Asymptote LaTeX articles integration trigonometry

I have downloaded asymptote as well as ghostview.  When I try to have it do anything, it says I need to instal ghostscript.  What is the easiest way to do this?

You will find Ghostscript [url=http://www.cs.wisc.edu/~ghost/doc/AFPL/index.htm]here[/url]. If you are using Windows then the download file will have w in its name, for example gs854w32.exe. It's simple to install.

GSView puts an easier to use graphical interface onto Ghostscript which is why the latter has to be installed. It looks like the Asymptote instructions in the Wiki assumed it had already been installed for LaTeX use, which is clearly not always the case.

When I followed the mirror link from that web page, it sent me to the page in the attached screenshot.  The sourceforge link eventually led to a similar situation.  Which file(s) do I need to download for it to work?

Edit: While I am asking about $\LaTeX$, quotation marks are always formatted as ending quotes.  How do I tell $\LaTeX$ to format a quote as a beginning quote.  I.e., John said, "The quote on the left is a beginning quote, while the one on the right is an ending quote."

So you aren't using Windows then as I have already said above that [quote="stevem"]If you are using Windows then the download file will have w in its name, for example gs854w32.exe.[/quote]Also links such as http://sourceforge.net/project/showfiles.php?group_id=1897 tell you that most of the files are source files for Linux etc to compile themselves and that, as I said, gs854w32.exe is the (32 bit) Windows binary.

So what operating system are you using?

For quotes use the backtick ` typed twice for opening quotes (probably on the left-hand side of your keyboard) and a single quote typed twice for closing quote, so you get $\text{``these are quotes''}$

[quote="stevem"]So you aren't using Windows then ... [/quote]
Oops.  I am using windows.  So that's the only file I need to download, then?

Edit: I downloaded the file and followed the instructions from the Wiki.  However, when I try to run Asymptote on their example.tex file, error messages appear, as in the attached screenshot.

Yes, gs854w32.exe is the installation file for Ghostscript on Windows and is all you need.

Which example file? Should it be example.asy not example.tex? See the second Problem on the [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help]Asymptote Wiki Help[/url].

I just edited my earlier message with an update on the problem.

Edit: I just managed to fix the problem on my own--I had saved the asymptote packages in the wrong folder.

I'm still having problems.  When I try to write asymptote code within a $\LaTeX$ document, the asymptote window just opens and closes immediately.  I can write an asy document in $\LaTeX$ and the resulting image compiles fine in Asymptote, but I can't find a way to insert the resultant image into a $\LaTeX$ document.  By the way, is it possible to make a hyperlink in $\LaTeX$?

Have you followed [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration]Asymptote: Advanced Configuration[/url]  including the installation of Ghostscript and using the Wizard in TexnicCenter so it can find Ghostscript - you should have a LaTeX=>PS=>PDF option there?

Simple links can use the url package where you write \url{http://www.artofproblemsolving.com} and can be assured that hyphens won't be inserted into long URLs and it can deal with characters such as ~

If you want many more options use the [url=http://www.tug.org/applications/hyperref/manual.html]hyperref package[/url] where you can colour the link and give them titles rather than just the URL plus loads of other possibilities.

I was sort of thinking more along the lines of hyperlinking to another document rather than to a web page.  Is there a way to do that?

The hyperref package should still do that since you are essentially using HTML code and links can be to URLs or documents.

[quote="roadnottaken"]I'm still having problems.  When I try to write asymptote code within a $\LaTeX$ document, the asymptote window just opens and closes immediately.[/quote]
I am still having this problem, and yes, I did make sure to follow *all* of the directions.  I can make the diagram as a separate .asy file and include it as an image, but this is not quite ideal (I'd rather keep the entire document in one file if possible).

We need to keep the DOS window open so that it can report any errors. Suppose you want to compile test.asy which is in c:\somewhere. Then go to Start->Run. Type cmd and press OK. This should give you a DOS window. Then type [code]"C:\Program Files\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" c:\somewhere\test[/code]If there are any spaces in the path for test.asy then you must use quotes.
Press Enter and see what information you get. If everything works then you'll get a pdf, but if not then you may get something useful.

That window is exactly the same as TexnicCenter uses, but has the advantage that it stays open to read error mesages.

If that works but TexnicCenter doesn't then re-check that Tools->Customize->Tools->Asymptote says:
[code]"C:\Program Files\Asymptote\asy.exe"
-batchView -tex "pdflatex" -align C -f "pdf" %tc[/code]

When I write a .asy file in LaTeX, I can run asymptote on it to produce the image.  The problem comes when the code for the .asy file is part of the code for another program.  This is when the window flashes and closes immediately.

As far as I'm aware there is no window when asymptote code is included. If I compile one of the examples:
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Hello.
I like to make pics with Asymptote like this one:
\begin{figure}[h]
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(Circle((0,0),1),yellow,black);
  fill(Circle((-.3,.4),.1),black);
  fill(Circle((.3,.4),.1),black);
  draw(Arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}
\par It makes me happy,
since I can still type my normal LaTeX stuff around it:
\(\displaystyle\int_0^{\pi}{\sin{x}}\,dx=2\)
\end{document}[/code]it creates, amongst other files, a .asy one which says:
[code]access settings;
if(!settings.multipleView)
 settings.batchView=false;
settings.tex="pdflatex";

eval(quote{
defaultfilename='LaTeX1_1';
size(345.0pt,0,Aspect);

include graph;
  size(1inch);
  filldraw(Circle((0,0),1),yellow,black);
  fill(Circle((-.3,.4),.1),black);
  fill(Circle((.3,.4),.1),black);
  draw(Arc((0,0),.5,-140,-40));
});[/code]and in the log file I get
[code](C:\localtexmf\tex\latex\asymptote\asymptote.sty
\AsyStream=\write3
\AsyTestStream=\read1
\c@asy=\count88
)[/code]So can you try compiling the code and tell me what you get in .asy and .log files?

When I tried to run asymptote on the following line
[code]draw(Arc((0,0),sqrt(1998),0,90));[/code]
it came back with the error message
[code]-: 1.6: no matching variable 'Arc'[/code]
Why is this?  Except for the specific values, it is copied from the example program they give for using asymptote in $\LaTeX$.

I presume that's because you haven't got Asymptote working yet inside the tex file. Stick to the same example and give the errors you get in the log and in the DOS window when run from the command line. Does it complain that it can't find asymptote.sty for example?

No, this was in an asy file.  When I write an asy as a separate file in LaTeX and run asymptote on it, it normally gives me the pdf file, but it has a problem with this specific line of programming.

So are you saying the values in the example you changed worked?

I'm saying that they did not work, but that when I run asymptote on a .asy file typed in TeXnicCenter, it works as long as my syntax is correct.

Edit:  I figured out what was going on.  Because the example used capital letters, asymptote was trying to read it as a variable, which, naturally, had not been defined.  When I changed if from "Arc" to "arc", the program recognized it as a command, and ran properly.

I can get that error message by saving the file as for example test.asy.tex, but if I save as test.asy it's fine. In TexnicCenter it's essential that is says All files when you save it otherwise it will add .tex to the name of the file.


analytic geometry
number theory proposed
number theory

Consider the real number axis (i. e. the $x$-axis of a Cartesian coordinate system). We mark the points $1$, $2$, ..., $2n$ on this axis. A flea starts at the point $1$. Now it jumps along the real number axis; it can jump only from a marked point to another marked point, and it doesn't visit any point twice. After the ($2n-1$)-th jump, it arrives at a point from where it cannot jump any more after this rule, since all other points are already visited. Hence, with its $2n$-th jump, the flea breaks this rule and gets back to the point $1$. Assume that the sum of the (non-directed) lengths of the first $2n-1$ jumps of the flea was $n\left(2n-1\right)$. Show that the length of the last ($2n$-th) jump is $n$.

Grobber has asked me for the solution of the problem. Since I don't have an own solution, I forward the one communicated to me by Barry Wolk:

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

[quote="Barry Wolk"]

First you modify the problem by asking for the maximum possible total distance jumped, instead of asking for this distance to equal n(2n-1). To get a maximum, two consecutive jumps cannot be in the same direction, since (e.g.)
x1, x2, 9, 5, 2, y1, y2, ... gives a smaller total than
x1, x2, 9, 2, y1, y2, ... , 5 .
So the max occurs at some x1, x2, ... that satisfies x1=1, x1<x2, x2>x3, x3<x4, etc. And (take n=5 as typical) the total distance jumped becomes
  -x1+2(x2+x4+x6+x8)-2(x3+x5+x7+x9)+x10.
To maximize this with x1=1, take
{x2, x4, x6, x8} = (as large as possible) = {7,8,9,10},
and take
{x3, x5, x7, x9} = (as small as possible) = (2,3,4,5}.
This leaves x10=6, which is the result to be proved. And in the general case, the total distance for these solutions becomes exactly n(2n-1), thereby solving the original problem.

[/quote]

[quote="Darij Grinberg"]

You wrote:

[quote="Barry Wolk"] First you modify the problem by asking for the maximum possible total distance jumped, instead of asking for this distance to equal n(2n-1). To get a maximum, two consecutive jumps cannot be in the same direction, since (e.g.)
x1, x2, 9, 5, 2, y1, y2, ... gives a smaller total than
x1, x2, 9, 2, y1, y2, ... , 5 .[/quote]

Why? I don't understand how this applies to all cases.

[/quote]

[quote="Barry Wolk"]

[quote="Darij Grinberg"]You wrote:

[quote="Barry Wolk"] First you modify the problem by asking for the maximum possible total distance jumped, instead of asking for this distance to equal n(2n-1). To get a maximum, two consecutive jumps cannot be in the same direction, since (e.g.)
x1, x2, 9, 5, 2, y1, y2, ... gives a smaller total than
x1, x2, 9, 2, y1, y2, ... , 5 .[/quote]

Why? I don't understand how this applies to all cases.[/quote]

This really is a trivial point. The general case would be a permutation of {1, 2, ... , 2n} of the form
x1, x2, ..., xp, a, b, c, y1, y2, ... , yq
with x1=1, p+q=2n-3, a>b>c or a<b<c. The total distance jumped here is
  j=|x1-x2|+|x2-x3|+...+|a-b|+|b-c|+...|y_(q-1) - yq|.
Since |a-b|+|b-c|=|a-c|, just compare this j with the total distance jumped in the permutation
x1, x2, ..., xp, a, c, y1, y2, ... , yq, b
and you easily see that the latter sequence gives a larger total distance.

So in the permutation that gives the maximum j, you cannot have two consecutive jumps in the same direction.

[/quote]

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

[b]PS.[/b] Thanks, Barry, for the solution, and sorry for forwarding personal correspondence. But your solution is the nicest one to this problem I have seen.

  Darij


geometry
incenter
cyclic quadrilateral
geometry solved

[u][b]The author of this posting is : romano[/b][/u]
____________________________________________________________________

Let a quadrilateral ABCD . AC intersects BD at I . $I_1$ is the center of the inscribed circle of triangle AIB . $I_2$ is the center of the inscribed circle of triangle BIC . $I_3$ is the center of the inscribed circle of triangle CID . $I_4$ is the center of the inscribed circle of triangle AID . Prove that $I_1I_2I_3I_4$ is a cyclic quadrilateral if and only if $ABCD$ is a cyclic quadrilateral.

[u][b]The author of this posting is : grobber[/b][/u]
____________________________________________________________________

After some experimenting with Dynamic Geometry software, this doesn't seem true  :? . Maybe you wanted $I_1,I_2,\ldots$ etc. to be the incenters of $ABC,BCD,\ldots$? (I'm just guessing here, so sorry if it's a stupid question :) )

Actually, if you denote the incenter of $XYZ$ by $I_U$, where $\left(X,Y,Z,U\right)$ is any permutation of $\left(A,B,C,D\right)$, then it can be shown that $\angle BAD + \angle BCD - \pi$ and $\angle I_BI_AI_D + \angle I_BI_CI_D - \pi$ have the same sign (both $>0$, both $0$, or both $<0$). However, this might have no connection to what you wanted to post :) .

[u][b]The author of this posting is : pestich[/b][/u]
____________________________________________________________________

Romano wrote:

Prove that I1I2I3I4 is a cyclic quadrilateral if and only if ABCD is a cyclic
quadrilateral.

It should have been ABCD is circumscribed (has an incircle).

Pestich.

[u][b]The author of this posting is : romano[/b][/u]
____________________________________________________________________

Sorry , this problem is edited :
Let a cyclic quadrilateral ABCD . Prove that I1I2I3I4 is a cyclic quadrilateral if and only if ABCD is a circumscribed quadrilateral.

[quote="romano"]Let a cyclic quadrilateral ABCD . Prove that I1I2I3I4 is a cyclic quadrilateral if and only if ABCD is a circumscribed quadrilateral.[/quote]

A solution of this can be found on http://www.mathlinks.ro/Forum/viewtopic.php?t=21758 .

  Darij


A sheet of notebook paper weighs 0.005 ounces per square inch. A box contains 24 notebooks. Each notebook has 100 sheets of 10-inch by 6-inch paper. For each notebook, the back cover and the front cover together weigh one-half ounce. What is the total weight of the notebooks in the box, in ounces?

Every notebook has $ .005\cdot100\cdot10\cdot6\equal{}30$ ounces of paper, and $ .5$ ounces of cover. $ 24(30)\plus{}24(.5)\equal{}720\plus{}12\equal{}\boxed{732}$

For each notebook, the back cover and the front cover together weigh one-half ounce. Isn't one-half = 1.5?


let $a,b,n$ be natural number prove that there are $\frac{(a+1)(b+1)-2}{2}$ number such that we can't show with this form $ax+by$(x,y are natural)

I think the formula is false.
The coin exchange problem of Froebenius is solved here :
http://www.mathlinks.ro/Forum/viewtopic.php?p=264029#p264029

Have a look at pages 30-32.

Pierre.

The largest number which cannot be written as $xa+yb,\ xy\in\mathbb N$ ($\mathbb N$ contains $0$) is $ab-a-b$, and among the numbers $0,1,\ldots,ab-a-b$, half can be written as a natural linear combination of $a,b$, if I remember correctly.

I believe a generalization was posted by sam-n a long time ago.

I remember once prooving such fact:

if $a,b,c$ are pairwise coprime then $2abc-ab-bc-ac$ is the largest number which is not of the
form $xbc+yca+zab$ where $x,y,z$ are non-negative integers.


The Classroom Math forum is for middle school students to discuss the math from their classes.

[b]This forum should only be used by middle school students and teachers.[/b]  (Others are free to read, but please leave the discussion to middle school students.)

Middle schoolers should use this board to answer challenge questions posted by board moderators, administrators, and other users. They can also use this board to ask questions they have about their math studies.

Older students should leave the challenge questions for the younger students - answer the challenge problems that are appropriate for your age and experience level only! You are free to help out students who have questions.

We expect the math in this forum to be topics you might find in a middle school classroom.  Typical classes in middle school classrooms include Pre-Algebra, Algebra 1, and very basic Geometry. If you wish to discuss contest problems, use MATHCOUNTS for middle school contest problems.

If you are uncertain which forum to place your question in, always place it in the more advanced forum of those you are considering.

Please refrain from titling a post as "easy" or "hard."  If it is in the right forum, it is [i]appropriate[/i] regardless.

Please do not post messages that are meant to "teach" others about forum etiquette.  Messages such as "do not revive old topics" should not be placed on the forums.

If you feel someone is abusing the forum, please contact a moderator or administrator.

The following are examples of problems that are appropriate for this forum.  You can answer these problems in other threads if you like, but please do not make a thread of solutions here.

[b]Arithmetic[/b]:

Simplify the fraction $ \frac{44}{180}$.

Convert $ \frac{19}{25}$ to a percent.

Perform the indicated arithmetic:
\[ 1.02\plus{}5.3\minus{}0.13\plus{}2.771.\]
[b]Algebra[/b]:

Find $ x$ where $ 3x\plus{}7 \equal{} 16$.

Find $ (a, b)$ where $ a\plus{}2b \equal{} 11$ and $ b\minus{}a \equal{} 4$.

If $ 3^{x}\equal{} 81$, find the value of $ 2^{x}$.

Factor $ 2x\plus{}6$, $ x^{2}\plus{}2x\plus{}1$.  More advanced factoring belongs in High School Basics

[b]Geometry[/b]:

Find the area of a right triangle that has hypotenuse of length 37 and one of whose legs is 12.

The area of a circle is $ 32\pi$.  Find the length of its diameter.

[b]Counting[/b]:

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

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

[b]Probability[/b]:

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

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

[b]Number Theory[/b]:

Find the greatest common divisor of 60 and 105.

Find the least common multiple of 40 and 90.

Problems that use this level of curriculum, but a greater level of problem solving creativity should be posted in [url=http://www.artofproblemsolving.com/Forum/index.php?f=299]The Basics[/url] or the [url=http://www.artofproblemsolving.com/Forum/index.php?f=132]MATHCOUNTS[/url] forums.

To add one very important guidelines to what rrusczyk has said:

- Please keep subject material to topics that your principal would find acceptable for discussion at your middle school.

If your principal would object to the posting of a particular topic in a school newsletter, the topic has no place on this forum.

Questions and solutions not meeting these guidelines will be deleted.

And another thing:

Please only post one problem per post  :D

Some of you are posting multiple problems in one post and that makes it difficult for the person answering the post see all of the problems in front of them.  If you have to make 5-6 topics in a row, then do so.  :)  ;)  :P

I'll second robinhe's suggestion for another reason: some posters in various forums have grouped problems of highly varying difficulty.  This makes it difficult or impossible to move threads into the forums in which they fit best.

For now, we are locking marathon threads and asking that students discontinue creating them.  The reasons are that (1) many of the problems are growing too difficult for the level of the forum, (2) the depth of discussion of the solutions is low, and (3) it's hard to search/identify the problems in any way.  Please keep posts to one problem per thread (except where another problem relates to the original problem discussed).

i am an eight grader and i was just wondering...in one of my posts...(medium easy hard prob) the question had an integral in it ($\int$) although knowledge of the integral was not necessary for the question because it cancelled out...

although one poster asked me what the integral was and i explained... is that ok?

Just a friendly reminder to review the forum guidelines.  Simply scroll to the top and read the first few messages.  We have had numerous messages of late falling outside the guidelines.

Reminder of this guideline which is often ignored:

Please do not post messages that are meant to "teach" others about forum etiquette.  Messages such as "do not revive old topics" should not be placed on the forums.

Clutters the forums, and moderators have to delete the messages.  Moderators will tell the user as needed.  In general, we don't really have a rule against reviving old posts.

[quote="gauss1181"]Why is it that? Some people just need reminders..especially if they're new.

I think it's ok.[/quote]

It's best if you just pm the person who is at fault and remind them nicely not to revive old posts.

[quote="gauss1181"]What's so bad about reviving old topics?[/quote]

It iritates people...

I posted a reply in this forum yesterday simply to say that I thought the poster had posted a great question ('distance problem')! This seems to have been deleted. Just wondering why.

I have a question and I don't know where to post it

If you're in a higher grade but you're not very good at math, can you still post here in problem solving threads and such?

Of course, you can post at whatever level you feel comfortable with. :)

I posted a problem 3 days ago and no one has answered it should I post the answer?

The level of these problems are way harder than they are supposed to (2nd post). Is the problem difficulty change supposed to happen or what?

The post was made 8 years ago. Things are pretty much bound to change in that time.

But the topics jumped from simple equations and arithmetic to modular congruences and stuff. I wouldn't imagine that big of a change could happen in 8 years. But I guess MC and the AMC's changed a lot in 8 years...

Also one thing I have noticed is the increase in trigonometry in this forum. Trigonometry belongs in the high school intermediate (or basics if it is simple 30-60-90 trig) but definitely not in the middle school forum.

More complex math has started appearing in this forum, some stuff that I think should be considered high school.

I think we really need to stress this. Looking at this current forum right now, there are A LOT of advanced topics that do NOT belong in this forum and have been moved. People really need to pay more attention to stickies and other important posts.

I think that the major difference is that many people that use the forums are way ahead of the "Middle School Classroom" curriculum, but still post things that are not especially challenging for them but for us....extremely challenging.

Also to add, there have been some people posting some commercials. Not to mention any names, but I don't think you are supposed to do that on the forum. Are you?

Eh, those things are not people but rather "spambots", i.e. accounts purposely made to spam advertisements in the hopes that they will effectively advertise something, whatever it may be.  Except they usually don't and are deleted somewhat quickly.  Don't worry about them so much - AoPS will take care/make fun of them on its own.

Good post and I personally appreciate it

Tip: use the thanks button to thank posts.


(For those of you who have lost your packets, the AHSME problems in activity 14 are also [url=http://www.math.uncc.edu/~hbreiter/AHSME/1997st.pdf]here[/url].)

This is a transplant from the old forum.  Here are the answers we have arrived at so far, with a voting system of sorts (19 had two votes for D and one for E; 24 had two votes for C and one for B).

14. B
15. D
16. D
17. A
18. E
19. D
20. A
21. C
22. E
23. D
24. C

Solutions

[b]18[/b] (from Sara S.)
i called the sum S and # of terms n:
S/n = 22, (S+10)/n = 24, solve: n=5, S=110

mode is 32, smallest is 10, and m-4 is in it, so list is:
10, m-4, m, 32, 32 = 110
2m + 70 = 110
m=20, E

[b]22[/b] (from Kevin)
it wasnt exactly quick, but heres how i did it:
first, look at the possible answers (i love multiple choice).
none of them are bigger than 11, so A has 11 more than E (too lazy to write out their names). if B has 19 more than A then there's too much money, so B has 19 less than A, and A has 8 less than B. you need to add/subtract 7,5, and 4 to make 8, so 7+5-4. once you figure out how much they have relative to each other, finding the exact amount is easy.

Here's how I did 22.
[hide]we have a+b+c+d+e=56 and
a-b=  :pm: 19
b-c=  :pm: 7
c-d=  :pm: 5
d-e=  :pm: 4, and
e-a=  :pm: 11
Note the left hand side of the latter 5 equations add up to 0, so the right hand side must also. so we have
a-b= 19
b-c= -7
c-d= -5
d-e= 4, and
e-a= -11.
A little bit of substitution does the trick, I believe the answer was E. 10[/hide]

23. [hide]
If you can sort of visualise it, you've got a cube that's got a tetrahedron cut out of it. Our cube has a side of length 1, and our tetrahedron has a base area of 1/2, and a height of 1. Thus the volume of the tetrahedron is 1/6, and the volume of our folded figure is 5/6.[/hide]

Also, if anyone wants me to, I'll post the rest of the problems to this problem set, since Mr. Feng copied the same page twice.

[quote="captcha000"]Also, if anyone wants me to, I'll post the rest of the problems to this problem set, since Mr. Feng copied the same page twice.[/quote]

You can, but they're also just on the last page of the following PDF
http://www.math.uncc.edu/~hbreiter/AHSME/1997st.pdf


A customer at a 7-11 store selected 4 items to buy, and was told that the cost was  7.11, He was curious that the cost was the same as the store name, so he enquired as to how mathe figure was derived. The shop asiis said that he had simply multiplyed the prices of the four individual items. The customer protested and said that the four prices should have been added, not multiplied. But when he did the result was exactly the same: 7.11 What were the 4 invidual prices?

This is a very nice problem, which involves a lot of work! I remember it took me something like 10 hrs to solve it one year ago...

Answer: [hide]the prices are 3.16, 1.25, 1.50, 1.20 [/hide]


These two spinners are divided into thirds and
quarters, respectively. If each of these spinners
is spun once, what is the probability that the
product of the results of the two spins will be an
even number? Express your answer as a common fraction.

[asy]size(5cm,5cm);

draw(Circle((0,0),1));

draw(Circle((3,0),1));

draw((0,0)--(0,1));

draw((0,0)--(-0.9,-0.47));

draw((0,0)--(0.9,-0.47));

draw((2,0)--(4,0));

draw((3,1)--(3,-1));

label("$3$",(-0.5,0.3));

label("$4$",(0.5,0.3));

label("$5$",(0,-0.5));

label("$5$",(2.6,-0.4));

label("$6$",(2.6,0.4));

label("$7$",(3.4,0.4));

label("$8$",(3.4,-0.4));

draw((0,0)--(0.2,0.8),Arrow);

draw((3,0)--(3.2,0.8),Arrow);[/asy]

The probability that the product will be odd is $ \frac{2}{3} \cdot \frac{1}{2}\equal{}\frac{1}{3}$.  Thus, the probability that the probability will be even is $ \frac{2}{3}$.


How many non empty subsets of a set can you make containing only certain elements? Like if you wanted to find how many non-empty subsets of {1,2,3,4,5,6,7} there are that contain only prime numbers. Is it 2^(number of wanted elements)-1?

Yeah, imagine that you are just finding subsets of $ \{2,3,5,7\}$, since no other terms can be there.

So it would be 4C4+4C3+4C2+4C1=2^4, but no empty subsets, so 15?


Try to find as many solutions to this puzzle as possible.

Place 8 peices on a chessboard (8x8 square), so that no 2 peices are in the same row, coloumn, or diagonal. (Basically, place 8 queens so that no Queen can take any other Queen.)

Finding one solution and then rotating or reflecting it DOES NOT MAKE ANOTHER SOLUTION

So far, i got 9 different ways to do it :P

Well, this is the well-known [url=http://mathworld.wolfram.com/QueensProblem.html]Queens' problem[/url]. :)

See also : http://www.math.utah.edu/~alfeld/queens/queens.html .


[quote="hxy09"]find all $ f$: R\{0} $ \rightarrow$ R\{0}
       $ f(x) \plus{} f(y) \equal{} f(xyf(x \plus{} y))$  with $ x \plus{} y \neq 0$[/quote] Can you please post a solution because I don't find this as an easy func equation.

$ f(x)\equal{}\frac1x,\forall x\ne0$
Suppose $ f(a)\equal{}b\ne\frac1a$ for $ a,b\ne0$.
Then we can let $ x\equal{}a\minus{}\frac1b\ne0$ and $ y\equal{}\frac1b$ and we will get $ f(\frac1b)\equal{}0$. Contradiction!

[quote="xxp2000"]$ f(x) \equal{} \frac1x,\forall x\ne0$
Suppose $ f(a) \equal{} b\ne\frac1a$ for $ a,b\ne0$.
Then we can let $ x \equal{} a \minus{} \frac1b\ne0$ and $ y \equal{} \frac1b$ and we will get $ f(\frac1b) \equal{} 0$. Contradiction![/quote]
Right!!!A similar solution:
$ yf(x\plus{}y) \neq 1$
yields $ f(x)\equal{}\frac {1}{x}$


[quote="mr. math"]i need help with this question. is there any rule for it?

what is the units digit of 7 to the power of 79?

 :alien:  :rotfl:[/quote]

What's up with the alien and rotfl?

[hide="look for patterns"]$7^{1}=7$
$7^{2}=49$
$7^{3}=343$
$7^{4}=2401$
$7^{5}=16807$

Yes the pattern goes ${7,9,3,1,7,9,3,1,7,...}$
The answer is $\boxed{3}$[/hide]

EDIT: D'OH!

Now would be an excellent time to use modular arithmetic. \[7\equiv7\pmod{10}\\7^{2}\equiv9\pmod{10}\\7^{3}\equiv3\pmod{10}\\7^{4}\equiv1\pmod{10}\\7^{5}\equiv7\pmod{10}\\\vdots\] As you can see, every 4 terms, the sequence repeats.  Thus, the 80th term is the same as the 4th term, and the 79th term is the same as the 3rd term, which is $\boxed{3}$

Totally Zealous, you might want to check your answer (not the sequence, what you get from using the sequence)...

The general rule is that whenever we want to find the last digit of something, it will repeat every four times.  (Due to the fact that we're looking for mod 10, and $\phi 10 = 4$).  So we can repeatedly subtract four from the exponent (or just subtract the largest multiple of 4 that fits):

$7^{79}$ implies $7^{79-76}=7^{3}$.

So we want the last digit of $7^{3}=343$, which is obviously $\boxed{3}$.

If we want the last digit of any x^y, we can use this method of repeatedly subtracting 4 from the exponent.

(I'm not sure I made sense, any clarifications?)

So this is the rule, right: $a^{k}\equiv a^{k \mod \phi m}\pmod m$.

$7^{79}\equiv7^{79\mod\phi10}\pmod{10}=7^{79\mod4}=7^{3}\equiv3 \pmod{10}$


Hola a todos. Me preguntaba si podrian darme alguna sugerencia para que pueda resolver este problema...

[b]Problema[/b]

[i]Encuentra el entero positivo mas grande $n$, para el cual existe un unico entero $k$ tal que:[/i]

$\frac{8}{15}< \frac{n}{n+k}< \frac{7}{13}$

[color=blue]Me llama la atención que el valor de $n$ es muy cerrado, ya que la desigualdad corresponde a $\frac{104}{195}<\frac{n}{n+k}<\frac{105}{195}$. Cierto es que, si aumentamos el valor del denominador, no se apreciará tanto, pero la cota para $\frac{n}{n+k}$ es llamativa.
Por ahora, no encuentro un método de resolución efectivo. Apenas estoy echando un vistazo. Tengo una lista de enunciados que recibí como tarea, y debo dedicarme a ellos. Más escribo para que este foro reanude su actividad, y que alguien con más tiempo pueda ayudarte mejor.[/color]

[quote="R.G.A.M."][color=blue]Me llama la atención que el valor de $n$ es muy cerrado, ya que la desigualdad corresponde a ...[/color][/quote]

Entiendo  :) . Gracias por tu respuesta  :lol: .

[quote="R.G.A.M."][color=blue]Me llama la atención que el valor de $n$ es muy cerrado[/color][/quote]

Yo no entiendo esto. ¿Qué es un valor muy cerrado? No suena muy coherente.

 :wink:

[color=blue]Me refería a que estaba muy acotado el valor de la expresión.
No necesitas prestar mucha atención, sólo divagaba para que alguien intentara resolverlo pues, como mencioné antes, tengo una tarea bastante complicada que requiere mucha atención de mi parte.[/color]


inequalities
trigonometry
function
logarithms
inequalities unsolved

Help please, the book I have tells me to solve the problem using Jensen's inequality:

Let $ a,b$ and $ c$ be positive real numbers. Demonstrate that

\[ a^ab^bc^c\ge (abc)^{\dfrac{a\plus{}b\plus{}c}{3}}\]

also how do you prove that

If $ \alpha, \beta$ and $ \gamma$ are angles in any acute triangle then:

$ \sin \alpha \plus{}\sin \beta \plus{} \sin \gamma > 2$

I don't know about jensen, but what about this

assume w.l.o.g that $ a\ge b \ge c$

therefore $ a \minus{} b \ge 0$ and $ a\ge b$ implies $ a^{a \minus{} b}\ge b^{a \minus{} b} \Leftrightarrow a^a b^b \ge a^bb^a$

Similarly $ a^ac^c\ge a^cc^a$ and $ b^bc^c \ge b^cc^a$

Multiplying all these together gives

$ a^{2a}b^{2b}c^{2c}\ge a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b}$

$ a^ab^bc^c \ge (abc)^{\frac {a \plus{} b \plus{} c}{3}}$

EDIT: actually this is wrong, sorry

[quote="keyree10"]To use Jensen's, consider the function $ f(x) \equal{} x\ln x$[/quote]

Thanks, how did you find this?

[quote="tonypr"]Help please, the book I have tells me to solve the problem using Jensen's inequality:

Let $ a,b$ and $ c$ be positive real numbers. Demonstrate that
\[ a^ab^bc^c\ge (abc)^{\dfrac{a \plus{} b \plus{} c}{3}}
\]
also how do you prove that

If $ \alpha, \beta$ and $ \gamma$ are angles in any acute triangle then:

$ \sin \alpha \plus{} \sin \beta \plus{} \sin \gamma > 2$[/quote]
$ \Leftrightarrow a.lna \plus{}b.lna\plus{}c.lnc \geq \frac{1}{3}(a\plus{}b\plus{}c)(lna\plus{}lnb\plus{}lnc)$
$ a\geq b\geq c\Rightarrow lna\geq lnb\geq lnc$
using cheybyshev :
Done !

[quote="tonypr"]also how do you prove that
If $ \alpha, \beta$ and $ \gamma$ are angles in any acute triangle then:
$ \sin \alpha \plus{}\sin \beta \plus{} \sin \gamma > 2$[/quote]
$\sin \beta \plus{} \sin \gamma >1+cos  \alpha \implies$
$ \sin \alpha \plus{}\sin \beta \plus{} \sin \gamma >1 \plus{}\sin \alpha \plus{}cos  \alpha=1+\sqrt{2}sin(\alpha \plus{}\frac{\pi}{4}) > 2$.
or
$ \sin \alpha \plus{}\sin \beta \plus{} \sin \gamma >1 \plus{}\cos \alpha \plus{}cos  \alpha \plus{}cos\gamma >2$.


function
combinatorics proposed
combinatorics

a) Find the value of $\sum_{n=1}^{\infty}\frac{\phi(n)}{2^n-1}$;

b) Show that $\sum_k {m\choose k}{{n+k}\choose m}=\sum_k {m\choose k} {n\choose k} 2^k$ for $m,n\geq 0$;

c) Using the identity $(1-x)^{-\frac 12}(1-x)^{-\frac 12}=(1-x)^{-1}$ derive a combinatorial identity!

d) Express the value of $\sum (2^{a_1}-1)\ldots (2^{a_k}-1)$ where the sum is over all $2^{n-1}$ ways of choosing $(a_1,a_2,\ldots,a_k)$ such that $a_1+a_2+\ldots +a_k=n$, as a function of some Fibonacci term.

[quote="amir2"]just a question:
Is $\phi(n)=\varphi (n)$?[/quote]
Yes it is.

See Theorem 1 from my post from: [url]http://www.mathlinks.ro/viewtopic.php?p=1101215#1101215[/url].

[quote="BaBaK Ghalebi"]I know this is an old thread,but I'm looking for a solution for part [b]b)[/b],I thought asking it here is better than posting it again,so does anyone have a solution for part [b]b[/b]? :maybe:[/quote]

[i]Double Counting[/i] :

LHS can be interpreted in the following way : Choose $ k$ elements from a set $ M$ with $ |M|\equal{}n$ elements in $ \binom {m}{k}$ distinct ways. Consider another set $ N$ (with $ |N|\equal{}n$ elements) disjoint from $ M$. Choose $ m$ elements from the union of the set $ N$ and $ k$ elements previously chosen. One can do this in $ \binom {n\plus{}k}{m}$ distinct ways. It follows that the number of ordered pairs of sets $ (X,Y)$ where $ X \subset M$ , $ Y \subset N \cup X$ with $ |X|\equal{}k$ , $ |Y|\equal{}m$ is equal to $ \binom {m}{k}\binom {n\plus{}k}{m}$. If the result is extended to $ k$ which varies from $ 0$ to $ m$, the number of all pairs $ (X,Y)$ will be equal to $ \sum_{k\equal{}0}^m \binom {m}{k} \binom {n\plus{}k}{m}$.

This problem can also be solved by first choosing the sets $ Y \cap M$ and $ Y \cap N$. If $ |Y \cap N|\equal{}j$ , then $ 0 \leq j\leq m$ and it follows that $ |Y \cap M|\equal{}m\minus{}j$. Thus we can choose $ Y$ so that $ |Y \cap N|\equal{}j$ and $ |Y \cap M|\equal{}m\minus{}j$ in $ \binom {n}{j}\binom {m}{m\minus{}j}$ different ways. It remains to choose the set $ X$ so as to satisfy $ Y \cap M \subset X \subset M$,

since $ Y \subset  N \cup X$ and thus $ Y \cap M \subset (N \cup X) \cap M\equal{}(N \cap M) \cup (X \cap M)\equal{}\null \cup X\equal{}X$. But the set $ X$ can be chosen in $ 2^j$ ways, because for each of the $ j$ elements of the set $ M\ (Y \cap M)$ there exist exactly two possibilities: it belongs or does not belong to the set $ X$. It follows that the number of pairs $ (X,Y)$ with the stated property is equal to $ \sum_{j\equal{}0}^m \binom {n}{j} \binom {m}{j}2^j$,
and this observation completes the proof of [b](b)[/b].


Write it as $(n-1)(n+1) = 5 \cdot 2^m$
$(n-1, n+1) = 2$ so if $2^t||n-1$ then $2||n+1$ and vice versa.
So we have possibilities:
$n-1=2$
$n+1=2$
$n-1=10$
$n+1=10$
From that we see that only $n=9$ works.

$(n, m)=(9, 4)$

we have
1 + 5.$2^{m}$ = $n^{2}$;
5.$2^{m}$ = $n^{2}$ - 1 = (n+1)(n-1);
$2^{m}$ = $\frac{(n+1)(n-1)}{5}$
so n is odd and (n+1)(n-1) has only one prime 5 and the rest are all 2, wich means (n+1)(n-1) ends in 0, but 5 must divide one of them, so we have: n + 1 = 10 or n - 1 = 10, wich lead us to the unique solution (m,n) = (4,9)

[b][hide=Solution:]
We have,
$$5.2^m=(n+1)(n-1) \implies n-1=10 \implies n=11 $$
Or,
$$ n+1=10 \implies n=9$$

So, $(m,n)=(4,9)$ only suffices.

[hide=Another solution]
Move $1$ to the other side and factor to get $5\cdot 2^m=(n+1)(n-1)$. Since $\gcd(n+1, n-1)=2$, both $n+1$ and $n-1$ must be even. Thus, either $n-1$ or $n+1$ must be in the form of $2^p$ for $p<m$. Then we have two cases.

Case 1: $n-1=2^p$

We have
\begin{align*}
2^p(2^p+2)&=5^m\\
2^{p+1}\left(2^{p-1}+1\right)&=5\cdot2^m.
\end{align*}Since $2^{p-1}$ is odd, we know that $p=3$ and $m=4$. Then, we have $n=9$, so $(m,n)=(4,9)$ is a solution.

Case 2: $n+1=2^p$

Similarly to case 1, we get $2^p-1=5$, which yields no integer solution.

We have $\boxed{(m, n)=(4, 9)}$ as the only solution.
[/hide]

$$1 + 5 \cdot 2^m = n^2 \implies n^2-1=5 \cdot 2^m \implies (n-1)(n+1)=5 \cdot 2^m$$
Clearly, either $n-1 = 2^{m-1}$ or $n+1=2^{m-1}$[hide=.]Just to rigorize this: $\gcd(n-1,n+1)=\gcd(2,n-1)$ by the Euclidean algorithm, so since $n$ is odd, $\gcd(n-1,n+1)=2$. [/hide] In the first case, $n+1=10 \implies (n = 9,m=4)$. In the second case, $n-1=5 \implies n = 6$, which yields no solutions.

I’m gonna Try to take a proper crack at this instead of plugging the solutions

[b][u]CLAIM:[/u][/b] $m$ is an even integer and $n$ is an odd integer

[hide=Proof]
Using $mod(4)$ we realise that $n \equiv 1(mod 4)$ and $m \equiv 0(mod m)$ since $1$ is $1 \equiv 1(mod 4)$
[/hide]

Now Algebraic manipulations would lead us to find that
$$5.2^m = (n-1)(n+1)$$
Since this has to be true and $n$ is odd ,
Let $n=2k+1$ for some $k \in I^+$
Therefore
$$5.2^m = (2k)(2k+2)$$
$$\implies 5.2^m = 4(k)(k+1)$$
Since $m$ is even we already have $4| m$ for all $m \in {2,4,6,8.....}$
$5=k$ or $5= k+1$
Therefore
$$n+1 = 2.(k+1)$$
$$\implies n+1 = 10$$

Therefore
$ n = 9$ comes out as a solution and correspondingly $m= 2$ also comes out as solution

[b][u]CLAIM:[/u][/b] There are no more solutions to the above.

[hide=Rather Dodgy proof:]
Other solutions are not possible since the other pairs present do not satisfy the above given conditions and are not relatively prime,as shown above.
[/hide]

This means that $5 \cdot 2^m = (n+1)(n-1)$. Now assume $n \neq 4, 9$. Then both $n-1$ and $n+1$ are something multiplied by a power of $2$. One must only be a power of two, and the other one must be $5$ multiplied by a power of $2$. Therefore, we must find positive integers $y, z$ satisfying $2^y+2=5 \cdot 2^z$. But the RHS is a multiple of $4$ when $z \geq 2$, so we only need to test $z = 0,1$. Out of these two, only $z = 1$ works, giving $y = 3$. Then the only solution is $n + 1 = 10, n - 1 = 8 \implies \boxed{(n,m) = (9, 4)}$.

[quote=Arne]Find all positive integers $m$ and $n$ such that $1 + 5 \cdot 2^m = n^2$.[/quote]

We have $(n-1)(n+1)=5 \cdot 2^m$. We clearly have $\gcd(n-1,n+1)=1,2$. If the gcd is 1, then neither factor is divisible by 2, implying that $m=0$, which gives no solutions for $n$. Thus the gcd is 2. Evidently, this implies that both factors are divisible by 2, and exactly one is divisible by 4. If the factor divisible by 5 is also divisible by 4, then that factor is at least 20 while the other factor is 2, contradiction. Thus the v_2 valuation of the factor divisible by 5 is 1, so it is equal to 10. Checking, the only such solution is $(m,n)=(4,9)$, and no other solutions exist.

[hide=Solution]
It is clear that $n\equiv 1,4 \pmod{5}$. We can then write that $n=5a+1, 5a+4$. Let's first examine the first case. Notice that we can then get that $2^m=(5a+2)(5)$. We have to have $a$ to be even, so denote this as $2k$. Plugging this in again gives us that $2^{m-2}=(5k+1)k$, which doesn't have any solutions. Now, we go onto the last case. It goes through a similar process, but this one gives some solutions. The pair that works is $\boxed{(m,n)=(4,9)}$.[/hide]

[quote=Arne]Find all positive integers $m$ and $n$ such that $1 + 5 \cdot 2^m = n^2$.[/quote]
Solved [hide=here][[https://artofproblemsolving.com/community/c6h351572p1895582]]

Nice one!
First, of all, $(n-1)(n+1)=5\cdot 2^m$, now, if $n$ is even, we have a contradiction. Now, if $n,$ is odd, then, both factors, are divisible by $2,$ and one of them is divisible by $4,$ it is a contradiction, if the factor, that is divisible by $5,$ is also a factor of $4.$ Now, we have that, we have that, we test all values less than $10,$ if we look at the powers of $2.$ Hence, we see, that $\boxed{(4,9)}.$

We first note that $m=0$ and $m=1$ yield $n^2=6$ and $n^2=11$ respectively, so there exists no integer solutions in these cases. Thus, $m\geq 2$ which gives us that $n>4$. Now, note that we have
\[5 \cdot 2^m = n^2-1=(n-1)(n+1)\]
and at most one of $n-1$ and $n+1$ is divisible by 5. Thus, we have two cases.\\

[b]Case 1 :[/b] $5 \mid n+1$. This means, that $2$ is the only prime dividing $n-1$ and so $n-1=2^a$. By our bound on $n$, this gives us that $a \geq 2$. But, notice that,
\[n+1=2^a+2=2(2^{a-1}+1)\]
which means $v_2(n+1)=1$ (since $a-1\geq 1$). Thus, we have
\[m=v_2(5 \cdot 2^m) = v_2((n-1)(n+1)) = a+1\]
Thus, $a=m-1$. Plugging this in we have $n=2^{m-1}+1$ and further,
\[2^{m-1}(2^{m-1}+2)=2^m(2^{m-2}+1)\]
which requires, $2^{m-2}+1=5$ which gives us that $m=4$ and in turn $n=9$.\\

[b]Case 2 :  [/b]$5 \mid n-1$. This means $n+1=2^a$ and we have $a \geq 3$. Then, by a similar $v_2$ argument as before, we obtain that $a=m-1$. So, $n=2^{m-1}-1$. This gives,
\[2^{m-1}(2^{m-1}-2)=2^n(2^{m-2}-1)\]
which requires $2^{m-2}-1=5$ and thus, $2^{m-2}=6$ which clearly has no solutions. Thus, the only pair that works is $(4,9)$ which can easily be checked.


The formula for the sum of the first $ n$ number is $ \frac{n(n\plus{}1)}{2}$

Plugging in, $ S\equal{}\frac{2009 \cdot 2010}{2}\equal{}2009 \cdot 1005\equal{}2009000\plus{}10045\equal{}2019045$ so A

[quote="moldovan"]The sum $ 1 \plus{} 2 \plus{} ... \plus{} 2009$ equals:

$ (A)$ $ 2019045$
$ (B)$ $ 2018045$
$ (C)$ $ 2009045$
$ (D)$ $ 1019045$
$ (E)$ $ 2019005$[/quote]

Let the sum be $ S.$ And:

$ S \equal{} 1 \plus{} 2 \plus{} \cdots \plus{} 2009$
$ S \equal{} 2009 \plus{} \cdots \plus{} 2 \plus{} 1$

Adding the lists gives us $ 2S \equal{} 2009\cdot2010 \Rightarrow S \equal{} 2009\cdot1005 \equal{} 2019045 \equal{} \boxed{A}$

Of course, more generally, $ 1 \plus{} 2 \plus{} \cdots \plus{} n \equal{} \sum_{k \equal{} 1}^{n}k \equal{} \frac {n(n \plus{} 1)}{2}$


For each positive integer $n\leq 49$ we define the numbers $a_n = 3n+\sqrt{n^2-1}$ and $b_n=2(\sqrt{n^2+n}+\sqrt{n^2-n})$. Prove that there exist two integers $A,B$ such that
\[ \sqrt{a_1-b_1}+\sqrt{a_2-b_2} + \cdots + \sqrt{a_{49}-b_{49}} = A+B\sqrt2. \]

nice.. very nice  :D
please, give me a solution

Make the guess that
$a_k-b_k=(a\sqrt k+b\sqrt{k+1}+c\sqrt{k-1})^2\\
=a^2k+b^2(k+1)+c^2(k-1)+2ab\sqrt{k^2+k}+2ac\sqrt{k^2-k}+2bc\sqrt{k^2-1}$.
Comparing coefficients, we find that $ab=ac=-1$ and $bc=\tfrac12$, and $a=\sqrt2$, $b=c=\tfrac{1}{\sqrt2}$ works. So $\sqrt{a_k-b_k}=\sqrt{2k}-\sqrt{\tfrac{k+1}{2}}-\sqrt{\tfrac{k-1}{2}}$ (making sure to check that it actually works and is positive). Now, watch it telescope!

Zabelman, one question.  You say "make the guess that"---but what would lead you to make such a guess in problems of this sort?  Your solutions to problems in the forum often involve you saying something like "make the educated guess that" or "hypothesize so and so," but some people just aren't so incredibly smart that they can get flashes of insight like that (keep in mind you were the ONLY person in the U.S. to get 150 on the AMC 12A this year).  What was your motivation in this problem?  And can you give any tips, in general, on how to search for such seemingly random "guesses" for how to solve problems?  Do you have some kind of system?

I do have a process: hope. The thing we're trying to evaluate involves $\sqrt{a_k-b_k}$, so I hope this comes out to something manageable. Next, the value of $a_k-b_k=3k+\sqrt{k+1}\sqrt{k-1}-2\sqrt{k}\sqrt{k+1}-2\sqrt{k}\sqrt{k-1}$ involved parwise products of $\sqrt{k}$, $\sqrt{k+1}$, and $\sqrt{k-1}$, hence the guess. Many problems are solved simply by hoping the right thing.

Here's what I did:
First I factored, $b_n=2\sqrt{n}(\sqrt{n+1}+\sqrt{n-1})$. Then I noticed that $\sqrt{n+1}\sqrt{n-1}=\sqrt{n^{2}-1}$ which is part of $a_{n}$. So, I say $r=(\sqrt{n+1}+\sqrt{n-1})$ and square it to get $r^{2}=2n+2\sqrt{n^{2}-1}$. Then I said $\displaystyle a_n=\frac{4n+r^2}{2}$ and noticed $4n=(2\sqrt{n})^2=s^{2}$. From these it immediately follows that $\displaystyle a_n=\frac{s^2+r^2}{2}$, $b_n=rs$, and $\displaystyle a_n-b_n=\frac{s^2-2rs+r^2}{2}\Rightarrow \sqrt{a_n-b_n}=\frac{s-r}{\sqrt{2}}=\frac{2\sqrt{n}-(\sqrt{n+1}+\sqrt{n-1})}{\sqrt{2}}=\frac{(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})}{\sqrt{2}}$
So,
$\displaystyle\sum^{49}_{n=1}\sqrt{a_n-b_n}=\displaystyle\sum^{49}_{n=1}\frac{(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})}{\sqrt{2}} \\
\displaystyle=\frac{1}{\sqrt{2}}\left(\sum^{49}_{n=1}(\sqrt{n}-\sqrt{n-1})-\sum^{49}_{n=1}(\sqrt{n+1}-\sqrt{n})\right)
\\
\displaystyle=\frac{1}{\sqrt{2}}\left(\left(\sum^{49}_{n=1}\sqrt{n}-\sum^{49}_{n=1}\sqrt{n-1}\right)-\left(\sum^{49}_{n=1}\sqrt{n+1}-\sum^{49}_{n=1}\sqrt{n}\right)\right)
\\
\displaystyle=\frac{1}{\sqrt{2}}\left(\left(\sum^{49}_{n=1}\sqrt{n}-\sum^{48}_{n=0}\sqrt{n}\right)-\left(\sum^{50}_{n=2}\sqrt{n}-\sum^{49}_{n=1}\sqrt{n}\right)\right)
\\
\displaystyle=\frac{1}{\sqrt{2}}\left(\left(\sqrt{49}-\sqrt{0}\right)-\left(\sqrt{50}-\sqrt{1}\right)\right)
\\
\displaystyle=-5+4\sqrt{2}$


[quote="staretak"][quote="SorrowMan"][quote="Anto"]Esto H = { z in C | Re(z) > 0 }.
Iparxei olomorfiki sinartisi f : H -> H etsi oste f(1)=1 kai f(2)=4 ?

Stergios[/quote]

Olomorfiki ennoeis fantazomai olomorfi , (dld paragogisimi se olokliro to H) ... I 3*z-2 den einai mia tetoia sinartisi??[/quote]

Basika me to olomorfiki o Stergios olomorfi ennoei...elpizw na ton sinxoreseis, alla ta ellinika tou Stergiou einai kaka   :rotfl:  :rotfl:

Tsekare mia an i sinartisi sou einai apo to H STO H...[/quote]

Sigxoremeno to paidi :p ... Ontws nai den proseksa oti thelei apo to H sto H , tespa tha tin ksanakitaksw kapoia alli stigmi , twra tis eriksa mia grigori matia aplos
\\


conics
ellipse
limit
calculus
calculus computations

For $a>0,\ b>0,$ let $r$ be the radius of the circle passing through the three points $(0,1),\ (a,b),\ (-a,b)$ on the ellipse $\frac{x^2}{4}+y^2=1.$ Find the limit value $\lim_{a\rightarrow 0} r.$


Prove that:
for $ i=1,2,\cdots,6$ and $ 0\leq x_{i}\leq1$,

$ \sum_{i=1}^{6}x_{i}\leq3+\sum_{cyc}x_{1}x_{2}$

Lokman GÖKÇE


calculus
derivative
algebra unsolved
algebra

Find all functions $f: \mathbb R \to [-1,1]$ such that for all $x,y \in \mathbb R$,
\[f(x\plus{}y)\equal{}f'(x)f'(y)\minus{}f(x)f(y).\]

I think problems involving derivative does not belong to the olympiad section.

[quote="reason"]Hi!!
find all f such that:

$ f(x \plus{} y) \equal{} f'(x)f'(y) \minus{} f(x)f(y)$

$ f: IR \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > [ \minus{} 1,1]$

thanx.[/quote]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\equal{}f'(x)f'(y)\minus{}f(x)f(y)$

If $ f'(x)\equal{}0$ $ \forall x$, we get $ f(x)\equal{}c$ and $ c\equal{}\minus{}c^2$, hence two trivial solutions $ f(x)\equal{}0$ $ \forall x$ and $ f(x)\equal{}\minus{}1$ $ \forall x$

We'll now consider that $ f(x)$ is not a constant function. So $ \exists c$ such that $ f'(c)\neq 0$, $ P(x,c)$ $ \implies$  $ f'(x)\equal{}\frac{1}{f(c)}(f(x\plus{}c)\plus{}f(x)f(c))$ and so $ f(x)$ is $ C_{\infty}$

Then, derivating $ P(x,y)$ along $ x$, we get $ f'(x\plus{}y)\equal{}f"(x)f'(y)\minus{}f'(x)f(y)$
Same, derivating $ P(x,y)$ along $ y$, we get $ f'(x\plus{}y)\equal{}f'(x)f"(y)\minus{}f(x)f'(y)$

Equating these two equations, we get $ (f"(x)\plus{}f(x))f'(y)\equal{}(f"(y)\plus{}f(y))f'(x)$

And so either $ f(x)\plus{}f"(x)\equal{}0$ $ \forall x$, either $ f'(x)\equal{}a(f(x)\plus{}f"(x))$ $ \forall x$ for some $ a$

The first case give us $ f(x)\equal{}\alpha\cos(x\plus{}\beta)$ and, plugging back in the original equation, we get $ f(x)\equal{}\minus{}\cos(x)$

The second case, after some longer calculus, and using the fact that $ f(x)\in[\minus{}1,\plus{}1]$ implies no bounded non constant solutions.

Hence the solutions :
$ f(x)\equal{}0$
$ f(x)\equal{}\minus{}1$
$ f(x)\equal{}\minus{}\cos(x)$

[quote="reason"]Thanx for the answer pco  :)

But I have a question: What do you mean by $ C_{\infty}$ ? :oops:

thanx again.[/quote]

I mean that the $ n^{th}$ derivative $ f^n(x)$ exists and is continuous for any $ n$


Now i want to send some exercises but i can't type mathematics formular. Can you show me how to type them?
   :(  :(  :(

Have you visited the LaTeX tutorial on the left side panel?

Can you please show me again? Please help me! OK? But don't laugh at me. Someone is in this trouble.

[quote="Valentin Vornicu"]If you are using the MathLinks style (the blueish one) you have to click the LaTeX Help button located above, 2nd circle from the left.[/quote]
Thank you very much! I went to LaTEX help after having read your reply.
Thanks,

Out of curiousity, what does the acronym LaTeX stand for, assuming it's an acronym?

There is original typesetting system TeX (I think "technical" and "text" are appropriate sources for this name), which was developed by Donald Knuth in 1977.
LaTeX is a "[i]format[/i]" over basic TeX's commands. LaTeX was created by Leslie Lamport ("La" in LaTeX) in the beginning of 80th.


algebra
difference of squares
special factorizations

I'm asked to completely factor the following:

$5x^{3}-6y^{3}+6x^{2}y-5xy^{2}$

The textbook answer is:

$(5x+6y)(x+y)(x-y)$

When I attempted the problem, I first factored out $5x$ which led me to $5x(x+y)(x-y)-6y(y^{2}+x^{2})$, but I don't how to get from there to the answer, assuming this is the proper way to start.

Can somebody help me with the proper steps to take to get this correct?

$(x+y)(x-y)=x^{2}-y^{2}$ <-- most useful identity that exists

[quote="pircdefense"][quote="kohjhsd"][quote="pircdefense"]I'm asked to completely factor the following: $5x^{3}-6y^{3}+6x^{2}y-5xy^{2}$

[...] led me to $5x(x+y)(x-y)-6y(y^{2}+x^{2})$[/quote]

Actually, it should be $5x(x+y)(x-y)-6y(y^{2}-x^{2})$[/quote]

Thanks for your reply. I don't understand that last minus sign. When I rearrange the terms I come up with my original intermediate step. Can you show me what you did to end up with that signage?[/quote]

[hide]Try expanding your original intermediate step.
\begin{eqnarray*}5x(x+y)(x-y)-6y(y^{2}+x^{2}) &=& 5x(x^{2}-y^{2})-6y(y^{2}+x^{2})\\ &=& 5x[x^{2}+(-y^{2})]+(-6y)(y^{2}+x^{2})\\&=&5x \cdot x^{2}+5x \cdot (-y^{2})+(-6y) \cdot y^{2}+(-6y) \cdot x^{2}\\ &=&5x^{3}-5xy^{2}-6y^{3}-6x^{2}y\\ &=& 5x^{3}-6y^{3}-6x^{2}y-5xy^{2}\end{eqnarray*}
But we started with $5x^{3}-6y^{3}+6x^{2}y-5xy^{2}$, so the sign of $6x^{2}y$ term should be corrected.[/hide]

[quote="kohjhsd"][/quote][quote="pircdefense"][quote="kohjhsd"][quote="pircdefense"]I'm asked to completely factor the following: $5x^{3}-6y^{3}+6x^{2}y-5xy^{2}$

[...] led me to $5x(x+y)(x-y)-6y(y^{2}+x^{2})$[/quote]

Actually, it should be $5x(x+y)(x-y)-6y(y^{2}-x^{2})$[/quote]

Thanks for your reply. I don't understand that last minus sign. When I rearrange the terms I come up with my original intermediate step. Can you show me what you did to end up with that signage?[/quote][quote="kohjhsd"]

[hide]Try expanding your original intermediate step.
\begin{eqnarray*}5x(x+y)(x-y)-6y(y^{2}+x^{2}) &=& 5x(x^{2}-y^{2})-6y(y^{2}+x^{2})\\ &=& 5x[x^{2}+(-y^{2})]+(-6y)(y^{2}+x^{2})\\&=&5x \cdot x^{2}+5x \cdot (-y^{2})+(-6y) \cdot y^{2}+(-6y) \cdot x^{2}\\ &=&5x^{3}-5xy^{2}-6y^{3}-6x^{2}y\\ &=& 5x^{3}-6y^{3}-6x^{2}y-5xy^{2}\end{eqnarray*}
But we started with $5x^{3}-6y^{3}+6x^{2}y-5xy^{2}$, so the sign of $6x^{2}y$ term should be corrected.[/hide][/quote]

Thanks for taking the time to answer my question. I'll work through some more problems to see if I understand it completely.

I'm not sure what kohjhsd did, but I'm pretty sure I have a feasible method to reach the "textbook" answer.

$ 5x^{3}-6y^{3}+6x^{2}-5xy^{2}$ begin by grouping. to do this, we must re-arrange the formula slightly.

$ 5x^{3}-5xy^{2}+6x^{2}-6y^{3}$

$ 5x(x^{2}-y^{2})+6y (x^{2}-y^{2})$

$ (5x+6y)(x^{2}-y^{2})$

We will now use what's called the difference of squares to reach our final answer:

$ (5x+6y)(x-y)(x+y)$

If you're don't understand why this factoring method works, then read below. given:

$ x^{2}-y^{2}$

now recall that $ (x+y)(x+y) = x^{2}+xy+xy+y^{2}$. However, we want to cancel out xy, so we do the following:

$ x^{2}-y^{2}= x^{2}+xy-xy+y^{2}$
$ x^{2}-y^{2}= x(x+y)-y(x+y)$
$ \boxed{x^{2}-y^{2}= (x-y)(x+y)}$


How do u think those are solved?
x-2y+2z=4
x-y   =-2
2x-3y+2z=2
3x-4y+a.z=0

And
A=  2 1  B=   2 -1
      3 1        -3 -1
Solve:
a)X+B=3(A-X)
b)AX-X=AB-B
c) YA+Y=AB+B

[b]policy[/b]

\[ \{\begin{array}{l}
x-2y+2z=4 \\
x-y=-2 \\
2x-3y+2z=2
\end{array}
\Rightarrow
\{\begin{array}{l}
x=f(z) \\
y=g(z)
\end{array}\]
So, $ 3x-4y+a z=3f(z)-4g(z)+a z=0$.

The first system is impossible, as we can observe by adding the symmetric of the second equation to the first and third equations.

For the second set:

$ A \equal{} \left[\begin{array}{cc} 2 & 1 \\
3 & 1 \end{array} \right]$, $ B \equal{} \left[\begin{array}{cc} 2 & \minus{} 1 \\
\minus{} 3 & \minus{} 1 \end{array} \right]$, and the first equation is equivalent to 4X = 3A-B, from which we get

$ X \equal{} \left[\begin{array}{cc} 1 & 1 \\
3 & 1 \end{array} \right]$.

The solution to the second equation is $ X \equal{} I \plus{} (I \minus{} A^{ \minus{} 1})B \equal{} \left[\begin{array}{cc} 8 & \minus{} 1 \\
\minus{} 15 & 1 \end{array} \right]$.

For the third equation, $ Y \equal{} (I \plus{} A)B(I \plus{} A)^{\minus{}1}$.


vector
function
search
calculus
calculus computations

Also how would I go about solving this question...

Sorry, id offer more but my mount has broked me.

A nine meter verticle antenna is erected on a planar hill whose shape is the graph of x + 2y + 2z = 11
with the base of the antenna at the point on the hill where x = 1 and y = 4. A wire is to be stretched from
the top of the antenna to a control room on the ground (ie on the hill),
and it is desired to make the wire as short as possible.

Where on the hill should the control room be located? give its x, y, z coordinates.

Those two expressions are the same direction- positive constant multiples are irrelevant.

As for your second post- please don't post the same question in more than one place.

sorry on that, I posted the second question first, then I couldnt find it, I even ran a search on the forum for myself as the author.  It never turned up, this happend emediately after I posted my second question, which happens to be the topic of this thread.  I figured I couldnt post 2 consecutive topics in under some time frame so I decided instead to post my first question as a reply to my second, could you link me to my first post, second question?

Just realised I ran my search to this specific forum subsection.  I ran another search to the full forum and found it, sorry for the confusion.

As to my first question on this post, the positive multiplier make the direction vector a unit vector, could you explain how the unit vector is irrelevant Thanks.

[quote]sorry on that, I posted the second question first, then I couldnt find it[/quote]
I moved it to "Calculus Computations and Tutorials", which is the right forum for this sort of classroom material. I'm about to move this as well.

[quote="Abdar"]The question is:  If I want to increase a function f most rapidly from a point P
which direction should I move?

(Grad)f(P) or

(Grad)f(P) / |(Grad)f(P)|[/quote]

The function f is a scalar field. The gradient operator is a vector operator. Thus

$\nabla f$ is a vector.  The [b]direction [/b]of any vector, and I believe this is the heart of your question, is specified by a unit vector equal to the vector itself divided by its magnitude.  In other words, $\nabla f(P) \text{ and }\frac{\nabla f(P)}{|\nabla f(P)|}$ point in the same direction. The magnitude, $|\nabla f(P)|$ merely tells you how fast $f$ is changing in that direction


function
algebra
floor function
calculus
calculus computations

1). Prove that the limit (-1)^[x] as x approaches infinity does not exist. Bracket indicates floor function

2).Suppose that the limit of f(x)=L as x approaches a and L is positive. Show that there exists an interval I containing a such that f(x)>0 and x E I

Thank you  :)

I will give a proof for problem number two by using the definition of limit
[hide="Hint"]
Because $ Lim(f(x))\equal{}L$ for $ x$ approaches $ a$, then take $ t\equal{}\frac {1}{2}L$, there exist $ d$ such that $ d\minus{}a<x<d\plus{}a$ that implies $ \frac {1}{2}L<f(x)<\frac {3}{2}L$, take $ d$ that I have mentioned above, then $ f(x)$ is strictly positive in the interval.
[/hide]

Thank You Mr. Qruni. . . You made me realize that all I need to use is the definition of limit. I thought that it will be proved in some other way :)

[hide="1."]for 1 define $ a_{n} \equal{} ( \minus{} 1)^{n}$
 if the limit exists and say is equal to $ L$,then the sequence $ a_{i}$ also converges to a limit $ L$.Then ever  subsequence of $ a_{i}$ should converge to limit $ L$ consider $ n \equal{} 2m$ and $ n \equal{} 2m \plus{} 1$ and obtain contradiction
The point is if $ f(x)$ as $ x \to a$ exists and is equal to $ L$ where $ a \in \mathbb{R}$ or is $ \infty$ then for ever sequence $ b_{n} \to a$ $ c_{n} \equal{} f(b_{n}) \to L$[/hide]


Suppose that $ b^c \equal{} 121$.  What is $ b^\frac{c}{2}$?

[quote="Kermitdafrog"]then u know its 11^2=121.[/quote]
Not necessarily. $ b$ and $ c$ could just as easily be $ 9$ and $ \log_{9}121$, respectively. All you can say for sure is $ b^c\equal{}11^2$. Thus, $ 11^{\frac{2}{2}}\equal{}\boxed{11}$.


How many arrangements of the numbers 1, 1, 2, 2, 3, 3, 4,4 are there in which no adjacent numbers are equal?

Please help! I think I am supposed to use the principle of inclusion-exclusion but I am somewhat stumped on this one. I'd appreciate any help you have to offer. :)

simply put
$\alpha_1$ happens when two $1$ are adjacent
$\alpha_2$ happens when two $2$ are adjacent
$\alpha_3$ happens when two $3$ are adjacent
$\alpha_4$ happens when two $4$ are adjacent
now we want to count $N(\bar{\alpha_1,\alpha_2,\alpha_3,\alpha_4})$
did you get it??


Let side $BC$ of $\bigtriangleup ABC$ be the diameter of a semicircle which cuts $AB$ and $AC$ at $D$ and $E$ respectively. $F$ and $G$ are the feet of the perpendiculars from $D$ and $E$ to $BC$ respectively.  $DG$ and $EF$ intersect at $M$.  Prove that $AM \perp BC$.

Orl, you could remember this one! ;)

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/9531?expand=1]Hyacinthos message #9531[/url]"][b]From:[/b] "orl_ml" (Orlando Doehring)
[b]Subject:[/b] AM is perpendicular to BC

[code]Let side BC of  triangle ABC be the diameter of a semicircle which
cuts AB and AC at D and E respectively. F and G are the feet of the
perpendiculars from D and E to BC respectively. DG and EF intersect
at M. Prove that AM is perpendicular to BC.[/code][/quote]

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/9550?expand=1]Hyacinthos message #9550[/url]"][b]From:[/b] Darij Grinberg
[b]Subject:[/b] Re: AM is perpendicular to BC

[code]Dear Orlando,

In Hyacinthos message #9531, you wrote:

>> Let side BC of  triangle ABC be the diameter
>> of a semicircle which cuts AB and AC at D and
>> E respectively. F and G are the feet of the
>> perpendiculars from D and E to BC respectively.
>> DG and EF intersect at M. Prove that AM is
>> perpendicular to BC.

Well, D and E are just the feet of the altitudes
from C and B, respectively.

Let S be the foot of the altitude from A, and let
H be the orthocenter of triangle ABC. Let the
line DG meet the line AS at M', and let the
line EF meet the line AS at M". We will show that
the points M' and M" coincide.

Since the point M' lies on the lines AS and DG,
the Menelaos theorem applied to triangle ABS
yields

  AM'   SG   BD
  --- * -- * -- = -1,
  M'S   GB   DA

and therefore

  AM'     DA   GB   AD   GB
  --- = - -- * -- = -- * --.
  M'S     BD   SG   BD   SG

Since EG || HS, we have GB / SG = EB / HE, so
that

  AM'   AD   EB   AD   DH   EB   AE
  --- = -- * -- = -- * -- * -- * --.
  M'S   BD   HE   DH   BD   AE   HE

The triangles BDH and BEA are similar, so

  DH   AE            DH   EB
  -- = --,    and    -- * -- = 1,
  BD   EB            BD   AE

so that our equation above simplifies to

  AM'   AD   AE     AD   AE
  --- = -- * -- = - -- * --.
  M'S   DH   HE     DH   EH

Now, this expression is symmetric in B and C;
hence, by analogy,

  AM"     AD   AE
  --- = - -- * --,
  M"S     DH   EH

and the points M' and M" coincide. Thus, the
lines DG, EF and AS meet at one point M' = M" = M,
and AM is perpendicular to BC.

There should be a simpler proof.

Thanks for this and all the other problems, many
of them are still too difficult or non-standard
for me.

  Friendly,
  Darij Grinberg[/code][/quote]

[quote="[url=http://groups.yahoo.com/group/Hyacinthos/message/9553?expand=1]Hyacinthos message #9553[/url]"][b]From:[/b] Vladimir Dubrovsky
[b]Subject:[/b] Re: [EMHL] Re: AM is perpendicular to BC

[code]Dear Darij,

here is a shorter solution
to Hyacinthos message #9531 --

> >> Let side BC of  triangle ABC be the diameter
> >> of a semicircle which cuts AB and AC at D and
> >> E respectively. F and G are the feet of the
> >> perpendiculars from D and E to BC respectively.
> >> DG and EF intersect at M. Prove that AM is
> >> perpendicular to BC.

Let D', E' be reflections of D, E in BC, N the intersection of DE' and D'E.
Obviously, N lies on BC and MN is parallel to DD' and EE', or perpendicular
to BC (because M and N divide DG and DE', resp., in the same ratio
DF:EG=DD':EE'). So it remains to show that, in your notation, N=S, the foot
of the altitude from A, or, equivalently, that angle NEC = angle ABC..
But angle NEC = angle D'EC = angle DE'C = angle DBC. QED

Best,
Vladimir Dubrovsky[/code][/quote]

  Darij

:roll: ,although it noway explians why result is true,it admits inelegent but short co ordinate sol,take A-0,1,B-b,oand C-c,o.

I have got another solution today, so I would like to post it here.
Sorry for reviving this old topic, but it is always better not to create a new topic! :)

Obviously $CD$ and $BE$ are the altitudes of the triangle $ABC,$ so let them intersect at the orthocentre $H$ of $\triangle ABC.$ Now join $AH,HM.$ Let $K=BE\cap DF, L=CD\cap EG.$
Note that $\angle BDF=\frac{\pi}{2}-\angle DBC=\angle LCG,$ and $\angle DFB=\angle LGC=\frac{\pi}{2}.$ Then $\triangle BDF\sim \triangle CGL.$ this leads to $\frac{GL}{GC}=\frac{BF}{DF}.$
Analogously we have $\triangle KBF\sim\triangle CEG,$ implying $\frac{KF}{FB}=\frac{CG}{GE}.$
These two jointly imply that $GC\cdot BF=GL\cdot DF=KF\cdot GE;$ so that we have
$\frac{EG}{DF}=\frac{GL}{KF}=\frac{EG-GL}{DF-KF}=\frac{EL}{DK}.$
On the other hand, note that $M=FE\cap DG\implies \triangle DMF\sim\triangle MEG,$ so that $\boxed{\frac{EG}{DF}=\frac{DM}{MG}}.$ But, $\triangle DHK\sim\triangle HEL\implies \boxed{\frac{EL}{DK}=\frac{DH}{HL}}.$
Using the last three relations, we note that $\frac{DM}{MG}=\frac{DH}{HL},$ which, from Thale's theorem, gives $HM\parallel EG.$ Since $EG\perp BC,$ we note that $HM\perp BC,$ ie $M$ lies on $AH.$ So $AM$ is the $A$-altitude of $\triangle ABC.\Box$
:)

$\square BCED$ is cyclic, $DE\cap BC=I$ and $A$ is on the polar of $I$ ([i][b]Brocard[/b][/i])
Draw a line through $M$ perpendicular to $BC$ and let it cut $DE$, $BC$ at $H$, $J$
$FJ/JG=DF/EG=FI/GI$, that is, ($F$,$G/J$,$I$)$=-1=$ ($D$,$E/H$,$I$)
Therefore, $HM$ is the polar of $I$ , $\overline{AHM}$ is the polar of $I$ and $AM\bot BC$

Dear Mathlinkers,
why not with Pappus theorem by considering the tangent to the circle at B and C...
Sincerely
Jean-Louis


prove that for $ a,b,c %Error. "belong" is a bad command.
\mathbb{R}$,and $ a^2 + b^2 + c^2 + abc = 4$ then $ ab + bc + ca - abc \leq 2$

$ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} abc\geq 2bc \plus{} a^{2} \plus{} abc$ (AM-GM)

$ (2 \minus{} a)(2 \plus{} a)\geq 2bc \plus{} abc$  $ \implies$  $ bc\leq 2 \minus{} a$

$ ab \plus{} bc \plus{} ca \minus{} abc\leq ab \plus{} 2 \minus{} a \plus{} ac \minus{} abc\leq2$  $ \iff$  $ ab \plus{} ac \minus{} abc \minus{} a\leq0$

$ \implies$  $ b \plus{} c \minus{} bc \minus{} 1\leq0$  $ \implies$  $ (b \minus{} 1)(c \minus{} 1)\leq 0$

if $ a,b,c\leq 1$  $ \implies$  $ (b \minus{} 1)(c \minus{} 1) \equal{} 0$

if $ a > 1,b,c\leq 1$ $ \implies$  $ (b \minus{} 1)(c \minus{} 1)\geq 0$

if $ a,b,c > 1$  $ \implies$  $ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} abc > 4$ contradiction!

if $ a < 1 ,b,c\geq 1$  $ \implies$  $ (b \minus{} 1)(c \minus{} 1)\geq 0$

we can choose $ b,c,(b \minus{} 1)(c \minus{} 1)\geq 0$  :wink:


The other portion of this puzzle is in Puzzles and Brainteasers, because it had more to do with words.

How many permutations of any length can you form from the letters in ARTOFPROBLEMSOLVING? That is, what is the sum of the number of ways to pick one letter, two letters, three letters, four letters...... nineteen letters from the letters in ARTOFPROBLEMSOLVING where order DOES matter and the letters with the same name are indistinguishable?

Please show your work.

Any solutions? Come on, this can't be too hard. Trouble is, though, that I don't know the answer. :blush:


integration
trigonometry
vector
calculus
geometry
geometric transformation
rotation

How can i calculate $ \displaystyle\int_0^\infty r^2 \ \mathrm{d} r \int_0^{2\pi} \ \mathrm{d} \phi \int_0^\pi \sin\theta \ \mathrm{d} \theta
\frac{e^{\minus{}\epsilon r}}{r}
e^{\minus{}\imath \bold{p} \cdot \bold{r}}$ where $ \epsilon >0$ is a constant and $ \bold{p}\in \mathbb{R}^3$ is a constant vector?

What is [b]r[/b] ?

May be the integral is $ \int\int\int_{{\mathbb R}^3} {e^{ - \varepsilon r}\over r} e^{ - \imath {\bf p\cdot x}}dx dy dz$ and [b]x=[/b](x,y,z), $ r = \sqrt {x^2 + y^2 + z^2}$?

In such a case you may do a rotation that puts [b]x[/b] parallel to $ {\underline e_1}$
and the integral becomes (the Jacobian is 1)
$ \int\int\int_{{\mathbb R}^3} {e^{ - \varepsilon r}\over r} e^{ - \imath p_1 r}dx' dy' dz'$ and then introduce polar coordinates.
Sorry if  this hint is useless

$ \bold{r} \equal{} (r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)$; thanks to your advice, making with a rotation $ \bold{p}$ be parallel to $ e_3$, i get $ \frac{4\pi}{p^2\plus{}\epsilon^2}$

If $ {\bf x} \equal{} (x,y,z)$ (cartesian coordinates) the integral is meaningless, if $ {\bf x} \equal{} {\bf x}(r,\theta,\phi)$ the notation is unclear, if $ {\bf x}$ is a constant (unlikely), the second exponential is constant

[quote="Clausius"]$ \bold{r} \equal{} \bold{x}$[/quote]

Key words: Newtonian potential.

The Fourier transform of $ \frac {1}{|x|}$ in $ \mathbb{R}^3$ is $ \frac {C}{|\xi|^2}.$ The form of the constant $ C$ will depend on where you put the $ 2\pi$ in your definition of Fourier transform.

If you define the Fourier transform by $ \widehat{f}(\xi) \equal{} \int_{\mathbb{R}^d}f(x)e^{ \minus{} 2\pi i\xi\cdot x}\,dx}$ then for $ 0 < b < d,$ we have that the Fourier transform of $ f(x) \equal{} |x|^{ \minus{} b}$ is $ \widehat{f}(\xi) \equal{} C|\xi|^{b \minus{} d},$ where $ C \equal{} \frac {\pi^{b \minus{} d/2}\Gamma\left(\frac {d \minus{} b}2\right)}{\Gamma\left(\frac {b}2\right)}.$

With the $ 2\pi$ where I have it, in your case, which is $ d \equal{} 3,b \equal{} 1,$ we get $ C \equal{} \frac {1}{\pi}.$ Your $ C$ will be different because you seem to be putting the $ 2\pi$ in a different place in your definition.

The key to the computation: rather than attempt the particular integral you wound up with, instead we write $ |x|^{ \minus{} b}$ as a "linear combination" of dilated Gaussians:

$ |x|^{ \minus{} b} \equal{} \frac {\pi^{b/2}}{\Gamma\left(\frac b2\right)}\int_0^{\infty}u^{b/2 \minus{} 1}e^{ \minus{} \pi u|x|^2}\,du.$

Considered in terms of the function of $ x,$ that integral doesn't converge in any nice normed space; consider this to be a case of weak (distributional) convergence. But then, you're asking about the distributional Fourier transform anyway. All of the functions we're talking about here are locally integrable but not globally integrable. (That - the lack of convergent integrals - may be what Arrigo-Sacchi is objecting to.)

So: just take the Fourier transform of that, and see what happens. In your case, with the $ 2\pi$ in a different place, you might want to use $ e^{ \minus{} u|x|^2/2}$ rather than $ e^{ \minus{} \pi u|x|^2}.$ Same idea, slightly varying arithmetic details.

Note: this idea (I've heard it called the "principle of subordination") is not mine; it's something I learned long ago.


[quote="NC7"]Given $a^2+b^2+c^2=1$.prove that
0$\le$abc +2(1+a+b+c+ab+bc+ca)[/quote]
If $abc\leq0$ then $abc +2(1+a+b+c+ab+bc+ca)=2(1+a)(1+b)(1+c)-abc\geq0.$
If $abc\geq0$ then$abc +2(1+a+b+c+ab+bc+ca)=(a+b+c+1)^2+abc\geq0.$ :)


geometry
3D geometry
sphere
geometry proposed

There is given a triangle $ABC$ in a space. A sphere does not intersect the plane of $ABC$. There are $4$ points $K, L, M, P$ on the sphere such that $AK, BL, CM$ are tangent to the sphere and $\frac{AK}{AP} = \frac{BL}{BP} = \frac{CM}{CP}$. Show that the sphere touches the circumsphere of $ABCP$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=40884 .

  darij


Show  that if x,y,z are positive integers then $ (xy\plus{}1)(yz\plus{}1)(zx\plus{}1)$ is a perfect square if and only if
$ (xy\plus{}1),(yz\plus{}1),(zx\plus{}1)$  are all perfect squares.

This has been posted many times before. See http://www.mathlinks.ro/viewtopic.php?t=150369 for example.


graph theory
combinatorics proposed
combinatorics

pro1.A complete graph has $2^{n}+1$ vertices.All edegs colored by n color.Prove that there are a odd cycle in graph which all edeg is the same color.
pro2.Given $X=\{1,2,...,n\}$.Let a family $m$ subset of set $X$.If this family such that any two element of $X$ then exist a set of family so it contains just one of two element.prove that $n\leq 2^{m}$.
pro3.Let k point on a line.Pass any two point $A,B$ on line then draw a circle which let $AB$ be diamater.Some circle colored by one of $n$ color such that any two circle be touch out then colored by 2 different color.
Have you problem ,similar?

$k\leq 2^{n}$ for problem 3.It is question.


In a circle $C$ with centre $O$ and radius $r$, let $C_1$, $C_2$ be two circles with centres $O_1$, $O_2$ and radii $r_1$, $r_2$ respectively, so that each circle $C_i$ is internally tangent to $C$ at $A_i$ and so that $C_1$, $C_2$ are externally tangent to each other at $A$.

Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent.

Assign weights $m_{A1} = OO_1 = r - r_1, m_O = O_1A_1 = r_1$ to $A_1, O$, respectively, so that these 2 weights balance at $O_1$, which is then assigned weight $m_{O1} = m_{A1} + m_O = r - r_1 + r_1 = r$. Assign weight $m_{O2}$ to $O_2$, such that the weights $m_{O1}, m_{O2}$ balance at $A$:

$\frac{m_{O2}}{m_{O1}} = \frac{O_1A}{O_2A} = \frac{r_1}{r_2}$

$m_{O2} = m_{O1}\ \frac{r_1}{r_2} = \frac{r r_1}{r_2}$

Assign weight $m_{A2}$ to $A_2$, so that the weights $m_O, m_{A2}$ balance at $O_2$:

$\frac{m_{A2}}{m_O} = \frac{OO_2}{O_2A_2} = \frac{r - r_2}{r_2}$

$m_{A2} + m_O = m_{O2} = \frac{r r_1}{r_2}$

Solving for $m_{A2}$,

$m_{A2} = m_O\ \frac{r - r_2}{r_2} = \left(\frac{r r_1}{r_2} - m_{A2}\right) \frac{r - r_2}{r_2}$

$m_{A2} \left(1 + \frac{r - r_2}{r_2}\right) = \frac{r r_1}{r_2} \cdot \frac{r - r_2}{r_2}$

$m_{A2} = \frac{r_1(r - r_2)}{r_2}$

Let $OA$ meet $A_1A_2$ at $P$. The weights $m_{A1}, m_{A2}$ now balance at $P$, i.e.,

$\frac{A_1P}{A_2P} = \frac{m_{A2}}{m_{A1}} = \frac{r_1(r - r_2)}{r_2(r - r_1)}$

Since we also have $\frac{OO_1}{A_1O_1} = \frac{r - r_1}{r_1},\ \frac{OO_2}{A_2O_2} = \frac{r - r_2}{r_2}$, it follows that

$\frac{OO_1}{O_1A_1} \cdot \frac{A_1P}{PA_2} \cdot \frac{A_2O_2}{O_2O} = \frac{r - r_1}{r_1} \cdot \frac{r_1(r - r_2)}{r_2(r - r_1)} \cdot \frac{r_2}{r - r_2} = 1$

By Ceva's theorem, the cevians $A_2O_1,\ OA \equiv OP,\ A_1O_2$ in the triangle $\triangle OA_1A_2$ are concurrent.

in this problem, we just apply the ceva theorem with triangle $OO_1O_2$, with A in $O_1O_2$, $A_1$ in $OO_1$, $A_2$ in $OO_2$

[quote="shobber"]Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent.[/quote]

This result is also true if the three circles are [b][color=blue]externally tangent[/color][/b].

That's what [b]neverstop[/b] means and this is true because $\frac{s-a}{s-b} \cdot \frac{s-b}{s-c} \cdot \frac{s-c}{s-a} =1$

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1992Problem2

Vo Duc Dien

oh nice problem!!
lemma 1:$O1,O2,A$ are on same line.
proof:we know that $\angle O2AO =90$ $  \angle O1AO =90$ so $O1,A,O2$ are on same line :)
lemma 2:$O,O2,A2$are on same line and $O,O1,A1$ are on same line.
proof:we know that $O2A2$ is perpendicular to line that tangent to circle$C$ and we know that $OA2$ is perpendicular to line that tangent to circle$C$ so $A2,O2,O$ are on same line :wink: !
proof of problem:
suppose $A2A1O2 = X$ and $O2A1O=Y$ and $OA2O1 = P$ AND $O1A2A1=Q$ we must proof:
$\frac{\sin X}{\sin Y} . \frac{\sin AOO1}{\sin AOO2} . \frac{\sin P}{\sin Q} =1$
we know that :
$\frac{\sin AOO1}{\sin AOO2} = \frac {r_1}{r_2} . \frac{OO2}{OO1}$
and we know that:
$\frac{\sin X}{\sin Y} . \frac{AA2}{A1O} = \frac {r_2}{OO2}$
and also it is right for $\frac{\sin P}{\sin Q}$and we know that $OA2=OA1$ so the problem is proof!
excuse me because im so lazy to right all of my solution but im sure that you know that my solution :wink:

Dear MLs

It is simple.

The concurrency point is this case is, obviously,
the Lemoine point (symmedian point) of  triangle AA1A2.

r1 and r2  are two red herrings, while  r  isn't.

This post is dedicated to both Mr. Referendum - G. Papandreou   and
Mr. M.I.T. -  Darij, not necessarily in this order.

Friendly,
M.T.

@armpist: very fine observation, although not quite 'obvious'; I think it can bring new good questions about this configuration.

Best regards,
sunken rock

@sunken rock

maybe the very first candidate should be the Sangaku of Gumma, even though there
are plenty of solutions by now.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=18266

It may lead to a new approach.

M.T.


induction
modular arithmetic
LaTeX
Divisibility Theory
number theory

Let $f(x)=x^3 +17$. Prove that for each natural number $n \ge 2$, there is a natural number $x$ for which $f(x)$ is divisible by $3^n$ but not $3^{n+1}$.

The following is from [b]Meru Alagalingam[/b], who for some strange reasons doesn't want to register here and told me to post it:

Let $p$ be a prime. Then we write $p^{k}\parallel n$ iff $p^{k}|n$ but $p^{k+1}\not|n$.
We will prove a stronger statement than requested in the problem: For each natural number $n$, $n\geq 2$, there is a natural number $x\not\equiv 0\mod 3$ for which $f\left(x\right)$ is divisible by $3^{n}$ but not by $3^{n+1}$.
And we will prove that by mathematical induction on $n$ whereas the basis case is trivial as $3^{2}\parallel f\left(1\right)$ with $1\not\equiv 0\mod 3$. So given any $x\not\equiv 0\mod 3$ for which $3^{n}\parallel f\left(x\right)$ we want to construct a $y\not\equiv 0\mod 3$ for which $3^{n+1}\parallel f\left(y\right)$ and if we succeed we're done:
\[a\not\equiv 0\mod 3\Longrightarrow 3^{b}a^{2}\equiv 3^{b}\mod 3^{b+1}\]
\[\Longrightarrow\left(a+3^{b-1}\right)^{3}=a^{3}+3a^{2}3^{b-1}+3a3^{2b-2}+3^{3b-3}\equiv a^{3}+3^{b}\mod 3^{b+1}\ \ (1)\]
 $3^{n}\parallel x^{3}+17$, i.e. you can distinguish two cases:
[u]Case 1: $x^{3}+17\equiv 2\cdot 3^{n}\mod 3^{n+1}$ [/u]
\[3^{n}\parallel x^{3}+17\Longrightarrow x\not\equiv 0\mod 3 \\ \Longrightarrow \left(x+3^{n-1}\right)^{3}+17\equiv x^{3}+3^{n}+17\equiv 2\cdot 3^{n}+3^{n}\equiv 0\mod 3^{n+1}\\ \Longrightarrow y: =x+3^{n-1}\]
[u]Case 2: $x^{3}+17\equiv 3^{n}\mod 3^{n+1}$ [/u]
\[\left(x+2\cdot 3^{n-1}\right)^{3}+17 = \left(\left(x+3^{n-1}\right)+3^{n-1}\right)^{3}+17\equiv\left(x+3^{n-1}\right)^{3}+3^{n}+17\\ \equiv x^{3}+3^{n}+3^{n}+17\equiv 3^{n}+3^{n}+3^{n}\equiv 0\mod 3^{n+1}\\ \Longrightarrow y: =x+2\cdot 3^{n-1}.\]
 In both cases $y\equiv x\not\equiv 0\mod 3$.
[u]Case 1: $3^{n+2}\not|y^{3}+17$[/u] and we are done.
[u]Case 2: $3^{n+2}|y^{3}+17$[/u] and $3^{n+1}\parallel z^{3}+17$ whereas $z: =y+3^{n}$:
 \[z\equiv y\equiv x\not\equiv 0\mod 3\]
 \[z^{3}+17=\left(y+3^{n}\right)^{3}+17=\left(y+3^{n-1}+3^{n-1}+3^{n-1}\right)^{3}+17\equiv y^{3}+3^{n}+3^{n}+3^{n}+17\equiv 0\mod 3^{n+1}\]
follows by applying (1) repeatedly, i.e. $3^{n+1}|z^{3}+17$ but (1) also yields $z^{3}+17=\left(x+3^{n}\right)^{3}+17\equiv x^{3}+3^{n+1}+17\not\equiv x^{3}+17\equiv 0\mod 3^{n+2}$, i.e. $3^{n+2}\not|z^{3}+17$.

[quote="ZetaX"]We will prove a stronger statement than requested in the problem: For each natural number $n$, $n\geq 2$, there is a natural number $x\not\equiv 0\mod 3$ for which [...][/quote] That's not really stronger, is it? If $x\equiv0\pmod3$ then automatically $3\not|f(x)$.

Here's my solution. Sorry, i don't know Latex type setting, so i've attached my solution as a word document.

[quote="tanrock"]Help me! I don't understand
[quote="ZetaX"]
[u]Case 1: $3^{n+2}\not|y^{3}+17$[/u] and we are done.
[u]Case 2: $3^{n+2}|y^{3}+17$[/u] and $3^{n+1}\parallel z^{3}+17$ whereas $z: =y+3^{n}$.[/quote][/quote]
me too .
please explain more :)

We will prove the problem by mathematical induction on $n$. The basis case, $n=2$, is successful when $x=1$.
Let $x_0$ be the number satisfying $3^n \parallel f(x)$. And we will find $a$ such that $3^{n+1} \parallel f(x_0 + a) $.
$(x_0 + a)^3 + 17 = a^2 ( a+ 3x_0 ) + 3ax^2 + (x^3 +17)$. Obviously, $3 \not | x_0$.
If $3^n | a$, $a^2 ( a+ 3x_0 ) + 3ax^2 + (x^3 +17) \equiv 3a + k \times 3^n \mod 3^{n+2}$ where $k<3$.
Then, when $a = 3^n -  k \times 3^{n-1}$, the proof finishes since $3a + k \times 3^n \equiv 3^{n+1} \mod 3^{n+2}$.

   Equivalent statement:Suppose $n\in \mathbb N$,and $n\ge 2$,prove:there are $m\in \mathbb N$ such that $3^n||m^3+17$.
   Induction:$n=2$,take $m=1$.
     Suppose the statement is true for $n(\ge 2)$,that is there exists $m\in \mathcal N$ such that $m^3+17=3^n(3q+r)$,$r\in \{1,2\},q\in \mathbb N$.Then $3$ does not divide $m$,implies $m^2\equiv 1(mod 3)$.
    For any positive integer $s$,we have $(m+3^{n-1}s)^3+17=m^3+17+3^nm^2s+3^{2n-1}ms^2+3^{3n-3}s^3\equiv 3^n(3q+r)+3^ns=3^n(r+s)(mod 3^{n+1})$.
     If $s=3-r$,let $x_1=m+3^{n-1}(3-r)$,easily $3^{n+1}|x_1$.If $s=6-r$,let $x_2=m+3^{n-1}(6-r)$,also $3^{n+1}|x_2$.We prove at least one of $x_1,x_2$ is not divided by $3^{n+2}$ is Ok.Assume sake of contradiction,then $0\equiv x_2^3+17=(x_1+3^n)^3+17=x_1^3+17+3^{n+1}x_1^2+3^{2n+1}x_1+3^3n\equiv x_1^3+17+3^{n+1}\equiv 3^{n+1}(mod 3^{n+2})$.Contradiction.

we prove it using induction. the base step for $n=1$ is obvious, now assume that we have proved it for $n=k,k>3$

and we prove it for $n=k+1$.

according to induction $\exists a : 3^k \mid a^3 +17 , 3^{k+1} \nmid a^3+17 \implies a^3+17=3^k \times r$ that $3 \nmid r$

now we prove that for some

$n : 3^{k+1} \mid (n\times 3^{k-1}+a)^3+17, 3^{k+2} \nmid (n\times 3^{k-1}+a)^3+17$

which is equivalent to proving that: $(n\times 3^{k-1}+a)^3+17 \equiv 3^{k+1} \mod 3^{k+2}$

$(n\times 3^{k-1}+a)^3+17=n^3 \times 3^{3k-3} + 3a(n\times 3^{k-1})(n\times 3^{k-1}+a) + a^3+17 \equiv$

$(a^3+17) +3a^2(n\times 3^{k-1}) \equiv r\times 3^k + 3^k\times (a^2n) \equiv 3^{k+1} \mod 3^{k+2} \iff r + a^2n \equiv 3 \mod 9$

 and we easily see because $a,r$ are constants and $3\nmid r,a \implies \exists b : ba^2 \equiv 1 \mod 9 \implies n \equiv (3-r)\times b \mod 9$


symmetry
algorithm
analytic geometry
geometry solved
geometry

Given $\triangle ABC$ find locus $M$ such that $MA+MB+MC=3a$ with given real positive number a.
Can any body help me to draw it, what is degree of it?
if $\triangle ABC$ is equilateral is this locus symmetry?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=59086 . Unfortunately, you can´t really say much about that locus...

  darij

Yes, through your topic we know $FA+FB+FC=\frac{1}{2}\sqrt{a^{2}+b^{2}+c^{2}+4\sqrt{3}S}$=k,
And this locus is only meaning if $MA+MB+MC=r \ge k$ and $C(k)$ is a curve througt $F$ and obtains $F$ is a exteme point !
but what is happen with $C(r), r>k$

But can any body help me to draw it, or can you help me to show algorithm to draw it with cabri, I can draw it in Cartesian coordinate with maple but maybe it isn't true !


trigonometry
geometry
incenter
symmetry
cyclic quadrilateral
geometry proposed

$P$ is a point in the interior of $\triangle ABC$, and $\angle ABP = \angle PCB = 10^\circ$.
(a) If $\angle PBC = 10^\circ$ and $\angle ACP = 20^\circ$, what is the value of $\angle BAP$?
(b) If $\angle PBC = 20^\circ$ and $\angle ACP = 10^\circ$, what is the value of $\angle BAP$?

I can't think of any other ways but Trig Ceva...... :?

[quote="k2c901_1"]Actually, the best solution I've seen has nothing to do with trigonometry.[/quote]
Can you show the construction?  :?

I'll wait for a few days, then I'll post it if no one furnishes the construction then. (Actually the construction is not mine...)

I think I have got the construction of (1)
[hide]
Constuct equilateral triangle with base $BC$. Then after a congruence and a cyclic quadrilateral proof, we will get $\angle{BAP}=30^o$.[/hide]

Construct as I mentioned in my last post.

Since $\triangle{BCD}$ is an equilateral triangle, so $DC=BC$, $\angle{BCD}=60^{o}$.

We already have: $\angle{PCB}=10^{o}$ and $\angle{ACP}=20^{o}$, so $\angle{ACB}=30^{o}$. So we also have $\angle{DCA}=30^{o}$.

From $AC=AC$, we shall have: $\triangle{DCA} \cong \triangle{BCA}$. Hence $\angle{CDA}= \angle{CBA} = 20^{o}$

Now, use the fact that $DB=DC$ and $PB=PC$, so we must have $DP$ bisect $\angle{BDC}$. Hence: $\angle{BDP}= \angle{CDP}=30^{o}$. Hence: $\angle{ADP}=10^{o}$, so $\angle{ADP}=\angle{ABP}=10^{o}$, which gives us A, P, B, D lie on the same circle. Hence: $\angle{BAP}=\angle{BDP}=30^{o}$

[quote="k2c901_1"]Nice construction![/quote]
Thanks.  :)

Can you show the construction you mentioned above?
Also, I am pretty sure the answer of number two is also $30^{o}$ and I have found out the triangle in (2) is congruent to the triangle in (1). But I am kinda stuck here......

As soon as I get through my notes... I saw a solution using the excenter of a certain triangle, but I don't seem to remember it. Can anyone help me?

For (a):
Choose $E$ on $AC$ so that $\angle EBC=30 ^\circ$. Notice that $A$ is the excenter of ?

Hope this is enough. I don't want to be too obvious; I hope that you can find ? yourself.

[quote="k2c901_1"]For (a):
Choose $E$ on $AC$ so that $\angle EBC=30 ^\circ$. Notice that $A$ is the excenter of ?
[/quote]
Use the same method I posted above, we can prove that: $EP$ bisects $\angle{BEC}$. Hence: $\angle{BEP}=\angle{AEP}=\angle{AEF}=60^{o}$. Hence $A$ is the ex-centre of $\triangle{BEP}$. So $\angle{AEP}=40^{o}$. Which gives: $\angle{PAC}=100^{o}$. Since $\angle{BAC}=130^{o}$, so $\angle{BAP}=30^{o}$.

Actually, the fact that $EP$ bisects $\angle BEC$ is trivial if you notice the symmetry between $B$ and $C$. (Note that $\angle EBC=\angle ECB$, and $\angle PBC=\angle PCB$.)

[quote="k2c901_1"]for (b):
I'm not going to say too much, but a continuation of the excenter argument in (a) shows that $P$ is the incenter of some triangle (not necessarily given in the problem!)[/quote]

Which one?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55281

\[{ \overline {\underline | \ How\ can\ I\ learn\ to\ draw\ on\ the\ computer,\ too\ ?\ | }} \]

Can anybody tell me which triangle in (b) is $P$ the incenter of?

Can anybody post a complete solution to part (b) of this problem?


quadratics
modular arithmetic
algebra
binomial theorem
number theory
number theory proposed

[b]Problem[/b][i] Let $F_{n}$ represent the $n^{\text{th}}$ Fibonacci number.  For prime $p>5$ prove that
\[F_{p}\equiv \left(\frac{p}{5}\right)\pmod p.\][/i]

Could you please tell me the source of this problem, [b]boxedexe[/b]? Some time ago I found it in a paper of Naoki Sato, but no source appeared for it... Thanks in advance! :)

It is from Ireland. But I don't know year.

[quote="Rust"]$F_{p}=\frac{q^{p}-r^{p}}{\sqrt 5 },q=\frac{1+\sqrt 5 }{2},r=1-q$, therefore $F_{p}=5^{(p-1)/2}(mod p)$. It is true for all $p>2$.[/quote]

Rust, can you give more explanation ? How can you convert the irrational into integer ? Thx..

Indeed, if you expand Binet's formula using the binomial theorem, you'll realize that no irrational terms remain-they all cancel. After doing this, since $\dbinom{p}{k}\equiv 0 \pmod{p}$, for every $0<k<p$, it follows that the only non-zero modulo $p$ term of the sum is $5^{(p-1)/2}$. On the other hand, since Fermat's little theorem gives us $2^{p-1}\equiv 1 \pmod{p}$, we can leave out the denominator of the expression, for it's congruent to 1 modulo $p$. Finally, upon noticing that Euler's criterion states that $\left(\frac{a}{p}\right) \equiv a^{(p-1)/2}\pmod{p}$, it follows that $F_{p}\equiv 5^{(p-1)/2}\equiv \left(\frac{5}{p}\right) \pmod{p}$.


quadratics
calculus
integration
irrational number
number theory
prime factorization

Normally, I have lots of trouble with problems like:

Show that $ n^2\plus{}2n\plus{}12$ is not divisible by 121 for all real n.

Obviously, $ 121\equal{}11^2$ so I could just prove that the equation is not divisible by 11. But, I have no idea how to proceed further. So, it would be nice if someone could gather up 5 or 6 problems that deal with divisibility and mods that are sort of similar to this one. Factoring will do also.

First, no, you cannot simply show that it is not divisible by 11. For example, when $ n \equal{} 10$, the quadratic evaluates to 132, which is divisible by 11 but not 121.

[hide="My approach to this problem"]
$ n^2 \plus{} 2n \plus{} 12 \equal{} (n \plus{} 1)^2 \plus{} 11$.

If 121 divides $ (n \plus{} 1)^2$ then certainly 121 will not divide $ (n \plus{} 1)^2 \plus{} 11$ (but 11 will).
If 121 does not divide $ (n \plus{} 1)^2$ then 11 does not divide it either, and so $ (n \plus{} 1)^2 \plus{} 11$ will not be divisible by 11, let alone 121.[/hide]

Yongyi said:
[quote]If 121 does not divide $ (n\plus{}1)^{2}$ then 11 does not divide it either[/quote]

umm... what about if for example  $ (n\plus{}1)^{2}$ is equal to 132. It is clearly not divisible by 121, but it IS divisible by 11??

I don't understand the question. Let $ n$ be the irrational number such that $ n^2 \plus{} 2n \equal{} 109$. Therefore, $ n^2 \plus{} 2n \plus{} 12 \equal{} 121$, which is divisible by $ 121$.

Unless you meant integral $ n$ and not real $ n$. If this is the case, then if $ 11$ divides $ n^2$, then $ 121$ must define $ n^2$ because the power of any prime in the prime factorization of a square must be even by definition.

which question you don't understand? mine or Poincare one?

Poincare's original question. It stated to prove the claim for all real $ n$, which doesn't make much sense (see my earlier post), which makes me believe that the claim should be proven for integral $ n$.

[quote="Chalao"]Yongyi said:
[quote]If 121 does not divide $ (n \plus{} 1)^{2}$ then 11 does not divide it either[/quote]

umm... what about if for example  $ (n \plus{} 1)^{2}$ is equal to 132. It is clearly not divisible by 121, but it IS divisible by 11??[/quote]

Notice that $ (n\plus{}1)^2$ is a perfect square. Therefore being divisible by 11 IMPLIES divisibility by 121.

I also think Poincare wants to prove it for integers.

My step-by-step solution

We have:

$ n^2 \plus{} 2n \plus{} 12 \equal{} (11)(11)(a)$ (which we want to prove is imposible for integer numbers)

$ n^2 \plus{} 2n \plus{} 1 \equal{} (11)(11)(a) \minus{} 11$

$ (n \plus{} 1)(n \plus{} 1) \equal{} (11)(11)(a) \minus{} 11$

$ (n \plus{} 1)(n \plus{} 1) \equal{} (11)(11a \minus{} 1)$

For the equation to have a solution we need n+1 be divisible by 11...... so $ n$ should be a number of the form 11x-1 (because   11x-1+1 is clearly divisible by 11)

then ... the same number  $ 11x \minus{} 1$ we choose should be divisible by $ (11a \minus{} 1)$   substituding $ n \equal{} 11x \minus{} 1$ in $ (n \plus{} 1)$ we get a number of the form $ 11x$ which $ (11a \minus{} 1)$ should divide, that clearly demonstrate there are not integer solutions. (because any number $ 11x$ is divisible by $ (11a \minus{} 1)$)

My original question was not how to solve the problem, but to acquire and solve more problems like it :P


[quote="Command"]Given $a$ and $b$ are both irrrational, is $a^{b}$ rational or irrational?[/quote]
Can be rational. Let $a=\sqrt 2 ,b=2\log_{2}(3)$, then $a^{b}=3.$

Yeah.
I forgot that property and didn't deploy it  thank you very much


Show that threre is a natural m so that
   a^(m-1) is congruent to 1 mod m, for every natural a that satisfies (a, m) = 1 and m is not a prime number.


Factor:

1) (x^3 + y^3)^2  +  x^5.y - 3.x^3.y^3 + x.y^5

2) a^3 + 3.a^2.b + 3a.b^2 + 2.b^3 - 6.b^2.c + 12b.c^2 - 8.c^3

Sorry for not being able to type in superscript...I do not have the programme to do so...  :(

You don't need a program; it's built in to the forum.  Click on the link that says "LaTex" over to the left for more explanation/help.

I will tex them for you this time, but make sure to learn it for the future.

$(x^3 + y^3)^2 + x^5y - 3x^3y^3 + xy^5$

$a^3 + 3a^2b + 3ab^2 + 2b^3 - 6b^2c + 12bc^2 - 8c^3$

thx to all u guys!   :lol:
ok i'll try to learn how to use latex too, so i won't trouble u guys again..  :P

[quote="break314dancer"]

So, using Pascal's triangle, you'd get:

$(a+b)^3+(b-2c)^3$, right?[/quote]

then how do i continue factorising from here? it seems so simplified, i cant factorise it any more... :oops:

[quote="shobber"][quote="kunny"]You can use the factorisation of $A^3+B^3$. ,mathical88 :)[/quote]
Yep.$A^3+B^3=(A+B)(A^2-AB+B^2)$.
But that only make the expression ugly.[/quote]

Or $(a+b)^3-3ab(a+b)$ but I don't know if that's more simplified or not


Suppose we have a deck of 10 cards - say the ace to ten of spades
(denoted 1,. . . , 10 for definitiveness). Permutations of the deck may be
regarded as elements of S(10).

Define s to be the ‘shuffle’ which hides the top card:
s =    1 2 3 4 5 6 7 8 9 10
  ..... 2 4 6 8 10 1 3 5 7 9 .

Let t be the ‘shuffe’ that leaves the first card unchanged:
t = 1 2 3 4 5 6 7 8 9 10
 ... 1 3 5 7 9 2 4 6 8 10.

Finally, let c be the ‘cut’:

c = 1 2 3 4 5 6 7 8 9 10
.... 6 7 8 9 10 1 2 3 4 5 .

(a) Write s, t, c, cs and scs using cycle notation, as a product of disjoint
cycles, and find the sign of each.

(b) For each of these basic and combined shuffes s, t, c, cs and scs, how many times must it be repeated before the cards are returned to their original positions?

damn..this latex thing is really complicated. I have made some changes to the problem..basically the numbers on each of the two rows of s,c and t are arranged like a rectangle.


cosider any 6 points on any given conic
construct any hamiltonian circuit through the given 6 points
extend the first segment and the 4th to intersect at P,
extend the 2nd segment and the 5th to intersect at Q,
extend the 3rd segment and the 6th to intersect at R,
P, Q and R are collinear
in other words locus of a sixth point given 5 pointssuch that P,Q AND R are collinear is the conic passing through the given 5 points

for a Pair of straight lines it amounts to the Papus's theorem

i have a beautiful proof using projections but would like to have some more ideas


$ A \equal{} \{\text{factors of 360}\}$, $ B \equal{} \{\text{perfect squares}\}$, and $ C \equal{} \{\text{odd numbers greater than 2}\}$. What number is in all three sets?

B is the most smallest set, so let's go with B.

if you look at B and C, you know the number is a square of an odd number.

so its 1, 9, 25, 49, 81, 121...

and the factor of 360 in the set is 9.

answer : 9

B is the most smallest set, so let's go with B.

if you look at B and C, you know the number is a square of an odd number.

So, it's 1, 9, 25, 49, 81, 121...

And, the factor of 360 in the set is 9.

Answer : 9

umm...
@qball, i'm not sure why you copied fantasy's answer and posted again
if it's just to add some punctuation, that's pointless and please refrain from doing so in the future

edit: to fantasy's post below me, oops  :blush: i'll fix that

[quote="vallon22"]umm...
@qball, i'm not sure why you copied iin's answer and posted again
if it's just to add some punctuation, that's pointless and please refrain from doing so in the future[/quote]

ehhh vallon its my answer  :D

1 doesn't work because it doesn't satisfy the restrictions for $C$.  $1$ is not greater than $2$.  :)

W00T 1337TH POST YEAH


a,b,c>0 prove that

(a/b+b/c+c/a)(a+b+c) >= 3 :sqrt: (a 2 +b 2 +c 2 )

I think a very hard inequality is the following

$(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})(a+b+c) \ge 3\sqrt{3(a^2+b^2+c^2)}$. ;)