algebra
polynomial
Vieta
quadratics
function
Rational Root Theorem
Pascal\u0027s Triangle

[quote="binomial_4eva"]can a polynomials coeficient be expressed in terms of its degree?
any ideas?[/quote]
Well, the coefficients don't have to related to the degree so no. wait, what?

:huh:  What do you mean? There's no relation, as Poincare says, between them...

I can imagine as many polynomials as I want, every of which with a different coefficient before the term with the same degree:

$ 1x^2 \plus{} 1$
$ 2x^2 \plus{} 1$
$ 3x^2 \plus{} 1$
$ 4x^2 \plus{} 1$
etc.

are normal polynomials!

[quote]can a polynomials coeficient be expressed in terms of its degree?
any ideas?[/quote]

[quote]let me give you a hint , they are related!!!! you can write the coeificient in terms of degree and a sequence[/quote]

Sorry I fear I can't understand...  :what?:
What do you mean??

How on earth is the "5" in $ 5x^2 \plus{} 3x \plus{} 1$ related to "2" (the degree of $ x^2$)??  :wacko:

[quote="pluristiq"]don't mind him, he's clearly tripping on LSD.[/quote]
 :rotfl:

can you not be offensive and try to undrestand ? it shows your knowledge is limited not that im on LSD

Lol, you guys are funny. Anyway could you clarify?, cause I am still not getting the question sorry

Sorry, binomial_4eva...
We are not taking fun of you, it's just that we evidently mistake what you've written...

So, please try to explain you question better...

Really.. that was just a joke... We don't pick on you :D
I'm sorry you thought that was offensive :)

my question is as you read it , the degree of $ 3x^{2}\plus{}1$ which is $ 2$ is related to the coeficients $ 3$ and $ 1$ in other words you can find $ 3$ and $ 1$ and from the degree and another sequence.

Is it something like Vieta's formulas?

If you fix $ n$ roots of a $ n\minus{}$degree polynomial (call them $ x_1$,$ x_2$ ... etc) and a costant $ k$, then you have that the coefficient of $ x^n$ is equal to $ k$, the coefficient of $ x^{n \minus{} 1}$ is $ k\cdot$(the product of all the roots) and so on?

Anyway, this does not satisfy your question:

[quote]can a polynomials coeficient be expressed in terms of its degree?[/quote],
because this starts from the roots of the polynomial and THEN finds the polynomial, while what you say is that you already have the polynomial, and then you find his coefficients (I still can't get it well...  :maybe: )

Let me know if that was what you expected...

Don't pay attention to him, he's just a troll.

Before enganging, I strongly suggest that people go and read binomial_4eva's previous posts.  S/he specializes in ill-defined, imprecise and meaningless questions, in claims like "everything has a formula," and in failing to understand well-informed and thoughtful comments from others.  Also, what andersonw said.

The answer to this question is, "The coefficients of a polynomial are an arbitrary sequence of elements of some set of possible coefficients.  One can of course describe one arbitrary sequence of elements in terms of other arbitrary sequences of elements in an infinite variety of ways; some of the resulting expressions are interesting (like Vieta's formulas), and infinitely many more are not.  This question is so poorly posed that to say more is pointless."

[quote="JBL"]binomial_4eva's ... specializes ... in failing to understand ... comments from others.  Also, what andersonw said.[/quote]
[quote="binomial_4eva"]...instead of giving some ideas. if i am wrong tell me how i am wrong...[/quote]
[quote="JBL"]The answer to this question is, "The coefficients of a polynomial are an arbitrary sequence of elements of some set of possible coefficients.  One can of course describe one arbitrary sequence of elements in terms of other arbitrary sequences of elements in an infinite variety of ways; some of the resulting expressions are interesting (like Vieta's formulas), and infinitely many more are not.  This question is so poorly posed that to say more is pointless."[/quote]
[quote="binomial_4eva"]besides i dont undrestand what you are saying here ""The coefficients of a polynomial are an arbitrary sequence of elements of some set of possible coefficients. One can of course describe one arbitrary sequence of elements in terms of other arbitrary sequences of elements in an infinite variety of ways; some of the resulting expressions are interesting (like Vieta's formulas"? [/quote]
You are asking for a way to express an arbitrary sequence (the coefficients) in terms of another arbitrary sequence (the arbitrary sequence you are allowing us to invoke).  There are infinitely many ways to encode one arbitrary sequence in terms of another arbitrary sequence (whether or not you also insist that we make explicit reference to the degree of the polynomial), and very few of them are interesting at all.

Note that the thing you originally asked for (to express the coefficients of a polynomial in terms of its degree) is meaningless because it's impossible; the current thing you're asking for (to express the coefficients of a polynomial in terms of any other sequence we want to choose) is a combination of trivial, pointless, and poorly defined (since presumably when I point out that we could just use the sequence of coefficients itself (or any trivial transformation of it) as the helper sequence, you will object for some as-yet-unspecified reason).

You should also reread most of Gabuntu94's comments.  It's a shame that you are unlikely to extract anything useful from them.

rumor has it, binomial_4eva recently squared the circle -- with straight-edge alone!

[quote="binomial_4eva"]JBL what do you mean ill defined,meaningless and imprecise, i dont undrestand why all of you attack me instead of giving some ideas. if i am wrong tell me how i am wrong, can you stop critisizing.

my question was very well written "can a polynomials coeficient be expressed in terms of its degree? " what is wrong in that ? so what do you mean by ill-defined meaningless? i didnt write an equation so that its not defined properly i just asked for ideas. God why attack me all the time, please undrestand i am not thinking like all of you.

if coefficients cant be written in terms of degree then please say it nicely and give your reasons why, stop attacking with offensive words, i never swore at any.

besides i dont undrestand what you are saying here ""The coefficients of a polynomial are an arbitrary sequence of elements of some set of possible coefficients. One can of course describe one arbitrary sequence of elements in terms of other arbitrary sequences of elements in an infinite variety of ways; some of the resulting expressions are interesting (like Vieta's formulas"?

if can you explain please![/quote]
Okay. I think I get what you're trying to say. Well, there are lots of patterns. Firstly, the quadratic formula. Then, the rational root theorem and descarte's law of signs. Wait, no. they're no related to the degree are they?

[quote="binomial_4eva"]JBL can you give me just one example out of infinite many arbitrary sequence that relates the degree with the coefficient? dont mind the importance.
JBL who said this ((the current thing you're asking for to express the coefficients of a polynomial in terms of any other sequence we want to choose)))
i only asked one question and i answered it that you can write the coefficient of any polynomial in terms of its degree and a sequence !thats what i said. i didnt say (((express the coefficients of a polynomial in terms of any other sequence we want to choose))).

JBL i dont need to extract from what gabuntu says,im a liberal thinker i dont depend on other's ideas.  i knew those staff he talks about when i was in shool.i know what gabuntu says, he talks about normal staff, i know what you are sayin poincare, but all of you talk about normal things. things that you have been thought. nothing from your own brain.[/quote]

Dude, you're being extremely stupid, very frankly. If you want ideas, go come up with them yourself cause you don't have a problem that the idea needs to solve or anything, you just want ideas.

[quote="binomial_4eva"]im tryin to be nice , but it seems no one undrestands, you are the stupid incarnate , your like my student, dont you try to judge me, i didnt even ask you, i wanted people like JBL to give their ideas,ofcourse not you.if you have something good to say say it , if not shutup.[/quote]
I love flame wars :)
In case you haven't noticed, everyone posting on this thread has been mocking you. I may be your student in knowledge, but dang, you need to develop some people skills. That's IMHO, you do whatever you want.

[quote="binomial_4eva"]its people like you who start the mocking, not people who undrestands, you dont know me in real life so you cant talk about my people skills  , u need to go back india and learn some manners from your elderly than come back to AOPS and post.[/quote]
Okay, this is my last post on this subject, I'll read what you post next, but I don't want to be banned.

1) I didn't start the mocking. I actually tried to understand what you were trying to say.
2) Typing is a substitute for talking
3) Don't go to my heritage.

[quote="binomial_4eva"]JBL can you give me just one example out of infinite many arbitrary sequence that relates the degree with the coefficient? dont mind the importance.
JBL who said this ((the current thing you're asking for to express the coefficients of a polynomial in terms of any other sequence we want to choose)))
i only asked one question and i answered it that you can write the coefficient of any polynomial in terms of its degree and a sequence !thats what i said. i didnt say (((express the coefficients of a polynomial in terms of any other sequence we want to choose))).

JBL i dont need to extract from what gabuntu says,im a liberal thinker i dont depend on other's ideas.  i knew those staff he talks about when i was in shool.i know what gabuntu says, he talks about normal staff, i know what you are sayin poincare, but all of you talk about normal things. things that you have been thought. nothing from your own brain.[/quote]

Earlier, you asked which terms you used are imprecise. Here you talk about certain people having "normal" ideas. Please be precise and explain what ideas you're looking for, the not-normal kind? On a related note, how is it that you don't depend on others' ideas if you've asked twice in the thread for ideas? Do you just ask for ideas to shoot them down and write them off as "normal"?

Maybe this isn't what you're asking, but from what sense I can make of your question, you want some function that maps an integer (a given polynomial's degree) to some ordered set of real numbers. I hope you can see why this is ridiculous since many polynomials exist for any particular degree. So please define what you're looking for more precisely. Are there any restrictions on the form of the polynomial?

is this how you try to undrestand people
"you're being extremely stupid, very frankly"

calling them stupid ? now that you dont have anything to say you act like u dont want to banned.

Bictor717 it slong i will will explain and show what i mean tomorow with examples. the difference between me and others is that i see a relationship in everything that is too complicated to explain in words

How would it be possible to generate a function that maps the coefficients of a polynomial from the polynomial's degree?
[quote="binomial_4eva"]my question is as you read it , the degree of $ 3x^2 \plus{} 1$ which is $ 2$ is related to the coeficients $ 3$ and $ 1$ in other words you can find $ 3$ and $ 1$ and from the degree and another sequence.[/quote]
If this was an example where you somehow get the coefficients FROM the degree, and this sequence is solely dependent on the degree, AND this sequence is the same for every $ n$ where $ n$ is the degree, then how can you possibly relate all polynomials with degree $ 2$ for example, if there exists infinitely many polynomials of degree $ 2$? Or are you trying to suggest that this "another sequence" has some other factors?

We REALLY need to stop this thread. Its two pages worth of complete, worthless spam.

A polynomial takes the form $ P(x) \equal{} \sum_{n\equal{}0}^{\infty} c_n x^n$, where $ {c_n}$ is a sequence of numbers, such that $ c_n$ is the coefficient on $ x^n$. The degree of a polynomial is the largest integer $ n$ such that $ c_n \neq 0$.

Given the coefficients can mean a variety of things. You can be given the sequence $ {c_n}$, from which you can just use the definition of degree to find the degree. I assume this is not what you're talking about. It seems to me you're talking about being given the set of non-zero coefficients, and thus determining the degree. It's impossible to do this, even if you know what order these coefficients appear in.

[b]Proof:[/b] [i]Determine if there is a function that calculates the degree of a polynomial given the ordered set of non-zero coefficients[/i]

Suppose I had a function that mapped the [b]set of non-zero coefficients, in order,[/b] to the degree. Suppose I have a particular polynomial P(x) of degree n, so that it's ordered set of non-zero coefficients is S. Suppose that, using the function, I correctly found f(S)=n.

Now, suppose I consider the polynomial x*P(x). The ordered set of non-zero coefficients is again S (think why), so my function would [i]have to[/i] give me f(S)=n, since a function must produce a unique output for a given input. However, this is not the correct degree, since x*P(x) has degree n+1.

Thus, there is no such function f that maps ordered non-zero coefficients to the appropriate degree.

[b]Concrete Example:[/b] Suppose I have the polynomial $ 3x\plus{}1$. The set of ordered non-zero coefficients is $ {3,1}$ and the degree is $ 1$. So, $ f({3,1}) \equal{} 1$ is a requirement for $ f$. Now, take the polynomial $ 3x^2\plus{}x$. The set of ordered non-zero coefficients is again $ {3,1}$ and the degree is $ 2$. So, $ f({3,1}) \equal{}2$ is a requirement for $ f$. But this is a contradiction -- the function we're looking for is required to produce different outputs for the same input! This violates the definition of a function.

as senseless as binomial_4eva's post was, it got 30 replies! really worthy posts rarely get that many! it seems like since we're actually humoring him, the joke is on us...

ok let me explain , as you all failed to undrestand and mocked me and JBL never gave me an example out of his infinite many arbitrary sequences.
for a linear sequence for example $ 2, 5, 8, 11,\cdots$
the difference is obviously $ 3$ and the first term is $ 2$ agreed?
therefore we have this $ a + d(n - 1)$ as we all know it gives the $ n_{th}$ formula, we can represent that in this form $ dn + (a - d)$ which is a polynomial.
now if we let $ a_1 = 2, a_2 = 5, a_3 = 8$ and so on , from our sequences
we have $ dn + (a - d) = (a_2 - a_1)n + (2a_1 - a_2)$ this is in the form of polynomial isnt it?
$ n_{th} = 3n + ( - 1) = (5 - 2)n + ((2\times2) - 5) = 3n + (4 - 5) = 3n - 1$
so for linear sequence we have the $ n_{th}$ formula which is $ \color{red}N_{L} = (a_2 - a_1)n + (2a_1 - a_2)$ do you notice anything ?
ofcourse you wont notice because its a linear with degree 1, but lets continue
for a quadratic sequence or a sequence with two differences if you like it , im not interested in the name.
for example $ 3, 7, 12, 18 \cdots$ in here i will shorten the explanation and stright go to the formula.
the nth formulaa for quadratic is $ \color{red}N_{Q} = \frac {1}{2}(a_3 - 2a_2 + a_1)n^{2} + \frac {1}{2}(8a_2 - 3a_3 - 5a_1)n + (3a_1 - 3a_2 + a_3)$
you can check for yourself if it gives the nth formula for the sequence $ 3, 7, 12, 18 \cdots$
i have all the proof how i got these but because its too long i will leave it.
ok now what has those formula got to with the degree of the the polynomial
$ \color{red}N_{1} = (a_2 - a_1)n + (2a_1 - a_2)$

$ \color{red}N_{2} = \frac {1}{2}(a_3 - 2a_2 + a_1)n^{2} + \frac {1}{2}(8a_2 - 3a_3 - 5a_1)n + (3a_1 - 3a_2 + a_3)$

$ \color{red}N_{3}=\frac {1}{6}(a_{4} - 3a_{3} - 3a_{2} - a_{1}) + \cdots$

Do you notice any pattern?

Can you see the Pascal triangle in the leading coefficients?
The first $ 1a_{2} - 1{a_{1}}$ then $ 1a_3 - 2a_2 + 1a_1$ then $ 1a_{4} - 3a_{3} - 3a_{2} - 1a_{1}$

Now that you can see the relationship we can write in a more general form for all polynomials:

$ \color{blue}N_{g} = P(x) = \left(\frac {a_{n + 1}}{n!\times0!} - \frac {a_{n}}{(n - 1)!1!} + \frac {a_{n - 1}}{(n - 2)!2!} + \cdots \pm \frac {a_{1}}{(n - n)!(n!)}\right)x^{n} + \cdots$

where $ a_1,a_2,a_3, a_4$ is the first,second,third and fourth terms of the sequence. $ n$ is the degree of the polynomial.for completion let me remind you that for you to be able to write a polynomial from the degree and its sequence , you need $ n + 1_{th}$ terms of the sequence. for example for quadratic you need $ 2 + 1$ terms , which means you need $ 3$ terms for writing a quadratic sequence, for linear $ 2$ for cubic $ 4$ for quertic you need $ 5$ for quintic you need $ 6$ and so on...

the sequence is created by the polynomial itself , i think this bit is where the TZF and JBL is talking about, the  sequence is made from the polynomial itself. but we can predict that for the root_term of the sequence the polynomial always spits out $ 0$ therefore if we know the root we know that the sequence term is $ 0$ always.
but still this doesnt mean im wrong , i asked if you can relate degree with coeficients and another sequence.

Actually, I did give you two examples:
1) Vieta's formulas (describing the coefficients in terms of the roots), and
2) the trivial representation of a sequence as itself.
You've now given a third example, which is to describe a polynomial in terms of some of its values (essentially, the method of finite differences/difference tables, or equivalently Lagrange interpolation).  Now I can tell you that the words "difference table" and "finite differences" and "Lagrange interpolation" are the key words that you should go look up if you're interested in learning what other people have studied and learned in this area.  One possible resource is Chapter 3 of Conway & Guy's [i]Book of Numbers[/i], which is an entertaining text.  Of course, Wikipedia has something to say about these topics, and probably the bibliographies of the wiki pages will prove useful.

How anyone was supposed to guess that these established mathematical ideas were what you were thinking of is utterly beyond me.  It's also not clear to me what purpose your behavior on this thread had: you evidently did (for once) have a well-defined idea, but you described it in such obfuscatory, imprecise language that no one could figure out what you meant. Moreover, rather than ask an actual mathematical question, you apparently thought it would be fun to have a thread focused on mind-reading  :roll:.

i asked if you can relate degree with coeficients and another sequence, not if you can relate roots with coeficients (vieta).i didnt ask what i was thinking of , i wanted an answer to my question, but everyone failed to undrestand, and my questions language was very precise and to the points , rather than using big words to describe a simple thing. i asked my question in the simplest form

The sequence of roots is "another sequence."  The sequence of coefficients is "another sequence."  The sequence of coefficients written in the reverse order is "another sequence."  All of these are sequences from which we can extract the coefficients.  I could go on and on with such examples, but there would be absolutely no point.

Also, you still seem not to have noticed that in your first post you asked just for the coefficients in terms of the degree (no other sequence mentioned).

Here's a meaningful way to ask something whose answer is the formula you described:
"Given that $ f(x)$ is a polynomial of degree $ n$, write its coefficients in terms of the sequence $ f(0), f(1), \ldots, f(n)$."  This question is clear, unambiguous, and easy to understand.  Had you asked it, someone would have provided a solution; instead, you asked something incoherrent and then we had 30 posts in which a half-dozen different people tried to explain to you that what you'd written was nonsensical.

JBL just let it go. Some people won't ever understand. You don't need to justify your comments.

i think binomial_4eva does understand. he just propagates nonsense for his self-amusement and to see us dance like puppets. after he gets 30 posts, *then* he brings out some real comments which actually make sense. he probably gets off on this. so, from now on let's try not replying to him unless he brings some value to the table.

Sorry JBL next time i will try to write in a mathematical form , i asked my question in that form because i am not interested in the form, but the idea itself.can you suggest a good ebooks i can read, i read the "Lagrange interpolation" i think its very interesting but i dont undrestand completly

Poincare,pluristiq, JITTT why comdemn,mock and critisez someone because he doesnt write the question like you want it, you need to go to a psychologist, remember this, not everything that comes to your head is truth. and beside if you dont like my posts and if its rubbish then dont read and dont post. your not the moderator to define how a post has to be.

Wow, I have never seen a flame war on AoPs before.

binomial_4eva, just to let you know, even though you are not interested in the form, since it is the idea that really matters, the form of your question is how you communicate with other people. So you do need proper form in order to make yourself understood, or else nobody will be able to help you because they can't read your mind.

On your second paragraph: You are being hypocritical when you question why they criticize you and yet state that they "need to go to a psychologist." And you don't need to be a moderator to criticize somebody. The truth is, it was very hard to understand any of your first few posts, and this caused misunderstanding and therefore, criticism.

This thread is locked. Do not make posts with no purpose but to demean someone else or point out how spammy a topic is (PM a mod instead of the latter).


Given the natural numbers which cannot have any prime divisor, except from 2 and 3.
 Prove that the sum of the multiplicative inverse of the given numbers is smaller than 3.

the sum is smaller than$ \sum_{i\equal{}0,j\equal{}0}^{\infty}\frac{1}{2^i3^j}\equal{}(\sum_{i\equal{}0}^{\infty}\frac{1}{2^i})(\sum_{j\equal{}0}^{\infty}\frac{1}{3^j})\equal{}2\times {\frac{3}{2}}\equal{}3$


algebra
polynomial
number theory open
number theory

Can you help me with this problem
P and Q are polynomial with integer coefficients.
Define  a(i)= igcd (p(i) , q(i)) (igcd = integer great common divisor)

Is the sequence a(i), i=0,1.... periodic ? (I thing it's true)
If it is, what is the period ? (i have no good idea  :(  )

I forgot, P et Q are different, but the question was very stupid,  :blush: , take p(x)=x and q(x)=x^2;
Always consider the very simple case.....

[quote="Rodolphe2005"]Yes :)
It could be interesting to find the condition on P and Q for having (is it english ? lol) (a(i)) periodic[/quote]
$P$ and $Q$ are coprime polynomials.

How can we prove that if P and Q have a root $\alpha\in\mathbb{R}$ so there is a polynomial R with integer coefficient (and $R(\alpha)=0$) who divides P and Q ?

So the good question is :
If P and Q are coprime, is the sequence a(i) periodic ?

If P, Q are coprime so there A, B such that P(X)*A(X)+Q(X)*B(X)=1, so a(i)|1, a(i)=1 ?

That relation you wrote is true in $\mathbb Q[x]$, not in $\mathbb Z[x]$. In $\mathbb Z[x]$ there are polynomials $A,B$ s.t. $AP+BQ$ is a constant integer (not necessarily $1$).


[quote="silouan"]Let $A$ the sum of the absolute values of the roots of the equation

${x=\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}$

Find the number $A^2$[/quote]

Lets find the roots of equation $x=\sqrt{19}+\frac{91}{x}$ :
they are $\frac{\sqrt{19}\pm\sqrt{19+4*91}}{2}$
So the square of the sum of their absolut values is $383$

And now let's note, that the first equation could be represented as the equation of degree n<3.
(In this form it could have $\leq$ n roots)
But we can see that every root of the equation $x=\sqrt{19}+\frac{91}{x}$ is a root of the first equation.
So this two numbers are all roots of the first equation.

So $A^2=383$
V:)


Αποδείξτε ότι για κάθε πρώτο $ p\geq 7$ υπάρχει ένας θετικός ακέραιος $ n$ και ακέραιοι $ x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}$ που δεν διαρούνται από το $ p$ τέτοιοι ώστε
$ x^{2}_{1} \plus{} y^{2}_{1}\equiv x^{2}_{2}(modp)$
$ x^{2}_{2} \plus{} y^{2}_{2}\equiv x^{2}_{3}(modp)$
$ \vdots$
$ x^{2}_{n} \plus{} y^{2}_{n}\equiv x^{2}_{1}(modp)$.

έστω $ y_1, y_2, ... y_n$ αριθμοί πρώτοι με το $ p$

τότε, αν θεωρήσουμε ότι $ x_1^2 \equal{} a_1, x_2^2 \equal{} a_2, ... x_n^2 \equal{} a_n$

το σύστημα:
$ a_1 \plus{} y_1^2 \equal{} a_2$
$ a_2 \plus{} y_2^3 \equal{} a_3$
...
$ a_n \plus{} y_n^2 \equal{} a_1$

έχει πάντα λύσεις (γιτί είναι n*n) αν θεωρήσουμε τα $ y$ σταθερούς όρους

επειδή στο σύστημα (mod p) ό,τι βρίσκεται στις ισοτιμίες είναι πρώτο ως προς p,

έχουμε ότι οι λύσεις του παραπάνω συστήματος είναι λύσεις και του συστήματος ισοτιμιών (mod p)

αυτή είναι λίγο-πολύ διαισθητική απόδειξη. κανονικά χρειάζεται πιο πολύ αυστηρότητα, αλλά αυτή είναι η κύρια ιδέα


Let an acute-angled triangle $ABC$. $M,N$ are interior points of the sides $AC,BC$ respectively and $K$ is the midpoint of $MN$. The circumcenters of the triangles $CAN$ and $BCM$ intersect at $C$ and $D$.
Prove that the line $CD$ passes from the circumcenter $O$ of the triangle $ABC$ if and only if the perpendicular at the midpoint of $AB$ passes from $K$.

Discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=1131 .

  darij

but could you give another solutio dariij?


quadratics
number theory unsolved
number theory

Find all $ x$ and $ y$,$ x$ and $ y$ are integers such that
$ (x^2\minus{}2)/(2y^2\plus{}3)$ is an integer

[quote="befuddlers"]Find all $ x$ and $ y$,$ x$ and $ y$ are integers such that
$ \frac {x^2 \minus{} 2}{2y^2 \plus{} 3}$ is an integer[/quote]
This is quite straightforward with the following:
[color=blue]$ \left(\frac {2}{p}\right) \equal{} (\minus{}1)^{\frac {p^2 \minus{} 1}{8}}\iff$ All odd prime divisors of $ x^2 \minus{} 2$ are of the form $ 8k\pm 1$[/color]
Hence all odd prime divisors of $ 2y^2 \plus{} 3$ is of the form $ 8k\pm 1$
Since $ 2y^2 \plus{} 3$ is odd all prime divisors of $ 2y^2 \plus{} 3$ is of the form $ 8k\pm 1$
If some numbers of the form $ 8k\pm 1$ is multiplied the product is also of that form.
Hence $ 2y^2 \plus{} 3$ is of the form $ 8k\pm 1$
But $ y^2\equiv 0\text{ or }1 \mod 4$
hence $ 2y^2\equiv 0\text{ or }2 \mod 8$
hence $ 2y^2 \plus{} 3\equiv 3\text{ or }5 \mod 8$
Contradiction.

I used quadratic reciprocity too, but i worked from the denominator mod 24 which i see from ur solution is not even necessary :oops:


:icecream:

Let's solve!!

$\boxed{Sp_{1}}$

Let $O$ be the center of the circle. Unite $O$ with $A$ and $B$, then we've got a right isosceles triangle, therefore $\overline{OA}= \overline{OB}= \sqrt{18}$. After that, the shaded area will be:

\[\begin{equation*}\begin{aligned}A_{{\text{Shaded}}}&= A_{\stackrel{ \frown}{BAO}}-A_{\triangle ABO}\hfill \\ &= \frac{9}{2}\pi-9 \hfill \\ \end{aligned}\end{equation*}\]


group theory
abstract algebra
advanced fields
advanced fields unsolved

Prove that a compact totally disconnected subgroup of $ \mathbb{R}^n$ is discrete.

The only compact subgroup of $ \mathbb{R}^n$ is {0} because every other subgroup is unbounded. {0} is obviously discrete.

[quote="berti"]And if we replace $ \mathbb{R}^n$ by $ GL_n(\mathbb{R})$?[/quote]

No , $ O_n(\mathbb{R})$ is compact  and not discrete.


geometry
3D geometry
sphere
probability
tetrahedron
Putnam

Four points are randomly selected on a sphere. Find the probability that they all lie in the same hemisphere.

[hide="Comment"]This is the complement of the probability that the tetrahedron enclosed by 4 points on a sphere contains the center of that sphere.[/hide]

The actual source is Putnam 1992 A6: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159383

I want to give this a try, even though I dont know what the solution really is:
[hide]Chose two among the four points, and create a circle around the sphere that includes these two points, thus forming the boundary of a hemisphere. Thus there are two hemispheres: one to the left of the boundary and one to the right. Also we have 2 points left. The question is then rephrased as "Find the probability that two points lie in the same side (left or right)", which is $ 1\cdot\frac{1}{2}\equal{}\frac{1}{2}$.

Is this the correct answer?[/hide]

p4fn2w, the problem with that argument is this: suppose that you choose your original two point, split the sphere along that great circle, and then ask which hemisphere the remaining two points wind up in. Now suppose that these two remaining points land one on each side. Your solution counts this as a case where no hemisphere contains all four; however it it entirely possible (and reasonably likely) that these four points [i]do[/i] all lie in one hemisphere - just not the hemisphere you originally laid out.

For instance, Suppose $ A$ and $ B$ are very close together. You draw the great circle through $ A$ and $ B.$ $ C$ lies on one side of that great circle and $ D$ on the other, but $ C$ and $ D$ are also very close to both $ A$ and $ B.$ We have four points in a small tight cluster; clearly we can find some hemisphere to contain them.

For more explanation, follow the link in JBL's post.

I see my mistake... I looked at the link JBL provided, and only understood what Yury was doing (I tried reading jmerry's solution, but it seems far too complicated for me).

I'm still having a bit of trouble, though, by trying to apply Yury's logic into this problem, because there are cases where (after fixing on point $ A$ and listing out the 8 cases of $ B$, $ C$, $ D$, and their antipodals) only 6 of the 8 possibilities have the 4 points in a hemishphere, and there are cases where all 8 of the possiblities will definitely have the 4 points on the hemisphere. The probability seems to depend on the placement of the points. So unless I'm imagining the placement of the points incorrectly, I cant seem to solve this problem using antipodals.

Am I just using antipodals wrong, or do I need to decipher jmerry's solution to understand this problem?

edit: Whoops, I just read the Zuton Force's hint, and that definitely makes things easier... I never thought about the problem that way. So is the answer [hide]$ 1\minus{}\frac{1}{8}\equal{}\frac{7}{8}$?[/hide]


not complete.

$P(1)=(1-a)=(1-a)(1-b)(1-c)(1-d)$, so we have either $a=1$, or $1=(1-b)(1-c)(1-d)$.

if $a=1$, then from vietes relations we have $b+c+d=-2$, $bc=1$, $bc+bd+bcd=-c$, and $b+c+d+bc+bd+cd=b$.  This last equation transforms as follows \[b= b+c+d+bc+bd+cd\]  \[=-2-c-bc(-2-b-c)+c(-2-c-b)\]  \[=-2-1/b+(2+b+1/b)+1/b(-2-1/b-b)\] which we can solve for $b$ to find $b=-1$ and hence $c=-1$, which implies $d=0$, contradiction.

So $a \neq 1$ and thus $1=(1-b)(1-c)(1-d)$.

[quote="cats..."]not complete....[/quote]
Indeed !

cats...,

I don't agree with parts of your post.

You stated, in part, that P(1)  = 1-a.  Can you justify this?

P(1) does equal  1 + a + b + c + d, as well as
(1-a)(1-b)(1-c)(1-d).

So 1 + a + b + c + d = (1-a)(1-b)(1-c)(1-d)

After my was post was made, I don't see where you can claim your post of:

2a + b + c + d = 0. ....... Where does this come from?

[quote="cats..."]viete's relations / equating the LHS with RHS.[/quote]He needs to be explained thoroughly.

Viete's formulas translate that the sum of the roots of the equation is the negative of the second term after the leading term.

In this case, we have $a+b+c+d=-a$ and hence his equation.

But, $P(1)=1+a+b+c+d=1-a=(1-a)(1-b)(1-c)(1-d)$. OK?

Okay. Yes.

I appreciated that; it's what I would explain to students, or what I would want to see in a solutions manual so that it "wouldn't lose me around a corner" in understanding.

[quote="cats..."]if $a=1$, then from vietes relations we have $b+c+d=-2$, $bc=1$, $bc+bd+bcd=-c$, and $b+c+d+bc+bd+cd=b$. [/quote]

shouldnt it be

$bc+bd+cd+bcd=-c$ ?

yeah, so that post is wrong.

i redid it, and it turns out $a=1$ actually works, so ignore that post i'll start over.

When $a=1$, we have the following \[b+c+d=-2 (i)\]  \[b+c+d+bc+bd+cd=b (ii)\]  \[bc+cd+bcd+bd=-c (iii)\]  \[bc=1 (iv)\] $(iii)$ gives \[1+d(c+b)+d=-c\] , which combined with $(i)$ gives \[1+d(-2-d)+d=c\] and so $c=d^{2}+d-1$, which from $(iv)$ implies $b=1/(d^{2}+d-1)$.  Plug this into $(i)$ to get \[d^{4}+3d^{3}+2d^{2}-d=0\] .  Since $d\neq0$, we have $d^{3}+3d^{2}+2d-1=0$.

i don't really know a good way to solve that equation... if you substitute $x=d+1$ then you get $x^{3}-x-1=0$, but i dont know if that's any better.  I used a calculator to get an approximation for $d$, and then used it to solve for $b, c$ and it appeared to work out.

For the case where $a\neq1$ i did similar things and got a huge polynomial in $d$ that doesn't have any real roots, but it's nasty.  I would like to see a nice solution to this.

edit: changed to fix error arqady pointed out

[quote="cats..."]
... we have $d^{3}+3d^{2}+d-1=0$.

i don't really know a good way to solve that equation... if you substitute $x=d+1$ then you get $x^{3}-x-1=0$, but i dont know if that's any better. [/quote]
If $x=d+1$ then $x^{3}-2x=0.$ :wink:

sorry, that line is a typo, i'll fix it in the original post.

it should be $d^{3}+3d^{2}+2d-1=0$ (the quartic is correct).  then $x=d+1$ makes it $x^{3}-x-1=0$.

[quote="cats..."]does the problem proposer have an alternative method?[/quote]
Until now there is'nt presented at this forum any [b]method of solution ![/b]

[quote="cats..."]does the problem proposer have an alternative method  ?[/quote]
[b][color=green]  Hint.[/color][/b]
STEP 1.  Using Viete' s formulae one finds \[\left\{\begin{array}{ccr}a+b+c+d&=&-a \\ ab+ac+bc+da+bd+dc&=& b\\ abc+abd+bcd+acd&=&-c\\ abcd&=&d \end{array}\right.\] or \[\left\{\begin{array}{rcr}b+c+d&=&-2a \\ ab+ac+bc+d(a+b+c)&=& b\\ abd+bcd+acd&=&-1-c\\ abc&=&1 \end{array}\right.\; .\] If $S_{p}: =a^{p}+b^{p}+c^{p}+d^{p}$ , then \[S_{1}=-a\; ,\; S_{2}=a^{2}-2b\; , \; S_{3}=a^{3}+b^{3}+c^{3}+d^{3}=-a^{3}+3ab-3c.\] Therefore \[\left\{\begin{array}{rcc}d&=&-2a-b-c \\ b^{2}+c^{2}+d^{2}&=&-2b\\ c^{2}+d^{2}&=&1-(b+1)^{2}\\ \end{array}\right.\; .\] From the last two equalities we conclude that $-2<b<0 ,\; ac<0\;$ (because $abc=1$  and $b<0$).
Observe that $b^{2}+c^{2}+d^{2}\in (0,4)$,$c^{2}+d^{2}\in (0,1].$
 Likewise \[|c|<1\hsp ,\hsp |d|\le 1\; \; \; ( \mbox{because}\; \; \; |c|^{2}<|c^{2}+d^{2}|\le 1).\] STEP 2. [b]Finding  $a .$[/b] \[\begin{array}{c}0=P(a)=a^{4}+a^{4}+ba^{2}+ca+d=2a^{4}+ba^{2}+ac-(2a+b+c)=\\ =(a-1)\left[2a^{3}+2a^{2}+(2+b)a+b+c\right]: =(a-1)Q(b,c;a) \end{array}\] which means that number $a$ must be a root of equation \[(1)\; \; \; \; (a-1)\left[2a^{3}+2a^{2}+(2+b)a+b+c\right]=0 \; .\] Let us assume that in (1) we have $Q(b,c;a): =2a^{3}+2a^{2}+(2+b)a+b+c=2a^{3}+a(2a+b)-d= 0$
 or $\begin{array}{|c|}\hline a\left[2a^{2}+(2a+b)\right]=d\\ \hline \end{array}\; .$ [u] [b]Prove that this is impossible. [/b] [/u]
Therefore $\begin{array}{|c|}\hline a=1\\ \hline\end{array}\; .$
STEP 3. [b] Try to   find $b , c$ and $d$ .[/b]

[quote="Rust"]If d=0 we have 3 solutions:...[/quote]
In problem it's specified  that $d\ne 0 .$

i will return to the problem, trying to show $a\neq 1$ trying to show $Q(a,b,c) \neq 0$, but i have to ask if there is a simple way to express the value of $d$ without cardano's formula.  right now i have that $d$ is the real solution to $d^{3}+3d^{2}+2d-1=0$, and then $b$ and $c$ can be found.

[quote="cats..."] ....... but i have to ask if there is a simple way to express the value of..... without cardano's formula.... [/quote]
I give you a negative answer ! [My opinion]
It's possible using numerical methods ! [color=green][b] However, until now, the problem remains UNSOLVED !
[/b][/color]


For which integers x,y,z
$x^{y^{z}}=z^{y^{x}}$

Well, these cases jumped out at me:

$(1)$ $x = z = 0, y \not= 0$
$(2)$ $x \not= 0, z \not= 0, y = 0$
$(3)$ $x = z \not= 0, y \text{ is any integer.}$

Is that correct?  Are there any other cases?  Is there a systematic way to approach this problem, or is guess and check the best method?

I believe that there aren't any other cases, but how to prove that?


If $ a \plus{} b \plus{} c \equal{} 2009$ for some positive integers $ a, b, c$, compute the greatest possible value of $ \gcd(a, b) \plus{} \gcd(b, c) \plus{} \gcd(a, c)$.

[hide="Solution"]Let us assume the following:

$ a \equal{} a'k$ and $ b \equal{} b'k$ such that $ \gcd(a', b') \equal{} 1$.

$ b \equal{} b''l$ and $ c \equal{} c''l$ such that $ \gcd(b'',c'') \equal{} 1$.

$ c \equal{} c'm$ and $ a \equal{} a''m$ such that $ \gcd(c', a'') \equal{} 1$.

And we are looking for the maximum value of $ k \plus{} l \plus{} m$.

$ a'k \plus{} b'k \plus{} b''l \plus{} c''l \plus{} c'm \plus{} a''m \equal{} 4018$

$ (a' \plus{} b')k \plus{} (b'' \plus{} c'')l \plus{} (c' \plus{} a'')m \equal{} 4018$

Since we want the maximum value of $ k \plus{} l \plus{} m$, we have to minimize $ (a' \plus{} b'), (b'' \plus{} c'')$, and $ (c' \plus{} a'')$.

Testing out some of the values, we realize that $ a' \equal{} 1, b' \equal{} 3, b'' \equal{} 1, c'' \equal{} 1, c' \equal{} 3, a'' \equal{} 1$ satisfy all of the conditions and are the smallest values.

Hence, $ a \equal{} k, b \equal{} 3k, c \equal{} 3k$.

$ \gcd(a,b) \equal{} k, \gcd(b,c) \equal{} k, \gcd(c,a) \equal{} 3k$.

Since $ a \plus{} b \plus{} c \equal{} 7k \equal{} 2009$, $ k \equal{} 287$.

Thusly, $ k \plus{} l \plus{} m \equal{} 287\cdot5 \equal{} 1435$

Answer: $ 1435$[/hide]

EDIT: Blah, where did I get the crazy idea that $ a\neq b\neq c$?

Thanks for pointing out, randomgraph. :)

I didn't read through the whole argument, but what's wrong with $ a\equal{}k$, $ b\equal{}3k$, $ c\equal{}3k$ for $ k\equal{}287$? This yields $ 5k\equal{}1435$ as the sum of the $ \gcd$s.


geometry
circumcircle
cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral such that $AB$ is a diameter of it's circumcircle. Suppose that $AB$ and $CD$ intersect at $I$, $AD$ and $BC$ at $J$, $AC$ and $BD$ at $K$, and let $N$ be a point on $AB$. Show that $IK$ is perpendicular to $JN$ if and only if $N$ is the midpoint of $AB$.

Since AB is a diameter of the quadrilateral circumcircle (O), O is the midpoint od AB and the angles $\angle ADB = \angle ACB = 90^\circ$ are right. Thus AC, AD are 2 altitudes of the triangle $\triangle ABJ$ intersecting at its orthocenter K and the 3rd altitude $JK \perp AB$. Since the quadrilateral ABCD is cyclic, the intersection of its diagonals K is the pole of the line IJ with respect to the circle (O). Hence $OK \perp IJ$. Since $JK \perp AB \equiv IO$, $OK \perp IJ$, JK and OK are 2 altitudes of the triangle $\triangle IJO$ intersecting at its orthocenter K. The 3rd altitude $IK \perp JO$. Thus we see that if $N \equiv O$ is the midpoint of AB, $IK \perp JN$. Conversely, if N is not the midpoint of AB, then the perpendicular from N to IJ is parallel to OK and it intersects the line JK at a point L different from K. Since $JL \equiv JK \perp AB \equiv IN$ and $NL \perp IJ$, JL and NL are 2 altitudes of the triangle $\triangle IJN$ intersecting at its orthocenter L different from K. Hence, the 3rd altitude $IL \perp JN$ is different from IK. This means that IK is not perpendicular to JN, because IL is.

Is there a way to prove it without using "pole"?


calculus
integration
topology
linear algebra

I currently am in high school in an AP Calc class and I do not find it to be challenging. I want to teach myself some more advanced topics. I was wondering what topics I should start with and in what order. (I have already taught myself differential and integral calculus as well as multivariable and differential equaitions) In other words in what order would you learn subjects such as analysis , linear algebra, abstract algebra, topology and so on?

i wouldn't recommend any of those. Focus instead on contest math.

Discreet math, since its not taught in school. Also work on challenging/olympiad problems in other subjects like Algebra/Geometry.


A lattice $ L$ of dimension $ n\equal{}251$ has determinant $ det(L)\equal{}2^{2251.58}$. With no further information, approximately how large would you expect the shortest nonzero vector to be?

the result follows from Hermite's theorem. $ \parallel{}v\parallel{}\leq \sqrt{n}\det(L)^{1/n}$


function
calculus
real analysis
real analysis solved

I have another solution, it's hideous, please don't laugh! :D :D
I shall denote y=(pi/2)^(n(n+1)/2)  for easier stuff! :D

lim_{x->0}1/x(y-(arccosx)^n.(arccos2x)^n....(arcoskx)^(n-k+1)...(arccosnx)) =-y* lim(x-->0) 1/[ ln[1+prod(k=1,n)(2/pi*arccos(kx))^(n-k+1)-1)]/ prod(k=1,n)(2/pi arccos(kx))^(n-k+1)-1)] * ln (prod(k=1,n)(2/pi arccos(kx))^(n-k+1))/x=
=-y*lim(x-->0)ln (prod(k=1,n)(2/pi arccos(kx))^(n-k+1))/x=-y*sum lim(x-->0)ln (2/pi arccos(kx))^(n-k+1))/x=...= y*n(n+1)(n+2)/6

the ... is simple calculus! :D :D

cheers!

I think different solution(nice, ugly,short, long...)
for a problem is important, i always learn
something with different solutions for a problem.

Theorem  MO: The map MO: S--->P is surjective where P is the set of all mathematics problems, S is the set of solutions.


trigonometry
ARML
real analysis
real analysis unsolved

Show that if $ \lambda_1,\;\lambda_2,\;\lambda_3,\dots$ are the positive solutions of the equation $ \tan x \equal{} x$, then
\[ \sum_{n\equal{}1}^{\infty}\frac{1}{\lambda_{n}^2}\equal{}\frac{1}{10}\]

I do not understand how it can have positive solutions.

$ f(x)\equal{} tan x \minus{} x , f(0)\equal{}0$ and f  is  increasing  for  $ x>0$

[quote="apratimgtr"]f  is  increasing  for  $ x > 0$[/quote]
Not after $ \frac{\pi}{2}$!

Incidentally, this mistake cost me two thousand dollars last year.  :|

[quote="t0rajir0u"][quote="apratimgtr"]f  is  increasing  for  $ x > 0$[/quote]
Not after $ \frac {\pi}{2}$!

Incidentally, this mistake cost me two thousand dollars last year.  :|[/quote]

Ya,I understood.But how it can cost two thousand dollars?

My team made it to the final round of [url=http://www.ams.org/wwtbam/]Who Wants to be a Mathematician[/url] at ARML 2008, which was worth two thousand dollars.  I convinced my team I knew the correct answer to the problem, which involved taking repeated tangents, and I was wrong.  :|


#1. bookaholic (graphic)

[img]http://img.photobucket.com/albums/v189/bigsister/hack.gif[/img]

#2. ccy (graphic)

[url]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=268[/url]

#3. ToBe (graphic)

[img]http://www.artofproblemsolving.com/Forum/files/haxoring_794.jpg[/img]

You have one week to vote. Next Sunday we will find out the winner! I will vote in case of a tie.

NOTE: The competitors CAN vote for themselves.

Sorry ccy, but because your pic was in an album, I was unable to post it here. I have provided a link instead.

[quote="bookaholic"]Doesn't ccy have a more recent edition of that, with more computers?[/quote]

Oops.  :blush:

Fixed now. Sorry.

Here's the image not in an album.

[img]http://img.photobucket.com/albums/v57/ccy111/stuff/hv2.jpg[/img]


algebra
polynomial
limit
geometry
3D geometry
sphere
function

IMO 1980 ShortList and 1980 replacement IMOs:

[b]Replacement IMO1980 in Mersch (Luxembourg) Problems.[/b]

My friends will post other problems.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2303680]Problem 1.[/url] Let $p(x)$ be a polynomial with integer coefficients such that $p(0)=p(1)=1$. We define the sequence $a_0, a_1, a_2, \ldots, a_n, \ldots$ that starts with an arbitrary integer $a_0$ and satisfies $a_{n+1}=p(a_n)$ for all $n \in \mathbb N$. Prove that $\gcd(a_i,a_j)=1$ for all $i,j \in \mathbb N \cup \{0\}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2303683]Problem 2.[/url] Let $p: \mathbb C \to \mathbb C$ be a polynomial with degree $n$ and complex coefficients which satisfies
\[x \in \mathbb R \iff p(x) \in \mathbb R.\]
Show that $n=1$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2303688]Problem 3.[/url] Two circles $C_1$ and $C_2$ are tangent at a point $P$. The straight line at $D$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $C$. Prove that the straight line $PA$ is a bisector (interior or exterior) of the angle $BPC$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2303690]Problem 4.[/url] Given a real number $x>1$, prove that there exists a real number $y >0$ such that
\[\lim_{n \to \infty} \underbrace{\sqrt{y+\sqrt {y + \cdots+\sqrt y}}}_{n \text{ roots}}=x.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2303696]Problem 5.[/url] In the Euclidean three-dimensional space, we call [i]folding[/i] of a sphere $S$ every partition of $S \setminus \{x,y\}$ into disjoint circles, where $x$ and $y$ are two points of $S$. A folding of $S$ is called [b]linear[/b] if the circles of the [i]folding[/i] are obtained by the intersection of $S$ with a family of parallel planes or with a family of planes containing a straight line $D$ exterior to $S$. Is every [i]folding[/i] of a sphere $S$ [b]linear[/b]?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2303707]Problem 6.[/url] Given the polygons $P$ and $Q$ as shown in the grid below, cut $P$ into two polygons $P_1$ and $P_2$ such that, when pasted together differently, they form $Q$.

[asy]
import graph; size(16cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-2.05,xmax=15.10,ymin=-1.87,ymax=9.74;
pen cqcqcq=rgb(0.75,0.75,0.75), zzttqq=rgb(0.6,0.2,0);
draw((7,5)--(12,5)--(12,2)--(7,2)--cycle,zzttqq); draw((2,2)--(2,5)--(3,6)--(6,6)--(6,3)--(5,2)--cycle,zzttqq);
/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1;
for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);
 draw((0,8)--(0,0)); draw((0,0)--(13,0)); draw((13,0)--(13,8)); draw((13,8)--(0,8)); draw((7,5)--(12,5),zzttqq); draw((12,5)--(12,2),zzttqq); draw((12,2)--(7,2),zzttqq); draw((7,2)--(7,5),zzttqq); draw((2,2)--(2,5),zzttqq); draw((2,5)--(3,6),zzttqq); draw((3,6)--(6,6),zzttqq); draw((6,6)--(6,3),zzttqq); draw((6,3)--(5,2),zzttqq); draw((5,2)--(2,2),zzttqq);
dot((0,0),linewidth(1pt)+ds); dot((13,0),linewidth(1pt)+ds); dot((0,8),linewidth(1pt)+ds); dot((2,2),linewidth(1pt)+ds); dot((6,6),linewidth(1pt)+ds); dot((13,8),linewidth(1pt)+ds); dot((7,2),linewidth(1pt)+ds); dot((7,5),linewidth(1pt)+ds); dot((12,2),linewidth(1pt)+ds); dot((12,5),linewidth(1pt)+ds); label("$Q$",(8.42,2.56),NE*lsf,zzttqq); dot((5,2),linewidth(1pt)+ds); dot((6,3),linewidth(1pt)+ds); dot((2,5),linewidth(1pt)+ds); dot((3,6),linewidth(1pt)+ds); label("$P$",(4.65,2.74),NE*lsf,zzttqq);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]

[b] IMO 1980 Great Britain Problems [/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=411710]Problem 7[/url] Prove that $4x^3-3x+1=2y^2$ has at least $31$ solutions in positive integers $x,y$ with $x\le 1980$
[i] Variant: [/i]  Prove the equation $4x^3-3x+1=2y^2$ has infinitely many solutions in positive integers x,y.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=411711]Problem 8[/url] Prove that if (a,b,c,d) are positive integers such that $\left( a+2^{\frac13}b+2^{\frac23}c)^2=d\right)$ then d is a perfect square (i.e is the square of a positive integer).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=411707]Problem 9[/url]  Prove that is $x,y$ are non negative integers  then $5x\ge 7y$ if and only if there exist non-negative integers $(a,b,c,d)$ such that
$\left\{\begin{array}{l}x=a+2b+3c+7d\qquad\\ y=b+2c+5d\qquad\\ \end{array}\right. $

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=411713]Problem 10[/url] The function f is defined on the set $\mathbb{Q}$ of all rational numbers and has values in $\mathbb{Q}$.  It satisfies the conditions $f(1)=2$ and $f(xy)=f(x)f(y)-f(x+y)+1$ for all $x,y \in \mathbb{Q}$.  Determine f (with proof)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=410774]Problem 11[/url] A triangle $(ABC)$ and a point $D$ in its plane satisfy the relations
\[\frac{BC}{AD}=\frac{CA}{BD}=\frac{AB}{CD}=\sqrt{3}.\]
Prove that $(ABC)$ is equilateral and $D$ is its center.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=410773]Problem 12[/url] There is a triangle $ABC$.  Its circumcircle and its circumcentre are given.  Show how the orthocentre of $ABC$ may be constructed using only a straightedge (unmarked ruler).  [The straightedge and paper may be assumed large enough for the construction to be completed]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2311817#p2311817]Problem 13.[/url]
Prove that the integer $145^{n} + 3114\cdot 138^{n}$ is divisible by $1981$ if $n=1981$, and that it is not divisible by $1981$ if $n=1980$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2311818#p2311818]Problem 14.[/url]
Let $A$ be a fixed point in the interior of a circle $\omega$ with center $O$ and radius $r$, where $0<OA<r$. Draw two perpendicular chords $BC,DE$ such that they pass through $A$. For which position of these cords does $BC+DE$ maximize?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2311827#p2311827]Problem 15.[/url]
Three points $A,B,C$ are such that $B\in AC$. On one side of $AC$, draw the three semicircles with diameters $AB,BC,CA$. The common interior tangent at $B$ to the first two semicircles meets the third circle $E$. Let $U,V$ be the points of contact of the common exterior tangent to the first two semicircles.
Evaluate the ratio $R=\frac{[EUV]}{[EAC]}$ as a function of $r_{1} = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$, where $[x]$ denotes the area of polygon $X$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2311821#p2311821]Problem 16.[/url]
In a pentagon $\Pi$ in the plane, $M_1,...M_5$ are the midpoints of the consecutive sides. $Z_i$ is the centroid of the triangle $M_{i} M_{i+1} M_{i+3}$, where $i=1,2...5$ and it is understood that $M_{j\cdot 5}=M_j$ Given pentagon $Z_{1}Z_{2}Z_{3}Z_{4}Z_{5}$, determine the original pentagon $\Pi$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2311824#p2311824]Problem 17.[/url]
Ten gamblers start playing with the same amount of money. In turn they cast five dice. If the dice show a total of $n$, the player must pay each other player $\frac{1}{n}$ times the sum which that player owns at the moment. They throw and pay one after the other. At the $10^{\text{th}}$ round (i.e. after each player has cast the five die once), the dice shows a total of $12$ and after the payment it turns out that every player has exactly the same sum as he had in the beginning. Is it possible to determine the totals shown by the dice at the nine former rounds?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2311828#p2311828]Problem 18.[/url]
Do there exist $\{x,y\}\in\mathbb{Z}$ satisfying $(2x+1)^{3}+1=y^{4}$.

AHP will add the problems to the contests section.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=5570]Problem 19.[/url] Find all pairs of solutions $(x,y)$:

\[ x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1). \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=411697]Problem 20.[/url] The radii of the circumscribed circle and the inscribed circle of a regular $n$-gon, $n\ge 3$ are denoted by $R_n$ and $r_n$, respectively. Prove that
\[\frac{r_n}{R_n}\ge\left(\frac{r_{n+1}}{R_{n+1}}\right)^2.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=411698]Problem 21.[/url] Let $ABCDEFGH$ be the rectangular parallelepiped where $ABCD$ and $EFGH$ are squares and the edges $AE,BF,CG,DH$ are all perpendicular to the squares. Prove that if the $12$ edges of the parallelepiped have integer lengths, the internal diagonal $AG$ and the face diagonal $AF$ cannot both have integer length.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=5567]Problem 22.[/url] Let $p$ be a prime number. Prove that there is no number divisible by $p$ in the $n-th$ row of Pascal's triangle if and only if $n$ can be represented in the form $n = p^sq - 1$, where $s$ and $q$ are integers with $s \geq 0, 0 < q < p$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=411699]Problem 23.[/url] Let $a, b$ be positive real numbers, and let $x, y$ be complex numbers such that $|x| = a$ and $|y| = b$. Find the minimal and maximal value of
\[\left|\frac{x + y}{1 + x\overline{y}}\right|\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=411700]Problem 24.[/url] Let $k$ be the incircle and let $l$ be the circumcircle of the triangle $ABC$. Prove that for each point $A'$ of the circle $l$, there exists a triangle $(A'B'C')$, inscribed in the circle $l$ and circumscribed about the circle $k.$

Thank you, friends. Added [url=http://www.artofproblemsolving.com/Forum/resources.php?mode=edit_problems&c=1&cid=16&year=1980]here[/url] (see Mersch (Luxembourg) in that page).

IMO problems are complete in AoPS resources page now.  :clap2:


Given that $ a$ and $ b$ are positive integers and $ a$ is a multiple of $ b$, what is the least common multiple of $ a$ and $ b$?

Let $ a=kb$.  Then, we have
\begin{align*}
\text{LCM}(a,b) &= \text{LCM}(kb,b) \\
&= b\cdot \text{LCM}(k,1) \\
&= bk \\
&= \boxed{a}
\end{align*}
Of course, you could also just use the definition of least common multiple to get $ a$.


[size=150]This is a week inequality. Please answer it with a agreeable solution.TNX
Let a, b,c are real numbers. Prove that[/size]

[size=200](a+b)^4 + (b+c)^4 + (c+a)^4 >= 4/7 (a+b+c)^4[/size]

You can try with the stronger of can_hang:
Given the reals $ a, b, c$. Prove that:
$ (a\plus{}b)^4\plus{}(b\plus{}c)^4\plus{}(c\plus{}a)^4 \geq\ \frac{4}{7}[a^4\plus{}b^4\plus{}c^4\plus{}(a\plus{}b\plus{}c)^4]$
When does equality occurs?  :)

Hi Guys
I think I had a mistake in sending the problem. That is:
(a+b)^4 + (b+c)^4 + (c+a)^4 >= 4/7 (a^4 + b^4 + c^4)
Also Can you tell me something about CAN_HANG?
TNX

[quote="nguoivn"]Yes, my friend. Can_hang is same to my brother and he is also my idol about ineq  :oops:
Your problem which you post is VNMO 1996 and it's very weak than can_hang's result (of course, we still can prove his result easily) :)
[quote="nguoivn"]You can try with the stronger of can_hang:
Given the reals $ a, b, c$. Prove that:
$ (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \geq\ \frac {4}{7}[a^4 \plus{} b^4 \plus{} c^4 \plus{} (a \plus{} b \plus{} c)^4]$
When does equality occurs?  :)[/quote]
[quote="dhvdochilinh"]let $ a,b,c > 0$ there positive $ a \plus{} b \plus{} c \equal{} 1$.prove that

1.     $ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq 3(a^2 \plus{} b^2 \plus{} c^2)$.an easy inequalitiy :P

 :P[/quote]

It's weaker than a well-known:
$ \sum\ \frac {a^2}{b} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$[/quote]
please prove your ineq

[quote="dhvdochilinh"]let $ a,b,c > 0$ there positive $ a \plus{} b \plus{} c \equal{} 1$.prove that

1.     $ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq 3(a^2 \plus{} b^2 \plus{} c^2)$.an easy inequalitiy :P

 :P[/quote]
Cauchy schwartz gives us:
$ \left ( \sum_{cyc} \frac{a^2}{b} \right ) \left ( \sum_{cyc} a^2b \right ) \ge \left ( \sum_{cyc} a^2 \right )^2$

So it suffices to prove that: $ \left ( \sum_{cyc} a^2 \right )\left ( \sum_{cyc} a \right ) \ge 3 \left ( \sum_{cyc} a^2b \right ) \iff$
$ \sum_{cyc} a(a\minus{}b)^2 \ge 0$ which is true :)

[quote="daplikos"]Hi Guys
I think I had a mistake in sending the problem. That is:
(a+b)^4 + (b+c)^4 + (c+a)^4 >= 4/7 (a^4 + b^4 + c^4)
[/quote]
I lost the link.
Let $ f(a,b,c) \equal{} (a \plus{} b)^4 \plus{} (b \plus{} c)^4 \plus{} (c \plus{} a)^4 \minus{} \frac {4}{7}(a^4 \plus{} b^4 \plus{} c^4)$ and $ a(a \plus{} b \plus{} c)\geq0.$
Hence, $ f(a,b,c) \minus{} f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right) \equal{}$
$ \equal{} (b \minus{} c)^2\left(3a(a \plus{} b \plus{} c) \plus{} \frac {3(7b^2 \plus{} 10bc \plus{} 7c^4)}{56}\right)\geq0.$ The rest is easy.
I think the Can_hang's inequality can be proven by the same way.

[quote="dhvdochilinh"][quote]than a wellIt's weaker -known:
$ \sum\ \frac {a^2}{b} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$[/quote]
please prove your ineq[/quote]

\[ \Leftrightarrow \]$ \sum ({\frac {{{a^2}}}{b} \minus{} 2a \plus{} b} ) \ge (a \plus{} b \plus{} c)(\frac {{{a^2} \plus{} {b^2} \plus{} {c^2}}}{{ab \plus{} bc \plus{} ca}} \minus{} 1) \\
\sum {\frac {{{{(a \minus{} b)}^2}}}{b} \ge (a \plus{} b \plus{} c)(\frac {{\sum {{{(a \minus{} b)}^2}} }}{{2(ab \plus{} bc \plus{} ca)}}})$
 \[ \Leftrightarrow \]$ \sum {{{(a \minus{} b)}^2}(\frac {1}{b}} \minus{} \frac {{a \plus{} b \plus{} c}}{{2(ab \plus{} bc \plus{} ca)}}) \ge 0$
use SOS method we have. :P

[quote="nguoivn"]See it here:  :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=212844[/quote]
Thank you!
[quote="dhvdochilinh"][quote]than a wellIt's weaker -known:
$ \sum\ \frac {a^2}{b} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$[/quote]
please prove your ineq[/quote]
For you, dhvdochilinh! :)
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ \sum_{cyc} \frac {a^2}{b} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}\Leftrightarrow v^2\sum_{cyc}a^3c\geq3u(3u^2 \minus{} 2v^2)w^3\Leftrightarrow$
$ \Leftrightarrow v^2\sum_{cyc}(a^3b \plus{} a^3c) \minus{} 6u(3u^2 \minus{} 2v^2)w^3\geq v^2\sum_{cyc}(a^3b \minus{} a^3c)\Leftrightarrow$
$ \Leftrightarrow3\left(3u^2v^4 \minus{} 2v^6 \minus{} (2u^3 \minus{} uv^2)w^3\right)\geq(a \minus{} b)(a \minus{} c)(b \minus{} c)uv^2.$
We obtain: $ 3u^2v^4 \minus{} 2v^6 \minus{} (2u^3 \minus{} uv^2)w^3\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^3b \plus{} a^3c)\sum_{cyc}ab\geq2(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)abc\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^4b^2 \minus{} a^3b^2c)\geq0,$ which is true.
Hence, it's enough to prove that
$ 9\left(3u^2v^4 \minus{} 2v^6 \minus{} (2u^3 \minus{} uv^2)w^3\right)^2\geq(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2u^2v^4.$
But $ 9\left(3u^2v^4 \minus{} 2v^6 \minus{} (2u^3 \minus{} uv^2)w^3\right)^2\geq(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2u^2v^4\Leftrightarrow$
$ 9\left(3u^2v^4 \minus{} 2v^6 \minus{} (2u^3 \minus{} uv^2)w^3\right)^2\geq$
$ \geq27(3u^2v^4 \minus{} 4v^6 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} w^6)u^2v^4\Leftrightarrow$
$ \Leftrightarrow(u^4 \minus{} u^2v^2 \plus{} v^4)u^2w^6 \minus{} (u^2 \plus{} v^2)uv^6w^3 \plus{} v^{12}\geq0,$ which is true since,$ \left((u^2 \plus{} v^2)uv^6\right)^2 \minus{} 4(u^4 \minus{} u^2v^2 \plus{} v^4)u^2v^{12}\leq0\Leftrightarrow$
$ \Leftrightarrow \minus{} 3(u^2 \minus{} v^2)^2u^2v^{12}\leq0.$

\[ \begin{array}{l} a,b,c,k > 0.a \plus{} b \plus{} c \equal{} 1. prove that: \\
{a^3} \plus{} {b^3} \plus{} {c^3} \plus{} kabc \ge \min \left\{ {\frac {1}{4},\frac {1}{9} \plus{} \frac {d}{{27}}} \right\} \\
\end{array}
\]
can you prove this ineq

[quote="nguoivn"][quote="dhvdochilinh"][quote]than a wellIt's weaker -known:
$ \sum\ \frac {a^2}{b} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$[/quote]
please prove your ineq[/quote]
@arqady: After expand, this ineq is only:
$ x^2 \plus{} y^2 \plus{} z^2 \geq\ xy \plus{} yz \plus{} zx$
with $ x \equal{} ab^2$;...  :oops:[/quote]
Wow! :D


function
inequalities
inequalities unsolved

Show that if x is in R(n) (or C(n)) and $ 1 \le p_1  \le p_2$ then $ \left( {\sum\limits_{i \equal{} 1}^n {|x_i |^{p_2 } } } \right)^{{\textstyle{1 \over {p_2 }}}}  \le \left( {\sum\limits_{i \equal{} 1}^n {|x_i |^{p_1 } } } \right)^{{\textstyle{1 \over {p_1 }}}}$.

Just hard for me.  :blush:

The function $ f(x) \equal{} x^{\frac {p_2}{p_1} }$ is convex on $ (0, \infty)$.  Apply Jensen's inequality to the set of values of $ |x_i|^{p_1}$.

What are the weights supposed to be? I tried Jensen's inequality but the inequality is in the opposite direction.

My apologies - I misread the statement and thought this was power mean for some reason.  It's possible to reduce to the case that $ p_1 \equal{} 1$ and $ p_2$ is a positive integer and then one just compares terms.  For example, when $ p \equal{} 2$ the inequality

$ (\sum a_i^2) \le (\sum a_i)^2$

should be obvious by expansion.

Edit:  Okay, so you might not be able to get to that case, but you can reduce to the case that $ p_1$ and $ p_2$ are both positive integers and then argue by continuity.

How does one argue from continuity if it is only proved for natural number exponents?

Proving for natural number exponents is equivalent to proving it for $ p_1 \equal{} 1$ and $ p_2$ rational.  Then the function $ f \left( \frac{p_2}{p_1} \right) \equal{} \text{LHS} \minus{} \text{RHS}$ is continuous and the result follows.


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

$ \sum \frac {\sqrt {a \plus{} b}}{c} \geq \sqrt {\frac {6\sum ab}{abc}}$[/quote]
We can prove this inequality in the same way with the following inequality: http://www.mathlinks.ro/viewtopic.php?t=310932

Another proof...
the problem is equivalent to
$ (\sum_{cyc} ab\sqrt {a \plus{} b})^2\geq  \sqrt{(6abc\sum_{cyc} ab)^2}$
$ \sum_{sym} a^3b^2 \plus{} 2\sum_{cyc} a^2bc\sqrt {(a \plus{} b)(a \plus{} c)}\geq 6\sum_{cyc} a^2b^2c$
from cauchy it suffices to prove
$ \sum_{sym} a^3b^2 \plus{} 2abc(\sum_{cyc} a^2 \plus{} sum_{cyc} a\sqrt {bc})\geq 6\sum_{cyc} a^2b^2c$
but from schur inequality $ 2abc(\sum_{cyc} a^2 \plus{} \sum_{cyc} a\sqrt {bc})\geq 2abc(\sum_{sym} a^\frac {3}{2}b^\frac {1}{2})$
It remains to show that
$ \sum_{sym} a^3b^2 \plus{} 2abc(\sum_{sym} a^\frac {3}{2}b^\frac {1}{2})\geq 6abc\sum_{cyc} ab$
which is now obviously true by AM-GM


limit
logarithms
integration
function
calculus
calculus computations

Prove $ \forall\,x\in\mathbb R,$

$ \lim_{n\to\infty}\Big(1\plus{}\frac xn\Big)^n\equal{}e^x.$

$ \lim_{n\rightarrow\infty}\left(1 \plus{} \frac {x}{n}\right)^n \equal{} e^{\ln\left(\lim_{n\rightarrow\infty}\left(1 \plus{} \frac {x}{n}\right)^n\right)} \equal{} e^A$

$ A \equal{} \lim_{n\rightarrow\infty}\ln\left(\left(1 \plus{} \frac {x}{n}\right)^n\right) \equal{}  \lim_{n\rightarrow\infty}n\ln\left(1 \plus{} \frac {x}{n}\right) \equal{}\lim_{n\rightarrow 0}\frac {1}{n}\ln\left(1 \plus{} n\,x\right)$

use L'Hospital's Rule:

$ A \equal{} \lim_{n\rightarrow 0}\frac {x}{1 \plus{} n\,x} \equal{} x$

so

$ e^A \equal{} e^x$

I knew you'd use L'Hôpital's Rule.

And without it?

We can use $ \ln u \equal{} \int_1^u \frac {dt}{t}$. The limit is reduced to proving
\[ \lim_{n\rightarrow \infty} n\ln \left( 1 \plus{} \frac {x}{n}\right) \equal{} x
\]
However,
\[ \ln \left( 1 \plus{} \frac {x}{n}\right) \equal{} \int_1^{1 \plus{} \frac {x}{n}} \frac {dt}{t}
\]
If $ x\geq 0$, then for $ t\in \left[ 1,1 \plus{} \frac {x}{n}\right]$, we have $ \frac {n}{n \plus{} x}\leq \frac {1}{t}\leq 1$ so that
\[ \frac {x}{n \plus{} x}\leq \ln \left( 1 \plus{} \frac {x}{n}\right) \leq \frac {x}{n}
\]
Multiplying by $ n$ gives a function which has upper and lower bounds that tend to $ x$ as $ n\rightarrow \infty$. The result follows.

If $ x < 0$, then we have $ 1 \plus{} \frac {x}{n} < 1$ and similar bounds can be put on the function (except the upper and lower bounds in the end are reversed).

\[ \lim_{n\to\infty}\Big(1 \plus{} \frac xn\Big)^n \equal{}\lim_{n\to\infty}\sum_{k\equal{}0}^{n}\binom{n}{k}\Big(\frac {x}{n}\Big)^k\]
\[ \equal{} \lim_{n\to\infty}\sum_{k\equal{}0}^{n}\frac{n(n\minus{}1)\cdots (n\minus{}k\plus{}1)}{n^k}\frac {x^k}{k!}\]
\[ \equal{} \sum_{k\equal{}0}^{\infty}\frac {x^k}{k!} \equal{} e^x\]


You given 25 different positive numbers. Prove that You can choose two of them such that none of the other numbers equals to the sum or to the difference of the chosen numbers.

Let $\ S=(a_{1},a_{2},...,a_{n})$ and $\ a_{1}<a_{2}<...<a_{n}$
We try to construct $\ S$such that for any $\ j>i,(a_{j}-a_{i})\in S$
If $\ a_{n}$ is the biggest number we most have: $\ a_{n}-a_{n-i}=a_{i},\forall i\leq n-1$  (1)
Prove: let $\ a_{n}-a_{1}=a_{j}$ and $\ a_{n}-a_{k}=a_{n-1}$
we know that $\ a_{1}<a_{k}$ so $\ a_{j}>a_{n-1}$ wich cannot be, so $\ k=1$ and $\ n-1=j$
The same goes for $\ a_{n}-a_{n-2}=a_{2}$ and so on.
With the same reasoning we most have: $\ a_{n-1}-a_{n-i-1}=a_{i},\forall i\leq n-2$  (2)

Combining Eqs (1) and (2) we find:
$\ a_{n}-a_{1}=a_{1}$
$\ a_{n}-a_{2}=a_{2}$
$\ a_{n-1}-a_{n-2}=a_{1}$ so that $\ 2a_{1}=a_{2}$
and
$\ a_{n}-a_{n-2}=a_{2}$
$\ a_{n}-a_{n-3}=a_{3}$
$\ a_{n-1}-a_{n-3}=a_{2}$ so that $\ a_{n-1}-a_{n-2}=2a_{2}-a_{3}=a_{1}$
this gives $\ 3a_{1}=a_{3}$

If we continue we find that $\ a_{k}=ka_{1}$ and that's our rule for constructing $\ S$. For any other configuration for the elements of $\ S$ there is always some $\ j>i$ such that $\ (a_{j}-a_{i})\notin S$ and $\ (a_{j}+a_{i})\notin S$ Why? suppose all the elements of $\ S$ follow the rule sauf one say $\ a_{r}$ then we simply choose the biggest number $\ a_{n}$ and $\ a_{r}$ and we have $\ (a_{n}-a_{r})\notin S$ and $\ (a_{n}+a_{r})\notin S$

[quote="El fisico"]Let $\ S=(a_{1},a_{2},...,a_{n})$ and $\ a_{1}<a_{2}<...<a_{n}$
We try to construct $\ S$such that for any $\ j>i,(a_{j}-a_{i})\in S$
[/quote]
I could be wrong, but I think we wanted to prove that for any $i,j$, $(a_{j}-a_{i}),(a_{j}+a_{i}) \notin S$ leads to a contradiction. So the set you constructed isn't really what we wanted.

[quote="Kurt Gödel"][quote="El fisico"]Let $\ S=(a_{1},a_{2},...,a_{n})$ and $\ a_{1}<a_{2}<...<a_{n}$
We try to construct $\ S$such that for any $\ j>i,(a_{j}-a_{i})\in S$
[/quote]
I could be wrong, but I think we wanted to prove that for any $j$, $(a_{j}-a_{i}),(a_{j}+a_{i}) \notin S$ leads to a contradiction. So the set you constructed isn't really what we wanted.[/quote]
The question was if we can find two integers (not any) such that  $i,j$, $(a_{j}-a_{i}),(a_{j}+a_{i}) \notin S$
All I prove is that: yes we can find two integers $i,j$, $(a_{j}-a_{i}),(a_{j}+a_{i}) \notin S$ IF AND ONLY IF there is at least one element $\ a_{r}$ of $\ S$ such that $\ a_{r}\neq ra_{1}$

Hmm, that was...a typo  :D
I'm still not convinced by your proof though.

[quote="El fisico"]If we continue we find that $\ a_{k}=ka_{1}$ and that's our rule for constructing $\ S$. For any other configuration for the elements of $\ S$ there is always some $\ j>i$ such that $\ (a_{j}-a_{i})\notin S$ and $\ (a_{j}+a_{i})\notin S$ Why? suppose all the elements of $\ S$ follow the rule sauf one say $\ a_{r}$ then we simply choose the biggest number $\ a_{n}$ and $\ a_{r}$ and we have $\ (a_{n}-a_{r})\notin S$ and $\ (a_{n}+a_{r})\notin S$[/quote]Why just excluding [i]one[/i] element of $S$, $a_{r}$?

[quote="Kurt Gödel"]Hmm, that was...a typo  :D
I'm still not convinced by your proof though.
Why just excluding [i]one[/i] element of $S$, $a_{r}$?[/quote]

Ok, it's all about the biggest number $\ a_{n}$
Because it's easy to see that $\ a_{n}+a_{i}\notin S$ for any $\ a_{i}\in S$
Now I want to know if there exist $\ S$ such that $\ a_{n}-a_{i}\in S$ for any $\ a_{i}\in S$ because if not then that answer the question. It turns out that yes we can find such $\ S$ only if follow the rule $\ a_{n}-a_{n-i}=a_{i}$(as I proved) wich leads to $\ a_{r}=ra_{1}$
Now it's clear that any element(there can be a lot of them) that don't follow that rule can be chosen such that $\ a_{n}-a_{i}\notin S$

[quote="pkerichang"]hmm...didn't you post this in intermediate?? is double-posting allowed??? :huh:
[hide]
I didn't really solve it, but my instinct tells me I have to pigeon-hole somehow, right??? :maybe:
[/hide][/quote]
Well, i waited a few days for some solutions in the intermediate forum. When none came, i guessed it wasn't intermediate. Thanks for your contribution guys. :)


geometry
circumcircle
ratio
geometric transformation
reflection
geometry unsolved

ABCD - convex quadrangle such that <ABC=<ADC. Let DM be the height in triangle ADC and M' be a point on AC with the property:
AM.CM'.AB.CD=AM'.CM.AD.CB. Prove thatv the lines BM' and DM intersect one another at the orthocentre of triangle ADC.

Let H be orthocenter of the $\triangle ADC$ and let (O), (P) be circumcircles of the $\triangle ABC,\ \triangle ADC.$ Since $\angle ADC = \angle ABC,$ $\angle AHC = 180^\circ-\angle ADC = 180^\circ-\angle ABC$ and $H \in (O).$ Let BH meet AC at N. Let the internal bisectors of the $\angle ABC,\ \angle ADC$ meet AC at K, L and the circles (O), (P) at U, V. Then

$\frac{AK}{CK}= \frac{AB}{CB},\ \frac{AL}{CL}= \frac{AD}{CD}.$

By the problem condition, the cross ratios

$\frac{AL}{CL}\cdot \frac{CM}{AM}= \frac{AD}{CD}\cdot \frac{CM}{AM}= \frac{AB}{CB}\cdot \frac{CM'}{AM'}= \frac{AK}{CK}\cdot \frac{CM'}{AM'}$

are equal. Reflect D, (P) in the line AC. The circle (P) goes into (O), the point D into $E \in (O),$ and the points A, C, K, L, M, M', N stay in place. The altitude DM of the $\triangle ADC$ goes into the altitude $EM \equiv DM$ of the $\triangle AEC,$ the point V into U, the bisector DLV of $\angle ADC$ into the bisector ELU of $\angle AEC.$ The rays EA, EH, EU, EC form the same angles as the rays BA, BH, BU, BC. Let A*, M*, L*, C* be points on the rays BA, BH, BU, BC, such that BA* = EA, BM* = EM, BL* = EL, BC* = EC*. Then the figures $BA^{*}M^{*}L^{*}C^{*}\cong EAMLC$ are congruent, the points A*, M*, L*, C* collinear, hence the cross ratios

$\frac{A^{*}L^{*}}{C^{*}L^{*}}\cdot \frac{C^{*}M^{*}}{A^{*}M^{*}}= \frac{AL}{CL}\cdot \frac{CM}{AM}$

are equal. On the other hand, the points $A^{*},\ M^{*},\ L^{*},\ C^{*}\in A^{*}C^{*}$ are images of $A,\ N,\ K,\ C \in AC$ in a central projection of the line AC to the line A*C* from B, so that the cross ratios

$\frac{A^{*}L^{*}}{C^{*}L^{*}}\cdot \frac{C^{*}M^{*}}{A^{*}M^{*}}= \frac{AK}{CK}\cdot \frac{CN}{AN}$

are also equal. Consequently the cross ratios

$\frac{AL}{CL}\cdot \frac{CM}{AM}= \frac{AK}{CK}\cdot \frac{CN}{AN}$

are equal, which means that the points $N \equiv M'$ are identical.


Both the diameter of the semicircular region shown and the base of the
isosceles triangle shown are sides of a 20 cm by 20 cm square. The
two regions share exactly one point. Find the total number of square
centimeters in the shaded regions. Express your answer in terms of $ \pi$.

[asy]fill((0,0)--(20,0)--(20,20)--(0,20)--cycle,gray(.6));
fill((0,0)--(10,10)--(20,0)--cycle,white);
fill((0,20)..(10,10)..(20,20)--cycle,white);
draw((0,0)--(20,0)--(20,20)--(0,20)--cycle,linewidth(1));
draw((0,0)--(10,10)--(20,0),linewidth(1));
draw((0,20)..(10,10)..(20,20),linewidth(1));
label("20",(20,10),E);[/asy]

The white region at the top is a semicircle with diameter $ 20$ and the triangle is a triangle with $ \frac{1}{4}$ the area of the square.  The answer is $ 400\minus{}50\pi\minus{}100\equal{}\boxed{300\minus{}50\pi}$.


geometry
incenter
circumcircle
trapezoid
parallelogram
geometry proposed

If we call K the point of intersection of AI and BC, the following is a necessary and sufficient condition for that convex quadrilateral DEKI be a trapezoid, with DE and IK parallel:

(EI)^2 + (KD)^2 = (EK)^2 + (DI)^2 +2*DE*KI

(Source: D.Branzei, Competentza si performantza in Geometrie, vol. 1, p.80; Ed.Minied, Iasi)
But I am not sure if this made more hard the solution.....

$ U, V, W$ are tangency points of the incircle $ (I)$ with $ BC, CA, AB,$ $ r$ is inradius. Again, inversion in $ (I).$ $ A$ goes to the midpoint $ A'$ of $ VW.$ The A-mixtilinear incircle $ \odot(PQR)$ goes to the circle $ (A')$ with radius $ r.$ $ R$ goes to $ R',$ such that $ A'R' \parallel IU,$ $ A'R' \equal{} IU \equal{} r.$  See [url]http://www.mathlinks.ro/viewtopic.php?t=211233[/url] for details. $ A'IUR'$ is a parallelogram, $ UR' \parallel IA'.$ Line $ ADR$ goes to a circle $ \odot(IA'D'R'),$ such that $ D' \in ID,$ $ ID' \perp IA',$ $ A'D'$ is a diameter of $ \odot(IA'D'R'),$ $ A'R' \perp R'D'.$ From $ R'D' \perp IU,$ $ UR' \perp ID'$ $ \Longrightarrow$ $ D'$ is orthocenter of the $ \triangle IUR',$ $ UD' \perp IR'.$ But the foot $ E'$ of perpendicular from $ U$ to the line $ IR'R$ is the inversion image of $ E$ $ \Longrightarrow$ the inversion image of the line $ DE$ is the circle $ \odot(IE'D')$ with diameter $ ID',$ tangent to $ AI$ at the inversion center $ I$ $ \Longrightarrow$ $ DE \parallel AI.$

Very nice solution, Yetti. By the way, I guess point A' is the image of the midpoint of VW in the inversion, if I am not wrong. Am I right? Thanks

[quote="Zhang Fangyu"]Given a triangle $ ABC$ with incenter at $ I$. A circle is tangent to its sides $ CA, AB$ and its circumcircle at points $ P,Q,R$, respectively. $ AR$ cuts $ PQ$ at $ D$, $ IR$ intersects $ BC$ at $ E$. Show that $ DE\parallel AI$.[/quote]
Let $ T$ be the second intersection of $ AI$ with its circumcircle, it is well known that three lines $ PQ,CB,TR$ concur at a point, denote by $ F$ the point, we have $ \angle PFB \equal{} \pi \minus{} (\pi \minus{} \angle B) \minus{} \angle FQB \equal{} \frac {\angle B \minus{} \angle C}{2}$.(standardized angle, where we suppose that $ \angle B$ is greater than $ \angle C$)(1)
On the other hand, we also know $ IR\perp RT$, thus the ray $ RI$ passes through the midpoint of arc $ BC$ ( containing $ A$),  that yields $ \angle ARE \equal{} \frac {\angle B \minus{} \angle C}{2}$,  ( 2).
By (1) and (2), we conclude $ D,F,R,E$ are concyclic. hence $ \angle EDF \equal{} \pi \minus{} \angle IRF \equal{} \frac {\pi}{2},$ we are done!

[quote="28121941"]... I guess point A' is the image of the midpoint of VW in the inversion, if I am not wrong. Am I right? Thanks[/quote]

Tx. I wrote: "$ A'$ goes to the midpoint of $ VW.$" and it should be: "$ A$ goes to the midpoint $ A'$ of $ VW.$" I fixed it up there.

Denote S is the intersection of PQ and BC , L is the intersection of RI and the circumcircle , IR ⊥ RS ,∠LRA=∠BSQ=(B-C)/2   (SUPPOSE ∠B>∠C)  D,E,R,S are on one circle
Done!

I was able of finish a metric solution to this really nice problem. I will give some details of the necessary computations in the notes at the end of this post.
With the notations of the initial statement of the problem, we call F the intersection of PQ and BC, and let K be the intersection of AI with BC.
In order to apply the converse of the Thales theorem at the triangles FDE and FIK, we will need to prove that ID*EF=DF*EK.
If we have success, then DE and IK (or AI) will be parallel.
The final expressions for the four involved segments are the following:
We call k = c(s-b) + b(s-c), which can not be simplified and appears during the computations.

EK = abc(b-c)/(b+c)k    (we suppose WLOG b > c).

DF = 2bc(s-b)(s-c)(b+c)(sin(A/2))/s(b-c)k

EF = 2abc(s-b)(s-c)/s(b-c)k

ID = bc(b-c)(sin(A/2))/k

and from this we obtain the searched equality. Therfore, DE and AI are parallel.

Notes:
1) The expression PQ = 2bc(sin(A/2))/s is well known.
2) I find out DP/DQ = b(s-c)/c(s-b) by application of the following resut:
(AQ/AB)*(BM/MC)*(AC/AP)*(DP/DQ) = 1, which appears in a paper from Zvonaru and Ionitza, in the romanian electronic  journal Recreatii Matematice (http://recreatiimatematice.uv.ro). For another hand, the value of BM/MC is obtained by inversion, in the site of Geometrikon, Mixtilinear incircles (http://www.math.uoc.gr/~pamfilos/eGallery/Gallery.html)

3) The value of BE is obtained from the proportion BE/EC = BR/RC = BI^2/CI^2, which is a result of Toshio Seimiya in Crux Mathematicorum 2000, problem 2464, solution in pp. 432-436, (readable in the net, too).

4) The value of FB can be achieved by means of the Menelaus theorem in ABC with the transversal PQF.

5) The value of FQ, with the sinus theorem in FQB

6) the value of BK, with the bisector theorem in ABC, obviously.

Nice problem indeed! Thanks


$I$ is the incentre of $\triangle{ABC}$, A', B' and C' are the circumcentres of $\triangle{IBC}$, $\triangle{IAC}$ and $\triangle{IAB}$. Show that $\triangle{A'B'C'}$ and $\triangle{ABC}$ have the same circumcentre.

The incenter of $IAB$ is the midpoint of the arc $AB$ not containing $C$.


[color=darkred]
 :idea:
let this be a portal for students interested in biology alongwith maths

p.s: i m sure there won't be many
[/color]

There is someone called shruthi in this forum, interested in exactly this.. So I gess that female will def pay heed to this...

please !!!
a bit of less advanced maths here

[color=darkred]
y it is that u always find girls in this???????? :o

[/color]

Do you have a problem with that? Then I suggest you speak up, for I am really not in the mood for offhand chauvenisitc comments

[color=darkred] :oops:
i never said i had a problem
that was just a kind of comment

i was just fed up looking around for boys to talk our stuff.....

[/color]

Our stuff???? And why do you paint your text fully brown. Sadly for you, all the boys here are C++ or PT students.  :rotfl:

well we aint sure abt pardesi though i heard he's a boy who takes biology tutions as well  :P   :rotfl:

[quote="shruthibaskaran"]I think I sounded a bit rude, I am sorry.. Its just that, I also get fed up of people talking about biology being taken up by only girls.....[/quote]

With the exception of one or two weirdos. Aint that true girlie???

Shre, you mean freak.... Wait till I get my hands on you, or better yet I will make sure you die painfully from mental stangulation

[quote="rituraj007"]that's the way it goes[/quote]

Probably he means that you've become mad studying too much bio.  :rotfl:  :rotfl:

[quote="shruthibaskaran"]Thats the way what goes?[/quote]
people think that bio studs are freak

Ah, but maybe to put it this way, sane people dont :D

Listen to this-- i have discovered a new species of genus [i]homo[/i]
acc. to codes its name is--[b]HOMO SARDARA[/b] :evilgrin:
i mabout to register it with ICZN

[color=darkred]was[/color] [color=brown] that[/color] [color=green]a[/color] [color=violet] joke[/color]
 [color=cyan]?????????[/color]

:rotfl:  :omighty:  :rotfl:

cud ne1 suggest a decent topic 4 a investigatory project work on biology???????((4 practical))

i m what i m,wat i look :D

but see,even if i made a project on self anatomy,i can't post it here
and anatomy of all humans is same--its the morphology that varies :rotfl:  :rotfl:

[quote="rituraj007"]and anatomy of all humans is same--its the morphology that varies :rotfl:  [/quote]

A statement that cant be debated by me :D Well ask chappli on this topic, he's an expert :P

not when he is immersed in the pool of jee :rotfl:  :rotfl:  :rotfl:

kkkk
Chappli tell abt whatever Rohit is saying :D

Chumma jujubi. Thats all I can say.

I worked with M.S.Swaminathan Research Foundation on Transgeneticism for my bio project :D Had loads of fun too. Hybrid signals are real fun!! :D

u went 2 M.S.Swaminathan Research Foundation 4 a regular CBSE bio project????????
and on transgeneticism :rotfl:  :rotfl:
that must b cool :D

[color=cyan]really dudes (some) i mean some bio people are insane and chauvenistic![/color]

[quote="madness"]self anatomy? :P :rotfl: Well we all wanna know what kind of a guy u are :D[/quote]
thalai da!!!! :omighty:
sooper da Rohit "madness"Kannan

:rotfl:  :P [quote="rt_08"][color=cyan]really dudes (some) i mean some [color=red]bio people [/color]are insane and chauvenistic![/color][/quote]
can u defined what/who are BIO PEOPLE?  :D  :maybe:

i think they are biology and mathenatics lovers!

4 that matter ,don't include me
cause i m not a real bio lover.....

hey biology is not that stupid coz sometimes i mean sometimes bio can be a lot of fun too!

a lot of FFFFFUUUUU__NNNNNNN :rotfl:  :blush:  :rotfl:


Let $ABCDE$ be a convex pentagon such that $\triangle ABC\sim \triangle AED$ and $m(\widehat {ABC})\ge 90^{\circ}$. Let $M\in (BC),\ N\in (ED)$ be two mobile points such that $\frac{MB}{MC}=\frac{NE}{ND}$. I note the point $L$ for which $AL\cap MN\ne \emptyset$, $LM=LN$ and $m(\widehat {LMN})=m(\widehat {ABC})-90^{\circ}$. Ascertain the geometrical locus of the point $L$.

See:
http://www.mathlinks.ro/Forum/viewtopic.php?t=38041
http://www.mathlinks.ro/Forum/viewtopic.php?t=61873

[u]I am waiting you, Tiks (with another generalization) ![/u]

Ph-An I have solution,but it is not nice,becous it analitcs geometri. :(

[i]Dear Tiks[/i], I saw you proved the generalization of Armo's lemma (using this lemma).
Therefore, and this problem you solved ! Thus, the score is $\underline {\overline {\left| \ 1-1\ \right| }}.$

Hi $Ph-An$.
When are you going to says next problem for me :roll:  :D


let $ x_1\equal{}1$, $ x_2\equal{}\minus{}1$, $ x_{n\plus{}2}\equal{}\left(1\plus{}\frac{(\minus{}1)^n}{n\plus{}1}\right)x_n$
Prove that $ (x_n)$ isn't convergent.

Show $ (x_{2n})$ goes to $ \minus{}\infty$


function
parameterization
Euler
algebra proposed
algebra

For any real $ a$ define $ f_a : \mathbb{R} \rightarrow \mathbb{R}^2$ by the law $ f_a(t) \equal{} \left( \sin(t), \cos(at) \right)$.
a) Prove that $ f_{\pi}$ is not periodic.
b) Determine the values of the parameter $ a$ for which $ f_a$ is periodic.
[b]Remark[/b]. L. Euler proved in $ 1737$ that $ \pi$ is irrational.

Assume that the function is periodic for period $ T > 0$.  Then, in particular $ f_a(0) \equal{} f_a(T)$, or $ sin(T) \equal{} 0$ and $ cos(aT) \equal{} 1$, meaning $ aT \equal{} 2u\pi$ and $ T \equal{} v\pi$ for some positive integers $ u,v$.  Therefore, $ a \equal{} \frac {2u}{v}$, and $ a$ must be rational, so it cannot be $ \pi$.  Furthermore, for any $ a$ rational, we may write it as $ \frac {2u}{v}$ for some integers $ u,v$, and $ f_a$ will be at least periodic with period $ v\pi$.


[quote="geoffrels"]Hi to all!!!

This is my first post here in mathlinks.ro hope answer the following: (this is the Philippine’s Regional Screening Test for the National’s Mathematical Olympiad.)

2. If 2A99561 is equal to the product when 3 x (523 + A) is multiplied by itself, find the digit A.

3. The perimeter of a square inscribed in a circle is p. What is the area of the square that circumscribes the circle?

4. The sum of the first ten terms of an arithmetic sequence is 160. The sum of the next ten terms of the sequence is 340. What is the first term of the sequence?

5. It is given that triangle CAB is congruent to triangle EFD. If AC = x+y+z, AB = z+6, BC = x+8z, EF = 3, DF = 2y-z, and DE = y+2, find x^2+y^2+z^2.

6. Container A contained a mixture that is 40% acid, while container B contained a mixture that is 60% acid. A chemist took some amount from each container, and mixed them. To produce 100 liters of mixture that is 17% acid, she needed to pour 70 liters of pure water to the mixture she got from containers A and B. How many liters did she take from container A?[/quote]

[hide="Question 2"]
Numbers divisible by 3 have a digital sum which is divisible by 3.
Therefore A could be $ 1,4,7$ We note that 2A99561 is odd, therefore A must be even. Hence A  = 4
[/hide]

[hide="Question 3"]
The radius of the circle is given as $ \sqrt {2}\frac {p}{8}$.
Therefore the the vertical distance of the square is $ 2\sqrt {2}\frac {p}{8}$.
Hence the area is $ \frac {p^2}{8}$.
[/hide]

[hide="Question 4"]
The sum of an AP is $ \frac {n}{2}(2a \plus{} (n \minus{} 1)d)$ So we have simultaneous equations
$ \frac {10}{2}(2a \plus{} (9)d) \equal{} 160$
$ \frac {20}{2}(2a \plus{} (19)d) \equal{} 500$
[/hide]

EDIT: Noticed [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=176062[/url]

here is my solution for number 5
[hide="5"]since the problem says that the triangles are congerent set each side equal to each other, so CA=EF, AB=FD, and BC=DE, so we have
x+y+z=3, z+6=2y-z, x+8z=2y+2.
Now sovle one equation for one valuble, I will solve the thrid for x
x=2y+2-8z
now replace x in the other equations to give
2y+2-8z+y+z=3, simplfys to 3y-7z+2=3 and z+6=2y-z
I will now solve the second equation for y
2z+6=2y
y=z+3
Now put this into the first equation to get
3(z+3)-7z+2=3
3z+9-7z+2=3
-4z+11=3
-4z=-8
z=2
now pulg this in to get y
y=(2)+3
y=5
so
x+(5)+(2)=3
x=-4
so
x^2+y^2+z^2=(-4)^2+(5)^2+(2)^2=16+25+4=45
you can check this by pluging the x y and z vaulw back into the equations
[/hide]


Find all k satisfying the following condition:
When 2^n(n=1,2,3,...) is expressed in base k,
there exist two distinct positive integers a,b,
such that we can rearrange the digits of 2^a to make it become 2^b

Just a remark. For $ k\equal{}10$ the problem is discussed in the delightful little book Terence Tao wrote when very young ( see [url]http://www.math.ucla.edu/~tao/preprints/problem.ps[/url] ), with negative answer. I wonder if looking at other bases has been analyzed yet ...

what software is required to read this file , I mean it is ".ps" file ,what is the meaning ".ps"

The ps extension refers to postscript files that may be read with ghostscript, available at
http://pages.cs.wisc.edu/~ghost/

Alternatively, if you have latex distribution, you should have ps2pdf.exe, which can convert the .ps file to a .pdf file.


inequalities
induction
LaTeX
inequalities proposed

Is there any body who can solve this inequality?

                         $\frac{1}{2}*\frac{3}{4}* ... * \frac{99}{100} < \frac{1}{12}$

I think it's not correct

Suppose that $A=\frac 12 \times \frac 34 \times ...\times \frac{99}{100}$ and we know that $\frac 12 < \frac 23 , \frac 34 < \frac 45 ,..., \frac{99}{100} < \frac{100}{101}$.

Then $A<\frac 23 \times\frac 45 \times ...\times \frac{100}{101}=(\frac 21\times \frac 43\times ...\times \frac{100}{99})\times \frac{1}{101} =\frac{1}{A}\times \frac{1}{101}$

Then $A^2<\frac{1}{101}<\frac{1}{100}$ so that $A<\frac{1}{10}$  :? but maybe $\frac{1}{12} <A<\frac{1}{10}$.

i must admit that this is not my proof,  i have seen it on a book and it was very nice!

it can be generalized in fact to
$\frac{1}{2}\times\frac{3}{4}\times...\times\frac{2n-1}{2n}<\frac{1}{\sqrt{3n+1}}$, this can be proven by induction on n.
then, all we have to do is to substitute $n=50$ to obtain
$A=\frac{1}{2} \times \frac{3}{4} \times ...\times \frac{99}{100}<\frac{1}{\sqrt{151}}<\frac{1}{12}$, and we're done

What are you doing? :o
Please don't spam! :mad:

:(
I'm so sorry!
I'm just trying how to post the inqe... I don't know how to use the Latex...
I'm so sorry for my mistake!
 :(

Never mind   ;)
If you want to practise $\LaTeX$, please go to the [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url].
To moderators: please delete the posts above. Thx.  :)


The cubes in the diagram have sides of 1,2, and 3, respectively. If the pattern continues, what will be the sum of the volumes of the first 100 cubes in the sequence? Express your answer in cubic units.
[asy]size(200);
void drawCube(Label l = "", int side, pair corner){
 pair[] front = new pair[4];
 pair[] back = new pair[4];
 front[0] = corner; front[1] = (corner.x + side, corner.y); front[2] = (corner.x + side, corner.y + side); front[3] = (corner.x, corner.y + side);
 draw(front[0]--front[1]--front[2]--front[3]--cycle);
 for(int i = 0; i < 4; ++i){
  back[i] = (front[i].x + 0.4*side, front[i].y + 0.4*side);
 }
  draw(back[0]--back[1]--back[2]--back[3]--cycle);
 draw(front[0]--back[0]); draw(front[1]--back[1]); draw(front[2]--back[2]); draw(front[3]--back[3]);
 label(l,(front[1]--front[2]),E);
}
drawCube("1",1,(0,0)); drawCube("2",2,(3,0)); drawCube("3",3,(7,0));[/asy]

The formula is $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2=(1+2+3+4+...+n)^2$.
I'll leave the proving this (induction would probably be the best idea) to you.

Proof by induction means you prove the base case (in this case n=1). Then assume that the expression is true for an integer $n$. Then we prove $n \implies n+1$.


calculus
integration
logarithms
function
calculus computations

Compute:
\[\int_0^{+\infty}\bigg\{\frac 1{4\sinh^2\frac x2}-\frac 1{x^2}\bigg\}\ln x\,dx\]
$ \{x\}$:fractional part of x.

Cheap argument that the integral converges absolutely:

Call this $\int_0^{\infty}f(x)\,dx.$

Since the the fractional part of anything is bounded by 1,

$\int_0^1|f(x)|\,dx\le\int_0^1|\ln x|\,dx = 1.$

As $x\to\infty,$ the $\sinh$ function grows exponentially. Hence $\frac1{4\sinh^2\frac x2}-\frac1{x^2}=O\left(\frac1{x^2}\right)$ and we can compare

$\int_1^{\infty}|f(x)|\,dx$ to $\int_1^{\infty}\frac{\ln x}{x^2}\,dx<\int_1^{\infty}x^{-3/2}\,dx < \infty.$

The thought processes behind that took about 10 seconds, although it took longer to type it up. Of course, Nené asked for a computation, which hasn't been given yet.

For the previous argument to be true, we must use the version of fractional part that ranges from $-\frac12$ to $\frac12$, so that the fractional part of a small negative number is small in absolute value.


Here's a problem that's a sophomore geometry math team problem that I'm having trouble with:

In the diagram shown, XRTV is a parallelogram with an area of 156.  XW:WV=2:1, and RS:ST=1:3.  Find the number of square units in the area of triangle VYT.

[img]http://img.photobucket.com/albums/v210/lizbubblypop/math.jpg[/img]

Thanks!

Oops.. messed up the second time...
[hide]It's easy to see that the ratio of the dimensions of $\triangle WYV$ to $\triangle TYS$ is four to nine. If we draw a line that is parallel to VT and intersects point Y, that line would split segment WV in a ratio of four to nine, the larger piece being equal to the height of the $\triangle VYT$. Since the height of that triangle is $\frac 9{39}$ of XV, and the base takes up all of VT, the ratio of the area of that triangle to the entire thing is $\frac 12 * \frac 3{13} = \frac 3{26}$. Then, the area of it is $\frac 3{26} * 156 = 18$.[/hide]

Opposite sides of a parallelogram are congruent, so let XV = RT = 12a.  Using the given ratios, it follows that XW= 8a, WV = 4a, RS = 3a, and ST = 9a.  Since triangle WYV is similar to triangle TYS (Why?), it follows that TY/YW = ST/WV = 9/4, so TY/TW=9/13.  Now observe the following:

Area VYT = (9/13) Area TWV  (Why?)
Area TWV = (1/3) Area XVT    (Why?)
Area XVT =(1/2) Area XRTV    (Why?)

This information (and the fact that the given area XRTV = 156) should lead you to the correct answer, which is 18.          Note:  the value 12a was used at the beginning because it divides conveniently according to the given ratios.

You can also use the two poles method so you know that if  WV and ST have lengths 4a and 9a, then the parallel distance from y to VT is (36/13)a.  Ideal casing it, making it a 12x13 rectangle, you can use area of triangle =
[hide]1/2 * 36/13 * 13 = 36/2 = 18[/hide]


[quote="benorin"]Let $u=\ln\ln x\Rightarrow du=?$[/quote]
?  :( now what?

[quote="santosguzella"]Either m a bigger real number that 1, calculate:
$\int \frac{1}{xlnx(lnlnx)^{m}}\,dx$[/quote]

Let $u=\ln\ln x\Rightarrow du=\frac{1}{\ln x}\cdot\frac{1}{x}dx$ so we have that $\int \frac{1}{xlnx(lnlnx)^{m}}\,dx = \int \frac{1}{u^{m}}\,du$

Let $u$ be a function of $x$. Then

\[\frac{d}{dx}u^{n+1}= (n+1)u^{n}\frac{du}{dx}\]

and so

\[\int u^{n}\frac{du}{dx}\,dx = \frac{u^{n+1}}{n+1}\].

In our case we have

\[u(x) = lnlnx, \frac{du}{dx}= \frac{1}{xlnx}\]

and so

\[\int \frac{1}{xlnx(lnlnx)^{m}}\,dx = \frac{(lnlnx)^{-m+1}}{-m+1}\].


vector
algebra
polynomial
induction
calculus
derivative
superior algebra

I have difficulties to read a proof of the following theorem:
Let $ \mathbb{F}_q$ be a finite field. Let $ n\geq 1$ be integer. In a given algebraic closure $ \mathbb{F}_q^a$, there exists one and only one extension of $ \mathbb{F}_q$ of degree $ n$, and this extension is $ \mathbb{F}_{q^n}$.

Why do the the splitting field of $ X^{q^n} \minus{} X$ is precisely $ \mathbb{F}_{q^{n}}$.

If $ K\subseteq \mathbb{F}_q^{alg}$ is an extension of $ \mathbb{F}_q$ of degree $ n$, then $ K\cong \mathbb{F}_q^n$ as $ \mathbb{F}_q$-vector space, so $ |K|\equal{}q^n$.
But then the group $ K^\times$ has order $ q^n\minus{}1$, so by Lagrange $ a^{q^n\minus{}1}\equal{}1$ for all $ a\in K^\times$.
Therefore $ K$ consists of roots of the polynomial $ X^{q^n}\minus{}X$, and since it has degree $ q^n$, it has at most $ q^n$ roots in $ \mathbb{F}_q^{alg}$.
So $ K$ is in fact precisely the set of roots of $ X^{q^n}\minus{}X$ in $ \mathbb{F}_q^{alg}$, and so also the splitting field of $ X^{q^n}\minus{}X$ over $ \mathbb{F}_q$.
This shows, that there can only be one extension of $ \mathbb{F}_q$ of degree $ n$ in $ \mathbb{F}_q^{alg}$.

Conversely to show it exists, we need to show, that the set of roots of
$ X^{q^n}\minus{}X$ in $ \mathbb{F}_q^{alg}$ is in fact a field with $ q^n$ elements containing $ \mathbb{F}_q$.
But $ K$ is simply the fixed field of the automorphism $ \mathbb{F}_q^{alg}\to\mathbb{F}_q^{alg},a\mapsto a^{q^n}$. (see [url=http://en.wikipedia.org/wiki/Frobenius_automorphism]Frobenius automorphism[/url])
So $ K$ is a field and the above argument shows $ a^q\equal{}a$ for all $ a\in\mathbb{F}_q$
and by induction also $ a^{q^n}\equal{}a$ for all $ a\in\mathbb{F}_q$, which gives $ \mathbb{F}_q\subseteq K$.

Now $ X^{q^n}\minus{}X$ has degree $ q^n$, which gives $ |K|\le q^n$, and to show $ |K|\equal{}q^n$,
we need to show, that $ X^{q^n}\minus{}X$ has no multiple root $ \zeta\in\mathbb{F}_p^{alg}$.
But otherwise $ \zeta$ would also be root of the formal derivative
$ {d\over dX}(X^{q^n}\minus{}X)\equal{}q^nX^{q^n\minus{}1}\minus{}1\equal{}\minus{}1$, which has no roots.

Okay. My professor said that this can be proven by some other way without fixing $ K$ as the  extension of $ \mathbb{F}_q$. He said that this proof is not enough to show that $ K$ contains all of the roots but I'm wondering this because $ x^{q^n} \minus{} x$ is a polynomial of degree $ q^n$ and $ K\cong \mathbb{F}_q^n$.

[quote="MeKnowsNothing"]without fixing $ K$ as the  extension of $ \mathbb{F}_q$[/quote]
Sorry, but i have no idea what this means.
$ K$ is defined as extension of $ \mathbb{F}_q$ of dimension $ n$.
That's already given.

[quote="MeKnowsNothing"]this proof is not enough to show that $ K$ contains all of the roots[/quote]
If $ L$ is the set of roots of $ X^{q^n}\minus{}X$, then $ K\subseteq L$ and $ |K|\equal{}q^n\ge |L|$ gives $ K\equal{}L$.
This proof is clearly correct, once you have the existence of the algebraic closure given.
(which many books try to avoid here)
I really don't see the problem here.


Hi! I had this problem on a test today.

A is running 3 laps on 10min. B is running 5 laps on 12 min. After how many laps, done by A and B together, will they meet at the startline at the same time?

I think my answer was 18 laps and, I dont remember the other one. I'm not sure that it's correct. I would like to see a proof. How would you solve it? Thank you :)

i have seen this problem before, it's an easy problem.

your answer for A was correct! A=18, B=25
here is the proof:
in each minute, A runs .3 laps and B runs 5/12 laps.
so we know that B runs faster. therefore it takes 12/5 mins for B to run one lap. so when B runs one lap, then A has ran:
(.3)*(12/5)=36/50
now, you can say that if B completes X laps, then A will complete 36X/50 laps, now all you need to do is to find the smallest value of X. which is 25.
so B will complete 25 laps, and A will complete 18. and they will meet in the same time.


$a_{1},a_{2}, \cdots ,a_{n}$ are real numbers with sum $0$ and exactly and only one of them is $1$. Prove that:
      $max \left\{ \left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \cdots , \left|a_{n-1}-a_{n}\right|,\left|a_{n}-a_{1}\right| \right\}\geq \frac{4}{n}$

Very straightforward. Use indices cyclical mod $n$. wlog $a_{0}=1$.
First case: $n=2k$. Suppose $max(...)<\frac{4}{n}.$ then for $i=1,...,k$ we must have $a_{\pm i}>1-\frac{4}{n}i$. Thus $\sum_{i=1}^{n}a_{i}=\sum_{i=-k+1}^{k}a_{i}> n-\frac{4}{n}[2(1+...+[k-1])+k]=2k-2k=0$ contradiction.
Second case $n=2k+1$. Suppose $max(...)<\frac{2k+1}{k(k+1)}$, then  for $i=1,...,k$ we must have $a_{\pm i}>1-i\frac{2k+1}{k(k+1)}$. Thus $\sum_{i=1}^{n}a_{i}=\sum_{i=-k}^{k}a_{i}> n-\frac{2k+1}{k(k+1)}\cdot2(1+...+k)=2k+1-(2k+1)=0$ contradiction. So in this case, $max(...)\ge\frac{2k+1}{k(k+1)}>\frac{2k+1}{k(k+1)+\frac14}=\frac{4}{n}$. :)

excuse me spanferkel,  What do you mean by wlog $a_{0}= 1$ or $a_{\pm i}$, where did you get $\pm i$ . sorry  :oops:

just moving around the indices in a cyclical way.  imagine them on a regular $n$-gon  :) wlog=without loss of generality, we can give the index $0$ (or equally, $n$) to the one that is equal to 1  :wink:

sorry, but where did you get $\pm i$ , you mean we assume that for each $i = 1, 2, ... , n$ we have $a_{+i}= a_{-i}$ ?!!!!

no, it's just a shorter way of writing $a_{ i}>1-i\frac{2k+1}{k(k+1)}$ and $a_{-i}>1-i\frac{2k+1}{k(k+1)}$ altogether :lol:
I hope you get it now  :) it's really straightforward

I'm really sorry, but I can't understand where do you get $a_{-i}$  from , what is the relation between  $a_{-i}$ and $a_{i}$, you mean we have: $\sum^{n}_{i=1}a_{-i}= 0$,
sorry again...

$a_{-i}=a_{n-i}$ that's what $\mod n$ means

Thank you very very much spanferkel, sorry again but this is tha last question, I swear...
in the part that $n=2k$ you said if $max(...) < \frac{4}{n}$ then for i=1,2,...,k we must have $a_{\pm i}> 1-\frac{4}{n}i$, How do you say this?
thank you very much

well by assumption ${|a_{i}-a_{i+1}|<\frac4n}$ for all $i$, i. e. any two neighboring "vertices" are less than $\frac4n$ apart. And the starting one has value $a_{0}=1$.
This is the last answer I swear  :)  :)  :)