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geometry proposed

Given are $\triangle ABC$ area $S$ with any $M$ onplane prove that
$\frac{b^{2}}{a}MA+\frac{c^{2}}{b}MB+\frac{a^{2}}{c}MC\ge 4S$ :)

[quote="evarist"]b^{2}.\frac{MA}{a}\geq b^{2}.\frac{\vec{MA}\vec{GA}}{GA.a}\geq [/quote]
Hey are you sure when you multiply two sides with a number you don't know is it positive ? :D

[quote="evarist"]...$b^{2}.\frac{\vec{MA}\vec{GA}}{GA.a}\geq \frac{3\sqrt{3}GA^{2}b^{2}}{a^{2}+b^{2}+c^{2}}$
   [/quote]
Explain for me that inequality!

[color=purple]Sorry Iam wrong gemath ! .This is not correct proof.I'll search a different solution .[/color]

[color=purple]Sorry for these above posts make you loss time.Here is the correct proof
  We always have$xa^{2}+yb^{2}+zc^{2}\ge \4\sqrt{xy+yz+zx}S$
 Thus$\frac{b^{2}}{a}MA+\frac{c^{2}}{b}MB+\frac{a^{2}}{c}MC$
    = $\frac{MA}{a}b^{2}+\frac{MB}{b}c^{2}+\frac{MC}{c}a^{2}\ge 4\sqrt{\frac{MA.MB}{ab}+\frac{MB.MC}{bc}+\frac{MC.MA}{ca}}S$
  Using the stronger inequality:$\frac{MA.MB}{ab}+\frac{MB.MC}{bc}+\frac{MC.MA}{ca}\ge 1$ we finish :D . [/color]


geometry
3D geometry
tetrahedron
vector
modular arithmetic
The Geometry of Numbers

Show there do not exist four points in the Euclidean plane such that the pairwise distances between the points are all odd integers.

If this existed, would we have a tetrahedron in space with odd integer sidelengths and volume $ 0$. I will show this is not possible.

Take one of the points as the origin and label the vectors to the other three points $ x,y,z$, then $ V \equal{} |x\cdot(y\times z)|$. So $ V^2 \equal{} \left|\begin{array}{ccc}x\cdot x & x\cdot y & x\cdot z \\
y\cdot x & y\cdot y & y\cdot z \\
z\cdot x & z\cdot y & z\cdot z\end{array}\right|$, and since $ 2a\cdot b \equal{} |a|^2 \plus{} |b|^2 \minus{} |a \minus{} b|^2$ we can express this as $ 8V^2 \equal{} \left|\begin{array}{ccc}2|a|^2 & |a|^2 \plus{} |b|^2 \minus{} |a \minus{} b|^2 & |a|^2 \plus{} |c|^2 \plus{} |a \minus{} c|^2 \\
|a|^2 \plus{} |b|^2 \minus{} |a \minus{} b|^2 & 2|b|^2 & |b|^2 \plus{} |c|^2 \minus{} |b \minus{} c|^2 \\
|a|^2 \plus{} |c|^2 \plus{} |a \minus{} c|^2 & |b|^2 \plus{} |c|^2 \minus{} |b \minus{} c|^2 & 2|c|^2 \end{array}\right|$.

But if we assume that $ |a|,|b|,|c|,|a \minus{} b|,|a \minus{} c|,|b \minus{} c|$ are odd integers, then of course (squares are 1 mod 8) Sarrus' rule gives $ 8V^2\equiv 4\pmod 8$, so $ V\not \equal{} 0$, contradiction.

Peter, for your formula for Volume, shouldn't the triple scalar product be divided by 6?

Yes, it does, however it is irrelevant here, since it only adds a factor of $36$, and the conclusion stays.


$ f$ is a function of class $ C^1.$

I've seen in some books that there is another C, like a greek one to denote a class function. What's that code?

You will have to show us what you mean since we have no idea what "some books" are or what "greek one" means.

One of these?

$ \mathcal{C}, \mathfrak{C}, \mathbb{C}$

I presume you mean that $ \mathfrak{C}$ is not quite right. Why won't you show us what you want eg by scanning an image, otherwise how else are we to guess what you are seeing?

Here's the symbol:

[img]http://img338.imageshack.us/img338/6941/dsfsdfph7.png[/img]

I didn't want to scan the book then I remember I could take a picture.

You can't do it on the AoPS forum but in a document use:
[code]\usepackage{mathrsfs}
...
\mathscr{C}^1[/code] which gives you [img]http://sixthform.info/gblog/pictures/97429fc452a49784bfef4808ff0fb056.gif[/img]

Hmm...

I tried that, because I was interested in whether the "S" and "D" that come with that might be the right symbols for the Schwartz space and the $ C_c^{\infty}$ functions of distribution theory.

And it didn't work, not for S, not for D, not for C. What I got were ugly oblique sans-serif characters - some sort of default font. The console had no messages or warnings that i can decipher, just this:

(/usr/local/texlive/2007/texmf-dist/tex/latex/jknapltx/mathrsfs.sty)

[Mac operating system, TeXShop version 1.43.]

Maybe some font hasn't worked. Is there anything relevant in the log? In a Windows system using the minimal document
[code]\documentclass{article}
\usepackage{mathrsfs}
\begin{document}
$\mathscr{C}$
\end{document}[/code]I get
[quote](C:\texmf\tex\latex\jknappen\mathrsfs.sty
Package: mathrsfs 1996/01/01 Math RSFS package v1.0 (jk)
\symrsfs=\mathgroup4
) (LaTeX1.aux)
....
LaTeX Font Info:    Try loading font information for U+rsfs on input line 4.
 (C:\texmf\tex\latex\jknappen\ursfs.fd
File: ursfs.fd 1998/03/24 rsfs font definition file (jk)
)[/quote]On a Mac you should get something like
[quote](/usr/local/texlive/2007/texmf-dist/tex/latex/jknapltx/mathrsfs.sty)
(./LaTeX2.aux) (/usr/local/texlive/2007/texmf-dist/tex/latex/jknapltx/ursfs.fd)
 [1{/usr/local/texlive/2007/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./LaTeX2.aux) )</usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cmr1
0.pfb></usr/local/texlive/2007/texmf-dist/fonts/type1/hoekwater/rsfs/rsfs10.pfb[/quote]

Assuming at you were compiling the minimal document I quoted, I'm a bit concerned that you are getting references to bold and different size fonts such as cmbx12.pfb, cmr7.pfb.
What happens if you use the following alternative package?
[code]\documentclass{article}
\usepackage{calrsfs}
\begin{document}
$\mathcal{C}$
\end{document}[/code]

Sorry, the previous wasn't quite a minimal document - it did have some boldface and amsmath and amssymb in it.

With the calrsfs minimal document:

[quote](/usr/local/texlive/2007/texmf-dist/tex/latex/calrsfs/OMSrsfs.fd) [1{/usr/local
/texlive/2007/texmf-var/fonts/map/pdftex/updmap/pdftex.map}] (./testcal.aux) )<
/usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cmr10.pfb></usr/local
/texlive/2007/texmf-dist/fonts/type1/hoekwater/rsfs/rsfs10.pfb>
[/quote]
And the same sans-serif default font.

If I change the \usepackage from calsrsfs to amssymb, I do get the letter $ \mathcal{C}.$

[quote="Kent Merryfield"]If I change the \usepackage from calsrsfs to amssymb, I do get the letter $ \mathcal{C}.$[/quote]The calrsfs package redefines \mathcal normally used by amssymb nd mathrsfs redefines \mathscr.
It looks like there's a problem with the rsfs fonts so something else is being used. Does using XeLaTeX instead of LaTeX help? XeLaTeX uses a different font mapping so may help.
The fact that you get cmr10.pfb in your log makes it looks as if the wrong font is being chosen.
Fonts are notoriously difficult to deal with but let's see if the two files ursfs.fd, pdftex.map are the same.
[b]usrs.fd[/b]
[code]%% ursfs.fd
%%
%% (c) Copyleft 1995, 1996, 1998, 1999 J"org Knappen
%% Licence: GNU public licence version 2
%%
\ProvidesFile{ursfs.fd}[1998/03/24 rsfs font definition file (jk)]
\DeclareFontFamily{U}{rsfs}{\skewchar\font127 }
\DeclareFontShape{U}{rsfs}{m}{n}{%
   <5> <6> rsfs5
   <7> rsfs7
   <8> <9> <10> <10.95> <12> <14.4> <17.28> <20.74> <24.88> rsfs10
}{}
\endinput

J"org Knappen, Barbarossaring 43, D-55118 Mainz, Allemagne
Author of the Book: Schnell ans Ziel mit LaTeX2e, Oldenbourg-Verlag
M"unchen, 1997, ISBN 3-486-24199-0.
% /ursfs.fd[/code]
[b]pdftex.map[/b] contains the lines:
[code]rsfs10 rsfs10 <rsfs10.pfb
rsfs5 rsfs5 <rsfs5.pfb
rsfs7 rsfs7 <rsfs7.pfb[/code]
Finally, check that the files /usr/local/texlive/2007/texmf-dist/fonts/type1/hoekwater/rsfs/rsfs10.pfb and /usr/local/texlive/2007/texmf-dist/fonts/type1/hoekwater/rsfs/rsfs10.pfm exist.

OK. I've got two computers - home and work. I dropped in to work this evening and tried these files there, they work.

(As for the question of whether mathrsfs is right for the distribution spaces -- I like the D. I'm not so sure about the S but it's better than anything else I've seen.)

The work computer has a more recent version of TeXShop than the home computer - maybe I should try re-installing.

Edit: Yes, downloading and installing the latest MacTex package did make a difference. It works on that computer now, too.

Edit again: so \mathscr has no lower case letters? \mathscr{ABCDEFGHIJKLMNOPQURSTUWXYZ} printed the whole alphabet with lots of swirls and flourishes, while \mathscr{abcdefghijklmnopqrstuvwxyz} printed nothing at all. No error message - just nothing at all on the page.

I'm glad you got it sorted. I was leaving the re-installing option until the end, but perhaps I should have mentionedit earlier.

[quote="Kent Merryfield"]Edit again: so \mathscr has no lower case letters?[/quote]That's correct. \mathcal and \mathbb aren't defined for lower case either giving all sorts of random symbols instead of the alphabet. The alternative \mathscr defined by \usepackage[mathscr]{eucal} only gives the odd symbol for lower case letters.

Using amssymb package (as this forum does) gives
\mathcal: $ \mathcal{abcdefghijklmnopqrstuvwxyz}$
\mathbb: $ \mathbb{abcdefghijklmnopqrstuvwxyz}$
but on the other hand
\mathfrak: $ \mathfrak{abcdefghijklmnopqrstuvwxyz}$


Given a positive integer $ m > 2$, show that there exist positive integers $ p$ and $ q$ such
that $ p < q$ and

$ \dfrac{1}{m} \equal{} \displaystyle\sum_{j\equal{}p\plus{}1}^{q}\frac{1}{j(j\plus{}1)}$

[quote="guru_bhai"]Given a positive integer $ m > 2$, show that there exist positive integers $ p$ and $ q$ such
that $ p < q$ and

$ \dfrac{1}{m} \equal{} \displaystyle\sum_{j \equal{} p \plus{} 1}^{q}\frac {1}{j(j \plus{} 1)}$[/quote]

$ \dfrac{1}{m} \equal{} \displaystyle\sum_{j \equal{} m \minus{} 1}^{m^2\minus{}m\minus{}1}\frac {1}{j(j \plus{} 1)}$


it seems a innocent inequalitie, but its not really so, i would say, a physics one. has it a beauty solution?

for x, y, z in [b]R[/b]
Show:

min{(x-y)^2,(y-z)^2,(z-x)^2} <= (x^2 + y^2 + z^2)/2

[quote="tchebytchev"]your problem is very similar to this : :wink:
[hide] http://www.mathlinks.ro/viewtopic.php?t=170209 [/hide][/quote]
Yes! This method works here. :wink:

[quote="arqady"][quote="El Chamo"]but hier can be x, y oder z negative[/quote]
Try it. It still works for your inequality for all $ x,$ $ y$ and $ z.$ :wink:[/quote]

I tried it. You need b > x, and that is not always true.
First see, that whit the number 5 is not true for all x, y and z, only whit 2, and the method not explain when you pick b < 0 (x>0 allways).
I don´t see how that work.

[quote="El Chamo"]
I don´t see how that work.[/quote]
OK. We need to prove that
\[ \min\{(x \minus{} y)^2,(y \minus{} z)^2,(z \minus{} x)^2\}\leq\frac {x^2 \plus{} y^2 \plus{} z^2}{2}
\]
Let $ x\leq y\leq z$ and $ z \equal{} x \plus{} u,$ $ y \equal{} x \plus{} v.$ Then $ u\geq v\geq0.$
Hence, we need to prove that $ \min\{v^2,(u \minus{} v)^2\}\leq\frac {3x^2 \plus{} 2(u \plus{} v)x \plus{} u^2 \plus{} v^2}{2}.$
Case1. $ v\leq u \minus{} v$
Id est, we need to prove that $ 3x^2 \plus{} 2(u \plus{} v)x \plus{} u^2 \minus{} v^2\geq0.$
But $ \frac {\Delta}{4} \equal{} (u \plus{} v)^2 \minus{} 3(u^2 \minus{} v^2) \equal{} 4v^2 \plus{} 2uv \minus{} 2u^2 \equal{} 2(2v\minus{}u)(u\plus{}v)\leq0.$
Case2. $ v\geq u \minus{} v.$
Hence, we need to prove that $ 3x^2 \plus{} 2(u \plus{} v)x \plus{} 4uv \minus{} u^2 \minus{} v^2\geq0.$
$ \frac {\Delta}{4} \equal{} (u \plus{} v)^2 \minus{} 12uv \plus{} 3u^2 \plus{} 3v^2 \equal{} 4u^2 \plus{} 4v^2 \minus{} 10uv \equal{} 2(2u \minus{} v)(u \minus{} 2v)\leq0.$
Done! :)

Very nice inequality, but what is so physical about it?

[color=blue][b]Theorem 1.[/b] If $ x$, $ y$, $ z$ are three reals, then

$ \min\left\{  \left(  y\minus{}z\right)  ^{2},\left(  z\minus{}x\right)  ^{2},\left( x\minus{}y\right)  ^{2}\right\}  \leq\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}$.[/color]

Before we prove this, we show:

[color=blue][b]Lemma 2.[/b] If $ a\geq0$ and $ b\geq0$ are nonnegative reals, then

$ \min\left\{  b^{2},\left(  a\plus{}b\right)  ^{2},a^{2}\right\}  \leq\frac {b^{2}\plus{}\left(  a\plus{}b\right)  ^{2}\plus{}a^{2}}{6}$.[/color]

[i]Proof of Lemma 2.[/i] Since the statement of Lemma 2 is symmetric in the variables $ a$ and $ b$, we can WLOG assume that $ b\leq a$. Then,

$ \min\left\{  b^{2},\left(  a\plus{}b\right)  ^{2},a^{2}\right\}  \leq b^{2}\equal{}\frac{b^{2}\plus{}\left(  b\plus{}b\right)  ^{2}\plus{}b^{2}}{6}$
$ \leq\frac{b^{2}\plus{}\left(  a\plus{}b\right)  ^{2}\plus{}a^{2}}{6}$

(since $ b\leq a$ yields $ \left(  b\plus{}b\right)  ^{2}\leq\left(  a\plus{}b\right)  ^{2}$ and $ b^{2}\leq a^{2}$). This proves Lemma 2.

Now to the [i]proof of Theorem 1:[/i] Since the statement of Theorem 1 is symmetric in the variables $ x$, $ y$, $ z$, we can WLOG assume that $ x\geq y\geq z$. Then, $ x\minus{}y\geq0$ and $ y\minus{}z\geq0$. Denote $ a\equal{}x\minus{}y$ and $ b\equal{}y\minus{}z$. Thus, $ a\geq0$ and $ b\geq0$. Besides, $ z\minus{}x\equal{}\minus{}\left(  \left(  x\minus{}y\right)  \plus{}\left( y\minus{}z\right)  \right)  \equal{}\minus{}\left(  a\plus{}b\right)$, so that $ \left(  z\minus{}x\right) ^{2}\equal{}\left(  \minus{}\left(  a\plus{}b\right)  \right)  ^{2}\equal{}\left(  a\plus{}b\right)  ^{2}$.

Since $ a\geq0$ and $ b\geq0$, Lemma 2 yields

$ \min\left\{  b^{2},\left(  a\plus{}b\right)  ^{2},a^{2}\right\}  \leq\frac{b^{2}\plus{}\left(  a\plus{}b\right)  ^{2}\plus{}a^{2}}{6}$.

Since $ a\equal{}x\minus{}y$, $ b\equal{}y\minus{}z$ and $ \left(  a\plus{}b\right)  ^{2}\equal{}\left(  z\minus{}x\right)  ^{2}$, this becomes

$ \min\left\{  \left(  y\minus{}z\right)  ^{2},\left(  z\minus{}x\right)  ^{2},\left( x\minus{}y\right)  ^{2}\right\}  \leq\frac{\left(  y\minus{}z\right)  ^{2}\plus{}\left( z\minus{}x\right)  ^{2}\plus{}\left(  x\minus{}y\right)  ^{2}}{6}$.

Hence,

$ \min\left\{  \left(  y\minus{}z\right)  ^{2},\left(  z\minus{}x\right)  ^{2},\left(x\minus{}y\right)  ^{2}\right\}  \leq\frac{\left(  y\minus{}z\right)  ^{2}\plus{}\left(z\minus{}x\right)  ^{2}\plus{}\left(  x\minus{}y\right)  ^{2}}{6}$
$ \leq\frac{\left(  y\minus{}z\right)  ^{2}\plus{}\left(  z\minus{}x\right)  ^{2}\plus{}\left( x\minus{}y\right)  ^{2}\plus{}\left(  x\plus{}y\plus{}z\right)  ^{2}}{6}$ (since $ \left(  x\plus{}y\plus{}z\right)^{2}\geq0$)
$ \equal{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{2}$,

and Theorem 1 is proven.

Clear solution, Darij!

Moreover, the following similar inequality is also true,

\[ \min\{(a\minus{}b)^4,(b\minus{}c)^4,(c\minus{}a)^4\} \le \frac{a^4\plus{}b^4\plus{}c^4}{2}.\]

Let's prove this!

I think the following inequality is true and much harder
\[ \min\{(a \minus{} b)^4,(b \minus{} c)^4,(c \minus{} a)^4\}\minus{}abc(a\plus{}b\plus{}c) \le \frac {a^4 \plus{} b^4 \plus{} c^4 }{2}.
\]

[quote="darij grinberg"]Very nice inequality, but what is so physical about it?
[/quote]

It´s geometrie-physic. I thought this problem so: $ P$ is a point in $ R^3$ whit components $ (x,y,z)$, then:

$ \min\left\{\left( y \minus{} z\right) ^{2},\left( z \minus{} x\right) ^{2},\left( x \minus{} y\right) ^{2}\right\} \leq\frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{2}$

is equivalent to

$ \min\left\{ d\left( y , z\right),d\left( z , x\right),d\left( x , y\right)\right\} \leq\sqrt {\frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{2}} \equal{} \frac {d(P,0)}{\sqrt {2}}$ ,
where $ d(. , .)$ is the euclideans distance. But it´s only an idea, can you find throught this way a solution. I no :maybe:

No matter, I found another solution in internet here:

http://www.math.ethz.ch/~gruppe5/www/Analysis_I_HS07/loesung4.pdf
...the last exercise.

it says:

Let $ S \equal{} \frac {x \plus{} y \plus{} z}{3}$ the gravity center and so on... you dont need understand germanic languages... just math! :|

you can try this... suppose wlog that $ x\leq y\leq z$. let $ y \equal{} x \plus{} a$ and $ z \equal{} x \plus{} a \plus{} b$, $ a,b\geq 0$. then, we have to prove that
$ \min \{a^2,b^2\}\leq \dfrac{3x^2 \plus{} 4ax \plus{} 2bx \plus{} 2a^2 \plus{} 2ab \plus{} b^2}{2} \equal{} \dfrac{2(x \plus{} a)^2 \plus{} (x \plus{} b)^2 \plus{} 2ab}{2}$

this follows because $ \min\{a^2,b^2\}\leq ab$, since $ a,b > 0$.


Find the maximum amount of $54x^2y^3(1-x-y)$ for $x,y>0$.

We must have $1-x-y>0$. Then use AM-GM inequality:
\[ 1=\frac x2+\frac x2+\frac y3+\frac y3+\frac y3+(1-x-y)\ge 6\sqrt[6]{\frac{x^2y^3(1-x-y)}{4\cdot 27}} \]


[quote="mdk"]Find k so that the solutions of x^3-3x^2 + kx + 8 = 0 are in arithmetic progression.[/quote]

$ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}3$

arithmetic progression ----> $ \frac{x_{1}\plus{}x_{3}}{2}\equal{}x_{2}$ ----> $ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}3x_{2}\equal{}3$  ---->

$ x_{2}\equal{}1$ ----> $ 1^3\minus{}3*1^2 \plus{} k*1 \plus{} 8 \equal{} 0$ ----> $ k\equal{}\minus{}6$

I feel dumb and am probably forgetting something obvious  :blush: , but how do you know $ x_{1}\plus{}x_{2}\plus{}x_{3}\equal{}3$?

This could help: [url]http://mathworld.wolfram.com/VietasFormulas.html[/url]

@zaya_yc: nice solution!

$ 3\equal{}t_1\plus{}t_2\plus{}t_3\equal{}t_1\plus{}(t_1\plus{}n)\plus{}(t_1\plus{}2n)\Longrightarrow t_2\equal{}1$
Now $ t_1t_2t_3\equal{}\minus{}8\equal{}(1\minus{}n)(1)(1\plus{}n)\Longrightarrow n\equal{}3$

So $ (t_1,t_2,t_3)\equal{}(\minus{}2,1,4)$ and $ k\equal{}t_1t_2\plus{}t_1t_3\plus{}t_2t_3\equal{}\minus{}6$

You know that the middle term is the average of the sum of all the terms (in an arithmetic progression).  So you know the middle term is $ \frac{3}{3}\equal{}1$.  So 1 is a root of the polynomial.  Use synthetic division to divide out a 1, and you see $ k$ has to be $ \minus{}6$.


Let be given [tex]a \in R[/tex] .
Sequence [tex](x_n),n=0,1,2,...[/tex] defined by
[tex]x_0=\sqrt{1996},x_{n+1}=\frac{a}{1+x_n^2},n=0,1,2...[/tex]
Find all [tex]a[/tex] such that sequence has limit finite

My general condition is $a \in [2,-2]\cup\{x_0^3 + x_0\}$
:cool: Do you know why ? :cool:

If $a \geq 0$ the all of the $x_n \geq 0$ and if $a \leq 0$ than all of the $x_n \leq 0$
So $\frac {x_{n+1}}{x_n}=\frac{a}{x_n+\frac{1}{x_n}}$ consequently we get that :
$\frac{x_{n+1}}{x_0}=\frac{a^n}{(x_1+\frac{1}{x_1})(x_2+\frac{1}{x_2})....}$ and
from this we get that $\frac{x_{n+1}}{x_0}<\frac{a^n}{\b^n}$ where $b > 2$ and if you take $n$ to
infinity and $a\subseteq[-2;2]$ we get that $x_n$ has limit.
I don't know how u can prove for $a=x^{3}_{0}+x_0$ and why there are no other solutions??

Go to :
http://www.mathlinks.ro/Forum/topic-24096.html
:cool:


M is a matrix

             | 1 1 2 |
    M  =  | 0 1 1 |
             | 0 0 1 |

Find M^n where n is a positive integer.

I cranked out the first powers and looked for a pattern.

                 | 1 2 5 |
    M^2  =  | 0 1 2 |
                 | 0 0 1 |

                 | 1 3 9 |
    M^3  =  | 0 1 3 |
                 | 0 0 1 |

                 | 1 4 14 |
    M^4  =  | 0 1   4 |
                 | 0 0  1 |

                 | 1 5 20 |
    M^5  =  | 0 1  5  |
                 | 0 0  1 |

                         | 1  n  n(n+3)/2 |
Hence:  M^n  =  | 0  1        n       |
                         | 0  0        1      |

Im supposed to find M^n where n is a positive integer. I found the pattern, but I'm not sure exactly what the question is asking and what my answer should be?

The idea is to come up with a general formula that involves n, like you tried to do in your last step. I'm not going to CHECK that it's correct, but you have the right idea.


The edges of a graph containing $ 6$ points and all $ 15$ edges connecting the points are colored in two colors. Prove that there exists a cycle of length $ 4$ of one color.

Here's another solution:

[hide]At least 8 edges are the same color, so it is sufficient to prove that in a graph with 6 vertices and 8 edges, there is a cycle of length 4.  Assume no such cycle exists.  Consider the vertex X with maximal degree.  Since the sum of the degrees of all vertices is 16, by Pigeonhole X has degree at least 3.  Let A be the set of vertices connected to X by an edge, and let B be the set of vertices not connected to X by an edge.  We proceed by casework based on the degree of X:

Case 1: X has degree 3.  If there are 2 edges connecting elements of A to other elements of A, then both of these edges must include some vertex Y since A only has 3 elements.  Suppose these edges connect Y to some vertices Z and W, then we have the cycle X-Z-Y-W, a contradiction.  Therefore there can be at most one edge connecting elements of A to other elements of A.  Since B only has 2 elements, there can be at most one edge connecting elements of B to other elements of B.

This leaves at least 3 edges connecting an element of A to an element of B.  Two of these must include the same vertex Y from B.  Suppose these edges connect Y to vertices Z and W in A, then we have the cycle X-Z-Y-W, a contradiction.

Case 2: X has degree 4.  If there are 3 edges connecting elements of A to other elements of A, then two of these edges must include some vertex Y since A only has 4 elements.  This creates a cycle of length 4, a contradiction.  Therefore there are at most 2 edges connecting elements of A to other elements of A.  B only has one element V, and we are left with 2 edges connecting V to elements of A.  This also creates a cycle of length 4, a contradiction.

Case 3: X has degree 5.  There are 3 edges left that must connect elements of A to other elements of A, and two of these must include some vertex Y.  Again we have a cycle of length 4, a contradiction.

Therefore, we always have a cycle of length 4.[/hide]


Suppose that $a,b,c$ are three positive real numbers and $abc=1$. Prove that:

$\sum_{cyc}\frac{a}{\sqrt{b+\sqrt{ac}}}\geq \frac{3}{\sqrt{2}}$

We have:
\[\sum\frac{a}{\sqrt{b+\sqrt{ac}}}\ge\frac{\left(\sum\sqrt a\right)^{2}}{\sum\sqrt{b+\sqrt{ac}}}=\frac{2\sqrt 2\left(\sum\sqrt a\right)^{2}}{\sum 2\sqrt 2\sqrt{b+\sqrt{ac}}}\\ \ge\frac{2\sqrt 2\left(\sum\sqrt a\right)^{2}}{\sum 2+b+\sqrt{ac}}=\frac{2\sqrt 2\left(\sum\sqrt a\right)^{2}}{\sum a+\sum \sqrt{bc}+6}\]
We will prove that: $\frac{2\sqrt 2\left(\sum\sqrt a\right)^{2}}{\sum a+\sum \sqrt{bc}+6}\ge\frac{3}{\sqrt 2}\quad (*)$
We have $(*)\Longleftrightarrow a+b+c+5\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\ge 18,$ which is true because: $\begin{cases}a+b+c\ge 3\sqrt[3]{abc}=3\\ 5\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\ge 5.3\sqrt[3]{\sqrt{ab}.\sqrt{bc}.\sqrt{ca}}=15\end{cases}$


There is a circle with diameter $76$  and inscribed inside the circle are three smaller circles with radius $r_1,r_2,r_3\in \mathbb{N}$ . If the diamater of the big circle passes through the center of all the three small circle with $r_1+r_2+r_3=38$ (all three small circles touch each other) . Given the total area of the three small circles are equal to half the area of the big circle . Find all possible $r_1,r_2,r_3$ .

[quote="al.M.V."]what??? the sum of the squares of r_1, r_2, r_3 is an integer, but 39^2/2 is not....so what's wrong? :?[/quote]

well .... it is $\frac{38^2}{2}$ though ....


Who's interested in an AI contest??  I can host one if a handful of people are interested.  as for the allowed languages, anything that works in STDIO is okay.
is the poll disabled?  I cant find it.  Anyways here are some of the topics:

Connect 4
Tron (aka Snake) (Two Players and/or multi-players)
4x4x4 Tic Tac Toe
Robocode (only java)
Other (let me know)

Tron is played on a grid, each turn every snake extends by one square.  if there are no squares left over, you lose.
Robocode is robots shooting bullets at each other.  see here:
http://www.alphaworks.ibm.com/tech/robocode

Dots and Boxes is always an interesting game involving a good amount of math...

I guess I'd give it a shot if you did it.

As for dots and boxes...hasn't it been solved?

I'd like to try it, but I won't be very good at it.

Dots and Boxes has not been solved.  :D  It is a very interesting game, with tons of math. There are, however, many very good programs already available developed to play the game, and they do lose to human opponents rather frequently.


calculus
integration
induction
algebra proposed
algebra

For every $m,n\in N$ denote
\[ f(m,n)=\sum_{k=1}^\infty(k^n(\frac{m}{m+1})^k)  \]
Prove that $f(m,n)\in Z$ and divisible by $m(m+1)$ for all $m,n$(of couse $m\neq 0$)

Am I wrong or $f(m,0) = m$ so that the desired result is not true?

Pierre.

The simple case with n=0 can be negligible.

Ah?  :|

Let $S_n(x) = \sum_{k \geq 1} k^nx^k.$
Then, for all $n$ and $x$, we have $S_{n+1} = xS_n'(x)$

Let $P(x) = \frac 1 {1-x}.$
An easy induction leads to $P^{(i)}(x) = \frac{i!} {(1-x)^{i+1}}$, where $P^{(i)}$ denotes the $i^{th}$ derivative.
Another easy induction leads to :
$S_n(x) = \sum_{i=1}^na_{i,n}x^iP^{(i)}(x)$ where the $a_{i,n}$'s are integers.

Since $f(m,n) = S_n(\frac m {m+1})$ it follows that $f(m,n) = \sum_{i=1}^na_{i,n}\cdot i! \cdot(m+1)m^i = 0 \mod [m(m+1)].$

Pierre.


Find $ a,b,c$  from the system $ ?$

$ a \plus{} b \plus{} c\equal{}1$
$ a^2\plus{}b^2\plus{}c^2\equal{}1$
$ a^3\plus{}b^3\plus{}c^3\equal{}1$

[quote="Obel1x"]Find $ a,b,c$  from the system $ ?$

$ a \plus{} b \plus{} c \equal{} 1...........(1)$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1...........(2)$
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 1............(3)$[/quote]

$ (1)^2\minus{}(2)$, we get $ ab\plus{}bc\plus{}ca\equal{}0$

$ (3)\equal{}(1) \times (2)$, we get $ ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)\equal{}0$
$ \Leftrightarrow ab(1\minus{}c)\plus{}bc(1\minus{}a)\plus{}ca(1\minus{}b)\equal{}0$
$ \Leftrightarrow 3abc\equal{}0$

So one of $ a,b,c$ , say $ a\equal{}0$.

then $ (b\plus{}c)^2\equal{}b^2\plus{}c^2$ gives $ bc\equal{}0$, so one of $ b,c$ is $ 0$ and another is $ 1$

$ (a,b,c)\equal{}(0,0,1),(0,1,0),(1,0,0)$ are the only solutions.

This was a problem in your regional olympiad , wasn't it ?

[hide="Here is my solution"]Easy to show $ ab\plus{}bc\plus{}ca\equal{}0$

$ \Rightarrow a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}(ab\plus{}bc\plus{}ca))\Leftrightarrow 1\minus{}3abc\equal{}1\Rightarrow 3abc\equal{}0\Rightarrow a\equal{}0\vee b\equal{}0 \vee c\equal{}0$ . Assume $ a\equal{}0$ .

$ (a\plus{}b\plus{}c)^3\minus{}(a^3\plus{}b^3\plus{}c^3)\equal{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)\Rightarrow (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0\Leftrightarrow a\equal{}\minus{}b\vee b\equal{}\minus{}c\vee c\equal{}\minus{}a$ .

Since i assumed $ a\equal{}0$ it is impossible $ b\equal{}\minus{}c$ . Assume $ a\equal{}\minus{}b\Rightarrow a\equal{}b\equal{}0$ and $ c\equal{}1$ .

Same for the permutations and i get the same result as stephencheng .[/hide]

From $ a\plus{}b\plus{}c\equal{}1, ab\plus{}bc\plus{}ca\equal{}0, abc\equal{}0$ it follows $ a,b,c$ are the roots of the equation $ x^3\minus{}x^2\equal{}0$, i.e. $ \{a,b,c\}\equal{}\{1,0,0\}$

Square the first equation and subtract the second to get $ ab\plus{}ac\plus{}bc\equal{}0$.
Take the product of the first and second equations and subtract the third equation to get $ a^2b\plus{}a^2c\plus{}b^2a\plus{}b^2c\plus{}c^2a\plus{}c^2b\equal{}0$.
Multiply the first equation by $ ab\plus{}ac\plus{}bc$, then subtract what we got above to get $ 3abc\equal{}0\equal{}abc$. Therefore at least one of the variables is zero. Call then $ x$ and $ y$. We have
$ x\plus{}y\equal{}1$
$ x^2\plus{}y^2\equal{}1$
Square the first equation and subtract the second to get $ xy\equal{}0$. Therefore one of the remaining variables must be 0, and the other is 1. Therefore the solutions to the given system are $ (1,0,0)$, $ (0,1,0)$, and $ (0,0,1)$.


[quote="Franz Joseph"]Prove that:
$\sin \theta+\sin (\theta+\alpha)+\sin(\theta+2\alpha)+.......+\sin(\theta+n\alpha)= \frac{\sin\frac{(n+1)\alpha}{2}\sin(\theta+\frac{n\alpha}{2})}{\sin\frac{\alpha}{2}}$[/quote]
Below we use the identity $2\sin{A}\sin{B}= \cos{(A-B)}-\cos{(A+B)}$

$\sin \theta+\sin (\theta+\alpha)+\sin(\theta+2\alpha)+.......+\sin(\theta+n\alpha)= \frac{\sin\frac{(n+1)\alpha}{2}\sin(\theta+\frac{n\alpha}{2})}{\sin\frac{\alpha}{2}}$

$\Leftrightarrow 2\sin{\frac{\alpha}{2}}[\sin \theta+\sin (\theta+\alpha)+\sin(\theta+2\alpha)+.......+\sin(\theta+n\alpha)]= 2\sin\frac{(n+1)\alpha}{2}\sin(\theta+\frac{n\alpha}{2}$
$\Leftrightarrow [\cos{(\theta-\frac{\alpha}{2})}-\cos{(\theta+\frac{\alpha}{2})}]+[\cos{(\theta+\frac{\alpha}{2})}-\cos{(\theta+\frac{3\alpha}{2})}]+[\cos{(\theta+\frac{3\alpha}{2})}-\cos{(\theta+\frac{5\alpha}{2})}]+...+[\cos{(\theta+(n-\frac12)\alpha)}-\cos{(\theta+(n+\frac12)\alpha)}] = \cos{(\theta-\frac{\alpha}{2})}-\cos{(\theta+(n+\frac12)\alpha)}$
$\Leftrightarrow 0=0$

Note that if $\alpha=0,$ the identity is still true in the limit sense.

$\lim_{\alpha \to 0}LHS = (n+1)\sin{\theta}$

$\lim_{\alpha \to 0}RHS = (n+1)\sin{\theta}$

This is a very elementary problem using complex numbers.  Just use the exponential form of a complex number, $e^{i\theta}$, and geometric sums.

Masoud Zargar


I want to know , what's the math name about such a special numbers:
A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1  a2  ...  ak.

For example, 6 = 1 + 2 + 3 = 1  2  3.

is there any website solve this number like pi=3.1415.......and what 's its name?

any assistance will be appreciated!

thanks, but i have a little question. For example: let a1=3,a2=3,a1>=a2>=3..   Ms=a1*a2=9, Ss=a1+a2=6, so k=9-6=3. a3=1.  but a1+a2+a3=7 a1*a2*a3=9.... what can i do??

I am sorry. I edit my post $k=M_{s}-S_{s}+s$.

Thanks,Rust, but can we improve our problem:

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 * 2 = 2 + 2
k=3: 6 = 1 * 2 *  3 = 1 + 2 + 3
k=4: 8 = 1  * 1 * 2 * 4 = 1 + 1 + 2 + 4
k=5: 8 = 1   *  1 * 2 * 2  * 2 = 1 + 1 + 2 + 2 + 2( not 1+1+1+3+3=3*3*1*1*1)
k=6: 12 = 1 * 1 * 1 * 1 * 2 * 6 = 1 + 1 + 1 + 1 + 2 + 6

[quote="mgb"]Numbers like this are called perfect numbers.[/quote]
N called perfect number if (and only if) $2N=\sigma(N)$.


group theory
abstract algebra
superior algebra
superior algebra solved

If H is a subgroup of G and [G:H] = 2 then the righ cosets of H in G are the same as the left cosets of H in G. Why?

Thanks for the help!!!!

The right cosets are $H$ and $xH$ for some (and thus any) $x\in G\setminus H$. This can be restaded as: The right cosets are $H$ and $G\setminus H$.
Repeat this process for the left cosets.

I am not sure what you said applies... I am still confused any other help would be greatly appreciated.

From the beginning:
Because $[G: H]=2$, there are 2 right cosets (this is just the definition).
I might have scrambled the definitions, but generally the right cosets have the form $Hx$ with $x\in G$.
The right cosets form a partition of $G$. The right coset corresponding to the identity $e$ of $G$ is $He=H$. Since there are only two right cosets and they form a partition of $G$, the other right coset (as a subset of $G$) must be $G\setminus H$.

The same holds for left cosets.

i guess what i dont understand why are the same... like i understand how to get them.. so the right coset would be e and G / H and the left coset would be e and then H / G right... or am i looking at it wrong?  or does the left coset have to be G / H as well??

....
Let $(G,\cdot)$ be a group and $H$ a subgroup of $G$. Then we define $(G/H)_l = \{x \cdot H | x\in G\}$ (left cosets) and we denote the elements of $(G/H)_l$ by $\widehat x$.

Now $(G/H)_r=\{Hx | x\in G\}$ (right cosets) and we denote the elements by $\widehat x$.

Generaly if we have a set $M$ and $\sim$ an equivalence relation on $M$ we define $M/\sim=\{x: x \in M\}$ with the property that if $x\in M$ then there is an unique $y\in M$ so that $x\sim y$.

In your case we have to equivalence relations $\sim_l$ (left) and $\sim_r$ (right).

Back to the problem.

Because $|G: H|=2$ it means that if $x\in G-H$ then $H\cup xH=G$ and $H\cup Hx=G$.
Because $H\cap xH=\emptyset$ and $H\cap Hx=\emptyset$ it follows that $xH=Hx$, so $H$ is normal in $G$ so the right and left cosets are the same ($(G/H)_l=(G/H)_r$).


Suppose $y$ varies directly as $x$ and inversely as $z$, and $y=12$ when $x=8$ and $z=2$. When $x=3$ and $z=6$, the value of $y$ is :


Circle O has radius 10, circle P is internally tangent to O and has a radius of 4.  Line QT is tangent to circle P at T, and line PT intersects circle O at points A and B. Q is on the larger circle with Q,O and P  colinear and Q isnt toching the smaller circle. Compute the product TA*TB.

[quote="chickendude"]is Q supposed to be O?

if Q is a new generic point, then T could be anywhere on the circle P, and that product will vary[/quote]

sorry Q, O, and P are colinear.  Theres actually a picture for this, but I dont know how to get it on here.  O is the center of the larger circle, P of the smaller circle. and Q is on the larger circle with Q,O and P colinear and Q isnt toching the smaller circle.

ok that's much better

here we go

[hide]
first, let the other point where line $QT$ intersects the circle be called R
Let the point diametrically opposed to $Q$ be called $N$

we see that $AT \cdot BT = QT \cdot TR$ by power of the point

$QT = \sqrt{16^{2}-4^{2}}= 8\sqrt{6}$ by the pythagorean theorem on triangle PQT

PQT is similar to NQR  (both are right triangles sharing angle Q)

from this we see that

$\frac{16}{20}= \frac{8\sqrt{6}}{8\sqrt{6}+TR}$

$TR = 2\sqrt{6}$

$AT \cdot BT = QT \cdot TR = 8\sqrt{6}\cdot 2\sqrt{6}= 96$
[/hide]

[quote="chickendude"]ok that's much better

here we go

first, let the other point where line $QT$ intersects the circle be called R
Let the point diametrically opposed to $Q$ be called $N$

we see that $AT \cdot BT = QT \cdot TR$ by power of the point

$QT = \sqrt{16^{2}-4^{2}}= 8\sqrt{6}$ by the pythagorean theorem on triangle PQT

PQT is similar to NQR  (both are right triangles sharing angle Q)

from this we see that

$\frac{16}{20}= \frac{8\sqrt{6}}{8\sqrt{6}+TR}$

$TR = 2\sqrt{6}$

$AT \cdot BT = QT \cdot TR = 8\sqrt{6}\cdot 2\sqrt{6}= 96$
][/quote]
chickendude, I get a differnet answer then you but that could be primarily that $QT = \sqrt{16^{2}-4^{2}}$ but that does not equal $8\sqrt{6}$, as it is $4\sqrt{15}$
Everything else you did seems right, but that would make $TR=\sqrt{15}$ making the answer $60$.


geometry
geometric transformation
Support
search
LaTeX

Many members admit that they have difficulties translating terms from the native language into English. I would suggest to start up such a page for some of the nations represented at mathlinks in order to help people getting started. In order to provide a small example, please provide the translation of something like that: equilateral, acute, obtuse, isosceles ,... What do the other people think about it ?

it's a good idea, maybe we should create a sub-forum for that? :)

I think it would just be better to have some kind of dictionary, that people can browse like "triangle" would result a list as well of acute-angled, congruent, same-shape-but-not-same-size (whatever it's called), etc.

like that they can find whatever they want. Cuz elseway you'd need sooo much, and what do people do that are alone from their country?

I guess a lot of such dictionaries already exist. This work need not be done twice, we should rather collect links here.

For german, there is [url=http://www.math.uni-goettingen.de/baule/wbuch.html]das mathematische Wrterbuch[/url], which is not bad and may be downloaded, but leaves out most of geometry.

Another one, covering german and bulgarian, and which is moreover user-extendable, is available at [url=http://www.uni-bonn.de/~manfear/mathdict.php]the University of Bonn[/url]. Sadly, I found no way to get it home.

In Vietnam , there is no way to find a book like that. All of us must use the knowledges teached by teachers to translate them.

for french people, there is a good dictionary in the site http://www.bibmath.fr.
otherwise the best way to learn math vocabulary is reading many books and essays.

Hi orl, all,

[quote="orl"] ... I would suggest to start up such a page for some of the nations represented at mathlinks in order to help people getting started. In order to provide a small example, please provide the translation of something like that: equilateral, acute, obtuse, isosceles ,... What do the other people think about it ?[/quote]

The recently launched site thesaurus.maths.org < [url]http://thesaurus.maths.org/mmkb/view.html?resource=index[/url] > "contains brief explanations of mathematical terms and ideas in English, Danish, Finnish, Hungarian, Lithuanian, Polish, Slovak." [From thesaurus.maths.org's homepage].   The people working on the site are trying to expand the number of mathematical terms and ideas contained on the site, as well as the number of languages that it currently supports.  From the site's homepage, they report they are currently working on integrating Spanish into the site .   One of the nice things about this site is, for example, that if you are Finnish and want to look up a term you may do so and get results in Finnish and English, plus any other languages the site supports that you may wish to view on the same web page.  Thus one may compare the English explination of a term to the Finnish explination of the term quite easily.

There is also a thing they call an "m-button" feature that one can use on some web pages.  When you click on your "m-button" a search field appears at the top of the web page that you are viewing alowing you to search for a term on the thesaurus.maths.org site.   I haven't tried this feature on [i]this[/i] site yet.  My guess is that it won't work.  I'll report back about this after I submit this post.  I don't want to loose this post to an error on the page because of clicking the "m-button"  :P

Yours truly,

mathfreak

Hi Everyone,

[quote="mathfreak"]There is also a thing they call an "m-button" feature that one can use on some web pages.  When you click on your "m-button" a search field appears at the top of the web page that you are viewing alowing you to search for a term on the thesaurus.maths.org site.   I haven't tried this feature on [i]this[/i] site yet.  My guess is that it won't work.  I'll report back about this after I submit this post. [/quote]

Here is my report.  The "m-button" did not work on this site :(   I got the following error message from the thesaurus.math.org site:

[quote]Error Report Your request failed to complete. Details follow:
[list]Exception in action handler
null[/list][/quote]

I believe that this probably has to do with conflicting Java code or something on the respective webpages.  I don't really know though because I only know basic HTML.  I would not be able to put something together as impressive as what Valentin Vornicu and the other administrators have put together here with my current level of computer and programing knowledge.

Anyway, check out [url]http://thesaurus.maths.org/mmkb/view.html[/url] if you happen to speak Danish, Finnish, Hungarian, Lithuanian, Polish, or Slovak.   With the aid of the site, you may be able to better understand the English mathematical terms used here at MathLinks Everyone.

Yours truly,

mathfreak

[quote="mathfreak"] I would not be able to put something together as impressive as what Valentin Vornicu and the other administrators have put together here with my current level of computer and programing knowledge.
[/quote]

:D I'm the only Admin, but I had some precious help with installing Latex from Stevem. Anyway, thanks a lot for the appreciation. It's not because of MathLinks that the button does not work, some other issues are at stake here.

Anyway, I expect them to upgrade their coding, as soon as feedback about bugs come in (that's usually what happens with new software).


(Great Britain 1) Consider the polynomial
$P(x) = a_{0}x^{k}+a_{1}x^{k-1}+. . .+a^{k}$
with $a_{0}$, $a_{1}$, . . . , $a_{k}$ being integers. Such a polynomial is said to be divisible by $m$ if $P(x)$ is a multiple for every integer $x$. Show that $a_{0}*k!$ is a multiple of $m$ if $P(k)$ is divisible by $m$. Prove also the following property: If $a_{0}$,$k$ and $m$are positive integers with $a_{0}*k!$ a multiple of $m$, then it is possible to find a polynomial divisible by m of which $a_{0}xk$ has the highest degree.

I'm having trouble finding good criteria for divisibility.

Does anyone have a proof?


geometry
rectangle
parallelogram
probability
MATHCOUNTS
AoPS Books

it probably would..............
just go view the table of contents of the book
see if you like it

[quote="jhredsox"]after doing practice chapters just to warm up for the chapter, i realized i really need help in geometry as in different shapes and such.

would intro to geo help for states in this subject?

i really want to go to nats, but u need top 4 in state (cd is unofficial) and i got 6th written at chapter

-jorian

i hope this doesn't reveal anything about the chapter

if so, delete it (just making sure - i tried not to mention it[/quote]
They usually don't use really complicated geometry subjects.

Know the properties of special quadrilaterals (rhombi, squares, rectangles, parallelograms, isosceles trapezoids).

Simple answer: Yes - will be good in the long run.

i just want to make sure i get top 4 at states b/c 4's a cool number.

jorian

[quote="bpms"][quote="7h3.D3m0n.117"]yeah they focus a lot on geometry...

[hide="rant"]
WHICH IS WHY OUR STUPID CURRICULM PUT ALG 2 FOR 8TH GRADE AND GEO FOR 9TH...even though MC ends after 8th grade....grr im blaming my math teachers for me not getting in to the countdown round...
[/hide][/quote]
Do you expect your teachers to hold your hand and teach you all of the math you need to know if you want to succeed in competitions? In my school they do geo in 8th grade, and I had taught it myself in 7th because I knew that I would need to know it if I wanted to make nationals.[/quote]

no in fact id rather have my math teachers not within 25 miles of me while im studying....

i'm just saying that my curriculm is horriblely against MC...i think it is far easier learning geometry and learning some probability parts from Alg 2 by urself than vise versa....

[quote="perfect628"]my school didn't let 7th graders take alg. 1, and i BEGGED, and they still didn't let me last year. then, this year, they let 7th graders take it. figures...[/quote]

yeah. that always suckk...if u want something u have to beg a few years before u the actual time  :D

[quote="7h3.D3m0n.117"]yeah they focus a lot on geometry...

[hide="rant"]
WHICH IS WHY OUR STUPID CURRICULM PUT ALG 2 FOR 8TH GRADE AND GEO FOR 9TH...even though MC ends after 8th grade....grr im blaming my math teachers for me not getting in to the countdown round...
[/hide][/quote]
our school does the same, i think they do that so that in 9th grade we take geometry the same time the honors students do.

[quote="ehehheehee"]And our school puts Geo in 9th and Algebra 2 in 10th (if you take honors).

Perfect628, The way I skipped my math grades is to come in the first day and ask for an exit exam (while nobody else is looking).  If you are lucky, you will get your exam by the end of the week, get an A and move on.  Meanwhile, the teacher will be asking you whether you are really, really, really sure you know your stuff.

I would think the AOPS texts on Geometry would help more on Mathcounts than the school could.  (My geo teacher skipped the chapter on power of the point because she said there was not much use to it).[/quote]

i believe most of the AoPS books help with mathcounts...

anyway, my parents won't let me skip Alg 2 even though i know all the material because they don't want me too....then they keep bugging me to study and excel in math......


integration
real analysis
real analysis unsolved

Is there any way(elementary) to prove that the sequence $ \frac{e^{\minus{}n}n^n}{(n\minus{}1)!\sqrt{n}}$ is bounded without using Stirling?

I think i finally found a pretty easy way to bound it:
\[ n! \equal{} \Gamma(n \plus{} 1) \equal{} \int_0^{\infty}e^{ \minus{} t}t^n dt \equal{} n^{n \plus{} 1}\int_0^{\infty}e^{ \minus{} n(\tau \minus{} ln(\tau))}d\tau, \ t \equal{} n\tau
\]
Let $ \tau \equal{} x \plus{} 1$ to get $ n! \equal{} n^{n \plus{} 1}e^{ \minus{} n}\int_{ \minus{} 1}^{\infty}e^{ \minus{} n(x \minus{} ln(1 \plus{} x))}dx\geq n^{n \plus{} 1}e^{ \minus{} n}\int_{0}^{\infty}e^{ \minus{} n(x \minus{} ln(1 \plus{} x))}dx\geq n^{n \plus{} 1}e^{ \minus{} n}\int_{ 0}^{\infty}e^{ \minus{} n\frac {x^2}{2}}dx$, as far as $ ln(1 \plus{} t) \minus{} t \plus{} \frac {t^2}{2}\geq 0 \ \forall t > 0$, finally $ n!\geq n^{n \plus{} 1}e^{ \minus{} n}\frac {1}{\sqrt {n}}\sqrt {\frac {\pi}{2}}$.


In how many different ways can a committee with 3 male and 3 female members be chosen from a group of 4 females and 6 males?

Or, following YOUR posting guidelines :P

First, we pick the $ 3$ females.  There are $ 4$ different ways to pick the person to exclude, so there are $ 4$ ways to pick the females.

Next, we pick the $ 3$ males.  There are $ 6 \times 5 \times 4\equal{}120$ ways to pick them, however, we have overcounted by $ 3!$ because we didn't divide for order.  Thus, there are $ 20$ ways to pick the males.

Finally, we multiply these to get $ 4 \times 20\equal{}\boxed{80}$.


find all integer$x,y$ such that:

$x^2-3y^2=1$

We can assume $x, y$ are non-negative. Note that $(1, 0)$ and $(2, 1)$ are solutions. Also, suppose $x^2-3y^2=(x+\sqrt{3}y)(x-\sqrt{3}y)=1$. Multiplying by $(2+\sqrt{3})(2-\sqrt{3})$, we see that $(2x+3y, x+2y)$ is another solution.

Now suppose $(x, y)$ is a solution. Write $x=2a+3b, y=a+2b$, then using the recursion backwards $a=2x-3y, b=2y-x$ is a solution, that is, $(2x-3y, 2y-x)$ is a solution. It's easy to check that these numbers are non-negative. Since $2x-3y+2y-x=x-y<x+y$, the sum $x+y$ of the solution is decreased unless $y=0$. Obviously it can't always decrease, so at some point $y=0$. But then $x=1$, so the solution belongs to the ones obtained by the recursion $(x, y) \rightarrow (2x+3y, x+2y)$. This gives all the solutions.

but it's also important to show that these are the only solutions,
anyway
if we have an equation $x^2-ky^2=t$ and we want to find all positive integer solutions we first find a single pair of positive integers $(x_0,y_0)$ such that $x_0^2-ky_0^2=1$.
If we find a solution $(x',y')$ for the equality $x'^2-ky'^2=t$ then it's easy to check that the pair
$(x'x_0+ky'y_0,x'y_0+y'x_0)$ is also a solution.
Let's put a linear transformation $f(x,y)=(xx_0+kyy_0,xy_0+yx_0)$ thus $f^{-1}(x,y)=(xx_0-yky_0,-xy_0+yx_0)$ so we have to check when $xx_0-yky_0 \leq 0$ or $-xy_0+yx_0 \leq 0$ in order to find the "begining" pairs $(x',y')$ that create classes of solutions and differ pairwise.
In the example above we have $x^2-3y^2=1$ so let's take a pair $x_0^2-3y_0^2=1$ - for instance
$(x_0,y_0)=(2,1)$ so if $(x,y)$ is our solution then $(2x+3y,x+2y)$ also satisfies our equation.
Going ''backwards" we receive a pair $(2x-3y,-x+2y)$ so we have to check when
$2x-3y \leq 0$ or $2y-x \leq 0$ but $2x \leq 3y$ is impossible so $x \geq 2y$ thus $3y^2+1 \geq 4y^2$, so
$y \leq  1$ - so the only beggining pair can be $(2,1)$, so indeed it's the only pair that generates all other solutions.


How many times in a day does the minute hand cross the hour hand?
(edit)At what times do they cross?

Remember that in a day the minute hand goes around 24 times while the hour hand goes around twice.

Oh, yeah, and for what time they cross there is an equation that works for every crossing.

[hide]The equation is 30x+30n=360x, or x=n/11, n being the hour at which we start. For example, at hour 0, x is time 0. At hour 1, x is 1/11 hours (if you look at a clock, you'll see that that fits the common-sense estimate of a bit more than 5 minutes.) Then you have to add n to x to get the time relative to time 0.

If you look at a clock, the hour hand travels at 30 dph (degrees per hour) and the minute hand travels at 360 dph. So if it's, for example, 3 o'clock, at the time the hands meet both hands have traveled x minutes, but the hour hand has traveled a distance of 30x degrees and the minute hand has traveled the same distance plus 3*30 degrees (since it started at 12.) Since the minute hand has also traveled 360x degrees, the two quantities are equal, leading to 360x=30x+3*30. Now just replace 3 by n, representing the hour. Then solve for x, giving x=n/11, and substitute whichever hour you want, from 0 to 11, for n. Then solve for x and add n.[/hide]


Find all triples of positive integers $(x, y, z)$ such that \[(x+y)(1+xy)= 2^{z}.\]

[quote="Peter"]Find all triples of positive integers $ (x, y, z)$ such that
\[ (x \plus{} y)(1 \plus{} xy) \equal{} 2^{z}.
\]
[/quote]
Supose that $ x \leq y$,then $ x;y$ are odd;$ z \geq 2$ and exist $ m < z$ such that$ x \plus{} y \equal{} 2^m$
and $ 1 \plus{} xy \equal{} 2^{z \minus{} m}$
We have $ (1 \plus{} xy) \minus{} (x \plus{} y) \equal{} (x \minus{} 1)(y \minus{} 1) \geq 0$
so $ 2^{z \minus{} m} \geq 2^m$ or $ m \leq \frac {z}{2}$
*If $ x \equal{} 1$ then $ y \equal{} 2^m \minus{} 1;z \equal{} 2m$
*If $ x > 1$ then $ x \geq 3$=>$ 2^m \equal{} x \plus{} y \geq 6$ so $ m \geq 3$
We have $ x^2 \minus{} 1 \equal{} x(x \plus{} y) \minus{} (1 \plus{} xy) \equal{} 2^mx \minus{} 2^{z \minus{} m} \equal{} 2^m(x \minus{} 2^{z \minus{} 2m})$
So $ 2^m|x^2 \minus{} 1$
Note that $ (x \minus{} 1;x \plus{} 1) \equal{} 2$=>$ 2^{m \minus{} 1}|x \minus{} 1$ or $ 2^{m \minus{} 1}|x \plus{} 1$
From $ x \leq y$  we have $ 0 < x \minus{} 1 \leq \frac {x \plus{} y}{2} \minus{} 1 \equal{} 2^{m \minus{} 1} \minus{} 2 < 2^{m \minus{} 1}$
So $ 2^{m \minus{} 1}|x \plus{} 1$
From $ 1 < x \leq y$ then $ x \plus{} 1 < x \plus{} y \equal{} 2^m$
=>$ x \plus{} y \equal{} 2^{m \minus{} 1}$ or $ x \equal{} 2^{m \minus{} 1} \minus{} 1$
=>$ y \equal{} 2^{m \minus{} 1} \plus{} 1$=>$ 4z \equal{} 3m \minus{} 2$

Let we have
$x+y=2^{a}=U,xy=2^{b}-1=V$ then $b\geq a$.
With Viet,we have x,y are root of equation
$t^{2}-Ut+V=0$ then we must have $k^{2}=2^{2(a-1)}-2^{b}+1$ or $(k-1)(k+1)=2^{b}(2^{2(a-1)-b}-1)$
note gcd(k-1,k+1)=2
hence $k=1$ or $k+1\geq 2^{b-1}\to k^{2}\geq 2^{2(b-1)}-2^{b}+1$ then $a\geq b$
All we have $a=b$ or $b+2=2a$
do next is easy

equation implies that $x+y=2^{z_0},xy+1=2^{z_1}$ WLOG assume that $x\le y$ now if $x=1 \implies (1+y)=2^{\frac{z}{2}}$

then one possible solution is

$(1,2^{z}-1,2z)$ so now because $x,y$ should be odd then $x\ge 3 \implies x+y \le xy+1 \implies x+y \mid xy+1 \implies$

$x+y=2^{z_0} \mid x^2-1 =(x+1)(x-1)$, it's easy to see $(x-1,x+1)=2$ now we have to check two case:

[color=#f00]I)[/color] $x+y \mid 2(x+1)$ now as $2(x+y) > 2(x+1)$ so $x+y=2(x+1) \implies y=x+2$

$\implies 2x+2=2^{z_0},xy+1=(x+1)^2=2^{z_1}$ so one other possible solution is $(2^{z}-1,2^{z}+1,3z+1)$

[color=#f00]II)[/color]  $x+y \mid 2(x-1) $ now as  $2(x+y) > 2(x-1)$ implies $x+y=2x-2 \implies y=x-2$

which is a contradiction with $y \ge x$ so no solution in this case

so the solutions of the equation are: $(2^{z}-1,2^{z}+1,3z+1)$ and $(1,2^{z}-1,2z)$


Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $ \$56$. The absolute difference between the amounts Ashley and Betty had to spend was $ \$19$. The absolute difference between the amounts Betty and Carlos had was $ \$7$, between Carlos and Dick was $ \$5$, between Dick and Elgin was $ \$4$, and between Elgin and Ashley was $ \$11$. How much did Elgin have?

$ \textbf{(A)}\ \$6\qquad
\textbf{(B)}\ \$7\qquad
\textbf{(C)}\ \$8\qquad
\textbf{(D)}\ \$9\qquad
\textbf{(E)}\ \$10$

A-B= $ \pm$ 19
B-C=$ \pm$ 7
C-D=$ \pm$ 5
D-E=$ \pm$ 4
E-A=$ \pm$ 11
But the sum of all five numbers must be 0. We find 19+4-11-7-5=0 works.
A-B=19
B-C=-7
C-D=-5
D-E=4
E-A=-11
Write everything in terms of E
E+11=A
E+4=D
E-1=C
E-8=B
Plugging into A+B+C+D+E=56 and solving gives E=10, which checks.


The math section on the SATs and PSATs I have no problem with, but the writing and reading sections are nightmares!  I scored a 56 on both the reading and writing section.  Does anyone have any suggestions?  Are there online SAT writing and reading classes(aops-like literature websites)?  Will doing tons and tons of SAT questions help?  because I know for math you need to go over the concepts not only do tons of math problems.  What concepts are there for reading and writing?
Thank you so much!

Just remember that people can wake up from nightmares and have a great day. Yes, you have time to get better.

P.S. Good luck to everyone taking the ACT today.

Well I do not really think there is much that is conceptual to do with reading.  I've heard the best ways to improve your score is to start reading magazines like Time, the New Yorker and the editorial page of the NY Times I guess is really good as well.

Other than that, get a big fat book and start doing reading sections -- they take 25 minutes a piece and probably about 7-8 per passage... so whenever u have a minute you could sit down and do one.  Then, just start building your vocab, look up any words you don't know in anything.

I think all of that should help.

As far as writing, there are a ton of grammatical rules and such that can be learned that are in most of the SAT Prep books.

it's good to keep in mind the vocabulary on the PSAT and the SAT that account for many points also. and a big contributor to your overall score

I find that just being engaged in my English and humanities classes automatically improves my reading and writing on SAT-type stuff. However, in this case it might be too late for that...

Well, I'm not going to college, but when I took the SATs two years ago, the main thing that got me down was the vocab. It definitely contributed to at least half of my missing points. (I got a...was it 620 or 660? I forgot.)

Reading scores improve steadily throughout the high school years for students who read a lot. You can read mostly about math, if math is what you like, but be sure to read more than you are assigned to read in school. I agree with Probability 1.01 that if you get involved in your school English lessons, you will get better as a reader and writer, even if the school lessons are rather lousy. A book I recommend about writing, because it is funny and thus easy to remember, is [url=http://www.amazon.com/How-Not-Write-Essential-Misrules/dp/039332723X/]How Not to Write[/url] by William Safire.

SparkNotes makes these vocab books that really helped for the reading section... they're normal books with common SAT vocabularly words in bold and defined.  I think they're called "Score-Raising Novels" or something similar.

Good Luck!


[quote="Honey_S"][b]Problem.[/b] Let $ a,b,c$ be positive real number such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \sqrt {\frac {a}{{b^3 \plus{} abc}}} \plus{} \plus{} \sqrt {\frac {b}{{c^3 \plus{} abc}}} \plus{} \sqrt {\frac {c}{{a^3 \plus{} abc}}} \ge \frac {{ab \plus{} bc \plus{} ca}}{{\sqrt 2 }}
\]
[b]P/s.[/b] For the stronger one I haven't got a solution for it
\[ \sqrt {\frac {a}{{b^3 \plus{} abc}}} \plus{} \plus{} \sqrt {\frac {b}{{c^3 \plus{} abc}}} \plus{} \sqrt {\frac {c}{{a^3 \plus{} abc}}} \ge \frac {{3}}{{\sqrt 2 }}
\]
:)[/quote]

Here is my solution for the first. How about the stronger one  :wink: . There is a solution for both of them with AM-GM by can_hang2007. :)


geometry
power of a point
projective geometry

Draw 2 tangent lines from the point $ (3,\ 1)$ to the circle $ x^2 \plus{} y^2 \equal{} 5$, denote 2 points of tangency by $ A,\ B$. Find the equation of the line passing through 2 points $ A,\ B$.

Let O be (0,0) and P be (3,1), let PO cut the circle at Q and R, with Q closer to P. By Pythag, $ OP \equal{} \sqrt {10}$. By power of a point, $ PQ \cdot PR \equal{} AP^2$, so $ (\sqrt {10} \minus{} \sqrt {5})(\sqrt {10} \plus{} \sqrt {5}) \equal{} AP^2$, so $ AP \equal{} \sqrt {5}$. So OAP is an isoceles right triangle, so OAPB is a square. So AB is the line that goes through the midpoint of OP and is perpendicular to it, which is $ y \minus{} \frac {1}{2} \equal{} \minus{} 3(x \minus{} \frac {3}{2})$, or $ y\equal{}\minus{}3x\plus{}5$.

Correct. :)

Let the points of tangency be $ A(x_1,\ y_1),\ B(x_2,\ y_2)$, by the formula of tangent line for a circle, the tangent lines at $ A,\ B$ are $ x_1x \plus{} y_1y \equal{} 5\ , x_2x \plus{} y_2y \equal{} 5$ respectively, since these lines intersects in the point $ (3,\ 1)$, we have $ 3x_1 \plus{} y_1 \equal{} 5,\ 3x_2 \plus{} y_2 \equal{} 5$, which means that 2 points $ A,\ B$ are on the same line $ \boxed{3x \plus{} y \equal{} 5}$.

Comment:

This line is called "polar line" with the point $ (3,\ 1)$ called "pole".


A matrix is not uniquely determined using solely the first value and its determinant; I'm relatively certain of this without actually having a good proof of it.


least common multiple
greatest common divisor

Agh, apparantly my brain is malfunctioning.  :blush:

I meant least common multiple.  With greatest common factor the problem is trivial as i_like_pie showed.

[quote="E^(pi*i)=-1"]The least common multiple of $72$, $500$, and $x$ is the same as the least common multiple of $392$, $384$, and $x$, where $x$ is a positive integer.  Find the minimum possible value of $x$.[/quote]
[hide]$72=2^{3}\cdot3^{2}$
$500=2^{2}\cdot5^{3}$

$392=2^{3}\cdot7^{2}$
$384=2^{7}\cdot3$

Take all the largest powers to get $x=2^{7}\cdot3^{2}\cdot5^{3}\cdot7^{2}=\boxed{7056000}$.[/hide]


Anyone play the game? It uses pure ASCII...

judging by the wikipedia article, it looks similar to pokemon mystery dungeon.

speaking of pokemon mystery dungeon...
[youtube]GDdNsomWEno[/youtube]
I AM MUDKIP

LOL

[quote="junggi"]speaking of pokemon mystery dungeon...

I AM MUDKIP

LOL[/quote]

watch out
he's-a gonna wat'r gun u

that's a bit boring...

I fought:
Umber hulk
Python
Couatl
Ettin Mummy
Ogre King

Spell it out in Nethack...

Yep.


I think problem 4 can be generalized . For example

 Every point in the numerical plane is colored either red, green, or blue.
Prove that there exist a rectangle in that plane such that all four of its vertices are the same color and that

$A.$ it's biggest side is smaller than 19 .
$B.$ it's smallest side is smaller than 4.

Yeah, that is true, but that would actually make it easier. If I had known that there was a 4X19 rectangle involved, I probably would have gotten that problem right.

You could have the entire plane, not just the lattice points, and demand an arbitrarily small rectangle. Of course there's no real difference.

[quote="Phelpedo"]I think you can generalize it to $n$ colors and an arbitrarily "thin
subset of the plane. You would just have to take $n^{4}+1$ parallel lines.[/quote]

Yes, I think we can.

Every point in the numerical plane is colored either n different colors.
Prove that there exist a rectangle in that plane such that all four of its vertices are the same color and that

A) it's biggest side is smaller than n*6+1 .
B) it's smallest side is smaller than 4.

I think it's not hard

all possible ways are
BBxy. . . .
BxBy. . . .
BxyB. . . .
xByB. . . .
xyBB. . . .
xBBy. . . .
x,y are some colors
B=B is one of n colors =>
after S=n*6+1 strings we have

the same combination

for example
    CxyC
    ........
     .......
        .
     .........
    CxyC  which be a rectangle with same colored vertices.


inequalities
superior algebra
superior algebra unsolved

Let $ s(n)$ be the sum of positive divisors of n. How can I show that $ s(n) < 2n\sqrt{n}$.

You can pair up $ d$ with $ \frac nd$, this gives that there at most $ 2 \sqrt n$ divisors. Now every divisor is at most $ n$, the result follows.

Here's a brute force solution.  It suffices to show the result where $ n$ is a power of $ p$.

Then $ 1 \plus{} p \plus{} p^2 \plus{} \ldots p^k < (k \plus{} 1)p^k < 2p^{k/2} p^k \equal{} 2p^k\sqrt {p^k}$, where the inequality $ k \plus{} 1 < 2p^{k/2}$ can be shown by induction.

...not that this adds very much, but since I worked it out I thought I might as well post it. :)

EDIT: Well, I guess this approach has the advantage that it lets you refine the bound to $ Cn^{1 \plus{} \epsilon}$ for all $ \epsilon > 0$.

EDIT 2: The first assumption I gave above is incorrect--I think this whole argument is bogus.

As it seems not to be clear, a little bit more: if $ d$ divides $ n$, then $ \frac nd$ divides $ n$ as well. Especially, one out two divisors is $ \leq \sqrt n$ (as either $ d$ ir $ \frac nd$ is). Thus the number of divisors is at most $ 2 \sqrt n$.

thanks, now I have also worked with this, and got more or less the same conclusion, but anyways thanks to make it clear:-)


conics
hyperbola
geometry
parallelogram
circumcircle
geometry proposed

ABCD is cyclic quadrileteral. Then six lines that pass through the midpoints of sides of ABCD and are perpendicular to opposite sides are concurrent ( diagonals are opposite sides). Prove that the point of intersection is center of right hyperbola that pass through points A,B,C,D

Let $H_A,H_B,H_C,H_D$ be the orthocenters of $BCD,CDA,DAB,ABC$ respectively. That hyperbola is the unique conic passing through $A,B,C,D,H_A,H_B,H_C,H_D$ (we disregard the annoying cases when the quadrilateral has right angles; those can be treated as limit cases of the general case). As is well-known, if two chords of a conic intersect in their mutual midpoint, then this midpoint must be the center of the conic.

Notice that $ADH_AH_D$ is a parallelogram, since $AH_D,DH_A\perp BC$, and they're both equal to twice the distance from $O$ (the circumcenter of $ABCD$) to $BC$. By the observation above, $AH_A\cap DH_D$ is precisely $P$, the center of the hyperbola under consideration. Clearly, since the perpendicular to $BC$ in the midpoint of $AD$ is parallel to $AH_D,DH_A$ and equidistant from these two lines, it must pass through the center $P$ of the parallelogram $ADH_AH_D$, and we're done (in the same way, we prove that all six lines perpendicular to a side through the midpoint of the opposite side pass through $P$).


Are there integers $x,y,z,t$ such, that $x^2-y^2=y^2-z^2=z^2-t^2\not=0$

No, it was posted before that there are no four squares in arithmetic progression.
http://www.mathlinks.ro/Forum/viewtopic.php?t=52021 (without solution given there, but a reference to the MLC: http://www.mathlinks.ro/Forum/contest/4th%20Edition/4th%20edition%20-%20round%201.pdf)
also http://www.mathlinks.ro/Forum/viewtopic.php?t=850 is interesting.


Hello everyone. Recently, I found myself wondering. Who is the best player on FTW nowadays?

My first instinct: Look at the current top 10 ratings. It's probably one of:

AIME15 2034
Mewto55555 1974
nt811581147 1931
FantasyLover 1897
miller4math 1893
EcstaticPotter 1874
stevenmeow 1870
Ttocs45 1862
BOGTRO 1851
ernie 1843

Then I looked at that list. Mew, AIME, and ttocs had all been involved in their fair share of scandals. Stevenmeow is relatively inactive. Etc. etc. (Note this isn't meant to slam AIME or Mew or ttocs, its only to say that this isn't their true rating, and that rating isnt the best indicator).

Highest ever rating fails to identify a top player as well.

So I looked at tournaments. Of all the people, Mewto, superllamel, fantasy, and RG were among the high placers. However, RG and super have been inactive, mew is, well, mew is mew, and fantasy is good, but is he the best?

Then it hit me. If I don't know who the best is, why don't I [b]find out?[/b] And once I know, shouldn't I post it so others aren't in the same dilemma?

And so, I create a new system of mini-tournaments. Now before TNCTC (The "New" CTC) gets on my back for skipping the queue (spelling?), I must point out, this is no ordinary tourney.

This is a system. Not a tournament. A series of games, to determine the champion. Sure sounds like a tourney, eh?

But no. I decided to borrow my idea from a game we almost all have been, are, or will be addicted to. No, not Starcraft!

This system calls for 12 volunteers.

A person who wishes to be FTW Grand Champion must first sign up. There will be a long list of waiting people. At various times, whenever the heck they want, they can choose to challenge one of the 8 "FTW Masters". These are 8 volunteers. If our hero, I will call him Bob, defeats a master, he gains a "FTWGC Badge". He then challenges another Master, and so on. If he ever loses to a Master, he loses 3 random badges.

Once he has all 8 badges, he moves on to challenge the Epic Four. These are 4 more volunteers, but who are of a very high caliber of FTW playing. He will play all of these in a row. If Bob loses to anyone of these, he will be stripped of 5 random badges, and must begin again in his badge collecting.

If, however, Bob defeats all of the Elite, I mean Epic Four, he will become FTW Grand Champion! If however, someone was Grand Champion before him, then he will have to face the current champion and defeat him to gain his current spot. If he loses to the champion, he loses 5 random badges. However, if he wins, he begins his reign as [b]FTW Grand Champion!!![/b]

A few things to note:

FTW Masters and Epic 4 may not become FTWGC. I'm sorry, that's life. However, you will still gain recognition as a good FTW player?

The first 12 people to PM [b]me[/b] with their current and highest ratings will be given the spots of Epic 4 and Masters. Obviously, the Epic 4 will be the best of those who sign up.

If you miss the original signup deadline, you can still be a Master or Epic 4.

First, you must PM me your desire. I will ask around among Masters if any of them are willing to give up their spot. If no one does, you may then choose any Master and attempt to unseat them by beating them in a beast of 9 cd. You may only challenge 1 Master a month in this manner. Once you have been a Master for 2+ weeks, you may choose to challenge an Epic 4 member. The same rules apply, however, you can challge E4 members once every two weeks.

All games for badges are best 3 of 5.

All games against the epic 4 are best 5 of 9.

All games against champion are best 7 of 13.

Signup to play and earn badges in-thread I guess.

No flaming, mods please don't lock, just remove flames/spam.

PM me with questions.

Tytia.

PS: Was I too verbose?

This reminds me of Pokemon :rotfl:

Hm, this "maxschindler" person reminds me of a certain user...:P

EDIT: Make it so that the Epic 4 and Gym Leader people need, say, at least 1800 and 1600 rating respectively, just to keep the levels up there.

Oh by the way, it doens't matter how active you are. You can challenge like once a week maybe if you want. The only people who need to be pretty active (once every day or two) are the Masters (and possibly the Epic Four).

[quote="AIME15"]People who I believe should apply for Epic 4 or Masters:

AIME15
Mewto55555
FantasyLover
miller4math
EcstaticPotter
stevenmeow
Ttocs45
BOGTRO
ernie
Jongao
isabella2296
mousy[/quote]

if this is cd no way miller's in

Don't worry about Masters yet.

Anyway, there has been some confusion. If you wish to become a master or part of the epic four, PM me.

If you show any interest at all in playing and trying to get badges and whatnot, just post that you are interested. You don't need to post ratings, I just want a general idea of how many people are doing this.

I guess I'm interested to pwn the so-called 'Epic 4'.

Rofl Pokemon :rotfl:

BTW I'm not signing up.

[quote="Meta_Knight"]
if this is cd no way miller's in[/quote]

miller4math is not bad at all in FTW cd, it's real cd he's bad at :rotfl:

A bit more clarification:

What I said about the one month rule between challenging Masters is [b]only if you want to become a Master[/b]. If you just want badges, you can challenge them as often as you want (within reason)!

[quote="maxschindler"]6 people thus far have applied to be Masters or members of the Epic 4.

In addition 5 people including myself have expressed interest in actually participating.[/quote]

Really? You or Mewto55555?

I see that you borrowed this from my ftw mmorpg.

I'll sign up for non-master non-epic4.

[quote="AIME15"]People who I believe should apply for Epic 4 or Masters:

AIME15
Mewto55555
FantasyLover
miller4math
EcstaticPotter
stevenmeow
Ttocs45
BOGTRO
ernie
Jongao
isabella2296
mousy[/quote]

aww
i feel left out now  :(

Yes, but I am not AIME, and I am accpeting the first 12 to apply. There are still a few slots left. Being a master is a great honor!

Hello!  I feel I have a good shot at this... 1962 highest rating, current 1942. :D

I WANNA JOIN!  LET'S HAVE FUN !!!!

should each of the "masters" have a special badge?

and just some advice, everyone participating in this as a regular person should post their name in a separate thread
when they defeat one of the "masters", the "master" should pm one of the mods to edit the person's post to include the badge
or, if the person is still allowed to edit their own post, they can do it themselves

does this make sense to anyone?


trigonometry
function
logarithms
inequalities
algebra unsolved
algebra

solve the following
\[ 2^x+2^{\sqrt{1-x^2}}=3  \]

 :)

$x=0$ and $x=3$
(I suppose you want a good expanation along with that...I think its acceptable just to see that $2^0+2^1=3$, but I suppose that you could prove that there are no other solutions.)

EDIT: Whoops, I should have typed $x=0$ and $x=1$.

I will show that $x=0$ and $x=1$ are the only solutions.
First note that $x \in [-1,1]$, as otherwise the square root takes complex values.
A natural substitution would be $x = \sin t$ where $t \in [-\pi/2,\pi/2]$.
Consider the function $f(t) = e^{a \sin t} + e^{a \cos t}$ where $a : = \log 2$.
It suffices to show that $f$ is strictly increasing on $[-\pi/2,0]$, strictly increasing on
$[0,\pi/4]$ and strictly decreasing on $[ \pi/4,\pi/2]$.
The first claim is obviously true,  because both $\sin t$ and $\cos t$ are strictly
increasing on $[-\pi/2,0]$. The third follows from the second via the substitution
$t \to \pi/2 - t$. For the second it suffices to check that $f'(t) = \cos t e^{a \sin t} -\sin t e^{a \cos t} \ge 0$,
i.e. $\tan t \le e^{a(\sin t - \cos t)}$, i. e. $\log \tan t \le a(\sin t - \cos t)$ for $t \in [0,\pi/4]$.
In the last inequality we have equality for $t = \pi/4$, so it suffices to check that
on the given interval we have
$\frac d {dt} \log \tan t \ge \frac d {dt} a(\sin t - \cos t)$, i. e. $\frac 1 {\tan t} \frac 1 {\cos^2 t} = \frac 2 {\sin (2t)} \ge a(\cos t + \sin t)$.
But on the given interval $\sin (2t)$ takes its maximal value $1$ for $t = \pi/4$
and $\cos t + \sin t$ takes its maximal value $\sqrt{2}$ for $t = \pi/4$, which again
can be seen e.g. by differentiation. So we are left to check
$2 \ge a \sqrt{2}$, which is obvious.  ;)

If I am not wrong I found a nice proof .
First of all we have   $-1\leq x\leq 1$ .
I will brake it in two cases .
   If $-1\leq x\leq 0$
The fuction $2^x$  is increasing so  $2^x\leq 2^0=1$
Also $\sqrt{1-x^2}$  is an increasing function when $-1\leq x\leq 0$ .So  $2^{\sqrt{1-x^2}}\leq 2$ .
So $2^x+2^{\sqrt{1-x^2}}\leq 3$  for $-1\leq x\leq 0$
   If $0\leq x\leq 1$ then  $2^x\geq 2$  and  because
$\sqrt{1-x^2}$  is  decreasing when $0\leq x\leq 1$  we have
$2^{\sqrt{1-x^2}}\geq 1$ .
So for $0\leq x\leq 1$ we have  $2^x+2^{\sqrt{1-x^2}}\geq 3$
So the only solutions are x=0 and x=1

Well, not quite:
If $0 \le x \le 1$ then $1 \le 2^x \le 2$, which is too weak to give a useful inequality approach.

You are right .....  :blush:
I think  it holds that for $0\leq x\leq 1$
$2^x+2^{\sqrt{1-x^2}}\geq 3$  but I have no proof  yet

Well, this is exactly what I have shown...  :P


algebra
binomial theorem
Pascal\u0027s Triangle

Compute the exact value of 1000000001^4 without a calculator.

I think this is a long way to do it:
[hide]We can write that as: 100000000+1, and let x=100000000.
We want to find $(x+1)^{4}$, so using the binomial theorem that is:
$x^{4}+4x^{3}+6x^{2}+4x+1$ and I guess if I wanted to, I could expand all of that, but who cares. Anyway, I get
100000000000000000000000000000000+4000000000000000000000000+60000000000000000+400000000+1, or if I added correctly: 1000000004000000006000000004000000001.
[/hide]

ok, now I'm sure there's a faster way.

Edit:Ya, there was.

It basically [i]is[/i] the binomial theorem. Just remember 11^4=14641 and add in the zeros in between to get: 1000000004000000006000000004000000001.

Edit: woops  :blush:

I think Anaxerzia meant $11^{4}$. Oh, I can't believe I missed that. I must have been half asleep.

You can use pascal's triangle to solve this problem. The fifth number(not the fourth number because the first number is the zeroeth power) is 14641. You can put all eight zeroes between the digits to get the answer.So the answer is 1000000004000000006000000004000000001 . This also works for other numbers with zeroes in the middle like 101,1001,10001 etc.

Thats pretty much what I did you know...


at the cost of being bugging u ppl with problems:).. i will go ahead with another one
here's the problems statement;
Q)

Harry Potter has n mixtures in front of him, arranged in a row. Each mixture has one of 100 different colors (colors have numbers from 0 to 99).

He wants to mix all these mixtures together. At each step, he is going to take two mixtures that stand next to each other and mix them together, and put the resulting mixture in their place.

When mixing two mixtures of colors a and b, the resulting mixture will have the color (a+b) mod 100.

Also, there will be some smoke in the process. The amount of smoke generated when mixing two mixtures of colors a and b is a*b.

Find out what is the minimum amount of smoke that Harry can get when mixing all the mixtures together.
Input

There will be a number of test cases in the input.

The first line of each test case will contain n, the number of mixtures, 1 <= n <= 100.

The second line will contain n integers between 0 and 99 - the initial colors of the mixtures.
Output

For each test case, output the minimum amount of smoke.
Example

Input:
2
18 19
3
40 60 20

Output:
342
2400

well smells of dp...or am i wrong??
please help
vasu

well well here is my solution
using dynamic programming
i constructed an array of size n*n;
and let dp[i][j] store the minimum value possible between the ith and jth index in the n mixturess..
so what we are looking for is dp[0][n-1]...using zero-based indexing.

i used dp..not sure if it is right
but it's getting wrong answer on the spoj forum
//code follows
#include<cstdio>
#include<iostream>
using namespace std;

int dp[101][101];
//below function calculates sum all values between indices i till j inclusive
int sum(int i,int j)
{
  int ret=0,k;
  for(k=i;k<=j;k++)ret+=dp[k][k];
  return (ret%100);
}

int main()
{
  int n;
  while(1)
    {
      if(!scanf("%d",&n))break;
      if(feof(stdin))break;
      int i,j;
      memset(dp,0,sizeof(dp));
      for(int i=0;i<n;i++)
	{
	  scanf("%d",&dp[i][i]); //storing it along the diagonal of the array
	}
      for(i=1;i<n;i++)for(j=0;j<n-i;j++)
	{
	  if(i==1)dp[j][j+1]=dp[j][j]*dp[j+1][j+1];
	  else
	    {
	      dp[j][j+i]=min(dp[j][j+i-1]+sum(j,j+i-1)*dp[j+i][j+i],dp[j+1][j+i]+sum(j+1,j+i)*dp[j][j]);
	    }
	}
      printf("%d\n",dp[0][n-1]);
    }
  return 0;
}

please help as i cant make out what's wrong with this

A couple of points:
1) When you get down to one colour, from what I can tell you are adding that colour to the total smoke. You don't get any more smoke once you get to one colour, if you have written the problem correctly.
2) I think your code isn't considering all options - when you want to mix together colours from position i to position j, it looks like you are only trying:
a) mix the colours from i+1 to j, then mix that with colour i
b) mix the colours from i to j-1, then mix that with colour j.
There are lots more possibilities - you could mix colours i to i+k, and colours i+k+1 to j, and then combine those two. The answer should be the minimum of those options, for each possible k.

I'm thinking it will also save a lot of time if you store the sums in an array too (calculated with DP again).

[edit - just submitted my way on sphere, and it works.]

Wow, recomputing sums is such a waste  :o

P.S. Just submitted my solution on SPOJ and got AC with 0.02 s runtime  :D

i am getting a wrong answer for quite sumtime now..so i am bugged...
guess u guys cud help
here is my code

#include<iostream>
#include<cstdio>
using namespace std;
#define INF 1000000000
/*array storing values of smoke obtained by using mixtures between i and j inclusive*/
int dp[101][101];
int n;//no of mixtures
int sums[101][101];/*array storing values of sums between i and j inclusive*/
//process of computing sums
void process1(void)
{
	int i,j;
	int sum=0;
	for(i=0;i<n;i++)
	{
		sum+=dp[i][i];
		sums[0][i]=sum;
	}
	for(i=1;i<n;i++)for(j=i;j<n;j++)sums[i][j]=sums[i-1][j]-dp[i-1][i-1];
}
//now the real thing bottom up dp to calculate values
int process2()
{
	int i,j,k;
	for(j=1;j<n;j++)for(i=0;i<n-j;i++)
	{
		if(j==1)dp[i][i+1]=dp[i][i]*dp[i+1][i+1];
		else
		{
			int min=INF,mm;
			for(k=i;k<i+j;k++)
			{
				if(k==i)
				{
					mm=dp[k+1][i+j]+(sums[k+1][i+j]%100)*dp[i][i];
				}
				else if(k==i+j-1)
				{
					mm=dp[i][k]+(sums[i][k]%100)*dp[i+j][i+j];
				}
				else
				{
					mm=dp[i][k]+dp[k+1][i+j]+(sums[i][k]%100)*(sums[k+1][i+j]%100);
				}
				if(mm<min)min=mm;
			}
			dp[i][i+j]=min;
		}
	}
	return dp[0][n-1];
}

int main()
{
	while(1)
	{
		memset(dp,0,sizeof(dp));
		memset(sums,0,sizeof(sums));
		if(feof(stdin))break;
		if(!(scanf("%d",&n)))break;
		for(int i=0;i<n;i++)scanf("%d",&dp[i][i]);
		if(n==1)printf("0\n");
		else
		{
		process1();
		int ret=process2();
		printf("%d\n",ret);
		}
	}
	return 0;
}

i hope it is commented enuf for u guys to understand as it is ugly..i am storing values of the mixtures along the diagonal...but when n=1 there wont be smoke..(as tripleM said) so i take care of that in the main()
for computing dp[i][i+j] i consider all values of k between i and i+j-1 inclusive
//with special consideration for k=i and k=i+j-1

can u guys spot something wrong
please help!!
thankss a lot anyways
 :D

well i advise you to use CODE option while posting your code. it would be better to understand. and i think you should use indentation to make your code more readable!

Well, he did indent it ;) It just doesn't show up unless you use code tags.
Anyway.. I've looked at your code for ages, and can't see a thing wrong with it.
Heres my (java) code if you want to compare (though I ordered the dp in a slightly different way).
[code]import java.util.*;
import java.io.*;
public class Harry {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(new InputStreamReader(System.in));
		while (sc.hasNext()) {
			int N = sc.nextInt();
			int[] nums = new int[N];
			for (int i=0; i<N; i++) nums[i] = sc.nextInt();
			int[][] sums = new int[N][N];
			for (int i=0; i<N; i++) {
				sums[i][i]=nums[i];
				for (int j=i+1; j<N; j++) {
					sums[i][j]=(sums[i][j-1]+nums[j])%100;
				}
			}
			int[][] ans = new int[N][N];
			for (int i=N-1; i>=0; i--) {
				ans[i][i] = 0;
				for (int j=i+1; j<N; j++) {
					ans[i][j] = Integer.MAX_VALUE;
					for (int k=i; k<j; k++) {
						ans[i][j] = Math.min(ans[i][j],ans[i][k]+ans[k+1][j]+sums[i][k]*sums[k+1][j]);
					}
				}
			}
			System.out.println(ans[0][N-1]);
		}
	}
}[/code]


geometry
calculus
analytic geometry
graphing lines
slope
trigonometry
function

A density curve is defined as a definite curve such that the area underneath is exactly equal to 1.

[b]Question:[/b] A particular density curve beings at the origin and has a slope of 1. What are the coordinates of the right endpoint?

Because my stat class does not require calculus, I was wondering if there was a non-calculus method used to obtain the answer to this question? I've already solved this problem (using calculus, however), and the answer is

[hide]
$ (\sqrt {2},\sqrt {2})$
[/hide]

[hide]Well the slope is 1 so your coordinates will be (a,a) for some value a... and then its a 45-45-90 right triangle?[/hide]

But in using $ \frac{ab \sin C}{2}$, I don't get the same answer as I do with calculus? (Could be my calculation error, but I went over it 3 times...)

Well, you have a right triangle with legs $ a$ and $ a$, so the area is $ a^2/2$ so $ a\equal{}\sqrt{2}$ so the coordinate is $ (\sqrt{2},\sqrt{2})$

Oh whoops, I accidently used the $ 45^{\circ}$ as $ \sin C$  :ewpu: .

Thanks.

The area under a linear function (a function with constant slope) that touches the origin (!) is trivial... Just imagine a right triangle with legs $ a$ and $ f(a)$... so the area is $ \frac {af(a)}2$... and since in this case $ f(a) \equal{} a$, we have $ \frac {a^2}2 \equal{} 1$, and solve to get $ a^2 \equal{} 2$ and $ a \equal{} \sqrt 2$.


[quote="itazzirun"]without using calculator, prove that cos(2pi/5)+cos(4pi/5)=-1/2[/code][/quote]
 I put A= $cos(\frac{2\pi}{5})+cos(\frac{4\pi}{5})$
 :[code]cos(\frac{2\pi}{5})+cos(\frac{4\pi}{5})=-\frac{1}{2}[/code]

$cos(\frac{2\pi}{5})+cos(\frac{4\pi}{5})=-\frac{1}{2}$

$2Asin(\frac{\pi}{5})=....$


calculus
derivative
function
limit
inequalities open
inequalities

a,b>0 and satisfied $ a \plus{} b \equal{} 1$ . Find the biggest value of
 $ A \equal{} \frac {1}{a^3 \plus{} b^3} \plus{} \frac {1}{ab}$

hello, i think, it doesn't exist a maximum, do you mean the  minimum?
Sonnhard.

[quote="AngelHolder"]You wrong! It exits maximum, for a=b[/quote]
Try $ a \to 0, b \to 1$ :) Then the max is $ \plus{} \infty$ :)

[quote="EastyMoryan"]Did you mean $ A \equal{} \frac {1}{a^3 \plus{} b^3}$? (Which has a max of $ 4$ when $ a \equal{} b \equal{} \frac {1}{2}$)

Let $ \alpha \equal{} 3 \plus{} \sqrt {3}$ and $ \beta \equal{} 3 \minus{} \sqrt {3}$

Since $ a \equal{} 1 \minus{} b$

The question is equivalent to

$ A \equal{} \frac {1}{(1 \minus{} b)^3 \plus{} b^3} \plus{} \frac {1}{(1 \minus{} b)(b)}$
$ A \equal{} \frac {1}{1 \minus{} 3b \plus{} 3b^2 \minus{} b^3 \plus{} b^3} \plus{} \frac {1}{b \minus{} b^2}$
$ A \equal{} \frac {1}{3b^2 \minus{} 3b \plus{} 1} \plus{} \frac {1}{b \minus{} b^2}$
$ \frac {dA}{db} \equal{} \frac { \minus{} 1(6b \minus{} 3)}{(3b^2 \minus{} 3b \plus{} 1)^2} \plus{} \frac { \minus{} 1(1 \minus{} 2b)}{(b \minus{} b^2)^2}$
Setting $ \frac {dA}{db} \equal{} 0$
$ (2b \minus{} 1)(3b^2 \minus{} 3b \plus{} 1)^2 \minus{} 3(2b \minus{} 1)(b \minus{} b^2)^2 \equal{} 0$, with no new restrictions on $ b$.
$ (2b \minus{} 1)[(9b^4 \plus{} 9b^2 \plus{} 1 \minus{} 18b^3 \plus{} 6x^2 \minus{} 6x) \minus{} 3b^2 \plus{} 6b^3 \minus{} 3b^4)] \equal{} 0$
$ (2b \minus{} 1)(6b^4 \minus{} 12b^3 \plus{} 12b^2 \minus{} 6b \plus{} 1) \equal{} 0$
$ (2b \minus{} 1)(\alpha b^2 \minus{} \alpha b \plus{} 1)(\beta b^2 \minus{} \beta b \plus{} 1) \equal{} 0$
The first bracket shows that there is a max/min/POI at $ x \equal{} \frac {1}{2}$.
The third bracket has complex roots, which I'm ignoring.
The second bracket has roots at $ b \equal{} \frac {1}{2} \pm \frac {\sqrt {2\sqrt {3}}}{6 \plus{} 2\sqrt {3}}$

I suppose I could go through the second derivative, but suffice it to say that the roots of the second bracket are local minimums (they are equal, and they're the minimums of this function (since it's bounded from $ (0,1)$, and the local maximum is when $ a \equal{} b \equal{} \frac {1}{2}$, and at that point $ A \equal{} 8$, but $ \lim_{b \to 0^ \plus{} } \equal{} \infty$ and $ \lim_{b \to 1^ \minus{} } \equal{} \infty$, meaning this function has no absolute maximum.[/quote]
By the power-mean ineq: $ \frac{a^3\plus{}b^3}{2} \ge \left ( \frac{a\plus{}b}{2} \right ) ^3 \equal{} \frac{1}{8}$ hence $ \frac{1}{a^3\plus{}b^3} \le 4$..


Hi all  :)  i want to ask about this theorem.

If $S$ is a set and let $a=\inf \, S$ and $b= \sup \, S$. if $(a,b) \subset S$ then $S$ is an intervaL.

 i've no idea how to start? :?  In my books it is stated without proof, and it is used to prove the characterization theorem. its look obvious but not a definition so must be proved isn't it?
sometimes something obvious is not easy to prove for me.  :P

thanks.

$a$ is a lower bound for $S$ and $b$ is an upper bound for $S$. Therefore $(a,b)\subset S\subset [a,b]$. There are only four possible sets $S$ with this property, and it is easy to check that all are intervals.


When the book says "Why?" or "How would you solve this by..." where could I find the answer?

Yeah, but how do you know if you got it right?

The point is to convince yourself that it makes sense. Often to make these things more clear, on problems that can be easily solved with a lot of tedious computations, you will be asked to prove it with no computation at all to demonstrate that you understand the concept.

its just to make you think some more and not just say, "oh yeah, i got the problem!", even though you may have been lucky and not fully understood it. etc etc

generally Tare, the book was written by mr R here and mr Sandor to make you think...

so, unless that "Why?" was included for you in the solutions book then you'll have to figure it out or post here for us to attempt to help...most of it is common sense though...

Best bet if you can't figure it out yourself is to post in the forum, as several others have done when they've had questions about the book.

You think it's good to make a section of the board called "From the Books" or something?

Well...if we advertise the book more here (in AoPS.com), then it might be good, but other than that it'd be pretty unpopular...or is it only me?


analytic geometry
geometry
incenter
complex numbers
geometry unsolved

Anyone?

  In the Holidays spirit I  relax the requirements and declare that trigonometric and/or

  coordinate methods are welcome here.

  Thank you.
  M.T.

Allow me to relax the conditions a bit more: you also can use complex numbers,
  Lagrange factors.

  Happy holidays to all.
  M.T.

[quote="armpist"]Given  are two triangles in general position.

  Construct a straight line splitting each in two equal areas.

  That's it.

  M.T.[/quote]

 I knew that we can prove this fact using Bolzano's theorem , but I do not know any euklidean construction.
 I will be expecting this interesting proof.

 [u]Babis [/u]

[quote="El Chamo"]a linie through the geometric centers of the triangles  :)[/quote]

Nonsense (I am hearing it not for the first time, who made people believe that every line through the centroid bisects the area??).

  darij

hello Darij

  El Chamo must be from a French speaking Swiss canton. They call not centroid
  but incenter as center of triangle. So do not get excited, there is something
  about a line through incenter connected with triangle area.
  Yet the solution clearly is not that simple. If it were, I would not
  have relaxed the conditions to give even slightly more than carte blanche.

  M.T.


algebra
polynomial
geometry
vector
superior algebra
superior algebra solved

Let $P_1,\ldots,P_m\in\mathbb Z[X_1,...,X_n]$. Prove that these polynomials have a common complex zero if and only if they have a common zero over finite fields of characteristic $p$ for infinetely many primes $p$.

If $P_1,....P_m$ does'nt have a common complex zero,then from hilberts nullstellensatz we get that the ideal they generate in $\mathbb{C}[X_1,....X_n]$ is the unit ideal,hence we get the ideal they generate in $\mathbb{Q}[X_1,.....,X_n]$ is the unit ideal and so there exists an integer $a$ which belongs to the ideal $<P_1,.....,P_m>$ in $\mathbb{Z}[X_1,....,X_n]$.From this we get that they ca'nt have a common zero (by reducing modulo primrs p) in finite fields of charactrestic $p$ s.t $p$ does not divide $a$.Therefore we get if they have a common zero in finite fields of characttrestic $p$ for infinitely many$p$,then they have a common complex zero.

I am still thinking about the other way  :roll:

I here used the fact that if $k \subset K$ are 2 fields and if$I$is an ideal of $k[X_1,....,X_n]$ ,then
$IK[X_1,....,X_n] \cap k[X_1,.....X_n] = I$,i have the proof for this with me.I believe its correct.

Hopefully u will atleast let me know if my approach is correct or not,sadly i found that i do'nt get replies to many of my posts  :( ,in that case i do'nt understand the purpose of this forum.

[quote="rohith"]
I here used the fact that if $k \subset K$ are 2 fields and if$I$is an ideal of $k[X_1,....,X_n]$ ,then
$IK[X_1,....,X_n] \cap k[X_1,.....X_n] = I$,i have the proof for this with me.I believe its correct.

Hopefully u will atleast let me know if my approach is correct or not,sadly i found that i do'nt get replies to many of my posts  :( ,in that case i do'nt understand the purpose of this forum.[/quote]

I think the approach is correct. At least I hope it is, because it is my approach as well :D
Can you post the proof for $IK[X_1,....,X_n] \cap k[X_1,.....X_n] = I$?

Please use don't and doesn't.

[quote="amfulger"]

Please use don't and doesn't.[/quote]

And write can't, not ca'nt.

You have to write the ' where you would say an o in a conversation : you will never say : I caont come, but you do say : I can not come-> I can't come  :)

hi
thanks for correcting my english,the first time when i read ur reply,i couldn't understand what u meant by saying "use don't and doesn't"  :lol: .Anyway regarding the proof (I saw this in my algebraic geometry class):
So let $k \subset K$  be two fields,Let $I$ be an ideal of $k[X_1,....,X_n]$.
Let $K=k\oplus{_{\alpha}} e_{\alpha}k$ as a $k$ vector space.
So $K[X_1,......,X_n]=k[X_1,....,X_n]\oplus{_{\alpha}}e_{\alpha}k[X_1,....,X_n]$
$\Longrightarrow IK[X_1,....,X_n] = \oplus{_{\alpha}}e_{\alpha}I$
$\Longrightarrow IK[X_1,....,X_n] \cap k[X_1,....,X_n] = I$

Assume $P_1,\ldots,P_m$ have a common complex zero. Then the ideal they generate in $\mathbb C[X_1,\ldots,X_n]$ is not $(1)$.

Then the ideal they generate in $\mathbb{\bar Q}[X_1,\ldots,X_n]$ is not $(1)$, where $\mathbb{\bar Q}$ is the algebraic closure of $\mathbb Q$ in $\mathbb C$. Now the polynomials $P_1,\ldots,P_m$ have a common zero in $\mathbb{\bar Q}$. Let this zero be $(a_0,\ldots,a_n)$.

Since $a_i\in\mathbb{\bar Q}$ for all $i$, they are all algebraic, hence $L=\mathbb Q(a_0,\ldots,a_n)$ is a finite algebraic extension of $\mathbb Q$, thus $L=\mathbb Q(a)$ for some algebraic number $a$.

Let $q_i$ be polinomials in $\mathbb Q[X]$ such that $q_i(a)=a_i$.
Let $R_i(X)=P_i(q_1(X),\ldots,q_n(X))$, $R_i\in\mathbb Q[X]$. $R_i(a)=0$ for all $i$, so the ideal generated by the $R_i$s is not $(1)$ in $\mathbb Q[X]$. Then this ideal is generated by a non-constant polynomial $R\in\mathbb Z[X]$ (note the we can assume that this polynomial has its coefficients in $\mathbb Z$). Let $S_i\in\mathbb Q[X]$ be defined by $R_i=S_i\cdot R$.

Let $p$ be a prime number which does not divide any of the denominators of any $R_i$, $q_j$ or $S_k$ and such that $R$ is not constant mod $p$. There are infinetely many such primes and for all of them it makes sense see all the polynomials above as polynomials over $\mathbb F_p$.
Let $a_p$ be a root of $R$ in a finite algebraic extension $K_p$ of $\mathbb F_p$ (such a finite extension exists). Then $R_i(a_p)=0$ for all $i$ and $P_i(q_1(a_p),\ldots,q_n(a_p))=0$ for all $i$, hence $(q_1(a_p),\ldots,q_n(a_p))$ is a common zero for all $P_i$ in $K_p$ which is a finite field of characterstic $p$.
================================================================================================
It can be proved that $R$ has roots in $\mathbb F_p$ for infinetely many primes $p$. With this, it can be proved that $P_1,\ldots,P_m$ have roots in $\mathbb F_p$ for infinetely many primes $p$ and the problem text can be improved to:
$P_1,\ldots,P_m$ have a common complex zero iff they have a common zero in $\mathbb F_p$ for infinitely many primes $p$.

hi,
thanks nice solution,btw presently i am not getting a proof for the problem,
If $f \in \mathbb{Z}[X]$,then the set {$p ; p/f(a);a \in \mathbb{Z}$}is infinite.
I believe u used this in the claim given in ur footnote.
I will try for a few more days and ask u.


Find all function $f: \mathbb{R}$ \ $\{\pm 1\}\to\mathbb{R}$ such that:
$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x-1}\right)=x$ $\forall x\in\mathbb{R}$ \ $\{\pm 1\}$

[quote="rjt"]Is this posted correctly? Because at the moment, if we let $x=2$ and $x=-2$:
$f(-\frac{1}{3})+f(5) = 2$
$f(5)+f(-\frac{1}{3}) =-2$[/quote]

I'm sory for my carfuless typing. The right problem is:
$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{1-x}\right)=x$ $\forall x\in\mathbb{R}$ \ $\{\pm 1\}$

[b]Solution.[/b]  Substitute $\frac{x+3}{1-x}$ and $\frac{x-3}{x+1}$ to obtain the following two equations.
\[\\ f(x)+f\left(\frac{x+3}{1-x}\right)=\frac{x-3}{x+1}\\ f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{x+3}{1-x}\]
Adding the above two and using the original equation, we have
\[2f(x)=\frac{x-3}{x+1}+\frac{x+3}{1-x}-x \]
And the conclusion follows.$\blacksquare$

[quote="hien"]The right problem is:
$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{1-x}\right)=x$ $\forall x\in\mathbb{R}$ \ $\{\pm 1\}$[/quote]

Becomes easier then. Let $g(x) = \frac{x-3}{x+1}$. Then $g(g(x)) = \frac{x+3}{1-x}$ and $g(g(g(x))) = x$. Therefore, \[\left\{\begin{array}{l}f(g(x))+f(g(g(x)) = x \\ f(g(g(x))+f(x) = g(x) \\ f(x)+f(g(x)) = g(g(x))\end{array}\right.\]It follows immediately that \[\begin{aligned}f(x) & = \frac{1}{2}\left(\left(f(x)+f(g(x))\right)+\left(f(g(g(x)))+f(x)\right)-\left(f(g(x))+f(g(g(x)))\right)\right) \\ & = \frac{1}{2}\left(g(g(x))+g(x)+x\right) = \frac{1}{2}\left(\frac{x+3}{1-x}+\frac{x-3}{x+1}+x\right)\end{aligned}\]

[quote="JBL"]... and $f$ can take any value at $x = \pm 1$.[/quote]

x=+-1 is not in the domain of the function [given]

You know, I specifically went back to the first post to check what the domain was, looked it over, and decided that it was all of $\bf R$.  Feh.


algebra
polynomial
function
algebra proposed

For a finite set $X \subset \mathbb{R}$, define $L_2(X) := \{x+x' \mid x,x' \in X, x \neq x'\}$ to be the multiset of all sums of two different elements of $X$. Prove the following classical result of Selfridge: if $A,B \subset \mathbb{R}$ are two different sets such that $L_2(A) \equiv L_2(B)$, then $|A|$ is a power of $2$. That is, if $X \subset \mathbb{R}$ and $|X|$ is not a power of $2$, then $X$ can be recovered from $L_2(X)$.

Enjoy!  ;)

--Vesselin

I'm sure it's been discussed before. The key idea is, I think, looking at $f_A(x)=\sum_{a\in A}x^a$ and $f_B$ constructed similarly for $B$ (after we assume that all the elements of $A,B$ are non-negative, which we can clearly do). Then some computations follow, which I don't have the time to write here right now :).

Of course, this idea works directly only when $A,B \subset \mathbb{Z}$ are sets of integers. And it's quite easy, actually: if $A,B \subset \mathbb{N}$ satisfy $L_2(A) \equiv L_2(B)$, then $f_A(x)^2 - f_A(x^2) = f_B(x)^2 - f_B(x^2)$, i.e.
\[
f_A(x)^2 - f_B(x)^2 = f_A(x^2) - f_B(x^2).
\]
Because $A \not\equiv B$, $f_A(x) - f_B(x)$ is a non-zero polynomial; taking $k$ to be the multiplicity of $x-1$ in the factorization of this polynomial, we obtain from the above that $|A| = f_A(1) = 2^{k+1}$.

But what about the case when $A,B \subset \mathbb{R}$ do not necessarily consist of integers?  ;)

Then I think that we can just aproximate all the numbers sufficiently well with integers and we are done. If this does not work (though I doubt) we can work with analityc functions instead of polynomials.

[quote="harazi"]Then I think that we can just aproximate all the numbers sufficiently well with integers and we are done. [/quote]

 :? Maybe I don't see obvious things, but could you please explain how can we reduce the $\mathbb{R}$ case to the $\mathbb{Z}$ case so easily?  :?

We know that if we have n real numbers $x_1,...,x_n$ and an $ r>0$ then we can find $ a_1,...,a_n$ integers and $p$ integer such that $|a_i-px_i|<r$ for all i. I imagine that simple estimations will prove that practically if $x_1,...,x_n$ and $ y_1,...y_n$ with the same sets of double sums, then the corresponding sets of doubles sums for the corresponding $a_i$ will be equal. Though, I am not very sure, since I cannot check now.


induction
modular arithmetic
arithmetic sequence
number theory proposed
number theory

prove that for every odd prime p and for every positive integer k there are infinitely many primes q such that $q^{p-1}=p^{k}*h+1$  (h is natural) :(  :(  :(

let $a>1$ be an integer coprime to $p$

by Eular-Fermat theorem,
\[a^{\phi(p^k)} \equiv 1 (mod p^k)\]

since $\phi(p^k)=p^k-p^{k-1}$, therefore if we set $b=a^{p^{k-1}}$, we have
\[b^{p-1} \equiv 1 (mod p^k)\]

consider the arithmetic progression $b+n\cdot p^k \; \; (n=1,2,3,...)$, suppose that $q=b+n \cdot p^k$ is a prime from the arithmetic progression, then
\[q^{p-1} = (b+n \cdot p^k)^{p-1} \equiv b^{p-1} \equiv 1(mod p^k)\]

therefore the result is true for every prime in the arithmetic progression $b+n\cdot p^k \; \; (n=1,2,3,...)$, by Dirichlet Theorem, there are infinitely many such primes because $b$ is coprime to $p^k$

This is a direct application of Dirichlet's Theorem. We know that there is an $x$ s.t. $p^k|x^{p-1}-1$ (we can construct such solutions by induction on $k$, for example). After that, we know, by Dirichlet's Theorem, that there are infinitely many primes $q\equiv x\pmod {p^k}$, because $(x,p^k)=1$.

Edit: I hadn't seen the post above when I started to write mine :).