Asymptote LaTeX FTW articles geometry function

AIME15 PMed me about this, so I thought I might as well post it here for future reference.

My procedures are *cough* messily *cough* outlined in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1105243#1105243]this forum topic[/url]. I'll try to outline cleanly here:

-- Download the asymptote [b]1.43[/b] source from [url=http://sourceforge.net/project/platformdownload.php?group_id=120000&sel_platform=1345]here[/url] (NOTE: this is a NEWER version than the one I used in the post above). I'll assume you've downloaded to the Desktop.

-- Follow directions in the INSTALL file --- I've adapted it for this version:

1. Open Terminal. It's located in /Applications/Utilities/Terminal.app

2. Run: [i]cd ~/Desktop[/i]

3. [i]gunzip asymptote-1.43.src.tgz[/i]

4. [i]tar -xf asymptote-1.43.src.tar[/i]

5. [i]cd asymptote-1.43[/i]

6. Put [url]http://www.hpl.hp.com/personal/Hans_Boehm/gc/gc_source/gc-7.1.tar.gz[/url]
in the current directory.

7. [i]./configure[/i]

8. [i]make all[/i]

9. [i]sudo make install[/i]

That should have installed Asymptote. For usage, let's say you have a [b]asylatex.tex[/b] that contains an asy image on the Desktop.

Use whatever LaTeX editor to compile the file. It'll return an error that looks something like this:
[code]Package asymptote Warning: file asylatex_1.pdf does not exist on input line 33.[/code]

Terminal again. Run:
[code]cd ~/Desktop
asy asylatex[/code]

That will run asymptote on the file [b]asylatex.asy[/b], producing [b]asylatex_1.pdf[/b].

Now compile [b]asylatex.tex[/b] again, and with any luck, it should compile with no errors.

[b]EDIT[/b]: I just noticed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1241236#1241236]this post[/url] that provides a way better alternative to manually running asy every time.

And with any luck, I made no errors in the short tutorial above. :|

Hope it helped! :)

In step two, i got this error: gunzip: asymptote-1.43.src.tgz: No such file or directory
Is it my machine?

Did you download the source package to the Desktop?

Then you must've forgotten to [i]cd Desktop[/i].. :)

A terminal session, by default, starts out in your home directory.

But if i forgot that, it wouldn't have said:

eugene-chens-imac:Desktop Eugene$ \$$
 that's the thing in front of where you type.

huh :huh: that's weird.. It's not a typo, is it? I can't really think why it wouldn't work..

Well if you can see it on the Desktop, you can also extract it using Mac's archive tool. Double-clicking on the package will probably do that (speculating - I'm not in OS X right now).

OK, i double clicked it and it's open now...what do i do now?

Extract it until you get a folder (not an archived package) called asymptote-1.43 on your desktop. Then continue from step 5, so that you're in the directory [b]~/Desktop/asymptote-1.43[/b]

Got this: -bash: cd: asymptote-1.43: No such file or directory

Got it after step five.  Did what you said before.

Uhh could you go to the FTW room so it's easier for us to talk?

hmm you either have a different asymptote version, or you're in the wrong directory.. The latter is more probable, though.

Could you try this: in Terminal, run [i]cd /Users/[b]yourusername[/b]/Desktop/asymptote-1.43[/i]

and see if you get to the directory?

EDIT: yup I'm there.

No, I was already in the directory.  It was after step five when i got that.

hmm, just wondering, AIME15, did you get the Xcode tools and get asy to work? (note to self: I might have to include this in the howto).

No.  On step seven, I got this:

checking for gc-7.1.tar.gz... (cached) no

Please put the Boehm garbage collector tar.gz file in the current directory.
FOR EXAMPLE, USE THE COMMAND:

wget http://www.hpl.hp.com/personal/Hans_Boehm/gc/gc_source/gc-7.1.tar.gz

So i put that in, but it said like bash - no command wget or something like that

Same thing when i tried sudo wget

 :maybe:

You missed step 6. but ah yes that was what I was afraid of---Mac systems do not have wget.

What wget does:

[i]wget some_file_on_the_web[/i]

that just downloads the file to the current directory. So for example, if you're compiling asymptote in ~/Desktop/asymptote-1.43/ , the file will be downloaded there.

Since macs don't have that, you can just manually download the file to that directory and run the configuration again.

EDIT: or you can just install wget: [url]http://wget.darwinports.com/download/[/url]

Well I tried putting it on Desktop (where asymptote 1.43 is located) and also tried putting it INSIDE asymptote 1.43, but neither of them worked.

Put it [i]inside[/i] asymptote-1.43.

And in terminal, when you run ./configure, make sure you're [i]inside[/i] that same directory. Your terminal should look something like the attached picture.

But I am already...

eugene-chens-imac:asymptote-1.43 Eugene$

Hmm...

BTW, i used to use ubuntu :D

hmm.. could you run [i]ls[/i] (that's an L, not an I) in that directory and post the output? Just to make sure the file really is in the directory..

and yeah Ubuntu FTW!

Yeah it's in there :(

gc-7.1 right?

this si everything in there:

BUGS		camperror.d	envcompleter.d	knot.o		record.d
ChangeLog	camperror.h	envcompleter.h	lex.yy.cc	record.h
Compile		camperror.o	envcompleter.o	lex.yy.d	record.o
Delaunay.cc	castop.h	errormsg.cc	lex.yy.o	refaccess.cc
Delaunay.d	coder.cc	errormsg.d	lexical.h	refaccess.d
Delaunay.h	coder.d		errormsg.h	locate.cc	refaccess.h
Delaunay.o	coder.h		errormsg.o	locate.d	refaccess.o
GUI		coder.o		errors		locate.h	runtime.cc
INSTALL		coenv.cc	errortest.asy	locate.o	runtime.d
LICENSE		coenv.d		examples	main.cc		runtime.h
Makefile	coenv.h		exp.cc		main.d		runtime.in
Makefile.in	coenv.o		exp.d		main.o		runtime.o
README		common.h	exp.h		mathop.h	runtime.pl
ReleaseNotes	config.h	exp.o		memory.h	settings.cc
TODO		config.h.in	fftw++.cc	mod.h		settings.d
absyn.cc	config.log	fftw++.d	modifier.h	settings.h
absyn.d		config.status	fftw++.h	name.cc		settings.o
absyn.h		configure	fftw++.o	name.d		simpson.cc
absyn.o		configure.ac	fileio.cc	name.h		simpson.d
access.cc	constructor.cc	fileio.d	name.o		simpson.o
access.d	constructor.d	fileio.h	newexp.cc	stack.cc
access.h	constructor.o	fileio.o	newexp.d	stack.d
access.o	dec.cc		flatguide.cc	newexp.h	stack.h
angle.h		dec.d		flatguide.d	newexp.o	stack.o
application.cc	dec.h		flatguide.h	pair.h		stm.cc
application.d	dec.o		flatguide.o	parser.cc	stm.d
application.h	doc		fpu.h		parser.d	stm.h
application.o	drawclipbegin.h	frame.h		parser.h	stm.o
array.cc	drawclipend.h	fundec.cc	parser.o	symbol.cc
array.d		drawelement.h	fundec.d	patches		symbol.d
array.h		drawfill.cc	fundec.h	path.cc		symbol.h
array.o		drawfill.d	fundec.o	path.d		symbol.o
arrayop.h	drawfill.h	gc-7.1		path.h		table.h
asy-keywords.el	drawfill.o	genv.cc		path.o		tests
asy-list.pl	drawgrestore.h	genv.d		path3.cc	texfile.cc
asy.ico		drawgroup.h	genv.h		path3.d		texfile.d
asy.rc		drawgsave.h	genv.o		path3.o		texfile.h
asymptote.nsi	drawimage.h	getopt.c	pen.h		texfile.o
asymptote.spec	drawlabel.cc	getopt.h	picture.cc	transform.h
autogen.sh	drawlabel.d	getopt1.c	picture.d	triple.h
base		drawlabel.h	guide.cc	picture.h	types.cc
bbox.h		drawlabel.o	guide.d		picture.o	types.d
builtin.cc	drawlayer.h	guide.h		pipestream.h	types.h
builtin.d	drawpath.cc	guide.o		primitives.h	types.o
builtin.h	drawpath.d	guideflags.h	process.cc	util.cc
builtin.o	drawpath.h	inst.h		process.d	util.d
callable.cc	drawpath.o	install-sh	process.h	util.h
callable.d	drawverbatim.h	interact.cc	process.o	util.o
callable.h	entry.cc	interact.d	program.cc	varinit.cc
callable.o	entry.d		interact.h	program.d	varinit.d
camp.l		entry.h		interact.o	program.h	varinit.h
camp.tab.cc	entry.o		item.h		program.o	varinit.o
camp.tab.d	env.cc		keywords.cc	psfile.cc	vm.h
camp.tab.h	env.d		keywords.pl	psfile.d	wce
camp.tab.o	env.h		knot.cc		psfile.h	xasy
camp.y		env.o		knot.d		psfile.o	xasy3
camperror.cc	envcompleter.cc	knot.h		record.cc	xstream.h

ah I see. It's already extracted, but it shouldn't be. Just download the tar.gz to that directory [i]without[/i] extracting it; leave it in archived form. :)

[quote="mac and asy noob(aime15)"]Hm...how do i do that?[/quote]

:blush:[/quote]

lol that's okay. :lol: erm when you download [url=http://www.hpl.hp.com/personal/Hans_Boehm/gc/gc_source/gc-7.1.tar.gz]the file[/url], a dialog should pop up asking whether you'd like to save or open it. Simply choose to [b]save[/b] it to asymptote-1.43.

WOOT!  it worked!  can you give me a sample asy file to test on texshop?  (becuz im an asy noob :( )

[quote="undefined117"]haha okay. Make sure you've actually installed it by running steps 8 and 9, though. :)

Very basic asy file:
[code]
\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{asymptote}

\begin{document}

\begin{asy}
size(300);

draw((0,0)--(1,1),blue);
\end{asy}

\end{document}[/code]

Sorry er I forgot most default asymptote commands. I use [url=http://www.artofproblemsolving.com/Forum/latexrender/asy/cse5.asy]cse5[/url] (from AoPS) a _lot_. You should really download it, as well as the olympiad package. I'll explain how to install it (it's not nearly as painful as installing asymptote has been) if you'd like.[/quote]

YES! IT WORKED! ok, how can i install cse5 and olympiad?  thanks so much

if you can help, please help quick :blush:

my internet goes down in 8 minutes

The packages need to be placed in the directory [b].asy[/b] in your home directory. Namely, in ~/.asy

The dot in front of the name makes it a hidden directory. Run the following commands (the dollar signs are not part of the commands, mind you) in Terminal to show the hidden directories (WARNING: this will restart Finder):

[code]$ defaults write com.apple.finder AppleShowAllFiles TRUE
$ killall Finder[/code]

You should then be able to see [b].asy[/b] in your home directory. Put the files cse5.asy and olympiad.asy inside the directory [b].asy[/b].

In terminal, run [i]sudo texhash[/i] <-- this will update the asymptote "database" of packages.

Then you can try compiling this document:

[hide="kinda long code, may take a moment to compile"][code]
\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{asymptote}

\begin{document}

\begin{asy}
import cse5;
import three;
size(400);
currentprojection=perspective((50,35,30));
defaultpen(linewidth(.8));

triple O=(0,0,0);
triple AA=(0,0,1), AB=(1,0,0), AC=(0,1,0), AD=(-1,0,0), AE=(0,-1,0), AF=(0,0,-1);
triple BA=(1,1,1), BB=(-1,1,1), BC=(-1,-1,1), BD=(1,-1,1), BE=(1,1,-1), BF=(-1,1,-1), BG=(-1,-1,-1), BH=(1,-1,-1);
triple DA=(0,0,2), DB=(2,0,0), DC=(0,2,0), DD=(-2,0,0), DE=(0,-2,0), DF=(0,0,-2);
triple EA=(0,0,3), EB=(3,0,0), EC=(0,3,0), ED=(-3,0,0), EE=(0,-3,0), EF=(0,0,-3);
triple FAA=(3,-3,3), FAB=(3,-3,0), FAC=(3,0,3), FAD=(0,-3,3), FBA=(-3,3,3), FBB=(-3,3,0), FBC=(0,3,3), FBD=(-3,0,3), FCA=(3,3,-3), FCB=(3,0,-3), FCC=(3,3,0), FCD=(0,3,-3), FDA=(-3,-3,-3), FDB=(0,-3,-3), FDC=(-3,-3,0), FDD=(-3,0,-3);

D(AA--AB--AF--AD--cycle--AC--AF--AE--AB--AC--AD--AE--cycle);
DPA(BA--BB--BC--BD--BA^^BE--BF--BG--BH--BE^^BA--BE^^BB--BF^^BC--BG^^BD--BH,red);
DPA(BA--BC--BH--cycle--BF--BC^^BF--BH,green);
D(EA--EB--EF--ED--cycle--EC--EF--EE--EB--EC--ED--EE--cycle,orange);
DPA(FAA--FBA--FCA--cycle--FDA--FCA^^FBA--FDA);
\end{asy}

\end{document}[/code][/hide]

Yay it all worked!!! Thanks for everything undefined!!!!!!

dang now my excuse for not having geometry hw is invalid :(

Macs do have curl; you can use "curl -O url" to download url.

I've never used that, but it should do the same thing as [i]wget[/i].

hehe so instead of downloading manually, you could've just typed "curl -O http://www.hpl.hp.com/personal/Hans_Boehm/gc/gc_source/gc-7.1.tar.gz" :dry:

OK.
Is there a way to make it do *asy asyfile* automatically instead of going to terminal every time?

worthawholebean mentioned a script to automate it in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1241236#1241236]the thread[/url] you created in the LaTeX forum.

Copy the code into TextEdit and save it somewhere (like your Desktop) as [b]asypdftex[/b]. Then you can drag and drop the file to /usr/local/texlive/2007/bin/i386-darwin/ (it'll ask for the password). Then run this in Terminal:
[code]sudo chmod +x /usr/local/texlive/2007/bin/i386-darwin/asypdftex[/code]
Then, like worthawholebean said, configure stuff in TeXShop.

Erm...how do i write a script?  What do i make the file extension?

It says "No such file or directory" after i do chmod.  I triple-checked, and the file is there...

Did you drag-drop the file to that directory? Also notice that this is [i]not[/i] in your home directory. There is no tilde (~) anywhere in /usr/local/.../i386-darwin/

i know.  Does it have to be drag-drop?  Because i just copied....

And i didn't put tilda.

ehh that's kind of weird.. OS X has chmod functionality.. hmm. I'll assume you saved a script [b]asypdftex[/b] on the [b]Desktop[/b]. Try copying this to Terminal:
[code]cd ~/Desktop
sudo cp asypdftex /usr/local/texlive/2007/bin/i386-darwin/
sudo chmod +x /usr/local/texlive/2007/bin/i386-darwin/asypdftex[/code]
if that doesn't work, well.. I don't know. :(

Sorry for reviving, but I recently had to restore my iMac because there was so much clutter...

Installing again.

At the "make all" step, it all compiled fine till:

[code]
c-7.1/include -o drawfill.o -c drawfill.cc
g++ -Wall -ansi -DHAVE_CONFIG_H -D_FILE_OFFSET_BITS=64 -DUSEGC  -g -O3  -I . -Igc-7.1/include -o path3.o -c path3.cc
g++ -Wall -ansi -DHAVE_CONFIG_H -D_FILE_OFFSET_BITS=64 -DUSEGC  -g -O3  -I . -Igc-7.1/include -o drawpath3.o -c drawpath3.cc
g++ -Wall -ansi -DHAVE_CONFIG_H -D_FILE_OFFSET_BITS=64 -DUSEGC  -g -O3  -I . -Igc-7.1/include -o drawsurface.o -c drawsurface.cc
drawsurface.cc: In member function ‘virtual void camp::drawSurface::render(GLUnurbs*, double, const camp::triple&, const camp::triple&, double, bool)’:
drawsurface.cc:260: error: invalid conversion from ‘GLvoid (*)(...)’ to ‘GLvoid (*)()’
drawsurface.cc:260: error:   initializing argument 3 of ‘void gluNurbsCallback(GLUnurbs*, GLenum, GLvoid (*)())’
make: *** [drawsurface.o] Error 1
[/code]

and it stopped.

Yes, I'm having the same issue.

[code]c-7.1/include -o drawsurface.o -c drawsurface.cc
drawsurface.cc: In member function ‘virtual void camp::drawSurface::render(GLUnurbs*, double, const camp::triple&, const camp::triple&, double, bool)’:
drawsurface.cc:260: error: invalid conversion from ‘GLvoid (*)(...)’ to ‘GLvoid (*)()’
drawsurface.cc:260: error:   initializing argument 3 of ‘void gluNurbsCallback(GLUnurbs*, GLenum, GLvoid (*)())’
make: *** [drawsurface.o] Error 1
[/code]

I'll post if I find a fix.


How many 9-digit positive integers are there in which no digit occurs 3 or more time consecutively? (The integers are allowed to begin with 0's, but no more than 3 of them, of course.)

Besides the fact that I can make a backtracking program to solve this problem, because 9 is rather small, I'll give a nasty solution to this one that gives a recursive method to compute the desired number:
For a number n, denote by f(n,i,j) the number of good (that verify the conditions of the problem) sequences (integers) of n terms (digits) that end up in j of i. In here 0 \leq i \leq 9 and 0 \leq j \leq 2.
We have f(n,i,2)=f(n-1,i,1) for all i in [0..9].
And f(n,i,1)= \sum {k<>i} f(n-1,k,1)+ \sum {k<>i} f(n-1,k,2).

The number of good sequences for n is  \sum {i in 0...9 and j in 1..2} f(n,ij).

If it is your gratest wish that I give you the exact number of 9-digit or m-digit for some other (but small) m, I will write a computer program to compute the number by the method I posted here.

I don't know what amfulger thought about it, but...
Desired number can be one of  the following forms : ab... or aab...
So it is absolutely clear that recurrence is $f_n=9f_{n-1}+9f_{n-2}$, $f_1=10$, $f_2=100$. Therefore, using standart method, we can find its closed form.
Answer is $f_{10}=9282568410$.


LaTeX
number theory proposed
number theory

Let $a_{1}, a_{2}, ..., a_{n}$ be positive integers such that:

\[\begin{array}{c}a_1\mid{2^{a_2}-1}\\a_2\mid{2^{a_3}-1}\\\cdot\\\cdot\\\cdot\\a_{n-1}\mid{2^{a_n}-1}\\a_n\mid{2^{a_1}-1}\end{array}\]

Show that all $a_i$'s are equal to 1

Take $p$ to be the smallest among all the primes which divide $a_i,\ i\in \overline{1,n}$. Assume $p|a_1$. In this case, $p|2^{a_2}-1$ and $p|2^{p-1}-1$ because $p>2$, since $a_i$ are odd. All of this shows that $p|(2^{a_2}-1,2^{p-1}-1)=2^{(a_2,p-1)}-1=1$, which is a contradiction, and the conclusion follows.

Nice!, my solution is not as elegant as yours. I show that being $p_i$ the smallest prime dividing $a_i , i\in{1,...,n}$, then $p_1>{p_2}>{\cdots}>{p_n}>{p_1}$ and the conclusion follows :D:D:D
(Thx valentin for the trick  :) )

just simply use $>$.

Wow, i gave a solution about a latex problem!!! :D

Pierre.

[quote="Leonardo"]Thx, :blush:  I was looking for the symbol like 15 minutes. After I made the post I notice my stupidness....:D[/quote] another small trick, you can edit your own post, so that you can correct it if necessary :) (like in this case) so go ahead, try it :P

and Pierre, right on ;) you could also start using \displaystyle after the dollar sign to make the inline equations look nicer (when having sums and fractions and stuff). for example $ \sum_{k=1}^n$ is different from $\displaystyle \sum_{k=1}^n$ (another latex tip: pause your mouse over a formula -> its code will be revealed to you :D).


geometry
cyclic quadrilateral
geometry unsolved

Convex quadrilateral $ ABCD$ is inscribed in circle $ (O,R)$.Let $ L\equal{}AC \cap BD,Q\equal{}AB \cap CD$.Let $ H$ be the orthocenter pf triangle $ ADQ$.Let $ M,N$ be the midpoints of $ AC,BD$.Prove that $ MN$ is perpendicular to $ PH$

$ MN\perp HL$ iff $ LH$ and $ LO$ are the isogonal lines wrt $ \angle ALD$.
Let $ E, F$ be the projections of $ H$ on $ BD, AC$.
We will prove that $ MENF$ is cyclic quadrilateral:
Since $ AJFH$ is cyclic we get $ \angle FJH \equal{} \angle FAH \equal{} 90^o \minus{} \angle ACD \equal{} 90^o \minus{} \angle ABD \equal{} \angle BDJ$
$ \Rightarrow FJ$ passes through $ N$
Similarly $ ER$ passes through $ M$
So $ \angle LFN \equal{} \angle JFA \equal{} \angle JHA \equal{} \angle RHD \equal{} \angle RED \equal{} \angle LEM$
$ \Rightarrow MENF$ is cyclic $ \rightarrow$ QED

1.QH⊥AD  QH1⊥BC  =>  ∠HQH1=∠DPB
2.⊿QAD∽⊿QCB  => QH/QH1=AD/BC
3.∠DPB=∠DAE=∠HQH1    AE/AD=BC/AD=QH1/QH    =>  ⊿QHH1∽⊿ADE
4.notice AH1⊥AE   =>  HE⊥QA  =>  H,D,E are collinear  => HE//MN⊥HH1
Ps: actually, the orthocenters of  ⊿DPB, ⊿ACP and ⊿DBQ, ⊿ACQ are ⊥ MN respectively.


modular arithmetic
induction
number theory
relatively prime
number theory proposed

Let a1, a2, . . . be a sequence such that a1 = 43, a2 =
142, and a(n+1) = 3a(n) + a(n−1)for all n>1. Prove that
(a) an and a(n)+1 are relatively prime for all n > 1;
(b) for every natural number m, there exist infinitely many natural
numbers n such that a(n) − 1 and a(n+1) − 1 are both divisible by
m.

(a) We have $ a_{n\minus{}1}\equal{}a_{n\plus{}1}\minus{}3a_n$. If $ d|a_{n\plus{}1}$ and $ d|a_n$, then $ d|a_{n\minus{}1}$. Continuing this, $ d$ must also divide $ a_{n\minus{}2},\ldots,a_2,a_1$. But $ \gcd(43,142)\equal{}1$, so $ d\equal{}1$. Therefore $ a_n$ and $ a_{n\plus{}1}$ are relatively prime.

(b) Define $ a_{n\plus{}1}\equal{}3a_n\plus{}a_{n\minus{}1}$ for all integers $ n$. We have $ a_0\equal{}a_2\minus{}3a_1\equal{}13$, $ a_{\minus{}1}\equal{}a_1\minus{}3a_0\equal{}4$, $ a_{\minus{}2}\equal{}a_0\minus{}3a_{\minus{}1}\equal{}1$, $ a_{\minus{}3}\equal{}a_{\minus{}1}\minus{}3a_{\minus{}2}\equal{}1$. Consider this sequence modulo $ m$. There are finitely many possible values for $ (a_n,a_{n\plus{}1})\pmod{m}$, more exactly there are $ m^2$ possible values. So there exist two pairs $ (a_i,a_{i\plus{}1})$ and $ (a_j,a_{j\plus{}1})$ which are congruent modulo $ m$. Since this sequence is defined recursively, then $ (a_{\minus{}3},a_{\minus{}2})\equiv(a_{j\minus{}i\plus{}1},a_{j\minus{}i\plus{}2})\pmod{m}$. Let $ j\minus{}i\equal{}x$. By simple induction, we have $ (a_{\minus{}3},a_{\minus{}2})\equiv(a_{kx\plus{}1},a_{kx\plus{}2})\pmod{m}$ for all positive integer $ k$. Since $ a_{\minus{}3}\equiv a_{\minus{}2}\equiv1\pmod{m}$, then $ a_{kx\plus{}1}\equiv a_{kx\plus{}2}\equiv1\pmod{m}$. Therefore $ a_{kx\plus{}1}\minus{}1$ and $ a_{kx\plus{}2}\minus{}1$ are divisible by $ m$ for all $ k$, whence the result follows.


My post count reads 585, but when I click on "Find all posts by Tare" on my profile, it says "Search found 504 matches"

Did someone lower my post count directly or is the "**********'s post number" problem happening again (only this time it's less than the total post)?

This is a result of posts being culled by the natural paring of the Etc. forum we set up (any post that has been sitting there unanswered for over a week gets culled).  Your post count is unaffected by this 'natural' paring, it only gets reduced if a moderator has to manually clip posts.

So "culling" is like "pruning" mentioned earlier? and what is "paring"? is it different from "culling" or "pruning"?
This isn't in the FAQ, so can you add this?

Those words are all mean the same thing in relation to the board.  I'll see what I can do about the FAQ.

Thanks!

I have one last question. You said we have more than enough space [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=308]here[/url]. Exactly how many posts (or file size) can this forum support?

Currently, around 50K (posts).  Eventually, much more.  We'll be moving to a dedicated server sometime either this fall or next spring.

Then what was the reason for deleting the logs of the Etc. forum?

It allows us a little more time before having to go to a dedicated server.

Recall that we [i]had[/i] about 10,000 posts (although the post count is only < 9000) a few days ago.  There's still probably quite a bit of time.  However, the sooner you move, the fewer files there are to move, right?


You do know that 1/7 = .142857 repeating, right? That's why it's so cool, I guess, because you can use it to estimate 1/7 to 6 or 12 significant figures and annoy people.

But on basis of arbitrary favoritism, 247 is obviously the best number since there's no reason why it's cool.

Another cool thing about it, which is I guess based on the previous things, but is fun to annoy your friends with anyway, is the following problem:

Bozo the clown saw a six digit number as the price of a car. The last digit was 2. He noticed that if he were to move the 2 to be the first digit (and not the last), it would be three times the size of the original number. What is the original number?

The solution is obvious based on the previous posts, but it is a fun little problem anyway.
Avi Arfin

42 is the best.  Or maybe 5/12 is because, um, to "any dice problem where you don't know the answer, guess 5/12... it should work!"  Oh well

A cool set of numbers I like is \[\frac{1}{98},\frac{1}{998},\frac{1}{9998},\frac{1}{99998}, etc.\] $\frac{1}{9998}=0.00001000020000400008000016000032\ldots$
Just add one more space for every extra digit in $99\ldots8$.
They are really just fun to annoy your friends with. :P  :lol:

[quote=daermon]my favorite number is 3 :jump:[/quote]

Reason?


What is the sum of the 25 smallest perfect squares which are not perfect cubes?

The first few sixth powers are (squares and cubes) $ 1,64,729,\ldots$.

Thus, we want to find the sum of $ (1^2 \plus{} 2^2 \plus{} \cdots \plus{} 26^2\plus{}27^2\plus{}28^2) \minus{} 1^2\minus{}8^2\minus{}27^2 \equal{} \frac {28\cdot29\cdot57}6 \minus{} 794 \equal{} \boxed{6920}$.

Edit: Thanks for reminding me that $ 1^3$ is a perfect cube, mewto55555. :wink:  :P

Um waht about 1^2? and then we have to worry about 27^2


Let $ x$ and $ y$ be positive integers with no prime factors larger than $ 5$.
Find all such $ x$ and $ y$ which satisfy
$ x^2 \minus{} y^2 \equal{} 2^k$
for some non-negative integer $ k$.

[hide="What I tried"]I let $ x\equal{}2^a3^b5^c$ and $ y\equal{}2^d3^e5^f$. then we have:
$ 2^{2a}3^{2b}5^{2c}\minus{}2^{2d}3^{2e}5^{2f}$
It was pretty easy to prove that if $ x$ has a factor of $ 3$, then $ y$ cannot, and same for if $ x$ has a factor of $ 5$. After that, Im lost.
[/hide]

Hints and suggestions are appreciated. I would really like to know how to solve these general types of problems

Hints (the fewer you look at, the better  :) ):

[hide="First hint"]Without loss of generality, (using your notation), $ \text{min}\,(a,d) = 1$. In other words, only one of $ x,y$ needs to be even.

[hide="Second hint"]Difference of squares. The factors of $ 2^k$ must also be powers of $ 2$.

[hide="Third hint"]Note that $ x - y, x + y$ are both powers of $ 2$, say $ 2^a$ and $ 2^b$, and solve for $ x$ and $ y$. Now, apply Mihalescu's Theorem.

$ (x,y) = (3 \cdot 2^k, 2^k), (5 \cdot 2^k, 3 \cdot 2^k)$ for $ k \in \mathbb{Z}_{\ge 0}$.

Suppose $ 2^m \parallel \text{gcd}\,(x,y)$; then one can factor out $ 2^{2m}$ from $ x^2 - y^2$, so it follows that without loss of generality, we can make at most one of $ \{x,y\}$ even. Now, $ x^2 - y^2 = (x - y)(x + y) = 2^k$, so it follows that $ x - y = 2^a, x + y = 2^b$ for $ b > a$. Summing the two equations gives $ x = \frac {2^a + 2^b}{2} = 2^{a - 1} \cdot (2^{b - a} + 1)$, and subtracting gives $ y = \frac {2^b - 2^a}{2} = 2^{a - 1} \cdot (2^{b - a} + 1)$. Let $ k = b - a$; then, we are given that the prime factors of $ 2^k \pm 1$ must be either $ 2,3,5$. For $ k = 0$, we obtain $ y = 0$, which contradicts the hypothesis of positive integers. For $ k > 0$, obviously $ 2 \nmid 2^k \pm 1$, and furthermore, if $ 3,5 | 2^k \pm 1$, then $ 3,5 \nmid 2^k \mp 1$. It follows that either $ 2^k - 1 = 5^c$, $ 2^k + 1 = 3^d$, or $ 2^k + 1 = 5^c, 2^k - 1 = 3^d$. By Mihalescu's Theorem, it follows that $ k = 1,2$. Remembering the we factored out powers of $ 2$ in our first statement, the answer set follows.

[/hide][/hide][/hide]

Evidently, Im tackling problems way over my head :) . Thanks for the hints, though.

We can still solve the problem without Mihailescu, though. I will just build off azjps' solution.

[hide]The situation $ 2^k \minus{} 1 \equal{} 5^c$ is false mod $ 4$ unless $ k \equal{} 1$, $ c \equal{} 0$. So in the other case we have $ 2^k \minus{} 1 \equal{} 3^d$. Mod $ 4$, we find that unless $ k \equal{} 1$, $ d \equal{} 0$, $ d$ is odd, so $ 2^k \minus{} 1 \equal{} 3 \cdot 9^m$ for some $ m$. But this is false mod $ 8$, unless $ k \equal{} 2$, $ m \equal{} 0$. Thus, $ k \equal{} 1$ or $ k \equal{} 2$.[/hide]


I set up the AMC 12 Problems/Solutions pages so that problems and solutions could be easily added.  One problem I just ran into while beginning to post the problems themselves is writing the answer choices (A, B, C, D, E).  I'm not knowledgeable about the wiki formatting codes, so I'm not sure how to write the answer choices.  Hopefully there is a way to write them similar to how we have it in the Contests/Resources pages.  Any ideas or suggestions?

Also, is it possible to center equations in math mode, or just to center text?  Thanks!

The HTML tag text you want centered centers the text, but often it looks weird on the wiki...

To center equations in math mode use <math> .... </math>

edit:

Also, look at what I did for problem 1: http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12A_Problems#Problem_1

If you like that you can use
[code]
<math>
\mathrm{(A) \ }
\qquad \mathrm{(B) \ }
\qquad \mathrm{(C) \ }
\qquad \mathrm{(D) \ }
\qquad \mathrm{(E) \ }
</math>
[/code]

as a solution template.

That's a pretty good template.  It can be hacked pretty easily to create more or less space when needed.

That answer template works nicely, but I'm having problems with the centered math.  Sometimes centered math stuff gets displayed below or above where it should be, covering up other text.  Is this just my computer, or is this happening to other people as well?  I've been avoiding centered math just for this reason.

Could you try an recreate an example and post the link here?  I don't think I've ever experienced that.

Take a look at the answer choices [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12A_Problems]here[/url].

On my computer, starting on problem 12, the answer choices are a little below where they should be.  It gets really bad as you go farther down.  I didn't even center anything beyond problem 20.

What browser are you using?  I don't see the problem in Firefox or IE6.


Determine a triangle for which the three sides and an altitude are four consecutive integers and for which this altitude partitions the triangle into two right triangles with integer sides. Show that there is only one such triangle.

I couldn't find an elegant way to solve this problem. Does somebody have a better way to do it? (not casework)

[hide=" Casework Method"]
Denote the lengths as $x,$ $x+1,$ $x+2,$ $x+3.$ You know that the altitude must be less than at least two of the sides, since it's a leg in two right triangles. Hence, the altitude has length $x$ or $x+1.$

Then form equations with permutations of the lengths. The case that happens to work is when the altitude has length $x,$ the side it's drawn to has length $x+2,$ and the others have length $x+1, x+3.$ By some pythagorean theorem/difference of squares, we find that $x=12.$

Indeed, this method's a mess...  :blush:
[/hide]


function
floor function
induction
calculus
integration
algebra unsolved
algebra

A function $f: \N\rightarrow\N$ is circular if for every $p\in\N$ there exists $n\in\N,\ n\leq{p}$ such that $f^n(p)=p$ ($f$ composed with itself $n$ times) The function $f$ has repulsion degree $k>0$ if for every $p\in\N$ $f^i(p)\neq{p}$ for every $i=1,2,\dots,\lfloor{kp}\rfloor$. Determine the maximum repulsion degree can have a circular function.

[b]Note:[/b] Here $\lfloor{x}\rfloor$ is the integer part of $x$.

There is a typo in the problem.  It should read "The function has repulsion degree $ k > 0$ if for every $ p$, $ f^{i}(p)\neq p$ for $ i \equal{} 1,2,...,\lfloor kp\rfloor$."

Solution coming later today by yours truly (unless somebody's faster).

Obviously $ f(1) \equal{} 1$, and we shall assume that it is possible to find a circular function $ f$ with repulsion degree $ k\geq\frac{1}{2}$.

Because $ \lfloor 2k\rfloor \equal{} 1$ by hypothesis, $ f(2)\neq2$, and because of circularity, $ m > 2$ exists such that $ f(2) \equal{} m$ and $ f(m) \equal{} 2$.  If $ m\geq4$, then $ \lfloor km\rfloor\geq 2$, in contradiction with $ f(f(m)) \equal{} f(2) \equal{} m$, hence $ m \equal{} 3$ necessarily.

For any positive integer $ q$, define $ C_{q}$ as the set of different numbers $ p$ such that $ f^{i}\left(2^{q}\right) \equal{} p$ for some $ i$.  Because of circularity, $ C_{q}$ is also the set of integers $ p$ such that $ f^{i}(p) \equal{} 2^{q}$ for some $ i$, and the cardinal $ n_{q}$ of $ C_{q}$ is the smallest number such that $ f^{n_{q}}(p) \equal{} p$ for all $ p\in C_{q}$, being obviously $ 2^{q}\geq n_{q}$.

Assume that $ p\in C_{q}$ exists such that $ p\geq2^{q\plus{}1}$.  Then, $ \lfloor kp\rfloor\geq2^{q}$, or $ f^{i}(2^{q})\neq2^{q}$ for all $ i \equal{} 1,2,...,2^{q}$, which is absurd because of circularity.  Therefore, all $ p\in C_{q}$ are less at most $ 2^{q\plus{}1}\minus{}1$.  In particular, $ 2^{q}$ cannot belong in $ C_{q'}$ for any $ q' < q$, or all $ C_{q}$ are disjoint.

We will now show by induction that $ C_{q}\equal{}\left\{2^{q},2^{q}\plus{}1,2^{q}\plus{}2,...,2^{q\plus{}1}\minus{}1\right\}$.  The result is proved for $ q \equal{} 1$.  Assume the result true for $ 1,2,...,q\minus{}1$.  Then, any element in $ C_{q}$ is not lower than $ 2^{q}$, since all positive integers less than $ 2^{q}$ are either 1, or belong to some $ C_{q'}$ with $ q' < q$.  The largest element in $ C_{q}$ is then at least $ 2^{q}\plus{}n_{q}\minus{}1$, or $ n_{q}\minus{}1\geq\lfloor\frac{2^{q}\plus{}n_{q}\minus{}1}{2}\rfloor\geq2^{q\minus{}1}\plus{}\frac{n_{q}}{2}\minus{}1$, and we can conclude that $ n_{q}\geq2^{q}$, yielding $ n_{q}\equal{} 2^{q}$.  Since no element of $ C_{q}$ may be smaller than $ 2^{q}$ or larger than $ 2^{q\plus{}1}\minus{}1$, then $ C_{q}\equal{}\left\{2^{q},2^{q}\plus{}1,2^{q}\plus{}2,...,2^{q\plus{}1}\minus{}1\right\}$, qed.

Since the largest element of $ C_{q}$ is $ 2^{q\plus{}1}\minus{}1$, and $ i \equal{} n_{q}\equal{} 2^{q}$ is the smallest $ i$ such that $ f^{i}\left(p\right) \equal{} p$ for any $ p\in C_{q}$, then $ 2^{q}\minus{}1\geq\lfloor k\left(2^{q\plus{}1}\minus{}1\right)\rfloor > k\left(2^{q\plus{}1}\minus{}1\right)\minus{}1$, or $ k <\frac{2^{q}}{2^{q\plus{}1}\minus{}1}\equal{}\frac{1}{2}\plus{}\frac{1}{2\left(2^{q\plus{}1}\minus{}1\right)}$.  When $ q$ takes on increasing values of integers, the second term tends to zero, so $ k$ cannot be larger than $ \frac{1}{2}$.

Note that a circular function with $ k \equal{}\frac{1}{2}$ may be constructed by setting $ f(p) \equal{} p\plus{}1$ if $ p\plus{}1$ is not an integral power of 2, and $ f(p) \equal{}\frac{p\plus{}1}{2}$ if $ p\plus{}1$ is an integral power of 2.

A function $f: N \rightarrow N$ is circular if for every $p \in N$ there exists $n\in N,\ n \le p$ such that $f^n(p)=p$ ($f$ composed with itself $n$ times) The function $f$ has repulsion degree $k>0$ if for every $p\in N$ $f^i(p)\neq{p}$ for every $i=1,2,\dots,\lfloor{kp}\rfloor$. Determine the maximum repulsion degree can have a circular function.

[b]Note:[/b] Here $\lfloor{x}\rfloor$ is the integer part of $x$.


algebra
functional equation
algebra unsolved

The title explains what question I'd like to ask;

what happens to the Cauchy's Functional Equation $ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$

IF we are given the following condition?

$ f(f(x)) \equal{} x$.

The Cauchy Equation simply says, that $ f$ is $ \mathbb{Q}$-linear.
$ f\circ f \equal{} \text{id}_\mathbb{R}$ says $ f$ has minimalpolynomial $ |x^2 \minus{} 1 \equal{} (x \minus{} 1)(x \plus{} 1)$,
so only eigenvalues $ \pm 1$ and the decomposition $ \mathbb{R} \equal{} \ker (f \minus{} \text{id}_\mathbb{R})\oplus \ker (f \plus{} \text{id}_\mathbb{R})$. (given by $ x \equal{} {x \plus{} f(x)\over 2} \plus{} {x \minus{} f(x)\over 2}$)

Conversely for any decomposition $ \mathbb{R} \equal{} V\oplus W$ with $ \mathbb{Q}$-subspaces $ V,W$ of $ \mathbb{R}$
we get a solution $ f: \mathbb{R}\to \mathbb{R},v \plus{} w\mapsto v \minus{} w$ with $ v\in V,w\in W$,
and every decomposition gives a different $ f$. (as $ V \equal{} \ker (f \minus{} \text{id}_\mathbb{R})$ and $ W \equal{} \ker (f \plus{} \text{id}_\mathbb{R})$)

Now if $ B$ is any $ \mathbb{Q}$-basis of $ \mathbb{R}$ (which exists by [url=http://en.wikipedia.org/wiki/Zorn%27s_Lemma]Zorn's Lemma[/url])
then any disjoint union $ B \equal{} B_1\sqcup B_2$ gives another decomposition $ \mathbb{R} \equal{} V\oplus W$,
where $ V$ is $ \mathbb{Q}$-spanned by $ B_1$ and $ W$ is $ \mathbb{Q}$-spanned by $ B_2$.


parameterization
analytic geometry
geometry
function
probability
algebra unsolved
algebra

Hello,

I've been trying to figure out an index for the economic health of our language center and am going slightly berzerk.  These are the parameters:

- I begin with two percentages.  One (Student Weeks, or SW) tells me the increase in enrollment during a given period (a month) relative to what was expected.  For example, a SW of 11% indicates that there are 11% more students than were originally expected for that month, so in a case like this I would probably have to spend more on books and maybe hire another teacher.  The other is Costs and Margins (CM) , which indicates the change in expenses for that same time period relative to what was originally budgeted.  For example, -3% indicates that we spent 3% less than what we budgeted.  Following the examples above, if your SW were 11% and your CM was -3%, this means that you had more students than expected (11% more) yet you managed to bring your budget under by 3%, which is very good.  You were a good manager that month.  Conversely, if your SW were 11% (same amount) but your CM was 15%, this is not good--your expenses went higher by 4%.  Note that these numbers can go into both positive and negative zones, with limits of 100 & -100.

- What I wish to produce is an index of the center's financial health for that month that is ranged between -1 and 1.  0 would always be the middle point, or breaking even, e.g. if your SW are -4% and your CM are -4% too, then you did your job acceptably because you managed to reduce your expenses at exactly the same rate as your number of students decreased.  The further away you move from the 0 point, your index (we call it the Center Budget Index, or CBI) should approach 1 or -1.  Seems straightforward, but unfortunately it's not.

- If one imagines a set of coordinates with SW as the vertical axis and CM as the horizontal axis, we can quickly position the 0 value area on the diagonal that moves from bottom left through top right (that's the line along which both SW and CM have the same value.)  After that, things become way cloudier for me!  I had tried figuring out what the angle created by the two coordinate points (SW and CM) was, and setting the CBI value based on the angle, but as elegant as that was, it didn't quite work, for the reasons described next.

- If for example we move into the upper left quadrant of the coordinates, and create a diagonal line from the top left corner (which is values SW = 100 & CM = -100, because SW and CM being a percentage change +-100 are the maximum values) down to the 0,0 coordinate, you can immediately see how my angle theory breaks down.  First, it is fairly clear that if your SW are 100 and your CM are -100, you couldn't possibly have done better as a manager.  You are at the extreme top left point of the possible coordinates; the amount of students in your school is way up, and yet you brought your costs down enormously (this would probably never happen in a real-life scenario, but it helps illustrate my point).  You deserve a CBI of 1, the best possible score.  To stay at the extreme edges, neither a SW of 99 and CM of -100 or a SW of 100 and a CM of -99 are quite as good, so the value should decrease along the margin.

- Staying on the diagonal line I created in the upper left quadrant, I'll move down along it closer to 0,0.  Let's say my SW are 25 and my CM are -25.  Like 100 and -100, the values are proportionate.  However, we cannot say that this is the best we could do.  In fact, not even close.  The best you could do is, of course, CM -100 (regardless of the value of SW, really).  In such a case (25 / -100), you did quite well because you cut your costs a lot despite having an increase in students, but it's not as astounding as 100 / -100.  However, would you deserve a 1, because you did the best possible (-100 CM) anyway?  My instinct says yes, but I'm not completely sure.  This is the conundrum.  And eventually, can these values be assigned in a logical, mathematical way that can be reduced to a formula?

If you've got any ideas, please throw them my way.  My expertise is language, not mathematics!  Furthermore, I'd love to plug all this as a formula in excel, if it's possible.

Many thanks.

Try this formula:
$ \frac{4}{\pi} \cdot \arctan(c(SW\minus{}CM))$
where $ c$ is some constant that allows you to adjust how responsive the index is to small differences between $ SW$ and $ CM$.

The best justification I can give is the graph of $ \arctan$ does what you want, and I can't see why anything more than the difference between $ SW$ and $ CM$ is signficant.

Thanks for your posting.  I'm not sure I understand what the value of 'c' should be, though.  Ultimately, what I need is a value between -1 and 1 where 1 is the 'best' and -1 is the worst.

Forgive me, I was using the wrong tangent function in excel--using the coordinates function instead of a number function.  Now I do get a value between -1 and 1.  Still trying to figure out the value of 'c'.

UPDATE: This formula somewhat different from yours is what seems to work best for me in Excel:  =(4/PI())*(ATAN((PI()/10)*(B5-B6)))/2  Don't ask me why I was just tinkering around.  I'd love to know why, actually!

Oh, crap. I meant $ \frac{2}{\pi}$ instead of $ \frac{4}{\pi}$. That's why you have a divided by two at the end. Other than that, it's the same formula with $ c \equal{} \pi/10$.

Another function that is sometimes used in similar circumstances is
\[ \frac{{1.5}^x \minus{} 1}{{1.5}^x \plus{} 1},
\]
where $ x \equal{} SW \minus{} CM$ in your problem.  The base $ 1.5$ can be changed to fit your application.  This function is sometimes called the [url=http://en.wikipedia.org/wiki/Logistic_function]logistic function[/url].  For example, it is used in the usual chess rating systems to compute the probability that a player with one rating will beat a player with another rating.


Let [tex]n \in N^*[/tex]
Find the least number M and largest N such that forall real namber [tex]x_1;x_2;..;x_n[/tex],holds the following
[tex]N(x_1^2+x_2^2+...+x_n^2)\le \sum\limit_{k=1}^n(x_1+x_2+..+x_k)^2\le M(x_1^2+x_2^2+...+x_n^2)[/tex]

i think you mean positive reals: else take x_i = 0 and we have M=N=0.

Otherwise, take x_1 .. x_{n-1} as tending towards zero.  Then everything tends towards $x_n^2$... no clue how to proceed apart from this.


In the plane are given 100 lines such that no 2 are parallel and no 3 meet in a point. The intersection points are marked. Then all the lines and k of the marked points are erased. Given the remained points of intersection for what max k one can reconstruct the lines?

[i]Proposed by A. Golovanov[/i]

Can anyone hepl me with a link at wich i can find the pb form tuymaada (yakutsk) contest 2003 and 2002?

[quote="ciprian"]No solution? No one likes this pb? :?  :([/quote]
Puedes encontrar los problemas de Tuymaada en:
(Review this URL:)

[url=http://www.guas.info/competit/tuyme.htm]www.guas.info/competit/tuyme.htm[/url]

I found these problems in PDF format but I don't remember where!!!

Enrique.

I'm not sure if this is right but in this case three points define a line. Every point denotes an intersection between two lines so if three lines are collinear, there must be a line passing through it. Since two lines share a point each point "contributes" one point to where the line passes through. Since there are 100 lines we need 150 points (since each point contributes points to two lines and each line needs 3) in order to define 100 lines.


dar har khane az yek jadval n*n yeli az adad {1,2,.....,n} neveshte shode ke har adad daghighan n bar amade ast sabet koid satr ya sotuni vojud darad ke hadeaghal $ \sqrt {n}$adad motmayez dar on vojud darad

[quote]ba borhane kholf farz konid dar har satro sotoon,har adad kamtar az $ \sqrt n$ bar oomade bashe...
hala biaid tedade joft a'adade mokhtalef ke too ye satr ya sotoon hastano beshmorid,farz konid in te'dad barabare $ S$ bashe...
khob hala deghat konid ke chon too har satro too har sotoon har adad kamtar az $ \sqrt {n}$ bar oomade,pas too har satro sotoon kamtar az $ \binom{\sqrt {n}}2$ ta azin jofta darim,pas dar majmoo kamtar az $ n\times n\times\binom{\sqrt {n}}2$ ta azin jofta darim yeni:

$ \boxed{S<n^2\binom{\sqrt {n}}2} (I)$

khob hala yebaram biaid intori beshmorid:
ye khoonaro dar nazar begirid,in khoonehe ba $ 2n \minus{} 2$ ta khoone dge ham satro ham sotoone,hala deghat konid ke tebghe farze masale har adad daghighan $ n$ bar oomade,pas az beine in $ 2n \minus{} 2$ ta khoone dge,haddeaghal $ n \minus{} 1$ tashoon ba in adade fargh daran...
pas darim:

$ \boxed{S\geq n\binom{n}2}(II)$

khob hala az $ (I),(II)$ natije mishe:

$ \frac {n^2(n \minus{} 1)}2 < \frac {n^2(\sqrt {n})(\sqrt {n} \minus{} 1)}2$

$ \iff n < 1$

ke inam tanaghoze,pas satr ya sotooni vojood dare ke haddeaghal $ \sqrt n$ adade motamayez dashte bashe...

[/quote]


i got these error messages:

! File ended while scanning use of \frac .

! Emergency stop.

and only the first 2 pages of my file showed up in the dvi file (where's the rest of it?)

You probably forgot a closing } or similar and TeX kept looking for it ('scanning') until it gave up and stopped. Just before it says ![i] File ended while scanning use of \frac[/i] does it say anything about a Runaway argument? The following line might help locate the problem.

If you can't see it easily then try and compile part of the document and see if that works. Use a binary search - compile half the document and that will tell you which half the problem is in, then repeat until you find the offending line.


linear algebra
matrix
combinatorics unsolved
combinatorics

Let $A_{1},A_{2},...,A_{n}$ be subsets of $\{1,2,...,n\}$. For any different $i,j\in \{1,2,...,n\}, i\in A_{j}\leftrightarrow j$ is not in $A_{i}$, and $i$ is not in $A_{i}$ for all $i$. Moreover, there exists a constant $k$ such that $n(A_{i}\cap A_{j})=k$ for all different $i,j$.

Prove or disprove that the numbers of elements in each set $A_{i}$ are all equal.

[quote="Zorro"]what does [quote="Summerburn"]$n(A_{i}\cap A_{j})=k$[/quote] mean??[/quote]

The number of elements in the set $A_{i}\cap A_{j}$ is equal to $k$.

Let $d_{i}: = \{A_{j}| i\in A_{j}\}$. We have $A_{i}+d_{i}=n-1$.
we have $\sum{d_{i}\choose 2}= k{n \choose 2}$ so $\sum A_{i}^{2}= (k+1){ n \choose 2}$. we have also $\sum A_{i}={n \choose 2}$ and $A_{i}\geq k+1$, so $\sqrt{\frac{\sum A_{i}^{2}}{n}}=\frac{\sum A_{i}}{n}$, thus $A_{i}$ must be equal
i hope i didn't make any mistake

[quote="Zorro"]
so $\sum A_{i}^{2}= (k+1){ n \choose 2}$. [/quote]

I think it should be $2k+1$, not $k+1$.  :)

yes:/
so we have $\frac{n}{4}\leq k \leq \frac{n}{2}$ and we have still nothing

Hey, I'm still interested.

Can anybody solve it?

now we have $2k+1\geq \frac{n-1}{2}$, and $A_{i}\geq k+1$ which gives us $\frac{n-3}{2}\geq k \geq \frac{n-3}{4}$
but I think there must be $A_{i}\geq 2k+1$ so we would have $\frac{n-1}{2}\geq 2k+1 \geq \frac{n-1}{2}$
(I cant find a counterexample)

Hmm... this has been unsolved for a long time!

consider set $ A_n$ and assume wlog that $ A_n \equal{} \{1,2,...,a_n\}$. Consider $ B$ matrix such that $ b_{ij} \equal{} 1$ if $ j \in A_i$ and $ 0$ otherwise.
We will count sum of entries $ b_{ij}$ with both $ i,j \leq a_n$. Since $ |A_i \cap A_n| \equal{} k$ we have that this sum is equal $ a_n\cdot k$. On the other hand $ i \in A_j \leftrightarrow j \notin A_i$, so $ a_{ij} \plus{} a_{ji} \equal{} 1$ and $ a_{ii} \equal{} 0$ thus this sum must be equal to $ \frac{a_n (a_n\minus{}1)}{2}$. Finally we have $ a_n\cdot k \equal{} \frac{a_n (a_n\minus{}1)}{2}$, which gives $ a_n \equal{} 2k\plus{}1$. So every set has the same number of elements


I run http://www.ExpertBee.com where people who have a question pay for someone to answer it, and someone posted a query about two problems related to ElectroMagnetic Radiation, and no one has answered.
"Why does the sky overhead appear blue on sunny days? Why do sunrises and sunsets appear red? Why do clouds look white or grey?

 From the Conservation of Energy law, what can happen to EMR energy incident on the Earth? What do you understand by the term 'spectral reflectance'?

A good answer will have a step by step explanation! Please put in your profile how you know!"
Here is a link to the query:
http://www.expertbee.com/auction-query_i12894.html
If you can help him, answer it and make a little money.
Thanks

I can tell about red. It is due to refraction. There is such quality of light if you know. Different frequency lights refract in different ways. The red light appeared to be the most refracted. About the color of light I knew but I dont remember. I'll think a little bit.

sorry man I have just noticed not refraction but disperstion.

you know, i think these questions can be just answered through wikipedia...

if anyone knows how to shut down expertbee.com, please do so!

they have been scamming people for like a year now jeez


induction
inequalities
inequalities solved

The result immediately follows from

$1+\frac{x^{2}}{y}\geq\frac{\left(1+x\right)^{2}}{1+y}$.

I will post a solution.

We use induction on n.
- For n = 2, the desired inequality is equivalent to (x_1 - x_2) 2 (x_1 + x_2)  \geq 0.

- Now suppose that the inaquality holds for a fixed value of n \geq 2.
Let x_1,...,x_(n+1) be any positive real numbers. Since the desired inequality is cyclic, we may assume that x_(n+1) = max{x_i}.
Then :
(1+ (x_1) 2 /x_2 )...(1+ (x_(n+1)) 2 /x_1 )
= (1+ (x_1) 2 /x_2 )...(1+ (x_n) 2 /x_1 )*(1+ (x_n) 2 /x_(n+1) )(1+ (x_(n+1)) 2 /x_1 )/(1 + (x_n) 2 /x_1)
\geq (1+x_1)...(1+x_n)*(1+ (x_n) 2 /x_(n+1) )(1+ (x_(n+1)) 2 /x_1 )/(1 + (x_n) 2 /x_1)  (from the induction hypothesis)

Thus it suffices to prove that :
(1+ (x_n) 2 /x_(n+1) )(1+ (x_(n+1)) 2 /x_1 )/(1 + (x_n) 2 /x_1)  \geq (1+x_(n+1)).
Expanding, this last inequality is equivalent to
(x_1 - x_(n+1))(x_n - x_(n+1))(x_n + x_(n+1))  \geq 0, which clearly holds from our choice of x_(n+1).

Pierre.


Let $\alpha$ be given positive real number, find all the functions $f: N^{+} \rightarrow R$ such that $f(k + m) = f(k) + f(m)$ holds for any positive integers $k$, $m$ satisfying $\alpha m \leq k \leq (\alpha + 1)m$.

it seems very hard and nice;anybody have solution

hi,

here is my solution :

1-we will find all fonction g such that $g(k+m) = g(k)+g(m) (1)$.

lets assume that : $g(1) = \beta$such that $\beta$ is a real

number .

$g(1)=x \Rightarrow  g(2)=2x \dots$ by recurence ,we can prove that $g(x) = ax$

2- $g$ verifies also the second relation .

so $f(x) = g(x)= \beta x$
 ;) .

[quote="toetoe"]
$g(1)=x \Rightarrow  g(2)=2x \dots$ by recurence ,we can prove that $g(x) = ax$
[/quote]

why ? are you sure it is true?

[quote="toetoe"]hi,

here is my solution :

1-we will find all fonction g such that $g(k+m) = g(k)+g(m) (1)$.

lets assume that : $g(1) = \beta$such that $\beta$ is a real

number .

$g(1)=x \Rightarrow  g(2)=2x \dots$ by recurence ,we can prove that $g(x) = ax$

[u][i][b]2- $g$ verifies also the second relation .[/b][/i][/u]

so $f(x) = g(x)= \beta x$
 ;) .[/quote]

your solution is wrong !

http://www.artofproblemsolving.com/community/q1h448277p2524052 ;)
Almost same

[hide = Solution]

We claim that all solutions are $f(x) \equiv cx,$ for $c \in \mathbb{R}.$ Firstly, it's immediate to verify that these functions $f$ work.

Now, we will show that they are the only such functions which work. Let's select some arbitrary large positive integer $m$, and consider the interval $((m+1)\alpha, m(\alpha + 1))$. Select some integer $k$ such that $k , k+1$ are both within this interval. Then, by the condition of problem, we know that $f(m+1+k+1) = f(m+1) + f(k+1), f(m+1+k) = f(m+1)+f(k), f(m+k+1) = f(m) + f(k+1)$, and $f(m+k) = f(m) + f(k).$

Therefore, this gives that $f(m+k+2) - f(m+k+1) = f(k+1) - f(k) = f(m+k+1) - f(m+k).$

Let's call the natural number $t$ $\mathbf{good}$ if $f(t)$ is the average of $f(t-1), f(t+1)$, so that we know that any number in $(m + \alpha (m+1) + 1, m(\alpha +1) + m)$ is good, for all $m$ sufficiently large. However, it's clear that all sufficiently  large numbers are contained in an interval of that form, and so hence we conclude that all sufficiently large numbers are good, say $t \ge C$ are all good.

Then, we know that $f(C), f(C+1), f(C+2), \cdots$ is an arithmetic progression, and so if we apply the condition with $m, k \ge C$, we know that $f \equiv cx$ for some $c \in \mathbb{R}$ when $x \ge C.$

Now, it simply remains to use downwards induction to show that $f \equiv cx, \forall x \in \mathbb{R}.$ (Just set $m =k$ in the condition of the problem)

$\square$

[/hide]

Quite late to post a solution, but the problem is amazing so I'll do it anyways  :D

Call an integer $n$ [i]separable[/i] if there exist two integers $a,b$ such that $a+b=n$ and $f(a+b)=f(a)+f(b)$. We will prove all sufficiently large integers are separable: it is enough to have $\alpha \leq \frac{a}{b} \leq \alpha +1 \iff \alpha+1 \leq \frac{a+b}{b} \leq \alpha +2 \iff \frac{1}{\alpha +1} \leq \frac{b}{n} \leq \frac{1}{\alpha +2}$. Start with $\frac{1}{n}$ and add $\frac{1}{n}$ to it in every step, and as long as $\frac{1}{n}$ is smaller that the difference between $\frac{1}{\alpha +2}$ and $\frac{1}{\alpha +2}$ we can't jump over the interval $(\frac{1}{\alpha +2}, \frac{1}{\alpha +1})$ so such integer is separable by the problem condition.

We use the above claim to conclude that $f(n)$ can be represented as a linear combination of $f(1), f(2), ..., f(k)$ for some fixed $k$ (the $f$-s of $1$ through $k$ being just themselves). Now consider the smallest subset of these such that the linear representation using those is [b]unique [/b](sorry for probable abuse of terminology), and write $f(n)=g_1(n)f(a_1)+...+g_m(n)f(a_m)$ where $\{a_1, a_2, ..., a_m\} \subset \{1,2,...,k\}$. By the uniqueness of the linear representation, we have that the problem condition carries over to the $g_i$ functions (you get the reasoning by thinking of $f$ as a vector field). But we have the functions $g_i$ are from the integers to rationals (I am skipping a bit of reasoning but it should make sense), and every rational can be represented as a linear combination of other rationals, and combine that with the first paragraph we get that $g_i$s have to be linear combinations of just one $g_i(something)$, so it has to take form $g_i(n)=nc_i$ where $c_i$ is constant.

So we conclude we can represent $f$ in a comfortable way: $f(n)=n(c_1f(a_1)+...+c_mf(a_m)) = nD$ where $D=const.$, this being true for all $n>k$(although we can be done here it is really hard to explain why).  Now for some final touches, divide the problem into two cases: $\alpha \leq1 $ and $\alpha >1$. In the first case we have $f(2n)=2f(n)$ and from here just choose any $n$ and really big $k$ so that $2^kf(n)=f(2^kn)=2^knD \implies f(n)=nD$. For the second case, choose an integer $\alpha \leq a \leq \alpha+1 \implies a>1$. We have $f((a+1)n)=f(n)+f(an)$ and choose $n\in (k/2, k]$ to expand the $f(n)=nD$ to number less that $k$ in an obvious manner, and so we are done $\blacksquare$

I again apologize to anyone who might read this for the terrible explanation and terminology, my aim was to make it fell natural :D


Is there a family of subsets of [i]N[/i] with a cardinality of continuum which is totally ordered with inclusion?[/code]

Yes. Hint: think of $ \mathbb Q$ instead of $ \mathbb N$ (these sets work the same for your purposes, since there is a bijection between them).

For each $ y\in\mathbb{R},$ let $ E_y\equal{}\{x\in\mathbb{Q}\mid x<y\}.$

If $ y<z,$ then $ E_y\subset E_z$ and $ E_y\ne E_z.$ That's a total order.

This works since you can show $ \mathbb{Q}$ as being in a one-to-one correspondence with $ \mathbb{N}.$


hallo!
 Fände es toll wenn wir hier über die aufgaben der 2. runde diskutieren könnten. Hier sind meine Ergebnisse:

Aufgabe1: $ \minus{}\sqrt{2}$,$ 0$ ,$ \sqrt{2}$
Aufgabe2: widerspruchsbeweis
Aufgabe3: Symmetrie, Spiegelung sämtlicher ebenen
Aufgabe4: $ 6 \cdot 4^{n\minus{}3}$

Bin gespannt auf eure ergebnisse...

Bitte warte noch bis die Aufgaben auf der [url=http://www.bundeswettbewerb-mathematik.de/aufgaben/aufgaben.htm]offiziellen Web-Seite[/url] sind. Es gibt dazu auch einen Beitrag auf dem [url=http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=92728&start=200]Matheplaneten.[/url]


geometry trigonometry function calculus derivative

The largest side of a triangle is $100$, and its largest angle is double that of another angle. What is the triangle's maximum area?

If I didn't make some mistakes, then I think the maximum area of the triangle is $\frac{a^{2}}{4}\cdot\sqrt{-9+6\sqrt 3}$, where $a=100\ .$

[quote="Altheman"][quote="cincodemayo5590"][quote="Virgil Nicula"]If I didn't make some mistakes, then I think the maximum area of the triangle is $\frac{a^{2}}{4}\cdot\left(3\sqrt 2-\sqrt 6\right)$, where $a=100\ .$[/quote]

How did you find your answer? (I am trying one method, but I am stuck on maximizing a trigonometric expression)[/quote]

would that expression happen to be

$\frac{a^{2}}{2}\cdot \frac{\sin x\sin 3x}{\sin 2x}$

with $36<x<60$[/quote]

Yes.  :P

[quote="cincodemayo5590"][/quote][quote="Altheman"][quote="cincodemayo5590"][quote="Virgil Nicula"]If I didn't make  mistakes, then I think the maximum area of $S$ the triangle is $S=\frac{a^{2}}{4}\cdot \sqrt{-9+6\sqrt 3}$, with $a=100\ .$[/quote]How did you find your answer? (I am trying one method, but I am stuck on maximizing a trigonometric expression)[/quote]Would that expression happen to be $\frac{a^{2}}{2}\cdot \frac{\sin x\sin 3x}{\sin 2x}$ with $36<x<60$[/quote][quote="cincodemayo5590"]Yes.  :P[/quote][color=darkblue]Here is  how I calculated the maximum value of the area of the triangle $ABC$, i.e. $S=\frac{a^{2}}{4}\cdot f(x)$, where the function is $f: \left[\frac{\pi}{5},\frac{\pi}{3}\right)\rightarrow \mathbb R\ ,\ f(x)=\frac{\sin 3x}{\cos x}\ .$ Obtain $f(x)=\frac{\sin x\cos 2x+\sin 2x\cos x}{\cos x}=\tan x\cos 2x+\sin 2x \ .$ From the remarkable formulas $\tan x=t\Longrightarrow$ $\sin 2x=\frac{2t}{1+t^{2}}$ and $\cos 2x=\frac{1-t^{2}}{1+t^{2}}$, using the substitution $\tan x=t\in \left[\tan\frac{\pi}{5},\sqrt 3\right)$ obtain $g(t)=f(\arctan t)=\frac{3t-t^{3}}{1+t^{2}}$ and derivatives - the maximum point is $t_{0}=\sqrt{2\sqrt 3-3}$ and $g(t_{0})=\sqrt{-9+\sqrt 3}$a.s.o.
[b]Remark.[/b] $\cos 2t_{0}=\frac{1-t^{2}_{0}}{1+t^{2}_{0}}=\frac{\sqrt 3-1}{2}$, i.e. $A=\arccos\frac{\sqrt 3-1}{2}\ .$

Know somebody solve this problem without derivatives ? Sorry, I didn't succeed.[/color]


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a}{1+ab^{2}}+\frac{b}{1+bc^{2}}+\frac{c}{1+ca^{2}}\ge \frac{3\sqrt[3]{abc}}{1+abc}.\]

Let $ LS$ be the left side of the inequality and $ p=abc$. Then,

$ LS=\sum\frac{ac}{c+pb}\ge \frac{(\sum a\sqrt{c})^{2}}{\sum a(c+pb)}=\frac{(\sum a\sqrt{c})^{2}}{(1+p)\sum ab}$.

The proof reduces to show that $ abc=1$ implies
$ \frac a{b}+\frac b{c}+\frac c{a}\ge \sqrt{3(a^{2}+b^{2}+c^{2})}$.
I think this is a known inequality.

[quote="NguyenDungTN"]
+ $ a \ge b \ge c$
Similarly, we have $ \frac{a-b}{1+ab^{2}}+\frac{b-c}{1+bc^{2}}+\frac{c-a}{1+ca^{2}}\ge 0$[/quote]
Unfortunately, this is not true: try $ a=2, b=1$ and $ c=1/10$.

[quote="10maths_tp0609"][quote="hungkhtn"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a}{1+ab^{2}}+\frac{b}{1+bc^{2}}+\frac{c}{1+ca^{2}}\ge \frac{3\sqrt[3]{abc}}{1+abc}. \]
[/quote]

Let $ a=\frac{kx}{y}$, $ b=\frac{ky}{z}$ $ c=\frac{kz}{x}$
The inequality is equivalent to:
$ \sum_{cyc}\frac{k^{3}x^{2}}{y(k^{2}x+z)}\ge \frac{3k^{3}}{1+k^{2}}$
Applying Cauchy-Schwarz inequality:
$ LHS \ge \frac{k^{3}(x+y+z)^{2}}{(k^{2}+1)(xy+yz+zx)}\ge \frac{3k^{3}}{1+k^{2}}$[/quote]

Dear Brother, I think your solution is not true. Please check again,

My solution is using VASC's famous result

\[ (x^{2}+y^{2}+z^{2})^{2}\ge 3(x^{3}y+y^{3}z+z^{3}x),\]

but anyway, VASC's solution to this one is very nice. So It is not necessary to post my own here.


linear algebra
matrix
linear algebra unsolved

Here is a very beautiful result: suppose $A_1,...,A_n$ are $2x2$ real matrices such that for all $x_1,...,x_n$ real numbers, the matrix $x_1A_1+...+x_n A_n$ is diagonalisable. Prove that there is a matrix $Q$ such that all matrices $QA_j Q^{-1}$ are symmetric.
  Non posed question: does the result remain true in higher dimensions? What is the maximal dimension of a subspace of diagonalisable matrices (it would not wonder me to see this at oral 2006).

Do you have a nice proof, harazi? All I did was to perform some rather boring computations (not long or anything like that, but nothing out of the ordinary) :(.

Unfortunately, no. I also made some computations with 2 and 3 matrices and got the result. I posted the problem so that probbaly comes up with a nice idea that could lead to a generalization in higher dimensions.

I did reduce it to the case when we have only two matrices as follows:

The space generated by our matrices must clearly have dimension $<4$. Suppose the dimension is $3$, and assume the space does not contain the identity matrix. This means that every $2\times 2$ matrix has the form $A+xI$, where $x$ is a real and $A$ is diagonalizable, which is absurd (all such matrices are diagonalizable), so if the dimension is $3$, then the space contains $I$, we can complete $\{I\}$ to a basis, and we can restrict our attention to the other two matrices in the basis.

Let $A_1,\cdots, A_n\in M_n(K)$  where $K$ is any field and $characteristic(K)\geq n$. Suppose that, for every $(x_i)\in K^n$, $\sum_ix_iA_i$ is diagonalizable over $K$.
Ref [1] shows that if $n=2$, then $A_1A_2=A_2A_1$. It is easy to deduce that the$(A_i)$ pairwise commute.
 Thus the $(A_i)$ are silmultaneously diagonalizable over $K$.
Of course, when $K=\mathbb{R}$, the diagonal matrices are symmetric matrices.

[1] T.S. Motzkin, O. Taussky, Pairs of Matrices with Property L. II, Trans. Amer. Math. Soc. 80 (1955) 387-401,

[quote=harazi] What is the maximal dimension of a subspace of diagonalisable matrices [/quote]

I can answer this (over $\mathbb R$): $\frac{n(n+1)}2$

Indeed, symmetric matrices are an example, and each subspace of $M_n(\mathbb{R})$ of strictly greater dimension contains a non-zero antisymmetric matrix.


geometry
cyclic quadrilateral
geometry solved

Let $ABC$ be an acute triangle. The altitude from $B$ meets the circle with diameter $AC$ at points $P$,$Q$, and the altitude from $C$ meets the circle with diameter $AB$ at $M$,$N$. Prove that the points $M$,$N$, $P$,$Q$ lie on a circle.

Let the orthocentre be $H$ and let $AD \perp BC$ where $D \in BC$. Hence $D$ lies on both of the circles. Since $MN$ and $PQ$ all pass through $H$, hence by the power of point theorem, we have $PH \cdot QH=DH \cdot AH=MH \cdot NH$, therefore $M$, $N$, $P$, $Q$ lie on the same circle.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=42613 .

  darij

Let D, E, F, the traces of altitudes AH, BH, CH ( where H is the orthocenter of ABC ), on BC, AC, AB, respectively.

     From cyclic quadrilateral BFEC and the right triangles AQC, ANB,
( where Q, N, outwardly to ABC ), we have that:

     (AQ)² = (AE)·(AC) = (AF)·(AB) = (AN)², =>  AQ = AN.

     So, the circucircle of M, N, P, Q, is centered at vertex A, of ABC
( because of AM = AN and AP = AQ ). This is also a simple result, but useful for the proof of the result that the intersection point A', of the segment lines QM, NP, lies on the sideline BC of ABC.

     If now, we denote similarly the points B', C' ( by the intersection points R, S, of the segment line AD, from the circle with diameter BC ), on AC, AB respectively, then there is another interesting and more difficult result, that the lines AA', BB', CC', are concurrent.

     In this configuration, of a given triangle with it’s altitudes and their intersection points from the circles with diameters it’s sidelines, we can find many results, some of them interesting.

     Kostas Vittas.

[color=darkblue][b]A similar problem.[/b] Let $H$ be the orthocenter of the triangle $ABC$ and let $M\in AB$, $N\in AC$ be any points so that $H\in MN$.
Denote the points $U\in HB$, $V\in HC$ so that $MU\perp MN$ and $NV\perp MN\ .$ Prove that $UV\parallel BC\ .$[/color]

We denote as E, F, the traces of the altitudes BH, CH of ABC, on the sidelines AC, AB, respectively.

     It is easy to prove that the quadrilateral MFNV is cyclic ( from the perpendicularities of the lines MN, NV and AB, CF ). So we have that: (HM)·(HN) = (HF)·(HV),  (1).

     Similarly from the cyclic quadrilateral MENU, we have that: (HM)·(HN) = (HE)·(HU),  (2).

     From the cyclic quadrilateral BFEC, => (HE)·(HB) = (HF)·(HC),  (3).

     From (1), (2), (3),  =>  (HU) / (HB) = (HV) / (HC),  =>  UV // BC and the proof is completed.

     Kostas Vittas.


Let $ a_i > 0$ for $ i \equal{} 1,2,\cdots,n$. Show that
\[ 2\prod\limits_{i \equal{} 1}^n\sqrt [2n]{a_i} \plus{} \prod\limits_{i \equal{} 1}^n\sqrt [n]{(\sqrt {a_i} \minus{} 1)^2}\le \prod\limits_{i \equal{} 1}^n\sqrt [n]{a_i \plus{} 1},\]

Sorry, this is a known result. So don't waste your time to it. Sorry. :blush:


An interesting problem; try to find a nice solution without brute forcing it.

Compute the sum of the distinct prime factors of $292677$.

if $n= \prod^{k}_{i=1}p_{i}^{\alpha_{i}}$ ( n is a positive interger)
then the sum of its positive divisors is $\sum^{k}_{i=1}\frac{p_{i}^{\alpha_{i}+1}-1}{p_{i}-1}$

[hide]i dont know if it's "brute forcing" or not  :P [/hide]

Note that he asked for the sum of its [b]prime[/b] factors.  The problem is to factor without brute forcing.

[quote="cosinator"]Well, I notice that $292677=30000-7323=3(10000-2441)=3(10^{4}-7^{4})$

(if my arithmetic is right)[/quote]Hmm, interesting.  Not my solution though.  [hide="Hint"]$541^{2}$[/hide]

Well technically that is brute forcing since the general solution to the problem (of prime factorization) looks for differences of squares also. However, it isn't really because it's specially tailored to this problem :P

[quote="10000th User"][quote="cosinator"]Well, I notice that $292677=30000-7323=3(10000-2441)=3(10^{4}-7^{4})$

(if my arithmetic is right)[/quote] $2441\neq7^{4}$ and 292677 is a 6-digit number[/quote]

Darn my mental arithmetic!


factorial
calculus
integration
Support
inequalities
number theory open
number theory

Solve for $ a,b$ integers:

$ (a \plus{} b)! \equal{} a!b!\cdot 2008$
If there are more than $ 1$ solutions, write all, and show the minimal solution (for $ a,b$ minimal).

Thank you.
And please, stick to the elementary methods.

[hide]$ b\not \equal{} 0$ or $ a\not \equal{}0$

If $ b\equal{}0$ then $ a!\equal{} a! 2008 \implies 2008\equal{} 1$ ,it is impossible . And so impossible for $ a\equal{}0.$

If $ b\equal{}1$ ,then $ (a\plus{}1)!\equal{}a! 2008  \implies  a\plus{}1\equal{}2008 \implies a\equal{}2007 . (a,b)\equal{}(2007,1) or (1,2007).$

 Of course ,$ b\equal{}2 ,b\equal{} 3 ,b\equal{}4,$........$ a$ also has certainly corresponding possible integral vaules. You may find it.[/hide]

[hide="constraint for a"]WLOG $ a\le b$,
$ (a \plus{} b)! \equal{} a!b!\cdot 2008$
$ (a\plus{}1)(a\plus{}2)\cdots(a\plus{}b)\equal{}2008 b!$
$ a^b \cdot b! <(a\plus{}1)(a\plus{}2)\cdots(a\plus{}b)\equal{}2008 b!$
$ a^a < a^b <2008$
$ a < 5$[/hide]

[quote="vinskman"]

 Of course ,$ b \equal{} 2 ,b \equal{} 3 ,b \equal{} 4,$........$ a$ also has certainly corresponding possible integral vaules. You may find it.[/hide][/quote]

I don't think that for every b there exists a. Maybe mszew is right.

$ a^b \cdot b! <(a\plus{}1)(a\plus{}2)\cdots(a\plus{}b)$

msew,
How do you support?
Please show it.

[quote="PKMathew"]$ a^b \cdot b! < (a \plus{} 1)(a \plus{} 2)\cdots(a \plus{} b)$

msew,
How do you support?
Please show it.[/quote]

$ a^b \cdot b! < (a \plus{} 1)(a \plus{} 2)\cdots(a \plus{} b)$

$ \text{when } b\equal{}4 ,a\equal{}3$

$ a^b\equal{} 81, b!\equal{} 24 , a^b b!\equal{}81. 24\equal{}1944$

$ (a\plus{}1)(a\plus{}2)(a\plus{}3)(a\plus{}4)\equal{} 4.5.6.7\equal{}840$

The suggested inequality does not hold in such condition!!

[quote="mszew"][hide="constraint for a"]WLOG $ a\le b$,
$ (a \plus{} b)! \equal{} a!b!\cdot 2008$
$ (a \plus{} 1)(a \plus{} 2)\cdots(a \plus{} b) \equal{} 2008 b!$
$ a^b \cdot b! < (a \plus{} 1)(a \plus{} 2)\cdots(a \plus{} b) \equal{} 2008 b!$
$ a^a < a^b < 2008$
$ a < 5$[/hide][/quote]
Sorry about the mistake, confused + and .

Here comes with the proper edition:
$ (a \plus{} 1)(a \plus{} 2)\cdots(a \plus{} b) \equal{} 2008 b!$
$ \frac{(a \plus{} 1)}{1}\frac{(a \plus{} 2)}{2}\cdots\frac{(a \plus{} b)}{b} \equal{} 2008$
$ (\frac{a}{1}\plus{}1)(\frac{a}{2}\plus{}1)\cdots(\frac{a}{b}\plus{}1) \equal{} 2008$ because $ a\le b$
$ (\frac{a}{1}\plus{}1)(\frac{a}{2}\plus{}1)\cdots(\frac{a}{a}\plus{}1) \le(\frac{a}{1}\plus{}1)(\frac{a}{2}\plus{}1)\cdots(\frac{a}{b}\plus{}1) \equal{} 2008$
Noticing that $ 2 \le (\frac{a}{i}\plus{}1)$ for $ 1\le i \le a$ then $ 2^a< 2008$
$ a<11$


geometry
trigonometry
AMC
AIME
circumcircle
trig identities
Law of Sines

Let $ ABC$ be a triangle.$ D$ is midpoint of $ BC$
Suppose that $ \angle BAD\equal{}\angle ACB;\angle DAC\equal{}15^o$
Caculate $ \angle ACB\equal{}?$

Unfortunately, I don't remember my triangle geometry too well, but doesn't the fact that $ D$ is a bisector mean that $ \angle BAD\equal{}\angle DAC$?

Nope...

[b]Example:[/b] Consider the triangle with vertices A(0,0), B(1,0), C(1,2k), and midpoint of BC D(1,k).

As k increases indefinitely, <BAC approaches a right angle. However, <BAD does not approach 45 degrees, it approaches 90 degrees as well.

[hide="Hint"]
[asy]pointpen = black; pathpen = black + linewidth(0.7);
pair A=(0,0), B=(5,0), C= B + (5*2^.5)*expi(75 * pi/180), D=(B+C)/2;
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);
D(A--D(MP("D",D,NE)));
D(anglemark(B,A,D,20));D(anglemark(A,C,B,20)); D(anglemark(C,B,A,20)); D(anglemark(C,B,A,14));
MP("15^{\circ}",A,(8.5,7.2)); MP("\theta",C,(-2.5,-4));[/asy]

Let $ \theta \equal{} \angle ACB$. Notice that $ \triangle DAB \sim \triangle ACB \Longleftrightarrow \frac {BD}{AB} \equal{} \frac {AB}{BC} \equal{} \frac {AB}{2BD} \Longrightarrow \sqrt {2} BD \equal{} AB$. By the Law of Sines on $ \triangle ABC$, we have
\[ \frac {\sin \theta}{AB} \equal{} \frac {\sin (\theta \plus{} 15^{\circ})}{2BD} \Longrightarrow \frac {\sin (\theta \plus{} 15)}{\sin \theta} \equal{} \frac {2BD}{AB} \equal{} \sqrt {2} \Longrightarrow \sqrt {2} \sin \theta \equal{} \sin (\theta \plus{} 15)
\]
[/hide][hide="Answer"]
$ \theta \equal{} 30^{\circ}$.
[/hide]

[hide="Comment"]I found the trig solution as well, but I feel like there's got to be something synthetic. There was a similar AIME problem (2004 I think) with a quadrilateral that involved constructing an angle bisector but I couldn't do that here.[/hide]

From $\angle BAD=\angle BCA$ it yields that $AB$ is tangent to the circle $\odot ACD$, i.e. $AB^2= BD\cdot BC\iff 2AB^2=BC^2$.
Take $E$ the midpoint of $AC$, $DE\parallel AB\implies\angle ADE=\angle BAD=\angle BCA$, or $AD$ tangent to the circle $\odot CED$, so $AD^2=AE\cdot AC\iff 2AD^2=AC^2\ (\ *\ )$.
Construct the right-isosceles triangle $\triangle AOC$, $AO=OC$, $AC$ separates $O$ and $D$. With $(\#)$ it follows that $AO=AD$ and, with $\angle CAD=15^\circ\implies \angle OAD=60^\circ$ and $\triangle OAD$ is equilateral. It follows that $OA=OD=OC$, i.e. $O$ is the circumcenter of $\triangle DAC$, and, comsequently $m(\widehat{ACD})=\frac {m(\widehat{AOD})}{2}=30^\circ$.

Best regards,
sunken rock

[u]Nice your proof[/u], [b]Sunken Rock[/b] ! [u]See[/u] [b]PP7[/b] from [url=http://www.artofproblemsolving.com/blog/69176][color=darkred][b]here[/b][/color][/url].

one more different solution...

What about applying m-n theorem...!! :)

let the bisected parts equal m

so,

$-2mcot(\theta + 15^0) = mcot\theta - mcot15^0 ---- (1)$ and,
$-2mcot(\theta + 15^0) = -mcot(2\theta + 15^0) - mcot\theta ---- (2)$

(1) - (2) gives,

$cot(2\theta + 15^0) + 2cot\theta  - cot15^0 = 0$

or, $2cot\theta + \frac{1-tan2\theta.tan15^0}{tan2\theta + tan15^0} - cot15^0 = 0$

on further simplifications.. the eqn. reduces to..

$tan^2\theta(6-\sqrt{3}) - 2tan\theta + \sqrt{3}-2 = 0$

and then obviously, a piece of cake ...

$tan\theta = \frac{1 \pm 2\sqrt{(\sqrt{3}-1)^2}}{6-\sqrt{3}}$ (on furthur simplifications after applying shree Dharacharya's rule..) :P

finally ignoring the -ve sign, gives

$tan\theta = \frac{1}{\sqrt{3}}$ or $\theta = 30^0$


A person throws darts at a dartboard. They have $5$ turns. They can get $1$, $2$, $4$, or $6$ points per turn. What is the smallest score that is impossible for them to get?

There is a long, brute-force method to do this, but I want to see an short, elegant method since you need to go as fast as possible on sprint round problems.

I remember this problem was posted a long time ago, I don't think there's a non-brute force way to do it.

Wasn't this a target on this year's chapter?

[quote="Ubemaya"]yeah it was.

#7 I think.[/quote]

#8, 2006...

Check the mathcounts website, they post their solutions to the chapter and state problems.

Yeah, they really help...but if they had nationals....

[hide]
8. Eli throws five darts at a circular
target as shown and each one lands
within one of the four regions.
The point values of the red (outer),
blue, green and yellow (inner) areas
are 1, 2, 4, and 6 points,
respectively. We are asked to find
the least score greater than five
points that is not possible when the
point values of the five darts are
added together.
What values can we make with only
5 darts?
6 = 1 + 1 + 1 + 1 + 2
7 = 1 + 1 + 1 + 2 + 2
8 = 1 + 1 + 2 + 2 + 2
9 = 1 + 2 + 2 + 2 + 2
10 = 2 + 2 + 2 + 2 + 2
11 = 4 + 1 + 2 + 2 + 2
12 = 4 + 2 + 2 + 2 + 2
13 = 4 + 4 + 1 + 2 + 2
14 = 4 + 4 + 2 + 2 + 2
15 = 4 + 4 + 4 + 1 + 2
16 = 4 + 4 + 4 + 2 + 2
17 = 4 + 4 + 4 + 4 + 1
18 = 4 + 4 + 4 + 4 + 2
19 = 4 + 4 + 4 + 1 + 6
20 = 4 + 4 + 4 + 2 + 6
21 = 4 + 4 + 1 + 6 + 6
22 = 4 + 4 + 2 + 6 + 6
23 = 4 + 1 + 6 + 6 + 6
24 = 4 + 2 + 6 + 6 + 6
25 = 1 + 6 + 6 + 6 + 6
26 = 2 + 6 + 6 + 6 + 6
27 = ?
27 is the first one that we can’t
make. Note that there is a limit to
the checking. We can never get to
a point value of 31.
Note: A pattern emerges
making the counting easier. Every 1
can change to a 2 to create the next
number; then every 2,2 can change
to a 4,1 for the next number; every
4,1 can change to a 4,2 for the next
number, every 4,2 can change to a
6,1 for the next number; and every
6,1 can change to a 6, 2 for the next
number. Seeing this pattern makes
the listing of the possibilities easier.
[b]27 Ans.[/b][/hide]

-jorian

The awesome solution by Ravi B
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=449079&%2A#449079


I think that Greg would win. There's a chance that Neal would be fast enough (I wasn't at Nationals this year, I don't know just how fast he is) on easy problems to win, but I'm sure that Greg is more intelligent, and he would definitely win if the problems were a little more difficult than standard Countdown problems.

This is also assuming same age; Greg would obviously have the advantage if they faced each other at the ages they are now. :P

NOT ANYMORE *clicks vote button* Oh wait, I already voted  :?

They'll be facing off enough in high school. Let's watch for them then :) .

Greg easily. Neal is fast, but Greg is AMAZING

[quote="nat mc"]If Neal Wu and Gregory Gauthier faced off in countdown, who would win.  I think Neal would.  He is super fast.  Gregory was fast but he was too precise to be fast enough.[/quote]

Did you see last year's Countdown? I'm quite sure Greg would win- he was fast AND precise. I think only one person buzzed in before him-once- and got it wrong. And Neal wasn't THAT fast or consistent on every problem (i.e. last one). Next year might be a different story though.

Well, for the last one, why would he be stupid and risk the championship. If you're opponent got it wrong wouldn't it make sense to wait it out and check your answer a lot?

That was only an example. But still, a lot of times competitors are so sure of their answers that they really don't need to check- and in this case, the problem was pretty easy.

[quote="jasonho2004"]<Editor's note to above> I am neither Neal Wu nor Greg Gauthier. I sucked in chapter, state, and especially nationals. You don't even want to know my score for Nationals.[/quote]

Bah. You spent too much time being an un-pimp :P .

[quote="jasonho2004"]<Editor's note to above> I am neither Neal Wu nor Greg Gauthier. I sucked in chapter, state, and especially nationals. You don't even want to know my score for Nationals.[/quote]
Let me guess.  15?


Find all ordered pairs of positive integers $ (a,b)$ such that $ \binom{a\plus{}b}{a}$ is prime.

[hide="Solution"][b]Lemma:[/b]  $ {n\plus{}1 \choose k\plus{}1} \equal{} \frac{n\plus{}1}{k\plus{}1} {n \choose k}$.

Suppose $ {n\plus{}1 \choose k\plus{}1} \equal{} p$ is prime.  Then $ (k\plus{}1)p \equal{} (n\plus{}1) {n \choose k}$.

[b]Case:[/b]  $ p | n\plus{}1$.  If $ k \neq 0, n$ then $ {n \choose k} \ge n \ge k\plus{}1$, so it follows that $ p \equal{} n\plus{}1$.  Then $ {n \choose k} \equal{} k\plus{}1$, which occurs only when $ k \equal{} 1$.  But $ {p \choose 2}$ is prime only when $ p \equal{} 3$.  If $ k \equal{} 0, n$ then $ {p \choose 1}$ is always prime.

[b]Case:[/b]  $ p | {n \choose k}$.  But we know that $ n \plus{} 1 > k \plus{} 1$, so this is absurd. [/hide]

[hide="Uses a similar idea"]Prove by induction or by the factorial formula that for $ n \geq 4$, if $ 2 \leq k \leq n \minus{} 2$, then $ {n \choose k} > n$. Use the factorial formula to show $ {n \choose k}$ can't have prime factors higher than $ n$.[/hide]


Stie cineva informatii in legatura cu pregatirea de la Bucuresti?

Am primit acum f putin timp un email de la domnul Gologan. Incerc sa anunt pe toata lumea.
[quote] Bacalaureatul olimpicilor 19 mai-29 mai
Pregatire I OIM:
6-13 mai Bucuresti; 2 baraje pe 12-13 mai cu cate 4 probleme, lotul se injumatateste
Pregatire II OIM:
10-19 iunie. Baraj geometrei (3 probleme) 11 iunie, baraj tip OIM  16-17 iunie cu corectare tip OIM intre voi (pe copiile xerox facute cu voi de fata)  si apoi cu coordonarea de catre profesori cu elevii de fata.
2-7 iulie tabara cu Ungaria la Debretin sau Poiana Pinului cu lotul de 6 plus +1 + cei 3 ce merg in Iacutia.
[/quote]


Let $ x\in \mathbb R$. Find the minimum and maximum values of the expresion:

$ E\equal{}\dfrac{(1\plus{}x)^8\plus{}16x^4}{(1\plus{}x^2)^4}$

Let $ x \equal{} \tan \theta$, we have $ E \equal{} (\cos \theta \plus{} \sin \theta)^8 \plus{} 16\sin ^ 4 \theta \cos ^ 4 \theta$, and let $ t \equal{} \sin 2\theta$,

we obtain $ E \equal{} (1 \plus{} t)^4 \plus{} t^4$ with $ \minus{} 1\leq t\leq 1$.

Just to point out that $2^4 + 1^4 = 17$, not $33$.


[quote="ueamath"]A mass [i]m[/i] is suspended from the ceiling by a spring of natural length [i]l[/i] and stiffness [i]k[/i]. When the mass is displaced a distance [i]z[/i] from its equillibrium position, the equation governing its subsequent motion is given by
$(z'' + (k/m)z =0$ (*)

The potential energy in the spring is $kz^2/2$ . The kinetic energy of the mass is $m(z')^2?\"$> mULTIPLY EQUATION (*) by z@ and integrate withe respect to time to show that

Potential Energy + Kineetic energy + constant
YOUR HELP WILL BE APPRECIATED

Thanks[/i][/quote]

$z'' + \frac{k}{m}z =0$
$z'*z'' + \frac{k}{m}z*z' =0$
$\frac{z'^2}{2} + \frac{\frac{k}{m}z^2}{2} =0$
$\frac{mz'^2}{2} + \frac{kz^2}{2} =0$

Done.


Determine all real solutions of this equation :

$ 2(x^4\minus{}2x^2\plus{}3)(y^4\minus{}3y^2\plus{}4)\equal{}7$

[quote="dgreenb801"][hide]That's $ 2[(x^2 \minus{} 1)^2 \plus{} 2][(y^2 \minus{} \frac {3}{2})^2 \plus{} \frac {7}{4}] \equal{} 7$
Each of the brackets is at least 2 and 7/4. Equality holds when $ x \equal{} 1$, $ y \equal{} \sqrt {\frac {3}{2}}$.[/hide][/quote]

$ x\equal{}\pm 1$ and $ y\equal{}\pm \sqrt{\frac{3}{2}}$
so and these solutions:
$ (\minus{}1,\sqrt{\frac{3}{2}}),(1,\minus{}\sqrt{\frac{3}{2}}),(\minus{}1,\minus{}\sqrt{\frac{3}{2}})$


geometry
3D geometry
quadratics
calculus
integration
modular arithmetic
LaTeX

Find all $x;y \in Z$ such that:
$54x^{3}+1=y^{3}$

[hide]Besides the trivial (0,1), there aren't any other solutions.
Look at this mod 54.  It has an 18 series cycle, which means that y=18z+1.
Now subtract one and factor the difference of cubes and replace y with z.
After complete simplification, $x^{3}=108z^{3}+18z^{2}+z$.
Now, for this expression to equal a cube, $18z^{2}+z$ must equal $(a^{3}-108)z^{3}$.
To solve, we set it as z($(a^{3}-108)z^{2}-18z-1)=0$ and solve for the quadratic since we know z=0 works.
Via the quadratic formula, $\frac{9\pm\sqrt{a^{3}-27}}{a^{3}-108}=z$, but the minus conjugates are irrelevant for the least is -1/9 and tends to 0.
$a$ must be an integer so x=az.  $z$ is an integer because $x^{3}$ is integral and since the only possible non-integer denominators are the factors of 18 except 1 and plugging them into the equation concerning z and x gives that $108z^{2}+18z+1$ must be a factor of the denominator.  For 9 and 18, it obviously fails, and for 2, 3, and 6, the resulting integer isn't a multiple of the denominator.  Thus, z is an integer.
Now, seeing that a=3 gives z=-1/9, while a=4, 5 are irrational and a=6 is less than one, we conclude that no a is possible, which means that no z is possible except when z=0, for at that point, the quadratic will yield 0/0 because $a=108^{1/3}$.[/hide]
My solution isn't elegant.  Does anyone have a better idea?

[hide="a start?"]
Rewrite as
\[54x^{3}=(y-1)(y^{2}+y+1)\]
Now observe that $y^{2}+y+1$ is odd, so 2 divides $y-1$. Also, $y^{2}+y+1\equiv (y-1)^{2}\pmod{3}$. Since either $y-1$ or $y^{2}+y+1$ is divisible by 3, both are. Finally, $x^{2}+x+1$ is never divisible by 9, so 9 must divide $y-1$. We end up with
\[x^{3}=\left( \frac{y-1}{18}\right) \left( \frac{y^{2}+y+1}{3}\right)\]

But any common divisor of $y-1$ and $y^{2}+y+1=(y-1)^{2}+3(y-1)+3$ must divide 3, so $\frac{y-1}{9}$ and $\frac{y^{2}+y+1}{3}$ are relatively prime. Therefore, both factors are perfect cubes (even in all of $\mathbb{Z}$, since $(-1)^{3}=-1$).

We now have, for some integers $a,b$,
\[\frac{y-1}{18}=a^{3},\ \frac{y^{2}+y+1}{3}=b^{3}\]
\[\Rightarrow (18a^{3}+1)^{2}+(18a^{3}+1)+1=3b^{3}\]
\[\Rightarrow 108a^{6}+18a^{3}+1=b^{3}\]
Now I'm stuck... :maybe:
[/hide]

[quote="sylow_theory"]Is there a nice solution to the equation,
$2a^{3}+1 = b^{3}$

Because we can write,
$2(3x)^{3}+1 = y^{3}$[/quote]

Well, you could do it if $a$ is prime  :P ...

$2a^{3}= (b-1)(b^{2}+b+1),$

and let $b-1 = 1, 2, 2a, \cdots, 2a^{3}$ and see which ones give valid answers.

I think we can prove that (x;y)=(0;1) is the only solution. :maybe:

Or another thing we could do is let $y^{3}= A^{2}, x^{3}= B^{2},$ so we have $54B^{2}+1 = A^{2}\Leftrightarrow A^{2}-54B^{2}= 1,$ which is a special case of the Pell Equation. If anyone could find a small value for $A,B$ apart from $(1, 0),$ then we could generate infinitely many values of $(A, B).$ Then we would just have to find the values for $(A^{2},B^{2})$ which are perfect cubes.

@scorpius119: How do you center LaTeX?

[quote="Karth"]Then we would just have to find the values for $(A^{2},B^{2})$ which are perfect cubes.[/quote]

They're far from guaranteed to exist... I don't really think this approach is viable.  The similarity to a Pell's equation is tantalizing, but no corresponding easy field extension exists.


[quote="tonypr"]Im not sure if this is too hard for classroom math but here's a question with mods

Find the remainder of $ 2^{100}$ when divided by $ 101$.[/quote]

$ a^{\phi(x)}\equal{}1 \mbox{ (mod } x)$ is true if $ a\nmid x$
$ \phi$ denotes the amount of numbers less than $ x$ that are relatively prime to x.
Now in this case, $ 2\nmid 101$, and 101 is prime.
Therefore, $ \phi(101)\equal{}100$
Which is basically what the formula says.
So, $ 2^{100}\equal{}1 \mbox{ (mod } x)$


ratio
number theory solved
number theory

Determine the least possible value of the largest term in an arithmetic progression of seven distinct primes.

Are you sure it's seven and not six? It looks like we might have to give a numerical example, and the examples get rather large for seven numbers :?

I am sure it is seven..
and also I don't understand the problem meaning because my english is not good enough,can you first explain the problem to me?

the least possible value is $907$, given by the progression $7+150k, k=0,1,...,6$. Obviously the common ratio $d$ must be divisible by $2.3.5=30$. If the first term $a_1$ is not 7, then the common ratio must be also divisible by 7 (if not, one of the seven numbers will be a multiple of 7, contradiction) and therefor $d\geq 210$, so $a_7\geq 7.210 \geq 907$. Then, with $a_1=7$ we see that $d\geq 150$ because we have:
\[7+6.30=7+3.60=7+2.90=187=11.17\]
\[7+2.120=247=19.13\]
And with $d=150$ we get $7,157,307,457,607,757,907$.. all primes!!

btw, i think this one should have been posted in the pre-olympiad section


[hide]the sum so far is 426, and she needs a total of 4 * 145 or 580 points.
580 - 426 is [color=brown]154 points[/color].[/hide]

[quote="MCrawford"]In three bowling games, Alice scores 139, 143, and 144.  What score will Alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]

[hide="Solution"](139+143+144+X)/4=145
426+X=580
X=154

Alice needs a 154 on the 4th game to have an average of 145.[/hide]

[quote="MCrawford"]In three bowling games, Alice scores 139, 143, and 144.  What score will Alice need in a fourth game in order to have an average score of 145 for all four games?[/quote]

[hide]$\frac{139+143+144+x}{4}=145$

$x=154$[/hide]


[quote="happyme"]The midpoints of the sides of a square are connected to form a second square. The midpoints of the sides of the second square are then connected to form a third square, and successive side midpoints are connected in the same way to form a fourth, fifth, and sixth square. The area of the original square is 64 square inches. Determine the number of square inches in the area of the sixth square.

I'm gonna stop my problem spree while I learn how to do fractions in LaTeX.  :ninja:[/quote]

[hide]The area becomes $\frac{1}{2}$ each time the next square is formed.

Therefore the sixth square should have area of $64 \cdot \frac{1}{2^{5}}=\boxed{2}$ square inches.[/hide]

Fractions in $\LaTeX$: [code]$\frac{x}{y}$[/code] which will display as $\frac{x}{y}$.

Oh...
I was putting a forward slash, instead of a backslash.


[hide]240000
I'm assuming its in the order of
[u]A[/u]  [u]B[/u]  [u]C[/u]  [u]D[/u]  [u]E[/u]
where A,B,C,D are numbers and E is a letter.
for A, you have digits 0,1,2,3,4,5,6,7,8,9; so thats 10 choices
for B, C, and D, you have the same scenario as A, so 10 choices for each letter
for E, there are 24 choices. Originally, there were 26(since there are 26 letters in the alphabet), but since the problem eliminated Letters "I" and "O", thats 26-2= 24
so its 10*10*10*10*24=240000[/hide]

Thanks.

Does anybody have a link to a website which goes through combinatorics?

This problem was posted in another thread awhile back.

[quote="Adam4444"]"License plates are issued that contain four digits followed by one letter. If the letters O and I cannot be used, how many different license plates are used?"

How would you solve this problem? And how should one think, in general, when facing these kind of problems?[/quote]
That would just be 10*10*10*10*24=240,000 possible license plates.

Andrewliu alreadygave the answer and solution;there's no need to post it again.
If you want to learn more about combinations and such, check the Wiki or order Intro to Counting and Probability, it's a good book.

I think gauss was just saying that the solution was a bit overcomplicated or something...

It's not complicated at all. That's exactly what you need to do to answer the problem. It only looks long because it explains everything in a detailed way.

andrewliu actually provided the reasoning for where the numbers multiplied together come from.  gauss just showed the calculation and assumed that it's obvious where the numbers came from.

Ahh I see.
You're getting a bit active...

Yeah, Ummm, I don't think this is combinations; shouldn't it be permutations(since order does matter) OR is there a difference between combinations and combinatorics? IDK!!! :?:

[b]Combinatorics: [/b]Combinatorics is a branch of pure mathematics concerning the study of discrete (and usually finite) objects. It is related to many other areas of mathematics, such as algebra, probability theory, ergodic theory and geometry, as well as to applied subjects in computer science and statistical physics. Aspects of combinatorics include "counting" the objects satisfying certain criteria (enumerative combinatorics), deciding when the criteria can be met, and constructing and analyzing objects meeting the criteria (as in combinatorial designs and matroid theory), finding "largest", "smallest", or "optimal" objects (extremal combinatorics and combinatorial optimization), and finding algebraic structures these objects may have (algebraic combinatorics).

[b]Combinations: [/b]In combinatorial mathematics, a combination is an un-ordered collection of unique sizes. (An ordered collection is called a permutation.) Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important (two lists with the same elements in different orders are considered to be the same combination). Also, the elements cannot be repeated in a combination (every element appears uniquely once); this is often referred to as "without replacement/repetition". This is because combinations are defined by the elements contained in them, s the set {1, 1, 2} is the same as {2,1,1}. For example, from a 52-card deck any 5 cards can form a valid combination (a hand). The order of the cards doesn't matter and there can be no repetition of cards.

(from wikipedia)

How would you solve the same problem for number plates which only contains two numbers from 1-3. I just don't get this with combinations.

Can you clear that up?  It sounds like the number of ways to choose 2 numbers from 3, which would be $ \binom{3}{2}\equal{}3$.

I think he means that the license plate can have 2 numbers, and each can be 1-3

To solve problems like this, in general you would think:

How many ways can I place the first digit? Well, it can be from 0 - 9, so 10 ways.
This would repeat until...
How many ways can I place the last letter? Well, there are 26 letters, but I can't use 2 of them. So there are 24 ways.

This is also known as constructive counting.


search
induction
modular arithmetic
AMC
USA(J)MO
USAMO
greatest common divisor

Prove that there exsits two positive integer x,y so that:
                n^x=x+my
for all positive integer m,n

Hi pigfly,

Is there an error in your problem?

Let us say there exists two such integers $x$ and $y$.
Then $n^x = x + m y $ is valid for all positive integers $m$ and $n$.
Put $n = m = 1$, then we have $1 = x + y$.
Put $n = 1$ and $m = 2$, then we have $1 = x + 2y$.
The above two imply that $y = 0$, and hence, $x = 1$.
But then $n^1 = 1$ is not true for all positive integers $n$.

Thus, there exist no such integers $x$ and $y$.

Kindly clarify this.

**********************************************************************************
Sambit

I guess the question is to prove that: \[\forall m,n \in \mathbb{N}_0, \exists x,y \in \mathbb{N}_0: n^x=x+m\cdot y\]

We search x for having n^x = x (m)
x=k*m+i with k in N and i in |[0;m-1]|
n^(km+i) = i (m)
Let a=n^m
a^k * n^i = i (m)
We can set that n ^ m = 1, if no we just need to take n' = n/pgcd(n,m) and m'=m/pgcd(n,m) and at end we multiply by pgcd(n,m) for refind n and m
So it exists b such that (it's correct ?) n^i * b = 1 (m)
Now we need to find k such that a^k = b*i (m)

If m is prime, Vect(a)=Z/mZ so for all i it exist k for a^k = b*i (m)

If m is not prime I doesn't find, for the moment, a solution

i think we cant that $n^m \equiv 1 mod m$. this is for example not possible if m is prime and n is not 1 mod m, since by little fermat, $n^m \equiv n mod m$. furthermore, we cannot cancel the greatest common divisor out of m,n, look at the equation(though this wouldn't anyway).

Peter

"n ^ m = 1"
I wanted to say gcd(m,n)=1
But it's right I have done some errors, nice problem !

Well, it’s not hard to show for the case when $m=p^t$, with $p$ prime. It’s really easy for $t=1$, and then we use induction on $t$:

We only need to prove it in the case $p\not |\ n$. Assume $p^t|n^a-a\Rightarrow p^t|n^{a_k}-a_k$, where $a_k=a+p^t(p-1)k$. However, $n^{a_k}-a_k\equiv n^a-a+kp^t\pmod {p^{t+1}}$, and it’s clear that we can select $k$ such that the last expression is $0\pmod{p^{t+1}}$.

For the case $p|n$, just take $a$ (what you called $x$) equal to $p^t=m$ and we get $p^t|n^a-a$ because $a\ge t$.

Maybe a similar inductive argument works for the case when $m$ isn’t a prime power, but I haven’t tried that.

This problem is a consequence of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=110644#110644]this[/url] one.

First of all, I upologize for some mistake during posting up the problem. I'm thankful to PeterVDD for understanding my ideas . Now,I had the solution.We will prove the sequence which is defined by $n_1=n;n_{i+1}=n^{n_i}
,...(mod n)$ is eventually constant (1)(This is also Grobber's problem:http://www.mathlinks.ro/Forum/viewtopic.php?t=15713 ).Then , the problem was solved. Proof by induction
+) m=1: it is easy(m=2:Problem 3 ,20 th USAMO )
+) suppose the problem corrects with every number <m . We prove that it corrects with m.
Lemma:If (1) corrects with h,q such that gcd(h,q)=1 then it corrects with hq
Indeed, (1) corrects with h,q$\to\exists c,d,k_0:
n_k$ :equiv:$ c$(mod h)
$n_k$ :equiv: $d$(mod q)$  \forall k$ :ge: $k_0$
According to the Chinese' theorem  remainder which there exist only b(mod hq) such that:   b :equiv: c(mod h); b :equiv: d(mod q)$\to n_k$ :equiv: b( mod hq)$\forall k$ :ge: $k_0$.This lemma is proved
If n=$\displaystyle\prod_{i=1}^t p_i ^{a_i}$ with t>1$\to p_i{a_i} <m \to$ from lemma we have the solution
If n=$p^a, p\in P,a\in N*\to\varphi(m)=p^{a-1} (p-1)<m$
then$\exists k_0:n_k$ :equiv: t(mod$\varphi(m)$)$\forall k\geq k_0$
-)gcd(n,m) <> 1 we have $n_a$ is divisible by m $\to n_k \equiv 0(mod m)\forall k$ :ge: a
-)gcd(n,m) = 1:$n^{\varphi(m)}$ :equiv: 1(modm)(euler's theorem)
$\forall k$ :ge: $k_0:n_k=h+\varphi(m) q$
deduce that $n_{k+1}$ :equiv: $n^h$(mod m). Hence, $n_k$ :equiv: $n^{k_0}$(modm) $\forall k$ :ge: $k_0$,the sequence is eventually constant .In every case we have solution of (1).  So, problem 3 is solved :P


Let $a,b$ be real　numbers such that $a>-1,\ b>-2.$ Find the minimum of the following equality. \[2b+\frac{2}{a+1}+\frac{2a+2}{b+2}.\]

Let $\alpha=\frac{1}{a+1}$, $\beta=b+2$. Then we wish to minimise $2\left(\alpha+\beta+\frac{1}{\alpha\beta}\right)-4\ge6\sqrt[3]{\frac{\alpha\beta}{\alpha\beta}}-4=2$, with equality if and only if $\alpha=\beta=\frac{1}{\alpha\beta}$, or $\alpha=\beta=1$, which gives $(a,b)=(0,-1)$.


We have f(x)=$ x^3$+$ ax^2$+$ bx$+c, b<0 and ab=9c/ Prove then f(x) have 3 roots

i assume you mean real root.  :wink:  but
let a=1, b=9, c=1. then equation dfoesn't have 3 real roots. :huh:  :wink:

Someone can write D for $ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$?

no need to write D for f(x).
write D for $ f'(x)$.
then cleary $ a^2 \ge 3b$, since b<0 & $ a^2\ge 0$.
so there are two critical points of  f(x).
let they be $ m_1,m_2$. then $ m_1 * m_2 \equal{} b<0$.

we then simply have to prove $ f(m_1)f(m_2)<0$

which is easy i believe.

I think you not true, because you don't use, that $ ab\equal{}9c$, I [b]forgot[/b] D of $ f(x)$....

PS My English is no good, I have 4points of 5 points on English lesson on Russia :blush: and I don't know, how write  #forgot#

ith_power, you are true)))
$ f(m)f(l) < 0$
$ f'(m) \equal{} 0$ and $ f'(l) \equal{} 0$
We have $ f(x) \equal{} f'(x)(x/3 \plus{} a/9) \plus{} 2/3x(b \minus{} \frac {a}{3})$
$ f(m) \equal{} \frac {2}{3}m(b \minus{} \frac {a}{3})$
$ f(l) \equal{} \frac {2}{3}l(b \minus{} \frac {a}{3})$
$ f(m)f(l) \equal{} \frac {2}{3}l(b \minus{} \frac {a}{3})\frac {2}{3}m(b \minus{} \frac {a}{3}) \equal{} \frac {4}{9}ml(b \minus{} \frac {a}{3})^2$
But $ ml \equal{} b/3 < 0$, because $ b < 0$
$ f(m)f(l) \equal{} \frac {4}{9}ml(b \minus{} \frac {a}{3})^2 < 0$
Contradaction


geometry
trapezoid
combinatorics unsolved
combinatorics

Find the least positive integer $ n$ with the following property: Paint each vertex of a regular $ n$-gon arbitrarily with one of three colors, say red, yellow and blue, there must exist four vertices of same color that are the vertices of some isogonal trapezoid. Here an isogonal trapezoid is defined as a convex quadrilateral with a pair of parallel sides, and a pair of equal length but not parallel sides.

[hide]There exist some $ m>n$ such that $ m$ has no such property.[/hide]

Already posted :
http://www.mathlinks.ro/Forum/viewtopic.php?t=184424

Pierre.


geometry
circumcircle
advanced fields
advanced fields unsolved

I have the following problem and no solution... please, help me. :)

In Euclidean three-dimensional space we have three distinct points A, B, C. Find the set of lines which are equally distanced from these three points.

For example, in plane there are three such lines, each of them joining midpoints of two sides of triangle ABC. But in space it seems rather complicated.
Do you know something related to this problem, maybe even the solution?

mazo

If the three points are collinear, each line parallel to the line through them is equidistant from the points and no others are. Otherwise, if $\ell$ is not parallel to $AB$, $BC$, or $AC$, there is exactly one line in the direction of $\ell$ that works; project to a plane perpendicular to $\ell$ and use the circumcenter of the projected points. Finally, if $\ell$ is parallel to $AB$, then any line parallel to $\ell$ in the plane halfway between $AB$ and $C$ works.


Functional Analysis
real analysis
real analysis unsolved

Show that if $ \mathbf{C}$ is compact and $ \{ \mathbf{M}_n \}$ tends strongly (pointwise) to $ \mathbf{M}$, then $ \mathbf{C}\mathbf{M}_n$ tends uniformly (in norm) to $ \mathbf{CM}$.

Any idea?  :oops:

The simplest idea that you can employ when you cannot prove something for a long time is to try to find a counterexample ;) In this case it is very easy: let $ H$ be a Hilbert space, let $ e_n$ be the orthonormal basis in $ H$, and let $ e$ be any non-zero vector. Take $ M_nx\equal{}(x,e_n)e$ and choose $ C$ so that $ Ce\ne 0$. What is true, is that $ M_nC$ tends to $ 0$ in norm, but that you can, probably, easily prove yourself...