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Tuong gi chu trang hay ve Toan thi cung nhieu. http://www.jmilne.org/math/ . Trang nay la cua J.S. Milne , trong do co chua cac bai bao , sach ve Ly thuyet so, Hinh hoc dai so so hoc, AG, ANT,...

Uhm... Hungkhtn cũng đã đính post bài trên National Forum nhưng mấy hôm trước bận quá quên béng đi mất.

Hầu hết các bài post trên ML về BDT toán của Hs Việt Nam, mà không có sân nhà cho Việt Nam thì thật tiếc. DDTH không còn là DDTH nữa, mà người VIệt nam yêu BDT còn rất nhiều, nên ko thể không có 1 forum cho BDT được. Tất nhiên về chất lượng của Forum thì Hungkhtn là admin và điều hành tất cả mọi việc bây giờ nên các bạn có thể yên tâm về điều đó 100%  :lol:

Chỉ tiếc caí tên miền hơi khó nhớ quá... nhưng nó mới xuất hiện đc 1 tuần mà cứ xem số post và chất lượng của nó thì sẽ thấy VI Forum chẳng kém gì ML cả.

Anh Tuấn có thời gian thì vào đó post bài nhé. VI Forum rất cần người tận tuỵ như anh.  :lol: Hungkhtn cũng chào đón tất cả các thành viên Vietnamese mathlinkers khác nhé  :oops:  ở VI Forum không như ML mà mấy anh em việt nam ta cứ lập nick giả để vote 6 cho mình hết. Còn Hungkhtn lúc nào cũng bị một kẻ tiểu nhân có nick giả vote là spam  :rotfl: so trivial... tiếc là cũng là người VN ta...

Bye, hen

[quote="hungkhtn"]

Hầu hết các bài post trên ML về BDT toán của Hs Việt Nam, mà không có sân nhà cho Việt Nam thì thật tiếc. [b]DDTH không còn là DDTH nữa[/b][/quote]
Noi  linh tinh!  :mad:

[quote]Anh Tuấn có thời gian thì vào đó post bài nhé. VI Forum rất cần người tận tuỵ như anh.  :lol: [/quote]
Ben do toan thien tai, chang dam dau  :rotfl:
P/S: Cai van cua chu anh khong ngui duoc, nghe tinh tuong , nen co thang no ghet, no vote cho la phai :D Anh cung bi 1 thang ghet, nhung ko phai la anh tinh tuong :P Bay gio anh di vote cho chu 6* goi la an ui  :D

:)) neu ko trung y tuong thi` ko can di chung con duong.

DDTH ko con la dien dan toan hoc ngay xua vi ca may thang troi co bai post nao dau. Cai day e tuong a bit chu...  :lol:

[quote="hungkhtn"]:)) neu ko trung y tuong thi` ko can di chung con duong.
[/quote]
Mắc lỗi chính tả! :P

[quote]

DDTH ko con la dien dan toan hoc ngay xua vi ca may thang troi co bai post nao dau. Cai day e tuong a bit chu...  :lol:[/quote]
Đúng thế thật chú ạ :D Sao thế nhỉ? ở đây có ai làm Mod của ddth cho biết tí thông tin được không?  :maybe:

Trở lại vấn đề chính :P . Đây là một trang khác http://www.grupoteorema.mat.br/links.html . Trên trang này có rất nhiều link hay về Olympic Toán! :)

em là mod đây nhưng em chả biết gì cả  :P  :rotfl:
Mọi người vào VI Forum chơi vui hơn nhiều ... được cái spam ko bị xóa bài  :D

[quote="N.T.TUAN"]Vừa ngó lại VI thấy bây giờ đẹp hơn nhiều, có điều tên miền nhìn chán quá, chẳng muốn vào lại. Đây là một trang khác về Toán, thấy bảo cũng hay http://mathnfriend.org/ . Nó là một forum kiểu như DDTH nhưng trẻ hơn nhiều, box bdt bên đó cũng mạnh lắm! :)[/quote]

Bay gio ben mathnfriend da to chuc 'Thi dau BDT' de canh tranh voi web bdt cua anh Hung. Moi nguoi vao xem va cho y kien nhe!

Đây là forum của Serbia http://imocompendium.com/phpBB/ . Trang này mới đấy!  :wink:

[color=purple]Lam` sao doc dc ha anh Tuan ? Toa`n tieng' Serbi  :D
@ Lau ro`i ko gap anh anh van khoe chu' ? [/color]

Trang đó dùng tiếng Nga thì phải? Nhưng có mục bằng tiếng Anh đấy! Dạo này anh rất khoẻ, cảm ơn chú! :)

Em là new member nhưng cũng có mấy trang :
diendan3t.net >>> Toán THCS và Tiểu Học (của anh Hieuchuoi@)
VÀ toanthpt.net >>>Toán THCS và THPT của thầy Phú Khánh
Nếu có ai giới thiệu rồi thi xoá bài này đi nhá .

[quote="NguyenDungTN"][quote="N.T.TUAN"]Hồi trước bên DDTH có mục này thấy cũng hay, bây giờ lập lại :D Mọi người khi post cái link thì bỏ chút thời gian nói qua về nó nhé! :)

Xin được bắt đầu với http://www.ch06k.com/forum/index.php . Đây là một forum về Bất đẳng thức, và chỉ về B D T thôi, nó cũng trẻ. :D[/quote]

Trang BDT của anh Phạm Kim Hùng đã chính thức đổi thành : http://batdangthuc.li/forum/index.php
Mong mọi người vào ngó qua và cho ...vài bài post[/quote]

Khong phai la trang web cua Anh  :blush:  Anh chi bat dau xay dung no thoi, chu bay gio trang web la cua tat ca moi nguoi  :wink: Hi vong moi nguoi se giup no phat trien. Mot web BDT tot cua ng` Vietnam do ng` Vietnam xay dung la uoc mo lon nhat va... cuoi cung cua hungkhtn doi voi BDT.

Database cua Forum cu van con` nhung co 1 so loi ky thuat nen hien tai no chua up dc len Forum moi.

Chào mừng bạn đến với mathlinks. Còn sách olympiad bạn có thể xem ở trên mục tài nguyên của http://diendantoanhoc.net hoặc http://trungtuan.wordpress.com  và 1 số trang kể ở trên kia  :)

Forum mới về Toán http://www.math10.com/forum/ .

[quote="quangpbc"]Cảm ơn anh Tuân, trang này rất hay . à quên còn có 1 trang nữa http://trungtuan.wordpress.com/  ở đây có thể tìm dc một số cuốn sách hay ,những vấn đề cũ mà hấp dẫn và còn để 1 số người kô có tự trọng vào đó chép đáp án THTT ,chẳng hạn như nmtvn,toanhoc....Em dẫn chúng đến blog của anh rồi đó anh Tuân ạ [/quote]
 :rotfl: đừng nói chúng như thế chơ  với lại bài trong THTT dạo này toàn bài dễ  :P Công nhận blog của anh Tuân hơi bị được  :D

Cảm ơn lời khen tặng(Khi người khác khen, ta hãy nói ''Cảm ơn'') :P

Một trang khác http://mathschallenge.net/index.php?section=home . Trang này mình chưa tìm hiểu được nhiều.

[quote="N.T.TUAN"]New forum http://www.olimpiade.org/Forum/portal.php .[/quote]
Hic! Anh cho trang tiếng Hi Lạp thế này thì bọn em ngồi khóc!

Trang đó có mục download, chắc sẽ kiếm được chút ở đó, mà trang đó của Indonesia
 cơ mà?  :lol:

Chữ gì mà nửa Anh nửa Lào vậy nhỉ :wacko:  hay là  tiếng Lào kiều bên Anh nhỉ  :rotfl:

http://en.wikipedia.org/wiki/Portal:Mathematics sẽ thích hợp hơn http://en.wikipedia.org/wiki/Main_Page.

Anh Tuấn ơi, anh có cuốn này về concrete Math thì buzz cho em nhé ^^ ! :lol: Cám ơn anh trước ạ ! :blush:

Forum moi ve Toan http://ant.edu.ms/forum/index.php . Minh la 1 trong cac Admin.  :roll:

Tôi thấy trang này có nhiều thứ quá. Nên tập trung chính vào Toán học thì hay hơn.

Toan thi ben ngoai co roi. Forum Toan tieng Viet thi co day, vi du http://mathscope.org/forum/index.php  :lol:


I got a strange email, and I know several people on this forum got one as well. It said:

glads_777k@yahoo.com,

My name is Glads. i am a female I was impressed when i saw your profile today
at(http://www.mathlinks.ro)and will like to establish a long lasting
relationship with you. In addition,i will like you to reply me through my e
mail box (glads_777k@yahoo.com).This is because i don't know the
possibilities
of remaining in forum for a long time. please If you are interested in knowing
more about me please contact me directly to my email adderss box so i will send
you some of my pictures imediately,
Thanks waiting to hear from you .
Glads.

      glads_777k@yahoo.com

Did anyone else get this?

I got this. Is it because I included my email address in my profile. I'm assuming everyone who got the email also did too. I don't see how "Glads" could have gotten emails otherwise. It's probably a bot, anyhow.

It appears to be an attempt to fish out live e-mail addresses.  Strongly recommend you throw it in a spam folder.

Don't show your email address in your profile.  If you need to post it somewhere, post it like:

MyEmailAddress @ DomainName.com

so it doesn't get caught by spambots.

Intriguing how they sent a message regarding "relationships" and "pictures" to people on a MATH forum  :rotfl:

yes , i got that guy's email too and the content is exactly what you show here .

wired , why he does that .

I didn't get this email, but I get plenty of these type emails, if it has an attachment, DON'T download it!  If it asks for a reply don't reply, cause next thing you know they will be sending Virus through it.  It's most likey a online predator trying to get private info, if you use gmail, then go to "Settings", click "Filters" and enter the sender's email, then choose "Trash it" after clicking next.

  I have strong suspicions of "her" even being "female". :?:

I didn't get this email, but I got one a while ago from someone named Loveth Johnson? Did anyone else get that? It was the same type of email Izzy got.

Since I put my email on AoPS, I have gotten about a dozen emails saying "You have just won 1,500,000 euros. Send us the following form to claim your prize. Name... Address... Telephone..." and it goes on. My email has only been on my profile for a few weeks. Do you think it will stop if I take it off? Or is it permanently on "the list"? I am taking it off right now.

This has happened before, numerous times. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=872496180&t=272342]Another example.[/url]

I think that if AoPS/ML puts your email on a way that spam bots can't detect, like, say, type it on "Paint" and save as an image and post it on.  Thats what Facebook does, So only humans can read it.

Spam bots can read even images like that. Unless we make your email into a captcha it won't work. And also it won't protect you from Spam in your email. If you find a way to fix the Spam problem with emails then you're on your way on becoming a billionaire ;)

[quote="Bugi"]I think that this is out of AoPS/ML problem. Just delete, or somehow forbid them. Hotmail has its spam folder, Gmail - do as yunhua98 said, and so on...

This should be locked now![/quote]
Yeah, but people always find a way around them, by having multiple emails, etc.

[quote]Yeah, but people always find a way around them, by having multiple emails, etc.[/quote]

  Then keep on blocking them!

  and also, Gmail can set settings so that certain sujects or content could make it spam too, you can custimize this, but the default settings will always be on.  Here's a way i use:  Firstly, set up a gmail account, second, click Settings>Forwarding and IMAP (or POP3, can't remember which)>Choose Forward messages to: blank, and put in your primary email address.  That way, gmail helps you filter spam, and only the ones that make it to the inbox get forwarded.  Also, when you sign up for stuff, put in the filtering account, not your primary one, currently, my filter account has over a thousand spam messages, but it doesn't effect me.  I get about one spam per month on my primary account, and I sign up for LOADS of things!

P.S.  If you already have a gmail account, still follow the above steps and make another, do this especially if you sign up to many websites.  i.e. Facebook, Myspace, Twitter, etc.


Determine the number of binary strings of length n that don't contain the pattern 001.

For $n>3$, the pattern can occur in $n-3$ places ($001$, $0010$, etc. aren't valid numbers).  Each of the remaining digits, of which there are $n-4$, has $2$ possible numbers.  So there are
$(n-3)*2^{n-4}$
binary numbers containing this sequence with $n$ digits.  So there are
$2^{n-1}-(n-3)*2^{n-4}=8*2^{n-4}-(n-3)*2^{n-4}=(11-n)*2^{n-4}$
binary numbers not containing this sequence for every number with $n$ digits.

I think you are overcounting the numbers that contain the pattern more than once.

Shoot you're right...
Counterexample to my previous counting method: $001001001$ would be counted $3$ times.


[quote="ridgers"]i am an albanian boy in 9 class. I won the olympiad in my city and at 15 march ,so after 2 weaks, i have albanian olympiad. I am not so prepared because i haven't worked so much . I want to know which formulas should i learn. Which formulas appears mostly in questions?? If you can you can say me the names. thank you all.[/quote]
Do you speak russian?
I dont think formulas can help you to win. :wink:

[quote="ridgers"]no i don't speak  russiam. i know english,albanian, turkish and italian[/quote]

Great!

Very easy and useful:    $ \frac{a}{b}\plus{}\frac{b}{a}\geq2$
I suppose (sure) you know this. :)

one other useful identity useful to factorisations is:
if $ a\plus{}b\plus{}c\equal{}0$ then $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$
(try to proove it)

[quote="DIMITRIS T"]one other useful identity useful to factorisations is:
if $ a \plus{} b \plus{} c \equal{} 0$ then $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3abc$
(try to proove it)[/quote]

Very Easy. :lol:
$ a\plus{}b\plus{}c\equal{}0$, so $ (a\plus{}b)^2\equal{}c^2$.
Then: $ a^3\plus{}b^3\plus{}c^3\equal{}3abc,(a\plus{}b)(a^2\minus{}ab\plus{}b^2)\plus{}c(c^2\minus{}3ab)\equal{}0$,
$ (a\plus{}b)(a^2\minus{}ab\plus{}b^2)\plus{}(a^2\plus{}2ab\plus{}b^2\minus{}3ab)c\equal{}0$,
$ (a\plus{}b\plus{}c)(a^2\minus{}ab\plus{}b^2)\equal{}0$

Are you agree?
 :wink:

Ok. Prove that:
$ a_1^2\plus{}a_2^2\plus{}\cdots\plus{}a_n^2 \geq\frac{1}{n}$, if
$ a_1\plus{}a_2\plus{}\cdots\plus{}a_n\equal{}1$.
 :wink:

[quote="YURI"]Ok. Prove that:
$ a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2 \geq\frac {1}{n}$, if
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$.
 :wink:[/quote]
Ι found  2 solutions:
1)by b-c-s inequality (n=1+1+...1 n times)
2)by using andreescu inequality for the fractions $ \frac{(a_1)^2}{1},...,\frac{(a_n)^2}{1}$
all these for n [u]natural[/u] number

i finished the olympiad today but i didn't do very well. i will write the questions in another topic tomorrow!

[quote="DIMITRIS T"][quote="YURI"]Ok. Prove that:
$ a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2 \geq\frac {1}{n}$, if
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$.
 :wink:[/quote]
Ι found  2 solutions:
1)by b-c-s inequality (n=1+1+...1 n times)
2)by using andreescu inequality for the fractions $ \frac {(a_1)^2}{1},...,\frac {(a_n)^2}{1}$
all these for n [u]natural[/u] number[/quote]

Well, actually, it’s very easy problem. Prove that
$ (a_1b_1\plus{}a_2b_2\plus{}\cdots\plus{}a_nb_n)^2 \leq(a_1^2\plus{} a_2^2\plus{}\cdots\plus{}a_n^2) (b_1^2\plus{} b_2^2\plus{}\cdots\plus{}b_n^2),$
$ a_i, b_i \geq 0, i\equal{}1,2,\cdots,n.$ :wink:

[quote="YURI"][quote="DIMITRIS T"][quote="YURI"]Ok. Prove that:
$ a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2 \geq\frac {1}{n}$, if
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$.
 :wink:[/quote]
Ι found  2 solutions:
1)by b-c-s inequality (n=1+1+...1 n times)
2)by using andreescu inequality for the fractions $ \frac {(a_1)^2}{1},...,\frac {(a_n)^2}{1}$
all these for n [u]natural[/u] number[/quote]

Well, actually, it’s very easy problem. Prove that
$ (a_1b_1 \plus{} a_2b_2 \plus{} \cdots \plus{} a_nb_n)^2 \leq(a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2) (b_1^2 \plus{} b_2^2 \plus{} \cdots \plus{} b_n^2),$
$ a_i, b_i \geq 0, i \equal{} 1,2,\cdots,n.$ :wink:[/quote]
this as the bcs inequality and it holds for every $ a_i,b_i$ [u]real[/u] numbers
see a proof there [url]http://planetmath.org/encyclopedia/ProofOfCauchySchwarzInequalityForRealNumbers.html[/url]

[quote="DIMITRIS T"][/quote][quote="YURI"][quote="DIMITRIS T"][quote="YURI"]Ok. Prove that:
$ a_1^2 + a_2^2 + \cdots + a_n^2 \geq\frac {1}{n}$, if
$ a_1 + a_2 + \cdots + a_n = 1$.
 :wink:[/quote]
Ι found  2 solutions:
1)by b-c-s inequality (n=1+1+...1 n times)
2)by using andreescu inequality for the fractions $ \frac {(a_1)^2}{1},...,\frac {(a_n)^2}{1}$
all these for n [u]natural[/u] number[/quote]

Well, actually, it’s very easy problem. Prove that
$ (a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 \leq(a_1^2 + a_2^2 + \cdots + a_n^2) (b_1^2 + b_2^2 + \cdots + b_n^2),$
$ a_i, b_i \geq 0, i = 1,2,\cdots,n.$ :wink:[/quote][quote="DIMITRIS T"]
this as the bcs inequality and it holds for every $ a_i,b_i$ [u]real[/u] numbers
see a proof there [url]http://planetmath.org/encyclopedia/ProofOfCauchySchwarzInequalityForRealNumbers.html[/url][/quote]
Yes. :D
I have an interesting problem.
Find all real $ x$ such that $ \sqrt{5-x}=x^2-5.$:wink:

People, this is not a thread for posting random exercises that come into one's mind. Every exercise belongs into a separate topic unless there are good reasons why several exercises should go together (e. g. if they use exactly the same tactic to prove). The original question was what formulas one should learn to be prepared for the algebra problems of an olympiad, and there are several ones I could name here:

[b]Binomial formula.[/b]

If $ a$ and $ b$ are two elements of a field $ K$, and $ n$ is a nonnegative integer, then $ \left(a\plus{}b\right)^n\equal{}\sum_{k\equal{}0}^n\binom{n}{k}a^kb^{n\minus{}k}$.

[b]Some nice formulae for three numbers.[/b]

If $ a$, $ b$, $ c$ are three elements of a field $ K$, then
$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}\left(a\plus{}b\plus{}c\right)\left(a^2\plus{}b^2\plus{}c^2\minus{}bc\minus{}ca\minus{}ab\right)$;
$ \left(b\plus{}c\right)\left(c\plus{}a\right)\left(a\plus{}b\right)\plus{}abc\equal{}\left(a\plus{}b\plus{}c\right)\left(bc\plus{}ca\plus{}ab\right)$;
$ \left(a\plus{}b\plus{}c\right)^5\minus{}\left(a^5\plus{}b^5\plus{}c^5\right)\equal{}5\left(b\plus{}c\right)\left(c\plus{}a\right)\left(a\plus{}b\right)\left(a^2\plus{}b^2\plus{}c^2\plus{}bc\plus{}ca\plus{}ab\right)$;
$ \left(b\minus{}c\right)^2\plus{}\left(c\minus{}a\right)^2\plus{}\left(a\minus{}b\right)^2\equal{}2\left(a^2\plus{}b^2\plus{}c^2\minus{}bc\minus{}ca\minus{}ab\right)$;
$ \left(b\minus{}c\right)^4\plus{}\left(c\minus{}a\right)^4\plus{}\left(a\minus{}b\right)^4\equal{}2\left(a^2\plus{}b^2\plus{}c^2\minus{}bc\minus{}ca\minus{}ab\right)^2$.

[b]Some nice formulae for $ n$ or $ 2n$ numbers.[/b]

If $ a_1$, $ a_2$, ..., $ a_n$ are $ n$ elements of a field $ K$, then

$ \left(\sum_{k\equal{}1}^na_k\right)^2\equal{}\sum_{k\equal{}1}^na_k^2\plus{}2\sum_{1\leq i<j\leq n}a_ia_j$;
$ n\sum_{k\equal{}1}^na_k^2\minus{}\left(\sum_{k\equal{}1}^na_k\right)^2\equal{}\sum_{1\leq i<j\leq n}\left(a_i\minus{}a_j\right)^2$.

If $ a_1$, $ a_2$, ..., $ a_n$, $ b_1$, $ b_2$, ..., $ b_n$ are $ 2n$ elements of a field $ K$, then

$ \left(\sum_{k\equal{}1}^na_k^2\right)\left(\sum_{k\equal{}1}^nb_k^2\right)\equal{}\left(\sum_{k\equal{}1}^n a_kb_k\right)^2\plus{}\sum_{1\leq i<j\leq n}\left(a_ib_j\minus{}a_jb_i\right)^2$.

[b]Lagrange interpolation.[/b]

Let $ x_{1}$, $ x_{2}$, ..., $ x_{n}$ be $ n$ [i]pairwisely distinct[/i] elements of a field $ K$ (for instance, $ K\equal{}\mathbb{Q}$ or $ K\equal{}\mathbb{R}$ or $ K\equal{}\mathbb{C}$). The [i]Lagrange interpolation polynomials[/i] for these $ n$ elements $ x_{1}$, $ x_{2}$, ..., $ x_{n}$ are $ n$ special polynomials $ N_{1}$, $ N_{2}$, ..., $ N_{n}$ of degree $ n \minus{} 1$ over $ K$ which are defined as follows:

$ N_{i}\left(X\right) \equal{} \prod_{1\leq j\leq n;\ j\neq i}\frac {X \minus{} x_{j}}{x_{i} \minus{} x_{j}}$ for every $ i\in\left\{1,2,...,n\right\}$.

These polynomials satisfy $ N_{i}\left(x_{i}\right) \equal{} 1$ for every $ i\in\left\{1,2,...,n\right\}$, and $ N_{i}\left(x_{j}\right) \equal{} 0$ for every $ i,j\in\left\{1,2,...,n\right\}$ such that $ i\neq j$.

The name-giving property of these Lagrange interpolation polynomials is the following one: For every polynomial $ S$ of degree $ \leq n \minus{} 1$, we have $ S \equal{} \sum_{i \equal{} 1}^{n}S\left(x_{i}\right)N_{i}$.

This formula allows us to compute the explicit form of the polynomial $ S$ if we know the values $ S\left(x_1\right)$, $ S\left(x_2\right)$, ..., $ S\left(x_n\right)$. In other words, we can thus compute the coefficients of a polynomial of degree $ \leq n\minus{}1$ if we know the values of this polynomial at $ n$ distinct numbers.

[b]Vanishing finite differences.[/b]

Let $ P$ be a polynomial in one variable $ X$ over a field $ K$. Let $ n\geq 0$ be an integer. Then,

$ \sum_{k\equal{}0}^n\left(\minus{}1\right)^k\binom{n}{k}P\left(X\minus{}k\right)$

is a polynomial of degree $ \leq \deg P\minus{}n$ in $ X$. In particular, if $ n>\deg P$, then the polynomial

$ \sum_{k\equal{}0}^n\left(\minus{}1\right)^k\binom{n}{k}P\left(X\minus{}k\right)$

is identically zero.

[b]Binomial inversion.[/b]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=141592 and http://www.mathlinks.ro/Forum/viewtopic.php?t=112581 for two equivalent formulations of this.

Someone should write about the Viete formulae, Abel summation and whatever else can be added...

  darij


Clearly, $a,b,c,d,e,f$ are nonnegative numbers. Suppose, for instance, that $b\le d$. Then $c+d+e+f\le e+f+a+b$ and this implies $a\ge c$. Similarly, we obtain $f\le b,e\ge a,d\le f$, therefore, we must have $a=b=c=d=e=f=0$or $\frac14$.

Denote $ a,\dots,f$ by $ a_0,\dots,a_5$, with indices modulo 6. Note that the numbers $ a_i$ are positive.
Let $ S \equal{} a\plus{}\dots\plus{}f$. Then $ \sqrt{2}\sqrt[4]{a_i} \plus{} a_i \plus{} a_{i\minus{}1} \equal{} S$.
Subtracting those equalities for $ i$ and $ i\minus{}1$, we get
$ a_{i\minus{}2} \minus{} a_i \equal{} \sqrt{2}(\sqrt[4]{a_i} \minus{} \sqrt[4]{a_{i\minus{}1}})$.

Thus, if $ a_i < a_{i\minus{}1}$, then $ a_{i\minus{}2} < a_i$. But then $ a_{i\minus{}1} > a_{i\minus{}2}$, so with the same logic we get $ a_{i\minus{}3} > a_{i\minus{}1}$. Going on, we see that  $ a_{i\minus{}k}$ is alternatively greater than the greatest or smaller than the smallest of the numbers $ a_i, a_{i\minus{}1}, \dots, a_{i\minus{}(k\minus{}1)}$. Since we'll eventually come back to $ a_i$, this is a contradiction.

So all the numbers are equal; the value is easily seen top be 0 or 1/4.

[hide] Without lost of generality $ f\leq e\leq d\leq c\leq b\leq a$
$ 4b \equal{} (c\plus{}d\plus{}e\plus{}f)^{4}$  $ 4c \equal{} (d\plus{}e\plus{}f\plus{}a)^{4}$ so
$ (c\plus{}d\plus{}e\plus{}f)^4\leq (d\plus{}e\plus{}f\plus{}a)^4$ so $ 4b\leq 4c$ but $ c\leq b$
therefore $ b\equal{}c$ and for $ 4a,4d,4e,4f$ $ a\equal{}d\equal{}e\equal{}f$ and $ a\equal{}b\equal{}c\equal{}d\equal{}e\equal{}f$
so $ 4a\equal{}(4a)^4$ $ a\equal{}0 or \frac{1}{4}$
[/hide]

Sorry yildobekir, your solution is wrong in the fact that WLOG $ f \le e \le d \le c \le b \le a$...since the equations are not symmetric but cyclic
In cyclic equations...we can only assume that one element is maximal...suppose $ a$...
Then we can show that $ a \equal{} b \equal{} c \equal{} d \equal{} e \equal{} f$ using $ a \ge b,c,d,e,f$

Similarly, enescu should first assume $ d$ as the maximal element before taking $ d \ge b$

This question is an extended version of a INMO-1996 problem,
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341887&sid=d903588743a45ae2589116e2c4e29f66#p341887


trigonometry
real analysis
real analysis unsolved

Define a sequence ${a_n}$ by $a_0=1$, $a_{n+1}=\sin{a_n}$. Prove that the sequence $na_n^2$ is bounded.

just find the first term of an equivalent of $a_n$ with the usual method (already discussed on the forum)


How two people can split something up and both feel like they got more than half ?

It sounds as if your headline is referring to an algorithm.

My try at this is that you have a pie with at least another part of a pie positioned on top of it a certain way.
You make one cut cutting both the pie and the other pie part so that two people end up with more than half of a pie.

I'm leaning that this is not what you wanted.

Is this the problem of valuation, where one person prefers one half of the good to the other, so between them they are getting more than they think is half?

Unfortunately there is no way to guarantee that they both feel like they got more than half.  If they both veiw the pie in the same way, and assign the same value to the peices as eachother if one person gets more than half then the other feels like he got less than half.

However you can guarantee that both people feel like they got at least half the cake, the classic way is for one of them to make a cut which he thinks is half, and for the other one to choose which peice he thinks is larger. This method however forces the first person to get a peice he thinks is exactly half.

Here is a modification to the above idea which guarantees that each person feels like they got at least half but doesnt force one to have half.  Person A cuts the cake into what they feel is even, person B then decides which he thinks is bigger.  Person A then cuts the "bigger" peice again into what he feels is fair, and person B chooses which one of those peices he gets and person A gets the other.  Now to deal with the "smaller" peice from the first cut, person B cuts and person A chooses which peice he wants.  Again, this gives a 50-50 cut if the people have identical preferences, but in all practical cases this yeilds a division where both people feel they got more than half.

what if the person cut the pie sort of diagonally so it looks like you have a lot?

[quote="LeoYard"]This is not about fair division.
In the well-known 2-person, 1-cut cake-cutting procedure, “I cut, you choose.”
The algorithm satisfies two desirable properties:
(1) Envy-freeness: Each person thinks that he or she receives at least a tied-for-largest piece and so does not envy the other person.
(2) Efficiency (Pareto-optimality):There is no other allocation that is better for one person and at least as good for the other person.

But in the cut-and-choose does not satisfy a third desirable property:
(3) Equitability: Each person’s subjective valuation of the piece that he or she receives is the same as the other person’s subjective valuation.

The posted problem, while it does not satisfy equitability in an absolute sense,
does satisfy it in a relative sense, which we call proportional equitability: After ensuring that each person receives exactly 50% of the cake, it gives each person the same proportion of the cake that remains, called the surplus, as he or she values it.[/quote]

That's more like reverse psychology.. kind of...isn't it?


algebra
polynomial
combinatorics unsolved
combinatorics

[b]Problem 6.[/b] Let $p>2$ be prime. Find the number of the subsets $B$ of the set $A=\{1,2,\ldots,p-1\}$ such that, the sum of the elements of $B$ is divisible by $p.$

[i] Ivan Landgev[/i]

See  http://www.mathlinks.ro/Forum/viewtopic.php?t=15112  ;) . It is the same because you can find the number of $l-element$ subsets of $A$ with the condition for every $l$(consider the polynomial $P(x)=\left(x-\epsilon\right)\left(x-\epsilon^2\right)...\left(x-\epsilon^{p-1}\right)$, $\epsilon$ is $p^{th}$ root of unity!).

Indeed, the problem is well known! It's strange that it appeared in an IMO TST...  :ninja:

Yes, I think that it was on Sankt-Peterburg Olympiad in 2004 also.

[quote=bilarev]http://www.mathlinks.ro/Forum/viewtopic.php?t=15112[/quote]
This link isn't working.
Can anyone post the solution of the problem.

[quote=megarnie]The link is now [url]https://artofproblemsolving.com/community/c6h15112[/url][/quote]

thx

Also British MO 2019 P3, and Indonesian First Stage TST for IMO 2021


inequalities
trigonometry
inequalities unsolved

Hint: Construct a polygon with ${x_{i}}$ and ${\frac{2\pi}{n}}-\alpha_{i}$

Really it is very hard, can you post all your solution ?

More Hint:

Construct a polygon $A_{1}A_{2}\cdots A_{n}$ which satisfy:

(1) $OA_{i}=x_{i}$

(2) $\angle A_{i-1}OA_{i}=\frac{pi}{n}+\alpha_{i}$

You will need:

(1) the Relation between $S$ and $L^{2}$

(2) CS ineq.

Here $S$ means $S[A_{1}A_{2}\cdots A_{n}], L=A_{1}A_{2}+\cdots+A_{n}A_{1}$


i just now noticed this, but i can't get to the wiki from the mathlinks thing. is there a link somewhere?

Well, it is called AoPsWiki for a reason

Valentin promised it to us here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=97646

conceive

Valentin is back only for a couple of days and has probably had more urgent things to do. :)


[quote="xxxyyyy"]Find the largest and smallest $18$ digit numbers which are perfect squares and which contain all numbers from $1$ to $9$ exactly twice.

One way is to use brute force method, but is there any other method using which this can be solved?[/quote]

no takers yet?

[quote="Potato Theory"]Where did you get this problem?[/quote]

saw it somewhere on internet, dont remember now.


[quote="DaReaper"]does anyone know a website with old math team tests/practice math team tests. Not AMC or Mathcounts. it would be best if the tests were like the 25 question + 3 tiebrakers at the end test.
(BTW, this is for ALABAMA geometry math team [i.e. Vestavia tourny, grissom tourny, hoover tourny, etc.) thus, any help would be greatly appreciated  :lol:[/quote]
My best guess would be to ask this question in the Alabama forum. One of the people from vestavia might provide the links for it..
But to my knowledge I haven't seen any vestavia tests online.


If Arccos x+Arccos 2x+arccos 3x=pi, and the value of x is a solution to
ax^3+bx^2+cx-1=0,
Find a, b, and c.

Move one of the inverse cosine functions to the right side and take the cosine of both sides. Simplify, and you have your answer.


It is classical (and posted multiple times, http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=nilpotent+trace&t=64870) that if $ M\in \mathcal{M}_n(\mathbb{C})$, and $ \mbox{tr}(M^k)\equal{}0$ for $ k\equal{}1,\ldots , n$, then $ M$ is nilpotent.
1. $ M\in \mathcal{M}_n(\mathbb{C})$, and $ \mbox{tr}(M^k)\equal{}0$ for $ k\equal{}1,\ldots , n\minus{}1$ and $ \mbox{tr}(M^n)\not\equal{}0$, what can we say about $ M$?
2. $ M\in \mathcal{M}_n(\mathbb{C})$. $ i_1, \ldots , i_{n}$ are distinct integers between $ 1$ and $ n$, and $ \mbox{tr}(M^{i_k})\equal{}0$ for $ k\equal{}1,\ldots , n\minus{}1$ and $ \mbox{tr}(M^{i_n})\not\equal{}0$, what can we say about $ M$?

I think that it has already been discussed, but here is one answer for the 1st one:
$ M$ is not nilpotent, hence has at least $ r \geq 1$ distincts non zero eigenvalues. Writing the condition on traces, one can see (Vandermond matrices are invertible) that in fact $ r\equal{}n$ : $ M$ has $ n$ distinct eigenvalues $ \lambda_1, \lambda_2, \ldots, \lambda_n$. But one can say much more because $ (X \minus{} \lambda_1)(X\minus{}\lambda_2)\ldots(X\minus{}\lambda_n) \equal{} X^n \minus{} a$. This shows that $ M$ is similar to $ diag(\lambda \omega, \lambda \omega^2, \ldots, \lambda \omega^n)$ where $ \lambda \in \mathbb{C}^*$ and $ \omega \equal{} e^{\frac{2 i \pi}{n}}$.

Need must have a solution !
[quote]1)that if $ M\in \mathcal{M}_n(\mathbb{C})$, and $ \mbox{tr}(M^k) \equal{} 0$ for $ k \equal{} 1,\ldots , n$, then $ M$ is nilpotent.[/quote]
  Cayley Hamilton Theorem $ M^n \plus{} a_1M^{n \minus{} 1} \plus{} ... \plus{} a_kM^k \plus{} ... \plus{} a_{n \minus{} 1}M \plus{} a_nI_n \equal{} 0$ ($ a_0 \equal{} 1$)
 I Use Newton formula to find $ a_k ,\forall k \in${$ 1,2,..,n$} With $ S_k \equal{} Tr(M^{k}) \equal{} \lambda_1^k \plus{} \lambda_2^k \plus{} .. \plus{} \lambda_n^k$ ,$ \lambda_i ,\forall i \in$ {$ 1,2,..,n$} are eigenvalues of $ M$
         We have $ \forall k \in${$ 0,1,...,n \minus{} 1$} $ \sum_{i \equal{} 0}^{k}{a_iS_{k \minus{} i}} \equal{} (n \minus{} k)a_k$ [b](1)[/b]
By $ S_0 \equal{} n,S_k \equal{} 0 ,\forall k \in$ {$ 1,2,..,n \minus{} 1$} then we have $ a_kS_0 \equal{} (n \minus{} k)a_k$ or $ ka_k \equal{} 0$ then $ a_k \equal{} 0 ,\forall k\in$ {$ 1,2,..,n\minus{}1$} then $ M^n \plus{} a_nI_n \equal{} 0$ from here $ Tr(M^n) \plus{} na_n \equal{} 0$ or $ a_n \equal{} 0$ >Hence $ M^n \equal{} 0$
[quote]2. $ M\in \mathcal{M}_n(\mathbb{C})$. $ i_1, \ldots , i_{n}$ are distinct integers between $ 1$ and $ n$, and $ \mbox{tr}(M^{i_k}) \equal{} 0$ for $ k \equal{} 1,\ldots , n \minus{} 1$ and $ \mbox{tr}(M^{i_n})\not \equal{} 0$, what can we say about $ M$ ?[/quote]
One example $ i_k \equal{} k$
 The same $ a_1 \equal{} a_2 \equal{} ... \equal{} a_{n \minus{} 1} \equal{} 0$ and We have $ M^n \plus{} a_nI_n \equal{} 0$ from here $ Tr(M^n) \plus{} na_n \equal{} 0$ or $ a_n \equal{} \frac {Tr(M^n)}{n} \neq 0$,$ \det(M) \equal{} \frac {( \minus{} 1)^nTr(M^n)}{n}$ so we have $ M^n \plus{} \frac {Tr(M^n)}{n}I_n \equal{} 0$ , $ M \in GL_n{(\mathbb{C})}$ and $ M$  is diagonalizable !
 You can finish cases remain . I hope you will finish it !! :wink:


$n$ vectors are on the plane. We can move each vector forward and backeard on the line that the vector is on it. If there are 2 vectors that their endpoints concide we can omit them and replace them with their sum (If their sum is nonzero). Suppose with these operations with 2 different method we reach to a vector. Prove that these vectors are on a common line

Hello,

You should clarify your wording. What do you mean by "if their sum is not zero" ? Do you mean that we then keep the two initial vectors or we just eliminate them without replacing them by their sum (which is just one point) ?
Also can you please define what you call two "different" methods ?

Thanks,

--
Julien Santini

We have some allowed operations in each operation we can add two vectors that their sum is not zero vector (with problem conditions); And we can move a vector on a line that it's on it. You see we can do these allowed operations in many methods to reach to a single vector. You must prove that all these vectors are on the same line

I wonder why you don't understand what I said. I think everything is clear!!

[quote=Omid Hatami]
I wonder why you don't understand what I said. I think everything is clear!![/quote]

The grammar, for starters.

I think this problem reduce to showing the operation become associative (eg. $n=3$), and this we can represent as a triangle geometry problem. After associativity then rest is simple :)

The sum of the vectors is invariant. Right???
Or I'm wrong?


analytic geometry
geometry
geometric transformation
invariant
The Geometry of Numbers

Prove that on a coordinate plane it is impossible to draw a closed broken line such that [list][*] coordinates of each vertex are rational, [*] the length of its every edge is equal to $1$, [*] the line has an odd number of vertices.[/list]

[quote="Peter"]Prove that on a coordinate plane it is impossible to draw a closed broken line such that [list][*] coordinates of each vertex are rational, [*] the length of its every edge is equal to $ 1$, [*] the line has an odd number of vertices.[/list][/quote]

We can rephrase it as follows:
[list]
[*] coordinates of each vertex are integers,
[*] the length of its every edge is a fixed natural number,
[*] the line has an odd number of vertices.[/list]

Well, it is only a idea... I'm trying to use modulo 4 to take some things over.
The possible pairs for values mod4 is (0,0),(1,0) and (0,1) ($ \delta_x^2\plus{}\delta_y^2\equal{}N^2$).
We can exclude the case (0,0) WLOG (if N is even, then $ \delta_x,\delta_y$ are even too).
Now, we can think in how change the mod4 coordinates of broken-line vertices.

That's a good start. As you remark, by Pythagoras' theorem we have $ a^2 \plus{} b^2 \equal{} c^2$, implying mod 4 that $ c$ is odd iff $ a\plus{}b$ is odd.

Assume such a configuration exists, and take one with minimal fixed sidelength $ L$. Then neceessarily $ L$ is odd (otherwise rescaling with factor $ 1/2$ gives a smaller configuration with integer coordinates & lengths).

Start at a point, after translation we may assume both coordinates are even. Now every time we go along a side, by the first observation, the sum $ x \plus{} y$ switches in parity. So, after an odd number of edges, we can never end back on our starting point.

Nice invariant huh? :)


I just compiled something, the report on the bottom says

[code]
LaTeX-Result: 0 Error(s), 0 Warning(s), 1 Bad Box(es), 1 Page(s)
[/code]

Or however many pages. This is the first time I got a bad box. What is a "bad box"? Doesn't sound good... Although it looks just fine...

If you go to the warrning you'll probably find an underfull or overfull hbox which means that LaTeX had trouble fitting things on a line nicely. See [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url] page 20.
They are the least important warnings - good to remove but not something to stress about.


How can you tell if a matrix has an inverse or not?

[quote="deej21"]I think it has no inverse if it's determinant is 0.  I wouldn't put money on it, but that's what I initially remembered when I saw the question.[/quote]

100% right. A [b]square[/b] matrix has an inverse iff it is non-singular. To be non-singular obviusly is not to be singular, and to be singular means determinant equal to 0. If the matrix is not square, there are not really inverses, but sometimes are pseudoinverses, but that is another story.

The following are equivalent about an nxn matrix A:

- A is not invertible.
- A is singular.
- det A = 0.
- rank A < n.
- A is not row-equivalent to the nxn identity matrix I.
- 0 is an eigenvalue of A.

And probably a few others too, which I can't remember. So there are numerous ways to check.

[quote="TheBB"]
- A is row-equivalent to the nxn identity matrix I.
[/quote]

You should may be edit the one I am quoting, because it is false!. The identity itself is row-equivalent to itself, and it is invertible!! It should be, not row-equivalent!!

Oh yeah, the textbook I was reading from was listing the converses (which also are true, by the way). Got mixed up there, sorry. :)


Perfect Square
modular arithmetic
function
number theory

Assume that $a,b$ are integers and $n$ is a natural number. $2^na+b$ is a perfect square for every $n$.Prove that $a=0$.

Clearly both $4(2^na + b)$ and $2^{n+2} a + b$ are perfect squares for all $n$. Their difference is fixed: $3b$. If $a \neq 0$ (then $a>0$, for otherwise we have negative perfect squares for large $n$), then for sufficiently large $n$, next largest square to $2^{n+2} a + b$ will be greater than $4(2^na + b)$, causing a contradiction.

Btw, this is Polish 2001, problem B1. Please change your source to reflect that.

Could you explain this solution well? I think I wasn't able to understand it...

why contradiction? [if you don't put why that contradicts...it seems little hard to understand...]

thanks

It's a contradiction because we have that $4(2^na+b)$ is a square for all natural $n$ and so if next square to $2^{n+2}a+b$ is larger than $4(2^na+b)$ then if follows that $4(2^na+b)$ isn't a square. Hence contradiction. Is it clear now?

He says that we have $2^{n+2}+b=x^2, 2^{n+2}+4b=y^2$ with some $x,y$, so $y^2=x^2+3b$.
Now when $b>0$ ( I think he forgot to mention), for $x \geq \frac{3}{2}b$ (which is possible to archieve if $a>0$) this gives $(x+1)^2=x^2+2x+1 > x^2 +3b = y^2 > x^2$ or short $x^2<y^2<(x+1)^2$ stating that there is a square between two consecutive squares, being clearly impossible.
For $b<0$ the same argumentation works with $(x-1)^2<y^2<x^2$.
So we are left with $b=0$, but then just set $n=0$ and $n=1$ to get  that $a$ and $2a$ are perfect squares, again giving $a=0$.

My proof is different from others:

Wlog we can let $b$ to be odd.(The other cases are trivial to handle)
Choose an arbitrary odd prime $p$ such that $2$ is a primitive root of $p$.Then $2^{p-1} \not\equiv 1\pmod{p^2}$.

[u][b]Lemma[/b][/u]:There exists ${x \in \{0,1,2,\cdots,p-1}$ such that $2^xa \equiv -b\pmod{p}$

[i]Proof[/i]:Consider the set of integers $2^ka+b,k=0,1,2,\cdots,p-1$.If none of the integers in the set are divisible by $p$,then by PHP there exists $0 \le i<j \le p-1$ such that $2^ia+b \equiv 2^ja+b \pmod{p}$ or $2^{j-1} \equiv 1\pmod{p}$.This contradicts the fact that $2$ is a primitive root of $p$.So the claim holds.

Let $x_0$ be a solution of the congruence.Then
$p \mid 2^{x_0}a+b$ and $p \mid 2^{x_0+p-1}a+b$ (The second one follows from FLT)
$\Rightarrow p^2 \mid 2^{x_0}a+b$ and $p^2 \mid 2^{x_0+p-1}a+b$ (Since both are perfect squares)
$\Rightarrow p^2 \mid 2^{x_0}a(2^{p-1}-1)$
$\Rightarrow p \mid a$ (Since $2$ is a primitive root of $p$)

Since there are infinitely many primes for which $2$ is a primirive root,there are infinitely many prime divisors of $a$.Thus $a=0$

"there are infinitely many primes for which 2 is a primirive root" why?, as wikipedia says Artin's conjecture has not been proven for any value yet.

[hide=" @sayantanchakraborty"] You might want to revise your proof. Whether or not there are infinitely many primes $p$ for which 2 is a primitive root is actually an open problem. [/hide]
[hide=" My solution"]
Note that $4(2^na+b)$ and $2^{n+2}a+b$ are both squares and their difference is constant ($3b$). Now choose a $n$ such that $4 \sqrt{2^na+b} + 1 > 3b$ and note that $2^{n+2}a+b$ lies between $(2\sqrt{(2^na+b)})^2$ and $(2\sqrt{2^na+b}+1)^2,$ contradiction. $\blacksquare$
[/hide]

[quote="AnonymousBunny"][b]Lemma:[/b] There exists no unbounded function $f(x)$ from the natural numbers to the natural numbers such that $f(n)+c$ is a square for all $n \in \mathbb{N}$ for a fixed positive constant $c.$
[b]Proof:[/b]Assume the contrary. Choose a $n$ such that $2f(n)+1>c.$ Note that $(f(n)+1)^2 = f(n)^2 + 2f(n)+1 > f(n)^2+c.$ However, we also have $f(n)^2+c>f(n),$ so we have just found a square between two perfect squares, contradiction. $\blacksquare$[/quote]
Your Lemma is clearly wrong (I will let you find yourself the error(s) in the proof), since the function $f(x) = c^2x^2-c$ is unbounded, and $f(n)+c = (cn)^2$, always a perfect square.

Whoops, I see. That was completely stupid. :oops: Thanks for pointing it out!

Maybe this can make solution to the previous problem even more clear. Once we get $x^2=y^2+3b$ for some non-negative integers $x$ and $y$, we then have $(x-y)(x+y)=3b$. If $b \neq 0$ then $x \neq y$. Then we just note that both $x$ and $y$ tend to $\infty$ as $a$ does implying that $x+y \rightarrow \infty$ so does the product $(x-y)(x+y)$ (depending on the sign of $b$) and it eventually becomes larger (if $b>0$) or smaller (if $b<0$) than $3b$.

Can someone check my solution?
(The OP has said $n$ to be a natural number but $n$ is non negative in the actual problem)
Let $a\neq 0$. $b$ is clearly not $0$.
Let $(a,b)= g$ and consider $g\neq 1$.Let $p$ be a prime divisor of $g$.
Clearly $p|2^na+b$ so $p^2|2^na+b\implies p|2^n\cdot\frac{a}{p}+\frac{b}{p}$
$p$ cannot divide both $\frac{a}{p}$ and $\frac{b}{p}$ as then the gcd will not be $1$.
If $p$ divides even one of them, then $p$ will also have to divide the other (because $p|2^n\cdot\frac{a}{p}+\frac{b}{p}$)
Contradiction.
So. $\left(\frac{a}{p},p\right)=\left(\frac{b}{p},p\right)=1$
Now, for $n=0$, $p\mid\frac{a}{p}+\frac{b}{p}$ and for $n=1$, $p\mid 2\cdot\frac{a}{p}+\frac{b}{p}$

Subtracting, we get $p\mid\frac{a}{p}$, contradiction.
So, $(a,b)=1$
Let us select a prime $p$ such that $p\mid a+b$
We know that there exists $n'$ such that $2^{n'}\equiv 1\pmod p$ and $2^{n'}\not\equiv 1\pmod {p^2}$
$\implies 2^{n'}a+b\equiv a+b\pmod p$
Therefore $p^2|2^{n'}a+b$ and $p^2|a+b\stackrel{\text{subtracting}}{\implies} p^2|a(2^{n'}-1)$
So we have $p|a$ and $p|a+b$. Contradiction. (as $(a,b)=1$)
So $a$ has to be equal to $0$.

[quote="Ashutoshmaths"]
$p$ cannot divide both $\frac{a}{p}$ and $\frac{b}{p}$ as then the gcd will not be $1$.
[/quote]
Why? If $p$ divides both $\dfrac{a}{p}$ and $\dfrac{b}{p},$ all we get is $p^2 \mid g$. I fail to see how that is a contradiction.

Also, I don't think you're allowed to put $n=0$ (I'm not sure what you mean by the "original problem"; I'm pretty sure the OP intended $n$ to be positive).

[quote=billzhao]Clearly both $4(2^na + b)$ and $2^{n+2} a + b$ are perfect squares for all $n$. Their difference is fixed: $3b$. If $a \neq 0$ (then $a>0$, for otherwise we have negative perfect squares for large $n$), then for sufficiently large $n$, next largest square to $2^{n+2} a + b$ will be greater than $4(2^na + b)$, causing a contradiction.[/quote]

What is the inspiration for looking at the difference of $4(2^na + b)$ and $2^{n+2} a + b$? Thanks a lot!

Another solution:
Suppose that $a\neq0$ then $a>0$ and for large values of $n$ the number $2^n.a+b$ is negative.
From the hypothesis, there exist a sequence of positive inntegers $(x_n)_{n\ge1}$ such that $x_n=\sqrt{2^n.a+b}$ for all $n$.
Easy to see that $\text{lim}(2x_n-x_{n+2})=0$, implies the existence of a positive integer $N$ such that $2x_n=x_{n+2}$ for all $n\ge N$.
But $2x_n=x_{n+2}$ is equivalent to $b=0$. Then $a$ and $2a$ are both perfect squares, which impossible for $a\neq0$.
This show that $a=0$. ;)

[url=http://artofproblemsolving.com/community/c6h1433593p8105886]Here[/url] is a different approach.

[quote=ZetaX] (which is possible to archieve if $a>0$) how do you get this?.[/quote]

[quote=ZetaX], for $x \geq \frac{3}{2}b$ (which is possible to archieve if $a>0$)[/quote]
How did you get this?


How fast should you be doing the national practices?

I mean for sprints and target.  I HIGHLY doubt I'll get to cds.

I have not gotten any practice nationals, but I should have gotten them by next week.

My Ideals:  1.5 min per sprint.

                1.899999999999999999999999 min per target.  Apprx 1 min 48 secs.

Is there any ways to increase accuracy and speed?

PLUS:  Should I even attempt the last page? 25 ---> 30?

No idea how fast you should go, but I suggest skipping around. I noticed that on some nationals, there are actually some (relativley) easy problems on the last page.

Well what I would do is try going really fast. If there is one that is even semi-hard, skip it. You should have about 10 minutes left at end by doing this, so if you have greater than 10 problems that you haven't done, then do them in order from easiest to hardest. if you have about 4-5 that you haven't done, I reccomend checking all of the ones that you aren't sure of.

In target, do the easiest one first, check it, then do the hardest one.

i actually like the last page.

the first 25 are usually about tricks and preventing mistakes.

The last page is usually the conceptual stuff, so you'll get it if you know how to do it.  There's not as much risk of making stupid mistakes.

Basically just make sure you can finish all the problems in 40 minutes and also make sure you can check over all your work at least once.

that's helpful advice, peijin.
and how do you recommend doing that?

You might be able to find some on Google but otherwise I believe some AoPSers have some past national tests. You can also do some of the questions that appear on the MC forum here; as many are of national level.

you should try to finish the first 10 probs in like 5 minutes then maybe 1.5 minutes about for the rest and then some checking

for target... it's good to be able to finish both problems in like 3 minutes and have 3 minutes leftover to chech both of them a billion times

For target, I usually try to finish the easier one in like a minute, spend about 3 minutes on the other, and spend the rest of the time checking.

Oh, but if it were me, I would just do 1-20. Skim over 21-30, see if there's any I could do, do those, then go back and check over 1-20.
And if I were to have additional time, I would go back and try the ones I didn't do from 21-30.

There's really no secret to taking the test - you can either get the answer or you can't...

do it backwards if you want a confidence boost

backwards? the point on making the difficulty of the problems rise throughout the sprint is to get the students warmed-up to tackle the harder problems (25-30). Doing the hardest problems first won't do any good :wink:

For States i usually carefully do problems 25-30 and skim the last few problems. It worked :D

I discover that on nats the last several problems aren't really much harder than the others though. so maybe you should do all the problems in the given time. I'm still working on this :)

But if you want a confidence boost, like the other poster mentioned, then you can do the hard ones first and just breeze through the others.
You'd probably win tiebreakers too, but I wouldn't do it unless you're [i]really[/i] good.

I don't know. Let's say you can get at least 25 on sprint most of the time.
Wait, that totally falls apart if it's a really hard year. Argh, I'm all confused now. Just disregard what I said.

Hm...the highest score on national sprint last year was, I think, 26, so you can easily make countdown with 22 or 23; suppose you aim for 22.  Basically, aim to get the first 15 right.  This shouldn't be hard - just don't make stupid mistakes.  Then, you only need 7 out of the next 15 - not too bad right?

Countdown...well, it depends on the problem and who you are up against...

You got the 26, right?
It shouldn't be too hard, unless you're someone like me who just gets wiped out by harder problems.

unless you are aiming for the top 10, i suggest you do the first couple of pages, then pick problems in the last couple, but then again, this comes from a guy who never went to nationals and does horrible in national tests... :P

you should be able to hit the first ~20 in about 20-ish minutes (if anything takes you more than two minutes, skip it immediately) for top twelve standards
20-30 in the next 15 minutes is a good estimate, then go back and check over your work/do the things you haven't done

Meh, for Chapter Target, I had the answers before time started.
:huh:
Basically got the answers while the proctor was reading the problems to us...
:P
Anyways, for Nats, just try to do as many problems as you can, and don't waste your time on problems you probably can't do.

yeah chapter target was ridiculously easy this year, 'specially the first four or so

Your proctors read the Target Round questions to you? Kinda weird... because mine didn't.

[quote="ZhangPeijin"]
Basically got the answers while the proctor was reading the problems to us...
:P
[/quote]

yeah...that's weird. The proctor read the questions then started time? or started time then read?

they're allowed to do that? All we got was the 6 minutes.

[quote="ra5249"]for our chapter, they read out loud but didn't let us pick up our pencils, then they started time.[/quote]

I'm pretty sure the extra time wouldn't have been needed though :|

yeah, honestly, given the difficulty level of this year's target, it probably wouldn't have made a difference...

Well they started time after they finished reading, but they read pretty fast.
I wasn't sure whether or not I was allowed to use my pencil, so I played it safe and tried to the problems mentally.

[quote="ra5249"][quote="CountdownKing"]It doesn't matter if they read it, because it's the same for everyone in the chapter. As long as your state doesn't have wild card teams like mine, and not all the chapters read the problems. Still, I've never heard of that.

I have some questions for Allen Yuan.

I've never been to Nationals, so I don't know what scores are good. What did you get last year and what did Justin get?[/quote]
the scores were like 41, 40, 39, scores between 36 and 38 (36 was cd cutoff) 7-way tie for 35, and so on[/quote]

wow....thats 7 people killing themselfs for making stupid mistakes....


ratio
modular arithmetic
AMC
AIME
arithmetic sequence
geometric sequence

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30.  Find the sum of the four terms.

[quote="sen"][quote="E^(pi*i)=-1"][hide]

The four numbers are $a, a+k, a+2k, \frac{(a+2k)^2}{a+k}$.  We see that $a+30=\frac{(a+2k)^2}{a+k}$, so that $(a+30)(a+k)=(a+2k)^2$.

Expanding and simplifying yields $30a+30k=3ak+4k^2$, or $3a(10-k)=2k(2k-15)$.  We see that $10-k$ and $2k-15$ have the same sign.  If they are both positive, $k<10$ and $k>7$.  The only solutions are 8 and 9.  If both are negative $k>10$ and $k<8$, impossible.  For $k=8$ we get $a=\frac{8}{3}$, which is impossible since all the terms are integers.  For $k=9$ we get $a=18$.  Evaluating the four terms we get 18, 27, 36, and 48.  The answer is 18+27+36+48=129.

[/hide][/quote]

But what if $a$ and $k$ are not of the same sign?[/quote]

$a$ and $k$ have to be of the same sign because the sequence is $increasing$ and is of $positive$ integers.

[quote="E^(pi*i)=-1"][hide]

The four numbers are $a, a+k, a+2k, \frac{(a+2k)^{2}}{a+k}$.  We see that $a+30=\frac{(a+2k)^{2}}{a+k}$, so that $(a+30)(a+k)=(a+2k)^{2}$.

Expanding and simplifying yields $30a+30k=3ak+4k^{2}$, or $3a(10-k)=2k(2k-15)$.  We see that $10-k$ and $2k-15$ have the same sign.  If they are both positive, $k<10$ and $k>7$.  The only solutions are 8 and 9.  If both are negative $k>10$ and $k<8$, impossible.  For $k=8$ we get $a=\frac{8}{3}$, which is impossible since all the terms are integers.  For $k=9$ we get $a=18$.  Evaluating the four terms we get 18, 27, 36, and 48.  The answer is 18+27+36+48=129.

[/hide][/quote]Here's something that makes the calculation slightly faster. (just slightly :))

[hide]Note that since $3a(10-k)=2k(2k-15),$ you must have $k|3$ or $2k-15|3\implies k|3.$ Hence, $k|3$ and you only need to test $k=9$ instead of $k=8,9.$[/hide]

Click on the smiley below.

[hide=" :) "]Let the sequence be: $a-x, a, a+x, \frac{(a+x)^{2}}{a}$. We have that:

$\frac{(a+x)^{2}}{a}-(a-x) = 30 \Leftrightarrow x^{2}= 3a(10-x)$

We know that $a,x$ must be integers, so we know that $0 < x < 10, 3|x$, which gives us the possibilities that $x=3,6,9$.

[b]Case 1: x = 3[/b]

This case clearly does not work, as this value of $x$ introduces a prime factor of $7$ on the right hand side.

[b]Case 2: x = 6[/b]

This case does not work, as $a = 3$ and $a-x>0$ by our given conditions.

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

We cross our fingers and hope that this case does work, and fortunately, our finger-crossing method works. This case implies that $a = 27$.

From the above, we conclude that our sequence is: $18, 27, 36, 48$. Adding these values yields the value of $129$.[/hide]

[hide="heh"]
well let's see the numbers are $x, x+c, x+2c, \frac{(x+2c)^{2}}{x+c}$

and since $x+30=\frac{(x+2c)^{2}}{x+c}$

it's pretty easy to see that $x =(\frac{c}{3})( \frac{30-4c}{c-10})$

and we know that both $30-4c$ and $c-10$ gotta be either negative or positive, and we can tell that only $c = 8, 9$ work.

$c=8$ fails (lol c8) but $c=9$ produces $x=18$, and there we are.

$\boxed{129}$
[/hide]

[hide="Another approach"] Let the common ratio of the geometric part of the series be $\frac{a}{b}$, where $gcd(a, b) = 1$.  Then we can write the geometric portion of the series as

$kb^{2}, kab, ka^{2}$

The first term must therefore be

$kb^{2}-(kab-kb^{2}) = k(2b^{2}-ab)$

Now we are only subject to the requirements $a > b, 2b > a$ and

$ka^{2}-k(2b^{2}-ab) = k(a^{2}+ab-2b^{2}) = 30$

Well, one way to satisfy the first condition comes when $b = 2, a = 3$.  This gives

$a^{2}+ab-2b^{2}= 9+6-8 = 7 \not | 30$

Hmm.  The next possibility is $b = 3, a = 4$.  This gives

$a^{2}+ab-2b^{2}= 16+12-18 = 10 | 30$

So $k = 3$ and we have the familiar series.  The important thing to stress with this solution is that it is easy to generate many such series if the last condition is removed. [/hide]

Note that
$ \frac{(x\plus{}2y)^2}{x\plus{}y}\minus{}x\equal{}30$
is equivalent to
$ y(3\plus{}\frac{y}{x\plus{}y})\equal{}30$
But $ \frac{y}{x\plus{}y}$ is between 0 and 1
So y is strictly between 30/4 and 30/3
So we need to check whether y equals 8 or 9.
We find y=9 x=18 works.

[hide="Solution"]

Let the terms be $a-d, a, a+d, \frac{(a+d)^2}{a}$.

Then the sum is $3a+\frac{a^2+2ad+d^2}{a}$ or $4a+2d+\frac{d^2}{a}$.

Also, $\frac{(a+d)^2}{a}-a+d=30$, so $\frac{(a+d)^2-a^2+ad}{a}=30$, or therefore $\frac{3ad+d^2}{a}=30$.

This also gives $3d+\frac{d^2}{a}=30$.

Therefore, our desired answer is the same as $4a+30-d$.

Now, since $d$ is positive, and when multiplying by $a$ and taking $\pmod{3}$, we get $d\equiv 0\pmod{3}$, we do casework on $d$.  $d=3$ gives $9+\frac{9}{a}=30$ or $a=\frac{1}{3}$ contradiction.  $d=6$ gives $\frac{36}{a}=12$ or $a=3$, but $a>d$ since the first term is positive, so contradiction.  Therefore, $d=9$ giving $\frac{81}{a}=3$ or $a=27$.

This gives the last term (to check for all conditions) to be equal to $\frac{(27+9)^2}{9}$ which is clearly an integer, so hence the answer is $4*27+30-9=\boxed{129}$.  [/hide]

[hide="Solution"]Denote by $d$ the common difference in the arithmetic progression. The first three terms are then $a$, $a+d$, and $a+2d$. The last term is equal to the third term multiplied by the ratio of the third term to the second term, or $\frac{(a+2d)^2}{a+d}$. Thus, we have the equation $$\frac{(a+2d)^2}{a+d}-a=30,$$ which upon simplification and rearrangement becomes $$3ad+4d^2=30a+30d.$$ We will solve for $a$ in terms of $d$ and use the fact that all terms are integers to deduce $d$. This gives $$a=\frac{4d^2-30d}{30-3d}.$$ Hence, $4d^2-30d$ is divisible by $30-3d$. The quantity $30-3d$ must also then divide $4d^2-30d+d(30-3d)=d^2$. This tells us that $d^2$ is divisible by $3$, so $d$ is divisible by $3$. Since all terms are positive, $d$ must be positive, giving us the possible choices $3$, $6$, and $9$. But $4d^2-30d$ must also be positive, so $d>\frac{15}{2}$. Therefore, $d=9$, and plugging this into our expression for $a$, we find that $a=18$. The sequence is $18$, $27$, $36$, $48$, and its sum is $\boxed{129}$.[/hide]

Let's denote the sequence by $a, b, c, d$. We have $a+c = 2b, bd = c^2, d-a = 30$. We are trying to find $a+b+c+d$. Also, we are given that the terms are positive integers. We now split into two cases.

We know $d-a = 30$. Then $a = d-30$ and $2b = c+d-30$, so $2bd = d(c+d-30) = 2c^2$, which means that $dc+d^2-30d = 2c^2$. Solving as a quadratic in $d$, we get discriminant $9c^2-60c+900$. Denote $9c^2+60c+900 = p^2$, and solving as a quadratic in $c$ gives discriminant $p^2-800$. Letting $p^2 - 800 = x^2$ gives $(p-x)(p+x) = 800$, and checking possibilities gives $p = 102, x = 98$ so $c = 36, d = 48, a = 18, b = 27$ and the answer is $\boxed{129}$.


[color=blue][i]Οι παρεγγεγραμμένοι κύκλοι παρουσιάζονται συχνά σε διαγωνισμούς. Ευκαιρία να θυμηθούμε τις βασικές μετρικές σχέσεις:[/i][/color]

Έστω [b]Δ[/b] ένα σημείο στην πλευρά [b]ΒΓ[/b] ενός τριγώνου [b]ΑΒΓ[/b]. Θεωρούμε τους παρεγγεγραμμένους κύκλους $ C_1,C_2$ των τριγώνων [b]ΔΑΒ,ΔΑΓ [/b] που εφάπτονται στα τμήματα [b]ΒΔ,ΔΓ [/b]και την άλλη , εκτός της [b]ΑΔ[/b] , κοινή εσωτερική εφαπτομένη των $ C_1,C_2$  που τέμνει την [b]ΒΓ[/b] στο [b]Τ[/b]. Να αποδειχθεί ότι ο παρεγγεγραμμένος κύκλος του τριγώνου [b]ΑΒΓ[/b] που αντιστοιχεί στη γωνία [b]Α [/b]εφάπτεται με την[b] ΒΓ [/b]στο [b]Τ[/b].

 Μπάμπης

Έστω ότι η $ BC$ εφάπτεται στον $ C_1$ στο $ Q$ και στον $ C_2$  sto $ S$ .
Τότε εύκολα βγάζουμε ότι $ DQ\equal{}TS$
Όμως  $ DQ\equal{}\frac{AB\plus{}BD\minus{}AD}{2}$ και επιπλέον
$ TC\equal{}TS\plus{}SC\equal{}\frac{AB\plus{}BD\minus{}AD}{2}\plus{}\frac{AD\plus{}DC\minus{}AC}{2}\equal{}\frac{AB\plus{}BC\minus{}CA}{2}\equal{}t\minus{}b$

όπου $ t$ η ημιπερίμετρος του $ ABC$ , το ζητούμενο . :)


I was looking at the solution for a problem about sets and didn't understand one part of the solution. It said something like

... and, letting |S| denote the number of elements in set S and U denoting the union of sets sign, 2^101 + 2^|C| = 2^|A U B U C| . If |A| = |B| = 100 and |C| > 0, then the only possible solution to this equation is |C| = 101 and  |A U B U C| = 102.

However, I don't understand why that is the only solution given the above equation. Does anyone happen to know?

It's actually for a number theory reason. When can the sum of two powers of 2 be another power of 2? In other words, if 2a + 2b = 2c, where a, b, and c are integers - then what does that say about a, b, and c?

[quote]then what does that say about a, b, and c?[/quote]

Dunno. a=b, b=c, or a=c would obviously work, but I dunno about other solutions.

Actually a has to equal b. To see why, suppose a < b. Then factoring out 2a, you get 2a(1 + 2b-a) = 2c.

But that means 1 + 2b-a is a power of 2, which is only possible when b-a = 0. Therefore a = b, and c = a+1.


Consider $ C^0([a,b],\mathbb{R})$ and let $ C^{0,\alpha}$ be the usual Holder space with the norm , where $ \parallel f\parallel_{\alpha}\equal{}\parallel f\parallel_{\infty}\plus{}\sup_{x\neq y}\frac{|f(x)\minus{}f(y)|}{|x\minus{}y|^{\alpha}}$. .  Suppose that $ 0<\delta<\gamma<1$.  Let $ U\equal{}\{f\in C^{0,\gamma}: \parallel f\parallel_{\gamma} \leq C\}$. Given that there is some $ K>0$ such that $ [f]_{\delta}\leq K(\parallel f\parallel_{\infty})^{1\minus{}\frac{\delta}{\gamma}}([f]_{\gamma})^{\frac{\delta}{\gamma}}\forall f\in C^{0,\gamma}$, prove that $ U$ is compact in $ C^{0,\delta}$.

It look like we have to use the Ascoli's theorem
is $ 0<\alpha<1$?

You can say that.  This restricted is given to $ \delta$ and $ \gamma$.

EDIT: I just finished solving it.


Hi guys,

if you have a 3d space curve s(t), then is the value $ \hat{s}\cdot\nabla\hat{s}$ equivalent to the normal vector, $ \frac{\frac{d\mathbf{T}}{dt}}{\parallel{}\frac{d\mathbf{T}}{dt}\parallel{}}$ where $ \mathbf{T} \equal{} \frac{d\mathbf{s}}{dt}$?

also, is $ \hat{s}\cdot\nabla\times\hat{s}$ equivalent to the binormal vector, or torsion, $ \mathbf{B} \equal{} \mathbf{T}\times \mathbf{N}$?

it seems to me they should be equivalent, but i'm not exactly sure.

Huh? Those all look like meaningless malformed expressions to me. $ \nabla$ is not a meaningful operation on vector-valued functions of a scalar. The curl isn't, either. You don't get vectors out of dot products.

isn't the gradient of a vector a second-order tensor? also i'm sorry I meant the magintude of the normal and binormal vectors.

those expressions actually come from physics - if you really want you can find them in brizard and hahm reviews of modern physics 79, 2007. i was very confused when i first encountered them too - i thought at first that $ \hat{s}\cdot\nabla\times\hat{s}$ always = 0, but i worked it out and it doesn't for some reason. sorry, i'm not very good at linear algebra - i've never formally taken it

For a function [b]of a scalar[/b], there is only the derivative. All other differential operators come from the possibility of differentiating with respect to more than one variable.

Also, here's an expression for the magnitude of the normal $ N$: $ 1$. That's also the magnitude of the binormal $ B$.

what about for a function of a vector? i think that's what they meant

Then that's a completely different question, which is not (entirely) about curves. In what you posted, $ t$ is the only independent variable.

yes i'm sorry - the t was all my doing. s should be a function of 3 variables.

So now the other side makes no sense. Go back to the drawing board.


geometry
analytic geometry
geometry proposed

In the plane is given a right isosceles triangle with smaller sides equal 1. It is allowed to replace one vertex of the triangle with its symmetrical wrt to one of the two other vertices of the triangle. As a result of such operations we obtain a triangle with side lengths $a,b,c$ with $c$ being the longest one. Prove that $a+b-c\leq2-\sqrt{2}$.

If $ABC$ is the triangle and $\angle C=\pi/2$ then set $C(0,0),A(1,0),B(0,1).$ Now use the fact that after a transformation we obtain a triangle with vertices with integer coordinates and area $\frac{1}{2}.$


I found this rather interesting :
Suppose the harmonic series converges to $h$ ... then by rearranging the terms we get $h-1=h$ which is contradiction.
http://mathworld.wolfram.com/BernoullisParadox.html

How exactly does one rearrange the terms?

For that paradox you have to assume that $h$ is finite. Otherwise the statement of the "paradox" wouldn't make sense anyway.

[quote="leibniz"]$\sum_{k=1}^{\infty}\frac{1}{k}=+\infty$ ;)[/quote]

How exactly does this help me understand what I asked?

Guys, spix is asking how you can rearrange $h = \sum_{k=1}^{\infty}\frac{1}{k} \text{to get } h = h - 1$.


I care, can you prove that formula?
Thanks.

[quote="bpms"][quote="ragnarok23"]A circle with radius 8 has two perpendicular diameters drawn, $AB$, and $CD$ as shown. Four congruent circles are each inscribed in the quarter circles formed. Then a smaller circle is inscribed between the other four circles, such that it is tangent to each circle. Find the area of this circle (shaded region). Express your answer in terms of $\pi$ and in simplest radical form.[/quote]
Another problem I would call nice.
[hide]The formula for a quartercircle is $\frac{r}{\sqrt{2}+1}$ (if anyone cares I can prove it), so the radius of the quarter circle is $8(\sqrt{2}-1)$ and the diameter is $16(\sqrt{2}-1)$. Note that subtract this from the radius of the largest circle gives you the radius of the smalles circle. So the radius of that is $24-16\sqrt{2}$. Squaring and multiplying by pi we get $1088\pi-768\pi\sqrt{2}$ or the factored form $64\pi(17-12\sqrt{2})$.[/hide]
Hopefully no careless mistakes.[/quote]
for your second sentence, don't you have to subtract that twice since there are two quarter circles on each side of the smallest circle?

[quote="13375P34K43V312"]Proofs don't really belong in the MC forum...[/quote]
An explanation of a formula is perfectly fine though.
What we mean by proofs not allowed in this forum is, for example, problems that begin with "Prove that..."; in other words, solutions that require rigorous proofs.

This problem is crushed by pythagorean.
For why the formula works, draw the picture, take the dot in the center of the inscribed circle, draw both altitudes to the radii of the larger circle to get a square. Draw the relevant diagonal of the square, and draw a radius from the dot to the point on the circumference such that it makes one segment. There you go.
If this is to vague I can post a picture, but I would prefer not to because it would be difficult...

Ahhh, cool. I wish I didn't suck at geometry.

I used to, but it's easier when you're required to learn it. It's kind of boring, so I generally don't do it when I don't have to.

in my (limited) experience with circles inside circles, the pythagorean theorem is usually the best way to go  :D


inequalities
algebra
function
domain
inequalities unsolved

Hi!

This problem is a simple aplication of Holder´s Inequality.
Do you know this inequality?

I've heard of this inequality. But what if it is not in the domain of knowledge required? Is it still able to be resolved? Must I do it from the scratch like proving Holder's or can I exploit other inequalities? Thanks a lot.

[quote="HK Angel"]The question:
If a, b, c, d, e, f are positive real numbers, prove that
(a3 + d3) ^ 1/3  *  (b3 + e3)^ 1/3  *  (c3 + f3)^1/3 >=  abc + def
---------------
It seems obvious that the AM>=GM>=HM inequality has to be used. However, the right part appears strange to me. It appears that the AMGMHM inequality has to be applied twice but then how many terms (2 or 3?) are involved is quite confusing. Any clue on how to start on this problem or any good hints? Thank you.[/quote]

$ (a3 \plus{} d3) ^ 1/3  *  (b3 \plus{} e3)^ 1/3  *  (c3 \plus{} f3)^1/3 >\equal{}  abc \plus{} def$   ---> $ (a^{3}\plus{}d^{3})(b^{3}\plus{}e^{3})(c^{3}\plus{}f^{3})\geq(abc\plus{}def)^{3}$----> $ (a^{3}\plus{}d^{3})(b^{3}\plus{}e^{3})(c^{3}\plus{}f^{3})(abc\plus{}def)\geq(abc\plus{}def)^{4}$---->
$ (a^{3}\plus{}d^{3})(b^{3}\plus{}e^{3})(c^{3}\plus{}f^{3})(abc\plus{}def)\geq(\sqrt{a^{3}b^{3}}\plus{}\sqrt{d^{3}e^{3}})^{2}(\sqrt{abc^{4}}\plus{}\sqrt{def^{4}})^{2}\equal{}((\sqrt{a^{3}b^{3}}\plus{}\sqrt{d^{3}e^{3}})(\sqrt{abc^{4}}\plus{}\sqrt{def^{4}}))^{2}\geq((\sqrt{\sqrt{a^{3}b^{3}}\sqrt{abc^{4}}}\plus{}\sqrt{\sqrt{d^{3}e^{3}}\sqrt{def^{4}}})^{2})^{2}\equal{}(abc\plus{}def)^{4}$


$x,y,z>0$ such that $x+y+z=1$, prove that
\[xy\sqrt{x+y}+yz\sqrt{y+z}+zx\sqrt{z+x}\le \frac{1}{2}\sqrt{xy+yz+zx-xyz}\]

We have
$ \sqrt{xy\plus{}yz\plus{}xz\minus{}xyz}\equal{}\sqrt{(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\minus{}xyz}\equal{}\sqrt{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$
This inequality becomes
$ \sum xy \sqrt{x\plus{}y}  \leq \frac{1}{2} \sqrt{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$
$ <\equal{}> \frac{\sum xy \sqrt{x\plus{}y}}{\sqrt{(x\plus{}y)(y\plus{}z)(z\plus{}x)}}  \leq  \frac{1}{2}$
$ <\equal{}> \sum \frac{xy}{\sqrt{(y\plus{}z)(z\plus{}x)}} \leq \frac{1}{2}$
By AM-GM we have
$ \sum \frac{xy}{\sqrt{(y\plus{}z)(z\plus{}x)}}  \leq  \sum \frac{1}{2} xy  ( \frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x})$
$ \equal{}  \frac{1}{2} [  (\frac{xy}{x\plus{}z}\plus{}\frac{yz}{x\plus{}z})\plus{}( \frac{xy}{y\plus{}z}\plus{} \frac{zx}{y\plus{}z})\plus{}(\frac{yz}{x\plus{}y}\plus{}\frac{zx}{x\plus{}y}) ] \equal{} \frac{x\plus{}y\plus{}z}{2}\equal{} \frac{1}{2}$
Done!


Human beings have discovered an inhabitable planet and soon after, they find 10 more inhabitable planets. Of these 11, only 5 are deemed "Earth-like" in its resources and the rest are deemed "Mars-like" since they lack many important resources. Assume that planets like Earth take up 2 units of colonization, while those like Mars take up only 1. If humanity mobilizes 12 total units of colonies, how many different combinations of planets can be occupied if the planets are all different from each other?

Why don't you post this in another forum? The high school forums are much more active than the Alcumus forums.


Show that in any set of eleven integers there are six whose sum is divisible by $6$.

This problem written in that way, isn't interesting, it can be done 'manually'.
From the other hand, this is a special case of a strong thing : Olson's theorem. Thinking generally, we can prove much more than this task requires.

[hide]Now I remember that it posted before http://www.mathlinks.ro/Forum/viewtopic.php?t=57546[/hide]

a general version --
it is well known that ,given (2n-1) positive integers. show that , it is possible to get n from them , such that, their sum is divisible by n.
put n=6 and done!


inequalities
trigonometry
inequalities solved

okay.. strange .. when I pressed post.. an idea stroke me and I came with this solution: (hope it's good)

let H be the orthocentre of the traingle. Then we have H inside ABC and HC+HB<AB+AC => 2R(cosB+cosC)<2r(sinB+sinC) which gives:

CosB+cosC<sinB+sinC and then sum (cosB+cosC)<sum(sinB+sinC) which gives what we want!!

cheers! if anybody has a kooler idea give a horn! bye bye :D

that's pretty cool - i don't think that it can get anymore cooller than that :)

[quote="galois"]hey sheer coincidence i was trying this very problem when i made the last post.and by using what grobber proved( using a very elegant argument)
sin a + sin b + sin c >2>1 + 8sin(a/2)sin(b/2)sin(c/2)>1 + 4sin(a/2)sin(b/2)sin(c/2)=cos a + cos b + cos c
this shows that this inequality is very weak
cheers[/quote]

[size=150]sin a + sin b + sin c >2>1 + 8sin(a/2)sin(b/2)sin(c/2)>1 + 4sin(a/2)sin(b/2)sin(c/2)=cos a + cos b + cos c
How to prove it???thanks![/size]

[quote="Lagrangia"]

let H be the orthocentre of the traingle. Then we have H inside ABC and HC+HB<AB+AC => 2R(cosB+cosC)<2r(sinB+sinC) which gives:

[/quote]
I thinks RHS shoud be $2R(\sin B+\sin C)$
but LHS
if one of them greater than $\frac{\pi}{2}$
$HC+HB$ shoud be  $\mathbb{-}2R(cosB+cosC)$

[quote="galois"]hey sheer coincidence i was trying this very problem when i made the last post.and by using what grobber proved( using a very elegant argument)
sin a + sin b + sin c >2>1 + 8sin(a/2)sin(b/2)sin(c/2)>1 + 4sin(a/2)sin(b/2)sin(c/2)=cos a + cos b + cos c
this shows that this inequality is very weak
cheers[/quote]
$\sin a + \sin b + \sin c >2\text{ is true if all angle are not exceed }\frac{\pi}2$

Example
\[ \sin 10^\circ+\sin 60^\circ+\sin 110^\circ=1.97937...\not >2 \]
Tummy

[quote="Lagrangia"]galois with his problem: inequality involving sines reminded me of a problem:
 prove that in any acute angle triangle:  sum (cosA)< sum (sin A) holds! g'luck![/quote]
$\sum\cos{A}\le \frac{3}{2}< 2<\sum \sin{A}$ .

[b]
Note:
1) The inequality  $\sum\cos{A}\le \frac{3}{2}$  appears in  J.M.Child Math.Gaz., 23(1939)138-143.
2) $2<\sum\sin{A} \le \frac{3}{2} \sqrt{3}$ in  acute triangles seems to be observed by O.Bottema , Euclides 20(1954/55) 114-116. [/b]

Another variant of proposed inequality :  if $A,B,C\in \left(0,\frac{\pi}{2}\right]$ then
\[ \cos{A}+\sqrt{2}\left(\cos{B}+\cos{C}\right)< \sin{A}+\sin{B}+\sin{C} \; . \]
Note: the LHS is maximum for $A=\frac{\pi}{2}$ and $B=C$.


Let $ a,b,c\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that:
$ 5(a\plus{}b\plus{}c)\plus{}\dfrac{3}{abc}\geq 18$

I'm sorry, I'm sure this problem was posted before. It looks so familiar...

Let $ p\equal{}(a\plus{}b\plus{}c)$  and  $ q\equal{}ab\plus{}bc\plus{}ca$  Then

  $ p^2\equal{}2q\plus{}3  , p>\sqrt{3}$ and $ (ab\plus{}bc\plus{}ca)^2\geq 3abc(a\plus{}b\plus{}c)$, denote $ \frac{1}{abc}\geq\frac{3p}{q^2}$

 Thus, is suffices to prove that $ 5p\plus{}\frac{9p}{q^2}\geq 18$

 Since

$ 5p\plus{}\frac{9p}{q^2}\equal{}5p\plus{}\frac{36p}{(p^2\minus{}3)^2}\minus{}18\equal{}\frac{5p^2\minus{}18p^4\minus{}30p^3\plus{}108p^2\plus{}81p\minus{}162}{(p^2\minus{}3)^2}\equal{}\frac{(p\minus{}3)(5p^3\plus{}12p^2\minus{}3p\minus{}18)}{(p^2\minus{}3)^2}$

  we still have to show that $ 5p^3\plus{}12p^2\minus{}3p\minus{}18\geq 0$

  It is true because $ p>\sqrt{3}$ and then $ 5p^3\plus{}12p^2\minus{}3p\minus{}18\equal{}p^2(5p\plus{}12\minus{}\frac{3}{p}\minus{}\frac{18}{p^2})>p^2(5\sqrt{3}\plus{}12\minus{}\sqrt{3}\minus{}6)>0$

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:
$ 5(a \plus{} b \plus{} c) \plus{} \dfrac{3}{abc}\geq 18$[/quote]
setting $ F(a,b,c) \equal{} 5(a \plus{} b \plus{} c) \plus{} \dfrac{3}{abc}$

$ F(a,b,c) \minus{} F(a,\sqrt {\frac {b^2 \plus{} c^2}{2}},\sqrt {\frac {b^2 \plus{} c^2}{2}}) \equal{}$
$ (b \minus{} c)^2(\frac {3}{bc(b^2 \plus{} c^2)} \minus{} \frac {5}{b \plus{} c \plus{} \sqrt {2(b^2 \plus{} c^2)}})$

we can be prove easy $ F(a,b,c)\ge F(a,\sqrt {\frac {b^2 \plus{} c^2}{2}},\sqrt {\frac {b^2 \plus{} c^2}{2}})$

by Mixing Vriable we need prove $ F(\sqrt {3 \minus{} 2t^2},t,t)\ge 0$


Source: Georgia GHP Problem Solving Set I

Let A and B be two non-concentric circles with radii of different lengths. A and B have two points of intersection, P and Q. Let W and X be points on the locus of A, and let Y and Z be points on the locus of B. Show that if line segments WPY and XQZ are parallel, then WXYZ is a parallelogram.

(An example drawing included)

Let $\angle WPQ=x$, then $\angle PQZ=x$, also

$\angle WXZ=180-x=\angle PYZ$ by subtended arcs

by symmetry, $\angle XWY=\angle XZY$,

so $WXYZ$ is a parallelogram since both pairs of opposite angles are equal


function
complex analysis
algebra
rational function
complex analysis unsolved

Let $f$ be holomorphic function in the unit disk $|z|<1$ and continious on it's boundary $|z|=1$ and $|f(z)|=1$ for all $|z|=1$. Prove that $f$ -rational function

I don't have a rigorous answer.  But I think the idea is that we can subtract out some Blaschke products and then we have a set on which $f=0$ which contains a limit point.

[quote="Kent Merryfield"][quote]But I think the idea is that we can subtract out some Blaschke products[/quote]
Change the word "subtract" to "divide."[/quote]
Yes, I think the sketch of the solution is approxamately the following:
If our function doesn't have zeros inside the circle it must be constant(it follows from the maximum modulus principle applied firsly for $f$ and then for $\frac{1}{f}$):if we denote these zeros by $0,\alpha_1,\cdots,\alpha_k$ (0 has $m$ multiply), then we can divide our function by Blashke product with these zeros.Thus, ibtained function doesn't have any zeros inside the disk $|z|\leq 1$, and have identical modulus on the circle, therefore it must be constant.


inequalities
real analysis
real analysis unsolved

In geometric terms, the quantity $r=\frac{y_{0}-x_{0}}{\sqrt{2}}$ is the distance from the point $(x_{0},y_{0})$ to the line $y=x$. The inequality says that any point whose distance to $(x_{0},y_{0})$ is less that $r$ lies above the line $y=x$. Geometrically this is obvious. If you need an algebraic proof, then use the Cauchy-Schwarz as follows:
\[|1\cdot (x-x_{0})+(-1)\cdot(y-y_{0})|\le \sqrt{1^{2}+(-1)^{2}}\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<y_{0}-x_{0}\]
hence $y-x>y_{0}-x_{0}-|(x-x_{0})-(y-y_{0})|>0$


conics
ellipse
calculus
integration
trigonometry
geometry
real analysis

What is the length of the ellipse (deduced by integral calculus) ?

[quote="Daniel Stanica"]What is the length of the elypse ?[/quote]
If you meant "ellipse", I think this is a joke. Otherwise, define "elypse"...

It's easy to find an integral that gives the length: if the axes are $a$ and $b$, we can parametrize the ellipse as $x=a\cos t$, $y=b\sin t$. This leads to $\int_0^{2\pi}\sqrt{a^2\sin^2 t+b^2\cos^2 t}\,dt=\int_0^{2\pi}a\sqrt{1+\left(\frac{b^2}{a^2}-1\right)\cos^2 t}\,dt$.

Unfortunately, this is not an elementary integral unless $a=b$ and the ellipse is a circle. It is at least very easy to approximate; even the simplest quadrature methods are extremely accurate.

I know some quadrature methods, but I need an elementary solution. This problem is from romanian textbook (for secondary school).

[quote="Daniel Stanica"]I know some quadrature methods, but I need an elementary solution. [/quote]
Well, as jmerry said, there isn't one. Maybe there is a misprint in the textbook and the authors actually meant to ask about the area, not the length.


Who's taking 10B?
12B

How you guys prepping?
Want some hints or random blabbings from me?
LoL

Aim: DarkQuantumDy

i'm taking it

i thinking of preparing and learn as much as possible here =)

i'm taking it aannnnnnnnnnnnnnnnnnnnnnnnnnnnnnnndddddddddddddddddddddddddddddddddddddddddd my life's is on it i'm betting my soul lol. i am pissed off at the AMC-12 didn't know you weren't suppose to guess i am pissed off at how im wasting time so i plan to prepare for AMC by ibnstalling so good o,ld fashion discipline.

[b]Since I did not make the 100 requierement on the AMC I am going to go the life of abstience. No MSN MESSENGER, USELESS FORUMS ON AOPS, UESLESS BROWING, HANGING OUT WITH FRIENDS, MOVIE THEATER, BEER (lol), I AM GOING TO ALL THE WAY TO USAMO HOPE ME LUCK I AM FOLLOWING THE STEPS OF KING NEAL aka SYNTAX ERROR.[/b]

wish me luck fubu.

good luck to you too churchill. is everyone just doing old tests? it seems like the most logical way to prepare.

beans

the more i prepare the worse i do. thus, i will not do a single problem to practice for amc 10.

Hehe ... it's extracurricular.  If you want to tell colleges you've done it, you can.  It's not going to affect your GPA :)

[quote="Unanswered"]my first time i'm taking the AMC.....will this be put on your school record ?[/quote]

While it may not go on your school record, some colleges and universities ask for AMC scores on their applications and most good universities are interested in hearing about your score.


\[ \begin{array}{l}
 1\left\{ {\begin{array}{*{20}c}
   {2\sqrt {x^2  \plus{} y^2 }  \plus{} xy \equal{} 1}  \\
   {\frac{x}{y} \plus{} \frac{y}{x} \equal{} 1}  \\
\end{array}} \right. \\
2 \left\{ \begin{array}{l}
 x \plus{} xy \plus{} y \equal{} m \\
 xy(x \plus{} y) \equal{} 3m \minus{} 8 \\
 \end{array} \right. \\
 \end{array}
\]

We can find x, y of this sytem of equation $ \begin{cases}2 \sqrt {x^2 \plus{} y^2 } \plus{} xy \equal{} 1\\\frac{x}{y} \plus{} \frac{y}{x} \equal{} 1\end{cases}$
Because sytem of equation $ \Longleftrightarrow \begin{cases}2 \sqrt {x^2 \plus{} y^2 } \plus{} xy \equal{} 1\\x^2\plus{}y^2\equal{}xy \end{cases}$

$ \Longleftrightarrow \begin{cases}2 \sqrt {xy} \plus{} xy \equal{} 1\\x^2\plus{}y^2\equal{}xy \end{cases}$

$ \Longleftrightarrow \begin{cases} \sqrt {xy} \equal{} \sqrt{2}\minus{}1 \\x^2\plus{}y^2\equal{}xy \end{cases}$

$ \Longleftrightarrow \begin{cases} xy \equal{} 3\minus{}2\sqrt{2} \\x\plus{}y\equal{}\pm \sqrt{3}(\sqrt{2}\minus{}1)\end{cases}$

easy to find $ (x; y)$.

We can find $ m$ to symtem of equation has roots from $ 2$ by Let $ \begin {cases} S\equal{}x\plus{}y \\ P\equal{}xy \end{cases}$ with $ S^2\minus{}4P \ge 0$


I think the prime number is a number that only can be divided by one and itself.
But I'm wondering why $ 1$ isn't a prime number  :?:

A prime has two factors, $ 1$ and itself. $ 1$ only has one factor.

It can be divided by $ 1$ and by itself also  :wink:
What the one factor that you mean?

A prime number has, by definition, two [b]distinct[/b] positive integral factors.
$ 1$ is the only positive integer that divides $ 1$, so $ 1$ cannot be prime.
However, a composite number has, by definition, at least [b]two prime factors[/b]. The only number that divides $ 1$ is $ 1$, and $ 1$ is not prime, so $ 1$ is not composite.

So $ 1$ is neither prime nor composite.

mmm. So we can say it must be two different numbers. Since $ 1$ have only $ 1$ so its not a prime number.
Thanks a lot guys  :)


If the equation (x + y)^2 = 49 is graphed, how many points (x, y) will the graph
pass through where x and y both have positive integer values and x > y

A. three B. four C. five D. six E. seven

What you could do for this problem is to square root both sides because it simply works. Then you could see that there are 3 points:
(6,1)(5,2)(4,3) So the answer is A.

$ (x \plus{} y)^2 \equal{} 49$

$ x \plus{} y \equal{} 7$  (we exclude -7, since x and y are positive)

Easily enough, we can list the pairs:

$ (6,1), (5,2),$ and $ (4,3)$, so the answer is $ \boxed{A) \ 3}$.


dude ok I would like to say that

djiang is kevin hu

and Kevin Hu is me

you know why? he made djiang as a joke and I made Kevin Hu as a joke to get back at him. It's completely harmless, only a joke. I don't know why you're so fired up about this...

sorry if this wasn't allowed, we won't do it again.

[quote="not_trig"]sorry if this wasn't allowed, we won't do it again.[/quote]It's not allowed.  Don't do it again.

[quote="gauss1181"]Some people are using multiple accounts on the site.

dingzhou and (^_^) (already resolved)
Lyndon Ji and mathcrazed
cognos599 and ozzinator99
ra5249 and THE RA
Aryth and Kevin Hu[/quote]

I think THE RA is like bobby/some texas people
I'm certain Lyndon Ji is an imposter
and damien just clarified the Kevin Hu thing

i think someone used one account for the forum/classes and another for FTW only

ternary0210 is spamming the NC forum A LOT, and I suspect that it is an alternate account. Please, please ban it.

[quote="gauss1181"]Why are ppl using multiple accounts to fool their identities?[/quote]

So their original account doesn't get blamed?  :wink:

[quote="isabella2296"][quote="gauss1181"]Seriously. Why are ppl using multiple accounts to fool their identities?[/quote]

Probably because some people want to protect their original account like 7h3 said, and also if they want to preserve a good rating and still want to play. Both are unjustified reasons.[/quote]
then why don't you play as a guest?

[quote="myyellowducky82"]their other account should get blamed too.[/quote]

The other account does get blamed. Moderators and admins can run IP checks to see which accounts are related.

And this problem isn't restricted to FTW, you know.  :wink:

If you use one account on one computer and another on a different computer is that bad?

Your question is a bit ambiguious. You should only have one account. You shouldn't make a new account for every computer. (At least that's how I interpreted it).

EDIT: isabella beat me.

they can only track your ip adress if you use a different internet connection

How can admins tell the difference between siblings on the same IP address, and someone with multiple accounts?

I wonder whether this thread is becoming a discussion of "how to create multiple accounts and not get caught". :roll: The only sure way is to have a truly split personality and use one account for Dr. Jekyll and one for Mr. Hyde. If two posters post from the same IP address, attend the same fora, use the same style and language, chat/quarrel with the same people, etc., it doesn't take too much brainpower to start suspecting that it is the same person.

The last remark: despite all the policies, etc., nobody really cares about possible multiple accounts as long as the behavior is consistent with the netiquette rules and the posts are top quality. You can create 2 accounts, make a lot of meaningful and interesting posts from each of them, and nobody will ever bother to try to figure out whether it is the same person or not (though it beats me why on Earth you would need 2 accounts in this case). The cases that get prosecuted are those of bad spammers that would be banned anyway, just, maybe, a couple of weeks later. So, it isn't like somebody goes over fora and looks for similar posts (both admins and mods have much more interesting things to do than that). It rather works like this: a spammer is banned/warned; another account is created or suddenly becomes active with an identical IP and posting style; it gets deleted by admins on the ground of "multiple account violation" even if the amount of spam from the new account is not yet sufficient to warrant the ban; if the spammer is persistent in creating new aliases, the IP itself is blocked, and so on.

The moral of the story is simple: [color=red][size=134]Do not spam and you'll have no trouble[/size][/color] (in particular, many of the above questions will become moot)

[quote="fedja"]
The moral of the story is simple: [color=red][size=134]Do not spam and you'll have no trouble[/size][/color] (in particular, many of the above questions will become moot)[/quote]

Exactly.  And on that note, I'll lock this thread,


[u][b]Team: Total Score[/b][/u]
Philips Exeter A: 171
Lehigh Valley Fire: 169
New York City A: 163
North Carolina A: 162
Georgia A: 161
Texas Gold: 160
Chicago A: 159
TJ A: 157
Montgomery A: 156
Michigan Reals: 155

[u][b]Team: Score before Relays[/b][/u]
Philips Exeter A: 153
Lehigh Valley Fire: 153
New York City A: 149
North Carolina A: 146
Michigan Reals: 141
Montgomery A: 140
Georgia A: 137
TJ A: 137
Texas Gold: 132
Chicago A: 131

[quote="nat mc"]Dang Indiana Gold had 130 before the relays and finished with 140 which was good enough for 11th[/quote]

only half correct. yes, Indiana Gold had 140, but it wasn't good enough for 11th; they came in 18th
 	Alabama 	149
                Upstate NY A  148
                Eastern Mass A 145
                Florida ALpha  145
	AAST Mu A   143
	Southern California A 143
	Washington Gold  143
                Indiana Gold  140
(SFBA and Colorado Granite were next, at 137 -- )

Does anyone have any results for the B Division?

I think we had a power score of 27 and a team score of 16, but we had good individual scores that gave us the win, by one point.

Wow...if we (SFBA A team) hadn't screwed up on #1 on the Team Round, and hadn't bombed the Relay and Power rounds, we probably would've gotten something like 10th. (Assuming a I74/R22/T32/P28 or something)...in reality, we got I74/R14/T28/P21, I think.

[quote="mysmartmouth"]Does anyone have any results for the B Division?
[/quote]

top five teams in B division:
South Carolina Doom  116
 Massachusetts  115
 San Diego A  113
 North Carolina B 112
 Southern California B  112
then a gap of 18 points to 6th place:
 Vermont All Stars  94
...

[/quote]TMA mysmartmouth wrote:
Does anyone have any results for the B Division?

top five teams in B division:
South Carolina Doom 116
Massachusetts 115
San Diego A 113
North Carolina B 112
Southern California B 112
then a gap of 18 points to 6th place:
Vermont All Stars 94
...

wait.... at Penn state they announced that VT got fifth place.  Did they count So. CA and No. CA as 4th place and just gave 5th to VT?

Well, they told us we won by one point, so I think Mass was in the B Division, assuming those scores are correct.

Anyways, I didn't realize how close these scores were...every single point from every single person was needed...without any one person, we probably wouldn't have even placed.

dang 3-7 are really close  :read:

go Georgia A!  :surf:

your info on "before the relays" is incorrect--NYC A had 139 before the relays and got 24 on them to get to 163.

Wow, one point separated each place from 3 to 7.

Ok I guess when our coach said we were right behind Michigan he meant in our site.

I'm really happy for Lehigh Fire. Although my opinion could be slightly skewed...

Minnesota B was 8th and Kentucky Gold was 9th in division B.

I'm kinda bummed that our team (alabama) only got 20 on the team...we missed #1 and the easy base 9 problem. Oh well.

You're bummed? I have no sympathy for teams that can get 24 points on the relays. You were in 4th place at Las Vegas, 12 points behind Washington, before the relays.

Really? I figured our relay score was mediocre, at best. I guess we lucked out in terms of the other teams not doing that great, I guess..

Our relay teams were stacked... our scores were 0 0 2 4 8.

[quote="jmerry"]So- was your 8-point relay a team of three middle-schoolers, Ubemaya? You may have discovered the secret to relay success: Mathcounts training.

It's all about speed.[/quote]
actually it was Ubemaya on #1, krustyteklown on #2 and Adrian Sanborn on #3.

The average top-12 team got 20 points on relays. The average of all of the 117 teams got 6.6 points on relays. Here are the top 9 teams ranked just by relay score:

$\begin{array}{cccc}\text{Team}&\text{R1}&\text{R2}&\text{Total}\\ \text{Texas Gold}&18&10&28\\ \text{Chicago A}&16&12&28\\ \text{NYC A}&16&8&24\\ \text{Georgia A}&16&8&24\\ \text{Alabama}&18&6&24\\ \text{TJ A}&10&10&20\\ \text{Exeter A}&12&6&18\\ \text{AAST }\mu\text{ A}&10&8&18\\ \text{Iowa Power}&10&8&18\end{array}$

Our A team got 10 + 2 = 12. Washington Gold had a 6 point lead on us and we had a 6 point lead on you going into the relays; Washington got 2+4=6.

["You" here refers to Elemennop and Alabama.]

our relay team had two 7s and a 6, and we got a 2  :(

[quote="samath"][quote="Kent Merryfield"]Our A team got 10 + 2 = 12. Washington Gold had a 6 point lead on us and we had a 6 point lead on you going into the relays; Washington got 2+4=6.
[/quote]

woohoo what a high score >.<

wow. my relay team had 3 7's, and we got a relay score of 0 ...[/quote]

Um, I don't think the individual score has an effect on the relay score because the relay is about speed

I'm just saying, we had a strong relay team, I expected to do better >.<

Yeah, Sam, I expected you to do better too.   :P

But then again, the way I missed Individual #2...  :oops:

We now know two things about how to win ARML.

1. You must wear pink shirts. (Anyone want to guess how many pink shirts we will see next year?)
2. A Georgia coach must score your team. I had North Carolina last year and a different Georgia coach had Exeter this year. This really sucks for Georgia since we can't score our own team.

Tom

It's kinda sad how TX Au scored 28 on relays but TX Ag only scored 4.

Yay my relay for scoring our only relay points! (4 on R2)

[quote="jmerry"]The complete team results are now available [url=http://www.arml.com/results.htm]here[/url].[/quote]
Am I missing something :maybe: ?  That only gives results up to 2006.

Try refreshing the page - you've probably checked it within the last day or so, and the new posting is quite recent.

[quote="worthawholebean"][quote="xxazurewrathxx"]Wow yeah, Florida really sucked it up on relays. We only had maybe 2 of the teams get any points at all.

Anyway, we'll be better next year if we can get all of our top students to go. We only had 6 of our top 15 go this year. It's a problem because we can't get any funding  :([/quote]
Considering Ignite and I were the best on our team, and we are definitely not best in the state, we had very few of the really good people come.[/quote]

i resent that statement

i am the #1 english speaking person in my state

we cna disqualify phumpong watanaprakornkul, jeffrey rosen, andall those illegals/robots

wow Indiana Gold could have been top 7 or 8 (1 behind Chicago before relays) but we screwed up the relays so bad. Mine got 4 pts on the first one (4/10 for our team total)

omg i just saw results.

WOOOO MAN 4 NEW RECORDS THIS YEAR!!

-Lowest power score ever in over 30 years of competing for A team
-Lowest team score ever in over 30 years of competing for A team
-First ever last place
-First ever time someone from EO Smith has gotten a 0 on individual, gotten all his relay questions wrong, gotten all his team questions wrong, etc.

Man I wonder if we can finally go back to B division in next year. I mean I know we won B division by a huge margin in 2003 and did ok in 2004 but srsly, we're lowering the floor and forcing all those poor teams into division A.

PS. WAWB: I also resent you saying that you and I were the best on the team. That's an extremely arrogant statement to make. Maybe the reason we did so bad on the team/power this year was because of CERTAIN condescending attitudes toward other team members.

WOW

Chicago A [my team] got owned on the team. How did that happen???

Georgia A had a 2 point lead over Phillips Exeter Red before the power round was scored and we ended up 10 points behind them. I guess we know what we need to work on for next year.

We (Exeter) got pwned on relays, and we were very lucky, because we lost to pretty much everyone in the top ten except the one that mattered--Lehigh Valley.

Our score distribution for the five teams was 0 0 4 6 8.  I guess that means we grouped our teams correctly...

[quote="Kent Merryfield"]It may be more of a question of which division Massachusetts 1E was in. Can someone explain the relationship between the 2006 teams of Mass A and Mass B and this year's teams of Eastern Massachusetts A, Massachesetts 1E, and Massachusetts 2B?[/quote]

Hmm, I believe we've always went A, E, B, with E and B in the B division, I never knew about adding the 1 and 2 infront of the E and B. But it was intended that E>B this year, not sure about what exactly happened this year (but I believe the same was intended last year).

I just updated the [url=http://www.artofproblemsolving.com/Wiki/index.php/ARML_historical_results]results page of the wiki[/url]. Someone should look that over to make sure I got it right.