True or false and why: \[\lfloor \sqrt{\lfloor x \rfloor}\rfloor = \lfloor \sqrt{x}\rfloor,\] for all non=negative $x$.

Put $x=n+\alpha$ where $n=\lfloor x\rfloor$ and $\alpha=\{x\}$. Then we have to prove or disprove

$\lfloor\sqrt{n}\rfloor=\lfloor\sqrt{n+\alpha}\rfloor$

Now put $n=k^{2}+m$ where $k,m$ are integer and $0\leqslant m\leqslant 2k$. That gives $\lfloor\sqrt{n}\rfloor=k$, and $n+\alpha=k^{2}+m+\alpha\leqslant k^{2}+2k+\alpha<k^{2}+2k+1$. Therefore $\lfloor\sqrt{n+\alpha}\rfloor=k$, which proves our initial equality.

[u]Generalization[/u]:
$\left[ \sqrt[k]{n}\right] = \left[ \sqrt[k]{[n]}\right]$, $\forall k \in \mathbb{Z}, k \ge 2; n \ge 0$.

[hide="Proof"]
Let $s \le \sqrt[k]{[n]}< s+1, s \in \mathbb{Z}$. We have $s^{k}\le [n] < (s+1)^{k}\ (1)$ and we need to prove $s^{k}\le n < (s+1)^{k}$. The left inequality is trivial ($s^{k}\le [n] \le n$). We have from (1) $[n]+1 \le (s+1)^{k}$, but $n < [n]+1$, hence the second inequality.
[/hide]


How many 8-digit numbers are there such that exactly 7 digits are all different?

$\textbf{(A)}\  {{9}\choose{3}}^2 \cdot 6! \cdot 3 \qquad\textbf{(B)}\  {{8}\choose{3}}^2 \cdot 7! \qquad\textbf{(C)}\  {{8}\choose{3}}^2 \cdot 7! \cdot 3 \\ \qquad\textbf{(D)}\  {{7}\choose{3}}^2 \cdot 7! \qquad\textbf{(E)}\  {{9}\choose{4}}^2 \cdot 6! \cdot 8$

[hide="Solution"]We can select the first digit by $ {{9}\choose{1}}$.

For the rest $ 7$ digits, we will select $ 6$ different numbers. $ {{9}\choose{6}}$.

Now we choose the repeated digit. The repeated digit is either one of the $ 6$ digits or the first digit.
If it is the first digit, all $ 7$ digits are different. So $ 7!$.
If it is the one of the $ 6$ digits, choose the repeated digit $ {{6}\choose{1}}$, then scramble them. $ {{6}\choose{1}}\frac {7!}{2!}$.

So the result is $ {{9}\choose{1}}{{9}\choose{6}}\left( 7! \plus{} {{6}\choose{1}}\frac {7!}{2!} \right) \equal{} {{9}\choose{3}}^{2}\cdot 6!\cdot 3$.

$ \boxed A$

[/hide]


[hide]
Clearly, we  can see that $\frac{x^{2}}{3}=5x^{3}\Longrightarrow 1=15x\Longrightarrow x=\frac{1}{15}$.[/hide]

what if it was log((x^4)/3)=log(5x^3)?

what are you doing with the exponants to get 1/15 from before

The thing here is that $\log$ is a one-to-one function. That means, $\log a = \log b \implies a = b$

$\log \left( \frac{x^{4}}{3}\right) = \log(5x^{3}) \implies \frac{x^{4}}{3}= 5x^{3}\implies x = 15$


At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $ 48$ free throws. How many free throws did she make at the first practice?

$ \textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Geometric progression, so on the first practice it was: $ \frac{48}{2^{5\minus{}1}}\equal{}3$

She made $ x$ free-throws on her first practice.

First day: $ x$
Second day: $ 2x$
Third day: $ 4x$
Fourth day: $ 8x$
Fifth day: $ 16x$

From the problem, $ 16x\equal{}48\Leftrightarrow x\equal{}3$

Answer: Letter [b](A)[/b]

To speed this one up, you can just use the answers as a guide.  Letter $ (E), (D),$ and $ (C)$ are all way too big, so now all we have to look at is $ (A)$ and $ (B)$.  After a quick check, we see that $ \boxed {(A)}$ is definitely the answer $ (3 \bullet 2^{4}\equal{}48)$.

Simplest way:

Day5:48
Day4:24
Day3:12
Day2:6
Day1:3

So the answer is A.


Hi

please allow me a question about cutting the numbers of a given Latex document by half.

Is there anyway to either
(1) take an existing PDF document that contains hyperlinks (coming from pdflatex with hyperref used) and produce a “2-up” version of it that still contains suitable hyperlinks, or
(2) take an existing .tex file that produces a formated and hyperlinked document and somehow tweak the frontmatter, so that the finally produced document looks exactly the same as the original (! page numbers contain the same content, headings, breaks, etc. look the same), only that now 2 pages appear on one final page in landscape?

Background of my question is that I would like to take my highly hyperlinked PhD thesis and create a 2-up version (that means a 2 pages per final page in landscape) of the thesis for online viewing. I personally prefer such representation for quick flipping through massive content with all common monitors being landscape mode, and the increasingly found 16:9 instead of 4:3 format further increasing this effect.

I did play around a bit with the “2up” and “booklet” packages, but could not get the trick done. The best I could do is use the “pdfpages” package, but all hyperlinks are lost. Just for your interest, I put the code futher below. It is already useful for printing, but hyperlinks are missing for online viewing. Does anyone know how to do the trick but WITH working hyperlinks?

Thank you a lot,
D.

--

[code]% creates a version of "thesis.pdf" with 2up = 2 pages per final page in landscape, but no hyperlinks
\documentclass[letter]{article}
\usepackage[final]{pdfpages}        % only works with pdflatex; inserts pages from external PDF document

\begin{document}
\includepdf[
    pages=-,            % pages=- ... all pages
    nup=1x2,            % nup=1x2 ... number of logical pages vertical and horizontal
    openright=true,     % first page printed on the right
    noautoscale=true,   % no automatic scaling, manual scaling with "scale"
    scale=.76,
    delta=0mm -25mm,    % (horizontal) and vertical distance between pages
    landscape
        ]{thesis.pdf}
\end{document}[/code]

I don't think the pdfpages route will work judging by [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/36ac2b991fc614fa/06e1a995e1865847]maintaining interactivity with pdfpages[/url], unless things have changed in the last 4 years. I suppose recompiling would work but, as I expect you've found, it would take a lot of work getting the page numbering etc correct.

I wonder whether you could try some of the pdf programs out there. Some may print out in 2-up form and also print to pdf. Whether this would maintain the hyperlinks would depend on the program. Most have free trials for you to try. You may also find an answer in an Adobe forum.

Thanks for your ideas:

I posted the request in an Adobe forum. The response seems to be discouraging:
http://www.adobeforums.com/cgi-bin/webx/?13@37.3BUshdpJp6c@.3bc4207b/0


topology
real analysis
real analysis theorems

Let $X$ be a metric space with metric $d$ and let $d_1$ be the metric defined by

$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}.$

Show that the two metric spaces $(X,d)$ and $(X,d_1)$ have precisely the same open sets.

There is a hint to the exercise, namely to show the two spaces have the same open balls with one exception.

Any ideas anyone?

basically everything you need, you said it yourself

all you have to do now, is show that if you make d small enough, d1 will go below any desired number

and the converse


Do there exists positive integer X such that:
          $ X^3=\overline{2003.....}$

There's a well-known theorem stating that unless $a$ is a power of $10$, $a^n$ can begin with any combination of digits. Just take $a=8$, for example. There is $n$ s.t. $(2^n)^3=(2^3)^n=8^n=\overline{2003\ldots}$.

I think pigfly means you should prove this "well known theorem". ;)  :)


[quote="Altheman"]i was wondering:
how well do i have to do to qualify for the aime?
how is the usamo index calculated with the usamts score?

hopefully it is not that high because i am only sure about 3 of the 5 problems, and the other 2 i have weak solutions, and i would like to get to the aime this year[/quote]

It's usually around 60's I heard.
You get basic AMC-12 score: 100.

This was said to me as well but concentrate on AMC itself. That will help you more. I just do USAMTS for fun.

As I understand it, you'll qualify for the AIME based on a cut-off point determined after the first three rounds.  I don't remember what it was last year (68 maybe?), but out of 75 points, you'd better have made as few mistakes as possible if you want to qualify for AIME through USAMTS.  Answering 3/5 will lose you 10 points which unfortunately just might be enough to put you below the cut-off.  However, it is variable and I personally find this year's problems so far to be significantly more difficult, which will naturally lower the cut-off.  You may be just fine; it's hard to say at this point. :)

However, it is still recommended that you take the AMC.  You can qualify for the AIME through the USAMTS, yes, but you won't get those points you need from the AMC if you're planning on trying for USAMO.  If you have no desire to make USAMO, then I guess it's not a problem, but I imagine everybody on this site would like to make it that far (if they haven't already), so take the AMC and do well.  And in all honesty, it's much easier to qualify for the AIME through the AMC than USAMTS.  Sure, you get ~30 days to solve the problems, but they're considerably more complex.

I don't know exactly how one's USAMO index is calculated with the USAMTS exactly (it may not factor in at all); all I know is that it's advantageous to take the AMC and not rely solely on the USAMTS.

Edit:  Oops, Silverfalcon beat me to the response.  His answer is probably more accurate anyway, I'd listen to him. ;)

You get an AMC-12 score of 100 points if you qualify for the AIME through the USAMTS.  Last year the cut-off was 68, but I'm not sure what formula is used to determine the cut-off.  As Silverfalcon said, concentrate on the AMC for qualification for the AIME, because a 100 on the AMC-12 is going to be a pretty big handicap when it comes time to calculating your USAMO index.

i will take the amc 10 and 12, but it would be nice to have already qualified

Are we even permitted to discuss questions?

[quote="TheAmazingOne"][quote="h_s_potter2002"]Well, if you're in 10th grade or lower, then it doesn't really matter that you get an AMC score of 100.  But I'd still take the AMC's.[/quote]

I qualified for the AIME in 9th grade. I took the AMC-12, though, not the AMC-10. But it's definitely possible to qualify for AIME through the AMC in 10th grade or lower...[/quote]
What he means is that if you're in 10th grade or below, your USAMO index is irrelevant - whether you qualify is soley based on whether your AIME score exceeds the floor value.  So, whether you get a 100 or 150 on the AMC, you still have an equal shot as a 9th or 10th grader to make USAMO.

[quote="TheAmazingOne"]Really? Why? So they wouldn't recognize the fact that someone got 150 on the AMC-12 just because the person is in 9th or 10th grade? Or is that only for the AMC-10?[/quote]
Actually, the grading system is beneficial - it's harder to qualify for USAMO using the index than it is simply by hitting the AIME floor value.  I believe this system was put into place to allow more 9th and 10th graders to qualify for the USAMO - and you can still get recognized for betting a really high AMC score, it just doesn't factor into whether you qualify for USAMO.

[quote="TheAmazingOne"]Oh. Then I guess I better try to take advantage of that this year =P[/quote]
Exactly. :)

Welcome to the AoPS, by the way! :welcomeani:

[quote="fuu-rei"]Wait, but it says on the USAMTS site's FAQ: "If you qualify via the USAMTS, your USAMO index is calculated based on a formula of your USAMTS score and AIME score. If you qualify via both the USAMTS and AMC, your USAMO index is whichever is greater by the two ways of calculating it. (Thus it is to your advantage to qualify via both ways.)"

But if anyone who qualifies via the USAMTS gets the AMC score of 100 to calculate the USAMO index, then the index using the AMC score would always be the higher one, right?[/quote]
Yes (or equal, if you got exactly 100).  So it's to your advantage to qualify both ways because if you have a bad day and bomb the AMC, you can still get to the AIME using the USAMTS; on the other hand, if you do well on the AMC, then your USAMO index will be better than if you had just relied on the USAMTS.

[quote="K81o7"]If you wouldn't qualify for the USAMO using your AIME score alone (10th grade or below), but if you used your index, you would, then do you qualify?

For example, if you needed either an index or 210, or an AIME score of 8, and you got a 7 on the aIME and 140 on the AMC, would you qualify?[/quote]
Interesting question - my guess is that you have to meet the floor value AND have the appropriate USAMO index if you're not a 9th or 10th grader, though I'm not sure about that.

[quote="h_s_potter2002"][quote="K81o7"]If you wouldn't qualify for the USAMO using your AIME score alone (10th grade or below), but if you used your index, you would, then do you qualify?

For example, if you needed either an index or 210, or an AIME score of 8, and you got a 7 on the aIME and 140 on the AMC, would you qualify?[/quote]

I'm pretty sure it's not possible for the index to be 210, and the floor to be 8 if you're an 11th or 12th grader who got a 140 on the AMC and a 7 on the AIME.  The floor value is determined by the lowest AIME score of the top 160 11th and 12th graders.  So the situation you just described is impossible (If that person is an 11th or 12th grader).  At least, I'm pretty sure it is.[/quote]
Also, anyone who is good enough to get a sufficiently high AMC score to create the situation above would probably score higher than the floor value on the AIME.

Perfect squares or perfect scores...:)

AMC is much easier than AIME, which probably explains most of it.

Actually, I'm studying in Taiwan, but I have US citizenship.  Is it still possible for me to qualify for AIME through USAMTS?

Perhaps it's still safer to go by AMC; I've got in since 8th grade.  But it's just good to be certain  :P .

[quote="lowest_reynolds"][quote="h_s_potter2002"]Actaully, at least half of the perfect scorers on the AMC didn't make the USAMO last year.[/quote]

That's really interesting.  Now is this referring to AMC 10 or AMC 12 perfect scores?  Or both?  How is it that one can achieve a perfect score, and yet still miss the USAMO (despite the fact that one would probably need an AIME score of oh say...9).  Maybe the AMC is also luck on one person's part?[/quote]
If you got a perfect score on the AMC 12, you would only need like a 7  (or an 8 on a really easy year) on the AIME to qualify for the USAMO.


geometry
incenter
geometric transformation
reflection
geometry proposed

Let $ I$ be the incenter of $ \triangle ABC$. The incircle touches the triangle in points $ A', B', C'$. ($ A' \in BC, B' \in AC, C' \in AB$). Let $ l$ be an arbitrary line through $ I$ and let $ A'', B'', C''$ be the reflections of $ A',B',C'$ in $ l$. Prove that $ AA'', BB'', CC''$ are concurrent.

None other than a Steinbart point sent to the line at infinite.  :roll:

Reflections $ A'',B'',C''$ of $ A',B',C'$ across the straight line $ \ell$  obviously lie on the incircle $ (I).$ Thus, by Steinbart's theorem, $ AA'',BB'',CC''$ concur if and only if $ A'A'',$ $B'B'',$ $C'C''$ concur. Indeed the lines $ A'A'',B'B'',C'C''$ meet at the infinite point of  $\perp \ell.$

Consider triangle $ ABC$ with the points of tangency of the incircle with the sides as $ A', B', C'$ respectively. We know that $ AA', BB' , CC'$ are concurrent at the Nagel point of triangle $ ABC$.. Now all of $ A"A', B"B', C"C'$ are parallel due to reflections upon the line which contains the diameter of the incircle as its segment. So let $ A"A', B"B', C"C'$ meet at infinity.  That will be the spiral similarity center for $ ABC, A"B"C"$. ( they are congruent due to reflections).  So by property of transformations, even $ AA", BB", CC"$ concur.


symmetry
complex numbers
complex analysis

let $ z_j = e^{\frac {2\pi i j}{n}}$ be the $ n$ roots of unity.  define $ c_j = |1 - z_j|$ to be chord lengths from 1 to $ z_j$'s.  Show that $ \prod_{k = 1}^{n - 1}c_k = n$

[hide]
This is a direct consequence of the fact that for any two complex numbers $ z$, $ w$,

$ |zw| \equal{} |z| |w|$,

and the definition of the roots of unity:

$ \sum_{k \equal{} 0}^{n \minus{} 1} z^k \equal{} \frac {z^n \minus{} 1}{z \minus{} 1} \equal{} \prod_{k \equal{} 1}^{n \minus{} 1} (z \minus{} \zeta_n^k)$.
[/hide]

Seems to me because of symmetry, this would be true using any of the roots:
\[ \prod_{k\ne j} |w_j \minus{} w_k| \equal{} n
\]
If so, what about the set $ \{w_n\}$ of n roots of any $ |z_c| \equal{} 1$? That is, would not this also be true:
\[ \prod_{k\ne j} |w_j \minus{} w_k| \equal{} n
\]
where the set $ \{w_n\}$ is the $ n$ roots of $ z_c$.
What about for any set of $ n$ points $ |w_n| \equal{} 1$?  When is the following true:
\[ \prod_{k\ne j} |w_j \minus{} w_k| \equal{} n
\]
would the last be true only in the case where the $ w_i$ are n roots of some $ |w_c| \equal{} 1$?
The Mathematica code below solves $ z^n\minus{}w\equal{}0$ for $ n\equal{}5$ and in this particular case $ w\equal{}i$, then forms the product  $ \displaystyle\prod_{k\ne j} |r_j\minus{}r_k|$ for each root $ r_j$.

[code]In[152]:=
n = 5;
w = Exp[Pi*(I/2)];
roots = z /. Solve[z^n == w];
Table[Simplify[Times @@ (Abs[roots[[num]] - #1] & ) /@
     Drop[roots, {num}]], {num, 1, 5}]

Out[156]=
{5, 5, 5, 5, 5}

[/code]


inequalities
number theory unsolved
number theory

[quote="marko avila"]find the smallest positive integer wich can be written as the sum of N positive integers all with the same digit sum for N=2002 and N=2003. i tried something with mod9 but no luck i tried inequalities no luck either.[/quote]

for N = 2002 and N = 2003 *simultaneously*?  or are these separate problems...?

[quote="marko avila"]find the smallest positive integer wich can be written as the sum of N positive integers all with the same digit sum for N=2002 and N=2003. i tried something with mod9 but no luck i tried inequalities no luck either.[/quote]
Of course the solution is a multiple of $2002$ and $2003$ then the minimal solution (non zero) is $2002*2003$ and is equal to $2002$ copies of $2003$ and $2003$ copies of $2002$, so solution is $2002*2003$.

RobertuX


Solve the ecuation

$a^2\frac{(x-b)(x-c)}{(a-b)(a-c)}+b^2\frac{(x-c)(x-a)}{(b-c)(b-a)}+c^2\frac{(x-a)(x-b)}{(c-a)(c-b)}=1$ knowing that

$a,b,c\in R$ ant $a\ne b\ne c\ne a$.

Rewrite $a^2 \frac{(x-b)(x-c)}{(c-a)(a-b)}+b^2 \frac{(x-c)(x-a)}{(a-b)(b-c)}+c^2 \frac{(x-a)(x-b)}{(b-c)(c-a)}=-1$.

Clearing the denominator,

$a^2 (b-c)(x-b)(x-c)+b^2 (c-a)(x-c)(x-a)+c^2 (a-b)(x-a)(x-b)$

$=-(a-b)(b-c)(c-a)$

$\Longleftrightarrow -(a-b)(b-c)(c-a)x^2=-(a-b)(b-c)(c-a)$, since $(a-b)(b-c)(c-a)\neq 0\Longleftrightarrow x=\pm 1$.

Nice :D.  Better that the solution I found in my book :P


advanced fields
advanced fields unsolved

Let $ A$ be an ordered set. For each $ \alpha$ in $ A$, $ X_\alpha$ is a finite set. $ X_\alpha \subset X_\beta$  whenever $ \alpha \le \beta$. Prove that the union $ \cup\{X_\alpha | \alpha \in A\}$ is a countable set, i.e., a monotone union of finite sets is countable.

Suppose $ \kappa$ and $ \mu$ are infinite cardinalities. We say $ \kappa$ can jump to $ \mu$ if there is a monotone union of sets such that each is of cardinality $ \kappa$ and that the union is of cardinality $ \mu$. Any infinite cardinality can jump to the next cardinality, i.e., $ \aleph_{p}$ can jump to $ \aleph_{p+1}$. Is there other cases?

[quote="pleurestique"]$ A$ is a poset? (partially ordered set) or is it totally ordered?[/quote]
It is totally ordered.

Let $ A'$ be the maximal subset of $ A$ such that no two $ X_\alpha$ with $ \alpha\in A'$ coincide. (it exists by Zorn lemma). Clearly, every set $ X_\alpha$ with $ \alpha\in A$ coincides with some $ X_\alpha$ with $ \alpha\in A'$. Hence, we can replace the union by that over $ A'$. Now, since any two sets $ X_\alpha$ with $ \alpha\in A'$ are comparable and different, the mapping $ A'\ni\alpha\mapsto\text{card\,}X_\alpha\in\mathbb Z_{+}$ is an injection, so $ A'$ is at most countable. The rest is clear.

i was thinking: there's a totally ordered set $ A'\subset A$ (with the induced order, of course) which is dense in $ A$, in the sense that $ \sup A' \ge \alpha$ for any $ \alpha\in A$.
and now you can construct an ordered map $ j$ from $ A'$ (that we suppose identified with some ordinal $ \lambda$) to $ \kappa^{+}$ (the next cardinal) in this way:
$ j(0)=\kappa$;
$ j(\alpha+1) = \min\{\beta\in\kappa^{+}\mid |\beta\setminus j(\alpha)| = |X_{\alpha+1}\setminus X_\alpha|\}$ (which is well-defined since $ \kappa^{+}$ is well ordered and $ 2j(\alpha)\in\{\beta\in\kappa^{+}\mid |\beta\setminus j(\alpha)| = |X_{\alpha+1}\setminus X_\alpha|\}$);
$ j(\alpha) = \min\{\beta\in\kappa^{+}\mid |\beta\setminus \bigcup_{\gamma<\alpha}j(\gamma)| = |X_\alpha\setminus \bigcup_{\gamma<\alpha}X_\gamma|\}$ (which is well-defined for similar arguments).
clearly this map satisfies the identity $ |\bigcup_{I}X_\alpha| = |\bigcup_{I}j(\alpha)|$ for any subset $ I\subset A'$, which gives us what we were searching for: we can jump from $ \kappa$ to $ \kappa^{+}$ and nothing more.

i think a similar argument shows also the first part, which has already been proved by fedja :)
just, i don't think you need AC, this time. (someone check this, please :P )

When we take the union, the cardinality of the union will be same as the cardinality of the X_c where   c > x , for any x belonging to A. Cardinality of X_c  is finite.


function
algebra
polynomial
real analysis
real analysis unsolved

I need help with this exercise problem on a version of the SW theorem.

For an algebra $ A$ of continuous real-valued functions on a compact space $ X$ such that $ A$ separates points of $ X$, show that either $ \bar{A}\equal{}C(X)$ or there exists $ p \in A$ with $ \bar{A}\equal{}\{ f \in C(X): f(p)\equal{}0 \}$.

[The version of the theorem that was already proven is where the algebra is also assumed to contain the constant functions and then the algebra is dense in $ C(X)$.

The hint given was: If $ 1 \in \bar{A}$, then we are done. If there is $ f \in A$ which is nowhere zero, then $ 1 \in \bar{A}$. If for each $ x \in X$ there exists $ f \in A$ with $ f(x) \neq 0$, then there is $ g \in A$ with $ g>0$ everywhere.]

[hide="Additional hints..."]
If a function is non-zero at $ x$, then it's non-zero at a neighborhood of $ x$. If for every point there's a function that doesn't vanish at that point, you get an open cover. Now, the space is compact... what can you do? Then, how can you manufacture a function which is never zero from what you have?
[/hide]

[quote="chungyc"]I've managed to get that function $ g$ that is positive everywhere but I can't quite figure out why the existence of such a function implies that $ 1 \in \bar{A}$.[/quote]
On any compact interval of the positive real line (in particular, an interval containing the range of $ g$), the function $ \sqrt t$ is the uniform limit of polynomials $ p_n$ with zero constant term. The sequence of functions $ x\mapsto p_n(g(x))$ will converge in $ \bar{A}$ to the square root of $ g$. If you repeat taking the square root often enough, you'll get uniformly close to the function 1.

[quote="chungyc"]Is it possible to approximate $ 1/g$ by such polynomials as well?[/quote]
Yes. The closure of the range of g is a compact set K that does not contain 0. The function f given by $ f(t)\equal{} t^{\minus{}1}\ (t\in K)$, and $ f(0)\equal{}0$, is continuous on $ K\cup\{0\}$. By the (Stone–)Weierstrass theorem, $ f$ can be uniformly approximated by polynomials $ f_n$. Then $ f_n(g)\in A$, and $ f_n(g)\to g^{\minus{}1}$.

[quote="chungyc"]Can these polynomials $ f_{n}$ be taken to have zero constant term? If so, how is it done?[/quote]
Yes, that was the point of requiring that $ f(0)\equal{}0$. If $ f_n$ is uniformly within $ \varepsilon/2$ of $ f$ then in particular $ |f_n(0)|<\varepsilon/2$. So if we subtract the constant $ f_n(0)$ from $ f_n$, we'll have a polynomial with zero constant term that is still within $ \varepsilon$ of $ f$.


$x<\frac{log3}{log\frac{1}{a}}$ is correct??
tkz
[]´s

[quote="shobber"]Solve the inequality: ${|}\frac{a^x+1}{a^x-1}{|} < 2$ with $(0<a<1)$.[/quote]
$\frac{a^x+1}{a^x-1} = \frac{a^x-1+2}{a^x-1} = 1 + \frac{2}{a^x-1}$
$\frac{-3}{2}<\frac{1}{a^x-1}<\frac{1}{2}$, you can take it from here, the only problem is that if x is negative, or positive, it is different

I think we can divide cases for $x$ which is $x>0$ and $x<0$ and solve it there

For $x>0$ we can solve and get $x>-\frac{\log 3}{\log a}$

For $x<0$ we get $x<-\frac{\log 3}{\log a}$


Four cirlces are placed inside a quadrilateral $ABCD$ such that each circle touches two adjacent sides and also two other cirlces. If a circle can be inscribed un the given quad, prove that some two of the four circles have equal radii.

http://www.mathlinks.ro/Forum/viewtopic.php?t=24332
http://www.mathlinks.ro/Forum/viewtopic.php?t=39209

  darij


analytic geometry
graphing lines
slope
AMC
USA(J)MO
USAMO
geometry

I'm not sure if we can begin yet, but its safe to say I did [b]much[/b] better than last year. :lol:

Yes, I also did much better than I expected!!! :D  :D

So, when can we start discussing the problems?

- Nawahd Werdna

I also did much better than last year (probably roughly a 25 point increase as a matter of fact) but that's not saying much because my score last year was TERRIBLE  :blush:. By AoPS standards (and consequently, by my new standards) I didn't do too great, but by the original standards I set out to achieve in January, I'm right on the money. My main problem was more in contest-savvy then my actual knowledge IMHO.

Meaning you got 90+? Maybe not good by rem standards but by contest standards, I'm sure you did awesome. It's only Euclid anyways!

[quote="max_tm"]Meaning you got 90+?[/quote]
Yeah... still not good. :(

i thought it was really easy compared to past contests (#9 was a joke)

only hard ones were 8 b and 10b

but terrible thing is

i wasted too much time on 8b i think and i didn't have time to do the rest

Yeah 8b) seemed like it was going to be an undertaking (never got around to taking  second pass at it) and I didn't even get a chance to read 10  :(, I have serious time management issues  :P.

I had about a hour for 10b.  For 40 or so minutes, I messed around with Menelaus and got a really intimidating algebraic expression to minimize.  Then, I suddenly got the flash of insight that I should [b]try to incorporate part a. [/b]I quickly saw a way to do the problem, wrote down what steps there were, and jumped into the calculations.  Of course, I ran out of time.

This Euclid was easily pwnable.  I think there will be many 100s.

I think that I did pretty well.. I got everything 8b and 10b, but I had some progress on 8b.....  :roll:  :blush:

I spent most of my time on 10b though... it seemed to be a very strange question!!!

I think 10b has a loophole: a line through M with positive slope will cause point Q to lie on the "other" side, which would allow for arbitrarily small (positive) perimeters...

10b) was interesting, but I ran out of time. I did manage to do a part solution for 8b) though. hopefully, they'll give me part marks. :P
lol, opposite of you Nawahd Werdna. :D

I don't see your reasoning for that; if you make PQ have positive slope, although AQ might get very small, AP will get much larger (I think this possibility was forseen and intended in the solution).

EDIT: ^ to nneonneo

What do you mean, Bobby???
I did 10b, but got a pretty strange answer, so I guess it's all wrong...

Yeah, thanks guys. I guess you don't have to be a robot to get 90+ on Euclid though i coulda done it had i gotten 7b. i know it was easy cuz my friend got 59 and got it and showed me how to do it.

To Jingyuan Mo: I was looking on the UBC list too. Good job man!!! You did so well! Hey, are you on st. george's school reach team or do you know anyone on that team? Just wondering, that's all. :D

[quote="carpo"][quote="bryant124"][quote="gmoney"]HI GUYS!!! I GOT 86 ON THE EUCLID CONTEST AND I AM GOING TO UBC'S EUCLID MATH DAY!!! IS ANYONE ELSE GOING???

GARY HOU :)[/quote]

lol...ur so excited...i tink i might be going...but there isnt a lot to do there...the good ting is i can skip a day of school...thats cool...me no like school anymore...

AAAAANYWAYS...morning=tour...lunch=lunch...afternoon=?...wut do we do in  the afternoon?...ubc is sooooo far from my home...

btw ubc has 2 of my 3 favorite letters in the alphabet... :D[/quote]

Afternoon is a problem-solving session in a steaming-hot lecture hall.[/quote]

grrrrrrrr...steaming-hot...but its ok...i can take it

@gmoney: i live like 10 steps away from richmond(not really...but very close)...richmond center rocks...i buy almost all my clothes there...its much better than oakridge...i never heard of ur school tho...btw im not good at math so i dont have any secrets... :blush: but its ok...i'll get math at some point in my life... :wink:

[quote="JingyuanMo"]UBC list is not that accurate...[/quote]

What do you mean?

[quote="JingyuanMo"]some scores have been changed...actually,some ppl went from 94 to 93...[/quote]

oh, I see. well, i didnt change, ubc said i got 86 and so did waterloo. come to think of it, i did hear some ppl dropped as much as 4 or 5 pts.

waterloo website has you with a prize so i know that YOU ARE SMART.

[quote="gmoney"][quote="bryant124"][quote="gmoney"]hey bryant124, you must be somewhat good at math since you're going to euclid math day. :)

is your name bryant or are you a basketball fan? i'm like the biggest sports fan in the world.  :wink:[/quote]

me a kobe bryant fan...

but seriously...me no math at all...me got owned on euclid...me lucky me got 90... :([/quote]

at least you probably got 7b. i cant believe i didnt get it, it was so easy.  :wallbash: that's probably why i got 86 and not 90 cuz i got 26/30 on part b. where'd you lose your marks?

and im a phoenix suns fan cuz i like steve nash. i hope you hate the san antonio spurs cuz i hate them for beating my suns...

it's strange that i dont even know your real name... i guess i'll find out on euclid day...

and im new here, how do you not double post?[/quote]

grrr............i dont kno wut i got...i was dreaming my whole way thro euclid...lol...

i dont like spurs...i mean...just look at their team color...black and white...thyr like the most boring team in the league...i dont really hate the suns for beating lakers...lakers dont really have a chance this year anyways...and steve nash is CANADIAN!!hurraY!!!

grrr...lol...my name is J.Sun...yes...SUN...u'll probably meet me...if i go to ubc... :o  :o  :o

and u can just delete ur post to not double post...i tink...

[quote="bryant124"][/quote][quote="gmoney"][quote="bryant124"][quote="gmoney"]hey bryant124, you must be somewhat good at math since you're going to euclid math day. :)

is your name bryant or are you a basketball fan? i'm like the biggest sports fan in the world.  :wink:[/quote]

me a kobe bryant fan...

but seriously...me no math at all...me got owned on euclid...me lucky me got 90... :([/quote]

at least you probably got 7b. i cant believe i didnt get it, it was so easy.  :wallbash: that's probably why i got 86 and not 90 cuz i got 26/30 on part b. where'd you lose your marks?

and im a phoenix suns fan cuz i like steve nash. i hope you hate the san antonio spurs cuz i hate them for beating my suns...

it's strange that i dont even know your real name... i guess i'll find out on euclid day...

and im new here, how do you not double post?[/quote][quote="bryant124"]

grrr............i dont kno wut i got...i was dreaming my whole way thro euclid...lol...

i dont like spurs...i mean...just look at their team color...black and white...thyr like the most boring team in the league...i dont really hate the suns for beating lakers...lakers dont really have a chance this year anyways...and steve nash is CANADIAN!!hurraY!!!

grrr...lol...my name is J.Sun...yes...SUN...u'll probably meet me...if i go to ubc... :o  :o  :o

and u can just delete ur post to not double post...i tink...[/quote]

HOLY CRIPES!!! Are you JULIAN SUN? if you are, good job man! i was reading comc list, you're top 3 in bc and you're only in gr. 10 according to ubc list. good job man! :thumbup:

and i figured out why you're not so excited bout euclid day, cuz a lot of churchill ppl are going, im excited cuz im the first person in the history of my school to go!

and i dont mind lakers. i like andrew bynum and i think he'll be a good compliment to kobe in the future. lakers are good at finding young talent, aren't they?

go utah jazz. lol. :play_ball:

[quote="gmoney"][/quote][quote="bryant124"][/quote][quote="gmoney"][quote="bryant124"][quote="gmoney"]hey bryant124, you must be somewhat good at math since you're going to euclid math day. :)

is your name bryant or are you a basketball fan? i'm like the biggest sports fan in the world.  :wink:[/quote]

me a kobe bryant fan...

but seriously...me no math at all...me got owned on euclid...me lucky me got 90... :([/quote]

at least you probably got 7b. i cant believe i didnt get it, it was so easy.  :wallbash: that's probably why i got 86 and not 90 cuz i got 26/30 on part b. where'd you lose your marks?

and im a phoenix suns fan cuz i like steve nash. i hope you hate the san antonio spurs cuz i hate them for beating my suns...

it's strange that i dont even know your real name... i guess i'll find out on euclid day...

and im new here, how do you not double post?[/quote][quote="bryant124"]

grrr............i dont kno wut i got...i was dreaming my whole way thro euclid...lol...

i dont like spurs...i mean...just look at their team color...black and white...thyr like the most boring team in the league...i dont really hate the suns for beating lakers...lakers dont really have a chance this year anyways...and steve nash is CANADIAN!!hurraY!!!

grrr...lol...my name is J.Sun...yes...SUN...u'll probably meet me...if i go to ubc... :o  :o  :o

and u can just delete ur post to not double post...i tink...[/quote][quote="gmoney"]

HOLY CRIPES!!! Are you JULIAN SUN? if you are, good job man! i was reading comc list, you're top 3 in bc and you're only in gr. 10 according to ubc list. good job man! :thumbup:

and i figured out why you're not so excited bout euclid day, cuz a lot of churchill ppl are going, im excited cuz im the first person in the history of my school to go!

and i dont mind lakers. i like andrew bynum and i think he'll be a good compliment to kobe in the future. lakers are good at finding young talent, aren't they?

go utah jazz. lol. :play_ball:[/quote]

bingo!!!...but i actually do suck at math...i cant even explain a math problem to ppl...its frustrating...its just like telling them a triangle has 3 sides and most of them r gonna go like "huh?y?"...and its not their problem...and i cant write on a straight line...rrrrr...irritating...

altho i dont wanna admit it...spurs r probably gonna win this year...unless pistons can stop them in finals... :(

[quote="bryant124"][/quote][quote="gmoney"][/quote][quote="bryant124"][/quote][quote="gmoney"][quote="bryant124"][quote="gmoney"]hey bryant124, you must be somewhat good at math since you're going to euclid math day. :)

is your name bryant or are you a basketball fan? i'm like the biggest sports fan in the world.  :wink:[/quote]

me a kobe bryant fan...

but seriously...me no math at all...me got owned on euclid...me lucky me got 90... :([/quote]

at least you probably got 7b. i cant believe i didnt get it, it was so easy.  :wallbash: that's probably why i got 86 and not 90 cuz i got 26/30 on part b. where'd you lose your marks?

and im a phoenix suns fan cuz i like steve nash. i hope you hate the san antonio spurs cuz i hate them for beating my suns...

it's strange that i dont even know your real name... i guess i'll find out on euclid day...

and im new here, how do you not double post?[/quote][quote="bryant124"]

grrr............i dont kno wut i got...i was dreaming my whole way thro euclid...lol...

i dont like spurs...i mean...just look at their team color...black and white...thyr like the most boring team in the league...i dont really hate the suns for beating lakers...lakers dont really have a chance this year anyways...and steve nash is CANADIAN!!hurraY!!!

grrr...lol...my name is J.Sun...yes...SUN...u'll probably meet me...if i go to ubc... :o  :o  :o

and u can just delete ur post to not double post...i tink...[/quote][quote="gmoney"]

HOLY CRIPES!!! Are you JULIAN SUN? if you are, good job man! i was reading comc list, you're top 3 in bc and you're only in gr. 10 according to ubc list. good job man! :thumbup:

and i figured out why you're not so excited bout euclid day, cuz a lot of churchill ppl are going, im excited cuz im the first person in the history of my school to go!

and i dont mind lakers. i like andrew bynum and i think he'll be a good compliment to kobe in the future. lakers are good at finding young talent, aren't they?

go utah jazz. lol. :play_ball:[/quote][quote="bryant124"]

bingo!!!...but i actually do suck at math...i cant even explain a math problem to ppl...its frustrating...its just like telling them a triangle has 3 sides and most of them r gonna go like "huh?y?"...and its not their problem...and i cant write on a straight line...rrrrr...irritating...

altho i dont wanna admit it...spurs r probably gonna win this year...unless pistons can stop them in finals... :([/quote]

yea, i know what you mean cuz i tutor ppl math and i have trouble too. i could probably write on a straight line if i could draw one. but i cant, thats probably why i lose so many marks on my math tests.

you are a brave soul for being first to admit spurs would probably win, i think so too but im starting to like deron williams and carlos boozer of the jazz and i HOPE they will come thru. but i disagree on pistons, i think it'll be cavs. itll be close tho, cavs in 7. lebron 43 pts. lol

OMG, im so tired, i gotta study for my biology test on tues, i havent started yet, i'll spend all of tomoro studying, good night.

[quote="gmoney"][/quote][quote="bryant124"][/quote][quote="gmoney"][/quote][quote="bryant124"][/quote][quote="gmoney"][quote="bryant124"][quote="gmoney"]hey bryant124, you must be somewhat good at math since you're going to euclid math day.

is your name bryant or are you a basketball fan? i'm like the biggest sports fan in the world.  :wink:[/quote]

me a kobe bryant fan...

but seriously...me no math at all...me got owned on euclid...me lucky me got 90... :([/quote]

at least you probably got 7b. i cant believe i didnt get it, it was so easy.  :wallbash: that's probably why i got 86 and not 90 cuz i got 26/30 on part b. where'd you lose your marks?

and im a phoenix suns fan cuz i like steve nash. i hope you hate the san antonio spurs cuz i hate them for beating my suns...

it's strange that i dont even know your real name... i guess i'll find out on euclid day...

and im new here, how do you not double post?[/quote][quote="bryant124"]

grrr............i dont kno wut i got...i was dreaming my whole way thro euclid...lol...

i dont like spurs...i mean...just look at their team color...black and white...thyr like the most boring team in the league...i dont really hate the suns for beating lakers...lakers dont really have a chance this year anyways...and steve nash is CANADIAN!!hurraY!!!

grrr...lol...my name is J.Sun...yes...SUN...u'll probably meet me...if i go to ubc... :o  :o  :o

and u can just delete ur post to not double post...i tink...[/quote][quote="gmoney"]

HOLY CRIPES!!! Are you JULIAN SUN? if you are, good job man! i was reading comc list, you're top 3 in bc and you're only in gr. 10 according to ubc list. good job man! :thumbup:

and i figured out why you're not so excited bout euclid day, cuz a lot of churchill ppl are going, im excited cuz im the first person in the history of my school to go!

and i dont mind lakers. i like andrew bynum and i think he'll be a good compliment to kobe in the future. lakers are good at finding young talent, aren't they?

go utah jazz. lol. :play_ball:[/quote][quote="bryant124"]

bingo!!!...but i actually do suck at math...i cant even explain a math problem to ppl...its frustrating...its just like telling them a triangle has 3 sides and most of them r gonna go like "huh?y?"...and its not their problem...and i cant write on a straight line...rrrrr...irritating...

altho i dont wanna admit it...spurs r probably gonna win this year...unless pistons can stop them in finals... :([/quote][quote="gmoney"]

yea, i know what you mean cuz i tutor ppl math and i have trouble too. i could probably write on a straight line if i could draw one. but i cant, thats probably why i lose so many marks on my math tests.

you are a brave soul for being first to admit spurs would probably win, i think so too but im starting to like deron williams and carlos boozer of the jazz and i HOPE they will come thru. but i disagree on pistons, i think it'll be cavs. itll be close tho, cavs in 7. lebron 43 pts. lol

OMG, im so tired, i gotta study for my biology test on tues, i havent started yet, i'll spend all of tomoro studying, good night.[/quote]

lol...u study for tests?i never do...thats y i always get stuck on concepts...but its ok...science isnt about concepts...just like math...(yawn)...

weeeeeeeeee...pirates coming out...but if i go to the release then im probably gonna get home by 1 in the morning...sigh...

o...btw is there a deadline the ubc forms?...cuz i cant find them...in my messy room... :rotfl:

WHAT??? you havent got ubc forms in yet? i hope you're joking cuz they were due may 14. you could probably ask them for an extension tho.

i dont study for tests, i CRAM for tests. but its biology and im going to engineering so i dont really need it so i dont really care if i fail. but then again, i didnt study much in gr 10, a naturally smart person can get thru w/o studying til gr 12, cuz thats when all hard stuff come out.

im not that big on movies, tho, i know everyone else is, it's hard for me to sit in a theatre for three hours cuz i would hafta go to washroom and i hate sitting still, im always moving around and fiddling


#1. dinoman (animation)

[url]http://www.artofproblemsolving.com/Forum/files/time_flash_animation_212.swf[/url]

#2. ToBe (graphic)

[img]http://artofproblemsolving.com/Forum/files/kid_staring_at_clock_175.png[/img]

#3. -Terminator- (graphic)

[img]http://artofproblemsolving.com/Forum/files/paradox_129.jpg[/img]

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

NOTE: The competitors CAN vote for themselves.


I suppose members of high school class of '05 are mostly busy now applying to college class of '09 at various colleges. Good luck.

P.S. See the "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]Books on College Admission[/url]" and "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]Preparing for College Math Study[/url]" sticky threads for more information about getting into college, and how to study once in college.

Yeah, I'd also like to wish all of you good luck, just try not to set the bar too high, I'll have to follow next year.

Good Luck Class of '05

actually, i'd like to say good luck to the people that are going to college.

[quote="ComplexZeta"]How come? I don't find college a luck-oriented thing.[/quote]

Good point  :lol:

How about "We wish that the college admissions staff does not totally screw you over"?

[quote="ComplexZeta"]How come? I don't find college a luck-oriented thing.[/quote]

I think there is some luck involved though since the decision is somewhat subjective also.

I feel so uncomfortable now that the application is out of my hands. I mean, the course of my life will be determined within a matter of weeks-- it is just scary to think about it.

Welcome back to the new AoPS forums. Thanks for your patience waiting to get back online while the company fixed the boards. I suppose by now (Friday 10 December 2004) a few applicants for early action/early decision have heard results, and more will this weekend. Good luck to all of you in finding a suitable school for next school year.

Good luck to the people in the equatorial zone and southern hemisphere too, who apply to university on a different schedule.

Agreed.

The biggest thrill of my life possibly was opening the package containing my acceptance from Princeton.

All of the AoPSers who applied early should have heard of decisions by now, and by the end of March they will know how they fared in regular admission. Good luck to all. Have a happy New Year.


conics
parabola
analytic geometry
complex numbers

[quote="bos1234"]$ z$ satisfies $ |z\minus{}1| \equal{} Im z\plus{}1$. Sketch the locus of the point P representing z in the Argand diagram and write down its Cartesian equation.

This is the books soln.

Let $ A(0,1)$, $ P$ represent $ i$, $ z$ respectively. Then $ |z\minus{}i| \equal{} AP$, and $ Im z\plus{}1$ is the distance from P to the line $ y \equal{}\minus{}1$. Hence the locus of P is a parabola with focus $ A(0,1)$ and directrix $ y \equal{}\minus{}1$, with equation $ x^{2}\equal{} 4y$

[url=http://img165.imageshack.us/my.php?image=91967523js6.png][img]http://img165.imageshack.us/img165/1893/91967523js6.th.png[/img][/url]

How did they get this part : $ Im z\plus{}1$ is the distance from P to the line $ y \equal{}\minus{}1$[/quote]

[hide]The imaginary part of a complex number is Imz.  This is also the "y-coordinate" that people would normally associate with.  So the distance from P to y=-1 is Imz+1.[/hide]


geometry
real analysis
real analysis unsolved

What should be the length $W$ for yeloow area be half of circle area?


my school already signed up for the AMC A, but im trying to get it to sign up for the B too.

a) is it ok to have 2 separate registration forms?

b)  Are we in late registration time now?  http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/04bro1012/04bro1012html/04bro1012-regb.html   says we are, but http://www.math.unl.edu/~amc/registration/start12.php says discount registratin ends at the end of january.  Which is right?
thanks.

1 more question: if I take it at another school, is it official (i.e count towards my index)

im pretty sure it counts, but how can you take it at another school?

well, i mean, for AP tests and such, if your school doesnt offer them you can go to other schools to take them.  I know someone who took an AMC at another school, but it didnt count for some reason, that's why i wasnt sure if it counts.

I'd really like to take the B test (and have it count), but im pretty sure  it wont be through my school.

can anyone else verify what Neal said is true?  If you take the test at another school than your own, it can count officially?

Why not e-mail Steve Dunbar of the AMC?  He'll know.  Or anyone else from the AMC.

yeah you'd think that wouuld've been my first move, but i did this first, then i figured i wouldnt email him cause he also reads this forum.

Anyways, i sent an email, i'll post the response here for others who are wondering.

My school does not register for the B test, so I'm going to take it in another school. I think it still counts toward the index, but there will not be school awards since you're not a student of the school where you take the test.

im pretty sure it counts too because i've heard people say that their school did not offer it so they were going to take it somewhere else. besides, the AMC administrators are smart enough to realize that some schools wont offer it. though you might need to pay your own registration... i took it at my high school last year when i was in 8th grade so i guess that sort of counts, considering i never went to that high school... maybe?

Cats... asked

 > a) is it ok to have 2 separate registration forms?

Yes.

> b) Are we in late registration time now?
>www.unl.edu/amc/b-registration/b1-archive/2003-2004/04bro1012/04bro1012html/04bro1012-regb.html > says we are, but >www.math.unl.edu/~amc/registration/start12.php says discount
> registratin ends at the end of january. Which is right?

Actually, both are correct.  The first website is simply an image of the paper information and registration brochure that we send out to all schools.  If schools register with the paper form, either by mail or fax, the "Dec 20 - Jan 15" registration fee is required.

The second site is our on-line Web-mediated registration form.  It requires much less handling and is much easier for us.  Therefore, we give schools using that Web registration form a "discount" registration fee to encourage them to sign up, even now.  We can easily reset the deadlines in the program that generates the page in order to automatically display what we want, and charge the right amount.  We have emailed all prior exam managers alerting them to the discount.

Steve Dunbar
AMC Director

Mr. Dunbar, can you respond to my emails please.

Thanks.


function
quadratics
algebra
floor function
number theory
modular arithmetic

Show that for any rational  $x$  the following equality doesn't hold: $\{ x^2\}+\{x\}=1.$

${x}$ gives the fractional part of $x$ so, for example, ${3.14159}=.14159.$

[quote="nat mc"]Doesn't -7/4 work? Every rational x for -2 < x <= -:sqrt:3 works like -17/9 That is greates integer function right?[/quote]

$\{\frac{-7}{4}\}=\frac{-7}{4}-[\frac{-7}{4}]=\frac{-7}{4}-(-1)=-\frac{3}{4},\{\frac{49}{16}\}=\frac{1}{16}\Rightarrow \{\frac{-7}{4}\}+\{\frac{49}{16}\}\ne 1$ :? .$[]$ is the greatest integer function,$\{\}$ is the fractional part and $x=[x]+\{x\}$.

Usually, when dealing with the so-called "fractional part," it is best to convert the statement into one with the floor function by using {x}=x-floor(x).  Either way, you can turn the whole thing into a problem with modular arithmetic, at least when the numbers are rational like you said.  So you are saying x^2-floor(x^2)+x-floor(x)=x^2+x-floor(x^2)-floor(x)=1 has no rational solutions.  I propose that you subdivide the rationals into categories based on the value of their floor function and solve it by showing that in each of those categories, there is no solution.  But, that's just an idea, it's probably dumb.

I think
[hide]
If ${x} = a$ then $0 \le a < 1$.
$a^2 + a = 1 \Rightarrow a^2 + a - 1 = 0$

which does not yield any rational solutoins.
[/hide]

The solution to this problem is very easy,just put $x=\frac{p}{q}$.

i think if you set $ {x\}=x-[ x] }$ then it'll be solved.

[hide]Suppose the contrary.
It is clear that $x$ is not an integer number (we have $\{x^2\}+\{x\}=0$ for integer numbers). The condition  $\{x^2\}+\{x\}=1$ means that $x^2+x$ is an integer number :!:. So $x^2+x=n$. But if this quadratic equation has a rational root, then it should be an integer number! So $x$ is an integer. Contradiction.[/hide]

Well in other words,
x^2+x=an integer (apart from 0), and x is not an integer.
x^2+x=x(x+1).  Since both x and x+1 are rational, the product of the two numerators in those two has to be a multiple of the two denominators.  Let x be expressible in simplest form as a/b.  Then we have a(a+b)/ab is an integer.  Therefore, b goes into a+b, and so b goes into a.  That right there is a contradiction.


In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $ v$, $ w$, $ x$, $ y$, and $ z$. Find $ y \plus{} z$.

$ \textbf{(A)}\ 43 \qquad
\textbf{(B)}\ 44 \qquad
\textbf{(C)}\ 45 \qquad
\textbf{(D)}\ 46 \qquad
\textbf{(E)}\ 47$

[asy]unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}

label("$25$",(0.5,0.5));
label("$z$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
label("$w$",(2.5,2.5));[/asy]

i have different answer

[hide="easy guess and check"]
ok so consider subtracting an amount from each of the squares, making it a regular 1-9 magic square.

lets say we're subtracting a.

by simple logic, we know that a=16 or 17 (if a is less than 16, 25 is larger than 9, and if a is larger than 17, 18 becomes nonpositive(?))

however, if a=16, 21 becomes 5, and so a can't be 16 because 5 is in the middle of regular magic square.

so we know that a=17

lets subtract then.

[asy]unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}

label("$8$",(0.5,0.5));
label("$z'$",(1.5,0.5));
label("$4$",(2.5,0.5));
label("$1$",(0.5,1.5));
label("$x'$",(1.5,1.5));
label("$y'$",(2.5,1.5));
label("$v'$",(0.5,2.5));
label("$7$",(1.5,2.5));
label("$w'$",(2.5,2.5));[/asy]
(value of v,x,y,z doesn't matter after subtracting 17 so i just denoted them as v', x', y', and z')

and the regular magic square is

[asy]unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}

label("$8$",(0.5,0.5));
label("$3$",(1.5,0.5));
label("$4$",(2.5,0.5));
label("$1$",(0.5,1.5));
label("$5$",(1.5,1.5));
label("$9$",(2.5,1.5));
label("$6$",(0.5,2.5));
label("$7$",(1.5,2.5));
label("$2$",(2.5,2.5));[/asy]

so y-17=9, y=26
z-17=3, z=20

y+z=20+26=46

answer: 46

[/hide]

phew that was a long solution

It's 66. w is obviously 19, so we have 18 to 25 in our square, so we have 17 or 26, so 21 or 22 is in the center but 21 is used, so 22 is the center-one, sum is 66, so y=20,z=26, answer 46.


function
induction
algebra proposed
algebra

Let $\displaystyle M$ be a set composed of $\displaystyle n$ elements and let $\displaystyle \mathcal P (M)$ be its power set. Find all functions $\displaystyle f : \mathcal P (M) \to \{ 0,1,2,\ldots,n \}$ that have the properties

(a) $\displaystyle f(A) \neq 0$, for $\displaystyle A \neq \phi$;

(b) $\displaystyle f \left( A \cup B \right) = f \left( A \cap B \right) + f \left( A \Delta B \right)$, for all $\displaystyle A,B \in \mathcal P (M)$, where $\displaystyle A \Delta B = \left( A \cup B \right) \backslash \left( A \cap B \right)$.

What do you mean by $\phi$ in the first condition? This variable is not defined here.

Thanks. I think, $\emptyset$ is more common :-)

Let $X,Y$ be two disjoint subsets of $M$. Taking $A = X \cup Y$ and $B = X$ we get that
\[f \left( X \cup Y \right) = f \left( X \right)+f \left( Y \right) . \]
By induction we get that for any pairwise disjoint subsets $X_{1},X_{2},\ldots,X_{k}$ we have that
\[f \left( \bigcup_{i=1}^{k}X_{i}\right) = \sum_{i=1}^{k}f \left( X_{i}\right) . \]
Hence, $f \left( \emptyset \right) = 0$ and $f \left( \left\{ x_{1}, \ldots, x_{m}\right\}\right) = \sum_{i=1}^{m}f \left( \left\{ x_{i}\right\}\right)$, which, by (a), is $\geq \sum_{i=1}^{m}1 = m$.
Therefore, $f(M) \geq n$. But we know that $f(M) \leq n$, so equality must hold. That happens iff we have equality everywhere, i.e. $f \left( \left\{ x_{i}\right\}\right) = 1$ for all $i \in \overline{1,n}$, so $f(X) = |X|$.
Conversely, this function satisfies all conditions.


calculus
derivative
function
calculus computations

Find the derivative $ y'''\equal{}\frac{d^{3}y}{dx^{3}}$ of parametric function

$ x\equal{}e^{\minus{}t},\quad
y\equal{}t^{3}$.


inequalities
real analysis
real analysis unsolved

Let {x_n} :
 $ x_0 = 1 $
 $ x_{n+1}= \frac{a}{1+x_n^2} $
 Find a such that : {x_n} converge

For real $a$, let $y=y(a)$ be the unique real solution of the cubic equation $x=\frac{a}{1+x^2}.$ The sequence converges if $\frac{|2ay|}{(1+y^2)^2}<1.$ So we're looking for conditions on $a$ which make that inequality true.

My general  condition is $a \in [2,-2]\cup\{x_0^3 + x_0\}$
  :cool:

Since romano specified that $x_0=1$, the extra value of $a$ in diogene's answer is 2, and he has already included it in the interval $[-2.2].$

I haven't worked out the proof myself, but I've played with it numerically, and it looks like diogene is right, with slower convergence for $|a|$ closer to 2. Furthermore, if $|a|>2$, the sequence tends to a stable 2-cycle, more quickly for larger $a.$

Of course I'm right , and without numerical play ! :D
$x_{n+1} = f(x_n)$ where $f$  is bounded monotonic and continuous function.  The equation $x = f(x) $ has only 1 real  solution $L$.
The equation $x=f(fx)$ can have only  1  or 3 distinct real solutions (try to prove it :D ) .
So any accumulation point of $\{x_n\}$  is  solution of  $x=f(fx)$.  The possible cases for $g(x)= f(f(x))-x$ are :
1) $ |a| \leq 2 \Longleftrightarrow   $ 1 solution,$L$, so $x_n \rightarrow  L$
2) $ |a| > 2  \Longleftrightarrow  $ 3  solutions : $u_0=\frac{a-\sqrt{a^2-4}}{2}=f(u_1)  <  L=f(L)  <  u_1=\frac{a+\sqrt{a^2-4}}{2}=f(u_0)$
$x_0 = L  \Longleftrightarrow   a = x_0^3 + x_0 ,  x_n=x_0$ : trivial convergence
For the other sub-cases the accumulation points set is $\{u_0,u_1\} \Longleftrightarrow  x_n$ does'nt  converge.
 :cool:  Do you follow me ?  :cool:

[quote]So any accumulation point of $ \{x_n\}$ is solution of $x=f(f(x)).$[/quote]
This is the one step that's not obvious. But I see the argument. Because $f$ is decreasing (for positive $x$), if we have more than two accumulation points, and because one accumulation point must be mapped to another another accumlation point, then the largest such point must map to the smallest, the second largest to the second smallest, and so on. Hence we're restricted to 2-cycles (or convergence).

If $|a|>2$ the terms of the sequence tend in alternating fashion to the two points $u_0$ and $u_1$.

The equation $f(f(x))=\frac{a(1+x^2)^2}{a^2+(1+x^2)^2}=x$ factors as $(x^2-ax+1)(x^3+x-a)=0$. There are Diogene's 1 or 3 real roots.


[quote="Ester"]First let me start of by saying HI :D
...okay, now that the simple pleasantries have been made, let me ask my question.

My pre-chem. teacher, has consistantly asked the class simple but amusing questions (which for some reason or another has everyone completely "stumped"). The lastet being:

If you release a balloon (from sea level) that contains 1 mole of Helium gas, pretend that the skin of this balloon is made of a material that can be stretched or contracted to any size. In other words, the balloon can shrink to a zero size and expand to infinte size. Form a hypothesis as to what is going to happen to the volume of such a balloon when you release it from sea level and allow it to rise to a height of 150 km (outter space).
1) Use the atmospheric data in conjunction with Boyle's and Charle's law to calculate the volume of a balloon at intervals of 10 km up to 150 km.
2) Use the Ideal Gas law to predict the volume of the balloon as a function of its altitude (from sea level to 150 km in 10 km intervals).
3) Calculate the % deviation between the volume calculated in 1 and the volume calculated in 2.
4) Plot the size of the balloon as a function of atmospheric height.
5) How does your original hypothesis differ from the conclusion (the mathematical step in your data)?

I would really appreciate the help  :?
...if not, then it's still okay, no hard feelings   ;)[/quote]

You need to do your own homework, it's simple application of the ideal gas law.  Why don't you start by showing your work/ideas for 1)

[quote="GCT"]

You need to do your own homework, it's simple application of the ideal gas law.  Why don't you start by showing your work/ideas for 1)[/quote]

:huh: Okay maybe it's my fault for not explaining the situation correctly.....this is [b]not[/b] homework.
This is a question he posed to the class at the end of the semester. He said that if we can figure this out, then we'll be a step ahead for the first module of AP Chem.

Anywho, from what i understand, the Ideal Gas Law states:
PV=nRT (R being the universal gas constant)
But i still end up being stuck because i am trying to find the volume & temp. And i dont know where to start.
Maybe that is why i am asking for help :|

[quote="pkothari13"]as for exact numbers, you may have to go out and find that yourself.

to illustrate a similar example, pretend you have a balloon. now when you put it in the bottom of the pool, the balloon shrinks in size.  however, as you bring it closer and closer to the surface, the balloon slowly increases in size.

use this more intuitive example with your knowledge of environmental factors at higher altitudes to get your answer.[/quote]

thank you for your advice :)

But the further up the balloon goes, the temp. & pressure changes.....
hummmmm :|
Let me go play with the numbers again.
Thanx for the slight bit of help pkothari13

[quote="Ester"]
with further research I figured out that the 1 mole of Helium, the balloon contains, would equal 22.41 L at S.T.P.
But the further up the balloon goes, the temp. & pressure changes.....
[/quote]

now before you go any further, do some more research as to what S.T.P is, because it's not just anywhere where you will have those conditions.

and you are right about the temp and pressure changing... but how exactly does it change? and how do these changes affect volume??

this is where you use PV = nRT.  You already know n and R will be constant (n = 1 mol and R = .08206 atm*L/mol*K), so you can reduce it down to PV = kT, where k is just a constant and has no affect in the relative scheme of things.

i think it might be best if you get a better knowledge of gas laws in general. try [url=http://dbhs.wvusd.k12.ca.us/webdocs/GasLaw/KMT-Gas-Laws.html]this[/url]
site for a general overview of it.

Thanx again for the extra help pkothari13 ;)
...But i just thought of something, wouldnt the balloon only rise as high as its own density :?:
I say this because i wouldnt be suprised if this is some sort of trick question from my teacher :oops:

have you ever let a balloon into the sky and watch it fly away?  it'll only rise until it is so stretched that it pops.  howeverrrrrrrrr, in your case, your balloon is pretty durable, so we won't have to worry about that  :)

Oops, i forgot about that specific  part of the problem :oops:

However, I just figured out that adiabatic expansion would play a part of the entire problem. To bad I know nothing about that :?

adiabatic = no heat flowing in or out of the system.

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Thermal/ImportantThermalProcess.html

that gives you a good overview of certain thermal processes, including those with contstant heat and pressure. note that heat isn't the same thing as temperature.


hello

can anyone find the next hardy-ramanujam number after 1729?

This is a challenge. :)  :)

Don't double post. :mad: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=320524[/url]

the next number is 4,104 just search on google


algorithm
superior algebra
superior algebra unsolved

Hallo guys, can you tell me which group is  $ < x,y | x^4 \equal{} y^4 \equal{} 1, x^2 \equal{} y^2, xy \equal{} y^3x >$ ?

I used the Todd-Coxeter Algorithm and I got two possible groups: the Klein four-group and the Quaternions, but I don't know how to decide which one is.

Danken!

[quote="LydianRain"]Your guess should be the quaternions, as this group certainly "looks" nonabelian.

Easiest proof is probably just to show that the quaternions are consistent with the presentation. (that is, $ Q_8$ contains $ x$ and $ y$ satisfying those relations and generating $ Q_8$)  Presentations are "universal" in the sense that if a group is consistent with the presentation, it must be a quotient of the presented group.  In particular, the presented group has at least $ 8$ elements.[/quote]

why do you say that the group looks nonabelian?

The quaternions are consistent with this presentation, but the Klein four-group is also consistent.   For the former, take $ x\equal{}i$ and $ y\equal{}j$ and for the latter $ x\equal{}(1,0)$ and $ y\equal{}(0,1)$

Read what I wrote again.  Yes, both are consistent, implying that both are quotients of the group in question.  In particular, the group in question has to be at least as big as either of them.  And $ |Q| \equal{} 8>4 \equal{} |K|$, so the group has at least $ 8$ elements, and isn't the Klein 4-group.

I say that it "looks" nonabelian because there is no obvious way to prove from the relations that $ x^2\equal{}y^2\equal{}1$, which would need to hold if it were abelian.

Well, that last relation $ xyx^{ \minus{} 1} \equal{} y^{ \minus{} 1}$ definitely makes it nonabelian, unless $ y^2$ is the identity. Since we were given $ y^4 \equal{} e$, we should guess that $ y$ probably has order $ 4$ rather than $ 2$.


A wrecker's iron ball eight inches in diameter weighs 120 pounds. How many pounds would a similar iron wrecking ball twelves in diameter weigh?

well, the volume expands cubically(??), so:

(12/8)^3*120=
405 pounds.


Find all $f: R\rightarrow R$ such that

$f(xy)+f(xz)-f(x)f(yz)\geq 1$ for all $x,y,z$.

Let $x=y=z=1$, it is easy to see that $f(1)=1$. Let $x=y=z=0$, we get the same way $f(0)=1$. If $y=z=1$ then $f(x)\ge 1$; if $y=z=0$ then $f(x)\le 1$, so $f(x)\equiv 1$.

$f(0)+f(0)-f(0)f(0)\geq 1\implies f(0)=1$
$f(1)+f(1)-f(1)f(1)\geq 1\implies f(1)=1$.
$y=z=1\implies f(x)\geq 1,\forall x\in \mathbb{R}$.
$y=z=\frac{1}{x}, x\neq 0\implies f(x)f\left(\frac{1}{x^{2}}\right)\leq 1$ So $f(x)=1, \forall x\in \mathbb{R}$


How can a 1 x 3 rectangle be dissected into three pieces that can be rearranged to form a square?

Thanks.

[quote="MeKnowsNothing"]http://www.jimloy.com/puzz/dissect.htm[/quote]

Thanks for providing the above link; however, using the notation there, $ a \equal{} 3b$, and so at least when hand-drawn, the dissection doesn't seem to work, because the two diagonal lines seem not to intersect inside the rectangle.

Does this work? One other mathematician gave me this answer and I'm too tired to check the solution.

Measure $ \sqrt {3}$ from one corner (call this corner $ A$) along the long side of the rectangle, and call that point $ B$. Measure $ \sqrt {3}$ from the opposite corner $ (C)$ along the opposite long side, and call that point $ D$. Then cut out the right triangle with long side $ AB$, right angle at $ A$, and short side of length $ 1$. Now drop a vertical line from the point $ D$ and cut along that line, resulting in a small right triangle with one side of length $ 1 \minus{} \sqrt{3}$. Now rearrange those three pieces.

[quote="MeKnowsNothing"]
Measure $ \sqrt {3}$ from one corner (call this corner $ A$) along the long side of the rectangle, and call that point $ B$. Measure $ \sqrt {3}$ from the opposite corner $ (C)$ along the opposite long side, and call that point $ D$. Then cut out the right triangle with long side $ AB$, right angle at $ A$, and short side of length $ 1$. Now drop a vertical line from the point $ D$ and cut along that line, resulting in a small right triangle with one side of length $ 1 \minus{} \sqrt {3}$. Now rearrange those three pieces.[/quote]

That's correct, but I believe you mean to say that the small right triangle has sides of length $ 3 \minus{} \sqrt {3}$ and $ \sqrt {3} \minus{} 1$.

Thanks.


How can I change the size of my title (made \title{Untitled} ) and possibility change it's position, font type and color?

I know that change default thins in $ \text{\LaTeX}$ is a lot of hard work, so the simplest solution would be ommit the comand \title and make a page that look like a initial page?

1. The simplest way is to use
[code]\title{\color{red}\Huge \textit{This is the title}}
\author{Me}
\maketitle [/code]
2. If you also want to move the title around then
a) either use the titlepage environment without \maketitle
[code]\begin{titlepage}
\vspace*{2in}
\begin{flushright}
{\color{red}\Huge \textit{This is the title}} \\
Me
\end{flushright}
\end{titlepage}[/code]
b) or redefine the maketitle command
[code]\makeatletter
\renewcommand{\maketitle}{%
\begin{titlepage}
\vspace*{2in}
\begin{flushright}
\@title \\
\@author
\end{flushright}
\end{titlepage}}
\title{\color{red}\Huge \textit{This is the title}}
\author{Me}
\makeatother

\maketitle[/code]

Truly a nice solution. Thanks. I'm not too good in latex. If you have a link to a book about latex have thinks advanced and basic (Once I got one really good, but forgotten where I downloaded it) I would be glad to see it.

I have another question.

I use a Portuguese package to make my books. It's called AbnTeX (crudely ABNT is a set of rules to "how to format properly")

So instead of normal book, my book have the code

[code]\documentclass[a4paper,11pt,oneside]{abnt}[/code]

But one thing that I really dislike is the Table of Contents. The space between the chapters are really big, and some others annoying things, like, instead of "chapter 1 - ..." is just "1 - ..."

If the solution for those basic problems are in a ebook, just appoint the ebook and I try to learn it by it.

Thanks.

In general, try [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos Learning more about LaTeX[/url]. For the table of contents try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=titletoc]titletoc package[/url].


Let A={1,2,3..n}.If X denotes the subsets of A containing exactly three elements ,then find $ \sum_{P\in X}{min(P)}$

They ask you to compute the sum of the least elements in all triplets that can be made off the numbers $ 1,2,\ldots,n$.
[hide="Hint"]Try counting the triplets having $ k$, $ 1\leq k\leq n\minus{}2$.[/hide]


sabet konid bishomar addade avval beforme 4n+3 vojud darad?

(tamrine gosasteie pish!) :blush:

albate miga uni ke tu aksare ketabha neveshte nadoroste!akhe akharesh az inke p l 3 ro natige migire ke tanaghoze darhali ke momkene p hamun 3 bashe!
mage na?!vali esbate binahaiat addade avval doroste chon 1 avval nist...

man vase in ye esbat daram vali motmaen nistam ke dorosto hesabio kamel bashe:
khob farz kon tedade aadade avval be forme $ 4k \plus{} 3$ mahdood basheo majmooe $ \{p_1,\ldots,p_k\}$ majmooe hame in aadad bashe,khob hala dota halat darim:

$ \boxed{\textrm{halate avval: } K \equal{} 2t}$

dar in halat adade zir ro dar nazar begir:

$ N \equal{} p_1p_2\ldots p_k \plus{} 2$

khob az oonjai ke $ k$ ye adade zoje darim:

$ p_1p_2\ldots p_k\equiv 1\pmod 4$

$ \Rightarrow N\equiv 3\pmod 4$

khob pas $ N$ khodesh ye adadie be forme $ 4n \plus{} 3$ pas ye amele avval be forme $ 4k \plus{} 3$ bayad dashte bashe (cherake dar gheyre in soorat hame amelash be shekle $ 4k \plus{} 1$ mishano oonmoghe khode $ N$ ham majboore be shekle $ 4k \plus{} 1$ beshe)

pas ye adade avval be esme $ q$ vojood dare ke $ q\equiv 3\pmod 4$ va $ q\mid N$,amma be rahati natije mishe ke $ q\not\in\{p_1,p_2,\ldots,p_k\}$ chera ke dar gheire in soorat age masalan $ q \equal{} p_i$ oonmoghe khahim dasht:

$ p_i\mid N\Rightarrow p_i\mid 2\Rightarrow p_i \equal{} 2$ ke inam tanaghoze cherake $ p_i\equiv 3\pmod 4$

pas $ q\not\in\{p_1,p_2,\ldots,p_k\}$ va hamchenin $ q\equiv 3\pmod 4$ ke inam tanaghoz dare ba farze avvaliamoon...

--------------------------------------

$ \boxed{\textrm{halate dovvom: } K \equal{} 2t \plus{} 1}$

too in halat biaid adade zir ro dar nazar begirid:

$ N \equal{} p_1p_2\ldots p_k \plus{} 4$

hala chon $ k$ ye adade farde darim:

$ p_1p_2\ldots p_k\equiv 3\pmod 4$

$ \Rightarrow N\equiv 3\pmod 4$

khob hala ba hamoon estedlali ke too halate ghabl kardim natije mishe ye $ q$ vojood dare ke avvalan $ q\equiv 3\pmod 4$ va sanian $ q\mid N$ amma bazam mese halate ghabl age $ q \equal{} p_i$ oon moghe natije mishe $ p_i\mid 4$ ke bazam tanaghoze pas too in halatam $ q\not\in\{p_1,p_2,\ldots,p_k\}$ pas too in halatam be tanaghoz miresim...

[quote="nodet"]mamnoon agha babak vali nemidunam chera tuie pc man formulha besoorate latex dide mishe?! :maybe:
chi kar konam man ke chizi nemifahmam! :oops:[/quote]

nemidoonam chera,vali ide kollie esbato mitooni befahmi:

bebin farz kon tedadeshoon mahdood bashe masalan k ta basheo majmooe {p_1,p_2,...,p_k} hame in aadade avvale be forme 4k+3 bashan...

khob hala age k zoj bood bia adade zir ro dar nazar begir:

N=(p_1p_2...p_k)+2

bebin chon k zoje pas p_1p_2...p_k be peimane 4 barabare 1 mishe pas N be peimane 4 barabare 3 mishe,khob hala in adade N khodesh bayad ye amele avval be forme 4k+3 dashte bashe (cherake dar gheire in soorat hame amelash be shekle 4k+1 khahand boodo dar natije khode N ham be shekle 4k+1 mishod...)

khob pas ye adade avval be esme q vojood dare ke q|N va q=4k+3
khob hala be rahati mishe did ke q ozve majmooe {p_1,p_2,...,p_k} nis chon age ozve in majmooe bashe masalan q=p_i oonmoghe az p_i|N natije mishe p_i|2 ke inam tanaghoze.
pas ye adade avval be shekle 4k+3 peida shod ke too majmooamoon nabood ke inam tanaghoze...
--------------------------------------------
pas too halati ke adade k zoj bashe masale halle,hala farz kon adade k fard bashe oonmoghe bia adade ziro dar nazar begir:

N=(p_1p_2...p_k)+4

deghat kon chon k farde pas adade p_1p_2...p_k be peimane 4 ba 3 hamneheshte pas adade N ham be hange 4 ba 3 hamneheshte...
khob hala mese halate ghabli ye adade avval mesle q vojood dare ke q|N va q=4k+3 khob hala mesle estedlale ghabli adade q nemitoone ozve majmooe {p_1,p_2,...,p_k} bashe chon age q=p_i oonmoghe natije mishe p_i|2 ke inam tanaghoze,pas too in halatam be tanaghoz residim...

ideye in rahe halam mese rahe hale agha babake,faghat yekam kutahtare fek konam...
farz konid $ p_1,p_2,...,p_k$ hameye adade avale be surate $ 4k\plus{}3$ bashan.gharar dahid:
$ S\equal{}4p_1p_2...p_k\minus{}1$ be vozuh S daraye amele avali be surate 4k+3 ast.farz mokonim in amel $ p_i$ bashad.pas darim$ p_i|4p_1p_2...p_k\minus{}1$   ,   $ p_i|4p_1p_2...p_k$ pas $ p_i|\minus{}1$ ke tanaghoz ast...


Prove that $ \displaystyle\frac{1}{1\cdot 2\cdot\sqrt{3}}\plus{}\frac{1}{2\cdot 3\cdot\sqrt{4}}\plus{}\frac{1}{3\cdot 4\cdot\sqrt{5}}\plus{}\cdots\plus{}\frac{1}{n\cdot(n\plus{}1)\cdot\sqrt{n\plus{}2}}>\frac{n}{2(n\plus{}1)},$ for all $ n\in\mathbb{N^*}$

Really?

$ n \le 22 : \frac{1}{1\cdot 2\cdot\sqrt{3}}\plus{}\frac{1}{2\cdot 3\cdot\sqrt{4}}\plus{}\frac{1}{3\cdot 4\cdot\sqrt{5}}\plus{}\cdots\plus{}\frac{1}{n\cdot(n\plus{}1)\cdot\sqrt{n\plus{}2}}>\frac{n}{2(n\plus{}1)}$
$ n \ge 23 : \frac{1}{1\cdot 2\cdot\sqrt{3}}\plus{}\frac{1}{2\cdot 3\cdot\sqrt{4}}\plus{}\frac{1}{3\cdot 4\cdot\sqrt{5}}\plus{}\cdots\plus{}\frac{1}{n\cdot(n\plus{}1)\cdot\sqrt{n\plus{}2}}<\frac{n}{2(n\plus{}1)}$
by Mathematica

Where to [b]free[/b] download Mathematica ?

No. Mathematica is shareware.

Do you know Maxima?
Maxima is freeware :)


quadratics
algebra
polynomial
function
symmetry
algebra proposed

[quote="alex2008"]Let be $ f: \mathbb{R}\rightarrow \mathbb{R}\ ,\ f(x) \equal{} ax^2 \plus{} bx \plus{} c$ . Determine $ a,b,c\in \mathbb{Z}$ such that $ f(f(1)) \equal{} f(f(2)) \equal{} f(f(3))$ .[/quote]

are you asking to determine just  one triple?it seems that you are asking for just one triple...but it can´t be that, because then the problem is obvious:
$ a\equal{}b\equal{}c\equal{}0$ saitsfies it.
sorry if i misunderstood the problem :blush:

If $ a\equal{}0$, then $ f(x)\equal{}bx\plus{}c,$ and $ f(f(1))\equal{}b^2\plus{}bc\plus{}c, f(f(2))\equal{}2b^2\plus{}bc\plus{}c,$ so $ b^2\equal{}0 \Rightarrow b\equal{}0$. We see that $ (a,b,c)\equal{}(0,0,c)$ is a solution for all $ c\in \mathbb{Z}$.
Now assume $ a\neq 0$. Then $ f(x)$ is a quadratic polynomial with exactly 2 complex roots. We cannot have $ d\equal{}f(i)\equal{}f(j)\equal{}f(k)$ for some distinct $ i,j,k$, since then the function $ g(x)\equal{}f(x)\minus{}d$ would have 3 roots, a contradiction. So we must have some two of $ f(1), f(2), f(3)$ be equal. Note that all three cannot be equal, since then we would have $ f(x)$ be a constant function.

We now split our work into cases:
Case 1: $ f(1)\equal{}f(2)$.
We have $ a\plus{}b\plus{}c\equal{}4a\plus{}2b\plus{}c \iff b\equal{}\minus{}3a$, so $ f(x)\equal{}ax^2\minus{}3ax\plus{}c$, and $ f(3)\equal{}c$. Then we have $ f(f(1))\equal{}f(f(3)) \iff f(a\plus{}b\plus{}c)\equal{}f(c) \iff f(c\minus{}2a)\equal{}f(c)$. Since $ a\neq 0, x\equal{}c\minus{}a$ must be the vertical line of symmetry for $ f(x)$. But since $ f(1)\equal{}f(2)$, we know that this line is $ x\equal{}1.5$, so we have $ c\minus{}a\equal{}1.5$, a contradiction, as $ a,c \in \mathbb{Z}$.

Case 2: $ f(1)\equal{}f(3)$.
Similar to above, we have $ a\plus{}b\plus{}c\equal{}9a\plus{}3b\plus{}c \iff b\equal{}\minus{}4a$, so $ f(x)\equal{}ax^2\minus{}4ax\plus{}c, f(2)\equal{}\minus{}4a\plus{}c$. Then $ f(f(1))\equal{}f(f(2))$ gives $ f(\minus{}3a\plus{}c)\equal{}f(\minus{}4a\plus{}c)$, so $ \minus{}3.5a\plus{}c\equal{}2 \iff c\equal{}4\plus{}7a$. With this, we can find $ f(1)\equal{}4a\plus{}4, f(2)\equal{}3a\plus{}4$, and this tells us $ f(4a\plus{}4)\equal{}f(3a\plus{}4)$, implying $ 3.5a\plus{}4\equal{}2 \iff a\equal{}\minus{}\frac{2}{3.5}\equal{}\minus{}\frac{4}{7}$, a contradiction.

Case 3: $ f(2)\equal{}f(3)$.
Once again similar to the above, we have $ 4a\plus{}2b\plus{}c\equal{}9a\plus{}3b\plus{}c \iff b\equal{}\minus{}5a$, $ f(x)\equal{}ax^2\minus{}5ax\plus{}c, f(1)\equal{}\minus{}4a\plus{}c$. $ f(f(1))\equal{}f(f(2)) \Rightarrow f(\minus{}4a\plus{}c)\equal{}f(\minus{}6a\plus{}c) \Rightarrow \minus{}5a\plus{}c\equal{}2.5$, contradiction.

Thus, the only solutions are $ (0,0,c)$ for integer $ c$.

[quote="Brut3Forc3"]If $ a \equal{} 0$, then $ f(x) \equal{} bx \plus{} c,$ and $ f(f(1)) \equal{} b^2 \plus{} bc \plus{} c, f(f(2)) \equal{} 2b^2 \plus{} bc \plus{} c,$ so $ b^2 \equal{} 0 \Rightarrow b \equal{} 0$. We see that $ (a,b,c) \equal{} (0,0,c)$ is a solution for all $ c\in \mathbb{Z}$.
Now assume $ a\neq 0$. Then $ f(x)$ is a quadratic polynomial with exactly 2 complex roots. We cannot have $ d \equal{} f(i) \equal{} f(j) \equal{} f(k)$ for some distinct $ i,j,k$, since then the function $ g(x) \equal{} f(x) \minus{} d$ would have 3 roots, a contradiction. So we must have some two of $ f(1), f(2), f(3)$ be equal. Note that all three cannot be equal, since then we would have $ f(x)$ be a constant function.

We now split our work into cases:
Case 1: $ f(1) \equal{} f(2)$.
We have $ a \plus{} b \plus{} c \equal{} 4a \plus{} 2b \plus{} c \iff b \equal{} \minus{} 3a$, so $ f(x) \equal{} ax^2 \minus{} 3ax \plus{} c$, and $ f(3) \equal{} c$. Then we have $ f(f(1)) \equal{} f(f(3)) \iff f(a \plus{} b \plus{} c) \equal{} f(c) \iff f(c \minus{} 2a) \equal{} f(c)$. Since $ a\neq 0, x \equal{} c \minus{} a$ must be the vertical line of symmetry for $ f(x)$. But since $ f(1) \equal{} f(2)$, we know that this line is $ x \equal{} 1.5$, so we have $ c \minus{} a \equal{} 1.5$, a contradiction, as $ a,c \in \mathbb{Z}$.

Case 2: $ f(1) \equal{} f(3)$.
Similar to above, we have $ a \plus{} b \plus{} c \equal{} 9a \plus{} 3b \plus{} c \iff b \equal{} \minus{} 4a$, so $ f(x) \equal{} ax^2 \minus{} 4ax \plus{} c, f(2) \equal{} \minus{} 4a \plus{} c$. Then $ f(f(1)) \equal{} f(f(2))$ gives $ f( \minus{} 3a \plus{} c) \equal{} f( \minus{} 4a \plus{} c)$, so $ \minus{} 3.5a \plus{} c \equal{} 2 \iff c \equal{} 4 \plus{} 7a$. With this, we can find $ f(1) \equal{} 4a \plus{} 4, f(2) \equal{} 3a \plus{} 4$, and this tells us $ f(4a \plus{} 4) \equal{} f(3a \plus{} 4)$, implying $ 3.5a \plus{} 4 \equal{} 2 \iff a \equal{} \minus{} \frac {2}{3.5} \equal{} \minus{} \frac {4}{7}$, a contradiction.

Case 3: $ f(2) \equal{} f(3)$.
Once again similar to the above, we have $ 4a \plus{} 2b \plus{} c \equal{} 9a \plus{} 3b \plus{} c \iff b \equal{} \minus{} 5a$, $ f(x) \equal{} ax^2 \minus{} 5ax \plus{} c, f(1) \equal{} \minus{} 4a \plus{} c$. $ f(f(1)) \equal{} f(f(2)) \Rightarrow f( \minus{} 4a \plus{} c) \equal{} f( \minus{} 6a \plus{} c) \Rightarrow \minus{} 5a \plus{} c \equal{} 2.5$, contradiction.

Thus, the only solutions are $ (0,0,c)$ for integer $ c$.[/quote]
I think your solution is not enough (Just thinking  :D) , the roots are more general, I figure out they are: $ (a,b,c)\equal{}(2t,\minus{}8t,7t\plus{}2)$, for $ t$ is some interger.

I realize my error... for case 2, it should be $ \minus{}3.5a\plus{}c\equal{}2 \iff 2c\equal{}7a\plus{}4$. This is only possible if $ a$ is even, so we let $ a\equal{}2t$, and it gives the roots that you give above!

[quote="mathVNpro"][quote="Brut3Forc3"]If $ a \equal{} 0$, then $ f(x) \equal{} bx \plus{} c,$ and $ f(f(1)) \equal{} b^2 \plus{} bc \plus{} c, f(f(2)) \equal{} 2b^2 \plus{} bc \plus{} c,$ so $ b^2 \equal{} 0 \Rightarrow b \equal{} 0$. We see that $ (a,b,c) \equal{} (0,0,c)$ is a solution for all $ c\in \mathbb{Z}$.
Now assume $ a\neq 0$. Then $ f(x)$ is a quadratic polynomial with exactly 2 complex roots. We cannot have $ d \equal{} f(i) \equal{} f(j) \equal{} f(k)$ for some distinct $ i,j,k$, since then the function $ g(x) \equal{} f(x) \minus{} d$ would have 3 roots, a contradiction. So we must have some two of $ f(1), f(2), f(3)$ be equal. Note that all three cannot be equal, since then we would have $ f(x)$ be a constant function.

We now split our work into cases:
Case 1: $ f(1) \equal{} f(2)$.
We have $ a \plus{} b \plus{} c \equal{} 4a \plus{} 2b \plus{} c \iff b \equal{} \minus{} 3a$, so $ f(x) \equal{} ax^2 \minus{} 3ax \plus{} c$, and $ f(3) \equal{} c$. Then we have $ f(f(1)) \equal{} f(f(3)) \iff f(a \plus{} b \plus{} c) \equal{} f(c) \iff f(c \minus{} 2a) \equal{} f(c)$. Since $ a\neq 0, x \equal{} c \minus{} a$ must be the vertical line of symmetry for $ f(x)$. But since $ f(1) \equal{} f(2)$, we know that this line is $ x \equal{} 1.5$, so we have $ c \minus{} a \equal{} 1.5$, a contradiction, as $ a,c \in \mathbb{Z}$.

Case 2: $ f(1) \equal{} f(3)$.
Similar to above, we have $ a \plus{} b \plus{} c \equal{} 9a \plus{} 3b \plus{} c \iff b \equal{} \minus{} 4a$, so $ f(x) \equal{} ax^2 \minus{} 4ax \plus{} c, f(2) \equal{} \minus{} 4a \plus{} c$. Then $ f(f(1)) \equal{} f(f(2))$ gives $ f( \minus{} 3a \plus{} c) \equal{} f( \minus{} 4a \plus{} c)$, so $ \minus{} 3.5a \plus{} c \equal{} 2 \iff c \equal{} 4 \plus{} 7a$. With this, we can find $ f(1) \equal{} 4a \plus{} 4, f(2) \equal{} 3a \plus{} 4$, and this tells us $ f(4a \plus{} 4) \equal{} f(3a \plus{} 4)$, implying $ 3.5a \plus{} 4 \equal{} 2 \iff a \equal{} \minus{} \frac {2}{3.5} \equal{} \minus{} \frac {4}{7}$, a contradiction.

Case 3: $ f(2) \equal{} f(3)$.
Once again similar to the above, we have $ 4a \plus{} 2b \plus{} c \equal{} 9a \plus{} 3b \plus{} c \iff b \equal{} \minus{} 5a$, $ f(x) \equal{} ax^2 \minus{} 5ax \plus{} c, f(1) \equal{} \minus{} 4a \plus{} c$. $ f(f(1)) \equal{} f(f(2)) \Rightarrow f( \minus{} 4a \plus{} c) \equal{} f( \minus{} 6a \plus{} c) \Rightarrow \minus{} 5a \plus{} c \equal{} 2.5$, contradiction.

Thus, the only solutions are $ (0,0,c)$ for integer $ c$.[/quote]
I think your solution is not enough (Just thinking  :D) , the roots are more general, I figure out they are: $ (a,b,c) \equal{} (2t, \minus{} 8t,7t \plus{} 2)$, for $ t$ is some interger.[/quote]

Maybe here is another approach:
Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$. Then $ f(1) \equal{} a \plus{} b \plus{} c$, $ f(2) \equal{} 4a \plus{} 2b \plus{} c$ and $ f(3) \equal{} 9a \plus{} 3b \plus{} c$. Then $ f(f(2)) \minus{} f(f(1)) \equal{} a[(4a \plus{} 2b \plus{} c)^2 \minus{} (a \plus{} b \plus{} c)^2] \plus{} b[(4a \plus{} 2b \plus{} c) \minus{} (a \plus{} b \plus{} c)]$
$ \equal{} (3a \plus{} b)[a(5a \plus{} 3b \plus{} 2c) \plus{} b]$
$ \equal{} (3a \plus{} b)(5a^2 \plus{} 3ab \plus{} 2ac \plus{} b)$
$ f(f(3)) \minus{} f(f(1)) \equal{} a[(9a \plus{} 3b \plus{} c)^2 \minus{} (a \plus{} b \plus{} c)^2] \plus{} b[(9a \plus{} 3b \plus{} c) \minus{} (a \plus{} b \plus{} c)]$
$ \equal{} (8a \plus{} 2b)[a(10a \plus{} 4b \plus{} 2c) \plus{} b]$
$ \equal{} 2(4a \plus{} b)(10a^2 \plus{} 4ab \plus{} 2ac \plus{} b)$
Suppose that $ f(f(1)) \equal{} f(f(2)) \equal{} f(f(3))$. If $ b \equal{} \minus{} 3a$,
Then we must have $ 0 \equal{} 10a^2 \plus{} 4ab \plus{} 2ac \plus{} b \equal{} \minus{} 10a^2 \minus{} 12a^2 \plus{} 2ac \minus{} 3a$, whence $ c \equal{} a \plus{} \frac {3}{2}$. In this case, $ a$ and $ c$ cannot both be integers. However, if $ b \equal{} \minus{}4a$,
then $ 0 \equal{} 5a^2 \plus{} 3ab \plus{} 2ac \plus{} b \equal{} 5a^2 \minus{} 12a^2 \plus{} 2ac \minus{} 4a$, whence $ c \equal{} \frac {7a}{2} \plus{} 2$. So we can get integer coeﬃcients by taking $ (a,b,c) \equal{} (2t, \minus{} 8t, 7t \plus{} 2)$ for some integer $ t$.


Έστω τετράγωνο ΑΒΓΔ με διαγώνιο ΑΓ και έστω και δύο μετρικά τρίγωνα ΕΖΗ=ΚΛΜ=ΑΒΓ.
Να αποδειχθεί μετρικά ο τύπος εύρεσης εμβαδού τριγώνου βΧυ:2 ή αλλιώς ότι το ΑΒΓ είναι το μισό του ΑΒΓΔ.

Τι θα πει μετρικά?? :blush:

Μετρικά θα πει να εφαρμόσουμε επιθετικά το μέτρο επί του μετρούμενου, καθώς προβλέπει η θεωρία μετρήσεως. Από την επίθεση αυτή, θα εξαχθεί ένας αριθμός που θα αποτελεί το μέτρο του τετραγώνου. Σαν μέτρο χρησιμοποιούμε το ακριβές ΑΒΓ το οποίο έχουμε 2 φορές σαν ΕΖΗ και ΚΛΜ. Αυτό το κάνουμε διότι τα σχήματα αυτά καθαυτά δεν μετακινούνται επί του επιπέδου παρά μόνο σαν ομόλογα ή εικονικά. Τα μετρικά τρίγωνα όμως δεν έχουν τέτοια δέσμευση και ελεύθερα μετακινούνται ώστε να γίνει η επίθεσή τους επί του μετρούμενου.
Εν προκειμένω θα πρέπει να επιθέσεις τα δύο μετρικά τρίγωνα (ΕΖΗ=ΚΛΜ) επί του ενός τετραγώνου ΑΒΓΔ για να διαπιστώσουμε αν το τετράγωνο είναι διπλάσιο του ΑΒΓ, αφού καλείσαι να επιθέσεις τα 2 ίσα με το ΑΒΓ τρίγωνα ΕΖΗ και ΚΛΜ. Αν είναι διπλάσιο, τότε ο τύπος εύρεσης εμβαδού τριγώνου βΧυ:2 είναι ορθός, αφού θα αποδεικνύεται ότι το τετράγωνο που είναι βΧυ αποτελείται από 2 τρίγωνα (τα χωράει ακριβώς για να το πω αλλιώς) και επομένως το κάθε τρίγωνο, είναι το μισό τετράγωνο όπως λέει ο τύπος.
Ελπίζω να σου εξήγησα με επάρκεια τι ζητάω και να σου υπέδειξα πως να το κάνεις σύμφωνα με τη θεωρία μετρήσεως.


Mr Johnson had lived in the same house for many decades. One day Mr Johnson returned home to find it engulfed in flames. The fire department was on the scene, but they made no attempt to save Mr Johnson's house. Even though he is in tears, Mr Johnson says nothing.

What is going on?

Mr. Johnson, like the factions in Iraq, belongs to a different religious sect than do the firefighters, and they know he is different from their own own religion.  Mr. Johnson does not
say anything because he realizes that 1) the firefighters actually set the fire, and 2) he is next (to die) if he verbally
objects about it.

[quote="mathnerd314"][quote="E^(pi*i)=-1"][hide]
Mr. Johnson is a bird.  The firefighters are there to contain the forest fire, not to save Mr. Johnson's nest.  Since Mr. Johnson is a bird, he obviously doesn't say anything.
[/hide][/quote]

[hide]1. Why would a bird be named Mr. Johnson?
2. Why would a bird be in tears?[/hide][/quote]

mathnerd314, what is wrong with the name Mr. Johnson? and a bird can be in tears, as it has emotions! lol. well, it is a biotic creature, so yea..u know?

Clues:

[hide]1. Mr. Johnson is not a bird. He is a person (man).

2. The house could have been saved if the fire department had acted. Mr Johnson IS crying because his home is burning down.

3. As it was rightly pointed out in the first reply, the firefighters destroyed Mr. Johnson's house. The reason for destruction was however different from the responses received so far.[/hide]

Questions:

Did Mr Johnson do something to cause his house to be burnt down?

Can Mr. Johnson talk at all?

Question 1: Did Mr Johnson do something to cause his house to be burnt down?

Answer: Yes.

Question 2: Can Mr. Johnson talk at all?

Answer: Yes.

Does Mr. Johnson have a wife?

Also, Is Mr. Johnson politically involved with his country?

And, where does he live(country?)

ok, this is kinda silly...

[hide="probably not the answer :blush: "]is it because he left the house, as he had been living in teh same house for so long, and when he left, they burnt it? i dunno....this is too far fetched[/hide]

i could easily make one
[hide] mr. johnson is a werewolf and they don't cry so when he went through a lycanthropic time the firemen noticed and burned his house down and then he came back to being human and wept[/hide] :(


There are 5 patches on a coat of area $ 5$.  Each patch has area at least $ 2.5$.  Show that there exists an overlap between 2 patches of area at least $ 1$.

For crying out loud, do a search!

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1701477842&t=154180

Here's another proof from the earliest days of AoPS (my first post!):
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=24486#24486[/url].

[quote="unimpossible"]If you want a solution just check the Intermediate Counting book.[/quote]

Or the class.


real analysis
inequalities
real analysis unsolved

Let $f,g\in L^{p}$, $1/p+1/q=1+1/r$, $p,q,r\geq 1,f,g\geq 0$. Prove that
\[\parallel f*g\parallel _{r}\leq \parallel f\parallel _{p}\parallel g\parallel _{q}.\]


Is there a way to label only some of the equations in \eqnarray? I can only get it to label all or none.

First, you should know that eqnarray is now deprecated and has been replaced by the align environment from the amsmath package.  For information on why, see [url=http://tug.org/pracjourn/2006-4/madsen/]Avoid eqnarray![/url]

To label only some of the equations in an align environment, use \notag at the end of the equation, before the \\.  For example,

[code]\begin{align}
a&=b\\
c&=d\notag\\
\end{align}[/code]

It doesn't seem to work properly in the forum, but it should work in a document.

[quote="roadnottaken"]It doesn't seem to work properly in the forum,[/quote]It does work, but you put in an extra \\ by mistake giving the (2) after a blank line:
\begin{align}
a&=b\\
c&=d\notag
\end{align}


Find all sets $ A,B\in\mathbb{R}$ such that their union is $ \mathbb{R}$, their intersection is empty and if $ x,y\in A$ or $ x,y\in B$, then $ x+y\in A$ or $ x+y\in B$ respectively.

You need be more careful with your Latex (at least after Submit, check it turned out as you wished). By looking at the code you used, you meant

if $ x,y \in A$ then $ x \plus{} y \in A$, and if  $ x,y \in B$ then $ x \plus{} y \in B$.

Also, you should use $ A, B \subset \mathbb{R}$, not the other sign, since these are subsets, not elements.

[quote="bakerbakura"]Find all sets $ A,B\subseteq\mathbb{R}$ such that their union is $ \mathbb{R}$, their intersection is empty and if $ x,y\in A$ or $ x,y\in B$, then $ x \plus{} y\in A$ or $ x \plus{} y\in B$ respectively.[/quote]

I'm interested in your solution because there are infinitely many solutions and I did not succeed up to now finding a general form for them.

Beside trivial solutions $ (\mathbb R,\emptyset)$ $ ,(\mathbb R^\plus{}\cup\{0\},\mathbb R\minus{})$ $ ,(\mathbb R^\minus{}\cup\{0\},\mathbb R\plus{})$, Here is a method to generate infinitely many solutions :

Starting with a solution $ (A_0,B_0)$, let $ h(x)$ any bijective solution of Cauchy's equation $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$. Then $ (h(A_0),h(B_0))$ is a solution too.

Starting with $ ,(\mathbb R^\plus{}\cup\{0\},\mathbb R\minus{})$, for example :
Using continuous bijective solutions of Cauchy equation, we get the two trivial solutions $ ,(\mathbb R^\plus{}\cup\{0\},\mathbb R\minus{})$ and $ ,(\mathbb R^\minus{}\cup\{0\},\mathbb R\plus{})$
Using AC and non continuous bijective solutions of Cauchy equation, we get infinitely many non trivial couples $ (A,B)$

[u][b]Could you, please, give us your general solution [/b][/u]?

Please, bakerbakura, since you are active in mathlinks forum, could you give us the solution to this problem.

Thanks.

[quote="bakerbakura"]I found something wrong with my solution but it does work if we modify the problem as follows:

Given real numbers $ a < b$, find all sets $ A,B\subseteq\mathbb{R}$ such that their union is $ \mathbb{R}$, their intersection is empty, $ x\in A$ for all $ x\in(a,b)$ and if $ x,y\in A$ or $ x,y\in B$, then $ x \plus{} y\in A$ or $ x \plus{} y\in B$ respectively.

Sorry about that  :oops:[/quote]

Thanks for your answer.

And no worries.

[hide="Here is my solution"]Note that if $ n\in\mathbb{N}$ and $ x_1, x_2, \dots, x_n \in A$, then $ x_1\plus{}x_2\plus{}\dots\plus{}x_n\in A$. Putting all the $ x_i$ equal, we get that if $ x\in A$, then $ nx\in A$. The same applies for $ B$. Also, if $ nx\in A$, then $ \notin B$, since that would imply $ nx\in B$. Therefore if $ x\in A$, then $ \frac{x}{n}\in A$. So, if $ p,q\in\mathbb{N}$, then $ x\in A$ implies $ px\in A$, which implies $ \frac{p}{q}x\in A$, and the same applies for $ B$.
Also note that if $ x$ is a non-zero real number, then $ \{\frac{p}{q}x | p,q\in\mathbb{N}\}$ is dense on the real numbers. We now split the problem into three cases:

[b]1.    $ a<0<b$[/b] Since for any real number $ x$,  $ \{\frac{p}{q}x | p,q\in\mathbb{N}\}$ is dense on the real numbers, there is a pair $ (p,q)$ of natural numbers such that $ \frac{p}{q}x\in(a,b)$. Then $ \frac{p}{q}x\in A$, so that $ x\equal{}\frac{q}{p}\times\frac{p}{q}x\in A$. Therefore $ A\equal{}\mathbb{R}$, and so $ B\equal{}\emptyset$.

[b]2.    $ 0\leq a<b$[/b] By using a similar argument to the above, we have that $ x\in A \forall x>0$. Now if $ x\in A$ for some negative $ x$, then for all positive $ y$, $ x\plus{}y\in A$, i.e. $ z\in A\forall z>x$. Then for any negative $ c$, there is a positive rational number $ p$ such that $ pc\in(x,0)$, which implies that $ c\in A$. Then $ 0\equal{}x\plus{}(\minus{}x)\in A$, so that $ A\equal{}\mathbb{R}$, and $ B\equal{}\emptyset$. Otherwise, $ d\in B\forall d<0$. This gives the solutions $ A\equal{}\mathbb{R}^{\plus{}}$, $ B\equal{}\mathbb{R}^{\minus{}}\cup{0}$ and $ B\equal{}\mathbb{R}^{\minus{}}$, $ A\equal{}\mathbb{R}^{\plus{}}\cup{0}$.

[b]3.    $ a<b\leq0$[/b] Similarly, we obtain the solutions $ A\equal{}\mathbb{R}^{\minus{}}$, $ B\equal{}\mathbb{R}^{\plus{}}\cup{0}$ and $ A\equal{}\mathbb{R}^{\minus{}}\cup{0}$, $ B\equal{}\mathbb{R}^{\plus{}}$.[/hide]


What do you think? I didn't put this in the game forum because this is no joke. Did you even think that this was physically possible, that is hurdling a seven footer? I mean this is against one of the best teams in the world except they to didn't know that Vince Carter could violate the laws of physics?

[youtube]xp6qJ7LZNlg[/youtube]

uh, there's a good number of people that can jump that high (or higher), and carter's an amazing player, so i'm not surprised that he's able to jump that high. I dont' get why it's so surprising, a guy that practices basketball all day for years is able to do amazing things?

a lot people can jump high, including me(34 in vert) but doing it in a game at the world level is different , hurdling guy 7'2'' is nothing short of awesome. Also it's Carter, not carter.

[quote="mdk"]a lot people can jump high, including me(34 in vert) but doing it in a game at the world level is different , hurdling guy 7'2'' is nothing short of awesome.[/quote]

ok, yea, but he's a guy practices for these types of condition, so "game at world level" for him isn't any sort of barrier that stops him from performing what he can normally do anyways

[quote] Also it's Carter, not carter.[/quote]

You typed the exact same thing twice..? what are you tryign to say

since when do people practice jumping over seven footers from like 9 feet out. Do you understand that this has been dubbed the best dunk of all time?

just cause he doesn't practice that particular jump doesn't mean he doesn't practice jumping or work on atheltic ability in general, a lot of which translates twoards the ability to jump from 9 feet out over hagrid the giant

ok its the best dunk of all time. thats nice. i just dont think it's that hard to believe.

that's probably because it's been 7 years but when it first happened everyone was like WOW!


linear algebra
matrix
algebra
polynomial
induction
linear algebra unsolved

Prove that for every positive integer n there exist a square matrix $ A$ nxn such that $ A^3 \equal{} A \plus{} I$. Prove also that if $ A^3 \equal{} A \plus{} I$, then detA>0.

Over what?  $ \mathbb{Z}$?  (False when $ n \equal{} 1, 2$.)  $ \mathbb{R}$?

I think that $ A\in \mathcal{M}_n(\mathbb{R})$.
You can take $ A\equal{}aI_n$ , where $ a$ is the real root of the equation $ a^3\minus{}a\minus{}1\equal{}0$, such that $ A^3\equal{}A\plus{}I$
$ A^3 \equal{} A \plus{} I_n$ $ \Rightarrow$ $ A^4 \minus{} A^2 \plus{} I_n \equal{} A \plus{} I_n \equal{} A^3$.But $ A^4 \minus{} A^2 \plus{} I_n \equal{} \left(A^2 \minus{} \frac {1}{2}I_n\right)^2 \plus{} \left(\frac {\sqrt {3}}{2} I_n\right)^2$. And now, use [url=http://www.mathlinks.ro/viewtopic.php?t=174477]this[/url].The fact that $ \det A \neq 0$ follows from $ A(A^2 \minus{} I) \equal{} I$.

Another possible approach could be the following:
The polynomial $ X^3 \minus{} X \minus{} 1$ doesn't have any real root in the interval $ ( \minus{} \infty,0]$, so the eigenvalues of $ A$ must be strictly positive if they are real.Since the characteristic polynomial of $ A$ has real coefficients it follows that the eigenvalues of $ A$ are positive reals or if they are not real they must be complex conjugates.Therefore
\[ \det A \equal{} \prod_{k \equal{} 1}^p \lambda_k \overline{\lambda_k} \cdot \prod_{k \equal{} 1}^{n \minus{} 2p}x_k
\]
where $ \lambda_k$ and $ \overline{\lambda_k}$ are the complex eigenvalues of $ A$ and $ x_k$ are the real eigenvalues of $ A$.Since $ \lambda_k \cdot \overline{\lambda_k} > 0$ and $ x_k > 0$ the conclusion follows.

Sorry I didn't specify, the matrix is over reals. My aproach was induction for first part and second part as well. It is not difficult but I just wanted to see different solutions since I just started to learn liner algebra at university.


Let $ a,b,c,d>0$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$.Prove that:
$ \frac{ab}{a\plus{}b\plus{}cd}\plus{}\frac{bc}{b\plus{}c\plus{}da}\plus{}\frac{cd}{c\plus{}d\plus{}ab}\plus{}\frac{da}{d\plus{}a\plus{}bc} \le \frac{4}{3abcd}$

[quote="CSS-MU"]Let $ a,b,c,d > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ \frac {ab}{a \plus{} b \plus{} cd} \plus{} \frac {bc}{b \plus{} c \plus{} da} \plus{} \frac {cd}{c \plus{} d \plus{} ab} \plus{} \frac {da}{d \plus{} a \plus{} bc} \le \frac {4}{3abcd}$[/quote]
Ít can be solved easyli by AM-GM.
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$ so we have:
$ ab\plus{}bc\plus{}cd\plus{}ca \le 4$ and $ abcd \le 1$
$ \sum \frac{ab}{a\plus{}b\plus{}cd} \le \sum \frac{ab}{3\sqrt[3]{abcd}} \le \frac{ab\plus{}bc\plus{}ca\plus{}da}{3abcd} \le \frac{4}{3abcd}$

[quote="mathstarofvn"][quote="CSS-MU"]Let $ a,b,c,d > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ \frac {ab}{a \plus{} b \plus{} cd} \plus{} \frac {bc}{b \plus{} c \plus{} da} \plus{} \frac {cd}{c \plus{} d \plus{} ab} \plus{} \frac {da}{d \plus{} a \plus{} bc} \le \frac {4}{3abcd}$[/quote]
Ít can be solved easyli by AM-GM.
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$ so we have:
$ ab \plus{} bc \plus{} cd \plus{} ca \le 4$ and $ abcd \le 1$
$ \sum \frac {ab}{a \plus{} b \plus{} cd} \le \sum \frac {ab}{3\sqrt [3]{abcd}} \le \frac {ab \plus{} bc \plus{} ca \plus{} da}{3abcd} \le \frac {4}{3abcd}$[/quote]
Correct.Another solution? :wink:

My solution:
By CS, we have:
 $ \frac {1}{a + b + cd} \leq \frac {1}{9}(\frac {1}{a} + \frac {1}{b} + \frac {1}{cd})$
So that:
 $ \sum\frac {ab}{a + b + cd} \leq \frac {1}{9}[2(a + b + c + d) + \sum\frac {ab}{cd}] = \frac {1}{9}({\frac {2abcd(a + b + c + d) + \sum a^2b^2}{abcd})}$
By AM-GM and CS:
 $ abcd(a + b + c + d) \leq \sqrt {4(a^2 + b^2 + c^2 + d^2)} = 4$
 $ \sum a^2b^2 \leq \frac {(a^2 + b^2 + c^2 + d^2)^2}{4} = 4$
 $ \Rightarrow \frac {1}{9}(\frac {2abcd(a + b + c + d) + \sum a^2b^2}{abcd}) \leq \frac {4}{3abcd}$