inequalities
inequalities proposed
n-variable inequality

Let $0<a\leq 1$ be a real number and let $a\leq a_{i}\leq\frac{1}{a_{i}}\forall i=\overline{1,1996}$ are real numbers. Prove that for any nonnegative real numbers $k_{i}(i=1,2,...,1996)$ such that $\sum_{i=1}^{1996}k_{i}=1$ we have $(\sum_{i=1}^{1996}k_{i}a_{i})(\sum_{i=1}^{1996}\frac{k_{i}}{a_{i}})\leq (a+\frac{1}{a})^{2}$.

Under the given condition, for all $n\in \mathbb{N}$, stronger inequality holds:

$\left(\sum_{i=1}^{n} k_ia_i \right)\left(\sum_{i=1}^{n} \frac{k_i}{a_i} \right) \le \frac{1}{4}\left(a+\frac{1}{a}\right)^2$

[i]Proof:[/i]
From the condition, we have: $(a-a_i)\left(a-\frac{1}{a_i} \right)\ge 0 \Leftrightarrow a_i+\frac{1}{a_i}\le a+\frac{1}{a} $

Now, from am-gm, it follows:

$\left(\sum_{i=1}^{n} k_ia_i \right)\left(\sum_{i=1}^{n} \frac{k_i}{a_i} \right) \le \frac{1}{4}\left[ \sum_{i=1}^{n} k_i\left(a_i+\frac{1}{a_i}\right) \right]^2 \le \frac{1}{4}\left[ \sum_{i=1}^{n} k_i\left(a+\frac{1}{a}\right) \right]^2 = \frac{1}{4}\left(a+\frac{1}{a}\right)^2$

Another way to get $ a_i+\frac{1}{a_i}\le a+\frac{1}{a} $ is to notice we have $a\leq a_i\leq 1$ and that $f(x) = x+\frac {1} {x}$ is decreasing on $(0,1]$.

Also, for good measure, we have by Cauchy-Schwarz the reverse inequality
$\left(\sum_{i=1}^{n} k_ia_i \right)\left(\sum_{i=1}^{n} \frac{k_i}{a_i} \right) \geq \left (\sum_{i=1}^{n} k_i\right )^2 = 1$.


If $ f(x),g(x)$ are continuous when $ x\in [a,b]$

Prove: $ \max{\{f(x),g(x)\}}$ is continuous when $ x\in [a,b]$ :oops:

Start by taking three cases: choose $ x_0 \in [a, b]$.  If $ f(x_0) > g(x_0)$ then in a neighborhood of $ x_0$, ....  If $ g(x_0) > f(x_0)$, exactly the same thing happens with $ f$ and $ g$ switched.  And if $ f(x_0) \equal{} g(x_0)$ then for every $ \varepsilon > 0$ there exist ....  (Fill in the blanks ;).)

max$ (f,g)\equal{}\frac{1}{2}(|f\plus{}g|\plus{}|f\minus{}g|)$ which it is continious fuction as both $ |f\plus{}g|$ and $ |f\minus{}g|$ are continious .


In trading firms employing both quants and traders, how do salaries (plus bonuses) compare?

You should also ask yourself if you really are ready to sacrify your private life for those jobs ... I know some traders and quantitative analysts, I can tell you they work their ass, like for some of them a good 80 hours a week ... much of the burden you need to take after your work hours, meaning your private life is significantly restricted.
That is just to say guys, don't just think about the bucks.

The hours may be like that in early investment bank jobs, but they aren't like that at most hedge funds I know of.  I worked a fair amount at Shaw, but I work way more now on AoPS.  I would say the average work week at places like Shaw or Jane Street is closer to 55 hours, maybe lower (unless things have changed dramatically).

As for pay differential between quants and traders, that varies widely by company, and by strategy within that company.  Whichever group is most important in generating profit is generally the one that gets paid best.  So, a black-box quantitative strategy is going to pay quants more, and one in which information and relationships are most important is going to pay traders more.


suppose that the set A is denumerable, the set B is denumerable, and the sets A and B are disjoint; show that the union A u B is denumerable.

Can someone help with that.

You need to find a bijection between $A \cup B$ and $\mathbb{N}$, for instance. Why not send the elements of $A$ on the even integers, and the elements of $B$ on the odd integers ? ($0$ is supposed even, according to Bourbaki's conventions).

By definition, you have a bijection between $A$ and the $\mathbb{N}$. But you also know how to construct a bijection from $\mathbb{N}$ onto the even natural numbers ($n \mapsto 2n$). Now consider the composition of the two ... and do the same for $B$ with the odd integers.

i think it is supposed to be done without bijection.

Any other way to do it?

You cannot do it "without bijection", since countability is defined in terms of bijections (equipotence). Unless you have a different unusual definition, in which case you need to tell us ...

Ok well I still cannot solve it, can you help?


i have know an American maths magazine.it has a website http://www.maa.org/pubs/monthly.html
but i don't how to download them.there's no links on it.

Scientific American is also a good journal.there're maths problems for every month.i don't how to download them,either.

maybe there're some more good journals.you can tell me.thanks

Some journals are *gasp* only published in print! And you have to pay for them!!

For all of the old issues of several math magazines, try [url=http://www.jstor.org/]JSTOR[/url].

Here i prepared a list of journals which are free of cost...zome of them are in other languages...i also mentioned it

[url]https://drive.google.com/file/d/17dpeIXSFTZrfrkCrLOpxXKt4Mkuw3Ujc/view?usp=drivesdk[/url]


$a, b, c, d > 0$
Prove that
$\sum_{cyc}a^{7}b^{3}c \ge \sum_{cyc}a^{5}b^{4}cd$

Remark: the purpose of this question is to investigate the validity of an inequality similar to Muirhead except for symmetric sums. See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103761.
If you have a general approach for this sort of problem, discuss it in the other thread. This thread is for this particular inequality, the solution of which was not obvious to me... (in other words, AM-GM didn't work ;) )


1) Show that any positive integer can be represented as sums and differences of powers of 3.

2) a. Does the number 0.1234567891011121314..., which is obtained by writing successively all the integers, represent a rational number?
b. Does the number 0.011010100010100..., where a_n = 1 if n is prime, 0 otherwise, represent a rational number?

hmm, I definately recognize some of the problems you've been posting.  Larson's [i]Problem Solving through Problems[/i] perhaps?

well, a while ago i posed question 2 as a challenge problem, you can find it here

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11473

For 1) :
 Let $n > 0$ be given. Then $U_n = 1 + 3 + \cdots + 3^n = \frac {3^{n+1} - 1} 2$.
Thus, each integer of the form $x = \sum_{i=0}^n a_i3^i$, where $a_i \in \{-1,0,1 \}$ belongs to the set $I_n = [-U_n, U_n ]$.
Note that $|I_n| = 3^{n+1}$, and that since we have $3$ choices for each of the $a_i$'s there are $3^{n+1}$ number of the form above. Thus, if we prove that these $3^{n+1}$ choices leads to pairwise distinct numbers, it would follow that each integer which belongs to $I_n$ can be written with desired form. Now, each integer is in $I_n$ for some sufficientely large $n$, so that the desired result would follow.

But, if $ \sum_{i=0}^n a_i3^i =  \sum_{i=0}^n b_i3^i$, where $a_i,b_i \in \{-1,0,1 \}$ for all $i$.
If $a_i = b_i$ for all $i$ we are done. Otherwise, let $j$ b the minimal subscript for which $a_j \neq b_j$. Then $(a_j - b_j)3^j = \sum_{i > j} (b_i - a_i)3^i$ (if necessary, note that a sum indexed over the emptyset is equal to $0$). It follows that $(a_j - b_j)3^j = 0$ mod $[3^{j+1}]$, so that $a_j - b_j$ is divisible by $3$. But $-2 \leq a_j - b_j \leq 2$ thus $a_j - b_j = 0$ which contradicts that $a_j \neq b_j$, and we are done.

Pierre.

For 3) : Suppose, for a contradiction, that this number is rational. Thus, its decimal representation becomes periodic. Let $T$ be the length of one of its period.
Since there are infinitely many primes and infinitely many composite numbers, the periodical pattern contains at least one $0$ and one $1$.
But, it is well-known that we may find infinitely many sequences of $T+1$ consecutive  positive integers which are all composite. Thus, at least one of these sequence will occur after the decimal expansion will be become periodic. In that case, the periodical pattern contains only $0$. A contradiction.

Thus the given number is irrational.

Pierre.

I think you misunderstood 2 a, because it is not rational (it is not difficult to reason that it cant be periodic)

anyway, for 1 I remember seeing a very nice proof using Generating Functions, but I dont remember how to do it (it was something like multiplying (1+x^3+x^-3)*(1+x^9+x^-9)... with some trick) anyone know how to do it in this fashion?

Interesting. Champernowne's number (0.12345678910111213...) is not just irrational: it was proven to be transcendental by Mahler in 1961. However it seems to be equivalent to the said rational number if it is rewritten as

9
:Sigma: (n*10^-n) +
n=1

:inf:
:Sigma: (n+9)(10^-(2n+9))
n=1

why is this?

I suppose that you could prove that the 3rd is irrational using the prime number theorem.  The digits are getting further apart, so it can't be periodic.

[quote]Show that any positive integer can be represented as sums and differences of powers of 3.
[/quote]
Sorry if it was already posted, (did not look at all the responses) but [b]just write the number in the base of 3   (using digits  0,1,-1) [/b].  (BTW the representation is unique )

[hide="BTW Interesting story"] In old days (In india) grocers   only used to have weights of  $1,3,9,27$ "sers"  and using these they were able to weigh any weight between 1 and 40 ( 40 sers used to make 1  "mon")
For example if they wanted to weight  $11$ they will put $9,3$ on one side and $1$ on the otherside of the balance with the goods to be weighed....
:)

(I also remember  when "new" weight's  were introduced ( "ser" became a "kilogram" - (they were approximatly equal anyway)  "mon" went away and 100 "kilogram" was  called "quintal" all they would have needed an additional  $81$ piece to reach all they way up to  $121$ ...:)[/hide]

ZennyK wrote:2.a) The decimal represents the rational number
[hide] 60499999499/490050000000 [/hide]
Hint: the decimal can be expressed as a sum of two different sums.

that's wrong isn't it? that number goes 96 97 99 00 01 02 03

so it skips 98 and after that it doesn't write the 1 for the hundreds.

The number in 2a is not repeating and therefor not a rational right?

[quote="diophantus"]What's to prove.[?][/quote]

30=1

n(30)=n

[quote="diophantus"][quote="firefox"]1) Show that any positive integer can be represented as sums and differences of powers of 3.
[/quote]
What's to prove.  Due to the existence of base three, any integer can be written as $a_0*3^0+a_13^1...a_n83^n$ where $0 \ge a_i \le 2$ If a_i=0 simply don't write that term in the sum. If $a_i=1$ then that term is part of your sum, and if $a_i=2$, set $a_i$ to 1 and increase $a_{i+1}$. It's a matter of definition.[/quote]

You're only allowed to use each power of three once.

yes but for any n, 2 times 3^n is the same as 3^(n+1)-3^n.  That's why the existence of base 3 proves your statement.


trigonometry
algebra
polynomial
induction
calculus
calculus computations

One(trivial) :  $P_n(x)$ is a polynomial of degree $n$ and the coefficient of $x^n$ is $1$.

Two(easy) : prove by induction that $P_n(2cos(t))=\frac{sin((n+1)t)}{sin(t)}$

Three(conclusion) :  $P_n(x)$ has $n$ distinct roots $x_k=2\cos \frac{k\pi}{n+1},k=1...n\ $,  so  $\ P_n(x)=\prod_{k=1}^{n}(x-2\cos \frac{k\pi}{n+1}) $

 :cool:


calculus
derivative
analytic geometry
advanced fields
advanced fields unsolved

I am so confused about these Einstein Summation Notations.

For example, how do you show that

$ \mathcal{L}_X g_{ab} = \nabla_{(a}X_{b)}$
where $ \nabla_a$ is the Levi-Civita connection preserving the metric $ g_{ab}$, and $ \mathcal{L}_X$ is the Lie derivative.

This has nothing to do with the summation convention. The convention says that in expressions  like $ a_{k}b_{k}$, with one index repeated, there is an implicit summation with respect to that index: $ \sum_{k}a_{k}b_{k}$.

While I agree with neworder, balto's question *could* mean something like, "I understand how to prove this statement, but I don't know how to prove it using Einstein notation manipulations (i.e. how to prove it using coordinates)."

Anyway, the desired formula follows almost immediately from the "Einstein notation" formula for the Lie derivative of a tensor (see Wikipedia, for example, to find this formula).  What is less obvious perhaps is how the "Einstein notation" formula for the Lie derivative follows from the usual mathematical definition of the Lie derivative.

Actually, I think I don't know how to represent the condition: "$ \nabla_a$ preserves the metric $ g_{ab}$" in coordinates.

and yes, I know an Einstein notation for $ \mathcal{L}_X$, so the above formula is supposed to be expand in terms of these notation, but I have no idea how to simplify it.

Anyway, can anyone write a complete prove (using Einstein notation one, not the coordinate free one)?

In coordinates, you'd just write $ \nabla_c g_{ab}\equal{}0$.  This follows from using the coordinate formula for $ \nabla$ of a tensor in terms of Christoffel symbols, plus the formula for Christoffel symbols.

To prove the formula you want:

$ L_x g_{ab}\equal{}X^c \nabla_c g_{ab} \plus{}(\nabla_a X^c)g_{cb}\plus{}(\nabla_b X^c)g_{ac}\equal{}\nabla_a (X^c g_{cb})\plus{}\nabla_b (X^c g_{ac})\equal{}\nabla_a X_b\plus{}\nabla_b X_a$, where we used the fact $ \nabla_i g_{jk}\equal{}0$ three times in the second equality.


Simplify $\sqrt{n+\sqrt{2n-1}}+\sqrt{n-\sqrt{2n-1}}$, so that there aren't any "nested radicals".

Generalization: Find any conditions that allow us to express $\sqrt[n]{a+\sqrt[m]{b}}+\sqrt[n]{a-\sqrt[m]{b}}$, so that there aren't any "nested radicals".

Prove your results.

[hide=" easy part"]Just square it. Whatever you get after simplifying, put it under a square root.
$\sqrt{4n-2}$, for $n\ge 1$.
$\sqrt{2}$, for $\frac12\le n<1$.[/hide]


Find the point $ P$ in the interior of an acute-angled triangle $ ABC$ for which the value $ BL^2 \plus{} CM^2 \plus{} AN^2$ is a minimum, where $ L$, $ M$, $ N$ are the feet of the perpendiculars from the point $ P$ onto the sides $ BC$, $ CA$, $ AB$, respectively.

$BL^2+CM^2+AN^2=CL^2+AM^2+BN^2\Rightarrow BL\cdot CL+CM\cdot AM+AN\cdot BN=\frac{a^2+b^2+c^2-(BL^2+CM^2+AN^2+CL^2+AM^2+BN^2)}2$ must be maximized. Lucky for us there's a point where each of the products $BL\cdot CL,CM\cdot AM,AN\cdot BN$ is maximized, namely $O$ (each product is the product of two segments with constant sum, so each of them is maximized when the two segments are equal).


vector
function
algebra
polynomial
linear algebra
linear algebra unsolved

Let $ V$ be an infinite dimensional vector space over a field $ F$ with basis $ \left\{v_{i}\right\}$ and define $ v_{i}^{*}: V\rightarrow F$ by

\[ v_{i}^{*}\left(v_{j}\right)\equal{}\begin{cases}0 & i\neq j, \\ 1 & i\equal{}j.\end{cases}\]
Show that $ \left\{v_{i}^{*}\right\}$ does not span $ \textrm{Hom}\left(V,F\right)$.

I'm assuming that we need to find a function that cannot be written as $ \sum c_{i}v_{i}^{*}$, where only a finite number of $ c_{i}$ terms are nonzero.  Can anyone give me a push in the right direction?  Thanks!

Consider $ L(v)\equal{}L\left(\sum_jx_jv_j\right)\equal{}\sum_jx_j.$

Or, to phrase the same example another way: let $ V$ be the vector space of all polynomials. Consider a polynomial $ f$ as a function on $ F$ and define $ L(f)\equal{}f(1).$


Prove that $ sinA\plus{}sinB\plus{} sinC <\equal{}\sqrt{3}(cosA\plus{}cosB\plus{}cosC)$

[quote="nguoigo"]Prove that $ sinA \plus{} sinB \plus{} sinC < \equal{} \sqrt {3}(cosA \plus{} cosB \plus{} cosC)$[/quote]

Try $ A \equal{} B\rightarrow\frac {\pi}{2}$ and $ C\rightarrow0^ \plus{} .$ :wink:


Hi, new TeX user here.

Is there any way to elongate the vertical bar in the following equation?

x(t)|_{t=0} = 0

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/7be7fd8eaa2b57075c08c35c947495b8.gif[/img]

Thanks.

The usual way is to use \left & \right. Since you only want one | you use \left| and \right. with the . indicating no symbol, as in x(t)\left|_{t=0}\right. = 0 which gives $x(t)\left|_{t=0}\right. = 0$
If you want it still bigger you can use \big as in x(t)\big|_{t=0} = 0 which gives $x(t)\big|_{t=0}= 0$

Or a little more big x(t)\Big|_{t=0}=0 which give: $x(t)\Big|_{t=0}=0$.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a $ PID$, let $ B$ be a torsion $ R$-module and let $ p$ be a prime in $ R$.
Prove that if $ pb\equal{}0$ for some nonzero $ b \in B$, then $ Ann(B) \subset (p)$.

$ Ann(b)$ is an ideal of $ R$ which contains $ p$. we have to show that actually it equals $ (p)$. this follows from $ b \neq 0 \Rightarrow Ann(b) \neq R$ and that $ (p)$ is a maximal ideal.


Hi!TRAN QUOC HOAN o day nhieu dan minh lam

[quote="Kapi"][b]K09[/b] and[b]N.T.TUAN[/b] do you do your problem very well? :)  (who is [b]Tran Quoc Hoan[/b] ? :D )[/quote]

Thank you!

[quote="TyphoonDC"]Em cũng vào ML dc lâu gòi,nhưng ít vàu đây :D[/quote]

Hinh dai dien cua chu ac nhi? :D  :P  :D

Chào các bạn! Rất vui gặp các bạn ở đây.

Em cung chao thay a.Co phai thay rat than voi anh HA DUY HUNG khong.

[quote="math_friends"]yea.!! them mot nguoi` viet nam nua~!! yea. this is Toan Nguyen who loves to face with mathematic problems.. and let me tell you,this is a real good site. Nguoi Vietnam minh` smart qua' ha. :D [/quote]
em a`, nguoi` vi.t nam minh` ta^m` thu*o*ng` tho^i

smart qua' ?!?  :rotfl:

Rat vui duoc gap cac ban  av cac thay o tren 1 dien dan Tieng Anh nay!! :D Minh la 1 thanh vien moi.
  Han hanh duoc gap thay Nam Dung ,em da duoc doc 1 cuon sach cua thay.do la cuon Boi Duong Doi Tuyen VN 2003 di thi Quoc Te.Cac bai toan va cach giai ma thay suu tam rat hay.Xin cam on thay. :lol: Thay con viet quyen sach nao khac ko a??

Chao cac ban mot lan nua,

1. Dung la toi co quen than voi thay Ha Duy Hung. Toi rat thich lam viec voi cac thay giao tre, vi ho nang dong va luon huong ve phia truoc.

2. Rat thu vi la cuon tai lieu boi duong DT 2003 cua toi duoc cac ban don nhan. Do se la mot dong luc lon cho toi. Toi co y dinh viet 1 cuon sach voi su dau tu nghiem tuc hon, nhung day van la mot mon no cua toi voi cac ban hoc sinh. Rat tiec la thoi gian luon la thu ma toi thieu nhat.

3. Cac ban can trao doi gi, can tai lieu, cu email cho toi nhe.

Namdung

Troi a the thi con gi bang,em cam on thay lam. da may lan e m dinh gui thu cho thay nhung cu ngai mai.Mot lan nua em cam on thay

Co phai Ko9 muon noi den Ha Duy Hung ,cuu hoc sinh Tran phu,Hai Phong,tung dat giai 3 bao THTT?? :roll:

Anh Ha Duy Hung dang day tai khoi chuyen toan DHSP Ha Noi
Nhan tien hoi cac bac luon, khong biet ket qua thi IMO nam nay the nao roi
okie!?

[quote="nguyenquockhanh"]khong biet ket qua thi IMO nam nay the nao roi
okie!?[/quote]
Tớ cũng mong biết kết quả của các anh "[i]đem chuông đi đánh tai người[/i]" như thế nào rồi.
Mong trúng tủ thật nhiều nhiều cho thiên hạ điếc tai ...

To de nghi ban Keira xoa bai post cua ban badgirl di (neu khong thi nho Valentin ). Neu day la noi dua thi minh cho rang badgirl, ban da dua thai qua va khong dung luc dung cho roi, con neu la nghiem tuc thi xin nhac ban mot lan nua rang: hay can trong voi loi noi ban, dung phat bieu thieu suy nghi nua. :mad:
A xoa xong bai post cua badgirl thi xoa luon ca bai cua nntrkien di nha, day toan la nhung bai ko dang co tren dien dan.

Peace out nao.

Ket qua day nay:

VIETNAM
Score P1 P2 P3 P4 P5 P6 Final Medal
VNM1 5  7  0  7  7  4  30  Silver
VNM2 0  1  7  2  7  2  19  Bronze
VNM3 0  7  0  1  7  0  15  Bronze
VNM4 1  1  6  1  7  0  16  Bronze
VNM5 7  7  6  2  7  2  31  Silver
VNM6 7  7  7  2  7  2  32  Silver
TOTAL 143

3 bac, 3 dong. Thap hon han moi nam( nam ngoai hinh nhu 4 vang phai khong nhi?). Neu xet ve dong doi thi kem hon ca Singapore.  :( ( dang ngoi nghi, nam nay lo thang ban Sin di thi Olympic Toan quoc te no hoi sao ket qua VN kem nhu vay, chang biet tra loi the nao nua) Ma hinh nhu Brazil cung co huy chuong vang, China thi co toi 4 ban perfect score. Nhu vay bao de nam nay kho cung khong phai( hinh nhu co phan de hon nua, vi cut-off point cao hon nam ngoai), vay thi tai sao ket qua cua Viet Nam lai kem nhu vay?( tren thuc te, neu nhin vao ket qua chon doi tuyen vong 2 cua ta thi co the thay rang, diem cua cac thanh vien doi tuyen thi vong 2 cao hon han moi nam). Chang le do thieu thoi gian tap trung? Hay do du luan nam nay khong tot, dan den viec doi tuyen mat tinh than khong tap trung thi cu duoc? Hay vi nam nay chat luong doi tuyen khong duoc nhu moi nam? Xin moi nguoi cho biet y kien.

hic hic sao topic nay lai lac sang van de nay
Thoi toi cha binh luan gi vi chungta chua hieu ro nguyen nhan ,them nua ta khong phai la nguoi trong cuoc.Cac ban hay de sau di

Vang, minh dat cau hoi cung chi mong co ban nao trong cuoc( co mot so ban co nick o day, phai khong nhi?) co the giai dap thac mac, hoac co ban nao quen biet voi "nguoi trong cuoc" co the tra loi cung duoc vay. Cung nhu hoi VN that bai o Tiger Cup, minh thay moi nguoi cung rat tich cuc mo xe, phan tich nguyen nhan giam sut cua doi tuyen co ma. Roi khi doan the thao khong danh duoc huy chuong o Olympics 2004, cung da co khong it bai phan tich nguyen nhan that bai. Doan Olympic Toan Quoc te du gi cung co vai tro dai dien cho VN, thanh bai cua doan cung chinh la the dien quoc gia. Mac du chung ta van co huy chuong, nhung so voi binh dien chung thi co mot thuc te khong the phu nhan, do la thanh tich cua ta kem han. Vay thi cung rat nen phan tih, mo xe chuyen nay lam chu? Chi la minh khong biet ro lam nen khong dam phat bieu y kien gi nhieu, hy vong moi nguoi se giup minh giai dap thac mac nay nhe.  :)

thầy ơi!!!nếu em không lầm thì em sắp được học thầy roài!!em nghe đồn thầy hiền và đẹp trai lém,phải không ạ??? :blush:

học toán mà cần hỏi thầy đẹp trai nữa sao? thầy analysis cũng hiền và đẹp trai lắm nè ... hay là học thầy đi  :rotfl:

Dạ thầy analys ui!!!em thấy dân toán thì ai cũng đẹp trai hết mà thầy!thầy dạy trường nào vậy ạ??? :oops:  :blush: !!!

sao em thấy dân toán ai cũng đẹp trai sao??? nghe được câu này thầy thiệt là vui quá ... lòng đầy tự tin  :rotfl: nhưng thầy thấy dân toán phái nữ thì không được đẹp cho lắm... !?
thầy đang dạy part-time ở usa   :)

Thầy ở xa wá,làm sao mà em học được bây giờ hả thầy.em thấy con gái học toán không đẹp là đa số,nhưng mà ai đẹp thì đẹp lắm thầy ui,hix hix,lớp của em chỉ vẻn vẹn có 8/41 nữ,nhưng mà..... :(  :(

nhưng mà..... toàn là xấu không chứ gì... và trong đó có em đúng không :rotfl:
ờ em không học thầy được thì coi như may mắn  :)

hahaha!!!Có 1 điều thầy nhầm to oài thầy ui!!!em là con trai kia mà,cho nên mới khen những người học toán đẹp trai chớ thầy!!! :rotfl:

trời!... vậy mà thầy tưởng em là con gái khen dân toán phái nam. em là con trai thì khen cũng huề vốn thôi. làm thầy vui cười quá trớn
nhưng thôi, dù sao thì thầy cũng đã vui, vui trong hiểu lầm cũng là vui
thầy không hiểu em là con trai sao hỏi thầy ND đẹp trai làm gì ?!!

Dạ lớp em có 3 thầy dạy toán:thầy Thanh Dũng,thầy Nam Dũng và thầy Khánh Trình,mà em đã được học 2 thầy roài,chỉ còn có thầy Nam Dũng nữa thôi ạ,thầy của em mà,cho nên em thấy thía thì phải khen chứ ạ!!! :blush: mà sự thật là đa số dân toán đều đẹp trai hết!!!thầy cũng thấy như thế mà,phải không ạ?? :roll:

lạc quan mà nói thì ... thì đa số dân toán phái nam đều đẹp trai, trong đó có thầy  :rotfl: haha
thầy đồng ý với em về điểm này... thầy nghe nói chương trình giáo dục bên vn còn nhiều "bùa phép" lắm. có phải là em tính khen để lấy lòng? hay là em khen hối lộ?  :wink:

dạ đâu có đâu,tại em có thừa tự tin là mình đẹp trai,cho nên mới lôi cả dân toán vào í mà,mà trò đẹp trai thì tất nhiên thầy còn đẹp trai hơn ấy chứ ạ!!! :lol:

ko biet email cua thay Nam Dung la gi nhi?

Bác Nam Dũng mất ngủ rồi nhá! :P Bạn nào muốn biết e-mail của bác ấy thì nhấp chụôt vào nick namdung là xong! :D

[color=purple]Anh analysys co phai? nguoi` VN ko? Tai sao anh lai deo cai' co` Nguy?
   Da vay o? cai' chu~ ki' lai co' them cai' dong` chu~ ki' " Vong quoc"
   Anh lam` vay thi` dung` nhan minh` la` nguoi`VN ! [/color]

Thôi, chú không phải nói thêm chủ đề này nữa, anh em lại hục hặc nhau :D Câu hỏi na ná kiểu này đã xuất hiện nhiều trong box này rồi, và cũng đã được analysis trả lời.

[quote="evarist"][color=purple] Chung` nao` em chua dc giai thich thi` em kung~ van~ cu' hoi? anh Tuan' a Em ko hiu? sao vay ma` cung~ jam' vao` box nay` nhan minh` la` nguoi` VN.Em ma` gap ong anh nay` ngoai` doi` co' le~ se~ bup nhau mat' (neu' anh ta tuyen truyen` tu tuong? fan? quoc' the' nay`) !
          Ong anh cho y' kien' ti' chu' analysys ? [/color][/quote]
Chú vào đây mà đọc này. Có câu trả lời trong đó đấy! :P
http://www.mathlinks.ro/Forum/viewtopic.php?t=50552

[quote="evarist"][color=purple] Chung` nao` em chua dc giai thich thi` em kung~ van~ cu' hoi? anh Tuan' a Em ko hiu? sao vay ma` cung~ jam' vao` box nay` nhan minh` la` nguoi` VN.Em ma` gap ong anh nay` ngoai` doi` co' le~ se~ bup nhau mat' (neu' anh ta tuyen truyen` tu tuong? fan? quoc' the' nay`) !
          Ong anh cho y' kien' ti' chu' analysys ? [/color][/quote]
lâu rồi tôi không có vào đây,

trả lời câu thứ nhất: tôi dùng lá cờ vàng là vì nó biểu tượng cho một nước vn tự do dân chủ

trả lời câu thứ hai: chữ ký của tôi dùng để cảm thán cho thân phận đất nước cũng như còn người vn

có lẽ evarist khó chấp nhận được những gì tôi nói ... nhưng không vì thế mà những điều tôi nói là sai

nếu còn gì thắc mắc thì cứ hỏi tiếp

Hình như trong nội quy của các diễn đàn đều có điều là không thảo luận về các vấn đề chính trị nhỉ?

Đúng! Các mod nên đóng cái này lại được rồi.

Em nghi ca`n del het' cac' bai` cua lao analysys de canh cao'

Đã lâu lắm rồi analysis không lên forum này, có del bài cũng chẳng ảnh hưởng tới ai.

[quote="analysis"]ha mấy em này định chơi cái trò bịt miệng; học sinh mà đạo đức suy đồi như vậy  :([/quote]
Chúng nó trẻ con mà, đừng bận tâm :) . À này, không post những cái gì liên quan đến chính trị đâu đấy, nếu analysis post liên quan đến chính trị thì Mods không tha đâu :P . Có cái topic được đánh dấu trong box này, vào đó giới thiệu tí đê , nhể? :)

Em nào dám chơi trò bịt miệng. Chẳng qua lâu quá không thấy ông anh lên, mà nhiều người đòi close cái topic nên nói thế thôi.
See my signature


This is a pretty fun game. I don't know if it's already been on here or not, but what the heck.

Simply post a completely hypothetical statement that can be replied with a "True or False" by the next poster, who will then post a different statement for the next poster and so on.

Basically Poster A says "Next poster likes cake." Poster B will answer "True. Next poster likes dogs," which will be answered by Poster C.

Try to make the statements intriguing and not boring like my example.

I will start:

The next poster likes to watch sports.

...
theres a thread similar to this one in the forum


Does there exist an infinite collection of sets such that the intersection of every two distinct sets in the collection is nonempty, but the intersection of every three distinct sets in the collection is empty?

[hide="Solution"] Yes.  Let $ a_1, a_2, ...$ be an enumeration of the elements in some universal set and let $ S_1, S_2, ...$ be a collection of initial empty sets.  Let $ f : \mathbb{N} \to \mathbb{N}^2$ be a bijection.  For every positive integer $ k$, place the element $ a_{2k \minus{} 1}$ in set $ S_k$ and the element $ a_{2k}$ in sets $ S_{f(k)_1}, S_{f(k)_2}$.  [/hide]


Could you write it more clearly? What confused me is that there are 5 "(" and only 4 ")" :D

[quote="Moubinool"]Prove that  the infinite product

(1+1/p(1).sin(p(1).pi/2))(1+1/p(2).sin(p(2).pi/2))...(1+1/p(n).sin(p(n).pi/2)) ...... = 2/ \pi
where p(n) is n-th prime number[/quote]

Let A=(1+1/p(1).sin(p(1).pi/2))(1+1/p(2).sin(p(2).pi/2))...(1+1/p(n).sin(p(n).pi/2)) ...
It is easy to see that A=\pi p-prime,odd(1+(-1)(p-1)/2/p).

Let B=\pi p-prime,odd(1-(-1)(p-1)/2/p).

It is easy to see that 1/AB=\sum n-odd1/n2=\pi2/8. [b][1][/b]
1/B=\sum n-odd(-1)(n-1)/2/n=\pi/4. [b][2][/b]

[1] is well known and [2] is known as Gregory's formula.
From [1] and [2] we get A=2/\pi.


I qualified for the MAML level 2 with a score of 116. Do you have any advice for preparing? I have done one practice test(05-06), but really didn't do well. Do they give partial credit?

Well, if you were going to prepare, I'm not sure the night before is the best time...Yes they do give partial credit, but only if you actually make some progress. I'm not sure exactly how strict they are. Just budget your time well and if you spend a while on something and don't get anywhere, you should probably move on. (This is mainly just generic test-taking stuff though--math prep is useless at this point)

Good luck, and (if you're going to the Canton site) see you tomorrow.

Deerfield. It'll be fun(I did some prep in the past few weeks, but only discovered this forum today). The one thing I have noticed is induction seems to be really useful for these proof's, so I might practice some induction-based proof's.

proof's?

Don't make stupid mistakes. For example, don't copy down recursions incorrectly, then proceed to prove the incorrect recursions. That is all.


What is the volume of a spherical shell with an outer diameter of 8 cm and an inner diameter of 6 cm?

[quote="Jhonny Ruiz"][hide="Formula"]$V=\frac{4\pi}3 (R^{3}-r^{3})$[/hide][/quote]

Correct, but first, we need to find the radii.

[hide="In terms of diameter,"]$V = \frac{\pi}{6}(D^{2}-d^{2})$[/hide]
[hide="Solution"]Plug in my formula to get:

$V = \frac{\pi}{6}(64-36)$

$V = \frac{14\pi}{3}$[/hide]


Let $ k$ be a positive integer. Find all polynomials with real coefficients which satisfy the equation

$ P(P(x)) \equal{} [P(x)]^k$

[hide]Let $ \deg(P(x))\equal{}n$.  Since we must have $ \deg(P(P(x)))\equal{}\deg([P(x)]^k)$, we have $ n^2\equal{}kn$.  Thus, $ n\equal{}0$ or $ k\equal{}n$.

If $ n\equal{}0$, then let $ P(x)\equal{}c$ where $ c$ is real.  Then the equation gives $ c\equal{}c^k$ or $ c\equal{}0,1,\minus{}1$.

If $ k\equal{}n$ then we have $ P(P(x))\equal{}[P(x)]^k$.  Both sides of the equations are polynomial in $ P(x)$.  Since the two are equal, corresponding coeficients are equal.  That is, $ P(P(x))\equal{}[P(x)]^k\Rightarrow P(x)\equal{}x^k$.

Thus, $ P(x)\equal{}0,1,\minus{}1,$ or $ x^k$ for integer $ k$.

[/hide]

$ \forall x\in P(\mathbb{R}): \ p(x)\equal{}x^k.$
so if $ P(\mathbb{R})\ is\ infini$ we have $ \forall x\in \mathbb{R}: \ p(x)\equal{}x^k$
if $ P(\mathbb{R})\ is\ fini$ then  $ \exists c\in\mathbb{R},\forall x\in \mathbb{R}: \ p(x)\equal{}c$ and $ PoP\equal{}P^k$ gives $ c\equal{}c^k$

$ P(P(x))\equal{}[P(x)]^k\Rightarrow P(x)\equal{}x^k$

huh?

[quote="(^_^)"]$ P(P(x)) \equal{} [P(x)]^k\Rightarrow P(x) \equal{} x^k$

huh?[/quote]
you konw that if $ p(x)\equal{}x^k$ have uinfinit solutions then $ \forall x: \ p(x)\equal{}x^k$
then we have $ p(a)$ is solution for $ p(x)\equal{}x^k$ and we have infini values for $ p(a)$ if $ deq(P)>0$


Does the below looked just a little bunched up to you?
If I should put spaces between the lines, how do I do it?

\begin{align*}
\cot x + \tan x & = \frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\\
&= \frac{\cos^2 x + \sin^2 x}{\sin x \cos x}\\
&= \frac{1}{\sin x \cos x}\\
&= \csc x \sec x\\
\end{align*}

You could add [\jot] at the end of each line:

\begin{align*}
\cot x + \tan x & = \frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\\[\jot]
&= \frac{\cos^2 x + \sin^2 x}{\sin x \cos x}\\[\jot]
&= \frac{1}{\sin x \cos x}\\[\jot]
&= \csc x \sec x
\end{align*}

or, if you want a different amount of space then add a vertical space eg \\[2ex] (an ex is the height of x in the current font and you can vary the number of them (including decimal values)). You can use different spacing for each line.

\begin{align*}
\cot x + \tan x & = \frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\\[2ex]
&= \frac{\cos^2 x + \sin^2 x}{\sin x \cos x}\\[2ex]
&= \frac{1}{\sin x \cos x}\\[2ex]
&= \csc x \sec x
\end{align*}

By the way you don't want a \\ on the last line as that will add an unwanted blank line.


calculus
derivative
conics
parabola
quadratics
analytic geometry
graphing lines

I was having some trouble with a problem. The problem reads:
Sketch the graphs of y = x^2 and y = -x^2+6x-5, and sketch the two lines that are tangent to both graphs. Find equations of these lines.

I've graphed the two parabolas and drew the tangent lines. I've found the derivatives of each equation as well. I relabeled the points where the tangent line is tangent on each of the parabolas using Xsub1 and Xsub2. Am I going about doing this the correct way? Can someone give me a little help, thanks.

-Paul

if y=x^2 then y'=2x.Let m=2x i.e. x=m/2,and therefore y=(m^2)/4.

Equation of tangent at (m/2,(m^2)/4) to the curve y=x^2 is

y-((m^2)/4) = m(x-(m/2))

Solving for the point of intersection of the above tangent with the curve

y=-x^2+6x-5

we get a quadratic in x which should have only one root if tangent of this curve is required to be same as that of y=x^2

x^2+(m-6)x+5-((m^2)/4)=0

Putting Discriminant = 0 (since it has only one root)

we get m=2,4

The required equations of tangents are y=2x-1,y=4x-4

How do you get m=2,4 from:

x^2+(m-6)x+5-((m^2)/4)=0

and once you have 2,4 I understand thats the slope so y=2x,y=4x but where do you plug back into to get the -1 and -4?

Discriminant of the quadratic in x

=(m-6)^2 - 4(5-(m^2)

=m^2-6m+8

put m^2-6m+8=0 solve for m

and then put these values of m in the equation of the tangent

y-((m^2)/4) = m(x-(m/2))

to obtain the required result .

Hope it is clear now.


One day, my friend randomly gave me the following sequence:

If $a_n$ is even, then $a_{n+1}=\frac{a_n}{2}$
If $a_n$ is odd, then $a_{n+1}=3a_n+1$

($a_1$ is a [u]positive integer[/u].)

For instance, if we start with $a_1=3$, the sequence is $3,10,5,16,8,4,2,1,4,2,1,4...$.

I tried different numbers for $a_n$, and it seems that the sequence eventually ends up with $4,2,1,4,2,1...$.

Can anyone prove (or disprove) that the sequence always ends up with the 4-2-1 cycle regardless of the number for $a_1$?

This is a famous problem, called the "3n + 1 problem."  It is currently unsolved.

An interesting variation, and an indication of why this is hard- try the same iteration with negative initial values.

This is the famous Collatz Conjecture. No proof exists and I think it was Erdos who said that "the world is not yet ready for this type of math," or something similar.

Yeah, it was Erdos. Some info about it:
So far, no one has either proved/disproved it

The conjecture is that eventually it goes to 1, but this means that eventually it goes to some power of 2. It actually has to be a power of 4 because 3x+1 is never 2 raised to an odd power (proof?). According to mathword, if negatives are included (you are allowed to start at -3, for instance), they do not all go to 1. there are at least 4 known loops that occur in the negative numbers.

Other problems have been asked such as the 5x+1 problem, but none has proved as interesting.

27 is a ridiculously long chain.


any knows any other formula like vieta's formula,to do with polynomials, roots, coefficient or power?

thnx t0rajir0u, any other thereom or formula like this, any1?


Let $ n,k\in\mathbb{Z}$ greater than $ 1$ and $ \gcd(n\plus{}k^n,k)\equal{}1$
Prove that at least one of $ n\plus{}k^n$ and $ n.k^{k^n\minus{}1}\plus{}1$ is not prime

Have a look here :
http://www.animath.fr/tutorat/dossier_03042sol.pdf

see problem #6.

Pierre.


trigonometry
inequalities
inequalities unsolved

If $a,b,c,d$ are real numbers and $a,b,c,d=[-\frac{\pi}{2},\frac{\pi}{2}]$, $sina+sinb+sinc+sind=1$; $cos2a+cos2b+cos2c+cos2d\geq \frac{10}{3}$.
Prove that $a,b,c,d=[0,\frac{\pi}{6}]$.
 :wink:


calculus
integration
logarithms
real analysis
real analysis unsolved

After substitution $ u \equal{} xy$ the integral equals $ \int_0^1 \int_0^y \frac {1 \minus{} \frac {u}{y}}{( \minus{} \ln u)^\frac52} \, \frac {du}{y} \, dy$ .

Swap integration order $ \int_0^1 \int_u^1 \frac {\frac {1}{y} \minus{} \frac {u}{y^2}}{( \minus{} \ln u)^\frac52} \, dy \, du \equal{} \int_0^1 \frac {u \minus{} 1 \minus{} \ln u}{( \minus{} \ln u)^\frac52} \, du$ and replace $ u$ by $ e^{ \minus{} t}$ .

$ \int_0^\infty \frac {e^{ \minus{} t} \minus{} 1 \plus{} t}{t^\frac52} \, e^{ \minus{} t} \, dt \equal{} \sum_{k \equal{} 2}^\infty \frac {( \minus{} 1)^k}{k!} \, \int_0^\infty t^{k \minus{} \frac52} \, e^{ \minus{} t} \, dt$$ \equal{} \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k \, \frac {\left(k \minus{} \frac52\right)!}{k!} \minus{} \frac {\left( \minus{} \frac52\right)!}{0!} \plus{} \frac {\left( \minus{} \frac32\right)!}{1!}$

$ \equal{} \left( \minus{} \frac52\right)! \, \sum_{k \equal{} 0}^\infty {\frac32 \choose k} \minus{} \frac {\sqrt {\pi}}{3}\, 4 \minus{} \frac {\sqrt {\pi}}{3}\, 6 \equal{} \frac {\sqrt {\pi}}{3}\, 4\,\cdot \,2^{\frac32} \minus{} \frac {\sqrt {\pi}}{3}\, 10 \equal{} \frac {\sqrt {\pi}}{3} \left(\sqrt {128} \minus{} 10\right)$


quadratics
number theory
number theory unsolved

I believe this one comes from a book call Elementary Number Theory, by Keneth Rosen.

The problem:

http://www.geocities.com/jacobin2001/mathproblem.jpg

the law of quadratic reciprocity:

http://www.geocities.com/jacobin2001/Quadraticrecip.jpg

Part 1: set $a=\frac{p-q}{4}$ iff $p \equiv q \mod 4$ and $a=\frac{p+q}{4}$ otherwise.
In the first case, it follows $\left( \frac{-q}{p}\right) = \left( \frac{p-q}{p}\right) = \left( \frac{4a}{p}\right) = \left( \frac{4a}{q}\right) = \left( \frac{p-q}{q}\right) = \left( \frac{p}{q}\right)$.
The result follows easily, and the same works for the other case we deal with.

2. Part: their hint holds you away from defining the Jacobi symbol and using that it still fulfills the same properties. But at least that hint makes says everything that has to be said ;)


Let $ P_1(x)\equal{}x^5\plus{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e$. Find the polynomial $ P_2(x)$ whose roots are n times those of $ P_1(x)$.

[hide="answer, you can simplify"]
Consider $ P_{1}(\frac{x}{n})$ where x is a root of our equation. In order for $ k\equal{}\frac{x}{n}$ to be a root, k has to be n times x, that way when we divide by n, we are still taking $ P(x)$, which is 0. So, expand and stuff.[/hide]

Let $ r_1, r_2,\ldots,r_5$ be the five roots of $ P_1(x)$. $ P_1$ can thus be written as \[ P_1(x) \equal{} (x\minus{}r_1)(x\minus{}r_2)\ldots (x\minus{}r_5)\]. $ P_2$ is then \[ P_2(x) \equal{} (x\minus{}2r_1)(x\minus{}2r_2)\ldots (x\minus{}2r_5)\]. Multiplying out $ P_1$ values for $ a,b,c,\ldots,e$ can be found in terms of $ r_1,r_2,\ldots,r_5$. When $ P_2$ is multiplied out values for $ a,b,\ldots,e$ can be plugged in, yielding \[ P_2 \equal{} x^5\plus{}2ax^4\plus{}4bx^3\plus{}8cx^2\plus{}16dx\plus{}32e\]

I think it's a little more revealing to write

$ P_2(x) \equal{} n^m P_1\left( \frac {x}{n} \right)$,

don't you?  :)  If nothing else, it opens up much more straightforward techniques for dealing with related problems:

- Find the polynomial whose roots are $ k$ greater than those of $ P_1$.
- Find the polynomial whose roots are those of $ P_1$ squared.
- And so forth.

well... you may be right... anyway it isn't a great jump from $ \sum{n^{m\minus{}k}a_kx^k}$ to $ n^mP_1(\frac{x}{n})$  :lol:


[b]Δίνεται τρίγωνο $ABC$ και τα ύψη του $AD,BE,CZ$ που τέμνονται στο $H$.Έστω $M$ το μέσο της $EZ$ και $P$ η τομή των $ZE$ και $BC$.H $PH$ τέμνει τις $AB,AC$ στα σημεία $K,L$.Να αποδείξετε ότι η $AM$ και οι κάθετες από τα $K,L$ προς τις $AB,AC$ αντίστοιχα, συντρέχουν.[/b]

                                                        [u][i][b]Nick Rapanos[/b][/i][/u]

Μια πραγματικά φοβερή [u]Ραπάνια [/u]  γεωμετρία  :D
Λοιπόν νομίζω την έλυσα (Τρίτη διαφορετική λύση?)
Έστω ότι η κάθετη από το $L$ στην $AC$ τέμνει την $AB$ στο $Q$ και ότι η κάθετη από το $K$ στην $AB$ τέμνει την $AC$ στο $N$ και έστω ότι οι $QL$ και $NK$ στο $G$ .To $G$ είναι το ορθόκεντρο του $AQN$ .Έστω $O$  το μέσο της $BC$ .
To τετράπλευρο $ZBCE$  είναι εγγράψιμο άρα οι $ZE,\ BC$ είναι αντιπαρράλληλες και αφού $AM$ διάμεσος στο $AZE$ η $AM$ θα είναι συμμετροδιάμεσος στο $ABC$ . Έστω ότι η $AM$ τέμνει την $QN$  sto $S$ .
Από γνωστό θεώρημα (αποδεικνυεται με συζυγή αρμιονικά) η $PL$  είναι κάθετη στην $AO$ .Τώρα είναι
$\angle{OAC}=90-\angle{ALP}=\angle{KLQ}=\angle{KNQ}$
Όμως $\angle{KNQ}=90-\angle{KQN}$
Όμως αφού $AS$ συμμετροδιάμεσος του $ABC$ είναι
$\angle{SAQ}=\angle{OAC}$
Άρα $\angle{QAS}+\angle{AQN}=\angle{KNQ}+\angle{KQN}=90$
Άρα η $AS$  είναι ύψος του $AQN$ άρα περνά από το $G$   :wink:

Αν κάποιος θέλει να βάλλω την απόδειξη για το θεώρημα που χρησιμοποιώ να μη διστάσει .
Νίκο ελπίζω να είναι σωστή η λύση

Silouane,

eisai sostos!Vevaia to theorima pou xrisimopoieis den einai poly gnosto kai eukoloapodeikto gia to eury koino.Episis an kapoios ftasei sto theorima- i askisi teleionei ekei! (mporo na sou eksigiso to pws..)
Gia auto loipon tha se parakalousa an mporeis na valeis tin apodeiksi tou theorimatos.

Nick

Θεώρημα
Έστω $\triangle ABC$, και το μέσο [b]M[/b] της  BC,και το ορθόκεντρο [b]H[/b] που έχει τις προβολές του  $D\in BC,\ E\in CA,\ F\in AB$  και οι τομές  $L\in BC\cap EF,\ S\in AM\cap LH$.
Δείξτε ότι : $AM\perp LH$

Απόδειξη (υπάρχεικαι μία εκπληκτική με πολικές αλλά βάζω την πιο normal)
Είναι γνωστό ότι η τετράδα  (L,D,B,C) είναι αρμονική (h.d.), και τότε είναι γνωστό ότι DB.DC=DL.DM. Έτσι ,

$(L,D,B,C)-h.d.\Longleftrightarrow DB\cdot DC=DL\cdot DM\Longleftrightarrow DA\cdot DH=DL\cdot DM$ $\Longleftrightarrow \frac{DA}{DM}=\frac{DL}{DH}\Longleftrightarrow$
$\tan \widehat{AMD}=\tan \widehat{LHD}\Longleftrightarrow \widehat{AMD}\equiv\widehat{LHD}\Longleftrightarrow HL\perp AM.$
Πώς σας φαίνεται?

[quote="silouan"]Είναι γνωστό ότι η τετράδα  (L,D,B,C) είναι αρμονική (h.d.), και τότε είναι γνωστό ότι DB.DC=DL.DM. [/quote]  ??????? ΠΩΣ ΤΟ ΞΕΡΟΥΜΕ ???
ΕΙΜΑΙ ΑΣΧΕΤΟΣ ΑΠΟ ΓΕΩΜΕΤΡΙΑ ΚΙ ΘΕΛΩ ΝΑ ΜΟΥ ΠΕΙΤΕ

ρε κοίτα τι σημαίνει αρμονική τετράδα ? :wink:
Μετά από αυτό παίρνεις λίγο Ceva λίγο Menelao και βγήκε.

Συγχαρητήρια στον φίλο μου Νίκο Ράπανο για τις τόσο καλές ασκήσεις γεωμετρίας που θέτει κι συγχαρητήρια στον φίλο μου Σιλουανό που τις λύνει κάτι που δείχνει την τεράστια ικανότητα κι τον δυοσας περιμένω κι άλλες γεωμετρίες του Νίκου
 :D  :D  :D
Χαιρετώ

Ας δούμε μία γενίκευση του προβλήματος και μία άλλη προσέγγιση για τη λύση του.

ΠΡΟΤΑΣΗ. - Δίδεται τρίγωνο ABC και έστω AD, BE, CZ, τρεις σεβιανές του, συντρέχουσες σε τυχόν σημείο Η, στο εσωτερικό του. Έστω Μ το μέσον του EZ και P, το σημείο τομής των EZ, BC. Φέρνουμε την ευθεία PH και έστω K, L, τα σημεία τομής της με τις πλευρές AB, AC αντιστοίχως. Αποδείξτε ότι η ευθεία AM και οι δια των K, L, παράλληλες ευθείες προς τις ευθείες CZ, BE αντιστοίχως, τέμνονται στο ίδιο σημείο.

ΑΠΟΔΕΙΞΗ – (στο σχήμα μου είναι AB = 6.8, AC = 7.7, BC = 5.8). - Έστω Η', το σημείο τομής των EK, ZL. Τα τρίγωνα HBC, H'KL, έχουν τα σημεία E, Z, P, τομής των πλευρών τους ανά μία, κείμενα επί ευθείας και άρα, σύμφωνα με το θεώρημa του Desarques, είναι προοπτικά. Οι ευθείες επομένως, BK, CL, HH', τέμνονται στο ίδιο σημείο και αποδεικνύεται έτσι, ότι το σημείο Η' κείται επί της ευθείας ΑΗ.

Από το μέσον Μ του EZ, φέρνουμε παράλληλες ευθείες προς τις ευθείες BE, CZ αντιστοίχως, και έστω E', Z', τα σημεία τομής των, με τις πλευρές AC, AB.

Σύμφωνα με την παρακάτω βοηθητική πρόταση, έχουμε ότι E'Z' // KL και άρα προκύπτει ότι AZ' / AK = AE' / AL, (1).

Έστω Ν και Ν', τα σημεία τομής της ευθείας ΑΜ, από τις δια των K, L, παράλληλες ευθείες προς τις ευθείες CZ, BE. Αρκεί να αποδείξουμε ότι Ν ≡ Ν'.

Από Z'M // KN, => AZ' / AK = AM / AN, (2) και από Ε'M // LN', => AE' / AL = AM / AN', (3).

Από (1), (2), (3), => AM / AN = AM / AN', => AN = AN', => N ≡ N' και η πρόταση έχει αποδειχθεί.

ΒΟΗΘΗΤΙΚΗ ΠΡΟΤΑΣΗ. – Δίδεται τρίγωνο ABC, τυχόν σημείο P, στο εσωτερικό του και έστω D, E, F, τα σημεία τομής των πλευρών του BC, AC, AB αντιστοίχως, από τις ευθείες AP, BP, CP. Από το μέσον Μ του EF, φέρνουμε παράλληλες ευθείες προς τις ευθείες DE, DF αντιστοίχως, οι οποίες τέμνουν τις πλευρές AC, AB, στα σημεία E', F'. Αποδείξτε ότι E'F' // BC.

ΑΠΟΔΕΙΞΗ – (στο σχήμα μου είναι AB = 7.8, AC = 9.8, BC = 12.0). - Δια των E, F, φέρνουμε παράλληλες ευθείες προς τις πλευρές AB, AC του ABC αντιστοίχως, και έστω O, N, τα σημεία τομής των, με τις ευθείες CF, BE. Θα αποδείξουμε ότι NO // BC.

Από FN // AC, => PN / PE = PF / PC, => PN = (PE)·(PF / PC), (1).

Από EO // AB, => PO / PF = PE / PB, => PO = (PF)·(PE / PB), (2).

Από (1), (2), => PN / PO = PB / PC, (3), => NO // BC, (4).

Έστω K, L, τα σημεία τομής της ευθείας DE, από τις ευθείες FN, CF, αντιστοίχως. Θα αποδείξουμε ότι το σημείο Ν, είναι μέσον του FK.

Σύμφωνα με το θεώρημα του Μενελάου, στο τρίγωνο FKL,  και για διατέμνουσα την ευθεία NPE, έχουμε την ισότητα

(NF / NK)·(EK / EL)·(PL / PF) = 1, (5).

Για να είναι NF = NK (N, μέσον του FK), από την (5), συμπεραίνουμε ότι αρκεί να αποδειχθεί ότι ισχύει EK / EL = PF / PL, (6).

Από FK // AC, => EK / EL = CF / CL, (7).

Από (6), (7), προκύπτει ότι αρκεί να αποδειχθεί ότι ισχύει PF / PL = CF / CL, (8).

Η (8) όμως, προκύπτει ως αληθής από το πλήρες τετράπλευρο CDPEAB. Πράγματι, όπως είναι γνωστό, η διαγώνιά του CP, τέμνεται από τις άλλες δύο του διαγώνιες, DE, AB, στα σημεία F, L, συζυγή αρμονικά των P, C. Δηλαδή αποδεικνύεται ότι ισχύει η (8) και άρα FN = NK. Δηλαδή το N, είναι μέσον του FK και επομένως έχουμε ότι η ευθεία ME' // DE, διέρχεται δια του σημείου Ν. Ομοίως αποδεικνύεται ότι και η ευθεία MF' // DF, διέρχεται δια του σημείου Ο.

Από FN // AC, EO // AB και επειδή το M είναι μέσον του EF, προκύπτει ότι MN = ME' και MO = MF' και άρα το τετράπλευρο E'F'NO, είναι παραλληλόγραμμο.

Επομένως E'F' // NO // BC και η πρόταση έχει αποδειχθεί.

Κώστας  Βήττας.

ΥΓ. – Nick Rapanos, ωραία πρόταση για να ξεσκουριάζει το μυαλό. Silouan, ωραία απόδειξη για όσους απολαμβάνουν τη γεωμετρία. Σας ευχαριστώ θερμά και τους δύο και σας αφιερώνω την παραπάνω απόδειξη (πρωτοεμφανιζόμενη νομίζω).

Η γενίκευση και η λύση σας είναι πραγματικά τρομερές .
Ειδικά τα σημεία που χρησιμοποιείται πρώτα το CDPEAB ως πλήρες τετράπλευρο και στη συνέχεια η χρήση του Θεωρήματος Desarques είναι απίθανα.
Το κακό είναι ότι πρέπει να φάω πολλά καρβέλια για να γράψω κάποτε τέτοια λύση..... :D
Σας ευχαριστούμε

Πραγματικά συγκλονιστικές λύσεις κι συμφωνώ μαζί σου ΣΙλουανέ
Όλα είναι στην προπονηση

Σοβαρα χωρίς πλάκα είσαι ο μόνος που μπορείς να το κάνεις (εκτός από τον Νίκο τον Ράπανο)

Αμβρόσιε έχεις δίκιο, όλα είναι στην προπόνηση. Για τον κοινό νου, που δεν πιστεύει στην παρθενογέννεση, αλλά ούτε και στην επιφοίτηση του αγίου πνεύματος (μιας και εδώ μιλάμε για τη γεωμετρία), αυτό που του μένει, είναι το να ασκηθεί όσο γίνεται περισσότερο, μαθαίνοντας από τους άλλους (αυτούς που έχουν προηγηθεί, αυτούς με τους οποίους μαζί πορεύτηκε, αλλά κυρίως από τους νέους που ορμητικά τον προσπερνάνε).

Νομίζω ότι τελικά, η λύση γεωμετρικών (αλλά και γενικότερα) προβλημάτων, είναι θέμα πετυχημένης διαχείρισης πληροφοριών, για τις οποίες πιθανόν να έχεις συνεισφέρει και ο ίδιος. Όταν διαβάζουμε την απόδειξη κάποιου άλλου, μπορεί να εμπνευστούμε κάποια ιδέα χρήσιμη για τη λύση άλλου προβλήματος. Όταν κάποιος άλλος διαβάζει τη λύση μας, μπορεί να δει κάτι που μας ξέφυγε και να πάει το θέμα παραπέρα.

Αυτός είναι ένας καλός λόγος νομίζω, για να δημοσιεύονται όλες οι απόψεις για τα διάφορα προβλήματα. Το ζητούμενο είναι να εμπλουτίσουμε όλοι τις γνώσεις μας και ας προσπαθήσουμε, στο μέτρο των δυνάμεών του ο καθένας, να κάνουμε αυτό το φόρουμ ακόμα πιο δυνατό.

Σας ευχαριστώ και τους δύο, για τα καλά σας λόγια.
Κώστας Βήττας.

Φίλε Κώστα με βρίσκεις απόλυτα σύμφωνο σε όλες σου τισ σκέψεις ειδικά στην γεωμετρία στην οποία έχω αδυναμιες κι θέλω πραγματικά να βοηθήσω όλουςε μας
ευχαριστώ πολύ για την συμμετοχη σου εδώ
Φιλικά

Ιδού ένα πρόγραμμα καταστροφής της Γεωμετρίας: Πάρε κατάλληλο σύστημα συντεταγμένων και μετά βάλε τη Mathematica ή τη Maple ή την Derive ή τη Matlab να κάνει τις πράξεις, απαλειφές κτλ.  ...

symfono apolytos...alla etsi exoume ti lysi-oxi ti goiteia!

Δεν χρειάζεται να βάλουμε κάποιο πρόγραμμα να κάνει πράξεις, απαλοιφές κλπ, αν το ζητούμενο είναι να δούμε για παράδειγμα, ότι τρεις ευθείες τέμνονται στο ίδιο σημείο.

Με οποιοδήποτε σχεδιαστικό πρόγραμμα και αν κατασκευάζουμε τα σχήματά μας, αυτό γίνεται αυτόματα, γιατί είναι βασισμένο σε κάποιο κρυφό καρτεσιανό σύστημα συντεταγμένων και εκτελεί μόνο του όλες τις απαραίτητες πράξεις τις αναλυτικής γεωμετρίας που ξέρουμε.
(Χωρίστε ένα κύκλο σε επτά ίσα μέρη και θα μείνετε έκπληκτοι για την ακρίβεια χάραξης).

Έτσι έχουμε τη βεβαιότητα ότι οι ευθείες είναι συντρέχουσες, ή και όχι. Αυτό είναι ένα καλό βοήθημα σε σχέση με παλιότερα, που τα σχήματα γίνονταν με το χέρι.

Νομίζω όμως ότι το ζητούμενο είναι άλλο. Με την απόδειξη προσπαθούμε να εξηγήσουμε γιατί συμβαίνει αυτό και να καταλήξουμε σε κάποιο συμπέρασμα, με μία αναντίρρητη διαδοχή συλλογισμών, που δεν επηρεάζεται από τη θέληση κανενός (ανθρώπου ή και θεού ακόμα).

Άρα εξοικειωνόμαστε στην τεκμηριωμένη διατύπωση συμπερασμάτων, για τη λύση των διαφόρων προβλημάτων στη ζωή μας, όχι απαραίτητα μόνο μαθηματικών.

Όταν δεν καταφέρνω να λύσω ένα γεωμετρικό πρόβλημα, προσπαθώ να μάθω περισσότερα και να γίνω καλύτερος. Ξέρω ότι πάντα θα υπάρχουν άλυτα προβλήματα για μένα, που κάποιοι άλλοι καταφέρνουν να τα λύσουν. Δεν μου φταίει η γεωμετρία και δεν μπορώ να διανοηθώ την καταστροφή της, γιατί με βοηθάει να ξεπερνάω τα κάθε λογής δόγματα.

Κώστας Βήττας.

φίλε κώστα έχεις απόλυτο δίκιο
θα ήταν πολύ χρήσιμο για δύσκολα γεωμετρικά προβλήματα η βοηθεια της τεχνολογίας για την σχεδιαση κι μόνο μετά πρέπει να χρησιμοποιησουμε τα παραδοσιακα τεχνασσματα κι αξίες για την λύση
χαιρετώ

να υποθέσω ότι είστε υπέμαχοι της κλασικής γεωμετρίας?
εγώ πάντως βρίσκω πολύ γοητευτικό τη μετατροπή των σχημάτων σε εξισώσεις και το αντίστροφο
φυσικά ο καθένας έχει το δικαίωμα να γοητεύεται από διαφορετικά πράγματα...
α, και δε θα το έλεγα πρόγραμα καταστροφής της γεωμετρίας αλλά αναβάθμισης αυτής... :lol:


Given numbers $ a,b,c$ and $ d$ no two equal,solve the system
$ x\plus{}ay\plus{}a^2z\plus{}a^3w \equal{} a^4$
$ x\plus{}by\plus{}b^2z\plus{}b^3w\equal{} b^4$
$ x\plus{}cy\plus{}c^2z\plus{}c^3w \equal{} c^4$
$ x\plus{}dy\plus{}d^2z\plus{}d^3w \equal{} d^4$

Consider the polynomial $ P(t) \equal{} t^4 \minus{} wt^3 \minus{} zt^2 \minus{} yt \minus{} x$

It is easy to see $ x,y,z,t$ are four roots of $ P(t)$.

By Vièta's theorem, we have:
$ a \plus{} b \plus{} c \plus{} d \equal{} w,ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} \minus{}z,abc \plus{} bcd \plus{} cda \plus{} dab \equal{}  y,abcd \equal{} \minus{}x$

Since, with given $ a,b,c,d$, the system is solved. :blush:


(not exact wording) A triangle of area 1 is folded from its top vertex. The fold is parallel to the base. Find the minimum area of the resulting figure.

on the opposite side???..so at the edge of the other side i believe

[quote="13375P34K43V312"][quote="The Zuton Force"]Where does the vertex end up?[/quote]

Wherever. The problem asks for you to find the maximum area of all the places where the vertex ends up given that the fold is parallel to the base.[/quote]

Eh? The area can't be more than $1$, so you're better off not folding at all... maybe minimum area?

That makes more sense  :)

[hide="Hint"]It is the same for all triangles[/hide]

[hide="Solution"]
So we may as well pick an easy one. Let's use a right triangle with base 2 and height 1.

Clearly if the vertex is contained within the triangle, its maximized when the vertex is placed on the right angle, so we need to explore only what happens if we allow the vertex to go outside the triangle.

Let $k$ be the portion of the altitude that is "missing".
The remainder of the altitude is $1-k$, and the distance that the folded section juts out of the original triangle is $k-(1-k)=2k-1$

We want to maximize the overlapped region.
This has area equal to the area "lost" from the folded part minus the area "gained" from the part that juts out below the base:

$A=(2k)(k)/2-(2(2k-1))(2k-1)/2=k^{2}-(2k-1)^{2}=-3k^{2}+4k-1$

The maximum value of this occurs at the vertex, $k=\frac{-4}{2(-3)}=\frac{2}{3}$

Sticking this back into our equation for $A$, we achive $A=\frac{1}{3}$ to be the area of the overlapped region.

So, the minimum area of the triangle is $1-A=\boxed{\frac{2}{3}}$
[/hide]

ya and Zuton Force did it for when b=2 and h=1...so i guess thats right..

The thing is that it isn't constant for all triangles...so I don't see how there's a single answer.

[quote="paladin8"][quote="Elemennop"]I got that the minimum area was $1-\frac{b}{6h}$, where $b$ and $h$ are the base and height, respectively.[/quote]

I don't think that works. For any triangle, if you fold the vertex to the opposite side you get $\frac{3}{4}$ (the overlapped portion has half the height and half the base of the whole triangle).[/quote]

The minimum doesn't occur when the vertex is on the opposite side.

[b]@Elemennop:[/b]

I'm very interested to see your solution. :yup:

However, I believe it should be the same for all triangles of area 1, because all triangles can be divided into two right triangles along the altitude, and the method i presented produces the same result for right triangles.

By the way, I assumed that the vertex was chosen optimally, since choosing an acute vertex of an obtuse triangle would not lead to very much overlap.

Additionally, I'm sure I can find a triangle in which the base is 6 times the height, and I sure hope the minimum area doesn't become 0  :wink:

(in fact, we know the minimum area must be at least $1/2$, because if it is any less, then there must be double overlap somewhere, which can't happen with one fold)

[hide]Let  be the base,  be the height,  be the base of the fold, and  be the base of the triangle formed by the additional paper jutting out past the base once it's folded. Let  be the height of the fold, and  be the height of the triangle formed by the additional paper jutting out.

First off, we know that . Also, note that , since area is .

Now, we want to maximize the lost area in order to minimize the actual area; the lost area is  By similar triangles, we have . Also, . Solving for  and , we have , and . Plugging this into  and cleaning it up a bit, we have  The minimum of  occurs at , so plugging this in yields  so the minimum is .

Oops.

So yea Zuton's right.

[/hide]


Let $ a;b;c$ be positive real numbers.Prove that:
 $ \sum (\frac{a}{b})^2\plus{}3(\frac{ab\plus{}cb\plus{}ca}{a^2\plus{}b^2\plus{}c^2})^4 \ge 6$

[quote="mathangel"]How can you find these numbers ?[/quote]

You try some values, trying to give a countrexample, so that the originial ineq is false
like $ a \equal{} b \equal{} c$ or $ a \rightarrow \plus{} \infty$  :wink:


Insert three fractions between $ \frac{1}{4}$ and $ \frac{1}{3}$ so that all 5 of the fractions are in arithmetic sequence. Write the sum of the three inserted fractions as a fraction in lowest terms.

$ \frac{1}{3}\minus{}\frac{1}{4}\equal{}\frac{1}{12}$, so each fraction is separated by $ \frac{1}{48}$.  That is, the fractions are $ \frac{13}{48}$, $ \frac{14}{48}$, and $ \frac{15}{48}$, and their sum is $ \frac{42}{48}\equal{}\frac{7}{8}$.


algebra
polynomial
complex numbers
algebra unsolved

$ \omega_1$, $ \omega_2$, $ \omega_3$ are 3 complex numbers such that
\[ \omega_1 \plus{} \omega_2 \plus{} \omega_3\equal{}1\]
\[ \omega_1^2 \plus{} \omega_2^2 \plus{} \omega_3^2\equal{}3\]
\[ \omega_1^3 \plus{} \omega_2^3 \plus{} \omega_3^3 \equal{} 7\]
Determine, with proof, the value of $ \omega_1^{12} \plus{} \omega_2^{12} \plus{} \omega_3^{12}$.

[hide="My Idea"]I was thinking that maybe one could let the three complex numbers be the roots of a cubic equation, and then try to solve for the coefficients, but this became very tedious.[/hide]

Did you try using [[Newton's Sums]] to find the coefficients of such a polynomial?


calculus
integration
trigonometry
function
calculus computations

Evaluate :

$ \int_0^{\pi/6} {\sin x\plus{}\cos x\over 1\plus{}\sin x\cos x} dx$

let the function inside the integral be $ f(x)$.   now find f(pi/6-x)    .  then put  under the integral.

[quote="PKMathew"]Evaluate :

$ \int_0^{\pi/6} {\sin x \plus{} \cos x\over 1 \plus{} \sin x\cos x} dx$[/quote]

$ I \equal{} 2 \ \int {\sin x \plus{} \cos x\over 2 \plus{} \sin 2x} dx$

$ \sin 2x \equal{} 1 \minus{} (\sin x \minus{} \cos x)^2$

Thus, I = $ 2 \ \int {\sin x \plus{} \cos x\over 3 \minus{} ( \sin x \minus{} \cos x)^2} dx$

$ \sin x \minus{} \cos x \equal{} t \ \Rightarrow \ ( \sin x \plus{} \cos x ) dx \equal{} dt$

$ I \equal{} 2 \ \int \frac{dt}{3 \minus{} t^2}  \equal{} \frac{1}{\sqrt{3}} \ ln|\frac{t\plus{}\sqrt{3}}{t\minus{}\sqrt{3}}| \plus{} c$

$ \equal{} \frac{1}{\sqrt{3}} \ ln|\frac{\sin x  \minus{} \cos x \plus{}\sqrt{3}}{\sin x \minus{} \cos x \minus{}\sqrt{3}}| \plus{} c$

now apply limits from 0 to $ \pi / 6 $   :D

[b]PKMathew wrote:[/b]

[quote]Evaluate :

$ \int_{0}^{\pi/6}{\sin x\plus{}\cos x\over 1\plus{}\sin x\cos x}dx$

[/quote]

As the numerator contains put  $ \ sin x \plus{} cos x$ so put $ \ sin x  \minus{}  cos x \equal{} t$  as $ \sin x\minus{}\cos x \equal{} t\ \Rightarrow\ (\sin x\plus{}\cos x ) dx \equal{} dt$

therfore the integral reduces to $ I \equal{} 2\ \int\frac{dt}{3\minus{}t^{2}}$ ,now solve and substitute the limits.

Your solution is same as Nora.91's one. :wink:

sorry kunny at first i jus din notice that i am sorry :P  :P


Hello. How are the people who make the APMO decided? Also, how is the IMO team decided? Is it from the CMO, USAMO or APMO? Or is it from all three? Thanks. :)

I think the top 30 or so students on the COMC plus people who go to Winter Camp are invited to the APMO (though I'm not sure of the exact number). The IMO team is decided from all 3 but so far I think everyone that has made it has received honourabe mention or better on CMO so that probably matters the most.


oh recently I 'm not able to see any post that is written in latex they are shown as an original code do u know what should i do? :blush:

Try the support forum -- that's where such questions are likely to be answered.

Go to your profile.  There's a line there that says 'Always allow BBCode'.  If it says no, change it to yes.

if that didn't solve the problem it may be an issue with your browser not allowing pictures. Do you see the emoticons, such as this one:  :?  ?


modular arithmetic
number theory unsolved
number theory

find all tuples(x,y,z) such that y is prime and z not divisible y and 3 satisfy:
x^3-y^3=z^2

Hold on, I am probably horribly wrong on this......

[hide="but........"]Rearranging, we have
$x^3 = y^3 + z^3$

This has no solutions because of Fermat's Last Theorem.[/hide]

I believe there is a typo, it might be that the right equation is $x^3 - y^3 = z^2$: not sure!

[quote="cezar lupu"]Anyway, about this Fermat's last theorem, I heard that a russian named Alexander Ilikin gave an elementary proof.
Is this true? :?  :o[/quote]

Thousands of people have been claiming they proved it in an elementary way.
Actually, none of them did prove it :D

I'm sorry but this is a recent proof. I heard he sent the proof to the Russian Academy to verify it, but no answer so far.... :?

That doesn't change my opinion. I don't think an elementary proof exists.

And, by the way, this is not the right place for such a discussion. If you really want to discuss Fermat's last theorem, rather visit the Round Table section or something similar :)

Note that $x^3-y^3 = (x-y)(x^2 + xy+y^2)$.
Note also that $(x^2+xy+y^2) - (x+2y)(x-y) = 3y^2$
$\Rightarrow \gcd (x-y, x^2 + xy+y^2) = \gcd (x-y, 3y^2)$.
Since $y \nmid z$ and $3 \nmid z$, we have $3y^2 \nmid z^2$. And since $(x-y) \mid z^2$, it follows that $\gcd (x-y, 3y^2) = 1$. Thus $x-y$ and $x^2+xy+y^2$ are both perfect squares.

Let $x-y = a^2 \Rightarrow x^2 + xy +y^2 = {(a^2+y)}^2 + (a^2+y)y + y^2 = a^4 + 3a^2y + 3y^2$ is a perfect square. $\Rightarrow 4(x^2 + xy + y^2) = {(2a^2 + 3y)}^2 + 3y^2$ is a perfect square. Letting $4(x^2 + xy + y^2) = m^2$ and ${(2a^2 + 3y)}^2 = n^2$, we have $m^2 = n^2 + 3y^2 \Rightarrow (m+n)(m-n) = 3y^2$. At this point, we can split into $3$ cases to consider. (Remember that $y$ is prime, so its only positive divisors are $1$ and $y$.)

Case 1: $(m-n, m+n) = (1, 3y^2)$
We have $2n = 3y^2-1 \Rightarrow 4a^2 = 2n-6y=3y^2-6y-1 \Rightarrow {(2a)}^2 \equiv -1 \pmod{3}$ (Contradiction)

Case 2: $(m-n, m+n) = (y, 3y)$
We have $n = y < 2a^2 + 3y = n$ (Contradiction)

Case 3: $(m-n, m+n) = (3, y^2)$
We have $4a^2 = {(2a)}^2 = 2n - 6y = y^2 - 6y - 3 < {(y-3)}^2$. Note also that $y^2 - 6y - 3 > {(y-4)}^2 \forall y \geq 10$. Since $y$ is prime, it suffices to check the cases $y = 2, 3, 5, 7$. Checking, we find that $y^2 - 6y - 3$ is a perfect square only for $y=7$.

Note that it suffices to check these $3$ cases for $m-n < m+n$, so the cases $(m-n, m+n) = (y^2, 3), (3y, y), (3y^2, 1)$ can immediately be omitted.

Indeed, checking, we find that when $y=7$, the corresponding $(x, y, z) = (8, 7, 13)$ is a solution. Thus this is the only possible solution.


Let $p$ denote a prime number of the form $4k+3$ and $x,y$ be positive integers and $p$ divides $x^2-xy+\frac{p+1}{4}y^2$. Prove that there exist integers $u,v$ such that
$x^2-xy+\frac{p+1}{4}y^2 = p(u^2-uv+\frac{p+1}{4}v^2)$.

Note that $x^2-xy+\frac{p+1}{4}y^2 = \left\| x+ \frac{1+i \cdot \sqrt{p}}{2} y \right\|=\left\| x+\zeta y \right\|$ with $\zeta=\frac{1+i \cdot \sqrt{p}}{2}$.
Since $\| \cdot \|$ is multiplicative, we see that the numbers of desired type are closed under multiplication.
Since $p=1^2-1 \cdot 2 + \frac{p+1}{4} \cdot 2^2$, we get the result.

Very nice solution,Zetax.
I think it can also be proved in a regular way.


prove that for every set $X$ of at least $4k^2$ distinct residue classes modulo a prime $p$ ,there is an integer  $a$ such that the set $\{\ ax(mod p) : x\in X \}\$intersects every interval in $\mathbb{Z}_p$ of length at least $p/k$.

it seems that just Prof Alon is able to solve this one.any try?


function
algebra
functional equation
algebra unsolved

Find all functions $f,g$:$R \to R$ such that $f(x+yg(x))=g(x)+xf(y)$ for $x,y \in R$.

Partial solution (if $g(0) \neq 0$). [hide]$x=0$ $\to$ $f(yg(0))=g(0)$ $\to$
case 1)
$g(0)=k \neq 0$ $\to$ $f(t)=k$ for every $t= \frac{y}{k} \in \mathbb{R}$ ($f$ constant).
We substitute $f(t)=k$ in the functional equation,
$k=g(x)+xk$ $\to$ $g(x)=k(1-x)$, and $g(0)=k$ for every $k$.
So, in this case, the only solutions are:
$f(x)=k$, $g(x)=k(1-x)$, for every $k \in \mathbb{R^*}$,

case 2)
$g(0)=0$
...[/hide]
I'll keep working  :)

I'll assume $g(0)=0$ (in a previous post I solved otherwise).

Fact 1: $f(x)=g(x)$ for every $x$.
[hide="Proof"]$y=0$ $\to$ $f(x)=g(x)+xf(0)$ $\to$ $g(x)=f(x)-f(0)x$.
In particular, if $x=0$, $g(0)=f(0)$, and we know $g(0)=0$, so $f(x)=g(x)$.[/hide]
Fact 2: Either $z=0$ is the only zero of $f(z)=g(z)=0$, or $f(x)=g(x)=0$ for every $x$.
[hide="Proof"]If $f(z)=g(z)=0$,
$x=z$ $\to$ $0=zf(y)$ for every $y$.[/hide]
[I edited it to make it easier.]

After my last two posts, the functional equation that still remains to be solved is $f(x+yf(x))=f(x)+xf(y)$, and the only zero is at $f(0)=g(0)=0$.

For every $x \neq 0$, $f(x) \neq 0$ and we can define $h(x)= \frac{-x}{f(x)}$.
$y=h(x)$ $\to$ $f(0)=f(x)+xf(h(x))$ $\to$ $f(h(x))= \frac{1}{h(x)}$.

It would be helpful to prove that there's only one possible value for $c$ with $f(\frac{1}{c})=c$.
In that case, $h(x)=\frac{1}{c}=\frac{-x}{f(x)}$ and the only solutions left would be $f(x)=-cx$.
And checking with the functional equation,
$-cx+c^2xy=-cx-cxy$ $\to$ $c^2=-c$ $\to$ $c=0$ or $c=-1$.
The only new solution would be $g(x)=f(x)=x$.
Unfortunately, our assumption is wrong, because with this function there's not only one possible value for $c$ with $f(\frac{1}{c})=c$. There are two, $c=-1$ and $c=1$. So, we're back to the beginning  :(

Let's solve the case when $g(x)=f(x)$
It obvious that $f(x) \equiv 0$ is a solution.Assume that there exists $s \in R$ such that $f(s) \neq 0$.
$y=0, x=s \Longrightarrow f(s)=f(s)+sf(0) \Longrightarrow f(0)=0$
Let $f(t)=0 \Longrightarrow y=s, x=t \Longrightarrow f(t+sf(t))=f(t)+tf(s) \Longrightarrow t=0$.So $f(t)=0$ iff $t=0$
$x=y=-1 \Longrightarrow f(-1-f(-1))=0 \Longrightarrow f(-1)=-1$
$y=1 \Longrightarrow f(x+f(x))=f(x)+xf(1)$ $(1)$
$y=-1,x=1 \Longrightarrow f(1-f(1))=f(1)-1$
$1-f(1)+f(1-f(1))=0 \Longrightarrow f(1-f(1)+f(1-f(1)))=0 \Longrightarrow$ from $(1)$ $\Longrightarrow 0=f(1-f(1)+f(1-f(1)))=f(1-f(1))+(1-f(1))f(1) \Longrightarrow f(1)=1$
$x=1 \Longrightarrow f(1+y)=1+f(y)$
$y \Longrightarrow \frac{1}{f(x)} \Longrightarrow f(x+1)=f(x)+xf(\frac{1}{f(x)})$
$\Longrightarrow f(x)+1=f(x)+xf(\frac{1}{f(x)}) \Longrightarrow f(\frac{1}{f(x)})=\frac{1}{x}$
$y=x, x \Longrightarrow \frac{1}{f(x)} \Longrightarrow f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$
$f(\frac{1}{f(x)})+1=\frac{1}{x}+\frac{1}{f(x)}.x \Longrightarrow 1=\frac{1}{f(x)}.x$
So $f(x)=x$.

[quote="cefer"]$y=x, x \Longrightarrow \frac{1}{f(x)}\Longrightarrow f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$
[/quote]

If I have understood your substitutions correctly,you should got
$f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}*f(x)$ in place of
$f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$,shouldn't you?

$f(x+yf(x))=f(x)+xf(y)$
I solved this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=125594]here[/url].
It's quite complex.

[quote="cefer"]Let's solve the case when $g(x)=f(x)$
It obvious that $f(x) \equiv 0$ is a solution.Assume that there exists $s \in R$ such that $f(s) \neq 0$.
$y=0, x=s \Longrightarrow f(s)=f(s)+sf(0) \Longrightarrow f(0)=0$
Let $f(t)=0 \Longrightarrow y=s, x=t \Longrightarrow f(t+sf(t))=f(t)+tf(s) \Longrightarrow t=0$.So $f(t)=0$ iff $t=0$
$x=y=-1 \Longrightarrow f(-1-f(-1))=0 \Longrightarrow f(-1)=-1$
$y=1 \Longrightarrow f(x+f(x))=f(x)+xf(1)$ $(1)$
$y=-1,x=1 \Longrightarrow f(1-f(1))=f(1)-1$
$1-f(1)+f(1-f(1))=0 \Longrightarrow f(1-f(1)+f(1-f(1)))=0 \Longrightarrow$ from $(1)$ $\Longrightarrow 0=f(1-f(1)+f(1-f(1)))=f(1-f(1))+(1-f(1))f(1) \Longrightarrow f(1)=1$
$x=1 \Longrightarrow f(1+y)=1+f(y)$
$y \Longrightarrow \frac{1}{f(x)}\Longrightarrow f(x+1)=f(x)+xf(\frac{1}{f(x)})$
$\Longrightarrow f(x)+1=f(x)+xf(\frac{1}{f(x)}) \Longrightarrow f(\frac{1}{f(x)})=\frac{1}{x}$
$y=x, x \Longrightarrow \frac{1}{f(x)}\Longrightarrow f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$
$f(\frac{1}{f(x)})+1=\frac{1}{x}+\frac{1}{f(x)}.x \Longrightarrow 1=\frac{1}{f(x)}.x$
So $f(x)=x$.[/quote]

Oops!I have a bad mistake, but here is a solution (i think here isn't any mistake :lol: )
From above we got
$f(\frac{1}{f(x)})=\frac{1}{x}\Longrightarrow f(x)=1$ iff $x=1$
Replacing $y$ with $\frac{y}{f(x)}$ we get
$f(x+y)=f(x)+xf(\frac{y}{f(x)})$   $(1)$
$y \Longrightarrow y+1$
$f(x+y+1)=f(x)+xf(\frac{y+1}{f(x)})$
$f(x+y)+1=f(x)+xf(\frac{y+1}{f(x)})$ $(2)$
From $(1)$ and $(2)$
$1+xf(\frac{y}{f(x)})=xf(\frac{y+1}{f(x)})$
$1+xf(\frac{y}{f(x)})=xf(\frac{y}{f(x)}+\frac{1}{f(x)})$ $(*)$
Applying $(1)$
$1+xf(\frac{y}{f(x)})=x(f(\frac{y}{f(x)})+\frac{y}{f(x)}f(\frac{\frac{1}{f(x)}}{f(\frac{y}{f(x)})}))$
$\frac{f(x)}{xy}=f(\frac{1}{f(x)f(\frac{y}{f(x)})})$  $(**)$
Replacing $x$ with $\frac{1}{f(x)}$ and using the fact that $f(\frac{1}{f(x)})=\frac{1}{x}$
$\frac{f(x)}{xy}=f(\frac{x}{f(xy)})$
Replacing $y$ with $\frac{f(x)}{x}$ we get
$1=f(\frac{x}{f(f(x))}) \Longrightarrow f(f(x))=x$
In $(**)$ taking $y=\frac{f(x)}{x}$ we get $1=f(\frac{1}{f(x)f(\frac{1}{x})}) \Longrightarrow f(x)f(\frac{1}{x})=1$
$x \Longrightarrow f(x)$ in $(*)$
$1+f(x)f(\frac{y}{x})=f(x)f(\frac{y}{x}+\frac{1}{x})$
$\frac{1}{f(x)}+f(\frac{y}{x})=f(\frac{y}{x}+\frac{1}{x})$
$f(\frac{1}{x})+f(\frac{y}{x})=f(\frac{y}{x}+\frac{1}{x})$
$f(x)+f(yx)=f(x+yx) \Longrightarrow f(a+b)=f(a)+f(b)$
Using this and main functional equation
$f(x)+f(yf(x))=f(x)+xf(y) \Longrightarrow f(yf(x))=xf(y) \Longrightarrow f(xy)=f(x)f(y)$
We got that $f$ is additive and multiplicative so $f(x)=x$


Euler
geometry
circumcircle
radical axis
geometry proposed

Given triangle $ ABC$. Denote $ A', B', C'$ the midpoints of $ BC, CA, AB$, respectively. Prove that the radical center of $ (A, AA'), (B,BB'), (C, CC')$ lies on [b]Euler line[/b] of triangle ABC.

Let $ H,O,N$ denote the orthocenter, circumcenter and nine-point center of $ \triangle ABC.$ Let $ P$ be the midpoint of $\overline{HN}.$ By Stewart's Theorem for the cevian $ AP$ in $ \triangle AHO$

$ AP^2 \equal{} \frac {_1}{^4}R^2 \plus{} \frac {_3}{^4}AH^2 \minus{} \frac {_3}{^{16}}OH^2 \equal{} \frac {_1}{^4}R^2 \plus{} \frac {_3}{^4}(2OA')^2 \minus{} \frac {_3}{^{16}}OH^2$

$ AP^2 \equal{} \frac {_1}{^4}R^2 \plus{} 3(R^2 \minus{} \frac {_1}{^4}a^2) \minus{} \frac {_3}{^{16}}OH^2 \equal{} \frac {_{13}}{^4}R^2 \minus{} \frac {_3}{^4}a^2 \minus{} \frac {_3}{^{16}}OH^2$

Let $ \mathcal{P}_a$ denote the power of $ P$ with respect to $ (A,AA').$ Then

$ \mathcal{P}_a \equal{} {AA'}^2 \minus{} AP^2 \equal{} \frac {_1}{^2}(b^2 \plus{} c^2) \minus{} \frac {_1}{^4}a^2 \minus{} \frac {_{13}}{^4}R^2 \plus{} \frac {_3}{^4}a^2 \plus{} \frac {_3}{^{16}}OH^2$

$ \Longrightarrow \ \mathcal{P}_a \equal{} \frac {_1}{^2}(a^2 \plus{} b^2 \plus{} c^2) \minus{} \frac {_{13}}{^4}R^2 \plus{} \frac {_3}{^{16}}OH^2$

Since this expression is symmetric in terms of $ (a,b,c),$ we conclude that midpoint $ P$ of $\overline{HN}$ has equal power with respect to circles $ (A,AA'),$ $(B,BB'),$ $(C,CC').$

Here is my solution for this problem
[b]Solution[/b]
Let $G$ be centroid of $\triangle$ $ABC$; $D$, $E$, $F$ be reflections of $A'$, $B'$, $C'$ through $A$, $B$, $C$; $D'$, $E'$, $F'$ be midpoints $EF$, $FD$, $DE$; $T$ be NPC center of $\triangle$ $DEF$
We have: $\dfrac{\overline{GA}}{\overline{AD}}$ = $\dfrac{\overline{GA}}{\overline{A'A}}$ = $\dfrac{2}{3}$
Similarly: $\dfrac{\overline{GB}}{\overline{BE}}$ = $\dfrac{2}{3}$, $\dfrac{\overline{GC}}{\overline{CF}}$ = $\dfrac{2}{3}$
So: $\dfrac{\overline{GA}}{\overline{AD}}$ = $\dfrac{\overline{GB}}{\overline{BE}}$ = $\dfrac{\overline{GC}}{\overline{CF}}$ = $\dfrac{2}{3}$ or $AB$ $\parallel$ $DE$, $BC$ $\parallel$ $EF$, $CA$ $\parallel$ $FD$
Then: $T$ lies on Euler line of $\triangle$ $ABC$
Let $I$ be midpoint of $E'F'$; $J$, $K$ be reflections of $C'$, $B'$ through $BC$, $L$, $M$, $N$ be orthogonal projections of $T$ on $EF$, $FD$, $DE$
We have: $\dfrac{\overline{AC'}}{\overline{DE}}$ = $\dfrac{1}{2}$ . $\dfrac{\overline{AC'}}{\overline{DF'}}$ = $\dfrac{1}{2}$ . $\dfrac{\overline{GA}}{\overline{GD}}$ = $\dfrac{1}{5}$ and $\dfrac{\overline{IA}}{\overline{ID}}$ = $\dfrac{\overline{D'A'}}{\overline{D'D}}$ = $\dfrac{\overline{D'A}}{\overline{D'G}}$ . $\dfrac{\overline{D'G}}{\overline{D'D}}$ = $\dfrac{3}{5}$ . $\dfrac{1}{3}$ = $\dfrac{1}{5}$
Hence: $\dfrac{\overline{AC'}}{\overline{DE}}$ = $\dfrac{\overline{IA}}{\overline{ID}}$ = $\dfrac{1}{5}$ or $I$, $C'$, $F$ are collinear
Similarly: $I$, $B'$, $E$ are collinear
We also have: $B'C'$ $\parallel$ $BC$ $\parallel$ $EF$ and $B$, $C$ are midpoints of $B'E$, $C'F$ so: $J$, $K$ $\in$ $EF$ and $J$ $\in$ ($C$; $CC'$), $K$ $\in$ ($B$, $BB'$)
But: $T$ is NPC center of $\triangle$ $DEF$ and $I$ is midpoint of $E'F'$ so: $TI$ $\perp$ $EF$ or $I$, $T$, $L$ are collinear
We have: $\dfrac{\overline{LJ}}{\overline{LE}}$ = $\dfrac{\overline{IC'}}{\overline{IE}}$ = $\dfrac{\overline{IB'}}{\overline{IF}}$ = $\dfrac{\overline{LK}}{\overline{LF}}$
So: $\overline{LK}$ . $\overline{LE}$ = $\overline{LJ}$ . $\overline{LF}$ or $P_{L / (B; BB')}$ = $P_{L / (C; CC')}$
But: $LT$ $\perp$ $BC$ then: $LT$ is radical axis of ($B$; $BB'$), ($C$; $CC'$)
Similarly: $MT$, $NT$ are radical axises of (($C$; $CC'$), ($A$; $AA'$)), (($A$; $AA'$), ($B$; $BB'$))
Therefore: $T$ is radical center of ($A$; $AA'$), ($B$; $BB'$) and ($C$; $CC'$) or radical center of ($A$; $AA'$), ($B$; $BB'$) and ($C$; $CC'$) lies on Euler line of $\triangle$ $ABC$


A regular hexagon is inscribed in a circle. What is the ratio of the length of a side of the hexagon to the minor arc that it intercepts?

(1) $\frac{\pi}{6}$           (3) $\frac{3}{\pi}$
(2) $\frac{1}{2}$         (4) $\frac{6}{\pi}$

is it

[hide]

Clearly the radius "$r$" equals the length of the hexagon and the arc length is $\frac{1}{6}$ of the total circumference therefore

Arclength $=\frac{2\pi r}{6}=\frac{\pi r}{3}$

Ratio $\implies\frac{r}{\frac{\pi r}{3}}\implies\boxed{\frac{3}{\pi}}$[/hide]

If this was the AMC, you could guess very easily the correct answer. The answer will be slightly smaller than $1$, since the side length is slightly shorter than the intercepted arc. The only one close is $\frac{3}{\pi}$.

Thanks for the important mathematical data.


inequalities
function
calculus
quadratics
analytic geometry
graphing lines
slope

Given that $x$ and $y$ are real numbers such that $x+y=4$ find the minimum value of $x^2+y^2$.

[hide]$(x+y)^2 = 16$
$x^2 + 2xy + y^2 = 16$
$x^2 + y^2 = 16 - 2xy$.

From AM-GM, we have
$\frac{x^2+y^2}{2} \ge \sqrt{x^2 y^2}$.
$x^2+y^2 \ge 2xy$

Equality in the above inequality holds when $x^2 = y^2$. (This will maximize $2xy$ and will minimize $x^2 + y^2$.)

Thus, $x = y$, and since $x + y = 4$,
$x = 2$, and $y = 2$.

$x^2 + y^2 = 2^2 + 2^2 = 8$.[/hide]

Edit:  The "hide" function does not seem to be working here - what is wrong?

[quote="kunny"]$\sqrt{x^2y^2}=xy?$[/quote]

well it's true when x and y are both positive (or negative) - if one's positive and the other's negative it's a fairly trivial case

Yeah that's the almost the exact solution I have. But kunny does have a good point. You can't apply AM-GM when unless the numbers are positive or 0 (I think), so a proof of your solution would have to consider the case where one of $x$ or $y$ is less than $0$. Here is what I found for that case.

Suppose one of $x$ or $y$ is less than $0$. Without loss of generality, suppose $x<0$. Then $y=4-x>4-0=4$ so $y^2>16>8$. We already found two numbers that give $x^2+y^2=8$ so clearly the minimum could not occur when one of the number is less than $0$.
And of course they can't both be negative.

[hide]
You can use calculus aswell
$x^2 + 2xy + y^2 = 16$
We're trying to maximize 2xy,
$y = 4 - x$
$xy = 4x - x^2$
$4 - 2x = 0 \rightarrow x = 2 \text{ is a maximum.}$
$x^2 + 2(2)(2) + y^2 = 16$
$x^2 + y^2 = 8$
[/hide]

Or just substitute $y=4-x$ into $x^2+y^2$, getting $x^2+(4-x)^2=x^2+16-8x+x^2=2x^2-8x+16$ and then just use the formulae for the minimum/maximum of a quadratic function.

Yeah, AM-GM only works when $x,y \geq{0}$. To prove it start with $(x-y)^2 \geq{0}$ like Kunny said. Then reorder this to get $\frac{x^2+y^2}{2} \geq{xy}$. Now let $u=x^2$ and $v=y^2$ to get the desired inequality $\frac{u+v}{2}\geq{\sqrt{uv}}$. By definition, $u,v \geq{0}$.

sweet, i just got kunny's hint. the answer is the square of the smallest radius a circle can have that is centered at the origin and intersects the line $x+y=4$

edit: actually, its easier to say the square of the shortest distance from the line $x+y=4$ to the origin.

[hide]Draw a circle and find tangents with slopes of -1.[/hide]

there's another solution. just plug in for $x+y$, divide by 2 and use the trivial inequality on $(x-y)^2$.

You can use Cauchy-Schwalz inequality.

$(a^2+b^2)(x^2+y^2)\geq (ax+by)^2.$

[quote="kunny"]You can also use inner product of vector.

Rewrite $x+y=(1,1)\cdot (x,y),$ ...... :lol:[/quote]

This is equivalent to the Cauchy solution  ;)

Edit:  Geometric solution.  $x + y = 4$ is a line perpendicular to the line $y = x$, so its minimal distance to the origin occurs at $x = y = 2$.  :)

Yes. When you see the eqation the sum of squares, you will imagine this as the square of a distance and regard $ax+by$ as the inner product of two vectors $(a,b),\ (x,y).$ These informations give a hint for using Cauchy-Schwarz inequality.

Geometric solution has posted in the former posting, t0rajir0u. ;)

I think some replies have gotten a little over-complicated:

$x+y=4$
$y=4-x$

So $x^2+y^2=x^2+(4-x)^2$
$=2x^2-8x+16$
$=2(x^2-4x+8)$

Completing the square gives
$2[(x-2)^2+4]$
So the minimum of this expression occurs when $x=2$, and this value is $8$

coffeym, you should see others solitions before you replies. ;)

Umm, i don't see completing the square in any of the above solutions.


real analysis
trigonometry
function
group theory
abstract algebra
calculus
derivative

I have two questions.

1. Find the limits superior of $\{a_n\}$. Justify your answer.
                        $a_n=\sin n,\;n\in\mathbb{N}$

2. Use the definition of uniform continuity to show which of the following function is
uniformly continuous.
                          $f(x)=x\sin \frac{1}{x},\;x\in(0,\pi)$.

For the first one you can use that the set $M=\{n+2\pi m\,/\, n,m\in\mathbb{Z}\}$ is dense, then $\sin(M)$ must be dense in $[-1,1]$, so the limsup u wanted is $1$.

About the density of $M$, first note that it is an additive subgroup of $\mathbb{R}$, and it can be proved that such subgroups are of the form $\{cn/n\in\mathbb{Z}\}$  for some $c$ or dense.

$|f(x_1)-f(x_2)|\leqq 2|x_1-x_2|$, so $f$ is Lipschiz continuous, stronger than uniformly continuouity.

No, it's not Lipschitz.
On the other hand, it is uniformly continuous- any function which extends to a continuous function on a compact set is. Use a derivative estimate away from zero, and the fact that $|f(x)|\le |x|$ near zero.


inequalities
algebra
polynomial
number theory unsolved
number theory

Let  $ a_1<a_2<...$ be a sequance  integers  such  that  $ gcd(a_i,a_j)\equal{}1$
and  \[ a_{i\plus{}2}\minus{}a_{i\plus{}1}\ge a_i\minus{}a_{i\minus{}1}\].

  Prove  that   \[ \sum  \frac{1}{a_i}<\infty\]

Define $ b_i \equal{} a_{2i}$. Since $ \sum \frac {1}{a_i} < a_1 \plus{} 2\sum\frac {1}{b_i}$, it suffices to prove that $ \sum \frac {1}{b_i}$ converges. If $ i \ge 3$,

$ \begin {cases} a_{2i} \minus{} a_{2i \minus{} 1} \ge a_{2i \minus{} 2} \minus{} a_{2i \minus{} 3} \\
a_{2i \minus{} 1} \minus{} a_{2i \minus{} 2} \ge a_{2i \minus{} 3} \minus{} a_{2i \minus{} 4} \end{cases} \Rightarrow a_{2i} \minus{} a_{2i \minus{} 2} \ge a_{2i \minus{} 2} \minus{} a_{2i \minus{} 4}$

$ \Rightarrow b_{i} \minus{} b_{i \minus{} 1} \ge b_{i \minus{} 1} \minus{} b_{i \minus{} 2}$

Now we see that, for $ n > 2$, we can't have $ b_{i \plus{} 2n} \minus{} b_{i \plus{} 2n \minus{} 1} \equal{} b_{i \plus{} 2n \minus{} 1} \minus{} b_{i \plus{} 2n \minus{} 2} \equal{} ... \equal{} b_{i \plus{} 1} \minus{} b_i \equal{} n$. In fact, looking the sequence $ b_i, b_{i \plus{} 1}, ..., b_{i \plus{} 2n}$ modulo $ n \minus{} 1$, we have $ 2n \plus{} 1$ consecutive numbers, so at least two zeros. Therefore we have two numbers divisible by $ n \minus{} 1$, so not coprimes - contradiction. Case $ n \le 2$, the difference $ n$ can also occur only a finite number of times. So we can assume that the sequence increases by $ 3$ exactly $ 6$ times, then increases by $ 4$ exactly $ 8$ times, then increases by $ 5$ exactly $ 10$ times... (we have proved it doesn't happen, but this is the worst case). Also, we can ignore the first terms (terms whose differences are $ < 3$); this does not change convergence.

I'll just sketch the end. $ 1 \plus{} 2^2 \plus{} 3^2 \plus{} 4^2 \plus{} ... \plus{} n^2 > \frac {1}{10}n^3$ (the formula for this sum is a polynomial of third degree in $ n$). So the $ 2n$ consecutive terms with difference $ n$ are at least of magnitude $ \frac {n^3}{10}$. As there are $ 2n$ of these terms, the sum of their inverses can be overstimated by $ 2n*\frac {10}{n^3} \equal{} \frac {20}{n^2}$. Thus, the convergence of this series is reduced to the convergence of $ \sum \frac {1}{n^2}$, which is well known.

All right?


$ 7x^3\minus{}8ax^2\equiv 0 \pmod 3$
$ \minus{}8ax^2\plus{}9x\plus{}3a \equiv 0 \pmod 7$

Is there an elegant way to solve it? The first congruence is easy; from the secondo one I get $ a\equal{}1$, but don't know if it's right, hence to continue.

There are many solutions. First equavalent to $ x(x\plus{}a)\equiv 0\mod 3$, second to $ a\equiv x\equiv 9\mod 7$ or $ 3(a^2\minus{}1)\equiv (ax\minus{}2)^2\mod 7$.
Second had solutions
$ a\equiv x\equiv 0\mod 7$,
$ a\equiv 1,x\equiv 2\mod 7$,
$ a\equiv 2,x\equiv 3, x\equiv 6\mod 7$,
$ a\equiv 5,x\equiv 1, x\equiv 4\mod 7$,
$ a\equiv 6,x\equiv 2\mod 7.$

I solve the first the second samilar.
$ \Longleftrightarrow x^2(7x-8a)\equiv 0 (\mod 3)$
$ \Longleftrightarrow x\equiv 0(\mod 3) \text{or} 7x\equiv 8a (\mod 3)$
The second has to case :
 [u]Case 1[/u] $ a\equiv 0 (\mod 7)$
  [u]Case 2[/u] Consider $ \Delta=3(27+24a^2)$
So $ (\frac{\Delta }{7})=1$
With this solution then use chinese remainder theorem.


I think electrophilic addition but I dont understand how the unsaturated chains are produced.

the answer a is quite obvious looking at the options , but what about the mechanism ???

That product is not the only one. Can anyone provide the structure of the other product that is obtained when styrene is treated with acid?


1. $ P(Q(x))\equal{}\frac{Q(x)}{3x\plus{}1}$ then what is the remainder of $ P(3x^2\minus{}5x\minus{}2)$ when divided by $ x\minus{}6$ ?

2. $ P(x)\equal{}2x^3\plus{}7x^2\minus{}6x\plus{}1$ if $ P(x)$ can divide $ x^2\minus{}mx\minus{}1$ then $ m\equal{}?$ ?

[hide]2)
Since we are dividing a cubic by a quadratic we know that our answer will be a linear function.
Without any calculations it's easy to see that the quotient has to be $ 2x\minus{}1$.

So we have that $ 2x^3\plus{}7x^2\minus{}6x\plus{}1\equal{}(x^2\minus{}mx\minus{}1)(2x\minus{}1)\equal{}2x^3\minus{}x^2\minus{}2mx^2\plus{}mx\minus{}2x\plus{}1$
We pair up the $ x^2$ terms: $ 7\equal{}\minus{}2m\minus{}1\rightarrow m\equal{}\minus{}4$
Or also work with $ x$ terms: $ \minus{}6\equal{}m\minus{}2 \rightarrow m\equal{}\minus{}4$[/hide]

[quote="Poincare"][quote="RisingFlame"][hide="1"]$ P(x) \equal{} (x \minus{} 6)G(x) \plus{} R$
$ P(3x^2 \minus{} 5x \minus{} 2) \equal{} P((3x \plus{} 1)(x \minus{} 2)) \equal{} \frac {(3x \plus{} 1)(x \minus{} 2)}{3x \plus{} 1} \equal{} x \minus{} 2$
$ x \minus{} 2 \equal{} (x \minus{} 6)G(x) \plus{} R$
Substituting $ x \equal{} 6$ gives $ R \equal{} 6 \minus{} 2 \equal{} \boxed{4}$.[/hide][/quote]

You could just use synthetic division and the remainder theorem...[/quote]

...which is essentially what I did.


Just curious, does anyone know who the National Scholarship winners were for 2008? I heard Alex got the 25k but declined, and 2 ppl from my school got 14k but I don't know any beyond that.

so where did these scholarship winners go instead?

I went to a dual degree program at the University of Pennsylvania (Huntsman Program in International Studies and Business), which comprises a liberal arts degree and a degree from the Wharton School in 4 years. I was also accepted to Princeton but decided to take the business route at Wharton.

I felt bad rejecting the National Scholarship, since it is a great chance to study at a good university (they offered me a spot in the Honours Business/Mathematics program or something like that) and hope that somebody else benefited from it.


Let $ a;b;c \ge\ 0$, $ c \equal{} max$ or $ min$ {$ a;b;c$}. Find condition of $ M,N$ such that :

$ M.a^2\plus{}M.b^2\plus{}N.c^2 \ge\ (2M\minus{}N)ab\plus{}N.ac\plus{}N.bc$

[quote="onlylove_math"][quote="arqady"][quote="onlylove_math"]Let $ a;b;c \ge\ 0$, $ c \equal{} max$ or $ min$ {$ a;b;c$}. Find condition of $ M,N$ such that :

$ M.a^2 \plus{} M.b^2 \plus{} N.c^2 \ge\ (2M \minus{} N)ab \plus{} N.ac \plus{} N.bc$[/quote]
If $ 0\leq N\leq4M$ then your inequality is true for all $ a,$ $ b$ and $ c.$[/quote]
Thanks you but $ M ,N \ge\ 0$ my inequality is obvious  :)[/quote]
Yes,because you are waiting a condition of SS technique
Your ineq becomes $ M(a\minus{}b)^2\plus{}N(a\minus{}c)b\minus{}c)\ge0$ and if $ M,N\ge 0$ it is true
I think it is still an open question :lol:


Let us have a theory, "All monkeys like bananas". We wish to prove this theory.

Let us assume it is not provable and false. But if it is false, then there must be a counterexample. Then we use this counterexample to prove that it is false. Therefore it is provable.

Let us assume it is not provable. Therefore it must be true. However, now we have a proof that it is true, and thus it is provable.

Therefore it must be provable no matter if it is false or not.

Does this work?

No: a universally quantified statement can be false, true and provable or true and unprovable.  Your framework doesn't distinguish between the latter two because it doesn't distinguish between "there (provably) is no counterexample" and "I haven't laid hands on a counterexample" and "there (unprovably) is no counterexample."

It is true that "not provable" implies "true"; this does not constitute a proof of any particular statement whose truth-value is in question.


geometry
geometric transformation
rotation
linear algebra
matrix
topology
rotation matrix

Prove that the no. of independent qt.s in the rotation matrix from a certain orthornmal basis to other in n dimensional space is $ \frac{n(n\minus{}1)}{2}$

$ A\in{O(n)}$ iff $ \phi(A)\equal{}0$ where $ \phi: A\in\mathcal{M}_n(\mathbb{R})\rightarrow{A^TA\minus{}I_n}$.
$ d\phi_A: H\in\mathcal{M}_n(\mathbb{R})\rightarrow{H^TA\plus{}A^TH}\in\mathcal{M}_n(\mathbb{R})$.
$ ker(d\phi_A)\equal{}A(SS_n)$ where $ SS_n$ is the set of skew-symmetric matrices.
Thus $ O(n)$ is a manifold s.t. $ dim(O(n))\equal{}dim(SS_n)\equal{}n(n\minus{}1)/2$.
$ SO(n)$ is a connected component of $ O(n)$; thus its dimension is the same.


inequalities
trigonometry
inequalities proposed

[quote="Vasc"]Let $a,b,c$ be non-negative numbers.
(a) If $a+b+c=3$, then
$ \frac{a}{a+bc}+ \frac{b}{b+ca}+  \frac{c}{c+ab} \geq \frac{3}{2}$;
(b) If $ab+bc+ca=3$, then
$ \frac{1}{1+a^2}+ \frac{1}{1+b^2}+  \frac{1}{1+c^2} \geq \frac{3}{2}$.[/quote]

 :rotfl:  :rotfl:  :rotfl:
For b). Assume $x=ab$, $y=bc$ and $z=ca$ to obtain a).

For b, is this proof ok?
WLOG, let c>=b>=a, then
bc>=(ab+bc+ca)/3=1
Then,
\[
\frac{1}{{1 + b^2 }} + \frac{1}{{1 + c^2 }} \ge \frac{2}{{1 + bc}}
\]
because it is equivalent to
\[
(b - c)^2 (bc - 1) \ge 0
\]
Then,
\[
\begin{array}{l}
 \frac{1}{{1 + a^2 }} + \frac{2}{{1 + bc}} \ge \frac{3}{2} \\
  \Leftrightarrow 2(3 + bc + 2a^2 ) \ge 3(1 + a^2 )(1 + bc) \\
  \Leftrightarrow a^2  + ab + ac \ge 3a^2 bc \\
  \Leftrightarrow \frac{{a + b + c}}{3} \ge abc \\
 \end{array}
\]
This is true because
\[
\frac{{a + b + c}}{3} \ge \sqrt[3]{{abc}} = \frac{{ab + bc + ca}}{3}\sqrt[3]{{abc}} \ge abc
\]
I hope it is correct, because i have never proved any inequality proposed by you..  :?

Congratulations, siuhochung for your very nice solution for this very nice ineq :)

vasc, can you kindly show us your proof? :D

[quote="Vasc"]I used Schur's inequality to show that

$(\sum {ab})^2-3abc \sum {a} \ge abc(3+3abc-2\sum{ab})$.

But your solution is nicer.[/quote]

Thanks for your solution!
Moreover I think hungkhtn's method or your method's better than mine because my solution came out by luck.. and you two's methods are more general and can be applied in other cases. Anyway, thank you very much for your solutio nagain!

Siuhochung, don't diminish your great proof.

You have been really fantastic  :)  :)  :)

and remember that luck help genial minds ;)

After such beautiful proofs, my solution will probably pass unobserved, but here is a nice fact: the problem is equivalent to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3171]the well-discussed problem[/url] $ \sum{\frac{m_{a}\cdot m_{b}}{ab}\geq\frac{9}{4}}$ in any triangle. I found this very nice and interesting.

This Harazi brilliant solution is very unexpected for me. :?

So this nice inequality has  well four  interesting solution. ;)

That is all,.......for now!! :)

Harazi, can you please explain to me how to transform the original inequality into the equivalent form

$ \sum{\frac{m_{a}\cdot m_{b}}{ab}\geq\frac{9}{4}}$

Thank you very much.

Ok, excuse me for the very late answer, I simply forgot about this problem. Suppose the inequality is not true. Then you can find $ u,v,w$ such that $ u^2+v^2+w^2=3$ and $k<1$ such that $\frac{1}{1+a^2}=\frac{ku^2}{2}$ and so on. Substitute the values you found for $a,b,c$ in the given condition and you will find that $\sum{\frac{\sqrt{(2-u^2)(2-v^2)}}{uv}}<3$. Now, apply the inequality with the medians in the triangle with sides $ u,v,w$.

Why does Harazi say that $u,v,w$ are sides of a triangle?

Or better how to prove that

$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{1}{\sqrt{1+c^2}}$

for $ab+bc+ca=3$

I am not sure if this one can be true :?

Thank you.

[quote="Vasc"]I used Schur's inequality to show that

$(\sum {ab})^2-3abc \sum {a} \ge abc(3+3abc-2\sum{ab})$.

But your solution is nicer.[/quote]
  Dear Vasc , How can you apply Schur's inequalities for this ineq ?

It suffices to show that from Schur's ineq. we get Cauchy's ineq.  :P
Indeed, adding Shur's ineq
$x^3+y^3+z^3+3xyz \ge xy(x+y)+yz(y+z+zx(z+x)$
and
$x(y-z)^2+y(z-x)^2+z(x-y)^2 \ge 0$,
we get
$x^3+y^3+z^3 \ge 3xyz$,
which is just Cauchy.

Here is my view on how to prove Vasc's inequalities from post #1. I will only verify the inequality [b](a)[/b], since inequality [b](b)[/b] is equivalent to [b](a)[/b], as Myth showed in post #2.

With the notation $\sum$ for cyclic sums, the inequality [b](a)[/b] becomes $\sum\frac{a}{a+bc}\geq\frac32$. On the other hand, the condition a + b + c = 3 yields $\frac{a+b+c}{3}=1$, and thus

$\frac{a}{a+bc}=\frac{a\cdot\frac{a+b+c}{3}}{a\cdot\frac{a+b+c}{3}+bc}=\frac{a\left(a+b+c\right)}{a\left(a+b+c\right)+3bc\right)}$
$=\frac{2ca+2ab-bc}{2\left(bc+ca+ab\right)}+\frac{3bc\left(a-b\right)\left(a-c\right)}{2\left(bc+ca+ab\right)\left(a\left(a+b+c\right)+3bc\right)}$.

Hence, the inequality $\sum\frac{a}{a+bc}\geq\frac32$ is equivalent to

$\sum\left(\frac{2ca+2ab-bc}{2\left(bc+ca+ab\right)}+\frac{3bc\left(a-b\right)\left(a-c\right)}{2\left(bc+ca+ab\right)\left(a\left(a+b+c\right)+3bc\right)}\right)\geq\frac32$.

But

$\sum\left(\frac{2ca+2ab-bc}{2\left(bc+ca+ab\right)}+\frac{3bc\left(a-b\right)\left(a-c\right)}{2\left(bc+ca+ab\right)\left(a\left(a+b+c\right)+3bc\right)}\right)$
$=\sum\frac{2ca+2ab-bc}{2\left(bc+ca+ab\right)}+\sum\frac{3bc\left(a-b\right)\left(a-c\right)}{2\left(bc+ca+ab\right)\left(a\left(a+b+c\right)+3bc\right)}$
$=\frac{3\left(bc+ca+ab\right)}{2\left(bc+ca+ab\right)}+\frac{3abc}{2\left(bc+ca+ab\right)}\cdot\sum\frac{\left(a-b\right)\left(a-c\right)}{a\left(a\left(a+b+c\right)+3bc\right)\right)}$
$=\frac32+\frac{3abc}{2\left(bc+ca+ab\right)}\cdot\sum\frac{\left(a-b\right)\left(a-c\right)}{a^2\left(a+b+c\right)+3abc\right)}$.

Hence, if we succeed to show that $\sum\frac{\left(a-b\right)\left(a-c\right)}{a^2\left(a+b+c\right)+3abc\right)}\geq 0$, then we are done. But this is easy: WLOG we can assume that $a\geq b\geq c$, and then we obviously have $a^2\left(a+b+c\right)+3abc\geq b^2\left(a+b+c\right)+3abc\geq c^2\left(a+b+c\right)+3abc$, and thus $\frac{1}{a^2\left(a+b+c\right)+3abc}\leq\frac{1}{b^2\left(a+b+c\right)+3abc}\leq\frac{1}{c^2\left(a+b+c\right)+3abc}$, and thus, by the Vornicu-Schur inequality ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212]http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9[/url] Lemma 1), we get $\sum\frac{1}{a^2\left(a+b+c\right)+3abc}\left(a-b\right)\left(a-c\right)\geq 0$, what is equivalent to $\sum\frac{\left(a-b\right)\left(a-c\right)}{a^2\left(a+b+c\right)+3abc\right)}\geq 0$. And the inequality is proven.

  Darij

(Olympic Revenge 2013)Let $a,b,c,d$ to be non negative real numbers satisfying $ab+ac+ad+bc+bd+cd=6$. Prove that\[\dfrac{1}{a^2+1} + \dfrac{1}{b^2+1} + \dfrac{1}{c^2+1} + \dfrac{1}{d^2+1} \ge 2\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2916452
Let $a,b,c$ to be non negative real numbers satisfying $bc+ca+ab=3$. Prove that\[\dfrac{3}{2}\leq\dfrac{1}{a^2+1} + \dfrac{1}{b^2+1} + \dfrac{1}{c^2+1} <2.\]
\[\dfrac{1}{a^2+1} + \dfrac{1}{b^2+1} + \dfrac{1}{c^2+1}  \ge\dfrac{3}{ 2}\Leftrightarrow \frac{1}{2-\cos A}+ \frac{1}{2-\cos B}+ \frac{1}{2-\cos C} \ge 2 \]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=518350


[1] Use z = a + bi to show that z(under two bars) = z (under no bar).

[2] Use z = a + bi and w = c + di to show that
z times w(both letters under a bar) = z(under a bar) times w(under a bar).

Thanks!

[hide="1"]  If $z=a+bi$, then $\overline{z}=a-bi$. It should be fairly obvious to prove from this.  :) Two $-$s make a $+$...
[/hide]
[hide="2"] $zw=(ac+(ad+bc)i-bd)=(ac-bd)+(ad+bc)i$. Hence $\overline{zw}=(ac-bd)-(ad+bc)i$.

$\overline{z}=a-bi$ and $\overline{w}=c-di$
Hence $\overline{z}\cdot \overline{w}= (a-bi)(c-di)=(ac-bd)-(ad+bc)i=\overline{zw}$.[/hide]

Thanks but the prove is not obvious as you said.

um...
what isn't obvious about it?  the + turns into a - with the first bar and back to a + with the second.

So, the bar makes all the difference, right?

[quote="ckck"][quote="13375P34K43V312"]For the millionth time, can you use $\LaTeX$?

$\overline{z}$
$\overline{\overline{z}}$[/quote]
But what do the bars mean?[/quote]
The bar of something means the conjugate of something.  The conjugate of something means that if the conjugate was added or multiplied to the original number, the product and sum are real.  For example: If $z=a+bi, \overline{z}=a-bi$ because $(a-bi)(a+bi)=a^{2}-(b^{2})(i^{2})=a^{2}+b^{2}$, which is real, we also have that a+bi+a-bi=2a, which is real.