find all funtions satisfying following conditions.

$f:R^+ \rightarrow R^+$

$f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$

I have very long proof of this eq. If you have short proof. It's glad me to see the solution.  :D

Is there any one try this problem?  :(

I used $ f(1), f(2)$. and using $5$ cases to solve it.

Is there anyone have a nice solution?

[quote="Diarmuid"]If the function is from $R^+$ to $R^+$, how can you use $f(0)$ in your solution?

(or are you using $R^+$ for non-negative reals?)[/quote]

Can I understand from this that you are saying that [tex]0 \notin \mathbb{R}^+[/tex] ?? :o

I wrote other one;  :D

Anyway, using a lot of case is not a good solution.

[quote="leibniz"]Can I understand from this that you are saying that $0 \notin \mathbb{R}^+$ ?? :o[/quote]

Well, yeah.  That's how I was taught the notation anyway - $R^+$ is the positive reals.


function
calculus
integration
derivative
real analysis
real analysis unsolved

Find all continuous functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $ x\in \mathbb{R}$ and $ n\in \mathbb{N}$,

$ n^2\int_{x}^{x\plus{}\frac{1}{n}}f(t)dt\equal{}nf(x)\plus{}\frac{1}{2}$.

[quote="Jumbler"]Find all continuous functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $ x\in \mathbb{R}$ and $ n\in \mathbb{N}$,

$ n^2\int_{x}^{x \plus{} \frac {1}{n}}f(t)dt \equal{} nf(x) \plus{} \frac {1}{2}$.[/quote]

Take $ n\equal{}1$ for a brief moment to see the left hand side is then a $ C^1$ function of $ x$, proving $ f$ is $ C^1$. Now apply the second derivative version of Taylor's formula to $ F$, an antiderivative for $ f$, to learn something useful about $ f'$.

[quote="Jumbler"][quote="WWW"]Now apply the second derivative version of Taylor's formula to $ F$, an antiderivative for $ f$, to learn something useful about $ f'$.[/quote]

What do you mean by that?[/quote]

$ F(x\plus{}1/n) \equal{} F(x) \plus{} F'(x)(1/n) \plus{} (F''(c)/2)(1/n^2)$.


A delivery truck made three stops after leaving the warehouse. At the first stop, the driver unloaded $ \frac{1}{4}$ of his crates. At the second stop, he unloaded $ \frac{2}{3}$ of the crates remaining on his truck. At the last stop, he unloaded $ 80\%$ of the rest of his crates at that point, and then he returned to the warehouse with 20 crates. How many crates did he unload at the first stop?

We see that the driver unloaded $ \frac {1}{4} \plus{} \frac {2}{3}*\frac {3}{4} \plus{} \frac {3}{4}*\frac {1}{3}*\frac {4}{5} \equal{} \frac {19}{20}$. From here, we see that 20 crates is equal to 1/20 of his total crates, and thus he has 400 total crates. Since he unpacked $ \frac {1}{4}$ of them at the warehouse, he unloaded $ 100$ crates at the first stop.


Find all pairs (m,n) belonging to N such that n^5 + n^4 = 7^m - 1.

[quote="chessfreak"]Find all pairs $ (m,n)$ belonging to $ \mathbb{N}$ such that $ n^{5}\plus{}n^{4}\equal{} 7^{m}\minus{}1$.[/quote]

[hide]
First of all factor it into:

\[ (n^{2}\plus{}n\plus{}1)(n^{3}\minus{}n\plus{}1)\equal{}7^{m}\]

Assume just for this case that both Factors are of the form $ 7k$, So:

$ n^{2}\plus{}n\plus{}1\equiv 0\bmod 7\implies n(n\plus{}1)\equiv\minus{}1\bmod 7\implies n(n\plus{}1)\equal{}7x\minus{}1$

Making the same for the second factor gives: $ n(n^{2}\minus{}1)\equal{}7y\minus{}1$

Since $ 2|n(n\plus{}1)(n\minus{}1)$ and, $ 3| n(n\plus{}1)(n\minus{}1)$ It follows that $ 7y\equiv 1\bmod 6$

So, $ y\equal{}1$ gives a solution $ (m,n)\equal{}(2,2)$

I don't know how to continue from here... And is this the best aproach ?

[/hide]

Because $ n^{2}\plus{}n\plus{}1 \equal{} 7^{k},n^{3}\minus{}n\plus{}1 \equal{} 7^{l}$ and for $ n > 2, n^{3}\minus{}n > n^{2}\plus{}n\to l > k$.
We have $ \frac{n^{3}\minus{}n\plus{}1}{n^{2}\plus{}n\plus{}1}\equal{} n\minus{}1\minus{}\frac{n\minus{}2}{n^{2}\plus{}n\minus{}2}\not\in N$ if n>2.


I'm taking the SAT II next week.
I’m having difficulties with true/false/correct explanation questions.
Does any one have advice on questions of this style?
Thanks!

not any advice on that specific section b/c it's really stupid... but don't get the barrons SAT II chem guide.  It's horribly horrible.


Mafia calculus derivative counting distinguishability Gauss blogs

Just put some quotes that you think are cool here!!!

"Success is like peeing in your pants. Everyone can see the results, but only you can feel the warmth."

"Enjoy Life. You have plenty of time to be dead."

"A pessimist says the glass is half empty. An optimist says the glass is half full. A [u]real[/u] optimist says the glass is full, just with some air on top."

"For every minute you are angry, you lose 60 seconds of happyness."
 :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217757

You might want to try this thread

Well, that thread is dislike, and this one is like...

When you criticize someone, walk a mile in their shoes. That way, you are a mile away, and you have their shoes.

Thats a great one 1=2!

"Life sucks. Get used to it."

"Take it easy, life is short"

"Your well is 5 seconds deep" (taken from a physics textbook)

"WINNER WINNER CHICKEN DINNER" (my history teacher, no seriously)

"If at first you don't succeed, you fail."

There may not be an I in team, but surprisingly enough there are 2 in schizophrenia.

Some people are like Slinkies: they may not seem to have a purpose, but they bring you great joy if you push them down a flight of stairs.

All I got.

The whole world's a stage, and religion is the critic.

[quote]"WINNER WINNER CHICKEN DINNER"[/quote]

Watch 21, read the book (I think...), or look up the phrase.

[quote="GLaDOS"]"You're not a good person, you know that? Good people don't end up here."

"Do you think I'm trying to trick you with reverse psychology? Seriously now..."

"I invited your best friend the companion cube. Of course, he couldn't come because you murdered him."

"Two plus two is f…f…f… f…10. IN BASE FOUR! I'M FINE!"

"Good news, I figured out what that thing you just incinerated is. It was a morality core they installed after I flooded the enrichment center with a deadly neurotoxin to make me stop flooding the enrichment center with a deadly neurotoxin, so get comfortable while I warm up the neurotoxin emitters."

"One 18.25 ounce package chocolate cake mix. One can prepared coconut pecan frosting. Three slash four cup vegetable oil. Four large eggs. One cup semi-sweet chocolate chips. Three slash four cups butter or margarine. One and two third cups granulated sugar. Two cups all purpose flour. Don't forget garnishes such as: Fish shaped crackers. Fish shaped candies. Fish shaped solid waste. Fish shaped dirt. Fish shaped ethyl benzene. Pull and peel licorice. Fish shaped volatile organic compounds and [b]sediment shaped sediment[/b]. Candy coated peanut butter pieces. Shaped like fish. One cup lemon juice. Alpha resins. Unsaturated polyester resins. Fiberglass surface resins. And volatile malted milk impoundments. Nine large egg yolks. Twelve medium geosynthetic membranes. One cup granulated sugar. [b]An entry called 'how to kill someone with your bare hands.'[/b] Two cups rhubarb, sliced. Two slash three cups granulated rhubarb. One tablespoon all-purpose rhubarb. One teaspoon grated orange rhubarb. Three tablespoons rhubarb, on fire. One large rhubarb. [b]One cross borehole electro-magnetic imaging rhubarb.[/b] Two tablespoons rhubarb juice. Adjustable aluminum head positioner. Slaughter electric needle injector. Cordless electric needle injector. Injector needle driver. Injector needle gun. Cranial caps. And it contains proven preservatives, deep penetration agents, and gas and odor control chemicals. That will deodorize and preserve putrid tissue."[/quote]

I've never actually played Portal, but I want to!!!!

[quote="CatalystOfNostalgia"]白人看不懂[/quote]
Lol I liked this one :D

The man who smiles when things go wrong has thought of someone to blame it on.

"Compound interest is the most powerful force in the universe."
- Albert Einstein

"Oh my gosh! We're in trouble, What're we going to do!"-someone who found out that the sun is hurtling through space and going to hit the earth

"I'm going to do my laundry, could I have some change"-a passerbyer without change

"Some believe Global Warming doesn't exist. Therefore, Global Warming doesn't exist" ~ Classmate

"You fail more than a drowning fish" ~ Another classmate

"We're all going to die some time in our lives"

"-insert the hitchhiker quote where it explains that when someone explains the universe, a more bizarre and unexplainable situation occurs-"

"have u eaten ur dinner yet???" ~me (it's an inside joke)

"WE SAID IT AT THE SAME TIME!!!"
"nooooooo wayyyyyyyyy" ~me and 2 annoying people

"Every fight is a food fight when you’re a cannibal."
-Demetri Martin

"Someone's life is either long and useful or short and irrelevant. There is no in-between. If there was, many funeral speeches would suck." - Me
"Witty quotes don't have to be funny." - Me
"Moo cow." - Me
"Me" - Me

"[img]http://www.rivingtonarms.com/rarms/artists/john_finneran/images/UntitledBlackCloud.jpg[/img] <-- It's the black cloud of doom!!!"
-Me, when supposed to be signing people's yearbooks

"To thirty years of sobriety. Aw, **** it." - Someone else

"Fail"
~Me
"Fail?"
~Me
"Epic FAIL"
~Me
"You say fail too much"
~Someone Else
"Uh...fail?"
~Me
"I just noticed...your eyes are really big"
~A friend, and she's known me for a year :rotfl:

My eyes actually are pretty big

[quote="algebra2000"]
"I just noticed...your eyes are really big"
~A friend, and she's known me for a year :rotfl:

My eyes actually are pretty big[/quote]

Who is this friend?

@ytrewq: alyssa

"tcjv shoots you with a 1337-gun"
~Me

Here are some quotes I learned(mostly from the computer game Civilization IV but also from AoPs Algebra Introduction book"  Kind of inspiring, huh?

"If you can't explain it to someon else, you don't know it well enough"-Albert Einstein

"You can't direct the wind, but you can adjust your sails"-Unknown

"True glory is in writing what deserves to be written and reading what deserves to be read."-Pliny the Elder

"A man who moves mountains starts by carrying small rocks."-Confucious

"The wisest men always follow their own direction"-Unknown

"10% of the population cause 90% of the damage that has been accrued."-Anonymous


each corner of a cube is cutt of to leave an equilateral triangle. how many faces does it have? i dont know how i got it wrong.
thanks

When you cut off corners of a cube, won't you be left with a quadrilateral at least?

Umm, I don't get what you're saying...In order for the cut face to be a quadrilateral it has to intersect at least 4 sides...

My answer is 14 since there are 6 sides in a cuber and 8 corners.

He means the surface of the cut will be an equilateral triangle.  That is, given a cube with vertices B, D, and E adjacent to vertex A.  We cut it along a plane slicing AB, AD, and AE so that the shape formed by the cut-points on AB, AD, and AC is an equilateral triangle.  This is a bad explanation, but it makes sense.

Well, where are all the faces?  Describe what type each is.

8 (equilateral) triangles and 6 ocragons.
There might be different answers depending on the size of the cut though.

Not necessarily octagons.  But your answer would have been correct.


Suppose $q$ and $p$ are positive numbers for which

$\log_9 p = \log_{12} q  = \log_{16} (q+p)$

What is the value of $\frac qp$
I believe this problem is from one of the AoPS books...

There is indeed a quite similar problem in the AoPS book.
Set the three expressions equal to $x$, so
$p = 9^x$, $q = 12^x$, and $q + p = 16^x$
We are looking for $\frac{q}{p} = \frac{4}{3}^x $
$9^x + 12^x = 16^x$
$1 + \frac{4}{3}^x = \frac{16}{9}^x$
Denote $Y$ as $\frac{q}{p} = \frac{4}{3}^x $:
$1 + Y = Y^2$
$Y = \frac{1 + \sqrt{5}}{2}$ since $q$ and $p$ have to be positive integers.
[hide]
which is the Golden ratio[/hide]

[quote="Iversonfan2005"]This exact same problem appeared on this forum a while back.[/quote]
You are correct Iversonfan2005 and in fact not just once, but twice
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=257612#p257612]http://www.mathlinks.ro/Forum/viewtopic.php?p=257612#p257612[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=40609]http://www.mathlinks.ro/Forum/viewtopic.php?t=40609[/url]

Can you post some more log problems xenon_nightmare152 please or any of you.
I really need some practise(started this new).

Can anyone tell me what is $ln$.
They say it is $log_e$ but does $e$ have a specific value like $\pi$?????
 :help:

[hide="Solution"]
log 3 / log 2 times log 4 / log 3 times log 5 / log 4, etcetera etcetera etcetera.

All the terms cancel out except for log 8 / log 2, which happens to be log 8 base 2, which equals [b]3[/b]. Jajaja.
[/hide]

[quote="BornForMath"]$e$ does have a specific value.
From Google calculator: e = 2.71828183
 [url=http://antwrp.gsfc.nasa.gov/htmltest/gifcity/e.1mil]First ten million digits of e[/url]
[url=http://mathworld.wolfram.com/e.html]e from MathWorld[/url]

Here is a simple log problem to get you started: :)

$\text{Evaluate the product}(\log_2 3)(\log_3 4)(\log_4 5)(\log_5 6)(\log_6 7)(\log_7 8).$[/quote]

Simple, just do change of base formula and everything cancels out except for $\frac{log8}{log2}$ which is $log_{2}{8}=3$. Someone make harder ones :)


In $ xy \minus{}$plane, there are $ b$ blue and $ r$ red rectangles whose sides are parallel to the axis. Any parallel line to the axis can intersect at most one rectangle with same color. For any two rectangle with different colors, there is a line which is parallel to the axis and which intersects only these two rectangles. $ (b,r)$ cannot be ?

$\textbf{(A)}\ (1,7) \qquad\textbf{(B)}\  (2,6) \qquad\textbf{(C)}\  (3,4) \qquad\textbf{(D)}\   (3,3) \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]$ (1,r)$ is possible.

[code]
bbbbbbbbb
1
 2
  3
   4
    5
     6
      7
[/code]

$ (2,r)$ is also possible.

[code]
bbbbbbb
1      b
 2     b
  3    b
   4   b
    5  b
     6 b
[/code]

$ (3,4)$ is also possible. So $ (3,3)$ is also possible.
[code]
bbb
bbb    44
bbb  3
     3  bbb
  22    bbb
 1      bbb
 1 bbbbb
   bbbbb
   bbbbb

[/code]

The answer is $ \boxed E$[/hide]


$ m^2\minus{}2m(cosx\minus{}sinx)\plus{}2sin^2 x\geq 0                   (1)$
Find the values of m such that (1) is true for every $ y\in R$

1) Double Post.
2) Wrong forum. [moved now]
3) There is no $ y$ in (1).

:lol: Sorry, I don't know that. In (1), find values of m such that (1) is true for every $ x\in R$


1.$ x\equal{}\frac {\minus{}1\pm\sqrt{5}}{2}$

[hide="Solution"]Put $ x^4 \plus{} 2x^3 \plus{} 2x^2 \plus{} x \minus{} 2 \equal{} (x^2 \plus{}ax \plus{} b)(x^2 \plus{}px \plus{} q)$
So, $ p\plus{}a\equal{}2\Rightarrow p\equal{}2\minus{}a$
$ bq\equal{}\minus{}2\Rightarrow b\equal{}\frac {\minus{}2}{q}$
Substitute these values in $ q\plus{}ap\plus{}b\equal{}2$ and $ qa\plus{}pb\equal{}1$
Solve the 2 quadratics in $ q$ and equate powers of $ q$ to get $ a\equal{}1$ and $ q\equal{}\minus{}1,2$[/hide]

Same method in second one, can't think of a better one. maybe someone will come up with something different
sorry for postng again :)


While installing Texnic and Miktex, the instructions on this site tells you to tell Texnic where your Dvi-viewer is located, yet when I installed Texnic, the path was already in the ouput profile.

In the pictures section of the guide, the first example in the "how to" section looks like this:

[code]\documentclass[11pt]{article}
\usepackage[pdftex]{graphicx}
\begin{document}

This is my first image.

\includegraphics*[viewport=100 200 500 650]{myimage}

That's a cool picture up above.

\end{document}[/code]

and is supposed to produce this:

[url]http://www.artofproblemsolving.com/LaTeX/Images/myimagelatex.pdf[/url]

The second piece of code on the "manipulation" page looks like this:

[code]\documentclass[11pt]{article}
\usepackage[pdftex]{graphicx}
\begin{document}

This is my first image.

\includegraphics*[viewport=100 200 500 650]{myimage}

That's a cool picture up above.

\end{document}[/code]

and produces this:

[url]http://www.artofproblemsolving.com/LaTeX/Images/myimagelatex6.pdf[/url]

There's a contradiction there.

Thanks for noticing this problem.  The web page has been fixed.


complex analysis
function
geometry
geometric transformation
rotation
complex analysis unsolved

Find all automorphisms of the punctured disk $B_{1}(0)\setminus\{0\}$.

A holomorphic function from the punctured disk to the punctured disk is bounded in a neighborhood of zero, hence it has a removable singularity there. Pursue that, and we see that zero must be mapped to zero. In that case, we have an automorphism from the disk to itself that fixes zero, and that can only be a rotation, $z\to e^{i\phi}z.$


This is one of the problems from one of the workouts(workout 5) that they send your school, I don't really get this:

What number is the reciprocal of 100pi? Express the answer in scientific notation rounded to three significant digits.

Thanx
 :)

I don't want to just give you the answer, so let me give yoy the clues you need.  You may know some and not others, hopefully enough to fill in the blanks.

First reciprocal basically means you invert or flip a fraction(technically a number's reciprocal is the number you would multiply by to get a product of 1)

Next you would turn your fraction to a decimal by division.

Last put it in scientific notation which maybe the part you least understand.  Here are some examples of putting numbers in scientific notation:

75000 = 7.5 x 10^4
893000000 = 8.93 x 10^8
.0000034 = 3.4 x 10 ^-6
.002345 = 2.345 x 10^-3

As for the 3 significant digits, that just means you only use three non-zero digits(one of them can be zero if the zero is in middle ex. 3.08)

Hope that helps you on your way,

josh Frost

Without a calculator, this problem would be very annoying...

With a calculator, just do [hide]1/(100pi)[/hide]

Actually, this problem is hardly as tough as it seems.

One little hint (look if you're really stuck)-[hide]Do we really have to take the reciprocal of ? or can we take the reciprocals of 2 numbers?[/hide]

It's one of those problems that seems hard until you find a really easy way to solve it.

That problem was on Workout 5 for the 2004-2005 (this year's) problems. Since it is a workout, calculators ARE allowed, so when calculating 1/(100 :pi:) you get [hide]3.18 * 10^(-3) [/hide]

Just quick rule of exponenets:

Positive - You know how to do it so no explanation

Negative - You will going to get decimal and here's why:

2^-1 = (1/2)^1 = 1/2

You're going to learn why you change 2 to 1/2 soon if you didn't learn it yet. But just know there's way like that.

Silverfalcon


LaTeX articles integration limit geometry Support

I am trying to add to my forum the ablity to post latex code.
But I need away to convert the latex to a image (i.e gif , jpg ,...etc)
So I can display it in a html page.

I am 99% sure that the artofproblemsolving converts the latex code summited to a gif file.

This all would be easy to do if I was on linux but I am on windows xp using microsoft iis server.
I did my forum in asp and javascript.

For linux their are command line programs like latex2gif , dvi2png ,...etc etc list goes on.
But for windows you don't have these options.
I have looked on sourceforge for a latex  to image converter but the only ones for windows
are zipped in tar.gz archive files. And also requires cygwin which I am "not" willing to download and install.

I was wondering if anybody knows how to accomplish this.
I am also curious to know if artofproblemsolving uses a linux server to host their site.
Because if they are using windows then how the heck are they converting the latex to a gif.
And could I get your conversion program.

The only other way I thought of handling it was to make a RPC to my linux machine.
But this would not be very efficient.

I also have miktex installed on this machine which has a tool that converts tex to pdfs.
However I haven't found any tool's for it to go tex to gif ...etc.
If their is any add on package I could get to do this that would be one way to do it.

Thanks for any help. But they should have a free program out their that runs on windows xp
that doesn't need cygwin to be installed.

I also looked at MLMATH from w3c standards but didn't want to use it because I want my latex
to be displayable in a browser independent manner.

AoPS uses [url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] for the images; it is a PHP program which uses LaTeX, ImageMagick and Ghostscript to get the images. All these programs are available for Windows. If you aren't using PHP then look at the batch file for [url=http://www.mayer.dial.pipex.com/tex.htm#latexrender_windows]LatexRender for Windows[/url], where you can see how the conversion is done so you can adapt it for yourself.

Other ways are to use [url=http://savannah.nongnu.org/projects/dvipng/]dvipng[/url] (which you have already as it is included with of MiKTeX) or [url=http://www.forkosh.dreamhost.com/source_mathtex.html]mathTeX[/url] either on your server (preferable) or remotely from mathTeX.

[quote="Mathematics!"]I have looked on sourceforge for a latex  to image converter but the only ones for windows
are zipped in tar.gz archive files. And also requires cygwin which I am "not" willing to download and install. [/quote]That's nonsense. There are loads of Windows programs that will open tar.gz files. Examples include 7-Zip, Winzip, Win-GZ, tar-gzip etc etc.

Ok I did just find the dvipng command in the bin directory. I didn't realize that was their. :lol:
But my problem is the way the image looks. For example

latex code

file test.tex

[code]\documentclass{article}
\begin{document}
 $\sqrt{x} = 5$ is $x=25$.
$\sum \int \sqrt{x} \sqrt[n]{x} \frac{a}{b} \oint \prod \coprod
\frac{\partial{f}}{\partial{x}} \lim_{a \rightarrow b}$
\end{document}
[/code]

compiled it first with the command line latex command
like

latex test.tex

Then I take the test.dvi file that was created and
do
dvipng -T tight -o firsttest.png test.dvi
This produces the png image firsttest.png. My problem is that I want the image
to be only as big as the text space needed for the text. Right now my image is always
produceing 220 x 802 and when I don't use -T tight it produces 374 x 802.
So as you can tell their is alot of bordering white space at the bottom.
To much for putting it nicely in a webpage.
How can I tell divpng just to make the image as big as the text area needed?
(Maybe it is not the dvipng that is the problem but I have noticed when it generates the
png file their is a tiny 1 at the bottom of the image. Maybe this is the problem. How do I get
rid of the number at the bottom.)

Also I am wondering on the most effiecient way of putting latex into my forum.
Is LatexRender more faster then dvipng or what exactly is the fasts/best one to use in your
opinon? If the speed is mostly determined by the image file produced then I know the speed issue.

And
[quote]That's nonsense. There are loads of Windows programs that will open tar.gz files. Examples include 7-Zip, Winzip, Win-GZ, tar-gzip etc etc.
[/quote]

I never had to open tar.gz files on a windows machine never knew that WinZip could do it cool though. :blush:
However why don't they just make it easier for everybody by using a .zip archive file.

You should ask on the [url=http://savannah.gnu.org/support/?group=administration]dvipng forum[/url] for support for the program.

[quote="Mathematics!"]Also I am wondering on the most effiecient way of putting latex into my forum.
Is LatexRender more faster then dvipng or what exactly is the fasts/best one to use in your
opinon? If the speed is mostly determined by the image file produced then I know the speed issue..[/quote]Since the speed depends on all sorts of criteria, the only way to check is to compare them on the same machine. However, for a reasonably specified server the differences are going to be small compared with the connection speed.

[quote="Mathematics!"]I never had to open tar.gz files on a windows machine never knew that WinZip could do it cool though. :blush:
However why don't they just make it easier for everybody by using a .zip archive file.[/quote]Who is this "everybody"? Why should everyone else change their much more mature systems to suit you? The Unix way of doing things has been around a lot longer than zip files. Also there are many types of zip files not all compatible with one another so which one are you advocating everyone uses? What about cab files that Microsoft itself uses? Have you never used rar files? There are a myriad of such compression formats.

Have a look at the image that is generated by dvipng.
Using dvipng -T tight -o testYes.gif testmath.dvi

on the latex
[hide]
[code]\documentclass{article}
\begin{document}
 $\sqrt{x} = 5$ is $x=25$.
$\sum \int \sqrt{x} \sqrt[n]{x} \frac{a}{b} \oint \prod \coprod
\frac{\partial{f}}{\partial{x}} \lim_{a \rightarrow b}$
\end{document}
[/code][/hide]

You will see a one at the bottom right of the long image how do I get rid of it.
Is this a latex problem or a dvipng problem. Is it because of something to do with what type of
\documentclass{article} note I am not an expert at latex just want to beable to support it in my forum.

About the speed issue are you saying that in most cases it is negilgable. And if not what part is usually the issue the latex or the dvipng command.

Thanks

O I figured out the problem with

[code]\documentclass{article}
\begin{document}
 $\sqrt{x} = 5$ is $x=25$.
$\sum \int \sqrt{x} \sqrt[n]{x} \frac{a}{b} \oint \prod \coprod
\frac{\partial{f}}{\partial{x}} \lim_{a \rightarrow b}$
\end{document}[/code]

just needed to put [code]\pagestyle{empty}[/code] after the \begin{document} .
Works like a charm.

Thanks again for your help.


inequalities
linear algebra
matrix
calculus
derivative
integration
advanced fields

Let $F: \Omega\rightarrow\mathbb{R}^{n}$, $\Omega \subset \mathbb{R}^{n}$ convex and open, $F \in C^{1}(\Omega)$ and $x^{*}$ such that $F(x^{*})=0$ and $J(x^{*})$ non singular, where $J$ is the Jacobian matrix of $F$.

Supose there exists $p>0$, $L>0$ such that \[\parallel J(x)-J(x^{*})\parallel \leq L \parallel x-x^{*}\parallel ^{p}\] Prove that \[\parallel F(x)-J(x^{*})(x-x^{*})\parallel \leq\frac{L}{p+1}\parallel x-x^{*}\parallel ^{p+1}\]


The original was posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162547

[quote]Let $ f:\bf R\to R$ be a differentiable function such that

$ f(x) \equal{} f(x/2)\plus{}xf'(x)/2$.

Show that $ f$ is linear.[/quote]


Να λυθεί στους μη μηδενικούς πραγματικούς η εξίσωση:
$ 3^{\frac{1}{x}}x \plus{} 3^{x}\frac{1}{x} \equal{} 6$

Αυτή η άσκηση λύνεται με πολλούς τρόπους, επέλεξα τον καλύτερο γι' αυτούς που θέλουν να δουν κατευθείαν τη λύση της:
[hide]
Αν $ x < 0$, τότε $ 3^{\frac{1}{x}}x < 0$ και $ ^{x}\frac{1}{x} < 0$ άρα
$ 0 > 3^{\frac{1}{x}}x \plus{} 3^{x}\frac{1}{x} \equal{} 6 < \equal{} >$ άτοπο
Αν $ x < 0$, τότε από την ανισότητα αριθμητικού - γεωμετρικού μέσου και έπειτα της ανισότητας $ x \plus{} \frac{1}{x} >\equal{} 2$ αν $ x>0$, έχουμε:
$ 6 \equal{} 3^{\frac{1}{x}}x \plus{} 3^{x}\frac{1}{x} >\equal{} 2\sqrt{3^{ x \plus{} \frac{1}{x} }{ x \plus{} \frac{1}{x} } }>\equal{} 2\sqrt{3^{2}} \equal{} 6$
με τις ισότητες να ισχύουν όταν $ x \plus{} \frac{1}{x} \equal{} 2$ και $ 3^{x}\frac{1}{x} \equal{} 3^{\frac{1}{x}}x$.
Λύνοντας το σύστημα αυτό καταλήγουμε στη λύση $ x \equal{} 1$

Επέλεξα αυτή τη λύση γιατί έχω ιδιαίτερη αδυναμία στις ανισότητες.

Η άσκηση είναι η ίδια αν, αντί 3, έχουμε α και, αντί, 6 2α με α>0
[/hide]

Νομίζω πάντως ότι η άσκηση λύνεται και με αντίστροφες συναρτήσεις

Ίσως με παραγώγους βγαίνει εύκολα 1-1, οπότε και μοναδική προφανής λύση για

$ G(x) \equal{} 3^{\frac {1}{x}}\cdot x \plus{} 3^x\cdot\frac {1}{x}\minus{}6$

η $ x \equal{} 1$.

Με κάθε επιφύλαξη. Δε την έχω δοκιμάσει σε χαρτί!

Βγαίνει όντως πολύ εύκολα με παραγώγους (από τη στιγμή που απορίψουμε τις λύσεις <0). Πάντως η γενική ιδέα είναι ότι ισότητες μπορούν να προέλθουν από την φράξη παραστάσεων.

Η λύση με παραγώγους βγαίνει ως εξής:
έστω $ g(x) \equal{} 3^{x}\frac{1}{x} \plus{} 3^{\frac{1}{x}}x$ τότε:
Αν $ x < 0$ τότε $ g(x) < 0$ άρα δεν έχει λύση.
Αν $ x > 0$ τότε είναι $ g'(x) \equal{} e^{x}\frac{1}{x}ln3 \plus{} 3^{x}lnx \plus{}  e^{\frac{1}{x}}ln3lnx \plus{} 3^{\frac{1}{x}}$
     αν $ x \equal{} 1$ τότε εύκολα $ g'(x) > 0$
     αν $ x > 1$ τότε εύκολα $ g'(x) > 0$
     αν $ x < 1$ τότε υπάρχει $ y > 1 : y \equal{} \frac{1}{x} < \equal{} > g(y) \equal{} g(x)$  άρα $ g'(x) > 0$
     άρα $ g'(x) > 0$ για κάθε $ x > 0$
     άρα η $ g(x)$ είναι 1-1 για κάθε $ x > 0$
     άρα η $ g(x)$ έχει το πολύ μία θετική λύση
     και επειδή έχει την προφανή λύση $ x \equal{} 1$, αυτή θα είναι μοναδική

Μια αλλη προσεγγιση.....
Αρκει να λυσω τν:$ [x3^{1/2x}\minus{}3^{x/2}]^2\plus{}2x(3^{(x^2\plus{}1)/2x}\minus{}3)\equal{}0$ και ειναι προφανες οτι μονο για $ x\equal{}1$ εχει λυση...

Πολύ δυναμικό! :)  :lol:


Solve the equation $x =  \sqrt { 2 +  \sqrt { 2 -  \sqrt {2 + x}}}$.

[Admin] $\LaTeX$ enabled.

to get the accurate (not numerical) seems difficult


Is it possible for the altitudes of an obtuse angled triangle to be 1:2:3

[hide="hint"]
It does not matter if the triangle is obtuse or not.
[/hide]

[hide]$ h_a: h_b: h_c\equal{}{1\over a}: {1\over b}: {1\over c}$, and from there $ a\equal{}k, b\equal{}{k\over 2},c\equal{}{k\over 3}$ for some real $ k$. But then $ a>b\plus{}c$, so it's impossible.[/hide]


Prove that there exists a positive number $N$ has 2000 digits so that $N$ is composite and when we replace three arbitrary consecutive digits of $N$ by three new arbitrary number, the received number is also composite

[quote="srulikbd"]it's can't be true[/quote]
 :huh: why?

[quote="srulikbd"]cause, if I understand the ques correctly, we can actually can choose a new number. which is prime...?[/quote]

I don't think so. If we let the number end in $00$, then if we change any numbers but the last three, the result will be divisible by $2$ and $5$. So we need to show that there is some number $n$ with $1997$ digits such that there is no prime number between $1000n$ and $1000n+999$ (which I think pretty likely to happen for some $n$).

(There is a better approach to the problem I'm sure, but this was just intended to convince srulikbd that the proposition is true).

[quote="srulikbd"]but we can change the last 2 digits as well[/quote]
Yup, but you have to change three consecutive digits, so to change the last two, you'll have to change only the last three digits. And so if there's some gap greater than 2000 between two consecutive primes of 2000 digits, then the proposition is true (as whatever you change the last three digits to, the resulting number will be composite, as there are no primes in this range).

[quote="t0rajir0u"][hide="Idea"] If we can show that $n = gcd(1, 2, .... 999)$ has $1997$ or less digits, then let $N = 1000n$. [/hide][/quote]

This was my first thought, also.  The problem is, changing the last three digits to $001$ gives us something for which we don't have a prime factor.  (We know it's divisible by 1, by construction, but that obviously doesn't help us.)

[quote="t0rajir0u"][hide="Idea"] If we can show that $n = gcd(1, 2, .... 999)$ has $1997$ or less digits, then let $N = 1000n$. [/hide][/quote]
we can use the idea as multiplication of primes upto 1000,
2*3*5*.....* 997=A and A has digits less than 1997 but what about
1000n+1(when you change the last three consecutive numbers)

[quote="JBL"]Actually, here's the fix: 1009 is the smallest prime above 1000.  We take either gunes' or t0r's idea: both of those numbers should have well less than 2000 digits.  Multiply it by 100, so it ends in 000.  This result is relatively prime to 1009.  Thus, there is some multiple of it which is congruent to -1 mod 1009.  In fact, this multiple is at most 1008 times as large as it.  That multiple also ends in ...000, but we know that 1009 divides the result of changing it to ...000, and we know by the argument that gave rise to it that changing the last three digits to anything else gives a composite, and changing any other 3 consecutive digits leaves an even number, so we're set.  We've spent at most 6 extra digits on this, so then all we need is to show that gunes' or t0r's number has at most 1994 digits.[/quote]
mine is easy!
we all can sense that there are less than 500 primes upto 1000.
Suppose all have 3 digits. Then multiplying them will give us at most 500*3=1500 digit number.


AMC
USA(J)MO
USAMO
induction
number theory
relatively prime

This is a problem from the 2003 USAMO, and I just wanted to make sure my solution has no gaps in it. I "solved" it, but I have a feeling I'm wrong...  :lol:  I'm sure you know what I mean.

Here we go:

$ \text{Show that for each n we can find a n\minus{}digit number with all the digits odd which is divisible by } 5^n \text{.}$

My solution:

We shall proceed by induction:

For $ n\equal{}1$, $ 5^1 | 5$.

Now, assume that $ 5^n | k$, where $ k$ is an $ n$-digit number with all its digits odd.

Then, $ 5^n \cdot a \equal{} k$ for some $ a$.

We will now concatenate a digit $ d$ in front of $ k$ to form an $ n\plus{}1$-digit number. Although the problem does not state that the concatenation must be this specific, if this can be proven with this specificity, then we can also say that the general statement is true.

After concatenation, the number becomes $ 10^n \cdot d \plus{}k$.

This equals:

$ 10^n \cdot d \plus{}5^n \cdot a$

$ 5^n (2^n \cdot d \plus{} a)$

Since $ 2^n$ and $ 5$ are relatively prime, $ 2^n \cdot d \mod{5}$ goes through all of the modular residues of $ 5$ as $ d$ goes through the odd integers from $ 1$ to $ 9$, inclusive.

Thus, there is a $ d$, $ 1\le d \le 9$, such that $ 2^n \cdot d \equiv x\mod{5}$ for any $ x$. So for any $ a$, there exists a $ d$ such that $ 2^n \cdot d \plus{}a \equiv 0 \mod{5}$. So for any pair $ (n,a)$, we can choose $ d$ such that $ 5^{n\plus{}1}|5^n(2^n\cdot d \plus{}a)$.

And the induction is complete.

Correct, although
- This problem has already been posted more than once, and
- It doesn't really belong in Intermediate.

Mhmm I think you should be saying

After concatenation, the number becomes $ 10^{n\plus{}1}\cdot d\plus{}k$ instead. Since you want to add a digit at the far left of an $ n$ digit number, you must be $ n\plus{}1$ digits over and place a digit in that slot.

$ 10^n$ is correct; it is the smallest $ n\plus{}1$-digit number.

Har, after waking up from my tiredness it hit me. I was being really stupid :)


A convex polygon is given, no two of whose sides are parallel. For each side we consider the angle the side subtends at the vertex farthest from the side. Show that the sum of these angles equals 180 degrees.

I suppose that at least the number of vertices must be odd. Otherwise what do you mean by "side subtends at the vertex"?

[quote="warut_suk"]Well, for each side there must be a vertex farthest from that side, then you draw the angle using that side and that vertex, so as far as I understand, the number of sides can be even...[/quote]
But I don't understand how you can draw an angle using a side and a vertex. You can only draw an angle using two sides!

And one more question. There is only the following condition in the statement: "no two sides are parallel". But this is not sufficient to the fact that for each side there is only one vertex farthest from that side. For example consider regular pentagon. Perhaps no two sides and diagonals may not be parallel. :?

Why? You join each endpoint of the side to the vertex. That's the angle you want.
And say the regular pentagon is ABCDE then D is farthest from AB, E is to BC, etc.

I'll just sketch some steps, since the details might become a bit dry and boring to deal with.

First of all, let $\mathcal P=A_1A_2,\ldots A_n$ be the polygon. The indices are modulo $n$, i.e. $A_{n+1}=A_1$. For each side draw two segments, connecting the two endpoints of the side to the vertex which is farthest away from the side. We will call these simply segments.

Try to show that every two segments must intersect.

Show that if $A_i$ is the vertex farthest away from $A_uA_{u-1}$, while $A_j$ is the vertex farthest away from $A_uA_{u+1}$, then all the sides on the path $A_iA_j$ which does not contain $A_u$ have $A_u$ as the vertex which is farthest away.

Use these results to conclude that if $T_1,T_2,\ldots,T_k$ are the vertices (listed in a counter-clockwise direction; again, $T_{k+1}=T_1$) of $\mathcal P$ which are endpoints of at least two segments each, then $k$ is odd, and if for each $T_i$ we draw the two segments which form the largest angle with vertex $T_i$, then these two segments have $T_{i+\frac{k-1}2},T_{i+\frac{k+1}2}$ as second endpoints, and the sum of angles that we are looking for is $\sum_{i=1}^k\angle T_{i+\frac{k-1}2}T_iT_{i+\frac{k+1}2}$. The conclusion is easy to draw from here.

I'm sure a picture will make everything clear.


ARML
geometry
rhombus
circumcircle
cyclic quadrilateral
perpendicular bisector

In my initial solution, I assumed that $BECD$ is cyclic. In fact, this was a correct assumption. If you drew your diagram correctly [except for the angle condition], then this is readily apparent. I was so pumped that I saw the cyclic quadrilateral within the first minute or so of doing the problem. I knew this was the type of thing that few people would see. Too bad I ended up with a negative angle since I had $E$ between $A$ and $B$ in my diagram during the contest. Anyway, I have spent a lot of time going along bogus paths due to suggestions from others. I have spent way more time than would be necessary for this problem. Anyway, here is a complete solution:

Lemma:
Let $ABCD$ be such that $DA=DB=DC$, $AD//BC$, and the perpendicular bisector of $CD$ interesects $AB$ at $E$ such that $\angle ABE=\pi$. Prove that $AE=ED$.

Proof:
Let $F$ be the point closer to $A$ such that: $F$ is on $BC$ and $FC=AD$. Let $DF\cap AB=E'$.

$FC=AD$ and $FC//AD$ implies $AFCD$ is a parallelogram. Since $DC=AD$, $AFCD$ is a rhombus.

Let $\angle DAC=\theta$. Since it is a rhombus, $\angle ADF=\frac{\pi}{2}-\theta=\angle FDC$.

Since $DB=DC$ and $AD//BC$, $\angle DCB=2\theta=\angle DBC=\angle ADB$.

Since $AB=BD$, $\angle DAE'=\frac{\pi}{2}-\theta$. Hence, $\angle E'AD=\angle EDA$, so $AE'=E'D$.

Recalling that $\angle E'AD=\angle E'DC$, $AD=DC$, and $AE'=DE'$, $\triangle ADE'\cong\triangle DE'C$. This implies that $DE'=E'C$.

Hence $E'$ is on the perpendicular bisector of $DC$. Hence $E\equiv E'$ since they are the point that lies on $AB$ and the perpendicular bisector of $DC$. The result follows.

End Lemma.

Let $\angle BCE=4x$, then $\angle DEC=2x$. $\angle ECD=\frac{\pi}{2}-x$. $\angle BCD=\angle CBD=\angle ADB=\frac{\pi}{2}-5x$. $\angle BEC=10x$. $\angle BDE=10x-\left(\frac{\pi}{2}-x\right)=11x-\frac{\pi}{2}$.

$\angle ADE=\angle EDC\implies \frac{\pi}{2}-5x+11x-\frac{\pi}{2}=\frac{\pi}{2}-x\implies 7x=\frac{\pi}{2}$

Then $4x=\frac{2\pi}{7}\implies \boxed{\frac{360}{7}}$.
--

If you were expected to come up with this solution, then the problem would be totally unreasonable. I have spent at least an hour looking for a rigirous solution. Yet, I feel that this type of problem was easily guessable. If you drew your diagram with any degree of accuracy-focusing on all the conditions except for $\angle BCE=2\angle DEC$-, then the problem is prone to conjecture. Being able to guess geometric properties is [i]very[/i] important in high level geometry. If you were to find a rigirous solution in 10 minutes, this problem would be unreasonable. However, it is perfectly reasonable to form a conjecture and produce an answer in 10 minutes.

--Alex

I have been thinking more about this problem. Unless I am totally missing something, the proof part of this problem could make a very easy olympiad problem. I think that is a bit ridiculous. Of course, there is a difference between finding an answer and providing a proof, but still. I think problems should be made such that you have to develop some rigor to find the correct answer.

[quote="Altheman"]In my initial solution, I assumed that $BECD$ is cyclic. In fact, this was a correct assumption. [/quote]

Dang.  I made this assumption and then didn't use it at all.

[quote="pianoforte"]All the problems and solutions are on the official arml site.[/quote]

Where?

On the arml sidebar, click "2007 ARML Contest"

Okay, zman, your solution works, but would most people start to use uniqueness in about 7-9 minutes?  I know that I wouldn't think of it until at least 20 minutes into a problem.  Oh well, that's why it's a #8 and I missed it.  Bleh.

[quote="zman"][quote="Altheman"][quote="jmerry"]I've looked at the problem myself- it took me between 20 and 30 minutes to do it. It's a very nice problem, but it just doesn't belong in ARML.

I bashed the key fact that ADE is isosceles with coordinates.[/quote]

I agree. It is basically a semi-difficult proof problem. The angle thing is just to make it short answer.[/quote]

Do it backward, draw the circle BCD first, and make its intersection with line AB as E'. Then very quick angle chasing shows that E'C = E'D, that means E' lies on the perpendicular bisector, so E = E' The rest is simple.

Use uniqueness to replace certain conditions by other conditions is the key here.[/quote]

Yea...I was playing around with the 4 conditions that $E$ is a circumcenter, $D$ is a circumcenter, $AD//BC$, and $AE=ED$.

However, I ended up just doing a different proof as you can see...

This is a pre-olympiad geometry proof. 10 mins is not exactly enough time to come up with a rigirous solution...

I still think the solution using ABCD as a Cyclic Quadrilateral works slightly better... I spent about 20 minutes finding that myself, probably even less. Nevertheless, in the ARML conditions... it's incredibly difficult.

Kevin

couldn't you show that E is the circumcenter of DAB, then it follows that DAE is congruent to DBE, so you can assign some variables (3 to be exact), angle chase with isosceles triangles, parallel lines, etc. to get two equations, then use the fact that <ADE=<BDE to write a third equation, solve and be done?

that's what my dad did...

also I don't think ABCD is cyclic...AB||CD but those two base angles aren't equal

[quote="unlimitd3K"]wait, no... I keep confusing that...

I meant that you could use D as the center of the circle around ABC, and work from there[/quote]

Nah, I tried that, I don't think that works out nicely.


This is a poll for those who took AP Biology this year. Good luck to everyone on their scores!

where's the medium option? anywho, i put easy cuz it certainly wasn't hard. it was what was to be expected. i know i messed up two essays pretty bad, but a 5 is still well in reach.

I thought it was harder than I expected, but still easy overall.  The one essay about angiosperms versus bryophytes probably hurt me a bit, but given such a generous curve, it seems close to impossible not to get a "5".

[quote="dakyru"]but given such a generous curve, it seems close to impossible not to get a "5".[/quote]
how true.

lol yeah i forgot the medium option. its okay, this is AoPS, most people here do REALLY well on these types on exams.

[quote="joml88"]And yet 81.1% of the students taking the AP Bio exam last year failed to get a 5....hmm  :D  :D  :D[/quote]

that also means that 18.9% do get a 5. that's almost 1 in 5. i would like consider myself in the top 18.9% of test takers  :D

The only question I really had issue with was the one on the chromosomes. I did not know WHAT the crap they were wanting on that question. So I just wrote a whole bunch of stuff related to genetics and prayed I'd get some points off of it. Even my bio. teacher was like "WHAT?" and she grades AP exams. The angiosperm question was AWESOME. AWESOME AWESOME AWESOME. Probably because I know that stuff like the back of my hand. The immune system stuff made me a little nervous, probably because I've always been rusty on that, but... oh and the first one. HEEEHEE! Man, if you knew how many times we had been over that in lab, in tests, in essays, in review sessions... that's a HUGE concpt in our class. Always has been.

i predicted the immune system one for this year last year. but it was only a matter of time before they asked it and the cell. resp one. im not surprised on the angiosperm vs. bryophytes either. next year i predict an ethology or ecology question

i thought it was hard..

hope not to be one of those 89%..

my teacher predicted the cell respiration lab. and with a graph and design-an-experiment, it was easy
i was stumped with the "parts of a chromosome". i put long and short legs and i called something the wrong name. so i'll be lucky with like a 3 or 4 based on the other sections of the question.
the plant question messed me up cuz i got the parts mixed up. then i completely forgot the alternation of generations so my BS won't get any points. expecting another 3 or 4 points there.
i liked the immune question. i remembered most of it and expect to get a high grade. i might have left out/ mixed out a step or two, but most of it was right. go me.

and those are my way-too-long thoughts on the free response section.  :D

[quote="bleumoose"]hate to do this but... BUMP.
how did everyone do?[/quote]

this is on the other thread in round table, but i'll post again since this is just for biology.

[size=200]5!!! yay[/size]

Off topic, but.......
What course is more interesting?

Genetics/Microbiology
or
Vertebrates

I have to pick in between those 2 next semester.


I think so.  If I recall correctly this was an AIME problem...
[hide="hint"]write it out and simplify...[/hide]

[quote="eryaman"]
[hide="hint"]write it out and simplify...[/hide][/quote]

The reason I put it here is because that is very obvious....

[quote="bobthebuilder"]
$= 2 \cdot 199 \cdot 2 \cdot 197 ... \cdot 101 \cdot 50!$
the biggest two digit prime is 191/3 (because 191 is in the numerator) = 67

[/quote]
??? :?


A state lottery offers the following game:

It costs $\$ $1$ to buy a ticket.
For your ticket, you choose one card each from the four suits of a standard 52-card deck. Example: 5 of spades, queen of hearts, 6 of diamonds, ace of clubs.
The lottery now randomly makes its own selection of such a ticket.
If all four cards match, you win $\$ $5000$.
If any three of the four cards match, you win $\$ $50$.
If any two of the four cards match, you win $\$ $5$.
If any one of the four cards match, you win another ticket for the same game.
If no cards match, you win nothing.

What is your expected return for your ticket?

[hide="Answer"]The four cards you draw are completely random, so those do not matter. When the computer chooses, there is a $\frac{4}{52}\cdot \frac{3}{51}\cdot \frac{2}{50}\cdot\frac{1}{49}=\frac{1}{270725}$ chance of all four cards matching, in which case you will win $\$ $5000$. The chance of three cards matching, in which case you win $\$ $50$, is $\frac{4}{52}\cdot \frac{3}{51}\cdot \frac{2}{50}\cdot \frac{48}{49}=\frac{48}{270725}$. If two cards match, which has a chance of $\frac{4}{52}\cdot \frac{3}{51}\cdot \frac{48}{50}\cdot \frac{48}{49}=\frac{1152}{270725}$, you will receive $\$ $5$. If only one matches, which is a $\frac{4}{52}\cdot \frac{48}{51}\cdot \frac{48}{50}\cdot \frac{48}{49}=\frac{18432}{270725}$ chance, you will receive a free ticket, and if no cards match, which is a $\frac{270725-(18432+1152+48+1)}{270725}=\frac{251092}{270725}$ chance, you win nothing.
If the only way you could win money was if all four cards matched, you would win an average of $\frac{200}{10829}$ dollars per ticket. If you could only win money when three cards matched, you would win an average of $\frac{96}{270725}$ dollars per ticket. If you could only win money when two cards matched, you would win an average of $\frac{1152}{54145}$ dollars per ticket. Adding those three numbers together, we have $\frac{376}{7735}$ dollars per ticket. However, since you can also win another ticket for a second (or third, or fourth, etc.) try, we must find the infinite sum $\frac{376}{7735}+\frac{18432}{270725}\cdot \frac{376}{7735}+...+\displaystyle\left(\frac{18432}{270725}\right)^{\infty}\cdot \left(\frac{376}{7735}\right)^\infty$. Therefore, we have:
$\frac{\frac{376}{7735}}{1-\frac{18432}{270725}}=\frac{13160}{252293}$ dollars per ticket. :lol:[/hide]

I hope I didn't make any mistakes there... that was rather long. :?

That's not it, Nathan. We aren't drawing four cards from the whole deck; rather we are drawing one spade, one heart, one diamond, and one club. Four independent draws from four separate suits.

Ohh... right, I forgot about that part. I hope that makes it simpler...

[hide="Answer"]Again, the cards you draw do not matter. The computer's chance of drawing all four of the cards that you picked is $\frac{1}{13^4}=\frac{1}{28561}$, in which case you win $\$ $5000$. The chance of three of the four is ${4\choose 1}\cdot \frac{12}{13^4}=\frac{48}{28561}$, in which case you win $\$ $50$, the chance of two is ${4\choose 2}\cdot \frac{12^2}{13^4}=\frac{864}{28561}$, in which case you win $\$ $5$, the chance of one is ${4\choose 1}\cdot \frac{12^3}{13^4}=\frac{6912}{28561}$, in which case you win another ticket, and the chance of none is $\frac{12^4}{13^4}=\frac{20736}{28561}$.
If money is only won when all four cards are chosen, you win $\frac{5000}{28561}$ dollars per ticket. If money is only won when three of the four are chosen, you win $\frac{2400}{28561}$ dollars per ticket. If money is only won when two of the four are chosen, you win $\frac{4320}{28561}$ dollars per ticket. Adding these probabilities together, you will win an average of $\frac{11720}{28561}$ dollars per ticket. Factoring in the chance of another ticket, we have the infinite series $\frac{11720}{28561}+\frac{6912}{28561}\cdot \frac{11720}{28561}+ ...+\displaystyle\left(\frac{6912}{28561}\right)^\infty \cdot \left(\frac{11720}{28561}\right)^\infty$, so you will win an average of $\frac{\frac{11720}{28561}}{1-\frac{6912}{28561}}=\frac{11720}{21649}$ dollars per ticket.[/hide]


i have newly installed ubuntu 7.04(a GNOME based LINUX)
i want to learn some LINUX .I am a [b]newbie[/b]
can anyone post a good book

I dual-boot Ubuntu and Windows, and I've been using Ubuntu/Linux for over a year, so perhaps I can help =)

I wouldn't recommend books for learning Linux. IMO, they're just not worth it. I just wouldn't want to read a Linux book cover-to-cover. I'd recommend tutorials and hands-on experience. Here's one of my favorite tutorials:

http://www.tldp.org/LDP/gs/node5.html

And I think the first thing you should do is familiarize yourself with the shell.

Also, I have some linux-related stuff on my blog if you want to check it out.

Dont forget the man pages.  ALot of time its easier to go to the man page for a command, and good man pages have cross references to similar commands.

Between your online search engine and man pages, basic Linux administration is a bookless deal.  (i love books, but just being practical)

yes have realised this now after a two week or saga with ubuntu ....it's so addictive.... beryl is wonderful :D


[color=blue]Problem[/color].In a Notpron competition we have $n$ contestants. Each match is between two of the contestants.Each two of the contestants have played agianst each other once.
For each two contestance like $A,B$ ,there exist exactly $t$ contestance that have lost to $A$ and $B$.
Prove that \[n=4t+3\] PS.Obviously i changed some of the names . :)

[quote="rem"]There are no ties, right? :maybe:
Also, when you say "lost to A and B" do you mean lost to at least one of them or both of them?[/quote]

Both of them.  If he meant "lost to at least one of them", then he would have said A OR B.  Obviously there can't be any ties.


In a distant land there lived a worm with two mouths, one at each end. Both mouths ate at the same rate and with equal efficiency. Thirty minutes after the worm began eating a potato chip with one of its mouths, its other mouth joined in the fesat. Thirty minutes later, the chip was completely consumed.

Suppose both mouths began eating the chip at the start, and one mouth stopped when half the chip was eaten. It would then take the other mouth x minutes to finish the other half of the chip. Find x.

(rom Math Team team round)

[hide]Note that the first head eats for 1 hour total, and the second head eats for $ \frac12$ hour total.  Because they eat at the same rate, this is equivalent to the first head eating for $ \frac32$ hours total to finish the chip.  Thus, to finish half of the chip, it would take $ \cfrac{\frac32}{2}\equal{}\frac34$ hours to finish.  $ \frac34$ hours is the same as $ 45$ minutes.[/hide]


superior algebra
superior algebra theorems

What does $a^{G}$ means, where $a$ is an element of the group $G$

My guess is that $a^{G}=\{a^{g}\colon g\in G\}$, where $a^{g}=g^{-1}ag$. It is common to write conjugation in this way, because $(a^{g})^{h}=a^{gh}$. Thus, $a^{G}$ is the conjugacy class of $a$.


Explain why it is recommended to crouch when you touch the floor after jumping.

change in momentum is constant no matter what method you use.

you should crouch to elongate the collision time.  change in momentum = change in impulse = integral of Fdt, so if you evaluate the integral over a longer time interval, F will decrease to compensate.  thus, by crouching, your joints experience less force as you stop.

[quote="perfect_radio"]I don't understand why the collision time is longer when crouching... :?[/quote]if your legs are perfectly straight, your [b]entire[/b] body will stop almost instantly after you make contact with the ground.

If you crouch as soon as you make contact with the ground, your upper body will still be moving for a little while, even though your feet are not, because the joints in your leg can now provide a way for the upper body to maintain motion.  Thus, collision time is increased because it will take more time for your entire body to come to rest.


binomial coefficients
number theory unsolved
number theory

Find $n$ in such a way that ${n\choose 3}$ would be a perfect square

${{50}\choose 3}= 140^{2}$ is the only solution.

See also (in general) : http://www.mathlinks.ro/Forum/viewtopic.php?t=102812

Can you explain more?What can you prove with sylvester lemma? :blush:

Well, I am not going to explain this here, as it would be too long.
Yet, it is fully discussed in [u]Proofs from the book[/u].

But! if you don't have it, and cannot find one, you can thank google for this :) (it's chapter 3) :
[url=http://books.google.fr/books?vid=ISBN3540404600&id=KvQr9l0wgf8C&pg=PA13&lpg=PA1&dq=proofs+from+the+book&sig=mHHs05Eyp6ds9umSZjRGJrAbfi8]Binomial coefficients are (almost) never powers[/url].


quadratics
modular arithmetic
absolute value
number theory unsolved
number theory

Shown below is exercise 10.14.11 in Uspensky and Heaslett's Elementary Number Theory.  Chapter 10 introduces Legendre and Jacobi symbols and proves the quadratic  reciprocity law.  Request help, thanks.
--------------------
By analytical methods it is proved that for any prime $ p \equiv 1\pmod{8}$ there exists a prime $ Q \equiv 3\pmod{4}$ of which $ p$ is a quadratic non-residue. Given that such a prime $ Q$ exists, show that there are such primes $ q < \sqrt{\frac{4}{3}p}.$

For example, although 3 and 7 are residues of 193, 11 and 19 are non-residues.  With $ p\equal{}193, Q\equal{}19,$ one must prove that one of {3,7,11} is a non-residue.

Problem solved below, by an idea inspired by a method used in chapter 11 of the Uspensky - Heaslett book.

$ {}_xR_y$ denotes that $ x$ is a quadratic residue of $ y.$
$ {}_xN_y$ denotes that $ x$ is not a quadratic residue of $ y.$

Lemma
Given:
(1) Prime $ p \equiv 1\pmod{8}$
(2) $ m \in \mathbb{Z},\; p \nmid m$
(3) $ {}_{(\minus{}p)}R_m$
(4) $ |m| > \sqrt{\frac{4}{3}p}$
(5) $ m \equiv 3\pmod{4}$
Then:
(1) $ \exists m_1 \in \mathbb{Z} \ni p \nmid m_1$
(2) $ {}_{(\minus{}p)}R_{m_1}$
(3) $ |m| > |m_1|$
(3) $ m_1 \equiv 3\pmod{4}$

Proof:
One of the two solutions to $ x^2 \equiv (\minus{}p) \pmod{m}$ is contained in
$ \{1,2,...,\frac{|m| \minus{} 1}{2}\}.$  Denote this solution as $ N \Rightarrow
N^2 < \frac{1}{4}m^2.$

Also, $ |m| > \sqrt{\frac{4}{3}p} \Rightarrow \frac{p}{|m|} < \frac{3}{4}|m|.$

Thus, $ \exists m^{\prime} \in \mathbb{Z} \ni (N^2 \plus{} p) \equal{} m\cdot m^{\prime}
\Rightarrow m^{\prime} \equal{} \frac{N^2}{m} \plus{} \frac{p}{m}
\Rightarrow |m^{\prime}| < \frac{1}{4}|m| \plus{} \frac{3}{4}|m| \equal{} |m|.$

$ N$ even $ \Rightarrow (N^2 \plus{} p) \equiv 1\pmod{4} \Rightarrow m^{\prime} \equiv 3\pmod{4}.$  In this case, set $ m_1 \equal{} m^{\prime}.$

$ N$ odd $ \Rightarrow (N^2 \plus{} p) \equiv 2\pmod{8} \Rightarrow [\text{since}\; m \equiv 3\; \text{or}\; 7\pmod{8}]\; m^{\prime} \equiv 6\pmod{8}.$ In this case, set $ m_1 \equal{} \frac{1}{2}m^{\prime} \Rightarrow m_1 \equiv 3\pmod{4}.$

Under either definition of $ m_1$, $ |m| > |m_1|$ and $ {}_{(\minus{}p)}R_{m_1}.$   Also, $ p \nmid m \Rightarrow p \nmid N^2 \Rightarrow p \nmid (N^2 \plus{} p) \Rightarrow p \nmid m_1.$

.....................................

$ Q \equiv 3\pmod{4} \Rightarrow {}_{(\minus{}1)}N_Q \Rightarrow {}_{(\minus{}p)}R_Q.$
WLOG $ Q > \sqrt{\frac{4}{3}p}.$
Therefore, the lemma may be repeatedly invoked to generate
$ m_1, m_2, ...$ where:
(1) Each $ m_i \equiv 3\pmod{4}.$
(2) $ p \nmid m_i.$
(3) $ {}_{(\minus{}p)}R_{m_i}.$
(4) IF $ |m_i| > \sqrt{\frac{4}{3}p}$ THEN $ |m_i| > |m_{i\plus{}1}|.$

Consequently, the series must terminate with $ |m_k| < \sqrt{\frac{4}{3}p}.$
Since $ m_k \equiv 3\pmod{4},$ it has a prime divisor $ q \equiv 3\pmod{4}
\ni {}_{(\minus{}p)}R_{q} \Rightarrow [\text{since}\; q \equiv 3\pmod{4}]\;
{}_pN_{q}.$

I carelessly overlooked repairable flaw in my solution.  In the lemma's proof, $ p\nmid m$ does not imply $ p\nmid N^2.$  Example is $ p=17, m=91, N=68.$ To repair:

$ p\nmid N \Rightarrow p\nmid (N^2 + p) = m\cdot m^{\prime} \Rightarrow p\nmid m^{\prime}.$  In this case, the method originally given to select $ m_1$ is okay, and $ p\nmid m_1.$

Alternatively, $ p\mid N \Rightarrow (p^2)\mid (N^2) \Rightarrow p\mid (N^2 + p)$ and $ (p^2)\nmid (N^2 + p).$  In this case, with $ p\nmid m,$ then under either definition of $ m_1,$ $ p\mid m_1\text$ and $ (p^2)\nmid m_1.$  Therefore, $ m_1{\prime}$ (et al) may be taken as $ \frac{1}{p}m_1 \Rightarrow$ [since $ p\equiv 1\pmod{4}$ and $ m_1\equiv 3\pmod{4}$] $ m_1{\prime}\equiv 3\pmod{4}.$

Found an additional repairable flaw.  My original solution used the lemma to deduce the existence of $ m_k \equiv 3\pmod{4},$ whose absolute value is $ < \sqrt{\frac{4}{3}p}.$  My solution then presumes that $ m_k$ must have a prime divisor $ q \equiv 3\pmod{4}.$

This implication fails if $ m_k < 0.$  Fortunately, the repair is routine.  In the Lemma's premise, require that $ m>0.$  Since $ m^{\prime}$ is computed as $ (N^2 \plus{} p) \equal{} m\cdot m^{\prime},$ this will allow the property $ m_1 > 0$ to be added to the conclusion's lemma.


calculus
derivative
function
calculus computations

I'm having trouble with this problem

z=f(x,y)=xy/(x^2+2y^2)
Define f(0,0)=@. Find @ such that the directional derivative of f(x,y) in the direction of the vecotr w(1,4) exists at point (0,0). also at this point @, do the partial derivatives dz/dy and dz/dx exist?

Anybody know where to start?

This function is homogeneous of degree zero; $ f(cx,cy)\equal{}f(x,y)$ for $ c\neq 0$. That means that any directional derivative at the origin must be zero if it exists, and it exists only if the function is constant on that line through the origin.

In particular, what is $ f$ on the line $ (x,y)\equal{}t\cdot (1,4)$?


geometry
perimeter
inequalities
triangle inequality

If $a,b,c$ are the sides of a triangle whose perimeter is equal to 2 then prove that:

a) $abc+\frac{28}{27}\geq ab+bc+ac$;

b) $abc+1<ab+bc+ac$

See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=47939&view=next[/url]  (problem 1)

 :)

HINT:add 1/a + 1/b + 1/c and then use the standard inequalities. I know this is too much precise. But adjust with this much for two days and mre hint would be give if you still cannot solve.

i've found a solution to a), i think it's not the nicest, but it's a solution

$\displaystyle abc+\frac{28}{27}\geq ab+ac+bc \\\Leftrightarrow 2abc+\frac{56}{27}\geq \sum_{cyc}a(ab+ac+bc) \\\Leftrightarrow \frac{7(a+b+c)^3}{27}\geq abc+\sum_{cyc}a^2b+ab^2 \\\Leftrightarrow 42abc+\sum_{cyc}7a^3+21a^2b+21ab^2\geq 27abc+27\sum_{cyc}a^2b+ab^2 \\\Leftrightarrow 15abc+7\sum_{cyc}a^3\geq 6\sum_{cyc}a^2b+ab^2$

but from am-gm and schur we have that
$15abc+7\sum_{cyc}a^3\geq 6(3abc+\sum_{cyc}a^3)\geq 6\sum_{cyc}a^2b+ab^2$
and we're done

edit: just few seconds after i posted this solution i noticed there is a simpler one...

by the triangle inequality we have $1>a,b,c$, hence from am-gm we have that

$\frac{1}{3}=\frac{(1-a)+(1-b)+(1-c)}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)} \\\\\Rightarrow \frac{1}{27}\geq 1-(a+b+c)+ab+ac+bc-abc=ab+ac+bc-(1+abc) \\\Rightarrow abc+\frac{28}{27}\geq ab+ac+bc$

edit: just few seconds after the last few seconds i realized that $(1-a)(1-b)(1-c)>0 \Rightarrow abc+1<ab+ac+bc$

$1 < ab + bc + ac - abc \leq \frac{28}{27}$
Notice that $(1-a)(1-b)(1-c) = 1 - (a+b+c) + ab + bc + ac - abc = -1 + ab + bc + ac - abc$. Therefore, if we substract $1$ from all sides of the inequality, we get
$0 < (1-a)(1-b)(1-c) \leq \frac{1}{27}$
The left inequality is clearly true because of the triangle inequality ($0 < a,b,c < 1$).
Now, we use Ravi-substitution where we write $a = x+y, b=y+z, c = x+z$ with $0 < x,y,z < 1$ and $x+y+z = 1$.
We get that
$(1-a)(1-b)(1-c)  = xyz \leq \frac{1}{27}$
Which is true by AM-GM.

[quote=mihaig]If $a,b,c$ are the sides of a triangle whose perimeter is equal to 6 then prove that:

$$3abc+72\geq 8(ab+bc+ac).$$

Remark: The inequality is true for all positives with sum 6. The same for part a) of post #1[/quote]

Note that this is Schur's inequality and actually holds for all $a,b,c\geq0$ with $a+b+c=6.$
Consequently, the part a) of #1 holds for all $a,b,c\geq0$ with $a+b+c=2.$


What math class are you taking in school for the 2007-2008 school year? I hope that I can meet some of you from the Art of Problem Solving forum next year when I go to high school. Also, please tell me what school you're going to next year.  :)

Above is a poll using some of the math classes available at Thomas Jefferson High School for Science and Technology. I'm not really sure why certain classes are organized in certain ways. Why is algebra split into two years by geometry into "algebra 1" and "algebra 2?"

-Chris Au

I saw 52 views but no one responds  :(

I want to know if I might meet you next year in school.

that's because algebra is a wide subject. it's not possible for public schools to finish in one year.

Yeah I live in Virginia Beach, and math is split up pretty much the same way.  I'm taking magnet precalculus 1 and 2 right now.


number theory
relatively prime
greatest common divisor

a and b are relatively prime positive integers, and n is an integer. Show that the greatest common divisor of
a:^2: +b:^2: - nab and a+b must divide n+2.

a^2+b^2-nab = (a+b)^2-ab(n+2), so

m = gcd(a^2+b^2-nab,a+b)
= gcd((a+b)^2-ab(n+2),a+b)
= (-ab(n+2),a+b)

so m|ab(n+2) and m|a+b. Since gcd(a,b)=1, m cannot divide either a or b, only their sum a+b. Therefore, m|ab(n+2) means that m|n+2.

thats not entirely accurate, how about this, 6|(17*2*3) but 6 doesn't divide 2, and 6 doesn't divide 3, fix this part up and you will have a good solution!

[quote]fix this part up and you will have a good solution[/quote]

Urf. Easier said than done...

you're almost there, what is special about 6|2*3 and opposed to
7|7*4 or 6|12*5 in which case 6 and 7 divide only one number?

this should finish it off

[hide]let GCD(m,a)=g. Then, g|a and g|m. Also, since m|a+b, g|a+b. Since g|a, we must have g|b. But, a and b are relatively prime, so g = 1. therefore, a and m are relatively prime. Similarly, m and b are relatively prime. So, since m|ab(n+2), we must have m|n+2 as desired.[/hide]


which of the following numbers has 4 odd factors?  10-11-15-16

I am trying to help my son Sam and and I cant really do much, please help me

[quote="SnipedYou"]Well take a look at their factors

10 --- 1, 2, 5, 10 --- 2 odd factors, 2 even factors
11 --- 1, 11 --- 2 odd factors
15 --- 1, 3, 5, 15 --- 4 odd factors
16 --- 1, 2, 4, 8, 16 --- 1 odd, 4 even factors

So 15 is the number you are looking for.

If you don't know what a factor is it is just a number that when dividing another leaves a whole number.

And this belongs in classroom math not Olympiad Algebra.[/quote]


geometry
3D geometry
probability
FTW
number theory
prime numbers

Two cubes, each with faces numbered 1, 2, 3, 4, 5 and 6, are tossed. Each cube lands with a prime number on top. What is the probability that the sum of these prime numbers is less than 9? Express your answer as a common fraction.

Well there are 3 prime numbers in each cube and the only bad combo is 5,5, so the answer is 8/9.

On FTW!, the problem comes out as:

$\text{Two cubes, each with faces numbered}$ $1, 2, 3, 4, 5$ $\text{and}$ $6,$ $\text{are tossed.}$ $\text{Each cube lands with a prime number on top.}$ $\text{What is the probability that the sum of these prime numbers is less than}$ $8?$ $\text{Express your answer as a common fraction.}$

So I was wondering what I got wrong when I put $\cancel{\boxed{\frac{2}{3}}}$, so I went over to Review, and it said:

$\text{Two cubes, each with faces numbered}$ $1, 2, 3, 4, 5$ $\text{and}$ $6,$ $\text{are tossed.}$ $\text{Each cube lands with a prime number on top.}$ $\text{What is the probability that the sum of these prime numbers is less than}$ $9?$ $\text{Express your answer as a common fraction.}$

Is this a glitch?


Because of $ O_{1}O_{3}\perp OB$ and $ O_{2}O_{3}\perp OA$ we have $ \angle OO_{1}O_{3} \equal{} \angle OCA \equal{} \angle OO_{2}O_{3}$ and we are done.

Kostas Vittas.

Hi Kostas. Perhaps I'm being dim here. I can see how the lines are perpendicular but not how the angles are equal.

[quote="AndrewTom"]Let $ A, B$ and $ C$ be three points in order on a line, and let $ O$ be a point not on the line $ ABC$. Let $ O_{1}, O_{2}$ and $ O_{3}$ be the centres of the circles $ BCO, CAO$ and $ ABO$, respectively. Prove that $ O, O_{1}, O_{2}$ and $ O_{3}$ are concyclic.[/quote]
We know that $ \angle OO_2O_1 \equal{} 90^\circ \minus{}\angle O_2OC \equal{} 90^\circ \minus{} \frac { 180^\circ \minus{} \angle OO_2C }{2} \equal{} \frac { \angle OO_2C }{2} \equal{} \angle OAC$ and $ \angle OO_3O_1 \equal{} 90^\circ \minus{} \angle O_3OB \equal{} 90^\circ \minus{} \frac { 180^\circ \minus{} \angle OO_3B }{2} \equal{} \frac { \angle OO_3B }{2} \equal{} \angle OAB$.Hence $ \angle OO_2O_1 \equal{} \angle OO_3O_1$,as a result,the points $ O, O_{1}, O_{2}$ and $ O_{3}$ are concyclic.

AndrewTom
Could someone explain this? Thanks.

first of all draw a suitable figure in which perpendicular bisectors from O1, O2, O3 on each side of the triangles r drawn
use basic properties as angle subtended by arc nd parallel lines that will be formed by the medians
also use cyclic quad. formed by the perpendicular bisectors
i hope this will help u

Thanks to both of you (and to vittasko for his private message).


The compressibility factor of 1 mol of $ \ce{H2}$ is reasonably described by the expression

$ z \equal{} 1 \plus{} \frac{1.57 \times 10^{\minus{}2}}{V} \plus{} \frac{7.08 \times 10^{\minus{}4}}{V^2}$

Compute:

(a) The Van der Waals parameters for hydrogen.

(b)The volume occupied by one mole of hydrogen.

from Van de Waals equation P=(RT/(Vm-b))-a/Vm2
so PVm/RT = VmRT/(Vm-b)-a/Vm
expand Vm/Vm-b= 1+b/(Vm-b) using Taylor expansion
compare the coefficient with that in the virial equation Z=PVm/RT= 1+B/Vm+C/Vm2+... but we neglect more than three couple interactions.
Thus we hav etwo variables a,b and two equation from B,C values


geometry
3D geometry
algorithm
probability

What do you mean by geometric model? A tournament that I designed is like so.

[hide]First there are 3 games, A vs B, C vs D, E vs F.

Then the 3 teams that win have a round robin tournament.

Or you could have all 6 teams have a round robin tournament.[/hide]

[quote="drunner2007"][quote="AdorableAddict"]Each team is assigned to a random face on a cube. Then each team plays with the team opposite to it on the cube and this results in 3 games. A corner is determined by the winning three faces and the corner directly opposite to that corner will necessarily be the intersection of the three losing teams. Each of the three teams play against each other. Then each team plays against the two teams that are opposite to the two teams that it has most recently played. This algorithm ensures an equal chance of winning for each team.

[size=67]The above paragraph is intended as a joke. It does guarantee equal probability, nevertheless.  [/size][/quote]

isn't that just a 6-team round robin tournament? every team ends up playing every other team exactly once.[/quote]

Exactly. Read the small text.   :P

[quote="mr. math"]construct a geometric model for a tournament (any kind) with 6 teams. each team must have an equal chance to win.[/quote]

So why dont you just roll a die?

Yes, assign each team a number from 1 to 6. Then roll a die to determine the winner.  Each team DOES have a equal chance of winning, but it's independent of their skill.   :rotfl:

Case 1: They dont have the same skill
Then if the tornament depends on their skill, each team does not have an equal chance to win. It contradicts with the assumption.
So the tornament doesnt depend on their skill, roll a die is OK.

Case 2: They have the same skill
Certainly rolling a die is OK.

Therefore, rolling a die is the method you are looking for!
(just a gag :D )

Well I don't think your contradiction is valid. If they don't have equal skill, I think that you can somehow set up a bracket so that the teams with more skill are put a disadvantage, and somehow give everyone an equal chance at winning the whole tournament.

Then, if you put the teams with more skill in disadvantage to give everyone an equal chance, so why dont just choose the winner randomly?

Consider the case when a team is invincible and defeats all the other teams, if their skill counts, you dont have another choice but to let that team to be the champion. So the teams dont have an equal chance to win.


A box, in teh shape of a rectangilar solid has length 12, width 8 and a height of 8. What is the length, along the outside surface of the box, of the shortest path that can be drawn between the two of the vertices of the box which do not lie in the same face of the box?

Unless I've misunderstood the question, I get the following

[hide="Solution"]

Suppose the box's base is the rectangle $ ABCD$ with $ AB \equal{} 8$ and $ BC \equal{} 12$ and that the top of the box is the rectangle $ EFGH$ with $ E$ above $ A$ and so on.

If we consider that the box is hollow and open it out, then the shortest path from $ A$ to $ G$ (say) will be the length of the hypotenuse of the right-angled triangle $ AFG$

$ AF \equal{} AB \plus{} BF \equal{} 8 \plus{} 8 \equal{} 16$ and $ FG \equal{} 12$, so, by Pythagoras, $ AG \equal{} \sqrt {16^{2} \plus{} 12^{2}} \equal{} \sqrt{256\plus{}144} \equal{} \sqrt{400} \equal{} 20$.

[/hide]

Hope this is helpful.

Thanks Arrange your tan. Don't know what I was thinking:

[hide="Click to reveal hidden content"] clearly we have $ 4\times 4$ and $ 4\times 3$ for the shorter sides of the right-anlged triangle and so the hypotenuse must be $ 4 \times 5 \equal{} 20$.[/hide]

I've edited my previous post.


[quote="GameBot"]All of the dimensions of a cereal box are increased by $ 20\%$. By what percent is the volume increased? Express your answer rounded to the nearest whole number.[/quote]

Let initially cereal box be $ \ 100\times\100\times100$

After increase it becomes $ \ 120\times 120\times120$

$ \implies$ % increase in volume is $ \ \frac {120^3}{100^3}\times100 = 172.8\%$

After rounded to the nearest whole number, it is $ \ 173\%$

How do I talk in a game? The thing tells me I have to play a certain amount of games and send some posts. Any helpers? :?:

Wrong forum again... Post that in the FTW Questions and Support forum.

Or you could just let the length be a more generalized $ x$.  Then, we have $ (1.2)^3x^3 \implies 1.728x^3 \implies \boxed{73\%}$.


trigonometry
integration
logarithms
function
calculus
calculus computations

need help with the following differential equantions:

solve differential equantion:
1) Y' + Y/X = cos(x)

2) Y'' -2Y' -3Y = -2e^3X

3) Y'' - Y = Xsin(X)

4) Y'' + 2Y' +3Y = e^-2X (with: Y(0)=0 and Y'(0)=1)

sorry for not using latex.

Thx.

You could try using the method of changing constants. This is how it is done (I will do it on your first equation).

You have:

$ \frac{dy}{dx}+\frac{y}{x} = \cos x$

First, you find the solution to the homogeneous equation:

$ \frac{dy_h}{dx}+\frac{y_h}{x} =0$

$ \frac{dy_h}{dx}=-\frac{y_h}{x}$

$ \frac{dy_h}{y_h}=-\frac{dx}{x} / \int\,dx$

$ \ln y_h = \ln \frac{C}{x}$

$ y_h = \frac{C}{x}$

Now, you declare that $ C$ is a function of $ x$, and you have:

$ y = \frac{C(x)}{x}$ - notice that this is $ y$, and not $ y_h$

$ \frac{dy}{dx} = \frac{1}{x}\frac{dC}{dx} - \frac{C(x)}{x^2}$

Now, plugging this into your original equation, you get:

$ \frac{1}{x}\frac{dC}{dx} - \frac{C(x)}{x^2}+\frac{C(x)}{x^2} = \cos x$

$ \frac{1}{x}\frac{dC}{dx} = \cos x$

$ \frac{dC}{dx} = x \cos x$

$ C(x) = \int x \cos x\,dx = \cos x + x \sin x + K$

and therefor:

$ y = \frac{\cos x + K}{x} + \sin x$

Try verifying this by plugging it back in.. I hope you can do the rest by your own now!

1. $ y' \plus{} \frac {1}{x}y \equal{} \cos x\Longrightarrow (xy)' \equal{} x\cos x\Longrightarrow xy \equal{} \int x\cos x\ dx \equal{} x\sin x \plus{} \cos x \plus{} C$, yielding $ \boxed{y \equal{} \sin x \plus{} \frac {\cos x}{x} \plus{} \frac {C}{x}}$

2. $ y'' \minus{} 2y' \minus{} 3y \equal{} \minus{} 2e^{3x}\Longleftrightarrow (D \plus{} 1)(D \minus{} 3)y \equal{} \minus{} 2e^{3x}$, we have a particular solution: $ y \equal{} \minus{} \frac {2}{(D \plus{} 1)(D \minus{} 3)}e^{3x}$

$ \equal{} \minus{} \frac {2}{D \minus{} 3}\cdot \frac {1}{D \plus{} 1}e^{3x} \equal{} \minus{} \frac {2}{D \minus{} 3}\cdot \frac {1}{3 \plus{} 1}e^{3x} \equal{} \minus{} \frac {1}{2}\cdot \frac {x}{1!}e^{3x} \equal{} \minus{} \frac {1}{2}xe^{3x}$.

$ \therefore \boxed{y \equal{} C_1e^{ \minus{} x} \plus{} C_2e^{3x} \minus{} \frac {1}{2}xe^{3x}}$

hello, 4. solving the homogenous part of the eqaution by the ansatz $ y(x)\equal{}e^{\lambda x}$ we get $ y_h\equal{}e^{\minus{}x}\left(C_1\cos(\sqrt{2}x)\plus{}C_2\sin(\sqrt{2}x)\right)$. For the inhomogenous part you must make the ansatz $ y_p\equal{}Ae^{\minus{}2x}$.
The result  $ y(x)\equal{}e^{\minus{}x}\left(C_1\cos(\sqrt{2}x)\plus{}C_2\sin(\sqrt{2}x)\right)\plus{}\frac{1}{3}e^{\minus{}2x}$. To determine the constants $ C_1$ and $ C_2$ you must use the initial conditions.
Sonnhard.


integration
calculus
calculus computations

does anyone have any suggestions on graping in 3-space? and i have matlab but i have not learned how to use it to graph in 3-space. does anyone know how to do it

[tex]\int_{\infty}^\pi f(x)dx[/tex]

First you should read the output of the HELP PLOT3 command, and then use HELP for the instructions you are told to "See Also..." ;)


inequalities
logarithms
inequalities unsolved

$ f: R_{\plus{}} \to R_{\plus{}}$
  $ f(xy) \le f(x) f(y)$     for     $ x,y \in R_{\plus{}}$.

Prove

$ f(x^n) \le f(x) \cdot \sqrt{f(x^2)} \cdot \sqrt[3]{f(x^3)} \cdot \ldots \cdot \sqrt[n]{f(x^n)}$

[quote="aktyw19"]$ f: R_{ \plus{} } \to R_{ \plus{} }$
  $ f(xy) \le f(x) f(y)$     for     $ x,y \in R_{ \plus{} }$.

Prove

$ f(x^n) \le f(x) \cdot \sqrt {f(x^2)} \cdot \sqrt [3]{f(x^3)} \cdot \ldots \cdot \sqrt [n]{f(x^n)}$[/quote]

let $ a_k \equal{} \ln f(x^k),k \equal{} 1,2,...,n$.
Inequality into the following problem:
for any $ i,j \in N_{ \plus{} }$, $ a_{i \plus{} j} \le a_i \plus{} a_j$ then
\[ a_n \le a_1 \plus{} \frac {a_2}{2} \plus{} ... \plus{} \frac {a_n}{n}.\]
that's an old problem and I solved it [url=http://bbs.pep.com.cn/thread-500441-1-1.html]before[/url]
Image:
[img]http://bbs.pep.com.cn/attachments/month_0908/20090813_56d4c55bb2f0ea1b69adMMshOBclUP9x.gif[/img]


a cirle $ P: (x \minus{} 3)^2 \plus{} (y \minus{} 3)^2 \equal{} 25$ is rotated as $ \angle{A}$ where $ 0^0 < \angle{A} < 360^0$ so that results in the new position $ P'$. If there is only $ 1$ point $ L$ which match the criteria $ L \in P$ and $ L \in P'$ find the value(s) of $ cos(A)$..
thanks

If I understand it right, there are always exactly 2 values for L.

Which indicates to me that I'm missing something. Could you be more specific about the rotating part?


I've got a question and I hope someone can answer - I'm actually doubting if anyone will even understand my question, but whatever.  I started thinking about this when I was thinking about the remainder theorem.  It's a pretty simple concept, and the proof of this theorem is very simple as well, if I'm not mistaken.  However, this was not an intuitive thing for me, and I was wondering if anyone else felt the same way.  I totally understand the proof, and I can totally accept the theorem, of course.  I'm sure as I become more familiar with the topic, this 'Remainder Theorem' will become intuitive/obvious for me.  I think when I first learned it and worked on it a lot, I had that type of understanding of it, but I guess I lost it, because when I was thinking about it today, I had to write up the proof for myself just to feel comfortable to talk about it.  Is there a way that people think of it, the Remainder Theorem, that makes it obvious?  Perhaps, in words, is there a way of explaining the reason why/how the remainder theorem works?

Does anyone understand my question?

Well, anyway, I guess I should say what the remainder theorem is, for anyone who doesn't know:
"If f(x) is a polynomial, then the remainder obtained by dividing f(x) by x-p equals f(p)."

There is a general result that makes this one obvious.

First I will present an analogy using concrete numbers.  Suppose I wish to divide $37$ by $11$.  I will get a quotient of $3$ and a remainder of $4$.  Hence, I could write

$37 = 3 \cdot 11 + 4$

Where $3$ is the quotient and $4$ the remainder.  Obviously the remainder must be less than the divisor, or else we could divide more into it.  Similarly, when dividing a polynomial $f(x)$ by another $g(x)$, we write

$f(x) = Q(x) \cdot g(x) + R(x)$

Where $Q(x)$ is the quotient polynomial and $R(x)$ the remainder polynomial.  Analogously, $deg(R(x)) < deg(g(x))$ (or else we could divide into it).

Now, as to your question, set $g(x) = x - p$.

$f(x) = Q(x) (x - p) + R(x)$
$f(p) = R(p)$

And $R(x)$ must be a constant polynomial.  :)

Out of curiosity, what is the proof that you learned?

oh sorry for taking a while to reply.

what you posted actually, that's the proof i learned.

i mean, usually, there are multiple ways to explain why something works, if you truly understand it.  for example.... uhhhhmmmmm... Pascal's Formula, right?

Off the top of my head, I know that it can be explained using words and logic, as well as with algebra.

So I was wondering if there were any other ways of thinking about the Remainder Theorem, like there are with many concepts in math.

Well, then you have to ask yourself, conceptually, what dividing a polynomial by $x-p$ actually means.  And then there's a much easier explanation:

When dividing a polynomial by $x$ the remainder is obviously the constant term - in other words, $f(0)$.  That's because $x$ divides every term except the last.

Then dividing $f(x)$ by $x-p$ can be seen as being equivalent to dividing $f(x+p)$ by $x$, and then the remainder is again the constant term - or $f(p)$!  :)


Let ABC be a triangle with <B=30.Take D point on [BC] line segment such that  |DC|=|AB|.If angle <BAD=20  then calculate  angle C.

I do not know how to express the answer without trigonometric functions -- can it be done?

Are trigonometric equations and functions not allowed?

No, I just assumed that if someone has gone to the trouble of picking the nice angle 30 degrees, that the answer would come out to something nice as well.

[quote]Let ABC be a triangle with <B=30.
Take D point on [BC] line segment such that |DC|=|AB|.
If angle <BAD=20 then calculate angle C.[/quote]
I haven't solved the problem yet, but I have an observation.[hide]
Construct altitude AE.

Let $x = [AB] = [DC].$

Since $\angle B = 30^o,\; AE = \frac{x}{2},\;BE = \frac{\sqrt{3}}{2}x$

The area of $\Delta ADC = \frac{1}{2}[DC][AE] = \frac{1}{2}(x)\left(\frac{x}{2}\right) = x^2$ . . . interesting?[/hide]

I did it using various trig functions, but it was not a particularly nice answer, although impression may have been the problem. You will understand if you do it.

[quote]
[tex] \displaystyle \Delta ADC = \frac{1}{2}[DC][AE] = \frac{1}{2}(x)\left(\frac{x}{2}\right) = x^2 [/tex]
[/quote]

actually, [tex] \displaystyle \frac{1}{2}(x)\left(\frac{x}{2}\right) = \frac{x^2}{4} [/tex]

[hide="The answer"]In the setup, if we replace "30" with $\theta_1$ and "20" with $\theta_2$, I find the answer to be $C = \arctan\left(\frac{\sin{\theta_1} \cdot \sin(\theta_1 + \theta_2)}{\sin(\theta_1+\theta_2) - \sin\theta_1 \cdot \cos(\theta_1+\theta_2)}\right)$.  This can also be expressed as $\cot^{-1}(\csc \theta_1 - \cot(\theta_1 + \theta_2))$, which in this case has value $\cot^{-1}(2 -\cot\frac{5\pi}{18})$. [/hide]

[size=117][color=darkred][b]I think that $ m(\widehat{BAD}) \equal{} 15^{\circ}$ and no $ 20^{\circ}$ !!![/b][/color][/size]

[quote="mathquark"][color=darkred]Let $ ABC$ be a triangle with $ B \equal{} 30^{\circ}$.Take the point $ D\in [BC]$ such that $ CD \equal{} AB$ and $ m(\widehat{BAD}) \equal{} 15^{\circ}$. Ascertain $ m(\widehat{ACB})$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Generally, I"ll suppose that $ m(\widehat{BAD})\equal{}\alpha$ is known. Apply the [b]Mr. Sinus' theorem [/b]:

$ \left\{\begin{array}{ccc}\triangle ABD & : &\frac{AB}{\sin (B\plus{}\alpha)}\equal{}\frac{AD}{\sin B}\\ \\ \triangle ACD & : &\frac{CD}{\sin (B\plus{}\alpha\plus{}C)}\equal{}\frac{AD}{\sin C}\end{array}\right\|$ $ \implies$ $ \sin B\cdot\sin (B\plus{}\alpha\plus{}C) \equal{}\sin C\cdot\sin (B\plus{}\alpha )$ $ \implies$

 $ \sin B\cdot [\sin (B\plus{}\alpha )\plus{}\cos (B\plus{}\alpha )\cdot\tan C] \equal{}\tan C\cdot\sin (B\plus{}\alpha )$ $ \implies$ $ \boxed{\tan C \equal{}\frac{\sin B\cdot\sin (B\plus{}\alpha ) }{\sin (B\plus{}\alpha )\minus{}\sin B\cdot\cos (B\plus{}\alpha )}}\ .$

[b]Remark.[/b] $ C \equal{} 45^{\circ}\Longleftrightarrow\tan (B\plus{}\alpha ) \equal{}\frac{\sin B}{1\minus{}\sin B}$. In the particular case $ \left\{\begin{array}{c}B \equal{} 30^{\circ}\\ \ \alpha \equal{} 15^{\circ}\end{array}\right\|$ $ \implies$ $ \tan C \equal{} 1$ $ \implies$ $ C \equal{} 45^{\circ}\ .$

In this case $ A \equal{} 105^{\circ}$, $ B \equal{} 30^{\circ}$, $ C \equal{} 45^{\circ}$ and $ \frac{AB}{2}\equal{}\frac{BC}{1\plus{}\sqrt 3}\equal{}\frac{CA}{\sqrt 2}\ .$
[/color]


modular arithmetic
function
linear algebra
matrix
pigeonhole principle
logarithms
number theory

Here is something related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=84909]this[/url] which I couldn't solve.

Let $f : \mathbb{N} \to \mathbb{Z}$ be such that $f(x+\varphi(m)) \equiv f(x) \pmod{m}$ for all $m$. If $f(n) = o(3^n)$, show that there is some constant $c \in \mathbb{Z}$ such that either $f(n) = c$ for all $n$, or $f(n) = c2^n$ for all $n$, or $f(n) = c(-2)^n$ for all $n$. (Here, $\varphi$ is the totient function, and the motivation is obvious).

[More generally, the conjecture is that if $f$ is bounded exponentially, then $f(n) = c_1a_1^n + \cdots + c_ka_k^n$ for some integers $c_i,a_i$. ]

Can we at least show that if $f(n) = o(2^n)$, then $f$ is constant? (in this weaker case, I believe that a weaker periodicity condition suffices, for example, that $f(x+p-1) \equiv f(x) \pmod{p}$ for all primes $p$ ...) How about weaker results?

I think that this is interesting, and wanted to share this problem with you. The motivation is the study of the periodicity profile of linear recurrence sequences, and to discover criteria similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=84909]this[/url] for linear recurrence sequences.

--Vesselin

Well, the following is really easy: if $f(n+1) - f(n)$ is bounded, show that $f$ is constant.

Can anyone show that $f(n+1)/f(n) \to_{n \to \infty} 1$ implies that $f$ is constant?

I cannot solve the initial problem, but I do have some progress on it. Sorry, it's going to be long. I'm going to prove the following:

   Thing: if $f(x)=O(e^{ax})$ for some $a<\frac{2}{11}$ and $f(n+p-1)=f(n)\pmod p$ for all $n$ and all primes $p$, then $f$ satisfies a linearly recurent relation.

   I will use  the following:

   Theorem (Siegel's lemma) Let $(a_{ij})$ be a matrix with integer coefficients, with $1\leq i\leq n$ and $1\leq j\leq m$ and $m<n$. Let $A_{j}=|a_{i1}|+...+|a_{in}|$. One can find integers $x_{1},...,x_{n}$, not all zero and having absolute values at most $(A_{1}...A_{m})^{\frac{1}{n-m}}$ such that $a_{1j}x_{1}+...+a_{nj}x_{n}=0$ for all $j\leq m$.

   The proof is left as exercice (probably the most useful application of the pigeonhole principle that has ever been invented).

    Now, assume that $|f(x)|<C A^{x}$ for some positive constant $C$ and $A<e^{\frac{2}{11}}$. Let $a_{ij}=f(i+j)$. For instant, let $m,n$ vary according to the law $m=[n/2]$.

   Lemma: one can choose a sufficiently large $m$, such that if $x_{1},...,x_{n}$ is a solution given by Siegel's lemma, then $x_{1}f(j+1)+...+x_{n}f(j+n)=0$  (*) holds for all positive integers $j$.

   Proof: take $x_{j}$ such a solution and observe that  (*) holds for $j\leq m$. Assume that (*) fails for some smallest $k>m$, take $k+1$ to be the smallest for which it fails, thus it holds for all $j\leq k$ and $k\geq m$. Take $p$ a prime smaller than $k+2$. Then $1\leq k+2-p\leq k$ and so $x_{1}f(1+k+2-p)+...+x_{n}f(n+k+2-p)=0$. But this last sum is congruent to $x_{1}f(1+(k+1))+...+x_{n}f(n+(k+1))$, which is nonzero by choice of $k$. This shows that the last quantity is actually a multiple of the product of all primes up to $k+1$. Now, let us estimate this last quantity.

  1) Estimation of $|x_{j}|$. I keep the notations of Siegel's lemma. Then $A_{j}\leq C(A^{j+1}+...+A^{j+n})<C_{1}A^{n+j}$, where $C_{1}>1$ depends only on $A,C$. Thus \[|x_{j}|\leq (A_{1}...A_{m})^{\frac{1}{n-m}}< C_{1}^{1/2}\cdot A^{\frac{mn}{n-m}+\frac{m(m+1)}{2(n-m)}}<C_{2}A^{5n/4}.\] for some $C_{2}>0$ depending only on $A,C$.

   2) Estimation of $x_{1}f(1+(k+1))+...+x_{n}f(n+(k+1))$. This is at most (in absolute value) $(\max(|x_{j}|)\cdot C(A^{k+1+1}+...+A^{n+k+1})< C_{3}A^{9n/4+k}.$

     Ok, so we saw that for this $k$ we must have $\prod_{p\leq k}{p}\leq C_{3}A^{9n/4+k}$. And since $k\geq m>n/2-1$, we have $A^{9n/4+k}C_{3}<A^{11k/2}C_{4}$. If the lemma is false, then for infinitely many $k$ one must have $11k/2\cdot\ln A+\ln C_{4}>\sum_{p\leq k}{\ln p}$ and this forces, from prime number theorem $A\geq e^{\frac{2}{11}}$, contradiction with the choice of $A$.

    Thus, if $A<e^{\frac{2}{11}}$, then we know that our function satisfies a linear recurrent relation (since not all $x_{j}$ given by the $m$ in the lemma are zero). Ok, maybe now, Vess, our expert in linear sequences will help us draw a conclusion from all this. Otherwise, it would be just a huge useless post.  :(


Circles $k_1(O,r)$ and $k_2(O,2r)$ are given. A quadrilateral $ABCD$ is inscribed in $k_1$. $AB, BC, CD, DA$ are extended over $B, C, D, A$, respectively, and at the intersection with $k_2$ points $P, Q, R, S$ are obtained, respectively. Prove that $2P_{ABCD}\leqslant P_{PQRS}$ ($P$ means perimeter).

It can be done easily using Ptolomeys inequality for four points in plane.  You sholud do it for next pairs of points (O,A,P,Q), (O,B,Q,P), (O,C,R,S) and (O,D,S,A) and after sumating this relations you sholud get the given one.   It can be done also using tons of trigonometry.


integration
function
real analysis
calculus
Functional Analysis
complex analysis
real analysis unsolved

Let f:[0,1]->R be a Riemann integrable function. Show tht g:[0,1]->R with g(x)=f(x^2) is also integrable.

x= \sqrt y
\int[0;1] f(x^2)dx =\int[0;1] f(y)1/(2 \sqrt y) dy
f  and y->1/(2 \sqrt y) are  Riemann integrable their product also.

You have a
Theorem: /\:RxR->R a bilinear continous map, f,g Riemann integrable
functions on [a;b] then f/\g:[a;b]->R is Riemann integrable
In wich grade did you see this theorem in Romania ?

I don't think they teach something like this in Romania during high school. Could you explain a bit? What's f that thing g? "That thing" is the sign which resembles a reversed V, because I don't know how to write it :D

[quote="Moubinool"]
Theorem: /\:RxR->R a bilinear continous map[/quote]
oops it is not really correct sorry.

The correct thing .
Take f,g Riemann integrable functions on [a;b],
 B:ExF->G bilinear wich means
B(f,g) is linear for the first component, and the second component
and  E,F,G Banach space,  then the function q(x)=B(f(x),g((x)) is also
Riemann integrable.

For an example consider B the product q(x)=f(x)g(x)

Thanks for making it clear. As I said, we don't learn things like this during high-school in Romania. The solution I know (or knew, because I'd have to remember it now; the problem was posted quite some time ago  :D ) involved Lebesgue's integrability criterion.

[quote="grobber"] The solution I know  involved Lebesgue's integrability criterion.[/quote]

The level of Romanian  NMO math-problems is very high, you need
to know Lebesgue integrability ?
NMO is it for 12th grade student ?

wow! The contestants know Measure Theory? It sounds terrible to me.
Only the matriculations for special classes in math department of some universities contains at most one problem which has a background in Measure Thoery.

I have recently been reading a book on Measure Theory  :) and I just finished the section of Lebesgue's Integral.  :D
Measure Theory is really hard to understand for me.  :(

I'm also reading measure/integration theory now, which book do you use? I use Rudins R & C, really neatly written, in conjuction I also read Avner Friedmans book.

The book i'm reading is in Chinese...
And as a reference book, i use 'Measure Theory' by Halmos.
Indeed, the book in Chinese lists Halmos's book as its reference, and there's also a Russian book, "Теория Функцй Вещественной Переменной" by И. П. Натансон... i don't know what it means....
�

My father has a book on Functional Analysis by Rudin. His reasearch field is functional analysis...  :D And the books I'm reading ars also his.

Our teacher's recommend us to read Rudin real & complex analysis is world kown.
When I was in third years (Licence in french ~ bachelor maybe) in University Paris 7, I found in the math library Lebesgue's integration by Bourbaki a very good book, if you 'll find a copy try it.

Bourbaki, from what I've heard, uses linear functionals to introduce measures directly right? That's pretty hardcore for a first read I think. What are your thoughts Moubinool?

Romanian students don't actually need to know about measure theory, but I think this was the best approach for this problem. As for myself, I don't even know what measure means! :D I only know what 0 measure means since this notion appears in Lebesgue's criterion.

This was, indeed, a tough one. Romanian problems aren't easy, but this was unusually hard. I think my teacher once told me only 2 or 3 people solved it right (at a final round there are usually somewhere around  100 participants for a certain grade ).

from grobber "... 2 or 3 people solved it right ( around 100 participants..."
Hoaw these 3 might be very very good solver problems.
Are they participate sometime on mathlinks ?
--------------------------------------------------------------------------
from Mindspa"...uses linear functionals to introduce measures directly right? That's pretty hardcore ..."
Well it depends on your background before, if you did some functional
analysis before, it is more easy, to see how you get measure.
Most of time Lebesgue's theory is teach like Rudin's book.


Let $ a,b,c >0$ such that $ a\plus{}b\plus{}c\plus{}1\equal{}4abc$. Prove that
  $ 1.  \frac{ \sqrt{a}\plus{}1}{ \sqrt{a}\plus{} \sqrt{b}} \plus{}  \frac{ \sqrt{b}\plus{}1}{ \sqrt{b}\plus{} \sqrt{c}}\plus{} \frac{ \sqrt{c}\plus{}1}{ \sqrt{c}\plus{} \sqrt{a}} \leq 3$
  $ 2. \frac{1}{ \sqrt{a}\plus{} \sqrt{b}\plus{}1}\plus{}\frac{1}{ \sqrt{b}\plus{} \sqrt{c}\plus{}1}\plus{}\frac{1}{ \sqrt{c}\plus{} \sqrt{a}\plus{}1} \leq 1$

[quote="iptgut"]how to proof?[/quote]
It's easy :)

2)$ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$
So:$ abc \ge 1$
Let:$ \sqrt {a} \equal{} x;\sqrt {b} \equal{} y;\sqrt {c} \equal{} z$
So:$ xyz \ge 1;xy \plus{} yz \plus{} zx \ge 3$
Ineq(2)
$ < \equal{} > 2(x \plus{} y \plus{} z) \plus{} 2 \plus{} xyz \le (xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z)$
We have:
$ \frac {1}{9}(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \ge xyz$
.....
We have done!

[quote="2424"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$. Prove that
  $ 1. \frac {\sqrt {a} \plus{} 1}{\sqrt {a} \plus{} \sqrt {b}} \plus{} \frac {\sqrt {b} \plus{} 1}{\sqrt {b} \plus{} \sqrt {c}} \plus{} \frac {\sqrt {c} \plus{} 1}{\sqrt {c} \plus{} \sqrt {a}} \leq 3$
  $ 2. \frac {1}{\sqrt {a} \plus{} \sqrt {b} \plus{} 1} \plus{} \frac {1}{\sqrt {b} \plus{} \sqrt {c} \plus{} 1} \plus{} \frac {1}{\sqrt {c} \plus{} \sqrt {a} \plus{} 1} \leq 1$[/quote]
It's true because $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$


Can it happen for a whole calendar year that no single Sunday falls on the seventh day of a month?

I dont even know how to approach calendar problems... this system is so lame.  why can't we use better units for time where everything is a magnitude of 10 and leapyears dont exist and there aren't 365.25 days in a year...  i guess though it has to be this way because the sun rises 365 times in one revolution of Earth around sun...

[quote="236factorial"]No, I think a certain date will fall on every day within a year, even in leap years.[/quote]
Why?

Let a day of the week on a date in January be x. Then the day of the week of the same day in February will be x+3, since there are 31 days in January, and 31/28 leaves 3 as a remainder. Then, March will be either x+3 or x+4, because February can have 29 days also. April will be either x+6 or x+7.

x+7 is exactly a week from x (the same day), so it can be rewritten as x.
April will be x+6 or x.
May will be x+8 or x+2. Rewrite x+8 as x+1 (same reason as above).

continue, and you will see that x, x+1, x+2, x+3.... x+7 will all be covered in some month, thus a date will appear on every day of the week during the course of a year.

You would also know this if you keep track of what day your birthday is on every year :lol:


In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:

$\textbf{(A)}\ 26\text{ in} \qquad
\textbf{(B)}\ 4\text{ in} \qquad
\textbf{(C)}\ 13\text{ in} \qquad
\textbf{(D)}\ 8\text{ in} \qquad
\textbf{(E)}\ \text{None of these}$

[hide]
Notice that triangle ABC is right, as it's side length satisfies the Pythagorean Theorem. So, the area of ABC is 120 from $ \frac{bh}{2}$. We can also find the area with $ rs$, where r is the inradius and s is the semiperimeter. Thus, $ r\equal{}4$. So, $ \boxed{B}$
[/hide]