i'm new, so don't shoot me if this in the wrong section (i checked them all, and think it suits here  :? ) or if someone else came up with this question before... (just move it then)

sin1°+sin3°+...+sin359°=?

sin359 = -sin1
sin 357 = -sin3

so the equation becomes..

$\displaystyle \sin 1 + \sin 3 + \sin 5 + ... + -\sin 5 + -\sin 3 + -\sin 1$
$ = 0$

and now:

sin²(1°)+sin²(3°)+...+sin²(359°)=?

Ive solved that before so ill let the others answer... :)

I just realized that you are looking for odd degree values only!! Well, my answer is wrong but the idea is correct.

[quote="Philpoleon"]Well, my answer is wrong but the idea is correct.[/quote]
yeah, the idea behind it is indeed correct... but the answer isn't :)

by the induced formula $\sin{x}=-\sin{(360-x)}$ and $\sin{(90-x)}=\cos{x}$...

[quote="Mamat"]180? :?[/quote]
just a small miscalculation, i guess...


this was my first, and unfortunately last, year

[quote="13375P34K43V312"]This is my first year (8th). I actually think I did surprisingly well, according to my standards. I am guessing around 15, not too shabby.[/quote]
I don't think we can say what we think we got...

This is my third year, and I started in the 10th grade.  I'm hoping to get a perfect record of score improvements from year to year!

I think I ran across it in eighth grade, but it was practically over, and I gradually forgot about it.  Then I got AoPS books for my birthday at the beginning of ninth grade, and in them was a USAMTS flyer, and I decided to participate.

Attempted it last year (10th grade) but got too busy. I'm doing it this year though.  :lol:

[quote="solafidefarms"]Some people have expressed suprise at how long I've done USAMTS, and I'm wondering if I'm really that precocious :). What grade was the first grade you participated in USAMTS?

Billy[/quote]
Um, just to make a point, AoPSers are going to have a much lower average starting grade than non-aopsers...

My second year and I'm in ninth. I've gotten a lot better, though :thumbup:

1st year, (10th Grade).

I can come up with proofs in my mind, but I suck at wording them into a formal proof.

[quote="solafidefarms"][quote="SplashD"][quote="solafidefarms"]Some people have expressed suprise at how long I've done USAMTS, and I'm wondering if I'm really that precocious :). What grade was the first grade you participated in USAMTS?

Billy[/quote]

How long did you do USAMTS?[/quote]

This is my 3rd year, I've done it since 8th grade.

@jclarke, I have gotten 15 the first round both last year and my 8th grade year. Also, I tend to overestimate my score. (usually :))[/quote]

Oh. I don't think 8th is [i]that[/i] early. And I see from the polls that I'm still the only one that started in 7th.  :D

[quote="SplashD"][/quote][quote="solafidefarms"][quote="SplashD"][quote="solafidefarms"]Some people have expressed suprise at how long I've done USAMTS, and I'm wondering if I'm really that precocious :). What grade was the first grade you participated in USAMTS?

Billy[/quote]

How long did you do USAMTS?[/quote]

This is my 3rd year, I've done it since 8th grade.

@jclarke, I have gotten 15 the first round both last year and my 8th grade year. Also, I tend to overestimate my score. (usually :))[/quote][quote="SplashD"]

Oh. I don't think 8th is [i]that[/i] early. And I see from the polls that I'm still the only one that started in 7th.  :D[/quote]
What was your score?

I first heard about it at the very end of last year, so this is the first year I have tried it.

edit: I'm in 9th grade

I heard about usamts last year in my math class, but I was too lazy to go to the website.
this year, I finally decided to check it out, cept I was too lazy and didnt do the probs. I might start halfway in, but dunno

me=9th

I found the USAMTS through MathCounts and this year (9th grade) is my first year participating. :D


There are two people, A, and B, and there is a pile of $ a$ votes for A and $ b$ votes for B. It is given that $ a< b$. Now, the pile of votes is randomly shuffled, and every "turn" one vote is taken from the pile and the vote is recorded on a tally chart. The vote is not placed back into the pile.
Find (with proof) the probability that at least once during the tally there will be equal numbers of votes for A and B. (e.g. the same number of votes has been counted for both candidates)

Thank you.  :D  :D

[quote="Sage!"]Read about dominated sequences.[/quote]

I searched online, but could not find any sources. Perhaps you could refer me to some sources... thanks.

I found an article written by Polish mathematician in English.

Look at the [b]Observation 2[/b] : $ d(k,n) \equal{} \frac {n \plus{} 1 \minus{} k}{n} {{n \plus{} k} \choose {k}}$

Link : http://arxiv.org/PS_cache/math/pdf/0510/0510027v1.pdf

Another proof (unfortunately only in Polish): http://abstynent.eastgroup.pl/wiki/Twierdzenie_o_ci%C4%85gach_zdominowanych

I pretty sure that now you'll be able to count the expected probability.


Singapore Mathematical Olympiads 2001
Open Section #4

Each of the 48 small squares of an $8 \times 6$ rectangular board is coloured either black or white, and the number of black squares in each row and each column is even. In how many ways may the rectangular board be coloured?

More generally: in an $m \times n$ checkerboard, each square is filled with one of the numbers $1, 2, \ldots, k$ in such a way that the sum of the numbers in each row and each column is divisible by $k$.  In how many ways can this be done?

[hide]
Place numbers the top-left $m-1\times n-1$ portion of the checkerboard first in anyway that you want. There are $k(m-1)(n-1)$ ways to do this. This uniquely determines the rest of the board. We are able to deduce the right column (except for the bottom-right square). The bottom row can then be determined from the rest of each column. This was made such that the sum of all numbers in each column is divisible by k and the sum of all numbers in each row is divisible by k except for the bottom row. We can prove that this will indeed work:

Let $a_{1},a_{2},...,a_{m}$ be the sums of the numbers in each row and $b_{1},b_{2},...,b_{n}$ be the sums of the numbers in each column.

$\sum_{i=1}^{m}a_{i}=\sum_{i=1}^{n}b_{n}$
$\sum_{i=1}^{m}a_{i}\equiv \sum_{j=1}^{n}b_{j}\mod k$
Since we know all $a_{i}\equiv b_{j}\equiv 0$ for all $1\le i\le m-1$ and $1\le j\le n$
$a_{m}\equiv 0\mod k$

So each row and column is divisible by k.

The answer is $\boxed{k(m-1)(n-1)}$

The answer to this specific question is $2(8-1)(6-1)=\boxed{70}$
[/hide]

Err the answer is much larger than 70.. originally, the question said the answer was $N$ and we were told to find $\log_{2}N$.

Your proof is good, except that you've botched the actual counting part twice now.  It should be $k^{(m-1)(n-1)}$, and so in the original problem $2^{35}$.

wow i have no idea what i was thinking that second time

That's correct now.. the general problem insight is helpful, thanks.  :)


probability
number theory
least common multiple

This is the general case, when aliens abduct someone from Spokane once every $a$ days, someone from Seatlle, once every $b$ days, and someone from Moses Lake once every $c$ days, what is the probability that on a given day, someone is abducted from Spokane and Seattle, but not from Moses Lake?

I'm stuck trying to solve this:

Someone is abducted from Sponkane and Seattle every $\textrm{LCM}(a,b)$ days. The probability is $\frac{1}{\textrm{LCM}(a,b)}$. We don't want people abducted from Moses Lake. The
probability that they are abducted from Moses Lake is $\frac{1}{c}$, so the probability of them being not being abducted from
Moses Lake is $1-\frac{1}{c}$
Aren't the probabilities independent? If not, what would I do?

You are first looking for $\sum_{k\in\mathcal{S}}k\cdot\textrm{lcm}\left(a,b\right)$.  Of course, you are summing over \[\mathcal{S}=\left\{k\in\mathbb{Z}_{\textrm{lcm}\left(a,b,c\right)}\ | \ k\cdot\textrm{lcm}\left(a,b\right)\not\equiv 0\bmod c\right\}.\] The desired probability is then \[\textrm{P}=\frac{\sum_{k\in\mathcal{S}}k\cdot\textrm{lcm}\left(a,b\right)}{\textrm{lcm}\left(a,b,c\right)}.\]


function
calculus
derivative
limit
real analysis
real analysis unsolved

Suppose that f:R -> R is a function satisfying |f(x) - f(y)| <= c (x-y)^a for all x, y where c and a are constants a > 0.  Prove that f is a constant function.

I would start out like this but i'm not sure if it's right, please give me hint thanks in advance
the only way for f is a constant is f ' = 0

f ' (x) = |f(x) - f(y)| / (x-y)^a <= c
as x -> y , f(x) = f(y) = 0 so f ' is 0
but the problem doesn't say anything about differentiability, can we assume that or we have to prove? if so, how?

If $a > 1$ then your approach works. It follows that $f$ is differentiable. So you do not have to assume that.

If $a\leq 1$ then $f$ does not have to be a constant (this condition is called the Lipschitz condition). In fact, every differentiable function with a bounded derivative satisfies this property for $a = 1$, and some $c$. Consider, for example, $f(x)=x$.

BTW I presume that ythe LHS is $c |x - y|^a$ (not $c (x - y)^a$).

If $a>1$ we have

\[ |f'(x)| \leq \left|\lim_{y\to x} \frac{f(y) - f(x)}{y-x}\right| \leq \lim_{y\to x} \left|\frac{f(y) - f(x)}{(y-x)^a}\right| \cdot |y-x|^{a-1}\leq \lim_{y\to x} c \cdot |y-x|^{a-1} = 0. \]


MATHCOUNTS
abstract algebra
logarithms
function
floor function
algebra

I came across this question in a high school varsity competition, but I guess it's more or less a MathCounts level question:

How many digits does 3^1004 have?

Please explain your solution as well, because I might have gotten this question wrong. :blush:

theres some module thing way to solve this i just cant htink of it

[quote="abacadaea"][quote="xpmath"]Actually, it is the least number greater than $ 1004\cdot\log_{10}{3}$. I think.
It's nice formula.[/quote]

you sure? why not $ \log_{3}{1004}$[/quote]

xpmath is right. think about it this way, how can you determine the number of digits in 100? the method is take the floor function of the log 10 of the number plus 1.
so with $ 3^{1004}$, we have $ \log_{10} {3^{1004}} \Longrightarrow 1004 \cdot \log_{10} {3}$ take the $ \lfloor \lfloor$ function of that and add 1.

hope that helps abacadeaae

Um, you wouldn't be able to use a calculator or any other reference for the actual competition.  How would you do it without a calculator?

BTW: It's multiple choice, so I guess you could guess the answer:

(A) 440
(B) 460
(C) 480
(D) 500
(E) 520

I catually have the first few logarithms memorized (only those of 2,3, and 5).
$ \log_{10}3\equal{}0.4771$.

[quote="Math Geek"]Um, you wouldn't be able to use a calculator or any other reference for the actual competition.  How would you do it without a calculator?
[/quote]

how do you know this is a sprint round problem?

It ISN'T sprint, it's from a high school varsity competition.

$ \log_10{3}\approx 2$, but a little less.  Multiply by 1004 and you get $ \approx 500$.  But it's a little less, so I'd go with C)480

[quote="archimedes1"]Try not to get in the habit of using logs too much, unless you have to.  For example,
Find the number of digits in $ 2^100$ without using logarithms..[/quote]
i think you mean

Try not to get in the habit of using logs too much, unless you have to.  For example,
Find the number of digits in $ 2^{100}$ without using logarithms

i can do this one. $ 2^{10}\equal{}1.024*10^3, 2^{20}\approx 1.024*10^6, 2^{30}\approx 1.024*10^9$
an so on
so 31?

Because the number of digits in a positive integer $ n$ is $ \lfloor \log_{10} n \rfloor \plus{} 1$. To see why, notice that this number increases by 1 only when you reach a new power of 10, an event which will add a new digit.

Ah, that's pretty cool. I didn't realize why you were using log 10, but it's pretty obvious now that I've made the connection :P Thanks fro the help.


Find the quadrilateral of maximum area inscribed in a circle of radius R.

Answer: square with area $ 2R^2$
Too easy problem... It shouldn't be in geometry usolved problems section

I know it's classical,but what is the simplest solution to it?

$ S_{ABCD} \equal{} \frac{1}{2} AC \cdot BD \sin{\widehat{AC,BD}}$

Since $ ABCD$ inscribes a circle, $ AC, BD \leqslant 2R$ and $ \sin{\widehat{AC,BD}} \leqslant 1$, the maximum value of the area is $ 2R^2$.


function
probability
algebra
polynomial
domain
linear algebra
matrix

Suppose [tex]f(x,y): [0,1] \times \mathbb{R} \mapsto \mathbb{R}[/tex]  is an analytic function of both x and y in their domains. and for each fixed x, [tex]\max_{y} f(x,y)[/tex] exists, but it is possible to have multiple maximums. Let [tex] X[/tex]  be a random variable with uniform distribution over [tex] [0,1][/tex] . So [tex] f(X,y)[/tex]  is a random function of y.
Can we conclude that [tex] P[f(X,y) \text{has multiple maximums}][/tex]  is either 0 or 1?
Any contra-examples or any clues for the proof?

If I understood the question right, $f(x,y)=(x-1/2)y^2-y^4$ is a counterexample: it has a single maximum at $0$ for $0<x<1/2$ and two symmetric maxima for $1/2<x<1$, so the probability in question is $1/2$.

Thanks! That is really a good counterexample. Acturally, I am not dealing with all anaytic functions, but a subclass of anaytic functions. So maybe I should add more contraints to the functions. For my problem, the function can be written in this form:
[tex]f(x,y)=g(x,y)-y[/tex], where [tex]g(x,y)[/tex] is anaytic and [tex]-y[/tex] is not cancelable. I don't know if this will rule out all cases where the probability is greater than 0.

Maybe there was counterexamples too.

Thanks.

[quote="remgenesis"] For my problem, the function can be written in this form:
[tex]f(x,y)=g(x,y)-y[/tex], where [tex]g(x,y)[/tex] is anaytic and [tex]-y[/tex] is not cancelable. [/quote]
I'm afraid it is very hard to formally define what "not cancelable" means. Indeed, if in the above example you replace $y$ by $y+1$, it'll become a polynomial in $x$ and $y$ with a strictly negative coefficient at $y$ by which you can then divide to get a function of your form ($g(x,y)$ will then be a polynomial without pure $y$-term). So can you just tell exactly what class of functions you are looking at?

The functions I am looking at are:
[tex]f(x,y)=\log(1+g(x,y))-y[/tex], where
[tex]g(x,y): [0,1] \times \mathbb{R}^+ \mapsto \mathbb{R}^+[/tex]  is an analytic function of both x and y in their domains.

Not sure if there is still ambiguity.

BTW. I don't have the *exact* form of functions I am looking at. I am doing some engineering research. I have some conclusions on some specific functions. and I just hope the conclusion can be generalized as widely as possible.

Alas, I don't think this particular class is any better than the class of all analytic functions. You still have the possibility of the same effect only the explicit formulae for counterexamples become uglier... I also noticed that you changed the domain in $y$ from $R$ to $R^+$, but it also hardly makes any difference. Sorry, but it seems that no really meaningful general statement of the kind you want is possible.  :(  On the other hand, I'll be happy if somebody will prove me wrong  :)

Thanks. It seems i need to reformalize my problem...
Now, let me go back to a primitive problem:
Let $f(x)$ be a anaytic function on [0,1], and $X$ is a uniform random variable over [0,1].
How to prove that $P[f(X)=0]$ is either 1 or 0?  I suppose this claim is correct.

or a similar one. $f(x)$ is anaytic on $\mathbb{R}$ and is non zero at least at one point, then how to show that $f(x)$ only has countable zeros.

I have a feeling my problem in the orginial post is in some sense related to the problem in this post.

Thanks a lot. Forgive me if this is a naive question:
How do your proof extend to the case where the domain of $f(x)$ is $\mathbb{R}^n$. I think the number of zeros in this case may be uncountable. But the set of zeros should have some zero  measure (not sure which measure to use).

How to show this?

Just get back to this problem with a concrete expression:
Let $a_1,...,a_K$ be any given postive value.  $\mathbf{H}$ is a $N\times K$ random matrix with some joint density distribution, denote its i-th column vector by $\mathbf{h}_i$. Function $f: \mathbb{R}_+^K \mapsto \mathbb{R}$ is given by

$f(\mathbf{p})=(\sum_{i=1}^K a_i) \log \det (\mathbf{I}+\sum_{i=1}^K p_i \mathbf{h}_i \mathbf{h}_i^T)
-\sum_{j=1}^K a_j \log \det (\mathbf{I}+\sum_{i \neq j} p_i \mathbf{h}_i \mathbf{h}_i^T)-\sum_{i=1}^K p_i$

Note the function is defined on  $\mathbb{R}_+^K$, that is $p_i \geq 0$.
The function is non-concave, so it might have multiple (global) maxima.
The claim is the probability of $f(\mathbf{p})$ to have multiple maxima is zero. (Note $\mathbf{h}_i$ are random variables.)

Is the claim true? if so? how to prove? Thanks!

First, let me ask a couple of questions to see whether I understood you right:

1) are you interested in global maxima only? (i.e., if your function has several local maxima that are strictly less than the unique global maximum, you do not count that as multiple maxima)

2) is the joint density of elements of $H$ absolutely continuous with respect to the $NK$-dimensional Lebesgue measure? (i.e., are we just to show that the maximum is unique for almost every matrix $H$?)

Just wanted to make sure that there is no misunderstanding before trying to solve the problem  :) .

[quote]1) are you interested in global maxima only? (i.e., if your function has several local maxima that are strictly less than the unique global maximum, you do not count that as multiple maxima)[/quote]Yes, I am interested the global maxima only, not the local ones. since the function is non-concave, it usually has multiple local maxima for most H.

[quote]2) is the joint density of elements of $H$ absolutely continuous with respect to the $NK$-dimensional Lebesgue measure? (i.e., are we just to show that the maximum is unique for almost every matrix $H$?)[/quote]
Yes, like Gaussian, for example. Sorry i didn't make it clear in the last post.

Thank you very much!

I have an idea :idea: :). It suffices to prove that for a fixed real-analytic $F(p)$ growing as $o(\|p\|)$ and for a.e. set of coefficients $b_1,\ldots,b_n>0$, the expression $F(p)-b_1p_1-\ldots b_np_n$ has a unique global maximum in $\mathbb R^k_+$. This seems almost obvious but the formal proof escapes me. Can anybody else do it?

Thanks.
Sorry I don't have any ideas. I was hoping that the problem could be reduced to the number of zeros of an anaytic function. But I think you must know that much better than me.

OK, I solved it and the answer is "Yes, the probability of multiple global maxima is $0$". The key lemma is the geometric statement I posted on [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=203963[/url] as an open problem. I can prove it now but I do not like my proof since it uses a very heavy artillery: Sard's theorem for Lipschitz functions. I prefer to leave that lemma as a challenge to see whether somebody will come up with a more elementary proof. So the case is (temporarily?) closed. :lol:  Any questions?

The question is: I still don't know your solution  :P

If you don't like the way you proved that lemma, just post the rest part please.

Thanks alot for your continuous assistance!

[quote="remgenesis"]If you don't like the way you proved that lemma, just post the rest part please.
[/quote]
The rest is easy. The main idea is to notice that if one changes $h_i$ to $\frac 1{\sqrt b_i}h_i$ and makes the change of variable $p_i\mapsto b_i p_i$, then the question of uniqueness of the maximum of $f$ with the matrix consisting of the columns $\frac 1{\sqrt b_i}h_i$ and the symmetric linear term $-\sum p_i$ becomes equivalent to that of uniqueness of the maximum of the modified function $f$ with the matrix consisting of the columns $h_i$ (without $b$'s) and the asymmetric linear term $-\sum b_i p_i$. So, it is enough to prove that, for fixed $h_i$ and almost every choice of $b_i>0$ the global maximum of this new function is unique. But this is equivalent to the question whether the random plane orthogonal to the vector $(b_1,\dots,b_n)$, when moved so as to just touch the graph of a fixed function (the determinant part of $f$) touches the graph at just one point. Hence the geometric problem about compact convex sets (compact, because we can estimate the distance from the point of maximum to the origin in terms of $b_j$ and $h_j$ and convex, because for a plane to touch a compact set at one point is the same as to touch its convex hull at one point).

Based on your proof, I am trying to give a more general claim:
Let function $F: \mathbb{R}^K_+ \times \mathbb{C}^{N \times K} \mapsto \mathbb{R}_+$.
And $F$ satisfies $F(D^{-1}p,HD^{1/2})=F(p,H)$, where D is any diagonal matrix with positive diagonal elements,
and $F(p)/ \|p\| \to 0$ as $\|p\| \to \infty$.
Then function $f(p)=F(p,H) - \sum_i \lambda_i p_i$, where $\lambda_i$'s are fixed positive real numbers, has unique global maximum for almost every $H$.

are these condition sufficient? Did I miss some conditions? Do I really need $F$ to be an analytic function?

[quote="remgenesis"]...are these condition sufficient? Did I miss some conditions? Do I really need $F$ to be an analytic function?[/quote]
The statement is fine as you put it. In principle, you can generalize further (for instance, it is not so important that $H$ is multiplied by $D^{1/2}$: any other reasonable transformation would be fine too, but I do not see why one should generalize just for the sake of generalization, so I would leave the statement as you put it now unless you have some other particular application in mind).  As to analyticity, it has never been used in the proof. Continuity is enough. On the other hand, if you know that the function is analytic (or just sufficiently smooth), you probably can say more about the size of the set of exceptional $H$ than just that the measure is $0$ but I didn't try to think in this direction.

Yeah, I have some application in mind. So this generalization if pretty much enough for my problem.

Thanks for your comment!

[quote]then the question of uniqueness of the maximum becomes equivalent to that of uniqueness of the maximum of the modified function $f$ with the matrix consisting of the columns $h_i$ (without $b$'s) and the asymmetric linear term $-\sum b_i p_i$. So, it is enough to prove that, for fixed $h_i$ and almost every choice of $b_i>0$ the global maximum of this new function is unique. [/quote]
And when you proved the the zeros of an analytic function on $\mathbb{R}^n $has zero measure, you said:
[quote="fedja"]Just take a point where $f$ is not $0$ and observe that the intersection of every line through that point with the zero set of $f$ is at most countable. This is enough to conclude that the zero set has zero measure.[/quote]

They seems following from the same philosophy. Both are saying that intead of considering the whole space, it is sufficient to only consider a line in the space, which has lower dimension.
They seems right to me intuitively. But I am just not very clear how to rigorously get these and which known Theorem are applied?

Isn't it just integration in polar coordinates?  ;)

[quote="fedja"]
Hence the geometric problem about compact convex sets (compact, because we can estimate the distance from the point of maximum to the origin in terms of $b_j$ and $h_j$ and convex, because for a plane to touch a compact set at one point is the same as to touch its convex hull at one point).[/quote]

What does "compact" here mean? since $\|p\|$ is not bounded. how the set be compact?  :?

We know that the function $F(p,H)$ is much less than $\|p\|$ for large $\|p\|$, so the values of the cost function at $p$ with large $\|p\|$ are large negative, i.e., less than the value at $0$. So, the global maximum has to be achieved somewhere not too far from the origin (you can get an a priori bound in terms of $H$ and $\lambda$'s) and the values of $p$ with large $\|p\|$ can be just disregarded, i.e., we may think that the maximum is taken over $p$ with $\|p\|\leqslant M$ where $M$ is sufficiently large.

Another way to explain it is that you just consider the global maximum over $\|p\|\leqslant M$ for all $M=1,2,3,\dots$ and consider the event that such global maximum is multiple. Each such event corresponds to a compact set and, therefore, we can use the geometric lemma to conclude that it has probability $0$. But the event that the global maximum over the entire $\mathbb R_+^K$ is multiple is contained in the union of these events.

[quote="fedja"]That is a completely standard theorem from Complex Analysis that can be found in any decent textbook. The proof goes like this: if a zero $x_0$ of $f$ is not isolated, then the Taylor series of $f$ at $x_0$ is identically $0$. Hence $f$ identically vanishes in some neighborhood of $x_0$. Let now $U$ be the set of points $x$ such that $f$ vanishes identically in some neighborhood of $x$. Obviously $U$ is open. Since any $x$ on the boundary of $U$ is a non-isolated zero of $f$, $U$ is also closed. Thus $U$ must be the entire domain of $f$ (provided that it is connected, of course). So, if $f$ has at least one non-isolated zero, it vanishes identically. But any set of isolated points is at most countable.[/quote]

I do not understand when you say "Since any $x$ on the boundary of $U$ is a non-isolated zero of $f$, $U$ is also closed". For me the fact that $x$ is a non-isolated zero does not imply that $f$ vanishes identically in some *neighborhood* of $x$, because any neighborhood of $x$ could contain points that are not zeroes of $f$...

Can you bridge the gap ?

Read the definition, idoyen.
[quote="fedja"]Let now $U$ be the set of points $x$ such that $f$ vanishes identically in some neighborhood of $x.$[/quote]
Another argument for this: since $f$ can be represented locally by power series, $U$ is the intersection of the sets $U_n$, where $U_0=\{z: f(z)=0\}$, $U_n=\{z: f^{(n)}(z)=0\}$. Each of these sets is closed by continuity, so their intersection is closed.

[quote="jmerry"]Read the definition, idoyen.[/quote]
[quote="fedja"]Let now $U$ be the set of points $x$ such that $f$ vanishes identically in some neighborhood of $x.$[/quote]

We have that $f$ vanishes at $x$, but not necessarily in a  *neighborhood* of x.

In fact, I think the argument is that $f$ is equal to its Taylor series about $x$ in a neighborhood of $x$.
And since as you said $f^{(n)}(x) = 0$ for all $n$ when $x$ is at the boundary of $U$, we have that $f$ vanishes identically in a *neighborhood* of $x$.

Thanks


Define [b]A[/b] = {1, 2, 3, 4, 5, 6}.

How many functions f : [b]A[/b] --> [b]A[/b] satisfy f(f(x)) = x for all x in [b]A[/b]?

Generalize.

[i]Bosnia - Herzegovina Mathematical Olympiad 2002[/i]

Let a_n the number of functions f from {1,2,...,n} with values in the same set such that f(f(x))=x. Of course, a_1=1 and a_2=2. Now, if f(n)=n, then we can choose the function f in a_n-1 ways ( just look at its restriction on {1,2,...,n-1}) and if f(n)<>n then we can choose the function in (n-1)a_n-2 ways (f(n) is chosen in n-1 ways and for each such values we can define f in a_n-2 ways). So, a_n=a_n-1+(n-1)a_n-2.

By other arguments (pairing), we can show that

a_n = Sum(k=0..[n/2]) n!/(k!(n-2k)!2^k) (1)

So, we have a nice algebraic problem: let a_n defined as, so that a_n = a_(n-1) + (n-1)a_(n-2) (2). I think algebraically difficult to prove (2).

BTW, in Japanese MO (199?) there was following problem:
Define A = {1, 2, 3, 4, 5, 6,7}.

How many functions f : A --> A satisfy f(f(f(x))) = x for all x in A?

Namdung


you can view/download the problems by pushing [url=http://www.mathlinks.ro/Forum/resources.php?cid=24&year=2003]this link[/url].

yeah :D - i've just got back from it, which means it's over :)
i've got 2nd prize
Belarus and Hungary are the winning countries

Congratulations Valentin Vornicu! Were there other Romanian students?

yeah, these were the Romanian students (besides me) which got prizes:
Liviu Paunescu 1st prize
Razvan Iagar 2nd prize
Adrian Iuga 2nd prize
Daniela Petrisan 3rd prize
Ovidiu Preoteasa 3rd prize
Radu Remus 3rd prize
Raluca Tanase 3rd prize
Mihai-Aurel Fulger 3rd prize

Thanks for puting me the last in the list of  3rd prize winners. I think this is a clear case of last but not least.

sorry - the order is just the order that came up into my mind while writing this down in a hurry - no offence ment :)

I have a question:

Did Matei Adrian compete?

:D

i don't think so ... and anyway if he did, he didn't to very well ...

I was just wondering if he competed because he is from CAmpulung Muscel and he is a very good mathmatician.. he won third prize at the national stage when he was 11th grade! and I thought that he did compete at th IMC 2003 ... never mind ...thank you Valentin Vornicu!

thank you!!

bye bye

He is at the "politehnica" university in Bucharest!

[url= https://artofproblemsolving.com/community/c4376_2003_imc] problems  IMC 2003 [/url]
https://artofproblemsolving.com/community/c4376_2003_imc

[url= https://www.imc-math.org.uk/?year=2003&item=results] web site of IMC [/url]
https://www.imc-math.org.uk/?year=2003&item=results


this is killing me save me please please!!!!!!!!!!!!!!!!!!!!!!!!!!!!! download the attachment pdf file of hard math :blush:

[hide]1) it's $t_n=1+\frac{1}{t_{n-1}}$ since most of the continued fraction can be expressed as $t_{n-1}$.

4) If a sequence $t_n$ has a limit as $n\rightarrow \infty$ then $t_n\rightarrow t_{n-1}$.

If $\phi$ is the value of the continued fraction, then write $\phi=1+\frac{1}{\phi}$ and solve using quadratic formula.

The solution with $\phi>1$ (which must be true) is $\frac{1+\sqrt{5}}{2}$.

(Similarly $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=2$ because it satisfies $x=\sqrt{2+x}$ and $x>0$.)[/hide]

Homework help is usually discouraged on the forums. This really isn't the place to cheat on your next assignment, and if you don't understand it, ask about the concepts. For example, You could ask: How do I solve problems involving continued fractions, and we could discuss general methods. But, all in all, you can't learn if people supply you with all the answers, and the next homework might end up the same way.

Yeah, these kinds of problems are EXTREMELY common

Please, please, please don't post your problems in attachments.  It inconveniences everyone.  Instead, spend about a minute typing your problem directly into your post.

Hint: Remember the meaning of infinite.  Here, i'll give you a similar example.
Suppose $n=2+\frac{2}{3}+\frac{2}{9}+...$ and we want to find $n$.  Then
$3n=6+2+\frac{2}{3}+...=6+n \implies n=3$.
Now apply this idea...


Να βρεθούν όλες οι τριάδες πραγματικών που ικανοποιύν την παρακάτω ισότητα
2(b^4)+2a(b^3)+(a^2)(b^2)+(c^2)(b^2)+2cb+2=ac(b^2)

$ 2b^4 \plus{} 2ab^3 \plus{} a^2b^2 \plus{} c^2b^2 \plus{} 2cb \plus{} 2 \equal{} acb^2 \Leftrightarrow$
$ (2b^2)^2 \plus{} 2ab(2b^2) \plus{} a^2b^2 \plus{} a^2b^2 \minus{} 2(ab)(bc) \plus{} c^2b^2 \plus{} c^2b^2 \plus{} 2(2)(cb) \plus{} 2^2 \equal{} 0 \Leftrightarrow$
$ (2b^2 \plus{} ab)^2 \plus{} (ab \minus{} cb)^2 \plus{} (cb \plus{} 2)^2 \equal{} 0 \Leftrightarrow$
$ 2b^2 \plus{} ab \equal{} 0(I), ab \minus{} cb \equal{} 0(II), cb \plus{} 2 \equal{} 0(III)$

$ (III) \Rightarrow b \not \equal{} 0$
άρα $ (II) \Leftrightarrow a \equal{} c \Rightarrow ab \plus{} 2 \equal{} 0 \Leftrightarrow 2b^2 \equal{} 2 \Leftrightarrow b \equal{} \mp 1$

άρα $ (a, b, c) \equal{} ( \minus{} 2, 1, \minus{}2)$ ή $ (a, b, c) \equal{} (2, \minus{} 1, 2)$

Εχεις ενα αριθμητικό στο τέλος οι τριάδες είναι (2,-1,2) και (-2,1,-2) αλλά είσαι σβέλτος...

Δίκιο έχεις, το διόρθωσα. Άλλη μια ίδιου επιπέδου:

Να βρεθούν $ a, b, c$ τέτοια ώστε:
$ 2a^2 \plus{} 2b^2 \plus{} c^2 \plus{} 2ab \minus{} 2ac \plus{} 2b \plus{} 1 \equal{} 0$

Ειναι ταυτότητες.  Ετσι παίρνουμε

$ a^2 \plus{} c^2 \minus{} 2ac \plus{} b^2 \plus{} a^2 \plus{} 2ab \plus{} b^2 \plus{} 2b \plus{} 1 \equal{} 0\longleftrightarrow {(a \minus{} c)}^2 \plus{} {(a \plus{} b)}^2 \plus{} {(b \plus{} 1)}^2 \equal{} 0$ ετσι εχουμε (a,b,c)=(1,-1,1)

Και, για να γίνει ακόμα πιο δύσκολο, αν είχα ακόμα στο πρώτο μέλος τον όρο $ \plus{} 4ac$;

Αν εννοείς να προσθεσεις θα βγει απλώς $ {(a\plus{}c)}^2$

Ναι, εννοούσα να προσθέσω, αλλά έκανα τυπογραφικό λάθος έιναι $ \minus{} 4bc$


when $n-butyl,t-butyl ether$ was made to satnd with conc. sulphuric acid then a heavy hydrocarbon was formed after a long time what can it be and why

This seems to be an interesting problem. But what do you mean by heavy hydrocarbon? How heavy? Is it aliphatic or aromatic?

well it's aliphatic but if i give out how heavy i think the problems loses it's charm  :wink: the critical word is heavy rather very heavy

If the answer is a polymer, then "very heavy" should be more appropriate. In my opinion, the problem was not very well stated, because a lot of things could happen in solution. That's why in problems like this some kind of clue is often provided. Also, the context of the problem also gives some hints about the answer. Anyway, from where did you take this problem? Or you created it by yourself?

well i wrote "very heavy" in my third post..i got this in morrison boyd .it has lesss but great questions

[quote="Pardesi"]i got this in morrison boyd[/quote]

What chapter and page, if you don't mind?

Yes, and in the chapter of alkene reactions. The problem also says "...with an high boiling point" which is also relevant. Of course, now knowing all this the answer is straightforward.


Suppose the interior of the unit circle is divided into two equal areas
by an arc $ C$ (i.e., $ C$ is a non-self-intersecting path) with end points on
the circle. Prove that the length of $ C$ is at least $ 2$.

Find a way to construct such arc if its length is given (which is at least $ 2$).

There are different meanings for the word "arc" in mathematics.

Just so you know that I know (and others may know who aren't
speaking up) that in this context, an arc is any smooth curve
joining two points.**  For this problem, you must have added
the extra restriction that your arcs are not to cross themselves,
because my source doesn't say either way.

** My source:  Wolfram MathWorld


[quote="Naru"]Erm, pyoron >_>

You see, I just saw how to make kilometers into meters and and meters into centimenter :3
The same for liters grams and all those other shiz XD

The thing is, Im like... slow .__.
Are there any tips on how to like... make the process faster? D:

Help :<[/quote]

wiki it
I know 100 cm in a meter, 2.54 cm in an inch, about 1.61 km in a mile.
I'm not sure for liters/grams/etc
I think it's like 2.2 lbs for a kg

If you don't have the direct conversions in front of you, use dimensional analysis.  Of course you'll have to know other conversions but the method is great.


Evaluate $\sin^2 20^\circ + \cos 10^\circ \sin 40^\circ$.


[quote="nguoivn"]Given $ a, b, c \geq\ 0.$
Prove that: $ \sqrt []{1 \plus{} \frac { a^2 \plus{} 3b^2}{c^2 \plus{} ca}} \plus{} \sqrt []{1 \plus{} \frac { b^2 \plus{} 3c^2}{a^2 \plus{} ab}} \plus{} \sqrt []{1 \plus{} \frac { c^2 \plus{} 3a^2}{b^2 \plus{} bc}} \geq\ 3\sqrt []{3}$
 :)[/quote]
After using AM-GM, we just need to prove that
\[ (3a^2\plus{}b^2\plus{}c^2\plus{}bc)(3b^2\plus{}c^2\plus{}a^2\plus{}ca)(3c^2\plus{}a^2\plus{}b^2\plus{}ab) \ge 27abc(a\plus{}b)(b\plus{}c)(c\plus{}a)\]
Applying AM-GM Inequality again,
\[ (3a^2\plus{}b^2\plus{}c^2\plus{}bc)(3b^2\plus{}c^2\plus{}a^2\plus{}ca)(3c^2\plus{}a^2\plus{}b^2\plus{}ab) \ge 24\sqrt{3} abc \sqrt{(a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)}\]
It remain to show that
\[ 8\sqrt{(a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)} \ge 3\sqrt{3} (a\plus{}b)(b\plus{}c)(c\plus{}a)\]
which is trivial because $ x^2\plus{}xy\plus{}y^2 \ge \frac{3}{4} (x\plus{}y)^2$ for all $ x,y$.


function
real analysis
real analysis unsolved

[quote="the game"]Can any 1 here use cantor's little theorem & digonalisation theorem to prove baire's theorem & godel's 2nd incompleteness theorem.[/quote]

Well this question is here for days & heres no reply.
Cmon guys any ideas..

Could you please define all the terms you have used? For instance, I've never heard about "Cantor's little theorem", even though I'm sure I know what you mean. Anyway, it's better to have all the terms defined.

Ok heres the cantor's little theorem:
Let A,B be sets & f:AxA->B such that all functions g:A->B are representable by f.Then every function  :theta: B->B has a fixed point.

"all functions g:A->B are representable by f"  :?

[quote="grobber"]I don't know what that means either, I was wrong :). However, I have found [url=http://www.ias.ac.in/resonance/Aug2004/Aug2004p8-17.htm]this article[/url] online. I'm going to read it (it treats precisely the subject here: Cantor's Little Theorem and his diagonalization), and, hopefully, it will shed light on this.[/quote]

Well i certainly hope you have seen it.But the article you are seeing is from where I got the idea.I regularly take this magazine. I am waiting for your reply.


algebra
polynomial
linear algebra
linear algebra unsolved

If $T:V\longrightarrow V$ is nilpotent, prove that there exists $y\in V,y\neq0$ such that $Ty=0$, that is, $0$ is an eigenvalue of $T$.

(I know this is pretty easy, but I want to see a variety of solutions.)

Hi.

A linear map [tex]f:V\to V[/tex] is nilpotent iff its characteristic polynomial equals [tex]x^n[/tex], where [tex]n[/tex] is the dimension of [tex]V[/tex]. Since the zeros of the characteristic polynomial are the eigenvalues of [tex]f[/tex], we're done.

Another Solution: let [tex]f:V\to V[/tex] be a linear map and [tex]\lambda\in K[/tex] an eigenvalue of [tex]f[/tex]. Then, as we can easily proof, [tex]\lambda^k[/tex] is an eigenvalue of [tex]f^k[/tex]. For a nilpotent linear map [tex]f[/tex] with [tex]f^n=0[/tex], this yields [tex]\lambda^n=0[/tex], whence [tex]\lambda=0[/tex], since [tex]K[/tex] is a field.

Hanno

In fact, if the eigenvalues of $A$ are $\lambda_i$, then the eigenvalues of $f(A)$ (where $f$ is a polynomial) is $f(\lambda_i)$

[quote="towersfreak2006"](I know this is pretty easy, but I want to see a variety of solutions.)[/quote]
Let's get really simple-minded.
$T$ being nilpotent means that for some $m,\,T^m=0.$

Let $x$ be any nonzero vector. The sequence $\{x,Tx,T^2,T^3x,\dots\}$ starts out nonzero but is eventually zero. Let $y$ be the last nonzero element of the sequence. Then $y\ne0$ but $Ty=0=0\cdot y.$ Hence, $y$ is an eigenvector for the eigenvalue $0.$

[quote="Hanno"]Another Solution: let $f: V\to V$ be a linear map and $\lambda\in K$ an eigenvalue of $f$. Then, as we can easily proof, $\lambda^k$ is an eigenvalue of $f^k$. For a nilpotent linear map $f$ with $f^n=0$, this yields $\lambda^n=0$, whence $\lambda=0$, since $K$ is a field.
[/quote]
How do you know an eigenvalue exists?

HI JMerry.

In my second proof, I don't know. It's only a proof for the fact that if there's an eigenvalue, then it must be 0.

Hanno


geometry
circumcircle
inradius
cyclic quadrilateral
power of a point
radical axis
geometry proposed

The cirumradius and the inradius of triangle $ ABC$ are equal to $ R$ and $ r$; $ O$, $ I$ are the centers of respective circles. External bisector of angle $ C$ intersect $ AB$ in point $ P$. Point $ Q$ is the projection of $ P$ to line $ OI$. Find distance $ OQ$.

[i]Author: A.Zaslavsky, A.Akopjan[/i]

Let $ I_a, I_b, I_c$ be excenters of triangle $ ABC. O'$ is the circumcenter of triangle $ I_aI_bI_c$. We have O is [b]Nine-point circle[/b]'s center and I is the othorcenter of triangle $ I_aI_bI_c$ so $ O', O, I$ are collinear and $ OO'\equal{}OI\equal{}\sqrt{R^2\minus{}2Rr}$
Since $ I_aBAI_a$ is cyclic quadrilateral then $ PA.PB\equal{}PI_a.PI_b$ which follows that P lies on radical axis of (O) and (O'). But $ PQ\perp OI$ then PQ is the radical axis of two circles.
Therefore $ QO'^2\minus{}R^2_{I_aI_bI_c}\equal{}QO^2\minus{}R^2$
$ \Leftrightarrow (QO\plus{}OO')^2\minus{}4R^2\equal{}QO^2\minus{}R^2$
$ \Leftrightarrow OO'^2\plus{}2OO'.OQ\minus{}3R^2\equal{}0$
$ \Leftrightarrow OQ\equal{}\frac{3R^2\minus{}OO'^2}{2OO'}\equal{}\frac{2R^2\plus{}2Rr}{2\sqrt{R^2\minus{}2Rr}}\equal{}\frac{R^2\plus{}Rr}{\sqrt{R^2\minus{}2Rr}}$


Compute the following sum,
\[ \sum_{k \equal{} 1}^{n}\left\lfloor\frac {n \plus{} 2^{k \minus{} 1}}{2^k}\right\rfloor
\]

[quote="tdl"][hide="Hint"]Use $ [x \plus{} \frac12] \equal{} [2x] \minus{} [x]$[/hide][/quote]

woah. how did you get that identity?

It is a well-known identity!
You can prove it by consider $ 2$ cases: $ \{x\} < \frac12$ and $ \{x\}\geq\frac12$ (In this $ \{x\}$ is fraction part of $ x$)

thnx for the hint.
That pretty much wraps the problem.


who has seen the movie?  I dont get whats going on.

I've seen it; made sense, somewhat. It's some of that modern "artistic" stuff which, depending on your view of it, is either amazing, or requires no talent whatsoever to do and isn't really art. Normally I fall into the latter camp, but that director is interesting. He made Requiem for a Dream too.

I personally enjoyed Pi.  It involves a guy who ultimately stops looking for...[i]the ultimate[/i].  Interpret that however you like.  I felt like the artistic elements were well used in the telling of the story.

I didn't even know there's a movie called Pi out...guess it hasnt gotten all too much advertising

It came out several years ago.  It's an independent film.  I lived in New York at the time and it got some (unusual) advertising there, but I don't imagine it got a lot of advertising elsewhere.

Yep, I saw it in [i]Math and Physics in the Arts[/i] class a few weeks back.

Mmm, bzzzzzz (drilling noise).

I'll look it up...probably a very mathematical one...

i liked Pi quite a bit... the story itself is rather interesting, but i think the directing was much more...worthy of praise. it is told so frantically, the cinematography is employed so well, that it is worth a look just for the directing. also, check out Requiem for a Dream, by the same director.


algebra
polynomial
ratio
inequalities
calculus
integration
number theory unsolved

Show that x^8 + x^7 - x^5 - x^4 - x^3 + x + 1 cannot be a square whenever x is greater than 1.
( the polinomial is actually the ratio x^10-x^5+1 / x^2-x+1 ).

The following is my solution,I don't know whether it is right or not,which I know is that it is really an ugly solution. :?
We can prove that for $x \geq 2$ the inequality holds :
$\sqrt{x^8+x^7-x^5-x^4-x^3 +x+1} +0.289 <  x^4+\frac{x^3}{2}-\frac{x^2}{8}-\frac{7x}{16}<\sqrt{x^8+x^7-x^5-x^4-x^3 +x+1}+0.432  $
Now consider that $\sqrt{x^8+x^7-x^5-x^4-x^3 +x+1}$ is an integer,since $16(x^4+\frac{x^3}{2}-\frac{x^2}{8}-\frac{7x}{16})$is also an integer,clearly the only integer in the inteval $ (16 \times 0.289,16 \times 0.432)$ is 5 and 6.
So $x^8+x^7-x^5-x^4-x^3 +x+1=(x^4+\frac{x^3}{2}-\frac{x^2}{8}-\frac{7x}{16}-\frac{5}{16})^2$ or $(x^4+\frac{x^3}{2}-\frac{x^2}{8}-\frac{7x}{16}-\frac{6}{16})^2 $
For the first case,we have $\frac{3x^4}{64}-\frac{51x^3}{64}-\frac{69x^2}{256}+\frac{93x}{128}+\frac{231}{256}=0$,it is easy to show that the equation doesn't have integral roots.
The second case is similar.
The solution is too ugly.I hope someone will give a much more beautiful solution. :)

That is the 30th cyclotomic polynomial. So we need to show that $\Phi_{30}(x) = y^{2}$ has no solution. Let $a  = x - y$ and $f(x) = \prod_{d|n, d < n} \Phi_{d}(x)$. We have $x^n - 1 = (x-a)^{2}f(x)$. There is a double root lemma where, if $x^n - 1$ has a double root $\bmod p$, then $p|n$. Since the equality makes the congruence valid for all p, we have $\forall p (prime), p|30$, which is a contradiction. Q.E.D.

[quote=rchokler]Clearly $x^8+x^7-x^5-x^4-x^3+x+1>x^8$ since $x^5+x^4+x^3<3x^5<4x^5\leq x^7$ for all $x\geq 2$.

Also, $(x+1)^8>x^8+x^7-x^5-x^4-x^3+x+1$ since the difference is $7x^7+28x^6+57x^5+71x^4+57x^3+28x^2+7x$ with all positive coefficients.

Done.[/quote]

That only is to prove that it's not an 8th power, but doesn't suffice to prove it's not a square.


geometry
inradius
circumcircle
trigonometry
trig identities
Law of Sines

Points $ P$ and $ Q$ are on hypotenuse $ AB$ of right triangle $ ABC$, $ PB \equal{} CB$, and $ QA \equal{} CA$. If the diameter of the inscribed circle is $ 10$, compute the diameter of the circumscribed circle of triangle $ CPQ$.

[hide="I got"]$ 10\sqrt2$[/hide]

Please revise the statement, since such as it is, it implies $ P\equiv Q$ (the midpoint of $ AB$). Or expressly state that either $ P$ and $ Q$ or both are on the LINE $ AB$, but outside the SEGMENT $ AB$.

No... It says $ PB\equal{}CB$ and $ QA\equal{}CA$ (not $ PB\equal{}CP$ and $ QA\equal{}CQ$).

[quote="Yongyi781"]No... It says $ PB \equal{} CB$ and $ QA \equal{} CA$ (not $ PB \equal{} CP$ and $ QA \equal{} CQ$).[/quote]

Ah, sorry  :blush: I was too quick...

[quote="J.Mayers"]I got it too.
Letting PQ=$ \psi$, and $ AC \equal{} a$ and $ BC \equal{} b$, we get that QB=$ b \minus{} \psi$ and $ AP \equal{} a \minus{} \psi$.  Since the radius of the inscribed circle of $ \triangle{ABC}$ is $ 5$, we get that $ \psi \equal{} 10$ (using the fact that the inradius of a right triangle is given by $ I_{R} \equal{} \frac {a \plus{} b \minus{} c}{2}$.  Letting $ < CPQ \equal{} \alpha$ and $ < CQP \equal{} \beta$, we get that $ < PCQ \equal{} 180 \minus{} \alpha \minus{} \beta \equal{} \alpha \plus{} \beta \minus{} 90$ (From isosceles triangles AQC and BPC).  Hence, $ \alpha \plus{} \beta \equal{} 135^{\circ}$. Thus, $ < PCQ \equal{} 45^{\circ}$.  Drawing the circumcircle with center $ O$ of PQC, we see then that $ \triangle OPQ$ is an isosceles right triangle with hypotenuse $ 10$.  Therefore, its legs (which are the circle's radii) have lengths of $ 5\sqrt {2}$.  The diameter is twice this, or $ \boxed{10\sqrt {2}}$.[/quote]

I did a similar way, but you definitely have more proof, and the solution is very nice. Thanks for posting (and for also letting me know about the length of an inradius in a right triangle).

[quote="dgreenb801"][quote="mathwizarddude"] If the diameter of the inscribed circle is $ 10$
[/quote]
The inscribed circle of $ \triangle ABC$ or $ \triangle CPQ$?[/quote]
I don't know (but I take it to be ABC)... this is from the Intermediate Geometry Marathon (no one bothered to solve it there) and I just wanted to check if my ans is correct... (I did almost exactly the same thing as J.Mayers except I used the extended law of sines (actually pretty much the same thing)...
hmm, is this solvable if it's triangle CPQ???  :?:


Hi guys ;) ,her is the problems from an Armenian olimpiad,in it I have taken [b]first[/b] prise :first: ([b]notebook computer[/b]  :P  :D )

 [b]05.02.2006[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=426603#426603]Problem 1.[/url]
Solve the equation in $Z^{+}$
$y^{2}x^{4}+zy^{2}=4x+5z+x^{2}y^{2}z^{2}$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=426605#426605]Problem 2.[/url]
Let $ABC$ the triangle, $D$ the point in $BC$ that $BD*AB=DC*AC$.
$M$ and $N$ is midpoints of segments $AD$ and $BC$ respectively .We know that
$\angle{MBC}+\angle{MCA}=\angle{MBA}+\angle{MCB}$.Prove that circumcentre  of triangle $BMC$ lie on $MN$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73931]Problem 3.[/url]
Prove that if $n\geq 28$ and $0<x<\frac{\pi}{2n}$
than $\frac{sin2x}{sinx}+\frac{sin3x}{sin2x}+...+\frac{sin(n+1)x}{sinnx}<2ctgx$(we know that $\pi<3.15$).

[quote="jmtlf"]Does Armenian math olimpiad have only three problem?[/quote]
This isn't  Armenian National Olimpiad,it will be in march and I will post
the problems of that contest too,ther shoud be 6 problems like other NMO-s.In this olimpiad juri desided to give us only tree problems :(  :) ,first two isn't difficult,but laste one havn't been solved in that contest(it is still unsloved in Olimpiad Section  :(  ;) ).

[quote="Tiks"][quote="Valentin Vornicu"]What olympiad is this then, if it's not the National Olympiad? Regional Olympiad?[/quote]
This is the Olimpiad between   the  Armenian  grammar schools,but it is usually more harder than National Olimpiad :( ,today has finished first day of All Armenian Olimpiad,but the problems was too easy(it took me an hour to solve them all :blush: ) ,so I think I shouldn't post that problems in Mathlinks :? .[/quote]

Just go ahead and post them please.


Let $ X\equal{}\prod_{t\in T} X_t$

[b]Definition.[/b] We say that $ A\subseteq X$ depends on $ I\subseteq T$ (we write $ A\sim I$) iff for any $ x\in A$ and $ y\in X$ if $ x|_I\equal{}y|_I$ then $ y\in A$.

Show that

(1) $ A\sim I$ iff $ A\equal{}\pi^{\minus{}1}_I\pi_I(A)$ where $ \pi_I: X\to\prod_{t\in I} X_t$ is a projection,
(2) if $ A\sim I$ and $ A\sim J$ then $ A\sim I\cap J$.

Well (1) should be straightforward.
for (2) just show that for every $ A$ there is a unique minimal $ \mathcal{I}_A$ so that $ A\sim \mathcal{I}_A$. It satisfies: if $ \mathcal{I}_A\subset M\subset T$ then $ A\sim M$
and then its easy to show that $ \mathcal{I}_A\subset I$ and $ \mathcal{I}_A\subset J$ so that $ \mathcal{I}_A\subset I\cap J$ implying $ A\sim I\cap J$

Thanks for Your answer but I'm having some problems with (2): when I try to find $ \mathcal{I}_A$ it is tempting to define it as an intersection of all $ I$ such that $ A\sim I$ and then I'm back to my original problem (or maybe even worse one...).


geometry
circumcircle
ratio
geometry proposed

Given a triangle $ ABC.$  $ M$ is a point on the $ AC$. $ \omega_{1}, \omega_{2}$ is a circumcircle of triangles $ ABM$ and $ CBM$.
Bisector of $ \angle AMB$ intercet $ \omega_{1}$ in $ C_{1}$; bisector of $ \angle CMB$ intercet $ \omega_{2}$ in $ A_{1}$. $ L$ -- midpoint of $ AC$.

Prove that $ \angle C_{1}LA_{1}\equal{}90^{\circ}$.

Use inversion center at $ M$ with radius $ MA.MC$

[quote="Elementaryyy"]Use inversion center at $ M$ with radius $ MA \cdot MC$[/quote]
Consider the negative inversion with center $ M$ and power $ \overline{MA} \cdot \overline{MC}.$ Let $ B'$ be the inverse of $ B.$ Circles $ \omega_1,\omega_2$ are taken into the lines $ B'C,B'A$ respectively and angle bisectors of $ \angle(BM,AC)$ are double $ \Longrightarrow$ $ A_2 \equiv MA_1 \cap B'A$ and $ C_2 \equiv MC_1 \cap B'C$ are the inverse of $ A_1,C_1.$ Let $ N \equiv A_2C_2 \cap BB'$ and $ L' \equiv A_2C_2 \cap AC.$  From the harmonic cross ratio $ (A_2,C_2,N,L'),$ it  follows that $ (A,C,M,L)$ is harmonic as well $\Longrightarrow$ $ \overline{ML} \cdot \overline{ML'} \equal{} \overline{MA} \cdot \overline{MC}$ $ \Longrightarrow$ $ L$ and $ L'$ are inverse under the referred inversion. Since $ A_2,C_2,L'$ are collinear, then $ \odot(A_1C_1L)$ goes through $M.$

[quote="plane geometry"][hide]AC1A1' and BC1A1 are concurrent[/hide][/quote]

I did not understand your solution, but I think you  this:

Let $ A'_1$ point such that $ CA_1AA'_1$ is a parallelogram.
You can prove that $ \angle C_{1}AA'_{1}\equal{}\angle C_{1}BA_{1}$
Then, because $ AA'_1\equal{}A_1C$  and $ BC_1\equal{}AC_1$, the triangles $ A'_1AC_1$ and $ C_1BA_1$  are congruents
Then, $ A'_1MA_1$ is isosceles, and K is the midpoint and finish.


Note that $ f(e^{\frac {2\pi i}3}) \equal{} 0$, so $ x^2 \plus{} x \plus{} 1$ is a factor.

Dividing, we get $ x^5 \plus{} x \plus{} 1 \equal{} \boxed{(x^2 \plus{} x \plus{} 1)(x^3 \minus{} x^2 \plus{} 1)}$ (the latter of which is reducible over the rationals).

Wait, can you explain your method more fully? I got that it goes through Euler's Identity, but after that, not so much.

$ x^2 \plus{} x \plus{} 1 \equal{} (x \minus{} e^{\frac {2\pi i}3})(x \minus{} \overline{e^{\frac{2\pi i}3}})$. Since $ e^{\frac {2\pi i}3}$ is a root, its conjugate must be as well (since the coefficients of the polynomial are real). Then it's just polynomial division.


calculus
integration
trigonometry
Euler
real analysis
real analysis unsolved

Dear everyone.

I have two problems, but they are hard for me to solve. :blush:

$ 1) \int{\sqrt {4x^{2} \minus{} 4x \plus{} 10} dx}$
$ 2) \int_{0}^{\infty}{\sin{(x^{n})} dx}$

please help me with your answers. thanks :)

For the former, complete the square and do the trigonometric or hyperbolic substitution. It's ugly. Keep slogging.

For the latter, is that really the question you want to ask? I see no reason for the values to be nice in general.

$ \sqrt {4x^{2} \minus{} 4x \plus{} 10} \equal{} \sqrt {(2x \minus{} 1)^{2} \plus{} (3)^{2}}$

now use integration formula of  $ \sqrt {x^{2} \plus{} a^{2}}$

@jmerry :  are u getting the second one ?

$ \int_0^\infty \sin x^n \mbox{ }dx \equal{} \frac{1}{n}\Gamma\left(\frac{1}{n}\right)\sin\frac{\pi}{2n}$.

(Consider $ \int_0^\infty e^{\alpha x^n}\mbox{ }dx \equal{} \alpha^{\minus{}\frac{1}{n}}\Gamma\left(1 \plus{} \frac{1}{n}\right)$, which is seen to be true by a simple change of variables. You can obtain the corresponding integral for the cosine from this identity as well.)

hello, for the first integral you can use one of the substituions of Euler
$ \sqrt{ax^2\plus{}bx\plus{}c}\equal{}\pm x\sqrt{a}\plus{}t$ if $ a>0$.
Sonnhard.

Dear everyone.

thank all of you for sincere answers.
though I don't know what you said perfectly, I will try again. :)


inequalities
geometry
inequalities proposed

Prove:
27 (x+y)^2 * (x+z)^2 * (y+z)^2 >= 64 xyz (x+y+z)^3.
dont use jensens (by setting x+y+z=1 and log-ing both sides to make it really easy), its too easy. dont bother muirhead bashing it either because im sure it works.
Also try the general version or prove it false:
(n^n) [(a1+a2)(a1+a3) ... (a_(n-1)+a_n)]^2 >= [4^(n,2)] a1*a2 .. *a_n * (a1 + a2 .. + a_n)^n.
the bashing argument would get very bashy in this case and jensen's would simplify fail to work. note i haven't solved it yet.

[quote="epitomy01"]Prove:
27 (x+y)^2 * (x+z)^2 * (y+z)^2 >= 64 xyz (x+y+z)^3.
dont use jensens (by setting x+y+z=1 and log-ing both sides to make it really easy), its too easy. dont bother muirhead bashing it either because im sure it works.
[/quote]

What about using triangle geometry?

Let $a=x+y, b=y+z, c=z+x$, then the inequality is equiv. to $27R^2\ge (a+b+c)^2$ which is easy to prove.


Let $ \{a_n\}_{n \in \mathbb{N}_0}$ be a sequence defined as follows: $ a_1\equal{}0$, $ a_n\equal{}a_{[\frac{n}{2}]}\plus{}(\minus{}1)^{n(n\plus{}1)/2}$, where $ [x]$ denotes the floor function. For every $ k \ge 0$, find the number $ n(k)$ of positive integers $ n$ such that $ 2^k \le n < 2^{k\plus{}1}$ and $ a_n\equal{}0$.

Source: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=229046]Oliforum contest, round 2[/url]

Let $ n \equal{} x_0 \plus{} x_1*2 \plus{} ... \plus{} x_k*2^k$, $ x_i$ are 2-digits $ x_i \equal{} 0,1$ of n. Then by induction $ a_n \equal{} \sum_{i \equal{} 0}^{k \minus{} 1}( \minus{} 1)^{x_i \plus{} x_{i \plus{} 1}}$.
Therefore if k is odd, then $ N(k) \equal{} 0$, if k =2m, then $ x_i \equal{} x_{i \plus{} 1}$ for m means of n and $ x_k \equal{} 1$. Therefore $ N(2m) \equal{} \binom{2m}{m}$.
If $ N(k,l)$ is number of $ 2^k\le n<2^{k\plus{}1}$ suth that $ a_n\equal{}l\minus{}k,0\le 0<2k$, then $ N(k,2m\plus{}1)\equal{}0$, $ N(k,2m)\equal{}\binom{k}{m}.$


Josh was born at noon on July 4, 1991. Angelica was born at
noon on September 10, 1991. How many days older is Josh
than Angelica?

The number of days in July is 31, and the number of days in August is also 31:

$ (31\minus{}4)\plus{}31\plus{}10\equal{}\boxed{68}$

hmmm that way is a little confusing.

Theres a 2 month difference between sept. and july.

Both of those months have 31 days. So

31x2=62

then add 6 more days because 10-4=6.


quadratics
inequalities
inequalities unsolved

[quote="behdad.math.math"]for real numbers a,b,c,d prove that:
(a+b+c+d)^2=<3(a^2 +b^2 +c^2 +d^2) + 6ab[/quote]
It's a quadratic inequality of $ a.$ :wink:

Expanding is best choose
But CS ineq also works here
$ LHS\le (1 \plus{} 1 \plus{} 1)((a \plus{} b)^2 \plus{} c ^ 2 \plus{} d^2) \equal{} RHS$ :lol:


If $ 2(2^{2x})\equal{}4^{x}\plus{}64$, then x is equal to:

A) – 2
B) – 1
C) 1
D) 2
E) 3


function
integration
calculus
real analysis
real analysis unsolved

Let f be a function, defined on R, such $ \displaystyle f\left( x \right) = \int\limits_0^x {\frac {{2{e^{ - {t^2} + t}}}}{{1 + {e^t}}}dt}$. Prove that exist at least one ξ real number, such $ \displaystyle xf'\left( \xi \right) = \int\limits_0^x {{e^{ - {t^2}}}dt}$, for every x real number.

Set $ g(t) \equal{} \frac{2e^{\minus{}t^{2} \plus{} t}}{1 \plus{} e^t}$. Then g is continuous on the finite interval [0,x].
Then by the MVT for integrals, there exists $ \xi \in [0,x]$ such that $ g(\xi) \equal{} \frac{1}{x \minus{} 0}\int_{0}^{x} g(t) dt$. It is not hard to show that f is a differentiable function with $ f'(y) \equal{} g(y)$ whenever $ 0 <y< x$.

Hence we have $ xf'(\xi) \equal{} \int_{0}^{x} g(t) dt$.

:)
$ f'(t)\equal{}\frac{2{e^{\minus{}{t^{2}}\plus{}t}}}{1\plus{}{e^{t}}}\equal{}\frac{2}{1\plus{}e^{\minus{}t}}e^{\minus{}{t^2}}$
its range contains the range of $ e^{\minus{}{t^2}}$

[quote="lazyman"]:)
its range contains the range of $ e^{ \minus{} {t^2}}$[/quote]
And how this helps?  :maybe:

:)
$ \frac {1}{x}\int^{x}_{0}e^{ \minus{} t^2} \equal{} e^{ \minus{} {\xi_1}^2}$,
then we may find a $ \xi_2$ such that $ f'(\xi_2) \equal{} e^{ \minus{} {\xi_1}^2}$.
Does this help?

Yes helps, though solution seems complicated and the exercise has a simplier one


migan natayej amade hast vali besoorate kod hast vali ta shanbe be asami tabdil mishan va kamelan amade mishan!!!!dige bastegi dare ke bedan ya inke bezaran bad az emtehanae nahaiii

bacheha chand nafaretoon motmaenin ke hatman ghabool mishin begin ta man taklife khodamo befahmam!!!!!

pas har ki nomrasho bege.

man misham 23 (minimum) ke ghabool nemisham.

Agar rahe hal be javab berese nomre 7 minus, khahad bood va be ezaye har gaame asasyie esbat ke zekr nashode bashe haddeaghal 1 nomre kam mishe. Vali age faghat ye seri ahkam benivisi baste be mofid boodaneshoon 0 plus khahid shod. Dar zemn baraye rahe hal hayi ke tedade ziyadi az oon rahe hal rafteand barem bandi mizarand.

[quote="saeid"][quote="Aragorn"]natayeje nojoom oomad[/quote]

ye 5,6 sali az donya aghabi.[/quote]
khob aghaye jolo tar az donya natayeje physics 25 miad roye site

[quote="saeid"][quote="Aragorn"]khob aghaye jolo tar az donya natayeje physics 25 miad roye site [/quote]

baz ham az donya kheily aghabi man alan asamiye ghabool shodegan physic ro ham daram.[/quote]
baz ham az donya kheily joloee  chon natayej momkene ta 25 ham nayad  :rotfl:

[quote="mohamadhosein"]fekr konam gharare bargeye 130 nafare aval dobare tashih beshe.[/quote]
are doroste, vali alan dige tashihe oona ham tamoom shode. farda ba'd az zohr, natayej miad roo site.

goftam shayad (parinaz ,fatemeh ,.... ) on ha ham montazerand  in ham vaseye oon ha.

[quote="amir2"]goftam shayad (parinaz ,fatemeh ,.... montazerand) on ha ham montazerand  in ham vaseye oon ha.[/quote]
man age montazere nataiejam vase ine ke mikham bebinam kia too doran, vagarna dar morede khodam kamelan motmaennam :huh:

[quote="fatemeh"]man age montazere nataiejam vase ine ke mikham bebinam kia too doran, vagarna dar morede khodam kamelan motmaennam [/quote]

ghabool daram shoma khafanid.

vali man ham goftam montazerid dalilesho nagoftm.

[quote="samanfarhat"]bacheha bikhodi montazer nabashin.
nataiej be tore ghat'a 30 om miad  :)[/quote]
vayyyyy chera 30???? :o  man ke ta oon moghe mimiram natayeje adabi ham oomade

[quote="first mathematician"]ba inke ye edde 42 mishan, vali balatarin nomreie marhale2 hodoode 35e  :rotfl:  :rotfl:[/quote]

az koja midooni?

aghaye hatami hadeaghal shoma begid natayejo key midan akhe  har kasi baraye khodesh yek roozi mige!!!

Tebghe akharin akhbari ke man daram 10 tir dore shoroo mishe. Dar zemn natayeje riazi kamelan amade ast va montazer hastand ke natayeje computer ham amade beshe (Chon har kas hadde aksar ye olampiad mitoone sherkat kone va bayad olaviatha taeen beshe.)

Dar zemn ba tavajjoh be manabe olampiad computer oona hanooz dar hale tashihe soalat hastand va banabarin NATAYEJ BE ZOODI ELAM NAKHAHAD SHOD

Albate be gofteye masooline bashgah (ke mosalaman nabayad behesh etemad konid) yekshanbe natayej miyad.

Va dar zemn khabare dige in ke be ehtemale ziyad man dar dore (hamchon sale pish) dars midaham va hamchenin (mesle sale ghabl) az tarrahane soalha ham khaham bood.

khosham miyad ke har kasi baraye khodesh yek roozi mige !!!baba bikhail dige felan felan ke fekr konam bayad heyroon bashim!!!!!!!

dirooz yani panjshanbe be chand nafar az bachehaye allame helli ke tooye shobeye alef ta r emtehan midadan az bashgah zang zadand,be kesi zang nazadan az shoma?

[quote="samanfarhat"][quote="first mathematician"]ba inke ye edde 42 mishan, vali balatarin nomreie marhale2 hodoode 35e  :rotfl:  :rotfl:[/quote]
man ham to ro mishnasam ham dna.
ta chizi ro motmaen nabasham nemigam.
hala khodetoono koshtin az bas goftin.
man kamelan motmaennam.
tahala chand bar goftin man ham chizi nagoftam.
bas konin.[/quote]
man manzooram to nisti, kheilia hastan. to ham yekishooni. chera be khodet migiri?
"man kamelan motmaennam." ----> chi ro kamelan motmaenni?

doroste dna,mikhastam bedoonam ke dirooz be kesi ke tu shobeye alef ta r tehran emtehan dade zang nazadan?
chon be chand nafar zang zadan va azashun karname khastan,dar hali nagoftan chera.fekr konam chon tu shobeye ma karnameharo jam nakardan.

be hich kodoom az dovomi hayi ke too madreseye allame helli boodan karname nadade boodan , alan be chand ta az bache haye madreseye ma zang zadan goftan karname hatoono biarin ! ehtemalan kasayi ke beheshoon zang zadan hatman ghabool shodan ! (albate ye ehtemale ) DNA (!)  be to zang nazadan ?

[color=red]lotfan chert o pert nevisi ro baas konin, vagarna topic delete khahad shod. ;)  ;)[/color]

[quote="mohamadhosein"][quote="fatemeh"][quote="mohamadhosein"]yani shanbe ham natayej ro nemidan? :([/quote]
shanbe nataiej ma'loome, yekshanbe miad roo site :D[/quote]
motmaeni emrooz miyad?[/quote]

Fekr konam faghat ye joke bood. ;)  ;)

shaydam ta shab natayej biyad.mesle marhale aval. :roll:

[quote="Omid Hatami"]Hamoon tor ke mibinid, be ellate inke komiteye (tanbal) computer natayej ra elam nakard, bedoone elame natayeje computer (va dar nazar nagereftane olaviatha) natayeje rizazi elam shod.[/quote]
dar vaghe aya emkanesh hast ke ye tedad az in bachehayi ke ghabool shodan davat nashan dore? :?:


From a group of 10 students an executive board is chosen that has one president, one vice-president, one secretary and one treasurer. If each person can fill at most one of the positions, how many different executive boards are possible?


let Un be a sequence defined by
U0>=0,U_{n+1}=(1+Un)²+sqrt(Un²+2Un)
prove that limUn=+00

[quote="Farenhajt"]It must be $U_{0}>0$, since for $U_{0}=0$ you get $U_{n}=1$.

As for the proof, use induction to show $U_{n}>n$[/quote]
Why?
If $U_{0}=0$, then $U_{1}=1$, $U_{2}=4+\sqrt 3 >2$,...


we have the angle $Q$ , circle $C$ and point $P$ and lenght $k$ .
draw a line from $P$ which that $AB/CD=k$

[color=red][[b]Moderator edit:[/b] This problem cannot be solved using compass and ruler. This was shown for the case when P lies on one edge of the angle Q and when k = 1 in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=38828 .][/color]


[quote="Jason-Kountou"]Sorry but I think this is false, try $ a^2 \equal{} 1.7, b^2 \equal{} 0.8, c^2 \equal{} 0.5$.[/quote]

you are right. i have make a mistake.


geometry
perimeter
inequalities
triangle inequality
geometry unsolved

In a unit square one hundred segments are drawn from the centre to the sides dividing the square into one hundred parts (triangles and possibly quadruilaterals). If all parts have equal perimetr $p$, show that $\dfrac{14}{10} < p < \dfrac{15}{10}$.

[quote="Rushil"]In a unit square one hundred segments are drawn from the centre to the sides dividing the square into one hundred parts (triangles and possibly quadruilaterals). If all parts have equal perimetr $p$, show that $\frac{14}{10}< p < \frac{15}{10}$.[/quote]

Obviously each of the 100 segments have length atmost $1/\sqrt{2}$

Consider a triangle with lengths $a, b, c$ and let's suppose $a, c$ are the segments while $b$ is on the side of the square. By triangle inequality we have $a+b > c$ which gives $a+b+c > 2c$ so $p > 2/\sqrt{2}$ and $p > \sqrt{2}$ and $p > 1.4$.

Now $100 p = 2 * 100 (segment)+4$. We count each segment twice hence the factor $2$ and $4$ is because we count the sides of the square exactly once. Since each segment is $\leq  1/\sqrt{2}$ we get
$p \leq  2/\sqrt{2}+4/100$
$p \leq  1.414+0.04$
$p < 1.5$

So, $1.4 < p < 1.5$


topology
function
calculus
integration
Support
limit
Ring Theory

Let $ X$ be an infinite, compact, Hausdorff space. Denote by $ C(X)$ the real valued continuous functions on $ X$. Is $ (0)$ a decomposable ideal?

You mean non-trivially decomposable or something? $ (0)$ is itself a primary ideal

If by decomposable you mean intersection of two strictly larger ideals, then yes, it is: any infinite compact Hausdorff space $ X$ can be written as the union of two proper closed subsets $ A$ and $ B$ (let $ A$ be the closure of some small open set $ U$, and $ B\equal{}X\setminus U$), and consider the ideals of functions which vanish on $ A$ and respectivly $ B$; their intersection is $ (0)$.

Me too, I was thinking about the meaning of "decomposable" in this context. The exercise is quoted from the reference given above. There an ideal is called  decomposable iff it has a primary decomposition (which might not be a minimal one).

[quote="Third Edition"][quote="Kalle"]$ (0)$ is itself a primary ideal[/quote]I don't think so: If it was, it would be prime, since it is radical. But $ C(X)$ does not seem to be an integral domain. ;)[/quote]
Ohps

I should have mentioned.. sorry... by decomposable I mean intersection of primary ideals (finitely many).

I don't know why I was under the impression that Atiyah and MacDonald called "decomposable" an ideal which is a non-trivial intersection of two ideals :?. In fact, they call that "reducible". Anyway, with the proper definition given in Kuba's last post, the answer is 'no':

Assume $ (0)$ is decomposable. Passing to radicals, we can assume it's a finite intersection of prime ideals $ \frak p_i,\ i \equal{} \overline{1,n}$. The prime ideals of $ C(X)$ are in one-to-one correspondence with the maximal [url=http://planetmath.org/encyclopedia/NonPrincipalFilter2.html]filters[/url] of closed sets: given such a filter $ \mathcal U$, the corresponding ideal $ \frak p_\mathcal U$ is the set of functions vanishing on some member of $ \mathcal U$, and conversely, given $ \frak p$, the corresponding filter consists of those closed subsets $ F$ of $ X$ with the property that every function vanishing on $ F$ belongs to $ \frak p$. Denote the filters corresponding to $ \frak p_i$ by $ \mathcal U_i$ respectively.

Now write $ X$ as a union of $ n \plus{} 1$ closed subsets $ F_i,\ i \equal{} \overline{0,n}$, no $ n$ of which cover $ X$. This should be easy enough to arrange, given that $ X$ is infinite. Because the sets $ F_i$ cover $ X$ and the filters $ \mathcal U_i$ are maximal, each filter contains some set $ F_i$. After a renumbering, we can assume that $ F_i$ is a member of $ \mathcal U_i$ for $ i \equal{} \overline{1,n}$. Now let $ f$ be a non-zero function vanishing on $ F_1\cup\ldots\cup F_n$ - it exists because $ F_1,\ldots,F_n$ don't cover $ X$. $ f$ then belongs to the intersection $ \bigcap\frak p_i$, contradicting the assumption that this interection is $ (0)$.

[quote="grobber"]The prime ideals of $ C(X)$ are in one-to-one correspondence with the maximal [url=http://planetmath.org/encyclopedia/NonPrincipalFilter2.html]filters[/url] of closed sets: given such a filter $ \mathcal U$, the corresponding ideal $ \frak p_\mathcal U$ is the set of functions vanishing on some member of $ \mathcal U$, and conversely, given $ \frak p$, the corresponding filter consists of those closed subsets $ F$ of $ X$ with the property that every function vanishing on $ F$ belongs to $ \frak p$. Denote the filters corresponding to $ \frak p_i$ by $ \mathcal U_i$ respectively.[/quote]Do you need some property of the topological space to establish this correspondence? I suspect it does not hold for every topological space, in general. My proof heavily relies on the fact that $ X$ is paracompact.

I should make a correction to that statement. I called those filters "maximal", and I shouldn't have. The property I need in order for the correspondence to hold and the proof to work is this:

(*) Whenever the closed set $ A\in\mathcal U$ is a union of closed sets $ A \equal{} B\cup C$, one of $ B,C$ is a member of $ \mathcal U$.

I was misled by the fact that in general, a filter on a set is maximal (i.e. an ultrafilter) precisely when it satisfies (*) (after dropping the word "closed" everywhere, of course). When we deal with closed sets this is no longer the case. In fact, it should be clear that a filter of closed sets on a compact Hausdorff space $ X$ is maximal in the sense of not being contained in any larger filter of closed sets precisely when it's principal, i.e. the collection of all closed sets containing some point $ x\in X$.

Sorry about the inconvenience. I hope everything works now.

[quote="grobber"]... and conversely, given $ \frak p$, the corresponding filter consists of those closed subsets $ F$ of $ X$ with the property that every function vanishing on $ F$ belongs to $ \frak p$.[/quote]Apart from the fact that we are not dealing with filters, but with systems of closed sets fulfilling the filter axioms, I am not very sure about the quoted part. I doubt that the correspondence is well-defined because it is not evident to me that the finite intersection property of the "filter" corresponding to a prime ideal is fulfilled.  :maybe:

Yes, they're not exactly filters, but surely, everybody understood what I was talking about. The formulation "system of closed sets satisfying the filter axioms" would have been a nightmare. :)

Suppose every function vanishing on $ A$ belongs to $ \frak p$, and the same for $ B$ (both $ A$ and $ B$ are closed). Consider now a function $ f$ vanishing on $ A\cap B$. It can be written as a sum of functions $ g$ and $ h$ vanishing on $ A$ and respectively $ B$: the function $ \tilde g: A\cup B\to\mathbb C$ by $ \tilde g \equal{} 0$ on $ A$ and $ \tilde g \equal{} f$ on $ B$ is continuous, because the two branches agree on the intersection of the closed sets $ A$ and $ B$. Now extend $ \tilde g$ to $ g: X\to\mathbb C$ ([url=http://planetmath.org/encyclopedia/TietzeExtensionTheorem.html]the Tietze Extension Theorem[/url]), and put $ h \equal{} g \minus{} f$. As $ g$ and $ h$ vanish on $ A$ and $ B$, they belong to $ \frak p$, and hence so does $ f \equal{} g \plus{} h$.

Notice that this works for any ideal; the primality of $ \frak p$ is reflected in the property (*) in my previous post.

Edit:

Thanks Third Edition for noticing the typo; I had written $ \tilde g: A\cap B$ instead of $ \tilde g: A\cup B$.

Here's something sketchy.

Let's assume that $ (0)$ is decomposable, i.e. it is the intersection of finitely many primary ideals. Taking a minimal decomposition and passing to radicals ($ (0)$ is a radical ideal), we can assume wlog that $ (0)$ is the intersection of uniquely determined, finitely many prime ideals. It is known that they are annihilator ideals $ Ann(f_i)\equal{}\{g\in C(X)|gf_i\equal{}0\}$ with some $ f_i\in C(X)$, $ i\equal{}1,\ldots,n$ and that the set $ D$ of zero-divisors in $ C(X)$ is precisely the union of the annihilator ideals. Defining $ Z_i$ to be the zero-set of $ f_i$, this means that $ Z_1,\ldots,Z_n$ is a covering of $ X$ where none of the $ Z_i$'s can be dropped and every zero-divisor of $ C(X)$ is annihilated by one of the $ f_i$'s.

Exploiting extensively the paracompactness of $ X$, this implies that

1. the $ Z_i$'s are pairwise disjoint,
2. each of the $ Z_i$'s is a singleton

... altogether a contradiction.

[hide="Sketch of proof for 1. and 2."]To see 1. construct non-negative functions $ \varphi_i \in C(X)$ whose support is contained in $ Z_i$, but disjoint from every $ Z_j$ ($ j\neq i)$. Then show that $ \sum_i\varphi_i$ cannot be annihilated by any of the $ f_i$'s. For 2. it suffices to assume that $ X\equal{}Z_i$, such that $ C(X)$ is an integral domain. Suppose that $ X$ contains at least two points $ x_1,x_2$, then construct two non-negative continuous functions $ \varphi_i$ with $ \varphi_i(x_i)>0$ and disjoint support. Then $ \varphi_1\cdot\varphi_2\equal{}0$ leads to a contradiction.[/hide]

@grobber: could you also indicate why your maps between prime ideals and maximal filters of closed subsets are inverse to each other? I can't manage this since the prime ideals don't have to be closed (besides, the closed ones are maximal and in bijection to $ X$).

Unfortunately I was "a bit" hasty, and I got it all wrong. If that statement about the structure of prime ideals were true, it would imply, in particular, that every prime ideal of $ C([0,1])$ which contains a function vanishing only in $ 0$ contains all functions vanishing at $ 0$. This is not true: let $ f: [0,1]\to\mathbb(0,\infty)$ which decreases really fast towards $ 0$. Say, $ \lim_{x\to 0}\frac {f(x)}{x^n} \equal{} 0$ for all natural numbers $ n$. Then the ideal generated by $ f$ contains none of the functions $ f_n(x) \equal{} x^n$. Now let $ \frak p$ be an ideal containing $ f$ and which is maximal among the ideals containing none of the functions $ f_n$. It's easy to check that $ \frak p$ is prime, it contains one function vanishing precisely in $ 0$ (namely $ f$), but it doesn't contain all such functions (it doesn't contain the power functions $ f_n$).

So basically, everything I wrote in this topic before this message is nonsense :). However, I think my proof can still be repaired fairly easily. Starting with an arbitrary prime ideal, perform those two constructions which turned out not to be inverse to one another. You'll get another prime ideal, smaller than the one we started with, and of the desired form (i.e. obtained from a filter). Hence minimal prime ideals are of that form, and that's all we need in order to make the proof work: if $ (0)$ is a finite intersection of primes, then it's a finite intersection of minimal primes.


A boy pushed a circular table to the corner of a room so that it touched both walls. He then marked an spot on the extreme edge of the table with ink. The spot was exactly 8 inches from one wall and 9 inches from the other. What was the radius of the table?

[hide="Solution."]
Let the center of the table with radius $ r$ lie at $ (0, 0)$ and the spot marked be $ (x, y)$.  We know that the wall is $ 8$ inches away in one direction and $ 9$ inches away in the other, so:

$ x \plus{} 8 \equal{} r$
$ y \plus{} 9 \equal{} r$

Also, since the point lies on a circle:

$ x^2 \plus{} y^2 \equal{} r^2$

Substituting the the values of $ x$ and $ y$ into the last equation:

$ (8 \minus{} r)^2 \plus{} (9 \minus{} r)^2 \equal{} r^2$
$ r^2 \minus{} 34r \plus{} 145 \equal{} 0$
$ (r \minus{} 29)(r \minus{} 5) \equal{} 0$

Since $ 5$ is too small to be the correct answer, the radius must be $ \boxed{29}$.[/hide]

[hide="Contest-Style Approach"]It's approximately $ \frac{1}{1\minus{} \cos 45 ^{\circ}} \cdot \frac{8\plus{}9}{2} \approx 3.4 \times 8.5 \equal{} 28.9$. Verify that $ r\equal{}29$ works.[/hide]

I missed this before, but Begoner, there is no reason to discard the solution 5 - it is totally valid, it just won't be on the side of the table that is closer to the walls.


geometry
perimeter
calculus
integration
inequalities
triangle inequality

[hide]By the triangle inequality, we see that the triangle must have sides of 3,3, and 2. The altitude to the base with side 2 is $\sqrt8$ by the Pythag theorem, so the area is $\sqrt8$.[/hide]

[hide]
You realize that $8 = 4*2$ which simplifies to $2\sqrt2$[/hide]

I took the same approach as 4everwise and I also got $2\sqrt{2}$.

if you've got integral sides and perimeter, u might as well use heron's formula. i found it much easier with these sides.

[quote="4everwise"]A triangle with integral sides has perimeter $8$. The area of the triangle is

$\text{(A)}\ 2\sqrt2 \qquad \text{(B)}\ \frac{16}{9}\sqrt3 \qquad \text{(C)}\ 2\sqrt3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 4\sqrt2$[/quote]
[hide]
Only possible sides are 3,3,2.
Semiperimeter=4
$\sqrt{4(4-3)^2(4-2)}=\sqrt{8}=2\sqrt{2}$
Answer is A
[/hide]


probability
geometry
3D geometry
combinatorics unsolved
combinatorics

there are 27 squares with side 1. we will color 4 faces in orange, and let the two non-colored faces have one side in common.

Now, when we have 27 squares with side 1 into a square with side 3, find the probability of the large square having all of it's faces orange.

Thanks.

[quote="bjh7790"]squares[/quote]  A square is a polygon, i.e., a two-dimensional object.  You're trying to ask about [i]cubes[/i].

Please give your posts more informative titles -- this one could have been titled "Assembling colored cubes," for example.

This question is not really olympiad level and could just as easily been placed in the Preolympiad forum.  The positions of each cubelet are independent, so compute the probability for a corner square, edge square, face square and central square and then multiply appropriate powers together.


Prove that $ 2^{n}\minus{}1\in\mathbb{P}\implies n\in\mathbb{P}$.

[hide="Hint"]
$ (2^a-1)|(2^b-1)\Leftrightarrow a|b$
[/hide]

[quote="t0rajir0u"][quote="bertram"][hide="Hint"]
$ (2^a - 1)|(2^b - 1)\Leftrightarrow a|b$
[/hide][/quote]
I think many are familiar with the left direction, but it seems to me that the right direction is less obvious and deserves justification (although it is not necessary for this problem).
[hide="The relevant lemmas"] [b]Lemma 1:[/b]  $ x | y \Leftrightarrow \gcd(x, y) = x$ (this much should be clear, but it deserves to be stated; the proof should be easy and is left as an exercise)

[b]Lemma 2:[/b]  $ \gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1$

[hide="Proof"] Let $ d = \gcd(2^a - 1, 2^b - 1)$.  Then $ 2^a \equiv 2^b \equiv 1 \bmod d$.  By Bezout's Theorem, there exist integers $ x, y$ such that $ ax + by = \gcd(a, b)$.  Consequently, since $ d$ is odd,

$ 2^{ax + by} \equiv 2^{\gcd(a,b)} \equiv 1 \bmod d$

So $ d | 2^{\gcd(a,b)} - 1$, and clearly $ 2^{\gcd(a,b)} - 1 | d$.  The conclusion follows. [/hide] [/hide][/quote]

My proof is not as beautiful as yours,  but it's shorter. :P
[hide]
$ a|b\Rightarrow (2^a-1)|(2^b-1):$ In general, $ x^k-1=(x-1)\sum_{i=0}^{k-1}x^i$ (prove this by factorng the second equation). Now just set $ x=2^a, k=\frac ba\in\mathbb{N}$.
$ (2^a-1)|(2^b-1)\Rightarrow a|b:$ Set $ b = qa+r$ with $ q, r\in\mathbb{N}\cup\{0\}$ and $ 0\le r<a$. Then, by the above statement, $ (2^a-1)|(2^{qa}-1)\Rightarrow (2^a-1)|(2^{qa+r}-2^r)\Rightarrow (2^a-1)|(2^r-1)$ and $ r<a$ implies $ r=0\Rightarrow a|b.$
[/hide]

[hide="Another approach"]Prove the contrapositive: If $ n$ is composite, then $ 2^n\minus{}1$ is composite.

Let $ n\equal{}ab$, where $ a,b>2$.  Let $ x\equal{}2^a$ and $ y\equal{}1$.  Therefore $ x^b\minus{}y^b\equal{}(x\minus{}y)(x^{b\minus{}1}\plus{}x^{b\minus{}2}y\plus{}\ldots\plus{}y^{b\minus{}1})$.  Both terms are clearly greater than 1, hence $ 2^n\minus{}1$ is composite, and we're done.[/hide]

In general, we have that $ (a^{b}\minus{}1,a^{c}\minus{}1)\equal{}a^{(a,c)\minus{}1}\minus{}1$, right? Or does it only hold for prime a?

[quote="t0rajir0u"]Well, I think you meant $ (a^b \minus{} 1, a^c \minus{} 1) \equal{} a^{(b, c)} \minus{} 1$, but yes.[/quote]
Yes, I meant (b,c). So when you say yes, that means it only holds for prime a?

The argument I gave holds for all $ a$.  In fact, I believe it holds over a principal ideal domain.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=54800 , it hold's in arbitrary rings.


calculus
derivative
trigonometry
function
calculus computations

Now suppose you are asked to calculate the derivative of a trig function within a trig function such as:

y=cos(tan((1+2x^2)^.5))

Do you know the chain rule? \[ \left(f \circ g\right)'(x) = f'\left(g(x)\right) \cdot g'(x). \]

not sure how you apply it to this one though


Is it possible to put a Venn Diagram on $ \text{\LaTeX}$?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211639]How to draw Venn diagrams using Latex?[/url]


There are 30 students in Mrs. Taylor's kindergarten class.  If there are twice as many students with blond hair as with blue eyes, 6 students with blond hair and blue eyes, and 3 students with neither blond hair nor blue eyes, how many students have blue eyes?

[quote="zephyrrs"]This problem is solved easiest with a Venn Diagram.
Draw a Venn Diagram with 2 circles overlapping each other. Label one of the larger portions blond hair and the other one blue eyes.
If 3 kids have neither blond hair or blue eyes, the 3 parts of the Venn Diagram add up to equal 27.
If there are 6 kids that have both blond hair and blue eyes, there are 21 kids with either blond hair or blue eyes but not both.
It says that there are twice as many kids with blond hair than blue eyes, so the number of students with blue eyes is 21/3=7[/quote]

Don't forget to count the 6 students with blue eyes and blond hair as parts of both groups. Using your result of 7 plus the six would make 13 total students with blue eyes and 14 plus six or 20 with blond hair. So a 2 to 1 ratio would not exist as stated in the original problem.

Go back to your Venn diagram, a good strategy by the way, and let x represent the number of students with blue eyes but non-blond hair. Next represent blond hair, non-blue-eyed students in terms of the same x. Find the value of x by writing and solving an equation  that relates the 2 to 1 relationship.

Good luck!

[quote=zephyrrs]Oh I thought the 2 to 1 ratio didn't include the ones with both  :blush:
I was lazy and used guess and check instead, which wasn't too bad as I got the answer on the second try.
There are 16 people with only blond hair, and 5 people with only blue eyes. 16+6=22, 6+5=11
16+6+5+3=30 people total.[/quote]

I think that's incorrect, sorry for like the 8 year revive  :D

[hide = Solution]If there were $x$ blue eye people, there would be $2x$ blond people. That means there are $x - 6$ blue-eyed non-blond students. So that means that there are $2x-6$ blond non-blue-eyed students. This would all total to be $30$
$x - 6 +2x - 6 + 6 + 3 = 3x - 3 = 30$. Solving this for $x$ you get $x = 11$(Sorry for the hasty solution)[/hide]

I solved this differenly and got a different answer, where did I go wrong?

Total number of students = 30
3 have neither blue eyes or blond hair, so 27 have one of the two
since the ratio is 2:1, I assumed total number with blond hair = 18 and total number with blue eyes = 9
hence blond hair but no blue eyes 18-6 = 12
and blue eyes but not blond hair 9 - 6 = 3
My answer was 9 have blue eyes.


I dont really understand these questions, any help would be appreciated.

Determine whether each of these statements is true or false.

1.  null set is an element of {null}
2.  null set is an element of {null, {null}}
3.  {null} is an element of {null}
4.  {null} is an element of {{null}}
5.  {null} is a subset of {null set, {null}} (I thought the null set was a subset of every set?)
6.  {{null}} is a subset of {null. {null}}
7.  {{null}} is a subset of {{null}, {null}}

I don't really understand the difference between {null} and null set.  In my book, it says {null} is the singleton set, and has 1 element.  But doesnt the null set only have 1 element too?

i think the null set has 0 elements, or {}. the set {null} has 1 elements, that would be "null".

Yes, tetrahedron got it.  The null set, also known as the empty set, is a set without any elements.  The set {null} is a set which contains exactly one element, which happens to be another set, which happens not to contain any elements itself.

i think i get it.

So would these be right?

1. null set is an element of {null} -  True
2. null set is an element of {null, {null}} - True
3. {null} is an element of {null}  -  False     ?
4. {null} is an element of {{null}} - True
5. {null} is a subset of {null set, {null}} - True
6. {{null}} is a subset of {null. {null}}  - False
7. {{null}} is a subset of {{null}, {null}} - False

out of curiousity, where did you find these problems?

it's from a textbook for my discrete math class.  "Discrete Mathematics and its Applications" by Kenneth Rosen.  I think its supposed to be pretty good... lots of people seem to use it.  I didnt really like those questions though...


The Tastee Treats Ice Cream Shoppe presently offers its ice cream in cuboid containers with dimensions of $ 18$cm x $ 12$cm x $ 10$cm. The packaging firm with which they work indicates to them that a cylinder
with a radius of $ 7.5$ cm and a height of $ 12.2$ cm would decrease the surface area of the container and would be cheaper. To the nearest square centimeter, what is the positive difference between the numbers of square centimeters in the surface areas of the two containers? Use 3.14 as an approximation of $ \pi$.

The surface area of the cuboid container is $ 2lw \plus{} 2wh \plus{} 2hl \equal{} 1032$.

The surface area of the cylinder is $ 2\pi r^2 \plus{} \pi rh \equal{} \pi r(2r \plus{} h)$, which the problem wants us to approximate as $ \approx3.14r(2r \plus{} h) \equal{} 640.56$.

The positive difference is $ |1032 \minus{} 640.56| \equal{} 391.44\approx\boxed{391}$.

Extension: What 3-D object would be the most efficient in material usage (i.e. greatest volume to surface area ratio)?

[hide="Extension answer"]
a sphere[/hide]

I got a different answer to the initial problem.  The surface area of the rectangular prism would be $2(18\times12+12\times10+18\times10)=1032$.

The surface area of the cylinder would be $2(7.5^2\pi)+2(7.5)\pi\times12.2\approx{536.429}$.

The positive difference is thus $1032-536.429=\boxed{496\text{ cm}^2}$.

Can someone explain why the answer is $104$?

[hide="sol"]surface area of the original container is $2(18\cdot12+12\cdot10+10\cdot18)=1032$

surface area of the cylinder is the lateral surface area + the two circles

lateral surface area (just means surface area minus the bases): circumference $\cdot$ height [hide="why?"] if you unroll the cylinder, you'll notice that the lateral part is just a rolled up rectangle with one base as the circumference as the circle and the other as the height of the cylinder[/hide] $\implies (2\cdot\pi\cdot7.5)\cdot12.2=183\pi$

area of the bases: $\pi\cdot7.5^2$ then multiply that by 2 to get $112.5\pi$

adding the two together, we get $295.5\pi\approx927.87$

$1032-927.87\approx\boxed{104}$
[/hide]

[hide="errors made in previous posts"]

surface area is $2{\pi}r(r+h)$, not ${\pi}r(2r+h)$

i have no clue on how JSGandora got his answer, he had the right equation


videos
LaTeX
quadratics
Pythagorean Theorem
geometry
algebra
quadratic formula

The ultimate rule game

U MUST folow the rules before, and you must post all rules before.

Rule 1: All Rules may NOT mention Animals

Rule 1: All Rules may NOT mention Animals
Rule 2: All Rules must have the word "iLord" in them.

:D

Rule 1: All Rules may NOT mention Animals
Rule 2: All Rules must have the word "iLord" in them.
Rule 3: All Words Must Begin With A Capital Letter (I Don't Like ILord's Rule)
Rule 4: All Rules After This One Must Have A Word Starting With The Letter Q, Like Quirkly, Which Describes ILord.
Rule 5: A Very Very Very Confusing Quote Tricked Me Into Making This Rule, Which Is Probably Pleasing ILord. Each Rule After This Must Have At Least Two Sentences.
Rule 6: IlOrD MuSt DiE FoR ThIs RuLe To Be In AfFeCt AlL WoRdS MuSt AlTeRnAtE CaPiTal LeTtErs.

Ok. New game.

Rule 1: All rules must contain the number 1337.

Rule 1: All rules must contain the number 1337.
Rule 1337: All numbers in every rule up to rule 133[b]7[/b] must be replaced with 1337. Also, 1337 must be used at least 3 times.

1=2 broke his own rule: the number "3" was not replaced with "1337".

Rule 1: AlL WoRdS MuSt AlTeRnAtE CaPiTaL LeTtErS.

ye did not capitalize l at the end of capita[b]L[/b]

u EDITED!!

EDITING IS PROHIBITED!!!!
that just makes the whole game screwed up.

Rule 1: No rules following this one may state rules about capitalizing letters that are not gramatically capitalized.

AND THE SCORES????
meh, I'm too lazy to count it, so:

Rule 1: No rules following this one may state rules about capitalizing letters that are not gramatically capitalized.
Rule 2: All rules must start with the letter "A"

Rule 1: No rules following this one may state rules about capitalizing letters that are not gramatically capitalized.
Rule 2: All rules must start with the letter "A"
Rule 3: A rule following this one must include a number in the rule.

dang i stiffled up the thread

(at least i have for over 4 days.... which is amazing)

Rule 1: There must be no coloring in rainbow.
Rule 2: Nobody can post another rule anymore.
Rule 3: All rules must break rule 3 and not break this rule.

Ok. Enough with the oxymorons and the impossible posts.

(Rule 3 breaks Rule 2)

New Game:

Rule 1: At any given time, it must be possible for anyone to make another post.

6 is not a perfect square, so that breaks your own rule.

Rule 1: All rules must contain a perfect square, such as 4, not including the rule number.

A perfect square is not the same thing as a perfect number.

So we're still going with the old game? (Before dragon96 ruined it... :mad:  :wink: )

yep i guess........

Rule 63: You may not post rule numbers that are consecutive (so you can't post rule 4 after rule 3, you would have to post something besides 4)

you broke rule 3 so...

Rule 1: At any given time, it must be possible for anyone to make another post.
Rule 3: The rule numbers must be triangular numbers.
Rule 6: The rule numbers must also be perfect numbers from now on.
Rule 28: No rules can use a number bigger than 30. :rotfl:

AIME15 broke rule 3, theprodigy didn't start a new game, so I will.

Rule 1: At any given time, it must be possible for anyone to make another rule that will not break the previous rules.

Rule 1: At any given time, it must be possible for anyone to make another rule that will not break the previous rules.
Rule 2: [size=200]No more than[/size] [color=orange][size=59]3 tags may be used in any post.[/size][/color]

Rule 1: At any given time, it must be possible for anyone to make another rule that will not break the previous rules.
Rule 2: No more than 3 tags may be used in any post.
Rule 3: Each rule number must be the sum of all the rule numbers before it.

Uhh...this is like 3 months old.....

Can we continue though?  It wuz fun....

If you want to revive a rule game, revive [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227502]this one[/url].  It's more recent.


algebra
polynomial
modular arithmetic
search
Rational Root Theorem
algebra unsolved

Find all integers $ k$ such that for infinitely many integers $ n \ge 3$ the polynomial
\[ P(x) =x^{n+ 1}+ kx^n - 870x^2 + 1945x + 1995\]
can be reduced into two polynomials with integer coefficients.

[hide]First, note that it is obvious that $ k$ is not divisible by $ 5$, since if it were, Eisenstein's criterion would assert that the polynomial is irreducible in $ \mathbb{Z}[x]$.  If $ P(x)$ is reducible, then we have

$ P(x) \equal{} (a_kx^k \plus{} \cdots \plus{} a_1x \plus{} a_0)(b_jx^j \plus{} \cdots \plus{} b_1x \plus{} b_0)$

where $ k \plus{} j \equal{} n \plus{} 1$.  Notice that since $ a_0b_0 \equal{} 1995$ which is not divisible by $ 25$, we must have only one of $ a_0,b_0$ divisible by $ 5$.  Without loss of generality, assume $ a_0 \equal{} 5a_0^\prime$.  We prove that $ j \equal{} 1$:

We use the extremal principle. Let $ i$ be the least integer such that $ a_i$ is not divisible by $ 5$.  Then, consider the coefficient of $ x^i$ in the expansion.  It is $ a_ib_0 \plus{} a_{i \minus{} 1}b_1 \plus{} \cdots$.  Since all $ a_g$ for $ g < i$ are divisible by $ 5$, we must have that the coefficient is congruent to $ a_ib_0\pmod{5}$.  Since neither $ a_i$ or $ b_0$ are divisible by $ 5$, the coefficient of $ x^i$ is not divisible by $ 5$.  Looking back at $ P(x)$, the only coefficients not divisble by $ 5$ are $ kx^n$ and $ x^{n \plus{} 1}$.  So, $ i \equal{} n$ or $ n \plus{} 1$.  Obviously, $ i\ne n \plus{} 1$ since then $ k \equal{} n \plus{} 1\Rightarrow j \equal{} 0$ making $ b_jx^j \plus{} \cdots \plus{} b_1x \plus{} b_0$ constant.  Thus, we must have $ i \equal{} n$  Then, by similar reasoning, we must have $ k \equal{} n\Rightarrow j \equal{} 1$.

Thus, we must have $ b_1x \plus{} b_0$ divide $ P(x)$.  This implies that $ P(x)$ has a rational root.  By the Rational Root Theorem, the only possible rational roots for $ P(x)$ are divisors of $ 1995$.  Suppose, $ d|1995$ is a root of $ P(x)$ for infinitely many $ n$.  Then if $ |d| > 1$, then for large enough $ n$, we would have $ |d^{n \plus{} 1} \plus{} kd^n| > |870d^2 \minus{} 1945x \minus{} 1995|$ implying $ P(d)\ne0$.  Thus, we must have $ d \equal{} \pm1$ for infinitely many $ n$.  If $ d \equal{} 1$, then we get $ 1 \plus{} k \minus{} 870 \plus{} 1945 \plus{} 1995 \equal{} 0\Rightarrow k \equal{} \minus{} 3071$.  If $ d \equal{} \minus{} 1$, then we must have $ 1 \minus{} k \minus{} 870 \minus{} 1945 \plus{} 1995 \equal{} 0$ or $ \minus{} 1 \plus{} k \minus{} 870 \minus{} 1945 \plus{} 1995 \equal{} 0$ for which there are solutions $ k\equal{} \minus{}819, 821$.

Thus, the only $ k$ is $ k \equal{} \minus{} 3071,\minus{}819,821$.[/hide]

Actually, from Extended Eisenstein's Criterion, using $ p \equal{} 5$, we know that $ P(x)$ has an irreducible factor of degree at least $ n$
So if $ P(x)$ is reducible, then $ P(x)$ must be in the form $ (a_nx^n \plus{} \cdots \plus{} a_1x \plus{} a_0)(b_1x \plus{} b_0)$, that is P(x) has a rational root.
Then the rest is the same as cosinator's solution.

Small typo though... should be $ |870d^2\minus{}1945d\minus{}1995|$...

[quote="ishfaq420haque"]could you please tell what is extended einstein"s  citerion?
i know only einsteins criterion but what is its extended form??? :([/quote]
Here: [url]http://www.mathlinks.ro/viewtopic.php?search_id=2118740037&t=150952[/url]
Next time, try to use Search function.


i use kile. everything when fine till a new page was required though on compiling system showed no error but my document doesn't have a new page

I wouldn't have thought Kile was the cause of the problem, as it just passes the text onto the LaTeX compiler. Please post a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] so it can be tested by others.


let $ x,y,z$ be nonnegative real numbers. prove that:
$ \sum \sqrt{x^2\plus{}xy\plus{}y^2}\sqrt{y^2\plus{}yz\plus{}z^2} \geq (\sum x)^2$.

[quote="reza1370"]let $ x,y,z$ be nonnegative real numbers. prove that:
$ \sum \sqrt {x^2 \plus{} xy \plus{} y^2}\sqrt {y^2 \plus{} yz \plus{} z^2} \geq (\sum x)^2$.[/quote]
For any nonnegative $ x,y,z$ we have triangle $ ABC$ with point $ 0$ inside $ \triangle ABC$ , so that :
$ OA \equal{} x$ , $ OB \equal{} y$ , $ OC \equal{} z$ and $ \angle AOB \equal{} \angle BOC \equal{} \angle COA \equal{} \frac {2\pi}{3}$.

Point $ O$ for $ \triangle ABC$ is $ Torricelli$ Point.

Let $ AB \equal{} c ,BC \equal{} a,CA \equal{} b$ then : $ \sqrt {x^2 \plus{} xy \plus{} y^2} \equal{} c$ , $ \sqrt {y^2 \plus{} yz \plus{} z^2} \equal{} a$ , $ \sqrt {z^2 \plus{} zx \plus{} x^2} \equal{} b$
\[ \boxed{(x \plus{} y \plus{} z)^2 \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \plus{} 2\sqrt {3}S}
\]
So we must prove : $ ab \plus{} bc \plus{} ca \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \plus{} 2\sqrt {3}S \Leftrightarrow \boxed{r \plus{} 4R \ge \sqrt {3}s}$ which is well known  :)
( $ ab \plus{} bc \plus{} ca \equal{} r^2 \plus{} s^2 \plus{} 4rR$  ,  $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2s^2 \minus{} 2r^2 \minus{} 8rR$   )

I am very interested in your geometric solution! That is such a beatiful solution
I contribute my solution as followed
$ \sqrt{x^2\plus{}xy\plus{}y^2}\sqrt{x^2\plus{}xz\plus{}z^2}
\equal{}\sqrt{\left(x\plus{}\frac{y}{2}\right)^2\plus{}\left(\frac{\sqrt{3}y}{2}\right)^2}\plus{}\sqrt{\left(x\plus{}\frac{z}{2}\right)^2\plus{}\left(\frac{\sqrt{3}z}{2}\right)^2}
\ge \left(x\plus{}\frac{y}{2}\right)\left(x\plus{}\frac{z}{2}\right)\plus{}\frac{3yz}{4}$
which is the application of Cauchy-Schwartz.
Similarly, we implement the 2 other expressions. Adding them up and we get the desired result!


prove that for a,b,c>0
 ( let a+b+c=p , ab+bc+ca=q ,abc=r)

 p^2 +r +2 >=p +3q

 by the way happy new year to Iranian and some other country  :lol:

We need to prove $ (a \plus{} b \plus{} c)^2 \plus{} abc \plus{} 2\ge a \plus{} b \plus{} c \plus{} 3ab \plus{} 3bc \plus{} 3ca$
$ (a \plus{} b \plus{} c)^2 \plus{} abc \plus{} 2 \equal{} \sum_{cyc}(\frac {1}{2}a^2 \plus{} \frac {1}{2} \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \sum_{cyc}\frac {1}{2}a^2 \plus{} abc \plus{} \frac {1}{2}$
$ \ge a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \sum_{cyc}\frac {1}{2}a^2 \plus{} (\frac {1}{2}abc \plus{} \frac {1}{2}abc \plus{} \frac {1}{2})$ by AM-GM
$ \ge a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \sum_{cyc}\frac {1}{2}a^2 \plus{} \frac {1}{2}*3(abc)^\frac {2}{3}$ by AM-GM
$ \equal{} a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \frac {1}{2}((a^\frac {2}{3})^3 \plus{} (b^\frac {2}{3})^3 \plus{} (c^\frac {2}{3})^3 \plus{} 3(abc)^\frac {2}{3})$
$ \ge a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \frac {1}{2}(\sum_{sym}(a^\frac {2}{3})^{2}b^\frac {2}{3})$ by Schur
$ \equal{} a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \frac {1}{2}(\sum_{cyc}(a^\frac {2}{3})^{2}b^\frac {2}{3} \plus{} (b^\frac {2}{3})^{2}a^\frac {2}{3})$
$ \ge a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} \frac {1}{2}(\sum_{cyc}2ab)$ by AM-GM
$ \equal{} a \plus{} b \plus{} c \plus{} 3(ab \plus{} bc \plus{} ca)$
QED