Sorry if this got posted before... im new....

Let  A and B  be two irrational numbers such that 1/A + 1/B  = 1

Prove that the two sequences flr(nA) and flr(nB) are disjoint and their union covers all natual numbers. flr = floor value = integer part of the number. eg
flr(:sqrt:2) = 1,
flr(2.999) = 2

It was already posted. Latest post with solution is [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Beatty&t=22782]here[/url]

How about; prove the sequence:

flr(n + sqrt(n) + 1/2)

misses exactly the squares?


geometry
number theory
number theory solved

This can be found in Valentin's book:
For a given n let M={1<=x<=n|(x,n)=(x+1,n)=1}. If n is odd prove Product {x is in M} x = 1(mod n).

Yeah, it's in Romanian. I wonder when somebody will translate it to English... It's a really good book, you know.

I will translate it into English, of course :)

but I'm working on improving it (especially the grammar errors :D :D :D )

it's scheduled to January 2004 (if nothing else comes up :D).

I want to read Valentin's book.

Suppose (x,n)=(x+1,n)=1. Let y be number from {1,2,3,...,n} such that xy=1(mod n). It is clear y is unique, x<>y (because n is odd ==> x<>1) and (y,n)=1. Moreover x(y+1)=x+1(mod n) ==> (y+1,n)=1.

That's all.

I think your book will attract even more readers when it will be translated into English.

At the following web site by Hoojoo Lee you may find further topics for your book:

http://myhome.personaldb.net/ideahitme/

geometry (no syllabus):

http://myhome.personaldb.net/ideahitme/imogeometry.html

number theory:

http://myhome.personaldb.net/ideahitme/imonumbertheory.html
http://myhome.personaldb.net/ideahitme/syllabusnt.html#synumbertheory

algebra:

http://myhome.personaldb.net/ideahitme/imoalgebra.html
http://myhome.personaldb.net/ideahitme/syllabusag.html

Combinatorics:

http://myhome.personaldb.net/ideahitme/imocombinatorics.html
http://myhome.personaldb.net/ideahitme/syllabusct.html

Hey, Myth, that's the same as my proof. Pretty short and neat, huh? :D


A number is called perfect if the sum of its divisors, except itself, is equal to the original number. What is the sum of the two perfect numbers between 2 and 30?

The two perfect numbers are 6 and 28, so the sum is 34.

Just to add on:

The first few perfect numbers are 6, 28, 496, and 8128.

You can find some formulas on the Wikipedia article.


[quote="fleeting_guest"]Is there a way to search one/all of the blogs for some piece of text?[/quote]At this point no, but it is planned.


limit
logarithms
real analysis
real analysis unsolved

Hi,
Proove that : $\displaystyle\lim_{x\to 0^{+}}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$

Thanks.

The geometric mean is the 0-power mean. This is a standard result.

Take logarithms, then expand $a^x=e^{x\ln a}$ and $b^x=e^{x\ln b}$ as power series.

Try to read post #10 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30544]this thread[/url].

Thanks, I have just solving it. But with another way:

I have replaced $\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with $e^{\frac{1}{x}\ln\left(\frac{a^x+b^x}{2}\right)}$

And ${\frac{1}{x}\ln\left(\frac{a^x+b^x}{2}\right)}$ is the $f'(0)$ with : $f: x\rightarrow {\ln\left(\frac{a^x+b^x}{2}\right)}$.

Thats all!

Thanks again!


MIT
college
Stanford
Harvard
summer program
PROMYS
AMC

Has anyone who is just a well rounded student (didn't win any states awards or invented anything) get into really good collgeges, say MIT or Stanford.

you know, colleges look at more than just your awards and scores  :wink:

With sufficient evidence of said well-roundedness and general academic talent, plus some good essays, certainly.  Speaking from experience, there are some kids at Stanford with those standout achievements (science fair winners, IMO medalists, etc), but most people seem to meet your description of yourself (I'm assuming it's yourself).

Let me repeat the part about good essays.  That helps a lot, or so I'm told.

[quote="Sly Si"]

Let me repeat the part about good essays.  That helps a lot, or so I'm told.[/quote]

Yes, yes it does. There are going to be plenty of people out there with the same stats as you (this applies to almost everyone) that will be applying to the same college. Writing a good essay(s) will help the college determine who you are as a person and then can "distinguish" you from all the other people with the same stats as you, giving you the edge...

Yes, I have a friend that went to Harvard this year, and one that went to MIT last year; both of them were really hard workers and really intelligent (the MIT guy was a PROMYS kid), but neither of them had made USAMO or anything. They just have drive.

There are all kinds of people here at MIT, from what I can tell.  You have your hardcore scholars, you have your partiers, and everything in between.  The people I know best so far are those on the cross country team.  They seem to be the well rounded type, in general.

However, I think it's reasonable to assume that there are a lot more well-rounded students applying than people who are prodigies in one area.  So if you're in the well-rounded field, you'll have to cook up a stellar application.

I was fairly well-rounded, but I also had math accomplishments so I stressed those on my application.

wow u must be really smart. if i may ask what did you get on the AIME?  What clubs did you join during high school? also Do colleges look at your accomplishments in your senior year?

thanks :)

[quote="kimby_102"]Do u guys think that a summer program such as PROMYS plays a significant role ??[/quote]

YES!

In fact, it also shows that you have a passion for one of your ECs (in this case, math). A lot of people join thousands of clubs and stuff just so they can boost up their ECs, but colleges like it if you demonstrate an interest (a "passion") for one or two particular things, so yea it looks very good.

In 11th grade I got an 11 on the AIME.  The next year I had a pretty bad day (I tried to get away with minimal answer checking...not a good idea) and I got a 7.

If you go to PROMYS or any other math camp, I really hope you're going because you want to and not because you think you have to.  Otherwise that would be a lot of math/mathy people for you to endure.  Doing camps over the summer is always better than sitting at home the entire summer.  If possible, I would try to find a way to do summer research or be a summer research intern.  You'll usually pick up a little bit of cash and you can figure out just what it's like to do research.  It also gives you a glimpse into that field.

Most people at MIT are math, science, or computer nerds of some type (probably all actually; nerd means they are really good at it or they are really interested in it and, in most cases, both).  So  doing nerdy math/science/computer things will usually appeal to the admissions officers.

By the way, helping put content into the AoPSWiki also looks good on college transcripts (it also counts as community service!)  :wink:

Contrary to popular belief, being a good *student* is still the most effective measure for getting into college, where a good student is defined as one who gets good grades in challenging classes and has good standardized test scores.  Of course, to get into a *very* good school, one has to be a *very* good student.

[quote="joml88"]
By the way, helping put content into the AoPSWiki also looks good on college transcripts (it also counts as community service!)  :wink:[/quote]

How can that be recorded though?

I'm asking since I'm considering putting content on to help out, and I figure if I'm goign to do it anyways, I might as well find out ahead of time how it's recorded.

I think you can do it just like regular community service where you record the amount of time yourself.

If you feel like you need some 'proof' that you did it you can ask one of the admins to sign or give a statement of approval concerning your community service claim.  I can't guarantee that they will do this for you but if you ask nicely they'll probably do it for you.  Also, this is probably common sense, but only ask them to sign if you've done a reasonable amount of work.  I can't really give a hard number but if you've corrected 2 typos in the wiki, don't ask (and hopefully you aren't filling your college app with that).  I'd be willing to sign too but if you feel that I am too young (I'm a college freshman) I understand.


One day Max says to Liz, ``Out of the 25 people taking either English or French, you and I are the only two taking both.'' Liz, being mathematically inclined, responds by pointing out that there are exactly twice as many people in the English class as there are in the French class. How many people are taking English but not French?

Because Max and Liz are taking two classes, there are 27 kids in French and English combined.

Since there are 18 kids in English, the answer is 18-2 or $ \boxed{16}$.


calculus
integration
trigonometry
calculus computations

[b]1)[/b] Let $ I_n\equal{}\int_{0}^{1} x^n\sqrt{1\minus{}x}dx$
Prove that
$ I_{n\plus{}1}\equal{}\frac{2n\plus{}2}{2n\plus{}5}I_n$
[b]2)[/b] Let $ f: R\to [0,1]$ be a continuous function. Prove that
$ \int_0^{\pi} xf(sin x) dx \equal{} \frac{\pi}{2}\int_0^{\pi} f(sin x) dx$

1)
$ I_{n \plus{} 1} \equal{} \int_0^1 x^{n \plus{} 1} \sqrt {1 \minus{} x} \, \mbox{d}x$
By integration by parts we get
$ u \equal{} x^{n \plus{} 1} , \quad dv \equal{} \sqrt {1 \minus{} x} dx$
$ I_{n \plus{} 1} \equal{} \left[ x^{n \plus{} 1} \cdot \left( \minus{} \frac {2}{3} \right) (1 \minus{} x)^{3/2} \right]_0^1 \plus{} \frac {2 (n \plus{} 1)}{3} \int_0^1 x^n (1 \minus{} x) \sqrt {1 \minus{} x} \, \mbox{d}x \equal{}$
$ \equal{} \frac {2(n \plus{} 1)}{3} \left( I_n \minus{} I_{n \plus{} 1} \right)$
Now solve eq. for $ I_{n \plus{} 1}$ and you'll get the answer.

For the Second

$ \int_0^{\pi } x f(\sin (x)) \, dx\equal{}\underbrace{\int_0^{\pi } (\pi \minus{}x) f(\sin (x)) \, dx}_{x\to \pi \minus{}x}$

$ \int_0^{\pi } x f(\sin (x)) \, dx\equal{}\int_0^{\pi } \pi  f(\sin (x)) \, dx\minus{}\int_0^{\pi } x f(\sin (x)) \, dx$

$ \int_0^{\pi } x f(\sin (x)) \, dx\equal{}\frac{1}{2} \pi  \int_0^{\pi } f(\sin (x)) \, dx$


Each point in the hexogonal lattice is one unit from its nearest neighbor. How many circles of radius one contain at least two points of the lattice?
\begin{tabular}{c}
$ \bullet \hskip .2in \bullet$\\
$ \bullet \hskip .2in \bullet \hskip .2in \bullet$\\
$ \bullet \hskip .2in \bullet$
\end{tabular}

\begin{tabular}{c}
\bullet \hskip .2in \bullet\\
\bullet \hskip .2in \bullet \hskip .2in \bullet\\
\bullet \hskip .2in \bullet
\end{tabular}

More than 6 months later I have learned $ \text{\LaTeX}$ yay! I still am having trouble with the problem though.

We have $7$ circles from each of the lattice points.

Add $6$ more since we can connect 2 of the points with a center that is outside of the hexagonal lattice.

$\boxed{13}$


A larger box is $ 25\%$ longer, $ 20\%$ wider and $ 33\frac{1}{3}\%$ taller than a smaller box. The volume of the larger box is what percent of the volume of the smaller box?

Another way is to use an easy value, such as 10.

If it was 25% longer, than our large box has a length of 12.5.

It's 20% wider, so 12.

It's $ 33\frac{1}{2}$% taller, so $ \frac{100}{3}$

$ 12.5 \times 12 \times \frac{40}{3} \equal{} 2000 \div 1000 \equal{} 2$ so $ \boxed{200\%}$

Answer would be 200% Let me know if it helped!

Hi!

Please allow me to redirect you to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=503&t=234229]this[/url] sticky where the criteria for a proper solution was discussed before the time of its locking.
Simply posting an answer is considered spam by a large chunk of the AoPS community.

I no, Im sorry i just learned that yesterday so im trying to provide a solution now to.


Do you guys think a score of 213 is sufficient in NSEC? also what about 110 in NSEP. I am actually in class 10 so i hope it is ok. please also post the highest marks in nsec you know. :maybe:  :(

Are you a alien??????????
How could a Xthie score 213 in chem :o
Even the best of XIthies and XIIthies can't score above 200
or 110 in physics
You are definitely selected in NSEC and at a rate like this you'll clear INCHO as well :omighty:


Can anyone here solve,
$ x^{4}+x^{2}+x+1=0$ for complex numbers.

Im not saying it has a "good" solution, its something I ran into when trying to find the poles of a complex function in another forum.

${{x}^{4}}+{{x}^{2}}+x+1=({{x}^{2}}+\alpha x+\beta )({{x}^{2}}-\alpha x+\gamma )\Rightarrow $
\[\begin{cases} \beta +\gamma -{{\alpha }^{2}}=1\\ \alpha \gamma -\alpha \beta =1 \\\beta \gamma =1\end{cases}\]
\[\Rightarrow \begin{cases}\gamma +\beta ={{\alpha }^{2}}+1\\ \gamma -\beta =\frac{1}{\alpha }\end{cases}\Rightarrow \begin{cases} \gamma =\frac{{{\alpha }^{2}}+1+\frac{1}{\alpha }}{2}\\ \beta =\frac{{{\alpha }^{2}}+1-\frac{1}{\alpha }}{2}\end{cases} \]
 Substituting this values in the relation $ \beta \gamma =1$ gives
${{({{\alpha }^{2}}+1)}^{2}}-\frac{1}{{{\alpha }^{2}}}=4$ . Now Assume that ${{\alpha }^{2}}=u$. Then we have
${{u}^{2}}+2u+1-\frac{1}{u}=4$, that is ${{u}^{3}}+2{{u}^{2}}-3u-1=0$. Finally solve a cubic equation.


Congrats to all who made it! And good luck to those who missed out as well :)
Please post your expected marks and results as suggested by mustafa :)

Me-
Forgotten all marks! Will post when I remember :P
NSEA- IN
NSEP- OUT
NSEC/B- Did not give

Me-
NSEC- IN  (May be 184)
NSEP- IN   (May be 142)
NSEA/B- Not given  :(

btw Gaurav's marks - NSEA ([url=http://www.mathlinks.ro/viewtopic.php?p=1694698#1694698]may b 88[/url])

Congrats savy and Mihir
Also, thanks for telling me my marks  :rotfl:

NSEP-not selected(marks~110)
NSEC-not selected(marks~165) :mad:
NSEA-selected(marks=93) :lol:
Congratulations to all AOPSians and mathlinkers selected
To those not selected(including me for NSEC especally :rotfl: ) - Don't lose heart and all the best for future exams.

[quote="diptarka"]Chem:in: Score 213
Physics:out Score:110
help me for InChO as I am in Class X
Happy New year in advance to all.[/quote]
Same to you diptarka.
And PLEASE! You don't need to remind us that you are a tenth standard genius for anybody to help :P

Chem: In: 169 (sorry abhitayal)
Phy: In: 153 (objective)
NSEA/B: not given

@Asish
congrats on selection in both NSEP/C
and a really great score in NSEP :omighty:

Want to kill myself for my luck
missed the cutoff by inches :wallbash_red:

@Ashish
Congrats!!
Really really really excellent score in physics!!

@Abhitiyal
OH!!
That's really bad luck :(

@Diptarka
Same to you- Happy new year

hi

NSEA(sr) ; OUT
Marks; if i tell then i wud be disqualified frm the forum :(

if u still wanna know then........

[hide]either u have to be mod and promise me not to disqualify me[/hide]

[hide]or look below[/hide]
[hide]further[/hide]

[hide] still interested........[/hide]

[hide] i said its low [/hide]

[hide] ok its below 60[/hide]

[hide] i give it up [/hide]

[hide] it IS [/hide]

[hide] 52 [/hide] :(

Lol I have 70 odd in 12th, so its cool! :P

physics out 105
chemistry out 155
astronomy in 91

hi guys,
chemistry - IN (~173)
physics - IN (~133 objective 0 subjective)
astro - IN (~105)  :)
think jst made the cut in NSEP & C  :lol:
do we get to know our exact marks and cutoffs or ranks???

First and foremost congrats to all those who got selected in various olympiads. Do well in the next round and heres wishing you all the very best.

[quote="abhitayal"]:wow: Did anyone noyice that Sahal Kaushik has been selected in NSEP/C/B/A ie all 4 and he cleared RMO as well
This young boy is doing wonders
I don't think anybody till date has been selected in all 5 in the same year.

[hide]Hope you know who is Sahal Kaushik :rotfl:
If not see this
http://timesofindia.indiatimes.com/city/delhi/Home-tutored-Dwarka-boy-a-whiz/articleshow/4548895.cms[/hide][/quote]
There was one person my batch mate(on AOPS ofc :P) pradeep george mathias who got selected in 5 which does not include NSEB(bcs he was not the bio stud) but got thro' in Informatics (ZIO and IOI and eventually rep india). So def not the first. But great talent nonetheless.

@abhitayal
how is it possible to get selected in NSEB and NSEA both in same year?? :?:
they are at same time and one has to select any one of them??

pl rectify if i'm not correct!!

Well sound impossible but he did it(or two guys with same name and same roll no got selected from the same centre which is absurd). Perhaps he wrote astro in 1st 1hr and bio in the next 1hr . Or did both together. I don't know, but geniuses like him can do anything.

Lol maybe he could write with both hands and use half a brrain for each paper like the guy in 3 idiots  :rotfl:

some good news from my side...I got selected too
But not in Phy, Chem or Astro....but in Biology
Looks like no one in Bio here

List of INO centres, Schedule for INOs and procedural rules are put up at [url=http://olympiads.hbcse.tifr.res.in/announcements]HBSCE announcements site[/url]
Centre allotment to be put on 11th  :D

Thanks! I'll have to travel to Kanpur it seems. Weird that they didn't keep it in the state capital. We had it in the science city every year till now.

[url=http://olympiads.hbcse.tifr.res.in/announcements/inocentres]INO Centre allotment[/url] put up at HBCSE. {lol it's 2 days before the scheduled date :P (Monday)}.

BTW how're your preparations going on?  :huh:

@nuclear
science city = [hide="? ?"][hide="Lucknow?"]why so ? ?[/hide][/hide]

Hello guys,

I've received my INPhO selection letter cum Hall ticket yesterday.
Haven't yet received INChO Hall-ticket.

Anyone else?

Was selected only for INAO
Have not received any letter

I am also worried haven't got my incho hall ticket yet.
i hate the fact that a snail can beat the indian postal service in a race.

Please post on the INO 2010 forum regarding this. I mean NSEs are history by now and so we should let the dead past bury its dead and move to the present INO's.
p.s:any advice regarding INCHO . I am doing past papers but i am sinking into panic mode a bit too fast and ending up ruining all of it.Plus i change answers a bit too fast and the small box plus banning pencil written answers gets on my nerves.i feel the time too is limited.
do you think i should invest in a pen eraser?


[quote="harazi"]Please, do not repet more than 3 times problems already posted. I think this one was posted at least 4 times.[/quote]
sorry;but i don't know this one was posted

No problem, you are new, this was intended for the old ones who post the same problems several times.


If a,b are natural numbers such that
a/b=1-1/2+1/3-1/4+...-1/1318+1/1319, prove that 1979 divides a.


trigonometry
analytic geometry
graphing lines
slope
function
calculus
derivative

Let $ f(x)\equal{}sin(x)\minus{} kx$
a) Find the positive values of k so that f has only one strictly positive root. $ 0.128373<k<1$
b) Find the positive values of k so that f has three strictly positive roots. $ 0.070913<k<0.128373$

Can someone show me the idea to solve this problem?

Draw the picture. Draw the graph of $ y\equal{}\sin x$ and draw lines of various slopes through the origin to see where the intersections must lie. This helps in understanding what is going on.

There is a borderline case between the equation having one positive root and three positive roots in which it has two positive roots, the second one a double root.

For $ f(x)\equal{}0$ to have a double root, we must have $ f(x)\equal{}0$ and $ f'(x)\equal{}0$ simultaneously.

$ f'(x)\equal{}0$ gives $ \cos x\minus{}k\equal{}0$ or $ k\equal{}\cos x.$ Put this into $ f(x)\equal{}0$ to get the equation
\[ \sin x\minus{}x\cos x\equal{}0.\]
This can be rewritten as $ x\equal{}\tan x.$ In that form, we can see that it has a solution that's only slightly less than $ \frac{3\pi}2,$ and a solution that's slightly less than $ \frac{5\pi}{2},$ and so on. The solution near $ \frac{3\pi}2$ lies in the wrong quadrant - it would have a negative $ k.$ So be move on to the next quadrant. Using Newton's method with $ \frac{5\pi}2$ as a starting point we get $ x\approx 7.725251837.$ But then we had that $ k\equal{}\cos x$ for this $ x,$ so for this borderline case, we get $ k\approx \cos(7.725251837)\approx 0.128374554.$

The solution just before $ \frac{9\pi}2$ would give us the place where $ f(x)\equal{}0$ has exactly 4 roots, the last one a double root. For this we get $ x\approx 14.06619391$ and $ k\approx \cos(14.06619391)\approx 0.070913459.$

These numbers are approximations, obtained by numerical techniques. That much seems necessary.

I didn't understand this double root. Could you explain it better?

Just look at this picture:
[asy]size(300);

path P;
for(real h=0;h<16.5;h+=0.1) P=P--(h,2*sin(h));
DPA((0,0)--(16.5,0)^^(0,-3.5)--(0,3.5), black);
D(P,blue+linewidth(1));
path Q=D((0,0)--(9.5,2.43),green+linewidth(1));
D("A",IP(P,Q,2),N);
path R=D((0,0)--(16.5,2.33),magenta+linewidth(1));
D("B",IP(P,R,4),N);
D((12.5,2.2)--(0,0));[/asy]
The lines you are interested in for the question b), say, have the slopes between that of the green line and that of the magenta line. Note that both the green and the magenta line are tangent to the graph of $ y\equal{}\sin x$. The tangency points correspond to the double roots of the equation (both the function $ f(x)\equal{}\sin x\minus{}kx$ and its derivative $ f'(x)\equal{}\cos x\minus{}k$ vanish at those points)

The methods described will also find points $ C,D,E,\dots$ where further points of tangency occur, and hence ranges of $ k$ which yield 5 roots, 7, roots, 9 roots, ... But all of the computations will be numerical.


quadratics
number theory unsolved
number theory

Prove that the equation: $ x^2\plus{}3xy\minus{}2y^2\equal{}122$ doesn't have solutions in $ \mathbb{Z}^2.$

use quadratic reciprocity:
$ (2x\plus{}3y)^2\minus{}17y^2\equal{}488 \implies \left(\frac{12}{17}\right)\equal{}1$.
one can easily prove that we have $ \left(\frac{12}{17}\right)\equal{}\minus{}1$.
and we are done.

[quote="imedeen"]could you plain why $ \left(\frac {12}{17}\right) \equal{} 1$  :blush:
i don;t understand the your $ (...)$ symbol[/quote]
[url]http://en.wikipedia.org/wiki/Legendre_symbol[/url]


inequalities
trigonometry
algebra solved
algebra

Let $z\in \mathbb {C} \backslash \mathbb{R}$ such that there is $a\in \mathbb{R}$ and $n\in \mathbb{N}$ such that $z^n+nz=a$. Prove that $|z|=1$

I suppose  the correct sentence is $|Z| > 1$  :P   Note $Z = |Z|*(cos(t) + i*sin(t)) $
The basic idea is that $|sin(n*t)| < n*|sin(t)|$, if  n(natural) > 1 and t is not multiple of $\pi  $ .....
:cool:

Yes, you are right.
It was a simple problem after all, it was just me who couldn't see it right.

I believe the problem isn't so simple.
[tex] z^{n} = |z|^{n} ( \cos (n \alpha ) +i \sin (n \alpha )) [/tex]  and  [tex]nz=n|z| ( \cos \alpha + i \sin \alpha )[/tex]
[tex] a \in \mathbb{R} [/tex] so [tex]i ( |z|^{n} \sin (n \alpha ) + |z|n \sin \alpha ) = 0[/tex]
[tex]|z|^{n-1}=-\frac{n \sin \alpha}{ \sin n \alpha}[/tex]
I haven't finished but I believe the problem is realy prove that |z| = 1

Yes. But from here write $|z|^{n-1}=\frac {n|\sin\alpha|}{|\sin n\alpha|}\geq\frac {n|\sin\alpha|}{n|\sin \alpha|}=1$, so $|z|\geq 1$.  Look at the inequality in Diogene's post.


is there an infinite sigma algebra with countably many members? i have spent 3-4 days over this but nothing yet

Show that an infinite sigma algebra contains an infinite sequence of pairwise disjoint sets.

A nice argument for this:

Let the nonempty elements $\sigma$-algebra $S$ be partially ordered by inclusion.

Case 1: Every chain has finite length. Then for all $A\in S$, there is some $B\subseteq A$ in $S$ such that $B$ is a minimal element with respect to this partial order. It is now easy to show that there are infinitely many of these minimal elements, since every element of the $\sigma$-algebra is a union of minimal elements..
Also, any two distinct minimal elements are easily shown to be disjoint.

Case 2: There is an infinite chain $A_1\supset A_2\supset A_3\supset\dots$. Then the relative complements $A_i\setminus A_{i+1}$ are disjoint nonempty elements of $S$.

The set of countable subsets of an infinite set is uncountable, and we are done.

Suppose $S$ is a countably infinite $\sigma$-algebra in $X$. For each $x\in X$ denote by $A(x)$ the intersection of all elements of $S$ containing $x$. Then $A(x)$ is an element of $S$, because it is the intersection of at most countably many elements in $S$. Also, $F = \{A(x): x \in X\}$ is a partition of $X$ (One can prove that if $z\in A(x)$ then $A(x)=A(z)$). $F$ is infinite (if it is finite, then each element of $S$ is a disjoint union of elements of $F$ and $S$ become finite, contradiction). So $\mathcal P(F)$, the power set of $F$, is uncountable and there is an obvious injection of $\mathcal P(F)$ into $S$. Hence the cardinality of $S$ is uncountable, a contradiction.


logarithms
limit
function
number theory
prime numbers

okay, note that this is problem has extremely horrible wording.

anyway, prove that in the (infinite, i guess?) set of natural numbers, there are more even numbers than primes.
it's kinda obvious to intuitively prove, but how would you rigorously do it?

is it even possible to show how one thing is bigger than another when they're both infinite, or is this a stupid question?

The answer to this question depends on your definition of "bigger."  If your definition is based on [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url], there are [b]exactly[/b] as many even numbers as prime numbers.

A definition more appropriate to the intuition you're looking at is [url=http://en.wikipedia.org/wiki/Natural_density]natural density[/url].  The even numbers have natural density $ \frac{1}{2}$ while the prime numbers have natural density $ 0$.

Is the natural density of prime numbers a consequence of calculating the limit of the approximate amount of prime numbers divided by the amount of natural numbers?

What I mean is, the approximate amount of prime numbers smaller than or equal to $ x$ is $ \dfrac{x}{\ln x}$. So the natural density of prime numbers is equivalent to calculating $ \lim_{x\to \infty} \dfrac{x}{\ln x} \cdot \dfrac{1}{x} \equal{}\lim_{x\to \infty} \dfrac{1}{\ln x} \equal{} 0$?

Yes, but you don't need to prove the PNT to show that the prime numbers have asymptotic density zero.

but if there are infinite prime numbers...how is their density still 0?

The idea is that if you take a very large interval, the proportion of prime numbers in that interval steadily decreases to $ 0$ as you make the interval larger.  See, for example, the table of values of the prime-counting function [url=http://en.wikipedia.org/wiki/Prime-counting_function#Table_of_.CF.80.28x.29.2C_x_.2F_ln_x.2C_and_li.28x.29]here[/url].

In other words, for small values the density of primes is very high.  However, the more digits there are in the numbers you're looking at, the less likely you are to find a prime.  Natural density is an asymptotic notion.


There isn't a score cut off.
If you're in a really beast chapter,
sometimes a [b]45[/b] won't even make it(for example santa clara valley)
If you're in a really fail chapter,
you should be able to make states with a 20(no example as it's derogatory.)
The cut-off is usually around 40(not entirely sure)
It all depends on which chapter you're in.

[quote="AceOfDiamonds"]If you're in a really beast chapter,
sometimes a [b]45[/b] won't even make it(for example santa clara valley)[/quote]

Travis got a 45 in SCV and didn't make it.

To the OP: Ask your coach - it depends from region to region.  I know of a student who scored 18 at the Chapter round, and qualified for States.

[quote="AceOfDiamonds"]There isn't a score cut off.
If you're in a really beast chapter,
sometimes a [b]45[/b] won't even make it(for example santa clara valley)
If you're in a really fail chapter,
you should be able to make states with a 20(no example as it's derogatory.)
.[/quote]

Actually in my chapter, you only need 17/46.  Yes, my chapter STINKS.  But my state, well, that's another story.................. :(

But unless you're unlucky enough to live in CA or TX, you should probably be fine with around a 35.

I live in Texas and I have the Greater Houston Chapter. I got a 44/46, but only 7th place.

If you live in Texas, making nationals is harder than getting to countdown in nationals  :) .

Well, I live in Oakland County in Michigan, and its a pretty beast chapter. At Chapters, someone got a 40/46 and didn't make states, soo...yeah.

I took a practice and got just about a 35

[quote="basketball9"]I am new to mathcounts so I wolud like to know around what score is the cutoff to make states. I know it is top team and top 2 non team members[/quote]

It varies a lot depending on chapter and the level of the test

The top team and 2 nonteam is the [i]minumum[/i]

For example my chapter this year the cutoff was 36.

One chapter there was only 1 school.

And the top 10 in my chapter individual and top 4 teams!!

(95 people)

In my chapter, a 27 would get you 6th. The thing is, the top two teams and eight individuals make it to state. So you could make it to state if you have a really strong team as well.

at my school, we don't really have a chapter... like only 5 ppl participate. :o


calculus
integration
trigonometry
function
complex analysis

Use the following sequence of transformations:

$x+1=y$ and $y=2t$ and your integral is equal to

$\int_{0}^{1}\frac{dt}{\sqrt[3]{(1-t)t^{2}}}=\beta(\frac{1}{3},\frac{2}{3})=\frac{\Gamma(1/3) \Gamma(2/3)}{\Gamma(1)}=\frac{\pi}{\sin \frac{\pi}{3}}=\frac{2\pi}{\sqrt{3}}$.

Interesting solution (I've never seen the Beta function before), I was thinking something more along the lines of complex integrations.  Anyway, how did you know that $\Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right) = \frac{\pi}{\sin \frac{\pi}{3}}$?

More generally, the following equality holds:

$\int_{-1}^{1}\left(2-2x\right)^{\alpha}\left(2+2x\right)^{\beta}\left(1-x^{2}\right)^{-1/2}dx=\frac{\Gamma(\alpha+\frac{1}{2})\Gamma(\beta+\frac{1}{2})}{\Gamma(\alpha+\beta+1)}2^{2\alpha+2\beta}$, valid for $\min{Re(\alpha), Re(\beta)}>-1$.


My number of blog views has not increased for quite a while. ilovepink told me hers isn't increasing either. Is something going on?

I viewed your blog and the view count went up. Looks like it's working. Note that the counter won't go up if you were the last person to view your blog.

I seem to be having a similar-ish problem with my view counter. I just did around 20 minutes of categorizing entries in my blog, which means lots of clicking, etc., but I noticed the view counter barely increased. I experimentally refreshed and nothing happened. When I did it again, it jumped up by 3.

EDIT: Appears to be working normally now.

[quote="levans"]Note that the counter won't go up if you were the last person to view your blog.[/quote]

This is to prevent people from autoclicking to get blog views.

Which in turn keeps the site from crashing. So be glad this feature is in place.

Yes, it can. A friend of mine, username tinytim, did so and got banned as a result. You can search the memberlist, he no longer exists. However, I believe the admins fixed this so now the site won't crash.


Is it possible to draw a hexagon with vertices in the knots of an integer lattice so that the squares of the lengths of the sides are six consecutive positive integers?

No. As the vertices are knots of an integer lattice, by Pythagoras the squares of the lengths of the sides are sums of two integer squares. If those lengths are six consecutive integers, at least one is divisible by 3, but not by 9, hence cannot by the sum of two integer squares.


probability
calculus
real analysis
real analysis unsolved

[quote="nmtruong1986"]Calculate the following expression:
\[ f(r,p)=\sum_{k=r}^{+\infty}C^{r-1}_{k-1}p^r(1-p)^{k-r}  \]
Where $r \in N, 0<p<1$ and $C^m_n=\frac{n!}{m!(n-m)!}$

Deduce the sum $\sum_{k=r}^\infty C^{r-1}_{k-1} \frac{1}{2^k}$[/quote]
The terms of the sum come from the [url=http://mathworld.wolfram.com/NegativeBinomialDistribution.html]negative binomial distribution[/url] (r-th success occurs in the k-th trial, if the success probability is $p$). Hence all these sums are 1.

Of course Negative binary distribution implies the result. But I look forward another solution, more calculus (analysis)


geometry
circumcircle
trigonometry
power of a point

In a triangle $ABC$, $M$ is the midpoint of $BC$. Given that $\angle MAC=15^{\circ}$, find the maximum possible value of $\angle ABC$.

We draw the circumcircle of $\Delta ACM$, with center $O$
Every point $X$ on this circle is such that $\angle CXM=15$°

We see that the maximum value of $\angle B$ is obtained when $BX$ touches the circle.
So we draw the touching line $BX$ with $X$ a point on the circle.

Now, power of a point shows that $BX=\sqrt{2} BM$

Law of sine: \[\frac{\sqrt{2} BM}{\sin{M}}=\frac{BM}{\sin{BXM}}=\frac{MX}{\sin{B}}\]
And \[\frac{XC}{\sin{M}}=\frac{BM}{\sin{15}}=\frac{MX}{\sin{C}}\]
And \[\frac{2 BM}{\sin{X}}=\frac{\sqrt{2} BM}{\sin{C}}=\frac{CX}{\sin{B}}\]

Maybe this helps, I think it should be solveable this way.  :)


[quote="Phelpedo"]Let $p$ be an odd prime and $P(x)$ a polynomial of degree at most $p-2$.
(a) Prove that if $P$ has integer coefficients, then $P(n)+P(n+1)+...+P(n+p-1)$ is an integer divisible by $p$ for every integer $n$.

(b) If $P(n)+P(n+1)+...+P(n+p-1)$ is an integer divisible by $p$ for every integer $n$, must $P$ have integer coefficients?[/quote]
a) We can mass binomial theorem in modulo p to show that it suffices to prove that P(0)+...+P(p-1) is divisible by p, at which it suffices to show that 1^k+...+p-1^k is divisible by p for 1<=k<=p-1. Multiply through by (primitive root of p) ^k and we know that since primitiveroot^k doesnt equal 1 mod p, then its conclusive that 1^k+..+p-1^k is. And we are done.

b) No. Q(n)+Q(n+1)+...+Q(n+p-1) can be divisible by 2p, if we grant P(n)(n^2+n)=Q(n) where P(n) is any polynomial. Since not all the coefficients are divisible by 2, then Q(n)/2 is a non-integer polynomial such that the desired sum is divisible by p.


:ddr:
Here is the Problem Solving Prompt:

   Find a number between 2,000 and 3,000 that leaves a remainder of 1 when divided by 2, 3, 4,       5, 6, 7, 8, or 9.

We consider the number which is divisible by the numbers (since we could then add $ 1$ to get our number). Just multiplying yields $ 9!\equal{}362880$. This is too large, so we consider divisibility rules. We must multiply by $ 5$ and $ 7$. Since $ 9$ is in our count, we have also satisified divisibility by $ 3$. And multiplying by $ 8$ accounts for $ 2$, $ 4$, and with $ 9$ accounts for $ 3$.
$ 5\times{7}\times{8}\times{9}\plus{}1\equal{}\boxed{2521}$.

[quote="gaussintraining"]We consider the number which is divisible by the numbers (since we could then add $ 1$ to get our number). Just multiplying yields $ 9! \equal{} 362880$. This is too large, so we consider divisibility rules. We must multiply by $ 5$ and $ 7$. Since $ 9$ is in our count, we have also satisified divisibility by $ 3$. And multiplying by $ 8$ accounts for $ 2$, $ 4$, and with $ 9$ accounts for $ 3$.
$ 5\times{7}\times{8}\times{9} \plus{} 1 \equal{} \boxed{2521}$.[/quote]

thank you. that's the same answer i got.


counting
distinguishability
geometry
3D geometry
prism
geometric transformation
rotation

How many distinguishable ways can the faces of a hexagonal prism be painted with 8 different colors ( no color used twice)?

Answer
[hide]3360[/hide]

Are rotations distinguishable? (i.e. does orientation matter?)

EDIT: never mind. the answer is $ \frac {8!}{2\cdot 6} \equal{} 3360$

$ 8!$ ways to color them counting rotations, divided by two to make up for up-down symmetry, and divided by six to make up for the rotational symmetry.

EDIT 2: hey, I recognize this as problem 7 from the intro to c+p post test.

Or you can have $ \binom{8}{2}$ ways to choose the colors on top and bottom, then $ \frac{6!}{6}$ ways to choose the 6 colors with rotational (but not up-down) symmetry, giving the exact same answer.


Do you have any diagrams? Those would be helpful.

The area of a trapezoid is $ \frac{1}{2}h(b_1\plus{}b_2)$...(I think)

So can you set up some simultaneous equations to solve???

Ok, I am assuming that trapezoid ABCD is an isoceles trapezoid. If it isn't, then the problem is quite unapproachable and has many possible solutions (if you were to find them).

Since we know that E and F are the midpoints of sides BC and AD, respectively, we know that isoceles trapezoids ABEF and CDEF have the same height. Since the area of a trapezoid is h(b1 + b2)/2 and the area of isoceles trapezoid ABEF is two times the area of isoceles trapezoid CDEF, we can set up the following equation:

AB + EF = 2(CD + EF), or AB + EF = 2CD + 2EF, which simplifies to AB = 2CD + EF

Now to set up the ratio AB/CD, we need only to write EF in terms of AB or CD. Since we know that E and F are the midpoints of sides BC and AD, respectively, we know that EF = (AB + CD)/2. Substituting (AB + CD)/2 in for EF in the equation AB = 2CD + EF yields:

AB = 2CD + AB/2 + CD/2, which simplifies to AB/2 = 5CD/2

Now we can write the ratio of AB/CD! Knowing that AB = 5CD, we know that AB/CD = 5.

[u]Answer[/u]: 5

Example: AB = 15 while CD = 3. Therefore, EF = 12. Using the trapezoid area formula, we find that ABEF is two times as large as CDEF.


geometry
circumcircle
inequalities
inequalities proposed

For any triangle of sides $a,b,c$ and circumradius $R$:

 $(a+b+c)R \ge \sqrt{ab^3+bc^3+ca^3}$;

[quote="nickolas"]just  take [tex]  x=a , y=b , z=x  [/tex]  to  O.Kooi's  inequality :
[tex]  x,y,z>0 [/tex] , ABC  triangle  then [tex](x+y+z)^2R^2 \geq xya^2+yzb^2+zxc^2[/tex][/quote]

nickolas, do you mean this ineq?
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x+2+x+y+2&t=15558

O Kooi's  inequality  was  posted  by manlio , i  cannot  find  the post now , but i will  keep  looking  :)
If  you  aply  O Kooi's  ineq  to  the  triangle  ACB  and  put  [tex]x=a , y=b , z=c [/tex]  we  have  the initial  inequality

Nickolas is right: The given inequality directly follows from
$\displaystyle (x+y+z)^2R^2 \geq yza^2+zxb^2+xyc^2$.

But what do you say about the following

$27R^4 \ge ab^3+bc^3+ca^3$ ,

$(a+b+c)^2R \ge \sqrt{3}(a^2b+b^2c+c^2a)$ ?

I think $27R^4 \ge a^3c+b^3a+c^3b$ follows from that inequality too by putting z=a, x=b, y=c and $sinA+sinB+sinC \le \frac{3\sqrt{3}}{2}$

You are right, too.
It remains the last inequality. :lol:


Note that

$\sum_{\text{cyc}}a^{2}\geq \sum_{\text{cyc}}\frac{a^{3}(b+c)}{a^{2}+bc}$

$\iff \sum_{\text{cyc}}\frac{a^{2}(a-b)(a-c)}{a^{2}+bc}\geq 0$

Assume $a \geq b \geq c$ and let $f(a) = \frac{a^{2}}{a^{2}+bc}$ and define $f(b), f(c)$ similarly. Then,

$\sum_{\text{cyc}}f(a)(a-b)(a-c) \geq (a-b)(b-c)[f(a)-f(b)]+(c-a)(c-b)f(c) \geq 0$

Now, Cauchy gives

$2\sum_{\text{cyc}}ab \sum_{\text{cyc}}a^{2}$

$\geq \sum_{\text{cyc}}a(b+c) \sum_{\text{cyc}}\frac{a^{3}(b+c)}{a^{2}+bc}$

$\geq \left( \sum_{\text{cyc}}\frac{a^{2}(b+c)}{\sqrt{a^{2}+bc}}\right)^{2}$


Quick question for everybody: I'm working on designing an ARML t-shirt, and I need a bunch of team members' favorite numbers.  So post 'em here and I'll put a selection of them on the shirt.  Preference might go to non-integers.

I'm not going to do ARML this year, but I'll "donate" some numbers too:

pi, e, 17, 42, 1337, 96.5 (KOIT), 23, i, i^i (around 0.2), e^pi, pi^e, sqrt(2)^sqrt(2)

Real numbers, complex numbers, or quaternions?

[quote="Kent Merryfield"]pythag011 - this is not your T-shirt. If you've got suggestions, come over to our side.[/quote]

I suppose that was meant towards Sly Si, not pythag011?

1337.  LOL reminds me of a math team i coached once, where our team t-shirts said
[school name]
MATH1337S

[quote="eddyliu"]69 for sure[/quote]
this

[quote="eddyliu"]69 for sure[/quote]

Because it's the square root of $ 4761$, and certainly for no other reason at all whatsoever.

052495 (my school's collegeboard code)

I'm surprised the golden ratio and its conjugate hasn't been mentioned
1.618..., 0.618...

4/3 because it appears in the formula for volume of a sphere
8/15 because it appears in the formula for volume of a 5D sphere
$ V \equal{} \frac {8}{15} \pi^2 r^5$

1961 for being the most recent year that looks the same upside down.

12345679 for being a factor of 111111111.

9.80665 so our team can accelerate.

$ e^{i\pi}$

[quote="r15s11z55y89w21"]So why do you like $ 69$?[/quote]
So why do you like all the numbers you listed?

You're probably better off not knowing.
Yeah, putting that on the shirt would be pretty unwise. :D

I second -40.

And 1729 because it is the smallest number that can be expressed as the sum of two cubes in two different ways.
http://www.mathpages.com/HOME/kmath028.htm

[quote="DataBox"]I love how SoCal can post on our forum, but we can't post on their forum  :mad:[/quote]

:love:

[quote="r15s11z55y89w21"]So why do you like $ 69$?[/quote]
huh what does that have to do with why 69 is inappropriate
i swear
i ilke 69 purely due to pg reasons

Perhaps, then, $ \sqrt{4761}$ can go on the shirt.  It has all the same arithmetic properties, but none of the inappropriateness that comes from writing it out explicitly in base 10.

[quote="DeltaX"]And 1729 because it is the smallest number that can be expressed as the sum of two cubes in two different ways.
http://www.mathpages.com/HOME/kmath028.htm[/quote]

what about -1729?  :)
of course, i'm pretty sure that isn't the smallest one either..

i think he means smallest positive number

Tim, how about zero, for that matter?  In fact, zero is the only number that's the sum of two cubes in infinitely many ways!

No. 42. But what do I know? I am just an immigrant.

You know what my favorite number is?

!necro.

lol that was really lame o.O


maybe this belongs to classroom math...
But anyways, to the problem

remember: a^2 - b^2 = (a+b)(a-b)
Apply that principle twice to get the answer.

a^4 - 16 = (a^2 + 4)(a^2 - 4) = (a^2 + 4)(a + 2)(a - 2)

Hmm. You can factor $ (a^2\plus{}4)$ to get $ (a\minus{}2i)(a\plus{}2i)(a\minus{}2)(a\plus{}2)$

[quote="hunter34"]Hmm. You can factor $ (a^2 \plus{} 4)$ to get $ (a \minus{} 2i)(a \plus{} 2i)(a \minus{} 2)(a \plus{} 2)$[/quote]
maybe he means over the reals?

Most likely, but gaussian integers are pretty cool.


MATHCOUNTS
geometry
rectangle
3D geometry
probability
ratio

1. given rectangle QRST with Q(-2,4) and R(2,-4), if diagonal QS has a length of 2rt65, find the area of the rectangle.

2. points A and B are selected on the graph of y=-x :^2: /2 so that triangle ABO is equilateral. find the length of one side of triangle ABO.

3. seven points are equally distributed on a circle. how many non congruent quadrilaterals can be drawn with vertices chosen from among the seven points?

4. write  :sqrt: (5 - 2rt6)) as the difference of the square roots of two integers.

5. a rectangular aquarium 16 inches long, 9 inches wide, and 10 inches tall is filled with water to a level of 8 inches. a solid steel cube with an edge of 3 inches is dropped into the aquarium. how many inches high is the new water level in the aquarium? express your answer as a decimal rounded to the nearest tenth.

6. john ordered several copies of his favorite rock star's latest album to give his friends. he ordered 4 CD's and some additional tapes. the price of a CD was double the prie of a tape. when the order arrived the number of CDs had been interchanged with the number of tapes. the bill was 50% more than he expected. how many tapes did John order?

7. two diff integers are randomly selected from the set of integeres greater than 0 and less than 9. what is the probability that they have no common prime factor? express your answer as a common fraction.

8. what integer n has the property that 5^96 is greater than n^72 and 5^96 is less than (n+1)^72?

9. after 1/5 of the girls left a school dance, the ratio of girls to boys was 2/3. then 44 boys left and the ratio of boys to girls was 2/4. how many students remained at the dance?

10. how many four digit integers greater than 5000 are there for which the thousands digit equals the sum of the other three digits?

1. [hide]Diagonal QS has a length of 2  :sqrt: 65. And side QR has length 4 :rt5: . So using the pythagorean theory, the other side of the rectangle has length 6 :rt5: . So the area of the rectangle is 120[/hide]

2. [hide]Let (x, -x^2/2) be A and (-x, -x^2/2) be B. Because the sides of an equilateral triangle are equal,  :sqrt: (x^2+((-x^2)/2)^2) = 2x. Solving gives x=2 :rt3: [/hide]

3. [hide]Because any selection of 4 points from the seven would give a quadrilateral different from the others, there are C(7,4)=35 different quadrilaterals.[/hide]

4. [hide]Let  :sqrt: (5-2 :sqrt: 6) be equal to   :sqrt: a -  :sqrt: b. Squaring both sides gives

a+b=5 and 2 :sqrt: ab=2 :sqrt: 6

Thus since a>b, a=3 and b=2 and :sqrt: (5-2 :sqrt: 6)= :rt3: - :rt2: [/hide]

Well heres some more:

5. [hide]Since the water level is 8 in., there is 8*16*9=1152 cubic inches of water in the aquarium. the volume of the cube is 27 cubic inches so there is a total of 1152+27=1179 cubic inches. Divide this by the base, 16*9 and, to the nearest tenth, you get 8.2[/hide]

6. [hide]Let x be the price of a tape and y be the number of tapes he ordered. Then the original bill should have been 4*2x+x*y and the bill that arrived had 4*x+y*2x. So 3/2*(8x+xy)=4x+2xy. Solving and cancelling out the x's gives y=16.[/hide]

aaaaaaand...

8. [hide]n^72 < 5^96 < (n+1)^72. Take 24th root and you get

n^3 < 5^4 < (n+1)^3. Simple calculations give you n=8.[/hide]

9. [hide]From the first sentence, we have

b/(4g/5)=2/3 so 3b=8g/5

From the second, (b-44)/(4g/5)=2/4 so 4b-176=8g/5

3b=4b-176, so b=176 and g=330.

So after 1/5 of the girls left, there are 264 girls left and 176 boys. After 44 boys leave, there are 396 people left.[/hide]

you would think that the answer to number 3 is 35 but its 4 bc when they say different i think they mean not congruent. what about number 7 and 10....by the way, thanks so much!

Different = non-congruent, at least in this context.

I deleted this message because of the following post...

7. [hide]The possible sets of two are (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8)

(2, 3), (2, 5), (2, 7)

(3, 4), (3, 5), (3, 7), (3, 8)

(4, 5), (4, 7)

(5, 6), (5, 7), (5, 8)

(6, 7)

(7, 8)

There are a total of 21 of these and C(8, 2)=28 different ways to choose two of them. So the probability is 21/28.[/hide]

10. [hide]For this problem, I just listed them out for numbers between 5000 and 6000, and between 6000 and 7000. They are 21 and 28, two triangular numbers. So the total number of of four digit numbers is 21+28+36+45+55=185.[/hide]


So I just installed Asymptote (I have a pretty good base of Latex), and I was wondering if there was an efficient way to draw the following figure (aside from plugging in the co-ordinates).

[quote="fedja"]...and using the [b]intersectionpoint[/b] command.[/quote]
Doesn't cse5 let you use [b]IP[/b] instead of [b]intersectionpoint[/b] (and [b]D[/b] instead of [b]draw[/b])?

[quote="fedja"]Certainly. The whole beauty of Asymptote is that you almost never need to do any coordinate computations. Here is your picture (click on it or use the quote button to see the code)
[asy]size(200);
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;
path P=circle(O,r);
C=intersectionpoint(B--(B.x+r,B.y),P);
draw(P,dashed);
draw(C--B--O--A--B);
dot(O); dot(A); dot(B); dot(C);
label("$O$",O,SW);
label("$A$",A,NE);
label("$B$",B,S);
label("$C$",C,SE);[/asy]
This is not the only possible way, of course. The main things I wanted to attract your attention to are using coordinates of the existing points to build new ones and using the [b]intersectionpoint[/b] command.[/quote]

I understand your code, but how come when I try compiling, it doesn't work.

[quote="fedja"]For that I need more information. What exactly is the error message you get?[/quote]

Okay, so (thanks to the Wiki) I have asy_log which reads:

[quote]1983_tex.asy: 10.1: invalid token '\'
1983_tex.asy: 10.2: syntax error
error: could not load module '1983_tex.asy'
error: could not load module '\par'
[/quote]

This is a completely idiotic question but did you remove both {asy} and {/asy} tags when copying and pasting the code to your file? Also are you running it as a part of TeX file or just as an asy file? I'm asking because I cannot reproduce your error on my computer so I assume that something is wrong with the way you copied the code or the way you inserted it into a TeX-file. Try to do what stevem always suggests: remove the lines from the code one by one (starting with the last one, of course) until you stop getting error messages. The last line you removed is most likely the one causing the trouble. If you still do not see what the problem is, post the minimal code that causes your error.

@i_like_pie Certainly it does and I always use it myself, but I cannot assume that happyme has either cse5 or the olympiad package :)


Phythagoras said that '[b]Numbers[/b] Rules[b] The[/b] Universe[b].[/b]'If that is so,how is it that sometimes mathematician[u] are branded to either hopeless[b] romantic[/b]
 poet[b],and prophets[/b] itself? Is 'Heisenberg's' Uncertainty[b]
 Theory[/b] pre-supposes the 'mathematical[/u] uncertainty[u]?'

                                            Respectfully,
                                             A.Eddington[/u]

Heisenberg's Uncertainty Principle only states that one cannot predict the exact position and momentum of an electron in an orbital at one time. In this case, there isn't certainty, but there would be in other situations.

I'm not sure what your question is asking exactly, though. Can you clarify?


I need some help on functions. I'm not sure when to tell if something is even or odd. Like this problem F(x)= g(x)- g(-x). I don't know if this is even or odd. I'm so confused @_@. I also need some help with problems that are f of g of x or stuff like that. If f(x)= 2x^3-5, then find g(x), such that f  o  g(x)= g  o  f(x). Thanks  :D

Knowing the definition of what it means to be even or odd would be a good starting point.  A function is even if $ f(x)\equal{}f(\minus{}x)$ and is odd if $ f(\minus{}x)\equal{}\minus{}f(x)$.  Plug in $ \minus{}x$ and see what happens.


What grade are you in and how much homework do you get?

just wondering...

7th|3 hrs :huh: It gets more though as we near competitions.  :|

whoa 3 hours a day? I get barely 15 minutes because its like no problems assigned...

8th grade

and NO math homework  :wacko: because I do it all in class while my teacher is lecturing on some stuff I learned two years ago.

probably like 1.5 hours to 2 hours on the other stuff (combined)

History: None lately, and none for the coming weeks apparently.
English: - 1.5 hours once per week (yay procrastination)
Math: Done during chem class
Chem: Done during math class (sometimes the other way around)
Spanish: It varies. ~10 minutes/day average.
Health: 5 minutes, every other week?

7th| 5 hours.... I have to make solutions, and I have to learn Latin sometime in 2007

8th, and like 1/2 hour. They intend for it to be like 2 or 3 hours, but its EASY. It actually takes me longer thatn 1/2 hour, as I procrastinate WAY too much, but still like none. In math, on the computer, I just play games all clasas and do the tests periodically.

[quote="bubka"]I have ignored my homework completely through most of the first sem - especially vectors and matrices - and now I have done awfully in the endsems - especially vectors and matrices. :([/quote]

lol don't your teachers mark you down for not doing homework?

Well. I'm in eighth grade and I get 0-11 hours of homework. :D Seriously, sometimes I get 0 because I get it all done in class, and one time I got 11 because of a big fat stupid group project, which my big fat group just lazily dumped on me. :(

My math teacher gives about two hours of homework each night. He's crazy. I think he seriously thinks he's the only teacher that gives homework, so he can give so much of it. Of course, it takes me about an hour to do it because it's so easy :P , but it's still... tedious.

[quote="bubka"]I have ignored my homework completely through most of the first sem - especially vectors and matrices - and now I have done awfully in the endsems - especially vectors and matrices. :([/quote]

are you in linear algebra?

[quote="13375P34K43V312"][quote="7h3.D3m0n.117"][quote="Lena"]university, i get 4 questions per class, but it takes me 2 weeks to solve them (several hours per day)[/quote]
wow really????? with AoPS it should take you 30 minutes lol[/quote]

AoPS isn't for posting homework problems and letting people do your homework for you.[/quote]

i know but posting similar problems (not on homework) and recieving instruction tips makes life easier

[quote="7h3.D3m0n.117"][/quote][quote="13375P34K43V312"][quote="7h3.D3m0n.117"][quote="Lena"]university, i get 4 questions per class, but it takes me 2 weeks to solve them (several hours per day)[/quote]
wow really????? with AoPS it should take you 30 minutes lol[/quote]

AoPS isn't for posting homework problems and letting people do your homework for you.[/quote][quote="7h3.D3m0n.117"]

i know but posting similar problems (not on homework) and recieving instruction tips makes life easier[/quote]

or just post your homework and find out how to do it :lol:

9th in precal - our class gives us an assignment every night that isnt due until the end of the unit, so I dont even do the homework until the day before it's due. takes me 1.5-2 hrs

im in 4th grade lol and i barely have hw

[quote="mathnerd314"][quote="7h3.D3m0n.117"]and NO math homework  :wacko: because I do it all in class while my teacher is lecturing on some stuff I learned two years ago.[/quote]

Bleh, my teacher makes us take notes and actually checks them at the end of the quarter. Also, we aren't given the assignment until the end of class.[/quote]

wow...we're expected to take notees...because she walks around and looks at people's papers. she does lets use use our notes sometimes during quizes but thats it.

Luckily, I'm homeschooled, so I don't have to take notes, ut I do have homework. I have to check my homework with a study buddy, though.


Hello, I read through the College FAQ quite a while ago and bookmarked http://www.ocf.berkeley.edu/~abhishek/chicmath.htm because it had a bunch of great information on a lot of textbooks for different areas of math. Now I can't seem to access the web page (or any other berkeley ocf webpage for that matter). I was wondering if anyone else was having this problem, or if there was a mirror, or even if someone had the text of the web page saved. I would be very grateful.

Thanks,
Tyler

Try typing the webaddress into the Internet Archive [url=http://www.archive.org/index.php]Way Back Machine[/url]

You could also check out the (poorly named) [url=http://hk.geocities.com/mathphyweb/puremath.htm]How to Become a Pure Mathematician (or Statistician)[/url]

Or (the no longer maintained) [url=http://www.ericweisstein.com/encyclopedias/books/]Eric Weisstein's Encyclopedia of Scientific Books[/url]

Thank you for the Way Back Machine, that did work, and will definitely be put into future use! I will also bookmark the other links you provided, I'm a firm believer that you can never have too much information at hand!


Proof for positive a,b,c
$ \frac{1}{a(1\plus{}b)}\plus{} \frac{1}{b(1\plus{}c)} \plus{} \frac{1}{c(1\plus{}a)}  \ge  \frac{3}{1\plus{}abc}$

$ \frac {1}{a(1 \plus{} b)} \plus{} \frac {1}{b(1 \plus{} c)} \plus{}$ $ \frac {1}{c(1 \plus{} a)}\ge\frac {3}{1 \plus{} abc}$ $ \Longleftrightarrow$ $ (ab \plus{} bc \plus{} ca) \plus{} (ab^2 \plus{} bc^2 \plus{} ca^2) \plus{}$ $ abc(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} a^2b^2c^2(a \plus{} b \plus{} c)\geq 2abc(a \plus{} b \plus{} c) \plus{} 2abc(ab \plus{} bc \plus{} ca)$.
But $ (ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} a^2b^2c^2(a \plus{} b \plus{} c)\geq$ $ 2abc\sqrt {(a \plus{} b \plus{} c)(ab^2 \plus{} bc^2 \plus{} ca^2)}\geq 2abc(ab \plus{} bc \plus{} ca)$(Cauchy-Schwarz) and $ (ab \plus{} bc \plus{} ca) \plus{} abc(ab^2 \plus{} bc^2 \plus{} ca^2)\geq$ $ 2\sqrt {abc(ab \plus{} bc \plus{} ca)(ab^2 \plus{} bc^2 \plus{} ca^2)}\geq$ $ 2abc(a \plus{} b \plus{} c)$(Cauchy-Schwarz).

inny,check out this link: http://www.mathlinks.ro/viewtopic.php?t=308365
there is a nice solution posted by abdek.

http://www.mathlinks.ro/viewtopic.php?t=313218

 :)

I do it by following way:
$ \sum\frac{1\plus{}abc}{a(1\plus{}b)}\equal{}\sum (\frac{1\plus{}a}{a(1\plus{}b)}\plus{}\frac{b(1\plus{}c)}{1\plus{}b})\minus{}3\geq 6\minus{}3\equal{}3$
QED.


inequalities
logarithms
calculus
inequalities unsolved

Let $ 1\leq p \leq 2$. Prove that $ \forall x_i, x_n\in \mathbb{R}, n \in \mathbb{N}$,
$ \frac{1}{n^{\frac{2-p}{2p}}}$ $ {( \sum_{k=1}^{n}|x_k|^p)} )^{\frac{1}{p}}$ $ \leq$ $ {( \sum_{k=1}^{n}{x_k}^2)}^\frac{1}{2}$ $ \leq$ $ {( \sum_{k=1}^{n}|x_k|^p )}^\frac{1}{p}$

Thanks a bunch :) This has been causing me trouble

Looks like an application of Hölder's inequality.

Could you please prove it fully ? I don't seem to be getting it by holder's though :(

I think it's not Hölder. We get
$ \sum_{i \equal{} 1}^n x_i^p\leq \left (\sum_{i \equal{} 1}^n(x_i^p)^{2/p}\right )^{p/2}\left (\sum_{i \equal{} 1}^n1\right )^\frac {2 \minus{} p}{p}$
so
$ \sum_{i \equal{} 1}^n x_i^p\leq \left (\sum_{i \equal{} 1}^nx_i^2\right )^{p/2}n^\frac {2 \minus{} p}{p}$.
Thus
$ \left (\sum_{i \equal{} 1}^n x_i^p\right )^{1/p}\leq \left (\sum_{i \equal{} 1}^nx_i^2\right )^{1/2}n^\frac {2 \minus{} p}{p^2}.$
But unfortunately the inequality $ \frac {2 \minus{} p}{p^2} < \frac {2 \minus{} p}{2p}$ is not true. :(

[hide]The first inequality is just $ M(2) \ge M(p)$.[/hide]

[quote="Fermat's Little Turtle"]Let $ 1\leq p \leq 2$. Prove that $ \forall x_i, x_n\in \mathbb{R}, n \in \mathbb{N}$,
$ \frac {1}{n^{\frac {2 - p}{2p}}}$ $ {( \sum_{k = 1}^{n}|x_k|^p)} )^{\frac {1}{p}}$ $ \leq$ $ {( \sum_{k = 1}^{n}{x_k}^2)}^\frac {1}{2}$ $ \leq$ $ {( \sum_{k = 1}^{n}|x_k|^p )}^\frac {1}{p}$

Thanks a bunch :) This has been causing me trouble[/quote]

For the second inequality, note:
As it is homogeneous we may suppose
$ \sum_{k = 1}^{n}|x_k|^p = 1$
Therefore
$ |x_k| \le 1$ for all $ k$
and thus
$ x_k^2 \le |x_k|^p$ for all $ k$.
From this we get
$ \sum_{k = 1}^{n}x_k^2 \le \sum_{k = 1}^{n}|x_k|^p = 1$
and the proof is complete.

[hide]We can prove that the sum decreases while $ p$ increases by showing that its logarithm decreases, i.e.,

$ \frac {d \frac {ln(\sum |x_k|^p)}{p}}{dp} \le 0$.

This is equivalent with

$ \frac {\frac {\sum |x_k|^pln|x_k|}{\sum |x_k|^p}p \minus{} ln(\sum |x_k|^p)}{p^2} \le 0$

$ \Longleftrightarrow (\sum |x_k|^p)ln(\sum |x_k|^p) \ge \sum p|x_k|^pln|x_k|$

But $ (\sum |x_k|^p)ln(\sum |x_k|^p) \ge \sum |x_k|^pln|x_k|^p \equal{}  \sum p|x_k|^pln|x_k|$ since $ lnx$ is an increasing function.

Is there any mistake?

P.S.: never press "shift" for more than 8 seconds. I've just discovered that, when you make it, bad things happen.

P.S. 2: Your solution is much better, auj! Sorry... I do everything analitically...[/hide]

One approach to both inequalities is to consider the following problem:

Find the maximum and minimum of $ x_1^p \plus{} \dots \plus{} x_n^p$ given $ x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2 \equal{} c$ and each $ x_i \geq 0$.

This can be solved by Lagrange multipliers.  The Lagrangian condition becomes that for every $ i$ either or $ x_i \equal{} 0$ or $ px_i^{p \minus{} 1} \equal{} 2 \lambda x_i$.  Note that the latter case actually includes the former.  Furthermore, unless $ c \equal{} 0$ (which is a trivial case) we have $ \lambda \neq 0$.

Solving, we see that for $ p < 2$ the possible maxima/minima are those points where all of the nonzero values of $ x_i$ are equal in magnitude, which up to permutations we can think of as $ x_1 \equal{} \dots \equal{} x_k \equal{} z$, $ x_{k \plus{} 1} \equal{} \dots \equal{} x_n \equal{} 0$.  Now just solve for $ z$ in terms of $ k$ to see which $ k$ give you the maximum and minimum values.  This should immediately get you both sides of the inequality.

(edit: I had mistakenly thought this was in the analysis section, since the comparison of those $ p^{th}$ roots of the sum of $ x_i^p$ (the so-called $ l_p$ norms show up a lot in that area.  I guess you were looking for a non-calculus solution).


Hello!! this problem was in the list of selection of IMO 2006 of Brazil and now I'am posting.
Show that if gcd(x;y)=1 and (5z-4x)(5z-4y)=25xy then 10z+x+y or (10z+x+y)/3 is perfect square

Come on, it's not kind to post wrong problems.

Take: $x=5,y=6,z = 154$
Now: $(x,y)=(5,6)=1$
$(5z-4x)(5x-4y) = (770-20)(25-24) = 750 = 25 \cdot 5 \cdot 6$

$10z+x+y = 1551 = 3 \cdot 11 \cdot 47$
which is far from being a square, it's squareless.

I took me two hours to find it was wrong  :rotfl:

I post something wrong. Now I edit the problem.........

Please someone, I think that anyone try to solve it, please try!!!

I will tranlate and post others problems of BRASIL TST!!!Thank you for your solution!!!

Hello!!Thanks for your solution and thanks for the solution that was made by Carlos Shine, Eduardo Tengan e Samuel Pavlos (they send a e-mail to me).
 :)
 :jump:


geometry
geometric transformation
homothety

Two circles (K) and (L) are exrternally tangent at $ T$. A chord $ AB$ of (K) , produced , touches(is tangent to) the second circle at $ C$. Line $ CT$ meets circle (K) at $ D$. Prove that $ DA\equal{}DB$.

 [u]Babis[/u]

[quote="bertram"][hide="Solution"]A stretch at $ T$ transforms one circle into the other, so $ \Delta M_KTD \sim \Delta M_LTC$. Then $ \angle M_KDT \equal{} \angle M_LCT \equal{} 90^\circ \minus{} \angle ACT$. This means that $ M_KD$ is orthogonal to $ AB$ which was to be proven.

Edit: Actually the stretching argument implies $ M_KD$ and $ TM_L$ are parallel and thus both orthogonal to $ AB$.[/hide][/quote]

Are $ M_L,M_K$ the antidiametric of $ C,D$ to the circles ? If yes , then this solution with transformation is nice and short :) .

 Babis

Red angles and green angles are equal, so blue angles are equal.  :)
 [geogebra]50636e1d0efd1c41127a2bd465453a5e12ccb690[/geogebra]

Nice proof and diagram, archimedes! That is much easier than mine. The last time I used the theorem where an angle with one ray as a tangent to a circle and the other as a chord is half the intercepted arc was about three years ago! Although I don't get the notation in Bertram's.

Thank you.  I am glad to have revived the theorem for you.   :)
Bertram's solution is very nice and short.  But I have never seen the notation before.
What it says is (I think) let $ K$ and $ L$ be the centers of the circles.  Now a homothety at $ T$ takes circle $ k$ to circle $ l$, so it takes $ K$ to $ L$ and $ D$ to $ T$.  Thus, segment $ DK$ is taken to segment $ TL$.  Now these segments are parallel due to the nature of a homothety.  but segment $ TL$ meets line $ AB$ at a right angle, so line $ DK$ also meets $ AB$ at a right angle, implying that $ DAB$ is isosceles. [geogebra]1137a9c9adfaa8730aaf098ef03a3cb88762442e[/geogebra]

The two circles are homothetic at $T$, $C$ and $D$ are homologue points, hence the tangent of circle $(K)$ at $D$ is parallel to $AB$, meaning $D$ is the midpoint of the arc $AB$, implying the conclusion.

Best regards,
sunken rock


analytic geometry
induction
number theory unsolved
number theory

Let $P_1$, $P_2$, $\ldots$, $P_{1993} = P_0$ be distinct points in the $xy$-plane
with the following properties:
(i) both coordinates of $P_i$ are integers, for $i = 1, 2, \ldots, 1993$;
(ii) there is no point other than $P_i$ and $P_{i+1}$ on the line segment joining $P_i$ with $P_{i+1}$ whose coordinates are both integers, for $i = 0, 1, \ldots, 1992$.

Prove that for some $i$, $0 \leq i \leq 1992$, there exists a point $Q$ with coordinates $(q_x, q_y)$ on the line segment joining $P_i$ with $P_{i+1}$ such that both $2q_x$ and $2q_y$ are odd integers.

suppose that any point on the considered segments doesn't have the property.
take $ x_{i},y_{i} $ the coordinates of $ P_{i} $.
if $ x_{i}+y_{i}+x_{i+1}+y_{i+1} $ is even then $ x_{i}+x_{i+1} $ and $ y_{i}+y_{i+1} $ are congruent mod 2 and that means both are even (this case isn't good because the midpoint will have both coordinates integers) or odd but this is a contradiction.
then all the sums of this type are odd.
$ P_{0} $ has $ x_{0}+y_{0} $ congruent with $ w $ mod 2 and we will say that $ P_{0} $ is red.
then $ P_{1} $ has  $ x_{1}+y_{1} $  congruent with $ 1-w $ mod 2 and we will say that $ P_{1} $ is blue.
using induction we obtain that $ P_{2t} $ are red and  $ P_{2t+1} $ are blue.
so $ P_{0}=P_{1993} $ and $ P_{1992} $ are of the same colour so the sum of their coordinates is even which is impossible.
this is the proof for the existence of $ Q $.


Convert phenol to o-methoxybenzoicacid in not more than 3 steps.

$1$ Sodium metal. Gives sodium phenolate.

 $2$ Kolbe reaction. $CO_{2}$, $125$degrees  $4-7$ atm Sodium
       phenolate to $ortho-hydroxy benzoate$

 $3$ $(CH_{3})_{2}SO_{4}$ in acid.


Έστω μια επιφάνεια Φ με παραμετρική παράσταση x(u,v) όπου (u,v)∈D⊂E². Να αποδειχτεί οτι σε δίκτυο ασυμπτωτικών γραμμών ισχύει: τ²(s)=-Κ(s) , οπου τ(s) η στρέψη της καμπύλης και K(s) η καμπυλότητα Gauss της επιφάνειας..

Μιας και δεν βλέπω ανταπόκριση στη λύση της άσκησης θα τη δώσω ο ίδιος για όσους θέλουν να τη δούν..
Λοιπόν, γνωρίζουμε ότι σε δίκτυο γραμμών καμπυλότητος ισχύει ότι $ \overrightarrow{N}(s)=\pm{\overrightarrow{b}(s)}$ όπου $ \overrightarrow{N}(s)$ το κάθετο της επιφάνειας και $ \overrightarrow{b}(s)$ το δεύτερο κάθετο της καμπύλης..και $ s$ φυσική παράμετρος.

Παίρνουμε λοιπόν την σχέση $ \overrightarrow{N}(s)=\pm{\overrightarrow{b}(s)}$ και την παραγωγίζουμε. Θα είναι λοιπόν: \[ \dot{\overrightarrow{N}}(s)=\pm{\dot{\overrightarrow{b}}(s)}\] Κάνοντας χρήση των τύπων Frenet θα έχουμε ότι \[ \dot{\overrightarrow{N}}(s)=\pm{(-\tau(s)\overrightarrow{n}(s))}\]
Στη συνέχεια υψώνουμε στο τετράγωνο και θα έχουμε: \[ (\dot{\overrightarrow{N}}(s))^2=(\pm{(-\tau(s)\overrightarrow{n}(s)})^2\]
αλλά \[ (\dot{\overrightarrow{N}}(s))^2=(\overrightarrow{N}_{u}\dot{u}(s)+\overrightarrow{N}_{v}\dot{v}(s))^2=III(du,dv)\] όπου $ III(du,dv)$ η τρίτη Θεμελιώδης μορφή της επιφάνειας.. Θα έχουμε λοιπόν:
\[ III(du,dv)={\tau}^2(s)\] αφού $ (\overrightarrow{n}(s))^2=1$ διότι $ s$ φυσική παράμετρος. Στη συνέχεια κάνοντας χρήση της ταυτότητας που συνδέει της Θ.Μ. τις επιφάνειας δηλαδή: ${ III-2HII+KI=0}$ όπου $ 2H$ η μέση καμπυλότητα της επιφάνειας και $ K$ η καμπυλότητα Gauss της επιφάνειας θα έχουμε: \[ 2HII-KI={\tau}^2(s)\]
Αλλά $ 2H=0$ διότι σε δίκτυο ασυμπτωτικών γραμμών έχουμε $ K_{N}=0$ και $ K_{N}=\frac{II}{I}$ όπου $ K_{N}$ κάθετη καμπυλότητα της επιφάνειας. Καταλήγουμε λοιπόν ότι: \[ -KI={\tau}^2(s)\]
Αλλά $ I=(\dot{x}(s))^2=1$ αφού $ s$ φυσική παράμετρος. Οπότε τελικά καταλήξαμε ότι: \[ -K(s)={\tau}^2(s)\]


Hi everybody,

Can anyone take the pains to tell me what in-built functions are provided in the linux environment to sketch a line segment and a circle???

Thanks in advance to whoever does

-Have a good day!!!!

Uhm.... why don't you work in Windows? :) I've never done any graphics in Linux but if I suddenly decided to try it, I'd check [url=http://www.google.com]google[/url] first. I'm pretty sure you can find a lot of tutorials on the web.

If you are familiar with openGL, it is a system independent library. You can use it for example in C++ under linux with the same functions as it was used in windows.

[code]

void MyDisplay(void)
{
      glClear(GL_COLOR_BUFFER_BIT);          //clear the background

      glBegin(GL_LINES);
	glVertex2i(100,50);
  	glVertex2i(150,200);
      glEnd();

      glFlush(); // send to display
}

[/code]

Just use the [url=http://libsdl.org]SDL[/url]. Sound, graphics, everything.


Vieta
algebra
polynomial
Rational Root Theorem

Suppose that the rational numbers a,b and c are the roots of the equation $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$. Find all such rational numbers a, b and c. Justify your answer.

Note that one solution is $ a\equal{}b\equal{}c\equal{}0$.

By vieta's formulas, $ \minus{}a\equal{}a\plus{}b\plus{}c \implies 2a\plus{}b\plus{}c\equal{}0$, $ \minus{}c \equal{} abc \implies ab\equal{}\minus{}1$ if $ c \neq 0$, and $ b \equal{} ab\plus{}bc\plus{}ac \equal{} bc\plus{}ac\minus{}1 \implies b\equal{}bc\minus{}c/b\minus{}1$.  Hence $ 0 \equal{} b^2(c^2\minus{}1) \minus{}b\minus{}c \implies b\equal{}\frac{1 \pm \sqrt{1\plus{}4c(c^2\minus{}1)}}{2(c^2\minus{}1)}$

I'll try to finish later

First, find all solutions for when $ c\equal{}0$. The original polynomial becomes $ x(x^2\plus{}ax\plus{}b)\equal{}0$. By Vieta's, $ b\equal{}ab, \minus{}a\equal{}a\plus{}b\iff \minus{}2a\equal{}b$. If $ b\equal{}0$, then $ a\equal{}\minus{}a \iff a\equal{}0$, so $ a\equal{}b\equal{}c\equal{}0$ is a solution. Otherwise, $ a\equal{}1$ from the first equation of Vieta's, then $ b\equal{}\minus{}2$ from the second. Then $ a\equal{}1, b\equal{}\minus{}2, c\equal{}0$ is another solution.

Now consider when $ c\not\equal{}0$. By Vieta's, $ \minus{}c\equal{}abc \Rightarrow ab\equal{}\minus{}1 \iff b\equal{}\minus{}\frac{1}{a}$. Then also by Vieta's, $ \minus{}a\equal{}a\plus{}b\plus{}c \Rightarrow 2a\minus{}\frac{1}{a}\plus{}c\equal{}0 \iff c\equal{}\frac{1}{a}\minus{}2a\equal{}\frac{1\minus{}2a^2}{a}$.

Too lazy for the rest, but you use $ b\equal{}ab\plus{}bc\plus{}ac$ with $ b$ and $ c$ plugged in so it is all in terms of $ a$, then make an ugly quartic. This quartic will have no rational solutions, so you can stop. Yay.

Well the case of $ c = 0$ has been well settled, so here's the rest.

[hide="The Rest"]Let $ c \neq 0$. As stated before, we get, from Vieta's:

$ \begin{array}{cccc} \text{(i)} & abc = - c & \implies & ab = - 1 & \text{(ii)} & - a = a + b + c = & \implies & 2a + b + c = 0 & \text{(iii)} & b = ab + bc + ca & \implies & b = c(a + b) - 1 \end{array}$

Multiply (iii) by $ a$ to get $ ab = c(a^2 + ab) - a \implies - 1 = c(a^2 - 1) - a = c(a + 1)(a - 1) - a$
$ \therefore \text{(iv)} \qquad a - 1 = c(a + 1)(a - 1)$. Thus, either $ a = 1$ or $ 1 = c(a + 1)$.

[b]Case 1:[/b] If $ a = 1$, then we get $ b = - 1$ using (i). Then, using (ii) we get $ c = - 1$ as well. So $ (1, - 1, - 1)$ is a solution.

[b]Case 2:[/b] If $ a \neq 1$, then (iv) becomes $ 1 = c(a + 1) \implies ac = 1 - c$. Then, going back to equation (iii), we use this to get $ b = - c + bc \implies (b - 1)(c - 1) = 1$

Now, we employ equation (ii) to get $ (b - 1)( - b - 2a - 1) = 1$. Finally, use equation (i) to get:

$ (b - 1)\left( - b + \frac {2}{b} - 1 \right) = 1 \implies \qquad b^3 - 2b + 2 = 0$

Use the rational root theorem to verify that there are no rational roots in $ b$ to this equation, and we are done.

[b]Solution Set:[/b] $ \boxed{(a,b,c) \in \{ (1, - 2,0), \; (0,0,0), \; (1, - 1, - 1) \}}$[/hide]


Suppose we have a sample of twodimensional r.v., $ { X_i \choose Y_i}, i\equal{}1,\dots,n$. From this sample we can obtain empiric distributions for $ X$ and $ Y$, we denote them by $ \mathbb{F}_X,\mathbb{F}_Y$. You know in fact that r.v. $ X$ and $ Y$ are not independent (suppose the correlation is positive). Our goal is to simulate this dependent sample (not knowing the true theoretical distribution).
This is how I see it. First using the empiric distribution we "normalize" our two dimensional sample, that is we transform our variables in the following way $ {\mathbb{F}_X(X) \choose \mathbb{F}_Y(Y)}\equal{}{ U_1 \choose U_2 }$. With this transformation the marginal sample are now from a uniform distribution, but still there's the dependence structure, but now evident from a perspective of uniform random variable. Now I make another transformation, using the inverse of a normal distribution $ \mathbb{F}^{\minus{}1}_{\mu,\sigma}$, we obtain the following sample $ { \mathbb{F}^{\minus{}1}_{\mu,\sigma}(U_1) \choose \mathbb{F}^{\minus{}1}_{\mu,\sigma}(U_2) }\equal{}{ Z_1 \choose Z_2 }$. We can choose $ \mu\equal{}0,\sigma\equal{}1$.
Now we are on a familiar truf and we can simulate a two dim. sample from a normal distribution (please note that the dependence structure is still there). We find the covariance of this normal sample, simulate a new sample and transform back to uniform dist, and back to our empiric distribution.

Now this all sounds great, but I have no theoretical justification that this will actually work, and that I won't lose the dependence structure along the way. I tried doing this with real data, but somehow I'm not convinced that I'm geting the right thing.

I would be grateful, if someone with more experience and knowledge would comment this procedure, or even better give a theoretic-based algorithm how to simulate a sample from an empiric multivariate distribution. Thanks in advance.

this is a very interesting question and a lot of work has been done on this subject. You could start by having a look at [url=http://en.wikipedia.org/wiki/Copula_%28statistics%29]this[/url].


hi,
Let $f: [\frac{1}{a},a] \longrightarrow \mathbb{R}$ , where $a \geq 1$ with the property: $f(x)+f(\frac{1}{x})=k$ , where $k \in \mathbb{R}$.
Calculate:
\[\int_{\frac{1}{a}}^{a}{\frac{f(x) \cdot (1+x)}{x\sqrt{x^{2}+1}}}dx.\]


[hide]Let's assume there is 52 weeks in a year (there isn't exactly, but that will make the problem difficult). You work 65-20 = 45 years. That's a total of 45*52=2340 hours. That's also 2340*40=93600 hours. If you add that on to the 3000 hours before you were 20, you get
96600 hours total. [/hide]

[quote="Anonymous52111"]Dear everyone,

I don't understand how to solve this problem when I found this. So is there any formula that you can use it to solve this? Also, need your help for show works that will help me to understand much better than nothing, hopefully. Thanks.

Mathematics: Calculate the total number of hours you will work in your lifetime if you begin working full-time when you are 20; retire when you are 65; work in average of 40 hours per week; and work a total of 3,000 hours of part-time jobs before you turn 20.[/quote]
[hide]
65-20 = 45 years
there are 52 weeks, so each year you work 52x40 = 2080 hours.
since you work for 45 years, thats 2080x45=93600 hours, but you also worked 3000 hours before you turn 20 so you add that on.

Then your answer is 93600+3000= 96600 hours

[/hide]

This is a badly put problem in terms of correct English.
Supposedly one shouldn't say if at 20.
I mean, it should be if before you turn 20 you work 3000 hours doing part time jobs. not begin at 20, you began working jobs at a young age, whether it be full time or not.

If I wanted to be stuck up, I wouldn't have to add in the 3000 hours.

younger>past 20> retired.

Sometimes math problems like this make me wonder if it is trick or not.

[hide][quote="Anonymous52111"]Dear everyone,

I don't understand how to solve this problem when I found this. So is there any formula that you can use it to solve this? Also, need your help for show works that will help me to understand much better than nothing, hopefully. Thanks.

Mathematics: Calculate the total number of hours you will work in your lifetime if you begin working full-time when you are 20; retire when you are 65; work in average of 40 hours per week; and work a total of 3,000 hours of part-time jobs before you turn 20.[/quote]

you dont need a formula, its just simple calculations, if you can translate teh word problem in to eqations.

you work full time for 65-20 = 35 years.  each year has 52 weeks so thats a total of 35*52=1820 weeks.  multiply times 40 for the hours, which is 72800, and add 3000 hours for the part time, and the final answer is 75800hours[/hide]

star99, he worked 45 years, not 35 years  ;)
[hide]
40*52=2080 is how many hours you work a year.
2080*45=93600 is how many hours you worked after 20.
93600+3000= $\boxed{96600}$ is the number of hours you worked in your life.[/hide]


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines

The lengths of the sides of a triangle are consescutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is

$\textbf {(A) } \frac 34 \qquad \textbf {(B) } \frac{7}{10} \qquad \textbf {(C) } \frac 23 \qquad \textbf {(D) } \frac{9}{14} \qquad \textbf {(E) } \text{none of these}$

I liked this problem. :)
[hide="Hint"]Law of Sines and Law of Cosines  :P [/hide]

[hide="Solution"]Let the small angle be $\theta$ and the large one be $2\theta$. The sides are $n$, $n+1$, and $n+2$.

By the law of sines, $\frac{\sin \theta}{n}=\frac{2\sin \theta \cos \theta}{n+2}$. Simplifying, $\cos \theta=\frac{n+2}{2n}$.

But, we know that $\cos \theta$ can also be calculated by the law of cosines. Thus,

$\cos \theta=\frac{n+2}{2n}=\frac{n^2-(n+1)^2-(n+2)^2}{-2(n+1)(n+2)}$. Solving for n, $n=4$ and our answer is $\frac 34$.[/hide]

[quote="4everwise"]Nobody?  :huh:[/quote]
You've solved it already :P What is there left to do?

[hide]
another way of finding the lenght without trigo

Suppose the triangle ABC with $AB=n$ , $BC=n+2$ , $AC=n+1$ . Bisect angle A and let the bisector line which intersect with BC be $M$ . Then $ABC$ is similar with $MBA$ . And also let $AM=MC=p$ , $BM=q$ for which $p+q=n$ .

From the triangle similarity we have $\frac{q}{n}=\frac{p}{n+1}=\frac{n}{n+1}$ . Using this with $p+q=n$ and simplify we should get $n=4$ . Then it is easy to get the cosine of smallest angle.

[/hide]

I wasted 45 mins trying to find my mechanical mistake but I just can't find it  :mad:
if someone could tell me where I went wrong, that would be great...
$ \cos \theta \equal{} (x \plus{} 1)/2(x \minus{} 1)$
$ (x \minus{} 1)^2 \equal{} x^2 \plus{} (x \plus{} 1)^2 \minus{} 2x(x \plus{} 1)\cos \theta$
$ (x \minus{} 1)^3 \equal{} x^2(x \minus{} 1) \plus{} (x \plus{} 1)^2(x \minus{} 1) \minus{} x(x \plus{} 1)^2$
$ x^3 \plus{} 3x^2 \minus{} 3x \minus{} 1 \equal{} x^3 \minus{} x^2 \plus{} x^3 \plus{} x^2 \minus{} x \minus{} 1 \minus{} x^3 \minus{} 2x^2 \minus{} x$
$ 3x^2 \minus{} 3x \minus{} 1 \equal{} \minus{} x \minus{} 1 \minus{} 2x^2 \minus{} x$
$ 5x^2 \minus{} x \equal{} 0$
$ x \equal{} 1/5$

So $ \cos\theta \equal{} \minus{} 3/4$???

EDIT:
Never mind, I finally found it: I screwed up in expanding (x-1)³... which should have been x³ - 3x² + 3x - 1...
took me an hour just to find this stupid mistake...


geometry
incenter
circumcircle
geometric transformation
homothety
perpendicular bisector
angle bisector

It is easy
PQ through the incenter of triangle ABC( for all triangle ABC) then $PQ=\frac{2}{3}BC=8$

[quote="Line"][img]http://img407.imageshack.us/img407/4634/maraton6yj0bb6.png[/img]

Let $\triangle ABC$ be equilateral inscribed in a circle $C$. Let another circle be tangent internally to $C$ in $T$ and tangent to sides $\overline{AB}$ and $\overline{AC}$ in points $P$ and $Q$. If $\overline{BC}=12$, then find $\overline{PQ}$[/quote]

PQ through center of trianggle ABC and parallel to BC

and how you know that PQ pass through circumcenter of ABC?

[b]Problem:[/b] Let $\omega$ be the circle tangent to $O$ and both $AB$ and $BC$ in the diagram.

Let $D$ be the intersection of $AB$ and the perpendicular bisector to $AI$ through $I$, where $I$ is the incircle of triangle $ABC$. Call the intersection of the angle bisector of $\angle A$ and the perpendicular to $AB$ through $D$, $W$. $D_{2}$ is as given in the diagram.

Prove that $\omega$ is centered at $W$, i.e. $O_{1}(W;\overline{WD})$ is $\omega$.
[b]Proof:[/b] It is equivalent to proving that $O_{1}$ is tangent to $O$, and that $O_{1}$ is tangent to $AB,AC$.
         Draw the parallel to $DW$ through $O$.
Clearly $DW$ is perpendicular to $AB$; since the parallel goes through $O$, it is a diameter of $O$.
Furthermore, $DW$ is obviously radius of $O_{1}$.
         This means there is a homothecy taking $O_{1}$ to $O$, i.e. they are tangent, part 1.
         Since $\triangle AEI\sim \triangle ADW$, there is a homothecy taking incircle $O_{I}$ to $O_{1}$, notice especially that $E$ is taken to $D$. Since the incircle is tangent to the two sides, so is, therefore, circle $O_{1}$, part 2, and tangent at $D, D_{2}$.

[img]6766[/img]

The construction proven valid, we have by construction that $I$ is in fact the midpoint of $DD_{2}$, since $\triangle DIA$ is right, $AI$ is angle bisector, $AD_{1}=AD_{2}$ by tangent law. $DD_{2}$ is further perpendicular to $AI$ at $I$, stronger than the claim from previous posts.

an other sol for above pro
 you can use lemma :Let circle $(O)$ ,circle $(O_{1})$ intouch at $M$ ,tanget from$M$(on $(O_{1})$ intersect $(O)$  at $B,C$ then $AM$ through midle point of arc $BC$.
 then you can prove that $B,D,T,I$( see figure of me@home ) are cyclic ( $T$ be intouch of $(O_{1})$ and $(O)$
(note :In my prove I construct circle$(O_{1})$ intouch $(O)$at$T$ ,I be intersection of bisector of $\widehat{ABC}$with $DF ( D,F$ be tangent of $(O_{1})$ and $(AB,AC)$ )
Now i think you can complete sol

Sorry, I can't really understand what you're saying... It seems like you define $T$ and $M$ as the same point :maybe:
Also, the only ray that bisects arc $BC$ is the angle bisector, so what exactly is $M$?

Sorry!!  :lol:
I will construct circle $(O_{1})$ intoch $(O)$ at $T$ and touch side $AB,AC$ at $D,F$ and I be intersection of bisector of $\widehat{ABC}$ and $DF$ then $I$ will prove that $I$ be incenter of triangle$ABC$, from four points  $A,B,D,I$ are concyclic ( It is same to Thebault theorem )


For $\{1,2,3,...,n\}$ and each of its nonempt subsets a unique alternating sum is defined as follow: Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers.  For example, the alternating sum for $\{1,2,4,6,9\}$ is $9-6+4-2+1=6$ and for $\{5\}$ it is simply $5$.  Find the sum of all such alternating sums for $n=7$.

This problem has been posted at least 5 times in AoPS, please use SEARCH or use the RESOURCES page.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61279]http://www.mathlinks.ro/Forum/viewtopic.php?t=61279[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=69996]http://www.mathlinks.ro/Forum/viewtopic.php?t=69996[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67421]http://www.mathlinks.ro/Forum/viewtopic.php?t=67421[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67193]http://www.mathlinks.ro/Forum/viewtopic.php?t=67193[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52496]http://www.mathlinks.ro/Forum/viewtopic.php?t=52496[/url]


[hide="General Case (edited; sorry for the mistakes)"]
We will solve the problem in the general case, for an $m \times n$ grid.  (I am assuming that this means that there are $m+1$ and $n+1$ lattice points, respectively, on each side.)

Any two lattice points with distinct $x, y$ coordinates form one of four (ordered) pairs of coordinates that can each define the same (and uniquely determined) rectangle.  The number of distinct pairs of points $(x_1, y_1), (x_2, y_2)$ with $x_1 \neq x_2, y_1 \neq y_2$ is given by $(m+1)m(n+1)n$, which we must divide by four.

Hence our final answer is $\boxed{ \frac{(m+1)m(n+1)n}{4} }$.

The original problem, $m = 113, n = 206$, gives $\boxed{ 137328561 }$.
[/hide]

[quote="DanK"]How many rectangles can be drawn on a $113\times206$ grid?[/quote]

[hide]$(1+2+3+...+113)(1+2+3+...+206)=137328561$
Each dimension has that many sides of that length so you multiply each number of lengths by each number of lengths in the other dimension.[/hide]

Edit: My answer is exactly half of yours...

When you're asking for the number of rectangles that can be drawn, does this mean that just the vertices have to be on the lattice points? Because in that case, we can have rectangles going diagonal across the grid and two lattice points alone aren't enough to uniquely define a rectangle.

The general formula is

\[ \frac{mn\left(m+1\right)\left(n+1\right)}{\mathbf{4}}  \]

different rectangles for an $m\times n$ grid.  If you take an $m\times 1$ grid, you have $1+2+\cdots +m=\frac{m\left(m+1\right)}{2}$ rectangles.  The same thing happens for an $n\times 1$ grid.  The product will give you the total amount of rectangles on an $m\times n$ grid.

ch1n353ch3s54a1l has the correct answer for the problem you guys were doing, although I was actually just going for $m$ and $n$ lattice points on each side...

It's actually $\frac{(m+1)m}{2}\times\frac{(n+1)n}{2}=\frac{(m+1)m(n+1)n}{4}$ if you do it your way t0rajir0u (what pianoforte did), which is how I did it.

Here's a fun note I found the other day, which is why I posted the problem:
$\frac{(m+1)m}{2}$ is the formula for the $mth$ triangular number, which also happens to equal $\displaystyle{{m+1}\choose2}$.  Therefore, the formula can be also be written as $T_{m-1}T_{n-1}$ or ${{m}\choose{2}}{{n}\choose{2}}$ (or I guess $T_{m}T_{n}$ or ${{m+1}\choose{2}}{{n+1}\choose{2}}$ if you do the $m+1$ and $n+1$ lattice points...)

[hide]
Its actually:
$\boxed{\frac{R(R-1)C(C-1)}{2*2}}$
where $R$ and $C$ are the number of rows and columns.
Plugging in the info for this problem gives:
$\boxed{133615720}$
[/hide]

I guess I was a little late with my clarification for t0rajir0u...

And no, diagonal rectangles are not allowed...

And yes, that's the one I was going for demji...

demji:  Are you sure?  If $R$ is the number of rows and $C$ is the number of columns then there are $(R+1)(C+1)$ lattice points.  Moreover, for my formulation of the problem there are $R+1, C+1$ choices for $(x, y)$ and for the case of an $m \times 1$ rectangle your solution is clearly wrong, as amcavoy showed.

DanK:  The formula can be written ${n+1 \choose 2} \times {m+1 \choose 2}$ because, from my formulation of the problem, choosing two $x$-coordinates and then choosing two $y$-coordinates defines a rectangle.  This can be extended arbitrarily to higher dimensions, or to lattice triangles.  :)  (It is not a coincidence that the $m^{th}$ triangle number can be written this way; look at amcavoy's post and try to find a combinatoric interpretation of $1 + 2 + ... + n$.)

i am sure, break it down just 1-d shapes, a line of r points
then you make lines to connect the dots, then slowly shorten the line by 1 and build in all shifts, it ends up being $1+2+3+..+r-1=\frac{r(r-1)}{2}$ then repeat for columns and multiply together.

[quote="ch1n353ch3s54a1l"]Wait...I have a question...If  it's $\frac{m(m+1)n(n+1)}{4}$ , how do you know those four lattice points form a rectangle? Because if there are $m(m+1)$ lattice points able to be chosen on one dimension, there are only a certain amount of pairs of $(n_1,n_2)$ points that actually form a rectangle...maybe i'm missing something[/quote]

I'm not choosing four lattice points, I'm choosing two x-coordinates and two y-coordinates, which form two lattice points.  If you can show that they must form a diagonal (which they must iff $x_1 \neq x_2, y_1 \neq y_2$) then they uniquely define a rectangle (I am assuming, here, that "rectangle" meant "rectangle with sides aligned to the axes," as the general problem is a little harder...)

Ex.:  I pick the points $(1, 5), (3, 6)$.  The rectangle is

$(1, 5)$
$(1, 6)$
$(3, 5)$
$(3, 6)$

:)

[quote="demji"]a line of r points[/quote]

$R$ rows (by "rows" I mean the spaces between the lines, like the ones you would write on on graph paper while filling in a table) produces $R+1$ points.  :)

if u do it that way then u would be right, but the numbers that are given are points so u would have to switch it to be R and R-1

That's my fault T0rajir0u, I just didn't write the problem clear enough.  I meant $m\times n$ points, $(m-1)\times(n-1)$ rows and columns...

And I really like the other stuff with the combinations and triangles and such.  I figured that out when I found the thing that gave me the idea for the problem...

Oh, right, I was thinking [i]m[/i] [b]sides[/b] x [i]n[/i] [b]sides[/b].


im  interested in a proof for pascals theorem using only menallaus theorem.  i searched for it, but turned up nothing. anyone know this solution?


linear algebra
matrix
trigonometry
Putnam
linear algebra unsolved

(O.Skaskiv) For arbitrary square matrix $ A$ define $ \sin A$ by means of series
\[ \sin A\equal{}\sum_{n\equal{}0}^{\plus{}\infty}{\frac{(\minus{}1)^{n}}{(2n\plus{}1)!}A^{2n\plus{}1}}.\]
Does there exist a $ 2\times 2$ matrix $ A$ such that $ \sin A \equal{}\begin{pmatrix}1 & 2007\\ 0 & 1\end{pmatrix}$ ?

This (or this with a different year) was a Putnam problem.

Lemma: $ \sin^{2}A\plus{}\cos^{2}A \equal{} I.$

You can prove this through $ e^{iA}\equal{}\cos A\plus{}i\sin A$ and $ e^{iA}e^{\minus{}iA}\equal{} I.$

Then in this case, $ \sin^{2}A \equal{}\begin{pmatrix}1 & 4014\\ 0 & 1\end{pmatrix}$

so $ \cos^{2}A \equal{}\begin{pmatrix}0 &\minus{}4014\\ 0 & 0\end{pmatrix}.$

That can't be: the square of a $ 2\times 2$ matrix cannot be nilpotent.


vector
induction
linear algebra
linear algebra unsolved

Let $V$ be a vector space (not necessarily finite dimensional) over some field $k$. Given a positive integer $n$, we say that a unital algebra $\mathcal A$ of $k$-linear operators on $V$ is [i]$n$-fold transitive[/i] if for every linearly independent $n$-tuple $x_{1},\ldots,x_{n}\in V$ and every $n$-tuple $y_{1},\ldots,y_{n}\in V$, there is a member $T$ of $\mathcal A$ which maps $x_{i}$ onto $y_{i}$ for all $i\in\overline{1,n}$.

Prove that a $2$-fold transitive algebra of operators is $n$-fold transitive for all positive integers $n$.

Let us first prove that there exists an operator $T\in \cal A$ that maps $x_{1}$ to $y_{1}$ and that maps $x_{2}, \dots, x_{n}$ to $0$. Since $\cal A$ is 2-fold transitive:
1) there exists an operator $S_{1}\in\cal A$ s.t $S_{1}(x_{1}) = y_{1}, S_{1}(x_{2}) = 0$.
2) there exists an operator $S_{2}\in\cal A$ s.t $S_{2}(y_{1}) = y_{1}, S_{2}S_{1}(x_{3}) = 0$.
...
$n-1)$ there exists an operator $S_{n-1}\in\cal A$ s.t $S_{n-1}(y_{1}) = y_{1}, S_{n-1}\circ\dots\circ S_{1}(x_{n}) = 0$.

Then the operator $T= S_{n-1}\circ\dots\circ S_{1}$ maps $x_{1}$ to $y_{1}$, and it maps $x_{2},\dots, x_{n}$ to $0$ as we wanted.

Now let $T_{i}$ be the operator that maps $x_{i}$ to $y_{i}$ and all other $x_{j}$ to $0$. Then the operator $T= \sum_{i=1}^{n}T_{i}$ maps $x_{i}$ to $y_{i}$.

[quote="Yury"]
2) there exists an operator $S_{2}\in\cal A$ s.t $S_{2}(y_{1}) = y_{1}, S_{2}S_{1}(x_{3}) = 0$.
[/quote]

Why is that? What if you have $S_{1}(x_{3})=y_{1}\ne 0$, say?

I hope I didn not misunderstand.

You are right  :D . There is a gap in the proof.

It is more convenient to prove the first part by induction. We are are proving that for every $x_{1}, \dots, x_{k}$ there exists an operator $T_{k}$ that maps $x_{1}$ to $y_{1}$ and all other $x_{i}$ $(2\leq i\leq k)$ to $0$. If $T_{k-1}$ maps $x_{k}$ to $0$ or a vector which is not collinear with $x_{1}$, we are done (see my original post). Otherwise, $T_{k-1}x_{k}= \alpha x_{1}$ where $\alpha\neq 0$. By rescaling $x_{k}$ we may assume that $\alpha =1$.
By the induction hypothesis there exists an operator $Q$ that maps $x_{1}-x_{k}$ to $y_{1}$ and $x_{2},\dots, x_{k-1}$ to $0$. Now, if $Q(x_{1})$ and $Q(x_{k})$ are linearly independent, we can use the original argument. So
\[Q(x_{1}) = \beta y_{1}, \; Q(x_{k}) = (\beta-1) y_{1}.\]
Then for $T_{k}= Q-(\beta-1) T_{k-1}$, we have
\[T_{k}(x_{1}) = \beta y_{1}-(\beta-1) y_{1}= y_{1},\]
\[T_{k}(x_{k}) = (\beta-1) y_{1}-(\beta-1) y_{1}= 0,\]
\[T_{k}(y_{i}) = 0-0 = 0, \,\, \forall\, 2\leq i \leq k.\]

We are done  :lol:


This problem was used for Mathlinks Contest Round 3 if i 'm not mistaken:
If $x_1,x_2,...,x_n$ are positive reals number s.t. $\sum\frac{1}{x_i}=1$ then :
$\frac{x_2^2+x_3^2+...+x_n^2}{n-1}\frac{x_1^2+x_3^2+...+x_n^2}{n-1}...\frac{x_1^2+x_3^2+...+x_{n-1}^2}{n-1}\geq (\frac{\sum x_i^2}{n})^{n-1}$
    And solution was based on Lagrange's one.Here is an application:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=help&t=39548
.But my knowledge doesnt allow me understanding it!Any elementary solution!!

The solution was already posted. (by harazi i think)

It used an inequality with radicals from Old & New inequality, which was proved by Suranyi, so it's very difficult.

Unfortunately, I cannot find the topic anymore.


calculus
integration
trigonometry
real analysis
real analysis unsolved

complicated integral  :?:
\[\int \frac{x^{2}}{(x\cos x+\sin x)^{2}}dx \]

Yes, complicated. The [url=http://integrals.wolfram.com/index.jsp]Integrator[/url] couldn't even give any sort of closed form for it.


Show that $ f(x)\equal{}x^4\minus{}x^2\plus{}1$ is irreducible over the rationals

$ f(x) \equal{} \Phi_{12}(x)$.  Are you familiar with the proof that cyclotomic polynomials are irreducible?  If not, $ f(x)$ clearly has no rational linear factor and pairs of its conjugate factors don't multiply to rational quadratics.


ratio
trigonometry
geometry
angle bisector
geometry unsolved

$ABC$ is a triangle
$E, F$ are points in $AB$, such that $AE = EF = FB$

$D$ is a point at the line $BC$ such that $ED$ is perpendiculat to $BC$
$AD$ is perpendicular to $CF$.
The angle CFA is the triple of angle BDF. ($3\angle BDF = \angle CFA$)

Determine the ratio $\frac{DB}{DC}$.

%Edited!%

[quote="Johann Peter Dirichlet"]$ABC$ is a triangle
$E, F$ are points in $AB$, such that $AE = EF = FB$

$D$ is a point at the line $BC$ such that $ED$ is perpendiculat to $BC$
$AD$ is perpendicular to $CF$.
The angle CFA is the triple of angle BDF.

Determine the ratio $\frac{DB}{DC}$.[/quote]
I can’t understand sentence: The angle CFA is the triple of angle BDF.
Maybe it means $\angle CFA=3\angle BDF$. But $\angle CFA<\angle DFA=2\angle BDF$(since $BF=DF$)$<3\angle BDF$.
Also I think, that you had a mistake: you had in mind $\angle CFA=3\angle CFD$(or anything of such kind). Then this problem becomes rather nice.
Here it’s solution:
1) Since $\angle EDB=90^o$, $DF=FB=EF=AE=x$.
2) Let $\angle FBD=\angle FDB$ be $2a$. Then $\angle AFD=4a$ and $\angle AFC=3a$, $\angle CFD=a$.
3) $\angle DCF=\angle CFA-\angle CBA=a=\angle CFD$ :arrow: $CD=DF=x$.
4) $\angle ADE=90^o-\angle CFD-\angle EDF=a$.
5) From the theorem of sins for $\triangle {ADB}$
$AD=AB\frac{\sin \angle ABD}{\sin \angle ADB}=3x\frac{\sin 2a}{\cos a}$.
6) From the theorem of sins for $\triangle {ADF}$
$AD=AF\frac{\sin \angle AFD}{\sin \angle ADF}=2x\frac{\sin 4a}{\cos a}$.
7) From 5),6) we get  $\cos 2a=\frac{3}{4}$. But $BD=2BF\cos 2a=\frac{3x}{2}$. Hence $\frac{DB}{DC}=\frac{3}{2}$.

=====================================================
I can’t understand sentence: The angle CFA is the triple of angle BDF.
Maybe it means $\angle CFA=3\angle BDF$. But $\angle CFA<\angle DFA=2\angle BDF$(since $BF=DF$)$<3\angle BDF$.
Also I think, that you had a mistake: you had in mind $\angle CFA=3\angle CFD$(or anything of such kind). Then this problem becomes rather nice.
Here it’s solution:
Since $\angle EDB=90^o$, $DF=FB=EF=AE=x$.
Let $\angle FBD=\angle FDB$ be $2a$. Then $\angle AFD=4a$ and $\angle AFC=3a$, $\angle CFD=a$.
$\angle DCF=\angle CFA-\angle CBA=a=\angle CFD$ :arrow: $CD=DF=x$.
=====================================================
The ray $[AD$ is the bisector of the angle $\widehat {CAB}$, i.e. $AC=AF=2x$
Therefore, $\frac{DB}{DC}=\frac{AB}{AC}=\frac{3x}{2x}=\frac 32$.
Very nice problem and very nice Kozlov's proof.
Above is a ''broadway'' in your proof, Kozlov ! [b]I hope you don't make angry.[/b]

[b]A easy generalization.[/b] If $AE=y,\ EF=FB=x$, then $\frac{DB}{DC}=\frac{y+2x}{y+x}\ .$

Well, the question is correct (well, I participate of this competition :P). The problem is the picture: the angle C is very obtuse!

Is there a meaningful connection between the ratio 3 : 2  in this problem

  and the same ratio in Darij's

    http://www.cip.ifi.lmu.de/~grinberg/TPSymmedian.pdf

    Theorem 8  on p.17

   or is it a remarkable  coincidence ?

   T.Y.

   M.T.

Consider the $\triangle{ABC}$ whit $\angle{C}$ obtuse. Clearly $EF=FB=DF$ because $\angle{EDB}=90^{\circ}$ and $F$ is the midpoint of the hypotenuse $EB$ at the right triangle $\Delta{EDB}$. Let's call $\alpha$ to be the $\angle{BDF}$. $3\angle{BDF}=\angle{CFA}=3\alpha$. As $DF=FB$ the $\triangle{DFB}$ is isosceles whit $\angle{BDF}=\angle{DBF}=\alpha$. The $\triangle{DEF}$ is isosceles whit $DF=EF$ and $\angle{DFE}=2\alpha$ because $\angle{FBD}=\angle{FDB}=\alpha$. Let $Q$ be the point of intersection of $AD$ whit $CF$. As $\angle{CFA}=3\alpha$ and $\angle{DFE}=2\alpha$ then $\angle{DFQ}=\alpha$, therefore $\angle{QDF}=90^{\circ}-\alpha$. And as $\angle{EDF}=\angle{DEF}=90^{\circ}-\alpha$ it is clear that $DF$ is the bisector external of the $\angle{ADE}$. Now, as $DF$ is the bisector external of the $\angle{ADE}$ we have for the Angle Bisector Theorem that $\frac{ED}{AD}=\frac{EF}{AF}=\frac{1}{2}$. Let $M$ be the midpoint of $ED$, it is clear that $FM$ is the bisector of $ED$ because $EF=DF$, we see that the triangles $\triangle{EMF}$, $\triangle{DMF}$ and $\triangle{DQF}$ are congruent because $\angle{EFM}= \angle{DFM}=\angle{DFQ}=\alpha$ and $\angle{EMF}=\angle{DMF}=\angle{DQF}=90^{\circ}$ then cleary $ED=2DQ$, and as $\frac{ED}{AD}=\frac{1}{2}$ then $AD=4DQ$.
Now, $F$, $C$ and $Q$ are collinear points on the sides $AB$, $BD$ and $DA$ of the $\triangle{ABD}$, then for the Menelaus's Theorem we have that $\frac{QD}{QA}\cdot\frac{AF}{FB}\cdot\frac{BC}{DC}=1$, then $\frac{1}{5}\cdot\frac{2}{1}\cdot\frac{BC}{CD}=1$ $\Rightarrow{\frac{BC}{DC}}=\frac{5}{2}$
Therefore adding $1$, $\frac{BC}{DC}+1=\frac{BC+DC}{DC}=\frac{BD}{DC}=\frac{5}{2}+1=\frac{7}{2}$ :D

@above, isn't angle DFQ = -a?

(because it is 2a-3a as angle DFE is the larger one)

My answer is the same as @2above, $7/2$.
[asy]
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label("\begin{tikzpicture}
\coordinate[label=below:$A$] (A) at (-4,0);
\coordinate[label=below:$E$] (E) at (0,0);
\coordinate[label=below:$F$] (F) at (4,0);
\coordinate[label=below:$B$] (B) at (8,0);
\coordinate[label=above left:$D$] (D) at (.8,2.4);
\coordinate[label=above right:$C$] (C) at (3.2,1.6);
\coordinate[label=above:$K$] (K) at (2.4,3.2);
\draw[thick] (A)--(K)--(F)--cycle;
\draw[thick,blue] (E)--(D)--(B)--cycle;
\draw[thick,red] (D)--(F);
\draw[thick,purple] (E)--(K);
\foreach \s in {A,B,C,D,E,F,K}\filldraw (\s) circle (1.5pt);
\end{tikzpicture}");
[/asy]
Let $\angle BDF=x\implies \angle CFA=3x$ and $CF\cap AD=K$. Because $\angle BDE=\angle AKF=\pi/2$, then $E$ and $F$ are circumcenters of $(ADF)$ and $(BDE)$, respectively. We have \[\angle EFD = 2\angle EBD = 2\angle BDF = 2x\implies \angle DFC=x=\angle CDF\implies CF=DC.\]
Let $DC=a$. From LoS of $\triangle CFD$, we have $DF = 2a\cos (x)\implies AE=EF=FB=DF=2a\cos (x)$. BBy LoS of $\triangle BFD$, we have $BD=4a\cos^2 (x)$. By simple angle chasing, we have $\angle BAD=\pi/2 - 3x$ and $\angle ADB=\pi/2 + 2x$. From LoS of $\triangle ABD$, we have
\[\frac{4a\cos^2 (x)}{\cos (3x)} = \frac{6a\cos (x)}{\cos (2x)} \implies 2\cos (2x)\cos(x)=3\cos (3x)\implies 4\cos^3(x) - 2\cos (x) = 12\cos^3(x) - 9\cos (x)\]
and we have $\cos^2(x) = 7/8$. Therefore,
\[\frac{DB}{DC} = \frac{4a\cos^2(x)}{a} = 4\cos^2(x)=\boxed{\frac{7}{2}}.\]