Asymptote LaTeX articles trigonometry function conics parabola

I'm running Ubuntu and have Asymptote successfully compiling to LaTeX documents, but I can't seem to get additional packages to work.

I've tried adding cse5.asy and olympiad.asy to /usr/share/asymptote, where all the other .asy files seem to be, but Kile returns an error saying

[code][Asymptote] finished with exit status 1[/code]

When I try to compile

[code]
\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{asymptote}

\begin{document}

\begin{asy}
    import cse5;
    unitsize(.5inch);
    pair A=(0,0), B=(3,3);
    D(A--B);
    D(A);D(B);
\end{asy}

\end{document]
[/code]

I get the .log, .asy, and .aux files, but not the .pdf it's supposed to output (nor any filename_1.pdf). Would appreciate any help.

[quote="undefined117"]I've tried adding cse5.asy and olympiad.asy to /usr/share/asymptote, where all the other .asy files seem to be, but Kile returns an error saying[/quote]
Try to put the files in the folder [b]~/.asy[/b] (you may be obliged to create this folder).
Which version of Asymptote do you use ?

Nope, I get the same error with files in ~/.asy :(

"asy" in terminal gives Asymptote version 1.40

Hmm.. is this an Asymptote error, not a LaTeX/compiler error?

Try putting [b]cse5.asy[/b] and [b]olympiad.asy[/b] into [b]/usr/local/share/asymptote[/b] instead of [b]/usr/share/asymptote[/b].

That's where I have them, and everything works fine. :)

Hmm.. that's where I originally thought I was supposed to put the asy files, since it was there in OS X. However, my system didn't have that directory.

I just created /usr/local/share/asymptote and moved the asy files into it, but Kile still returns the same error :(

[code][PDFLaTeX] asytest.tex => asytest.pdf (pdflatex)
./asytest.tex:0:No file asytest.aux.
./asytest.tex:21: file asytest_1.pdf does not exist on input line 21.
[PDFLaTeX] 0 errors, 2 warnings, 0 badboxes
[PDFLaTeX] Done!

[Asymptote] asytest.tex (asy)
[Asymptote] finished with exit status 1[/code]

EDIT: huh the import command works in commandline asy.. so this is a Kile problem?

[url=http://i29.tinypic.com/2mgnksk.png]moving asy files in terminal[/url]

[url=http://i29.tinypic.com/sylfk7.png]Asymptote in terminal[/url]

In Kile, go to [b]Settings >> Configure Kile >> Tools >> Build >> Asymptote[/b], and post what you have in the [b]Command[/b] and [b]Options[/b] boxes (I have [b]Command: asy[/b] and [b]Options: '%S'[/b]).

yep, those are my settings too..

Command: asy
Options: '%S'

[code]> ~/Documents/latex$ asy asytest
/usr/share/asymptote/graph.asy: 1897.26: cannot cast 'path[]' to 'int[][]'[/code]

what does that mean?

In my original post in this topic, I mentioned asymptote "exiting with status 1" when I included the cse5 package.. however, it seems that the import command is the culprit, no matter which package I import (such as graph).

[hide="the log file"]
[code]This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) (format=pdflatex 2008.6.9)  2 JUL 2008 17:36
entering extended mode
 %&-line parsing enabled.
**asytest.tex
(./asytest.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, loaded.
(/usr/share/texmf-texlive/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/share/texmf-texlive/tex/latex/base/size12.clo
File: size12.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
\c@part=\count79
\c@section=\count80
\c@subsection=\count81
\c@subsubsection=\count82
\c@paragraph=\count83
\c@subparagraph=\count84
\c@figure=\count85
\c@table=\count86
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
)
(/usr/share/texmf/tex/latex/asymptote/asymptote.sty
(/usr/share/texmf-texlive/tex/latex/graphics/graphics.sty
Package: graphics 2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR)

(/usr/share/texmf-texlive/tex/latex/graphics/trig.sty
Package: trig 1999/03/16 v1.09 sin cos tan (DPC)
)
(/etc/texmf/tex/latex/config/graphics.cfg
File: graphics.cfg 2007/01/18 v1.5 graphics configuration of teTeX/TeXLive
)
Package graphics Info: Driver file: pdftex.def on input line 90.

(/usr/share/texmf-texlive/tex/latex/pdftex-def/pdftex.def
File: pdftex.def 2007/01/08 v0.04d Graphics/color for pdfTeX
\Gread@gobject=\count87
))
\AsyStream=\write3
\AsyTestStream=\read1
\openout3 = `asytest.asy'.

\c@asy=\count88
)
No file asytest.aux.
\openout1 = `asytest.aux'.

LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 18.
LaTeX Font Info:    ... okay on input line 18.
LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 18.
LaTeX Font Info:    ... okay on input line 18.
LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 18.
LaTeX Font Info:    ... okay on input line 18.
LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 18.
LaTeX Font Info:    ... okay on input line 18.
LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 18.
LaTeX Font Info:    ... okay on input line 18.
LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 18.
LaTeX Font Info:    ... okay on input line 18.

Package asymptote Warning: file asytest_1.pdf does not exist on input line 70.

LaTeX Font Info:    External font `cmex10' loaded for size
(Font)              <12> on input line 76.
LaTeX Font Info:    External font `cmex10' loaded for size
(Font)              <8> on input line 76.
LaTeX Font Info:    External font `cmex10' loaded for size
(Font)              <6> on input line 76.

Package asymptote Warning: file asytest_2.pdf does not exist on input line 86.

Package asymptote Warning: file asytest_3.pdf does not exist on input line 128.

[1

{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] (./asytest.aux) )
Here is how much of TeX's memory you used:
 547 strings out of 95086
 6299 string characters out of 1183256
 52862 words of memory out of 1500000
 3775 multiletter control sequences out of 10000+50000
 6376 words of font info for 23 fonts, out of 1200000 for 2000
 28 hyphenation exceptions out of 8191
 25i,4n,19p,167b,110s stack positions out of 5000i,500n,6000p,200000b,5000s
</usr/sha
re/texmf-texlive/fonts/type1/bluesky/cm/cmr12.pfb></usr/share/texmf-texlive/fon
ts/type1/bluesky/cm/cmr8.pfb>
Output written on asytest.pdf (1 page, 11762 bytes).
PDF statistics:
 14 PDF objects out of 1000 (max. 8388607)
 0 named destinations out of 1000 (max. 131072)
 1 words of extra memory for PDF output out of 10000 (max. 10000000)
[/code][/hide]

It would be helpful if you showed us the relevant Asymptote code. Without it, it is hard to tell much. As to the import command, most packages call the graph package either directly or by calling some package that calls it. In principle, it may happen that you have a corrupted version of graph.asy. Try to reinstall this particular package and see what happens.

[quote="undefined117"][code]> ~/Documents/latex$ asy asytest
/usr/share/asymptote/graph.asy: 1897.26: cannot cast 'path[]' to 'int[][]'[/code]

what does that mean?[/quote]
You're probably sending an array of paths where a 2D array of integers is expected, since it says it can't convert from the former to the latter.

That, however, is the most I can make of it without the actual code. :)

hmm I don't think it's corrupted or anything, since it happens with graph, cse5, and olympiad - as far as I know, anytime I use the import command.

[hide="asy code from LaTeX website"][code]import graph;

real f(real x) {return sqrt(x);}
pair F(real x) {return (x,f(x));}

real g(real x) {return -sqrt(x);}
pair G(real x) {return (x,g(x));}

path p=(0,0)--graph(f,0,1,operator ..)--(1,0);
fill(p--cycle,lightgray);
draw(p);
draw((0,0)--graph(g,0,1,operator ..)--(1,0),dotted);

real x=0.5;
pair c=(4,0);

transform T=xscale(0.5);
draw((2.695,0),T*arc(0,0.30cm,20,340),ArcArrow);
fill(shift(c)*T*circle(0,-f(x)),red+white);
draw(F(x)--c+(0,f(x)),dashed+red);
draw(G(x)--c+(0,g(x)),dashed+red);

dot(Label,(1,1));
arrow("$y=\sqrt{x}$",F(0.7),N);

arrow((3,0.5*f(x)),W,1cm,red);
arrow((3,-0.5*f(x)),W,1cm,red);

xaxis("$x$",0,c.x,dashed);
yaxis("$y$");

draw("$r$",(x,0)--F(x),E,red,Arrows,BeginBar,PenMargins);
draw("$r$",(x,0)--G(x),E,red,Arrows,PenMargins);
draw("$r$",c--c+(0,f(x)),Arrow,PenMargin);
dot(c);
[/code][/hide]
------

[hide="another asy code"][code]import cse5;
unitsize(1cm);
draw((0,0)--(4,5));[/code]

It gave me the same 1897.26 error, even though no cse5 command was used.[/hide]

I'm going to try reinstalling asymptote to see if it helps.

EDIT: nope, reinstallation didn't help. :(

It is graph.asy for sure. Go to the corresponding line in graph.asy and you'll see
[code]
path[] segment(pair[] z, bool[] b, interpolate join=operator --)
{
  checklengths(z.length,b.length,conditionlength);
  int[][] segment=segment(b);
  return sequence(new path(int i) {return join(... z[segment[i]]);},
                  segment.length);
}
[/code]
The line int[][] segment=segment(b); is causing the problem. It looks a bit misterious but what it actually does is to call the segment function from math.asy (which is completely unrelated to the segment function being defined). Now look at the beginning of the graph.asy file. You should see
[code]
private import math;

import splinetype;
import graph_settings;
[/code]
If the line with import math; is missing or corrupted, you get the error you mentioned (the same will happen if math.asy is corrupted). cse5 and olympiad both call graph, so they have nothing to do with the error. The first step to trace the problem would be to replace =segment(b) with =math.segment(b) in graph.asy and see what error you get now.

The line with import math; is there. I copy/pasted the code you provided over the existing code just to make sure, and it still didn't work.

I also replaced "=segment(b)" with "=math.segment(b)" to no avail - I get the same error. :(

In what way could the math.asy file get corrupted? I found this in it:
[code]// Return the indices of consecutive true-element segments of bool[] b.
int[][] segment(bool[] b)
{
  int[][] segment;
  bool[] n=copy(b);
  n.delete(0);
  n.push(!b[b.length-1]);
  int[] edge=(b != n) ? sequence(1,b.length) : null;
  edge.insert(0,0);
  int stop=edge[0];
  for(int i=0; i < edge.length-1;) {
    int start=stop;
    stop=edge[++i];
    if(b[start])
      segment.push(sequence(start,stop-1));
  }
  return segment;
}[/code]

is there anything wrong with that?

okay nvm "import math;" doesn't return any errors, so it means it's not corrupted..

I upgraded from Asymptote 1.40 to 1.43 just now to make sure no files are corrupted.

hmm so now I have this:
[code]/*path[] segment(pair[] z, bool[] b, interpolate join=operator --)
{
  checklengths(z.length,b.length,conditionlength);
  int[][] segment=segment(b);
  return sequence(new path(int i) {return join(... z[segment[i]]);},
                  segment.length);
}*/[/code]
and I still get the error :(

but now the error says:
[code]/usr/local/share/asymptote/graph.asy: 1900.26: cannot cast 'path[]' to 'int[][]'[/code]
why did that number change from 1897.26 to 1900.26?

I'm pretty sure there aren't duplicate files.. I didn't touch any of the file structure except to add cse5.asy and olympiad.asy.

It just can't be. The only place in graph.asy where this particular error may arise is the one I mentioned and we just removed it  :o . OK, the single command import graph; should already cause the error, right? Now just do the usual thing for code debugging: comment parts of graph.asy file and see if the import graph; command causes the error in various cases (start with commenting the entire file; if the error is still there, it is a sure indication that you are editing the wrong file). This way you should be able to pinpoint the problem. Also, if you want, attach your version of graph.asy. I'll see if I can observe the behavior you described on my computer.

okay I commented out the entire file, and the error was still there. So I deleted:

/usr/bin/asy
/usr/bin/xasy
/usr/share/asymptote/
/usr/share/texmf/tex/latex/asymptote/
/usr/local/bin/asy
/usr/local/bin/xasy
/usr/local/share/asymptote/

hoping it would get rid of Asymptote, and reinstalled Asymptote with [url=http://linux.softpedia.com/get/Multimedia/Graphics/Asymptote-2112.shtml]this[/url], which installed everything above except the first three (they were there probably because I'd installed them with aptitude). I then moved asy and xasy from /usr/local/bin/ to /usr/bin/

Commenting out all of /usr/local/share/asymptote/graph.asy had no effect on the error message.

"asy asytest" in terminal as well as compiling in Kile gives the error, but import graph; in terminal Asymptote doesn't:
[code]:~$ asy
Welcome to Asymptote version 1.43 (to view the manual, type help)
> import cse5;
> D((0,0)--(3,3));
(0,0).. controls (1,1) and (2,2)
 ..(3,3)
> [/code]
and outputs out.eps like it should.

So I thought maybe this was a problem with /usr/bin/asy, but my text editor won't open it. :| agh.

graph.asy attached as txt file.

okay wth it suddenly started working.. :huh:

Thanks for the help, fedja - in the future, I'll have to remember the commenting out thing for code debugging :oops:

[quote="undefined117"]
Commenting out all of /usr/local/share/asymptote/graph.asy had no effect on the error message.
[/quote]
The site essentially died yesterday, so I decided to go to the bed rather than to wait for it to come up again  :sleep2:

Two comments:

1) Wasn't the error message about /usr/share/asymptote/graph.asy (without /local/?) If so, editing usr/local/share/asymptote/graph.asy would not be very useful...

2) I'm happy it started to work but I'm still puzzled about what caused the problem. My best guess is that usr/share/asymptote had normal graph.asy in it and some corrupted or duplicate version of math.asy (at least, I couldn't get the described behavior in any other way). But it is only a guess :(.

yeah it used to be just "/usr/share/asymptote/", but "/local/" got included when I installed asymptote version 1.43

I got rid of all of the previous asymptote installation as described in an above post (hopefully).

ehh yeah I'm really confused too.. though hopefully it won't break again :| sorry if I kept you up out of bed or anything. Thanks again for your help. :)

mmk sorry to bring this up again, but unfortunately, Asymptote is still very ill. :| Broke again.

Currently trying to compile:
[hide="asy code"][code]\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{asymptote}
\usepackage{graphicx}
\usepackage{caption}
\usepackage{multicol}
\usepackage{fullpage}
\usepackage{pslatex}

\everymath{\displaystyle}

\begin{document}

\begin{asy}
import cse5;
import graph;
size(300);

Label f;
f.p=fontsize(10);
xaxis(-2,2);
yaxis(-2,2);

real f(real x) {return x^2;}
D(graph(f,-2,2));
\end{asy}

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

[quote="fedja"]math.asy seems OK. That you get the same error after replacing segment(b) with math.segment(b) is very strange. Let's try this: put the entire definition of the segment() routine in graph.asy in comment marks /* */ and see if you get the same error now. This way we'll at least make sure if it is this place that malfunctions. Also make sure that the files graph.asy and math.asy are unique (if you named some other file math.asy it may cause trouble or you may be making changes in not the same graph.asy as the one that is called). Also, try adding import math; as the first line of your code.[/quote]
Upon replacing segment(b) with math.segment(b) like you suggested, I get:
[code]/usr/local/share/asymptote/graph.asy: 1900.23: no matching variable 'math.segment'[/code]
When I commented out the entire segment() routine, it started looping, spitting this out over and over again:
[hide="kinda long"][code]Loading cse5 from /home/yoos/.asy/cse5.asy
Loading graph from /usr/local/share/asymptote/graph.asy
Loading math from math.asy
Loading splinetype from /usr/local/share/asymptote/splinetype.asy
Loading graph_settings from /usr/local/share/asymptote/graph_settings.asy
Loading slopefield from /usr/local/share/asymptote/slopefield.asy
Loading plain from /usr/local/share/asymptote/plain.asy
Including plain_constants from /usr/local/share/asymptote/plain_constants.asy
Loading version from /usr/local/share/asymptote/version.asy
Including plain_pens from /usr/local/share/asymptote/plain_pens.asy
Including plain_paths from /usr/local/share/asymptote/plain_paths.asy
Including plain_filldraw from /usr/local/share/asymptote/plain_filldraw.asy
Including plain_margins from /usr/local/share/asymptote/plain_margins.asy
Including plain_picture from /usr/local/share/asymptote/plain_picture.asy
Loading simplex from /usr/local/share/asymptote/simplex.asy
Including plain_shipout from /usr/local/share/asymptote/plain_shipout.asy
Including plain_xasy from /usr/local/share/asymptote/plain_xasy.asy
Including plain_Label from /usr/local/share/asymptote/plain_Label.asy
Including plain_arcs from /usr/local/share/asymptote/plain_arcs.asy
Including plain_boxes from /usr/local/share/asymptote/plain_boxes.asy
Including plain_strings from /usr/local/share/asymptote/plain_strings.asy
Including plain_markers from /usr/local/share/asymptote/plain_markers.asy
Including plain_arrows from /usr/local/share/asymptote/plain_arrows.asy
Including plain_debugger from /usr/local/share/asymptote/plain_debugger.asy[/code]
btw I've configured Kile to run asy -vvv, which might be why I'm getting this long line of code showing in the loop.[/hide]

Adding import math; doesn't fix the problem either.

So it's the segment() routine is what's causing this problem? meh it's weird how no one else seems to be having this problem, based on a Google search. :(

Again, running just the asy code from terminal asymptote gets me the parabola without a problem.

This is a bit idiotic, but since I've seen it 3 times by now, I'll ask:

Didn't you, by chance, have a bright idea to name one of your TeX files with Asymptote code in them math.tex in the nearest past?

The segment routine is OK by itself. It misbehaves if the math package is missing or altered. The behaviour you described is completely consistent with having a file named math.asy in one of the directories that Asymptote searches before its own directory (since now editing the graph.asy file has the desired effect, I hope no more duplicate files graph.asy are around). The directory where you have your TEX file is the most suspicious. So, please, do search for files named math.asy and graph.asy on your computer and make it completely sure that they are unique. If duplicates do not show up anywhere, copy (but don't move) graph.asy and math.asy to the directory where you keep your TeX files. This should help if the duplicates are hiding in some directories not shown on the regular search.

If you are still not convinced that a wrong math.asy is really there, look at these three lines:

Loading graph from /usr/local/share/asymptote/graph.asy
Loading math from math.asy
Loading splinetype from /usr/local/share/asymptote/splinetype.asy

Asymptote tells you the full path to graph.asy but not to math.asy.

uh oh.. yes I had that bright idea. I never realized that asymptote would search the same directory as the .tex file for the math.asy and graph.asy and stuff.. I also never thought of the obvious fact that compiling math.tex would also produce math.asy, which would be a duplicate.

After deleting math.asy (and renaming math.tex) it works now. Sorry for my idiocy. :|

I should have guessed earlier but I just didn't expect that. I knew that the duplicate files must be somewhere but I was looking for some traces of previous installations and such rather than for this obvious possibility. OK, since you are the 4th person on AoPS to do it, I added the corresponding remark to the AoPSWiki Asymptote help section. :)


The Northern lighthouse flashes every $ 18$ seconds. The Southern lighthouse flashes every $ 15$ seconds. The two lighthouses flash together. How many seconds will it be before the next time they both flash together?

[hide="Solution"] Take the lcm of both: $ lcm(15,18)\equal{}\boxed{90}$. [/hide]


Each positive integer is colored red or green in such a way that the sum of any three (not necessarily distinct) equally colored numbers also has the same color. Find all possible colorings.

What do you mean by equally colored? I'm assuming it means that something those elements are colored the same.

WLOG, let $1\in R$ $1+1+1=3\in R$. More experimenting shows that all odd colors are in $R$.

Now consider where $2$ is. If $2\in R$, then all integers are in $R$.

If $2\in B$, then $6,10,16,22,\cdots\,6k+4,\cdots\in B$. If we put any even integer not in $B$ in $R$, then we will get a contradiction (If we have $6k$, then $1+3+6k=6k+4$. If we have $6k+2$, then $1+7+6k+2=6(k+1)+4$). Thus, all even integers must be in $B$.

So the possibilities are: All integers painted red, all integers painted green, even painted red and odd painted green, and odd painted red and even painted green.

This seems a bit trivial. Am I answering the right question?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find al groups with 30 elements and 8 automorphisms.

Well, a group of order $ 15$ must lie in the center, since the inner automorphism group is a subgroup of the automorphism group, and is isomorphic to $ G/Z(G)$. So $ G/Z(G)$ has order $ 1$ or $ 2$, hence is cyclic. This implies that $ G$ is abelian (i.e. $ G/Z(G)$ must have order $ 1$...you can verify that if $ G$ mod its center is cyclic, then $ G$ is abelian).

By the Fundamental Theorem of Finitely Generated Abelian Groups, $ G$ must be isomorphic to $ Z_{30}$. This has $ \phi(30) \equal{} 1 \cdot 2 \cdot 4 \equal{} 8$ automorphisms, so it is the unique group that satisfies the property.


Hello, this is my first real post.

Howcome $\displaystyle\frac{\ln \sqrt{e} }{e} = \displaystyle\frac{1}{2e}$?

I would appreciate it if someone would clarify this for me.

thanks

$\frac{ \ln \sqrt{e} }{e} = \frac{ \frac{1}{2} \ln e}{e} = \frac{ \frac{1}{2} }{e} = \frac{1}{2e}$

[hide="Derivation"]
Recall the common exponentiation rules.

1.  $a^b = a \times a \times ... \times a$ ($b$ times)
1b.  $a^1 = a$ (from 1; $b = 1$)
2.  $a^b a^c = a^{b+c}$ (repeated use of 1)
3.  $( a^b )^c = a^{bc}$ (repeated use of 2)
4.  $\frac{a^b}{a^c} = a^{b-c}$ (from 1)
5.  $a^0 = 1$ (from 4; $b=c$)
6.  $a^{-b} = \frac{1}{a^b}$ (from 4 and 5)

Now we define a logarithm thus:  Iff $a^b = c$, then $b = \log_a c$.  From this definition we can derive several logarithmic rules from the exponentiation rules above:

1.  $a^{ \log_a x} = x$ (equivalent to definition)
1b.  $\log_a a = 1$
2.  $\log_a xy = \log_a x + \log_a y$
3.  $\log_a x^c = c \log_a x$
4.  $\log_a \frac{x}{y} = \log_a x - \log_a y$
5.  $\log_a 1 = 0$
6.  $\log_a \frac{1}{x} = - \log_a x$

Now, the problem.

$\frac{ \ln \sqrt{e} }{e}$
$= \frac{ \ln e^{ \frac{1}{2} } }{e}$
$= \frac{ \frac{1}{2} \ln e}{e}$ (rule 3)
$= \frac{ \frac{1}{2} }{e}$ (rule 1b)
$= \frac{1}{2e}$
[/hide]

I see my err, I didnt recognize that $\ln e^{\frac{1}{2}}$ was equal $\log_{e}{e}^{\frac{1}{2}}$  :oops: which obviously would imply $\frac{1}{2}\log_{e}{e}$

thanks

By multipying each side by $e$, we see that $ln\sqrt{e} = \frac{1}{2}$. Using the fact that $n^{\frac{1}{2}} = \sqrt{n}$,  the expression is clearly true.


I'm stuck with one of my homework problems, could anyone help me please?
Find linear transformation U, T: F2 to F2 such that UT= T0 ( the zero transformation) but TU does not equal T0. Any input would be appreciated.

Try matrices with exactly one entry equal to 1, and all other entries equal to zero.

Once you find a pair that works, it works over any field $ F$; no other conditions required. (I assume that your F2 is the two-dimensional vector space $ F^{2}\equal{}F\times F$)

thanks for your suggestion. Here are some examples that I found:
A= 0 0
     1 0

B= 0 0
     0 1

C= 0 1
     0 0

AB= 0 matrix, but BA does not
BC= 0 matrix, but CB does not

How does it go from here?

A matrix represents a linear transformation; $ A$ gives the transformation $ v\mapsto Av$, where $ v$ is a column vector. That is the zero map if and only if the matrix is zero.


I just want a general idea of how long people have been here! :lol:

yeah, but who would want to go around and look who joined when?

That would take a long time. :wink:

An administrator could plot a histogram based on the entire AoPS population instead of the self-selected sample taken here. However, the entire population includes such a large number of [url=http://en.wikipedia.org/wiki/Dead_Souls]dead souls[/url] that the results would not be reliable anyway.

If you just look at the Member List and sort it by the "joined date", you can get the full histogram yourself in about 10 minutes (though I admit that the dead souls will be a problem).

All the pages are nothing but dead souls up until about page 100.

2+ years for me. But that doesnt mean I've been posting for 2+ years

I first visited AoPS about 2.5 years ago, but didn't sign up for a few months.  I went with 3 years so the poll will still be accurate for me over the next few months (and due to rounding).

Posting (written from memory, so may not be entirely accurate):
1st 6 months:  almost entirely in the classes forums that I took; maybe 100 total???
2nd 6 months:  posted in other parts of the forum; had 500+ posts
2nd year:  started posting a lot last summer; I know I got several hundred that summer, probably near 900 total for summer; had about 1300-1400 at the end of the year
Remaining time:  didn't post as much; only about 1600 posts now.

[quote="xpmath"]Almost 6 months here.[/quote]

Same. We registered on the exact same day.

I'm nearing my first year  :D  (and wow with almost 1000 posts already!). The first time I joined AoPS I made an account but didn't really comprehend the problems that were posted. Last year, chess64 got me to join (when I was much better at math) at one of my former middle school MC meetings.

Oddly enough, I joined on Friday the 13th of October.  :diablo:

[quote="D3m0n Shad0w"]About 6 months, I suspect. I had known about AoPS before, but I wasn't that good at math  :P[/quote]

I beg to differ.

[quote="D3m0n Shad0ws Join Date"]Joined: 12 Nov 2006[/quote]

Almost a year, if I calculate correctly. 10 months.

Really? That's odd..

But if your talking about how active we've been on, I remember 6 months.

I have joined this forum for more than two months, this is a great forum! :lol:

Almost half a year here, although I knew of the existence of AoPS for maybe 3 years.


topology
advanced fields
advanced fields unsolved

I think this question is too easy for this forum. But since I don't know how to do it, I really need some help

Is there a nonempty perfect set in $R$ which contains no rational point?

Thank you.

[quote="t0rajir0u"]The negative follows from the fact that the rationals are dense in the reals, doesn't it?[/quote]

Yes: a perfect set is a set equal to its set of limit points, i.e. a closed set with no isolated points.  Take the complement of a closed set and you get an open set: we need an open set in $\bf R$ which contains all the rationals.  [EDIT: the following sentence is just plain old wrong.]  But the only such set is $\bf R$ itself.

wait what? the only open set in $\mathbb{R}$ which contains all the rationals is $\mathbb{R}$? Hrm. Really? Fix some $\varepsilon > 0$ and enumerate the rationals as $\{q_{n}\}_{n=1}^{\infty}$; consider the set $S = \bigcup_{n=1}^{\infty}B(q_{n}, \varepsilon 2^{-n})$ which is open as it is the union of open sets and clearly contains all the rationals. By countable subadditivity of measures, we find that $m(S) \le \sum_{n=1}^{\infty}\varepsilon 2^{1-n}= 2 \varepsilon$, so it seems like we can get open sets in $\mathbb{R}$ containing the rationals with arbitrarily small measure...

That was a vastly more complicated counter-example than you needed.  "Take the reals and remove a discrete set of irrationals" would have worked fine.

Lemma: Any closed subset $E\subset \mathbb R$ can be represented as a disjoint union $E=P\cup C$, where $P$ is a perfect set and $C$ is countable*.
Proof: Choose a countable family $\{I_{j}\}$ of open intervals that form a basis of topology of $\mathbb R$. (If you don't know what this means, think of $\{I_{j}\}$ as open intervals with rational endpoints.) Let $U$ be the union of all $I_{j}$ such that $E\cap I_{j}$ is countable. The sets $C=E\cap U$ and $P=E\setminus U$ are the ones we want.

By virtue of this lemma, you only need to find an uncountable closed set with no rationals, like the complement of the open set constructed by [b]MysticTerminator[/b].

(*) "countable" for me = "at most countable".

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=set+cantor+translate&t=71744]using translates of the cantor set[/url]

Mlok's result ie Cantor-Benedixon thm holds in separable spaces as well


Hi,
Has the following  problem been discussed in the forum?

Problem: If $(a_{n})_{n\in N}$ is a sequence of positive real numbers which converges to $a$ then it is known that $x_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\rightarrow a$, namely its arithmetic mean also converges to $a$.
The following question is natural.
Study the limit

$\lim_{n\rightarrow \infty}n\left(a_{n}-a\right)$.
Didi

If you stated it the way you meant to, there's no way to get any useful information.

Simple example: $a_{n}=\frac1{\sqrt{n}},$ whose limit is $a=0.$

Kent,
You are right, however I was thinking at the case when $a_{n}$ is a sequence for which $\lim_{n\rightarrow \infty}n(a_{n}-a)$ is finite, as a matter of fact this is the interesting case.
Didi


[quote="PeterTheGreat"]I run a [url=http://home.planet.nl/~p.j.hendriks/ppvdw.htm]website[/url] with a weekly math puzzle.  Regarding my puzzle nr 171, about folding a paper hat and using it as a water bucket, I am not sure wether my solution (based on a cone) is really the optimum. Could anybody help me out? If you want to solve the puzzle, please be so kind as to do so by email. Thanks in advance![/quote]

its in spanish or something

Well, I thought you would notice the Union Jack. If you click there, you will find an English version of the puzzle............


geometry
trigonometry
ratio
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14792[/img]

 :(

Below $ (x)$ denotes the area of the figure $ x$

Suppose that we have found the point $ P$

Let $ D\in AP \cap BC$

For any point $ H$ on the ray $ (AP$ we have $ \frac{(AHB)}{(AHC)} \equal{} \frac{AH\cdot AB \cdot \sin \angle BAP}{AH\cdot AC \cdot \sin \angle CAP} \equal{} \frac{AB \cdot \sin \angle BAP}{AC \cdot \sin \angle CAP}$ = constant

It follows that $ \frac{(APB)}{(APC)} \equal{} \frac{(ADB)}{(ADC)}$ which is equal to $ \frac{BD}{DC}$

So it is $ \frac{(APB)}{(APC)} \equal{} \frac{BD}{DC}$ and for the same reason, $ \frac{(BPC)}{(APB)} \equal{} \frac{CE}{EA}$ and $ \frac{(APC)}{(BPC)} \equal{} \frac{AF}{FB}$

So it's enough to divide the sides $ BC,CA$ in ratios $ \frac{APB}{APC}$ and $ \frac{BPC}{APB}$ respectively, where $ BPC,APC,APB$ are the so-denoted given segments

Then we draw the segments $ AD$ and $ BE$ to find the point $ P$. We find $ F$ on the intersection of $ CP$ with $ AB$

Then from Ceva it follows that $ \frac{FB}{AF} \equal{} \frac{BD}{DC}\cdot\frac{CE}{EA} \equal{} \frac{APB}{APC}\cdot \frac{BPC}{APB} \equal{} \frac{BPC}{APC}$, the desired

how can i divide BC in ratio APB/APC ?

 :(


$ P(x)$ is a quadratic trinomial. What maximum number of terms equal to the sum of the two preceding terms can occur in the sequence $ P(1)$, $ P(2)$, $ P(3)$, $ \dots?$

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

[hide]Let $ P(x) \equal{} ax^2\plus{}bx\plus{}c$. We are looking for the maximum number of integers $ n \ge 2$ such that:

$ P(n\plus{}1) \equal{} P(n) \plus{} P(n\minus{}1)$

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

$ (a)n^2 \plus{} (2a\plus{}b)n \plus{} (a\plus{}b\plus{}c) \equal{} (2a)n^2 \plus{} (2b\minus{}2a)n \plus{} (a\minus{}b\plus{}2c)$

$ (a)n^2 \plus{} (b\minus{}4a)n \plus{} (c\minus{}2b) \equal{} 0$

This is a quadratic which can have at most $ 2$ integer solutions $ n \ge 2$.

$ (a,b,c) \equal{} (1,\minus{}1,4)$ yields $ n^2\minus{}5n\plus{}6 \equal{} 0 \Rightarrow n\equal{}2,3$.

With $ P(x) \equal{} x^2\minus{}x\plus{}4$, we have: $ P(1) \equal{} 4$, $ P(2) \equal{} 6$, $ P(3) \equal{} 10$, and $ P(4) \equal{} 16$.
So, $ P(3) \equal{} P(2)\plus{}P(1)$ and $ P(4) \equal{} P(3)\plus{}P(2)$.

Thus, the maximum number of terms that are equal to the sum of the two preceding terms is $ 2$. [/hide]


The triangle $ \triangle SFA$ has a right angle at $ F$. The points $ P$ and $ Q$ lie on the line $ SF$ such that the point $ P$ lies between $ S$ and $ F$ and the point $ F$ is the midpoint of the segment $ [PQ]$. The circle $ k_1$ is th incircle of the triangle $ \triangle SPA$. The circle $ k_2$ lies outside the triangle $ \triangle SQA$ and touches the segment $ [QA]$ and the lines $ SQ$ and $ SA$.
Prove that the sum of the radii of the circles $ k_1$ and $ k_2$ equals the length of $ [FA]$.

Call $ \triangle{SPA} \equal{} T_1$ and $ \triangle{SQA} \equal{} T_2$ the inradii and exradii are $ r_1$ and $ r_2$ respectively then $ [SPA] \equal{} \frac {SP\cdot FA}{2} \equal{} S(T_1)r_1$ and $ [SQA] \equal{} \frac {AF\cdot SQ}{2} \equal{} (S(T_2) \minus{} AQ)\cdot r_2$. Then we need to prove that: $ \frac {SQ}{2S(T_2) \minus{} AQ} \plus{} \frac {SP}{2S(T_1)} \equal{} 1$ remember: $ AQ \equal{} AP$.
The last expression is the same as: $ \frac {SQ}{SQ \plus{} SA \minus{} AP} \plus{} \frac {SP}{SP \plus{} SA \plus{} AP} \equal{} 1$ remember: $ SQ \equal{} SP \plus{} PQ$ conclusion $ \iff$ $ SP(\frac {1}{SQ \plus{} SA \minus{} AP} \plus{} \frac {1}{SP \plus{} SA \plus{} AP}) \equal{} 1 \minus{} \frac {PQ}{SQ \plus{} SA \minus{} AP} \equal{} \frac {SP \plus{} SA \minus{} AP}{SQ \plus{} SA \minus{} AP}$ $ \iff$ $ SP(\frac {SP \plus{} SQ \plus{} 2SA}{SP \plus{} SA \plus{} AP}) \equal{} SP \plus{} SA \minus{} AP$ $ \iff$ $ SP\cdot SQ \equal{} SA^2 \minus{} AP^2 \equal{} FS^2 \minus{} FP^2$ but this last is clearly true since $ PF \equal{} FQ$

Call I the center of $ k_1$ and J the center of $ k_2$ and their radius $ r_1$, $ r_2$; $ k_1$ touch AS on K and $ k_2$ touch AS on B and SQ on C. Call H the projection of A on CJ. Then $ \angle IJA \equal{} \angle JAB \minus{} \angle ASI \equal{} 90 \minus{} \angle IAK \minus{} \angle FAP \minus{} \angle ASI \equal{}$
$ \equal{} \angle FPA \minus{} \angle IAK \minus{} \angle ASI \equal{} \angle IAK \plus{} \angle ASI \equal{} \angle JIA$.
Then $ \overline{IA} \equal{} \overline{IJ}$ and $ \angle KIA \equal{} 180 \minus{} (90\minus{} \angle ASI) \minus{} \angle JIA \equal{} 180 \minus{} (90 \minus{} \angle ISC) \minus{} \angle IJA \equal{} \angle AJH$. So the triangles $ \triangle IKA$ and $ \triangle JHA$ are equal and $ \overline{JH} \equal{} \overline{IK}$. Then $ CH \equal{} r_1 \plus{} r_2$, but CHAF is a rectangle so $ \overline{CH} \equal{} \overline{AF}$.

[hide="figure"][img]http://img233.imageshack.us/img233/1527/geometria34ht7.png[/img][/hide]


Hey all.

My name is Jeremy Pham, a Canadian citizen soon to be living in the United States.  I've just registered with Intermedia Algebra, already impressed with the AoPs books I ordered a week earlier.   I'm a cost concious kind of guy...is that bad?  Anyways, is the Drop Option available only up until the 3rd class of the entire course or the 3rd class in my sequence?  I just registered a few seconds ago.

Thanks ya'll!  I'm new to this board, so it might take me a while to adjust to AoPs lingo.

You can only drop before the 3rd class of the course, as you have access to the transcripts from the first two classes.


Let $ ABC$ be a triangle, and let $ A^{\prime}$ be the midpoint of its side $ BC$. Let $ Y$ be the foot of the $ B$-altitude of triangle $ ABC$. Finally, denote by $ H$ the orthocenter of triangle $ ABC$.

Prove that the circle with diameter $ HA$ touches the line $ A^{\prime}Y$ at the point $ Y$.

Since I need this problem for reference (at http://www.mathlinks.ro/Forum/viewtopic.php?t=155711 ), here is a [i]solution[/i]:

[hide="Solution"]Let $ Z$ be the foot of the $ C$-altitude of triangle $ ABC$.

We will work with directed angles modulo $ 180^{\circ}$. First, since the point $ H$ is the orthocenter of triangle $ ABC$, it lies on the altitudes $ BY$ and $ CZ$ of this triangle. Thus, $ \measuredangle HYA=90^{\circ}$ and $ \measuredangle AZH=90^{\circ}$.

Since $ \measuredangle BYC=90^{\circ}$ and $ \measuredangle BZC=90^{\circ}$, the points $ Y$ and $ Z$ lie on the circle with diameter $ BC$. The center of this circle is the midpoint of the segment $ BC$, i. e. the point $ A^{\prime}$. Thus, by the central angle theorem, $ \measuredangle A^{\prime}YB=90^{\circ}-\measuredangle BZY$. In other words, $ \measuredangle A^{\prime}YB=90^{\circ}-\measuredangle AZY$. Now, $ \measuredangle AZH=90^{\circ}$, so this becomes

$ \measuredangle A^{\prime}YB=90^{\circ}-\measuredangle AZY=\measuredangle AZH-\measuredangle AZY=\measuredangle YZH$.

From $ \measuredangle HYA=90^{\circ}$, it follows that the point $ Y$ lies on the circle with diameter $ HA$. If $ t$ is the tangent to the circle with diameter $ HA$ at $ Y$, then the tangent-chordal angle theorem yields $ \measuredangle\left(t;\ HY\right)=\measuredangle YZH$. On the other hand, $ \measuredangle\left(A^{\prime}Y;\ HY\right)=\measuredangle A^{\prime}YB=\measuredangle YZH$. Thus, $ \measuredangle\left(t;\ HY\right)=\measuredangle\left(A^{\prime}Y;\ HY\right)$, so that $ t\parallel A^{\prime}Y$. Hence, the lines $ t$ and $ A^{\prime}Y$ must coincide (as they are parallel but have a common point - namely, the point $ Y$). Since we have defined $ t$ as the tangent to the circle with diameter $ HA$ at $ Y$, it thus follows that the line $ A^{\prime}Y$ is the tangent to the circle with diameter $ HA$ at $ Y$. In other words, the circle with diameter $ HA$ touches the line $ A^{\prime}Y$ at the point $ Y$. The problem is thus solved.[/hide]

[hide="Or easier"]$ \angle{A'YB}=\angle{A'BY}$, since $ A'$ is the midpoint of the hypothenuse of $ \triangle{BYC}$ and $ \angle{A'BY}=\angle{XAY}$, where $ X$ is the foot of the $ A$-altitude. Thus, $ \angle{XAY}=\angle{AYB}$, i.e $ A'Y$ is tangent to the circle with diameter $ AH$.[/hide]

Easier indeed, thanks.
[hide="typo?"]I guess that by $ \angle{XAY}=\angle{AYB}$, you meant $ \angle{XAY}=\angle{A^{\prime}YB}$.[/hide]

  Darij

Or note that $ D, X, A', Y$ are concyclic since they lie on Feuerbach circle so $ \angle DYA' = 90^\circ$ :P.


function
inequalities
real analysis
real analysis unsolved

Let $ X$ and $ Y$ be nonempty sets and let $ h: X\times Y\rightarrow \Re$ have bounded range in $ \Re$.

Prove that $ \sup_y \inf_x h(x,y)\le \inf_x \sup_y h(x,y)$

Can someone please help.Thank you very much. :P

[b]Intuition. [/b]It is helpful to think of this problem as a game.  Player X chooses the value of $ x$; player Y chooses the value of $ y$. The outcome of the game is $ h(x,y)$. Player X wants to minimize this function; player Y wants to maximize it. Clearly, it is better to make your move after your opponent did his move. Now let us look at the inequality
\[ \sup_y \inf_x h(x,y)\le \inf_x \sup_y h(x,y).\]
The left hand side equals the outcome of the game when player Y makes the first move; the right hand side equals the outcome when player X makes the first move. In the former case, player X has advantage; in the latter case player Y has advantage. Therefore, the left hand side is smaller (recall that X wants to minimize the expression).

[b]Formal Proof.[/b]
Denote $ L \equal{} \sup_y \inf_x h(x,y)$. By the definition of $ \sup$, for every $ \varepsilon > 0$, there exists $ y_0$ s.t.
\[ \inf_x h(x, y_0) \geq L \minus{} \varepsilon.\]
Note that for every $ x$, by the definition of $ \sup$: $ \sup_y h(x,y) \geq h(x, y_0)$. Thus
\[ \inf_x \sup_y h(x,y) \geq \inf_x h(x, y_0) \geq L\minus{} \varepsilon.\]
Since this inequality holds for every $ \varepsilon > 0$, we have
\[ \inf_x \sup_y h(x,y) \geq L \equal{} \sup_y \inf_x h(x,y).\]

Yury, very good explaination indeed, the solution is very elegent, thank you very much for helping. :P


[quote="buzzer11"]i got d?
carrying a box together wont help them, right?[/quote]
Its not d. And they don't carry boxes together, they carry them by themselves. :)

[quote="buzzer11"][quote="Inspired By Nature"]Quincy and Celine have to move 16 small boxes and 10 large boxes. The chart indicates the time that each person takes to move each type of box. They start moving the boxes at 9:00 a.m. The earliest time at which they can be finished moving all of the boxes is
(A) 9:41 a.m. (B) 9:42 a.m. (C) 9:43 a.m. (D) 9:44 a.m. (E) 9:45 a.m.
[code]
       Celine Quincy

small  2 min. 3 min.
box

large  6 min. 5 min.
box[/code][/quote]

hmmm... so if Celine starts with the small boxes and Quincy starts with the large boxes, in 32 minutes, Celine will be finished her small boxes, and Quincy will be in the middle of her seventh.

If Celine starts helping Quincy, then three minutes, Quincy will be done one, leaving three left, and in three minutes, Celine will be done one, and in two minutes, Quincy. The last box will be finished by Celine in 4 minutes. I get:

$32+3+3+2+4 = \boxed{44}$
$9: 44$?

whats wrong with what i did?[/quote]
Is that the fastest time? :evilgrin:
I think you should experiment with this a little more. :P

[hide="a different answer"] i got c. i had celine get two large boxes and 15 small boxes. quincy got 8 large boxes and 1 small box.[/hide]

[hide]1. Celine gets 15 small boxes, Quincy gets 6 large boxes, the time is 9:30.
2. Celine gets 2 large boxes, Quincy gets 2 large boxes and a small box, the time is 9:43.[/hide]

[quote="Totally Zealous"][hide]1. Celine gets 15 small boxes, Quincy gets 6 large boxes, the time is 9:30.
2. Celine gets 2 large boxes, Quincy gets 2 large boxes and a small box, the time is 9:30.[/hide][/quote]

You should how many small boxes there and 9:30 isn't even one of the answer choices.

[quote="Totally Zealous"]2 facts:
1. We want the person who moves a type of box the shortest to move as much of that type of box as possible.
2. Since speeds (and time) is different for each person, we want a time when both people have just fully moved a complete box and no more at the same time. At 9:30 Celine has just fully moved 15 small boxes and no more while Quincy has fully moved 6 large boxes and no more. Then we just play out the rest.
If that sounds confusing, it's like this:
2 knobs are turned at different speeds. They started out with an arrow on them each facing north at the same time. Find how many time before they both face north at the same time again.
Just replace knob and arrow with celine, quincy, and box.[/quote]
I think your knobs explanation was a bit more confusing. :D
Nice facts though. :)

[quote="Totally Zealous"]2 facts:
1. We want the person who moves a type of box the shortest to move as much of that type of box as possible.
2. Since speeds (and time) is different for each person, we want a time when both people have just fully moved a complete box and no more at the same time. At 9:30 Celine has just fully moved 15 small boxes and no more while Quincy has fully moved 6 large boxes and no more. Then we just play out the rest.
If that sounds confusing, it's like this:
2 knobs are turned at different speeds. They started out with an arrow on them each facing north at the same time. Find how many time before they both face north at the same time again.
Just replace knob and arrow with celine, quincy, and box.[/quote]

okay, so i did fact number one. I dont think i completely expanded fact number 2.

[quote="buzzer11"]how did u guys approach this problem?[/quote]

i just made a table and used common sense. i'm more of a visual person, i guess.


So besides Tim, who made it?

Note to Tim: there is no way anyone ought to be able to force you to go to GHP if you made this. Talk to an ARML coach, they should know what to do.

lol, I improved my score from last year by one point.  Hurray!

Nice job, Tim!  I think Oliver also made it, in which case, nice job Oliver!

Whoah, three exclamations in a row...

Anyways, any other MOP qualifiers?

Thanks (I was sorta lucky), and Kudos to Oliver. So far the highest score in Georgia I know is Oliver's 21. Anyone who got higher?

Tim I completely agree with everyone else. Go to MOSP. I'll talk to the GHP folks if necessary.

Tom

Congrats to those who made it.

I got a 9. A whopping 2 point improvement.

Actually, I didn't send in my S-III form, so I guess I won't be doing MOP.

Hmm, I feel bad cuz red MOP's cutoff was three points(?) lower this year, and yet I still got less than half of the cutoff score.  That sadifies me...

But then again, I'm Sitan, so that was expected.  :)

PS: I lost.

PPS: Hey Oliver, are you going to be doing RSI instead?  Or were you just forgetful?

The red cutoff last year was 10, so it's 2 points lower this year. (Incidentally, the blue cutoff last year was 20, which is also 2 points lower this year)

@Oliver: so you forgot to turn in that form? :(

Congrats Tim!  :)

(Where are you guys finding these cutoff scores?)

@Sitan: I couldn't apply to RSI, I'm not a permanent resident, and yes, I guess I'm forgetful. It's not a big deal though, I can still go to GHP :)

Whee so I got +1 on last year: 7. Still -5 from my first time .....

Just kidding. I got an invitation anyways, even though I didn't turn it in on time. MOP it is then.

I would have made it if I'd been a year younger. :D  Same score as last year.  Congratulations, Tim and Oliver!

Good job Tim and Oliver!  :clap2:

And Sitan its okee! ur still in ninth grade (freshie buddyz :D) and will probably make like black mop. you have 3 yrs left, after all. xD


geometry
circumcircle
trigonometry
incenter
geometric transformation
reflection
search

Given a scalene triangle $ ABC$.Let $ O$ be the circumcenter triangle $ ABC$. Point $ A'$ lies on ray $ AO$ such that $ \angle BA'A \equal{} \angle CA'A.$ Let $ A_{1},A_{2}$ be the feet of perpendiculars from $ A'$ to $ CA,AB$, respectively. Let $ H_{A}$ be the foot of the altitude of triangle $ ABC$ from vertex $ A$. Denote by $ R_{A}$ the circumradius of triangle $ H_{A}A_{1}A_{2}$. Define $ R_{B},R_{C}$ similarly. Prove that $ \frac {1}{R_{A}} \plus{} \frac {1}{R_{B}} \plus{} \frac {1}{R_{C}} \equal{} \frac {2}{R}$.( where $ R$ is the circumradius of triangle $ ABC$)

It is clear that $ A^{\prime}$ is the second intersection of the line $ AO$ with the circumcircle of triangle $ BOC$. Since the quadrilateral $ A^{\prime}A_{1}AA_{2}$ is cyclic, $ \angle{A^{\prime}A_{1}A_{2}} \equal{} 90^{\circ} \minus{} \angle{A_{2}A_{1}A} \equal{} \angle{OAB} \equal{} 90 \minus{} \angle{C}$. Hence, $ \angle{A_{2}A_{1}A} \equal{} \angle{BCA}$, and thus, the lines $ A_{1}A_{2}$ and $ BC$ are parallel. Now, since the triangles $ ABC$ and $ AA_{2}A_{1}$ are homothetic, the circumcenter $ O^{\prime}$ of $ AA_{2}A_{1}$ lies on $ AO$, and $ \frac {AH}{AH_{A}} \equal{} \frac {AO}{AO^{\prime}}$.
On other hand, $ AA_{2} \equal{} AA^{\prime}\sin{C}$, $ AB \equal{} 2AO\sin{C}$, and therefore $ \frac {AB}{AA_{2}} \equal{} \frac {2AO}{AA^{\prime}} \equal{} \frac {AO}{AO^{\prime}} \equal{} \frac {AH}{AH_{A}}$. Thus, the lines $ A_{2}H_{A}$, $ A_{1}H_{A}$ are perpendicular to $ CA$, $ AB$, respectively, and hereby, the triangles $ HBC$ and $ H_{A}A_{2}A_{1}$ are homothetic. This yields that $ \frac {R}{R_{A}} \equal{} \frac {AH}{AH_{A}}$, and similarly, $ \frac {R}{R_{B}} \equal{} \frac {BH}{BH_{B}}$, $ \frac {R}{R_{C}} \equal{} \frac {CH}{CH_{B}}$, because the triangles $ BHC$, $ CHA$, $ AHB$ have the same circumradius $ R$.
By this, we conclude that
\[ \sum_{cyc}{\frac {R}{R_{A}}} \equal{} \sum_{cyc}{\frac {AH}{AH_{A}}} \equal{} \sum_{cyc}{\frac {(AH_{A} \minus{} HH_{A})a}{2S}} \equal{} \frac {4S}{2S} \equal{} 2,
\]
which means that $ \frac {1}{R_{A}} \plus{} \frac {1}{R_{B}} \plus{} \frac {1}{R_{C}} \equal{} \frac {2}{R}$, as desired.

Can you post the remaining CMO problems?

Slightly different solution, using a bit of trig:
Consider the antipode $ X$ of $ A$ on the circumcircle of $ ABC$. Choose a point $ Y$ so that $ X$ is the incentre of $ BCY$. Then $ \angle XBC \equal{} \angle YBC \equal{} 90\minus{}B$, $ \angle XCB \equal{} \angle XCY \equal{} 90\minus{}C$, $ \angle BYX \equal{} \angle CYX \equal{} 90\minus{}A$ and so $ \angle BXY \equal{} A\plus{}B$, but $ \angle AXB \equal{} \angle ACB \equal{} C$, so $ \angle BXY \plus{} \angle AXB \equal{} 180$, so $ A,X,Y$ are collinear, so $ Y$ lies on $ AO$. Also, $ \angle BYO \equal{} \angle CYO$. Thus $ Y$ has the same properties as point $ A'$, so $ Y$ is $ A'$.
Consider isogonal conjugate $ Z$ of $ A'$. Since $ A'$ lies on $ AO$, $ Z$ lies on $ AH_{A}$. Thus $ H_{A}$ is one of the vertices of the pedal triangle of $ Z$. Let the feet of the perpendicular from $ A'$ to $ BC$ be $ A_{3}$. Using the well-known fact that the pedal triangles of two points, that are isogonal conjugates of each other, have the same circumcircle, it follows that $ A_{1}A_{2}A_{3}H_{A}$ is concyclic. So $ R_{A}$ is simply the radius of the circumcircle of the pedal triangle of $ A'$.
Since $ BA'A_{2}A_{3}$ is concyclic, $ \angle A_{2}A_{3}A' \equal{} \angle A_{2}BA' \equal{} B$, and similarly $ \angle A_{1}A_{3}A' \equal{}\angle A'CA_{1} \equal{} C$. Further,  $ \angle A_{2}A'A_{3} \equal{} 180 \minus{} \angle A_{2}BA_{3} \equal{} B$, and $ \angle A_{3}A'A_{1} \equal{} 180 \minus{} \angle A_{3}CA_{1} \equal{} C$. Thus $ A_{2}A_{3}A'$ and $ A_{1}A_{3}A'$ are iscoceles, so $ A'A_{2} \equal{} A_{2}A_{3}$ and $ A'A_{1}\equal{}A_{3}A_{1}$. Thus $ A_{3}$ is the reflection of $ A'$ in $ A_{1}A_{2}$, and the circumradius $ R_{A}$ of $ A_{1}A_{2}A_{3}$, is the circumradius of cyclic quadrilateral $ AA_{2}A_{1}A'$, which is clearly $ \frac{AA'}{2}$.
Sin Rule in $ ABA'$ gives $ AA' \equal{} \frac {c sin B}{cos A} \equal{} \frac {2R sin C sin B}{cos A}$, so $ \frac{R}{R_{A}} \equal{} \frac {cos A}{sin B sin C}$. Thus we need to prove that: $ \sum \frac {cos A}{sin B sin C} \equal{} 2$, or equivalantly $ \sum \frac {cos A sin A}{sin A sin B sin C} \equal{} 2$, or equivalantly $ \sum sin 2A \equal{} 4 sin A sin B sin C$, which is well-known.
Here's a proof: We must show that $ sin 2A \plus{} sin 2B \equal{} sin C (4 sin A sin B \minus{} 2 cos C)$. But $ LHS \equal{} 2 sin (A\plus{}B) cos (A\minus{}B) \equal{} 2 sin C cos (A\minus{}B)$ and $ RHS \equal{} sin C (4 sin A sin B \plus{} 2 cos (A\plus{}B)) \equal{} 2 sin C ( 2 sin A sin B \plus{} cos (A\plus{}B)) \equal{} 2 sin C cos (A\minus{}B)$, so we are finished.

See  http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1136495808&t=55902

  It shows one more time that the tandem Gergonne-Grinberg is unbeatable as far as
  nice solutions are concerned, but only when  they are hand in hand
  (including all the other hands).

  T.Y.
  M.T.

My solution is a bit different than another solutions:
Solution:
It is easy to understand that there is only one such point $ A'$ on the ray $ AO$.Otherwise,$ A'B\equal{}A'C$,or which is equivalent to $ BA\equal{}AC$.
We shall construct a such point.Let $ H_1\in AC$,$ H_2\in AB$,be the points,such that $ H_AH_1\perp AC$ and $ H_AH_2\perp AB$.Denote
$ A_2\in H_AH_1\cap AB$ and $ A_1\in H_AH_2\cap AC$.Let $ A'$ be a point,such that $ H_AA_1A'A_2$ is a parallelogram.
But then $ \frac{A'A_1}{A'A_2}\equal{}\frac{A_2H_A}{A_1H_A}\equal{}\frac{H_AH_2}{H_AH_1}$,hence $ A,O,A'$ are collinear.
Denote $ A'A_1\cap BC\equal{} J$ and $ A'A_2\cap BC\equal{}K$.It is easy to understand that quadrilaterals $ AH_AA_1J,ABA'J,ACA'K,AH_AH_2K$ are cyclic,thus $ \angle JH_AA_1\equal{}\frac{\pi}{2}\minus{}\angle B\Rightarrow \angle AJB\equal{}\angle B\plus{}\angle C\minus{}\frac{\pi}{2}\equal{}\angle AKH_A$,hence
$ \boxed{\angle BA'A\equal{}\angle CA'A}$.
It is obvious that $ \triangle H_AA_2A_1\sim\triangle KJA'$ with coefficient $ 2$,but since $ AJ\equal{}AK\equal{}AA'$,then circumradius of $ \triangle KJA'$,equal to $ AJ\equal{}\frac{h_a}{\cos{\angle A}}$.Hence $ R_A\equal{}\frac{h_a}{2\cos{\angle A}}$,the rest is the same as in epitomy's proof.

:lol: This solution can prove Ha is an orthocenter
1. triangle [color=red]A'BA2[/color] and triangle [color=red]ABHa[/color] are similar (see the angle A'BA2)
Then, triangle [color=red]ABA'[/color] and triangle [color=red]HaBA2[/color] are similar too, by 1.

....


construct a measurable subset E of [0,1] such that for every sub-interval I both $ E\cap I$ and $ I \minus{} E$ have positive measure.

( hint: think of  cantor type sets)


Never mind, my poll isn't working so forget about that...so who's your fav character for SSBB? I know this topic is very exhausting, but still I would like your input...mine: Link & Snake

Well, I'm only basing this on prev knowledge(i.e. old characters) but I anticipate using Kirby and Jigglypuff quite a bit. I'm really gonna have to try the new characters myself before I make an opinion.

Totally same as perfect, except replace kirby/jiggly with Cpt. Falcon (oh please, let there be knee of justice)/Marth/Zelda

Don't create multiple threads of the same topic.

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=190999&sid=2c0782ee8e20a07b6e819176ca12b3a8]Here[/url] is your discussion of SSBB.


calculus
derivative
inequalities
inequalities unsolved

find value minimum of expression:
$ A \equal{} \frac {ma^x}{(b \minus{} c)^{2m_1}} \plus{} \frac {nb^y}{(a \minus{} c)^{2n_1}} \plus{} \frac {kc^z}{(b \minus{} a)^{2k_1}}$
 where : $ x,y,x ,m_1,n_1,k_1 \in Z$  and $ m,n,k$  real positive numbers
  i don't  it right or wrong  ?
 but i think that  it very interested

this problem  easier :
$ \frac{a^2}{(b\minus{}c)^2} \plus{}\frac{b^2}{(c\minus{}a)^2}\plus{}\frac{c^2}{(a\minus{}b)^2}  \geq 2$

[quote="Hong Quy"]this problem  easier :
$ \frac {a^2}{(b - c)^2} + \frac {b^2}{(c - a)^2} + \frac {c^2}{(a - b)^2} \geq 2$[/quote]
use: $ \sum_{cyc} {\frac{ab}{(b-c)(c-a)}=-1}$
we have $ LHS\geq\ -2\sum_{cyc} {\frac{ab}{(b-c)(c-a)}=2}$

we have other solution :
$ f(a,b,c)  \geq (a-b,b-c,0)$
 hence $ c=0$ we have ;
${{ \frac[a^2}{b^2} +\frac[b^2}{a^2} \geq 2$

[quote="toanIneq"]
 $ LHS\geq\ - 2\sum_{cyc} {\frac {ab}{(b - c)(c - a)} = 2}$[/quote]
why? this does not follow from AM-GM

also why? $ f(a,b,c)\geq f(a-b,b-c,0)$
and can u infact reach the equalitiy or can u just get arbitrary close?

anyway i managed to prove the case where $ a,b,c$ are all positve, but i have no good ideas for two of them positive and one negative (cases two negative or all negative are identical to those above i belive) except derivatives or muirhead-schurr combination. i didn't try to do it that way but i did quite a lot of trying to find counterexamples with computer and the statement apears true for all real $ a,b,c$.

if $ a,b,c\geq 0$. wlog assume $ a\geq b\geq c$
denote $ a-b=x$ and $ b-c=y$
$ \frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b^2)}=\frac{(c+y+x)^2}{y^2}+\frac{(c+y)^2}{(x+y)^2}+\frac{c^2}{x^2}$
RHS is obiously smaler if $ c$ is smaller so
$ RHS\geq \frac{(y+x)^2}{y^2}+\frac{y^2}{(x+y)^2}> 2$ impossible to reach equallity but possible to get arbitrarry close by $ c=0$ and $ x$ very small positive number

nobody has ideal  about my problem, i think  it's easy

[quote="Jure the frEEEk"][quote="toanIneq"]
 $ LHS\geq\ - 2\sum_{cyc} {\frac {ab}{(b - c)(c - a)} = 2}$[/quote]
why? this does not follow from AM-GM
[/quote]

[color=green]Using $ x^2 + y^2 + z^2\ge - 2(xy + yz + zx)$ ( because it is equivalent to $ (x + y + z)^2\ge 0$. )[/color]

[quote="Jure the frEEEk"]
also why? $ f(a,b,c)\geq f(a - b,b - c,0)$
[/quote]

[color=green]Expanding it, plz.[/color]  :)

I have some ideas :
[u]1[/u] If $ 8\not | xyz$ then $ f$ has no minimum if we cansider $ a\to \minus{}\infty$ and $ b>c\to 0$
[u]2[/u] So only to check for the case $ 2|x,y,z$
We can suppose have 2 number is negative. $ a>b\geq 0$ (other case have same method )
If $ m_1>x$ then we have : $ inf {f}\equal{}0$
So we consider the case $ m_1\geq x_1$
[b]Case 1 [/b] $ c\geq 0$ then $ f(a,b,c)\geq f (a\minus{}c,b\minus{}c,0)$
Easy to find minimum of $ f(a\minus{}c,b\minus{}c,0)$ in special case  :)
[b]Case 2[/b] $ c<0$ I think it is the most hard case . (I'll try to solve in the next time ).

[quote="Hong Quy"]this problem  easier :
$ \frac {a^2}{(b \minus{} c)^2} \plus{} \frac {b^2}{(c \minus{} a)^2} \plus{} \frac {c^2}{(a \minus{} b)^2} \geq 2$[/quote]

A simpler solution:

Let $ A \equal{} \frac{a}{b\minus{}c}, B \equal{} \frac{b}{c\minus{}a}, C \equal{} \frac {c}{a\minus{}b}$

then we have $ (A\plus{}1)(B\plus{}1)(C\plus{}1) \equal{} (A\minus{}1)(B\minus{}1)(C\minus{}1)$

which means $ AB\plus{}BC\plus{}CA \equal{} \minus{}1$

then $ A^2\plus{}B^2\plus{}C^2 \equal{} (A\plus{}B\plus{}C)^2\plus{}2 \geq 2$.


An equilateral triangle has an area of $ 64\sqrt{3}$ $ \text{cm}^2$. If each side of the triangle is decreased by 4 cm, by how many square centimeters is the area decreased?

$ \frac{s^2\sqrt3}4\equal{}64\sqrt3\implies s^2\equal{}64\cdot4\implies s\equal{}8\cdot2\equal{}16$

$ \frac{(s\minus{}4)^2\sqrt3}4\equal{}\frac{12^2\sqrt3}4\equal{}36\sqrt3$

Our final answer is $ 64\sqrt3\minus{}36\sqrt3\equal{}\boxed{28\sqrt3}$.


superior algebra
superior algebra unsolved

[quote="wiosna"]Give an example of ring that is semisimple but isn't semiprime.[/quote]

There's no such example because every semisimple ring is semiprime. If you meant a semiprime ring which is not semisimple, then $ \mathbb{Z}$ would be an example.


Does n e 1 know the date and syllabus of  second round of JMO?

i don't think so that even first level paper was held this year

r u talking abt the one conducted by Dr.S.P Arya


trigonometry
calculus
derivative
real analysis
real analysis unsolved

\[ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{n}\cos\frac{x}{2^n}\] is uniformly continuous or not ?

Yes it is. Just differentiate and observe that the series of the derivatives converges absolutely and uniformly (so the sum of the series is differentiable) and its sum never exceeds $ 1$ in absolute value.


Let P be an interior point of triangle ABC and extend lines from the vertices through P to opposite sides. Let AP = a, BP = b, CP = c, and the extensions from P to the opposite sides all have length d. If a+b+c = 43 and d = 3 then find abc.

does anyone know a mass points solution to this? since i found this AIME problem in a mass points article...

Let $D, E, F$ be the intersections of $AP$ and $BC$, $AC$, $AB$, then use:
\[ \frac{PD}{AD}+\frac{PE}{BE}+\frac{PF}{CF}=1  \]

hmm wait is that a well known rule or something? or do we have to figure that out ourselves

It can be used as a lemma. An one-line proof follows the area ratios.


Find the mean of the seventeenth term of the arithmetic sequence $ \{1,5,9,13,\ldots\}$ and the eighteenth term of the arithmetic sequence $ \{1,9,17,25,\ldots\}$.


salam ferhad.
[quote="Ferhad"]I hope this year Azerbaijan team will got the all medals[/quote]
maybe you meant next year :wink:

[quote="fedya89"]HI i am Ferhad aliyev from Azerbaijan/Nakhchivan.I hope this year Azerbaijan team will got the all medals.I want it with my heart.From here SALAM to everybody.....................[/quote]

Salam Farhad. Welcome on Aops forum :) .In which grade are you in?
Are you interested in olympiad training?


The source of this is a lecture I've given in a differential equations class. We'll talk about a family of objects of different sizes falling through the atmosphere. The fun of the problems involves getting actual numbers; the numbers should be treated as crude approximations. (Anything more than 2 significant figures is overkill.)

Naturally, we will make many simplifying assumptions in our model. We assume that the only forces that act are gravity and air resistance, and that the object is falling straight down from initially being at rest. We assume that air resistance can be modeled linearly - that air resistance is a force proportional to the first power of the velocity. We further assume that our various objects are the same shape, the same density, and fall in the same orientation, and that the coefficient of air resistance is itself proportional to surface area, which is in turn proportional to the 2/3 power of the mass. We ignore that fact that the properties of the atmosphere vary with altitude and act as if the coefficient of resistance is the same at all altitudes. Each such falling body will have a terminal velocity - a velocity towards which it tends in the limit if it falls far enough. To create a marker point, we say that a 70 kg object from our collection tends toward a terminal velocity of 54 m/s. (That's supposed to be approximately a human with a terminal velocity of about 120 mph.)

For each of the following objects, ask the following questions:

a) What is its terminal velocity? And for reference in later parts of the problem, what is $ 90\%$ of terminal velocity?

b) How much time does it take, falling from rest, to reach $ 90\%$ of terminal velocity?

c) What height must it fall from to reach $ 90\%$ of terminal velocity by the time it reaches the ground?

Object 1: Mass = 70 kg. (And it's given that terminal velocity = 54 m/s.)

Object 2: Mass = 5 kg.

Object 3: Mass = 15 g.

Object 4: Mass = 4000 kg.

Object 5: Mass = 50000 kg.

(In the lecture, I referred to these five objects as "human," "cat," "mouse," "elephant," and "whale." Of course, the real animals are in fact of very different shapes, which would throw off the calculation by quite a bit. We are oversimplifying here. I also, in discussing the answers, converted the heights into feet and the velocities into mph; in your case, report the answer in whatever units you're comfortable with.)

Alrighty i can actually do and understand this one:
[hide]
lets make positve down, cuz i get odd results if i dont, for some reason.
$ F_{grav}=mg$
$ F_{resistance} \propto vm^{2/3}$
$ F_{ar}=kvm^{2/3}$ opposing motion.
Therefore, $ F_{net}=mg-kvm^{2/3}$
$ a=g-kvm^{-1/3}$
let $ x: =km^{-1/3}$
then we get $ \dfrac{dv}{dt}=g-xv$
$ \displaystyle \int \frac{1}{g-xv} dv = \int dt$
$ \displaystyle -\frac{1}{x} \ln{|g-xv|}=t+C$
note that $ a$ is in the downward direction: therefore, $ 0<a=g-xv$
$ \ln{g-xv}=-xt+C$
$ g-xv=Ce^{-xt}$
$ v=\dfrac{g-Ce^{-xt}}{x}$
setting $ v_0=0$, we get that C=g.
then $ \displaystyle \lim_{t \to \infty} v=\dfrac{g}{x}$
x=$ \dfrac{g}{v}=\dfrac{9.8}{54}=0.18$
But $ x=km^{-1/3}=0.18=0.24k$
$ k=.075$
a) $ v_{terminal}=\dfrac{g}{k} m^{1/3}=13m^{1/3}$
$ 90 %
$ of $ v_{terminal}=12m^{1/3}$
since $ v_{terminal}=\dfrac{g}{x}$, $ 90%
$ of $ v_{terminal}=$
$ 0.9 \dfrac{g}{x}=\dfrac{g-ge^{-xt}}{x}$
$ 0.9=1-e^{-xt}$
$ e^{\dfrac{g}{k}m^{1/3}t}=10$
$ 13m^{1/3}t=2.3$
b)$ t=\dfrac{0.18}{m^{1/3}}$
now, $ \displaystyle \Delta h  = \int_0^{t_{impact}} \dfrac{g-ge^{-xt}}{x}$
$ \displaystyle = \dfrac{g}{x} t - \dfrac{g}{x^2}e^{-xt} |^{t_i}_0$
$ =\dfrac{g m^{1/3}}{k} \cdot \dfrac{0.18}{m^{1/3}} - \dfrac{g m^{2/3}}{k^2} \cdot e^{-xt_i}-(-\dfrac{g m^{2/3}}{k^2})$
$ =\ln{10} + 0.9\dfrac{g m^{2/3}}{k^2}$
c)$ h=2.3+16m^{2/3}$

human: a) 54, b) 0.043, c)270
cat: a) 22, b) 0.10, c) 48
mouse: a) 3.2, b) .71, c) 3.3
elephant: a)200, b) 0.011, c) 4000
whale: a) 480, b) 0.0048, c) 21 000
(kms)
human: a) 120 mph, c) 890 ft = 0.17 mi
cat: a) 50 mph, c) 160 ft
mouse: a) 7.2 mph, c) 11 ft
elephant: a) 465 mph, c) 2.5 mi
whale: a) 1100 mph, c) 13 mi
[/hide]
pretty amazing.

Let the coefficient of air resistance be $ \eta$. The differential equation for the motion is $ m\ddot{x}\equal{}mg\minus{}\eta\dot{x}$, or in terms of $ v$,
$ m\dot{v}\equal{}mg\minus{}\eta v$

Solving this, we have $ v(t)$,

$ v(t)\equal{}\frac{mg}{\eta}\left( 1\minus{}e^{\minus{}\frac{\eta t}{m}}\right)$

For terminal velocity, as $ t\rightarrow\infty$,
$ v_T\equal{}\frac{mg}{\eta}$

now, $ \eta\equal{}km^{\frac{2}{3}}$, and hence,

$ v_T\equal{}\frac{gm^{\frac{1}{3}}}{k}$

Putting in the marker values, we get $ k\approx 7.5\times 10^{\minus{}1}$ in S.I. units

a) Terminal velocities:
$ 5.4\times 10^1\text{m/s}$
$ 2.2\times 10^1\text{m/s}$
$ 5.8\times 10^{\minus{}1}\text{m/s}$
$ 2.1\times 10^2\text{m/s}$
$ 4.8\times 10^2\text{m/s}$

b) $ t_{0.9}\equal{}\frac{m^{\frac{1}{3}}\ln 10}{k}$
$ 1.3\times 10^1 \text{s}$
$ 5.3\times 10^0 \text{s}$
$ 7.6\times 10^{\minus{}1} \text{s}$
$ 4.9\times 10^1 \text{s}$
$ 1.1\times 10^2 \text{s}$

c)$ x\equal{}\frac{mg}{\eta}t\plus{}\frac{m^2g}{\eta^2}e^{\minus{}\frac{\eta t}{m}}$
Putting $ t\equal{}t_{0.9}$,
$ 7.3\times 10^2\text{m}$
$ 1.2\times 10^2\text{m}$
$ 2.6\times 10^0\text{m}$
$ 1.1\times 10^4\text{m}$
$ 5.6\times 10^4\text{m}$

The variation in the distances seems quite large

EDIT: Sorry I didn't see [b]bowei[/b]'s earlier response

For the terminal velocities, I agree with both of you for 1, 2, 4, and 5, and with bowei for #3 (the mouse); shreyas_patankar should check his numbers for that one.

For the time to 90% of terminal velocity, I agree with shreyas_patankar (within roundoff error, anyway) in all five cases, including the mouse this time. I don't agree with any of bowei's times.

I don't agree with either of you about the height. My methods were generally the same as shreyas_patankar, but in this case, you should recheck the details.

Roughly speaking, here are my answers:

[hide]For the human reaches 90% of terminal velocity, or 110 mph, falling from 1300 or 1400 feet - one of the tallest buildings in the world. If he doesn't have a parachute, the impact will be fatal.

The cat reaches 90% of terminal velocity, or 45 mph, falling from about 230 feet (a 20-story building). The cat probably will survive the fall, but could be seriously injured.

The mouse reaches 90% of terminal velocity falling of a bookshelf. No matter how high you drop it from, it touches down lightly (at about 7 mph) and runs away. (That is, if there aren't any curious hawks in the neigbhorhhod.)

Drop the elephant from an airplane at 20,000 feet, and it takes 45 or 50 seconds to fall and reaches over 400 mph (90% of its terminal velocity), leaving a crater and a big mess.

As for the whale, go up to 50 miles high, and the whale will reach supersonic speeds. For you Douglas Adams fans, I believe the phrase is "Oh no, not again."[/hide]

:oops: There was a syntax error while punching the values in my calculator.
a)Terminal velocity of mouse:$ 3.2\times 10^0\text{m/s}$
c)I am still getting slightly different answers for the heights
man: $ 7.1\times 10^2\text{m}$
cat: $ 1.2\times 10^2\text{m}$
mouse: $ 2.5\times 10^0\text{m}$
elephant: $ 1.1\times 10^4\text{m}$
whale: $ 5.7\times 10^4\text{m}$

So if a sky diver's parachute doesn't work, he will smash down at 200 km/h which will certainly kill him, but not much worse than that.

Shreyas - look at this, quoted from your first post:

[quote]c) $ x \equal{} \frac {mg}{\eta}t \plus{} \frac {m^2g}{\eta^2}e^{ \minus{} \frac {\eta t}{m}}$ [/quote]
There's a problem with that. Try setting $ t \equal{} 0.$ That would give you $ x \equal{} \frac {m^2g}{\eta^2}$ when it should give you $ x \equal{} 0.$

So it's an initial condition problem. As another clue - try expanding your answer in a power series about $ t \equal{} 0.$ You should get $ x(t) \equal{} \frac12gt^2 \plus{} O(t^3).$

------

The biggest weakness of this model is that linear resistance - force directly proportional to velocity - isn't a very good approximation across all ranges of both size and velocity. If you want to make this model more sophisticated, the first place to look would be into the phenomenon of fluid resistance; much has been written, of course.

silly me

$ x(t)\equal{}\int v(t)dt\plus{}C$
Substituting initial condition $ x(0)\equal{}0$, we have $ x(t)\equal{}\frac{mg}{\eta}t\plus{}\frac{m^2g}{\eta^2}e^{\minus{}\frac{\eta t}{m}}\minus{}\frac{m^2g}{\eta^2}$

giving heights as
man: $ 4.2\times 10^2\text{m}$
cat: $ 7.2\times 10^1\text{m}$
mouse: $ 1.5\times 10^0\text{m}$
elephant: $ 6.2\times 10^3\text{m}$
whale: $ 3.3\times 10^4\text{m}$

The problem for the cat is a solved example in Halliday/Resnick/Walker's text. According to them, the cat actually fares better on higher jumps as it spreads out and lowers the terminal velocity.


Wow... What a test!

How did all the Michiganders do?

There were some bad questions I thought... What is a "uniform border"??
lol...

Predicted: 35  :maybe:

Predicted: 36

Yeah, that question was weird.

I think I only got 23-25. It was tough. :( . I did all MMPC's 1997-2003 again for practice and it was my lowest score. The firsrt 10 questions of this year were similar in difficulty to 11-20 of 2003. I did so bad, and I'd been scoring 30's on previous ones. Someone console me. :(

I say the qualifying score will be one less than last year; it'll be a 23, so I may just barely make it.  Any opinions on this? How are you guys preparing for Part II?

I am putting my money on a 20 or 21 for part II cutoff.

How did you guys find the test compared to other years- hard/ easy? What questions stumped you?
Good score, tetrahedr0n and ragging; you all are definitely in the running for silver and gold and are assured a bronze.
Again, how will you study for Part II?

HA! Assured bronze... I would love to think of that!  :D .  If I screw up on the part II, I'm pretty sure there is NO way I could get bronze actually.  I thought this year's was about the same as last year's test.  I'd say it was about moderate difficulty... I made too many stupid mistakes I thought :? .  Congrats to you too samus.

Questions 29 and 25 stumped me... I really had no quick ideas on what to do and not enough time to find one.

I think the cutoff will be 23 this year just like lasts...

Btw, here are the answers if you want to check.

1. b
2. a
3. c
4. d
5. c
6. b
7. c
8. a
9. d
10. c
11. a
12. b
13. e
14. b
15. d
16. b
17. a
18. c
19. b
20. e
21. c
22. d
23. e
24. b
25. c *or E... not sure*
26. d.
27. b
28. c
29. d
30. a
31. c
32. a
33. e
34. d
35. c
36. a
37. e
38. c
39. b
40. d

25. c

isn't it log base 2 of x?

meaning that x has to be greater than 0 - one of solutions wouldn't work and that the answer is E?

-richtao

Richard Tao... I know who you are!  ICAE all the way.... (you probably dont know who I am anyways :))  I put E too, but some people in Troy say that it's C so I just said it was C too....  How'd you do btw?

Lol Richard Tao... haha well 25. E and im almost positive because me and Dan Pan tried doing it for a while.

Btw i think i got a 35.. a disappointment

This test was definitely harder than last year's. The cut off will probably be 21 or 22. Professor Cheng was the cause of the hardness :-P. He screwed everyone over. Well good luck guys on Part 2....

Soon the questions will be online Max. Ragging has posted arguably the most difficult questions of the test in the Intermediate Forum.

This years test seemed to be harder than previous years. I also predict the cutoff to be around 20. Also: was last years cutoff really as high as 23, because I got a 22 and was invited.

And I want to know how a 35 is a dissipointment.
I'm happy with the 30 I think I got.

Yep... Last year's cutoff was 23... I still think this year's test was about the same as that of last year so I guess cutoff is 22 or 23.

I think 35 is pretty good... 30 is good as well! Congratulations everyone.

Oh my god... I sure hope 49 is bronze... I'm hoping there is a massive amount of 43 or something.  43.0 is so high for Top 100. lol

Lots of people seem to be around 41-42 though...

Btw, don't you think that the MMPC should have have its own forum in the Michigan community since it's obviously such a big topic of discussion?

Yes, it would be nice of you guys to all start posting in the local forum... we're falling behind MA.

Hey, this is Marc--I go to Midland High School.

I scored a 31 on part I, and I'm hoping for 25 on part II.
31 + (25*1.2) = 61.

Based on the considerably higher scores this year, I'm guessing that puts me about 20-30th. Does that sound realistic?

By the way, I have Mr. Redman as a professor. (I'm dual-enrolled at Delta to take Differential Equations). I'm surprised he didn't have the scores up more quickly. He is an [i]incredibly[/i] organized teached. He told us how many homework problems we'd have, what dates we would have tests, how many questions each test would have and how much they'd be worth, the total numebr of points possible...all on the first day  :)

What did it take to score gold, silver, and bronze last year?

boy_21, you are now part of the MI community. You can check out the MI forum at the bottom in the "National and Local communities" section. That way, we can spam more about MMPC.

So, ask Redman how I did   :D

Anyway, I'm saying 61 will be a ... 12th - 24th place.

And for last year, I don't remember anything other than silver cutoff which was 52.0 (my score) + 1.6 = 53.6. Gold was higher by another 7 or 8 points? I wish they didn't take down the list from the website. I can still find the results from two years ago, but not last year.

While your at it, put in a good word for a person named Philip. See if Hamilton can convince him to release the scores. :D

The chance of Redman telling us any of our scores or places is slightly less than the chance of 1/x ever reaching 0. :(

I only got about 36 last year (I completely bombed part II) which put me about 120th place. What about you guys?

You could try blackmailing Redman?

Anyway, last year, I had 34 on part I, 15 on part II (less than 2 years ago :() and I ended with 52, which put me in 13th. Of course, that gave me all the glory of a 50th place, when a person with 2 more points than me ranked 11th got extra money, pictures, attention :)

2 years ago, 26 on part I, 16 part II, and i was about 120 th as well.

I'm considering putting on the form for "awards and activities":

1-3) a bunch of real accomplishments/activities
4) successful extraction of a rubber stopper from a volumetric flask.

we should all do something funny for #4

Tension building... What's every body hoping for? I want a bronze, but I don't think I'll make it.

Hoping for silver. I'm just looking forward to seeing people I haven't talked to in a long time. +the lectures.

I am looking for the MMPC pdfs - part2 of years 2008-09, 2011  does anyone have them?

Only with problem wordings (I do not care if it has solutions)


A carpenter wants to make a circular table out of a 4' by 8' piece of wood by putting together 2 congruent semicircles.  He wishes for the table to be as large as possible.  Find the diameter of the table.

[asy]draw((0,0)--(8,0)--(8,4)--(0,4)--cycle,blue); path circ1, circ2; circ1=circle((2.5,0),2.5); circ2=circle((5.5,4),2.5); draw(circ1); draw(circ2);[/asy]

Ignore the extra circle parts outside of the blue rectangle.

[quote="buzzer11"][hide]
Create a right triangle by connecting the centers of the two triangles and extending the center of the top triangle down to meet the rectangle.
By the Pythagorean Theorem, and calling the radius of the circles $r$, the third side (the base) equals $\sqrt{4r^{2}-16}= 2\sqrt{r^{2}-4}$.

Then, this base, added to two radii, should equal 8.

$2\sqrt{r^{2}-4}+2r = 8$
$\sqrt{r^{2}-4}= 4-r$
$r^{2}-4 = 16-8r+r^{2}$
$8r = 20$
$r= \frac{5}{2}$.
[/hide][/quote]

Since its asking for diameter, then you would have to double that for the answer, right?

Oh, sorry.  :blush:
Didn't read the problem carefully.

So yes,

[hide="The answer would be..."]5[/hide]


$ a;b;c;d\geq0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$
prove that:
$ abc \plus{} bcd \plus{} cda \plus{} dab\leq\frac {1}{27} \plus{} \frac {1}{27}abcd$

[quote="peine"]$ a;b;c;d\geq0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$
prove that:
$ abc \plus{} bcd \plus{} cda \plus{} dab\geq\frac {1}{27} \plus{} \frac {1}{27}abcd$[/quote]
Try $ a\equal{}b\equal{}0.$ :wink:

This has appeared many times, e.g.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168355[/url]


Find the solution of $ \frac{\frac{d^2y}{dx^2}}{\left\{1\plus{}\left(\frac{dy}{dx}\right)^2\right\}^{\frac{3}{2}}}\equal{}\frac{1}{a}\ (a>0)$.

hello kunny, we will rewrite your equation as follows $ ay^{''}(x) = \sqrt {(1 + y^'(x))^3}$ setting $ y^'(x) = p$ we have ${ \frac {dp}{\sqrt {(1 + p^2)^3}}} = \frac {dx}{a}$.Integrating both sides with respect to $ x$ we have $ \frac {p}{\sqrt {1 + p^2}} = \frac {x}{a} + C_1$. From here we get ${ y^'(x) = \pm\sqrt {\frac {(\frac {x}{a} + C_1)^2}{1 - (\frac {x}{a} + C_1)^2}}}$. Integrating again with respect to $ x$ we get the searched function as
$ y(x) = \pm\frac {\sqrt { - (aC_1 + x)^2}\sqrt {a^2(C_1^2 - 1) + 2aC_1x + x^2}}{aC_1 + x} + C_2$
Sonnhard.


The product of the $ 9$ factors $ \left (1\minus{}\frac{1}{2} \right ) \left (1\minus{}\frac{1}{3} \right ) \left (1\minus{}\frac{1}{4} \right ) \ldots \left (1\minus{}\frac{1}{10} \right )\equal{}$

\[ \textbf{(A)}\ \frac{1}{10} \qquad
\textbf{(B)}\ \frac{1}{9} \qquad
\textbf{(C)}\ \frac{1}{2} \qquad
\textbf{(D)}\ \frac{10}{11} \qquad
\textbf{(E)}\ \frac{11}{2}
\]

This is a telescoping product.
$ \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{8}{9}\cdot\frac{9}{10}\equal{}\frac{1}{10}$

[hide]
   (1/2)*(2/3)*(3/4).....*(9/10)==(1/10)
   only numerator of first term and denominator of last will remain uncancelled :)
[/hide]

[hide="Solution"]
This is equivalent to $\frac{1}{2} \cdot \frac{2}{3} \cdot \ldots\cdot \frac{8}{9} \cdot \frac{9}{10}$.  In this [b]telescoping product[/b], notice that for all the fractions but the last one, the denominator cancels with the numerator of the next fraction.  Therefore, everything cancels but the numerator of the first fraction and the denominator of the last fraction, leaving us with $\boxed{\text{(A)} \frac{1}{10}}$
[/hide]

Don't add replies to these really old posts, it's kind of annoying because the posts just end up at the top of everyone's feed

[quote=combinatorialgeometry]Another way to do it would be that it's 9!/10! from just multiplying so it's 1/10[/quote]

[quote=mathwhiz16][quote=doitsudoitsu][quote=DeathLlama9]...why did you revive a 5.5 year old thread that had already been resolved?[/quote][/quote][/quote]

works perfectly in this situation


geometry
perimeter
inradius
area of a triangle
Heron\u0027s formula

Hello,

Can someone please state me what is the use of Heron's formula and how can I use in a simple example from you?

Please since we are talking about geometry can someone please till me how can I find the area of a triangle by just knowing the sides only?

Thanking you

Heron's:

$ s \equal{} \frac {a \plus{} b \plus{} c}{2}$
A-area
$ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$

Example: Find the area of triangle with sides 13,14,15

Apply Heron's to get 84.

Edited, thanks to Rofl..

Herons Fromula does give the area of any triangle by just knowing the side lengths.  The relation between the two is $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ where $ s$ is the semi-perimeter or $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$.  A simple example may be like: what is the area of triangle $ ABC$ if $ AB\equal{}6$, $ BC\equal{}7$, and $ AC\equal{}9$. We first compute $ s$ as $ \frac{6\plus{}7\plus{}9}{2}\equal{}11$  Then, we have that the area is equal to: \[ \sqrt{11(5)(4)(2)}\equal{}\boxed{2\sqrt{110}}.\]

Try the [url=http://www.artofproblemsolving.com/Wiki/index.php/Heron's_formula]Wiki.[/url]

I'm sure you can come up with some problems on your own. For [hide="example,"]
Find the inradius of a triangle with side lengths $ 13,14,15.$[/hide]

Correction for Bugi. The area is actually \begin{align*}\sqrt{\frac{39}{2}\left(\frac{39}{2}-12\right)\left(\frac{39}{2}-13\right)\left(\frac{39}{2}-14\right)}&=\sqrt{\frac{39}{2}\cdot\frac{15}{2}\cdot\frac{13}{2}\cdot\frac{11}{2}}\\&=\frac{39\sqrt{55}}{4}.\end{align*}

Heron's formula states that $ \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ gives the area of triangle ABC where $ s$ is the semiperimeter.

This formula is often useful when trying to find the altitude/height of the triangle. A simple example would be:

Triangle ABC has lengths as follows: $ AB\equal{}6, BC\equal{}7, CA\equal{}11$. Find the length of the altitude from $ A$ to $ BC$. You would set up an equation comparing two expressions for the area: Heron's and the basic one.

$ \sqrt{12(12\minus{}6)(12\minus{}7)(12\minus{}11)}\equal{}\frac{7h}{2}$

$ \frac{7h}{2}\equal{}\sqrt{360}$

$ 7h\equal{}2\sqrt{360}$

$ h\equal{}\frac{2\sqrt{360}}{7}$

$ h\equal{}\frac{12\sqrt{10}}{7}$

There are many other applications, but this is a very simple one.


Anyone here goes to Duke or plans to apply to Duke this year?

I'm applying to Duke--it's my first choice!


ratio
symmetry
arithmetic sequence
algebra proposed
algebra

Problem 1:
Find all systems [tex] (a,b,c) [/tex] off positive real numbers such that:
[tex] \frac{\ a+b}{bc}=\frac{\ b+c}{ca}=\frac{\ c+a}{ab} [/tex]
Problem 2:
We are given an infinite arithematic progression:
[tex] a_1,a_2.....a_n,.......[/tex]
which contains the numbers [tex]1[/tex] and [tex]\sqrt{2}[/tex].Show that any three numbers of this progression can not be in a geometric ratio.

You should place problems in distinct topics.

For 2.
Since 1 and $\sqrt 2$ lies in the same arithmetic progression, we can write $\sqrt 2=1+t\cdot d$, where $t$ is an integer number and $d$ is a difference of this progression. We see that $d$ is an irrational number.
Suppose now that $1+kd$, $1+nd$ and $1+md$ are in geometric ratio ($k,n,m$ are integers). It follwos $(1+nd)^2=(1+kd)(1+md)$, i.e. $d(n^2-mk)=m+k-2n$. We know that $d$ is irrational, so obtained equation means $m+k-2n=0$ and $n^2-mk=0$. It implies $(m-k)^2=0$. Thus $m=k$ and all three terms $1+kd$, $1+nd$ and $1+md$ are the same number. Contradiction.

For 1 :
Rewrite it as $a^2+ab=b^2+bc=c^2+ca.$

From the cyclic symmetry, we may assume that $a \geq b,c$.
Now, the first equality leads to $(a-b)(a+b)+b(a-c)=0$.
Since $b,c > 0$ is forces $a=b=c$, which conversely is a solution.

Pierre.

[quote="pbornsztein"]

From the cyclic symmetry, we may assume that $a \geq b,c$.

[/quote]
What is cyclic symmetry can u please explain that to me?

The problem remains the same when using the transformation $(a,b,c) \rightarrow (b,c,a) \rightarrow (c,a,b).$

Pierre.

[quote]Since [tex]b,c > 0[/tex] is forces [tex]a=b=c[/tex], which conversely is a solution. [/quote]

Thanx for ur kind reply but please explain the above mentioned statement in ddetail......I would be really obliged.Thanking in anticipation.
Salman.

[quote="pbornsztein"]
Since $b,c > 0$ is forces $a=b=c$, which conversely is a solution.
[/quote]

Hmm... try a bit.... :?
Since $b+c>0$ and $b>0$, since $a-c \geq 0$ and $a-b \geq 0$ then the given sum is non-negative and the only case for which it is $0$ is that $a-c=a-b=0$.

Pierre.


[quote="FMako"]What is lg?[/quote]
$\lg x=\log_{10} x$

[quote="ifai"]Let a, b be two different values of x that satify the equation
\[ (3^{\lg x})(x^{\lg 3})-2(3^{\lg x}+x^{\lg 3})+3=0  \]
Find a+b.[/quote]

using  $x = 3^{\lg _3 x}$ and  $\log _x y\,\log _y z = \log _x z$
we write $x^{\lg 3} = 3^{\lg x}$

Hence the given equation becomes
$t^2 - 4t + 3 = 0$
where $t = 3^{\lg x}$

So, $t=1, 3$
gives x=1, 10

hence the answer is 11.

Through properties of logs, $(3^{\lg x})$ and $(x^{\lg 3})$ are equal, so we set both equal to y.

The original equation turns into, $y^2-4y+3=0$

Solves into y=3,1

Plug back in, x=10,1

a+b=11

Notice that if we $\log$ everything, $\log{3^{\log{x}} \longrightarrow \log{(x+3)}}$ and $\log{x^{log{3}} \longrightarrow \log{(3+x)}}$ so $3^{\log{x}}=x^{\log{3}}$, let us call those terms $x$, hence we get the equation ${x^2-4x+3=0 \longrightarrow\ x={3,1}}$. Since $x=3^{\log{x}}$, we can say that $1=3^{\log{x}} \longrightarrow \log{x}=0 \longrightarrow \boxed{x=1}$, also $3=3^{\log{x}} \longrightarrow \log{x}=1 \longrightarrow \boxed{x=10}$.

I had to prove that logarithm property to myself.  Nice problem, I agree that the answer is [hide="answer"] $11$ [/hide]

[hide="neat log property"] ${a^\log_b{c}} = {(c^\log_c{a})^{\log_b{c}}} = {c^{\log_c{a} \cdot \log_b{c}}} = {c^\log_b{a}}$[/hide]

That is pretty neat it means you can easily solve a problem like: Compute $3^{\log_5{20^{\log_3{5}}}}$ without having to blink.

Good call 10kth User.
$\log_x{a \cdot b} = \log_x{a}+\log_x{b}$

[quote="10000th User"][quote="Iversonfan2005"]Notice that if we $\log$ everything, $\log{3^{\log{x}} \longrightarrow \log{(x+3)}}$ and $\log{x^{log{3}} \longrightarrow \log{(3+x)}}$.[/quote]
Your deduction is not correct because

$\log{3^{\log{x}}=(\log x)(\log 3)\neq\log{(x+3)}}$[/quote]

oops


inequalities
geometry
circumcircle
inequalities proposed

Is the following inequality true ?

     $((a+b+c)^2.(a^2+b^2+c^2))/(abc(ab+bc+ca+2a^2+2b^2+2c^2))>=(sqrt3)/2r$ by standard notation $r$

are these triangle sidelengths and r is the radius of the incircle?

[quote="Singular"]A stronger version is $\frac{(a+b+c)^2}{3abc} \ge \frac{\sqrt{3}}{2r}$.. this looks easier but I don't know if its true.. maybe that will help someone.[/quote]
this ineq is reversed

by the way, the initial inequality is true, but it is not strong. You can just used some eliminations using areas ($2S=(a+b+c)r=abc/2R$ and then $a^2+b^2+c^2 \le 9R^2$ (yes, circumradius) and finally arrive at $3 \sum a^2 \ge 2 \sum a^2 + \sum ab$ which is obvious.


Ok, probably the following problem is very classical, but I confess I haven't the slightest idea of a construction: find a measurable set of real numbers $A$ such that $A+A$ is not measurable!

This would be interesting to see ... anybody has got an idea ?

You may be interested by the replies to the relevant thread I initiated on sci.math about that topic.


Let $ x \in \mathbb{R}^{n}$, $ W_{x} \equal{} \{ A \subset \mathbb{R}^{n}\, |\, \exists r > 0, B(x,r) \subset A \}$
where $ B(x,r)$ is the open ball centered on $ x$ whose radius is $ r$ and $ \mathbb{R}^{n}$ the usual n-dimensional real topological space.

Is $ W_{x}$ a filter on $ \mathbb{R}^{n}$ ? an ultrafilter ?

[quote="transseries"]No.  $ W_x$ is s a filter, yes, but not an ultrafilter.  If $ A$ is a set so that $ x$ is on the boundary of $ A$, then neither $ A$ nor its complement belongs to $ W_x$.[/quote]
Indeed :)

Additional question :
what about $ V_{x} \equal{} \{ S \subset \mathbb{R}^{n}\, |\, x \in S \}$ ??

[quote="Alp"]Additional question :
what about $ V_{x} \equal{} \{ S \subset \mathbb{R}^{n}\, |\, x \in S \}$ ??[/quote]

Yes, $ V_x$ is an ultrafilter.  Filters of this form are called [i]fixed[/i] ultrafilters.


trigonometry
calculus
integration
function
real analysis
real analysis unsolved

$ {\text{Evaluate:  }}\int_0^\pi  {\frac{{x\sin x}}
{{1 \plus{} \cos ^2 x}}{\text{ }}dx}$

[quote="dubito"]$ {\text{Evaluate: }}\int_0^\pi {\frac {{x\sin x}} {{1 \plus{} \cos ^2 x}}{\text{ }}dx}$[/quote]
Write,
$ \int_0^{\pi} x \left( \frac {\sin x}{1 \plus{} \cos^2 x} \right) ~dx$
Let $ u \equal{} x$ and $ v' \equal{} \frac {\sin x}{1 \plus{} \cos^2 x}$ so $ u' \equal{} 1$ and $ v \equal{} \minus{} \tan^{ \minus{} 1} (\cos x)$.
Integration by parts gives,
$ \minus{} x\tan^{ \minus{} 1} (\cos x) \bigg|_0^{\pi} \plus{} \int_0^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \frac {\pi^2}{4} \plus{} \int_0^{\pi} \tan^{ \minus{} 1} (\cos x) ~ dx$.

But (look below for proof),
$ \int_0^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \int_0^{\pi/2} \tan^{ \minus{} 1} (\cos x) \plus{} \int_{\pi/2}^{\pi} \tan^{ \minus{} 1} (\cos x) dx$
Note,
$ \int_{\pi/2}^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \minus{} \int_0^{\pi/2} \tan^{ \minus{} 1} (\cos x) dx$

That means,
$ \boxed{ \int_0^{\pi} \frac {x\sin x}{1 \plus{} \cos^2 x} dx \equal{} \frac {\pi^2}{4} }$.

Proof of stated claim:
$ \int_{\pi/2}^{\pi} \tan^{\minus{}1} (\cos x) dx \equal{} \int_{\pi/2}^{\pi} \minus{}\tan^{\minus{}1}[ \cos (\pi \minus{} x)]dx$
Let $ \mu \equal{} \pi \minus{} x$ to get,
$ \int_{\pi/2}^0 \tan^{\minus{}1} [\cos \mu] d\mu \equal{} \minus{} \int_0^{\pi/2} \tan^{\minus{}1} (\cos \mu) d\mu$.

[quote="sylow_theory"][quote="dubito"]$ {\text{Evaluate: }}\int_0^\pi {\frac {{x\sin x}} {{1 \plus{} \cos ^2 x}}{\text{ }}dx}$[/quote]
Write,
$ \int_0^{\pi} x \left( \frac {\sin x}{1 \plus{} \cos^2 x} \right) ~dx$
Let $ u \equal{} x$ and $ v' \equal{} \frac {\sin x}{1 \plus{} \cos^2 x}$ so $ u' \equal{} 1$ and $ v \equal{} \minus{} \tan^{ \minus{} 1} (\cos x)$.
Integration by parts gives,
$ \minus{} x\tan^{ \minus{} 1} (\cos x) \bigg|_0^{\pi} \plus{} \int_0^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \frac {\pi^2}{4} \plus{} \int_0^{\pi} \tan^{ \minus{} 1} (\cos x) ~ dx$.

But (look below for proof),
$ \int_0^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \int_0^{\pi/2} \tan^{ \minus{} 1} (\cos x) \plus{} \int_{\pi/2}^{\pi} \tan^{ \minus{} 1} (\cos x) dx$
Note,
$ \int_{\pi/2}^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \minus{} \int_0^{\pi/2} \tan^{ \minus{} 1} (\cos x) dx$

That means,
$ \boxed{\int_0^{\pi} \frac {x\sin x}{1 \plus{} \cos^2 x} dx \equal{} \frac {\pi^2}{4} }$.

Proof of stated claim:
$ \int_{\pi/2}^{\pi} \tan^{ \minus{} 1} (\cos x) dx \equal{} \int_{\pi/2}^{\pi} \minus{} \tan^{ \minus{} 1}[ \cos (\pi \minus{} x)]dx$
Let $ \mu \equal{} \pi \minus{} x$ to get,
$ \int_{\pi/2}^0 \tan^{ \minus{} 1} [\cos \mu] d\mu \equal{} \minus{} \int_0^{\pi/2} \tan^{ \minus{} 1} (\cos \mu) d\mu$.[/quote]

 :yup:

Merry Christmas

Let $ I \equal{} \int_0^{\pi}\frac{x \sin x}{1 \plus{} \cos^2x}\,dx$. Making the substitution $ x \equal{} \pi \minus{} u$, dx = -du, we get

$ I \equal{} \minus{} \int_{\pi}^0\frac{(\pi \minus{} u) \sin(\pi \minus{} u)}{1 \plus{} [\cos(\pi \minus{} u)]^2}\,du \equal{} \int_0^{\pi}\frac{(\pi \minus{} u) \sin u}{1 \plus{} [\minus{} \cos u]^2}\,du \equal{} \int_0^{\pi}\frac{(\pi \minus{} u) \sin u}{1 \plus{} \cos^2u}\,du$

$ \equal{}> I \equal{} \int_0^{\pi}\frac{\pi \sin u}{1 \plus{} \cos^2u}\,du \minus{} \int_0^{\pi}\frac{u \sin u}{1 \plus{} \cos^2u}\,du \equal{}> I \equal{} \pi \int_0^{\pi}\frac{\sin u}{1 \plus{} \cos^2u}\,du \minus{} I$

$ \equal{}> 2I \equal{} \minus{} \pi \int_0^{\pi}\frac{\minus{} \sin u}{1 \plus{} \cos^2u}\,du \equal{}> I \equal{} \minus{} \frac{\pi}{2}[\arctan(\cos u)]^{\pi}_0$

$ \equal{}> I \equal{} \minus{} \frac{\pi}{2} \left(\minus{} \frac{\pi}{4} \minus{} \frac{\pi}{4}\right) \equal{}> I \equal{} \frac{\pi^2}{4}$.

There's nice lemma: Let $ f$ be a continous function on $ [0;1]$. Then:
$ \int_0^\pi xf(\sin x)dx\equal{}\frac{\pi}{2}\int_0^\pi f(\sin x)dx$


function
LaTeX
inequalities
inequalities unsolved
China

Good solution. :)
I think it's better that you learn $\LaTeX$.
It's easy.

[quote="bel.jad5"]function f(x)=(x^2+9)/(2*x^2+(3-x)^2) is concave so
f(a)+f(b)+f(c)<=3f((a+b+c)/3)=3f(1)=3*10/6=5[/quote]
Also my solution.  :)

This is the first problem of the second day. And the Contest level is not very high, so there must exist simple and straightfoward solutions like that.

[quote="bel.jad5"]function f(x)=(x^2+9)/(2*x^2+(3-x)^2) is concave so
f(a)+f(b)+f(c)<=3f((a+b+c)/3)=3f(1)=3*10/6=5[/quote]

bel.jad5, how did you prove that $f(x)$ is concave? :maybe:

[quote]bel.jad5, how did you prove that f is concave? [/quote]

just by mathematic intuition   :P

When we use Jensen's inequality, we should show that $f(x)$ is concave up or down. :wink:

[quote="bel.jad5"]function $f(x)=\frac{x^{2}+9}{2x^{2}+(3-x)^{2}}$ is concave so
$f(a)+f(b)+f(c)\leq 3f\left(\frac{a+b+c}{3}\right)= 3f(1)=3\cdot \frac{10}{6}= 5$[/quote]
Are you sure that it is concave?
$f''(x)= \frac{4\left(x^{3}+9x^{2}-27x+9\right)}{3\left(x^{2}-2x+3\right)^{3}}$. This expression is positive for $x\to 0$, because $f''$ is continuous in $0$ and $f''(0)=\frac49$, but negative for $x=1$: $f''(1)=\frac{-4}{3}$. Thus the function $f$ is neither convex nor concave on interval $(0,3)$.


$ T: A\rightarrow B$ is a linear transformation on normed spaces. The book claims that:

$ \parallel{}T\parallel{} \equal{} \sup\{|Ta|: |a| < 1\} \equal{} \sup\{|Ta|: |a| \le 1\} \equal{} \sup\{|Ta| : |a| \equal{} 1\} \equal{} \inf\{M: v\in A \implies |Ta| \le M|a|\}$

How is the last equality true? I can reason myself into accepting the first 3 qualities, but I don't see the 4th one, or specifically how the inf could be equal to the sup ( not that I 'm saying that they can't be, but in this case I don't see it).

Please use \sup in your posts.

Thank you.

Take a look at this article:

 http://en.wikipedia.org/wiki/Operator_norm

All the descriptions of the operator norm seem to take place in a bounded space, does this mean that a normed space is necessarily bounded? ( because it's not stated in the immediate definition). It seems like the bounded condition would have to happen in order for there to be a sup and inf, etc.

In the fourth one, divide by $ |a|$, then $ v: \equal{}\frac{a}{|a|}$ has norm 1 (BTW, there is something wrong with the notation in the fourth part, $ v$ should be $ a$). So, you take the smallest upper bound for $ |Tv|$ that is [i]uniformly[/i] valid for all $ v$. This is the same as taking the $ \sup$ over all $ v$ with $ |v| \equal{} 1$.

hm, is this the browder analysis book published by springer?  :D


calculus
integration
LaTeX
real analysis
real analysis unsolved

integral of sin(x^2) dx

i'm sorry i don't know how to use latex

This only makes sense if $ x$ is bounded on $ [0,\infty).$ (Anyway, this has been posted several times here, try to seek Fresnel Integral.)

The question of whether the antiderivative $ \int \sin(x^2)\,dx$ can be expressed in closed form is essentially equivalent to the question of whether $ \int e^{x^2}\,dx$ can be expressed in closed form.

Given that, the answer is well known: no. It cannot be expressed in closed form in terms of elementary functions. It is, of course, Riemann integrable over any bounded interval, and (as Field hinted) its improper integral over the whole line converges.


Can you please post a solution, Thanhliem (or anyone else, for that matter) ?

I'm very interested in seeing how you solved this difficult problem.

[quote="Thanhliem"]Solve the equation:
$ x_1^2+x_2^2=3x_3 $
$ x_2^2+x_3^2=3x_4 $
$ \cdots $
$x_{2004}^2+x_1^2=3x_2$[/quote]

$x_1,x_2,\dots ,x_n$ are real, complex, integers ?

I think $x_i$ are real numbers.

Let $x_i=\frac{3}{2}y_i$. We have $y_{i+2}=\frac{y_i^2+y_{i+1}^2}{2}$. It is obvious $y_i>0$ (otherwise we obtain trivial solution $y_i=0$ for all $i$).

Suppose there is $n$ such that $y_n\leq 1$, $y_{n+2}\leq 1$ and $y_n+y_{n+1}<2$. In that case $y_{n+2}<1$ and $y_{n+3}<1$. It follows that for all $k\geq 4$ we have $y_{n+k}<\max\{y_{n+2},y_{n+3}\}$ -- contradiction with our equations.
The same contradiction for $y_n\geq 1$, $y_{n+2}\geq 1$ and $y_n+y_{n+1}>2$.

Suppose not all of $y_n$ equals 1 tnen we have for all n:
1. $y_n<1$ and $y_{n+1}>1$
or
2. $y_n>1$ and $y_{n+1}<1$.

Suppose $y_n=a<1$ and $y_{n+1}=b>1$. Consider $y_{n+2}=\frac{a^2+b^2}{2}<1$ and $y_{n+3}=\frac{(a^2+b^2)^2+4b^2}{8}>1$.
We will prove $y_{n+1}-y_{n}>y_{n+3}-y_{n+2}$. Actually, this inequality is equivalent to $8(b-a)>(a^2-2a+b^2)(a^2+2a+b^2)$. Now using $a<1$ we obtain $a^2-2a+b^2<b^2-a^2=(b-a)(b+a)$; $a^2+b^2<2$ ==> $a^2+2a+b^2<4$. So we have to prove $2>b+a$. It is immediate conclusion from $a^2+b^2<2$.

Suppose WLOG $y_1<1$ and $y_2>1$. We have just proved that sequence $y_{2k}-y_{2k-1}$ is positive and decreasing. It follows that $y_{2k}-y_{2k-1}$ converges. But it implies (due to our equations) $y_{2k}-y_{2k-1}=\mbox{const}$ and it can't be decreasing sequence -- contradiction.

myth 's solution similary my solution. but Michael Lipnowski ,you can solution it .try again please :)

if x1,x2 ,..x2004 is  positive number .Michael Lipnowski! Can you solve it? if you can solve it then you post  your solution ! :?


The ratio of losses to wins for Kyle’s team is 3 to 2. If the team had played the same number of games, but had won twice as many of its games, what would the ratio of losses to wins have been? Express your answer as a common fraction.

[hide] We have the ratio $ \frac{3x}{2x}$ for some positive integer $ x$. Therefore, he played $ 3x\plus{}2x\equal{}5x$ games. Since he originally won $ 2x$ of them, he now wins $ 4x$ of them. Therefore, the desired ratio is $ \frac{x}{4x}\equal{}\boxed{\frac14}$.[/hide]


Let $ABC$ be a triangle and $BB', CC'$ its bisectors. Compute the angles of the triangle if $m(AB'C') = 50$ and $m(AC'B') = 70$.

[quote="Lagrangia"]Let $ABC$ be a triangle and $BB', CC'$ its bisectors.[/quote]
Do you mean bisector of its angles?

Lagrangia: Which contest do these problems come from? Do you have the solutions to all of them?

I'll take that as "YES". And angle $A$ is $60^o$  :)  :D  :)  :D

And the others are $\angle B=80^o$ and $\angle C=40$. This problem is trivial, i am to bored to write its solution. Just look at quadriliteral $AB'C'X$ where $X$ is incentre, that quadriliteral is cyclic etc.