality as well).[/quote]
Definitely true. SInce I'm only 13  and I have lack of experience, I mostly answered the questions which related to experience as "no, I usually do not approach things in a traditional manner." This significantly affected my score.

These statistics have me wondering if the activities participate in affect our personality type, or vice versa. There is quite a definite lean towards INTx in these fora.

Close match between INTJ, INTP, ENTJ, and ENTP, but INTJ won.


trigonometry
geometry
cyclic quadrilateral
trig identities
Law of Cosines

Quadrilateral MSFC is inscribed a circle. The length of $ MS$ is $ 9$, $ SF$ is $ 3$, $ FC$ is $ 4$, and $ MC$ is $ 7$. What is $ \cos{\angle{M}}$, expressed as a common fraction?

[hide="Solution"]Let $ x=cos<M$.
Draw diagonal $ SC$ of the quadrilateral.
In a cyclic quadrilateral, opposite angles sum to $ 180^{\circ}$.  Thus, $ m<M+m<F=180^{\circ}$
Note that $ cos<M=-cos<F$.
Thus $ cos<F=-x$.
Applying law of cosines to $ \bigtriangleup MSC, 81+49-2*9*7x=SC^{2}$
For $ \bigtriangleup FSC, 9+16+2*3*4x=SC^{2}$
$ 25+24x=130-126x$
$ 150x=105$
$ x=\frac {7} {10}$[/hide]


inequalities
function
inequalities unsolved

We must have $a_{k}> 0$ for $\forall k$ natural.

So let $f(a_{1},a_{2},a_{3},...,a_{n}) = \sum_{k=1}^{n}\sqrt{\frac{a_{k}^{2}+1}{2}}-\sum_{k=1}^{n}a_{k}+\lambda(\prod_{k=1}^{n}a_{k}-1)$.

To obtain the maximun value of this function we must have:

$\frac{\partial f}{\partial a_{k}}= 0$ for $\forall k$ natural.
Indeed, our restriction is: $\prod_{k=1}^{n}a_{k}-1 = 0$.

If we pu $t=a_{k}$ we´ll obtain a third degree equation of the form:
$2\lambda t^{3}-(1+\lambda^{2})t^{2}+2\lambda t-\lambda^{2}= 0$.
But, we can easily prove that this equation has only one positive root. So we must have $a_{1}=a_{2}=a_{3}=...=a_{n}=1$ and by this way we´ll get $\lambda=\frac{2-\sqrt{2}}{2}$. And if we compute the hessian matrice of this function we´ll obtain a diagonal matrice such that all the elements of the diagonal are equal to $\frac{2\sqrt{2}-3}{4}$. And clearly, we notice that all the eigenvalues of this matrice is negative and so this is a point of maximun.

So we get our inequalitie: $\sum_{k=1}^{n}\sqrt{\frac{a_{k}^{2}+1}{2}}-\sum_{k=1}^{n}a_{k}< 0$.


calculus
integration
floor function
function
continued fraction
real analysis
real analysis unsolved

Let $a, b$ be two positive real number. When does the integral
\[ \int_a^b \left\lfloor \frac{1}{\{ \sqrt{x} \}} \right\rfloor \mbox{dx}  \]
converge ? ;)

Just to be clear: may we assume that by $\{y\}$ you mean the fractional part of $y$? That is, $\{y\}=y-\lfloor y\rfloor$ ?

Yes, of course, sorry for not pointing it out ;)

It converges whenever $[a,b)$ contains no perfect squares, of course. The improper integral $\int_0^1\left\lfloor\frac1{\{\sqrt x\}}\right\rfloor\,dx$ converges as well.

If $[a,b)$ contains a square $k^2$, the integral diverges by comparison to $\frac{2k}{x-k^2}$ near $k^2$; as $x\to k^2^+$, $\left\lfloor\frac1{\{\sqrt x\}}\right\rfloor\cdot\frac{x-k^2}{2k}\to 1$. Near zero, the function is comparable to $\frac1{\sqrt{z}}$, which is integrable.

For prettiness, use \left\lfloor and \right\rfloor; it's not as redundant as it looks.

[quote="jmerry"]It converges whenever $[a,b)$ contains no perfect squares,[b] of course[/b]. [/quote]
It was really unobvious to me, so excluding the possibility that i"m dumb (which may be of course true) I would like to know wheter it is a consequence of a more general result (related to continued fraction perhaps) ?

[quote]The improper integral $\int_0^1\left\lfloor\frac1{\{\sqrt x\}}\right\rfloor\,dx$ converges[/quote]
Could you give more details, please ;)

If there are no perfect squares in the interval, then $\{\sqrt x\}$ is bounded away from zero, and the function we want to integrate is bounded. A bounded, piecewise continuous function is integrable. A perfect square at the right endpoint is just one bad point, and it doesn't cause any problems.

The details in all other cases are in my second paragraph. As $x\to k^2$ from above, $\frac{\left\lfloor\frac1{\{x\}}\right\rfloor}{\frac1{\sqrt{x}-k}}$ tends to 1, so the behavior of the function is essentially that of $\frac1{\sqrt{x}-k}=\frac{\sqrt{x}+k}{x-k^2}$, which we understand.

There's nothing about continued fractions here.

Thank you jmerry. I've got one last question.
I'm still missing the flaw in my argument for the divergence of $\int_0^1 \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx}$

We have
\[ \int_0^1 \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx} = \sum_{j = 1}^{\infty} \int_{\frac{1}{j + 1}}^{\frac{1}{j}} \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx}  \]
But $\frac{1}{m + 1} < x \leqslant \frac{1}{m} \Rightarrow \left\lfloor \frac{1}{\{ x \}} \right\rfloor = m$. So $\int_{\frac{1}{j + 1}}^{\frac{1}{j}} \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx} = j \cdot ( \frac{1}{j} - \frac{1}{j + 1} ) = \frac{1}{j + 1}$. And
\[ \int_0^1 \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx} = \sum_{j = 1}^{\infty} \int_{\frac{1}{j + 1}}^{\frac{1}{j}} \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx} = \sum_{j = 1}^{\infty} \frac{1}{j + 1} = + \infty  \]

I don't know where my argument gets wrong :(.

[quote="jmerry"]Are you sure that's the integral you care about?[/quote]

Yes (if I understand youre question well), because originally I got a completely different answer than yours ;)
I found that assuming $b \geq a$, $\int_a^b \left\lfloor \frac{1}{\{ \sqrt{x} \}} \right\rfloor \mbox{dx}$ converges iff $0 \leq b-a < 1$. The divergence of $\int_0^1\left\lfloor\frac1{\{\sqrt x\}}\right\rfloor\,dx$ was a crucial point of that "proof" (I can post it if you want). This is why I would like to know where it gets wrong.
Anyway, I hope this discussion is not getting boring to you   :roll:

You proved the divergence of $\int_0^1\left\lfloor\frac1{\{x\}}\right\rfloor\,dx$, which is not the same thing. That integral diverges, but we're looking at one with a square root. The key points for the square root occur at perfect squares, not every integer.

Oh ! Yes, of course :blush:
Actually I was speaking of $\int_a^b \left\lfloor \frac{1}{\{ \sqrt{x} \}} \right\rfloor \mbox{dx}$ when in I was thinking of $\int_0^1 \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx}$ . I didn't noticed that fact for 2 days !. I don't know how this is possible ;)
I'm really sorry.

(Actually this is what happens at my earlier post, I speak of the divergence of $\int_a^b \left\lfloor \frac{1}{\{ \sqrt{x} \}} \right\rfloor \mbox{dx}$ and actually thinking about $\int_0^1 \left\lfloor \frac{1}{\{ x \}} \right\rfloor \mbox{dx}$ ;))


[quote=KDS]Let $ a_1,a_2,...,a_n,b_1,b_2,...,b_n$ be the real numbers in the interval $ [1,2]$ with $ a_1^2\plus{}a_2^2\plus{}....\plus{}a_n^2\equal{}b_1^2\plus{}b_2^2\plus{}...\plus{}b_n^2$.Determine the minimum value of constant $ c$ such that:
$ \frac{a_1^3}{b_1}\plus{}\frac{a_2^3}{b_2}\plus{}...\plus{}\frac{a_n^2}{b_n}$≤$ c(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)$[/quote]
Let $a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n\in[m, \lambda m](m>0),$ and $ \sum_{i = 1}^{n} a^2_i=\sum_{i = 1}^{n} b^2_i .$ [url=http://www.artofproblemsolving.com/community/c6h505234p2838103]Prove that[/url]\[\sum_{i = 1}^{n} \frac{a^3_i}{b_i}\le\frac{\lambda ^4+1}{\lambda (\lambda ^2+1)}\sum_{i = 1}^{n} a^2_i .\]


Is this more simple way ? I can't check this solution  :(  .

Where do you use the condition : $a,b,c$ are the length of triangle

We show more generally:For any a,b,c>0 the inequality holds.


AMC
USA(J)MO
USAMO
AIME
ARML
trigonometry
AIME I

I have a very big goal for myself next year that is, making USAMO and therefore having 80% chance of going to MOP(since I will be 9th grader). Is it a goal beyond me for next year? How should I prepare?

The simplest as well as the most important advice: do lots of problems.

Also, keep yourself interested. People can degrade very quickly if they start to lose interest in the subject.

And if you want to make USAMO, work on hard problems. Don't spend your time working on problems that you already know how to do. Instead seek to work on problems that are right at your current limit, i.e. the last 5 problems on AMC-12s, challenging AoPS problems, AIME problems, ARML problems, USAMO #1 and #4 problems, etc.

By working on challenging problems you will usually learn something new with each one.

Everything here is a great suggestion.  I was in the same position as you were in the beginning of this year, and I made a plan for myself to make USAMO, and I did!  Here's what I did:

I didn't study at ALL for the AMC.  (This may not be a good idea if your school or whatever doesn't offer both the A and the B.  But if it does, I suggest studying for AIME only.)

STUDY your !@#$ off for the AIME.  Get every contest since 1989, both the real ones and the alternates, and do (untimed) as many questions as you possibly can.

If you can't do any of the first 10 questions on any AIME, look at the solution, and MAKE SURE you understand that.  You should be able to, but sometimes they get a little wordy.

Take note of the similar strategies needed on different AIMEs.

You can really study however you want for the AIME, but study it a lot.  If you do, you'll see you're doing better the more you take.  This should get you into USAMO.

Under no time pressure, I can usually do first 10 problems of AIME(and some  of the last 5 problems). But I didn't make AIME this year. Either I just got a lot better after start posting and working on AoPS problems  or wierd things happened to me on the days of AMC. I am really really afraid something like that happen to me again.

Doing what GreenTwilight suggests works.

However, I suggest supplementing it with some books like Engel or Zeitz or something, if you want to get at least one question on USAMO. AIMEs won't teach you how to do proofs, but proofs can help you learn concepts important on the AIME.

[quote="Syntax Error"][quote="beta"]I have a very big goal for myself next year that is, making USAMO and therefore having 80% chance of going to MOP(since I will be 9th grader).[/quote]

Why is there a greater chance to make MOP because you're a 9th grader?  A donation this year does not equate to a donation next year.[/quote]

As is very well known from 2002-2003.

[quote="Syntax Error"]Why is there a greater chance to make MOP because you're a 9th grader?  A donation this year does not equate to a donation next year.[/quote]

True. And even more so for people in later graduating classes. And yet it is still a good goal for a person of about eighth or ninth grade age to start seriously preparing for USAMO (which mostly means preparing to get past AIME). Some (few) people make USAMO at younger ages, and historically a few people have tested into MOP through the USAMO at ages no older.

Whether the donation is forthcoming in subsequent years depends on the vagaries of the stock market, but it also depends on how young people trained in expanded MOP programs do in future math competitions. If it looks like a good idea to train younger people in the MOP style, several people in AMC will doubtless talk that up to lots of funding sources.

Of course work is required whether or not a larger MOP is there, but I was mostly wondering if there were some sort of news about an expanded MOP that I did not hear about.

I probably shouldn't say I have that great of a chance because I am still not a permanent resident in US yet(I am in process). Who knows how long it takes for the stupid INS people to give me a green card?

If you've already gone in and talked to them, it shouldn't be long. (referring to beta's question about a green card)

In any case, I actually never formally "trained" to get into USAMO.

In fact, I haven't done any math this year at all (not even curricular math) except AoPS.

I just took a couple of classes last semester, and all of a sudden I noticed my AIME score had jumped up 6 points from last year (it was like...whoa...when'd that happen??) according to the AIME I test. (My official AIME II score wasn't as high as that though...for reasons discussed on other threads)

So yeah...if your looking for a bridge from scoring high on AMC 10 into scoring high on AMC 12 and AIME, I think your best opportunity is right here.

And let me also say that doing past AIME and AMC problems is definitely not a bad way to go, but I found that looking at some individual problem types in problems from all different competitions to be more beneficial for me. It's pretty hard to explain how or why and stuff like that, but sometimes after u've seen problems of the same type backwards and forwards inspiration just comes to you out of nowhere. Like u can skip #3 and while working on #8 you realize (in the middle of chewing a pretzel) that you know how to do #3 because Mr. Crawford made you derive a formula back in a Number Theory class 5 months ago that required the same logical analysis.

I'm going to have to disagree with GT's advice not to study the AMC. I didn't study the AMC at all, and I very nearly didn't qualify for AIME, what with my score being 101.5. Then, of course, I got a 7 on AIME and barely made USAMO. So basically, if you really want to make MOP, you'd have to study everything - AMC, AIME, and USAMO. Remember, even with perfect scores on the AMC and AIME, you'd have to make at least a 7 or so on USAMO almost any year just to qualify for MOP under regular circumstances...

[quote="probability1.01"]MOP usually costs around 18 or so points.  This year its stock must have dropped, so I think they gave a slight discount.[/quote]

Haha nice analogy ;)

Yeah I think that estimate is about right, although probably a little high, by one or two points.

[quote="Nukular"][quote="Fermatprime"][quote="Nukular"]You need quite more than that.

My friend had almost perfect AMC/AIME. Got 8 pts on USAMO. No qualification.

And this was a hard year. For everything.[/quote]

I'm talking 'bout the EASY years to qualify.[/quote]

What's an EASY year to qualify? They take the top 30 people, no matter what.

A hard year of contests means you should require fewer points to qualify.[/quote]

easy meaning precisely that...

[quote="fuzzylogic"]My friend who is an 8th grader got  7 on USAMO (basically got the first problem right).  Certainly better than some 9th graders. No qualification.[/quote]

Wow was that this year?  I thought this year's #1 wasn't particularly easy, although last year's was.

[quote="probability1.01"]
Wow was that this year?  I thought this year's #1 wasn't particularly easy, although last year's was.[/quote]

This year's 1, if u know some trig tricks...wasn't awfully difficult...unless ur like me and forgot that u can use trigs on quads...

[quote="beta"]Yeah, even i solved this year's #1[/quote]

LOL, and I doubt you had to do much to get full credit, considering they basically gave me a point (I think) for just writing "Law of Cosines..."

[quote="Magnara"][quote="Title"]How should I prepare for next year?[/quote]

Steroids.[/quote]
Yes! except you've gotta have them shipped early, or else they won't arrive in time.  :lol:
actually, i'm pretty interested in all this advice since i really bombed the amc tests this year


please find positive integer solutions to the following equation.

a^2 + b^2 + c^2 = a^2 ．b^2

[url]http://www.mathlinks.ro/viewtopic.php?p=884205#884205
[/url]Here is the solution.
Don't post double.


MATHCOUNTS AMC AIME geometry probability logarithms trigonometry

Can I host my own contest for those beginners hoping to jump to the intermediate level (such as people hoping to make mathcounts nationals or AIME)? :)

If so...I will post questions here and those who wish to participate PM me your answers along with any questions you may have concerning each problem. Each question will be worth a certain point value which will be stated along with the question. There will be a total of about 30 questions involving geometry, number theory, probability theory, combinatorics, algebra, and trigonometry.

Format for sending answers:

(1) Answer
(2) Concise solution (the idea you used to solve the problem)

The idea of this contest is to sharpen your problem solving skills and to help you approach problems using different methods which will save you time. Thus, I will award one bonus point to any solutions which I see as elegant and less time consuming than the "long" method.

Please do not discuss questions with one another until I declare them officially dead. Each set of questions will be open for approximately 1 week. After each set is over I will post the best solutions submitted for each problem along with the cumulative results almost everyday (if I'm not very busy). I hope to have many of you participate!

The first few questions I will post one by one until I see how much feedback this will receive.

PPPC
Problem (1)
Value: 4 points

Consider $\triangle ABC$ with sides of length $13$, $14$, and $15$. The sum of the lengths of the altitudes drawn to sides $AB$, $BC$, and $AC$ can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Note: A calculator is not needed for this problem! :lol:

Moderator Note:

This post has been moved from Getting Started to AMC since the contest involves the concepts of AMC-12 and AIME.

More rules would be posted by author.

Silverfalcon

PPPC
Problem 2
Value: 5 points

Find all pairs $(x,y)$ of positive integers such that $x \leq y$ and   $\frac{1}{x}+\frac{1}{y}=\frac{1}{6}$

PPPC
Problem 3
Value: 3 points

A number $d_1d_2...d_n$ is called a distinguished $n$-digit number if $d_1$ is not $0$ and the digits $d_1, d_2,..., d_n$ are all distinct.

What is the sum of the digits of the largest distinguished 8-digit number that is divisible by 9?

PPPC
Problem 4
Value: 5 points

In how many ways can the number $1,000,000$ be expressed as a prodcut of three positive integers? Factorizations which differ only in the order of the factors are not to be considered different.

PPPC
Problem 5
Value: 4 points

If a chord is selected at random on a fixed circle, the probability that its length exceeds the radius of the circle is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

PPPC
Problem 6
Value: 2

Find the next three numbers in the sequence $\frac{2}{3}, 1 \frac{5}{12}, 2\frac{1}{6}, 2\frac{11}{12}$.

PPPC
Problem 7
Value: 3

How many digits does $2^{2222}$ have? $(\log 2=0.30103$)

PPPC
Problem 8
Value: 3

Which number is greater $2^{300}$ or $3^{200}$ and why?

PPPC
Problem 9
Value: 4

Solve the equation: $\sin x \cos x=\sin40$

PPPC
Problem 10
Value: 3

Solve the equation $\log_{x-2}(2x^2-9x+8)=2$


AMC USA(J)MO USAMO quadratics calculus integration

What is the smallest integer n>1 such that the root-mean square of the first n integers is an integer?

The root-mean square of n integers is

$ \sqrt{\dfrac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}}$

Root mean square has each term in the sequence squared before you add them up.

\[ \sqrt{\frac{a_1^2 \plus{} a_2^2 \plus{} \ldots \plus{} a_n^2}{n}}\]

I'm guessing that was an oversight.

let $ b_{n}$ is the RMS of first $ n$ natural numbers .
clearly $ b_{n}^{2}\equal{}\frac{(2n\plus{}1)(n\plus{}1)}{6}$
thus we have $ 2n^{2}\plus{}3n\plus{}1\minus{}6b_{n}^{2}\equal{}0$
solve the quadratic's discriminant to obtain that to have integral solutions
$ 48b_{n}^{2} \plus{}1 \equal{}k^{2}$ for some $ k \in \mathbb{N}$
thsu we get a pell's equation with initial solution $ (x_{1},y_{1})\equal{}(7,1)$
$ k^{2} \minus{} 48b_{n}^{2}\equal{}1$
let $ (x_{i},y_{i})$ be the ith solution then clearly
$ (x_{i\plus{}1},y_{i\plus{}1})\equal{}(x_{i}^{2}\plus{}48y_{i}^{2},4x_{i}^{2}y_{i}^{2})$
we generate the sequence and check for the least number required


If $n\geq 5$ ,$m\leq [\frac{n}{2}]$ and $n|(n-m)!+m$ then $n$ is prime

Let $n,m$ be an integers that satisfy the condition.
Suppose, for contradiction, that $n$ is composite. Then, either it can be expressed in the form $n=ab$ with $a \neq b$ and $a,b \leq \frac{n}{2}$ or it is the square of a prime.

In the first case, $n-m>\frac{n}{2}$ and hence $n=ab|(n-m)!$. Then we must have $n|m$, contradiction as $m<n$.

In the second case, $n=p^{2}$, where $p>2$. If $p=3$ then $n=9$ and we can check the few cases necessary. Otherwise $p,2p < \frac{n}{2}$. Hence $2p^{2}|(n-m)!$ and we finish as above.

My idea was similar but my mistake was that I only worked with a $p$ ,prime $p|n$ and I couldn't obtain that contradition..anyway nice solution


find k is a positive number so that k min and k*23^2004+85^2007 is divisible by 45600   :blush:

A TurboPascal program yields $k\equiv 1475 (mod\ 45600)$. So the answer is 1475.


The points A(-5,5), B(2,6) and C(-2,2) are the vertices of a triangle in a coordinate plane. What is the number of square units in the area of $ \Delta ABC$?

hello, we have $ \overline{AC} \equal{} \sqrt {3^2 \plus{} 3^2} \equal{} \sqrt {18}$ ,$ \overline{BC} \equal{} \sqrt {32}$ and $ \overline{AB} \equal{} \sqrt {50}$, since $ 18 \plus{} 32 \equal{} 50$ has our triangle a right angle and the area is $ A \equal{} \frac {1}{2}\sqrt {18}\sqrt {32} \equal{} 12$ square units.
Sonnhard.


A train, traveling at constant speed, takes 20 seconds from the time it first enters a tunnel that is 300 meters long until it completely emerges from the tunnel. One of tthe stationary ceilng lights in the tunnel is directly above the train for 10 seconds. Find the length of the train.

[hide]The distance that the train travels going through the tunnel is equal to 300 meters plus the length of the train (The time starts when the front goes in and ends when the back goes out).  The distance the train travels while the light is directly above the train is the length of the train itself.  Since this takes 10 seconds, half of the time it took to go through the tunnel, this means that the length of the train is equal to the length of the tunnel.  The train is 300 meters long.[/hide]

I got 300 meters as my answer.
Call the train's length x and use the formula D=rt

I set up two equations:
1st equation:
300+x=r20 seconds
(300+x)/20=r

2nd equation:
x=r10 seconds
x/10=r

Then x/10=(300+x)/20
20x=3000+10x
10x=3000
x=300

Hi, uh, please don't revive threads that are this old (look at the upper right side of the post to see the date).
Thanks! :)


integration
calculus
calculus computations

A particle moves along the x-axis so that at any time $ t > 0$, its velocity is given by
$ v(t) \equal{} 4 \minus{} 6t^2$
If the particle is at position $ x\equal{}7$ at time $ t\equal{}1$, what is the position of the particle at time $ t\equal{}2$?

We are given $ \frac{dx}{dt} \equal{} 4 \minus{} 6t^2$. We want to find the general form of $ x(t)$.

 $ \frac{dx}{dt} \equal{} 4 \minus{} 6t^2 \implies dx \equal{} (4 \minus{} 6t^2)dt \implies \int dx \equal{} \int (4 \minus{}6t^2) dt \implies x(t) \equal{} 4t \minus{} \frac{6}{3}t^3 \plus{} C$.

We are told that $ x(1) \equal{} 7$, from which we get $ 7 \equal{} 4 \minus{} 2 \plus{} C \implies C \equal{} 5$.
We get the general form for the position as $ x(t) \equal{} 4t \minus{} 2t^3 \plus{} 5$.
You are asked to find $ x(2)$.

This should serve as an example for the rest of the problems of this kind.

You could also use $ x(t_2) \equal{} x(t_1) \plus{} \int_{t_1}^{t_2} v(t) \,dt$ [which is just a rearrangement of the Fundamental Theorem of Calculus]. This eliminates the need to find the constant of integration in the equation for position.


Let a "five-tastic" number be defined as any number which contains at least three digit 5's in a row. For example, 34555, 95555, and 55555 are five-tastic, but 50505 is not. How many 5-digit numbers are there which are "five-tastic"?

[hide]
For a string with three consecutive fives $ 9 \cdot 10 + 8 \cdot 9 + 9 \cdot 9 = 243$
four fives $ 8 + 9 = 17$
five fives $ 1$

$ 243 + 17 + 1 = 261$.
[/hide]

EDIT $ \text{\LaTeX}$ typo.
EDIT Forgot zero as possible digit  :| ....

EDIT: First-digit mistake fixed.

[hide="Solution."]
Case 1: There are only three fives.

There are $ 3$ ways in which we may position the consecutive digit fives. For one of these, the other two places can be filled in $ 9^2$ ways, and for the other two, since the first digit can't be zero, the other two places can be filled in $ 8\cdot 9$ ways, and thus this case occurs in $ 1\cdot9^2 \plus{} 2\cdot 8\cdot 9 \equal{} 225$ ways.

Case 2: There are four fives.

There are $ 4$ ways to choose which digit is [i]not[/i] a five, since any digit besides the middle can be a non-five digit. For three of these, we can fill the non-five digit in $ 9$ ways, and for one of these, the non-five digit can be filled in $ 8$ ways, and thus this case occurs in $ 3\cdot9 \plus{} 8 \equal{} 35$ ways.

Case 3: There are five fives.

Hmm... 1 way.

In total, there are $ 225 \plus{} 35 \plus{} 1 \equal{} \boxed{261}$ five-tastic numbers.
[/hide]

[quote="AstroPhys"]
[...]
four fives $ 8 \plus{} 8 \equal{} 16$
[...]
[/quote]

I claim that:

15555 55550
25555 55551
35555 55552
45555 55553
65555 55554
75555 55556
85555 55557
95555 55558
         55559
are all five-tastic numbers that have four fives in them; there are at least seventeen in the above list, so there can't be $ 16$ in total.

I meant a "string of consecutive" fives.

[quote="AstroPhys"]
For a string with three consecutive fives $ 9 \cdot 10 \plus{} 8 \cdot 9 \plus{} 9 \cdot 9 \equal{} 243$
[/quote]

 :| Where is this coming from? Five-tastic numbers with three fives would be as follows:

"--555" can be filled in $ 8\cdot9\equal{}72$ ways (no 0,5; no 5),
"-555-" can be filled in $ 8\cdot9\equal{}72$ ways (no 0,5; no 5),
"555--" can be filled in $ 9\cdot9\equal{}81$ ways (no 5; no 5),

so there are $ 72\plus{}72\plus{}81\equal{}225$ five-tastic numbers with three fives.

[quote="jvenezuela716"]"555--" can be filled in $ 9\cdot9 \equal{} 81$ ways (no 5; no 5)[/quote]

And why can not there be a five in the ones place there?  :huh:

[quote="AstroPhys"]
And why can not there be a five in the ones place there?  :huh:[/quote]

Ah, I just realized that we are counting the same thing, but with different cases. My cases are divided depending on how many total digit-fives there are in the five-tastic number, whereas your cases are counting how many fives appear consecutively. This makes sense since we ended up with the same answer.  :P


calculus
real analysis
real analysis unsolved

Let p denote partitions and f fibonacci seqence. For each positive intger k, prove that f(n)>p(n+k) hold for all +ve integers n but finitely many values.

Bomb


conics
greatest common divisor
number theory
least common multiple
quadratics
analytic geometry

It's back!

1'.) Given the following multiplication in base 8, find base 10 value of the number xy8: xy8*2=yx8 (x and y are digits and they are all 2-digit numbers, not x times y)
2'.) The greatest common factor of two whole numbers is 12 and their least common multiple is 2520. Find the smallest possible value for their sum.
3'.) For what value of k is the following equation true? 2000:^2:+2000:^3:=2001k
4'.) Jill has three times more money than Jack. If Jill gives Jack $15, Jill will now have twice Jack's new amount. How many dollars did Jill start with?
5'.) Given the conic C, {(x,y): x:^2:+y:^2:-x-4y-12=0}, find the length of the chord whose endpoints are where the y axis intersects C.

This week's is short since many of the problem are much harder...still, buenas suerte!

2) [hide]you can multiply the gcf and lcm to get the product of the two numbers.  Then we can determine the closest numbers to its root that multiplies to 30240 which is 168 and 180.  The sum of those two are 348.[/hide]

theone853 wrote:[hide]2)348
3)4000000
4)$135[/hide]

You got all these right, but not 1...there's only one answer for each of these. Scores are updated.

Addendum: Whoops, I meant 1 not 2...sorry theone

[hide]1) xy*2=yx, 1 will be carried over so y is odd and more than 3,testing 25*2=52, we get that xy base 8 is 25. So xy in base 10 is 21.

2) LCM*GCF=x*y, we have the factors 2^5*3^3*5*7, the way to get x and y the smallest is to make them closest to each other which is achieved when x and y are 168 and 180. Their sum will be 248.

3)adding them we get 8,004,000,000 thus k needs to be 4,000,000.

4) solving the equations we get that jack had 45 to begin with, thus jill had 3 times as much, or 135.

5. later.[/hide]

You got 1, 3, and 4. 2 is so close but you missed the very last addition...

P.S. oooh ml2 zooms ahead before theone853 has a chance to take the lead...

Assume you know number bases, here's an easy method for #1:  [hide]Many of you probably know how to "cast out 9's" in base 10.  In base 8, the equivalent is to "cast out 7's".  By applying "casting out 7's", it's easy to show that x+y = 0 (modulo 7).  So the possible values for xy are 70, 61, 52, 43, 34, 25, 16.  Trial and error gives the (base 8) solution.  Or you can trivally rule out 70, 61, 52, 43, 34, and 16, because the second octet must be (twice the first octet) or (one greater than twice the first octet).[/hide]

2. yes i added wrong, the actual answer is [hide]348[/hide]

[hide]

1) 21

2) 312

3) 4,000,000

4) 135

5) 8

When the circle crosses the y-axis, x=0. The terms involving x or x-squared can be removed, leaving a nice quadratic whose solutions are 6 and -2. The distance between (0,6) and (0,-2) is 8.

[/hide]

#2[hide] The two numbers in question are both multiples of 12. 2520/12=210, and the 2 closest whole numbers that multiply to be 210 are 14 and 15. So the 2 numbers that the problem talks about are 168 and 180, whose sum is  348[/hide]

5. [hide]we want the cordinates where we have (0,k) and (0,m)

let us look at our conic and see where that happens.. {(x,y): x^2+y^2-x-4y-12=0}, we plug in 0 for x and get y^2-4y-12=0, factoring to (y+2)(y-6)=0 so k and m are -2 and 6. The distance between (0, -2) and (0, 6) is 8.[/hide]

Whoops, I didn't notice this till now, I'll go fix it :oops:


trigonometry
algebra
polynomial
limit
algebra unsolved

Let a trig polynomial :(deg >0)
 $ f(x)= a_1 cosx + b_1 sinx + a_2 cos2x + b_2 sin2x + ... + a_n cosnx + b_n sinnx $
  ($ a_n^2+b_n^2>0 $)
 Prove that : $ f(x) $ gets both negative and possitive value

$g(x)=\sum\frac{a_i}{i}\sin{ix}-\sum\frac{b_i}{i}\cos{ix}$
Then $g(x)=g(x+2\pi)$
$g'(x)=f(x)$
If $f(x) \ge 0$ or $f(x)\le 0$ for all $x$ then:
$g(x)$ increasing $\rightarrow g(x)=const->f(x)=0$

$\rightarrow\sum{a_i\cos{ix}}=\sum{b_i\sin{ix}}=0$

$\sum{a_i\cos{ix}}=0 \rightarrow a_n=0$,because it's a polinomial of $\cos{x}$
$\sum{b_i\sin{ix}}=0\rightarrow b_n=0$,because $\frac{\sum{b_i\sin{ix}}}{\sin{x}}$ is a polinomial of $\cos{x}$

dear hungkhtn ,
 Why $ f(x) = 0 $ ===> $ \sum a_icosix = \sum b_isinix = 0 $ ?

Oh ,at First I'm in a hury to think that:
if $f(x),g(x)$ are two polinomial and $f(x)+\sqrt{1-x^2}g(x)=0$ for all $x \in [-1,1]$ then $f(x)=g(x)=0$

I think It's obvious,But now I find it's not easy.
Here is other solution of mine Which doesn't use this way.

Take $f_i(x)=a_icos{ix}+b_isin{ix}$

Then $f(x)=\sum{f_i(x)}$

because $f(x)=0$ then $f^{(k)}(x)=0$

$\rightarrow \sum i^{2k}f_i(x)=0$

Then $\sum (\frac{i}{n})^{2k}f_i(x)=0$ for all natural $k$

If $i < n$ then $\lim(\frac{i}{n})^{2k}=0$

Then $f_n(x)=0$
And the rest is easy.


quadratics
Gauss
number theory unsolved
number theory

Prove that if $ 2$ is quadratic residue mod $ p$ then $ p \equiv 1,7\bmod{8}$ . Condition imposed
-Assume that you have never seen [hide="this"]$ \left(\frac {2}{p}\right) \equal{} ( \minus{} 1)^{\frac {p^{2} \minus{} 1}{8}}$[/hide] and prove the statement without using it . :D:  :blush:
Sorry if this is dumb .... :(

That would be equivalent to re-discovering Gauss' proof, so ...

sorry for the trouble ,but can you tell me about which Gauss proof ( in the sense proof for what) you are talking about .

I think Gauss was the first to compute the Legendre symbol $ \genfrac {(} {)} {} {} {2} {p}$.

Yes but this $ \left(\frac{2}{p}\right) \equal{} (\minus{}1)^{\frac{p^{2}\minus{}1}{8}}$ is a more stronger result than what I asked ...........so gauss or some mathematician would have proved for the weaker case then only extended its validity to a stronger case ...so suppose you were born before Gauss or whatever and you were intent upon proving this .......how would you have done....sorry if I have misunderstood what you are trying to tell.


does anyone know of a good intermediate-olympiad level geometry books, besides challenging problems in geometry, geo revisited, and those other books listed on AoPS wiki?
and
how about any interm-olympiad level algebra books, besides any of titu's, challenging problems in alg, and those listed on AoPS wiki?

I think
1) Geometry : Geometric Transformations by I. M. Yaglom and some notes got from internet :D, example : Kedlaya,...
2) Some notes in Algebra got from internet.

i would like books, not online papers, and the transformation books is on AoPS wiki


A set of 25 blocks is arranged into a 5x5 square.  How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?
(source:AMC 10B 2007)

[hide="Solution"]There are $ 25$ ways of picking the first square. Now, we have $ 4$ rows and $ 4$ columns to pick the next square from, so there are $ 16$ ways of picking the second square. Similarly, there are $ 9$ choices for the third square. Therefore, there are $ 3600$ permutations. There are $ 3! \equal{} 6$ orders per combination, so there are $ \boxed{600}$ combinations.

[/hide]


$x_{0}>0$

$x_{n+1}=\frac{x_{n}}{1+nx_{n}^{2}}$

1/ Find an equivalence of $x_{n}$

2/ Find an asymptotic expansion with two terms

We show that Xn=1/n+f(X0)/(n^3)+o(1/(n^3))
where f(X0)<0.
i) (Xn) is a positive decreasing sequence; thus its limit is 0.
ii) One shows by recurrence that if n>=2 Xn<=1/n:
Let Fn(x)=x/(1+n*x^2); X2<=sup(F1)=1/2 ; if Xn<=1/n then
X(n+1)<=Fn(1/n)=1/(n+1).
iii) Xn~1/n:  Let Un=n*Xn (<=1); U(n+1)=(n+1)*Un/(n+Un^2) ;
U(n+1)-Un=(Un-Un^3)/(n+Un^2)>=0; (Un) is a positive increasing sequence which converges to l in ]0,1] ;
 if l<1 then U(n+1)-Un~(l-l^3)/n ,
the series (U(n+1)-Un) fails to converge and (Un) reaches infinity. Contradiction.
iv) Let Zn=1-Un  (limit(Zn)=0); we assume that X0<>1
(if not Xn=1/n);
 it remains to show  that Zn~ -f(X0)/(n^2) :
Z(n+1)=Zn*(Zn-1+n)/(n+1+Zn^2-2*Zn);
Z(n+1)/Zn =1-2/n+o(1/n);
ln(Z(n+1))-ln(Zn)~ -2/n therefore ln(Zn)~ -2*ln(n);
let s be in ]0,1[; for all sufficiently large n :
Zn<=1/(n^(2-s)). Thus Z(n+1)/Zn=(n-1+o(1/n))/(n+1+o(1/n))=
1-2/n+2/(n^2)+o(1/(n^2)); ln(Z(n+1))-ln(Zn)=-2/n+o(1/n^2); therefore
ln(Zn)= -2*ln(n)+g(X0)+o(1) and Zn~exp(g(X0))/(n^2).


We know that in the absence of air drag, maximum range of a projectile is for an angle of $ 45^0$ for similar velocities.

Similarly what can we say if air drag is also considered??? Maximum range will definitely be less than the maximum range in CaseI, but what can we say about the angle of projection compared to 45.

Air drag can be approximated by $ \vec{F}\equal{}\minus{}k\vec{v}$, where $ k$ is a constant.

Using this, we get that the angle of projection for maximum range is dependant on the initial speed of projection.

(You get it by solving the differential equations
$ \frac{dv_x}{dt}\equal{}\minus{}kv_x$
$ \frac{dv_y}{dt}\equal{}\minus{}(kv_y\plus{}gt)$
I presently can't recall the solution, but I know this result)

I think the angle of projection turns for maximum range for a given theta turns out to be less than 45.

See here concerning this topic

http://farside.ph.utexas.edu/teaching/336k/lectures/node43.html


I'm wondering, are the earlier AIME's, like the ones from the 1980s easier than the ones we have now?  I tried doing the AIME in 1986, and I got 4 problems right in less than 40 minutes, while I only got a 3 on the AIME I this year.  It might be that I'm getting better, but I don't think I could improve so much.

I hate to break it to you, but yeah, they are easier. There's no perfect conversion function, but guess around a 2-3 (I think) point difference in difficulty (which is quite a lot). When you get into the early 90s though, there are some tough AIMEs. Most will be quite a bit more difficult than this year's AIME I.

Yeah, it sucks how variable the AIME's are.  I've averaged from a 0 (1997) to a 11 (1984). :)
My ratings:
1983-1989: Extremely easy
1991-1997: Extremely hard
1998-2005: Easy in general, but with lots of variability.

I tried that one and got 4 problems in the first 15 minutes. :)

yea, definately easier

wonder what the index for USAMO was back then

Yeh like evryone said the 80s ones were easy
BUT
this years AIME I was just as easy
I thought I had made USAMO for sure... but the floor became sky high

[quote="qweretyq"]this years AIME I was just as easy
I thought I had made USAMO for sure... but the floor became sky high[/quote]
Yeah, I agree. This year's was relatively easy in terms of "thinking"/"problem solving", but I thought it required some messy calculations... (I missed 3 Qs by simple arithemetic mistakes)


Prove that
\[ \sin^2\frac {\pi}{7} \plus{} \sin^2\frac {2\pi}{7} \plus{} \sin^2\frac {3\pi}{7} \equal{} \frac 74.\]

You can use $ \cos ^ 2 \theta \equal{}\frac{1\plus{}\cos 2\theta}{2}$ or $ a^2\plus{}b^2\plus{}c^2\equal{}(a\plus{}b\plus{}c)^2\minus{}2(ab\plus{}bc\plus{}ca)$.

$ 7 \theta \equal{} n \pi \Rightarrow \sin^2 4 \theta \equal{}  \sin^2 3 \theta$

$ \Rightarrow 16 \sin^2 \theta \cos^2 \theta (2 \cos^2 \theta \minus{}1)^2 \equal{} \sin^2 \theta (3\minus{}4 \sin^2 \theta)^2$

Discarding the root $ \theta \equal{} \pi$, we get an equation in $ \cos^2 \theta$ as

$ 64 \cos^6 \theta \minus{} 80 \cos^4 \theta \plus{} 24 \cos^2 \theta \minus{} 1 \equal{} 0$

$ \cos^2 \frac{\pi}{7}, \cos^2 \frac{2\pi}{7}, \cos^2 \frac{3 \pi}{7}$ are hence the roots of the cubic

$ 64z^3 \minus{} 80 z^2 \plus{} 24 z\minus{}1 \equal{} 0$

It follows from Vieta's formulas that

$ \cos^2 \frac{\pi}{7}\plus{} \cos^2 \frac{2\pi}{7}\plus{} \cos^2 \frac{3 \pi}{7} \equal{} \frac{5}{4}$

Hence $ \sin^2 \frac{\pi}{7}\plus{} \sin^2 \frac{2\pi}{7}\plus{} \sin^2 \frac{3 \pi}{7} \equal{} \frac{7}{4}$

you can use ...

$ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}(a\plus{}b\plus{}c)^{2}\minus{}2(ab\plus{}bc\plus{}ca)$

\[ \sin{\varphi}\plus{}\sin{(\varphi\plus{}\alpha)}\plus{}\sin{(\varphi\plus{}2\alpha)}\plus{}\cdots\plus{}\sin{(\varphi\plus{}n\alpha)}\equal{}\frac{\sin{\left(\frac{(n\plus{}1)\alpha}{2}\right)}\cdot\sin{(\varphi\plus{}\frac{n\alpha}{2})}}{\sin{\frac{\alpha}{2}}}.\]

___________________________

 - kl -


If $\sqrt{3+\sqrt{5}}=\frac{\sqrt{a}+\sqrt{b}}{c}$, where $a$, $b$ and $c$ are positive integers, find the minimum possible value of $a+b+c$.

[hide]$(\sqrt{x}+\sqrt{y})^2=x+y+2\sqrt{xy}$

since there is a 2 in $2\sqrt{xy}$, we'll multiply $3+\sqrt5$ by 2

$x+y+2\sqrt{xy}=6+2\sqrt{5}$

so $x+y=6$ and $xy=5$, so $x=1$ and $y=5$ (or the other way around).

$\sqrt{6+2\sqrt{5}}=\sqrt{5}+1$
$\sqrt{3+\sqrt{5}}=\frac{\sqrt{5}+1}{\sqrt{2}}=\frac{\sqrt{a}+\sqrt{b}}{c}$

this is simplified to $\frac{\sqrt{10}+\sqrt{2}}{2}$ so a+b+c=14.
[/hide]

Sorry. I don't have LaTeX.

[hide]
sqrt(3+sqrt(5))=sqrt(6+2sqrt(5))/sqrt(2)=(sqrt(5)+1)/sqrt(2).

Rationalizing, we get

(sqrt(10)+sqrt(2))/2.
a+b+c=10+2+2=14[/hide]

[quote="DanK"]You don't need $\LaTeX$ to post in it on here.  You just put dollar signs around the stuff you want to be in $\LaTeX$...[/quote]And he will have to put backslash \ to square roots like \sqrt for it to work, and don't forget to Enable BBCode (which should be enabled on default unless you clicked the first box just below this message :D )


[quote="Silverfalcon"]Difficult questions:

This marathon is different with one in this same forum..

In here: The question must be advanced question from state sprint round or target.. There should be two questions posted per time..

What is the remainder when 2 to 87th power divided by 7?

A line segment drawn from a vertex of a unit square to point on the square forms two regions. The area of the smaller region is one-third of the area of the larger region. How many units are in the length of that segment? Express your answer as a common fraction in simplest radical form.[/quote]
2^3 (mod 7)=1.
2^87(mod 7)=1
This is because 87 is 3*29.
1^29=1
divide it up into four triangular pieces.
3 will be for the larger region, and one will be the smaller region.
the side length of 1 side is 1.  The side length of the other side is 1/2.
Using pythagorean theorum, the other side is  :sqrt: (3/2).  that is  :sqrt: 3/ :sqrt: 2.  which is  :sqrt: 6/2

what is the smallest integer n where 7+30n isnt a prime?

I must insure you that even though I'm not nearly good as great mathletes in this site, I am trying hard on every move I make..
And this topic for difficult questions and like to remind you that this is not that simple.. For the first question, yes, it's correct.. For second questoin, however, it's incorrect..

I'll then solve to your problem when you got right answer plus you must post two questions at a time, not one..

um are you sure?
i cant find the problem with my work.

by the way i only have access to 2003 ones and for some reason it started screwing up my compute so i cant post another one.

[quote="Silverfalcon"]divide it up into four triangular pieces.
3 will be for the larger region, and one will be the smaller region.
the side length of 1 side is 1. The side length of the other side is 1/2.
Using pythagorean theorum, the other side is  (3/2). that is  3/  2. which is  6/2

That's your work.. Yes, you can divide it into four triangular pieces and 3 will be for the larger region and the one for the smaller region.. And we know side length 1 by 1 of square... Let's find the area of the smaller region... It's one-fourth of total area, meaning its area is 1/4. So, we know its height, 1, right? What's its base then? 1 * x * 1/2 =1/4
                x = 1/2

So, let's use pythogrem theorem..
1^2+(1/2)^2=1+1/4=5/4  then we must sqrt that and the result is  :sqrt: 5/2... Your answer was correct but you screwed up in the pythogrem theorem part..[/quote]

Give the next problem


Given $8$ distinguashable rings, find the number of possible five-ring arragement on $4$ fingers.

[hide]First, there are ${8\choose 5}=56$ ways to choose the 5 rings.  Then, there are 4 fingers for each of 5 rings, for $4^{5}=1024$ arrangements.  Then there are 57344 ways to arrange five rings.[/hide]

I think I must have misinterpreted the question-must there be one ring on each finger or something?  Because AIME answers are 3-digit numbers, so....

The (original) question asked for the three left-most non-zero digits of the answer (in order to make it a 3-digit number).

But I think you still have to consider the possible orderings of the rings on each finger.

Also, this question is currently also being discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111866]here[/url].

samath you might want to check your work :wink:


[quote="mihir0492"]but what about SA?[/quote]
do you mean South Africa or South America

search for kvpy in google and you will get to know about different streams.....

Yes my Interview is in Allahabad on 12th Jan(SA stream)

Have the call letters arrived  :what?:

When were you informed about the same :?:

3-4 days ago I think. Was I not supposed to receive it?  :what?:
Btw you can check your result online at the KVPY site

Yes mate I have also qualified for the interview. I am a bit suprised that I haven't recieved the letter yet.

Though I am confident that I will not qualify for the scholorship(I am not upto the standard that a KVPY scholor requires) :sniff:  :(

Finally I got the letter today.
Well what is the meaning of getting a form filled by the Teacher?(in the letter) I mean most of
teachers will fill goody-goody things about the students.
Secondly is it neccesary to get the form filled by our own school teachers. Can we get the form filled by our tutors/coaching class instructers as well?

:rotfl: I've been trying to call the KVPY authorities over the SAME question. In any case I've given photocopies of my form to my coaching teachers so that they can write the answers and those can be copy pasted by a school teacher in the eventuality that a school teacher is compulsory. Also, seeing as we 2 might not be the only ones with this query considering that the standard of coaching teachers far exceed the school standards in many places, it might be advisable to send a collective request to the authorities to allow the filling of questionnaire by the teachers who know us best including our coaching teachers.

i wud say the english of skul teacher is much better
so just find a teacher on whom u hav a gud impression

I disagree. At coaching, we are taught by IITians who are faaaaaar more proficient in English. And a KVPY scholar is likely to have a good impression on most teachers. In any case any teacher would be happy to fill up such a respectable exam's form ;-)

WELL............
can all the ppl who hav cleared the written xam ov kvpy.........pl post there date and reporting tym for interview..........
probably we migth b able to see each othr......... :P

mine is on 10th jan.......9:00am........... :roll:

Arnav Welcome back :)

Mine is on 12 jan in dehli(JNU)

then u may find rupamanjari ghosh...

she is a superb interviewer  :D  :oops:

What do you mean by  'superb' ?

:noo:  :stink:

when are your interviews???

my friends are having their between 4th to 18th jan

best of luck for interviews :P

hopefully most of AoPSers make it thru

is it necesarry that only a science teacher fills the form, or even a maths teacher can fill it....n also that many of my teachers at my coaching have completed there Phds (n  have a good english too), so im thinking of taking more than one teacher recommendation form with myself...

n my mom also has a query, do we have to give the interview in our school uniform?

:rotfl: Not uniform I guess, but yes its advisable to go in a formal dress. Lol I haven't taken out my formals since the NTSE interview. It feels pretty wierd. I'm not wearing a tie and all....makes me feel suffocated!  :ninja:

[quote="mihir0492"]is it necesarry that only a science teacher fills the form, or even a maths teacher can fill it[/quote]
get a teacher whom u hav a deep  impression...an english teacher will also do it then :rotfl:  :rotfl:

and well u shudn't mind abt uniform..nething wud do
 i remember i  wore a t-shirt which had nthng formal abt it... :P


AMC
AMC 10
ratio
geometry
trigonometry
trig identities
Law of Cosines

Look at the amc math jam.  the solution is in there, using ratio of areas

[quote="Hexaditidom"]25.
In triangle ABC we have AB = 25, BC = 39, and AC = 42. Points D and E are on AB and AC respectively, with AD = 19, and AE = 14. What is the ratio of the area of triangle ADE to the area of the quadrilateral BCED?
a) 266/1521   b) 19/75   c) 1/3   d) 19/56    e) 1

This is the only problem on the test I haven't figured out, because I keep using the following method and I end up with an answer which is obviously false:

1) Use AB, BC, AC in law of cosines to find <A
2) Use AD, AE <A to in law of cosines to find DE
3) Use heron's theorem to get the area of ABC and ADE
4) Answer is ADE/(ABC-ADE)

Is this the right method? I don't care much for the answer, I just want to know how to figure this out.
Thanks[/quote]

Easiest way is to use the formula [ABC] = absinC/2 and compare since angle C is equal in both the small and large triangles.

hint: [hide]you can draw in a common altitude for the whole triangle and the small triangle[/hide]

if you just brute force it using your way (the law of cosines twice and then herons)
you get D as the answer

[quote="gotztahbeazn"]but im not sure wat everyone else is talking about the common altitude
for some reason I cant... find this altitude for both triangles[/quote]

sorry, i don't mean a common altitude, i mean two separate altitudes in which those altitudes would be parallel. then, you can use similar triangles to find the answer.

I was wondering....could you use herons formula, and brahmaguptas formula to solve this?? I think it would be a long process if you could.

[quote="Hexaditidom"]My teacher gave me the answer key, and the altitude method is how they solved it. Can someone show me why if you have a triangle ABC and if you have a point D on AB, that AD/AB = Area ADC/Area ABC?[/quote]
Well, assuming the height is = to the height of the larger triangle that is true.

[quote="tarquin"]

the base is 19/25 and the height is 1/3 so the triangle is 19/75 of the thing i think.

[/quote]

except you found the ratio of that to the whole triangle, not the rectangle. so it should be

$\frac{19}{75-19} = \frac{19}{56}$


superior algebra
superior algebra solved

A,B \in Mn(R) such that 3AB+I=2BA
Proove that det(AB-BA)=0
Cristinel Mortici
----------------------------------------------------------------------
If you have a different solution, post it thanks.
2C= 2(AB-BA)= -I-AB
C= -1/2.I-1/2.AB
detC=(-1/2)^n.det(I+AB)=(-1/2)^n.det(I+BA) but BA=3/2AB+1/2I
detC=(-1/2)^ndet(3/2I+3/2AB)=(-1/2)^n.(3/2)^ndet(I+AB)
so (-1/2)^n.det(I+AB)=)=(-1/2)^n.(3/2)^ndet(I+AB) =>det(I+AB)=0
detC=0
An obvious remark: trAB=-n since trC=0

M.Omarjee
STT math-teacher

[quote="Lagrangia"]Well, in his book Mr. Cristinel Mortici says that:
"Try using the equality det(I_n+AB)=det(I_n+BA)" (as a hint)

and I think that this is what Moubinool also uses in his proof. I don't have another proof but this is what he says in his book!

cheers! :)[/quote]

If A is invertible
det(I+AB)=detAdet(A^(-1)+B)=det(A^(-1)+B)detA=det(I+BA)

If A is not invertible, we use

Lemma: the set of invertible matrices is dense in the set of matrices
Proof: do it as an exercice

So there exist a sequence invertible matrices A_p such that lim A_p = A
We know for any p \in N, det(I+A_p.B)=det(I+B.A_p), the det is continous application, let's tend p->+oo , det(I+AB)=det(I+BA)

For orl
STT Sciences Techniques Tertiaires is section in high school for trading, management, accounting.


Let we return an old topic.
I think
If we pose $ u_n\equal{}\sqrt{k}.tanx_n$, and note $ \sqrt{k}\equal{}tana$ we have
$ tanx_{n\plus{}1}\equal{}\frac{1}{\sqrt{k}}.tan(a\plus{}x_{n\plus{}1}$

[quote="nmtvn"]Let we return an old topic.[/quote]

Let we not until 15 October, ok?

  darij


Let $n$ be a positive integer. Show that \[\sum^{n}_{k=1}\tan^{2}\frac{k \pi}{2n+1}\] is an odd integer.

I claim that for all $ n \in \mathbb{N}$, $ S(n) : \equal{} \sum_{k \equal{} 1}^n \tan^2 \frac {k\pi}{2n \plus{} 1} \equal{} n(2n \plus{} 1)$.
First of all, denote $ t_k : \equal{} \frac {k\pi}{2n \plus{} 1}$, where $ k$ ranges from $ 1$ to a fixed $ n$, and $ m \equal{} 2n \plus{} 1$.
Now, in the relation

$ \frac {\cos(mx) \plus{} i\sin(mx)}{(\cos(x))^m} \equal{} \frac {(\cos(x) \plus{} i\sin(x))^m}{(\cos(x))^m} \equal{} (1 \plus{} i \tan(x))^m \equal{}$

$ \sum_{k, 2 | k} ( \minus{} 1)^{k/2} \binom{m}{k} \tan^k(x) \plus{} i \cdot \sum_{k, 2 | (k \minus{} 1)} ( \minus{} 1)^{(k \minus{} 1)/2} \binom{m}{k} \tan^k(x)$,

equating the imaginary parts and rewriting the sum yield

$ \frac {\sin(mx)}{(\cos(x))^m} \equal{} \tan(x) \cdot \sum_{k \equal{} 0}^n ( \minus{} 1)^k \binom{2n \plus{} 1}{2k \plus{} 1} (\tan^2 (x))^k$.
Now assigning to $ x$ all the values of $ t_k$ causes the left-hand side to vanish, hence the distinct numbers $ \tan^2(t_k)$ are all the roots of

$ P \equiv \sum_{k \equal{} 0}^n ( \minus{} 1)^k \binom{2n \plus{} 1}{2k \plus{} 1} X^k \in \mathbb{Z}[X]$.

This shows that $ S(n)$ is equal, by Viete's relations, to $ \minus{} \frac {[X^{n \minus{} 1}] P}{[X^n] P} \equal{} n(2n \plus{} 1)$, which was to be proved (I denoted by $ [X^t] P$ the coefficient in $ P$ of the monomial of degree $ t$).
Consequently the problem is not correct - the sum is odd iff $ n$ is odd. I strongly suggest changing the statement into either "find a closed form" or "prove that the sum is equal to [...]" - any others represent just useless deviations.
It'd be interesting to see the source of this nice question as well.

[b]Remark[/b]. The same idea for evaluating such sums is used in the elementary proof of the fact that $ \zeta(2) \equal{} \pi^2/6$.


Putnam
combinatorics proposed
combinatorics

Well this porblem is really interesting ,Or i believe it is :

There are $n$ gas stations around a square .
Such that the sum of all of the fuels is only suffice for a round ,around the sqaure.

Prove that a car(with empty fuel tank) can ,begin the journey from one of the stations and make a full lap around the sqaure.

This has been discussed before. I believe there was a topic where a Putnam problem was discussed, very similar to this one.

Edit:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=gas+stations&t=44152[/url]


LaTeX
combinatorics proposed
combinatorics

Hi guys,

i was doing the following problem which is was quite simple:
Consider all five-digit numbers formed by using the digits 1, 2, 3, 4, and 5 once and only once in each number formed. What is the sum of all of these five-digit numbers?

This seemed quite trivial for first I thought of manually finding the sum knowing there were 120 different combinations. instead I decided to derive a general formula from the basis of 3 numbers instead of five, it was much easier and i cam to the following formula:
n![0.5(n+1)][10^(n-1)+10^(n-2).....10^(n-n)] where n > 0 and n-n is the last power. I know I should have used sigma notation but i dont have mathtype, so sorry...

Not only did this formula I came up with work with the 3 numbers, but it also worked with the 5 numbers. This might be a known formula but I felt kind of proud when I discovered it alone..can someone please tell me if this formula had existed in the past??? :(

Welcome to the forum!

This is, indeed, known: it's not a very difficult problem.  Yet if you solved it on your own without guidance, you should feel happy!  As for difficulty, this is probably around the level of Intermediate topics or so.  Keep doing problems!

The thing is, you need to have $ n<10$.  If you want to extend it to larger $ n$, you have to use a different base.  But the reason this formula works is as follows...

There are $ n!$ numbers
The average value of the digit in one digits place as you go over all of the numbers is $ \frac{1\plus{}2\plus{}...\plus{}n}{n}\equal{}\frac{n\plus{}1}{2}$, so the average of all of the numbers is $ \left(\frac{n\plus{}1}{2}\right)\sum_{i\equal{}0}^{n\minus{}1}10^{i}$

As for typing in mathematical notation, AoPS is equipped with LaTeX (see my typing above). Go to the LaTeX page and learn how to make things look nice, or just learn by observing people on the forum (that's what I did).  You have to type some LaTeX code, and then put dollar signs around it to make it parse.  So for example, the sum you wanted would be as follows

\sum\limits_{i=0}^{n-1}{10^i} $ \implies\sum_{i\equal{}0}^{n\minus{}1}{10^{i}}$

The best place on the forum to begin learning LaTeX is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 (a LaTeX FAQ).

If you don't have MathType and don't know LaTeX, you can also download TeXaide, the free version of MathType we make available on our website, http://www.dessci.com. You can use point and click and/or typing to create an equation and then copy it to the clipboard as LaTeX and paste it into your posting.

Paul Topping
Design Science, Inc.


calculus integration logarithms trigonometry derivative symmetry function

hello, the antiderivative of your first integral containes the PolyLog-function.
By a numerical methode i have got
$ \int_{0}^1\frac{arcsin(\sqrt{x})}{x^2\minus{}x\plus{}1}\approx0.949703126294009395263498491745$.
The second one can be solved by substituting $ x\equal{}t^{12}$ then we got
$ 12 \int_{0}^1\frac{(1\plus{}t^6\plus{}t^4\plus{}t^3)t^{11}}{1\plus{}t^2}dt\equal{}\\
12\int_{0}^1t^{15}\plus{}t^{12}\minus{}t^{10}\plus{}t^9\plus{}t^8\minus{}t^7\minus{}t^6\plus{}t^5\plus{}t^4\minus{}t^3\minus{}t^2\plus{}t\plus{}1\minus{}\frac{t\plus{}1}{1\plus{}t^2}dt$
After a long calculation we have the result
$ \frac{918991}{60060}\minus{}6\ln(2)\minus{}3\pi$.
Sonnhard.

hello, the third integral containes also the Poly-Log function. By a numerical method we get
$ \int_{1}^{\frac{1\plus{}\sqrt{5}}{2}}\frac{x^2\plus{}1}{x^4\minus{}x^2\plus{}1}\cdot\log\left(1\plus{}x\minus{}\frac{1}{x}\right)\approx 0.2721982612879502663125861122797$
Sonnhard.

It is possible to obtain exact values for both the first and third integral.
Integral no 3.
Rewriting $ \frac{x^2 +1}_{x^4-x^2+1}$ as $ \frac{1+\frac{1}{x^2}}_{x^2-1+\frac{1}{x^2}}$=$ \frac{1+\frac{1}{x^2}}_{1+(x-\frac{1}{x})^2$,  and after making the substitution  y=x - $ \frac{1}{x}$   we get dy =1+$ \frac{1}{x^2}$dx and the new lower and upper limits of integration 0 and 1 respectively.

Thus we have now to calculate $ \int_0^1\frac{log(1+y)}_{1+y^2}$dy

 Let y=tan$ \theta$.  This gives d$ \theta$ =  $ \frac{dy}{1+y^2}$ and new upper and lower limits of integration 0 and $ \pi/4$ respectively.

We now must find $ \int_{0}^{\pi/4}log(1+\tan\theta)d\theta$.  But this is easily dealt with using the fact that $ \int_0^af(x)dx$=$ \int_0^af(a-x)dx$ .The answer is $ \pi/8$.log2

$ I\equal{}\int_0^1 \frac{\arcsin \sqrt{x}}{1\minus{}x\plus{}x^2}\, dx$ is after substitution $ x\mapsto \frac{1}{1\plus{}x^2}$

$ \int_0^\infty \frac{2x\, \arctan\frac{1}{x}}{1\plus{}x^2\plus{}x^4}\, dx$ . And this is $ \int_0^\infty \frac{2x\, \arctan x}{1\plus{}x^2\plus{}x^4}\, dx$ after substitution $ x\mapsto \frac{1}{x}$.

Because of $ \arctan\frac{1}{x}\plus{}\arctan x\equal{}\frac{\pi}{2}$ we can write

$ 2I\equal{}\frac{\pi}{2} \, \underbrace{\int_0^\infty \frac{2x}{1\plus{}x^2\plus{}x^4}\, dx}_{\frac{2\pi}{3\sqrt{3}}}$ . So $ I\equal{}\frac{\pi^2}{6\sqrt{3}}$ .

Dr Sonnhard Graubner, what is this numerical method you keep using? Is it just a program's approximation? :S

From substitution $ x\mapsto \frac{1}{x}$ and $ \arctan x\plus{}\arctan \frac{1}{x}\equal{}\frac{\pi}{2}$ it follows

$ \int_0^\infty \frac{\arctan x}{1\plus{}2\cos\alpha\, \,x \plus{}x^2}\, dx\equal{}\frac{\pi}{4}\int_0^\infty \frac{1}{1\plus{}2\cos\alpha\, \, x \plus{}x^2}\, dx$

$ \equal{}\frac{\pi}{4} \left[\frac{1}{\sin\alpha}\, \arctan\frac{x\plus{}\cos \alpha}{\sin\alpha}\right] _0^\infty\equal{}\frac{\pi}{4}\, \frac{\alpha}{\sin\alpha}$ .

And similarly from substitution $ x\mapsto 1\minus{}x$ and $ \arcsin\sqrt{x}\plus{}\arcsin\sqrt{1\minus{}x}\equal{}\frac{\pi}{2}$ it follows

$ \int_0^1 \frac{\arcsin \sqrt{x}}{1\minus{}\left(2\sin\frac{\alpha}{2}\right)^2\, x\, (1\minus{}x)}\, dx\equal{}\frac{\pi}{4} \int_0^1 \frac{1}{1\minus{}\left(2\sin\frac{\alpha}{2}\right)^2\, x\, (1\minus{}x)}\, dx$

$ \equal{}\frac{\pi}{4} \left[\frac{1}{\sin\alpha}\, \arctan\left((2x\minus{}1)\, \tan\frac{\alpha}{2}\right)\right]_0^1\equal{}\frac{\pi}{4}\, \frac{\alpha}{\sin\alpha}$ .

[quote]But this is easily dealt with using the fact that $ \int_0^af(x)dx \equal{} \int_0^af(a \minus{} x)dx$[/quote]

I don't have the slightest clue how I would go about proving this statement. I've tried writing out both sides as anti-derivatives, riemann sums and even trying to apply squeeze theorem, but to no avail. My intuition and a bit of graphing also tells me that the f(x) and f(a-x) on [0,a] are symmetric about some axis (possibly a/2).

Any hints?

Edit: @boxede: letting x = a-x (in order to find the two functions's intersection points) yields x = a/2, which is the axis of symmetry I mentioned earlier... but then what?

Edit 2: Nvm, I think I got it.

$ \int_{0}^{a} f(a \minus{} x) \; dx \equal{} \minus{} \int_{a}^{0} f(u) \; du \equal{} \int_{0}^{a} f(u) \; du \equal{} \int_{0}^{a} f(x) \; dx$

Hint for prime factorization: what happens if you substitute a-x for x?

Integral 1-another solution.

The formula $ \int_b^af(x)dx$ =$ \frac{1}{2}(\int_b^af(x)dx + \int_b^af(a-x)dx$) ,  with f(x) =   $ \frac{sin^{-1}\sqrt{x}}{x^2-x+1}$ and   a=1 ,  b=0  means that integral no1 can be expressed as

${ \frac{1}{2}(\int_0^1\frac{sin^{-1}\sqrt{x} +sin^{-1}\sqrt{1-x}}{x^2-x+1}}dx)$.............(1)    (    f(1-x) =  $ \frac{sin^{-1}\sqrt{1-x}}{x^2-x+1}$     because $ (1-x)^{2} -(1-x) +1 = x^{2}-x +1$  ).

To simplify the numerator  in .....(1) ,we use the formula

$ sin^{-1}p + sin^{-1}q$ = $ sin^{-1}$($ p\sqrt{1-q^2}+q\sqrt{1-p^2}$).Taking p=$ \sqrt{x}$ and q= $ \sqrt{1-x}$ we get

$ \sin^{-1}\sqrt{x} +\sin^{-1}\sqrt{1-x}$ = $ \sin^{-1}$($ \sqrt{x}.\sqrt{x} +\sqrt{1-x}.\sqrt{1-x}$) =$ \sin^{-1}(x+1-x)$ =$ \sin^{-1}(1)$ =$ \pi/2$.

The integral ......(1) therefore becomes $ \frac{\pi}{4}(\int_0^1\frac{1}{x^2-x+1})dx$ which is a standard form.  Routine steps lead to the answer  $ \frac{\pi^2}{6\sqrt{3}}$

On a scale of negative four to four, how would you rate the difficulty of these problems with respect to a freshman at college? Just a bit curious about what you think. Be back with solution as soon as I can find one.

Well 2 weeks I enter my second year at Waterloo, and I havn't come across integrals like these. That is, including special functions like the Polylog or Li(x) (not that this one showed up). Hopefully that lets you interpret them yourself.


Source: 2001 GA Varsity Math Contest

42. CD is the diameter of a semicircle with center O. If AB=OD, find angle BAO.

[img]http://mars.walagata.com/w/bthings-2000/Var2001-42.JPG[/img]

[hide="solution"]
Construct radius $BO$. Since $OD=OE=OB$ and $AB=OD$, $AB=BO$. So we have 2 isosceles triangles $ABO$ and $BOE$. Set $\angle{BAO}=x$ and solve. You get
$x+(180-4x)+45=180$
$45-3x=0$
$3x=45$
So $\angle{BAO}=15$ :D [/hide]

[hide]Construct radius OB. $\triangle BAO$ and $\triangle BOE$ are Isosceles.
$\angle EBO=2\angle BAO=\angle BEO$. (exterior angle= sum of remote interior angles)
$3\angle BAO=45\implies \angle BAO=15$ degrees.[/hide]


If $p\ge 1$ and $a,b,c$ are positive real numbers, then \[\frac{1}{a\sqrt{a^{2}+pbc}}+\frac{1}{b\sqrt{b^{2}+pca}}+\frac{1}{c\sqrt{c^{2}+pab}}\le\frac{1}{\sqrt{1+p}}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)\]

$\frac{(p+1)^{2}}{a^{2}+pbc}\le \frac{1}{a^{2}}+\frac{p}{bc}$
summing up we find that:
$\sum \frac{p+1}{a^{2}+pbc}\le \sum \frac{1}{a^{2}}$
then it's easy with cauchy.
the inequality is true for all positive p.

[quote="Albanian Eagle"]$\frac{(p+1)^{2}}{a^{2}+pbc}\le \frac{1}{a^{2}}+\frac{p}{bc}$[/quote]

Much nicer than mine!


HAPPY NEW YEAR

and a great 2007! hopefully!  :lol: [/quote]


algebra
polynomial
inequalities
real analysis
real analysis unsolved

Assume $ f(x)\equal{}(x\minus{}a_{1})(x\minus{}a_{2})...(x\minus{}a_{n})$

where $ a_{i} \in \mathbb{R}$

If $ (x\minus{}a)^2 | f'(x)$

show that: $ (x\minus{}a) | f(x)$

That is obviously false: take $ f(x) \equal{} (x\minus{}1)^3 \plus{} 1$.

You meant if $ (x\minus{}a)^2 \mid f(x)$ then $ (x\minus{}a) \mid f'(x)$, which is one of the most trivial results on polynomials.

But don't we want the polynomial with all real roots? The one u posted as contradiction has 2 complex

Sorry, you're right. I apologize for skipping that part of the statement. As is it's true.

The intended proof probably goes along these lines. Let $ \displaystyle P(x) \equal{} \alpha\prod_{i\equal{}1}^n (x\minus{}a_i)$, with real roots $ a_i$, be such that $ (x\minus{}a)^2 \mid P'(x)$, hence $ (x\minus{}a) \mid P''(x)$. One has, for $ x\neq a_i$, $ \displaystyle \frac {P'(x)} {P(x)} \equal{} \sum_{i\equal{}1}^n \frac {1} {x\minus{}a_i}$, therefore, by differentiating, $ \displaystyle \frac {P(x)P''(x) \minus{} P'(x)^2} {P(x)^2} \equal{} \minus{}\sum_{i\equal{}1}^n \frac {1} {(x\minus{}a_i)^2} < 0$, hence $ P(x)P''(x) < P'(x)^2$.
Assume $ (x\minus{}a) \nmid P(x)$; then $ a\neq a_i$, and $ x\equal{}a$ in the above inequality yields $ 0 < 0$, absurd.

In fact we can prove slightly more: if $ (x\minus{}a)^k \mid P'(x)$ for some integer $ k \geq 2$, then $ (x\minus{}a)^{k\plus{}1} \mid P(x)$.
We may assume wlog $ a\equal{}0$ (otherwise we can work with $ Q(x) \equal{} P(x\plus{}a)$), and so $ x^k \mid P'(x)$.
With $ \displaystyle P(x) \equal{} \alpha\left(\sum_{i\equal{}0}^{n\minus{}1} c_ix^i \plus{} x^n\right )$ this implies $ c_i \equal{} 0$ for $ 1\leq i \leq k$, in particular $ c_1 \equal{} c_2 \equal{} 0$.
Assuming $ a_i\neq 0$, this implies $ \displaystyle \sum_{1\leq i \leq n} \frac {1} {a_i} \equal{} 0$ and $ \displaystyle \sum_{1\leq i < j \leq n} \frac {1} {a_ia_j} \equal{} 0$. But $ \displaystyle \sum_{1\leq i \leq n} \frac {1} {a_i^2} \equal{} \Big ( \sum_{1\leq i \leq n} \frac {1} {a_i} \Big )^2 \minus{} 2 \hspace{\minus{}7pt} \sum_{1\leq i < j \leq n} \frac {1} {a_ia_j} \equal{} 0$, whence $ \displaystyle \frac {1} {a_i} \equal{} 0$, absurd.
So $ P(0)\equal{}0$, hence $ c_0 \equal{} 0$, and thus $ x^{k\plus{}1} \mid P(x)$.


AMC AIME calculus geometry USA(J)MO USAMO perimeter

I was just wondering how much you guys study each day?
I mean...how many hours do you guys spend on homework, studying for tests and competitions, ...

Thx!

I think there's a post about this somewhere...

I study generally 3-4 hours each night, mostly because of AP biology.

In general, 1 hour/night/subject. But I'm in middle school, which will probably making the studying time a little different.

EDIT: HW usually takes 30 min/subject.

I do like <1 hour on school homework.
I didn't really do much competition math in the last few weeks...maybe like 2-3 hours per week in the week before state?

I'm trying to do like 1-2 hours on weekdays and like 3+ on weekends to practice for Nationals.

HW supposedly takes 3.5 hrs, but i finish it in like an hour. Subtract like 30 min for minesweeper in taht :D

I've been prepping for AIME and since my dismal fialure due to lack of problem reading and sleeping for like 2 hrs before AIME I don't really have much to do...

time to prep for next year

So I'm trying around 3 hrs on weekdays per day doing AoPS v2 and like 7-8 hrs on weekends

Hmm...
I guess it depends.
I'd like to spend more time, but thus far, that doesn't seem to be working so well.

Homework takes me...a while....But that's because I'm prepping for the bell curve system in high school. I [i]have[/i] to top everyone else. :lol: It takes me around 1-3 hours. Average is about 0.3-1.12 hours.

Math homework takes the majority of the time..[size=75]..ironically. I hate exercises. Then there's the whole draw everything neatly with diagrams and...yeah. :wallbash_red: Then I get yelled at the next day for not gdoing it neatly enough. Meh. It takes around a half hour. Unless we get assigned most of the problems plus a worksheet. Then it takes 45 mins to an hour. Why does it take so long?!?!....[/end rant][/size] Then I've got short, miscellaneous assignments from other subjects that take anywhere from 5-15 minutes.

The remaining time is spent on working through math problems. There's about a half an hour for this. However, lately, I've been letting my homework get a little sloppier to ensure more time for math homework.

On the weekends, I finish the long-term assignments (mostly on Saturday morning). I usually use the afternoons and evenings and Sunday morning to do math stuff and pinano/flute/piccolo/other instruments. Weekends are usually dominated by as much math stuff I can cram in.

[size=75]Unfortunately, I have been buried under homework lately, so there's been no math stuff as of late. :wallbash_red:[/size]

....

....Then there's eating and sleeping and computer/AoPS and fun stuff/free time.

Why do you ask, anyways? :huh:

ummm generally an hour, but if my english teacher feels like really giving us hw, about 2-3hours plus violin practice time... plus extra math time..... so pretty much 2-3 hours every night if i include everything i have to do..... :harhar:

Zero hours studying for tests.
I'm always distracted when doing homework, so I have no idea how long it takes. However, after 2AM I get undistracted, and whenever I start at 2, I end at 3:30 to 4:00, so I'm guessing 2 hours about...
Next year it might get a lot tougher (hoping to skip Chem H and go to AP, which is like the hardest school class there is).

probably about 6-8 hours per homework, going upwards to 12 hours for cs assignments. On the other hand, this is the only learning I do at all outside of class, and I only have assignments every couple weeks, so it isn't that bad.

I did around 8 hours (or more) during the Olympiad season (February - July). About 2-3 hours in the rest of the time (some days even nothing at all).

I did 9 hours college math 8 months straight (75% things I was not inscribed for), then crashed :P

now, I study whenever I feel like it, much more relaxed :)

I normally study for about 2 hours a day, give or take. Also, when I have a test I normally study for twice as long.

I usually do homework 1-2 hours per day, on weekdays about 1-2 hours of training for MC nats and going on this site. I use the rest for leisure time.
On weekends I try to do lots of stuff, sometimes invite a friend over, play the saxophone, and I generally go over the MC tests I have taken to see if I completely understand how to do all of it. I also do an additional 2-3 hours per day of MC training over the weekends. On Sundays, I usually have a soccer game, and I starting to ref games.

not that much b/c MC is over and i hardly have homework b/c i normally do it during the class or during the class before... :lol:. But I DO have to study for my SAT II and my Calculus stuff....

So about 1.5 hrs a day.

I spend less than an hour on all of the schoolwork combined...hmmm

1-3 hours of homework (but I get some done in class when I finish stuff), never study for tests (except the period before it, I just look through my notes).

Enriched Geometry/Algebra I (take both) - often tests, daily homework, but all easy - I have a ~105 in both
[insert state name] Studies - my worst class (94%, going down), lots of homework, brutal tests
Earth Science - easy, barely any homework, short, easy quizzes
Tech Ed - no homework, no tests, in-class hands-on activities
English - fairly easy tests, but lots of homework
US History - weekly homework, mostly in-class notetaking, middle level tests
Computer Apps/Gym - nothing to be said here

The rest of my time goes to studying extra math.

hm....

your life seems awfully busy PoP.

yet you still have time to go on AoPS.

Amazing.

Poor evilhamster.

Your social studies teacher seems almost as mean as my evil 4th grade eacher who assigned a gazillion book reports.  Elementary school was SO much harder than middle school(where I am right now).  You had to write in cursive, were expected to do neat work, and you are constantly given projects that you have to do at home with your own time.  And they barely even count for a grade!  BUT MY PARENTS ALWAYS CHECK THEM AND MADE ME SPEND SO MUCH TIME ON THEM!  :censored:

But in middle school, tests are pretty easy.  For example(in social studies) here are a few questions on the test for "culture"(notice the constrast from evilhamster's tests)

1. How does culture change?
A.  When people want it to
B.  through cultural diffusion
C.  through my long lectures
D.  I don't care

2.  What is the number 1 rule in my class?(Hint:  look at the board and see the "Respect" sign and the #1 next to it)
A. Chewing Gum
B. Respect
C. These tests are boring.
D. None of these

and so on..........

and polkadotman1, I assume that's a state standardized test.

[quote="polkadotman1"]Do you study for like MEAP or something (idk if you have MEAP in ur state)?[/quote]

1. What's MEAP?
2. No. Obviously.

[quote="maybach"]and polkadotman1, I assume that's a state standardized test.[/quote]

yes, yes, that is, and i hate it.

we have to do it becuase bush signed a no child left behind contract (i bet ernie agrees with me XD).

that's why we have to do it.

Am i answering jjfun1's question too?

But standardized tests are so easy!

I get like perfects on them every year...............

I know but i hate how the essays have to be so long.

I know what to write but it takes too long to write

Yeah.  Those graders always took points off my essays for no reason until this year, when I finally got 6/6.  It's ironic how students say how "hard" the math section was.  It looked something like this(calculators allowed)
1. What is 25% of 80?  Record your answer on the grid-in sheet.
2. What's the perimeter of a square with a side length of 4?
a. 8
b. 12
c. 16
d. 32

and so on........

[quote="polkadotman1"]
we have to do it becuase bush signed a no child left behind contract (i bet ernie agrees with me XD).
[/quote]
what happened to the "no child left twiddling their thumbs" contract?

Ah!

I didn't know that.

My sis said that we started having to do the MEAP when Bush signed the NCLB

[quote="Maybach"]Standardized tests have been around a LONG time before NCLB.

Now, they've just become more frequently used.[/quote]
but standardized test are getting increasingly dumber because of NCLB
as a result, americans are getting dumber ( by fourth grade, a student must master one-digit addition because that's when bush learned to do it.)
the end of the world is near...

haha.

i heard a rumor that bush only got into yale because his parents had to give yale like 5 million dollars or something.

he really isn't that smart.

Hey i am an engineer student. In my Exam days i study about 15 hour in a day.

i'm not in middle school yet!

like 1-2hrs a day :)  :)  :)


topology
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

A ring homomorphism $ f: A \to B$ induces a map $ ^af: Spec B \to Spec A$ under which a point $ p \in Spec A$ has an inverse image $ ^a f^{ \minus{} 1}(p) \equal{} \{P \in Spec B \ \mid \ P \cap A \equal{} p \}$ which is homeomorphic to $ Spec(B \otimes_A k(p))$. Setting $ C \equal{} B \otimes_A k(p)$, and $ S \equal{} A \minus{} p$, and defining $ g: B \to C$ by $ g(b) \equal{} b \otimes 1$, then we have: $ C \equal{} B \otimes_A (A/p) \otimes_A A_s \equal{} (B/pB)_S \equal{} (B/pB)_{f(S)}$

I don't quiet see the step where localization at $ S$ and $ f(S)$ coincides. And $ f(S) \subset B$, so the last thing is a $ B \minus{}$module again, weird! I need some clarification...

thanks in advance

B/pB is a B-module, so you can localize at f(S), but also a A-module via f, so you can localize at S. using explicit contructions, they clearly conicide.


convert 4 nitro toluene to 3 nitro toluene

[hide="My method"]
1. 4-nitro toluene + Sn/HCl --> 4-methyl aniline
2. 4-methyl aniline + $ \ce{Ac_{2}O}$ --> 4-methyl acetanilide
3. 4-methyl acetanilide + $ \ce{HNO_3}$ + $ \ce{H_2 SO_4}$ --> 4-methyl 2-nitro acetanilide
4. 4-methyl 2-nitro acetanilide + $ \ce{H_3 O^\plus{}}$ --> 4-methyl 2-nitro aniline
5. 4-methyl 2-nitro aniline + $ \ce{NaNO_2}$/$ \ce{HCl}$  followed by $ \ce{H_3 PO_2}$ -->
3-nitro toluene
[/hide]

yep. thats one of doing it. ne other methods??

[hide]1. KCN, ethanol

2. Schmidt rearrangement

3. DMDO.

Drawback: possibly not very high yield in first step.[/hide]


Let a,b,c>0 such that $ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \frac{b^2}{a\plus{}b^2}\plus{}\frac{c^2}{b\plus{}c^2}\plus{}\frac{a^2}{c\plus{}a^2}\geq \frac{3}{4}$

Note that
\[ \sum_{cyc}\frac{b^2}{a\plus{}b^2}\ge\frac{(a\plus{}b\plus{}c)}{a\plus{}b\plus{}c\plus{}a^2\plus{}b^2\plus{}c^2}\equal{}\frac1{1\plus{}a^2\plus{}b^2\plus{}c^2}.
\] It remains to prove that $ \frac1{1\plus{}a^2\plus{}b^2\plus{}c^2}\ge\frac34$, i.e., $ 1\ge a^2\plus{}b^2\plus{}c^2$, i.e., $ (a\plus{}b\plus{}c)^2\ge a^2\plus{}b^2\plus{}c^2$, which is obvious.

I think you re' wrong.
$ \frac{1}{1\plus{}a^2\plus{}b^2\plus{}c^2}\geq 3/4 \Leftrightarrow 3(a^2\plus{}b^2\plus{}c^2)\leq 1$. That is'nt correct.

It 's here
http://www.mathlinks.ro/viewtopic.php?p=1300078#1300078

[quote="mathwizarddude"][quote="joh"]$ \[ \sum_{cyc}\frac {b^2}{a \plus{} b^2}\ge\frac {(a \plus{} b \plus{} c)}{a \plus{} b \plus{} c \plus{} a^2 \plus{} b^2 \plus{} c^2}$[/quote]
How did you get the above inequality?[/quote]
we can get that by using Cauchy-Swars(I don't know how exactly it's written); and I think he meant (a+b+c)^2, not a+b+c


vector
algebra
polynomial
induction
combinatorics unsolved
combinatorics

Let $ \{(A_1, B_1), \ldots, (A_m,B_m) \}$ be a family of pairs of subsets of $ [n]$ such that $ |A_i| \equal{} a, |B_i| \equal{} b \quad \forall i,$ and
\[ A_i \cap B_j \equal{} \begin{cases} \neq \emptyset & \text{if } 1 \leq i < j \leq m, \\
\equal{} \emptyset & \text{if } 1 \leq i \equal{} j \leq m \end{cases}
\]
Then $ m \leq \binom{a \plus{} b}{a}.$ If $ n \geq a \plus{} b,$ the bound is the best possible.

This is not really due to Bollobás - see http://gilkalai.wordpress.com/2008/12/25/lovaszs-two-families-theorem/#more-856 part 4. However, the proof presented there is not really a proof (it works only in the case $ n\equal{}k\plus{}l$, in which the Lovasz two families theorem is still interesting, but the original problem becomes trivial). Anyone with a solution?

  darij

Choose distinct real numbers $ t_1, t_2, \dots, t_n$.  Take $ {\bf{v}}_k : \equal{} \left(1,t_k,t_k^2,\dots,t_k^a\right) \in \mathbb{R}^{a \plus{} 1}$ for $ k \equal{} 1,2,\dots,n$.  Thus, any $ a \plus{} 1$ of the $ {\bf{v}}_k$'s are linearly independent (i.e., the set of $ {\bf{v}}_k$'s is [i]in general position[/i]).

For each $ i$, consider $ \mathcal{V}_i : \equal{} {\mathrm{span}}\left\{{\bf{v}}_k \,|\, k \in A_i\right\}$.  Thus, $ \mathcal{V}_i$ is an $ a$-hyperplane subspace of $ \mathbb{R}^{a \plus{} 1}$.  Let $ \boldsymbol{\alpha}_i$ be a normal vector to $ \mathcal{V}_i$ (which is unique up to a constant factor).  We define, for all $ B_j$, the polynomial
\[ p_j\left(\mathbf{x}\right) : \equal{} \prod_{\mu \in B_j} \left(\mathbf{v}_\mu\cdot\mathbf{x}\right)\,,
\]
for all $ \mathbf{x} \equal{} \left(x_0,x_1,\dots,x_a\right) \in \mathbb{R}^{a \plus{} 1}$.  Hence, $ A_i \cap B_j \neq \emptyset$ if and only if, for some $ \nu \in [n]$, $ \nu \in A_i$ and $ \nu \in B_j$.  Equivalently, $ \boldsymbol{\alpha}_i \perp \mathbf{v}_\nu$ or $ p_j\left(\boldsymbol{\alpha}_i\right) \equal{} 0$.  On the other hand, $ A_i \cap B_j \equal{} \emptyset$ if and only if $ p_j\left(\boldsymbol{\alpha}_i\right) \neq 0$.  Consequently,
(1) $ p_j\left(\boldsymbol{\alpha}_i\right) \equal{} 0$ for $ 1 \leq i < j \leq m$, and
(2) $ p_j\left(\boldsymbol{\alpha}_i\right) \neq 0$ for $ 1 \leq i \equal{} j \leq m$.

What is implied by the above argument is that the polynomials $ p_j$'s are linearly independent.  To see this, suppose that
\[ P(\mathbf{x}) : \equal{} \sum_{\lambda \equal{} 1}^m y_\lambda p_\lambda(\mathbf{x}) \equiv 0\,,
\]
for some $ y_1,y_2,\dots,y_m\in\mathbb{R}$.  We have that
\[ y_1p_1\left(\boldsymbol{\alpha}_1\right) \equal{} P\left(\boldsymbol{\alpha}_1\right) \equal{} 0\,.
\]
Because $ p_1\left(\boldsymbol{\alpha}_1\right) \neq 0$, we have $ y_1 \equal{} 0$.  By induction, we can show that $ y_1 \equal{} y_2 \equal{} \dots \equal{} y_m \equal{} 0$, which proves our assertion.

Finally, $ p_1(\mathbf{x}),p_2(\mathbf{x}),\dots,p_m(\mathbf{x})$ are linearly independent and they belong to
\[ \mathrm{span}\left\{x_0^{r_0}x_1^{r_1}\cdots{x_a^{r_a}}\,|\, r_0,r_1,\dots,r_a \in \mathbb{N}_0 \wedge r_0 \plus{} r_1 \plus{} \dots \plus{} r_a \equal{} b\right\}\,,
\]
which has $ \binom{a \plus{} b}{a}$ dimensions.  This proves that $ m \leq \binom{a \plus{} b}{a}$.  We can show that this bound is the best possible for all $ n \geq a \plus{} b$ by checking only the case when $ n \equal{} a \plus{} b$.  This is quite easy.  Let $ M : \equal{} \binom{a \plus{} b}{a}$.  Take $ \left(A_1,A_2,\dots,A_M\right)$ to be any permutation of elements of $ \binom{[n]}{a} \equal{} \binom{[a \plus{} b]}{a}$, and $ B_j \equal{} [n]\setminus A_j \equal{} [a \plus{} b]\setminus A_j$ for $ j \equal{} 1,2,\dots,M$.  This collection $ \left\{\left(A_1, B_1\right),\left(A_2,B_2\right), \ldots, \left(A_M,B_M\right) \right\}$ clearly satisfies the condition.  (Trivially, in addition, if $ n < a \plus{} b$, no such collections exist.)

PS1: Here is a brief proof of that the $ \mathbf{v}_k$'s are in general position.  Let $ s_1,s_2,\dots,s_N$ be any $ N$ real numbers.  We have
\[ \det\left(\begin{array}{ccccc} 1 & s_1 & s_1^2 & \hdots & s_1^{N \minus{} 1} \\
1 & s_2 & s_2^2 & \hdots & s_2^{N \minus{} 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & s_N & s_N^2 & \hdots & s_N^{N \minus{} 1} \end{array}\right) \equal{} \prod_{1\leq i < j \leq N} \left(s_j \minus{} s_i\right)\,.
\]
PS2: Yes, $ [n] \equal{} \{1,2,\dots,n\}$.

$ [n] \equal{} \{1, 2, ..., n\}$?

(I am reading the proof now, so don't worry if I am saying something stupid)


Let's have some definitive statistics on this point.


function
quadratics
algebra unsolved
algebra

Find all monotne(not neccessarily strictly) functions $ f : R_ \plus{} \cup {0} \mapsto R$ such that
$ f(x \plus{} y) \minus{} f(x) \minus{} f(y) \equal{} f(xy \plus{} 1) \minus{} f(xy) \minus{} f(1)$ for all nonegative real numbers $ x,y$ and
$ f(3) \plus{} 3f(1) \equal{} 3f(2) \plus{} f(0)$

Define $ g(x)\equal{}f(x\plus{}1)\minus{}f(x)$ and we use $ c_i$ for constants.
(1) $ g(x\plus{}1)\equal{}g(x)\plus{}c_9,\forall x\ge0$
$ P(x,2): g(x\plus{}1)\plus{}g(x)\plus{}c_1\equal{}g(2x)$ (F1)
$ P(x,4): g(4x)\equal{}g(x)\plus{}g(x\plus{}1)\plus{}g(x\plus{}2)\plus{}g(x\plus{}3)\plus{}c_2$
$ g(4x)\equal{}g(2x)\plus{}g(2x\plus{}4)\plus{}c_3$ or $ g(2x)\equal{}g(x)\plus{}g(x\plus{}4)\plus{}c_3$
Compare to (F1), we get
$ g(x\plus{}4)\equal{}g(x\plus{}1)\plus{}c_5$ or $ g(x\plus{}3)\equal{}g(x)\plus{}c_5,\forall x\ge1$ (F2)

For any $ x>0$, we can find integer $ n$ such that $ 3nx\ge1$.
compare $ P(x\plus{}1,3n)$ and $ P(x,3n)$ and apply (F2), we can derive
$ g(x\plus{}3)\equal{}g(x)\plus{}c_5,\forall x>0$
$ P(x,6): g(6x)\equal{}\sum_{i\equal{}0}^5g(x\plus{}i)\plus{}c_6\\\equal{}2(g(x)\plus{}g(x\plus{}1)\plus{}g(x\plus{}2))\plus{}c_7\equal{}2g(3x)\plus{}c_8,\forall x>0$
or
$ g(2x)\equal{}2g(x)\plus{}c_8,\forall x>0$.

Compare to (F1),
$ g(x\plus{}1)\equal{}g(x)\plus{}c_9,\forall x>0$
The other condition is equivalent to $ g(2)\minus{}g(1)\equal{}g(1)\minus{}g(0)$, which guarantees the equality also holds for $ x\equal{}0$.
Hence, $ g(x\plus{}1)\equal{}g(x)\plus{}c_9,\forall x\ge0$ (F3)

(2) $ g(x)\equal{}ax\plus{}b,\forall x$.
$ P(x,n): g(nx)\equal{}\sum_{i\equal{}0}^{n\minus{}1}g(x\plus{}i)\plus{}f(1)\minus{}f(n)\equal{}\sum_{i\equal{}0}^{n\minus{}1}g(x\plus{}i)\minus{}\sum_{i\equal{}1}^{n\minus{}1}g(i)
\\\equal{}ng(x)\minus{}(n\minus{}1)g(0)$
or
$ g(nx)\minus{}g(0)\equal{}n(g(x)\minus{}g(0)$.

Now $ g(rx)\minus{}g(0)\equal{}r(g(x)\minus{}g(0)),r\in\mathbb Q$.
Let $ a\equal{}g(1)\minus{}g(0)$ and $ b\equal{}g(0)$, we have $ g(r)\equal{}b\plus{}ar$ and $ c_9\equal{}g(1)\minus{}g(0)\equal{}a$
Also,
$ g(x\plus{}r)\plus{}b\equal{}r(g(\frac xr\plus{}1)\plus{}b)\equal{}r( g(\frac xr)\plus{}a\plus{}b)\equal{}g(x)\plus{}ar\plus{}b\equal{}g(x)\plus{}g(r)$
Monotonicity of $ f$ implies either $ g(x)\ge0,\forall x$ or $ g(x)\le0,\forall x$.
It is not hard to show $ g(x)\equal{}ax\plus{}b,\forall x$.

(3) $ f$ is quadratic polynomial.
Put back to the f.e.,
$ f(x\plus{}y)\minus{}f(x)\minus{}f(y)\equal{}axy\plus{}b\minus{}f(1)\equal{}axy\minus{}f(0)$
Let $ h(x)\equal{}f(x)\minus{}(\frac a2x^2\plus{}f(0))$, we see $ h$ satisfy Cauchy equation and $ h(x)\plus{}\frac a2x^2$ is monotone, say non-decreasing.

$ h(r)\equal{}rh(1),\forall r\in\mathbb Q$.
For any $ x>0$, we find positive rational sequences $ r_n\uparrow x,R_n\downarrow x$ and
$ h(x)\ge h(r_n)\plus{}\frac a2r_n^2\minus{}\frac a2x^2$
and
$ h(x)\le h(R_n)\plus{}\frac a2R_n^2\minus{}\frac a2x^2$
Let $ n$ go to infinity, we get $ h(x)\equal{}xh(1),\forall x$.

Conclusion, $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ with $ ab\ge0$

[i]Sorry for my stupidity but I cannot understand the second step in xxp2000's proof

      From his post we have :  $ g(x \plus{} r) \plus{} b \equal{} g(x) \plus{} g(r) \ \forall \ x \geq 0 ; \forall \ r \in \mathbb{Q^ \plus{} }$

   But I fail to prove $ g(x) \equal{} ax \plus{} b \ \forall \ x \geq 0$ :(

           Can anybody help me ? , I'm 90% sure that this solution is correct because xxp2000 is very good at Algebra  :|
[/i]


function
limit
calculus
derivative
real analysis
real analysis unsolved

I'm not entirely sure this goes in computations and tutorials, but here goes:

Suppose $T(x)$ is a continous function differentiable in $a$. For every $n$ let $m$ be the integer satisfying $\frac{m}{2^n} \le a < \frac{m+1}{2^n}$. Let $I_n$ be the interval $[m2^{-n},(m+1)2^{-n}]$. This interval has length $2^{-n}$, which tends to zero as n gets large, and it contains $a$. Since $T$ is differentiable in a then
$d_n = \frac{T((m+1)2^{-n} - T(m2^{-n})}{2^{-n}}$
has the limit $T'(a)$ as $n\to\infty$.

Can someone please show me why this is so? I don't understand it.

Well, it was a correct topic for Computations and Tutorials. I still wonder why it was moved to Unsolved Problems. Basically, we need to prove that if $x_n\le a\le y_n$ for all $n$ and $y_n-x_n\to 0$ as $n\to\infty$, then $\lim_{n\to\infty}\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(a)$, provided that the derivative at $a$ exists. But, using the definition of the derivative, we can write $f(y_n)=f(a)+f'(a)(y_n-a)+o(y_n-a)=f(a)+f'(a)(y_n-a)+o(y_n-x_n)$ (recall that $0\le y_n-a\le y_n-x_n$!) and similarly for $f(x_n)$. Subtracting and dividing by $y_n-x_n$, we get $\frac{f(y_n)-f(x_n)}{y_n-x_n}=f'(a)+\frac{o(y_n-x_n)}{y_n-x_n}=f'(a)+o(1)$.

Thanks! I've never actually seen that definition of differentiable used before (it was only quickly mentioned), but I guess it will be later in multivariable calculus.


Find the $ y$-intercept point of the line $ 3x\plus{}5y\equal{}20$.  Provide your answer as an ordered pair.

Expressing the line in slope-intercept form, \begin{align*}5y&=-3x+20\\y&=-\frac{3}{5}x+4..\end{align*} The intercept is thus $ \boxed{(0,4)}.$

[hide="Alternately"]
We know that the $ x$-coordinate of the $ y$-intercept is $ 0$.  Substituting, and solving for $ y$
\[ 3(0) \plus{} 5y \equal{} 20\Longrightarrow y \equal{} 4\]
Thus, our answer is $ (0,4)$.  [/hide]

We should put this in slope intercept form. So y=-3/5x+4

The y intercept is when x is 0 that means that y=4

the answer is (0,4).


are one of the best bands ever.  Even if they're lyrics are a little repetitive.

One of the best bands [i][u]ever???[/u][/i]

Well, not ever.  Bu their still pretty good.

just what do you mean by saying they have repetitive lyrics?
 of course not.
 they have exaaaactly   :mad:  the same lyrics for everything :
  mathematicians at heart you see  :D
 $Track A = Track B = ...$


Quadrilateral $ ABCD$ is a rectangle, and its length is twice its width.  The perimeter of the rectangle is 60 units.  Points $ P$, $ Q$, $ R$ and $ S$ are midpoints of each side.  What is the area of quadrilateral $ PQRS$?

[asy]size(4cm,4cm);

draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0));

dot((1,0));
label("$R$",(1,0),S);

dot((2,0.5));
label("$Q$",(2,0.5),E);

dot((1,1));
label("$P$",(1,1),N);

dot((0,0.5));
label("$S$",(0,0.5),W);

label("$A$",(0,1),N);
label("$B$",(2,1),N);
label("$C$",(2,0),S);
label("$D$",(0,0),S);[/asy]

AD+AB=30, AB=2*AD, so AB=20, AD=10. PQRS is one half the area of ABCD, so the area is 200/2=100


function
integration
calculus
complex numbers
calculus computations

Find with proof the range of $ x$ within which

$ S_x\equal{}\frac{1}{1^x}\plus{}\frac{1}{2^x}\plus{}\frac{1}{3^x}\plus{}\frac{1}{4^x}\plus{}.....$ has limit.

[hide="answer"] x>1[/hide]

[quote="Stephen"][quote="stephencheng"]Find with proof the range of $ x$ within which

$ S_x \equal{} \frac {1}{1^x} \plus{} \frac {1}{2^x} \plus{} \frac {1}{3^x} \plus{} \frac {1}{4^x} \plus{} .....$ has limit.

[hide="answer"] x>1[/hide][/quote]

Why?[/quote]

Note that this is an [b]answer[/b], not a solution, which serves as a hint.

I am still waiting for your solution. :D

$ S_x$ is the definition of the [url=http://en.wikipedia.org/wiki/Riemann_zeta_function]Riemann zeta function[/url], if that helps.

Is anyone interested in this question? There is no solution in the link.

[hide="solution"]

It's easy to see that for $ x \le 1$ , $ S_x$ do not have limit.

Now for fixed $ x >1$ and positive integers $ n \ge 2$,

$ \int_{n\minus{}1}^{n} \frac{1}{y^x}dy > \int_{n\minus{}1}^{n} \frac{1}{n^x} dy\equal{}\frac{1}{n^x}$

$ \Rightarrow S_x\minus{}1 < \int_{1}^{2} \frac{1}{y^x}dy\plus{}\int_{2}^{3} \frac{1}{y^x}dy\plus{}\int_{3}^{4} \frac{1}{y^x}dy\plus{}.....$
$ \equal{}\int_{1}^{\infty} \frac{1}{y^x}dy$
$ \equal{}\frac{1}{x\minus{}1}$

Together with the fact that $ \sum_{n\equal{}1}^{m} \frac{1}{m^x}$ is strictly increasing as $ m$ increases gives $ S_x$ has limit for $ x>1$

[/hide]

Calculus problems belong in the calculus fora.  Also, to correct a minor error by an earlier commentor: this sum gives the Riemann zeta function when the real part of $ x$ is greater than 1.  When the real part of $ x$ is less than or equal to 1, this sum diverges, but the function $ \zeta$ is still defined on that region (except actually at the point $ x \equal{} 1$) and given by a different expression.  The key concept here is [i]analytic continuation[/i].

This is exactly the usual statement of the so-called $ p$-series test, which is as far as I know universally taught in introductory calculus courses.  Anyway, I'm not a big fan of the integral test; use [url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Cauchy condensation[/url].

Edit:  You also didn't actually establish that the indefinite integral you wrote down converged.


A square and an equilateral triangle are both inscribed on a circle of radius 1. One of the sides of the triangle is parallel to two sides of the square. What is the area of the region which they overlap?

It should be $ \frac{9\sqrt{2}+2\sqrt{6}-6\sqrt{3}}{6}$ (i think...)

This is way too computational heavy...
but you can find the side lengths of the square and triangle using 30-60-90...

then you can use 30-60-90 to find the areas of 3 triangles that are part of the equilateral triangle but do not intersect the square..then sutract those areas from the equilateral triangles...

the full solution withh be to latex heavy..

can anyone confirm that answer?

Just want to be sure im right because there was so much computation.

You can edit your previous post (within some fairly small window of time) using the "edit" button, thus avoiding double-posts.  Your answer is correct.