It was good overall, but the ending was dumb.   :(  :lol:

The book was a great time consumer - had me very into it.  The movie doesn't do it justice - left too much out.  For those of you who haven't read it and need something to do this summer, check it out from the library...

I loved the start of it but the middle was okay.  I thought the end part was a bit cheesy.

The book was somewhat interesting overall.........though the end seemed to be kind of abrupt.......

It was enjoyable (not really something to delve into though). The book that I liked, which had some parts in common to the Da Vinci Code, was the The rule of Four. I thought it was much more interesting. ;)

"The rule of four" ?? I've not really heard of it..... Who's the author of the book ??

It is a Princeton and a Harvard guy(they wrote it together). Their names are Ian Caldwell and Dustin Thomason.

[quote="José"]What's your opinion about the book????[/quote]
Pulp but not completely junk :(  :)

The book is very cleverly written I feel. Dan Brown has successfully passed of what is mostly rumour, unproved conspiracies etc. as truth to the largely unwitting population. He's put enough padding between each of his rather dubious claims that you do eventually get suckered in.

Until you watch the film and realise the thigns we are supposed to believe are true are so riduculously unfounded on hard evidence you realise it must be a work of fiction.

My judgement: Moderately enjoyable fiction novel.

(And I'm not even Christian...)

I haven't seen the film yet, but for those of you who don't know it (shame on you! ;) )

[url=http://en.wikipedia.org/wiki/Audrey_Tautou]Audrey Tautou[/url]
also played in [url=http://en.wikipedia.org/wiki/Am%C3%A9lie]Le fabuleux destin d'Amélie Poulain[/url]

and what do you think about the anger of The Vaticano??? THe book is unreal, it doesn't say true thinks no? or I'm wrong?

The one thing that slightly bothers me is that Dan Brown attempts to reveal the fact that Jesus had children as some "earth-shattering" event that the world has never witnessed before, however, biblical historians and other such professionals have known about this fact, written about it, and have spoken about it since the 3rd century.

I did enjoy the movie though, purely for it's entertainment value. I find all those little historical flashbacks cool.

The movie was bad. I almost fell asleep at one stage. :?

The problem is Hanks and Howard aren't really suited to the slow pace of a conspiracy thriller. It does lag an awful lot and frankly you could hardly care less about what happened to the characters by the second half.

[quote="shadysaysurspammed"]The movie made up iys own story,i mean deleted things HIGHLY MODIFIED stuff,like instead of the library,they found A Pope by a mobile thing,the ending was totally changed,onl one cryptex,BS movie and bad direction.Even tough the movie wuld have been longer if they went according to the book,it would still be worth watching,like LORD OF THE RINGS was long but still won oscars and goodies.

@Centy--10 billion people who read it and liked it were not christian....[/quote]

Please make some effort to post comprehensibly.

If puerile writing and moderately original storylines are your cup of tea, then I could see why you would find the Da Vinci code interesting, but really there is so much better stuff out there, even in the same genre.  Frankly I'm incredibly tired of hearing people discuss it, whether on the street or on CNN, and both the Da Vinci code apologists and critics make me incredibly angry by this point in time.

I have no plans to see the movie.

Earlier, a few people said the ending was bad (to the book).  I loved the book, but the ending did seem kind of...stupid.  However, if you really think about the book, the ending is one of the most clever endings I'v eever seen.  On a wild tangent, Harry Potter #7 will have to have a KILLER ending to live up to the series standard.

The book itself was fairly enjoyable, but Dan Brown is a pretty poor writer in my opinion.  His prose is horribly bland and simplistic.

The movie went to fast for anybody that hadn't read the book to understand.   :(
6/10

The book was amazing. It was gripping and extremely thought-provoking. But it's unfortunate that Dan Brown made the claim in the beginning of the book that all historical descriptions were factual; not even most true scholars would assert that their findings are 100% accurate, even with sources cited, which Brown sorely lacked. If you're going to be that presumptuous, have something to back up your assertions, especially those as controversial as his.

[quote="JBL"]Okay, ready:  WORK OF FICTION.  Criticizing the author of a [i]novel[/i] for a lack of historical correctness is very, very silly.[/quote]

Particularly when it's in his best (financial) interests to imply that the events of the novel have some basis in reality.

JBL, I don't know if you're familiar with the book, but in the preface (or whatever he titled the pre-section), Brown states explicitly that "All descriptions of [historical stuff] are accurate." So I'm criticizing specifically his claim and subsequent failure to substantiate that claim, particularly as his assertions are very controversial. It would seem that this sort of criticism is not really silly at all.

I don't care what he says in the preface.  Shame on anyone who takes seriously a claim about the historical accuracy of a work marketed as fiction.

I'm not sure why you would assume that an author of a fictional work is not qualified to make historical claims merely because his work is labeled fiction. That claim is certainly silly.

No, I didn't say he wasn't qualified, I said no one should take him seriously.  Which is true.  If he were making an historical argument, he should have presented it in the language of history -- either an article to a peer-reviewed historical journal or a non-fiction work of history with proper sourcing.  Anyone who wants to have his work presented as fiction automatically gives up the opportunity to make any historical claims (or claims of any other sort: my historical novel about the Big Bang is never going to be taken seriously by cosmologists, with good reason).

Furthermore, who is going to begin a novel by saying that it [i]isn't[/i] historically accurate?

well, whether it's marketed as fiction or not, people don't seem to realize that- many take it at face value, especially with his statement that everything is accurate!


function
real analysis
real analysis unsolved

Show an example of a function $ f: R \rightarrow R$ which is measurable, but $ \frac{1}{f}$ is not.

f measurable means

$ A_c \equal{} \{ x \in R: f(x)\equal{}c \}$ are all in your sigma algebra

so $ \{x \in R: \frac{1}{f(x)} \equal{} c \} \equal{} A_\frac{1}{c}$ must be in the algebra as well

You need to change those sets $ A_c$ kenn, that is not enough to be measurable. More generally, if $ g$ is Borel-measurable and $ f$ is measurable (on any measure space), then $ g \circ f$ is measurable


inequalities
trigonometry
geometry
angle bisector
geometric inequality
geometry proposed

Given quadrilateral $ABCD,M$ is point inside it let internal bisectors of angles $\angle APB=l_{a},\angle BPC=l_{b},\angle CPD+l_{c},\angle DPA=l_{d}$ prove that
$MA+MB+MC+MD\ge \sqrt{2}(l_{a}+l_{b}+l_{c}+l_{d})$

[color=purple]This is the special case of the following problem which is created by Lenhard in 1968 0r 1970
  Let $A_{1}A_{2}...A_{n}$ is a convex polygon in plane.M in the plane has the polygon.Let $MA_{i}=m_{i}, l_{i}$is the angle bisector of the angle $%Error. "hatA" is a bad command.
_{i}MA_{i+1}$ with n+1 = 1 .Prove that
 $\sum m_{i}\ge \sum l_{i}$
   [/color]

You posted a nonstrict  inequality! Why do you like repling my topic with insignificant post ?

Evarist's inequality is wrong though. Let $A_{1}A_{2}A_{3}A_{4}$ be a quadrilateral and $A_{1}A_{3}$ is perpendicular to $A_{2}A_{4}$ at $O$. Let $M$ be $O$.

[color=purple]Sorry gemath if i make you angry ! I read it in Geometric inequality of Bottema page 144 and I think it's true so i post it and i think it's not insignificant.Please test it's and don't cricitize me ! [/color]

Did you see poster of treegoner? Check it ! , I said it isn't strict because it should be
$\cos\frac{\pi}{n}\sum_{i=1}^{n}MA_{i}\ge \sum_{i=1}^{n}w_{i}$ with $w_{i}$ is biector length of $\angle A_{i}MA_{i+1},(i=\overline{1,n},n+1\equiv 1)$

[color=purple]Sorry gemath ! It has a mistake and must be as you said .I posted  miss.
         Thanks treegoner![/color]


Hi!

Let $b|2n^{2}$ and $b,n\in \mathbb{N}$

Show that $n^{2}+b$ can't to be a perfect square.

plz :bye:

Let $2n^{2}= kb$ for a positive integer $k$.
Then $X = n^{2}+b = \frac{n^{2}(k+2)}{k}$
a) If $(k,n^{2}) = 1$, then $k|k+2 \Rightarrow k=1,2 \Rightarrow X \in \{2n^{2},3n^{2}\}$ which can't be a perfect square.

b) If $(k,n^{2}) = p > 1$, let $k=ps, n^{2}= pt, (s,t) = 1$.
Then, $X = \frac{t(ps+2)}{s}\Rightarrow s=1,2 \Rightarrow X \in \{t(p+2),t(p+1)\}$.

b1)Consider $t(p+1) = \frac{n^{2}}{p}\cdot (p+1)$.
If $\frac{n^{2}}{p}$ is not a perfect square, then there must be $q$, a common prime factor of $n$ and $p$ so that the multiplicity of $q$ is odd in $\frac{n^{2}}{p}$. Since $X$ is a perfect square, then $q|p+1$ which is impossible since $(p,p+1) = 1$.
If $\frac{n^{2}}{p}$ is a perfect square, then $p$ is a perfect square and so does $p+1$. But no two positive perfect squares differ by $1$.
So, $t(p+1)$ isn't a perfect square.

b2)Consider $t(p+2) = \frac{n^{2}}{p}\cdot (p+2)$.
If $p$ is odd, then $(p+2,p) = 1$. By a similar (but not exactly the same) reasoning with b1), $t(p+2)$ isn't a perfect square.
If $p$ is even, let $p = 2r$. Then $t(p+2) = \frac{n^{2}}{k}\cdot (k+1)$ which is exactly the case b1).

Therefore, if $b | 2n^{2}$, then $n^{2}+b$ can't be a perfect square..  :D

It is obviosly. If odd prime $p|b$ and $n^{2}+b$ is perfect square, then $ord_{p}(b)$ is even. Therefore or $n=xy, b=x^{2},n^{2}+b=x^{2}(y^{2}+1$, or $n=xy,b=2x^{2},n^{2}+b=x^{2}(y^{2}+2)$. But never $y^{2}+1,y^{2}+2$ are perfect square for $y\in N$.

[quote="Rust"]It is obviously. If odd prime $p|b$ and $n^{2}+b$ is perfect square, then $ord_{p}(b)$ is even. Therefore or $n=xy, b=x^{2},n^{2}+b=x^{2}(y^{2}+1)$, or $n=xy, b=2x^{2}, n^{2}+b=x^{2}(y^{2}+2)$.
But never $y^{2}+1,y^{2}+2$ are perfect square for $y\in N$.[/quote]

Well, do you have any complete proof for your two statements ? I can' t understand how you can deduce them...  :maybe:

If $ord_{p}(b)$ is odd, then $ord_{p}(n^{2})>ord_{p}(b)\Longrightarrow ord_{p}(n^{2}+b) \-$ odd - contradition (because $ord_{p}(n^{2}+b)$ even). Therefore for all odd primes p|b we get $ord_{p}(b)$ is even. It mean $b=x^{2}$ or $b=2x^{2}$. For both case $x|n$ (or $n=xy$).


calculus
integration
calculus computations

Let $ f: \mathbb{R}^n\rightarrow R$ be defined by $ f(x) = |x|^{ - p}\chi_{B}(x)$, where $ B$ is unit ball in $ \mathbb{R}^n$ with center at the origin. Compute the integral of $ f$ over $ \mathbb{R}^n$.

There is a nice equality (for non-negative $ f$)
\[ \int_{\mathbb R^d} f(x) \lambda(dx) \equal{} \int_0^\infty \lambda(\{x: f(x) \ge t\}) dt.\]

Thanks,
Just I have proved this nice equality and I have solved the problem. :roll:


How long did you spend on round 1? I personally spent 1.5 hours, which is why I did so crappily (10 points approximately).

Ha ha, I didn't even turn in my solutions.  I just dropped out,

I had other things...

 :oops:

about 4 hours (over the course of two days) on 1,2a,3,4,5, and then about 3 hours (over the course of a week) on 2b and I still missed it yay

[quote="Phelpedo"][quote="bpms"][quote="SplashD"]I don't think the polls will show true results because some people don't answer all the questions.[/quote]
You mean like me :P.[/quote]

So...why won't those be the true results?[/quote]

People who answers all five questions will most likely spent more time on it than people who only attempted one and gave up on the others right away.

Around 5 hours to solve them, then 10+ hours to typeset my solutions. (I'm relatively inexperienced with typesetting using TexnicCenter)

[quote="bpms"][quote="K81o7"]About 30min-1 hour on everything but #4, and then about 1.5 hours on #4.[/quote]
Do you mean 30 min-1 hour on each problem or combined!!!!!!!!!![/quote]

I meant 30 min to 1hr for all.

But then again, that was only the time taken to solve the problems.  The solutions typing took a couple hours as well.

Does writing the solutions count? Because when I was writing the solutions, I would write like 1 sentence, spin around my room for 5 minutes in my rolly chair, write another sentence, etc.

hahah, i did the rolly chair approach too, I'm afraid some of the spin times were over 5 minutes tho....i expect the same this month

Um... 30 minutes on #1, 10 minutes on #2a, 1.5 hours (at least) trying to find nice solution to #2b (brute forced it...), 40 minutes on #3, 10 minutes on #4, and 1 hour on #5.

Plus 1 hour to write it up...

(5 hours?)

Wait, more than that

[quote="13375P34K43V312"]How did it take you 30 minutes for #1? It seemed pretty trivial to me.[/quote]
This is just rude.

[quote="13375P34K43V312"]How did it take you 30 minutes for #1? It seemed pretty trivial to me.[/quote]

how did you not get #4 and #5 AT ALL, they seem pretty trivial to me ;).

I am a parent. This will be my son's 4th year with USAMTS....

I do hope no one is discouraged by this thread. I remember my son's first year. He was in 8th grade, and had never participated in a math competition.

When he first saw the questions, he said he couldn't do them. Later that day, I tried to encourage him - I told him that math problems could be daunting at first, but sometimes if you think about them for a while, a solution will come. He shrugged and said he had already figured out two of them. The other three took many hours, with the last one only being solved the day before the deadline. That first year he put an amazing amount of time and effort into the test. He wrote computer programs to brute force some of the answers. He answered all the questions, except the geometry question in the 2nd round. I am sure he put, on average, 20-30 hours per round. This was the last year the test was administered by the NSA. We had not yet found AoPS. I am glad he never had to see how trivial the test was for some kids.

His second year he also put a fair amount of work into it - but no longer needed to use computer programs. It was the first year on AoPS, so he also learned Latex. I know as I read this forum, I was amazed at how quickly some of the older students were doing the problems.

Last year, as a sophomore, it took a lot less time.

This summer we were on a family trip. We stopped at a library to check our email, and discovered the first round was published. He solved two before we left the library about 20 minutes later. What with setting up camp, and a few forced hikes, he didnt finish all of them until the next day. The 2nd round came on a day he was pretty busy, but when I went in to talk to him in the evening, I noted there were tick marks next to each of the problems. He had solved them all.

Of course, he takes a while to Latex them all, but that also takes a lot less, because with experience comes shorter answers.

People should not be asking how long it takes, without also wondering how long they have been doing it, and what is their other math experience. Heck, you should not care at all. This is a great test - you compete only against yourself, and there is so much to learn from those questions. Stick with it! It is a process that can lead to much better outcomes of more competitive contests!

Question 1: 45 minutes of Number Theory  :rotfl:
Question 2: 1 hour of working to find that my solution was incorrect, another 2 hours finding a partially incorrect solution (because I interpreted the problem wrong).
Question 3: 1 and a half hours
Question 4: an hour of contemplating, then 15 minutes after I found a method
Question 5: an hour of contemplating and beginning an analytic solution, another hour to finish (not the analytic solution :D)

About 8 and a half hours, not including the many hours of $\LaTeX$ing and constructing diagrams. :lol:


I think this is an IMO problem and its solution is specified in Arthur Engel :)  :?:

[quote="befuddlers"]
Could somebody post a solution for this problem.I could not figure it out. :(
If ab+1 /a^2 +b^2, prove that a^2 +b^2/ab+1 is always a perfect square.a,b are whole numbers.[/quote]

As stated this makes no sense ...can you print the problem  clearly (and  correctly ,  (putting "(" at right places, and actually stating what is given etc...)?..


A square is 3 units on a side. A 1x1 square is cut from one of its corners, and this new figure is inscribed in a circle. Find the exact radius of the circle.

draw the perpendicular biscectors of the segments connecting the points, and find the intersection...but one thing that bothers me is that the other 3 verticies of the triangle are still the same so the circumcircle won't change if you cut off one of the corners

The new vertices of the figure aren't concyclic, since three points define a circle (so take the three unchanged points).  The radius should be the same as if it were still a square, which is $\frac{3 \sqrt{2} }{2}$... right?

Oops.  :blush:  Didn't think this one through. sorry.

So okay... How about, what would be the radius of a circle that was inscribed inside this figure?

well, it is tangent to the sides opposite to the cut, and the inner vertex is on the circle, also the center lies on the line connecting the inner cut vertex, and the opposite vertex, and all you have to do is find a point that is tangent to the sides and passes through that point...


In the year 1985, the date April 25th had a unique property.  When the date was written as 4-25-1985, the number for the month and day, 425, could actually equal the year 1985. In other words, in what number base does 425 equal 1985 in base 10?


LaTeX
inequalities
geometry
3D geometry
dodecahedron
videos
number theory

It's a self-contradictory statement, because when I say "I do not speak English", I am actually saying several words of English.

Post similar stuff for other languages. For example: "Yo no hablo espanol" (and of course there should be a tilde over the N, but I'm too lazy to find the special characters).

It's Opposite Day

Think about that one for a moment...

[quote="fredbel6"]I've never read a topic full of spam.[/quote]
I have never quoted a message.

I have never fallen asleep while I was reading pointless forum topics.

[quote="mathnerd314"]It's a self-contradictory statement, because when I say "I do not speak English", I am actually saying several words of English.[/quote]
Actually this is not self-contradictory? You could've copied this without understanding it.  And you don't have to speak, you only have to write.

[quote]
Post similar stuff for other languages. For example: "Yo no hablo espanol" (and of course there should be a tilde over the N, but I'm too lazy to find the special characters).[/quote]
Buy an AZERTY-keyboard.

[img]http://bblfish.net/tmp/2004_09_29/titanium_azerty_keyboard.jpg[/img]

I have never posted an image.

[hide]how can u hide stuf?[/hide]
[i]im still wondering how to italicize[/i]

I cannot believe that today isn't tommorow, yesterday isn't today and people have feelings  :o

I am inside the exterior of my abode watching the inside of my skull while juggling cartoon characters on my invisible friend.

Porque é que estou a perguntar porquê?...

I need to take my computer to the Apple Store.

Where is that darn keyboard?

I do not understand what I'm typing.

My plane to France crashed, my family died, and I'm a vegetable.  Be back on Wednesday!

anybody know a person named [b][color=green]Pingala[/color][/b].

i cant seem to find him!! :(

i dont even know how to type!! :rotfl:  :(  :P

That "Rating: 6.0" in your location is really corny, you know.

On a side note, such things can not possibly be corny, and therefore you should change it to avoid an anomaly that will destroy the universe.

I believed what Temperal just said. I also died while living because ships sink upwards.

A better version of the sentence I said earlier:

I remember very vividly that I was diagnosed with severe amnesia, caused by hypothermia, at the age of 5, and I still am not near recovering.

(If you are not stupid) There are two lines in this post that may deceive you.

_________________
This is not an automatic signature.

"i will never say never"

- Justin Beiber

Within the true sentences of this post, and only withing them, I don't talk about myself using reflexive pronouns.

Finally, I never post trying to be fun.

_________________
This is not an automatic signature.


inequalities
trigonometry
inequalities unsolved

Here is its genaral  http://www.mathlinks.ro/viewtopic.php?t=217783   :lol:

[quote="arqady"]I agree with you, tobeno_1! It's very easy, but it's nice too. :wink:
The generalization:
Let $ x_i\in\mathbb R,$ $ y_i\in\mathbb R$ such that $ \sum_{i \equal{} 1}^nx_i^2\leq1,$ $ n\in\mathbb N.$ Prove that
\[ \left(\sum_{i \equal{} 1}^nx_iy_i \minus{} 1\right)^2\geq\left(\sum_{i \equal{} 1}^nx_i^2 \minus{} 1\right)\left(\sum_{i \equal{} 1}^ny_i^2 \minus{} 1\right)
\]
[/quote]

Yes, very nice generalization.

Obviously, $ \sum{y_i^2}\leq 1$

$ RHS\leq (\frac{\sum{x_i^2}\plus{}\sum{y_i}^2\minus{}2}{2})^2 \equal{} (\frac{2\minus{}\sum{(x_i^2\plus{}y_i^2)}}{2})^2 \leq LHS$ :lol:

Nice proof, tobeno_1! Your better than mine:
Let $ \vec x \equal{} (x_1,x_2,...,x_n),$ $ \vec y \equal{} (y_1,y_2,...,y_n),$ $ |\vec x| \equal{} x,$ $ |\vec y| \equal{} y$ and $ (\widehat{\vec x, \vec y}) \equal{} \alpha.$
Hence, $ \left(\sum_{i \equal{} 1}^nx_iy_i \minus{} 1\right)^2\geq\left(\sum_{i \equal{} 1}^nx_i^2 \minus{} 1\right)\left(\sum_{i \equal{} 1}^ny_i^2 \minus{} 1\right)\Leftrightarrow$
$ \Leftrightarrow(xy\cos\alpha \minus{} 1)^2\geq(x^2 \minus{} 1)(y^2 \minus{} 1)\Leftrightarrow$
$ \Leftrightarrow(x^2\cos^2\alpha \plus{} 1 \minus{} x^2)y^2 \minus{} 2xy\cos\alpha \plus{} x^2\geq0.$
$ x^2\cos^2\alpha \plus{} 1 \minus{} x^2\geq0$ because $ x^2\leq1$ and
$ \frac {\Delta}{4} \equal{} x^2\cos^2\alpha \minus{} (x^2\cos^2\alpha \plus{} 1 \minus{} x^2)x^2 \equal{} \minus{} x^2(1 \minus{} x^2)\sin^2\alpha\leq0.$ Done! :)


Let be $ n\in \mathbb{N}^*$ and the functions $ f,g: \mathbb{R}\rightarrow \mathbb{R}$ with $ f(x)\equal{}\sum_{k\equal{}1}^n|x\plus{}k|\plus{}\sum_{k\equal{}1}^n|x\minus{}k|$ and $ g(x)\equal{}f(x)\plus{}|2x\minus{}1|\ ,\ (\forall)x\in \mathbb{R}$ . Prove that $ f(x)\ge n(n\plus{}1)\ ,\ (\forall)x\in \mathbb{R}$ and find the minimum value of $ g(x)$ .

[hide="solution"]
1)$ f( \minus{} x) \equal{} f(x)$,so $ f(x)$ is even function,thus we can assume $ x\ge 0$.

The minimum happens when $ x$ is an integer because $ f(x)$ is linear on each interval $ [i,i \plus{} 1]$,
and for $ x > 1$,the tangent is positive,hence $ f(x)$ is increasing for $ x > 1$.

therefore,$ f(x)\ge min\{f(0),f(1)\} \equal{} n(n \plus{} 1)$ as desired.

2)$ g(x)\equal{}f(x)\plus{}|2x\minus{}1|\ge n(n\plus{}1)$
equality occurs if and only if $ x\equal{}\frac{1}{2}$

hence $ n(n\plus{}1)$ is the minimum.
[/hide]


I'm pretty sure you missed the tryouts by a long shot.


AMC
AIME
MATHCOUNTS
USA(J)MO
USAMO
blogs
AMC 8

I hope this isn't too spammy, but if anyone could, I'd be wondering if there could be an entire AMC sequence run (AMC 8=>AMC 10/12A=>10/12B=>AIME=>USAMO). If I was smart enough I would start one, but...

I look forward to other people doing it  :D

I think it is kind of unlikely that this would happen, because in general, the people interested in taking/writing a mock AMC are not the same as the people interested in taking/writing a mock AIME or USMO.  I think most people preparing for olympiads would really rather just work on olympiad problems.  Aside from this, it would be a [i]lot[/i] of work, especially if it would include the AMC 8 and two versions of the other AMCs, and whether people would stay with it from beginning to end is another issue.  Note that the AMC has 30 to 40 people in committees, plus 100-odd more contributors, to make its annual competitions. :wink:   But if someone is willing to come along and prove me wrong, that's fine, too.

I doubt that a single person would have enough time to do it, but I'm sure if you got a small group of people it wouldn't be too hard. I'd be glad to contribute (I am considering writing another Mock AIME anyway after college applications are done).

I'm working on doing an entire mock mathcounts track. Hopefully it'll be out by January.

I have a full mock USAMO that I haven't released yet. I would be willing to save it for this if other people are willing to write mock other things.

BTW, check my blog for a current mock USAMO that ends Dec. 3 (day 1 solutions due Nov. 26).


$P$ is a point inside $\triangle ABC$ such that $\angle PAC + \angle PBC = \angle PAB + \angle PCB$. What is the locus of $P?$

The A-Apollonian circle of triangle ABC.

Beautiful problem! Xeroxia did you solve this problem? Can you post a solution?

Let $AP, BP$ and $CP$ intersetcs the circumcircle at $D,E$ and $F$ respectively.
We know that $\angle DAC + \angle EBC = \angle BAD + \angle FCB$.
$\angle DAC = \angle DFC$ and $\angle EBC = \angle EFC$. So  $\angle DAC + \angle EBC = \angle EFD$. Similarly $\angle BAD + \angle FCB = \angle FED$. So $DE=DF$.
$\triangle PAC \approx \triangle PFD$ $\Rightarrow$ $\frac {AP}{FP} = \frac {AC}{FD}$  $(..i..)$

$\triangle PAB \approx \triangle PED$ $\Rightarrow$ $\frac {AB}{ED} = \frac {AP}{EP}$  $(..ii..)$

$\triangle PBC \approx \triangle PFE$ $\Rightarrow$ $\frac {PF}{PB} = \frac {PE}{PC}$  $(..iii..)$

And we also know $ED = FD$   $(..iv..)$
Multiply $(..i..)$, $(..ii..)$, $(..iii..)$ and $(..iv..)$.

$\frac {AP}{FP} \cdot \frac {AB}{ED} \cdot \frac {PF}{PB} \cdot ED= \frac {AC}{FD} \cdot \frac {AP}{EP} \cdot \frac {PE}{PC} \cdot FD$.

Finally we get $\frac {AB}{BP} = \frac {AC}{PC} \Rightarrow \frac {AB}{AC} = \frac {PB}{PC}$. This means $P$ and $A$ is on same Apollonius Circle of points $B$ and $C$.

From the definition of point $P$. The locus is an arc which is the intersection of that circle and the interior of $\triangle ABC$.

Now we will show that every point $P$ on this arc satisfies $\angle PAC + \angle PBC = \angle BAP + \angle PCB$:

[b]Let A, B, C, A', B', C' be concyclic points. Then the lines AA', BB', CC' are[/b]              [b]conccurently iff there is the relation AB.CA'.B'C'=BC.A'B'.C'A.[/b]
With this wellknown above problem can to solve the IMO's problem.
Remark. ABCA'B'C'    ----->    AB.CA'.B'C'=BC.A'B'.C'A'
             ACBA'C'B'    ----->    AC.BA'.C'B'=CB.A'C'.B'A
             AB'CA'BC'    ----->    AB'.CA'.BC'=B'C.A'B.C'A  a. s. o.

In the last message I remark this wellknown problem: [b]If the point[/b] $A,B,C,A',B,C'$ [b]belong to the circle[/b] $C$ [b]then[/b]
$AA'\cap BB' \cap CC' \ne \emptyset\ (ABCA'B'C') \Longleftrightarrow AB\cdot CA'\cdot B'C'=BC\cdot A'B'\cdot C'A.\Longleftrightarrow$
$(ACBA'C'B')\ AC\cdot BA'\cdot C'B'=CB\cdot A'C'\cdot B'A.\Longleftrightarrow $
$(BACB'A'C')\ BA\cdot CB'\cdot A'C'=AC\cdot B'A'\cdot C'B.\Longleftrightarrow a.s.o.$

I return to the propose problem. From the relations $DE=DF,$
$(ABCDEF)\Longrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$ and
$(ACBDFE)\Longrightarrow AC\cdot BD\cdot FE=CB\cdot DF\cdot EA$ results
$\frac{AB}{AC}\cdot \frac{CD}{BD}=\frac{FA}{EA}$, i.e. $\frac{AB}{AC}=\frac{BD}{CD}\cdot \frac{AF}{AE}=\frac{BD}{AE}\cdot \frac{AF}{CD}=\frac{PB}{PA}\cdot \frac{PA}{PC}=\frac{PB}{PC}\Longrightarrow \frac{AB}{AC}=\frac{PB}{PC}$a.s.o.

Rewrite $\angle PAC + \angle PBC = \angle PAB + \angle PCB$ as
\[ 180^\circ - \angle APC - \angle ACP + \angle PBC = 180^\circ - \angle APB - \angle ABP + \angle PCB \iff \angle APC - \angle ABC = \angle APB - \angle ACB =: \theta \]

Let also $\beta : = \angle ABC$ and $\gamma : = \angle ACB$. Thus $\angle APC = \beta + \theta$ and $\angle APB = \gamma + \theta$.

Apply an inversion with center $A$ and any radius $k$. We denote $X'$ the image of point $X$ of this inversion. Hence $\angle AB'P' = \angle APB = \gamma + \theta$ and $\angle AB'C' = \angle ACB = \gamma$ and, consequently, $\angle P'B'C' = \angle AB'P' - \angle AB'C' = \theta$. Anagolously, $\angle P'C'B' = \theta$.

Thus the triangle $P'B'C'$ is isosceles and $P'B' = P'C' \iff {k^2\over AP\cdot AB} PB = {k^2\over AP\cdot AC} PC \iff {BP\over CP} = {BA\over CA}$, that is, $P$ belongs to the $A$-Apollonius circle of $BC$. Note the we can reverse everything, including the inversion, so the desired locus is indeed the double-open-ended arc of this Apollonius circle inside the triangle.


logarithms
function
real analysis
real analysis unsolved

For what values of $ \lambda$ will the solution $ x$ to
	\[ \begin{aligned}\dot x(t) &\equal{} x^{3}(t)\plus{}\lambda x^{2}(t)\\ x(0) &\equal{} 1\end{aligned}\]
	exist for all $ t\ge0$ and be uniformly bounded?

For $ \lambda\le\minus{}1$. You can solve for $ t(x)$ explicitly.

[hide="Solution of the d.e."]I will write the problem in a more familiar notation (for me): $ y' \equal{} y^{3}\plus{}\lambda y^{2}$, $ y(0) \equal{} 1$. This is a separable variable d.e.:

$ \frac{dy}{dx}\equal{} y^{3}\plus{}\lambda y^{2}\Rightarrow \frac{dy}{y^{3}\plus{}\lambda y^{2}}\equal{} dx$

$ \Rightarrow \left[\frac{1}{\lambda^{2}}\left(\frac{1}{y\plus{}\lambda}\minus{}\frac{1}{y}\right)\plus{}\frac{y^{\minus{}2}}{\lambda}\right]dy \equal{} dx$

$ \Rightarrow \frac{1}{\lambda^{2}}\ln\left(\frac{y\plus{}\lambda}{y}\right)\minus{}\frac{1}{\lambda y}\equal{} x\plus{}C$.

For y(0) = 1 we get $ C \equal{}\frac{\ln(1\plus{}\lambda)\minus{}\lambda}{\lambda^{2}}$, and so

$ x \equal{}\frac{1}{\lambda^{2}}\ln\left(\frac{y\plus{}\lambda}{y(1\plus{}\lambda)}\right)\plus{}\frac{y\minus{}1}{\lambda y}$.[/hide]

I hoped that someone make some conclusions without explicit construction of the solution.

Here is a nonconstructive proof that $ \lambda >\minus{}1$ is impossible.
[hide]
Since $ x(0) \equal{} 1$, we must have that $ x(t) > 0$ for $ t\ge0$ (by uniqueness of solution). Since $ f(x): \equal{} x^{3}\plus{}\lambda x^{2}> 0$ for $ x > 1$, $ \dot x > 0$ in $ (0,\infty)$ implying that $ x$ is strictly increasing. But $ f$ is strictly increasing on $ [1,\infty)$, hence $ \dot x$ is strictly increasing on $ [0,\infty)$, i.e. that $ x$ is strictly convex on $ [0,\infty)$. It remains to note that an increasing strictly convex function cannot be bounded from above. Thus this case is impossible.[/hide]

Okay, for $ \lambda<\minus{}1$ the solution remains in the strip $ 0<x<1$ (it cannot reach $ 0$ by the uniqueness theorem), in which $ x'<0$.
With $ \lambda\equal{}\minus{}1$ you have a stationary solution.


Do you have any WASL/USAMO conflicts? I have issues with the WASL on my own, even though I didn't qualify for the USAMO.

SInce I'm dropping out of high school this year, I question if I really need to take the WASL. I would much prefer spending the time on much needed AP examination prep.

Conveniently, my school has the WASL end at 9:30AM so I can just make the USAMO. I'll probably leave it 10-15 minutes early but nobody cares anyway.

As for your situation, I don't really know whether or not you need to take the WASL.

The WASL doesn't really serve any purpose...
skip it...its not like they can do anything to you for not taking it.

For our school, passing has become graduation requirement ---- AFTER our year!!!! So i am happy to skip it. :D

BTW, I am going to postpone my reading section due to USAMO. Heheee :D

Lol - problem is, I don't think that the administrators will take skipping the WASL very nicely so I don't think trying to negotiate is worth the effort. What I'm thinking about doing is going through the WASL really quickly (hopefully for less than half the time required) then telling the teacher I finished and if I can read a book or not. THen I'll pull out my AP exam prep. :D

most of the time during WASL time i spent doing nothing b/c they give you SOOOOOOOOOOOOOOOOO much time for SOOOOOOOOOOOO easy math. (im in elementary school  :lol: so i dont know what its like anywhere higher, but yeah...)

I take WASL in a 10 person class full of nerds this year, we all finish in 1/4 to 1/5 of the time and then we play hearts and/or watch British sitcoms. I would skip WASL if I were you, and not making USMAO sucks, I skrewed up the AMC 10 (98 I think) :( .


If a < b, find all solutions to $ a^b = b^a $

$(a,b) = (2,4)$ works. :D

2^4 = 16 = 4^2 :D  :D

And I think $\gcd(a,b) \neq 1$ for sure. Otherwise, they'ren't going to be equal..

[hide="From there.."]
Let $b = ka$ where $k$ is an integer > 1. Then:

$ka^a = a^{ka}$..

I don't know how to go from here.. Perhaps I have wrong approach? :? [/hide]

2 and 4, I just did it by reasoning. Are there any other solutions? :?

Ah! I thought of and solved this problem a while ago! The only answer is indeed $(2,4)$, but theres another one like it, if you agree with it: $(-2,-4)$.

Heres the way i solved it.

The given equation is equivelent to $a^{\frac{1}{a}}=b^{\frac{1}{b}}$

Consider the graph of $y=x^{\frac{1}{x}}$   It is a known fact that this graph peaks at $x=e$

Now, consider a horizontal line that cuts this graph twice, once on either side of $e$, the maximum. The x-values of these two co-ordinates satisfy the equation we want to solve. These are the only solutions to it, aswell. Since a and b are to be dfferent, they have to lie on either side of $x=e$, so that means we only have to check the values 1 & 2. a=1 yields b=1. a=2 yields b=4. The only solution! Voilá! Hope this is clear, ask if i need to go deeper, but think about it first!

Consider the graph of $y=\frac{\ln x}{x}$.

[quote="flip2004"]If $ 0 < a < b \; , $  then  \[ a=a(t):=\left(1+\frac{1}{t}\right)^t \; \; ,\; \; \; b=b(t):=\left(1+\frac{1}{t}\right)^{t+1}\; \; ,\; \;
 \; \; t>0 .\]
[/quote]
In the following $ e $ is Napier's constant and  $ {\mathbf P}:= \frac{e}{2}-1 \; ,\; {\mathbf Q}:=\frac{e}{24}\; . $ If $ \left( a(t),b(t) \right) $  are the solutions of equation $  a^b=b^a  $ with $0<a<b  $ (see  above), then I have observed the following inequalities which are valid for $ t>0 $
\[
 e-\frac{1}{t+1}-\frac{{\mathbf P}}{t}< a(t)<  e-\frac{1}{t+1}-\frac{{\mathbf P}}{t+1}\; \; \; \; ,\; \; \;
\; \;
e+\frac{1}{t}+\frac{{\mathbf P}}{t+1}< b(t) < e+\frac{1}{t}+\frac{{\mathbf P}}{t}\; .\]
\[ \exp\left(\frac{{\mathbf Q}}{(t+1)^2}\right)< \frac{a(t)^{b(t)}}{\displaystyle e^e}<
\exp\left(\frac{{\mathbf Q}}{t^2}\right)\; .\]
Perhaps are of interest.Regards, Alex

Wait a minute!

$ a = \sqrt{3}, b = \sqrt{3}^3 $

Does that mean there are infinitely many solutions?
For any integer n >= 2,
if
$ a = \sqrt[n-1]{n} $
and
$ b = \sqrt[n-1]{n}^n $
then $ a^b = b^a $ right?

of course there are infinitely many solutions in the reals.  I think integers was implied.


$n^{n}$

You can just try a value of $n$.  If $n=4$, then $4^{4}=256$ and $7\cdot 5\cdot 3=105$.

[quote="miyomiyo"]Which is bigger - $n^n$ or $(2n-1)(2n-3)(2n-5)...3*1$?
e.g. 1000 * 1000 * 1000 * ... * 1000
or    1999 * 1997 * 1995 * ... * 1[/quote]

[hide]
First find the ratio between successive terms of the sequence $n^n$:

$\frac{(n+1)^{n+1}}{n^n} = (n+1) \left( \frac{ n+1 }{ n } \right)^n = (n+1)\left( 1 + 1/n \right)^n$

And the ratio between successive terms of the second sequence is clearly $( 2n+1 )$.

As $n$ becomes larger and larger, the ratio for the first sequence approaches $(n+1)e = en + e$, which is bigger than $2n + 1$. So since the first sequence increases more rapidly, it is "bigger."
[/hide]

This is pretty much the same. The ratio between successive terms of $n^n$ approaches $e(n+1)$, which is greater than the ratio between successive terms of the second sequence, $2(n+1)$.


Christina Skelton and Sherrylyn Branchaw will host a Math Jam on linguistics this Friday night, starting at 7:30 PM ET (4:30 PM PT).  As you'll see, there's a great deal of problem solving in linguistics, which may explain why each year several members of the US team for the International Linguistics Olympiads are also top math students!  (I think many AoPSers will recognize some of the ILO winners from the past few years [url=http://en.wikipedia.org/wiki/International_Linguistic_Olympiad]listed here[/url].)

 Here's a description of the Math Jam:

[quote]Have you ever looked at two languages that share many similar-looking words, and come to the conclusion that they are probably related? There is a linguistic theory, called the Comparative Method, which describes how languages change over time, and, by extension, what you would expect to see if two languages had descended from a common ancestor. We'll let you discover the basic principles of the Comparative Method, and show how you can use this theory to demonstrate that languages-- like English, Greek, Latin, Sanskrit, and Hittite-- are related even when they diverged thousands of years ago and the similarities between them are not always apparent to the untrained eye.[/quote]

Here's a little more information about the hosts:

Christina Skelton (Osud) received a BA in Classics from the University of Texas at Austin with minors in geology and biology. She is currently a graduate student in the Indo-European Studies Program at UCLA. She believes in polymathy, and specializes in using techniques from computational biology to study the evolution of languages and writing systems. She received the $ \$20,\!000$ Mitchell Award for Academic Excellence from the University of Texas Co-Op in 2007 for her work using phylogenetic systematics to reconstruct the history of the pre-alphabetic Greek writing system Linear B, and received an NSF Graduate Research Fellowship in Linguistics in 2009. In her spare time, which many people are amazed to learn exists, she enjoys playing classical guitar, cooking, and sewing dresses for herself.

Sherrylyn Branchaw (atheneglaukopis) received a BS in math and BAs in linguistics and Classics from the University of Arizona, with a minor in computer science. She is now finishing up her PhD in Indo-European Studies (dead languages!) at UCLA, with an emphasis on English historical linguistics. She, too, believes in polymathy and is forever inflicting linguistics on mathematicians and mathematics and computer science on linguists.

ahh cool. too bad I'll be ... on my way to Princeton Math Competition :(

Wow Indo-European languages...  :roll:

Is this one of the few(?) math jams in AoPS that are not that related to math?
Same for me: I also go to PUMaC, so I can't attend the Jam.

This is really interesting to me.  I'm definitely putting it on my schedule.

Thank you!   :)

This is really cool, definitely going to be there  :lol:

OMG ALAN HUANG GOT 3RD LAST YEAR!!! I KNOW HIM!!! Hes a beast at math as well.

Yeah, I will be attending. Its interesting how linguistics is connected to math.

When I tried learning sanskrit, it was nothing like English, it's going to be interesting to use math-related skills to see how they are similar.

in fact,Sanskrit is quite scientific in the sense that it is written as it is pronounced.That's one advantage over English.Still some words are really tongue twisters!(I hardly know any Sanskrit)

I looked at the problems. They require extreme carefulness. I'll be there :D

I will be there. Sounds interesting. Thank you AOPS for doing this.

I know some Sanskrit slokas from Vedas and Geeta! Would like to know how people in ancient India used to memorize the whole scripture and preserved it from generation to generation. There must be some algorithm there !!!!

 :!:

cool really interesting my friends would lovve it

The transcript is up, so you can go read it!

Oh cool, I just checked and it wasn't up. I'll go check again. Still not as good as going to the actual math jam though  :(
Did you like it? (people that were there)

I liked it alot and I thought it was pretty interesting.

The transcript is [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=266]here[/url].


I have a great idea! We could list prime numbers! Won't that be fun  :o ...just kidding...not really. If you see any number missing or any number that doesn't belong, please say so.

One more thing, put between 5 to 10 prime numbers in your post so it won't be considered  :spam: and other people can post more.

I'll start:

2
3
5
7
11
13
17
19

This should be really easy, because I have all the primes to around 220,000 on my calculator! (guess how long it took to execute the programs to get those primes...)

67
71
73
79
83
89
97
101
103

[quote]This should be really easy, because I have all the primes to around 220,000 on my calculator! (guess how long it took to execute the programs to get those primes...) [/quote]

aww..not fair  :P
107 109 113 ....127 131 137 139

149
151
157
163
167

b-flat
What programs did you execute and can you post the codes possibly?

Sure, I created 2 programs, A and AA.  For the first list (999 numbers), you test to see if it is divisible by every odd number, and for lists afterward, test for all members of the first list (contains primes to 7907).

[hide=code for lists greater than 1]
PROGRAM:A
;for the first line, put in the next odd number greater than the last number of the latest list as x.
:999->dim(L;whatever the next list name is
:1->A
:Repeat A=1000
:prgmAA
:If V=1
:Then
:x->L;whatever the list may be(A
:1+A->A
:End
:x+2->x
:End

PROGRAM:AA
:DelVar V
:2->R
:While 1
:If fpart(x/LPrime(R))=0:Return
:R+1->R
:If $\sqrt{x}$<LPRIME(R-1
:Then
:1->V
:Return
:End
:End
[\hide]

I have to go, so I will let you figure out how to create the first list (basically, test divisibility by all odd numbers instead of elements of LPRIME)

173
179
181
191
193
197
199
211
223
227

Um, Jose and 3, did you read the rules?

You are supposed to post CONSECUTIVE primes, not just random primes...

401, 409, 419, 421, 432, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499

[quote="Three"]ind33d 3 is a prime[/quote]

:spam:

To b-flat: sorry!!!!!!!!

[quote="MysticTerminator"][quote="Three"]ind33d 3 is a prime[/quote]

dude I strongly suspect that you are correct[/quote]

Really, never would have guessed it, but hay, you learn something new every day  :D

I see that everyone is posting lots of prime numbers in order  :roll:

503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599

[quote="José"][quote="Iversonfan2005"][quote="José"]

:spam:

[/quote]

 :spam: ...wow, how ironic[/quote]

My post wasn't a :spam: ... yours YES[/quote]

mmm explain how this last post of yours (that I'm quoting) and iversonfan's differ in any meaningful way JUST CURIOUS

He put that I had spammed (wrong) to increase his account... so I showed him that he was spamming.........

what are you implying that I spam goodness me I am so insulted oh man whatever will I do now

[quote="José"][quote="Iversonfan2005"]:rotfl: Jose, obviously you didn't pick up why i said "wow, how ironic"

a) G&FF is meant for spam
b) you calling something spam, especially in this forum, IS spam. And I spammed too by saying the same thing.
c) you are a bigger spammer than a lot of people you show this to :spam:, especially on this forum.
d)look at Arnav's first post on this page.

And good job MysticTerminator, kudos for you :P . Good luck at ARML btw :) (you better win!)[/quote]

a)you DID spam
b)the same
c)the same
d)the same[/quote]
Will you guys give it up? Both of you are spamming, and I'm probably sure you will call this spam too. Anyways, you said something was spam, now just forget about it.
Look how off topic this has gone.  :rotfl:
Prime Number$\longrightarrow$ Spam discussion

2003  2011  2017  2027  2029  2039  2053
 2063  2069  2081  2083  2087  2089  2099
I do not think I will live longer... :)
my phone number is prime but i wont reveal it


continued fraction
number theory unsolved
number theory

a.  If [i]a and [/i]b are positive integers, show that the equality [i]e<[/i]a/b<87/32 implies that b>39

b.  Given the infinite continued fraction [1;3,1,5,1,7,1,9...], find the best rational approximation [i]a/b with denominator [/i]b<25

For part a use continued fraction of e.

Bomb


A botanist had a plant which was 30 cm tall. Each successive month the plant grew by an amount equal to $ \frac25$ of its previous month's height. How many centimeters tall was it at the end of the second month? Express your answer to the nearest tenth.

Since the plant added $ \frac25$ of its height every month it grew by a factor of $ \frac75$. Thus, we have $ 30(\frac75)\equal{}42 \implies 42(\frac75)\equal{}\boxed{58.8}$.


Two rockets fly directly at each other. One travels at a speed of 8500 miles per hour, and the other travels at a speed of 3500 miles per hour. How many miles apart will they be one minute before they collide?

They will be 12,000 feet away an hour before the collision.  Divide that by 60, to get an answer of [b]200[/b].


36, and I got stucked at about 5 minutes

Woo...103

Who knows/cares about all those little islands???

i got 47 , and mostly coz i didnt know how to spell some countries :(

[quote="Elemennop"]I only got 32, but I spent 8 minutes trying to figure out why it wouldn't take China[/quote]
I eventually figured out they wanted "People's Republic of China"  :wink:

[quote="PenguinIntegral"]Memorize the following song to pwn this game:
http://www.youtube.com/watch?v=XAaqJhLrUhM[/quote]

Asia?  San Juan?  Czechoslavakia?

I got 104. Last year I memorized every single country and its location for a test, and it turned out I got 191/100 because I wrote down zambia twice  :rotfl:

Apparently still somewhat fresh in my mind. I remember SAD SSM for the islands in the carribean, in order from north to south except I forgot what all those stand for. I think:

St. Lucia and the Grenadines (not an official country apparently)
Antigua and Barbuda
Dominica
forget
forget
forget
Then Bermuda off to the side

hmm...china works for me.

lol i never knew there was a Dominican Republic of Congo

anyways i got 65 right:

Afghanistan, Argentina, Australia, Austria, Belgium,
Botswana, Brazil, Bulgaria, Canada, China,
Cuba, Denmark, Dominica, Dominican Republic, Egypt,
Ethiopia, Finland, France, Germany, Greece,
Haiti, Hungary, Iceland, India, Indonesia,
Iran, Iraq, Ireland, Israel, Italy,
Japan, Jordan, Kuwait, Lebanon, Libya,
Macedonia, Mexico, Mongolia, Morocco, Netherlands,
New Zealand, Niger, Nigeria, North Korea, Norway,
Pakistan, Panama, Peru, Portugal, Romania,
Russia, Saudi Arabia, Singapore, South Africa, South Korea,
Spain, Sweden, Thailand, Turkey, Ukraine,
United Arab Emirates, United Kingdom, United States, Venezuela, Vietnam,

I got 142 cause I didn't knowhow to spell those countries in eastern europe :(

[quote="rnwang2"]
Who knows/cares about all those little islands???[/quote]
the people who live there and its neighboring islands and nations

omg

i couldn't spell kyrgyzstan. how is there a z wtf. and i couldn't spell kazakhstan until i figured out where the h went after about a minute of thinking. and i kept wondering why it wouldn't take "great britain," then i see "united kingdom" then i go like wtf

got 93
mind was not working
should have got many more

[hide]Afghanistan, Albania, Andorra, Argentina, Armenia,
Australia, Austria, Azerbaijan, Bangladesh, Barbados,
Belgium, Bolivia, Brazil, Brunei, Bulgaria,
Cambodia, Canada, Chad, Chile, China,
Cyprus, Czech Republic, Denmark, Egypt, France,
Georgia, Germany, Greece, Guatemala, Honduras,
Iceland, India, Indonesia, Iran, Iraq,
Ireland, Italy, Jamaica, Japan, Kazakhstan,
Kenya, Kuwait, Kyrgyzstan, Laos, Latvia,
Lithuania, Luxembourg, Madagascar, Malaysia, Maldives,
Mali, Mexico, Monaco, Montenegro, Mozambique,
Myanmar, Netherlands, New Zealand, North Korea, Norway,
Oman, Pakistan, Papua New Guinea, Paraguay, Peru,
Poland, Qatar, Romania, Russia, Saudi Arabia,
Serbia, Slovakia, Slovenia, South Africa, South Korea,
Sri Lanka, Sudan, Swaziland, Sweden, Switzerland,
Tajikistan, Trinidad and Tobago, Turkey, Turkmenistan, United Kingdom,
United States, Uruguay, Uzbekistan, Vanuatu, Vietnam,
Yemen, Zambia, Zimbabwe,[/hide]

I've spent 15 minues thinking about teh other 3 but can't think of them :(

i didnt know how to spell most of the countries :oops:

Oh, only 113. I forgot almost every island...I got almost all of Europe+S. America, though...peru and lithiuania...

Eh 131. The message is [size=200][b]NOT[/b][/size] too small. Please [size=200][b]DO NOT[/b][/size]make the message longer before submitting.

Without the use of wikipedia countries list, I got 126.

With wikipedia and some nifty keyboarding skills, I got all 192!

51 countries remain

You forgot:Albania, Antigua and Barbuda, Azerbaijan, Bahrain, Barbados,
Botswana, Burkina Faso, Cameroon, Cape Verde, Comoros,
Dominica, East Timor, Equatorial Guinea, Fiji, Gabon,
Gambia, Georgia, Ghana, Grenada, Guinea,
Guinea-Bissau, Kiribati, Kuwait, Kyrgyzstan, Liberia,
Malawi, Marshall Islands, Mauritania, Mauritius, Micronesia,
Nauru, New Zealand, Palau, Papua New Guinea, Republic of the Congo,
Saint Kitts and Nevis, Saint Lucia, Saint Vincent and the Grenadines, Samoa, San Marino,
Sao Tome and Principe, Seychelles, Sierra Leone, Solomon Islands, Togo,
Tonga, Trinidad and Tobago, Tuvalu, Vanuatu, Zambia,
Zimbabwe,
Named so far:
Afghanistan, Algeria, Andorra, Angola, Argentina,
Armenia, Australia, Austria, Bangladesh, Belarus,
Belgium, Belize, Benin, Bhutan, Bolivia,
Bosnia and Herzegovina, Brazil, Brunei, Bulgaria, Burundi,
Cambodia, Canada, Central African Republic, Chad, Chile,
China, Colombia, Costa Rica, Croatia, Cuba,
Cyprus, Czech Republic, Democratic Republic of the Congo, Denmark, Djibouti,
Dominican Republic, Ecuador, Egypt, El Salvador, Eritrea,
Estonia, Ethiopia, Finland, France, Germany,
Greece, Guatemala, Guyana, Haiti, Honduras,
Hungary, Iceland, India, Indonesia, Iran,
Iraq, Ireland, Israel, Italy, Ivory Coast,
Jamaica, Japan, Jordan, Kazakhstan, Kenya,
Laos, Latvia, Lebanon, Lesotho, Libya,
Liechtenstein, Lithuania, Luxembourg, Macedonia, Madagascar,
Malaysia, Maldives, Mali, Malta, Mexico,
Moldova, Monaco, Mongolia, Montenegro, Morocco,
Mozambique, Myanmar, Namibia, Nepal, Netherlands,
Nicaragua, Niger, Nigeria, North Korea, Norway,
Oman, Pakistan, Panama, Paraguay, Peru,
Philippines, Poland, Portugal, Qatar, Romania,
Russia, Rwanda, Saudi Arabia, Senegal, Serbia,
Singapore, Slovakia, Slovenia, Somalia, South Africa,
South Korea, Spain, Sri Lanka, Sudan, Suriname,
Swaziland, Sweden, Switzerland, Syria, Tajikistan,
Tanzania, Thailand, The Bahamas, Tunisia, Turkey,
Turkmenistan, Uganda, Ukraine, United Arab Emirates, United Kingdom,
United States, Uruguay, Uzbekistan, Venezuela, Vietnam,
Yemen,

so i got 141.

lol i couldn't spell azerbaijan... i kept saying "azerbajan"... i forgot there was a z in kyrgyzstan...

i'm surprised i remembered brunei and forgot new zealand lol.

Algeria, Andorra, Angola, Antigua and Barbuda, Armenia,
Azerbaijan, Bahrain, Bangladesh, Barbados, Belarus,
Belize, Bhutan, Bosnia and Herzegovina, Brunei, Burkina Faso,
Cambodia, Cape Verde, Central African Republic, Chad, Colombia,
Comoros, Costa Rica, Czech Republic, Democratic Republic of the Congo, Dominican Republic,
Ecuador, El Salvador, Equatorial Guinea, Eritrea, Estonia,
Gabon, Gambia, Georgia, Ghana, Grenada,
Guatemala, Guinea, Guinea-Bissau, Guyana, Haiti,
Honduras, Iceland, Israel, Ivory Coast, Kazakhstan,
Kenya, Kiribati, Kuwait, Kyrgyzstan, Latvia,
Lebanon, Lesotho, Liberia, Liechtenstein, Lithuania,
Macedonia, Marshall Islands, Mauritius, Micronesia, Moldova,
Montenegro, Morocco, Mozambique, Namibia, Nauru,
Nepal, New Zealand, Nicaragua, Oman, Pakistan,
Palau, Panama, Philippines, Republic of the Congo, Rwanda,
Saint Kitts and Nevis, Saint Lucia, Saint Vincent and the Grenadines, Samoa, San Marino,
Sao Tome and Principe, Serbia, Seychelles, Sierra Leone, Slovakia,
Slovenia, Sudan, Syria, Tajikistan, The Bahamas,
Togo, Trinidad and Tobago, Tuvalu, Uganda, Ukraine,
Uruguay, Vanuatu, Venezuela,
Named so far:
Afghanistan, Albania, Argentina, Australia, Austria,
Belgium, Benin, Bolivia, Botswana, Brazil,
Bulgaria, Burundi, Cameroon, Canada, Chile,
China, Croatia, Cuba, Cyprus, Denmark,
Djibouti, Dominica, East Timor, Egypt, Ethiopia,
Fiji, Finland, France, Germany, Greece,
Hungary, India, Indonesia, Iran, Iraq,
Ireland, Italy, Jamaica, Japan, Jordan,
Laos, Libya, Luxembourg, Madagascar, Malawi,
Malaysia, Maldives, Mali, Malta, Mauritania,
Mexico, Monaco, Mongolia, Myanmar, Netherlands,
Niger, Nigeria, North Korea, Norway, Papua New Guinea,
Paraguay, Peru, Poland, Portugal, Qatar,
Romania, Russia, Saudi Arabia, Senegal, Singapore,
Solomon Islands, Somalia, South Africa, South Korea, Spain,
Sri Lanka, Suriname, Swaziland, Sweden, Switzerland,
Tanzania, Thailand, Tonga, Tunisia, Turkey,
Turkmenistan, United Arab Emirates, United Kingdom, United States, Uzbekistan,
Vietnam, Yemen, Zambia, Zimbabwe,

This is my first try. I will try later. :)


I'm looking for a book that discusses tactics for solving Olympiad level geometry problems like those on current USAMO tests. Would the book on geometry in the aops store be difficult enough?

No way would the AoPS geo book be on par with USAMO level geo problems. I suggest geometry revisited.

That is really heavy theory and not problem solving for AOPS book I did not mean introduction to geo

I've read geometry revisited (some) and I would prefer a book focused on problem solving like WOOT is.

Geometry Revisited is excellent - you should try it :)

[quote="Agr_94_Math"]Hello123, I suggest you to have a look at Geometry Unbound of Kiran Kedlaya and practice geometry problems from the Olympiad Section Geometry forums. All the best.[/quote]
Thanks for it :)

See here: http://www.amazon.com/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190

Also New Problems in Euclidean Geometry by David Monk 2009, ISBN 978-1-906001-09-4; see here: http://www.mathcomp.leeds.ac.uk/publications/

problems in plane and solid geometry by parsolov is a really good book and it has all problems


function
topology
complex analysis
complex analysis unsolved

What are the closed ideals of the ring of entire functions in the complex plane, equipped with the topology of uniform convergence on compact sets?

For each set ${B=\{w_{j}},$ where either $j\in\{1,2,\dots,n\}$ or $j\in\mathbb{N}$ such that $B$ is a set of isolated points, then the set of all entire functions that vanish on $B$ is a closed ideal.

We can do more: consider the set $\{w_{j},d_{j}\},$ where $d_{j}\in\mathbb{N}.$ Then the set of all entire functions that have zeros of order $d_{j}$ at $w_{j}$ is an ideal and I'm pretty sure it's closed.

So is that all of them?


Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that

$ \frac1{\sqrt {(1 \plus{} a \plus{} b)(1 \plus{} b \plus{} c)}} \plus{} \frac1{\sqrt {(1 \plus{} b \plus{} c)(1 \plus{} c \plus{} a)}}$
                                                                                $ \plus{} \frac1{\sqrt {(1 \plus{} c \plus{} a)(1 \plus{} a \plus{} b)}}\le1$

From http://www.batdangthuc.net/showthread.php?t=428

Hint: ((1+a+b)(1+b+c))>=(2(b)^1/2 + (b)^-1/2)^2 after that let a=e^x ,b=e^y, c=e^z and x+y+z=0 and after that use Jensen's inequality  :wink: (i think my solution is not nice)

This problem is very easy  :wink:  But following inequality is not easy

Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that

$ \frac1{\sqrt {(1 \plus{} a \plus{} b^2)(1 \plus{} b \plus{} c^2)}} \plus{} \frac1{\sqrt {(1 \plus{} b \plus{} c^2)(1 \plus{} c \plus{} a^2)}}$
                                                                                $ \plus{} \frac1{\sqrt {(1 \plus{} c \plus{} a^2)(1 \plus{} a \plus{} b^2)}}\le1$


geometry
rhombus
ratio
geometry unsolved

Given a rhombus $ABCD$. A point $M$ is chosen on its side $BC$. The lines, which pass through $M$ and are perpendicular to $BD$ and $AC$, meet line $AD$ in points $P$ and $Q$ respectively. Suppose that the lines $PB,QC,AM$ have a common point. Find all possible values of a ratio $\frac{BM}{MC}$.
[i]S. Berlov, F. Petrov, A. Akopyan[/i]

[hide]Suppose $PB, AM, QC$ concur in $R$. Then $\frac{BM}{MC}= \frac{PA}{AQ}$. $APMC$ is a paralleologram so $PA = MC$. The same for $BMQD$, with $BM = DQ$. Then $\frac{BM}{MC}= \frac{MC}{2BM+MC}\Rightarrow 2x^{2}+x-1 = 0$ with $\frac{BM}{MC}= x$. [/hide]

'Another' solutions:
If $\frac{BM}{MC}=\frac{c}{b}$ and $b+c=a$; from $BC//PQ$ we have $\frac{c}{b}=\frac{b}{a+c}$
$b=2c$  :wink:


hi! is there any way of changing the color of the text of the header???

thanks!!!

use the package "color" and in the definition of your header, use the command "\textcolor{color}{text}"


I found it extremly difficult to write counting proofs clearly; for example

Show that

$ \sum_{k \equal{} 0}^{n} \binom{n}{k} \equal{} 2^n$.

When I first saw this problem I found the result obvious  since the left side just counts every subset of a set with $ n$ elements and it is obvious that every of $ n$ elements can or can't be in a certin subset so therefore the right side counts the same thing ; However almost imediatly i started thinking 'is this clear enough?'
'would ; if this would be a contest problem  and this wouldn't be the offical solution every grader(ok i know since it is a trivial task but if it would be harder)  understood this ?'

And i often have this kind thoughts (specialy when i have to show some 1-1 correspondence).So I'm asking; does someone feel the same way .. or are this just my crazy thoghts?

[quote="mgao"]I think that's perfectly clear.
AoPS books WANT you to prove it that way. It's called a committee-selection proof.[/quote]
yes and we should definitely accept and worship the aops books because they are the bible of maths

[quote="Wichking"]I found it extremly difficult to write counting proofs clearly; for example

Show that

$ \sum_{k \equal{} 0}^{n} \binom{n}{k} \equal{} 2^n$.

[/quote]I find it extremely difficult to figure out what can be used to prove what.
Could we start with the expansion of $ (a\plus{}b)^n$ and just say make $ a\equal{}b\equal{}1$?


A bowling contest consists of several series. Mary got 185 points in her previous series and thereby increased her average score per series from 176 to 177 points. How many points would Mary need in her next series to increase her average to 178?

$ \text{(A)}\ 184 \qquad \text{(B)}\ 185 \qquad \text{(C)}\ 186 \qquad \text{(D)}\ 187 \qquad \text{(E)}\ 188$

[hide]First, we solve for how many rounds Mary has bowled before scoring 185 points. Setting up an equation, we have
\[ \frac{176n\plus{}185}{n\plus{}1}\equal{}177 \\
\Longrightarrow 176n\plus{}185\equal{}177n\plus{}177 \\
\Longrightarrow n\equal{}8\]
Then, we use this value of $ n$ to solve for the amount of points Mary needs in her next series to increase her average to $ 178$:
\[ \frac{177\cdot 9\plus{}s}{10}\equal{}178 \\
\Longrightarrow 1593\plus{}s\equal{}1780 \\
\Longrightarrow s\equal{}187
\Longrightarrow \boxed{D}\][/hide]

[quote="123456789"][quote="Silverfalcon"][quote="unimpossible"]what is the Niels Henrik Abel Contest?[/quote]

Niels Henrik Abel Contest is Norwegian Math Contest very similar to AMC in United States. I was known to their problems for couple years but only recently did I really decide to add them to Resource sections.

Right now, I'm focusing on Round 2 most because that's when I like their problems most (AMC-12 to AIME level) and basically solving every problem up there. Unfortunately, all the solutions are in Norwegian and majority of problems are in Norwegian as well. I'll do my best to translate it (using translators of course) and as of now, four years of Round 2 are available in Resource Section (click Norway and then Round 2). I have added solutions to [b]every[/b] problem in the round so people can now have English solutions![/quote]Thanks! Do you know any other contests with these type level questions?[/quote]

South Africa has some problems (senior challenges are better; except Round 3 because they are proofs)

http://www.samf.ac.za/samo/questions.html

Canada has some but probably harder levels than South Africa (suitable for AIME)

http://www.math.ca/Competitions/COMC/

Hope that helps. By the way, they will all EVENTUALLY be added to the Resource section.


geometry
incenter
ratio
articles
angle bisector

Hope this isn't too easy :P :

Prove that in any triangle ABC, with the angle bisector from A meeting BC at E, and incenter I, that the ratio AI:IE = (b+c)/a, where a, b, and c, are the lengths of BC, AC, and AB respectively.

[hide]
$\frac{AI}{EI}=\frac{AB}{BE}=\frac{AC}{EC}$
Therefore we have:
$\frac{AI}{EI}=\frac{AB+AC}{BE+EC}=\frac{b+c}{a}$[/hide]

[quote="nolachrymose"][hide]Through mass points, we can let the weight on <A be a, on <B be b, and <C be C. Then the "weight" at E is b+c. It is clear AI:IE = b+c:a.[/hide][/quote]

What is "mass points"?

This article has a pretty good explanation: [url]http://mathcircle.berkeley.edu/mpgeo.pdf[/url].


Doesn't it depend on how fast the snail was crawling?  If it's traveling at, say, 100mph, it could go a lot farther than if it was traveling 1mph.  You didn't specify a speed so that's why I bring this up.

it is given that the snail moves 1 foot in 1 minute

[hide="r u mad"]are u mad can a snail ever have that pace?[/hide]

[quote="apratimdefermat"]A snail crawled along a straight line for 6 minutes while several people watched it in progress. Each of these people watched the snail for 1 minute , also the snail was always watched by at least one person.What is the max possible distance the snail could crawl during these 6 minutes?[/quote]

Xaenir, where does it say he went 1 foot per minute?

sorry i over looked but let's solve it with that condition


please explain these problems from aops:

1. two bicyclists are seven-eighths of the way through a mile-long tunnel when a train approaches the closer end at 40 mph. the riders take off at the same speed in opposite directions, and each escapes the tunnel as the train passes them. how fast did they ride?

2. two men started at a point on a circular 1-miles track walk in opposite directions with uniform speeds and meet in 6 minutes, but if they walk in the same direction, it requires one hour for the faster walker to gain a lap. what is the rate of the slower walker?

3. two dogs, each traveling 10ft/sec, run toward each other from 500 feet apart. as they run, a flea flies from the nose of one dog to the nose of the other at 25ft/sec. the flea flies between the dogs in this manner until it is crushed when the dogs collide. how far did the flea fly?

Do you mean how to do them or what they mean?  If you're asking what they mean, please give the specific part that you don't understand.  Otherwise, here's #1:

[hide]They go 30 mph.

Let's first go forward in time from the beginning.  Let us consider the situation when one biker is at the end of the tunnel from which the train approaches.  This is when the train first enters the tunnel.  How far has the other biker traveled at this point?  The first biker has gone 1/8 of the tunnel and escapes.  The second biker has gone 1/8 of the tunnel in the other direction, so he is currently 3/4 of the way (1/4 of the way from the train) because 7/8 - 1/8 = 6/8 = 3/4.  If this biker travels the 3/4 of the tunnel in the same amount of time the train takes to travel the whole tunnel, then he must travel at 3/4 the speed.  3/4*40 = 30.  [/hide]

More coming later if you want.

ya thanks thats exactly what i mean! please send more!

2. two men started at a point on a circular 1-miles track walk in opposite directions with uniform speeds and meet in 6 minutes, but if they walk in the same direction, it requires one hour for the faster walker to gain a lap. what is the rate of the slower walker?

Here's 2.

[hide] If  x  is the rate of the slower man (in miles/min) and y is the rate of the faster man (in miles/min), then 6x+6y=1.  For the second part of the problem, let the time passed be z. Then 60x+1=60y, because the slower man does one less lap than the faster one, which is one mile long. So, solving the equations simultaneously, we have:

6x+6y=1 becomes

60x+60y=10

-(60x+1=60y)

60y-1=10-60y

120y=11

y=11/120

6x+6(11/120)=1

6x+(11/20)=1

6x=9/20

x=3/40

So the rate of the slower man is 0.0196(repeating) miles per min, and the rate of the faster man is 0.075 miles per min. This becomes about 1.2857 miles per hour and 4.5 miles per hour, respectively. If I might add, they're not going very fast![/hide]

mlee06 wrote:[hide]4.5 miles per hour[/hide]

That's pretty darn fast for walking.

Well, maybe not  the 4.5 miler but the 1.3 miler  is. He must be admiring the landscape around him or  walking backwards or something.

The long way to solve Problem 3 is to add up the lengths of each segment that the fly travels.  That method involves an infinite geometric series.  The short way to solve the problem is to first figure out how many seconds it took for the two dogs to collide.

3.  [hide]This is a really fun problem.  When I first saw it I thougth that it was impossible but then I saw the trick and figured it out the easy way.

The whole essence of solving this problem relies on the way that you approach it.

We are trying to find how far the flea flied.  Using d=rt and plugging in the rate we can see that d=(25 ft/sec)(t).  We can find the distance if we find the total time that it is flying.  To do this we can find out how long it will take the dogs to collide. Since both dogs are traveling at 10 ft/sec towards each other we can say they are closing in on each other at a rate of 20 ft/sec. So using d=rt we get 500 ft = (20 ft/sec)(t), thus t = 25 sec.  Plug t in the first equation, d=(25 ft/sec)(25 sec)=625 ft.  [/hide]

3.[hide]First we figure out how long it takes untill the dogs meet. Since they both travel 10 ft/sec they will travel the same distance, 250 ft. It will take them 25 seconds to travel this distance. The flea would then be flying for 25 seconds at 25 ft/sec. The flea flew 625 feet.[/hide]

Edit: someone beat me to it.

6x+6y=1 becomes
60x+60y=10
-(60x+1=60y)

60y-1=10-60y
120y=11
y=11/120

i am quite confused by that. in the first part, how do you get from 60x+60y=10 to -(60x+1=60y)? where does the y come from?

I couldn't solve problems 1 and 2 but I sure got number 3!

Can anyone tell me on what test number 3 would appear?

Like would it be in the AMC test, the AIME, etc.

Also, would it be one of the more difficult problems in that test?

Thanks.

I'm not really into using the formulas and x and y, so here's how I did problem 3:

[hide]We know the dogs are 500 ft apart, and by the time they collide each dog will have traveled 250 feet towards the other dog.  Then, we can figure the time it will take for the two dogs to collide since we know 250 feet must be traveled by each dog first, and since each dog travels 10 ft/second, we simply divide 10 into 250 and get 25 seconds for the dogs to collide.  Now, the flee travels 25 ft/second, and since it takes 25 seconds for the dogs to collide, we just multiply 25 and 25 and get 625.  After typing this I suppose it would have been easier using the formula, but hey, it's good to know there are different ways to solve a problem.[/hide]

#3 is not the type of problem that would appear on a test, I think mainly because it's so well-known.  #1 and #2 would probably appear at around #17-18 on an AMC 10.

Reply to NightFlarer: [hide]You are basically doing the same thing that I showed.  You may not think you are using the d=rt formula but you are.  The way you did it was better than memorizing the formula.  You are basically just using the info you have and then canceling some units out (by mult. or division of other known info) until you have your desired unit.  This is a very helpful method in problem solving(probably even more helpful in chemistry).  In fact problem solving is all about this kind of stuff.  It pretty much discourages memorizing formulas unless you understand them.[/hide]

Btw, according to the book this question appeared on the 1992 MA :theta: .  So it would be on the AMC 10 or 12 like prob 101 said.  I disagree with what you said about it being so well known probability.  It may be well known to most of the people on this web site but that doesn't mean that the majority of people taking the AMC would know it.  I would not be surprised if I saw this question on future tests(not necessarily AMC but equivalent ones).  You have to remember that you know a lot more than the average person taking these tests so what is trivial to you is terrifying to the general public.


Suppose first you have a computer in your room.  It does not work on batteries, but one of its parts is a plug.

Suppose you can enlarge this computer to the size of the universe.  Explain why this computer will not be able to function.

funcia,

no, but my explanation DOES involve that four letter word
you used that begins with a "pl."

Please do not read my answer until you have tried to answer for yourself.
*SPOILER*
[hide]Because the computer now takes up the whole universe, there is nowhere to put the plug into.  (Disregard the reality of the physics -- masses, gravity, densities, nuclear reactions created and other stuff.  Disregard the reality of the new length and diameter of the plug's chord.  No person, alien, or robot could exist to move anything anyway is also true because the computer is taking up ALL the space of the universe.)[/hide]

[quote="Padmavathi"]I was gonna say that
[hide] Since the computer is the size of the whole universe, where will you find a plugpoint for it??[/hide]
Or
[hide] Assuming you've plugged it in..where's the ON switch??[/hide]
:rotfl:  :rotfl:[/quote]

Hopefully there is another friendly universe in the neighbourhood! :alien:

It's not called a "prediction" when you a referring to past events.

[quote="bpms"][quote="mathnerd314"]EXTENSION:

Suppose you have a very powerful computer. Suppose this computer can predict everything that has happened and will happen in the entire universe. (Like Deep Thought does in the first H2G2 book.)

Explain why THIS computer will not work, either.[/quote]
Can it predict what it will do?[/quote]

Well, since it is part of the universe, yes.

Yes, box of rocks, I know that "prediction" only applies to the future. I just got lazy typing.


modular arithmetic
binomial coefficients
Divisibility Theory

Show that the coefficients of a binomial expansion $(a+b)^n$ where $n$ is a positive integer, are all odd, if and only if $n$ is of the form $2^{k}-1$ for some positive integer $k$.

Firstly, suppose $n=2^k-1$.

Then $\binom{n}{m}=\frac{(2^k-1)(2^k-2)....(2^k-m)}{1*2*......*m}$.

For any $a,b$ with $b<k$, $a \equiv -(2^k-a) \mod{2^b}$, so in particular, if one is zero, the other is zero also. Therefore, comparing factors in the numerator and the denominator, all the powers of two will cancel out exactly, leaving an odd binomial coefficient, completing the sufficiency part of the proof.

Now, suppose $n=2^k-l$ with $1<l\leq2^{k-1}$.

Consider the binomial coefficient $\binom{2^k-l}{2^{k-1}-1}=\frac{(2^k-l)!}{(2^{k-1}-1)!(2^{k-1}-l+1)!}=\frac{(2^k-l)(2^k-l-1)....2^{k-1}}{(2^{k-1}-l+1)!}$

Factoring out, this becomes $\frac{2^{k-1}}{2^{k-1}-l+1}\binom{2^k-l}{2^{k-1}-l}$, which is clearly even by choice of l.

Other method is Lucas theory for the case $ p \equal{} 2$
$ n \equal{} b_k.2^k \plus{} b_{k \minus{} 1}.2^{k \minus{} 1} \plus{} ... \plus{} b_1$
$ m \equal{} c_{k}2^k \plus{} c_{k \minus{} 1}2^{k \minus{} 1} \plus{} .. \plus{} c_1$
Where $ b_i,c_i\in\{0,1\}$
Then $ C_n^m\equiv \prod_{i \equal{} 1}^k C_{b_i}^{c_i}(\mod 2)$
Imply that $ C_{b_i}^{c_i}\equiv 1 (\mod 2)$
So $ b_1 \equal{} b_2 \equal{} ... \equal{} b_k\equal{}1$ imply that $ n \equal{} 2^{k \plus{} 1} \minus{} 1$
This problem was be prove.

The binomial coefficients are of the form \[\binom{n}{i} |  0\le i\le n\]

In base $2$, let $n=\sum_{i=0}^{j}2^ia_i$ where $a_j=1$.  By Lucas's theorem if there exists an $m$ such that $a_m=0$ then we would have \[\binom{n}{2^m}\equiv \binom{0}{1}\pmod{2}\equiv 0\pmod{2}\]

Contradiction.  Therefore $a_m=1\forall 0\le m\le j$ and hence $n=\sum_{i=0}^{j}2^i=2^{j+1}-1$.  We are done.

If I'm not mistaken, is that not precisely the proof above it, with just $a_i$ replacing $b_i$, and using the same Lucas' theorem?

[hide="A27"]
Since $\dfrac{n-k}{k+1}\dbinom{n}k=\dbinom{n}{k+1}$, we must have, for any $k$, $v_2(n-k)=v_2(k+1)$. Let $2^x$ be the largest power of 2 less that or equal to $n$. We let $k=n-2^x$, giving $x=v_2(k+1)$, or $2^x|n+1$, $2^{x+1}\not|n+1-2^x$. The first statement gives $n=a2^x-1$, where $a$ is 1 or 2, since we must have $\dfrac{n}{2^x} < 2$. The second statement rules out $a=2$, giving $a=1$, so $n=2^x-1$ for positive $x$ $\boxed{}$[/hide]

$\dbinom{n}{r} =\frac{n!}{(n-r!)\times(r!)}$ is odd if and only if $v_2(n!)=v_2(r!)x+v_2(n-r!) \iff \sum_{i=1}^{\infty} [\frac{n}{2^i}]=\sum_{i=1}^{\infty} [\frac{n-r}{2^i}]+\sum_{i=1}^{\infty} [\frac{r}{2^i}] \iff$

$\sum_{i=1}^{\infty} ([\frac{n}{2^i}]-[\frac{n-r}{2^i}]-[\frac{r}{2^i}])=0 \iff \forall i \in N: [\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}]$

now assume$\dbinom{n}{r}$ s are all odd it mean $\forall r<n,\forall i\in N : [\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}]$

we know there exists $k$ such as that $2^{k-1}\le n <2^k \implies n=2^k-1-s$

for some $s\ge 0 \implies$ now if we put $i=k-1$ and $r=2^{k-1}-1$

$[\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}] \implies[\frac{2^k-1-s}{2^{k-1}}]=[\frac{2^{k-1}-s}{2^{k-1}}]+[\frac{2^{k-1}-1}{2^{k-1}}]$

because LHS is equal to one and $[\frac{2^{k-1}-1}{2^{k-1}}]=0 \implies [\frac{2^{k-1}-s}{2^{k-1}}]=1 \implies s=0$ so $n=2^k-1$

we other part is easy, let $n=2^k-1$ we should prove $\forall r<n,\forall i\in N : [\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}] \implies$

if $i \ge k$ then both sides will be zero so supose $i<k$

$\forall r<n,\forall i\in N : [\frac{2^k-1}{2^i}]=[\frac{2^k-1-r}{2^i}]+[\frac{r}{2^i}]$

let $r=2^i\times t+r'$ that $2^i>r'\ge 0,t \ge 0 \implies  [\frac{2^k-1}{2^i}]=[\frac{2^k-1-2^i\times t-r'}{2^i}]+[\frac{2^i\times t+r'}{2^i}]$

LHS is obviously $2^{k-i}-1$ and for RHS , $[\frac{2^i\times t+r'}{2^i}]=t$ and $[\frac{2^k-1-2^i\times t-r'}{2^i}]=-t+[\frac{2^k-1-r'}{2^i}]$

and it's easy to see $2^{k-i}-1 \le \frac{2^k-1-r'}{2^i} < 2^{k-i}$ so $[\frac{2^k-1}{2^i}]=[\frac{2^k-1-r}{2^i}]+[\frac{r}{2^i}]$

the coefs of $(a+b)^n$ are of the form $\binom{n}{t}$ for some integer $t \le n$ then we have
\[\binom{n}{t}=\frac{n!}{t!(n-t)!} \qquad \textbf{which is odd}\]
by the parity condition we have that
\begin{align*}
v_2(n!) & = v_2(t!)+v_2((n-t)!)\\
n-s_2(n)& = n-s_2(t)-s_2(n-t)\\
& \implies s_2(n)=s_2(n-t)+s_2(t)
\end{align*}
since the last equation holds for $n+1$ different variables we easily obtain that the binary
representation of $n$ is equal to $111\cdots 111$ which is obviously in the form $2^k-1$ as desired.


I had to work a considerable bit to get this one - check it out. If $ x\plus{}y\plus{}z\equal{}0$ and $ x,y,z\in [\minus{}1,2]$ then show that
                \[ \sum_{\text{cyclic}}\sqrt{\frac{(2\minus{}x)(2\minus{}y)}{(2\plus{}x)(2\plus{}y}}\geq 3\]

Replacing first $ x,y,z$ with $ a\minus{}1,b\minus{}1,c\minus{}1$, respectively, using then a new suitable substitution and, finally, the contradiction method,
the inequality becomes a known one:

[i]If $ p,q,r$ are nonnegative numbers such that $ pq\plus{}qr\plus{}rp\equal{}3$, then[/i]$ \frac {1}{1\plus{}p^{2}}\plus{}\frac {1}{1\plus{}q^{2}}\plus{}\frac {1}{1\plus{}r^{2}}\ge\frac {3}{2}$.

Do you know any nice proof of this last inequality:
$ pq\plus{}qr\plus{}rp\equal{}3$, then$ \frac{1}{1\plus{}p^{2}}\plus{}\frac{1}{1\plus{}q^{2}}\plus{}\frac{1}{1\plus{}r^{2}}\ge\frac{3}{2}$?

[quote="j555"]Do you know any nice proof of this last inequality:
$ pq\plus{}qr\plus{}rp \equal{} 3$, then$ \frac{1}{1\plus{}p^{2}}\plus{}\frac{1}{1\plus{}q^{2}}\plus{}\frac{1}{1\plus{}r^{2}}\ge\frac{3}{2}$?[/quote]
Let $ p\geq q\geq r.$ Then $ \frac{1}{1\plus{}p^{2}}\plus{}\frac{1}{1\plus{}q^{2}}\plus{}\frac{1}{1\plus{}r^{2}}\geq\frac{2}{1\plus{}pq}\plus{}\frac{1}{1\plus{}r^{2}}\ge\frac{3}{2}.$ :wink:

[quote="Vasc"]
[i]If $ p,q,r$ are nonnegative numbers such that $ pq \plus{} qr \plus{} rp \equal{} 3$, then[/i]$ \frac {1}{1 \plus{} p^{2}} \plus{} \frac {1}{1 \plus{} q^{2}} \plus{} \frac {1}{1 \plus{} r^{2}}\ge\frac {3}{2}$.[/quote]

I don't see any proof  :(

[quote="HenryS"]

I don't see any proof  :([/quote]
See my post. :wink:

[quote="Vasc"]the inequality becomes a known one:

[i]If $ p,q,r$ are nonnegative numbers such that $ pq \plus{} qr \plus{} rp \equal{} 3$, then[/i]$ \frac {1}{1 \plus{} p^{2}} \plus{} \frac {1}{1 \plus{} q^{2}} \plus{} \frac {1}{1 \plus{} r^{2}}\ge\frac {3}{2}$.[/quote]
This inequality is [b]Problem 2.26 - page 64[/b] in new book (where it is presented with two simple solutions, and one of them is by Cauchy-Schwarz only):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote="HenryS"][quote="can_hang2007"]

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.[/quote]

But there is no such book in my library.  :(

Could you send me an e-book version, dear can_hang2007 (the author of book?)?  :?:

At least, you should post the proof of this inequality for me, please  :wink:[/quote]
I cannot send you the e-book version due to copyright! Sorry.

If you want to have the book, then try to contact the publisher.


Prove that the arithmetic mean $a$ of $x_1,\dots, x_n$ satisfies
$(x_1-a)^2+(x_2-a)^2+\dots+(x_n-a)^2\leq \dfrac{1}{2} (|x_1-a|+\dots+|x_n-a|)^2$


Team A and team B are playing volleyball against each other, and the score is tied 21-21. League rules state that in order to win, a team must have over 21 points and win by at least 2 points. Team A has a probability of 3/5 to score any given point. What is the probability that they win the match?

Each point is an independent event, so its $\frac{3}{5}*\frac{3}{5}=\frac{9}{25}$

[b]NUH UH UH[/b]

COMMON SENSE
IT DICTATES
THAT THE PROBABILITY
IS OVER 1/2

THIS IS LIKE
A POEM OF SORTS

yeah, you're right. what did i do wrong?

Well you didn't consider the fact that A could win one point, B could win another point, and then A could win two more points. It's basically an infinite series. Have fun with that.

[hide]
Let a = Pr(A wins the whole game).

Consider the next two games.
A is either wins both games, loses both games, or wins 1 game and loses 1 game.

$a = \text{Wins both games}+\text{Wins 1 game, loses 1}* a+\text{Loses both games}* 0$
$a = \frac{3}{5}\cdot\frac{3}{5}+2\cdot\frac{3}{5}\cdot\frac{2}{5}a$
$a = 9/13$
[/hide]

[hide]Casework!

CASE I: A wins both rounds.  Probability: $\frac{9}{25}$.
CASE II: B wins both rounds.  Probability: $\frac{4}{25}$.
CASE III: A and B win one round each.  They cancel each other out, ignore.

So the answer should be $\frac{\frac{9}{25}}{\frac{9}{25}+\frac{4}{25}}$ or $\boxed{\frac{9}{13}}$.[/hide]

Apparently, Treething's Caps Lock key is glued down.


Would anyone mind posting their AIME questions here?  My teacher is strange and won't let me have my test booklet back.

That's odd. My teacher wouldn't let me give my questions back (not  that I would have wanted to).