function
algebra
domain
system of equations

Solve the system of equation:
$ \left\{\begin{array}{c}xx(y^2-4y+5)=6y(z^2-4z+5)=6z(x^2-4x+5)=6\end{array}\right$

Here's what I have (it agrees with rue):
[hide]The question can be rephrased as:
 $ \{\begin{array}{c}x(y - 2)^2 = \frac {6}{x} - 1 \
 \
(z - 2)^2 = \frac {6}{y} - 1 \
 \
(x - 2)^2 = \frac {6}{z} - 1\end{array}$

Note that $ x, y,$ and $ z$ are $ > 0$ and $ \le 6$ to make the RHS nonnegative since the LHS is always nonnegative.

We claim the only such values for the solution $ (x, y, z)$ are $ (3, 3, 3)$. Substituting this solution works, since $ 1 = 1$ is true.

Assume there are other values. Assume one of the variable, WLOG $ x$, is greater than $ 3$. This makes $ \frac {6}{x} - 1 < 1$, so $ (y - 2)^2 < 1 \Rightarrow y < 3$. Then $ \frac {6}{y} - 1 > 1$ and $ (z - 1)^2 > 1 \Rightarrow z > 3$. Finally, $ \frac {6}{z} - 1 < 1$ so $ (x - 2)^2 < 1 \Rightarrow x < 3$. But we assumed $ x > 3$, so we reached a contradiction.

Thus, no variables can be greater than 3.

Now assume one of the variable, WLOG $ x$ again, is less than 3. This makes $ y > 3$, $ z < 3$, and $ x > 3$ if we repeat the process. Again, we reached a contradiction as we assumed $ x < 3$ but we reached $ x > 3$. Thus no variables can be less than 3.

Therefore, the only solutions $ (x, y, z)$ are $ \boxed{(3, 3, 3,)}$.[/hide]

We can generalize and ask when the solution to the system

$ x \equal{} f(y)$
$ y \equal{} f(z)$
$ z \equal{} f(x)$

is $ x \equal{} y \equal{} z$ (equals a fixed point of $ f(x)$).  Here, $ f(x) \equal{} \frac{6}{x^2 \minus{} 4x \plus{} 5}$.  I remember a thread that addressed the general question, but I can't find it :|

t0rajir0u, here is the generalization of this problem.

[u]Theorem.[/u] Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be a continuous and decreasing function with $ x \equal{} f(y),y \equal{} f(z),z \equal{} f(x)$.  Then it follows that there exists one unique solution $ (x,y,z) \equal{} \{x_0\}^3$, where $ x_0$ is the fixed point of $ f$.

[u]Proof.[/u] Since $ f$ is decreasing, it is clear that $ f(x) \equal{} x$ has exactly one solution $ x_0$ (with the use of the intermediate value theorem).  If $ (x,y,z)$ is a solution of the given system, then $ f(f(f(x))) \equal{} x$.  But $ f\circ f\circ f$ is also decreasing and continuous.  Hence, there exists a unique fixed point for $ f\circ f\circ f$, which is $ x_0$.  Hence, $ x \equal{} y \equal{} z \equal{} x_0$. $ \Box$

This theorem (I believe it is correct, but inform me of any inaccuracies) can be applied to the problem by finding the fixed point of $ f(x) \equal{} \frac {6}{x^2 \minus{} 4x \plus{} 5}$ (I believe we have to partition the domain of $ f$), which is $ x \equal{} 3$.  Hence, $ (x,y,z) \equal{} (3,3,3)$.

Hopefully, the above findings are correct.

May I ask, what is the name of this theorem?

I do not think that this has an actual name because I just created it. :P However, as t0rajir0u wrote in his/her post, the system of equations has been studied.


function
limit
algebra
polynomial
functional equation
algebra proposed

find all function $ f: R \to R$ that $ \forall x,y \in R$:
\[ f(x\plus{}y^n)\equal{}f(x)\plus{}(f(y))^n\]
$ ( n\geq 2$

1/ $ x \leftarrow 0, y \leftarrow 0$ gives $ f(0)^n \equal{} 0$ or $ f(0) \equal{} 0$.

2/ $ x \leftarrow 0$ gives $ f(y^n) \equal{} [f(y)]^n$. Hence, we have $ f(x\plus{}y^n) \equal{} f(x) \plus{} f(y^n)$.

3/ If $ n$ is even, we deduce that $ f(x)$ is monotone increasing. Also, for any $ y>0$, we have $ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$ [1]. Therefore, solving this Cauchy functional equation gives $ f(x) \equal{} x$ for all $ x,y \in R^\plus{}$. Then base on the fact that $ f(y^n) \equal{} [f(y)]^n$, we deduce $ f(\minus{}y) \equal{} \pm f(y) \equal{} \pm y$ for all $ y>0$. Combining with [1] to arrive at the only solution $ f(x) \equiv x$

4/ When $ n$ is odd, we deduce $ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$ directly. Then letting $ h \equal{} y^n \rightarrow 0$, we have $ \lim_{h\rightarrow 0} f(x\plus{}h) \equal{} f(x)$ or $ f(x)$ is continuous. Again, Cauchy functional equation gives $ f(x) \equiv x$.

DONE! The only solution is $ f(x) \equiv x$.

[quote="mathangel"]1/ $ x \leftarrow 0, y \leftarrow 0$ gives $ f(0)^n \equal{} 0$ or $ f(0) \equal{} 0$.

2/ $ x \leftarrow 0$ gives $ f(y^n) \equal{} [f(y)]^n$. Hence, we have $ f(x \plus{} y^n) \equal{} f(x) \plus{} f(y^n)$.

3/ If $ n$ is even, we deduce that $ f(x)$ is monotone increasing. Also, for any $ y > 0$, we have $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ [1]. Therefore, solving this Cauchy functional equation gives $ f(x) \equal{} x$ for all $ x,y \in R^ \plus{}$. Then base on the fact that $ f(y^n) \equal{} [f(y)]^n$, we deduce $ f( \minus{} y) \equal{} \pm f(y) \equal{} \pm y$ for all $ y > 0$. Combining with [1] to arrive at the only solution $ f(x) \equiv x$

4/ When $ n$ is odd, we deduce $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ directly. Then letting $ h \equal{} y^n \rightarrow 0$, we have $ \lim_{h\rightarrow 0} f(x \plus{} h) \equal{} f(x)$ or $ f(x)$ is continuous. Again, Cauchy functional equation gives $ f(x) \equiv x$.

DONE! The only solution is $ f(x) \equiv x$.[/quote]
Wrong! Notice,if $ f(x)\equal{}\minus{}x$ is also true
I had a proof for it ,but sorry,I cant post it,I takes me 7 pages to write (at home) so I am sorry
I am waiting a short proof( the cases $ n\equal{}2,3,4$ are quite easy,but with $ n\ge 4$  :(

I believe mathangel's solution is rigorous up to 3/. I kinda doubt about the argument to lead continuity in 4/ because $ f(y)>1$ may lead $ f(y^n),f(y^{n^2}),...$ to explosion. Before getting an explanation from mathangel. I am thinking another way to handle odd $ n$.
To sum up, we have, for odd $ n$.
(1) $ f(x^n)\equal{}(f(x))^n$.
(2) $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.
From (1), we have $ f(1)$ solves $ x^n\equal{}x$. So we have either $ f(1)\equal{}1$ or $ f(1)\equal{}\minus{}1$. For the latter case, we replace $ f(x)$ with $ \minus{}f(x)$ and see (1) and (2) still hold. So we assume $ f(1)\equal{}1$ from now on.

Let $ x$ fixed and $ y\equal{}f(x)$ and $ m$ a big integer.
$ f((mx\plus{}1)^n)\equal{}(f(mx\plus{}1))^n\equal{}(my\plus{}1)^n\equal{}m^ny^n\plus{}1\plus{}\sum_{i\equal{}1}^{n\minus{}1}m^iy^i$.
Also
$ f((mx\plus{}1)^n)\equal{}f(\sum_{i\equal{}0}^nm^ix^i)\equal{}\sum_{i\equal{}0}^nf(m^ix^i)\equal{}\sum_{i\equal{}0}^nm^if(x^i)\equal{}m^ny^n\plus{}1\plus{}\sum_{i\equal{}1}^{n\minus{}1}m^if(x^i)$.
The RHS of above two formulae should be equal.
$ \sum_{i\equal{}1}^{n\minus{}1}m^i(f(x^i)\minus{}y^i)\equal{}0$.
We can treat the LHS as a polynomial of $ m$ and $ m$ can be any large integer. Hence, the only way to solve it is to let $ f(x^i)\equal{}y^i$ for all $ i\equal{}1,...n\minus{}1$.

So we have $ f(x)\equal{}x$. When $ n$ is odd, we have another solution $ f(x)\equal{}\minus{}x$.


calculus
integration
trigonometry
calculus computations

[color=darkred] Find the recurrent relations for  $ A_n \equal{} \int (1 \plus{} \sin x \plus{} \cos x)^n\ \mathrm {dx}$

and $ a_n \equal{} \int_0^{\frac {\pi}{2}}(1 \plus{} \sin x \plus{} \cos x)^n\ \mathrm {dx}$ , where $ n\in \mathcal N$ .

See   http://www.mathlinks.ro/Forum/viewtopic.php?t=186681  [/color]

[quote="Virgil Nicula"][color=darkred] Find the recurrent relations for  $ A_n \equal{} \int (1 \plus{} \sin x \plus{} \cos x)^n\ \mathrm {dx}$

and $ a_n \equal{} \int_0^{\frac {\pi}{2}}(1 \plus{} \sin x \plus{} \cos x)^n\ \mathrm {dx}$ , where $ n\in \mathcal N$ . [/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] I"ll use the identity $ (1 \plus{} \sin x \plus{} \cos x)^2 \equal{} 2(1 \plus{} \sin x \plus{} \cos x \plus{} \sin x\cos x)$ .

For $ n\in\mathcal N$ , $ A_{n \plus{} 2} \equal{} 2\cdot \int (1 \plus{} \sin x \plus{} \cos x)^n(1 \plus{} \cos x \plus{} \sin x \plus{} \sin x\cos x)\ \mathrm {dx}\implies$

$ \boxed{A_{n \plus{} 2} \equal{} 2A_{n \plus{} 1} \plus{} H_n}$ , where $ H_n \equal{} \int (1 \plus{} \sin x \plus{} \cos x)^n\sin 2x\ \mathrm {dx}\implies$

$ H_n \equal{} \int (1 \plus{} \sin x \plus{} \cos x)^n\left[1 \minus{} (\cos x \minus{} \sin x)^2\right]\ \mathrm {dx}\implies$ $ \boxed {H_n \equal{} A_n \minus{} L_n}$ , where

$ L_n \equal{} \int (1 \plus{} \sin x \plus{} \cos x)^n(\cos x \minus{} \sin x)^2\ \mathrm {dx}$ . Apply the integration by parts :

$ \overline {\underline {\left\|\begin{array}{ccc} u(x) \equal{} \cos x \minus{} \sin x & \implies & u'(x) \equal{} \minus{} (\sin x \plus{} \cos x) \\
 \\
v'(x) \equal{} (1 \plus{} \sin x \plus{} \cos x)^n(\cos x \minus{} \sin x) & \implies & v(x) \equal{} \frac {1}{n \plus{} 1}\cdot (1 \plus{} \sin x \plus{} \cos x)^{n \plus{} 1}\end{array}\right\|}}$

Therefore, $ L_n \equal{} \frac {1}{n \plus{} 1}\cdot (1 \plus{} \sin x \plus{} \cos x)^{n \plus{} 1}(\cos x \minus{} \sin x) \plus{} \frac {1}{n \plus{} 1}\cdot \left(A_{n \plus{} 2} \minus{} A_{n \plus{} 1}\right)$ .

$ \boxed {\begin{array}{c} A_0 \equal{} x \plus{} \mathcal C\ \ ,\ \ A_1 \equal{} x \plus{} \sin x \minus{} \cos x \plus{} \mathcal C \\
 \\
(n \plus{} 2)\cdot A_{n \plus{} 2} \equal{} (2n \plus{} 3)\cdot A_{n \plus{} 1} \plus{} (n \plus{} 1)\cdot A_n \minus{} (1 \plus{} \sin x \plus{} \cos x)^{n \plus{} 1}(\cos x \minus{} \sin x)\ ,\ n\in\mathcal N\end{array}}$ .

Analogously, $ \boxed {\begin{array}{c} a_0 \equal{} \frac {\pi}{2}\ \ ,\ \ a_1 \equal{} 2 \plus{} \frac {\pi}{2} \\
 \\
(n \plus{} 2)\cdot a_{n \plus{} 2} \equal{} (2n\plus{} 3)\cdot a_{n \plus{} 1} \plus{} (n\plus{}1)\cdot a_n \plus{} 2^{n \plus{} 2}\ ,\ n\in\mathcal N\end{array}}$ .[/color]


LaTeX geometry search algorithm articles

I recently downloaded the KSEG Geometry diagram making software posted on may topics across the Forum. Though it does seem interesting, where can I get peoper documentation for extensive use? ALso, how do you include the .seg files made by this program into $LaTex$???

I tried that first of all but haven't found anything useful. Can you plz help???

KSEG documentation: [url]http://www.mit.edu/~ibaran/kseg_help_en.html[/url]

Including pictures in LaTeX: [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php[/url]

Though the documentation you have suggested doesn't have much info on my specific question, through my search, I think I have found an algorithm for converting .SEG files to .EPS which can easily be inserted into LAtex

1) Make files in KSEG and "export it " to the .PNG Format.

2) Using the Advanced Batch Converter I found on net(try google - I don't remember site!!) convert .PNG to .EPS ( actually it has loads of options .jpeg, .bmp  ... really good!!)

Volla!!!!

This really isn't a good way to make an eps file.  In fact, you're killing the advantages of postscript.

[url=http://www.illc.uva.nl/]The Institute for Logic, Language and Computation[/url] of the University of Amsterdam provides some [url=http://www.illc.uva.nl/SupportandInfo/createpostscript.html#win]helpful information[/url].

This is what [url=http://www.mit.edu/~ibaran/kseg_help_en.html]the documentation[/url] meant by "print from KSEG to a .ps file."

--

However, if you prefer just to Export an image, here's what you should do:

Export to image.
PNG and check High Quality
Add this line to your $\LaTeX$ document's preamble (a handy default taken from TeXshop):

[code]\DeclareGraphicsRule{.tif}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}[/code]

Now you can add png files using \includegraphics!

I don't know. Your method doesn't seem to work. Where have I to add your statement , between \usepackage and \begin{document}?? Also, have I to write something instead of 'dirname' and 'basename'???
Thanks!!!
Do you know where I can get a full version, free program similar to the Advanced Batch Converter I got here ![url=http://www.batchconverter.com/]www.batchconverter.com/[/url]. It seems it can very well do the job!

That's probably because I forgot a piece of the code :blush:

Here's the correct code:

[code]\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}[/code]

Yes, anywhere before \begin{document} should work.

Thanks for correcting but it seems it still doesn't work. Check this out:-

[code]\documentclass[11pt]{article}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}
\usepackage{graphicx,fullpage}
\begin{document}
\begin{figure}
	\centering
		\includegraphics{C:/Documents and Settings/a/My Documents/Rushil's Documents/Advanced Maths/FYKOS 19.1/rushil.png}
	\label{Mechanics}
\end{figure}
\end{document}[/code]

It would help if you posted the error message.  "It doesn't work" is not very specific.

Off the top of my head, you might try moving the \DeclareGraphicsRule command to after the graphicx package inclusion.


limit integration Putnam function AMC USA(J)MO USAMO

(1)   Find \[ \lim_{x \longrightarrow \infty} \frac{x^2}{e^x} .  \]
(2)   Find \[ \lim_{k \longrightarrow 0} \frac{1}{k} \int_{0} ^{k} (1 + sin 2x)^{1/x } dx .  \]

The Putnam in 1938 was not what the Putnam has become today. The aims and methods of the test-writing committee have changed quite dramatically, as has the mix of students taking the competition.

No need to defend the Putnam! It's obvious the level changes drastically through the years - as is evident with IMO as well!

By the way, what did you mean by 'changing priorities' of the Putnam committee? Any secret motive behind the Putnam ?  ;)

No, I'm not suggesting secret motives. What I'm suggesting is that the very meaning and purpose of mathematical competition has shifted. One thing you will notice with both this very old Putnam and with very old AHSME's is that the questions on these contests were originally much closer to standard questions that might be asked in a standard course in the curriculum. At first they came closer to asking how well as student had learned what he had been taught, and less how well that student could function when faced with something new and unexpected.

One of the things that happened to both the Putnam and the AMC contests over the years is that they gradually exchanged and absorbed ideas from problem-solving traditions in other parts of the world, notably Eastern Europe.

[quote="Rushil"]Indeed, US has benefited a lot in Math and Physics systems of education as well as research from Eastern Europe and Russia! India, China and other Asian nations , though they haven't 'taught' any systems to US - they have actually 'learnt' from them , have started lately to influence research in US![/quote]

India etc  have not, so far, exerted noticeable influence on the methods and traditions of US research except as an endless supply of manpower with the concomitant effects on the research market.  The US and European models of research and academia are being exported to China, India and the rest as Western-educated PhD's become professors in their home countries.   The US itself is in emulation of the British and German systems of the 19-20th century, with the added American ingredients of industrial-scale funding, R & D, collaboration with business (and political correctness, affirmative action and some other sociological differences, as well).

That's much, much broader than the point I was trying to make. I wasn't talking about educational systems in general, or even about the whole of math education.  I was talking narrowly just about mathematics competitions.

As far as I can tell (and I could be wrong about this), both the Putnam and the AHSME began as homegrown American events, and at first they tended to draw directly from the standard curriculum - as I said above, testing what students were supposed to have learned.

Of course, there have been scholarly competitions for a long time, including the challenge competitions for academic position and reputation that figure so prominently (and sometimes counterproductively) in the 16th and 17th century history of mathematics in Italy and France, and including the legendary Cambridge Tripos. But those would seem to have little to do with the subject at hand.

The Putnam and the USAMO started as home-grown American events, but at some point, the authors and organizers had to become aware of the student mathematical competition systems of certain European countries (Hungary, for instance). That, and the post-WWII and continuing migration of European-educated mathematicians led to some of them becoming prominently involved with these, or newer, competitions.


the sum of the angle measures of any triangle is 180 degres. find the angle measures of a triangle if the second angle measures 10 degres less than twice the first, and the third angle measures 25 degrees more than the second

Ok, we have an algebraic equation.

$ x \plus{} (2x \minus{} 10) \plus{} (2x \plus{} 15) \equal{} 180$

$ 5x \plus{} 5 \equal{} 180$

$ x \equal{} 35$

The angle measures are $ \boxed{35,60,85}$


Let A be the product of the first 2009 primes. Let S be the sum of all the factors of A and N be the number of factors of A. Find the units digit of A+S+N.

The units digit of $ A$ is $ 0$ because $ 2*5 \equal{} 10$.

The units digit of $ S$ is $ 0$ because $ 19 \plus{} 1 \equal{} 20$.

The units digit of $ N$ is $ 2$ because the units digit cycle repeats every 4 powers.

Therefore the units digit of $ A \plus{} S \plus{} N$ is $ 2$.

[quote="12markkram34"]The units digit of $ S$ is $ 0$ because $ 19 \plus{} 1 \equal{} 20$.[/quote]

This I don't understand. Where did 19 come from?

Lol, I don't understand either

It should've been $ (1\plus{}2)(1\plus{}3)(1\plus{}5)$...

There's at least 1 2 in this product(1+3), and there's at least 1 5 in this product (1+19), so the units digit is 2*5=0

That's what the other person said. (1+2)(1+3)... 1+19=20, or a multiple of 10.

That's what I was doing. $ 19\plus{}1\equal{}20$ so the units digit of $ S$ is $ 0$.

I don't understand. Can you explain it more clearly, please?

$ A \equal{} 2\cdot3\cdot5\cdots\equiv10\cdot3\cdot7\cdots\equiv0\pmod{10}$,

$ S \equal{} (1 \plus{} 2)(1 \plus{} 3)(1 \plus{} 5)\cdots(1 \plus{} 19)\cdots\equiv\cdots20\cdots\equiv0\pmod{10}$, and

$ D \equal{} \underbrace{(1 \plus{} 1)(1 \plus{} 1)\cdots(1 \plus{} 1)(1 \plus{} 1)}_{2009} \equal{} 2^{2009} \equal{} 2^{2004}\cdot2^5\equiv2^{2004}\cdot2 \equal{} 2^{2005}$

$ \equiv2^{2001}\equiv\cdots\equiv2^1\equiv2\pmod{10}$, so

$ A \plus{} S \plus{} D\equiv0 \plus{} 0 \plus{} 2\equiv\boxed2\pmod{10}$.

*This stuff works because $ 2^5\equiv2\pmod{10}$.


Let $ ABC$ be a triangle with $ B\equal{}C$.Find $ \max\left( \frac{sinA}{cosB}\right)$

$ \frac {sinA}{cosB} \equal{} \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}\equal{} 2  \cos \frac{A}{2}$ then cause $ \cos x \le 1$ we have he max when $ A\equal{}0$, so the max is 2.

[quote="Virgil Nicula"]2 is sup, no max ![/quote]

if we consider the degenerate triangle we have the max  :lol:  ok...is only the sup -.-


trigonometry
inequalities proposed
inequalities

[quote="xzlbq"]Let ABC be a triangle,prove that:
$ \sum\frac {a}{w_a} \geq \frac {8}{3}\sum\frac {\cos\frac {B}{2}\cos\frac {C}{2}}{\sin{B} \plus{} \sin{C}} \geq \sum\frac {1}{\cos\frac {A}{2}}$
BQ[/quote]

But more strang:

$ \sum\frac {a}{w_a} \geq \sum\frac {(b \plus{} c)r_a}{(w_a \plus{} r_a)w_a} \geq \frac {8}{3}\sum\frac {\cos\frac {B}{2}\cos\frac {C}{2}}{\sin{B} \plus{} \sin{C}} \geq \sum\frac {1}{\cos\frac {A}{2}}$

a side in easy inequeliti:
$ \frac {a}{b \plus{} c} \geq \frac {r_a}{r_a \plus{} w_a}$
to prove is easy.

BQ


let a,b,c are possitive numbers  such as

  a+b+c=1

k is a possitive number

a) prove that:

$ \frac{a\minus{}bc\plus{}k}{a\plus{}bc\plus{}k}$ +$ \frac{b\minus{}ac\plus{}k}{b\plus{}ac\plus{}k}$ +$ \frac{c\minus{}ab\plus{}k}{c\plus{}ab\plus{}k}$ $ \ge$ $ \frac{6\plus{}27k}{4\plus{}9k}$

b)if k is a any real number.what will happen with the innequalities

this problem is false.
-if a=1/2,b=c=1/4,k=0.2 we have LHS<RHS
-if a=1/2,b=c=1/4,k=0.1,we have LHS>RHS

to trannguyenquy:to endeavour.


I think, for this forum, the stuff we are supposed to concentrate on is this..

[quote]Counting:

How many multiples of 12 are there between 100 and 200?

In how many ways can at least one student from a group of four be selected for class honors if any combination of student(s) can be selected?

Probability:

When two fair 6 sides dice are thrown, find the probability that the sum of the rolls is 9.

John flips a fair coin 3 times. Find the probability that exactly 2 of the flips come up heads.

Number Theory:

Find the greatest common divisor of 60 and 105.

Find the least common multiple of 40 and 90[/quote]

rather than geomerty or arithmatic or algebra...

I don't see any reason why arithmetic, algebra, and geometry don't count as Middle School classroom math.  Why do you think the forum should only focus on certain subject areas?

Anyone can do Addition, Subtraction ,Multiplication, Division..

Most people come here, because they want to know a way to solve without using much of addition,subtraction,multiplication, or division on certain problem.

What they want to know is a way to solve problems in a neat and magical ways...

I think the moment we get into what advanced math students call "neat ways to do math," we're talking about posts that should go in the basics.  That's exactly the point of this thread.  This forum is for regular ole' ordinary classroom math.  We're not going to tell students who want to talk about classroom algebra that they have no forum for this.

I'm going to split this thread soon to keep this post relatively stand-alone.


inequalities
geometry
3D geometry
tetrahedron
geometry unsolved

A tetrahedron T' has all its vertices inside the tetrahedron T. Show that the sum of the edge lengths of T' is less than 4/3 times the corresponding sum for T.


function
topology
limit
superior algebra
superior algebra unsolved

Suppose I have the poset $(2^{N}, \subseteq)$ where N is the natural numbers. I define a function from $2^{N}\rightarrow 2^{(2^{N})}$ by saying that $f(x) = 2^{x}$ where $x \in 2^{N}$(so this is the power set function). Show that this is not continous (I use the subset relation on $2^{(2^{N})}$ aswell

It's worth pointing out that the [url=http://en.wikipedia.org/wiki/Order_topology]order topology[/url] is being used.

Ive tried in vain to find a (infinite) chain that doesnt fullfill the continuity property, but I allways end up with something that does.

I suspect that this function is [b]sequentially continuous[/b], that it, $\lim_{n\to\infty}f(\alpha_{n})=f\left(\lim_{n\to\infty}\alpha_{n}\right)$ for any sequence. However, sequential continuity is weaker than continuity unless the space is [url=http://en.wikipedia.org/wiki/First-countable_space]first-countable[/url]. And the space of ordinals is not first countable: consider what happens near the first uncountable ordinal.

Okey thanks. But I believe there is another, more trivial, solution to the problem (we have not talked about those things,first-countable and so on)


A= { (x,y) | x>0 and 0<y<x^2}
show that every straight line through the origin contains an interval around the origin that lies completely outside A.

Every straight line through the origin....
so..y=ax, a in R...
and then????

And then just notice that $|y|=|a||x|\geqslant x^2$ if $|x|<|a|$. This works for $a\neq 0$. But for $a=0$, the entire line lies outside the region and the same is true for the vertical line $x=0$.


What is the area of a pentagon with vertices (in the coordinate plane) (1,1), (2,4), (3,9), (4,16), and (2009,2010)?

we could solve the question by dividing the figure into simpler figures
supposing p(1,1) was A
point (2,4)was B
point (3,9)was C
point (4,16) was D
point (2009,2010) was E

we could form a square with points on the origin,(2009,1)i.e F,(2009,2010)i.e G and (0,2010)i.e H

we know the area of the square (2009^2)
we know the area of tri.AEF (2009*2008/2)

next we draw a perpendicular from points A,B,C,D to the y axis
 finding the area of each divided portion should be easy but the outline is to make a rectangle and subtract the extra triangle (i could explain that in further detail if needed)
 the areas are-3.5,12.5,24.5
the figure we are left with could be divided into a rectangle and a triangle whose areas are (4*1994) and (2005*1994/2)

by subtracting the areas of tri.AEF,the divided portions, the rectangle and the triangles from the area of the square we get-

4036081-2017036-3.5-12.5-24.5-7976-1998985=[b]6043.5[/b]
am i right?

[quote="Turk"]Use Shoelace theorem...[/quote]

You need to determine the order of the points first - if they are in the given order, then it is not a pentagon :)

could someone tell me what the shoelace theorem is
i've never heard of it before :huh:


function
algebra
polynomial
calculus
integration
advanced fields
advanced fields unsolved

I've got a problem concerning Stone-Weierstrass' theorem (applied to continuous complex functions on the unit circle) - how to show that if I have an algebra that separates points etc., but doesn't contain the constant function, then it is not dense in C(S^1)? That is, if we don't have the constant function right from the start, we can't produce it in our algebra.

There is no way, it is just false. Take any countable dense set of random continuous functions (define them as you wish; just have some oscillating random noise in each) and generate an agebra by it. There is really no chance of getting the identically 1 function in that algebra but it is certainly dense.

OK, but how do you formally prove for any algebra that if it doesn't contain constant functions, then they are not in the closure of that algebra?

fedja already told you that this is wrong.

OK.

What if the algebra is generated by a single function, i. e. it consists of all "polynomials" $ \sum a_{n}f^{n}$?
(of course $ f$ must be nonvanishing)

Polynomials with zero free term are dense in $ C[1,2]$.

Is the same true for complex polynomials on the unit circle?

No. That's because the integral of any complex polynomial over the circle is $ 0$ and this integral is a continuous functional on $ C(\mathbb T)$.


logarithms
real analysis
real analysis unsolved

Proof that
$
\sum\limits_{n = 0}^\infty  {\frac{{\left( {4n} \right)!}}{{\left( {2n + 1} \right)!\left( {n!} \right)^2 2^{6n + 1} }} = \frac{{2\ln \left( {1 + \sqrt 2 } \right)}}{\pi }}
$

ok
you are the best!!!!!
We can see now the solution?

An hint , starting with
$\sqrt {{\frac {1-\sqrt {1-x}}{x}}}=\frac{1}{\sqrt {2}}\sum _{n=0}^{\infty }
{\frac {{4\,n\choose 2\,n}{x}^{n}}{{16}^{n} \left( 2\,n+1 \right) }}
$
put $x=sin(t)^2$, integrate from $0$ to $\pi/2$ and use Wallis

Please remark that
$\frac{2\ln(1+\sqrt(2))}{\pi}=\frac{arcsinh(1)}{\arcsin(1)}$


inequalities
triangle inequality
real analysis
real analysis unsolved

Let $ p,q\in \langle 0,1\rangle, p\plus{}q\equal{}1$, then \[ \|px\plus{}qy\|^2\leq p\|x\|^2\plus{}q\|y\|^2,\] for all $ x,y\in X$ ( $ X$ is a normed space ).

:lol:
or since  norma is convex down (hence $ \parallel{}.\parallel{}^{2}$ is also convex down (since $ \parallel{}.\parallel{} \geq 0$).) function.

there is mistake when using triangular inequality it will appear $ p^4$llxll$ ^2$

where is mistake? in your proof here is no mistake .


Given a graph of n nodes that are all interconnected with weighted edges, what's the best way to find the minimal subtree that contains m nodes?

I've found two ways to approach the problem.
1. Generate all combinations of the nodes to connect and attempt to connect these nodes via either Kruskal's or Prim's algorithm while disregarding the rest. Compare all trees created and find the smallest one.
2. For each node, create a subtree via Prim's algorithm that originates from this node and stop the algorithm after m nodes have been found.

Obviously, the second approach is faster, but does it always produce the correct answer?

It seems like the second algorithm doesn't quite work because expensive edges will block Prim even if there are cheap edges that can be reached once the expensive edge is chosen. Consider the graph

[code]
A     B
 \   /
   C
   |
   D
   |
   E
   |
   F
 /   \
G     H[/code]

where the weights are AC = BC = FG = FH = 3, CD = EF = 1, and DE = 4. Suppose we want to find the smallest subtree with 4 nodes. Whichever node you start from, you'll never choose DE because it has the highest weight and you can get subtrees of size 4 on either side of DE. On the other hand, CDEF is the best subtree (weight 6).

It's clear that the subtree we want will be a subgraph of some MST of the original graph, but I don't see any easy way of finding it.


Here are some induction problems that I had difficulty solving... Hope you can help me. Ive done many easy induction, but the following are quite hard.
1. $1^{3}+3^{3}+5^{3}+\ldots+{(2n - 1)}^{3}= n^{2}(2n^{2}- 1)$
2. $|a_{1}+a_{2}+a_{3}+\ldots+a_{n}| \leq |a_{1}|+|a_{2}|+|a_{3}|+\ldots+|a_{n}|$
3. Prove that $x^{2n-1}+y^{2n-1}$ is divisible by $x+y$.
4. Prove that $n^{3}+2n$ is divisible by $3$.
5. Prove that $(cos \theta+n\pi) = (-1)^{n}cos \theta$ for all $n \in N$.
6. Prove that $10^{n}+3 (4^{n+2})$ is divisible by $9$.

By the way, why is my minus sign replaced with an a with cap?

Note that you are raising the whole of $(2n-1)$ to the $3$. You should type ${(2n-1)}^{3}$. It may not be that big a deal though.

Hope this helps.

[quote="kimby_102"]i think there's an error on #4; consider $n=2$.[/hide][/quote]
Thanks for informing me... That should have been $n^{3}$. Anyway, I have already edited it. :)

[quote="guile"]
4. Prove that $n^{3}+2n$ is divisible by $3$.
[/quote]

$n^{3}+2n=n^{3}-n+3n=n(n-1)(n+1)+3n$
this is clearly divisible by 3 since the first part is 3 consecutive the product of 3 consecutive integers, and multiples of 3 occur every 3 consecutive integers, the other term is also divisible by 3 so the whole thing is divisible by 3

Some of these would be easier without induction =/

#5 is quite easy with any method.  [hide="Hint on the induction"] The inductive step is one use of the cosine addition formula. [/hide]

[quote="Franz Joseph"][quote="guile"]
6. Prove that $10^{n}+3 (4^{n+2})$ is divisible by $9$.

[/quote]

I don't think this holds for $n=1$.[/quote]

Well, it never holds  :lol:  ($10^{n}+3(4^{n+2}) \equiv 1 \mod{3}$...)

I guess there's a typo again.

[quote="guile"]3. Prove that $x^{2n-1}+y^{2n-1}$ is divisible by $x+y$[/quote]

[hide="Possible method"] Try to find a recurrence for $a_{n}= x^{2n-1}+y^{2n-1}$.  ;) [/hide]


MATHCOUNTS
probability
geometry
geometric transformation
reflection

Post here your user name, state you will be competing for, your state competition score, and grade.

I will start with the 9 that I found on the original state results thread.  It would be cool to get at least 10% (are there about 250 competitors at Nationals?) of the National competitors to log on and record their name & score.  If I make a mistake let me know & I will edit this post.

I also think it would be cool to have some way to show that we are an "AOPSer" when we get to the competition.  Think of ideas and post them here.  We can decide closer to the date what we will do.

Chinaboy, Oklahoma, 41, 8th
yifman12, Conneticut, 40, 8th
dark_shadow, California, 42
cold entree, North Dakota, 39
Fermatprime, Georgia, 39, 8th
topper, Nebraska, 35, 7th  I'm the bottom dweller so far!!!
ml2, Washington, 42
justian, Washington, 40
captcha000, Georgia, 38, 8th

228 people in nationals. and if you wanted to know, i'm in 8th grade.
So, does anyone have a state with a cool pin design?

Pin designs?? Last year, our pins were express-mailed to us and got to us in time...for the awards banquet that saturday night...this year I know the Georgia pins have arrived, our coach has them, but we haven't seen them yet.

BTW me and Fermatprime are both 8th graders. Any ideas how we can recognize our fellow AoPSers??

Sure. The question is, do we want it to be obvious to everyone, or an "inside" thing?

[quote="yif man12"]good point. the only thing is, my state had the worst pins every last year. anyone in nationals last year remember connecticut pins? By the way, did georgia even have pins last year?[/quote]

Long story. As captcha said, though, we did get them express-mailed to us in time for the awards banquet (where we tried to trade them with everyone else...ugh.)

Basically, we ended up getting tons of extra Hawaii pins (they had a surplus or something) and trying to trade them to the 5 people who still hadn't gotten theirs. It was a real mess...

so that's what happened with georgia. wait until you see the connecticut pins from last year. most pathetic stuff ever. i just went around trading to everybody, which killed the value of our pins. that got me a beating from my teammates

I just went and fetched my "necklace" from last year. "Conneticut: We're full of surprises." lol

[quote="yif man12"]Yup. We're an image of patheticy (is that a word?)[/quote]

The noun related to the adjective "pathetic" is "pathos." This root is from Greek, so the word formation looks a little strange to English eyes.

But I would hardly call it pathetic to make it to Mathcounts nationals.

LOL. Oh well, at least you had them (and it had writing.)

[quote="captcha000"]  BTW me and Fermatprime are both 8th graders. Any ideas how we can recognize our fellow AoPSers??[/quote]

Well, if you are all willing to look extremely stupid, i'll post my idea

[quote="yif man12"]hey, i already look stupid. a little more stupidity can't hurt. we could also just agree on a place to meet. i wonder where the evening event will be this year. last year, my teammates were plotting to throw me into the tank with the beluga whale[/quote]

last last year my teammates went around blowing spit bubbles and trying to find hot girls to hit on.  beluga whale?  in chicago?  oh wait, the shedd aquarium?

[quote="yif man12"]what state did you make it from? and are you gonna be here this year?[/quote]

Cali 2002, and seeing as i'm two years over the limit, prolly, no.  and if you hadn't already guessed, i'm a girl.

IL, 45, 8th grade

Maybe we could all wear our pins a certain way to distinguish AoPSers.

So that means that you had a higher score than Greg at states probability 1.01.  Good job.  Were you the state champ?

[quote]I will start with the 9 that I found on the original state results thread. It would be cool to get at least 10% (are there about 250 competitors at Nationals?) of the National competitors to log on and record their name & score. If I make a mistake let me know & I will edit this post. [/quote]

You missed me.

State: IN
Score: 41 (would have gotten a 43 if I knew how many letters were in the alphabet.  :-x )

Sorry, I guess I posted in the wrong place.  It shows that I'm new here.  I got a 45 in the state competition in Nevada.  2nd place, I believe, got a 43.

Scott

[quote="probability1.01"]Yeah, but as I have said before, state doesn't reflect nationals.  I haven't practiced at all yet...  :oops:.  I suppose I can procrastinate until after the USAMO.[/quote]
that's how it's like in my state. It's kinda like getting first place is a curse, especially if you get first place after countdown *coughJEFFcough*. The funny thing is, the top scorer in nationals from connecticut the last two years has gotten 4th at states.
Congradulations to everyone who made it to nationals.

For the past two years, 4th place in OK did the best in written at nats.

Hehe.

Oh, that would have been my 8th grade year.  I was the girl hawking "special edition flag pins" that my grandmother just happened to have left over from something or other.

[quote="rand()%0"]heh, when I did nationals from Missouri 4 years ago (in DC) we traded the little CNA pins they gave everyone, by calling ourselves the Communists of North America, and people actually believed they were getting a rare pin.[/quote]

I don't really want to tell you guys where exactly I'm from cause I'm really not all that smart when compare with you guys. But I'm from one of the midwest state.
Last year when I was in Chicago, I did so horrible that it would be a disgrace 4 me to tell you where I'm from.

Really, it's no disgrace.  IL is supposed to be good, and I only got 50th last year :-(.  I think I can do better this time around though.

[quote="ComplexZeta"]If you don't want people to know where you are from, you shouldn't say in your location.[/quote]

lol that's true...

don't be ashamed at all. just remember, there's tons of kids out there who would like your score.

hey uh probability, 50th place isnt that bad ... ur PLACING was less than half of mine...I heh..didnt reach the halfway mark

yeah. at least your name was on that list thingy of the top 57.

Hey, just getting to nationals is an accomplishment... and keep in mind that you can still be very successful in math without winning Mathcounts...

I guess so, but still.
Anyway how can you say that you're bad when you place 50th last year.
I was more like in the bottom 10%.
I know, disgraceful, disgraceful, disgraceful, disgraceful, disgraceful, disgraceful, disgraceful.................... :(

I didn't do mathcounts in middle school but I can still understand.  Just give it everything that you have.  Being nervous can be a good thing.   It keeps pressure on you and most of the time you do better than at normal times.  Just try not to be too nervous because that can mess you up big time.  I'm sure that you'll do well but if you don't there will always be the next thing.

[quote="k.lee"]Thanks for trying to cheer me up.
The national in washington DC is getting REAL close and i'm so scared that i will choke and let down my teammates.
At school, i'm like the smartest kid. But i'm with the most intelligent kids in America, i'm feel really stupid sometime.[/quote]
don't worry. i let down my team last year on the team round. i got two easy problems wrong including the first problem! our team ended up with 7 right on the team round and got 19th place as a team. if i didn't screw up, who knows how well we might've done? But did my teammates beat me up? no. and i still ended up having a good time. feeling stupid is natural. i felt stupid for the entire weekend at nationals last year. but still, it was fun.

[quote="Rep123max"]Trust me, nothing is worse than in 7th grade getting 29th at Texas without working and in 8th grade choking after practicing the whole year and getting 29th again with a score of 37 compared to a 7th grade score of 38. Just going to nationals is a true accomplishment.[/quote]

Wanna bet? How about getting told in 6th grade (or was it 7th? Help me out here, captcha.) that you were in 2nd place and on the team, then retracting that score and informing you that you were 5th place, and then placing 5th once again the next year?

I really do know someone that happened to...

[quote="captcha000"]I heh...didn't reach the halfway mark[/quote]

Wow. You're still not exactly shouting your placement last year from the rooftops, but after Nats last year (actually, about last week even) I thought I could just mention that number around you and you'd start acting all weird...

Which reminds me, I need to practice so I can do better than 46th...
(btw, captcha, if you're reading this, SSM) ;)

*throws piece of crumpled-up paper out 20th story window and it hits captcha in head*

Did anything ever come of the plan for everyone to recognize each other as AoPSers?

I was thinking that we could all maybe wear the TI pins or the CNA (lol communists of north america) pins since most people end up wearing those around their neck instead of on their shirt...?? i dunno btw the way fermat, i have nooooo clue wut u r talking about...

it'd be cool if they had an AoPS pin there because Mr. Rusczyk will be doing the broadcasting again.


vector
geometry
geometric transformation
group theory
abstract algebra
linear algebra
matrix

let E a finite vector space and L(E) the vector space of the linear operators from E to E. I'm looking for a prime ideal  I of L(E)... i don't know the exact translation but
I is: (*) a subgroup of  ( L(E) , +)
       (*) if v is in I and a is in L(E) thsn a*v is in I
       (*) if a*b is in I then a OR b is in I

any idea..

There aren't any, because there aren't any nontrivial two-sided ideals.

Consider any matrix A.

We show that the two-sided ideal containing A contains the identity. This can eb done by row reducing A. If A is nonsingular, no problem. If A is singular, we can add to the echelon form of A certain matrices obtained through row and column operations, so that the sum is nonsingular. Then we are done.

Matrices in fact are an example of a class of algebras nown as central simple F - algebras i.e. algebras with no proper nontrivial twosided ideals and whose centre is isomorphic to the field F over which they are constructed.

Let $I$ be a nonzero bilateral ideal of $\mathcal L(E)$.
Consider in $I$ a nonzero $n\times n$ matrix $A$ with coefficients in the field $k$ over which $E$ is a vector space.
Let $r$ be the rank of the matrix $A$.
There exist matrices $L$ and $R$ such that $LAR=\left(\begin{array}{cc}I_r&0_{r,n-r}\\0_{n-r,r}&0_{n-r}\end{array}\right)\in I$.
Let $D$ be the diagonal matrix with diagonal entries: \[ D_{i,i}=\left\{\begin{array}{cc}1,&\mbox{if }i=1\\ -1,&\mbox{if }i=\overline{2,r}\\0,&\mbox{if }i=\overline{r+1,n}\end{array}\right. \]
Then $I\ni LAR+LARD$ is the diagonal matrix with one on the $(1,1)$ entry and 0 in the rest. This matrix is similar to all diagonal matrices having 1 somewhere on the diagonal and 0 everywhere else. A matrix similar to a matrix in $I$ is obviously in $I$, so all the matrices above are in $I$. But then their sum is also in $I$. Their sum is $I_n$, so $I=(I_n)=\mathcal L(E)$.


trigonometry
function
limit
calculus
calculus computations

i'm having lots of trouble with limits of trig functions... here are some of the problems i need help with:

1) lim (as x->0) of (tan(x))/x

2) lim (as x->0) of (sin(2x))/(x)

3) lim (as x->x)  of (sin(x))

4) lim (as x->0) of (1-cos(x))/x

5) lim (as x->0) of (sin(x))/((x^2)+3x)

any help ASAP would be greatly appreciated.. thanks!

Use $\lim_{x\to0} \frac{\sin \left(ax\right)}{x} = a$ for most of them and $\lim_{x\to0} \frac{1-\cos x}{x} =0$.

You should look at a proof of these, but here's a good way to think about it:

As $x\to 0, \ \sin x = O(x)$ meaning that $\sin(x)$ is bounded by $y=x$.  So $\lim_{x\to0}\frac{\sin \left(ax\right)}{x} = \frac{ax}{x} =a$

As $x\to 0, \ \cos(x) = O(1) \ \Rightarrow \lim_{x\to0} \frac{1-\cos x}{x} = \frac{1-1}{x} = 0$

Also these questions should go in Calculus Computations and Tutorials, where you can find a great post (at the top) by Kent Merryfield about Big-O and little-o.

that doesn't really make much sense to me... could you perhaps show me what you mean by doing number 1 for me? my calculus teacher did not explain any bit of trig limits to our class, so i have not even the slightest clue at how to approach these.  :mad:

#1 $\lim_{x\to0} \frac{\tan x}{x} = \lim_{x\to0} \frac{\sin x}{x\cos x} = \lim_{x\to0} \frac{\sin x}{x} \cdot \lim_{x\to0} \frac{1}{\cos x} = 1$

wow that was fast thank you very much.. one thing though is there some theorem or rule that explains why the limit (as x->0) of 1/cos(x) = 1? i know there is a theorem stating the limit (as x->0) of sin(x)/x = 1, but what about 1/cos(x)? thanks!

haha wow thanks that makes me feel stupid... i got so caught up in these theorems and such that i completely forgot the most simple thing  :blush:

P.S. - is there any work to be shown to prove that the answer for the limit (as x->infinity) of sin(x) is that it oscillates between 1 and -1 (besides a graph)?

Think about $\sin x$ in the unit circle (and the fact that $\sin x$ is periodic) , $-1\le\sin x\le 1$.  So $\lim_{x\to\infty} \sin x$ does not exist.

alright thanks! with your help i managed to get through pretty much the whole assignment except for 2 problems. i need help finding the answers (and showing work) to:

1) limit (as x->infinity) of x*sin(1/x)

and #2 is multiple choice:

2) Suppose a function [i]f[/i] is defined for all real numbers. If, for any positive number ε, there exists a positive number δ such that |f(x) - 3| < ε whenever 0 < |x - 1| < δ, then it follows that:

A. limit (as x->1) of f(x) = 3
B. limit (as x->3) of f(x) = 1
C. limit (as x->0) of f(x) = 3
D. limit (as x->infinity) of f(x) = 3
E. limit (as x->0) of f(x) = 0

once again, thanks ahead of time for any and all who can help me  :D

$\lim_{x\to\infty} x\sin\frac{1}{x} = \lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}} = \lim_{x\to0} \frac{\sin x}{x} = 1$

or

As $x\to\infty$, $\sin\frac{1}{x}$ is the same as $\frac{1}{x}$ and then we have $\lim_{x\to\infty} \frac{x}{x} =1$

The second question is just the definition of a limit $\lim_{x\to1} f(x) = 3$

haha whoa you're amazing... thanks a bunch. as you can tell my high school calc teacher doesn't teach me very much... thanks for all the help though  :D


Please do solve it Peter!!! It looks easy for me  :)

well in fact I was awaiting stryker to answer so I could ask another question, the sum is then A/(A+B) + A/(A+C) + B/(B+A) + B/(B+C) + C/(C+A) + C/(C+B) right?

This may look obvious, but I never saw an alike sum notation before I came to this site

anyway, supposing it is, I'll try to solve it.

hm no that's not possible, then it's just plain 3

so it should be A/(A+B) + B/(B+C) + C/(C+A)  ?

could someone correct me if I'm wrong...

well then, looks wrong too, or at least very easy: between 2 and 1..

The sum is A/(A + B) + B/(B + C) + C/(C + A) and it looks easy indeed.

Sorry, it is a / (a+b) + b / (b+c) + c / (c+a).

Namdung is right; the constants are not so obvious.

Hmm, I thought I moved this to the "Unsolved Problems" section -.-, because I have not solved it yet. ^^;

[quote="Namdung"]Hi all,

I think is  S = a/(a+b) + b/(b+c) + c/(c+a). But this question is not so obvious as you said. Note that a, b, c are sides of triangle. We can prove that  4/3 < S < 5/3, but 4/3 and 5/3 are far of the best constants. So, let's try, friends. I have some better approximation but I dont have the exact sln.

Namdung[/quote]

oh ya they're sides of a triangle :D heh woops :blush:

I found that

1.47777 < Sum a/(a+b) < 1.52223

Can anyone prove this result?

Namdung

In fact, it is (9+8 \sqrt 2-5 \sqrt 5)/6 and (9-8 \sqrt 2+5 \sqrt 5)/6.

Namdung, I think your result can be found by substituting a = x + y, b = y + z, c = z + x and then assuming x + y + z = 1 and letting x = 0 then solving for the minimum / maximum of the resulting function.

I remember getting some number like 1.477 from that method, but you might have had another solution in mind.

wpolly, could you please provide the proof? Your result sounds fascinating.

[quote="StRyKeR"]If a, b, c are sides of a triangle, find the maximum N and minimum M such that

N <  a / (a + b) < M[/quote]
I've found a similar problem. It goes like this :
x,y,z are the lenghts of the sides of a triangle, and let
f(s,y,z)=abs((x-y)/(x+y) + (y-z)/(y+z) + (z-x)/(z+x))
Find the upper limit of f(x,y,z). The answer to this question is (8 \sqrt 2 - 5 \sqrt 5)/3 < 0.04446 which is similar to wpolly's result.


algorithm
continued fraction
number theory unsolved
number theory

Actually i want to know about pell's equation correctly. Different sources states this differently so i am getting puzzled. so i need help from you guys!

Firstly:
$x^2 - ay^2 = 1$
easiest one. we need to express sqrt(a) in continued fraction. and if the period is k then:

p_sk, q_sk where s is any number for k=even number.
p_sk, q_sk where s is even number for k=odd number.

Right?

Secondly:
$x^2 - ay^2 = -1$
we need to express sqrt(a) in continued fraction. and if the period is k then:

p_sk, q_sk where s is odd for k=odd number.
k=even number NO SOLUTION.

Right?

Now tell me what will be in case of $x^2-ay^2=b$ for 1) b>0 and 2)b<0 ?

I will be glad to get response from you! :D

[quote=""mahbub"]p_sk, q_sk where s is any number for k=even number.
p_sk, q_sk where s is even number for k=odd number. [/quote]

what do you mean with p_sk and q_sk?

they r related with continued fraction. better u consult a number theory book for that!

[quote="mahbub"]
Now tell me what will be in case of $x^2-ay^2=b$ for 1) b>0 and 2)b<0 ?[/quote]
1) and 2) are essentially the same.

If equation $x^2-ay^2=b$ has at least one solution, then it has infinitely many of them.

If $b\neq 1$ then this equation has a solution iff equation $x^2-ay^2=-1$ has a solution.

If $(x,y)=(p_0,q_0)$ is one solution of $x^2-ay^2=b$ and $(x_n,y_n)$ are solutions of $x^2-ay^2=1$ then
$p_{n+1}=p_0x_n+aq_0y_n$
$q_{n+1}=p_0y_n+q_0x_n$

Btw. I've never seen that somebody writes p_sk and q_sk for continued fractions.

I've written here (last post) what's sufficient to gain all solutions:
[url=http://www.mathlinks.ro/Forum/topic-40958.html]www.mathlinks.ro/Forum/topic-40958.html[/url]

But it doesn't make a difference...the method is the same for all $t$ when $x^2-by^2=t$

[quote="mahbub"]well actually i was asking to find out all the base solutions for general c (or t). Can u help me?[/quote]
AFAIK, there is no some universal algorithm for that. You must guess one solution and then you can easily find infinitely many of them.

but i can see some books where they r giving algorithomic all the base solutions using continued fraction but i am not sure whether they works correctly. because i tried that with x^2-5y^2=-4 (posted in this site a few days ago) and according to the method it says there is no solution. that's why i am asking here!


Why can I see these class message boards:

Intro to Geometry
Intermediate Trigonometry/Complex Numbers
Olympiad Problem Solving
AMC-10 Problem Series
AIME Problem Series

I'm not in these classes!

No clue, but they show 0 posts/topics to me

A little mistake when we set up the forums - those are for the fall classes; we forgot to make them hidden.  You shouldn't see them anymore.


The value of $ \frac{3}{a\plus{}b}$ when $ a\equal{}4$ and $ b\equal{}\minus{}4$ is:

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ \frac{3}{8} \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ \text{any finite number} \qquad
\textbf{(E)}\ \text{meaningless}$

$ a \plus{} b$ is $ 0$, so we have a zero in the denominator and the expression is undefined.
Answer = $ \boxed{E}$.


real analysis
limit
real analysis unsolved

$ a$ is a point of Lebesgue of $ E$ if $ \lim_{h\to 0}\frac{m(E\cap (x\minus{}h,x\plus{}h))}{2h}\equal{}1$

let $ E$ be a measurable set of $ \mathbb{R}$ and $ 0$ be a  point of Lebesgue  density of $ E$, prove that there is an infinite sequence of points $ (x_{k})_{k\in \mathbb{N}}$  with $ x_{k}\neq 0$ , $ x_{k}\to 0$ as $ k\to \infty$ and $ \{x_{k},\minus{}x_{k}\}\in E$

$ E\subseteq [0,h)$or $ E\subseteq (\minus{}h,0]$ or $ E\subseteq A\cup (\minus{}h,0]\minus{}(\minus{}A),(A\subset[0,h))$the limit will <1. so we can find$ \{\minus{}x_1,x_1\}\subset E$.then we repeat the analysis in $ (\minus{}x_1,x_1)$. the $ X_k$ can be sufficent small, so $ x_k \rightarrow 0$.


this equation looks like it shouldn't be that hard, but I can't figure it out.

z is a complex valued function of a real variable with
$ z^{''}\equal{}\frac{\minus{}z}{|z|^{3}}$
where $ z^{''}$ is the second deriviative of z and |z| is the magnitude of z

find z that satisfies $ z(0)\equal{}a$ and $ z^{'}(0)\equal{}b$

Let's just proove that the solution is hard:

The simple case that z is a real function and positive the equation would be

$ \frac{d^{2}z}{dt^{2}}\plus{}\frac{1}{z^{2}}\equal{}0$

To solve this equation we need to solve the huge below

$ \frac{\left(c_{1}^{3/2}z(t)^{3/2}\plus{}2\sqrt{c_{1}}\sqrt{z(t)}\minus{}2\sinh^{\minus{}1}\left(\frac{\sqrt{c_{1}}\sqrt{z(t)}}{\sqrt{2}}\right)\sqrt{c_{1}z(t)\plus{}2}\right)^{2}}{c_{1}^{3}\left(c_{1}z(t)\plus{}2\right)}\equal{}\left(t\plus{}c_{2}\right)^{2}$

if z is complex, the solution would be complex :rotfl:

the real solution isn't [i]that[/i] hard (at least assuming I did it right)
I got
$ z(t)\equal{}\left(\frac{2(ut\plus{}v)^{2}}{9u^{2}}\right)^{1/3}$

making a substituition of your solution we get $ \frac{1}{\sqrt[3]{\frac{(t u\plus{}v)^{2}}{u^{2}}}}\equal{}0$

And In ODEs we have 2 kind of solutions, A General solution and a partial solution. example.

$ z''''' \equal{} z'''\plus{}z''''\plus{}z''$  have a solution $ z\equal{}a$ with $ a$ constant

oops, it should have been

$ z(t) \equal{}\left(\frac{\minus{}9u^{2}(ut\plus{}v)^{2}}{2}\right)^{1/3}$
I think that works

making a substituition $ \frac{u^{4}\plus{}1}{\sqrt[3]{\minus{}u^{2}(t u\plus{}v)^{2}}}\equal{}0$

the original solution would have 2 constants, but this have only one

it's not general

it does have 2 constants: u and v

I think I've really got it this time
$ z(t) \equal{}\left(\frac{\minus{}2(ut\plus{}v)^{2}}{9u^{2}}\right)^{1/3}$

does anyone know how to solve it when it is complex?


trigonometry
quadratics
algebra unsolved
algebra

[quote="Rust"][quote="Lovasz"]$\tan \cot x = \cot \tan x$
$\iff \tan \cot x = \tan \left(\frac{\pi}{2}-\tan x\right)$
$\iff \cot x = \frac{\pi}{2}-\tan x$
Now solve this quadratic equaton.[/quote]
$\tan \cot x= \tan{(\frac{\pi}{2}-\tan x)} \Longrightarrow \cot x=\frac{(2k+1)\pi}{2}-\tan x, \ k\in Z$[/quote]
Yeah, just a simple mistake, , that's never mind. :D


analytic geometry
geometry proposed
geometry

Let equilateral triangle $ ABC$. $ M$ is inbuilt triangular .$ MA,MB,MC$ intersect $ BC,CA,AB$ in $ X,Y,Z$. $ MD\perp BC, ME\perp CA, MF\perp AB$ ($ D\in BC,E\in CA, F\in AB$. Prove that $ S_{DEF}\geq S_{XYZ}$

I think it very hard
Can you post your solution

I think it's very hard too.....post the solution....I am very curious about the proof...

call $ x: y: z$ the trilinear coordinates of M and let AB = 1
then we must prove that

$ \frac{3}{8}\ \frac{xy \plus{} yz \plus{} xz}{(x\plus{}y\plus{}z)^2} \ge \frac{xyz}{(x\plus{}y)(x\plus{}z)(y\plus{}z)}$

but $ \sqrt{\frac{xy \plus{} yz \plus{} xz}{3}} \ge \sqrt[3]{xyz}$

so it become $ \frac{(x\plus{}y)(x\plus{}z)(y\plus{}z)}{\sqrt[3]{xyz} (x\plus{}y\plus{}z)^2} \ge \frac{8}{9}$

that is true with equalyty if x=y=z, so if M is the center of the triangle


ceiling function
logarithms
inequalities
function
number theory
totient function
number theory proposed

Define the phi-length $ l(n)$ of a positive integer $ n$ to be the least positive integer $ l$ such that $ \phi^l (n) \equal{} 1$, where $ \phi(n)$ is the number of positive integers less than and coprime to $ n$.

Prove that:

$ m \leq l(n) \leq \lceil k \rceil$

Where $ k \equal{} \log_2 {n}$ and $ m$ is any positive integer satisfying

$ 2^m \cdot k(k\minus{}1)(k\minus{}2)\cdots (k\minus{}m\plus{}1) \leq n$.

The upper bound is somewhat strong, but can anyone find a better lower bound?

if $ n$ is an  even number and $ d$ an integer with the following property : $ n \le 2^d$ then we can easily prove that $ \phi(n) \le \phi(2^d) \equal{} 2^{d \minus{} 1}$ and since $ \phi(n)$ is always(except $ \phi(2) \equal{} 1$) an even number we can do again what we have done.so if we put $ d \equal{} \lceil k \rceil$ then $ n \le 2^d$ and so we get $ l(n) \le \lceil k \rceil$.
the same thing can be done for odd numbers since after the first step we get an even integer.
for the left side of the inequality i think a better bound can be found using the inequality
$ \phi(n) > \frac {n}{e^{\gamma} \log \log n \plus{} \frac {3}{\log \log n}}$ (u can find more information [url=http://en.wikipedia.org/wiki/Euler%27s_totient_function]here[/url] in the part inequalities(7))
note that the bound given by [i]alkjash[/i] can easily be proven using this : $ \phi(n) \times 2k \ge n$.Which is rather easy to prove...


Show that $1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$ can never be an integer.

This is Problem 5 on PROMYS Problem Set #0.  Is it appropriate to post here?

sorry I do not know where the problem is from I just found it in ACoPS so assumed that this was the appropriate place for it.

Ah.  Well, it's just a very common problem then :)

i was thinking about proving by contradiction but i am unable to make it work.  :?

Mathworld cites a proof by p-adic numbers, but I wonder if there is a simpler method.   :|

There is a theorem(I do not know its name) that states that between a prime $p$ and $2p$ there is another prime.
Therefore among numbers $1,2...n$ we can find a prime $p$ such that $p$ divides only one number from the set of $1,2,...n$.
Write the expression $1+\frac 12 +\frac 13 +... +\frac 1n$ as a big fraction we will get in the denominator $n!$ and in numerator $n$ numbers of which $(n-1)$ are divisible by $p$ and one is not. Therefore $p$ will not divide the numerator of the fraction but will divide the denominator. Therefore the expression will never be an integer.

rem:
[quote]
There is a theorem(I do not know its name) that states that between a prime $p$ and $2p$ there is another prime.
[/quote]
It is called Bertrand's postulate. By the way, your proof is amazingly elegant.

Another one: look at the highest power $2^s$ of $2$ occuring in the first $n$ numbers. Then this power will be the only one of that numbers divisible by $2^s$ since the next number divisible by it is already the next power of two.
Then the same reasoning as before works.

More generally: \[ \sum_{\substack{k=1 \\ \gcd(k,m)=1}}^n \frac{1}{k} \] is never an integer (except $n=1$).

Similar: \[ \sum_{k=1}^n \frac{1}{dk+a} \] is never an integer.

I got this problem from ACoPS and it said to look at the numerator and denominator when u write the sum as one fraction in simplest form.

Yes, that will give you ZetaX's solution involving powers of $2$.

This problem has been posted SO MANY TIMES. I remember giving a bunch of links the last time the problem was posted. Just search "Harmonic Series" and you'll find it. And I don't know why this topic was revived, either.


Circles C1 and C2 are tangent to each other at K and are tangent to circle C at M and N. External tangent of C1 and C2 intersect C at a and B.
AK and BK intersct with circle C at E and F if AB is diameter of C.
Prove EF and MN and OK are concurrent. (O is center of circle C)

Inversion with center $K$ and power equal to the power of $ K$ with respect to $ C_{0}$ transforms $ C_{0}$ into itself and $C_1,C_2$ into  two parallel lines $ \ell_1, \ell_2$ tangent to $ C_{0}$ through $ M',N'$ $\Longrightarrow$ $ M',N'$ are antipodal points on $C_0.$ Line $ MN$ is taken into the circle $\odot( KM'N')$ and  line $ EF$ is taken into the circle $\odot(KAB).$ Note  that $O$ has equal power WRT $ \odot (KAB)$ and $ \odot (KM'N'),$ due to $ \overline{OM'} \cdot \overline{ON'}\equal{}\overline{ OB} \cdot \overline{OA}.$ Hence, $ \odot (KAB)$ and $\odot  (KM'N')$ meet again on the line $KO.$ If $\odot (KM'N'),$ $\odot (KAB)$ and $ KO$ concur $\Longrightarrow$ $ MN,EF$ and $ OK$ concur as well.


How do you find the remainder of (7^100)/9
Sorry for lack of Latex it messed up when i tried...

[hide="Solution"]
Use Euler's Totient Theorem. (Look it up.)

We have that $7^{ \varphi(9) }= 7^{6}\equiv 1 \bmod 9$.

This means that $7^{100}= 7^{16 \cdot 6}\cdot 7^{4}\bmod 9 \equiv 1^{16}\cdot (-2)^{4}\bmod 9 \equiv \boxed{7 \bmod 9}$, i.e. the division leaves a remainder of 7. [/hide]

[hide]I would just look at in $\pmod{9}$
$7 \equiv-2 \pmod{9}$
$\therefore 7^{100}\equiv (-2)^{100}\pmod{9}$
We then notice that $(-2)^{n}$ where n is a positive integer has the pattern $-2, 4,-8, 6,-2, 4...$
This translates to $7^{n}\equiv-2 (\iff n \equiv 1 \pmod{4}), 4 (\iff n \equiv 2 \pmod{4}),-8 (\iff n \equiv 3 \pmod{4}), 7 (\iff n \equiv 4 \pmod{4}) \pmod{9}$
Since $n$ in this case is $100$, and $100 \equiv 4 \pmod{4}$
Then $7^{100}\equiv 7 \pmod{9}$
(The remainder is $7$)[/hide]
Thanks for pointing out my retarded mistake cinco, I'm gonna hope that the only reason I did that was because I was in a hurry, lol.

[quote="cincodemayo5590"][quote="aahmadi"][hide]I would just look at in $\pmod{9}$
$7 \equiv-2 \pmod{9}$
$\therefore 7^{100}\equiv (-2)^{100}\pmod{9}$
We then notice that $(-2)^{n}$ where n is a positive integer has the pattern $-2, 4,-8, 6,-2, 4...$
This translates to $7^{n}\equiv-2 (\iff n \equiv 1 \pmod{4}), 4 (\iff n \equiv 2 \pmod{4}),-8 (\iff n \equiv 3 \pmod{4}), 6 (\iff n \equiv 4 \pmod{4}) \pmod{9}$
Since $n$ in this case is $100$, and $100 \equiv 4 \pmod{4}$
Then $7^{100}\equiv 6 \pmod{9}$
(The remainder is $6$)[/hide][/quote]

Wait... what? $2^{4}=16 = 7+9 \equiv \boxed{7}\bmod 9$. I think that you were thinking in $\bmod 10$.[/quote]

yes, cincodemayo is right. this was taken from a past GA tech tournament, and the answer is 7. thanks a lot for your help...


i'm trying to make _{-1}^{0}\text{e}
typeset so that the 0 is above the -1 on the right side
but right now it keeps typesetting on the left side
which makes it look really stupid

how do i fix this?

$ {\minus{}1}^{0}\text{e}$

Is that what you wanted?

no. the -1 is supposed to be a subscript and the 0 a superscript with the 0 being directly over the -1. the problem is that the 0 is placing itself on top of the "-" instead of on top of the "1" making it look reeeally stupid.

YES OMG
could you post the source code for that? or is there some super awesome way of finding it that i don't know of?
thanks!

The empty group is your friend when you want subscripts and superscripts for nothing; \phantom can help to align 1 and 0 (though it is not the only way):
{}_{-1}^{\phantom{-}0}\text{e} produces $ {}_{\minus{}1}^{\phantom{\minus{}}0}\text{e}$


geometry
parallelogram
vector
function
algebra
domain
real analysis

I just can't figure out how to approach this one, I guess.  :blush:

[i]Show that in an arbitrary quadrangle, the midpoints of the four sides are the vertices of a parallelogram.[/i]

Knowing that the midpoints at least make a quadrilateral, I thought if I could show that the sum of two adjacent sides equals the diagonal from their vertex, I'd be able to prove it. So far, though, I've yet to make it that far. Using complex analysis is the "key," but perhaps another way to look at it would help.

You have complex numbers $a,b,c,d$ which sum to zero (your arbitrary quadrilateral, the sides are taken in order clockwise).  The line segments formed by the midpoints of the sides are represented by $b/2+a/2,c/2+b/2,d/2+c/2,-a/2-b-c-d/2$ (again in order clockwise).  Applying the identity $a+b+c+d=0$, the last side can be written as $a/2+d/2$.  From here, to show that it's a parallelogram, we need only show that the sum of the first and third sides are zero, and that the sum of the second and fourth sides are zero.  We have $(b/2+a/2)+(d/2+c/2)=1/2(a+b+c+d)=0$ as desired, and $(c/2+b/2)+(a/2+b/2)=1/2(a+b+c+d)=0$, so we're done.

[quote="blahblahblah"] $(c/2+b/2)+(a/2+b/2)=1/2(a+b+c+d)=0$[/quote]

did you  mean $(c/2 + b/2) + (a/2 + d/2) = ... = 0$ ?
but don't let me sound ungrateful! thank you [u]so[/u] much!

This is really just vector geometry- you didn't use complex multiplication, and that's what makes complex numbers especially nice for geometry. In a purely affine problem like this, I'd use generic vectors instead

The one thing that makes me want to get the complex numbers involved are the possibilties this affords for using the Discrete Fourier Transform (the Fourier transform on $\mathbb{Z}_N.$ Of course, I'm also thinking beyond the quadrilateral to a more general $N$-gon.

Brief description of the DFT: Let $\omega=e^{2\pi i/N}.$ Suppose that $f: \mathbb{Z}_N\mapsto \mathbb{C}.$ (In other words, $f(n)$ is a complex sequence periodic of period $N.$)

The DFT of $f$ is the function $\hat{f},$ also with domain $\mathbb{Z}_N.$

$\hat{f}(k)=\frac1N\sum_{n=0}^{N-1}f(n)\omega^{-kn}$

We recover the original function as $f(n)=\sum_{k=0}^{N-1}\hat{f}(k)\omega^{kn}.$

I'll let you worry about defining the convolution of two functions, and proving the theorem that the Fourier transform of the convolution of two functions in the pointwise product of the Fourier transforms of the two functions (possibly with a constant multiplication.)

Let $z(n)$ be the vertices of an $N$-gon. Define $w=Tz$ as the $N$-gon whose vertices are the midpoints of the edges of the original polygon. That is, $w(n)=\frac{z(n)+z(n+1)}2.$ (Interpret those indices $\pmod{N}.$)

[Note that $T$ is a convolution operator.]

Here's the problem: determine what properties $w$ must have in order for it to be possible that there is a $z$ such that $w=Tz.$ From the previous posts in this topic, if $N=4,$ the condition is that $w$ be a parallelogram. What is the condition for other values of $N,$ and does it make a difference whether $N$ is odd or even?

[quote="hallie_72"]
[i]Show that in an arbitrary quadrangle, the midpoints of the four sides are the vertices of a parallelogram.[/i]
[/quote]

This is known as Varignon's theorem. Let the quadrangle be $ABCD$. Starting from the side $AB$ let the midpoints of $ABCD$ be $K,L,M$ and $N$. Then, using homothety with homothety center as $A,B,C,D$ with coefficent 2 to the line segments $KN,KL,LM,MN$ we see that $KN\mapsto BD,KL\mapsto AC,LM\mapsto BD$ and $NM\mapsto AC$. Therefore $KN||BD||LM$ and $KL||AC||NM$. Thus $KLMN$ is a parallelogram.

Sorry, I forgot to use complex geometry, maybe next time. :(


What is the product of all factors of 6?

The factors of 6 are 1, 2, 3, and 6, so the product is $ \boxed{36}$.


limit
logarithms
real analysis
real analysis unsolved

Find the limit:

$\lim_{n\rightarrow \infty}\frac{e^{\frac{n^{2}}{2}}}{C^{1}_{n}C^{2}_{n}\cdots C^{n}_{n}}$,

where $C^{k}_{n}$ stands for the binomial coefficient.


1.) Prove that $ \phi(p^k) \equiv p^{k\minus{}1}(p\minus{}1)$

2.) Prove that if $ n>2$ then $ \phi (n)$ is even.

1)
Any number can be either multiple of $ p$(a prime) or relatively prime with $ p$.

The multiples of $ p$ less than $ p^k$ are $ p, 2p,..., (p^{k\minus{}1}\minus{}1)p$, in total there are $ p^{k\minus{}1}\minus{}1$.

So there are $ (p^k\minus{}1)\minus{}(p^{k\minus{}1}\minus{}1\equal{}p^k\minus{}p^{k\minus{}1}\equal{}p^{k\minus{}1}(p\minus{}1)$ numbers relatively prime with $ p$ less than $ p^k$


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero such that $ a\plus{}b\plus{}c\equal{}2$. Prove that
\[ \sqrt{\frac{a\plus{}b}{a^2\plus{}ab\plus{}b^2}}\plus{}\sqrt{\frac{b\plus{}c}{b^2\plus{}bc\plus{}c^2}}\plus{}\sqrt{\frac{c\plus{}a}{c^2\plus{}ca\plus{}a^2}} \ge 2\plus{}\sqrt{\frac{2}{3}}.\]
:)

P/s: Best wishes for you, my dear friend! :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero such that $ a \plus{} b \plus{} c \equal{} 2$. Prove that
\[ \sqrt {\frac {a \plus{} b}{a^2 \plus{} ab \plus{} b^2}} \plus{} \sqrt {\frac {b \plus{} c}{b^2 \plus{} bc \plus{} c^2}} \plus{} \sqrt {\frac {c \plus{} a}{c^2 \plus{} ca \plus{} a^2}} \ge 2 \plus{} \sqrt {\frac {2}{3}}.
\]
:)

P/s: Best wishes for you, my dear friend! :)[/quote]

Thak you very much :) , my dear friend, for this very nice inequality. I will try to solve it, but unfortunately I am a bad solver of inequality. :(

[quote="nguoivn"]Dear Manlio, it's not hard. Using Am-Gm,we can prove the general problem:
$ \sum\ (\frac {a \plus{} b}{a^2 \plus{} ab \plus{} b^2})^k \geq\ 2 \plus{} (\frac {2}{3})^k$
With all $ a, b, c, k > 0$ and $ a \plus{} b \plus{} c \equal{} 2$
Good luck  :)[/quote]
Yes, it is because if we assume $ c\equal{}\min\{a,b,c\}$, we will have
\[ \frac{a\plus{}c}{a^2\plus{}ac\plus{}c^2} \ge \frac{1}{a\plus{}\frac{c}{2}}\]
\[ \frac{b\plus{}c}{b^2\plus{}bc\plus{}c^2} \ge \frac{1}{b\plus{}\frac{c}{2}}\]
\[ \frac{a\plus{}b}{a^2\plus{}ab\plus{}b^2} \ge \frac{a\plus{}b\plus{}c}{\left( a\plus{}\frac{c}{2}\right)^2 \plus{}\left( a\plus{}\frac{c}{2}\right) \left( b\plus{}\frac{c}{2}\right) \plus{}\left( b\plus{}\frac{c}{2}\right)^2}\]
;) :)

My hint is similar to you, but I use:
$ \frac{a\plus{}b}{a^2\plus{}ab\plus{}b^2} \geq\ \frac{2\sqrt[]{2}}{3 \sqrt[]{a^2\plus{}b^2}} \geq\ \frac{2\sqrt[]{2}}{3 \sqrt[]{(a\plus{}\frac{c}{2})^2\plus{}(b\plus{}\frac{c}{2}^2}}$
 :)

[quote="can_hang2007"][quote="nguoivn"]Dear Manlio, it's not hard. Using Am-Gm,we can prove the general problem:
$ \sum\ (\frac {a \plus{} b}{a^2 \plus{} ab \plus{} b^2})^k \geq\ 2 \plus{} (\frac {2}{3})^k$
With all $ a, b, c, k > 0$ and $ a \plus{} b \plus{} c \equal{} 2$
Good luck  :)[/quote]
Yes, it is because if we assume $ c \equal{} \min\{a,b,c\}$, we will have
\[ \frac {a \plus{} c}{a^2 \plus{} ac \plus{} c^2} \ge \frac {1}{a \plus{} \frac {c}{2}}
\]

\[ \frac {b \plus{} c}{b^2 \plus{} bc \plus{} c^2} \ge \frac {1}{b \plus{} \frac {c}{2}}
\]

\[ \frac {a \plus{} b}{a^2 \plus{} ab \plus{} b^2} \ge \frac {a \plus{} b \plus{} c}{\left( a \plus{} \frac {c}{2}\right)^2 \plus{} \left( a \plus{} \frac {c}{2}\right) \left( b \plus{} \frac {c}{2}\right) \plus{} \left( b \plus{} \frac {c}{2}\right)^2}
\]
;) :)[/quote]

Putting $ x\equal{}a\plus{} \frac {c}{2}$ and $ y\equal{}b \plus{} \frac {c}{2}$ with $ x\plus{}y\equal{}2$ we have

$ \frac{1}{\sqrt{x}} \plus{}\frac{1}{\sqrt{y}} \plus{}\frac{1}{\sqrt{x^2\plus{}xy\plus{}y^2}} \geq 2 \plus{} (\sqrt{\frac {2}{3}})$

How to solve this one in a simple way?

Thank you very much.

Please think about the tecnique: catch point of incidence in Am-Gm ineq  :)
It's very simple but very useful in some problem. It'll help us not have to use calculator.
I have said that: Using only Am-Gm, we are easily to solve the problems as:
$ \frac {1}{a^4 \plus{} b^4} \plus{} \frac {1}{b^4 \plus{} c^4} \plus{} \frac {1}{c^4 \plus{} a^4} \geq\ \frac {40}{(a \plus{} b \plus{} c)^4}$
...

[quote="nguoivn"]Please think about the tecnique: catch point of incidence in Am-Gm ineq  :)
It's very simple but very useful in some problem. It'll help us not have to use calculator.
[/quote]

Sorry, I don't understand this tecnique :blush:

Please any hint to solve in simple way :

For $ x,y$ positive reals such that $ x \plus{} y \equal{} 2$ we have

$ \frac {1}{\sqrt {x}} \plus{} \frac {1}{\sqrt {y}} \plus{} \frac {1}{\sqrt {x^2 \plus{} xy \plus{} y^2}} \geq 2 \plus{} (\sqrt {\frac {2}{3}})$

Thank you very much. :)


:blush: let $ ABCD$ incribed a circle $ (O)$ and $ P$ be arbitrary on plane $ (ABCD)$.Let $ O_{1}$;$ O_{2}$;$ O_{3}$;$ O_{4}$ be centre of circle $ (PAB)$;$ (PBC)$;$ (PCD)$;$ (PDA)$ respectively .prove that the mid point of $ PO$;$ O_{1}O_{3}$;
 $ O_{2}O_{4}$ are collinear. :wink:

I remember this problem,it was solved by yetti,long time ago,beautiful solution...
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=124926[/url]


In grading USAMTS Problem 4/2/17, the "Duck Goose Goose" problem, I encountered a submission that mistakenly reversed the roles of duck and goose ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62987]maybe this one[/url]). After the grading session, I looked at the reversed problem in more detail afterwards, and discovered that it is much more difficult than the original. If you are familiar with the "Duck Goose Goose" problem, you might be interested in the reasoning required to solve the reversed problem.

Let me phrase the reversed problem as "Goose Duck Duck". A teacher has $n$ students stand in a circle and she tags them in the cyclic pattern "Goose Duck Duck." Each student tagged as goose leaves the circle and each student tagged as duck remains in the circle until possibly retagged as a goose. The teacher continues around the circle until only one student remains. For which values of $n$ is the $n$th student in the circle the one remaining at the end?

The first observation is that if $n$ is one greater than a positive multiple of three, the $n$th student will be at the goose tag of the Goose Duck Duck cycle and be tagged out. So we know $n$ is not one more than a multiple of three. The case $n=1$ is an exception, because the teacher does not make a round in that case.

The concept of rounds is useful for sorting the cases. A round is when the teacher goes around the circle one full revolution. The first round starts with the first student and ends with the $n$th student. Since the $n$th student remains through all rounds, all rounds end with the $n$th student, which makes it easy to define the rounds.

We sort the cases for $n$ by modulus with respect to three: $n=3m$, $n=3m+1$, or $n=3m+2$ for some integer $m$. Let us see how each case changes after a round.

$n=3m$. After a round the $n$th student is still at the last position and $2m$ students remain. The next round would start "Goose Duck Duck" (GDD for short).
$n=3m+1$. After a round the $n$th student is out.
$n=3m+2$. After a round the $n$th student is still at the last position and $2m+1$ students remain. The next round would start "Duck Goose Duck" (DGD for short) and keep up that 3-long pattern.

We need to examine the effects of the "Duck Goose Duck" pattern. Once again we sort mod 3.

$n=3m$ DGD. After a round the $n$th student is still at the last position and $2m$ students remain. The next round would start "Duck Goose Duck."
$n=3m+1$ DGD. After a round the $n$th student is still at the last position and $2m+1$ students remain. The next round would start "Goose Duck Duck."
$n=3m+2$ DGD. After a round the $n$th student is out.

It also helps to examine the first few cases:
1-student GDD leaves the first ($n$th) student standing without any rounds.
2-student GDD leaves the second ($n$th) student standing after one round.
3-student GDD leaves the third ($n$th) student standing after two rounds.
4-student GDD leaves the third (not $n$th) student standing after three rounds.
5-student GDD leaves the second (not $n$th) student standing after three rounds.
1-student DGD leaves the first ($n$th) student standing without any rounds.
2-student DGD leaves the first (not $n$th) student standing after one round.
3-student DGD leaves the first (not $n$th) student standing after two rounds.
4-student GDD leaves the fourth ($n$th) student standing after three rounds.
5-student DGD leaves the third (not $n$th) student standing after four rounds.

Each round leads to a smaller circle of students. Since each round eliminates at most one third of the students, any circle with 6 or more students must go through the above rounds listed on its way to having only one student remaining. We can backtrack from the cases listed above to recreate the large circles that will keep the $n$th student.

Rule 1. If after a round we have an even number of students, $2m$, a GDD pattern, and the $n$th student still in place, then it could have come only from a circle of $3m$ students on a GDD pattern.
Rule 2. If after a round we have an odd number of students, $2m+1$, a GDD pattern, and the $n$th student still in place, then it could have come only from a circle of $3m+1$ students on a DGD pattern.
Rule 3. If after a round we have an even number of students, $2m$, a DGD pattern, and the $n$th student still in place, then it could have come only from a circle of $3m$ students on a DGD pattern.
Rule 4. If after a round we have an odd number of students, $2m+1$, a GDD pattern, and the $n$th student still in place, then it could have come only from a circle of $3m+2$ students on a GDD pattern.

So we start at an end and work backwards. One end is the 1 student (the $n$th one of course) and a GDD pattern. Rule 2 says that comes from 1 student in a DGD pattern. This is impossible, because if we had 1 student, we would not have taken another round.

The other end is 1 student and a DGD pattern. Rule 4 says that comes from 2 students and a GDD pattern. Rule 1 says that comes from 3 students GDD. Rule 2 says that comes from 4 students DGD. Rule 3 says that comes from 6 students DGD. Etc. We end up with the chain: 1 DGD, 2 GDD, 3 GDD, 4 DGD, 6 DGD, 9 DGD, 14 GDD, 21 GDD, 31 DGD, 47 GDD, 70 DGD, 105 DGD, 158 GDD, 237 GDD, 355 DGD, 533 GDD, ...  That chain is the only possible chain.

To find the values for $n$, we take the GDD cases in that chain. That gives us $n$ = 2, 3, 14, 21, 47, 158, 237, 533, ...

Problem 4/2/17, "Duck Goose Goose," had the sequence 4, 7, 16, 25, 52, 79, 160, ..., which splits into the sequence 7, 25, 79, ... generated by $3^k - 2$ and the sequence 4, 16, 52, 160, ... generated by $2\times 3^k - 2$. Alas, the "Goose Duck Duck" solution has no easy way to describe its sequence. The only way to generate the sequence of valid $n$-values is to backtrack the GDD-DGD chain and then separate out the GDD sequence. So not only is the "Goose Duck Duck" problem harder to solve, but the answer cannot be expressed cleanly.

Erin Schram

[quote="pkerichang"]so in this case there will be no specific sequence that fits the solution???[/quote]
There is a specific sequence. So long as we have a method to generate the next element of the sequence, we have a sequence. However, there is no explicit formula for generating this sequence. Nor can we predict how much work it takes to generate the next element. The situtation is messy.

We like clear solutions for the USAMTS, because they are easier for the participants to write and for us graders to read. So we discard a lot of messy problems. I was just surprised that reversing Duck Goose Goose to Goose Duck Duck made the problem so much more awkward. Sometimes the people who invent problems for the USAMTS are astonishingly clever at finding a clear problem in a family of messy problems.

Erin Schram


What are sublevels?? I never heard of this term before. Do $ e^{\minus{}}1$ exist in these subshells?

[hide="Explanation"]The quantum mechanical description of the electronic structure of atoms is based on the concept of atomic orbitals (which are regions of space where the probability of finding one or two electrons is very high - tipically 95%) which in turn are described by a set of three quantum numbers: the principal quantum number $ n$, the azimuthal quantum number (or angular momentum quantum number) $ l$, and the magnetic quantum number $ m_{l}$. The principal quantum number labels the different [i]shells[/i] of the electron cloud of the atom according to its energy and size: a shell is the set of all possible orbitals having the same energy and size (i.e, roughly the same distance from the nucleous), and so they all have the same $ n$. The principal quantum number assumes only integer values: n = 1, 2, 3, ... . The azimuthal quantum number is used to differentiate the orbitals in a given shell according to their shape, and orbitals having the same shape are said to belong to the same [i]subshell[/i] (or sublevel). For a given $ n$, $ l$ can assume the values l = 0, 1, ..., n-1. Also, each value of $ l$ is given a specific designation (derived from spectroscopy): orbitals with l = 0 are called $ s$ orbitals, those with l = 1 are $ p$ orbitals, with l = 2 are the $ d$ orbitals, and with l = 3 are the $ f$ orbitals. For example, for the shell with n = 3, we have the $ s$ subshell, the $ p$ subshell, and the $ d$ subshell. The magnetic quantum number have the function of differentiate the orbitals of a given $ l$ according to their spatial orientation, and for a given $ l$ the quantum number $ m_{l}$ can assume the set of values -l, -l+1, ..., 0, ..., +l, so that for each $ l$ there are $ 2l\plus{}1$ possible values for $ m_{l}$ - which means that the total number of orbitals in a subshell $ l$ is $ 2l\plus{}1$. For example, for the subshell l = 1 (the p orbitals) there are a total of 2*1+1 = 3 orbitals (as we know, there are three p orbitals), each one receiving a different $ m_{l}$ (-1, 0, or +1) because each p orbital have a different orientation is space.
In determining the electronic configuration of an atom, electrons are then added to each of these orbitals, according to three principles: minimum energy, Pauli exclusion principle, and Hund's rule. Finally, to completely describe a given electron, a total of four quantum numbers are needed: the three described above, and an additional one, the spin quantum number $ m_{s}$, which describes the spin state of the electron.
[/hide]


inequalities
function
inequalities proposed

$ a,b,c >0$. Prove that !
$ \sum_{cyc} \frac{a}{(b\plus{}c)^2} \ge \frac{9}{4\sum a}$

[quote="bigbang195"]$ a,b,c > 0$. Prove that !
$ \sum_{cyc} \frac {a}{(b \plus{} c)^2} \ge \frac {9}{4\sum a}$[/quote]

$ \sum_{cyc} \frac {a}{(b \plus{} c)^2} \ge\frac {(a \plus{} b \plus{} c)^2}{\sum a(b \plus{} c)^2}$

only prove

$ \frac {(a \plus{} b \plus{} c)^2}{\sum a(b \plus{} c)^2}\geq \frac {9}{4\sum a}$

$ \Longleftrightarrow  4a^3\plus{}3a^2b\plus{}3ca^2\plus{}3ab^2\minus{}30abc\plus{}3ac^2\plus{}4b^3\plus{}3b^2c\plus{}3bc^2\plus{}4c^3\geq 0$

true  due to AM-GM Inequality.

[quote="bigbang195"]$ a,b,c > 0$. Prove that !
$ \sum_{cyc} \frac {a}{(b + c)^2} \ge \frac {9}{4\sum a}$[/quote]

[quote="bigbang195"]$ a,b,c > 0$. Prove that !
$ \sum_{cyc} \frac {a}{(b + c)^2} \ge \frac {9}{4\sum a}$[/quote]

Since the inequality is homogenous, we may assume $ \sum a = 1$ and thus $ a,b,c \in [0,1]$.

Rewriting the inequality:

$ \sum_{cyc} \frac {a}{(b + c)^2} \ge \frac {9}{4\sum a} \& \iff$
$ \sum \frac {a }{(1 - a)^2}\geq \frac 9 4$

because of $ \sum a = 1$

Consider the function $ f(x) = \frac {x}{(1 - x)^2}$

You can easily check that its convex on $ [0,1]$

By Jensen, we have:

$ \sum \frac {a }{(1 - a)^2}\geq 3\cdot f\left(\frac {\sum a} 3\right) = \frac 9 4$

[quote="bigbang195"]$ a,b,c > 0$. Prove that
$ \sum_{cyc} \frac {a}{(b \plus{} c)^2} \ge \frac {9}{4\sum a}$.[/quote]
The inequality is equivalent to
\[ \sum_{cyc} \frac {a}{(b \plus{} c)^2}\cdot \sum a \geqslant\frac{9}{4}\mbox{,}\]
but with Cauchy-Schwartz and Nesbit, we get
\[ \sum_{cyc} \frac {a}{(b \plus{} c)^2}\cdot \sum a \geqslant \left(\sum\frac{a}{b\plus{}c}\right)^2\geqslant\left(\frac{3}{2}\right)^2\equal{}\frac{9}{4}\mbox{,}\]
which proves the desired.

[quote="bigbang195"]$ a,b,c > 0$. Prove that !
$ \sum_{cyc} \frac {a}{(b \plus{} c)^2} \ge \frac {9}{4\sum a}$[/quote]
by chebyshev we obtain $ \sum_{cyc} \frac {a}{(b \plus{} c)^2} \ge \frac {1}{3}(\sum_{cyc} \frac {a}{(b \plus{} c)})(\sum_{cyc} \frac {1}{(b \plus{} c)}) \ge \frac {1}{3} \frac {3}{2} \frac {9}{(\sum b \plus{} c)} \equal{} \frac {9}{4\sum a}$

Nice!

4 different solutions to the same problem  :lol:

PS: it can also be solved using $ a\equal{}\frac 1 x, \ b\equal{}...$ and AM-GM/Muirhead

[u][b]My proof[/b][/u]

$ LHS \equal{} \frac {a^3}{[a(b \plus{} c)]^2} \plus{} \frac {b^3}{[b(c \plus{} a)]^2} \plus{} \frac {c^3}{[c(a \plus{} b)]^2}\stackrel{Marin\ Bancos's\ lemma}{\geq}$
$ \geq\frac {a \plus{} b \plus{} c}{a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b)}\cdot\left[\frac {a^2}{a(b \plus{} c)} \plus{} \frac {b^2}{b(c \plus{} a)} \plus{} \frac {c^2}{c(a \plus{} b)}\right] \equal{}$
$ \equal{} \frac {a \plus{} b \plus{} c}{2(ab \plus{} bc \plus{} ca)}\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\stackrel{Nesbitt}{\geq}\frac {a \plus{} b \plus{} c}{2(ab \plus{} bc \plus{} ca)}\cdot\frac {3}{2} \equal{}$
$ \equal{} \frac {3(a \plus{} b \plus{} c)}{4(ab \plus{} bc \plus{} ca)}\stackrel{(a \plus{} b \plus{} c)^2\geq3(ab \plus{} bc \plus{} ca)}{\geq}\frac {9}{4(ab \plus{} bc \plus{} ca)} \equal{} RHS$

REMARKS:
1. [b]The author of this problem is Darij Grinberg[/b], and it has been posted before.
2. From my proof you can notice that the initial inequality could be strengthened, using the following statement

For $ a,b,c > 0$, the following inequality holds

$ \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2}\geq\frac {3(a \plus{} b \plus{} c)}{4(ab \plus{} bc \plus{} ca)}$
3. I'd like to remember you my lemma

If: $ a_{i}\geq0,b_{i} > 0,i \equal{} 1,2,\dots,n,$ and $ n\in N, n\geq1, k\in R, k > 0 ,$ then:
$ \left.\frac {{a_{1}}^{k \plus{} 1}}{b_{1}^{k}} \plus{} \frac {a_{2}^{k \plus{} 1}}{b_{2}^{k}} \plus{} \cdots \plus{} \frac {{a_{n}}^{k \plus{} 1}}{b_{n}^{k}}\geq\frac {a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{n}}{b_{1} \plus{} b_{2} \plus{} \cdots \plus{} b_{n}}\cdot\left(\frac {{a_{1}}^{k}}{b_{1}^{k \minus{} 1}} \plus{} \frac {{a_{2}}^{k}}{b_{2}^{k \minus{} 1}} \plus{} \cdots \plus{} \frac {{a_{n}}^{k}}{b_{n}^{k \minus{} 1}}\right)\right.$


geometry
circumcircle
LaTeX
geometry proposed

Let ABC be a acute angled triangle. The bisectors of ang.A and ang.C intersects BC and AB at A' and C' resp. Also the bisector of B intersects the circumcircle(O) at P. AA' and CC' int. (O) at Q and R resp.Given that ang.P is bisected by BP at A' and C'. and CI=2AI. prove QA'-RC'=CI

[quote="communist"]Given that ang.P is bisected by BP[/quote]
$ \angle P$ ?!? Is that $ \angle RPQ$ or $ \angle APC$, or something else?!

[quote="communist"][color=darkred]Let $ ABC$ be an acute triangle with the incircle $ C(I,r)$ and the circumcircle $ w \equal{} C(O,R)$ . Denote

$ \left\|\begin{array}{c} D\in AI\cap BC \\
\ E\in BI\cap CA \\
\ F\in CI\cap AB\end{array}\right\|$ and the second intersections $ M$ , $ N$ , $ P$ of the circle $ w$ with the lines $ AI$ , $ BI$ , $ CI$

respectively. Suppose that  $ IC \equal{} 2\cdot IA$ and $ \widehat {IND}\equiv\widehat {INF}$ . Prove that $ QD \minus{} RF \equal{} IC$ [/color]. [/quote]

[u]Please[/u] [b]Communist[/b], learn quickly [b]LaTeX[/b] !

hello sir,
 i am truly inspired by all your solutions. u have a truly different approach on every problem. i actually had another solution for this. but thank you.


modular arithmetic
search
quadratics
algebra
difference of squares
special factorizations
number theory unsolved

Let's see...I believe I found, with proof, all solutions to $ p^{n}\minus{}q^{m}\equal{} 1$, $ p$ and $ q$ primes, once...

*rummages through notebooks*

All right, I have it.  Here's the case you mentioned.

[hide]
$ 3^{n}\equal{} 2^{m}\plus{}1$
$ 3$ has order $ n$ modulo $ 2^{m}$.  Therefore, $ n$ is a power of $ 2$.  Thus,
\[ 2^{m}\equal{}(3^{\frac{n}{2}}\plus{}1)(3^{\frac{n}{4}}\plus{}1)...(3^{2}\plus{}1)(3\plus{}1)(3\minus{}1)\]
However, $ 5|3^{2}\plus{}1$.  Thus, $ n\le 2$.  If $ n \equal{} 2$, we have that $ m \equal{} 3$, and if $ n \equal{} 1$, we have that $ m \equal{} 1$.  So $ (m, n) \equal{} (1, 1), (3, 2)$.
[/hide]

If you would like the full proof, I can put it here.  It takes up nearly ten times as much space as the above case, though.

[quote="mathemagics"][quote="K81o7"]

$ 3^{n}\equal{} 2^{m}\plus{}1$
$ 3$ has order $ n$ modulo $ 2^{m}$.  Therefore, $ n$ is a power of $ 2$.
[/quote]

The order of $ 3 mod 2^{m}$ is a divisor of $ \phi(2^{m}) \equal{} 2^{m\minus{}1}$ so you get that the order of $ 3 mod 2^{m}$ is a power of $ 2$. But (I think) you are not sure that $ n$ is the order because you don't know about the minimality of $ n$. Correct me if I am wrong. Do you have a proof for the above about the form of $ n$ ?

For example $ 5^{3}\equiv 1\pmod{4}$ and $ 3\nmid 2 \equal{}\phi(4)$, because the order of $ 5 mod 4$ is $ 1$.[/quote]

Well if $ 3^{n}\equal{} 2^{m}\plus{}1$, then $ 3^{n}$ is the first power of $ 3$, apart from $ 3^{0}$, which is $ 1\mod{2^{m}}$, right?  :wink:

If we use it correctly, the equation we start with is extremely powerful in this way.  Variations on this same idea are used to prove the general case I mentioned.

[quote="K81o7"]
Well if $ 3^{n}\equal{} 2^{m}\plus{}1$, then $ 3^{n}$ is the first power of $ 3$, apart from $ 3^{0}$, which is $ 1\mod{2^{m}}$, right?  :wink:

If we use it correctly, the equation we start with is extremely powerful in this way.  Variations on this same idea are used to prove the general case I mentioned.[/quote]

You are right!! I didn't mention that. Thanks a lot for your nice solution

[quote="K81o7"]Thus,
\[ 2^{m}\equal{}(3^{\frac{n}{2}}\plus{}1)(3^{\frac{n}{4}}\plus{}1)...(3^{2}\plus{}1)(3\plus{}1)(3\minus{}1)\]
[/quote]

?? you should check this.

$ 3^{N}\minus{}1\equal{}(3\minus{}1)(3^{N\minus{}1}\plus{}3^{N\minus{}2}\plus{}\cdots\plus{}3\plus{}1)$

I am not sure what you did.  Maybe I am missing something.

[quote="baz"][quote="K81o7"]Thus,
\[ 2^{m}\equal{}(3^{\frac{n}{2}}\plus{}1)(3^{\frac{n}{4}}\plus{}1)...(3^{2}\plus{}1)(3\plus{}1)(3\minus{}1)\]
[/quote]

?? you should check this.

$ 3^{N}\minus{}1 \equal{} (3\minus{}1)(3^{N\minus{}1}\plus{}3^{N\minus{}2}\plus{}\cdots\plus{}3\plus{}1)$

I am not sure what you did.  Maybe I am missing something.[/quote]

$ 3^{n}\minus{}1\equal{}(3^{\frac{n}{2}}\plus{}1)(3^{\frac{n}{2}}\minus{}1)\equal{}(3^{\frac{n}{2}}\plus{}1)(3^{\frac{n}{4}}\plus{}1)(3^{\frac{n}{4}}\minus{}1)\equal{}...$

Difference of squares over and over again.  We can do this and reduce down to $ 3\plus{}1$ because $ n$ is a power of $ 2$.

[quote="TTsphn"]You can use Pell's Equation . :lol:[/quote]
Post a full solution, but do not give such a 0-information-argument.
[This and the post above will be deleted tomorrow]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1781083571&t=61460]You can see the simillar problem here[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=156125]also see here[/url]

So, in case anyone wants, it, here is the proof of the generalization I mentioned...

[hide]
So we have the equation
\[ p^{n}\minus{}q^{m}\equal{}1\]
Clearly one of the two primes is $ 2$.  We're going to consider two cases.

Case 1: $ p\equal{}2$

$ 2^{n}\minus{}1\equal{}q^{m}$
\[ q^{m}\equiv\minus{}1\mod{2^{n}}\]
In fact, because of the equation we have we can give a stronger statement: that $ q$ has order $ 2m$ modulo $ 2^{n}$.  Because the group of reduced residue classes modulo $ 2^{n}$ has order $ 2^{n\minus{}1}$, we conclude that \[ m|2^{n\minus{}2}\]
Therefore, $ m$ is a power of $ 2$.  But clearly we cannot have $ 2|m$, because $ \minus{}1$ is a quadratic nonresidue modulo $ 4$, and is therefore a qnr mod $ 2^{n}$.  Thus, $ m\equal{}0, 1$.  If $ m\equal{}0$, then $ n\equal{}1$, giving a trivial case.  If $ m\equal{}1$, then we have the general case that $ 2^{n}\minus{}1$ is a Mersenne prime (so $ n$ is prime...I can't say much else).

Case 2: $ q\equal{}2$

$ p^{n}\equal{}2^{m}\plus{}1$

We now consider two subcases.

Subcase 1: $ m$ is odd

In this case, we have $ 3|p^{n}\implies p\equal{}3$.
$ 3^{n}\equal{}2^{m}\plus{}1$
$ 3$ has order $ n$ modulo $ 2^{m}$.  Therefore, $ n$ is a power of $ 2$.  Thus,
\[ 2^{m}\equal{}(3^{\frac{n}{2}}\plus{}1)(3^{\frac{n}{4}}\plus{}1)...(3^{2}\plus{}1)(3\plus{}1)(3\minus{}1)\]
However, $ 5|3^{2}\plus{}1$.  Thus, $ n\le 2$.  If $ n\equal{}2$, we have that $ m\equal{}3$, and if $ n\equal{}1$, we have that $ m\equal{}1$.

Subcase 2: $ m$ is even

$ p\minus{}1|p^{n}\minus{}1\equal{}2^{m}$, so $ p\minus{}1$ is a power of $ 2$.  We would let $ p\equal{}2^{k}\plus{}1$, but then if $ k$ were divisible by an odd integer $ r$, other than $ 1$, we'd have $ 2^{\frac{k}{r}}\plus{}1|p$.  So let $ p\equal{}2^{2^{k}}\plus{}1$.

So $ 2$ has order $ 2^{k\plus{}1}$ modulo $ p$.  We therefore conclude that $ 2^{k\plus{}1}|2m\implies 2^{k}|m$.
Let $ m\equal{}2^{k}x$.  Then

\[ 2^{2^{k}}\equal{}p\minus{}1,\left(2^{2^{k}}\right)^{x}\equal{}p^{n}\minus{}1\]
\[ \implies (p\minus{}1)^{x}\equal{}p^{n}\minus{}1\]
Perform binomial expansion on the left hand side.  Observing the $ p^{1}$ term, we find that $ p^{n\minus{}1}|x$.  If $ n>1$, this means that
\[ (p\minus{}1)^{p^{n\minus{}1}}|p^{n}\minus{}1\]
All I can say is that that doesn't happen too often...only when $ p\equal{}2$, and we covered that case already.
So all that remains is $ n\equal{}1\implies 2^{m}\plus{}1\equal{}p$.  Then again, $ m$ is a power of $ 2$, giving our general form to be $ p\equal{}2^{2^{k}}\plus{}1, q\equal{}2, n\equal{}0, m\equal{}2^{k}$.

\[ \mathbb{QED}\]
[/hide]

You can find the whole proof of the Catalan's Conjecture and elementary proofs of some special cases (one of the exponents is 2) here: http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf

[quote="mathemagics"]Hi,

Is there any elementary proof that the only solution (in the set of the natural numbers) of the equation $ 3^{x}\minus{}2^{y}\equal{} 1$ is $ x \equal{} 2$ and $ y \equal{} 3$ ? If you know more than one proof post them all.

Note: Of course this problem is a very little part of Catalan's Conjecture so that's why I ask for an elementary proof.
Thanks[/quote]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130018]See here[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130446]A generalization[/url]


Let $n\ge2$ be a positive integer with divisors $1=d_{1}<d_{2}<\cdots<d_{k-1}<d_{k}=n$.  Prove that $d_{1}d_{2}+d_{2}d_{3}+\cdots+d_{k-1}d_{k}<n^{2}$

it has a surprisingly (or not) short solution

Ok...If a solution is short u cant say anything about the difficult,and specially when it is an imo problem.

OK...here is the solution.
Suppose $S=d_{1}d_{2}+...+d_{k-1}d_{k}$ to solve the problem we are going to use a very simple identity, $d_{i}/n=1/d_{k+1-i}$ (this is a simple identity so i am not going to prove it)

 then $d_{i}d_{i+1}/n^{2}=1/ d_{k+1-i}d_{k-i}$so we have to prove that $S/n^{2}<1$

$S/n^{2}=(1/{d_{k}}d_{k-1}+...+1/d_{2}d_{1})$ so lets make it teliscopic... $S/n^{2}/leq (1/d_{k-1}-d_{k})+...+(d_{1}-d_{2})=1-1/d_{k}< 1$ so we are done.

Moon

Oups,..Last two terms should be
1/d_1-1/d_2


quadratics
number theory unsolved
number theory

$p$ is a prime ($p=4k+3$) and $x$ and $y$ are integers. Solve this equation :

$x^2+p^2=y^3$   :)

The solution is very much like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=428447#p428447]that one[/url].
The fact $p=4k+3$ implies that $p$ is prime in $\mathbb{Z}[i]$. And similar reasonning shows that $x+ip$ is a perfect cube.

If $x+ip=(a+ib)^3$ then for imaginary parts we have $b(3a^2 - b^2) = p$. Hence, $b$ divide $p$, i.e., $b=\pm 1$ or $b=\pm p$ among which only $b=1$ and $b=-p$ give solutions.

If $b=1$ then the prime $p$ must be of the form $p=3a^2-1$. Then $x=\pm a(a^2-3)$ and $y=a^2+1$.
For example, $a=2,\ p=11$ imply $x=\pm 2$ and $y=5$; $a=4,\ p=47$ imply $x=\pm 52$ and $y=17$ etc.

If $b=-p$ then $3a^2 - p^2 = - 1$ implying that the prime $p$ belongs to the sequence $q_0=1,\ q_1=2,\ q_{n+1}=4q_n - q_{n-1}$ (see also [url=http://oeis.org/A001075]A001075[/url]). This sequence contains only one prime of the form $4k+3$ which is $p=7$ implying $a=4,\ x=\pm 524,\ y=65$.

[quote="maxal"]

If $b=-p$ then $3a^2 - p^2 = - 1$ implying that the prime $p$ belongs to the sequence $q_0=1,\ q_1=2,\ q_{n+1}=4q_n - q_{n-1}$ (see also [url=http://www.research.att.com/~njas/sequences/A001075]A001075[/url]). This sequence contains only one prime of the form $4k+3$ which is $p=7$ implying $a=4,\ x=\pm 524,\ y=65$.[/quote]

Excuse me, can you explain me what does it means? Why p=7 is the unique prime of the sequence?
Thanks

[quote="maxal"][quote="Gauss_87"]Excuse me, can you explain me what does it means? Why p=7 is the unique prime of the sequence?[/quote]
Because all terms of the form $4k+3$ (at positions of the form $4s+2$) in that sequence are divisible by 7.[/quote]

Why at the position $4s +2$ ? And the other positions? There's a Theorem or a Proof of this? Where?

[quote="Gauss_87"]Why at the position $4s +2$ ? And the other positions? There's a Theorem or a Proof of this? Where?[/quote]
All numbers at positions $4s+1$ and $4s+3$ are even, and all numbers at positions $4s$ are of the form $4k+1$.

[quote="metehan"]p is a prime (p=4k+3) and x and y are integers. Solve this equation

    [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/e9f30de2234d32c185c8272d1586abf2.gif[/img]   :)[/quote]

Very nice problem !!!
Nasilsiniz abi! Bu ben Devran eger hatirlarsiniz benim hotmail kostakoz@hotmail.com! Bende IMO 2005 katilmistim Tajik Turk lisede kuyorum inshallah bizde bir kampiniza katilacagiz bu sne !!

[quote="Davron"]Nasilsiniz abi! Bu ben Devran eger hatirlarsiniz benim hotmail kostakoz@hotmail.com! Bende IMO 2005 katilmistim Tajik Turk lisede kuyorum inshallah bizde bir kampiniza katilacagiz bu sne !![/quote]

Please post in english!

aren't there any elementary proof???
i have solve for p=3
for a prime number=q such that q divides y.==>q divides (x^2+9) .in this condition if q=3  then x^2≡-9 (modulo 27) if q is not 3 then; -9 is quadratic residue in (mod q).this means (-1) is quadratic residue in modulo q.then q≡1 (mod 4)
it is known that 2 can not divide y.this means that y≡1 (mod 4).
write the equality such that:
x^2+1=y^3+2^3=(y+2).(y^2-2y+4)
y+2≡3 (mod4) then there is a prime number s.such that s divides (y+2) and s≡3 (mod 4).so s divides (x^2+1) from quadratic residular rules;x^2≡-1 (mod s) is impossible.contradiction...
but i can't solve in general case.