Three nonzero real numbers $ a,b,c$ are said to be in harmonic progression if $ \frac {1}{a} \plus{} \frac {1}{c} \equal{} \frac {2}{b}$. Find all three term harmonic progressions $ a,b,c$ of strictly increasing positive integers in which $ a \equal{} 20$ and $ b$ divides $ c$.

[17 points out of 100 for the 6 problems]

[hide="My solution"]Suppose $ c \equal{} bk$
$ \frac {1}{a} \plus{} \frac {1}{c} \equal{} \frac {2}{b}\iff \frac {1}{20} \plus{} \frac {1}{bk} \equal{} \frac {2}{b}$
(Multiplying by $ 20bk$)
$ bk \plus{} 20 \equal{} 40k$
$ \iff 20 \equal{} k(40 \minus{} b)$
$ 20$ can be factored as $ 1\cdot 20,2\cdot 10,4\cdot 5,5\cdot 4,10\cdot 2,20\cdot 1$
Each corresponding to the following solutions:
$ (b,c) \equal{} (20,20),\boxed{(30,60),(35,140),(36,180),(38,380),(39,780)}$
The first one doesn't give increasing sequence.[/hide]

how can 16,80 be an answer????????????

i think it said strictlly increasing?????????????? :(  :o

It is a typo...if you read the solution you realise it is a $ 36$ not a $ 16$ The solution is correct

By the way...For what grades is this test made?

Daniel

o ya i just didnt think...........
 :rotfl:

well d only ques i got totally crrect
 :P

[hide][quote="Akashnil"][hide="My solution"]Suppose $ c \equal{} bk$
$ \frac {1}{a} \plus{} \frac {1}{c} \equal{} \frac {2}{b}\iff \frac {1}{20} \plus{} \frac {1}{bk} \equal{} \frac {2}{b}$
(Multiplying by $ 20bk$)
$ bk \plus{} 20 \equal{} 40k$
$ \iff 20 \equal{} k(40 \minus{} b)$
$ 20$ can be factored as $ 1\cdot 20,2\cdot 10,4\cdot 5,5\cdot 4,10\cdot 2,20\cdot 1$
Each corresponding to the following solutions:
$ (b,c) \equal{} (20,20),\boxed{(30,60),(35,140),(16,80),(38,380),(39,780)}$
The first one doesn't give increasing sequence.[/hide][/quote][/hide]
Yes there is a typo. Thanks to Arnav...!!! and Daniel. I've edited it. But I don't think I did the same in the contest. And I don't think they would subtract more than 2 points if I did it.

[hide="@Daniel"]It is the first level of the Indian Math Olympiads called Regional Mathematical Olympiads. It is open for grades 8-12. But only 6 ppl are selected from grade 12 per region, while from the 18 regions of India they choose about 30 from each in total. Mostly this contest is for grade 11.[/hide]

My solution's idea

[hide]
by simple algebraic computation, we get b(20+c)=40c
===>  (40-b)(20+c)=800.
Now since a<b, we have b>20.
===> 40-b<20.
Now for tracking the solutions for the diophantine, we can go via 40-b since we know it ranges in natural no. factors of 800 form 1 to 20 and correspondingly find the value of 20+c..First take 1X800, then powers of two till 4 for 40-b, then powers of 5 then combination of 2 and 5.
U get 7 equations but out these two don't satisfy the property that b divides c.
Hence there are five solutions:
20,30,60     20,35,140 , 20, 36,180 , 20,38.380   ,  20, 39,780.
Is the idea nice?[/hide][/hide]

hey hello,
i did the same calculations as akash but just messed up with the calculations
any marks????????

Don't know.
This is a Diphantine. So they expect u to be proper in calculations.
Also, when u finally obtain solutions, u must have checked them???U won't get any major points if the solutions were wrong.

romil and skand just chill out buddies. u will get selected :D  :)  :lol:
btw aravind u just factorised in a different way right
i think akash's solution can get a full too!!!!!!!!
i did it the same way as akash............
what abt others

well i know that i won't be selected!!!!!!!!!!

@befuddlers
It may appear so, but the tracking is more organised and rigorous.

[quote="A K Bhattacharyya"]Can anyone tell me how much marks would I get if I could find out only one solution ?[/quote]

One solution to this problem or one solution for only one problem?

What's your name? The results are out here: http://www.isical.ac.in/~rmo/
It's all about getting chance. No point worrying about how many marks etc. Are you in there?

This was one of the easiest questions... solution :
1/a + 1/c = 2/b
Let c = bx.
So, 1/20 + 1/bx = 2/b
=(bx + 20)/20bx = 2/b
=(bx + 20)/20x = 2
= bx - 40x = -20
=x(b - 40) = -20
=x(40-b) = 20
20 = 1*20, 2*10, 4*5, 5*4, 10*2, 20*1
Which gives us the solutions -
(b,c) = (20,20), [b][u](30,60), (35, 140), (36, 180), (38, 380), (39, 780)[/u][/b]
The solution (b,c) = (20,20) is not possible as we need the numbers in increasing order.
So we have a total of 5 solutions, highlighted above.
P.S. Can anyone suggest a website for learing LaTeX? I really need help.


If $ l x l \plus{} x \plus{} y\equal{}10$ and $ x \plus{} l y l \minus{} y \equal{} 12$, find $ x \plus{} y$.

(A) -2
(B) 2
(C) $ \frac{18}{5}$
(D) $ \frac{22}{3}$
(E) 22

Thanks in advance!

looking at the second  equation, if y were positive or 0, then abs(y)=y, and that would mean x=12, which does not check in the first equation. then check the cases where y is negative, so abs(y)=-y.
then you get a system in two variables, so i would suggest solving for abs(x)=x, and then abs(x)=-x

y < 0 is right.

We can intuitively conclude that x >= 0, since x < 0 would give you:

-x + x + y = y = 10

which contradicts our previous statement.

[hide]
x - 2y = 12
2x + y = 10

y = -14/5, x = 32/5

x + y = [b]18/5[/b]
[/hide]


Set x>0 and let the sequence $(x_n)$ be defined by $x_1=x,x_2=x^{x_1},x_3=x^{x_2},...,x_n=x^{x_{n-1}},...$
For what values of x is the sequence convergent?

Feels like $x \le 1$ where $x \in \mathbb{R}$. I shall try to prove it, please hold on :)

Mike

Seems strange to me that it also converges for some small values larger than 1.
I used Mathematica to compute it...

I remember thinking about it a very long time ago, during a train ride, and reaching he conclusion that the answer is: all $0<x\le e^{\frac 1e}$, or something like that :). I'm not really sure though, and I don't have the time to check it.

[quote="grobber"] the answer is: all $0<x\le e^{\frac 1e}$[/quote]
Quite probable.

The answer given in the book is $(1/e)^e$≦$x$≦$e^{(1/e)}$.
But I still have no idea how to do it.

Monotone convergence theorem for $1 \le x \le e^{1/e}$

Less then this is harder!

The complete solution is posted by Vladimir Lubyshev and fedja here:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=533566&highlight=#533566[/url]


1. I seemed to have automatically changed into the MathLinks mode (no longer AoPS look).
2. I try to access the classroom, but it just opens a new window where I am logged out.  When I try to log in, I go back to MathLinks forum mode, and I have the same problem again.

[quote="probability1.01"]OK, well I think I have the classroom issue resolved, but I still don't know why I'm stuck in this MathLinks version :([/quote]Don't panic just yet :) You can easily exchange skins in two ways: 1) go to Options -> Board Layout -> Board Style and choose AoPS  2) Go to profile -> Go down the options -> Board Style -> AoPS. I prefere the first.


inequalities
inequalities solved
3-variable inequality
cyclic inequality
Rearrangement
AM-GM

First of all, Cauchy solves them both. Another idea is to use AM-GM:
 $2a^3+b^3 \ge 3a^2b$ for the first one and $3a^4+b^4 \ge 4a^3b$. Summing up analogue inequalities we get the desired result.

applying Muirhead's inequality this comes out trivial right?

[quote=Lagrangia]Show that for $a,b,c \geq 0$ we have $ \sum a^3 \geq \sum a^2b$ and $ \sum a^4 \geq \sum a^3b$.[/quote]

Rearrangement.


geometry
geometric transformation
dilation

Write the complete balanced equations for the following reactions:

Reaction of potassium with liquid ammonia

Combustion of nitromethane.

Combustion of methane in an atmosphere of fluorine,

$ \ce{K}(s) \plus{} am \longrightarrow \ce{K \plus{} }(am) \plus{} e^ \minus{} (am)$

The solution will have a deep blue color if dilute and a bronze appearance if concentrated (which are independent of the metal). However, these solutions will slowly decompose according to

$ 2\, \ce{K \plus{} }(am) \plus{} 2\, e^ \minus{} (am) \plus{} 2\, \ce{NH3}(l) \longrightarrow 2\, \ce{NaNH2}(am) \plus{} \ce{H2}(g)$

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

$ 2\, \ce{CH3NO2}(l) \plus{} \frac32 \, \ce{O2}(g) \longrightarrow 2\, \ce{CO2}(g) \plus{} \ce{N2}(g) \plus{} 3\, \ce{H2O}$

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

Don't you mean "reaction of methane in an atmosphere of fluorine"?

$ \ce{CH4} \plus{} \ce{F2} \longrightarrow \ce{CH3F} \plus{} \ce{HF}$


When will the next AMC 12 Problem Series course be offered?

Also, do you think it would be too much to take both this course and the AIME Problem Series at the same time?

We'll offer it either this summer or this fall.  Many students double up Problem Series courses, so it probably wouldn't be too overwhelming time-wise.

By the way, how much are Problem Series courses? It doesn't say on the enrollment page...

[quote="kyyuanmathcount"]Personally, I took AMC12, AIME, Trig/complex numbers, and intro to geometry simultaneously.[/quote]
But how much spare time did you have, aren't you in middle school?


Let b,a(n),c(n) be  positive integers such that :  b > 1,  (a(n+1)-a(n))  is an increasing sequence,  c(n) is a bounded sequence .

Prove that X =  :Sigma: (n=1,..., :inf:){c(n)/ba(n)} is  irrational .

 :cool:

A hint for anyone trying it:

[hide]Consider what that summation looks like when expressed in base b.[/hide]


geometry
geometric transformation
reflection

In how many ways can 8 people be seated around a round table?

I put 2520 at first...but the answer was 5040...I cant tell whether to correct for over counting for reflections on these problems or not.

You can't "reflect" a table, because you can't flip over a table.  Well, you can, but the people won't really be happy.  So the answer is $ \frac{8!}{8}\equal{}7!\equal{}5040$.

[hide=sol]Fix the person at the top to account for symmetry, $\boxed{7!}$ remaining ways.[/hide]


--------------------------------------------------------------------------------------------
[color=blue][b]Prove that:
$\left| {ax^2 + bx + c} \right| \leqslant 1,\forall x \in \left[ {0,1} \right] \Rightarrow \left| b \right| \leqslant 8$[/color][/b]
--------------------------------------------------------------------------------------------

[u][b]Solution:[/b][/u]

$\left| {ax^2 + bx + c} \right| \leqslant 1,\forall x \in \left[ {0,1} \right] \Rightarrow \left\{ \begin{gathered} \left| c \right| \leqslant 1 \hfill \\ \left| {a + b + c} \right| \leqslant 1 \hfill \\ \left| {a \cdot \frac{1} {4} + b \cdot \frac{1} {2} + c} \right| \leqslant 1 \hfill \\ \end{gathered} \right. \Rightarrow$

$\Rightarrow \left\{ \begin{gathered} \left| c \right| \leqslant 1 \hfill \\ \left| {a + b + c} \right| \leqslant 1 \hfill \\ \left| {a + 2 \cdot b + 4 \cdot c} \right| \leqslant 4 \hfill \\ \end{gathered} \right. \Rightarrow$

$\Rightarrow \left| b \right| = \left| {a + 2 \cdot b + 4 \cdot c - \left( {a + b + c} \right) - 3 \cdot c} \right| \leqslant$

$\leqslant \left| {a + 2 \cdot b + 4 \cdot c} \right| + \left| {a + b + c} \right| + \left| {3 \cdot c} \right| \leqslant 4 + 1 + 3 = 8 \Rightarrow \left| b \right| \leqslant 8$


Why can't $ a^{p\minus{}1} \equiv 1 (mod p^{2})$. Is there a theorem to this?

Can't it?  This certainly occurs when $ a \equal{} 1$.  The question you should really be asking, then, is "when is this true?"

sorry... i forgot to say it was for odd primes

oh wait never mind now i get it... the article i got it from stated a was a primitive root $ mod p^{2}$... oops


calculus LaTeX Asymptote integration symmetry trigonometry logarithms

The past few weeks I've been converting calculus notes into latex, but the more progress I make, the more I feel as though I'm repeatedly making bad formatting choices that I should correct before this gets any longer than 106 pages. Here's the main things I have in mind:

-In my textbook, important equations are contained within red boxes, with the equation number in the top-left corner. The equation number is also contained within its own little red box. I wouldn't mind using a similar format for boxing in important text and equations.

-I'd like to switch to Roman numerals so that it's possible to distinguish between chapter 1 and 2 equations. The command I tried in one of stevem's reccomended websites didn't work for me.

-In the align* command I'm constantly using, I'm not entirely show how to get text to flush right. Using any flush command outright causes a string of errors.

-Decrease the space between asymptote pictures and text. I'm thinking of possibly assigning equation numbers to pictures for easier reference.

-Using the \! command to separate the dx differential from the integrand instead of \; .

I am of course open to any other suggestions and how to go about implementing them.

[quote="Prime factorization"]In my textbook, important equations are contained within red boxes, with the equation number in the top-left corner. The equation number is also contained within its own little red box. I wouldn't mind using a similar format for boxing in important text and equations.[/quote]
Have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=empheq]empheq package[/url].

[quote="Prime factorization"]I'd like to switch to Roman numerals so that it's possible to distinguish between chapter 1 and 2 equations. The command I tried in one of stevem's reccomended websites didn't work for me. [/quote]
Not sure how you want roman numerals but if you use \numberwithin{equation}[chapter} then the equations will be numbered 1.1, 1.2, ... for chapter 1 then 2.1, 2.2, ... for chapter 2 etc.

[quote="Prime factorization"]In the align* command I'm constantly using, I'm not entirely show how to get text to flush right. Using any flush command outright causes a string of errors.[/quote]
Use \tag*{text}

[quote="Prime factorization"]Decrease the space between asymptote pictures and text.[/quote]
Ensure there is no space inside the graphics themselves.

[quote="Prime factorization"]Using the \! command to separate the dx differential from the integrand instead of \; .[/quote]
Why? It's usual to use \,
$ \int x^2 \;dx$ space is too big
$ \int x^2 \!dx$ not enough space
$ \int x^2 \,dx$ just right

Please give [i]small[/i] minimal examples if you need help. No-one will want to look through and compile long documents.

[quote]Not sure how you want roman numerals but if you use \numberwithin{equation}[chapter} then the equations will be numbered 1.1, 1.2, ... for chapter 1 then 2.1, 2.2, ... for chapter 2 etc.[/quote]

The command still doesn't work. Note that all my equation numbers are given by the \eqref{} command.

[quote]Ensure there is no space inside the graphics themselves.[/quote]

The following

[code]\begin{figure}[h]
\begin{center}
\includegraphics{calculus_2.pdf}
\end{center}
\end{figure}
\flushleft{\textbf{Symmetry}}[/code]

produces excess vertical space between my picture and the text below. I am fairly certain it is not the picture itself which is causing this.

[quote]Please give [i]small[/i] minimal examples if you need help. No-one will want to look through and compile long documents.[/quote]

See the new attachment below.

\numberwithin works very well but you have redefined the numbering so why not add the chapter to that number? Use for example:
[code]newcommand{\eqnnumber}{\stepcounter{equation}\tag{\Roman{chapter}.\arabic{equation}}}[/code]
[quote="Prime factorization"]
The following

[code]\begin{figure}[h]
\begin{center}
\includegraphics{calculus_2.pdf}
\end{center}
\end{figure}
\flushleft{\textbf{Symmetry}}[/code]

produces excess vertical space between my picture and the text below. I am fairly certain it is not the picture itself which is causing this.[/quote]The vertical space is caused by the center environment. Use \centering instead.

At the end of the document you do strange things with an array - first you double the separation then you put \\[8ex] at the end of the lines as well! On top of that you add a centering environment with lots of \displaystyle commands when the align environment is made precisely for that purpose. So chuck out the arraystretch command and use align inside a box. Or use the empheq package which makes life a little easier.
Either
[code]\centering\fbox{\parbox{6cm}{\small\begin{align*}
&\int \sin^{-1}(x) \, dx =  x\sin^{-1}(x) +  \sqrt{1-x^{2} } + C  \\
&\int \cos^{-1}(x) \, dx = x\cos^{-1}(x) -  \sqrt{1-x^{2} } + C  \\
&\int \tan^{-1}(x) \, dx = x\tan^{-1}(x) - \frac{1}{2} \ln( 1 + x^{2} ) + C  \\
&\int \cot^{-1}(x) \, dx = x\cot^{-1}(x) + \frac{1}{2} \ln (1 + x^{2} ) + C \\
&\int \csc^{-1}(x) \, dx = x\csc^{-1}(x) + |\sin^{-1}(x)| + C \\
&\int \sec^{-1}(x) \, dx = x\sec^{-1}(x) + |\cos^{-1}(x)| + C
\end{align*}}}[/code]
or
[code]\usepackage{empheq}
...
\small{\begin{empheq}[box=\fbox]{align*}
&\int \sin^{-1}(x) \, dx =  x\sin^{-1}(x) +  \sqrt{1-x^{2} } + C  \\
&\int \cos^{-1}(x) \, dx = x\cos^{-1}(x) -  \sqrt{1-x^{2} } + C  \\
&\int \tan^{-1}(x) \, dx = x\tan^{-1}(x) - \frac{1}{2} \ln( 1 + x^{2} ) + C  \\
&\int \cot^{-1}(x) \, dx = x\cot^{-1}(x) + \frac{1}{2} \ln (1 + x^{2} ) + C \\
&\int \csc^{-1}(x) \, dx = x\csc^{-1}(x) + |\sin^{-1}(x)| + C \\
&\int \sec^{-1}(x) \, dx = x\sec^{-1}(x) + |\cos^{-1}(x)| + C
\end{empheq}}[/code]which give virtually identical results.

Still having a couple problems

1) The following

[code]\small{\begin{empheq}[box=\fbox]{align*}
\int \tan(x) \, dx & = \ln|\sec(x)| + C & \int \cot(x) \, dx & = -\ln|\csc(x)| + C \\
\int \sec(x) \, dx & = \ln|\sec(x) + \tan(x)| + C & \int \csc(x) \, dx & = -\ln|\csc(x) + \cot(x)| + C
\end{empheq}} [/code]

has some of the expressions outside the box's borders.

2) I've looked everywhere and I really cannot find any information on labelling pictures. The best I can think of doing is using \eqref{} so that the picture's label corresponds with that of a theorem or formula. I've also tried to give pictures labels independent of any theorems/formulas but the \tag, \numberwithin and eqnnumber command all seem to fail unless I put them within a math command like align.

3) I'm still not sure how to contain both text and math expressions in a single box. If I do say

[code]
\fbox{
Johnny walks the dog.
\begin{align*}
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}
\end{align*}
Jimmy has a cat }
[/code]

I'll simply get error messages.

1) That's because it has spread into the margins. You can constrain it with a parbox of size that you want:
[code]\centering\fbox{\parbox{10cm}{\small\begin{align*}
\int \tan(x) \, dx & = \ln|\sec(x)| + C & \int \cot(x) \, dx & = -\ln|\csc(x)| + C \\
\int \sec(x) \, dx & = \ln|\sec(x) + \tan(x)| + C & \int \csc(x) \, dx & = -\ln|\csc(x) + \cot(x)| + C
\end{align*}}} [/code]

2) Labelling pictures is standard and is in virtually every manual. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url]
[code]\begin{figure}[ht]
\centering
\includegraphics{greek.png}
\caption{this is my picture}
\label{mypic}
\end{figure}
That was \ref{mypic}[/code]

3) \fbox only takes a single paragraph so you need to put the code inside a parbox. See 1) and previous post.

[quote="stevem"]1) That's because it has spread into the margins. You can constrain it with a parbox of size that you want:
[code]\centering\fbox{\parbox{10cm}{\small\begin{align*}
\int \tan(x) \, dx & = \ln|\sec(x)| + C & \int \cot(x) \, dx & = -\ln|\csc(x)| + C \\
\int \sec(x) \, dx & = \ln|\sec(x) + \tan(x)| + C & \int \csc(x) \, dx & = -\ln|\csc(x) + \cot(x)| + C
\end{align*}}} [/code][/quote]

Those sets of commands keep giving me a bad box no matter what the size of the parbox is.

Perhaps 10cm is too wide for your page and goes into the margins - reduce it to whatever works.

No size paramater will fix it, no matter how big or small. In fact, the nature of the bad box is strange. The smaller I make it, the larger my "overfull hbox" becomes. Sometimes it gets as large as 350+ pts, though I can never visually see any changes. Still, I wouldn't mind getting rid of this error message.

Also, I really need a shortcut command that contains fbox, parbox and the align* environment. I've tried programming it myself, but I keep getting error messages.

[quote="Prime factorization"]No size paramater will fix it, no matter how big or small. In fact, the nature of the bad box is strange. The smaller I make it, the larger my "overfull hbox" becomes. Sometimes it gets as large as 350+ pts, though I can never visually see any changes. Still, I wouldn't mind getting rid of this error message.[/quote]Make it larger. 15cm (assuming the document has a left and right margin of 2cm) works fine.

[quote="Prime factorization"]Also, I really need a shortcut command that contains fbox, parbox and the align* environment. I've tried programming it myself, but I keep getting error messages.[/quote]You have to use quite different ideas to use \fbox, \parbox and \align inside \newenvironment. \box has to be replaced by \savebox, \parbox by minipage and \align needs \align...\endalign.

But, as I've said before, the empheq package makes all this much easier. If you really want a macro for
[code]\small\begin{empheq}[box=\fbox]{align*}
\int \tan(x) \, dx & = \ln|\sec(x)| + C & \int \cot(x) \, dx & = -\ln|\csc(x)| + C \\
\int \sec(x) \, dx & = \ln|\sec(x) + \tan(x)| + C & \int \csc(x) \, dx & = -\ln|\csc(x) + \cot(x)| + C
\end{empheq}[/code]then use
[code]\newenvironment{alignbox}
   {\small\empheq[box=\fbox]{align*}}
   {\endempheq}[/code]so you can write
[code]\begin{alignbox}
\int \tan(x) \, dx & = \ln|\sec(x)| + C & \int \cot(x) \, dx & = -\ln|\csc(x)| + C \\
\int \sec(x) \, dx & = \ln|\sec(x) + \tan(x)| + C & \int \csc(x) \, dx & = -\ln|\csc(x) + \cot(x)| + C
\end{alignbox}[/code]

I guess I just don't see how empheq is going to create the box format I'm looking for, which is basically the same as in the picture I scanned in the attachment.

The best I can do on my own is as follows:

[code]\fcolorbox{red}{white}{\parbox{10cm}{
If $f$ is continuous on $[a,b]$, then there exists a number $c$ in $[a,b]$ such that:
\begin{align*}
 f(c) = f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx  \eqnnumber \label{mvtintegral1}
\end{align*}
Insert some more writing here. }}[/code]

problem is, only about half the box gets a red border, there are no horizontal or vertical margins to separate the text from the box, and I have no way of labelling the box as a whole, just the contents within it.

Empheq is an easy way of using fbox so works very similarly. However, you want something different which is a frame around a lot of text which also has mathematics plus a label. You can get the layout in the image you posted with the number at the top left plus a reference to it, using the following code:
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{framed}
\usepackage{color}

\begin{document}
\newcounter{framedeq}
\refstepcounter{framedeq}
\color{red}
\begin{framed} {\color{black}
\noindent \fbox{\theframedeq} \label{feq}\ To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}}
\end{framed}
\normalcolor
That was the framed text with number \ref{feq}

%Now we repeat the process by incorporating this code in a new environment called 'myframed'

\newenvironment{myframed}[1]{\refstepcounter{framedeq}\color{red} \begin{framed} \color{black}
\noindent \fbox{\theframedeq} \label{#1}\ }{\end{framed}\normalcolor}

%This is used as follows.

\begin{myframed}{feq2}
To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}
\end{myframed}
That is the framed text with number \ref{feq2}\\
\medskip

Notice how the numbering is incremented and we can refer to the boxes: the second box has number \ref{feq2} whilst the first box has number \ref{feq}. There are no bad boxes either.

\end{document}[/code]

I copied your code into a new document but the second box keeps coming out out black, not red.

The code below produces the box I got in the attachment.

[code]\documentclass{article}
\usepackage{amsmath,amssymb,framed,color}
\newcounter{framedeq}
\refstepcounter{framedeq}
\newenvironment{myframed}[1]{\refstepcounter{framedeq}\color{red} \begin{framed} \color{green}
\noindent \fbox{\theframedeq} \label{#1}\ }{\end{framed}\normalcolor}
\begin{document}

\begin{myframed}{feq1}
To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}
\end{myframed}
That is the framed text with number \ref{feq1}.

\end{document}[/code]

Two errors, one yours, one mine:

1) You don't need \refstepcounter{framedeq} on line 4 as it is already defined in the newenvironment, unless you really want to start the numbering at 2 instead of 1.

2) There's a missing \color{red} near the end of newenviroment. It should read:
[code]\newenvironment{myframed}[1]{\refstepcounter{framedeq}\color{red} \begin{framed} \color{green}
\noindent \fbox{\theframedeq} \label{#1}\ }{\color{red}\end{framed}\normalcolor} [/code]

Now I get a strange warning message: "pdfTeX warning: pdflatex: pop empty color page stack 0" and the result is unaffected (ie the same bad box as before).

Here is a complete document that has no errors or bad boxes. Check that is OK and then slowly add other parts of your document to find where the error is.
[code]\documentclass{article}
\usepackage{amsmath,amssymb,framed,color}
\newcounter{framedeq}
\refstepcounter{framedeq}
\newenvironment{myframed}[1]{\refstepcounter{framedeq}\color{red} \begin{framed} \color{green}
\noindent \fbox{\theframedeq} \label{#1}\ }{\color{red}\end{framed}\normalcolor}
\begin{document}

\begin{myframed}{feq1}
To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}
\end{myframed}
That is the framed text with number \ref{feq1}.

\end{document}[/code]

The warning stems from the \myframed environment itself and goes away if I remove that new color command you added.

You have probably made an error somewhere in your code. If you copy and paste the above complete document, then it will work.

I did just that and the warning remains.

Are you using pdflatex to compile? Try using the xcolor package instead of color:
\usepackage{amsmath,amssymb,framed,xcolor}

It's the first line of a new paragraph so will be indented like all new paragraphs are as you would see if you write
[code]That is the framed text with number \ref{feq1} \\
isn't it?[/code]
\noindent will remove any automatic indentation of that line.

Oh one last thing, how can I turn the label argument into an optional one? If I simply do

\begin{myframe}
bla bla
\end{myframe}

The first letter of text dissapears. However, this isn't an issue if I put \begin{myframe}{}.

The label is conditional in the folowing code. Note the addition of the ifthen package and that the label must now use [] not {} as in \begin{myframed}[feq1] since [] is used for optional arguments.
[code]\documentclass{article}
\usepackage{amsmath,amssymb,framed,xcolor}
\newcounter{framedeq}
\usepackage{ifthen}
\newenvironment{myframed}[1][]{\refstepcounter{framedeq}\color{red} \begin{framed} \color{green}
\noindent \fbox{\theframedeq} \ifthenelse{\equal{#1}{}}{}{\label{#1}}\ }{\color{red}\end{framed}\normalcolor}
\begin{document}

% labelled case
\begin{myframed}[feq1]
To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}
\end{myframed}
That is the framed text with number \ref{feq1} \\ isn't it?.

%No label in this case
\begin{myframed}
To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}
\end{myframed}
That is the framed text with no label.
\end{document}[/code]

You my prefer another idea and that is to dispense with arguments, optional or otherwise. A label called framedeq1, framedeq2 etc is added in all cases, where the number is the one in that is shown in the box. Then you can refer to this label or not as the case may be. It means that you write \begin{myframed} without any arguments.
[code]\documentclass{article}
\usepackage{amsmath,amssymb,framed,xcolor}
\newcounter{framedeq}
\newenvironment{myframed}{\refstepcounter{framedeq}\color{red} \begin{framed} \color{green}
\noindent \fbox{\theframedeq} \label{framedeq\theframedeq}\ }{\color{red}\end{framed}\normalcolor}
\begin{document}

\begin{myframed}
To evaluate the integrals (a) $\int \sin mx \cos nx \,dx$, (b) $\int \sin mx \sin nx \,dx$ or (c) $\int \cos mx \cos nx \,dx$, use the corresponding identity:
\begin{align*}
\text{(a)} \quad \sin A \cos B &= \tfrac{1}{2}[\sin(A-B) + \sin(A-B)] \\
\text{(b)} \quad\sin A \sin B &= \tfrac{1}{2}[\cos(A-B) - \sin(A-B)] \\
\text{(c)} \quad\cos A \cos B &= \tfrac{1}{2}[\cos(A-B) + \cos(A-B)]
\end{align*}
\end{myframed}
That is the framed text with number \ref{framedeq1}.

\end{document}[/code]

The new box environment (I called it bx) still has some major issues.

First off, it doesn't get rid of spacing issues caused by the align* environment.

[code] \begin{bx}
\begin{align*}
\lim_{x \to a} ( \; f(x) + g(x) \; ) = L + M
\end{align*}
\end{bx}
[/code]

creates a line of empty space above and below what's contained in the align* environment.

Next is spacing issues involving text that is too close to the label in the top-left corner of the box. For instance,

[code]
\begin{bx}
If $f(x) \leq g(x)$ for all $x$ in an open interval that contains a (except possibly at a), then:
\begin{align*}
L\leq M
\end{align*}
\end{bx}
[/code]

the word "possibly" is practically touching the "18" label above it.

Then there's this slight alignment issue with the text.

[code]
\begin{bx}
A function $f$ is continuous at a number $a$ if
\begin{align*}
\lim_{x \to a} f(x)= f(a)
\end{align*}
or more precisely
\begin{align*}
0 < |x-a| < \delta \Longrightarrow |f(x)-f(a)| < \epsilon
\end{align*}
\end{bx}
[/code]

The line "or more precisely" isn't in alignment with the line "A function f is continuous ...".

Though, the worst effect comes when a box gets caught between 2 pages. What happens is the box has a black frame on one page, then has an empty red frame on the next page. See the attachment.

[quote="Prime factorization"]First off, it doesn't get rid of spacing issues caused by the align* environment..[/quote]Of course it doesn't. The align environment puts that in deliberately wherever it is used. Use \ [... \ ] instead. If you must use align then add  a negative vspace.

[quote="Prime factorization"]the word "possibly" is practically touching the "18" label above it.

Then there's this slight alignment issue with the text.

The line "or more precisely" isn't in alignment with the line "A function f is continuous ...".[/quote]Put the text and equations in a minipage such as:
[code]\newenvironment{bx}{\refstepcounter{framedeq}\color{red} \begin{framed} \color{green}
\noindent \fbox{\theframedeq} \label{framedeq\theframedeq}\ \begin{minipage}[t]{10cm}}{\end{minipage}\color{red}\end{framed}\normalcolor} [/code]
[quote="Prime factorization"]Though, the worst effect comes when a box gets caught between 2 pages. What happens is the box has a black frame on one page, then has an empty red frame on the next page. See the attachment.[/quote]Adjust your layout at the end so that it does't happen or put the boxes into floats to let LaTeX move it onto the next page.


OK, so I got first at my school, and my school team always pwns(in a relative manner, really the averagge chapter score is about 20, but it beats the other schools) so I'm guaranteed to go to state.  So who else is going?

I know AwesomeToad, summai, etc. but who else?

Mark Sellke, known as [u]geoishard[/u]

He doesn't post though... so don't bother him, let me bother him

[quote="Maybach"]According to my math teacher, the state competition is March 13. Is that right?[/quote]
Yep it is


function
real analysis
real analysis unsolved

Prove that if $ n>1$, the function $ u\equal{}loglog(1\plus{}\frac{1}{|x|})$ belongs to $ W^{1,n}(U)$, for $ U\equal{}B^{0}(0,1)$.

[quote="F-o-u-r-i-e-r"]Prove that if $ n > 1$, the function $ u \equal{} loglog(1 \plus{} \frac {1}{|x|})$ belongs to $ W^{1,n}(U)$, for $ U \equal{} B^{0}(0,1)$.[/quote]

$ u(4/3)$ is not well-defined?


PT For positive reals $ a, b, c$;
$ \frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{a}}{b\plus{}c}\plus{}\frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{b}}{c\plus{}a}\plus{}\frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{c}}{a\plus{}b} \ge \frac{9\plus{}3\cdot\sqrt{3}}{2\cdot\sqrt{a\plus{}b\plus{}c}}$

 Hope you will enjoy it!!!!        :D

[hide="Big hint"]
It's homogenous, so normalize to a+b+c=1. Now it looks easier :) [/hide]

I have some terms I don't understand

What do you mean with "homogenous" , "normalize" and what is the "Brute force"

Read Topics in Inequalities (Hojoo Lee)
Homogenize - to get every term in the inequality to the same degree.
Example: abc=1
a+1 isn't homogenous, as a is first degree, and 1 is zero degree
But, $ a\plus{}(abc)^\frac13$ is homogenous, as both terms are of first degree. This is homogenizing.
Normalize - look at my posts at http://www.mathlinks.ro/viewtopic.php?t=252146
Brute force - doing problems without nice ideas, just multiply everything (in this case). .

[hide]
Since the inequality is homogeneous let $ a \plus{} b \plus{} c \equal{} 1$ and the inequality becomes

$ \sum \frac {1}{1 \minus{} \sqrt {a}} \ge \frac{1}{1\minus{}\sqrt{\frac{1}{3}}}\equal{}\frac {9 \minus{} 3\sqrt {3}}{2}$

Now just apply weighted jensen [/hide]

I don't think the explanations of homogeneous inequalities above are very clear (indeed, I haven't ever seen many clear explanations), so let me rephrase that. An inequality with variables $ a_1,a_2,\ldots,a_n$ is homogeneous if we can substitute multiples of $ a_1,a_2,\ldots,a_n$ - i. e. $ ba_1,ba_2,\ldots,ba_n$, where we call $ b$ the [b]scaling factor[/b]  - for the original variables and still have the inequality hold true - this happens when all the terms are of the same degree $ k$, since multiplying the original inequality by $ b^k$ yields the substitution. Hence in an homogeneous inequality, it's always possible to pick a scaling factor $ b$ so that $ \sum ba_i \equal{} 1$, so we can assume without loss of generality that $ \sum a_i \equal{} 1$. For this specific case, we can assume that $ a \plus{} b \plus{} c \equal{} 1$.

What about solving without homogenizations or normalisations??
Think about it, it's easier.
It can be solved with even very little knowledge. :D
All your proposals are correct, but I did not think of it in any such way! :)

Some IMO problems become easy with these methods. Try and IMO '01/2 inequality.

[hide]
Okay, I will try to solve it without nomaliztion or jensen...
Rearrange to get
$ \sum \left(\frac{a\plus{}b\plus{}c}{b\plus{}c} \left( 1 \plus{} \sqrt{\frac{a}{a\plus{}b\plus{}c}}\right)\right) \ge \frac{9}{2} \plus{} \frac{3\sqrt{3}}{2}$

Now it is well known that $ \sum \frac{a}{b\plus{}c} \ge \frac{3}{2}$, which can be proven by expanding and applying Muirhead/AM-GM. So $ \sum \frac{a\plus{}b\plus{}c}{b\plus{}c} \ge \frac{9}{2}$

Now it suffices to prove that

$ \sum \frac{a\plus{}b\plus{}c}{b\plus{}c}\cdot\sqrt{\frac{a}{a\plus{}b\plus{}c}} \ge \frac{3\sqrt{3}}{2}$

So assume wlog $ a\ge b\ge c$, which implies $ \frac{a\plus{}b\plus{}c}{a\plus{}b}\le \frac{a\plus{}b\plus{}c}{a\plus{}c}\le \frac{a\plus{}b\plus{}c}{b\plus{}c}$
So by Checbyshev

$ \sum \frac{a\plus{}b\plus{}c}{b\plus{}c}\cdot\sqrt{\frac{a}{a\plus{}b\plus{}c}} \ge \frac{1}{3}\left(\sum \frac{a\plus{}b\plus{}c}{b\plus{}c}\right)\left(\sum \sqrt{\frac{a}{a\plus{}b\plus{}c}}\right)$

Now it suffices to prove that
$ \sum\sqrt{\frac{a}{a\plus{}b\plus{}c}}\ge \sqrt{3}$

Which is equivalent to
$ x\plus{}y\plus{}z\ge \sqrt{3}$ given that $ x^2\plus{}y^2\plus{}z^2\equal{}1$

And this is true because $ (x\plus{}y\plus{}z)^2 \equal{} x^2\plus{}y^2\plus{}z^2 \plus{} 2(xy\plus{}yz\plus{}zx)$

And by rearangement $ xx^2\plus{}y^2\plus{}z^2\ge xy\plus{}yz\plus{}zx$
$ \Rightarrow (x\plus{}y\plus{}z)^2\ge 3(x^2\plus{}y^2\plus{}z^2)\equal{}3$

QED
I dont know if i'd call that easier :P
Now show us your proof Polta

[/hide]

Actually my method is somewhat similar to ocha's one.
To me my method is easy but there is one thing.
I will post my solution in my next post.....
I don't know if you will think the method to be easy/tough/long(whatever!), but I assure you, its [b]fundamental[/b]!!! :)

There are two inequalities which I use here in my solution.
[hide]
[hide="No. 1"]
If $ x,y,z$ are positive reals st $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$, PT:
$ P \equal{} \frac {x}{1 \minus{} x^2} \plus{} \frac {y}{1 \minus{} z^2} \plus{} \frac {z}{1 \minus{} z^2}\ge \frac {3\sqrt {3}}{2}$
....................................(1)
[hide="solution/proof"]
From the RMS AM ineq,
$ \frac {(x \plus{} y)^2}{2}\le x^2 \plus{} y^2 \equal{} 1 \minus{} z^2$ so that $ \frac {z}{1 \minus{} z^2}\ge \frac {2z}{(x \plus{} y)^2}$
Hence it follows that $ P \ge \frac {2x}{(y \plus{} z)^2} \plus{} \frac {2y}{(z \plus{} x)^2} \plus{} \frac {2z}{(x \plus{} y)^2}$................(*)

From CS,
$ (x \plus{} y \plus{} z)(\frac {x}{(y \plus{} z)^2} \plus{} \frac {y}{(z \plus{} x)^2} \plus{} \frac {z}{(x \plus{} y)^2})\ge (\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y})^2\ge (\frac {3}{2})^2 \equal{} \frac {9}{4}$
(from the nessbitt's inequality).

Further, from the RMS AM ineq,
$ (\frac {x \plus{} y \plus{} z}{3})^2\le \frac {x^2 \plus{} y^2 \plus{} z^2}{3} \equal{} \frac {1}{3}$
$ \Longrightarrow x \plus{} y \plus{} z\le \sqrt {3}\Longrightarrow \frac {1}{x \plus{} y \plus{} z}\ge \frac {1}{\sqrt {3}}$

Using this (*) gives:
$ P\ge \frac {2}{x \plus{} y \plus{} z}(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y})^2\ge\frac {2}{\sqrt {3}}\frac {9}{4} \equal{} \frac {3\sqrt {3}}{2}$
[/hide][/hide]
[hide="Solution"]
Let $ s \equal{} a \plus{} b \plus{} c$ and $ x \equal{} \frac {s}{a}$; $ y \equal{} \frac {s}{b}$ and $ z \equal{} \frac {s}{c}$ Then we see that the given ineq can be written as:
$ \frac {\sqrt {s} \plus{} \sqrt {a}}{b \plus{} c} \plus{} \frac {\sqrt {s} \plus{} \sqrt {b}}{c \plus{} a} \plus{} \frac {\sqrt {s} \plus{} \sqrt {c}}{a \plus{} b}\ge \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {s}}$

On multiplying both sides by $ \sqrt {s}$ we get :

$ \frac {s \plus{} \sqrt {sa}}{b \plus{} c} \plus{} \frac {s \plus{} \sqrt {sb}}{c \plus{} a} \plus{} \frac {s \plus{} \sqrt {sc}}{a \plus{} b}\ge \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {s}}$

Again, dividing both the numerator & the denominator of the LHS by s gives:

$ \frac {1 \plus{} \sqrt {x}}{y \plus{} z} \plus{} \frac {1 \plus{} \sqrt {y}}{z \plus{} x} \plus{} \frac {1 \plus{} \sqrt {z}}{x \plus{} y}\ge \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {s}}$
Observe that $ x \plus{} y \plus{} z \equal{} 1$.

thus, it is reqd to prove that:
$ \frac {1 \plus{} \sqrt {x}}{1 \minus{} x} \plus{} \frac {1 \plus{} \sqrt {y}}{1 \minus{} y} \plus{} \frac {1 \plus{} \sqrt {z}}{1 \minus{} z}\ge \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {s}}$

Now, from the CS,
$ (1 \minus{} x \plus{} 1 \minus{} y \plus{} 1 \minus{} z)(\frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z})\ge 9$
or,  $ (3 \minus{} (x \plus{} y \plus{} z))(\frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z})\ge 9$
or, $ (\frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z})\ge \frac {9}{2}$

It's finally time to use no. (1):
hence
$ \sum \frac {\sqrt {x}}{1 \minus{} x}\ge \frac {3\sqrt {3}}{2}$..........(2)

Adding (1) & (2) we get the inequality RTP!!! :D  :lol: [/hide]
What do you think about it? It perhaps was long and tiring to write in LaTeX, but it was fundamental![/hide]

your proof of no. 1 is incorrect... in particular the sign is flipped on $ \frac{z}{1\minus{}z^2}\ge \frac{2z}{(x\plus{}y)^2}$

also your substitution is confusing...are s and p supposed to be the same?

[quote="Bugi"]No. 1 is incorrect, try $ x \equal{} \frac {1}{\sqrt {2}},z \equal{} 0.0000\ldots1$ :wink:

And s and p are the same, they both stand for halfperimeter.[/quote]
No.1 is correct :
$ \frac{x}{1\minus{}x^2} \geq \frac{3\sqrt{3}}{2}x^2$ is equivalent to $ (sqrt{3}x\minus{}1)^2(2\plus{}sqrt{3}x) \geq 0$. Setting up two other similar inequalities with y and z then add them up give the result

"Normalize"  doesn't weeken inequalities . It only makes it look easier and simpler , therefore , if one can solve the problem using "Normalize" , s/he can also prove it without using "normalize" .

[quote="Bugi"]No. 1 is incorrect, try $ x \equal{} \frac {1}{\sqrt {2}},z \equal{} 0.0000\ldots1$ :wink:
[/quote]

Bad example. I thought that there was really $ \frac{y}{1\minus{}z^2}$ :rotfl:  :rotfl:  :rotfl:

[quote="Potla"][quote="Altheman"]your proof of no. 1 is incorrect... in particular the sign is flipped on $ \frac {z}{1 \minus{} z^2}\ge \frac {2z}{(x \plus{} y)^2}$

also your substitution is confusing...are s and p supposed to be the same?[/quote]

Thanks I see that no 1 cannot be done that way.
But in my opinion, HTA has given the correct solution.
The inequality in no 1 is absolutely correct.
However, thanks for your notice, I make these types of mistakes quite a lot.[/quote]

Hey what am I talking about??? :huh:

The inequality is ABSOLUTELY CORRECT!!!! :lol:

As you see, $ 0<1\minus{}z^2<1$ (strictly as z is POSITIVE)
So $ \frac {z}{1 \minus{} z^2}\ge \frac {2z}{(x \plus{} y)^2}$ is completely correct as a statement....

Why did I take so many days to realise it?!?! :wallbash:

$ \frac {(x \plus{} y)^2}{2}\le x^2 \plus{} y^2 \equal{} 1 \minus{} z^2$
$ \implies$ $ 1 \minus{} z^2\geq\frac {(x \plus{} y)^2}{2}$
$ \implies$     $ \frac {1}{1 \minus{} z^2}\leq\frac {2}{(x \plus{} y)^2}$
$ \implies$  $ \frac {z}{1 \minus{} z^2}\leq\frac {2z}{(x \plus{} y)^2}$

@Potla:you see the inequality is flipped
And it took you so many days bcoz what u say now is incorrect :rotfl:  and you have forgotten what is correct  .


if A is congruent to B in mod M, and P is congruent to Q in mod M, then why is A+B congruent to P+Q in mod M and AB congruent to PQ in mod M

That's not true.

Let $ A \equal{} B \equal{} 1$, and $ P \equal{} Q \equal{} 3$. $ A\plus{}B$ mod $ 3$ is $ 2$, but $ P\plus{}Q$ mod $ 3$ is zero. Similarly, $ AB$ mod $ 3$ is $ 1$, but $ PQ$ mod $ 3$ is zero.

What I think you meant is $ a\equiv b\bmod m$ and $ p\equiv q\bmod m$ implies that $ a\plus{}p\equiv b\plus{}q\bmod m$, $ ap\equiv bq\bmod m$.

To prove those, you'd write $ a\equal{}mk\plus{}b$ and $ p\equal{}mj\plus{}q$ and go from there.

No...what I wrote is what I meant.... it's in aops volume 1, pg 44 - I don't really understand their explanation though, so I was hoping someone could explain it differently

Oh, that's a well known typo in AoPS 1. What they meant to say was what I wrote I that earlier post.

heh ok now I know the well known typo...thats good, because I was trying to make sense of what they wrote and I couldn't

but yea your version definitely makes sense

Is the typo still not fixed in the latest edition?

I have the 7th edition, not sure if it's the latest, but the error is in it.


I found that

1! = 1 is a square and
1! + 2! = 3
1! + 2! + 3! = 9 is a square

so for n=1 and n=3 we have squares.

How can we determine if there are any squares when n > 4 ?

[quote="Jean-Louis"]When is 1! + 2! + 3! + ... + x! a square ?[/quote]
Similary, find all positive integers $x,y,z$ such that $1!+2!+...+x!=y^{z}$.


Find number of ways in which you can go from $ A$ to $ B$ [b]without[/b] going through $ C$ (refer diagram).

There are total $ \binom{5\plus{}8}{5}$ ways to go A--->B

There are total $ \binom{3\plus{}5}{3}$ ways to go A--->C

There are total $ \binom{2\plus{}3}{2}$ ways to go C--->B

$ \binom{5\plus{}8}{5}\minus{}(\binom{3\plus{}5}{3}\plus{}\binom{2\plus{}3}{2})\equal{}1287\minus{}(56\plus{}10)\equal{}1221$

[quote="Bachukas"]There are total $ \binom{5 \plus{} 8}{5}$ ways to go A--->B

There are total $ \binom{3 \plus{} 5}{3}$ ways to go A--->C

There are total $ \binom{2 \plus{} 3}{2}$ ways to go C--->B

$ \binom{5 \plus{} 8}{5} \minus{} (\binom{3 \plus{} 5}{3}\cdot\binom{2 \plus{} 3}{2}) \equal{} 1287 \minus{} (56 \cdot 10) \equal{} 727$[/quote]

Fixed. :)


trigonometry
function
calculus
calculus computations

wassup peeps !  :D

here is a question i just made up:

show

$ \sum_{k\equal{}1}^\infty\;\dfrac{\sin\;\left(\dfrac{k\pi}{3}\right)}{k^3}\;\equal{}\;\boxed{\dfrac{5\pi^3}{162}}$

From a familiar Fourier series, we know that

$ \sum_{k \equal{} 1}^{\infty}\frac {\sin(kx)}{k} \equal{} \frac {\pi \minus{} x}2$ for $ x\in(0,2\pi).$

Integrate that to get that

$ \minus{} \sum_{k \equal{} 1}^{\infty}\frac {\cos(kx)}{k^2} \equal{} C \minus{} \frac {(\pi \minus{} x)^2}4$

But this function must have average value zero over $ [0,2\pi].$ Computing the average value of the function gives us that $ C \equal{} \frac {\pi^2}{12}.$ Hence,

$ \minus{} \sum_{k \equal{} 1}^{\infty}\frac {\cos(kx)}{k^2} \equal{} \frac {\pi^2}{12} \minus{} \frac{(\pi \minus{} x)^2}4$ on $ [0,2\pi].$

As an intermediate check of our computation, we set $ x \equal{} 0$ in this expression to get that $ \zeta(2) \equal{} \frac {\pi^2}6.$

Integrate again:

$ \minus{} \sum_{k \equal{} 1}^{\infty}\frac {\sin(kx)}{k^3} \equal{} C \plus{} \frac {\pi^2x}{12} \plus{} \frac {(\pi \minus{} x)^3}{12}$

Again, this must average to zero. Computing the average gives us that $ C \equal{} \minus{} \frac {\pi^3}{12}.$ Multiplying by $ \minus{} 1,$ we get

$ \sum_{k \equal{} 1}^{\infty}\frac {\sin(kx)}{k^3} \equal{} \frac {\pi^3 \minus{} \pi^2x \minus{} (\pi \minus{} x)^3}{12}$ for $ x\in[0,2\pi].$

Evaluating this expression at $ x \equal{} \frac {\pi}3$ gives the desired result.


my belief is that the constant in the right hand side should be 9/2.
for now, that's all :)

You are wrong, Valentin!

Let a=b and c nearly to 0.

oh ... yeah right - didn't tought about that ... the \geq sign tricked me :) ... it most certainly resembles the following

\sqrt ( a/(b+c) ) + \sqrt ( b/(c+a) ) + \sqrt ( c/(a+b) ) > 2

at the above, the first time I saw it I though that 2 should be replaced with 3/ \sqrt 2 because the equality case wasn't fulfilled, but in the end I've managed to prove this ... in a rather pelicular way.


The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle.  The circles are tangent to one another and to the sides of the rectangle as shown in the diagram.  The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\frac{1}{2}\left(\sqrt{p}-q\right),$ where $p$ and $q$ are positive integers.  Find $p+q.$

[asy]
size(250);real x=sqrt(3);
int i;
draw(origin--(14,0)--(14,2+2x)--(0,2+2x)--cycle);
for(i=0; i<7; i=i+1) {
draw(Circle((2*i+1,1), 1)^^Circle((2*i+1,1+2x), 1));
}
for(i=0; i<6; i=i+1) {
draw(Circle((2*i+2,1+x), 1));
}[/asy]

[hide]If the circles have radius $r$, the length is obviously $14r$ (7 circles).  For the height, look at a circle in the top row and one directly below it in the bottom row, call their centers $A$ and $C$.  Look at the circles in the middle row tangent to circles $A$ and $C$, and call their centers $B$ and $D$.  Rhombus $ABCD$ is composed of 2 equilateral triangles with side lengths $2r$, implying the height of the rhombus is $2\sqrt{3}$.  That only goes from the center of a top circle to the center of a bottom circle, so add on $2r$ to get the complete height.

$\frac{14r}{2r+2r\sqrt{3}}= \frac{14}{2\sqrt{3}+2}\cdot \frac{2\sqrt{3}-2}{2\sqrt{3}-2}= \frac{28\sqrt{3}-28}{8}= \frac{1}{2}(7\sqrt{3}-7) = \frac{1}{2}(\sqrt{147}-7)$

$\Rightarrow p+q = 154$[/hide]


topology
real analysis
function
limit
inequalities
analytic geometry
advanced fields

Consider all sequences of numbers $ 1,2,3$. Is it possible to split the set of all such sequences into finitely many ($ 3$?) groups such that any two sequences from the same group coincide in infinitely many positions? :wallbash:

Do you know such a decomposition for sequences of 1s and 2s?

Yes, Joel. Let us call two sequences equivalent if they coincide outside finitely many places. Let us call them complementary if they coincide in finitely many positions. Then all equivalence classes split into complementary pairs. Now just [b]choose[/b] one element in each pair to form the first set and send the other pair elements to the second set. I know that you may be unhappy with using continuum AC for this seemingly elementary question, but it is inavoidable. We can view our sequences as elements of the product of countably many copies of $ \mathbb Z_2$ with the standard product topology and complete Haar measure. Now, if you have any measurable set $ A$ of positive measure, then $ A\minus{}A$ contains a neighborhood of the origin and, in particular, a sequence in which all but finitely many elements are $ 1$. But then $ A$ cannot be used as one of our groups and we conclude that if all groups are measurable, each must have measure $ 0$, so not only finite, but even a countable measurable partition is not feasible. Now, our space is a complete separable metric space and the measure is non-atomic, so it is equivalent to $ [0,1]$ with Lebesgue measure (the equivalence being given by binary representations of real numbers) and it is known that it is impossible to construct Lebesgue non-measurable sets in $ [0,1]$ without AC.

[quote="fedja"]Consider all sequences of numbers $ 1,2,3$. Is it possible to split the set of all such sequences into finitely many ($ 3$?) groups such that any two sequences from the same group coincide in infinitely many positions?[/quote]
Reformulation: Given two such sequences, draw an edge between them if they agree in only finitely many positions.  Then the question is whether this graph has finite chromatic number.  My guess is no.

Somehow your problem reminds me of [url=http://mathworld.wolfram.com/KnesersConjecture.html]Kneser's Conjecture[/url], which was settled by Lovasz using topology.  Basically, Lovasz's proof (and subsequent proofs) gives a topological lower bound on the chromatic number of any graph.  Maybe one of these lower bounds will be useful for your graph.  For one discussion of these topological bounds, see [url]http://arxiv.org/abs/math/0208072v3[/url].

Note: This post is just speculation.  Feel free to ignore it.

Sorry, I will now try to prove the opposite of what I conjectured in the last post.  Namely, I will try to prove that your graph is 3-colorable.

I will use the fact that if every finite subgraph of a (infinite) graph is 3-colorable, then the entire graph is 3-colorable.  That principle is sometimes called compactness, and probably uses the Axiom of Choice.

So, let's say we have a finite subgraph of your graph.  For every edge in this finite subgraph, there are a finite number of bad positions (positions where the two sequences agree).  Let's ignore the (finite) union of all those bad positions.  That leaves us with infinitely many good positions.  Let's focus on one of those good positions.  Then that position provides us with a 3-coloring of the vertices of the finite subgraph.  So we are done.

Note: I am writing this early in the morning, so I might have written nonsense.

No nonsense, Ravi; what you wrote makes perfect sense. Let me tell what I'm driving at. Take any continuous increasing function $ h: [0,\plus{}\infty)\to [0,\plus{}\infty)$ such that $ h(0)\equal{}0$, $ h(\infty)\equal{}\infty$. Now define $ h$-content $ M(E)$ of a subset $ E$ of a metric space $ X$ by
\[ M(E)\equal{}\inf\left\{\sum_j h(\operatorname{diam} E_j): E\subset\bigcup_j E_j\right\}\]
I wonder if for every increasing sequence of sets $ E_n$, we have
\[ M(\cup_n E_n)\equal{}\lim_{n\to\infty}M(E_n)\]
What I can prove is that if $ h(2t)\le Kh(t)$ for all $ t$ with some constant $ K$, then
\[ M(\cup_n E_n)\le K\lim_{n\to\infty}M(E_n)\]
and if you replace the sum of $ h$ of diameters and coverings by arbitrary subsets by the sum of $ h$ of radii and coverings by balls, then this last inequality still holds and cannot be improved without additional assumptions about the metric space, but the reason for this sharpness I know is fairly stupid and the deBruijn-Erdos theorem you attracted my attention to eliminates it for the diameter case. I asked a couple of experts (like Pertti Mattila) and they didn't know any results of this type for arbitrary metric spaces but, of course, nobody is omniknowledgable.

Fix a free ultrafilter $ U$ on $ \mathbb N$ and split the set of all sequences of numbers 1,2,3 into the following three groups $ A_1, A_2, A_3$: $ A_j$ consists of those sequences whose $ U$-limit equals $ j$. Then for every two elements of $ A_j$ the set of coordinates where both elements equal $ j$ is an element of $ U$, i.e. is infinite.

This solution implies by the way that the number of such partitions is $ 2^{continuum}$ (like the number of all free ultrafilters).


squaring the relation gives $ x_{n \plus{} 1}^2 \equal{} x_n^2 \plus{} \frac {1}{x_n^2} \plus{} 2$. summing this equation over $ n \equal{} 0$ to 999 gives $ x_{1000}^2 \equal{} 45^2 \plus{} \sum_{n \equal{} 0}^{999} \frac {1}{x_n^2}$. clearly this shows that $ x_{1000} > 45$. and since $ 45.1^2 \equal{} 2034.01$, it remains to show that $ \sum_{n \equal{} 0}^{999} \frac {1}{x_n^2} < 9.01$.

in the same way we got $ x_{1000} > 45$, we get $ x_{100} > 15$. by the fact that $ (x_n)$ is strictly increasing and $ x_0 \equal{} 5$, it then follows that $ \sum_{n \equal{} 0}^{999} \frac {1}{x_n^2} < 100\cdot \frac {1}{5^2} \plus{} 900\cdot \frac {1}{15^2} \equal{} 8 < 9.01$.


calculus integration limit logarithms function algebra domain

Evaluate \[1.\lim_{n\to \infty}e^{-na}n^{b}\]
\[2.\lim_{n\to \infty}n^{-a}\log n\] when a>0,b>0.
Shew that \[1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-...=\frac{1}{2}\log 2\]
If a be real the integral \[\int_{0}^{\infty}\frac{cos(ax)}{1+x^{2}}dx\]
is a continuous function of a.

The first limit is 0: "the exponential greater than 1 grows faster than any power of its exponent".

The second limit is also 0: "the numbers grow faster than any power of their logarythms".

Regarding the series: multiplicate both sides by 2,

$2-1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+... = ln2$

$\Leftrightarrow1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-... = ln2$

which is a true equality.

$\left|\frac{\cos(ax)}{1+x^{2}}\right|<\frac{1}{1+x^{2}}$ and $\int_{0}^{+\infty}\frac{1}{1+x^{2}}dx = \arctan x |_{0}^{+\infty}=\pi/2<\infty$, so $\int_{0}^{+\infty}\frac{\cos(ax)}{1+x^{2}}ds$  convergence uniformly. The last statement is follows from Weierstrass theorem. Then \[F(a)=\int_{0}^{+\infty}\frac{\cos(ax)}{1+x^{2}}ds\] is a continuous function

Next challenge: compute the function anik_andrew called $F(a),$ exactly. (This is possible.)

One comment on anik-andrew's proof: if you were to attempt the same argument for $F'(a),$ the argument would be insufficient, since $\int_{0}^{\infty}\frac{x\,dx}{1+x^{2}}=\infty.$ Indeed, if we consider $F(a)$ to be a function with domain $\mathbb{R},$ it is an even function that is continuous for all $a$ but which fails to be differentiable at some particular point.

I computed $F(a)$ with Mathematica for a few integer values. The answer seems to be $F(a) = \frac{\pi}{2 e^{|a|}}$.
Can this be proven without using any advanced techniques?

The two integrals

$\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx=\frac{\pi}{2}e^{-|a|}$,

and

$\int_{0}^{\infty}\frac{x\sin ax}{1+x^{2}}dx=\frac{\pi}{2}e^{-|a|}sign(a)$,

are called Laplace integrals.

I will calculate the first one. let $a>0$.

Let $y=\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx$. Then

$y'=-\int_{0}^{\infty}\frac{x\sin ax}{1+x^{2}}dx$.

Now add to the preceding integral the following one: $\frac{\pi}{2}=\int_{0}^{\infty}\frac{\sin ax}{x}dx$ (here we use the fact that $a>0$, and we get that

$y'+\frac{\pi}{2}=\int_{0}^{\infty}\frac{\sin ax}{x(1+x^{2})}$

It follows that $y''=y$ and hence $y(a)=C_{1}e^{a}+C_{2}e^{-a}$.

Now observe that $|y|\leq \int_{0}^{\infty}\frac{dx}{1+x^{2}}=\frac{\pi}{2}$, i.e., $y$ is bounded , hence $C_{1}=0$. Thus $y=C_{2}e^{-a}$. Now $y(0)=\frac{\pi}{2}$ and it follows that $C_{2}=\frac{\pi}{2}$.
The result follows.

[quote="didilica"]
Let $y=\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx$. Then

$y'=-\int_{0}^{\infty}\frac{x\sin ax}{1+x^{2}}dx$.
[/quote]
You can't just do this without care. The second integral converges very poorly. Why can you differentiate under the integral sign? I mean, the derivative isn't even continuous.

Well, if $a>0$ then the derivative is nothing else but $e^{-a}\frac{\pi}{2}$ which is continuous if $a>0$. However recall that I have choosen $a>0$.

In this case $\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx$ is uniformly convergent with respect to $a$.

To see this we apply the following  criteria.

Let $I(a)=\int_{0}^{\infty}f(x,a)g(x,a)dx$. Then
If $\int_{0}^{A}f(x,a)dx$ is uniformly bounded as a function of $A$ and $a$ i.e.,
$|\int_{0}^{A}f(x,a)dx|<K$, $K$ is a constant , $a\geqq_{0}>0$
and $g(x,a)\rightarrow 0$ when $x\rightarrow \infty$ uniformly with respect to $a$ then the integral $I(a)$ is uniformly convergent with respect to $a$.

Now for our case $f(x,a)=\cos ax$ and $g(x,a)=\frac{1}{1+x^{2}}$.
Observe that for $a\geq a_{0}$ we have

$|\int_{0}^{A}\cos ax dx|=|\frac{\sin aA}{a}|\leq\frac{1}{a}\leq \frac{1}{a_{0}}$ hence is uniformly bounded. The second condition is also satisfied, therefore our integral converges uniformly with respect to $a\geq a_{0}>0$.

[quote="didilica"]

In this case $\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx$ is uniformly convergent with respect to $a$.

[/quote]

If we want to show that \[\frac{d}{da}\left(\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx\right)=\int_{0}^{\infty}\frac{d}{da}\left(\frac{\cos ax}{1+x^{2}}\right)dx\] we need to show that \[\int_{0}^{\infty}\frac{d}{da}\left(\frac{\cos ax}{1+x^{2}}\right)dx\] is uniformly convergent with respect to $a$, not  $\int_{0}^{\infty}\frac{\cos ax}{1+x^{2}}dx$

The integral $\int_{0}^{\infty}\frac{x\sin ax}{1+x^{2}}dx$ is uniformly convergent with respect to $a$ for $a\geq a_{0}>0$ as an application of the same criteria.
As a matter of fact,

$\int_{0}^{\infty}\frac{x \sin \beta x}{\alpha^{2}+x^{2}}dx$ , $\alpha, \beta>0$,

is uniformly convergent with respect to $\beta$ for $\beta\geq \beta_{0}>0$.

For convergence of these integrals, there's always integration by parts: $\int_{0}^\infty \frac{x\sin ax}{1+x^{2}}\,dx=\left[\frac{1-\cos ax}{a}\cdot \frac{x}{1+x^{2}}\right]_{0}^\infty-\int_{0}^\infty \frac{1-\cos ax}a\cdot\frac{1-x^{2}}{(1+x^{2})^{2}}\,dx$ for nonzero $a$. The first term always evaluates to zero, while the second integral converges well enough to deduce continuity in $a$.

Continuity in $a,$ provided $a\ne 0$. This is the derivative of $F(a)=\frac{\pi}2e^{-|a|},$ and it it discontinuous at $a=0.$


This is a nasty trick :cool:

The derivative of x, with respect to x, is 2x.  However, suppose we write x as the sum of x x's, and then take the derivative:

Let f(x) = x + x + ... + x  (x times)

Then f'(x) = d[x + x + ... + x]  (x times)
  = d[x] + d[x] + ... + d[x]  (x times)
  = 1 + 1 + ... + 1  (x times)
  = x

This argument appears to show that the derivative of x, with respect to x, is actually x.  Where is the fallacy?

This is nasty.

I m not really sure but I ll take a try at it anyways.

I m pretty sure that the fallacy occurs when you break the x^2 up into linear x 's.   I can't explain exactly why but if this were possible then many more abnormalities would occur in different differentiation instances.

Sry, that was not at all rigorous.  This is really cool... I would really like to see the proper solution.

I don't get it... isn't x the definition of x.x the definition of x+x+x+...+x (x times)

please try again ;)

first, show me what you mean by $\pi^2=\pi + \pi + ... + \pi$, $\pi$(!) times. second, the number of addends is not constant, so we may not use $d[f+g]=d[f]+d[g]$, one can only generalize this only to a constant natural number of addends. anyway, really nasty that stuff.

Peter

ok my tricky is for naturals only

but you're right, the trick is that the amount of x'es does change while deriving :)

We had an amusing thread about this topic in AoPS about 10 months ago.  Here is the link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8790[/url].

I know this is perhaps a bit outdated, but Im having some trouble really grasping this.

Peter said that only if the number of addends is constant can we use the derivative. Why is that?

Nevermind, a friend explained it to me quite well.

You're treating x as both a variable and a constant...

The derivative of a constant is 0, while the derivative of a variable usually isn't.


When determining what questions go on the AIME I and AIME II do you consider swapping problems between the two if one has say too much geometry or algebra?

Umm...It's not like they randomize the problems on both...Perhaps they use a stratified random assignment to each test, I really don't know. But they do design the tests to be about the same difficulty, with (very) roughly the same subject matter. Also, they randomly choose which is A and which is B.

Our goal is to create interesting, challenging, and appropriate contests.  Like anyone who creates a test for a class, or math contest we do whatever it takes to create such a contest.

Steve Dunbar
MAA Director, American Mathematics Contests


function
real analysis
real analysis theorems

So, I'm teaching myself Real Analysis this summer and like all of you, I like to play around with some ideas. I was wondering if you can imply differentiability (smoothness) from Uniform Continuity. Why or Why not? (Counterexamples or intuitive explanations and maybe some hints as to constructing a rigorous proof)

This is not to be confused with Lipschitz Continuity, which states that if a function "f" on an open Interval I is differentiable, then the function is also is uniformly continuous.

No.  Lipschitz is a stronger condition than uniform continuity, and even that doesn't imply differentiability.  Given a point $ x\in\mathbb{R}$, let $ g(x)$ denote the distance between $ x$ and the nearest integer.  Define $ f(x)\equal{}\sum_{n\equal{}0}^{\infty}2^{\minus{}n}g(2^{n}x)$.  This function is uniformly continuous by the Weierstrass M-test, but is not differentiable at any point $ x\in\mathbb{R}$ because the sequence of difference quotients at any point successively differ by at least 1 every time.


calculus
integration
function
trigonometry
logarithms
calculus computations

Evaluate

\[\int^{\frac{\pi}{2}}_0 \frac{d\theta}{1+\sin \theta +\cos \theta }\]

This one is fairly easy, in my point of view.
Just apply trignometric substitution :  $t = \tan \frac{x}{2}$
The integral becomes
$\int_0^1 \frac{dt}{1+t} = \ln(1+t)|_0^1 = \ln 2$


geometry
circumcircle
ratio
trigonometry
geometric transformation
homothety
power of a point

Let the circle $ {\omega}_{1}$ be internally tangent to another circle $ {\omega}_{2}$ at $ N$.Take a point $ K$ on $ {\omega}_{1}$ and draw a tangent $ AB$ which intersects $ {\omega}_{2}$ at $ A$ and $ B$. Let $M$ be the midpoint of the arc $ AB$ which is on the opposite side of $ N$. Prove that, the circumradius of the $ \triangle KBM$ doesnt depend on the choice of $ K$.

I have corrected the problem isn't anyone intersted  :maybe:

Easily seen that r3 = r2*sqrt[(r2 - r1)/r2] and is independent of the choice of point K.

let the centres of circles $\omega_1, \omega_2$  be denoted $O_2,O_1$ respectively

[u][b]lemma 1[/b] [/u] $M,K,N$ are collinear
 since $O_1M \perp AB$ and $O_2K \perp AB$ we get $O_2K \parallel O_1M$
$\Longrightarrow \angle O_2KN=\angle O_1MN$ now since $O_1M=O_2N$ this gives us that $\angle O_1MN=\angle O_2NK$
this implies that
$\angle O_2NK = \angle O_2KN$
which is true, hence M,K,N are collinear

Let $A_1M$ intersect $AB$ at $T$ now from Pythagoras we have ,
$MB^2=MT^2 + TB^2= MT^2+\frac{AB^2}{4}$
$MK^2=MT^2 + KT^2= MT^2+(AK-AT)^2=MT^2 +(AK-\frac{AB}{2})^2
=MT^2 +AK^2 -AK.AB + \frac{AB^2}{4}=MB^2 - AK.KB = MB^2 - MK.KN$
$\Longrightarrow MK.MN=MB^2$

$\frac{MK}{MB}=\frac{MB}{MN}$

thus $\Delta KBM \simeq \Delta BNM$
now also observe that $\Delta O_2KN\simeq O_1MN$
this gives us that $\frac{MK}{MN}= \frac{r_2 -r_1}{r_2}$
now circumradii are in the same ratio of the sides if the two triangles are similar (quite evident from sine law)
$ \frac{R(KBM)}{R(BNM)}=\frac{MK}{MB}=\sqrt{\frac{MK}{MN}}=\sqrt{ \frac{r_2 -r_1}{r_2}}
\Longrightarrow R(KBM)= r_2.\sqrt{ \frac{r_2 -r_1}{r_2}}=\sqrt{r_2.(r_2-r_1)}$
which is thus independent of the position of K

Let the centres of $\omega_1,\omega_2$ with radii $r_1,r_2$  be $O_1,O_2$ respectively. Let $NK$ meet $O_2$ at $M'$. Then $\angle O_1KN=\angle O_1NK=\angle O_2NM'=\angle O_2M'N$ and therefore $O_1K||O_2M'$, which implies $O_2M'\perp AB$ since $AB$ is a tangent to $\omega_1$. So since $O_2$ is the centre of $\omega_2$ and $AB$ a chord, $M'$ is the midpoint of the arc $AB$, i.e. $M=M'$. So $N,K,M$ are collinear. Let $NA,NB$ meet $\omega_1$ at $A',B'$ respectively.

It is clear the homothety (enlargement) with centre $N$ that takes $\omega_1$ to $\omega_2$ takes the quadrilateral $NA'KB'$ to $NAMB$ (consider the common tangent to $\omega_1$ and $\omega_2$, by the alternate segment theorem, since $BN$ and $B'N$ make an equal angle with this tangent, $\angle NK'B=\angle NMB$ so $KB'||MB$). Note this homothety also takes the tangent to $\omega_1$ through $K$ to the tangent to $\omega_2$ through $M$, i.e. the tangent to $\omega_1$ through $M$ is parallel to $AB$. Let $C$ be an arbitrary point on this tangent to $\omega_2$, such that $C$ lies on the same side of $O_2M$ as $B$. By the alternate segment theorem, $\angle CMB=\angle MNB$. Also since $MC||AB$, we have $\angle CMB=\angle MBK$. Therefore $\angle MNB=\angle MBK$ and it is clear $\triangle MBN\sim\triangle MKB$, hence $\frac{MK}{MB}=\frac{MB}{MN}$.

Note that since $\triangle MBN$ and $\triangle MKB$ are in proportion, so is their circumradii. So if $R$ is the circumradius of $\triangle BKM$ then $\frac{R}{r_2}=\frac{MK}{MB}=\frac{MB}{MN}$. Now $\left(\frac{MK}{MB}\right)^2=\frac{MK}{MB}\cdot\frac{MB}{MN}=\frac{MK}{MN}$ meaning $\frac{MK}{MB}=\sqrt{\frac{MK}{MN}}$.

The homothety implies $\frac{NO_1}{NO_2}=\frac{NK}{NM}$. Then $\frac{MK}{MN}=\frac{MN-NK}{MN}=1-\frac{NK}{NM}=\frac{NO_1}{NO_2}=\frac{r_2-r_1}{r_2}$. So $\frac{R}{r_2}=\frac{MK}{MB}=\sqrt{\frac{MK}{MN}}=\sqrt{\frac{r_2-r_1}{r_2}}$ and it follows that $R$ is fixed.

[hide="Non-synthetic, using trigonometry"]
Basically we directly compute the circumradius of $\triangle BKM$, which will be denoted by $R$. Let $O$ be the circumcenter of $\triangle BKM$. Let $P$ be the midpoint of $AB$ (with $MP \perp AB$). Also let $r_1$ and $r_2$ be the radii and $O_1$ and $O_2$ be the two centers of $\omega_1$ and $\omega_2$ respectively. Finally, let $\angle KNO_2= \theta$.

As in the other solutions, a homethety centered at $N$ takes circle $\omega_1$ to circle $\omega_2$, so $M$, $K$, and $N$ are collinear. We immediately get that $\angle KNO_1=\angle O_1KN=\angle MNO_2=\angle O_2MN=\theta$ and $\angle MKB = 90 - \theta$. First note that by the definition of $O$, we have $\angle MOB = 2\angle MKB = 180 - 2\theta$. Drawing the perpendicular from $O$ to $BM$ gives $\angle OMB = \theta$ and
\[R= OM=\frac{BM}  {2 \cos \theta} \: \: (\star).\]
Now all we need to do is calculate the value of $BM$ in a different way. To do this, first we need to find $MP$ and $BP$.

Dropping perpendiculars, it easily follows that $NK=2r_1 \cos \theta$ and $MN= 2r_2 \cos \theta$, so $MK=2(r_2-r_1) \cos \theta $. From the right triangle $\triangle KPM$, we get $MP=MK \cos \theta = 2 (r_2-r_1) \cos^2 \theta$.

By power of a point for chords, $BP^2 = MP (2r_2-MP)$. So:

$BM^2 \&= MP^2+BP^2  $
$\: =  MP^2+ MP (2r_2-MP)$
$ \: \: = 2r_2 \cdot MP $
$=4 r_2 (r_2-r_1) \cos^2 \theta,$

giving $BM=2\cos\theta \sqrt{r_2(r_2-r_1) } $. Plugging this into $(\star)$ gives $R = \sqrt{r_2(r_2-r_1) }$, which clearly doesn't depend on $K$.

[/hide]

[hide="Solution"]
Denote $R$ as the circumradius of $\triangle{KBM}$, $r_1$ the radius of $\omega_1$, and $r_2$ the radius of $\omega_2$.

Consider the homothety centered at $N$ taking $\omega_1$ to $\omega_2$. Since $M$ is the midpoint of $\widehat{AB}$ and $K$ is the point of tangency of $AB$ with $\omega_1$, the homothety takes $K$ to $M$. Thus, $N,K,M$ are collinear, and we have
\[ \frac{r_1}{r_2}=\frac{NK}{NM}=\frac{NM-KM}{NM}=1-\frac{KM}{NM}\implies\frac{KM}{NM}=\frac{r_2-r_1}{r_2}\]
Let $\angle{MNB}=\angle{MAB}=\angle{MBK}=\theta$. Since $\angle{BMK}=\angle{NMB}$, $\triangle{BMK}\sim\triangle{NMB}$ by AA similarity. It follows that $BM/NM=KM/BM$.

Next, note that by Extended Law of Sines on $\triangle{KBM}$ and $\triangle{ABM}$, we have $KM=2R\sin\theta$ and $BM=2r_2\sin\theta$, so $KM/BM=R/r_2$. Thus, $NM/BM=r_2/R$, so $KM/NM=R^2/r_2^2$. As a result,
\[ \frac{r_2-r_1}{r_2}=\frac{KM}{NM}=\frac{R^2}{r_2^2}\implies R=\sqrt{r_2(r_2-r_1)},\]
which is not dependent on the choice of $K$, as desired. // [/hide]

Note that by Archimedes' Lemma (or a simple homothety argument), $NK$ bisects $\angle ANB$.  (This follows from the fact that the homothety sending $\omega_1$ to $\omega_2$ sends $AB$ to the tangent at $M$, which implies that $M$ is the midpoint of arc $\widehat{AB}$.) Therefore $\angle MBA=\angle MNA=\angle MAB$, and as such $\triangle MKB\sim\triangle MBA$.  Let $k$ be the ratio of similitude of the two triangles.  We wish to show that $k$ is constant, as since the circumradius of $\triangle MBN$ is constant the circumradius of $\triangle MKB$ will follow to be constant.

Note that $k=\tfrac{MK}{MB}=\tfrac{MB}{MN}$.  Hence $k^2=\tfrac{MK}{MB}\cdot\tfrac{MB}{MN}=\tfrac{MK}{MN}.$ Thus, we have $1-k^2=1-\tfrac{MK}{MN}=\tfrac{NK}{MN}.$  However, note that the homothety centered at $N$ sending $\omega_1$ to $\omega_2$ sends $K$ to $M$, and as such $\tfrac{NK}{MN}=\tfrac{r_1}{r_2}$.  Solving gives $k=\sqrt{\tfrac{r_2-r_1}{r_2}}$, which is fixed, as desired.  $\blacksquare$

The problem has been posted (at least) twice before!

Best regards,
sunken rock

Bear in mind that this thread started in 2007, sunken rock...  :)

[hide=Solution]Archimedes' Lemma tells us that $MA^2=MB^2=MK \cdot MN$. Let $O_1$, $O_2$, and $O$ be the centers of $\omega_1$, $\omega_2$, and the circumcircle of $\triangle{MKB}$ respectively. Since $N,O_1,O_2$ are collinear as a homothety centered at $N$ sends $\omega_1$ to $\omega_2$, $\angle{NO_1K}=\angle{NO_2M}=\theta$. Then in quadrilateral $O_1KMO_2$, $\angle{KO_1O_2}=180^{\circ}-\theta$, $\angle{O_1O_2M}=\theta$, $\angle{O_2MK}=\frac{180^{\circ}-\theta}{2}$, so $\angle{O_1KM}=90^{\circ}+\frac{\theta}{2}$ so $\angle{BKM}=\angle{O_1KM}-\angle{O_1KB}=(90^{\circ}+\frac{\theta}{2})-(90^{\circ})=\frac{\theta}{2}$, so $\angle{MOB}=\theta$.

If we let the radius of $\omega_1$ be $r_1$ and the radius of $\omega_2$ be $r_2$ and the radius of the circumcircle of $\triangle{MKB}$ be $R$, $MN \cdot MK=4r_2(r_2-r_1)\sin^2{\theta}$ and $MB=2R\sin{\theta}$ so $R=\sqrt{r_2(r_2-r_1)}$, proving that $R$ is independent of the position of $K$.[/hide]

It's possible to do this simply with trig. Let $\alpha = \angle BKN$; then we compute $$\frac{MB^2}{\sin^2 \alpha} = \frac{MN^2}{\sin^2 \alpha} = \frac{MN \cdot NK}{\sin^2 \alpha}.$$ Now notice that $MN = 2r_2 \sin \angle KO_1N$ and $NK = 2r_1 \sin \alpha$, so the expression, which is precisely equal to the desired circumradius, evaluates to a constant.


Three schools compete in a chess tournament, each school with five students. School $A$ with students $a_{1},a_{2},...,a_{5}$, and so on.

1) If a school has to send a player to the match, the least number remaining always goes. A player who loses is immediately eliminated. The first match of the tournament is $a_{1}$ against $b_{1}$

2) Suppose a player $x_{i}$ of school $X$ beats player $y_{j}$ of school $Y$. If there is a player remaining in school $Z$, then $Z$ will send a player to meet $x_{i}$. If not, then $Y$ will send a player instead. If there is no students left in both $Y,Z$, the competition ends.

3) Each time players $x_{i}$ of school $X$ wins, school $X$ gets $10^{i-1}$ points.

Let $P_{A},P_{B},P_{C}$ be the points of schools $A,B,C$ at the end. Let $N$ be the number of all possible ordered triples $(P_{A},P_{B},P_{C})$. Find the remainder when $N$ is divided by $8$.

It is easy to prove that number of all possible score is equal to number of all possible match order.
Let $f(a,b,c)$ is a number of all possible match order when a students are in the school A, b students are in the school B, c students are in the school C.
We can show that $f(a,b,c)=f(b, c, a-1)+f(a, c, b-1)$.
 -> $f(a,a,c)=2f(a,c,a-1)$ -> $f(a,a,c)$ is even.
$f(5,5,5)=2f(5,5,4)=4f(5,4,4)$
just calculate the parity of $f(5,4,4)$.
Answer is 4.

[quote="Little Gauss"]It is easy to prove that number of all possible score is equal to number of all possible match order.
Let $ f(a,b,c)$ is a number of all possible match order when a students are in the school A, b students are in the school B, c students are in the school C.
We can show that $ f(a,b,c) \equal{} f(b, c, a \minus{} 1) \plus{} f(a, c, b \minus{} 1)$.
 -> $ f(a,a,c) \equal{} 2f(a,c,a \minus{} 1)$ -> $ f(a,a,c)$ is even.
$ f(5,5,5) \equal{} 2f(5,5,4) \equal{} 4f(5,4,4)$
just calculate the parity of $ f(5,4,4)$.
Answer is 4.[/quote]

I really don't understand your solution. Could you(or someone) explain more?

[b]Claim: [/b] The number of all possible score is equal to the number of all possible match order.
[b]Proof of Claim:[/b] If $a_1$ or $b_1$ wins 10 matches, the final score is $(10,0,0), (0,10,0)$.
If not, the remaining students win less than 10 matches, so the scores cannot be the same. $\blacksquare$

Let $f(a,b,c)$: number of all possible match order when $a$ students are in the school $A$, $b$ students are in the school $B$, $c$ students are in the school $C$.
Dividing the case by who won the first game, we have $f(a,b,c) = f(b,c,a-1) + f(a,c,b-1)$.
Note that $f(n,n,n) = 4f(n,n-1,n-1)$.
Therefore, we just need to check the parity of $f(5,4,4)$.
Straightforward calculation gives that $f(5,4,4)$ is odd, so the remainder is $4$.


function
calculus
derivative
real analysis
real analysis unsolved

Let be $ f$ one function and let definired like: $ f: \mathbb{R}^2 \to \mathbb{R}$.
If $ f$ is a differential funtion in all $ \mathbb{R}$ and if is [i]only one[/i] $ x_o \in \mathbb{R}^2$, where $ f'(x_o)\equal{}0$, prove that $ x_0$ is minimum of this funtion.

fitim

Wrong. For example, consider $ f(x,y)\equal{}xy$. Its derivative is zero only at the origin. That point is a saddle point, not either a minimum or maximum.

And if the question were $ f: \mathbb{R}^2\to[0,\infty),$ then $ f(x,y)\equal{}e^{xy}$ would serve the same purposes as an example.


suppose that

f(x) = $3^x$, g(x) = $100^x$

let $a_1$ = 3, and $a_{n+1}$=f($a_n$) for a>0, a is integral

let $b_1$ = 100, and $b_{n+1}$=g($b_n$) for b>0, a is integral

prove that:

$b_n$<$a_{n+2}$<$b_{a+1}$

HINT: [hide]prove $b_n$<$5b_n$<$a_{n+2}$ instead of $b_n$<$a_{n+2}$ then prove $a_{n+2}$<$b_{a+1}$[/hide]


$ \boxed{1}$

Let $ f(x)\equal{}\left|\frac{3}{2}x\minus{}1\right|\ (0\leq x\leq 2)$.

(1) Draw the graph of $ y\equal{}f(x)$.

(2) Draw the graph of $ y\equal{}f(f(x))$.

(3) Solve the equation of $ x\ : f(f(f(x)))\equal{}x$.

$ \boxed{2}$

Let $ a,\ b$ be real numbers.

(1) Draw the existence region of $ (a,\ b)$ such that  the quadratic equation of $ x\  : x^2\plus{}ax\plus{}b\equal{}0$ has more than one solution in the range of $ \minus{}1<x\leq 1$.

(2) Find the condition of $ a$ such that  the quadratic equation of $ x\  : x^2\plus{}ax\plus{}2a\minus{}\frac{3}{4}\equal{}0$ has more than one solution in the range of $ \minus{}1<x\leq 1$.

Question 1 part 3 solution

[hide]Consider if f(x) is equal to x.  Solving for x gives us values of x at 2 and 2/5.  These are two of the solutions that will always create themselves regardless of the number of functions used.  Now, we set f(x) equal to 2 or 2/5 and ignore the 2 and 2/5 solutions.  We obtain new solutions as -2/3 and 14/15.  However, since we must take f(x) three times, the -2/3 solution is extraneous.  Now we just set f(x) = 14/15 and obtain new solutions at 58/45 and 2/45.

The solutions are 2, 2/5, 58/45, and 2/45.[/hide]

[quote="azn6021023"]Question 1 part 3 solution

Consider if f(x) is equal to x.  Solving for x gives us values of x at 2 and 2/5.  These are two of the solutions that will always create themselves regardless of the number of functions used.  [color=red]10pt[/color]Now, we set f(x) equal to 2 or 2/5 and ignore the 2 and 2/5 solutions.  We obtain new solutions as -2/3 and 14/15.  However, since we must take f(x) three times, the -2/3 solution is extraneous.  Now we just set f(x) = 14/15 and obtain new solutions at 58/45 and 2/45.

The solutions are 2, 2/5, 58/45, and 2/45.[color=red]6pt[/color][/quote]

Your mark is [color=red]16pt[/color] in 25pt.

Am I missing solutions or was my explanation not detailed enough?


Find the formula fot the sequence which contains all natural numbers except squares. And PROVE IT


What rational number does the repeating decimal 0.9999...equal?

1

Many of the problems you're posting are way too easy for the forums you put them in. This should be in Classroom.

Thanks for the steps SMART and the other reply is great, too.


Given that a and b are positive integers and the difference between the expression a^2+b and a+b^2 is prime. Find all possible values of a and b.

[hide="Solution"]
WLOG, $ a>b.$ So $ a^2+b>b^2+a.$ Taking the difference gives \begin{align*}(a^2-b^2)+(b-a)&=(a-b)(a+b)-(a-b)\\&=(a-b)(a+b-1).\end{align*} From here, we get \begin{align*}a-b&=1\\a+b-1&=p\end{align*} where $ p$ is a prime. Adding both equations gives us $ 2a-1=p+1,$ and from here we get $ 2a-2=p.$ Since the LHS is even, $ p$ must be $ 2.$ Continuing, $ a=2$ and $ b=1.$ [/hide]

Of course, because you WLOGed it, (1,2) and (2,1) are both solutions.

[quote="cgyao15"]WLOG, what would that mean?  :([/quote]

Without the Loss Of Generality.

So he could assume a>b and there would be no loss of generality, which is a good thing. If you lose generality, you could end up getting a partial answer or lose points on a solution/proof.


Let $ c$ be a fixed positive integer number. Find all positive integers $ a,$ $ b$ in term of $ c$ such that such that $ ab$ divides $ c(c^2\minus{}c\plus{}1)$ and $ (c^2\plus{}1)$ divides $ a\plus{}b$.

the only solutions are $ \{a,b\}\equal{}\{c,c^2\minus{}c\plus{}1\}$...


inequalities
parameterization
probability
binomial coefficients
combinatorics unsolved
combinatorics

Given $n$ chairs around a circle which are marked with numbers from 1 to $n$ .There are $k$, $k \leq 4 \cdot n$  students sitting on those chairs .Two students are called neighbours if there is no student sitting between them.  Between two neighbours students ,there are at less 3 chairs. Find the number of choices of $k$ chairs so that $k$ students can sit on those and the condition is satisfied.

It looks too easy for VTST.
Assume $k \geq 2$.
Let $S$ be the set of chairs occupied by students.
Let $a = \max (S)$ and $b = \min (S)$.  Let $L = n + b - a$ be the length of interval between them.

Sum over all possibilities according to the value of $L$.
$L \geq 4 $ by definition of the permitted arrangements.
$L \leq n - 4(k-1)$ because there are $k-1$ additional intervals of length at least 4 on the circle, between consecutive students.
All values of $L$ satisfying these inequalities are possible.

Given $L$, there are $L$ possible positions of $a,b$ = number of possible positions of edge $(n,1)$ within this interval.
The number of placements of the remaining $k-2$ students is
the number of sets $0 = x_1 < ... < x_k = n - L$ with $x_i - x_{i-1} \geq 4$.
Shortening each interval by 3, this is the number of ways to find $k-1$ positive numbers with sum equal to $n - L - 3(k-1) = {{n - L - 3k + 2} \choose {k-2}}$.

So the answer is $\Sigma_{L=4}^{L=n-4} L {{n - L - 3k + 2} \choose {k-2}} $.
There is a closed form for the sum in terms of one or two individual binomial coefficients, but I am too lazy to work it out (use Abel summation to "integrate" ${p-L} \choose q$ to something like ${p+1 - L} \choose {q+1}$ and "differentiate" $L \to 1$, write the result as a difference of two values of $x \choose {q+2}$; here $p,q$ are the parameters from the summation).

Solution 2:

The probability that a randomly chosen arrangement has a student at position 1 is $k / n$, due to the rotational symmetry.
The number of arrangements with a student at position 1 is ${n - 3k - 1} \choose {k-1}$.
So the answer is $ \frac {n}{k} { {n - 3k - 1} \choose {k-1}}$ .

Writing $n/k = (n-3k)/k + 3$,  this is also equal to ${{n - 3k} \choose k } +  3 {{n - 1 - 3k} \choose {k-1}}$

Solution 3:
The second term, 3*(binomial coefficient) is equal to the number of arrangements where there is a student at position 1,2,or 3.
The first term is the number of arrangements with no student in locations {1,2,3}.


geometry
angle bisector
geometry proposed

Quadrangle $ ABCD$ is inscribed in a circle with radius 1 in such a way that the diagonal $ AC$ is a diameter of the circle, while the other diagonal $ BD$ is as long as $ AB$. The diagonals intersect at $ P$. It is known that the length of $ PC$ is $ 2/5$. How long is the side $ CD$?

By angle-chasing it's easy to see that $ PD$ is an angle-bisector of $ \hat{ADC}$.Hence by Stewart's theorem applying to triangle $ ODC$ we have:
$ CD \equal{} \frac {PC.OD}{OP} \equal{} \frac {2}{3}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group, and define by $G^k$ the set consisting of the $k$-th powers of the elements of $G$.

If $G^m$ and $G^n$ are commutative subgroups, prove that $G^{gcd(m,n)}$ is commutative.

what do you mean , subgroup consisting of k th powers

the set you describe, of all kth powers, does not seem a group

take k =3 in S3
you get one trivial element and a transposition?

Well, $G^k=\{e^k=e,a^k,b^k,...\}$. I don't see a problem in it...

Indeed, not all groups have such a subgroup for each index :) then assume they have one. I altered the wording a bit for that.


Choose any number from the set of natural number. Find the probability that this chosen number is even.

It is intuitively that the probability is $ \frac{1}{2}$, but I can't help to make the solution since the sample space is infinite.  :blush: Sorry but I know my question is very stupid.

Why the probability of a singular number $ 2n$ is clearly $ 0$ but the set of all of them has probability $ \frac{1}{2}$, while sum of infinite $ 0$ is still $ 0$?

You are right in your doubts: the set of all natural numbers cannot be given the structure of a probability space so that each number is assigned the same probability (you just proved it as a matter of fact). What people usually mean when talking of the probability that a random natural number is even (or square-free, or whatever) is that you take the numbers from $ 1$ to $ n$, turn this finite set into a probability space in the obvious way, compute (or estimate) the probability $ p_n$ that a random number between $ 1$ and $ n$ satisfies your property and take the limit $ \lim_{n\to\infty}p_n$.

[quote="yevon"]fedja, we know that your limit tends to 1/2 when n tends to infinite. Does it the possibility that we want to compute? or we cant say anything about a probability in the set of natural numbers, because it is not a probability space?

I can give an exemple, please forgive me if it is so stupid.

A(i) is the set of i first odd numbers and 2i first even numbers. Certainly with each A(i) the probability to find an even is 2/3, and when i tends to infinity, A(i) tends also to the set of all natural numbers. So can we say that the probability is 2/3? No, of course no.

So can we conclude that the set of natural numbers is never a probability space? or sometimes, with some rule it can be one?[/quote]

My teacher mentions a lot about the probability of an infinite sample space without mentioning anything, so I got confused  :oops:  Is there a standard to compute the probability in an infinite sample space?

Strictly speaking, probability is assumed to be countably additive, so to call the limit I mentioned above "probability" is an abuse of language. Still, many people do it even in textbooks because it is too convenient to think of densities in probabilistic ways.

As to infinite countable sample spaces, the problem is not in the absense of probability space structures on such spaces (you can take any sequence $ p_n$ of non-negative numbers such that $ \sum_{n\equal{}1}^{\infty}p_n\equal{}1$ and create a probability space on the natural numbers by assigning probability $ p_n$ to number $ n$, for instance), but in impossibility to create a uniform distribution on such a sample space, i.e., to assign the same probability to each number.


If $f_{p}: \mathbb{R}^{n}\to \mathbb{R}$ is a sequence of measurable functions. Why is $\{x \in \mathbb{R}^{n}\ | \ the \ sequence \ (f_{p}(x)) \ converges \ in \ \mathbb{R}\}$ a Borelset?

try to use the definition of a cauchy sequence.

Let $X =\{x : f_{n}(x) converges \}$.
Then $X = \bigcap_{k \geq 1}\bigcup_{N \geq 1}\bigcap_{m,n \geq N}\{x: |f_{n}(x)-f_{m}(x) | < \frac{1}{k}\}$ by Cauchy criterion.


geometry
trigonometry
analytic geometry
rectangle

$1$. Two circles of radii $17$ and $23$ intersect such that their radii drawn to one of the points of intersection meet at right angles. Without using co-ordinate geometry or trigonometry, please calculate the difference of the areas of the non-overlapping portions of the circles.

$2$. Given that $f(x) = x^4 + x^3 + x^2 + x^1 + x ^0$, find the remainder when $f(x^5)$ is divided by $f(x)$

$3$. Show that  $1110 \times 1111 \times 1112 \times 1113 = 1235431^2 - 1$   in any system of numeration with a base greater than $5$.

$4$. A triangle has sides $a, b, c$, such that $a^2 + b^2 + c^2 = ab + bc + ca$. Without using coordinate geometry , prove that the triangle is equilateral.

$5$. a rectangular card is folded through one corner such that the adjacent corner falls on a side, thus forming $3$ right triangles with areas in an arithmetic progression. (i.e., $ABCD$ is the rectangle, it is folded such that $D$ lies on $AB$.) If the area of the smallest triangle is $3 cm^2$ , what is the area of the largest?

[quote="xxxyyyy"]
$4$. A triangle has sides $a, b, c$, such that $a^2 + b^2 + c^2 = ab + bc + ca$. Without using coordinate geometry , prove that the triangle is equilateral.

[/quote]

Multiply both sides by $2$ and shift evrything to L.H.S.You'll get

$(a-b)^2+(b-c)^2+(c-a)^2=0$

So $a=b=c$

or else use A.M-G.M and get

$a^2+b^2+c^2 \geq ab+bc+ca$ and equality holds when $a=b=c$

[quote="xxxyyyy"]$2$. Given that $f(x) = x^4 + x^3 + x^2 + x^1 + x ^0$, find the remainder when $f(x^5)$ is divided by $f(x)$

[/quote]

I'm getting $5$.Is it right?

[hide="2"]
Let $\omega$ be a non-unity fifth root of unity.  $f(\omega) = 0$.  Now, $f(\omega^5) = f(1) = \boxed{5}$.
[/hide]

still problem $3$ and $5$ are unsolved.

Others are correct.

:-)

[quote="vidyamanohar"][quote="xxxyyyy"]

$3$. Show that  $1110 \times 1111 \times 1112 \times 1113 = 1235431^2 - 1$   in any system of numeration with a base greater than $5$.
quote]

Call 1110 as 'y'.
Then we have to evaluate y(y+1)(y+2)(y+3).
It is equal to $y^4 + 6y^3 + 11y^2 + 6y$
$= y^4 + 6y^3 + 11y^2 + 6y + 1 - 1$

$= (y^2 + 3y + 1)^2 - 1$
It is not difficult to evaluate the value of  $(y^2 + 3y + 1)$ for y=1110, which is 1235431.

Hence the proof.[/quote][/quote]

How do u relate it to any base greater than $5$?

Anyone for number $5$?

[quote="xxxyyyy"][quote="vidyamanohar"][quote="xxxyyyy"]

$3$. Show that  $1110 \times 1111 \times 1112 \times 1113 = 1235431^2 - 1$   in any system of numeration with a base greater than $5$.
quote]

Call 1110 as 'y'.
Then we have to evaluate y(y+1)(y+2)(y+3).
It is equal to $y^4 + 6y^3 + 11y^2 + 6y$
$= y^4 + 6y^3 + 11y^2 + 6y + 1 - 1$

$= (y^2 + 3y + 1)^2 - 1$
It is not difficult to evaluate the value of  $(y^2 + 3y + 1)$ for y=1110, which is 1235431.

Hence the proof.[/quote]

How do u relate it to any base greater than $5$?

Anyone for number $5$?[/quote][/quote]
Let x is the base.  Then we write $y = 1110 = x^3 + x^2 + x$
substituting this value of y in $(y^2 + 3y + 1)$ we get
$y^2 + 3y + 1=x^6 + 2x^5 + 3x^4 + 5x^3 +4x^2+ 3x + 1$ which is equal to 1235431 in base 'x', (naturally x has to be greater than 5)

[quote="frt"][hide="1"]
Let $A$ and $B$ be the circles of radii $23$ and $17$ respectively, and let $P$ be the overlapping region.

$(A-P)-(B-P)=A-B=23^2\pi-17^2\pi =\boxed{240\pi}$

I didn't use the fact that the radii drawn to the intersection meet at right angles... :|
[/hide][/quote]
Well this explains the fact that the intersection has nothing to do with the difference :rotfl: Good job frt


LaTeX
trigonometry
real analysis
real analysis unsolved

Let f(x) = |sin x sin 2x sin 4x ... sin 2nx|. how do I prove that f(x) ≤ 2/√3 f(π/3)?

Use $ \text{\LaTeX}$, please! Did you mean $ \sin x\sin 2x\dots \sin 2nx=\sin x\prod_{k=1}^{n}\sin 2k x$ or $ \sin x\sin 2x\dots \sin 2^{n}x=\prod_{k=0}^{n}\sin 2^{k}x$?

Then it is false: one of the factors in $ f(x)$ is $ \sin 6x$, which vanishes at $ x=\frac \pi 3$. On the other hand, it is apparent that $ f$ is not identically $ 0$. :P

Actually, it's true that if $ f_{n}= \left| \prod_{k=0}^{n}\sin \left( 2^{k}x \right) \right|$, then $ \left| f_{n}(x) \right| \leq \frac2{\sqrt 3}\cdot f_{n}\left( \frac{\pi}3 \right).$ But it's not particularly interesting :maybe:


What is the maximum number of 33-cents stamps that can be purchased with 5 dollars?

$ \frac{5}{0.33}\equal{}15 \frac{5}{33}$. Thus, $ \fbox{15}$ stamps can be purchased with $ 5$ dollars.


[quote="diophantus123"]what is the final digit of 7^7^7^7^7^7^7[/quote]

working mod 10:
7^7=3
3^7=7
so the final digit is 7...

If this is 7^(7^(7^(7^(7^(7^7))))), then I get a different answer. Please point out my solution's error.

7^(7^(7^(7^(7^(7^7))))) (mod 10) =
7^(7^(7^(7^(7^(7^7)))) (mod 4)) (mod 10) =
7^(7^(7^(7^(7^(7^7))) (mod 2)) (mod 4)) (mod 10) =
7^(7^(1) (mod 4)) (mod 10) = 7^3 (mod 10) = [b]3[/b].
The mods 2 and 4 are because of Euler's Generalization of Fermat's Little Theorem.

Sayoonara

Well, this is just a question of how you interpret the power tower 7^7^7^7^7^7^7.

Peter calculated the last digit of

  (((((7^7)^7)^7)^7)^7)^7,

and JGeneson calculated the last digit of

  7^(7^(7^(7^(7^(7^7))))).

And both solutions are correct.

  Darij

when no ( ) are placed, we read from left to right, every time ;)

We learnt that in 7th grade, so I really wonder why everyone keeps missing this over and over again? :?

[quote="darij grinberg"]No, actually there is a special convention that power towers like $\displaystyle a^{b^{c^d}}$ have to be understood as $\displaystyle a^\left( b^\left(c^d \right) \right)$. In fact, I hate this convention, but many people use it. The usual argument in favor of this convention is that the other interpretation, namely $\displaystyle \left( \left( a^b \right)^c \right)^d$, could be written in a much easier way in the form $\displaystyle a^{bcd}$, and hence there would be no need to write $\displaystyle a^{b^{c^d}}$.

  Darij[/quote]

$7^{7^{7^{7^{7^{7^{7}}}}}}\not=7^{777777}$....  very hard with numbers...

FWIW -

a^b^c is "normally"   interprested [b]as a^(b^c)[/b] and not (a^b)^c.
(in almost every  standard reference I have seen)

Of course,  using "(" will make it more clear.

(And with that convention one will  interpret Peter's answer was for 7^7^6 and not 7^7^7^7^7^7^7 ) .

And  BTW
It does NOT  matter how high the tower is (as long is  greater than 1) the last digit is 3

C[b]an you prove this:

Last digit of 7^7 or 7^(7^7) or 7^(7^(7^7) .. etc is 3 [/b]

(In fact it will not matter much .. after the   third  7 ... Eg last digit of
7^7^7^anypositivenumber^anypostivenumber^ ..... is still 3.

(If the third  tower number is even the asnwer would be 7..does not matter what the other numbers after that in the tower are ....)

[quote="Gyan"](in almost every standard reference I have seen)[/quote]

does that include any non-American reference? ;)

[quote]Does that include ..[/quote] Yes. But you can check that for yourself.

no, i can't :P I can't know what references you have seen ;)

But you may be right, anyway the mathematiciens here all say that that is an american convention that is not applied here... so since I'm not american, no need to use it :D

There was an AMC problem. It said :
According to the standard convention for exponentiation

2^2^2^2 = 2^(2^(2^2)) = 65536

If the order in which the exponentiations are performed is changed how many other values are possible?

So if theres no parantheses it is assumed to be (2^(2^(2^2))).
I know that this is another American source but its just fyi.

Yes, that's what the mathematiciens here said to me as well. In america it's seen like that, in europe (and probably most of asia, not sure here) they use the other notation. So since this is an international board we'll have no other way out than putting parentheses around everything :P

Every time I've seen a power tower like the above, what you are calling the american standard is what has been used. But the problems I've seen have been from many different countries. I find it interesting that you say Europe use the other convention, which seems completely pointless as mentioned earlier.. there is no reason to write it in that way, so why not use the other way :P

I have never heard about "American" conventions for power towers. In Russia, one writes $a^{b^c}$ for $a^{\left(b^c\right)}$, just as in the U. S. A.; I don't know how it is done in Germany (I have read very few number theory books in German), but I think the convention is rather international. Still, this doesn't change the fact that this convention can be misleading unless it is explicitely stated before being used (e. g., a good problem book should have a list of such conventions).

  Darij

[quote="TripleM"]the other convention, which seems completely pointless as mentioned earlier.. there is no reason to write it in that way, so why not use the other way :P[/quote]

Just because of the logics: between signs of the same priority, we read from left to right.  You don't read 8-13-7 as 8-(13-7) either do you? ;)

If writing 8-13-7 could be done in a MUCH simpler way, then I probably would. But thats not true. Why write powers like that if you could just write a multiplication in the index?

Well if you read from right to left in powers, then why not in other signs? It may be easier sometimes, but it's less consistent and most of all less general.


inequalities
geometry
circumcircle
inequalities unsolved

i think it's not true let $a=1$  $b=c= \frac{51}{100}$
may be i'm wrong or ...  :rotfl:

nope, i think it's true... the fact is that i (i think, maybe somebody else has done this before) obtained this inequality from the iran 96 inequality...

recall this:

$(xy+yz+zx)\left(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right)\geq \frac{9}{4}$.

set $(x,y,z)=(\cot A, \cot B, \cot C)$

using the identity $\cot A\cot B+\cot B\cot C+\cot C\cot A=1$, and some easy trigonometric manipulation we can find these are equivalent...

i just was trying to guess whether we could prove this inequality in a different way...  :D

my apollogies if this has been discused or posted before.

[quote="campos"]i don't know whether this has been posted in the forum or not, but anyway i think this inequality is nice!

Let $a,b,c$ be the sides of a triangle with circumradius $R$, prove that

$\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}}\geq 9R^{2}$.[/quote]
Clearly ,the inequality isn't true.
Counterexample:
$a=3,b=3,c=3\sqrt{3}\Longrightarrow R=3$
$57>81 ?$

Maybe you mean something like $\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}}\geq 3R^{2}$?

:oops:  in the proof $x,y,z$ must be positive, so the inequality is right but we have to say that $ABC$ is an acute triangle.  :P in your case $C=120$

keep trying it!

[quote="campos"]

Let $a,b,c$ be the sides of a triangle with circumradius $R$, prove that

$\frac{a^{2}b^{2}}{c^{2}}+\frac{b^{2}c^{2}}{a^{2}}+\frac{c^{2}a^{2}}{b^{2}}\geq 9R^{2}$.[/quote]
I think, it's true for acute triangle. :wink:

i said that in the last reply, but didn't correct the first one... and it seems that i can't do it... it says i must ask a moderator to do so... do you have a proof arqady?

[quote="campos"]... do you have a proof arqady?[/quote]
Let $a^{2}=y+z,$ $b^{2}=x+z$ and $c^{2}=x+y.$ Then your inequality equivalent to
$\sum_{cyc}(x-y)^{2}(4x^{2}+7xy+4y^{2})xy+2\cdot\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)xyz\geq0.$

More is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130999]here[/url] a part of my solution!


Given that x is a positive integer and y= 5 to the power of x, what is the greatest value for y for which y < 30,000?

$ y < 30,000$
$ 5^x < 30,000$
$ 5^x < 5^4 * 48$
$ 5^{x \minus{} 4} < 48$
$ 25 < 48$
$ x \minus{} 4 \equal{} 2$
$ x \equal{} 6$
$ y\equal{}5^x\equal{}5^6\equal{}15,625$

A quick check for this would be if you have a calculator. You see that $ 5^6 \equal{} 15,625$ which is less than $ 30,000$. The next biggest power would be $ 5^7 \equal{} 78,125$ which is clearly more that $ 30,000$.

In conclusion, $ \boxed{y\equal{}15,625}$.