function
algebra
polynomial
quadratics
complex analysis
complex numbers
complex analysis unsolved

Let $f$ be an entire function such that for any compact interval $[a,b]$ the maximum value of $f$ over $[a,b]$ is attained at one of the endpoints. Then what is the form of $f$?

It's entire but it takes real values on the real line (you say "maximum")? I think it's a bit unclear.

I forgot the absolute value. In fact $|f|$ attains the maximal value. Sorry.

One more stupid question, and then I'm done :): the intervals $[a,b]$ are on the real line? The notation $[a,b]=\{ta+(1-t)b\ |\ t\in[0,1]\}$, where $a,b$ are points in $\mathbb R^n$ is also common.

Let's not exagerate, the problem is difficult enough with real intervals.  :P  All the intervals are of real numbers here.

I was looking through some old AMM issues, and I found the following problem:

AMM 6279 - proposed by Lee A. Rubel

Let $f$ be an entire function with the property that the maximum of $|f|$ over any closed segment $L$ is achieved in one of the endpoints of $L$. Prove that $f$ is either of the form $f(z)=A(z-B)^{n}$ or $f(z)=Ae^{Bz}$.

I believe this is the problem you're referring to. The use of the word "segment" suggests that we are, in fact, concerned with segments in the plane, and not just real intervals, as you said. Moreover, the conclusion would not hold if "segment" simply meant "real interval" (contemplate the counterexample $z\mapsto e^{z^{2}}$). Besides, if instead of your version with real intervals we consider this one, then the conditions are more restrictive, so the functions enjoying this property should logically be [i]easier[/i] to classify, not harder :). Anyway, the problem in the previous paragraph is what I'll try to solve.

There are two cases: either $f$ has a zero (in which case if it is non-constant, which we assume from now on, it has precisely one zero), or it doesn't.

[i]Case I[/i]: $f$ has a zero

After translating the plane, we can assume that $f$ has the form $z\mapsto z^{n}e^{h(z)}$, where $h$ is entire. Extracting the $n$'th root changes nothing, so suppose $f(z)=ze^{g(z)}$, where $g=\frac{h}{n}$. The condition on $f$ implies that for all $z,\ |f(tz)|$ increases as $t$ increases along the non-negative real axis. This means that for all non-zero $z$ and all $t\ge 1$ we have $|ze^{g(z)}|=|f(z)|\le|f(tz)|=t|ze^{g(tz)}|$. After dividing this by $|z|$, we get $|e^{g(z)-g(tz)}|\le t$. The LHS is a bounded entire function, so it must be constant, meaning that $z\mapsto g(z)-g(tz)$ is constant (and equal to $0$; just plug in $z=0$) as well. We have thus obtained $g(z)=g(tz),\ \forall z,\ \forall t\ge 1$. Clearly, this implies that $g$ is constant, so $f(z)=Az$. After undoing all the transformations we performed on $f$, we see that the most general form of $f$ in case I is indeed $f(z)=A(z-B)^{n}$.

[i]Case II[/i]: $f$ has no zeros

We have $f(z)=e^{g(z)}$ for some entire (non-constant) function $g$. Let $u$ be the real part of $g$. We have $|f(z)|=e^{u(z)}$, so $g$ has the property that for each segment in the plane, $u$ reaches its maximum in one of the endpoints of the segment. The aim is to prove that this implies that $g$ is a polynomial of degree one.

For every $\alpha\in\mathbb R$, the set $u^{-1}(\alpha)$ is the boundary of the convex region $u^{-1}((-\infty,\alpha))$ (so it's a convex curve which does not intersect itself). Choose $\alpha$ such that $g'$ does not vanish on the curve $u^{-1}(\alpha)$. Then $g$ is conformal on this curve, and since its restriction to the curve takes values on the vertical line $\{z\in\mathbb C\ |\ \mbox{Re}~z=\alpha\},\ g$ is one-to-one on $u^{-1}(\alpha)$. This means that every complex number with real part $\alpha$ is taken as the image of $g$ at most once. Now, Picard's Theorem says that a non-polynomial entire function takes all its values infinitely often, with only one possible exception, so $g$ must certainly be a polynomial. It can't have degree $\ge 2$, because a polynomial which is at least quadratic takes all its values more than once with only finitely many exceptions, so we're done: $g$ is a linear function.


I have seen people post screenshots of what they see for problems or other things. How do you do them? Do you have a special program? Is it online? Thanks.

Above the place where you write messages, there's a place where you can "Add an Attachment," so when you click on it, it will instruct you on how to put an attachment in. Do note that bitmap files don't work on AoPS.

I meant how to get the actual picture created. BTW, when I click Add Attachment, nothing happens (maybe it is my computer's problem).

There is a button on most keyboards left of "Scroll Lock" and usually above "Insert"...it looks like this:

$ \frac {\text{Print Scr}}{\text{Sys Rq}}$

I call it the "Syzerk" button.  :rotfl:

Go to an image-editing device (I use the "ghetto" Microsoft Paint) and save the screen. People usually crop of the area outside of the window, or hold the $ \text{Alt}$ button while pressing the print-screen command so that privacy is insured. When you add the attachment, you could then import the file to the website.

[quote="r15s11z55y89w21"] and save the screen.[/quote]

Wait, how do you "save the screen"? Does paint have an option?

Ok, put it this way:

1. press Print Screen (Print Scr or Prt sc, depending on the keyboard). This places the screenshot of your screen in clipboard (equivalent to "copy")
2. open MS Paint or any other program for picture previewing editing (IrfanView, Fireworks etc)
3. paste the picture from clipboard on the "canvas"
4. Now do with it whatever you like, it's avilable for editing, and you can save it just like you'd save any other picture. Only keep in mind that you really don't need a bitmap (bmp) file, it's too big and not supported as a type of attachment on some fora (including MathLinks), so save it as a jpg if you don't mind the quality loss.


geometry circumcircle inradius perimeter

[quote="Centy"]Show that for $\triangle ABC$ with side lengths $a,b,c$, circumradius $R$, inradius $r$ and semi-perimeter $s$: \[ s^2 + r^2 + 4Rr = ab + bc + ca  \][/quote]

just put everything interms of area and sides, then just sides, and it follows after some algebra


[quote="N0vak"]http://yfrog.com/7h8016aj

I got it to 18 2/3 by  prolonging the denominator to 60. But when I look at the key answer it says 7 2/3[/quote]

[hide="fraction"]

$ \frac 5 6\plus{}\frac {27} {10}\plus{}\frac {11}{15}\plus{}\frac{17}{5}$

$ \equal{}\frac{50\plus{}162\plus{}44\plus{}204}{60}\\
\equal{}\frac{460}{60}\\
\\
\equal{}\frac {23}{3}\\
\\
\equal{}7\frac 2 3$
[/hide]

[quote="PKMathew"][quote="N0vak"]http://yfrog.com/7h8016aj

I got it to 18 2/3 by  prolonging the denominator to 60. But when I look at the key answer it says 7 2/3[/quote]

[hide="fraction"]

$ \frac 5 6 \plus{} \frac {27} {10} \plus{} \frac {11}{15} \plus{} \frac {17}{5}$

$ \equal{} \frac {50 \plus{} 162 \plus{} 44 \plus{} 204}{60} \\
\equal{} \frac {460}{60} \\
 \\
\equal{} \frac {23}{3} \\
 \\
\equal{} 7\frac 2 3$
[/hide][/quote]

Thank you. You are my hero!
I didn't think of prolonging 2,7 to  27/10. Great.


function
calculus
derivative
real analysis
real analysis unsolved

Does someone know a function which is (totally) differentiable, but its partial derivatives are discontinuous. The classical example is \[f(x)=x^{2}sin(1/x), \ f(0)=0\] But I'm looking for some other examples. Or are they very rare? It doesn't have to be an example of a function $f: \mathbb{R}\to \mathbb{R}$

[quote]are they very rare?[/quote]
Of course, you can take any continuously differentiable function and add $x^{2}\sin(1/x)$ to it. But this is probably not what you had in mind. There is a good reason why elementary functions of this kind involve $\sin(1/x)$ or its equivalent. Here it is:

[i]Theorem. Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ is differentiable at every point of $\mathbb{R}$. Then $f'$ satisfies the conclusion of the Intermediate Value Theorem. In other words, for any $a<b$ and any number $t$ between $f'(a)$ and $f'(b)$ there exists $c\in [a,b]$ such that $f'(c)=t$.[/i]

This is not very hard to prove, actually. The theorem implies that for $f'$ to be discontinuous, it must oscillate rapidly near its points of discontinuity. And this is where $\sin(1/x)$ comes in.

Derivatives can get very complicated in higher dimensions. Buczolich recently (2002) constructed a differentiable function $f\colon\mathbb{R}^{2}\to\mathbb{R}$ such that $\nabla f(0,0)=0$ but $|\nabla f(x,y)|\ge 1$ for almost every point $(x,y)$. [It took him 15 years.] Very recently, Maly and Zeleny found a different construction based on the following game:

Let $D$ be an open disk in the plane. Player A chooses a point in $D$. Then player B chooses a line passing through that point. Then player A chooses a point on that line (and still in $D$). Player B chooses a line passing through the last point, etc etc. Player A wins if the sequence of points diverges; B wins if it converges.

[i]Theorem [Maly-Zeleny]. Player B can always win.[/i]

Can you believe this?? (The [url=http://www.karlin.mff.cuni.cz/~rokyta/preprint/2005-pap/2005-160.ps]proof[/url] is accessible to professional mathematicians, but probably not to amateurs.)

Nice. Thanks for looking up that information. I shall take a look at it in a few years ;)  It looks pretty complicated.


Hello,

We have integers $ a$ and $ b\quad (a,b>0)$. We are looking for smallest possible integer $ t>0$, such that:

\[ \bigvee_{x>0}a^{(x+t)}\equiv a^{x}\quad (mod\quad{10^{b}})\]

($ x$ is also integer)

Is there any easy way to find $ t$?

Best regards,

tytus

For a given $ b > 3$ and over all integers $ a$ (for $ x > b$), the answer is given by the Carmichael function $ \lambda(m)$, where $ m$ is the modulus.  In this case,

$ \lambda(10^{b}) = \frac{10^{b}}{20}$.

(I should also give the particular values $ \lambda(10) = 4, \lambda(100) = 20, \lambda(1000) = 100$. )

For a particular integer $ a$, it's an open problem.  Whatever $ t$ is, it divides $ \lambda(10^{b})$.

Hello,

OK, but is there really no way to find $ t$ in easy way? If $ b$ is big $ (b=100)$ trying all $ t$, such that $ t|\lambda(10^{b})$ can be really hard.

Some examples for $ b=4$ (if I'm not wrong  :blush: ):
$ a=43\quad t=100$
$ a=49\quad t=50$
$ a=57\quad t=20$

Of course we can see that each time $ t$ divides $ \lambda({10^{b}})=\lambda({10^{4}})=\frac{10000}{20}=500$, but it was not easy to find it (and we have to remember that $ b$ is only $ 4$).

[quote="tytus"]OK, but is there really no way to find $ t$ in easy way? If $ b$ is big $ (b=100)$ trying all $ t$, such that $ t|\lambda(10^{b})$ can be really hard.[/quote]

There is no formula, if that's what you're asking.  That's why it's an open problem.

There are algorithms.  The problem is known as the order-finding problem.  One method is to repeatedly square $ a$.  Modular squaring is fairly fast, and once you have two repeated values $ a^{2^{i}}, a^{2^{j}}$, you know that the order divides $ gcd(2^{j}-2^{i}, \lambda(10^{b}))$.

Interesting, there's a quantum algorithm for this problem, but that's probably not of much practical interest.

Hello,

I understand what does it mean "problem is open"... :wink:

But as there is no *easy* way to find $ t$, maybe I have done something wrong...

Original problem is as follows:
Given integers $ a$ and $ b$ $ (a,b>0)$ find such $ t$, that
\[ \bigvee_{x>0}10^{b}|(a^{(x+t)}-a^{x}) \]
($ x$ is integer)
(by $ b|a$ I mean $ a$ is divisible by $ b\rightarrow a\: mod\: b=0$)

I thought that it is like:
\[ \bigvee_{x>0}10^{b}|(a^{(x+t)}-a^{x}) \leftrightarrow \bigvee_{x>0}a^{(x+t)}\equiv a^{x}\quad(mod\quad 10^{b}) \]
Where am I wrong? And of course - is there any *easy* way to solve original problem?

Best regards,
tytus

[quote="tytus"]Given integers $ a$ and $ b$ $ (a,b>0)$ find such $ t$, that
\[ \bigvee_{x>0}10^{b}|(a^{(x+t)}-a^{x}) \][/quote]

Well, this doesn't say the smallest $ t$, so letting $ t = \lambda(10^{b})$ should be fine.  On the other hand, if $ gcd(a, 10^{b}) > 1$ it's possible that $ t$ doesn't exist when $ x = 1$.

Hello,

Sorry - my mistake - we are looking for the smallest possible $ t$ of course... :blush:

Best regards,
tytus


If $a,b$ are positive integers such that $a^{2}+b^{2}/(1+ab)$ is an integer, then show that $a^{2}+b^{2}/(1+ab)$
must be a square.


vector
calculus
articles
Support
geometry
geometric transformation
reflection

[url=http://technocrat.net/d/2007/4/3/17225]Link to article.[/url]

Thoughts?

My thought is that the link is either broken or under maintainance; I just get an HTTP 500 error.

[url=http://www.kuro5hin.org/story/2007/4/1/22185/22601]Another link to the same op-ed.[/url]

My main objection: mathematics is better compared to a foreign language rather than the native tongue. The author should have compared and contrasted ESL and math classes. I personally was taught English in a way very similar to what is described in the article. And I hated it. But the knowledge acquired in grade school helped a lot when I studied English later, in preparation for graduate school in the USA. (Somehow, I failed to consider this scenario when I was in grade school.) Score one for the force-feeding of knowledge to students who totally don't need it.

But I share the author's skepticism toward Real World Applications. Here's one example from "Vector Calculus" by Marsden and Tromba (a good textbook overall, by the way):

[quote]A restaurant is being built on the side of a mountain which is described by equation $x^{2}+y^{2}+z=4R^{2}$. The restaurant is to be cylindrical with radius $R$. Its vertical wall (to be built of glass) has equation $x^{2}+(y-R)^{2}=R^{2}$.
(a) What is the surface area of this wall?
(b) To be large enough to be profitable, the consulting engineer informs the developer that the volume of the interior must exceed $\pi R^{4}/2$. For what $R$ does the proposed structure satisfy this requirement?
(c) During a typical summer day, the environs of the restaurant are subject to a temperature field $T(x,y,z)=3x^{2}+(y-R)^{2}+16z^{2}$. A heat flux density $V=-k\nabla T$ through all sides of the restaurant produces a heat flux. What is this total heat flux?[/quote]
How kewl is this? I mean, it's all about business, profitability and stuff. :cool:

Actually I completely disagree with this article :)

I think that many students consider math classes as most practical and relevant classes, especially since they see that what they learned in Math has applications in Physics, Computer Science/Programming and other areas.

On the other hand, I have an impression that students in general do not consider English classes (or Russian classes in Russia) too exciting since what they study there is nuances of punctuation and grammar.

A quote (I think  from P AM Dirac)
In math  paper you write about something which no one knew  before in a language which [is clear enough that] every one can understand , while in English (poetry ?) its just the opposite..:)

Another reason why English is not considered useful
is because
1) English is a language. Languages are spoken/used in order to communicate. Once you get to a point of being able to communicate your ideas to someone else; you really won't have to learn much more.
2) Math provides a challenge to solve. English really provides nothing but easy challenges in grammar.
3) There are other ways to communicate ideas than language. Such as music (yeah i know. It's a corny example), math (could be... :roll: ), and others

[quote]Once you get to a point of being able to communicate your ideas to someone else; you really won't have to learn much more.[/quote]
But is this point often reached in secondary school? Can we expect a randomly chosen high school graduate to be able to write a clear and convincing 500 word article, or deliver an effective 10 minute speech?

true true
but doesn't that also have something to do with the audience :ninja:

The opinion of English here is the opinion of a whole bunch of kids who love the maths/sciences and particularly dislike English studies... biased much? Most of you are comparing doing grammar worksheets to doing contest level problems and then saying that after a certain point the grammar worksheets are useless but the contest problems will always be useful. A closer parallel would be that a grammar worksheet is the equivalent of solving a page full of linear equations that have no context or interesting quirks. All of us would find these equally tiresome. Go study some humanities courses or read some classic English literature (or Russian, or French, etc.) and come back here and tell me that learning English in school has less of a benefit than learning math. As for kids thinking math is the most practical and relevant class, go tutor some kids who struggle in math and see how they react to that proposition.

I didn't say math was practical or the most useful
I just said it brings a challenge
I haven't seen many English-related work that required me to think for 2 hours

I'm not bashing English either
I'm not too bad at it (Score of 620 on the New SAT's Writing section and a Score of 9 out of 12 on the actual essay)
I did prepare for it
but
I prepared for it by just writing and writing and writing...
It got boring
but I did it because I wanted a good SAT score (I took it in 7th grade for those who don't know)

It is true that I'm a bit bias toward math -_- :ninja:

No worries I wasn't directing that at you in particular, actually, you hadn't even posted in the thread yet while I was typing that up  :P . Think of it this way though, you practiced for the English on the SAT, how boring would it be to practice for the math on the SAT for 99% of the people on this board? My point is that there are equivalents of the difficulty of math that is done on this forum in the language studies, but the fact that most of us aren't very interested in that level of English means that we don't study at that level. It's like when artsies say that math is sterile, boring and repetitive, they haven't been exposed to our math. It's all a matter of opinion.

I agree with you 100% now :thumbup:

Question:
What would be equivalent to math done on this forum that's done in Language studies?
just curious
I know there is a physics olympiad (but I heard it's much easier to make than the math olympiad)

Hmm, I thought I'd come back to this thread even though it's old.
I want to make the point that after having gone to college, I consider the English language to be the least of my concerns.
For one, it's main usage is communication and argumentation.
Argumentation, as many professors never state in the second college semester of English, primarily belongs to philosophy.

[quote="max_tm"]I'm not particularly well versed in the humanities  :blush: but I'm taking the A-level English Literature exams this year (both years condensed into one  :mad:) and I'm making a C (to be fair, I am doing two years of it at the same time :P) average right now, whereas in my regular English class I can maintain a 90 average no problem. To get an A I'm sure you'd need to know, at a pretty intimate level, the philosophies of the writers in the particular literary movement of the particular passage that you're analyzing and the implications of that and the historical context etc. etc. I could see how it would be stimulating to somebody who is interested in that kind of stuff.[/quote]

That's called environmental determinism. I suspect your "teacher" did not "teach" you very well. You'll get a lot of that from people in college. I don't like English or any of the subjective things in college, because sometimes the teacher doesn't "teach" very well, and because you lack knowledge that the professor knows but never told, you end getting a bad grade. Welcome to hell and the wrongfulness that is human existence, life, and human ignorance. Enjoy the stay. You'll have to learn that you may be strong, but others are weak, and they'll down you regardless, leaving you wondering if you are bad or if your circumstances are bad. Hence, the time we live in allows ignorant people to be biased, thus lacking a sense of empirical order for the standardized system of education that has been created. Do you see how someone from the future could look back and now environmentally determine your situation?

Now, English is just a language, such as Spanish, Japanese, etc...
However, math belongs to chemistry, physics, and more.

Math vs. English = Math wins
Math vs. Language = ? wins

Math is a language.
It's a language of logic.

College English stops focusing on logic, which is grammar, and starts focusing on rhetoric (persuasion, coherence, psychological influence of emotions via usage of language).
So, if you don't know grammar, you might be screwed.

I truly consider English to be the least of my concerns.
You'll realize this once you determine your major.
And science is my major, not English.
And I swear to God, I'd off myself if my major was English. Because that doesn't bring in money.

Not only these things, but from a capitalistic perspective, weilding language only does so much good.
But building devices, discovering things, and finding cures, does so much more than some witty language usage could ever do.

So, I hope you'll all realize that you're right about how learning English is the most pointless dribble in school.
But, I also want you to know something.
You cannot get far without mastering English, too.

You don't have to like it, but master as much of it you can.
Why?

Because when I'm becoming a scientist, I'll have to write documents, express my finding, and persuade people to give me research money.
But I think the most important part is expressing findings and knowing how to communicate.
It's called technical writing. If you dislike Shakespeare (get a time machine and assinate the bast*** for all I care), then you'll like technical writing.

But one last thing...
I may not have written this so well without weilding words and language.

Eh, English and Math aren't the same things, we don't use math to communicate with others do we?
Anyways, English taught at our school doesn't have any real life applications either.

I hope I'm not sounding like an authority for anything below; I certainly haven't delved deep enough into English nor Math to become anywhere close to an authority on either but I recognize that both are complex.

[quote="Yury"]Actually I completely disagree with this article :)

I think that many students consider math classes as most practical and relevant classes, especially since they see that what they learned in Math has applications in Physics, Computer Science/Programming and other areas.[/quote]

Most people don't take physics and a very small minority take computer science while everyone takes English.

[quote="footballrocks41237"]Personally I think after a certain point English is useless.  When do you need to know words like sesquipedalian...  Once you have enough to go about life, you really don't need anymore unless you plan to become an English professor or something...  Math, on the other hand, is always useful.  From construction to rocket science, math is always involved.  Most computers use binary code, and to truly understand binary you need math.

Bottom Line: Math > English[/quote]

First, you said after a certain point in English, but then used all of mathematics to support your point. You restricted English yet did not support math. If you restrict mathematics, one could make the same argument. When are you going to use complex analysis in construction or rocket science?

Next, hopefully you are learning much more in English class than grammar and spelling. In fact, in my English class we haven't done grammar or spelling at all. (although I do go to a school where the only thing emphasized is essentially English but grammar and spelling are implied. Because what is the point of grammar and spelling? To communicate your ideas effectively. Why do you use commas? To tell the reader which part of the sentence is doing what to what word.) What you learn in English is practicing yourself clearly and concisely. How is that ever useless in life?

[quote="Aznwithabrain"]I haven't seen many English-related work that required me to think for 2 hours[/quote]

Have you ever written an analytical essay?

[quote="Aznwithabrain"]I'm not bashing English either
I'm not too bad at it (Score of 620 on the New SAT's Writing section and a Score of 9 out of 12 on the actual essay)[/quote]

Saying the SAT for English is a comprehensive evaluation in English is like saying the SAT for Math is a comprehensive evaluation in Math, I bet you'd disagree with the latter, but only because you are familiar with the math problems that require creativity as well as elegance.

[quote="illusiat"]creativity is what makes english harder than (school) math. in english u have to write essays and use ur creativity. In math, u can just mem stuff :D[/quote]

Solving high-level contest problems does require creativity.

The skill-sets required for English and math are pretty much the same when you immerse yourself deep enough into them. What are you doing in math? Taking a variety of theorems that you have learned from a wide range of areas of math and putting them together to prove a point. You have logical steps so a reader can follow you clearly. You need to learn theorems and the names of them to communicate what you're doing. What are you doing in English? Proving a point using little bits and pieces of text from a variety of places and connecting them clearly and effectively. Paragraphs are your stepping stones, building on each other or proving different aspects of what you're trying to say. Then, your conclusion is for exploring the implications of your point being true. You need to learn grammar and spelling to communicate your point effectively.

From my experiences, English and Math are taught in a way such that the high-scoring student does not necessarily learn thinking and cleverness, but rather, methods and expectations.

After Junior High, most of English focuses on the analytical (and the occasional argumentative or reflective) essay. Read a book rich with threads to analyze. Find threads. Weave into coherent picture.

Similarly, math classes are also goal-oriented. Take given information--an expression, a geometric figure with some given statements, a dataset--and derive something useful from it--the value of a variable, some fact about the figure, the standard error of the dataset.

But to get good grades in either subject, the student figures out the course's expectations, and that is what English, Math, and school, in general, teaches. When I read Henry IV, certainly, I appreciated the language, but I also thought about what topics the teacher would approve of in an analytical essay. When I write that essay, I recall the format that this particular teacher prefers and follow it. I might have become somewhat of a better reader, but my grades reflect

In math, I marveled at proofs and questioned the importance of various fundamental theorems, just as idealists would like. However, in the end, the good grade comes from knowing what types of problems will show up on the exam, and how to attack them. Math tests in school courses are either limited in scope or limited in depth: there can either be three difficult problems, but from only 1 chapter, or 60 quick and obvious problems that cover half a year. When scope is liimted, I know that the teacher will test only certain methods--if necessary, I could simply try every method. When depth is limited, the grade comes from recognizing the "trap" (if any), avoiding it, and answering correctly.

Yeah. Teachers are creatures of habit, and their habit simplifies students' lives without strengthening their minds. I suppose they could make their teaching like the local math league: in every unit exam, there can be exactly one problem from outside the unit. That might encourage flexibility, to some extent.


Solve the following equation in integers $ x,y$: $ x^2\plus{}6xy\plus{}8y^2\plus{}3x\plus{}6y\equal{}2$

$ x^2\plus{}6xy\plus{}8y^2\plus{}3x\plus{}6y\equal{}(x\plus{}2y)(x\plus{}4y\plus{}2)\equal{}2$ therefore we have 4 posibilities:
$ x\plus{}2y\equal{}1$ and $ x\plus{}4y\plus{}3\equal{}2$ leading to $ x\equal{}3,y\equal{}\minus{}1$
$ x\plus{}2y\equal{}2$ and $ x\plus{}4y\plus{}3\equal{}1$ leading to $ x\equal{}6,y\equal{}\minus{}2$
$ x\plus{}2y\equal{}\minus{}1$ and $ x\plus{}4y\plus{}3\equal{}\minus{}2$ leading to $ x\equal{}3,y\equal{}\minus{}2$
$ x\plus{}2y\equal{}\minus{}2$ and $ x\plus{}4y\plus{}3\equal{}\minus{}1$ leading to $ x\equal{}0,y\equal{}\minus{}1$

i hope i claculated them right :D

There is an applet which can solve such problems with details: http://www.alpertron.com.ar/QUAD.HTM

[quote="Jure the frEEEk"]$ x^2 \plus{} 6xy \plus{} 8y^2 \plus{} 3x \plus{} 6y \equal{} (x \plus{} 2y)(x \plus{} 4y \plus{} 2) \equal{} 2$ therefore we have 4 posibilities:
$ x \plus{} 2y \equal{} 1$ and $ x \plus{} 4y \plus{} 3 \equal{} 2$ leading to $ x \equal{} 3,y \equal{} \minus{} 1$
$ x \plus{} 2y \equal{} 2$ and $ x \plus{} 4y \plus{} 3 \equal{} 1$ leading to $ x \equal{} 6,y \equal{} \minus{} 2$
$ x \plus{} 2y \equal{} \minus{} 1$ and $ x \plus{} 4y \plus{} 3 \equal{} \minus{} 2$ leading to $ x \equal{} 3,y \equal{} \minus{} 2$
$ x \plus{} 2y \equal{} \minus{} 2$ and $ x \plus{} 4y \plus{} 3 \equal{} \minus{} 1$ leading to $ x \equal{} 0,y \equal{} \minus{} 1$

i hope i claculated them right :D[/quote]

There's a slight mistake. You should have factored as

$ (x \plus{} 2y)(x \plus{} 4y \plus{} 3)$


Let $ \triangle ABC$, known that: $ sin2A\plus{}sin2B\equal{}4sinA.sinB$. Prove that $ ABC$ is a right triangle.

[quote="thanhnam2902"]Let $ \triangle ABC$, known that: $ sin2A \plus{} sin2B \equal{} 4sinA.sinB$. Prove that $ ABC$ is a right triangle.[/quote]

[hide="stuck"]$ f(x) \equal{} \sin x$ is concave over $ (0,\pi)$, and $ 0 < \angle A ,\angle B, \angle C < \pi$. So
\[ f\left(\frac {2A \plus{} 2B}{2}\right)\ge \frac {f(2A) \plus{} f(2B)}{2} \equal{} 2f(A)f(B)
\]
So $ \sin(A \plus{} B)\ge 2\sin A \sin B$, or $ \sin C\ge 2\sin A \sin B$.

Also $ \frac {\sin C}{\sin A} \equal{} \frac {c}{a}$ so $ \frac {c}{a}\ge 2\sin B$.

Then $ c\ge 2a\sin B$ or $ \frac {c^2}{4}\ge \frac {ac\sin B}{2} \equal{} [ABC]$

Then $ c^2\ge 4[ABC] \equal{} 2ab\sin C$ or $ a^2 \plus{} b^2 \minus{} 2ab\cos C\ge 2ab\sin C$, or $ a^2 \plus{} b^2 \minus{} 2ab(\cos C \plus{} \sin C)\ge 0$ which implies $ 90\le C < 180$...

Then I'm stuck...I tried AM-GM so that $ (\sin A \plus{} \sin B)^2\ge \sin 2A \plus{} \sin 2B$ to try to show $ 0 < C\le 90$ but I couldn't get anywhere...=/[/hide]

[hide="Hint"]
Try to use properties of triangles and prove that one of the angles is $ 90^{\circ}$.
[/hide]

[hide="Incomplete Solution"]

$ \sin2A \plus{} \sin2B \equal{} 4\sin A \sin B$

By product to sum formula:

$ \sin2A \plus{} \sin2B \equal{} 2\cos(A\minus{}B) \minus{} 2\cos(A\plus{}B)$

And using sum to product formula on the left side and dividing by 2:

$ \sin(A\plus{}B)\cos(A\minus{}B) \equal{} \cos(A\minus{}B) \minus{} \cos(A\plus{}B)$

and thus we can easily check that this will always hold true when
$ A\plus{}B\equal{}\frac{\pi}{2}$

What's remaining is to prove the "only if" of the problem.

[/hide]


A few people have posted questions on this forum about whether they are "worthy" or smart enough to be mathematicians, so I guess I'm really not alone in sometimes feeling inadequate about my abilities.

I've reached the stage where I'm sure that it is what I want to do, and I've got my PhD and a good postdoc lined up,.   But still I feel like maybe I can't do it.  The feeling comes and goes.  But some days I feel so overwhelmed by how much there is to learn, and how slow I seem to be at understanding things and how much smarter everyone else seems to be.  Does anyone have any advice on how to cheer up on days like this?  Perhaps you have developed a point of view that makes it easier?

I can confess that I feel quite miserable about my abilities most of the time despite I've been in the business for 20 years and I cannot say that my record is totally pitiful (being a research mathematician is quite a depressing job; you can be certain that what you know is just a tiny part of what is there and most of the time you fail to solve the problem you really want to solve). The good way out is to do something useful that requires just a fraction of your current strength (but not too small fraction): teaching a course, solving some simpler problems, doing some consulting, hanging on AoPS, etc. If you feel really bad, work in the garden, clean the house, play tennis, whatever, just have yourself occupied with some meaningful or, at least, "pseudomeaningful" activity for a while that requires sufficient amount of concentration but little thinking. It won't be long until you feel an urge to think of math. again (if you really like it). Then a few hours of work, another realization of your total idiocy, and back to the beginning of this loop :lol:.

As to "everyone else being smarter", it is a meaningless sentence. Of course, Terry Tao can solve some problems you cannot but you may be pretty sure that the converse is true as well. Everybody has their strengths and weaknesses. Also, comparing yourself with other people is rather useless most of the time. You'd better compare yourself to what you've been a year ago; this tells much more about yourself. As the same Terry Tao put it in his blog (though the idea itself is a few millenia old), just try to choose tasks slightly above your current abilities all the time and you'll be fine.

I think this is very sound guidance for *anything*.  It certainly describes a lot of my experience in building AoPS, at the very least.

The right answer:  You should go work on math some more.  The only way to address our inadequacies is to face them head on and remove them.

My actual answer:  I go on AoPS and write rambling posts directed at high school students.  Also, this one's petty, but it works for me sometimes:  I'm sure you've met at least a few modestly successful (read: more accomplished than you are) mathematicians who just don't seem as smart as you are.  If they can make it, so can you.  You just have to work harder.  At least that's what I tell myself.

Thanks very much for your answers.  Just reading them made me laugh and cheered me up a bit.  I took fedja's advice and tried working on an easier problem today than the one I'd been doing.  Thinking about something else that I could actually understand made me feel a lot better.  I will try to remember that for next time! (Or else just come and read your answers again).


algorithm
combinatorics unsolved
combinatorics

The following procedure gives a subset with no three term AP of $ \{1,2,\dots, n\}$ : Identify the element which participates in most number of three term arithmetic progressions, remove it. Then proceed iteratively, removing a element at a time till one gets a subset without a three term AP.

Prove or disprove the following:

(i) The subset thus obtained is maximal.
(ii) All maximal subsets obtained this way have the same cardinality.
(iii) All maximal subsets can be obtained by this procedure.

Things that have no good reason to be true are usually false  :wink:

(i) is false: for $ n \equal{} 9$ we can get to {1, 2, 6, 7}, which is contained in {1, 2, 6, 7, 9}.

(ii) is false: for $ n \equal{} 9$ we can get to both of the maximal sets {1, 2, 7, 8} and to {1, 2, 6, 8, 9}.

(iii) is false: for $ n \equal{} 9$ we cannot get to the maximal set {1, 2, 5, 6}.

@ JBL: That was real quick, I was trying to find some counter examples myself and could not find them easily. How did you figure them out? I am curious to know.

I wrote a small computer program to
(1) generate all AP-free subsets of a given set, and
(2) tell me for a given set how many APs each element belongs to,
then I started checking by hand.  There just aren't that many things to check: your hand is forced quite a lot even for $ n \equal{} 6, 7, 8$.

Actually I initially misunderstood "maximal" as "of maximum size," and also only started working with part (i), so it's possible that one or more of the statements has a smaller counterexample than $ n \equal{} 9$.  For example, with $ n \equal{} 5$ this algorithm outputs {1, 2, 4, 5}, so (iii) fails (we don't get {2, 3, 5}).


algebra
polynomial
function
functional equation
algebra unsolved

Find all polynomials $ p(x)$ such that:

$ (x\minus{}16)p(2x)$ $ \equal{}$ $ 16(x\minus{}1)p(x)$

The solutions can be described by the set $ \{p(x)\equal{}c(x\minus{}2)(x\minus{}4)(x\minus{}8)(x\minus{}16)\,|\,c\in\mathbb{R}\}$.

[i]Proof[/i]: The above set is contained in the solution set. Obviously, $ p\equal{}0$ is the only constant function solution. So from now on assume $ p$ to be non-constant. Comparison of the leading coefficients in the functional equation yields: $ 2^{\deg p}\equal{}16 \Longrightarrow \deg p\equal{}4$. Let $ z_1,\ldots,z_4$ denote the $ 4$ not necessarily different, complex zeroes of $ p$. Comparison of zeroes implies $ \{16,\frac{z_1}{2},\ldots,\frac{z_4}{2}\}\equal{}\{1,z_1,\ldots,z_4\}\Longrightarrow \{z_1,\ldots,z_4\}\equal{}\{2,4,8,16\}$.


I don't remember every single word, but it is like this:

A number follows the rule showed below until it becomes a single digit.

496----216----12---2

and

999---243----24----8

so we can say both 496 and 999 has a perseverance of 3.

What's the smallest number that has a perseverance of 12?

I think about it for 2 hours but I can't find a solution without using computer programs, does anyone have an idea?

[quote="wuyonglin2004"]Does any have an idea?[/quote]Have some patience. You'll have to give it some time for other people to try and solve it.

Also, you forgot to put what the rule is. The pattern seems like each successive number in a 'perseverance' is the product of the digits of the previous number, right?

In the meantime, consider my sketchy idea: think about numbers that cannot have certain factors. If there is a prime larger than 10, you can't have , say '11' or '13', as a digit since we are working in base 10. Though, a 'perseverance' of 12 can be quite... computational, I guess.

Yeah, I had no clue what the rule was.
And by perserverance, that means how many times you have to apply the rule to get a single digit number?

perseverance of 12 and perseverance of 3 differs a lot. can you provide any rules?

10000th User, You are right, the rule is "each successive number in a 'perseverance' is the product of the digits of the previous number".

 The perserverance means how many times you have to apply the rule to get a single digit number

I quote from Richard K. Guy - Unsolved Problems in Number Theory.

[quote]
... Neil Sloane defines the $ {\bf persistence}$ of a number as the number of steps before the number collapses to a single digit (after multiplying its decimal digits). ... The smallest number with persistence $ 11$ is $ 277777788888899$. There is no number less than $ 10^{50}$ with persistence greater than $ 11$. ... etc.
[/quote]

Is that true? I can't find it in that book though, which catalog is it?

Try searching for [url=http://www.research.att.com/~njas/sequences/index.html?q=persistence&language=english&go=Search]persistence[/url] in the [url=http://www.research.att.com/~njas/sequences/index.html]OEIS[/url].

In my Second Edition, it is item $ {\bf F25}$, page $ 262$.

Thanks mavropnevma and JBL, I got it. But I still don't understand why my math teacher put this on a precalculus worksheet, maybe trying to find another Carl Friedrich Gauss, haha. :lol:


number theory
greatest common divisor
algebra
function
domain
calculus
integration

Let $ R$ be an integral domain then why are the following statements equivalent:
i) every finite, non-zero set of elements of $ R$ has an hcf
ii) every finite, non-zero set of elements of $ R$ has an lcm

I understand that the above two statements are duals of each other and I read somewhere that the elements of the ring form a lattice with gcd being join and lcd being meet but I have no idea how to proceed. Any help is appreciated.

if you mean hcf = gcd, think about the identity lcm(a,b) gcd(a,b) = ab.

Ok, thanks. I think it generalizes to more than two factors by induction so I only need to consider elements two at a time.


Hello everyone, I need some help with consumer math. I have been trying all day  to figure out these units and problems that go along with them. I dont go to a normal school i go to a place called virutal high school, so it is kind of hard for me to get help when i need it. What I am looking for is emails from people that are willing to help me out.
The program i use to access my math courses is called nova net. So what i would have to do take screen shots of the problems and send them to whoever's email. I hope this is not too much trouble and i would really appreciate it if i could get some help. Thanks for your time and hopefully someone can help.

Or if someone that wants to help uses Aol instant messenger i can show the problems that way which might be easier.

okay... even though i don't get what "virtual high school" is, it still sounds really cool... u can e-mail me at LuCy4EvA@gmail.com, or my AIM sn is Luver of Rain... but please tell me that u are from AOPS when you IM me or I'll probably block you  :D

hey u can email me.
i ll try to help as much as i can!
email address:princesspriya09@yahoo.com :D

Email: point.line.square.cube.hypercube@gmail.com

Though you might be better off looking the high school forums for help on consumer math.

I don't really like the idea of revealing the e-mail to the publics..

EVERYONE, HIDE YOUR EMAILS!!!

[hide="my email"]swkcorean@gmail.com[/hide]

what's virtual highschool?

and this consumer math?

[color=blue]oo...thanks Lucy...fixed now..[/color] :D

shinwoo, i think ur e-mail was suppose to be gmail.com instead of gmai


Nine horses are in a race and the top three winners, in order, are announced: Blitzer (1); Southern Cross (2); and Smoke House (3). My uncle Roger knows nothing about horses; What is the probability of him randomly picking three horses and winning the Trifecta? (i.e. the probability of picking those exact three horses in that exact order?)

Your uncle could have picked the trifecta in any of $ 9\cdot8\cdot7\equal{}504$ ways. (That's the permutations of 9 things taken 3 at a time.) Only one of them is a winner. Hence the probability of picking a winner is $ \frac1{504}.$

I normally don't download files, and I won't download this one.

I've attached my sample final.  I've made it through to number 12.  Can someone check it out to see how I did?  I've tried to look up and understand 13-16 but still having trouble.  Thanks in advance,
MA120 Statistics
HW Assignment #16				   Name __________________________________

                       Fill in the blank or indicate the correct answer (A, B, C, or D).

Use the following sample data to answer questions 1 through 5:

	                              11, 11, 16, 6, 10, 12, 15, 7, 11,

_c____1.   The mean, , of the data above is (to the nearest 1/10)
		a.  9.0		b.9.9		c. 11.0		d.  answer not listed

___c__2.    The median of the data above is
		a.  9.0		b.  9.9		c.  11.0		d.  answer not listed

___c__3.    The mode of the data above is
		a.  9.0		b.  9.9		c.  11.0		d.  answer not listed

4.  The Mean Absolute Deviation (MAD) of this data is   ______9.33___

5.  The standard deviation of the data above is ______3.240______

The table below shows the frequency distribution for 12 students on a test with a maximum score of 20. Use this sample data to answer questions 6 through 8:

Score      Frequency      Midpoint    Midpoint x Frequency       midpoint squared x frequency
1-5          1                    3                      3                                   9
6-10        2                     8                     16                                  128
11-15      4                     13                    52                                  676
16-20      5                      18                   90                                  1620

total      12                     42                    161                                 2433

          (i was given score and frequency only, i filled in the rest)

6.   What is the mean score, , of the data in the frequency distribution above __13.417_______
7.   The standard deviation of the scores in the frequency distribution above ___4.981________

8.   The median score lies in which score interval? B
		a.  1 – 5	b.  6 – 10	c.  11 - 15	d.  16 - 20

9.  Nine horses are in a race and the top three winners, in order, are announced: Blitzer (1); Southern Cross (2); Smoke House (3).  My Uncle Roger knows nothing about horses: What is the probability of him randomly picking three horses and winning the Trifecta?  (i.e. the probability of picking those exact three horses in that exact order?) B
		a.  	b.  	c. 		d.  answer not listed

___b__10.  Suppose that =, P(A and B) = .54, and . What is ?

		a.  1.0		b.  0.36	c.   0.6		d.  answer not listed

For questions 11 through 14 use the table immediately below. The table shows the number of drinkers and nondrinkers by gender among 220 people. Assume one person is chosen at random from the 220.

                Drinkers     Nondrinkers       Totals

Male            60                40               100
Female         40                80               120
Totals         100                120             220

a   11.  If a male is chosen at random, the probability that he is a drinker is

		a.  		b.    	c. 		d.  answer not listed

__d___12.  If a drinker is chosen at random, the probability that the person is male is
		a.  	 	b.  	 	c.  		d.  answer not listed

_____13.  If one person is chosen at random from the 220 in the table above, the 			events of “Drinker” and “Male” are

		a.   independent events;
		b.   dependent events;
		c.   mutually exclusive events;
		d.   none of the above

14.  If one person is chosen at random: What is the probability the person will be Female OR a Nondrinker?  P(F or N): ________

____15.  Fantasy Pleasure Yacht Tours seat 13 passengers.  From experience, they have learned that, on average, 85% of the passengers holding reservations for a cruise actually show up.  If they book 15 passengers for a cruise, what is the probability that 13 or fewer passengers holding reservations will actually show up for the cruise?

	a.  	less than 80% 	b.  between 80 and 95%, inclusive 				c. 	more than 95%	d.  cannot be determined from information given

16.  The average number of cars going through a toll booth is 6 per minute.  If the booth can handle 12 cars per minute without forming a line.  What is the probability there
will be a line at the toll booth?  P(x > 12): _______


[quote="Vasc"]Let $a,b,c$ and $p$ be non-negative real numbers. Prove that
$\sum_{cyc}{(a-pb)(a-pc)(a-b)(a-c) \ge 0}$.[/quote]

(i)While $k=1$, the origin inequality holds obviously.
(ii)While $k<>1$, there is no harm in setting $b=a+x,c=a+x+y$, where $x>=0,y>=0$.
Then the origin inequality is equivalent to

$(k-1)^2(xy+y^2+x^2)a^2+(k-1)(y+2x)(kx^2+kxy-2y^2)a-ky^2x^2+y^4-kxy^3+k^2y^2x^2+k^2x^4+3y^2x^2+2k^2x^3y+3xy^3>=0$

When $x=0$ or $y=0$, the above inequality holds obviously.
When $xy<>0$, the discriminant of the above inequality is

$\delta=-3y^2x^2(k+2)^2(k-1)^2(x+y)^2<=0$.

Thus, the proof of the origin inequality is completed.

[quote="Yu-Dong Wu"]
When $xy<>0$, the discriminant of the above inequality is

$\delta=-3y^2x^2(k+2)^2(k-1)^2(x+y)^2<=0$.

Thus, the proof of the origin inequality is completed.[/quote]
Your calculation is correct. A little simpler is as follows.
Setting $b=a+x$, $c=a+y$, and then $t=(1-p)a$ and $q=2+p$, the given inequality is equivalent to
$At^2+Bt+C \ge 0$,
where $A=x^2-xy+y^2$, $B=(x+y)(2A-qxy)$, $C=(x+y)^2A-qxy(x+y)^2+q^2x^2y^2$.
The discriminant is
$D=-3q^2x^2y^2(x-y)^2\le 0$.
The inequality is true for all real numbers $x,y,z,p$, of course.  ;)


Two circles C and C' in the plane intersect at points A and B. A straight line passing through B intersects the circles C and C' respectively at M and N,
with B between M and N. Let I be the midpoint of the arc AM of circle C not containing B, and J the midpoint of the arc AN of circle C' not containing B.
 If K is the middle of MN, prove that KI is perpendicular to KJ.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253906[/url]


[quote="1/(ln x)"]also, does anyone know how to balance homework, debate club, and AMC/math so that you don't have to go to bed at 1 in the morning?[/quote]

i wish :( :( :(

Some of them are trivialized by the fact that you are allowed to use a calculator on AMCs, but the others are good. :)

[quote="erik-the-red"][quote="Iversonfan2005"][quote="1/(ln x)"]does anybody know if it is worth it to do old AHSME problems? when i mean old i mean 1950-1980 AHSME's.

also, does anyone know how to balance homework, debate club, and AMC/math so that you don't have to go to bed at 1 in the morning?[/quote]

Yes, ofcourse it is, it's extra practice and I think old AHSME problems are harder than the new AMC[/quote]

AMC 10 or AMC 12?[/quote]

both...atleast that's what I believe

Yea, since they didn't have the AIME, so doing well on the ASHME automatically qualified you for the USAMO.


very obvoiusly $a_n=(\dfrac {n}{n+2})a_{n-1}$ this is my help to the person who wants to solve it  :D  :D

[quote="silouan"]For the sequence $a_n$ we have that $a_1=1$ and

$a_1+a_2+.....+a_n=\dfrac{(n^2+n)a_n}{2}$ for each $n\in N*$.

Find $a_n$ as a function of $n$[/quote]

As ehsan said, we easily get $a_n=\dfrac{n}{n+2}a_{n-1}$

{just pose $n-1$ to $n$.... }

Now write $\dfrac{a_2}{a_1}=\dfrac{2}{2+2}$
               $\dfrac{a_3}{a_2}=\dfrac{3}{3+2}$
                                ........
               $\dfrac{a_n}{a_{n-1}}=\dfrac{n}{n+2}$

and multiply all.....

Actually this is an easy  problem
 :)


I'm not sure if any of you here actually do the practice sets for the PCF, but I just took a look at the Fermat practice set, and noticed something quite weird.

http://cemc.math.uwaterloo.ca/english/contests/pcf_prep/archive/fermat_practice4.pdf

The question I am looking at is #9. It does not look too complicated (at least, not the way I envisioned it). However, I cannot seem to get an answer that matches the given choices. I would like to know whether I am just plain doing it incorrectly, or there is some flaw with the problem (though I do doubt it is the latter).

dont focus on pcf's; they are not that important

what you want to do is start practicing amc's and aime's to make usamo which is much better than getting 150 on a fermat

:lol:  Thanks for the advice. I'm not sure if I can improve AMC as much now that I am in the eleventh grade, either way, I guess AMC is worth studying more for.

If anyone has the time, please look over #9. If I can't get this right, I doubt I'd do well on the AMC.

Yeah, I think something is wrong. I get $ 12(2\minus{}\sqrt{3})$ which is closest to $ 3$ cm.

Getting 150 on Fermat is better than [b]just[/b] making USAMO, because you get an actual award and a cash prize and you go to the Waterloo math camp, which is really fun, whereas making usamo but bombing it kind of gives nothing. The number of ppl in gr 11 who make usamo is much bigger than the number of people who get 150 on fermat. However, it is important to know that usamo gives you much more olympiad experience and brings you to a higher level than in fermat. Both are really good accomplishments with fermat 150 giving a greater sense of accomplishment while making usamo -- bringing you to a higher level.

Another thing -- AMC is basically the same as Fermat so if you study for AMC you automatically study for Fermat too, which is good. Note doing Fermat practice tests is not the bets way to prepare; its better to do the old contests.

Good luck to you on AMC and Fermat!

Thank you very much for the advice, Rem. The probability of me obtaining a perfect on Fermat is pretty darn close to zero. So I'll probably shoot for the latter option. Either way, I'll just work through some old contests and some problems from waterloo books. When I look at it though, it seems that AMC 12 is much harder than Fermat. So I'm guessing it's more effective to do some old AMC papers than Fermat ones. Again, thank you very much, and good luck in making the IMO team again this year  :lol:

BTW, I got the same answer (24-12/3), so I guess my thinking isn't that messed up after all.

I think so too, bryant124. Maybe I'll send an e-mail to the CEMC to inform them about this little "mistake".

Yeah, it does make more sense, for if it was diameter, the asnwer wsould be $ 6(4\minus{}\sqrt{3}) \approx 14$


Where I can find description of Crank - Nicholson method with example(s) of solutions for some differential equations?

Please dont cite books, just internet sources.

Thanks for links.

hello, i hope this will help you
http://www.ripublication.com/aama/aamav2n2_5.pdf
Sonnhard.


Let $ \xi$ be a random variable with distribution function $ F_{\xi}$ and let $ m_e$ be a median of $ F_{\xi}(x)$, i.e. a point such that

$ F_{\xi}(m_e \minus{} ) \leq 1/2 \leq F_{\xi}(m_e)$.

Show that $ \inf_{ \minus{} \infty < a < \infty} E|\xi \minus{} a| \equal{} E|\xi \minus{} m_e|$

Here is a solution:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193757[/url].


quadratics
vector
floor function
linear algebra
linear algebra unsolved

Hey guys, plz help me with the last part of question 10E (linear algebra) of paper 3 in this:
http://www.maths.cam.ac.uk/teaching/pastpapers//2008/Part_IB/index.html

(This part should be done in around 5 minutes, as the previous parts take quite long already. So it should be sth obvious but uuurrrgh... :mad: i don't know...)

Thanks a lot

It's not obvious... the Tripos is supposed to be difficult (though mostly it isn't).

Prove that if $ k$ is an algebraically closed field (of characteristic $ \neq 2$, to be safe), then any quadratic space of dimension at least $ 2$ over $ k$ has a nonzero isotropic vector (i. e. a nonzero vector $ v$ such that $ \left < v,v\right > \equal{} 0$). (Feel free to assume $ k\equal{}\mathbb{C}$ here and below.)

Prove that if $ k$ is an algebraically closed field (of characteristic $ \neq 2$, to be safe), then any quadratic space of dimension $ n$ over $ k$ has an isotropic subspace (i. e. a vector subspace consisting of isotropic vectors only) of dimension $ \left\lfloor\frac {n}{2}\right\rfloor$. (Construct such a space recursively, each time adding a new isotropic vector orthogonal to the isotropic subspace already constructed. For how long will this work?)

Conclude by induction.

  darij


A triangle with side lengths 16, 30, and 34 is inscribed inside Circle A. It is also circumscribed about Circle B. What is ratio of Circle B to Circle A?

[hide]Interestingly enough, this triangle is a right triangle. The formula for the radius of the insrcired circle is (leg1*leg2)/(perimeter).. so in this case, the radius of the inscribed circle is 6.
The radius of the outer circle is the product of all the sides divided by four times the area. so the radius is 17.

so if the question asks for area:
36/289[/hide]

[quote="erdogankerem123"]I didn't realize it was a right triangle at first, but now that you know it's a right triangle, you can say that the radius of the larger circle is 17 right away, since the hypotenuse is the diameter.
btw, can you prove those two formulas you used?[/quote]

The proof for the second formula, $r_{i}=\frac{l_{1}*l_{2}}{p}$ is simple

[hide="Proof"]

First, divide the initial triangle, say $\triangle ABC$ into 3 different triangles. These triangles should each have a base of one of the sides of $\triangle ABC$ and the respective height should be the radius of the incircle.

The areas of the three triangles can be written as $\frac{AB * r}{2}+\frac{AC * r}{2}+\frac{CB * r}{2}= \frac{(AB+BC+AC) * r}{2}$

This expression, $\frac{(AB+BC+AC) * r}{2}$ is equal to the area of the initial triangle, $\triangle ABC$.

Also, $(AB+BC+AC)$ can be replaced with $p$.

This yields the expression for area as: $\frac{p * r}{2}$

The area of $\triangle ABC$ can also be written $\frac{l_{1}*l_{2}}{2}$

Substitution gives: $\frac{l_{1}*l_{2}}{2}= \frac{p * r}{2}$

Multiplying each side by 2 and dividing by $p$ gives: $r_{i}=\frac{l_{1}*l_{2}}{p}$

[/hide]

[quote="erdogankerem123"]Nice, I hadn't tried it yet. Have you tried proving the second one yet?[/quote]

Not yet; it seems much more complicated; maybe it is in conjuction with heron's formula...

It's wrong.  Because it's a 8-15-17 rt triangle, the the longest side is the diameter, and the right angle is up against the edge of the circle.  Think about it- if a circle has to go around a triangle, the longest line you can draw in/on the triangle must be the diameter because that's what the circle stretches around.  If the diameter is 17, the radius is 8.5.

It is similar to an 8-15-17, but all of the side lengths are doubled, so the radius is 17, not 8.5.

[quote="erdogankerem123"]Nice, I hadn't tried it yet. Have you tried proving the second one yet?[/quote]

I tried proving it... but the only thing I could get is
\[r_{c}= \frac{abc}{(a+b+c)(a+b)(b+c)(c+a)}\]
where $r_{c}$ is the circumscribed radius, $a,b,c$ are the triangle's sides. Does anyone know how tor prove this?

Uh..extended laws of sines..but I dont think you can use that in this forum..Actually (abc)/(4*Area of triangle ABC)= R from extended laws of sines..and you plug in herons formula for the area of triangle ABC and you get that (as the 4s will cancel out leaving you with your equation).

[quote="magixter"]Oh, okay thanks Art of OWNA. But could you PM me with what you mean by [i]extended[/i] law of sines?[/quote]

Im pretty sure its just like regular laws of sines but you have these ratios and you cna multiply them to get that equation (for reference it is in art of problemsolving vol.2)


algebra
polynomial
linear algebra
linear algebra unsolved

For the real number $\alpha$, $\alpha$ not zero and linear operators $F$ and $G$ from $R^{n}$ to $R^{n}$ is given that $F\circ G-G \circ F=\alpha F$.
a)prove that $\forall k\in N$  $F^{k}\circ G-G\circ F^{k}=\alpha kF$.
b)prove $\exists k\in N$ sych that $F^{k}=0$.

If a) and b) were simultaneously true, then $F^{k}=0 \rightarrow GF^{k}=F^{k}G=0 \rightarrow GF^{k}-F^{k}G=0 \rightarrow F=0$. If F is 0, then the thesis is trivial, otherwise a) and b) are inconsistent.

It's obviously a mistake in the statement- I think making that $\alpha kF^{k}$ fixes (a).

Let $p(x)$ be the minimal polynomial of $F$, and consider $p(F)G-Gp(F)$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

[quote="jjchen611"]A beautiful and difficult  integral,who can help me? :?:

Ascertain $\int \frac{1}{(sinx)^5+(cosx)^5} \ dx$[/quote]

Maybe it's too difficult to anyone :

The following definite integral is weaker,but I also can't ,who can solve or give an idea :?:  :!:

Find $\int_0^{\frac{\pi}{2}} \frac{1}{(sinx)^5+(cosx)^5} \ dx$[/quote]

[quote="spx2"]you can use the formula's here at the end of the page
[url]http://aleph0.clarku.edu/~djoyce/java/trig/identities.html[/url]
for sinx depending on tanx/2
and also cosx depepnding on tanx/2

after that make substitution t=tanx/2
and you will have some thing like 1/polynomial you will decompose
in simple fractions and solve de integral[/quote]

1.Thank spx2  very much, but I don't agree with you,

2.For this integral, it is impossible to decompose
in simple fractions and solve de integra,

3.If you disagree with  me, Suppose you do  the following weaker integral:

Find $\int_0^{\frac{\pi}{2}} \frac{1}{(sinx)^3+(cosx)^3} \ {\rm d}x$

$\int \frac{1}{(sinx)^3+(cosx)^3} \ {\rm d}x$

[quote="Virgil Nicula"]Spx2, with your method you don't solve this indefinite integral !

[hide="Here ia a shorter method."]$t=\phi (x)=\sin x-\cos x\ ,\ 0<x<\frac{\pi}{2}\Longrightarrow 2\sin x\cos x=1-t^2\ ,$

$\phi '(x)=\cos x+\sin x\ ,\ [\phi '(x)]^2+[\phi (x)]^2=2\ .$

$\sin ^5x+\cos^5 x=(\sin x+\cos x)\left[1-(\sin x\cos x)-(\sin x\cos x)^2\right]\ .$

$I\equiv\int \frac{1}{\sin^5x+\cos^5 x}\ dx=\int \frac{(\sin x-\cos x)'}{(\sin x+\cos x)^2\left[1-(\sin x\cos x)-(\sin x\cos x)^2\right]}\ dx\ .$

$\boxed {\ I=F\circ \phi\ }\ ,\ F=\int \frac{4}{\left(2-t^2\right)\left[4-2\left(1-t^2\right)-\left(1-t^2\right)^2\right]}\ dt=$

$\int \frac{4}{\left(2-t^2\right)\left(1+4t^2-t^4\right)}\ dt=\ \ldots\ldots$ a.s.o. (it is easily from here).[/hide][/quote]

Virgil Nicula solve the integral in a clever way, and open up a possible approach to ascertain the following general integral which is difficult to deal with :

Let $n$ a nonnegitive integer, to find:
$\int \frac{1}{(\sin{x})^{2n+1}+(\cos{x})^{2n+1} }\ dx$ :rotfl:  :rotfl:


A music store owner bought a CD for $ \$9.00$ and sold it for $ \$12.15$. What was the percent of markup?

That's exactly what I and the player I was against in the cd both put, and apparently it's not 35.

what was the answer listed? did you guys report the question?

[quote="knittingfrenzy18"]Well I know the decimal is 0.740740....I forgot how to convert it though.[/quote]
To convert it, we know that it repeats every $740$, so we get the fraction $\frac{740}{999}=\frac{20}{27}$

Ah thanks, but I think this problem wants a percent.

[quote="knittingfrenzy18"]Well I know the decimal is 0.740740....I forgot how to convert it though.[/quote]
How did you get that? There is no explanation to how you solved this problem.

Haha. :blush: I didn't do that right. I didn't read the question carefully. Darn.

Well, this problem written in those words posted by GameBot, has a solution of 35 as solved by nikeballa96. So as PhireKaLk6781 said, the answer 35 wasn't the answer to this problem in FTW, so, this problem was just wrote out wrong.


geometry
analytic geometry
trigonometry
integration
real analysis
real analysis unsolved

Hello!
In what system of coordinates I can solve this problem?
Find volume $ V$ of volume $ G \subset R^3$ which is limited by a parametrical curve

$ \left( {\frac {{x^2 }} {{a^2 }} \plus{} \frac {{y^2 }} {{b^2 }} \plus{} \frac {{z^2 }} {{c^2 }}} \right)^2 \equal{} \frac {z} {h}\exp \left( {\frac {{\frac {{x^2 }} {{c^2 }}}} {{\frac {{x^2 }} {{a^2 }} \plus{} \frac {{y^2 }} {{b^2 }} \plus{} \frac {{z^2 }} {{c^2 }}}}} \right).$

$ a,b,c,h$ - const.

Please!

Using generalized spherical coordinates

$ \begin{gathered} \left\{\begin{gathered} x \equal{} ar\cos \left( \phi \right)\sin \left( \theta \right), \hfill \\
y \equal{} br\sin \left( \phi \right)\sin \left( \theta \right), \hfill \\
z \equal{} cr\cos \left( \theta \right). \hfill \\
\end{gathered} \right.{\text{ }}r > 0,{\text{ }}0 < \theta < \pi ,{\text{ }}0 < \phi < 2\pi \hfill \\
V \equal{} \iiint {dxdydz \equal{} abc\int\limits_0^{2\pi } {d\phi } }\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern \minus{} \nulldelimiterspace} 2}} {\sin \left( \theta \right)d\theta } \int\limits_0^{R\left( {\theta ,\phi } \right)} {r^2 } dr,{\text{ }}R\left( {\theta ,\phi } \right) \equal{} \sqrt [3]{{\frac {c} {h}\cos \left( \theta \right)e^{\left( {\frac {a} {c}} \right)^2 \sin ^2 \left( \theta \right)\cos ^2 \left( \phi \right)} }}. \hfill \\
\end{gathered}$


topology
real analysis
real analysis solved

Let $X$ be a metric space such that every sequence in $X$ has a convergent subsequence. Show that $X$ must be separable.

This is called sequential compactness. For a metric space, this is equivalent to the fact that it's totally bounded and complete. From the total boundedness, we get that we can cover the space with balls of radius $\frac 1n$, for each $n$. Now, for each $n$, choose a point in each of the finitely many balls of radius $\frac 1n$ which cover the space. The set of points you get is a countable dense subset.


Please solve the problem
Q: For what values of ‘a’ will the tangents drawn to the parabola y2 = 4ax from a point, not on the y-axis, will be normal to the parabola x2 = 4ay.

What do you mean by "will be normal to"?

Let me try to restate the problem indialover gave:

For what values of $ {a}$ will the tangents drawn to the parabola $ {y^2 \equal{} 4ax}$ from a point, not on the y-axis, will also be tangent to the parabola $ {x^2 \equal{} 4ay}$.

[hide="hint"]
Solve both equations for y, what happens?

[/hide]

[hide="further hint"]

If you know calculus, perhaps find the implicit derivatives?[/hide]

[quote="castigioni"]The tangent will be y=-x-a[/quote]

Could you please provide a full solution?

We substitute on both equations the equation of the tangent line: y=mx+p .As the line is tangent the equation in x will have only one solution.With this condition we find the values of m and p.

wait isn't the normal line the line perpendicular to the tangent line that passes through the point of tangency?

I think this should be in a higher forum, because the best way to do it is to use derivatives and stuff.


Sorry  :oops: I edited it, now please everyone continuous for b).

[hide="b"]
Suppose one of the $a$'s is negative. WLOG let $a_{n}$ be negative since the inequality is symmetric in the $a$'s. Then we have
\[a_{1}+a_{2}+\cdots+a_{n}<a_{1}+a_{2}+\cdots+a_{n-1}\]
\[\leq \sqrt{(n-1)(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n-1}^{2})}<\sqrt{(n-1)(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2})}\]
a contradiction.
[/hide]


calculus
integration
function
probability and stats

How can be the following s.i. $ \int \limits^{T}_{0} W_{1\minus{}e^{\minus{}2\theta t}}dW_t$ calculated or simplified? $ W_{\cdot}$ is Wiener process.

Are you sure that the integrand is an adapted process ?

Cause you are looking into the future as soon as $ 1 \minus{} e^{ \minus{} 2\theta t} > t$ this is the case for example when $ \theta > 1/2$ in some interval near 0, and then your integral is not well defined (unless you use Skohokhod-Type integral).

Tarasenya: is this a question of your own or an exercice ? there are very simple stochastic integrals that cannot be simplified (expressed more simply as function of well-known laws)

this question appeared in my little investigation. :ninja:


$BC=\sqrt{4^2+6^2}$  where $BC$ is the hypotenuse.
That means $2\sqrt{13}$

[quote="subhash333"]If side 1 of a right triangle is 4 and the other side is 6, then what is the hypotenuse :|[/quote]

[hide]You use the Pythagorean Theorem. a^2+b^2=c^2
a= leg 1
b= leg 2
c= hypotenuse

4^2+6^2=c^2
16+36=c^2
[size=150]square root of 52[/size] or [size=150]3 times the square root of 13[/size][/hide]

[quote="subhash333"]If side 1 of a right triangle is 4 and the other side is 6, then what is the hypotenuse :|[/quote]

[hide]Hypotenuse of the right triangle=$sqrt[4^2+6^2]$
=$sqrt[52]$
=$2$ $sqrt[13]$[/hide]

sqrt(16+36)
sqrt(52)
Using the Pythagorean Theorum......how do you pronounce that?... :?

[quote="subhash333"]If side 1 of a right triangle is 4 and the other side is 6, then what is the hypotenuse :|[/quote]

[hide="hypotenuse"]$2sqrt[2]13$[/hide]

[quote="easyas3.14159..."][quote="moogra"][quote="subhash333"]If side 1 of a right triangle is 4 and the other side is 6, then what is the hypotenuse :|[/quote]
[hide]Um... 4^2+6^2=42
root 42 would be correct. Or 6 could be the hyp if the other side was root 22... [/hide][/quote]

[hide]$4^2+6^2=52, so 2sqrt13 is correct.$[/hide][/quote]

right i did it in my head.

all you do is that you use pytogerean thereom which is a squared+ b squared= c squared (c being the hypotenuse). so in your case 4*4= 16 and 6*6=36 and  36 plus 16 is 52. now to find c you find the square root of 52 simplies to be 2 times the square root of 13

[hide="answer"]$2\sqrt{13}$[/hide]
In latex it is typed
[hide="what I typed"]2\sqrt{13}[/hide]
with the $\$...\$$ around it.

[quote="easyas3.14159..."][quote="moogra"][quote="subhash333"]If side 1 of a right triangle is 4 and the other side is 6, then what is the hypotenuse :|[/quote]
[hide]Um... 4^2+6^2=42
root 42 would be correct. Or 6 could be the hyp if the other side was root 22... [/hide][/quote]

[hide]$4^2+6^2=52, so 2sqrt13 is correct.$[/hide][/quote]
you forgot to do the square root of 52...which is [hide]$2\sqrt{13}$[/hide]

[quote="xxm0rg07hxx"]it was a problem on a amc 8 test possibly.[/quote]

Yes, surely....

a squared + b squared = c sqaured...

16+36= the sqaure root of 52

[quote="easyas3.14159..."]We've exhausted this topic.

Side note: There are 367 known proofs of the pythagorean theorem.[/quote]

That's a lot!!

[hide]
Using the Pythagorean Theorem, you get $4^2+6^2=16+36-52$. You take the square root of it and get $\sqrt{52}=2\sqrt{13}$.[/hide]

[quote="Ignite168"]This topic is over 2 months old and has A LOT of solutions already.. Stop bumping it![/quote]

Yes... please mods, lock it.

[hide]Using the Pythagorean Theorem, we have the square root of 36+16, which 2*square root of 13[/hide]

Hey guys, next time when you answer can you please hide your answer? Thank you!! :)

This is a poorly worded question. They should say that the two LEGS are 4 and 6.


A serie: $a_1,a_2,....$ is defined such that

$a_1=56$ and $a_{n+1}=a_n - \frac{1}{a_n}$

n is an integer number and $n\geq1$

Show that there's an integer k, $1 \leq k \leq 2002$, such that $a_k<0$


function
graph theory
combinatorics proposed
combinatorics

Let a set $P$ be of 2005 distinct primes. and set $A$ is consisted of Products of 1002 primes in $P$.

set $B$ is consisted of Products of 1003 primes in $P$.

Prove that there exist one to one and onto function $f:A->B $.

(For all $a \in A$ $a|f(a)$ )

Construct a bipartite graph having the elements of $A$ in one class (call it class $A$ as well, for simplicity), and the elements of $B$ in the other class (class $B$), and in which $a\in A,b\in B$ are connected iff $a\subset b$.

Each vertex in $A$ has precisely $1003$ edges, and so does each vertex in $B$. Let $M\subseteq A$. There are $|M|\cdot 1003$ edges starting from vertices of $M$, and if they have at most $|M|-1$ endpoints in $B$, then we immediately get a contradiction. This means that the vertices of $M$ are connected to at least $|M|$ vertices in $B$, so the hypotheses in Hall's Marriage Theorem hold, meaning that we can find a perfect matching in this graph (which is exactly what the function we were looking for must be).


[quote="diophantient"]What does $ (x,y,z) \in {\mathbb{N}}^3$ mean  :blush: ?[/quote]
I think: $ x \in N, y \in N, z \in N$

It means that $ (x,y,z)$ is a triple of positive integers.

$ \mathbb{N}^3 \equal{} \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ is the set of [i]ordered[/i] triples $ (x, y, z)$ such that $ x, y, z \in \mathbb{N}$.  This notation is general.  For example, $ \mathbb{R}^2 \equal{} \mathbb{R} \times \mathbb{R}$ is the set of ordered pairs $ (x, y)$ such that $ x, y \in \mathbb{R}$ (the real numbers); this can be taken to mean the Euclidean plane.

The notation $ \mathbb{A} \times \mathbb{B}$ refers to a direct product of rings, which has some additional properties, but for our purposes it means ordered pairs $ (a, b)$ such that $ a \in \mathbb{A}, b \in \mathbb{B}$.

[hide="Further discussion of direct products"] The direct product of rings is itself a ring, where operations are defined coordinate-wise.  This allows us to restate the Chinese Remainder Theorem rather elegantly in the following manner:  For a given positive integer $ n$, let $ n \equal{} p_1^{e_1} p_2^{e_2} ... p_k^{e_k}$ be its prime factorization.  Then

$ \mathbb{Z}_n \cong \mathbb{Z}_{p_1^{e_1}} \times \mathbb{Z}_{p_2^{e_2}} \times ... \times \mathbb{Z}_{p_k^{e_k}}$

In other words, the integers modulo $ n$ are isomorphic to the ordered pairs $ (a_1, a_2, ... a_k)$ where $ a_i$ is the remainder of a given integer $ \bmod p_i^{e_i}$. [/hide]


The attached figure shows two blocks, each of mass $m$, suspended From the ends of a rigid massless rod of length $L_{1}+L_{2}$, with $L_{1}= 20$cm, and $L_{2}= 80$cm. The rod is held horizontally on the flcrum and then released. What are the magnitudes of the initial accelerations of the blocks?

Any help would be appreciated.   :)

Consider the force on the objects when release.

Can you be more specific?
Can anyone provide the entire solution? PLZ? :maybe:

im still learning latex. so here is my solution. hope someone can convert it into latex.
by definition of torque and its characteristics,
t(torque)= sigma(F*r)=I(moment of inertia)*alpha(angular acceleration)
I = sigma(m*r^2) = 0.68m
Sigma(F*r) = 0.6 mg (since the two forces are opposite in direction, one clockwise and one counterclockwise)
thus,
alpha*0.68m = 0.6mg
alpha = 8.647 rad/sec^2
for linear acceleration, multiply alpha by r
a_1 = 8.647*0.2 = 1.729 m/sec^2
a_2 = 8.647*0.8 = 6.918 m/sec^2

Thanks Tong! I never expected the solution to be such a simple one.

Solution in $LaTeX$:

By definition of torque and its characteristics,
$\tau$ (torque) $= \Sigma Fr = I$ (Inertia) $\cdot \alpha$ (angular acceleration) \[I = \Sigma mr^{2}= 0.68m\] Since the two forces are opposite in direction, one clockwise and one counterclockwise: \[\Sigma Fr = 0.6mg\]  \[0.68\alpha m = 0.6mg\]  \[\alpha = 8.647 rad/sec^{2}\] For linear acceleration, multiply $\alpha$ by $r$ \[a_{1}= 8.647 \cdot 0.2 = 1.729 m/sec^{2}\]  \[a_{2}= 8.647 \cdot 0.8 = 6.918 m/sec^{2}\]


[hide]This is too easy for Pre-Olympiad. Just prove with induction that $f(n)=2\frac{f(1)}{n(n+1)}$.
This is trivial:
Ok for $n=1$, then the induction step:
\[(n^{2}-1)f(n)= f(1)\left( \frac{2}{2.1}+\frac{2}{3.2}+...+\frac{2}{n(n-1)}\right) \]

\[\iff f(n)=2f(1)\left(\frac{1-\frac{1}{n}}{n^{2}-1}\right) \]
hence we are done. The value $f(2007)$ is now within reach.[/hide]


Albatrosss' friend Caddlebot grows Apple trees. When he grows 100 trees he gets 800 apples per tree. For every tree after 100 the number of apples per tree decreases by 10. Find an equation that gives the number of apples per tree given the number of trees.

[hide="Hint"]Find a term to describe the numebr of tree more than 100[/hide]

I don't know if this is a little to hard. If you get stuck definatly read the hint.

Sorry, this is a little too hard isn't it? :blush:

I figured out the little n^2 thing at the bottom.

You want to find an equation that gives the number of apples per tree.

Sorry this is probably to hard.

[hide="Solution"]Find out the reasoning behind each of the hints.[/hide]

[hide="Hint 1"]Let x be the number of trees.[/hide]

[hide="Hint 2"]Let x-100 be the answer to A+Math's hint.[/hide]

[hide="Hint 3"]Let -10 be in your equation because of "decreases by 10".[/hide]

[hide="Hint 4"]Let -10(x-100) be in your equation for the total number of apples.[/hide]

[hide="Hint 5"]Let -10x+1000 be in your equation for the total number of apples.[/hide]

[hide="Hint 6"]Let 800-10x+1000 be in your equation for the total number of apples.[/hide]

[hide="Hint 7"]Let 1800-10x be in your equation for the total number of apples.[/hide]

[hide="Hint 8"]Let x(1800-10x) be in your equation for the total number of apples.[/hide]

[hide="Hint 9"]Let 1800x-10x^2 be in your equation for the total number of apples.[/hide]

[hide="Hint 10"]Let y=1800x-10x^2 be your equation for the total number of apples.[/hide]


Four congruent rectangles are placed as shown. The area of the outer square is $ 4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
[asy]unitsize(6mm);
defaultpen(linewidth(.8pt));

path p=(1,1)--(-2,1)--(-2,2)--(1,2);
draw(p);
draw(rotate(90)*p);
draw(rotate(180)*p);
draw(rotate(270)*p);[/asy]$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 \plus{} \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4$

let's let the area of the small square be 4. that means the area of the big square needs to be 16, so it's a 4 by 4. draw that, draw in the rectangles, and find the ratio for a 3x1 rectangle. :P

Let the rectangle have side lengths a and b. Let a > b.
(a+b)^2 = 4(a-b)^2
(a+b)^2 - (2a-2b)^2 = 0
(3a-b)(3b-a) = 0

Since a>b, 3b-a<3a-b and 3a-b must be positive. Therefore:
3b-a = 0
a/b = 3

[hide="Maybe a little faster?"]
Let the small square have side length $ x$, and the rectangle have side lengths $ a$ and $ b$.  Then just by looking at the diagram,

$ 2x\equal{}x\plus{}2b\equal{}a\plus{}b$

so $ x\equal{}2b$, $ 2x\equal{}4b$, and thus $ a\equal{}3b$.  Answer is $ A$.
[/hide]

[hide]My solution:
Let $ x$ be the longer side of the rectangle, and $ y$ the shorter.  The larger square has side length of $ x\plus{}y$ and the smaller $ x\minus{}y$.  The squares have an area ratio of 4, therefore a side ratio of 2.  $ \dfrac{x\plus{}y}{x\minus{}y}\equal{}2\Rightarrow x\plus{}y\equal{}2x\minus{}2y \Rightarrow x\equal{}3y$ therefore the answer is (A) 3

[quote="ProNivel"]It's dumb, but i can tell the answer by looking at the picture... :)[/quote]

In this case it happens to be correct, but sometimes the text will contradict the picture; they aren't always drawn to scale.


induction
number theory proposed
number theory

Do there exist [b]a)[/b] 6; [b]b)[/b] 1000 different natural numbers, such that if a and b are any two of them, then the sum a + b is divisible by the difference a - b ?

  Darij

We will prove, by induction on $n$, that for any integer $n > 1$, there exists pairwise distinct positive integers such that for any two $a,b$ among them the number $a+b$ is divisible by $a-b$.

- For $n = 2$, just consider $1$ and $2$.

- Now, assume that for some given $n \geq 2$ we have integers $1 \leq a_1 < a_2 < \cdots < a_n$ such that $a_i+a_j$ is divisible by $a_i-a_j$ for all $i\neq j$.

Let $\displaystyle N = a_1 \cdot a_2 \cdots a_n \cdot \prod_{ 1 \leq i < j \leq n} (a_i-a_j)$.
Then, define $\alpha_0 = N$ and $\alpha_i = N+a_i$ for $i=1,2,\cdots, n$.
It follows that :
- If $1 \leq k \leq n$, we have $\displaystyle \frac {\alpha_k + \alpha_0} {\alpha_k - \alpha_0} = \frac {2N} {a_k} + 1$ is an integer.
- If $1 \leq i < j \leq n$, we have $\dsp \frac {\alpha_i + \alpha_j} {\alpha_i - \alpha_j} = \frac {2N} {a_i - a_j} + \frac {a_i  + a_j}{a_i - a_j}$, which is an integer from the induction hypothesis.

This ends the induction step and the proof.

Pierre.

Many years ago I tried to solve this problem, but without success.
And only now I realize that Pierre's approach is native for this problem.
Perfect solution, Pierre!


In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

[hide="Quick Solution"]Let's label the columns $a$, $b$, and $c$, respectively. We're simply looking for the number of distinct arrangements of $aaabbbcc$. Our answer is $\frac{8!}{3!3!2!}=\boxed{560}$.[/hide]

[hide]We use a string of letters to model this problem.  We can represent any way of shooting down all the targets with an arrangement of the sequence $AABBBCCC$, where $A$ is shooting in the column with three targets, and $B$ and $C$ are the other two columns, with three targets each.  This denotes the order of shooting down targets, and is just $\dfrac{8!}{3!3!2!}=\boxed{560}$.[/hide]

[hide="AIME #8???"]
Note that this problem is equivalent to finding the number of different orders of $3$ indistinguishable objects, $3$ other indistinguishable objects, and $2$ other indistinguishable objects.  We will represent the first type of object as $\text{A}$, the second type as $\text{B}$, and the last type as $\text{C}$.  Thus, we are trying to find the number of arrangements of $\text{AAABBBCC}$.  Now we have turned this problem into a very well-known, slightly trivial problem; the number of arrangements is simply $\frac{(3+3+2)!}{3!3!2!}=\frac{8!}{3!3!2!}=\boxed{560}$
[/hide]

Yes this is AIME #8 which just boils down to $\binom{8}{3, 3, 2}$

Yeah, I know, as shown by my sol. @C-bass

Note that we just have to pick a column and then it's predetermined which target we shoot. Then the amount of ways is just $\frac{8!}{3!3!2!}=\boxed{560}$.

[hide=Solution]
Let $X$ represent shooting a target from the first column, $Y$ represent shooting a target from the second column, and $Z$ represent shooting a target from the third column.

Note that for each column, there is only $1$ possible order in which the targets in that column can be shot down, so the desired answer is the number of ways to arrange the letters in the string $XXXYYYZZ,$ which is
\begin{align*}
\binom{8}{3,3,2} &= \frac{8!}{3!3!2!} \\
&= \frac{40320}{72} \\
&= \boxed{560}.
\end{align*}
$\square$
[/hide]

[hide=Solution]If the three columns are $x$, $y$, and $z$, then the answer is just the number of ways to permutate $xxxyyyzz$, which is $\frac{8!}{3!2!2!}=560$[/hide]


quadratics
algebra
quadratic formula
SFFT
special factorizations
algebra unsolved

Consider the system:
$x+y=z+u$
$2xy=zu$
Find the maximum value for the real constant $m$, such that $m\leq \frac{x}{y}$, for any solution for the system $(x,y,z,w)$, with $x\geq y$.

x-2xy+y = z-zu+u
x-2xy+y-1 = z-zu+u-1 = (z-1)(1-u)
Sorry couldn't resist Simon's favorite factoring trick!
Solution:
[hide]
Ok, for the problem itself, set z and u constant.
x = zu/(2y)

zu/(2y) + y = z+u
zu + 2y^2 - 2(z+u)y = 0
Use quadratic formula from here to solve for y. Then, find x in term of y. Then, it is easy for decipher the minimum value of x/y.[/hide]


linear algebra
matrix
induction
graph theory
linear algebra unsolved

Let $ A=(a_{ij})$be a $ n \times m$ matrix with non-negative entries such that each row and each column has at least one positive entry, and suppose that whenever $ a_{ij}> 0$, then the row and column sums through $ a_{ij}$ must be equal. Prove that $ A$ must be square, i.e. $ m=n$.

Let's rephrase it first.

Construct a bipartite graph $ G$ with bipartition $ A,B$ as follows: the points in $ A$ are the rows of the matrix, while those in $ B$ are the columns. Two points are connected iff the corresponding row and column meet in a positive entry. In addition, each edge has a weight equal to the coresponding positive entry in the matrix. Define the weight of a vertex $ x$ to be the sum of the weights of the edges adjacent to $ x$. Now the problem reads:

Let $ G=A\cup B$ be a bipartite graph with weighted edges. There are no isolated vertices, and each two vertices which are connected have equal weights. Prove that $ |A|=|B|$.

We do this by induction on the number of edges, the case when this number is $ 1$ being trivial. If $ x\in A$ has degree $ 1$, then it must be connected to a vertex $ y\in B$ which, in turn, is connected only to $ x$. Remove $ x$ and $ y$ from $ G$ and apply the induction hypothesis to conclude. If there are no vertices of degree $ 1$, then we can find a simple cycle $ C$ in $ G$. Let $ e$ be an edge in $ C$ with minimal weight. Remove $ e$, and subtract the weight of $ e$ from that of every other edge in $ C$, thus modifying the weights of the remaining edges and vertices in $ C$. We obtain a new graph which still has all the properties that $ G$ had, and whose number of edges is one less than that of $ G$. We can now apply the induction hypothesis again.

P.S.

It should have been pretty obvious that the problem fits better in the 'Combinatorics' section. The fact that a matrix is mentioned somewhere in the text is not reason enough to post a problem in the 'Linear Algebra' forum :).


probability function trigonometry geometry algorithm calculus logarithms

This is from my last homework assignment of the semester in my probability course.

I wish to make my computer draw a plot of 5000 dots randomly but uniformly distributed on the unit disk $\{(x,y)|x^{2}+y^{2}\le 1\}.$ I have a programming language capable of drawing dots on the screen, and equipped with a "random number generator" which simulates instances of a uniform $(0,1)$ random variable. In the "pseudo-code" that follows, "rand()" is a function that produces an instance of a uniform $(0,1)$ ran variable, and each successive call to this program is independent of all previous calls. We also assume a built-in value of $\pi.$ My attempt to write the program looks like this:

.... Initialize the seed for the random number generator
.... Set up the screen for plotting, and draw a unit circle for background
.... For $j=1$ to $5000$ do:
......... Let $\theta: =2*\pi*\text{rand}()$
......... Let $r: =\text{rand}()$
......... Let $x: = r*\cos(\theta)$
......... Let $y: = r*\sin(\theta)$
......... Plot the point $(x,y)$
.... Next $j$
.... End program

Does this program work as intended? Does it produce 5000 samples from a distribution that is uniform in area on the unit disk, or does it produce some other distribution? If it produces some other distribution, how can you fix it so that it does work as intended?

...

In giving hints for this problem, I invited my students to go ahead and execute the program, if possible.

You've already posted this (***New Year Resolution: "Get over" Mathlinks/AOPS***). But I'm not giving any links since I would unnecessarily deprive someone from the pleasure of solving this problem.

If the random number generator works as intended, you would have a roughly equal number of points at each radius, which makes it very crowded near the middle.

You should get a better result using $r=1-(rand()^{2})$
At least I am

[quote="The Zuton Force"]You should get a better result using $r=1-(\text{rand}()^{2})$[/quote]

That's not exactly correct. If you do that, you would have

$P(R\le a)=P(1-U^{2}\le a)$ for a uniform $(0,1)$ r.v. $U.$

$=P(U^{2}\ge1-a)=P(U\ge\sqrt{1-a})=1-\sqrt{1-a}.$

But what we really want is $P(R\le a)=a^{2}.$

Note that for small $a,$ $1-\sqrt{1-a}\approx\frac{a}2,$ which is, in the limit, much larger than $a^{2}.$ You're still piling too many points into a neigborhood of the origin; your density (like the original density) goes to $\infty$ at the origin.

Can you use $r : = \sqrt{1-\text{rand}()}$?

Then $P(r \le a) = P(\sqrt{1-U}\le a) = P(U \ge 1-a^{2}) = a^{2}$.

Yes, that works.

Of course, $r: =\sqrt{\text{rand}()}$ accomplishes the same thing  a little more simply.

Alternate algorithm:

FOR j=1 to 5000 DO
.... DO (start a loop)
......... Let X := 2*rand() - 1
......... Let Y := 2*rand() - 1
.... LOOP until X*X + Y*Y <= 1
.... Plot the point (X,Y)
NEXT j

How many times will the inner loop be executed each time? That's a geometrically distributed random variable with $p=\frac{\pi}4.$ The expected number of executions of that loop would then be $\frac{4}{\pi}$ times the number of points - not bad at all.

I invented this problem (as a homework problem for a calculus-prerequisite first probability class) quite some number of years ago. At the time, I had a computer with an 8 MHz CPU and no floating-point co-processor. On that machine, the second algorithm (pick a point according to a distribution that is uniform on a square, test it to see if it's in the circle) was noticeably faster than the first (which requires the computation of a square root, a cosine, and a sine for each point). Once I got a machine with an FPU, I stopped noticing the difference.

A couple more side comments: First the programming comment. Several of my students went ahead and executed what they claimed to be the algorithm. In a couple of cases, all the points lined up along a spiral. What mistake could that be?

Then the simulation comment. The following steps

$R: =\sqrt{-2\ln(\text{rand}())}$
$\theta: =2\pi\text{rand}()$
$X: =R*\cos\theta$
$Y: =R*\sin\theta$

generate in $X$ and $Y$ two independent instances of a random variable that has a standard normal distribution. (In this context, "standard" denotes mean zero and variance one.) In fact, it's THE way to generate normally distributed random variables if what you're supplied with is a uniformly distributed random number generator. Can you explain why this works?

[quote="Kent Merryfield"]I have? So soon I forget.[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=computer&t=31042


Can someone explain how to add radicals (square roots)?

Do you have any particular problems that you have trouble with?

In general, to simplify square roots...(example is easiest):
$ \sqrt {1687500} \equal{} \sqrt {2^2\cdot3^3\cdot5^6}$

$ \equal{} 2\cdot3\cdot5^3\sqrt3$ (do you see how one takes out the factors by groups of $ 2$?)

Adding up is simply combining like terms, after simplification.

For example,

$ 2\sqrt2 \plus{} 3\sqrt5 \plus{} 4\sqrt2 \plus{} 8\sqrt8$

$ \equal{} 2\sqrt2 \plus{} 3\sqrt5 \plus{} 4\sqrt2 \plus{} 16\sqrt2$

$ \equal{} 22\sqrt2 \plus{} 3\sqrt5$

Edit: Thanks for pointing out my stupid mistake... :blush:

How does $ 8\sqrt{4}$ simplify to $ 16\sqrt{2}$?  Shouldn't it be $ 16$?

Yes, math154 made an error.

Basically, to add radicals together, you need to have the same number under the radical, just like when you add two terms together, you need the same variable.

For instance, there's no way to simplify $ 3\sqrt{5} \plus{} 4\sqrt{7}$, but what if we have something like $ 3\sqrt{5} \plus{} 6\sqrt{5}$? This becomes $ 9\sqrt{5}$.

When you want to add two radicals together that don't have the same radical, first see if you can simplify them, like math154 showed.

And subtracting radicals is just adding negative numbers before the radical, like; -1rad2

$ \sqrt{x}\plus{}\sqrt{y}$ cannot be expressed as $ \sqrt[n]{k}$. There n - positive integer, k - rational number, if $ \sqrt{\frac{x}{y}}$ is irrational number.

[quote="Bachukas"]$ \sqrt {x} \plus{} \sqrt {y}$ cannot be expressed as $ \sqrt [n]{k}$. There n - positive integer, k - rational number, if $ \sqrt {\frac {x}{y}}$ is irrational number.[/quote]

The nth root of k is just k^(1/n). Very important to remember.


calculus
integration
function
calculus computations

I've just learned integration by parts method and I can't figure out how to evaluate this integral. All functions that I choose lead me to the same resul (if the integral is I i get I=I  :blush:) So evaluate  $\int\frac{x^{2}}{1+9x^{2}}dx$. A hint may be useful :)

we dont need Integration By Parts at all ... change it to

$\frac{1}{9}\int\left(1-\frac{1}{9x^{2}+1}\right)dx$ and there we see inverse trigo comes in  :)

Thank you! In fact I had to integrate $\frac{x^{3}}{1+9x^{2}}$ ..but your idea works in this case as well. :D

When the degree of the numerator is greater than or equal to degree of the denominator, try long division.


I updated my signature and everything.
My profile syas it's there, but my messages dont show it.

Please help.

Make sure you are checking the [b]Attach signature (signatures can be changed in profile)[/b] option when posting your message or have your profile set to automatically attach your signature. This can be set in Profile->Preferences->Posting a Message->Always attach my signature


Three of the sides of a quadrilateral circumscribed about a cricle are 9,15,and 8.  what is the smallest possible value of the fourth side?

is there a certain formula for this?

[quote="i_like_pie"][hide="Not sure..."]Since the lengths $ 8$, $ 9$, and $ 15$ can form a non-degenerate triangle, a side of length $ \epsilon$ can be added between any two sides of the triangle to form a quadrilateral.[/hide][/quote]
But then how will the circle touch the side of tiny length, if it was previously a vertex of the triangle?

So that's where I went wrong...

[hide="Real Solution"]The sums of the opposite sides of a circumscribed quadrilateral are equal (this is a theorem).

$ 8\plus{}9\equal{}15\plus{}x$

$ x\equal{}\boxed{2}$

If the side opposite $ x$ is smaller (either $ 8$ or $ 9$), the value of $ x$ will be bigger (either $ 14$ or $ 16$).[/hide]

In an optimization problem, you must not only prove that your answer is the optimal one, you have to prove that it is attainable. If it is not attainable, you have to strengthen your bound.

[hide="hint"]equal tangents[/hide]
[hide="hint 2"]prove that the sum of opposite sides must be equal.[/hide]

then how would you know which is right?

[hide]the answers that are of choice are $ 1,2,4,14,NOTA$[/hide]


Let $ t_n$ is a sequence such that $ t_n \equal{} \frac{n(n\plus{}1)}{2}$. Find all $ m,n$ sucht that
$ t_n\plus{}t_{n\plus{}1}\plus{}t_{n\plus{}2}\equal{}t_m$

n=1 and m=4

Setting r=m-n
the given equation can be rewritten as 2t_(n+1) + n+2 = rn + t_r

writing t_n = n(n+1)/2 in the above, we get
 (n+2-r)(n+2) = r(r-3)/2 this means the LHS is divisible by r. That means RHS is divisible by r^2. This is impossible unless r=3 (since, r(r-3) < r^2). Hence r=3. Working out for n, we get  n=1 and so, m=4


[quote="manlio"]For $a,b,c$ positive reals prove that

$a^2+b^2+c^2 + \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq 3 +2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})$[/quote]

There must be a typo here; a = b = c gives a counterexample.

  Darij

Perhaps it is

For $a,b,c$ positive reals prove that

$a^2+b^2+c^2 + \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq 3 +(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})$

but I am not sure as I have not copied it on a paper but I remember only it as I read. Sorry.


I believe this was one of the last questions in the recent AMC exams.I forgot the wording of the question but I still remember the numbers So I will word it as far as I remember

There's a family and each of them has an 8 ounce cup. On a table there is milk and coffee. Each person fills their 8 ounce cup with varied amounts of milk and coffee. One member of the family, fills her cup with 1/4 of the total amount of milk and 1/6 of the total amount of coffee. According to this, how many family members are there?

I have a very fast way to solve this problem so try to find it.

[hide]There is more than four(derived off him taking 1/4) and less than six(derived from him taking 1/6) So he must  have 5.[/hide] Interesting problem.[/hide]

I reasoned the way Scrambled did when it came up on the AMC a few years back, but there is a more elegant way.

Spoiler:

[hide]Suppose the member of the family who drank 1/4 of the total milk and 1/6 of the total coffee drank m ounces of milk and c ounces of coffee.  Then there are at total of 4m ounces of milk and 6c ounces of coffee.  Note also that 0 < m < 8 and 0 < c < 8.

Also, we know that m + c = 8 (since the milk and coffee had to add up to 8 oz.) and 4m + 6c = 8x, where x is the number of people in her family.  Multiplying the former equation by -4 and adding it to the second equation, we get -4m - 4c + 4m + 6c = -32 + 8x.  This simplifies to 2c = 8x - 32, or c = 4x - 16.  But 0 < c < 8, so 0 < 4x - 16 < 8.  Thus 16 < 4x < 24, or 4 < x < 6.  Thus x = 5.[/hide]