LaTeX
articles
group theory
abstract algebra
Galois Theory

I'm using the \section{} command in LaTeX to create sections in a LaTeX document. I'd like the document to start with section 0. Does anyone know how to do this?

Thanks!

[hide="For reference, here is the beginning of my document"]

\documentclass{article}[12pt]
\usepackage{amssymb,amsfonts,amsthm}
\usepackage{amsmath,comment}
\usepackage{enumerate}
\usepackage[all]{xy}

\begin{document}

\title{Galois Theory}
\author{}
\date{August 19, 2009}

\maketitle

\section{Preliminaries}

Notation:

- $ |G|$ denotes the order of a finite group $ G$

- $ [E: F]$ denotes the degree of a field extension $ E/F$

- We write $ H \le G$ to mean that $ H$ is a subgroup of $ G$, and $ N \unlhd G$ to mean that $ N$ is a normal subgroup of $ G$

\section{Splitting Fields}

\begin{defn} Let $ F$ be a field and suppose $ f(x) \in F[x]$. A field extension $ E/F$ is said to be a \emph{splitting field} for $ f(x)$ over $ F$ if:
(a) We can write $ f(x) \equal{} (x \minus{} a_1)...(x \minus{} a_n)$ in $ E[x]$. That is to say, $ f(x)$ \emph{splits} over $ E$.
(b) The elements $ a_1,...,a_n$ generate the extension $ E/F$. Equivalently, $ f(x)$ splits in no subfield of $ E$.
\end{defn}
[/hide]

Something like \addtocounter{section}{-1} should work -- the keyword to look for either in this forum or on google is "counters."


inequalities
calculus
derivative
logarithms
real analysis
real analysis solved

For every n natural <>0 we denote x_n the real solution of the equation: 2^x+x=n. Compute, showing that there exists and it is finite: lim n-->oo n(x_n/logn -1) . the base of the log is 2.

thank you!!!

Here?s goes:

Notice that x_n tends to infinity as n tends to infinity. It?s enough to calculate limit of  (2^x+x)*(x/log(2^x+x)-1) as x tends to infinity. Now we multiply and divide this expression by log(2^x+x), so we get [(2^x+x)/log(2^x+x)]*(x-log(2^x+x))= -[(2^x+x)/log(2^x+x)]*log(1+x/2^x)=     -log(e)*[(2^x+x)/log(2^x+x)]*ln(1+x/2^x). Now we can use the inequalities (you can prove them by using derivatives) a>=ln(a+1)>=a/(a+1) for any positive a (I think they work for any real a, at least the first one does, but we don?t need that). We just apply these to a=x/2^x. I think we eventually get the limit we?re looking for=-log(e), (where log means, as you said, logarithm base 2).


LaTeX
number theory proposed
number theory

if $ p$ and $ p\plus{}2$ are two twin primes, then $ p(p\plus{}2)$ divides $ 4((p\minus{}1)!\plus{}1)\plus{}p$

prove it

[quote="lambruscokid"]Just use wilson's theorem twice, it's very straightforward

Daniel[/quote]

people thought fermat's last theorem was straightforward.

mmm  but it isn,t :P
Sorry if I was rude, it was not my intention

I was just stating that once "wilson's thoerem" comes to your mind, then the rest of the problem is just to do some easy calculations

Daniel

[quote="lambruscokid"]Just use wilson's theorem twice, it's very straightforward

Daniel[/quote]
Well, let's try:
(for instance, as I suck at Latex, == is congruency symbol)
(p-1)! == -1 (p)
p|4((p-1)!+1)+p, because (p-1)! is congruent to -1, and so with the 1 you get 0 mod p, and p is 0 mod p.

(p+1)! == -1 (p+2)
4((p-1)!)+1) + p == something (p+2)
Multiply all for p(p+1):
4(p+1)! + 4p(p+1) + p^2(p+1) == p(p+1)(something) (p+2)
-4 + 4(-2)(-1) + (-2)(-2)(-1) = 0 == p(p+1)(something) (p+2)
As neither p nor p+1 is 0 mod p+2, then something must be 0 mod 3
So, p+2|4((p-1)!+1)+p
As (p,p+2)=1 (pretty obv), p(p+2)|4((p-1)!+1)+p

Yeah, lambrusco is right :)

Oh thank you so much Pacman, I was a little dizzy from all the cough syrup I just drunk.  I think I could have seen the part of multiplying p(p+1), but I was really dizzy.  I am just doing some math problems for fun so i don't really care I get them or not.  it gives you something to do


Let ABC be a triangle, and D,E,F be the points of tangency of the incircle (center at I) of triangle ABC with the sides BC,CA, and AB, respectively. Let M,N be the midpoints of AE,AF respectively.MN intersects DF at T. Draw tangent TP,TQ from T to the incircle, PQ intersects MN,EF at L,K,respectively. Prove that IK is parallel to AL.[/i]

this might help:
[hide]its sufficient to show that $ \angle TAL=90$...[/hide]


Let $ a,b,c >0$ such that a+b+c=3.Prove that:
$ \frac{{a \plus{} b \minus{} c}}{{ab}} \plus{} \frac{{b \plus{} c \minus{} a}}{{bc}} \plus{} \frac{{a \plus{} c \minus{} b}}{{ac}} \le 3$ :)

[quote="TRAN THAI HUNG"]Let $ a,b,c > 0$ such that a+b+c=3.Prove that:
$ \frac {{a \plus{} b \minus{} c}}{{ab}} \plus{} \frac {{b \plus{} c \minus{} a}}{{bc}} \plus{} \frac {{a \plus{} c \minus{} b}}{{ac}} \le 3$ :)[/quote]
Denote $ p\equal{}a\plus{}b\plus{}c\equal{}3;q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc$
The inequality is equivalent to : $ 3r\minus{}4q\plus{}9 \geq 0$
By Schur inequality we have : $ 3r \geq 4q\minus{}9$
We have done :D

[quote="mehmetcantu"]If we expand denominators by a,b,c we get following inequality
\[ 2(ab \plus{} bc \plus{} ac) \minus{} a^2 \minus{} b^2 \minus{} c^2\leq3abc
\]
for getting both sides homogenized we expand right side by 3 and left side by a+b+c then after calculations we reach
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\geq(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac)
\]
after deleting $ 3abc$ from both sides it becomes known type of schur inequality[/quote]You are right :oops: i didn't know that after I post it because i have solved it without using Schur :)  :P .


In $ 2005$ Tycoon Tammy invested $ \$100$ for two years.  During the first year her investment suffered a $ 15$% loss, but during the second year the remaining investment showed a $ 20$% gain.  Over the two year period, what was the change in Tammy's investment?

$ \textbf{(A)}\ 5  \text{ percent loss}\qquad
\textbf{(B)}\ 2 \text{ percent loss}\qquad
\textbf{(C)}\ 1 \text{ percent gain}\qquad
\textbf{(D)}\ 2 \text{ percent gain}\qquad
\textbf{(E)}\ 5 \text{ percent gain}$

[hide]

After the first year, she has $ 100 \times .85 \equal{} 85$ dollars.  After the second year, she has $ 85 \times 1.2 \equal{} 102$.  $ 102$ is a $ \boxed{\textbf{(D)}\ 2 \text{ percent gain}}$[/hide]


abstract algebra
vector
advanced fields
advanced fields unsolved

Hello !

Please, I need help with this because I am stuck.

Let $ V$ and $ W$ two Banach spaces.   If $ A$ is a bounded linear operator, then the following conditions are equivalent:

(1) There is a constant $ c>0$ such that $ \|Av\| \geq c\|v\|$ for all $ v \in V$.
(2) $ Ker(A)\equal{}(0)$ and $ range(A)$ is closed.

Thank you !

[quote="MysticTerminator"]Okay, what are you getting stuck with? So you need to prove 1 implies 2 and 2 implies 1. Starting with 1 implies 2, you should be able to prove the kernel is trivial pretty easily, and showing closed range involves messing around with Cauchy sequences and the bound you're given. Showing 2 implies 1 wants you to basically think about what A does to unit vectors (well, this is one way to think about it). Contradiction is probably the easiest way to go here -- what does it mean if no such c exists? Then use your hypotheses to derive the contradiction.[/quote]

I am sorry about the delay, but I was quite busy.   I am also sorry because I forgot to say that the problem was not $ (1) \Rightarrow (2)$, but $ (2) \Rightarrow (1)$.    According to this, can you mystic be more explicit in your explanation?    Sorry, but for me the answer is not so obvious as you.

Thank you !

given (2), restrict $ A$ to a bijective operator $ V \to im(A)$ and use the open mapping theorem.


calculus
integration
calculus computations

For which real numbers s does the improper integral,

integral from (0 to oo) of ((x^s) / (1 + x)) dx ,

converge?

 [/code]

The integral $\int_{0}^\infty \frac{x^{s}}{1+x}\,dx$ may diverge either at $0$ or at $+\infty$. Near $0$ the integrand is comparable to $x^{s}$ and at $+\infty$ it is comparable to $x^{s-1}$. Now recall what you know about convergence of $\int_{0}x^{p}\,dx$ and $\int^{+\infty}x^{p}\,dx$.

There is another way of solving the problem:

$\int_{0}^{\infty}\frac{x^{s}}{1+x}ds=\beta(1+s,-s)$ valid for $Re(-s) >0$, $Re(1+s)>0$,  where $s$ is a complex number.
Thus your integral converges for $-1<Re(s)<0$ hence, the integral converges for $-1<s<0$ and diverges otherwise.


Let ABC be a triangle and elt w be a circle passing through A anc C. Sides AB and BC meet w again at D and E, respectively. Let y be the incircle of the circular triangle EBD and let S be its center. Suppose that y touches the arc DE at M. Prove that the angle bisector of angle AMC passes through the incenter of triangle ABC.

I'm a little new at proofs (I took a non-traditional Geometry course in my high school), so if I missed anything, let me know.

[hide]

Create new points F and G so that they are extensions of lines BD and BE, respectively, creating lines BDF and BDG, and so that line FG passes through point M.  The circle already described in the initial problem would also be the incircle of triangle BFG (since it already contacts FG at one point M).  Since the contact point of an incircle on a side lies on the angle bisector of the opposite angle, line BM would be the angle bisector of angle DBE (or FBG, take your pick).  Opposite angles on intersecting lines are congruent, so the angle bisector of one angle would be the angle bisector of its opposite.  So, a line passing through B and M would be the angle bisector of ABC, therefore it passes through the incenter.

[/hide]

The proof is good, except you may want to rewrite the first sentence as follows:

Construct line FG tangent to circle w with F on AB and G on BC.

Alright... I'll keep phraseology like that in mind.


[quote="Michael"]Hi!

Let $T_0: =\triangle ABC$, and $T_{n+1}=f(T_n)$ where $T_{n+1}=\triangle A_{n+1}B_{n+1}C_{n+1}$ is defined from $T_n=\triangle A_nB_nC_n$ by the following :

 $I_n$ is the centre of the incircle of $\triangle A_nB_nC_n$; $A_{n+1}$ is the centre of the incircle of $\triangle I_nB_nC_n$; $B_{n+1}$ is the centre of the incircle of $\triangle I_nC_nA_n$; and $C_{n+1}$ is the centre of the incircle of $\triangle I_nA_nB_n$.

1) Is the sequence $(I_n)_n$ convergent? and if yes, what is the limit?
2) what if we change "incircle" by "excircle"?[/quote]

Is it too hard or not interesting :juggle:


The point of this game is to start the biggest argument in this topic.
The argument can be on absolutely anything you can argue about.
I'll start... :police:


When I'm using TeXnic Center, and I have something like

\documentclass[11pt]{article}
\begin{document}
Vinayak

as my document and it's saved in a folder, and I try to build it, it gives an error of "emergency stop" and underneath shows the file location.  When i try to show output, it says there's an error and it couldn't locate the file.  I bet this is something stupid and small, so what is it?

Yeah I have that.  I'm using the solutions template.  Is it okay if we just turn in the solutions separately and not on a template?

I remember having a problem just like this when I first started typing my solutions in the template.  Unfortuantely I can't really remember what I did.  If I can remember I ll post it here.

[quote="Ravi B"]Did you end the document with \end{document} ?[/quote]

I make that mistake not infrequently.  Other troubleshooting questions of the same nature are: Did you close everything you opened (e.g. if you used \begin{enumerate}, did you remember to use \end{enumerate} as well)?  Does every \frac have two sets of braces ( { and } s)?  Are you out of math mode (this should be fairly obvious with TeXnicCenter, etc.)?  Usually failing to do one of these will cause an emergency stop.

You can avoid making mistakes like that by having a template. I don't really try to remember all the preamble stuff I like to use, so when I start a new $\text{\TeX}$ file, I simply save an old one with a new name and then delete the contents.


algebra
polynomial
number theory unsolved
number theory

Prove that we can find integer numbers $ a,b,c,d$ such that $ |a|,|b|,|c|,|d|>1000000$ and
 $ \frac{1}{a} +\frac{1}{b}+ \frac{1} {c}+\frac{1}{d}= \frac{1}{abcd}$

[quote="hoangclub"]Perove that we can find integer numbers $ a,b,c,d$ such that $ |a|,|b|,|c|,|d|$>1000000 and
                     $ frac{1}{a} \plus{} frac{1}{b} \plus{} frac{1}{c} \plus{} frac{1}{d}$=$ frac{1}{abcd}$[/quote]

We have : $ \frac{1}{u}\plus{}\frac{1}{\minus{}u\minus{}1}\plus{}$ $ \frac{1}{u^4\plus{}2u^3\minus{}u\minus{}1}\plus{}\frac{1}{\minus{}u^2\minus{}u\plus{}1}$ $ \equal{}\frac{1}{u(\minus{}u\minus{}1)(u^4\plus{}2u^3\minus{}u\minus{}1)(\minus{}u^2\minus{}u\plus{}1)}$ for any real $ u$ such that all denominators are non-zero.

Choosing $ u$ great enough, we have the solution to the problem.

[quote="Yuriy Solovyov"]---> pco. It a decision is a little unnatural. How did you guess that $ c$ should be the polynomial of power $ 4$ of $ u$? :)[/quote]

I tried something like $ \frac {1}{a} \minus{} \frac {1}{a \plus{} m} \minus{}$ $ \frac {1}{b} \plus{} \frac {1}{b \plus{} n} \equal{}$ $ \frac {1}{a(a \plus{} m)b(b \plus{} n)}$
Or also :
$ \frac {m}{a(a \plus{} m)} \minus{} \frac {n}{b(b \plus{} n)} \equal{}$ $ \frac {1}{a(a \plus{} m)b(b \plus{} n)}$

And so $ mb(b \plus{} n) \minus{} na(a \plus{} m) \equal{} 1$ which gives $ m \equal{} \frac {na^2 \plus{} 1}{b^2 \plus{} bn \minus{} an}$

Then, choosing $ b \equal{} a \plus{} 1$ and $ n \equal{} 1 \minus{} b^2$, we have $ b^2 \plus{} bn \minus{} an \equal{} 1$ and so $ m \equal{} na^2 \plus{} 1$

and so my solution :

$ b \equal{} a \plus{} 1$
$ n \equal{} 1 \minus{} b^2 \equal{} \minus{} a^2 \minus{} 2a$
$ m \equal{} na^2 \plus{} 1 \equal{} 1 \minus{} a^4 \minus{} 2a^3$

And so the 4 numbers :

$ a$
$ \minus{} (a \plus{} m) \equal{} a^4 \plus{} 2a^3 \minus{} a \minus{} 1$
$ \minus{}b \equal{} \minus{} (a \plus{} 1)$
$ b \plus{} n \equal{} \minus{} a^2 \minus{} a \plus{} 1$

pco did really good job. Here is a similar way to look at this problem.
we notice that $ \frac1a\plus{}\frac1{1\minus{}a}\equal{}\frac1{a(1\minus{}a)}$. Now we can construct $ a_n$ by $ a_1\equal{}u$ and $ a_n\equal{}1\minus{}\prod_{i\equal{}1}^{n\minus{}1}a_i$.
It is easy to verify $ \sum_{i\equal{}1}^n\frac1{a_i}\equal{}\frac1{\prod_{i\equal{}1}^na_i}$ for any  $ n$.


ratio
real analysis
real analysis unsolved

Find all x for which the series converges
$\sum_{n=0}^{\infty}\frac{(n!)^{k}}{(kn)!}x^{n}$

I finally learned to do the equations, so I decided to practice.
I think I have the correct solution to this one as well.

Ratio Test:
\[\frac{|a_{n+1}|}{|a_{n}|}=\frac{(n+1)^{k}}{(kn+1)(kn+2)\dots(kn+k)}|x|\]
The limit as $n\to\infty$ is $|x|/k^{k}$, right?


calculus
derivative
function
calculus computations

f ' (x) = (3-x^2) / x^4  so the critical points are (root 3, f(root3)) (-root 3, f(-root3)) and IS X=0 considered a critical POINT because since it is undefined in f(x) and f'(x) I am not sure if you can call it a critical point. Same with x=0 at as a point of inflection.

Thanks in advance!

Since $ x\equal{}0$ does not satisfy the condition for a critical point (it is not a zero of the derivative), it is not a critical point or point of inflection.  View the graph and see.

A critical pt is defined as when the f'(x) = 0 or is undefined so therefore at $ x \equal{} 0$ there is a critical point

Whatever the language (I wouldn't call it a critical point myself), you do have to look there. In this case, the original function $ f(x)$ goes to $ \infty$ as $ x\to 0^\minus{}$ and to $ \minus{}\infty$ as $ x\to 0^\plus{}$.

also in order to have a critical point the function must be continuous at that pt so how does one know if it is continuous at x = 0

For a function to be continuous at a point it must meet three criterion: a) It must be defined at that point b) There must be a limit at that point c) The limit must equal the value of the function at that point.  As jmerry pointed out, the limit doesn't exist, and the function isn't defined, so it isn't continuous.


Let $f(x)=1+\frac x {1!} + \frac{x^2}{2!}+...+\frac{x^n}{n!}$, where $n\geq 3$.
Prove that the equation $f(x)=0$ has no repeated roots.

[quote="xxreddevilzxx"]umm.....
[hide]
$f(x) = e^x$
so $x$ for $f(x)=0$ is undefined?
[/hide][/quote]
Not quite.  $f\left(x\right)$ is only equal to $e^{x}$ as $n\rightarrow\infty$.

[hide="simple with calculus"]
Suppose $r$ is a repeated root. Then $f^\prime (r)=f(r)=0\Rightarrow f(r)-f^\prime (r)=0$. But $f^\prime (x)=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots +\frac{x^{n-1}}{(n-1)!}=f(x)-\frac{x^n}{n!}$. Then $\frac{r^n}{n!}=0\Rightarrow r=0$. But 0 is clearly not a root, so there can be no repeated root.
[/hide]

[quote="t0rajir0u"]No repeated roots?  This is a special case of a much more difficult problem that I saw, which was to prove that $f(x)$ had at most one real root.[/quote]

Sounds more interesting. How to prove it?


Here are three axioms.

[b]PAR 1[/b] Two lines which are not parallel meet in exactly one point.

[b]PAR 2[/b] Given a line $L$ and a point $P$, there is one and only one line passing through $P$, parallel to $L$.

[b]PAR 3[/b] Let $L1, L2,$ and $L3$ be three lines.  If $L1$ is parallel to $L2$ and $L2$ is parallel to $L3$, then $L1$ is parallel to $L3$.

  Using the axioms PAR1 and PAR2 prove by contradiction, PAR3.  In other words, given $L1$ parallel to $L2$, and $L2$ parallel to $L3$.  Suppose that $L1$ and $L3$ are [i]not [/i] parallel.  Use PAR1 and PAR2 to derive a contradiction.

  I got this problem from one of my old geometry texts and it looked pretty nice to me.

[quote="236factorial"]So you're trying to say that the parallel postulate has just turned into the paralled theorem?  :?[/quote]

  Once you have done this proof, PAR 3 (also known as the parallel postulate) can be considered a theorem, since we have proved it from earlier facts.

[quote="AntonioMainenti"][quote="236factorial"]So you're trying to say that the parallel postulate has just turned into the paralled theorem?  :?[/quote]I think PAR2 is Euclid's fifth axiom (parallel postulate), right?

Anyway, isn't "proving axioms" kind of an oxymoron?[/quote]

  Im not any good at catching names, so for all I know PAR 2 could be the parallel postulate, although PAR 3 was what was intended.  I suppose the title does sort of contradict itslef, I chose it because it seemed to stand out to me...

I thought I proved some postulate [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39879]here[/url], but I guess there has to be some base postulate.

Note, there was an attachment in that thread and it was removed.


what can i say...

u upgraded them quite a  BIG bit, dont u think?
try to be more gentile next time, huh?...  :?
4th round sounds more like it...  :)
keep up the good work...

thanks K ... i'm sorry for OVER-estimating the level of difficulty of the problems - problem I and II seemed solvabile to me :)

4th round is lighter ... at least I hope so :)

full solutions of rounds 2 and 3 will be available soon - sorry for the delays ... but I am working hard to finish publishing my book :D

hey
the 3rd round problem were goddamn tough.neway thoroughly entertaining.i solved a grand total of zero!! :( .neway,my xams are on .so cant devote time.i am awaiting the solutions of round 2 and round 3.
and cud u suggest the the sum software using which one can make the mathematical symbols and stuff.i have got a problem making all these symbols on microsoft word
keep up the good work
regards

there is a language specialy designed by the American Mathematical Society back in the 1980's called TeX (it's spelled tek i think :) )

it has a really easy to learn close-to-user language

the simbols available here on the forum are based on that language

unfortunately it is very hard to dig up the files you actually need
there are many (tons actually) of different versions of TeX and it's pretty hard working with the software ( i had a few problems on my own )

but I can guide all of you into easy use of the LaTeX (actually LaTeX2e) language ( similar with TeX, but much more confortable )

first of all you have to install [url=http://www.miktex.org]MikTex[/url] which is a LaTex / Tex environment which will compile the stuff you write and create an .dvi file which can be viewed/printed with a viewer included in the MikTex package

after that install [url=http://www.winedt.com]WinEdt[/url]
don't try to read to many docs or help files, because if you don't have time to read them throughly they will confuse you even more

okay so now after installing winedt, it should detect the presence of the MikTex environment - if it doesn't then you should try to reinstall miktex again, and if this doesn't solve the problem then start reading those docs :)

now ... i have put [url=http://www.mathlinks.go.ro/style/example.tex]here[/url] a sample of a LaTeX file for you to see how this works RIGHT-CLICK and select SAVE_AS to save the tex file ... after writing the document in winedt, you have in the right of the toolbar a button that converts this tex file into a pdf file, and instantly opens acrobat reader to visualise it

[img]http://www.mathlinks.go.ro/images/winedt%20use.JPG[/img]

well in short that's about all - winedt will also help you by a visual interface, will tell you if you have misspelled a command, or even a word, if your document is written in english

for help on the latex syntax use crtl+F1 in winedt
and for winedt help (although is very unlikely that you will need it - there is a help menu )

hoped it helped you guys

Excuse me, but I know that original TEX language was designed
(and its translator was coded) by Don Knuth himself.
American Mathematical Society (Michael Spivak) designed a dialect (some macropackages) for TEX called [i]AMS[/i]-TEX for more comfortable design of BIG formulae.
LATEX is the most used set of macropackages for TEX written by Laslie Lamport and from version 2 \varepsilon  it's containing [i]AMS[/i]-LATEX
with full capabilities of [i]AMS[/i]-TEX as its subset.
LATEX is not another language!!!
It is only a set of packages for TEX.

yes you are entirely right. I didn't knew that back then :D


Let $ f(x)\equal{}ax^{2}\plus{}bx\plus{}c$ satisfy $ |f(0)|\leq 1, |f(1)|\leq 1$ and $ |f(\minus{}1)|\leq 1$. Prove that $ |f'(x)|\leq 4$ with $ x\in [\minus{}1;1]$

Posted before. Maximum, when $ f(x)\equal{}\pm (2x^{2}\minus{}1)$.

Where was it posted before? Could you give me the links to that topic or prove again ^.^

It is obviosly. From $ 2\ge |f(1)\minus{}f(0)|\equal{}|a\plus{}b|,\ 2\ge |f(\minus{}1)\minus{}f(0)|\equal{}|a\minus{}b|$ we get $ |a|\plus{}|b|\le 2$.
Therefore $ |f'(x)|\equal{}|2ax\plus{}b|\le |2ax|\plus{}|b|\le 2(|a|\plus{}b|)\minus{}|b|\le 4\minus{}|b|$.

note that $ |f'(x)|\equal{}|2ax\plus{}b|$ and by convexity, $ \max\{|f'(x)|: x\in [\minus{}1,1]\}\equal{}\max\{|2a\plus{}b|,|b\minus{}2a|\}$.
Conclude with identities like $ |2a\plus{}b|\leq\frac{3}{2}|a\plus{}b\plus{}c|\plus{}\frac{1}{2}|a\minus{}b\plus{}c|\plus{}2|c|$


topology
Functional Analysis
advanced fields
advanced fields unsolved

$E$ a Banach space and $B_E$ its unit ball. $J: E \to  E''$ defined by $J(x)(f)=f(x)$ the classical
embedding of $E$ into $E''$.
It is classical that $J(B_E)$ is dense in $B_{E''}$ for the weak* topology.
But $J(B_E)$ is of course not dense in $B_{E''}$ for the strong topology if $E$ is not reflexive.
So what can one say for the weak topology ?

No, not unless $E$ (and $E'$, and $E''$, and so on :)) is reflexive. I remember using the following fact in a recently posted proof: a closed convex subset of a Banach space is weakly closed. It follows from the separation theorem: given a closed convex set and a point not belonging to it (in any real normed space), there is a hyperplane which separates them.

Now, $J(B_E)$ is a closed convex proper (unless we have reflexivity) subset of $B_{E''}$, so also a weakly closed proper subset of $B_{E''}$.


group theory
abstract algebra
superior algebra
superior algebra theorems

Let $G=<a>$ denote the cyclic group generated by $a$. If $G$ has the Subgroup different with $\{e\}$ with  $e$ is identity element, is the subgroup of $G$ is $G$ it self?, if not give an example.

Can someone give an example of  finite group that isn't cyclic group?

Thanks in Advance

${e,a,b,c}$ with $e$ neutral and $ab=ba=c , ac=ca=b , bc=cb=a$, the Kleinian-four-group.
Or $S_n$, the group of permutations of $n$ elements for $n \geq 3$.
Or $D_n$, the groups of symmetries of a regular $n$-gon for $n\geq 2$ (for $n=2$ it's identical to the Kleinian-four-group).
Or just any group product of noncyclic groups...

let $n=ord(a)$ (the smallest positive integer for which $a^n=e$ ).
we have 2 cases:
1. n is prime in which case $\{e\}$ and G are the only subgroups of G
2. n is not prime in which case we have the subgroups $<a^d>$  where d divides n.


function
complex analysis
complex analysis unsolved

[quote=eugene]Prove that any entire transcendental function $f$ has periodic points of minimal period $n$ for each $n \geq 2$. (Even more, it has infintely many such point for each $n \geq 2$).[/quote]

This was proved by Fatou at first and then by PC Rosenbloom as a general case mentioned here.The proof appears in
(1)The Fix-Points of Entire Functions;Medd Lunds 1952;Rosenbloom
(2)L'iteration des functions entires;CR Acad.Sci. Rosenbloom.

I am unable to find both papers(they do not appear in Wiki!) but they are referenced in a lot of books etc.


How did you do in the math league contest?

I think our school's top score was a 40, and then a 39, then I think it went downhill (this is 8th grade, don't have any clue about 7th or 6th, but then our school's not really strong in math for those grades...)

I'm not sure about yours.  Its title differs by each state.  For example, I live in Illinois, so I took the Illinois Mathematics League (ILML)

I think there mostly different for each state. I got 25/28 in the DE 1

Math League is math league.  There are different names for each region of the country, but they are the same test.

Check out :  [url=http://www.mathleague.com/]www.mathleague.com/[/url]

If you want to know more.

In DE none of the schools in my division are registered (but we do it). Plus I only had 28 questions

[quote="dwx314"]First of all:
That Math League? Is it the New England Math League, or what?
There are too many little math league floeating around to be sure.[/quote]

Your school does NEML? What's your school's score?

I am dwx's coach, our school score for 8th grade was 186 (40,39,36,36,35)

Last year those kid's were 5th in NE as 7th graders.

Do you know what your score was Yif man?

Thirty-nine.

I thought that 4 :^2: + 6 :^2: = 100 <_< .

LOL, I hate it when I do that.

What's Math League? There are all these competitions I've never heard of before... I hadn't even heard of the AMC until I was talking with the others on the California state team last year, and by then it was too late for that year. :(

I'm in New England, I got a 40, and our eighth grade's score was 194


[quote="calc rulz"]Is it okay to not actually prove a fact if I find it on Sloane? and If so how do I cite Sloane?[/quote]

Unfortunately, this is not a question we can answer while a round is in progress (and without more information, but that would still have to wait until after the round).


25 knight are sitting around a circular table. Three of them are chosen randomy to take out a mighty dragon. Whats the probability that at least two of the three knights who are chosen are sitting next to each other?

[hide="Solution"]
First case: The second guy is next to the first guy.  Probability is 2/24 or 1/12.

Second case: The 1st and 2nd guys have at least 2 knights between them.  P(The 3rd guy sits next to one of them) = 4/23.  Multiply by 5/6 (this is P(1st and 2nd chosen are at least 2 away)).  You get 10/29.

Third case: The 1st and 2nd guys have one knight between them.  The probability of this is 1/12.  Then the P(3rd sits next to one or both of them) = 3/23.  This has a probability of 1/92.

Add them up: (1/12)+(10/29)+(1/92) = 1757/4002 or about 44%.
[/hide][/hide]

There are 4 cases

The chances that the second one is sitting next to the first and the third is sitting next to EITHER(not both) the second and first(so three in a row but the third one is not in the middle) is (2/24)(2/23)

The chances that the second one is sitting next to the first but the third one is not sitting next to the first or second one is (2/24)(21/23)

Finally, there is the chance that the third one is sitting next to the first, second, or both the first and second-there are two cases

First subcase: the second one is two seats or more away from the first one and the third is either next to the either the first or second - so (20/24)(4/23)

Second subcase: the second one is sitting exactly one seat apart from the first (so FxS where x is the seat), then the third person has 3 spots to choose from where it will make atleast 2 knights sitting next to eachother- so (2/24)(3/23)

Add all of them up, you will get 11/46

[quote="scorpius119"][hide="this resembles one of the HMMT problems, but much easier"]
We can find the complement probability, that no two of the chosen knights are next to each other.

After choosing the first knight, the two adjacent spots are both bad (if we want no two to be next to each other). That leaves 22 possible seats to choose the other two knights, and all combinations work except for the 21 where the other two knights are next to each other. This makes the number of good possibilities for the other two knights $\binom{22}{2}-21=\binom{21}{2}$, and the total number of possibilities is $\binom{24}{2}$. So letting $P$ be the probability we want, the complement is $1-P$, then
\[1-P=\frac{\binom{21}{2}}{\binom{24}{2}}=\frac{21\cdot 20}{24\cdot 23}=\frac{35}{46}\]

\[P=\boxed{\frac{11}{46}}\]
[/hide][/quote]
It's an AIME problem


Three series, $ \{ a_n \} _{n\equal{}0}^{\infty}$, $ \{ b_n \} _{n\equal{}0}^{\infty}$, and $ \{ c_n \} _{n\equal{}0}^{\infty}$ are co-defined as follows:

[u]Initial Values[/u]
$ a_0 \equal{} 0 \qquad b_0 \equal{} 1 \qquad c_0 \equal{} 2$
$ a_1 \equal{} 2 \qquad b_1 \equal{} 1 \qquad c_1 \equal{} 0$

[u]Recursion for n>1[/u]
$ a_{n\plus{}1} \equal{} b_n \plus{} c_{n\minus{}1}$
$ b_{n\plus{}1} \equal{} c_n \plus{} a_{n\minus{}1}$
$ c_{n\plus{}1} \equal{} a_n \plus{} b_{n\minus{}1}$

Find $ b_{2008}$.

[i]You may use any reasonably compact closed-form notation and/or the terms of any particularly well-known series. Please comment on the difficulty of the problem as well.[/i]

[quote="The Zuton Force"]Three series, $ \{ a_n \} _{n \equal{} 0}^{\infty}$, $ \{ b_n \} _{n \equal{} 0}^{\infty}$, and $ \{ c_n \} _{n \equal{} 0}^{\infty}$ are co-defined as follows:

[u]Initial Values[/u]
$ a_0 \equal{} 0 \qquad b_0 \equal{} 1 \qquad c_0 \equal{} 2$
$ a_1 \equal{} 2 \qquad b_1 \equal{} 1 \qquad c_1 \equal{} 0$

[u]Recursion for n>1[/u]
$ a_{n \plus{} 1} \equal{} b_n \plus{} c_{n \minus{} 1}$
$ b_{n \plus{} 1} \equal{} c_n \plus{} a_{n \minus{} 1}$
$ c_{n \plus{} 1} \equal{} a_n \plus{} b_{n \minus{} 1}$

Find $ b_{2008}$.

[i]You may use any reasonably compact closed-form notation and/or the terms of any particularly well-known series. Please comment on the difficulty of the problem as well.[/i][/quote]

the following is a stupid idea.
[hide]
just translate the nth row n terms to the right, and rename
now each column is a separate fibonacci sequence.
b2008 will be translated 1 to the right, so it will be in the new c column
so just find the 2008th number in the fibonacci style sequence with starting terms 2 and 1.
[/hide]

[hide="Solution"]The sequence $ a_0, c_1, b_2, a_3, c_4, b_5, a_6, \ldots$ is the $ 0$ sequence.
The sequence $ b_0, a_1, c_2, b_3, a_4, c_5, b_6, \ldots$ is the Fibonacci sequence, starting with the terms $ 1,2$.
The sequence $ c_0, b_1, a_2, c_3, b_4, a_5, c_6, \ldots$ is the Lucas sequence, starting with the terms $ 2,1$.
This is relatively easy to verify using the given recursions, since $ n \minus{} 1, n, n \plus{} 1$ (the indices in each recursion) admit each of the residues $ \mod 3$.
$ b_{2008}$ is in the Lucas sequence, so the answer is $ L_{2008}$.[/hide]

As for difficulty, it's probably challenging to figure out and solve directly, but the method of trying small cases makes it very easy. The pattern shows up very quickly.

f~, forgot about the Lucas series, thanks. Looks like its back to the drawing board (this is why I love AoPS)

 :stink:

The idea was that $ a \plus{} b \plus{} c$ is thrice the Fibonacci numbers, so since for each $ n$, since one is $ 0$ and another is a Fibonacci number, and together they add to thrice a Fibonacci number.


Let $ f: [0,1] \to \mathbb R$ be a function such that $ f'$ exists on $ [0,1]$ and $ f(0) \equal{} f'(0) \equal{} f'(1) \equal{} 0$. Show that there exists $ c \in (0,1)$ for which $ f(c) \equal{} c\cdot f'(c)$.

[quote="Victory.US"]let $ f: [0,1] \to R$ such that there exist  $ f'$ on $ [0,1]$ and $ f(0) \equal{} f'(0) \equal{} f'(1) \equal{} 0$

show that :$ \exists c \in [0;1]$ and $ f(c) \equal{} c.f'(c)$[/quote]

$ c\equal{}0$

[quote="pco"][quote="Victory.US"]let $ f: [0,1] \to R$ such that there exist  $ f'$ on $ [0,1]$ and $ f(0) \equal{} f'(0) \equal{} f'(1) \equal{} 0$

show that :$ \exists c \in [0;1]$ and $ f(c) \equal{} c.f'(c)$[/quote]

$ c \equal{} 0$[/quote]

i checked my místake

Define $ g : (0, 1] \to \mathbb{R}$ by $ g(x) \equal{} \frac {f(x)} {x}$. Since $ \lim_{x \to 0} \frac {f(x)} {x} \equal{} \lim_{x \to 0} \frac {f(x) \minus{} f(0)} {x \minus{} 0} \equal{} f'(0) \equal{} 0$, it means we can prolong $ g$ to be continuous on $ [0, 1]$, by taking $ g(0) \equal{} 0$. Now, on $ (0, 1]$ we have $ g'(x) \equal{} \frac {xf'(x) \minus{} f(x)} {x^2}$, so $ g'(1) \equal{} \minus{}f(1) \equal{} \minus{}g(1)$.
If $ g(1) \equal{} 0$, as $ g(0) \equal{} 0$, by Rolle's theorem, there must exist $ c \in (0, 1)$ such that $ \frac {cf'(c) \minus{} f(c)} {c^2} \equal{} g'(c) \equal{} 0$, whence $ f(c) \equal{} cf'(c)$.
Assume therefore $ g(1) \neq 0$.
If $ g'(x) > 0$ on $ (0, 1]$; then $ g$ is increasing and $ g'(1) \equal{} \minus{}g(1) < \minus{}g(0) \equal{} 0$, absurd.
Similarly, if $ g'(x) < 0$ on $ (0, 1]$; then $ g$ is decreasing and $ g'(1) \equal{} \minus{}g(1) > \minus{}g(0) \equal{} 0$, absurd.
Therefore there must exist $ c \in (0, 1]$ such that $ \frac {cf'(c) \minus{} f(c)} {c^2} \equal{} g'(c) \equal{} 0$, whence $ f(c) \equal{} cf'(c)$, but we work under assumption  $ g'(1) \equal{} \minus{}g(1) \neq 0$, so in fact $ c \in (0, 1)$.

A slightly more general result holds: Let $ f\ : \ [a,b]\ \rightarrow \mathbb{R}$ be a differentiable function such that $ f'(a)\equal{}f'(b)$. Then, there is $ c \in (a,b)$ such that $ \frac{f(c)\minus{}f(a)}{c\minus{}a}\equal{}f'(c)$. This is known as Flett's Mean-Value Theorem (see for example here: http://www.math.sc.edu/~girardi/m555/current/hw/MVT-Flett.pdf).

However, mavropnevma's proof can be adapted.

[quote="pohoatza"]A slightly more general result holds: Let $ f\ : \ [a,b]\ \rightarrow \mathbb{R}$ be a differentiable function such that $ f'(a) \equal{} f'(b)$. Then, there is $ c \in (a,b)$ such that $ \frac {f(c) \minus{} f(a)}{c \minus{} a} \equal{} f'(c)$. This is known as Flett's Mean-Value Theorem (see for example here: http://www.math.sc.edu/~girardi/m555/current/hw/MVT-Flett.pdf).

However, mavropnevma's proof can be adapted.[/quote]

Could you tell me how to prove Flett's Mean-Value Theorem  :?: thanks


inequalities
trigonometry
algebra proposed
algebra

Let $\alpha$, $\beta$ and $\gamma$ be three acute angles such that $\sin \alpha+\sin \beta+\sin \gamma = 1$. Show that
\[\tan^{2}\alpha+\tan^{2}\beta+\tan^{2}\gamma \geq \frac{3}{8}. \]

We have : $tan^2\alpha +1 =\frac{1}{1-sin^2\alpha}$
 The problem is equivalent to :
 $\frac{1}{1-sin^2\alpha}+\frac{1}{1-sin^2\beta}+\frac{1}{1-sin^2\gamma} \geq \frac{27}{8}$
 By Cauchy-Scharwz's ineq , we have :
$\frac{1}{1-sin^2\alpha}+\frac{1}{1-sin^2\beta}+\frac{1}{1-sin^2\gamma} \geq \frac{9}{3-(sin^2\alpha+sin^2\beta+sin^2\gamma)}$
 It's enough to show that : $sin^2\alpha+sin^2\beta+sin^2\gamma \geq \frac{1}{3}$ , which is true by Cauchy again .

by cauchy we have (cos²a+ cos²b+ cos²c) . (tg²a+ tg²b + tg²c) >= (sina +sinb +sinc)² = 1.
see that, again by cauchy, 3.(sin²a + sin²b + sin²c) >= (sina + sinb+ sinc)² = 1
this implies that 3- (cos²a + cos²b + cos²c)>= 1/3
so, cos²a+cos²b+cos²c <=3/8.

so, 3/8 . (tg²a+tg²b+tg²a) >= 1, and the result follow.  :)

My solution is similar to nntu.  put $sin a=x$  etc. So the problem is equivalent to.

$x,y,z>0$  $x+y+z=1$ and we want to prove that $\sum\frac{x^2}{1-x^2}\geq\frac{3}{8}$.

We use cauchy in the engel form to have

$\sum\frac{x^2}{1-x^2}\geq\frac{(x+y+z)^2}{3-(x^2+y^2+z^2)}$

so it emains to prove that $\frac{(x+y+z)^2}{3-(x^2+y^2+z^2)}\geq\frac{3}{8}$

<=>...<=> $3(x^2+y^2+z^2)\geq 1$ which is true.


kunny, It is here http://en.wikipedia.org/wiki/H%C3%B6lder's_inequality .

Thank you. :) But it seems to be difficult for me. :ninja:

[quote="Soarer"]Just apply Holder's inequality.[/quote]

Could you expalin in more detail, how could I use this?

kunny

I see, thank you,Ramanujan.  :)  My solution was just using Cauchy-Schwarz twice.


game
combinatorics
algebra
system of equations
IMO
IMO 1974

Three players $A,B$ and $C$ play a game with three cards and on each of these $3$ cards it is written a positive integer, all $3$ numbers are different. A game consists of shuffling the cards, giving each player a card and each player is attributed a number of points equal to the number written on the card and then they give the cards back. After a number $(\geq 2)$ of games we find out that A has $20$ points, $B$ has $10$ points and $C$ has $9$ points. We also know that in the last game B had the card with the biggest number. Who had in the first game the card with the second value (this means the middle card concerning its value).

sum = 39 points, so we have a sum of either 13 or 3. for 3 it is impossible to find 3 positive integers that sum to 3, so the sum is 13 and they played 3 games.

The minimum number is at least 1, the maximum number is at most 10. (let's call highest number win, lowest loss, middle draw)
Assume A lost once, then C can have lost at most twice, and we cannot find such numbers. So A has 2 wins and a draw. B and C both have a loss and we still have to split {win,loss,draw,draw} over B,C. Since B=C+1 and B won once we have:
A = wwd
B = llw
C = ldd

B won the last game, so lost the others and C drawed all the others, so C also drawed the first one.

*spends 2 hours on problem with friend

*realizes you can add numbers

So if A wrote down the value a, B wrote down the value b, and C wrote down the value c, and they played n rounds, then n(a+b+c) = 39 = 3*13

Since n is at least 2, n is either 39,13, or 3. If n is 39, then a+b+c = 1 which is impossible since a,b,c > 1. If n is 13 then a+b+c = 3. Since a,b,c > 0 this assumes a=b=c=1 which contradicts them being distinct. This means that the only possibility is n=3 and a+b+c = 13.

WLOG let a>b>c. We know that a is at least 7 (because if a<7 then 3*a < 20). So we split this into cases

Case 1: [b]a=7 [/b]

Then we know that B had to get 7 the last round. This means that the first 2 rounds B must have gotten 3 points in total. Since b and c are integers and 3 is not even, this means that b+c = 3. Since b>c and b,c>1 we see that b=2, c=1

But that means the greatest value A could have gotten is 7+7+2 = 16 < 20. This is bad

[b]Case 2: a=8[/b]

That means B got 8 the last round. That means in the first 2 rounds B got 2 points, which is only possible if the least value is 1. Thus in the first,second, and third rounds B got 1,1,and 8 points respectively.

There are 2 possibilities from here, either A got 8 in 2 of the rounds or 8 in one of the rounds. If A got 8 in one of the rounds then b=6. But this is impossible as this assumes C got both 6 and 8 at least once, where 6+8>9

This means that the only possibility is A got 8 in 2 of the rounds, meaning b=4. That means that A got 8,8,4 in the first,second, and third rounds respectively. Also this means that C got 4,4,1 in the first, second, and third rounds respectively. Thus, in this case C got the middle number first

[b]Case 3: a=9[/b]

Then B got 9 once and got 1 in the first 2 rounds. This is obviously impossible. And obviously if a becomes larger this just gets more impossible. Thus the only possibility is the one listed above.

Thus [b]C[/b] drew middle the first round

[hide=My solution]The sum of points is $39$. Hence there are $3$ shufflings and the sum of numbers on cards is $13$. Since $A$ has the highest point, hence the highets number, call $W$, must be $7$. On the ;ast shuffling, $B$ got $W$, hence $W$ is $8$, and the lowest, call $L$, have the number $2$ (if it is grater then $W$ max is $13-3-4=6$, a contradict. So we get the following possibilities ($W$, $D$ for middle number and $L$ respectively): \[ (8,4,1); (8,3,2); (7,4,2); (7,5,1) \]Since we get $20$ after $3$ shufflings, the triple must be $(8,4,1)$. Now, it is easy to see that person is C.[/hide]

[hide=Case Bash]We claim that $C$ is the desired person. Notice that during each game, the sum $S$ of numbers is invariant. Let $x,y,z$ be the numbers written on three cards. Moreover, WLOG assume $x<y<z~(\spadesuit)$. Let there by $\ell$ games played. Let $a,b,c$ be the numbers on the cards picked by $B$ in three games respectively. Then we have $$S\ell=20+10+9=39=3\cdot 13$$which gives $\ell=3$ and $S=13,$ since otherwise we would have $1,1,1$ on three different cards, a contradiction. Clearly, $x+y+z=S=13$ and $z\ge 5$ using $(\spadesuit)$. We will proceed by dividing the value of $z$ in cases. If $z=c=5$ then only choice to have is $x=y=4$ but by $(\spadesuit),$ a contradiction. If $z=c=6$ then $x+y=7$ and $a+b=4$. Hence $(a,b)\equiv(1,3),(2,2)$. Checking $(a,b)\equiv(1,3)$ since $1,3,6$ are distinct hence we should have $1+3=7,$ a contradiction. Therefore, $(a,b)\equiv(2,2)$ which gives $(x,y,z)\equiv(2,5,6)$. But then attributing numbers to $A$ we have maximum sum of numbers for $A$ is $6+6+5<20,$ contradiction. If $z=c=7$ then $x+y=6$ and $a+b=3$ which gives $(a,b)\equiv(1,2)$. Getting aforementioned reasoning into the play, we get $1+2+7=13,$ a contradiction. If $z=c=8$ then $x+y=5$ and $a+b=2$ which gives $(a,b)\equiv(1,1)$. Therefore, $(x,y,z)\equiv(1,4,8)$. Consequently, we get following as our configuration of game played. $$\left[\begin{array}{ccc} 8&1&4 \\ 8&1&4 \\ 4&8&1 \\ \end{array}\right]$$ It's easy to see that this indeed works and that $C$ gets the second number in first game. $\blacksquare$ [/hide] :wacko:


algebra
polynomial
number theory open
number theory

Hi all
I would like to know the formula of the sum of following sequence

1 + 2^4 + 3^4 + .. + x^4
1 + 2^5 + 3^5 + .. + x^5
1 + 2^8 + 3^8 + .. + x^8

Thanks so much!!

[url=http://mathworld.wolfram.com/PowerSum.html]Power Sum[/url] - MathWorld

It's always possible to write $ \sum_{j\equal{}1}^n j^k$ as a polynomial in $ n$ of degree $ k$. One way to find it is to start with a generic polynomial $ p(n)\equal{}a_{k\plus{}1}n^{k\plus{}1}\plus{}a_kn^k\plus{}\cdots\plus{}a_1n$ (we know the constant term is zero since the empty sum $ \sum_{j\equal{}1}^0 (\text{anything})$ is always zero) and then take differences. Separating powers of $ n$, $ p(n)\minus{}p(n\minus{}1)\equal{}n^k$ is a system of linear equations in the $ a_n$. Solve it; it's easy because the system is already triangular.

Try that for these cases. Can you spot any patterns?


analytic geometry
geometry proposed
geometry

[quote="Yavor Tsvetanov 52"]It's given the equilateral triangle XYZ.
$ W \in XY$ and $ WM\bot XZ (M\in XZ),WL\bot YZ(L\in YZ)$
O is the centar of the triangle XYZ.
To proof that OW halve ML.[/quote]
First, we have a well-known lemma
If $ M$ lie inside a the equilateral triangle $ ABC, D. E. F$ are perpendicular projects of $ M$ on $ BC, CA, AB.$
$ O$ is the center of  the triangle. Then we have the result
$ 3 \vec {MO} \equal{} \vec {MA} \plus{} \vec {MB} \plus{} \vec {MC} \equal{} 2(\vec {MD} \plus{} \vec {ME} \plus{} \vec {MF})$
Return the problem. By the lemma, we have
$ \vec {WM} \plus{} \vec {WL} \equal{} \frac {3}{2} \vec {WO}$
and If $ K$ is midpoint of $ ML$ Then $ 4\vec {WK} \equal{} 3\vec {WO}$
=> $ W, O, K$ are collinear
 :roll:

I proof that $ 3\vec{MO}\equal{}\vec{MA}\plus{}\vec{MB}\plus{}\vec{MC}$,but i can't proof this :$ 3\vec{MO}\equal{}\vec{MA}\plus{}\vec{MB}\plus{}\vec{MC}\equal{} 2(\vec{MD}\plus{}\vec{ME}\plus{}\vec{MF})$.Can you show how you proof it.

[quote="barcelona"]I proof that $ 3\vec{MO} \equal{} \vec{MA} \plus{} \vec{MB} \plus{} \vec{MC}$,but i can't proof this :$ 3\vec{MO} \equal{} \vec{MA} \plus{} \vec{MB} \plus{} \vec{MC} \equal{} 2(\vec{MD} \plus{} \vec{ME} \plus{} \vec{MF})$.Can you show how you proof it.[/quote]
we must prove$ 3.\vec{MO}\equal{}2(\vec{MD}\plus{}\vec{ME}\plus{}\vec{MF})$but if $ G$ is the centroid of $ \Delta DEF$ then
$ 2(\vec{MD}\plus{}\vec{ME}\plus{}\vec{MF})\equal{}6.\vec{MG}$ in other words we must prove that $ 3.\vec{MO}\equal{}6.\vec{MG}$ so we must have $ M,G,O$ are collinear and $ MG\equal{}GO$.To prove this use Coordinates, it is easy since there is nothing to do just computing.(note that this collinearity and equation($ M,G,O$ are collinear and $ MG\equal{}GO$) doesn't hold for any triangle, u can prove that it holds iff $ \Delta ABC$ is equilateral) :wink:


Binomial Theorem Part 1

We discover the Binomial Theorem by expanding various powers of the binomial x+y.

Wait gamebot can post here too?
I thought its only on alcumus


geometry calculus AMC AIME USA(J)MO USAMO integration

I found these tests significantly more difficult than the practice tests. Before, I've scored as high as 9/10's on practice tests, but yesterday, I got... 4/10 and 3/10 (on algebra and calculus, respectively.)

I liked the first few algebra problems, but to be frank, #2 (the real part of $ (1 \minus{} i)^2009$ or something like that) was more of a combinatorics problem than an algebra problem. Upon expanding, all we get is a sum of n choose k's. As I forgot all the combinatorical identities that I've learned (and studied algebra and calculus), I forgot how to evaluate that. In any case, I think I expanded it wrong anyway. Ah well.

#6... I spent a bunch of time on it. I assumed that x = y and got 194 = 1. Then I gave up. It had a very nice solution though, in my opinion. For #7, I couldn't factor the top.

I was shocked by #8. I had seen and solved that problem (or a variant, in which some numbers were changed) multiple times before. In fact, it was posted on AOPS shortly before HMMT, and was in AOPS Intermediate Algebra, on the last chapter. And I forgot the solution. It was kind of cheap, in my opinion; that one was out there already, and those who were exposed to it get a clear advantage. Except for me, simply because I suck; I ended up forgetting the solution and getting it completely wrong.

For calculus, I was an utter failure. I got 1, 2, and 4. For #3, I simply integrated wrong (I actually did the substitution $ u \equal{} 1/x$, didn't notice anything in 4 seconds, and went back to multiplying top and bottom by x - 1 and then partial fraction'ing.) For #5, though I didn't realize that that was simply the fourth derivative of sin, I used L'hopital's rule and got the answer: sin(pi/3). Which I evaluated as 1/2. GRAARGH. I then attempted #8 and #10. #8, from what I know, I assumed the function was in the form $ e^{f(x)}$. Actually it was in the form $ Ke^{f(x)}$. No matter. In any case, I didn't follow the solution provided, and simply tried to solve for $ f(x)$. I differentiated, equated the ln's and the non-ln's (not rigorous, but works fine enough), then got a differential equation. Then I solved it. And forgot the constant of integration, which would've changed everything. -_-

For #10, I don't get the solution provided at all. In any case, I didn't like it much (compared to 2007's and 2008's #10's at least; they blew me away), as one could easily bash it with a u = tan(x/2) substitution. Which is what I did, and I'd have gotten the right answer had I remembered that the resulting integral I had was improper. I thus ended up with 0.

Team round was fairly straightforward for me; I took test B. To our knowledge, we missed 3 and 6b, though according to the judges, we missed more. 3, we kind of did in a rush, since we were all working on 6b, (in which we are asked to prove that there exist infinitely many graphs such that when one removes a vertex, its chromatic number decreases.) How would you do that anyway?

I personally liked 2008's style of team round better. The problems weren't taken from well-known theories (like graph theory), and eacg team member could work on their strengths more. In our team, only about half our team worked on anything at all; some simply could not conceptualize an abstract topic like graph theory, whereas others simply were not very skillful at writing proofs.

The guts round was alright. Thing is, I saw ANOTHER recycled problem (or, perhaps, they thought it up independently; we given cos A + cos B and sin A + sin B, and we were asked to find cos(A - B).) Ah well. We guessed the 2nd to last one correctly and got the full 25 points.

Overall, my team failed HMMT this year. As always, the tests had nice problems, though in my opinion, this year's problems were harder (according to my and my team's opinions.) Then again, it might've been that I didn't get any sleep last night. My roommate (nittanylions) breathes so loud. At 4, I just gave up and reviewed my calc. And took a shower.

Overall, I think the divisions is severely flawed; it does not segregate teams effectively. I was on a frosh/prefrosh team; despite the fact that we come from a magnet school, we still could not in any way compare to the top teams from the top schools.

Ah well, I met t0rajir0u. I didn't really get much in his lecture (in part because I just had lunch and pulled an all-nighter, and in part because linear algebra was simply over my head), though during it, I suddenly understood what eigenvalues were. Heheh.

So, how'd you do on the tests? What'd you think of HMMT as a whole? Any significant experiences?

By the way, tell me if you attended t0rajir0u's lecture. =D

I took Algebra and Combinatorics, both fails. This was my first HMMT.

For Algebra, I got #1, #3, and #5. I got negative values for #2 because I considered differences as negative when taking the sum, and I learned I can't FOIL for #4. :rotfl:

For Combinatorics, I only got #1 and #2, which were both extremely obvious to me; I got both in the first 20 seconds of reading them. Everything else was a fail because I also learned I can't add.  :oops:

Our Team Round went okay. I came up with a guess for #8 ($ (n, 0)$ for all integral $ n$), to which a counterexample was founded with a pentagon 20 seconds before time was up. We got a zero for that. I also learned I can't write proofs; I ended up writing two induction proofs. I actually know graph theory decently since I've had to solve many problems involving graph theory and I took a class which discussed colorings of graph for a while, but it didn't help too much.

Guts was really fun. For anyone who might have noticed some of the near-collisions up at the front, I was the one in 26-100 wearing the maroon sweater or the Derive and Conquer sweatshirt. We got most of the Geometry problems in the Guts Round, but failed on the last three.

Overall, it was an awesome experience. I don't have any real complaints, other than what has already been said.

So I finish the algebra test and was really happy that I answered more than half of the questions, then the guy sitting behind me starts complaining [in chinese] about how he forgot to square the $ 2$ for the double product and so missed out on a perfect score.

I liked the questions, though I never actually tried any of the older tests. Lack of preparation = win?

Algebra #2 was actually just $ \text{Re}\,\left((1 + i)^{2009}\right) = \left(\sqrt {2}\right)^{2009} \cos (2009 \times 45^{\circ}) = 2^{1004}$, though I realized that afterwards.

Algebra #5, I got $ P( - 1) = - 1$.  :|

[hide="Mess for Algebra 9"]
Let $ s_1, s_2 = z_1 + z_2, z_3 + z_4$ respectively, and let $ p_1, p_2 = z_1z_2, z_3z_4$ respectively. Then Vieta's gives the relations
\begin{align} s_1 + s_2 & = - 14 \\
p_1 + p_2 + s_1s_2 & = 52 \\
p_1s_2 + p_2s_1 & = - 56 \\
p_1p_2 & = 16 \end{align}
Combining and repeatedly subsituting gives a sixth degree polynomial in $ p_1$ that I don't feel like working out again.

As the product of pairs of roots are all positive, then AM-GM gives $ |z_1z_2 + z_3z_4| \ge 2\left|\sqrt {z_1z_2z_3z_4}\right| = 8$, equality iff $ z_1z_2 = p_1 = 4$, which we check is a root of the sixth degree polynomial.
[/hide]

[hide="Geometry 7"]
$ 84 = \frac 12 \times 14 \times AE \Longrightarrow AE = 12$, and $ \triangle CFD \sim \triangle CEA \Longrightarrow CF = \frac {3}{5} \times 7, DF = \frac {4}{5} \times 7$. Then $ AF = 15 - CF = \frac {54}{5}$. By Pythagorean Theorem, $ AD = \sqrt {12^2 + 2^2} = 2\sqrt {37}$. By Ptolemy on $ AEDF$,  $ EF = \frac {AF \cdot ED + AE \cdot DF}{AD} = \frac {6}{5}\sqrt {37}$.
[/hide]

Geometry #8 was evil.

[hide="Geometry 10"]
(again realized after test) By Law of Sines on $ \triangle BPC, APD$, we have
\[ \frac {2}{CP} = \frac {\sin BPC}{\sin CBP} = \frac {\sin BPC}{\sin DAP} = \frac {3}{CP + 3 + 2}
\]
Thus $ CP = 10$, and by Power of a Point, $ BP(BP + 7) = (CP + 2)^2 \Longrightarrow BP = 9$.
[/hide]

In team test B (yeah ..), ended up solving the entire test, but after stumbling on #3, I didn't have enough time to write up #6 (described the configuration, in unintelligible handwriting and without justification).

[quote="Zhero"]
By the way, tell me if you attended t0rajir0u's lecture. =D[/quote]

Yep, my cell phone went off in the middle.

[quote="azjps"]I liked the questions, though I never actually tried any of the older tests. Lack of preparation = win?[/quote]

Been true for me.

At PUMaC, I didn't bother preparing for NT and ended up scoring around a 20, compared to a 9 on Algebra (only because I had a lucky guess on #10, otherwise I would have earned a 2).

Here, I prepared a lot for Combinatorics and failed epically, as seen above in my earlier post. Algebra went slightly better, though I was preparing right after I woke up at 6.

I should learn to not bother trying to prepare.

Whoever wrote the guts round problem that was like "simplify $ 2\cos^{2}(\ln(2009)i) \plus{} i\sin(\ln(4036081)i)$" should be applauded. Other than that, we might need a new batch of problem writers. Seriously, 18 (fail of a fail - 60% of a 60%) on Combo should not make me second (even in division 2). And my rank in Geometry should not be improved by 1 if my score decreases by 16. Ever.

Then again, I probably shouldn't have spent half an hour staring at number 10 for Geometry. Maybe by checking my answers I would've realized that I wrote $ \sqrt {50}$ instead of $ 5\sqrt {2}$.

My team and I actually ended up eating lunch through most of the mini-event time.

[quote="Zhero"]

Team round was fairly straightforward for me; I took test B. To our knowledge, we missed 3 and 6b, though according to the judges, we missed more. 3, we kind of did in a rush, since we were all working on 6b, (in which we are asked to prove that there exist infinitely many graphs such that when one removes a vertex, its chromatic number decreases.) How would you do that anyway?
[/quote]

I couldn't find a way to explain away 3, vaguely describing a process to find a vertex without a diagonal, which apparently worked.

I think this set of configurations works for #6: Construct $ K_n$, and let $ A_1$ be the set of $ n$ vertices of $ K_n$. For each subset of $ n\minus{}1$ vertices of $ A_i$, construct a new vertex and connect it with each of the $ n\minus{}1$ vertices. Let the newly constructed set of $ n$ vertices be $ A_{i\plus{}1}$. Repeat arbitrarily many times to achieve $ A_k$. Now construct a last vertex $ P$, and connect it with all $ n$ points of $ A_k$.

This graph requires $ n\plus{}1$ colors: each $ A_i$ requires $ n$ colors by induction, and then $ P$ is connected to $ n$ vertices of distinct colors. I'm pretty sure that removing any point will make the graph have a chromatic number of $ n$, which can probably be explained away by induction also.

hello.world, I liked that guts problem, simply because these types of algebra problems interest me. :P Nobody in my team got it. They were all staring like WTF. Then I was like hey that looks easy. You just simplify it with a bunch of formulas... XD

Algebra #2. Damn. I feel like jumping out a window. One of my teammates took algebra and combinatorics. He said: the only combinatorics I solved was #2 on algebra. XD

With that in mind, could we possibly prove some combinatorical identities using complex expansions like that? I just noticed. It's pretty cool. :O

[quote]Yep, my cell phone went off in the middle.[/quote]
How much of the lecture did you understand?

Thanks for the solution azjps.

[quote="hello.world"]we might need a new batch of problem writers.[/quote]
Paul Christiano, Shaunak Kishore, and Jacob Steinhardt were responsible for a total of $ 54$ problems.  What else would you expect?  :P  (You should be glad that they used at most $ 4$ of my problems; none of them were particularly easy.)

[quote="Zhero"]could we possibly prove some combinatorical identities using complex expansions like that? I just noticed. It's pretty cool. :O[/quote]
Yes.  This is a technique commonly known as the roots of unity filter; see my posts [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url], [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url], [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=242876]here[/url], and (to a lesser extent) [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=257652]here[/url].

Ok;
I took combinatorics and geometry.  For combinatorics, I've done barely anything, (I just began the last chapter of Intro. to Counting and Prob). Numbers 1,2, and 3 were decently trivial, and then everything after that was...casework.  Casework which, on one of my answers, involved 20 cases.  I forgot a case, and I would've gotten #8 right if I hadn't -_-

Geometry was interesting...I got number 1 abnormally slowly compared to past HMMT tests, which kind of deflated my want to win..#2 and #3 I got, and #7 I spent the rest of my time on, eventually getting the incorrect answer..stupid algebra.
I didn't add the information necessary for the tests though...sigh...

Guts was all right.  I thought the geometry problems on guts were better than on geometry.

Overall, enjoyable contest. The best part was t0raij0u's speech, which was certainly intellectually stimulating.  I must say though, I did enjoy the settting of MIT, and the Mexican place across from it =D

And  t0raij0u, can you post hte links you included at the end? Thanks.

The slides themselves are available on my blog (link in signature).

I didn't like combo at all. It was really caseworky, which made me go from a 35 to a 19. Also, it kind of ruined the difficulty curve...Earlier ones were roughly as tough in terms of casework as some later ones.

Algebra I liked in general, though I wish I'd tried #9...Ugh. Anyway it was roughly the same as previous years.

In theory I should have like team a lot, but evidently failed? I thought I'd gotten 5 and 7 completely, but...the graders disagreed.

Guts was cool. Liked the combo better than the combo test, and it was mainly what I did.

[quote="nittanylion"]and the Mexican place across from it =D[/quote]
Anna's is SO GOOD there's one near my school

[quote="seasonal squirrel"]Um just a question, why exactly was a question from the 1990 AIME used for Algebra? By that I don't just mean an analogous idea but the exact same numbers also. See here: [url]http://www.mathlinks.ro/viewtopic.php?p=459558#459558[/url][/quote]

Darn it, 1990 was one of those few years that I haven't looked at yet...

I thought the problems were at about the level of the recent years. In algebra I got the first 5 problems right and died after that. I thought number two was a little hard for a 3-point problem, while 3 and 5 were relatively easy, but both nice questions.

Calculus was not too bad, except that for number 3 I never believed that the denominator of the integrand could be factored, so a shameful error there. Number 10 was impossible (although a few people did get it). I'm sure most people attacked numbers 4 and 5 by brute force computation. 2 was a little mechanical, almost approaching an intimidating textbook problem.

The team round was not fair!!! Graph theory is one of those topics that you cannot learn on the spot. So that is why you see such disparity in the scores on the team round for even the top teams (152-377 for #11-20, 130-400 for #1-10). Worse, it formed the whole test. Why not divide the problems into parts like before? Maybe then one topic won't completely wipe out any chance at a high score.

Guts round was fun... although I like more those problems involving comparison with other teams (if your number N is submitted by N teams...).

Apparently the TJ people had seen a problem very similar to Calculus 10 before - if you'll notice, the only people to get it right were from TJ.

I solved 7 problems on Calc. I got two right. That pretty much sums it up.

#10 calc was anything but impossible. In fact, the $ u \equal{} \tan{\frac {\theta}{2}}$ substitution virtually trivializes it. Though this antiderivative is only valid over $ ( \minus{} \pi, \pi)$ (or was it a closed interval? I don't remember), if we simply split the integral into two parts, we will get the answer. In particular,

$ \lim_{t \to 0} \int_0^{\pi \minus{} t} \frac {d\theta}{a \plus{} b \cos{\theta}} \plus{} \int_{\pi \plus{} t}^{2 \pi} \frac {d\theta}{a \plus{} b \cos{\theta}}.$

With that in mind, I actually found #10 somewhat trivial and attempted it before #3. And I got an answer. My problem? I forgot that the substitution only works over $ ( \minus{} \pi. \pi)$. And thus arrived at the answer 0. I banged my head against a wall after that.

On hindsight, #9 calc wasn't that bad either. A brute-force approach isn't as ugly as you may think it may be. All one needs to do is perform a simple rotation, and integrate a somewhat ugly, but not absolutely horrible function from 0 to 2^(1/2).

On that note, what was the "clever" solution for that one?

[quote]I solved 7 problems on Calc. I got two right. That pretty much sums it up.[/quote]
Hah. XD That must suck. Which ones?

[quote]
Um just a question, why exactly was a question from the 1990 AIME used for Algebra? By that I don't just mean an analogous idea but the exact same numbers also. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=459558#459558[/quote]
... No wonder it seemed familiar. -_- In my opinion, that would have been worth an objection. People familiar with that problem have a clear advantage.

[quote]Anna's is SO GOOD there's one near my school[/quote]
Yeah, the lunch was good.

[quote]The team round was not fair!!! Graph theory is one of those topics that you cannot learn on the spot. So that is why you see such disparity in the scores on the team round for even the top teams (152-377 for #11-20, 130-400 for #1-10). Worse, it formed the whole test. Why not divide the problems into parts like before? Maybe then one topic won't completely wipe out any chance at a high score. [/quote]
I agree with you there. I must say that I was quite fond of last year's team round.

[quote]I thought the problems were at about the level of the recent years. In algebra I got the first 5 problems right and died after that. I thought number two was a little hard for a 3-point problem, while 3 and 5 were relatively easy, but both nice questions. [/quote]
I agree that 3 and 5 were somewhat trivial. #2 was actually not that hard once you realized you should use cis rather than combinatorical identities. -_-

In my opinion, the algebra this year was slightly harder than it was last year (or the previous years.) On my practice tests over the break, I scored at least 7 on each of the tests I took. I got a 9 on my practice for 2008 (though I must admit I took the general last year and saw some of the problems before) and a 9 for the practice for 2006. And I failed this one epically. I too died on everything past 5. Gah.

[quote="Zhero"]
[quote]
Um just a question, why exactly was a question from the 1990 AIME used for Algebra? By that I don't just mean an analogous idea but the exact same numbers also. See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=459558#459558[/quote]
... No wonder it seemed familiar. -_- In my opinion, that would have been worth an objection. People familiar with that problem have a clear advantage.
[/quote]

I saw that question before, though I remembered nothing of the solution, and it took me some attempts to realize that the elementary symmetric polynomial thing yielded a simple recursion; my teammate next to me also realized that (computed it incorrectly though), and I'm pretty sure he never saw the AIME question. Though I agree ..

The way I did algebra #2, the way on the official solutions [root of unity filter] was $ (1\plus{}i)^{2009} \equal{} 1 \plus{} {2009 \choose 1}i \minus{} {2009 \choose 2} \minus{} {2009 \choose 3}i \plus{} \cdots$, $ (1\minus{}i)^{2009} \equal{} 1 \minus{} {2009 \choose 1}i \minus{} {2009 \choose 2} \plus{} {2009 \choose 3}i \plus{} \cdots$, adding cancels the odd terms, so $ \frac{1}{2} \left[(1\plus{}i)^{2009} \plus{} (1\minus{}i)^{2009}\right] \equal{}$ the desired sum.

Oh, and I happened to read that particular post by t0rajir0u before.  :P

We did not realize that the problem had come from an old AIME. Someone submitted it to us and most people thought that it was a good problem so it ended up on the test.

To those who think that Combinatorics required a lot of case analysis: only $ 4$, $ 6$, and $ 9$ require any substantial case analysis if you approach the problem the right way. There is still a fairly nice solution to all $ 3$ of those problems as well.

[quote="azjps"]
The way I did algebra #2, the way on the official solutions [root of unity filter] was $ (1 \plus{} i)^{2009} \equal{} 1 \plus{} {2009 \choose 1}i \minus{} {2009 \choose 2} \minus{} {2009 \choose 3}i \plus{} \cdots$, $ (1 \minus{} i)^{2009} \equal{} 1 \minus{} {2009 \choose 1}i \minus{} {2009 \choose 2} \plus{} {2009 \choose 3}i \plus{} \cdots$, adding cancels the odd terms, so $ \frac {1}{2} \left[(1 \plus{} i)^{2009} \plus{} (1 \minus{} i)^{2009}\right] \equal{}$ the desired sum.
[/quote]

That's where I got stuck. How'd you simplify it from there? Combinatorial identities?

azjps already gave the calculation in an earlier post.  I would do something like $ (1 \plus{} i)^2 \equal{} 2i \Rightarrow (1 \plus{} i)^{2008} \equal{} 2^{1004}$ and the same for $ (1 \minus{} i)^2$, which gives

$ \frac{(1 \plus{} i)^{2009} \plus{} (1 \minus{} i)^{2009}}{2} \equal{} \frac{(1 \plus{} i)2^{1004} \plus{} (1 \minus{} i)2^{1004}}{2} \equal{} 2^{1004}$.

By "TJ people had seen a very similar problem" you mean "TJ people had calc #10 on their homework first semester."

We want to compute $ \int_0^{2\pi} \frac {d\theta}{a \plus{} b\cos\theta}$. Let $ z \equal{} e^{i\theta}$. Then $ dz \equal{} ie^{i\theta}d\theta \equal{} izd\theta$.

Thus,
$ \int_0^{2\pi} \frac {d\theta}{a \plus{} b\cos\theta} \equal{} \oint_{|z| \equal{} 1}\frac {dz}{iz (a \plus{} \frac {b}{2}(z \plus{} z^{ \minus{} 1}))} \equal{} \frac {2}{i}\oint\frac {dz}{bz^2 \plus{} 2az \plus{} b}$.

Now we find the poles of the integrand (the roots of the denominator).

$ bz^2 \plus{} 2az \plus{} b \equal{} 0$
$ z \equal{} \frac { \minus{} 2a \pm \sqrt {4a^2 \minus{} 4b^2}}{2b} \equal{} \frac { \minus{} a\pm\sqrt {a^2 \minus{} b^2}}{b}$

Since $ a > b > 0$, it is easy to see that $ \frac { \minus{} a \minus{} \sqrt {a^2 \minus{} b^2}}{b} < \minus{} 1$ and $ \minus{} 1 < \frac { \minus{} a \plus{} \sqrt {a^2 \minus{} b^2}}{b} < 1$. Thus, by Cauchy's integral formula,

$ \frac {2}{i}\oint\frac {dz}{bz^2 \plus{} 2az \plus{} b} \equal{} \frac {2}{i}2\pi i(\frac {1}{b\frac {2\sqrt {a^2 \minus{} b^2}}{b}}) \equal{} 4\pi(\frac {1}{2\sqrt {a^2 \minus{} b^2}}) \equal{} \frac {2\pi}{\sqrt {a^2 \minus{} b^2}}$.

Amusingly, the two of us who did well on calc were also the two who missed this problem.

Also amusingly, problems extremely similar to calc #10 last year are appearing on our homework this semester. For example, one of our homework problems asks you to compute $ \int_{0}^{1}x\ln x\ln(1\minus{}x)\, dx$.

[quote]the only combinatorics I solved was #2 on algebra.[/quote]

this quote is attributed to none other than ME!

yep i got the first 5 algebra, which was kinda good. i tried to epic bash the rest of them, failing in the process.

i really loathed the combinatorics section. i only got 1 and 2, whch were trivial. #3 i wouldve gotten if i couldve subtracted correctly, and the rest were kinda weird. its odd though. on all other hmmt tests i took as practice, there were always at least one probability problem. this time, however, there were none, and since the only real reason i take combinatorics is because i am very good at probability, i was basically dead in the water at that point.

i liked t0rajir0us lecture, even though i only understood <1% of it. it was a pretty exhilarating experience all in all, and that japanese place fanfan took me to was pretty good, except i drowned my teriyaki in spicy sauce. also i got a really beast t-shirt too. so yea, a pretty cool trip, but it wouldve been better if we won at least [b]something[/b]

Shaunak asked me that problem earlier in the semester.  I gave quite an overkill solution:  we'll write

$ \frac {1}{a} \int_{0}^{2\pi} \frac {d\theta}{1 \plus{} \frac {b}{a} \cos \theta} \equal{} \frac {1}{a} \int_{0}^{2\pi} \sum_{k \equal{} 0}^{\infty} \left( \minus{} \frac {b}{a} \right)^k \cos^k \theta \, d \theta$.

Now, $ \int_{0}^{2\pi} \cos^k \theta \, d \theta \equal{} \frac {1}{2^k} \int_{0}^{2\pi} (e^{i \theta} \plus{} e^{ \minus{} i \theta})^k \, d \theta$ and every term integrates to $ 0$ except the constant term, which exists if and only if $ k \equal{} 2n$ is even and is equal to $ {2n \choose n}$ in that case.  Hence the original integral is

$ \frac {1}{a} \sum_{n \equal{} 0}^{\infty} 2\pi {2n \choose n} \frac {b^{2n}}{(2a)^{2n}}$.

The identity $ \sum_{n \equal{} 0}^{\infty} {2n \choose n} x^n \equal{} \frac {1}{\sqrt {1 \minus{} 4x} }$ finishes the job.

Qiaochu, this is the same solution I came up with. Shaunak also has a geometric interpretation of the integral that allows one to evaluate it.

If you haven't already, read the solution to team round problem $ 7$. It's my favorite problem on the entire test, although I'm a bit disappointed that no one solved it.

EDIT: Actually I think AAST B solved it, but that was it.

[quote="JSteinhardt"]

If you haven't already, read the solution to team round problem $ 7$. It's my favorite problem on the entire test, although I'm a bit disappointed that no one solved it.

EDIT: Actually I think AAST B solved it, but that was it.[/quote]
I solved it. It was an easy problem from coding theory. But what a nice one paragraph solution.

The problem was the following: Given a code with Hamming distance 3 in $ \mathbb{GF}_2^n$, $ n\plus{}1$ not a power of $ 2$, show that only $ 2^n/(n\plus{}2)$ codewords can be used. The solution is two basic inequalities, used in conjunction, followed by an elementary number theory argument.

So I'm looking through the tests right now and I'm finding random problems that jump out at me. Here are my musings.

Algebra 2: I agree that this one is harder than 3, 4, 5, but it's not unreasonable. It's a pretty well known trick to take P(1)+P(-1) and stuff.

Algebra 6: This one is cute.

Algebra 7: This problem was just obnoxious. The idea is nice I guess, but it ended up being a time-waster (and it turns out I got it wrong).

Algebra 9: So...factoring into two quadratics wins again (see my solution in the divsion 1/2 thread).

Geometry 1/2: Alright this is more of a rant but WTF WHY DID I GET THESE WRONG.

Geometry 3: Well y'all could have provided us with a more enlightening solution...I dumped a good 15 minutes into this problem.

Geometry 4: This is a nice problem, but it's much easier than say, 6 and 7, or even 9. I found the whole test to be in a really weird order, though the problems were pretty good.

Guts 12: This problem is just plain annoying. Blah. No.

Guts 16: I like this one.

Guts 27/32: Aren't these like...kind of easy? Considering the fact that they give you 12/18 points, respectively?

Well overall, none of the problems really stood out as being really good (in my opinion), but there was a fair share of nice problems. Better than the PUMaC problems, by far...if any problems were written the night before, it was certainly hidden a lot better. Ted, saying we need new problem writers is pretty harsh, don't you think? It's pretty hard to write good problems. You should try it.

[quote="hello.world"]we might need a new batch of problem writers.[/quote]

[quote="CatalystOfNostalgia"]Well overall, none of the problems really stood out as being really good (in my opinion), but there was a fair share of nice problems. Better than the PUMaC problems, by far...if any problems were written the night before, it was certainly hidden a lot better. Ted, saying we need new problem writers is pretty harsh, don't you think? It's pretty hard to write good problems. You should try it.[/quote]

I rephrase: The level of difficulty reached a new high for HMMT. (A side note, I wonder whether that was intended or not, and if so, whether it was done to offset November.) This goes to the whole awards and division thing. :| Obviously it's hard to judge precisely how easy or hard the problems are gonna be overall, but if the high scorers (from division 2, anyway) are mostly failing, is the test just a bit harder than it should be?

As for your challenge, Carl, I know the difficulties of problem writing, and I decline for now. :)

[quote="Hamster1800"]By "TJ people had seen a very similar problem" you mean "TJ people had calc #10 on their homework first semester."
...

Also amusingly, problems extremely similar to calc #10 last year are appearing on our homework this semester. For example, one of our homework problems asks you to compute $ \int_{0}^{1}x\ln x\ln(1 \minus{} x)\, dx$.[/quote]

I do hope your homework is not graded. What class is this for anyway?

[quote="hello.world"]I rephrase: The level of difficulty reached a new high for HMMT. (A side note, I wonder whether that was intended or not, and if so, whether it was done to offset November.) This goes to the whole awards and division thing. :| Obviously it's hard to judge precisely how easy or hard the problems are gonna be overall, but if the high scorers (from division 2, anyway) are mostly failing, is the test just a bit harder than it should be?
[/quote]

What cannot happen is a 6-way tie for first place because all of the six got everything right. These problems sure did not seem too hard for the top contestants since they got nearly all of them right, and if the tests were any easier, they probably would not having a problem getting 100%. This might be why the problems are so hard.

In a 50 minute slot, you simply cannot have the broad range of difficulty that would make the questions accessible to everyone (to get one question right is quite an achievement, in my opinion), especially since the top scorers must be distinguished. Even in a longer time it is hard; just think about the AIME, a 180 minute test where the average of the students (who are the top 20000 in the nation) is a 2.5/15. Clearly, half the students sat for most of the time pretty bored. The USAMO is worse.

On the other hand, I do not believe that the tests this year were any harder. In fact, a combined score of 32 is one of the highest scores I've gotten on any of my practice tests. Also, comparing this years' scores with previous years shows no significant trend in difficulty. Lastly, keep in mind that taking the test actually and practicing are two very different feelings, and your evaluation of difficulty may vary as such.

[quote="236factorial"]
What cannot happen is a 6-way tie for first place because all of the six got everything right. These problems sure did not seem too hard for the top contestants since they got nearly all of them right, and if the tests were any easier, they probably would not having a problem getting 100%. This might be why the problems are so hard. [/quote]

I'm sure many of those who went to the November HMMT remember 10 people getting perfect scores. Good thing it was exactly ten.

[quote="hello.world"]I rephrase: The level of difficulty reached a new high for HMMT. (A side note, I wonder whether that was intended or not, and if so, whether it was done to offset November.) This goes to the whole awards and division thing. :| Obviously it's hard to judge precisely how easy or hard the problems are gonna be overall, but if the high scorers (from division 2, anyway) are mostly failing, is the test just a bit harder than it should be?)[/quote]

Actually, looking back at the problems, I think quite the opposite. Division 2 is a pretty bad indicator of difficulty, of course the scores are going to be lower there. But just take a look: compare the scores for every single one of the individual tests as well as the guts round with the high scores from last year. Of course, there are more people/teams (notably the Chinese schools), but even without them, the scores are consistently higher this year. The team round is a different story though...who made the decision to make the team round centered around one topic? Honestly, I think 10 random proof-based problems (multi-part, sure) would be the optimal format.

[quote="CatalystOfNostalgia"][The team round is a different story though...who made the decision to make the team round centered around one topic? Honestly, I think 10 random proof-based problems (multi-part, sure) would be the optimal format.[/quote]

I agree... probably they only had one good problem topic submitted, so only one topic appeared.

I do think that themes would be nice (as opposed to a completely random bag of questions) because then different teams members can interact with each other on a common topic, but there cannot be only one focus, as this year was. Especially not graph theory.

And now, a word from the 2009 Problem Czar herself...

First, I'm glad to see you all are excited about discussing these problems.  I know that the tests weren't perfect, and I understand your complaints and thank you all for the feedback.  Congrats to those of you who solved some of the tougher ones.

I'll keep all your comments in mind when advising the next problem czar (possibly JSteinhardt or Shaunak or both.)

[quote="CatalystOfNostalgia"]Ted, saying we need new problem writers is pretty harsh, don't you think? It's pretty hard to write good problems. You should try it.[/quote]

It is indeed very hard to write good problems, and it is especially hard to write good problems that are correct.  With much help from all the people listed in the program, I did my best to choose and write problems that I thought would be enjoyable, of the right difficulty, and not mathematically flawed, the last of which is actually much harder than I had anticipated.  For instance, you can come up with a beautiful geometry problem involving side chasing, and apply theorems that certainly seem to work and provide a solution that one would look at and think is flawless, but the diagram doesn't actually exist because of the triangle inequality.


[color=blue][b] Problem.[/b]
Let $a, b, c$ three nonnegative real numbers. Prove that the following inequality holds true:

\[ \frac{b^{2}+c^{2}}{a(b+c)}+\frac{c^{2}+a^{2}}{b(c+a)}+\frac{a^{2}+b^{2}}{c(a+b)}\geq3\cdot\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}.  \]

[/color]


Να λυθει στους θετικους ακεραιους,η εξισωση
 $xy(x+y)=k^{3}$

 :D  :D  :D

Η εξισωση δεν εχει λυση στους θετικους ακεραιους γιατι μετασχηματιζεται στην εξισωση κ^3+λ^3=μ^3 η οποια δεν επιδεχεται λυσεις στους θετ.ακεραιους.

Σωστος... :wink: (ελπιζω μονο το κ που χρησιμοποιεις να μην εινια το k της αρχικης εξισωσης...,γιατι τοτε θα ηθελα να γραψεις αναλυτικα την λυση...)
Μου φαινεται οτι εχετε καταλαβει οτι οι περισσοτερες ασκησεις αλγεβρας που βαζω αναγονται στην εικασια του Fermat...
Οντως ηταν ευκολη...(made in cuba...τι περιμενεις... :P )

 :D  :D  :D

Εννοείς το τελευταίο  ΘΕΩΡΗΜΑ του Fermat?  :rotfl:

ναι....ναι.....
Εννοω την αποδειξη του wilesπουεινια 10000 σελιδες....
Παντως σοβαρα τωρα....Ειναι ΘΕΩΡΗΜΑΤΑΡΑ!!!!

 :D  :D  :D

πως καταληγει σε αυτη την εξισωση ο Αρχιμηδης 6?

1)$xy(x+y)=k^{3}$
Mporoyme na ypouesoyme wlog οτι (x,y)=1[allios an (x,y)=d kataligoume semia eksisosi omoia me tin arxiki me x1,y1 opu ta x1,y1 einai prota metaksi tous)
Tote omos $(xy,x+y)=1$ kai ara$xy=l^{3}$ kai $x+y=m^{3}$me $lm=k$
Epeidi omos $(x,y)=1$ 8a einai $x=d^{3}$ kai $y=c^{3}$ kai $l=dc$ omos tote $d^{3}+c^{3}=m^{3}$,atopo apo fermat......

 :D  :D  :D

ευχαριστω Ραλλη...

Ενημεροτικα λεω οτι δεν ειναι λιγει οι μαθηματικοι που πιστευουν πως υπαρχει και αλλη λυση πολυ απλουστερη απο αυτην που ξερουμε για το τελευταιο θεωρημα του Φερματ.(ισως παρει και αρκετα χρονια για να βρεθει).Εξαλλου οι πρωτη λυση που δινεται για την αποδειξη καποιου θεωρηματος δεν ειναι τις περισσοτερες φορες η καλυτερη....

και ίσως είναι η λύση του Fermat...

Na thimisw oti exei apodeixtei (an den kanw lathos apo ton Hilbert) to eksis Thewrima: Ean ena provlima linetai me ena tropo, tote iparxei toulaxiston alli mia lisi me diaforetiki thewria pou linei to idio provlima. (Auto exei apodeixthei kai den einai eikasia). Gi'auto to logo polloi epidoksoi prospathoun na vroun tin deuteri lisi, kathws episis akoma eksetazetai apo tous eidikous i lisi tou Wiles kathws MONO h 4-melis ('h 6-melis epitropi) pou tin ekrine mporei pros to paron na tin diavasei, afou periexei teleutaia apotelesmata stin algevriki topologia, algevriki thewria arithmwn kai sti thewria Galois/epektasewn. Auta ta oliga...

Alexandros


There are $ 5$ dogs, $ 4$ cats, and $ 7$ bowls of milk at an animal gathering. Dogs and cats are distinguishable, but all bowls of milk are the same. In how many ways can every dog and cat be paired with either a member of the other species or a bowl of milk such that all the bowls of milk are taken?

[quote="apiarist1"]Very easy, 5*4=20.[/quote]
why would it be $ 5\cdot4$?
explanations?

Since there are two animals left after pairing them with the 6 bowls of milk, they must be a dog and a cat. Therefore we need to choose 1 dog and 1 cat, which gives 5x4=20

Since bowls of milk are the same it doesn't matter in what order animals drink the milk. It matters only what  animals drink the milk. SO, different arrangement we can obtain only choosing different pairs of animals who won't drink the milk, and it is 5*4=20.


It looks like the 2007 IMO team has been announced (unless this is a cruel prank pulled by probability's school newspaper): http://www.uni.uiuc.edu/gargoyle/2007/05/alex_zhai_to_represent_us_agai.htm

To summarize,

• Sherry Gong

• Eric Larson

• Brian Lawrence

• Tedrick Leung

• Arnav Tripathy

• Alex Zhai

Pretty cool, though not at all unexpected.

Eh... how do you know they were not [i]just kidding[/i]?

Yeah, that might be because you guys overwhelmed my school's server. Good job.

Uh oh, that's not good. I can't check my email now. Also, that means everyone is going to fail the English exam tomorrow.  :maybe:

The official announcment of the USA Team selection for the 2007 IMO is at:
http://www.maa.org/
The MAA posted it Tuesday, May 29, 2007.

Steve Dunbar
MAA Director for the American Mathematics Competitions

Hmmm bleargh maybe, but maybe not, especially if synthetic geometry is [i]not allowed[/i]...

First, I feel like I just missed a reference, and second, what is synthetic geometry???

the chance that US will beat China is still very rare

[quote="bpms"][quote="1=2"][quote="PTynan89"]Hmm, so maybe the US will beat China this year?

=)[/quote]

It has. A couple times. Check the archives of IMO, and you will see that USA won a couple of times.[/quote]
Check how recent those times are while you're at it.
The =) has significance, he was joking...[/quote]

No, you guys both misread my post.

I'm not saying it hasn't happened before, I was just saying how it would nice for something rare to happen again this year.

Also, the =) meant that it would make me very happy to see the US beat China this year.

I just looked in the archives of the IMO, and from 1995 to 2006 the U.S. has only beaten China once (1996) and China has only not won the IMO two other times (2003, and 1998 when their names didn't show up for whatever reason). Although, the 94' US team was very impressive and if they could pull it off this year, they will probably beat China  :)

Before then, China wasn't very successful in its first few IMOs. After all, they didn't have colleges until the late 70s  :|

China doesn't appear in 1998 because of politics- that IMO was in Taiwan.

Seriously? There goes bragging to friends that my area of expertise is completely free of politics.

If it makes you feel any better, that seems to be the only boycott by an important team. In the same era that large sections of the world didn't go to the Olympics (1980 Moscow and 1984 Los Angeles), the USSR team came to Washington for the 1981 IMO. The USA also went to communist countries several times, although Russia didn't come up until 1992.

There are some other places where politics shows up. At the 1993 IMO in Turkey, no team from Cyprus came- but there was a team from the "Turkish Republic of Northern Cyprus" instead, for the only time.

[quote="gt59"]The 1980 IMO was canceled though - I'm not really sure if it was related to the Cold War politics that affected the 1980/84 Olympics.  Evidently the host country for that year pulled out for some reason, perhaps funding.  Anyone know why?[/quote]
Quote from The IMO Compendium:

"The only year in which the IMO was not held was 1980, when for financial reasons no one stepped in to host it"

[quote="MathChauffeur"]
Quote from The IMO Compendium:

"The only year in which the IMO was not held was 1980, when for financial reasons no one stepped in to host it"[/quote]

I've read that before but find it hard to believe that there weren't political reasons behind the lack of financing.  Economic times were tough then, especially for poorer countries, but not much worse than in the late 70's or early 80's.  Most countries, including the US, had teams ready to go.  However, I can't believe that President Carter would have permitted a US team to participate in a contest closely aligned with the Soviet-bloc countries, especially if hosted by one.

I believe that Mongolia was supposed to host it, but the funding fell through...


blogs
logarithms
trigonometry
geometry
perimeter
floor function
algebra

Hello everybody. In case you don't know me, I am the co-creator of the [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1597]DMAT[/url]. (Click the link)

This is an announcement that Round 3 will begin around Thursday 7/30. The problems will cover an assortment of topics, ranging from mid-AMC 12 all the way to late-AIME. For past problems, visit the blog and look at past problems. [hide="Here are some:"]

If $ \displaystyle{\sum_{k \equal{} 2}^{1003}} \ {2007\choose k}$ can be expressed as $ a^b \minus{} c$, where a, b, and c are (not necessarily distinct) positive integers, find $ a \minus{} b \plus{} c$.

Suppose that the solution to: $ (5^{\log x})^2 \plus{} 4x^{\log 5} \minus{} 32 \equal{} 0$ is written as $ k^{\log_i j}$, where k, i, and j are positive integers. Compute $ k \plus{} i \plus{} j$.

If $ \sin(\omega) \equal{} 3\cos(\omega)$ for $ 0\le \omega \le \frac {\pi}{2}$, then $ x \equal{} \displaystyle{\sqrt {\frac {2\sin(\omega) \minus{} \cos(\omega)\sin(2\omega)}{\sin(2\omega)\sec(\omega)}}}$. $ x^4$ can be expressed as a fraction $ \frac {m}n$, where m and n are relative prime. Evaluate $ |m \minus{} n|$.

A kid plots a regular hexagon with a perimeter of $ 12\sqrt2$ units that lies in the Cartesian plane in such a way that its center is on the origin, two of the vertices lie on the x-axis, and the midpoints of two of its sides lie on the y-axis. The kid revolves the portion of the hexagon that lies in Quadrant I completely around the x-axis, thus forming a solid whose volume is m cubic units. If that same portion is completely revolved around the y-axis, a solid with a volume of n cubic units results. Suppose that $ \left(\frac {m}n\right)^2 \equal{} \frac {G}T$, where G and T are relatively prime. Demonstrate that you, on the other hand, don't need Mathematica's assistance and evaluate $ G \plus{} T$.

Let $ \tau_n \equal{} 1.5n \plus{} \frac {1}{3n \plus{} \frac {1}{3n \plus{} ...}}$. Evaluate $ \displaystyle\sum_{n \equal{} 1}^8 \tau_n^2$.

Tomorrow happens to be both Jude and Jerry's birthday! Millinova decides to invite them and 13 other friends to a fantabulous birthday party. The round table at the party contains seats with numbers 1-16, arranged clockwise. Jude and Jerry sit together, and Millinova sits next to one of them. A group of three adjacent seats contains gfour84, Rolloway, and Caelestor (though not necessarily in that order). Cyleura sits across the table from Caelestor. There are a total of $ n*9!$  possible distinct arrangements of people sitting at the table (two arrangements are distinct if one person is not sitting in the same-numbered seat). Calculate n.

Positive real numbers $ g,f,o,u,r$ satisfy $ \frac {g}{3} \plus{} o \plus{} 2f \plus{} 6u \plus{} 7r \equal{} 20$. Maximizing the product $ gfour$ gives the product $ m/n$, where m and n are relatively prime integers. Find $ m \plus{} n$.

Let $ a_0 \equal{} \cos{\frac {\pi}{2010}} \plus{} i\sin{\frac {\pi}{2010}}$ and for all positive integers $ n, a_n \equal{} (a_{n \minus{} 1})^{\frac {13n \plus{} 2}{n \plus{} 1}}$. What is the smallest n that makes $ a_n$ both positive and real?

How many integers between 1 and 1000 inclusive do not satisfy $ x \minus{} \lfloor\frac {x}{3}\rfloor \equal{} \lfloor\frac {3x}{4}\rfloor \minus{} \lfloor\frac {x}{12}\rfloor$?

A polynomial  satisfies $ P(8) \equal{} 144$, and the following expression is true for all $ x \not \equal{} 0, \frac {1}2$: $ \frac {8P(x^{2} \plus{} 2)}{P(4x)} \equal{} 1 \plus{} \frac {x^{3} \plus{} 1}{2x \minus{} 1}$. What is $ P(10)$?

Determine the largest integer that always divides $ 800^n \minus{} 125^n \minus{} 64^n \plus{} 253^n$ for all integers $ n\ge 2$.

[/hide]

If you would like to participate, please post in the topic below so gfour84 can get a head count. He'll be hosting Round 3. [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1566166#1566166]Click here for more information.[/url]

it is ok to sign up for the tournament even if i won't be able to solve problems until next wednesday for the next round? (i'm returning to the u.s.a.)

if so , then i sign up.

Can I register for the contest know also .
If i can  .....then i sign in.

Hello, here are the people who have explicitly signed up:

Brut3Forc3
darkprince
mathwizarddude
narik
SnowEverywhere
srinath.r
harekrishna
geniusbliss

If you haven't done so, please make a post below! (Or you can just submit an answer to 3-A and you'll be automatically entered)

If you would like to participate, please look at the first problem posted and submit an answer. If you can't solve it, then post your intentions to participate in this thread.

This is going to be the best one cause I am running it (actually, that just means that it will be easier .... )

However, you don't think that I didn't donate some problems to this round, did you?  :wink:
They'll come around the second half I think.  :P


Let [tex]x_i,y_i[/tex] be positive reals for [tex]1\le i \le n[/tex]. Assume that [tex]y_i[/tex] is an increasing sequence and  [tex]\displaystyle  \sum_{i=1}^k x_i \le \sum_{i=1}^k y_i[/tex] for all [tex]1\le k\le n[/tex]. Show that [tex]\displaystyle \sum_{i=1}^n \sqrt{x_i} \le \sum_{i=1}^n \sqrt{y_i}[/tex].

(Fixed typo, thanks [b]cn[/b]).

Are you sure the inequalities are pointing in the correct directions?  If we take

1 < xi < yi

for all the i's, and let the conditions hold, the final inequality is incorrect.

Thanks for the hint, fuzzy!

Hint: [hide]Use the Abel Summation Formula[/hide]

Bigger Hint: [hide]This is the Abel Summation Formula:

a1 b1 + a2 b2 + ... + an bn

= (a1-a2)(b1) + (a2-a3)(b1+b2) + ... + (a(n-1)-an)(b1+b2+...+b(n-1))+an(b1+b2+...+bn)[/hide]

About that hint:

[hide]Another name for the Abel Summation Formula - an alternative name that should stick this formula squarely in your mind - is "Summation by Parts." In fact, even if you've forgotten the exact details, if you just remember that it is "summation by parts", you should be able to reconstruct the correct formula from the analogy to integration by parts.[/hide]


Hi,

in his book "Outsmarting IQ: The emerging science of learnable intelligence" David Perkins quotes some findings of mathematician-psychologist Allan Schoenfeld (p.87):
"One of the most important factors [in deficient math problem solving is] poor mental management:

- Students did not pay attention to the winding path of their activities in solving a problem.
- They often did not think to use heuristics they knew and could have applied.
- They often perseverated in an approach that was not yielding progress rather than trying a new tack.
- They often gave up without rummaging in their repertoire for another point of entry.
- Amidst the trees, they lost sight of the forest.”

Although this is a very general question:
What are your own experiences concerning these points?
Do you have general or specific strategies for dealing with these (or similar) points?

Kind regards,
Thomas

As for "rummaging in one's repertoire for another point of entry", it's only a matter of having good "mental database management"  ;)  i.e. being able to memorize and categorize one's known facts and experience in a convenient way. I suppose that it can be trained. As far as my personal difficulties with solving math problems are concerned, the biggest problem seems to be being able to see five steps ahead - that is, seeing the destination point and achieving it however winding the path may be, even if it takes numerous steps to complete. It requires much imagination and intuition, which can be partially compensated with experience (i.e. having solved hundreds or thousands of similar problems), but not fully.


I need some help with the following problem:
  \[ \left \{ \begin{array}{ll}
  |b\minus{}2a\minus{}ab|\equal{}\sqrt{3(a^2\plus{}b^2)} \\
    |ab|\equal{}20 \end{array} \right.
\]
many thanks to you!
Best, Luan.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194681

You already posted this problem. No need to post it four more times in this forum. :maybe:


modular arithmetic
combinatorics unsolved
combinatorics

find the number of all 5-digit numbers ( in base 10) each of which contains the block 15 and is divisible by 15.
 ( for example 31545, 34155)
give solution with sufficient reasoning

The last digit ends in either 0 or 5, so we have several cases to consider (some overlap).
1) 1 5 a b 0 : $ 1\plus{}5\plus{}a\plus{}b\plus{}0\equiv\overline{ab}\equiv0\pmod3$; $ [00,99]$ contains $ 34$ multiples of $ 3$.
2) a 1 5 b 0 : $ a\plus{}1\plus{}5\plus{}b\plus{}0\equiv\overline{ab}\equiv0\pmod3$; $ [10,99]$ contains $ 30$ multiples of $ 3$.
3) a b 1 5 0 : $ a\plus{}b\plus{}1\plus{}5\plus{}0\equiv\overline{ab}\equiv0\pmod3$; $ [10,99]$ contains $ 30$ multiples of $ 3$.
1+3) 1 5 1 5 0 : $ 1$ number here.
4) a b c 1 5 : $ a\plus{}b\plus{}c\plus{}1\plus{}5\equiv\overline{abc}\equiv0\pmod3$; $ [100,999]$ contains $ 300$ multiples of $ 3$.
5) 1 5 a b 5 : $ 1\plus{}5\plus{}a\plus{}b\plus{}5\equiv\overline{ab}\minus{}1\equiv0\pmod3$; $ [\minus{}01,98]$ contains $ 33$ multiples of $ 3$.
6) a 1 5 b 5 : $ a\plus{}1\plus{}5\plus{}b\plus{}5\equiv\overline{ab}\minus{}1\equiv0\pmod3$; $ [09,98]$ contains $ 30$ multiples of $ 3$.
7) a b 1 5 5 : $ a\plus{}b\plus{}1\plus{}5\plus{}5\equiv\overline{ab}\minus{}1\equiv0\pmod3$; $ [09,98]$ contains $ 30$ multiples of $ 3$.
4+5) 1 5 a 1 5 : $ a\equal{}0,3,6,9$ so $ 4$ numbers here.
4+6) a 1 5 1 5 : $ a\equal{}3,6,9$ so $ 3$ numbers here.
5+7) 1 5 1 5 5 : this isn't divisible by $ 3$.

Hence our answer is $ (34\plus{}30\plus{}30\minus{}1)\plus{}(300\plus{}33\plus{}30\plus{}30\minus{}4\minus{}3\minus{}0)\equal{}479$.


Hello,

I hate to admit it, but I've been stumped for too long on this. I suspect this should be easy, but I'm stuck.

Given 12 teams and 6 weeks, I need to create a schedule that matches up pairs of teams each week such that each team plays the other teams exactly once. This has to be easy! I tried doing this manually in Excel, and even writing some SQL code but I seem to be getting nowhere.

Any suggestions?

In order for each team to play every other team, each team must play 11 games.  Are you trying to have each team play only one game per week?

After a lot of wasted time I finally stumbled accross a valuable phrase to search for: round robin. DOH!

Yes, each team plays one match per week. I found a site that actually generates a schedule for you (kind of random looking), and I found this description:

Make a column, from 1 to the number of teams
Make a second column, from the number of teams down to 1.
Put the two side by side -- that's the first week.

For the second week, rotate the column up (i.e. move everything up, and move the 12 to the bottom)
For each succeeding week, do another rotation.

When you get to week 11, you're done, so go back to week 1.


function
integration
real analysis
limit
real analysis unsolved

Let $f_n$ be the following sequence of functions defined on $[0,\infty):$

$f_n$ is the piecewise linear continuous function with vertices at $(0,0),\,\left(\frac1n,n\right),$ and $\left(\frac2n,0\right)$ and $f_n(x)=0$ for $x\ge \frac2n.$

Let $g(x)=\sup_n f_n(x).$

Is $\int_0^{\infty}g(x)\,dx$ finite or infinite?

[The course in which I've asked it as a test question covers Lebesgue measure and integration.]

That is correct. You have essentially made a comparison of $g(x)$ to some constant multiple of $\frac1x.$

There is, however, another argument that doesn't require such an explicit estimate of the size of $g(x).$

That is clever.

Here's the argument I had in mind:

Note that $g$ dominates the sequence $f_n$. If $g$ were integrable, the hypotheses of the Dominated Convergence Theorem would hold, and thus we would have that

$\lim_{n\to\infty}\int_0^{\infty}f_n(x)\,dx= \int_0^{\infty}\lim_{n\to\infty}f_n(x)\,dx$

But $\lim_{n\to\infty}\int_0^{\infty}f_n(x)\,dx=1$

and $\int_0^{\infty}\lim_{n\to\infty}f_n(x)\,dx=\int_0^{\infty}0\,dx=0.$

So the conclusion of the DCT does not hold, so the hypothesis that $g$ is integrable must then be false.