algebra
polynomial
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

the basic tool is your brain ;). this is just an exercise to practise the definitions, and you should do this by yourself. anyway, some hints:

1) permute the canonical k-basis.
2) map y to 0.
3) every choice works ... (besides the identity ring hom.)

[quote="Lippilus"]I've just started this topic so any help is appreciated:
We can look at $ k[x,y]$ as if it were a $ k$-module, $ k[x]$-module, $ k[y]$ or $ k[x,y]$-module. Construct the maps(non-trivial) $ k[x,y]\to k[x,y]$ which are:
1)k-module homomorphism but not $ k[x]$, $ k[y]$ or $ k[x,y]$-homomorphism
2)$ k[x]$-module homomorphism but not $ k[y], \ k[x,y]$ m.h.
3)a ring homomorphism but not a $ k[x,y]$ m.h.
What are the basic tools for this kind of problems?[/quote]

Hello.
I take the risk of an answer, but I'm not familiar with such problems.
The primitive monomials $ x^{i}y^{j}$ form a basis of the $ k$-module $ k[x,y]$, thus a $ k$-homomorphism of $ k[x,y]$ into itself is freely defined by its values on the primitive monomials : for each  primitive monomials $ x^{i}y^{j}$, choose a polynomial $ f{_i,j}(x,y)$; to this family of choices corresponds the $ k$-homomrphism which applies $ \sum a_{i,j}x^{i}y^{j}$ on $ \sum a_{i,j}f_{i,j}(x,y)$.
Such a homomrphism is a $ k[x]$-homomrphism if and only its (free) value at $ x^{i}$ is $ x^{i}$ for each $ i$; it is a $ k[y]$-homomrphism if and only if its value at $ y^{j}$ is $ y^{j}$ for each $ j$; it is a  $ k[x,y]$-homomorphism if and only it is the identity.
I think this gives the answers to questions 1) and 2).
For 3), choose two polynomials $ f(x,y)$ and $ g(x,y)$, with either $ f(x,y) \not \equal{} x$ or $ g(x,y) \not \equal{} y$. Apply $ \sum a_{i,j}x^{i}y^{j}$ on $ \sum a_{i,j}f(x,y)^{i}g(x,y)^{j}$. This defines a ring homomorphism which is not the identity, and thus not a $ k[x,y]$-module homomorphism.
But I think there can be other ring homomorphisms which are not $ k[x,y]$-module homorphisms, if one doesn't specify that one considers only $ k$-module homomorphisms.
For example, if there is a ring-homomorphism $ h$ of $ k$ into itself which is different from identity, choose two polynomials $ f(x,y)$ and $ g(x,y)$ and apply $ \sum a_{i,j}x^{i}y^{j}$ on $ \sum h(a_{i,j})f(x,y)^{i}g(x,y)^{j}$. If I'm not wrong, it is a ring homomorphism different from identity, thus not a $ k[x,y]$-module homomorphism.
More generally, this remains true if $ h$ is a ring-homomorphism of $ k$ into $ k[x,y]$ (instead of a ring-homorphism of $ k$ into itself) which doesn't fix all elements of $ k$ and I think this completes the general solution. But what are the ring-homorphisms of $ k$ into $ k[x,y]$ ?
This latter case is a little delicate, thus.

Hoping that I didn't say stupidities,
Friendly,
P.

P.S. I had not seen the (more concise) answer of -oo-.


[quote="JBL"]Why would you assume this has a nice value?  http://www.research.att.com/~njas/sequences/A065442[/quote]

I thought I had narrowed down a different problem to computing this sum, but it seems that I made an error solving the problem.

[quote="i_like_pie"][quote="13375P34K43V312"]Compute $ \sum_{k \equal{} 1}^{\infty}\frac{1}{2^{k}\minus{}1}$.[/quote]
There's an infinite amount of Mersenne primes...?[/quote]

Uh, wtf? First of all, that isn't proven, and second of all, that doesn't tell us that the sum diverges, there are infinitely many powers of two but that sum obvoiusly converges.

Well obviously

$ 0<\frac{1}{2^{k}\minus{}1}\le\frac{1}{2^{k\minus{}1}}$

for $ k\equal{}1,2,...$

so the sum converges by the squeeze theorem.

[quote="13375P34K43V312"]I thought I had narrowed down a different problem to computing this sum, but it seems that I made an error solving the problem.[/quote]

What was the original problem?


[quote="zibi"]Find all $n$, for which:

\[ n^{n}+1 \ \ \ and \ \ \ (2n)^{2n}+1  \]

are prime numbers[/quote]
Let's use this easy lemma.
[b]Lemma.[/b]If $m\geq 3$ and $a^m+1$ is prime numbre,wher $a>1\in{N}$,then $m=2^k$.
Proof.If the lemma is not true then $m=2^tq$ wher $q$ is odd numbre graten 1.=>
$a^m+1=(a^{2^t})^q+1>a^{2^t}+1$ and $(a^{2^t}+1)|(a^m+1)$contradiqtion.
If $n\geq 3$
From lemma we have that $n=2^k$=>$n^n+1=2^{k2^k}+1$=>$k=2^s$.
Also  $(2n)^{2n}+1$ is prome =>$2^{(k+1)2^{k+1}}+1$is prime=>$k+1=2^r$
but one of numbers $k$ or $k+1$ is odd,contradiction.
=>$n\leq 2$=>$n=1$
Answer:$n=1$  ;)

[quote="Tiks"]
$a^m+1=(a^{2^t})^q+1>a^{2^t}+1$ and $(a^{2^t}+1)|(a^m+1)$contradiqtion.
[/quote]

Sorry, I don't see how this is a contradiction, could you explain?   :(

$a^m+1$ is supposed to  be prime  hence  cannot  hav a divisor besides  itself and  1 contradiction!!! :roll:

[quote="zibi"]Answer : $n=1$ , $n=2$  ;)  Tiks, you omitted $k=1$ but still your proof is correct[/quote]
You rigth,sorry :)


A natural number is said to be prefectly descending if in its decimal representation all the digits from the left to right are in the decresing order.
For eg. 541 and 98761 are such prefectly descending nos.
Find all such natural nos.
PS This is a very simple problem.

Is this your own problem? Thats good progress for a 8th Grader.  :)

Anyway, let me assume, I choose 'k' unique digits. Then using them I can get only one such number which is perfectly descending.

Answer is $ 2^{10} \minus{} 1$ (The -1 is for the zero)

Edit: This also includes the 1 digit no's 1-9. If you don't want them subtract 9. :D


How do I do an equation array on the Wiki? It doesn't seem to recognize

\begin{eqnarray*}
\end{eqnarray*}

Thanks in advance,
4everwise

Currently, there are aspects of LaTeX not supported by MediaWiki. Hopefully somebody in the office will be able to install a new package or create a workaround at some point.

There are two ways you can do the same thing as eqnarray.  One is by using a matrix:

<math> \begin{matrix} (x+1)^3 &=& (x+1)(x^2+2x+1) \\ &=& x^3 + 3x^2 + 3x + 1 \end{matrix} </math>

Or using a table:

{|
|$(x+1)^{3}$
|$=(x+1)(x^{2}+2x+1)$
|-
|
|$=x^{3}+3x^{2}+3x+1$
|}

The table has an advantage in that it allows for much more flexibility.  For example:

{| style="width:400px; height:100px"
|$(x+1)^{3}$
|$=(x+1)(x^{2}+2x+1)$
|-
|
|$=x^{3}+3x^{2}+3x+1$
|}

Look at my [url=http://www.artofproblemsolving.com/Wiki/index.php/User:Joml88/sandbox]sandbox[/url] to see what these look like.


[hide]1/4*x = 64
solve for x
multipy both sides by 4
you get
x = $256[/hide]

[hide][quote="236factorial"]Linda paid 64 dollars for a jacket that was on sale for 25% of the original price. What was the original price of the jacket?[/quote]

$\text{25 percent of the original price means:}$

$\frac{64}{\frac{1}{4}}$

$ = 64 \times 4$

$ = 256$

$\boxed{\boxed{\boxed{\text{The original cost was 256 dollars. }}}}$[/hide]


geometry
circumcircle
trigonometry
projective geometry
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle,
CB'B"A, AC'C"B, BA'A"C three squares erected externally on the sides of ABC,
and L, M, N respectively the points of intersection of A'B" and A"C', B'C" and B"A', C'A" and C"B'.
Prove that LMN is perspective with ABC.
Sincerely
Jean-Louis

Let $ P_a \equiv  B'C'' \cap BC$ and  define $ P_b$ and $ P_c$ cyclically. Lines $ C''B$ and $ B'C$ meet at the antipode $ M$ of $ A$ WRT the circumcircle of $\triangle ABC.$ Hence, $ MB \equal{} 2R \cos C$ and $ MC \equal{} 2R \cos B.$ By Menelaus' theorem for $\triangle MB'C'',$ we get

$ \frac {P_aC}{P_aB} \cdot \frac{BC''}{MC''} \cdot \frac{MB'}{CB'} \equal{} 1 \Longrightarrow \  \frac {P_aC}{P_aB} \equal{} \frac {CA}{AB} \cdot \frac{ AB \plus{} 2R \cos C}{ CA \plus{} 2R \cos B} \ \ (1)$

By similar reasoning, we get the expressions

$ \frac {P_cA}{P_cB} \equal{} \frac {CA}{BC} \cdot \frac{ BC \plus{} 2R \cos A}{CA \plus{} 2R \cos B} \ \ (2)$

$ \frac {P_bC}{P_bA} \equal{} \frac {BC}{AB} \cdot \frac{AB \plus{} 2R \cos C}{BC \plus{} 2R \cos A} \ \ (3)$

Now, multiplying the expressions $(1),(2)$ and $(3)$ together yields

$ \frac {P_aC}{P_aB} \cdot \frac{P_cB}{P_cA} \cdot \frac{P_bA}{P_bC} \equal{} 1 \Longrightarrow   P_a, P_b,P_c$ are collinear and the conclusion follows.


He started with more dollars, spent more pennies than dollars, and ended up with less dollars than pennies? Wait, I must be misunderstanding something.  :huh:

Not sure about this one.

But don't you find out how much 150% is of 99 pennies by timing

1.5 * 99 = 148,5 pennies

add those to the 99

148,5 + 99 = 247,5 pennies
--------------------------------------

I have no idea if its correct since i don't know the money system of the US

1 dollar is 100 cents right??

Is 1cent  a 100pennies??

[quote="math92"]Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies. When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had. After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had. With how much money did Sam leave home that day?[/quote]

Kuresu, cents and pennies are the same thing. :)

Note: This solution is incorrect, because the equations don't end up working together. But can anybody tell me what I did wrong? Thanks.
[hide]
D=# of dollars

P=# of pennies

leaving home: he had D dollars and P pennies, and D=1.5P

spent: 10 dollars, 78 pennies

after spending: he had D-10 dollars and P-78 pennies, and D=P-1. Therefore, by substitution, D-10=P-78-1, or D-10=P-79.

Now, we have two equations. D=1.5P and D-10=P-79.

Substitute D=1.5P into D-10=P-79, and we get 1.5P-10=P-79. Then, 0.5P=-69.

This is where I get messed up. What did I do wrong? :?
[/hide]

The problem is impossible if he spent 10.78 with 10 dollars and 78 pennies.  If he had at least 78 pennies, that means he had at least 117 dollars (150% of 78). Spending 10 of those means that he has at least $107 left, which can't be one less than his number of pennies at any time because he doesn't carry more than 99.


The symbol $|a|$ means $a$ if $a$ is a positive number or zero, and $-a$ if $a$ is a negative number.  For all real values of $t$ the expression $\sqrt{t^4+t^2}$ is equal to:
(A) $t^2$
(B) $t^2+t$
(C) $|t^2+t|$
(D) $t\sqrt{t^2+1}$
(E) $|t|\sqrt{1+t^2}$

[hide]We can factor this out into:

$\sqrt{t^2(t^2+1)}$
$\sqrt{t^2}\sqrt{t^2+1}$
$|t|\sqrt{1+t^2}$

The absolute value sign comes in because anything squared is always positive, and the sqrt function returns the postive root. The answer is E.[/hide]

hi ... first post
I am looking to do well in upcoming math competitions, but i wish to start from the bottom.  Usually the competitions i enter are all multiple choice anyway...no essays.

[hide]factor to [tex](x^2)(x^2+1)[/tex], then take the positive square root of  [tex]x^2[/tex] to get [tex]|x|\sqrt{x^2+1}[/tex][/hide]

btw for some reason it said i was logged in as siavash_z when i posted this first...might want to check on that mods

Jeremy - that login bug should be fixed now (please let us know if it isn't).

[quote="Silverfalcon"]This is pretty good AHSME question, Mr. Crawford.

Then derivation from this question:

Using the concept above, what does $\sqrt {t^7+t^6}$ equal to? How does this similar or different from above AHSME? Explain.

P.S. For those who would walk away from this because of word "explain," short sentence is fine.[/quote]

$\sqrt {t^7+t^6} = \sqrt {t^6(t+1)} = \sqrt{t^6}.\sqrt{t+1} = |t^3|.\sqrt {t+1}$

I'm not sure how it differs.. the modulus function still applies because $t^6$ is still always positive


Putnam inequalities geometry MIT college Princeton calculus \/closed

Come on, you know you guys want to make one for next year.  Mostly because I'm going to college next year, and your classes are wonderful.

The Olympiad Problem Solving class is packed with Putnam problems and problems appropriate for Putnam preparation.

Our other Olympiad level classes would also be helpful with the Putnam.  There are parts of the Olympiad Inequalities class that would help with the Putnam.  We also expect to eventually offer Olympiad classes in number theory, counting/probability, algebra, and possibly other areas.

Sorry to bring up an old topic, but I was wondering if you have changed your mind about offering a Putnam class (or something equivalent to university level competitions) sometime in the near future, especially because the individual olympiad classes are no longer offered?  Thanks.

The WOOT class would certainly qualify as Putnam training, unless you're a student who can already get half the problems on a typical exam.  While not all of the topics taught are typical of the Putnam (the Putnam has very little geometry), a great deal of them will be helpful and the focus on proof-writing certainly transfers well to the Putnam.

Well, I took the WOOT class last year, and I agree the problem solving and proof writing would help a lot on the Putnam, but I was hoping you might consider offering a class that focused on some of the more advanced topics such as analysis and abstract algebra, but I guess not.  Thanks anyway.

I was curious if you could start up those old subject math jams. I found many of them tremendously helpful, and possibly they could fufill the gap, if only a little bit, of the people who dont have acess to top notch math clubs like MIT, Princeton,etc. For example many people have been exposed to convergence tests, and typical power series manipulation in calc BC, but there are few books that introduce ingenuity and problem solving into these ideas like you guys do with everything else.

In general, we run so many classes and math jams now that these kinds of Putnam topic math jams will be very rare.

Perhaps you can talk some of the college students on this site with ample Putnam experience to host a math jam or two along these lines.  I will set it up in the database if you can find a willing teacher.

If you're looking for a good problem-solving resource on Calculus, use Paul Zeitz's Art and Craft of Problem Solving, which does have a section on Calculus and follows the same method of teaching as AoPS does. It may not necessarily be a vast, deep resource for more advanced mathematics, but it certainly is something  :wink:


let a,b,c,d be nonzero integers such that a is different from c, iff:

a/c=(a 2 +b 2 )/(c 2 +d 2 )

prove that a 2 +b 2 +c 2  is not a prime!

There's something that seems unclear to me: did you mean "if" instead of "iff", which can also be interpreted (and usually is) as "if and only if" ?

Sorry, i mean that  if a,b,c,d satisfy condition, then result.


$ M$ is an interior point on the side $ AC$ of $ \triangle ABC$. The incircle of $ \triangle ABM$ touches $ AC$ at $ T_1$ and its center is $ O_1$. The incircle of $ \triangle ACM$ touches $ AC$ at $ T_2$ and its center is $ O_2$. $ O_1O_2 \cap AM \equal{} {N}$. If $ |T_1T2| \equal{} 2|NM|$, then find $ \frac{|O_1T_1|}{|O_2T_2|}$.

Not $ \triangle ACM$. It must be $ \triangle BCM$. And it is not $ AM$, it is $ BM$. Sorry for the typo.

The correct form is:

$ M$ is an interior point on the side $ AC$ of $ \triangle ABC$. The incircle of $ \triangle ABM$ touches $ AC$ at $ T_1$ and its center is $ O_1$. The incircle of $ \triangle BCM$ touches $ AC$ at $ T_2$ and its center is $ O_2$. $ O_1O_2 \cap BM \equal{} \{N\}$. If $ |T_1T2| \equal{} 2|NM|$, then find $ \frac {|O_1T_1|}{|O_2T_2|}$.

[hide]Let $ D_1$ and $ D_2$ be the points where incircles touch $ BM$. Let $ T_1M \equal{} a$ and $ T_2M \equal{} b$, then $ NM \equal{} \frac {a \plus{} b}{2}$. $ \frac {ND_1}{ND_2} \equal{} \frac {\frac {|a \minus{} b|}2}{\frac {|a \minus{} b|}2} \equal{} 1$. $ \frac {|O_{1}T_{1}|}{|O_{2}T_{2}|} \equal{} 1$[/hide]


Will be there a class specifically on Trig?

I notice that there tend to be a lot of trig problems on most exams...

For trig, just make sure you know all of the identities and the problems usually just reduce to algebra (at least on AMC/AIME-level questions).

Yes, we expect to build a course on complex numbers and trigonometry this summer.   In addition to covering the standard trig/complex number problems, we'll be highlighting many non-obvious applications as well.

[quote="MithsApprentice"]Does that mean it'll be run next fall/winter?[/quote]

That will probably be the case, though we don't set the schedule that far in advance -- particularly with classes that we have not written yet.


advanced fields
advanced fields unsolved

Show that if S is convex then it ,bounded and open in R then S is an interval (a,b)

Let $a=inf S$, $b=sup S$. Because $S$ is bounded, $a,b$ are reals.
Because $S$ is open, $a,b$ are not in $S$.
Take $t\in (a,b)$.
Take $S\ni a_n\to a$, $S\ni b_n\to b$. There exists $m\in\mathbb{N}$ such that $a<a_m<t<b_m<b$. Because $S$ is convex $(a_m,b_m\in S) \Rightarrow t\in (a_m,b_m)\subset S$.

Hi,

Thank you for your answer but still two questions:

1- Are am and an are different

2-How do you prove there exist such a 't' between an and bn

Thanks

$a_n\to a$ means that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to $a$.
$a_m$ is a term of that sequence.

I take $t\in (a,b)$ arbitrarly. Such a $t$ exists because $(a,b)$ is not void because $a\neq b$.
Because $a_n\to a<t$, there exists $m_1\in\mathbb{N}$ such that $\forall n\geq m$, $a<a_n<t$. Obtain similarly a $m_2$ for $(b_n)_{n\in\mathbb{N}}$ and take $m=\max (m_1,m_2)$.

Any more questions?


pigeonhole principle
inequalities unsolved
inequalities

$ a,b,c$ are positive real numbers,prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 1\ge2(ab \plus{} ac \plus{} bc)$

In Romania, in Valcea, we have a contest that`s called "Mathematica-Modus Vivendi". In 2006, I think, that problem was proposed by Darij Grinberg for the VII grade.
The proof I know it`s using the pigeonhole principle... . It`s a very nice problem... :ninja:


Consider a set $ A$ all element of $ A$ are positive number The minimum element is number $ 1$ and the maximum element is number $ 100$  .Each element $ \in$ $ A$ (except number $ 1$) is sum of two number (they can same value) $ \in A$.
 Determine the set $ A$ such that $ |A|$ is minimum.

[color=red]Edited by TUAN.[/color]

I think it should be A={100,50,25,13,12,6,3,2,1}......
and 9 should be minimum

first 100 should be sum of two number which belongs to A
and of course it should have a number >=50
then should have a number>=25
then>=13 then >=7 then >=4 then >=2 and then 1
so it should at least has 8 elements
but {100,50,25,13,7,4,2,1} did't satisfy the request.
so it should has at least 9 elements.....
I think didn't prove it so clearly.....
let someone else try it again.....

and it is the first time I visit here and I want to make friends with you all
I'm a Chinese student and I'm sorry my English is a little poor....

Let's arrange all the elements of $ A$ so that $ a_{1}< a_{2}< ... < a_{n}$. We also have $ a_{1}\equal{} 1$, $ a_{n}\equal{}100$. We are trying to find the smallest $ n$ possible.

It's clear that $ 2a_{i}\geq  a_{i\plus{}1}$

By applying the lemma, we get:

$ 100\leq  2a_{n\minus{}1}\leq  4a_{n\minus{}2}\leq  ...\leq  2^{n\minus{}1}$

From the recursion we read off the following estimation: $ n\geq  8$.

Now we easily get $ a_{7}\equal{} 50, a_{6}\equal{} 25$. Because of $ a_{5}\plus{}a_{4}< 25$, $ a_{6}$ is not the sum of two numbers from $ A$. Therefore $ n > 8$.

For $ n\equal{}9$ there are various systems of solution.


[quote="tigergt"]Find the value of 1^(1/5) + 2^(1/5) + 3^(1/5) + ... + 2009^(1/5).[/quote]I can bet you you can't even find $ \sum_{k \equal{} 1}^{2009}\sqrt {k}$. There is no elementary form to the answer of this question. It becomes a heavy-computation problem (non-Olympiad type).

However, $ \sum_{k\equal{}1}^{2009} \left \lfloor \sqrt[5]{k} \right \rfloor$ is quite easy; maybe he meant this!

$ \sum_{k\equal{}1}^{2009} \left \lfloor \sqrt[5]{k} \right \rfloor \equal{} \sum_{k\equal{}1}^{31} \left \lfloor \sqrt[5]{k} \right \rfloor \plus{} \sum_{k\equal{}32}^{242} \left \lfloor \sqrt[5]{k} \right \rfloor \plus{} \sum_{k\equal{}243}^{1023} \left \lfloor \sqrt[5]{k} \right \rfloor \plus{} \sum_{k\equal{}1024}^{2009} \left \lfloor \sqrt[5]{k} \right \rfloor \equal{}$
$ 31\cdot 1 \plus{} 211\cdot 2 \plus{} 781\cdot 3 \plus{} 986\cdot 4 \equal{} 6740$, if that is what you mean.


For those who are looking for materials and lectures given in Berkeley math circle i just found a link that contains several lectures in html,pdf,ps and dvi formats .I hope you'll find it interesting :):
http://mathcircle.berkeley.edu/BMC3/

There are some nice articles. Thanks for the link  :)


Hi Guys,
Where do we find USABO resources? I tried googling it but nothing good came up.

CEE has been poor about giving out resources for USABO. Have they even released a single semifinal exam to those who didn't take it? Really, the best access to resources is through your teacher, if he or she has accrued exams from the past several years. (I don't have the time to scan the hundreds of pages of semis right now, unfortunately.)

I once found a set of old USABO practice questions (10 sets for a total of 120 questions or so). PM me if you can't find that on Google (advanced search for USABO, filetype:PDF).

The official IBO guide on the IBO page has the list of phyla you need to know. It also has the syllabus, which shows the approximate percentages of each subject material on the exam. USABO generally follows this syllabus, although it has not been so good about following the percentages.

Other than that, your best bets are always the books listed on the USABO site: Campbell, Alberts, Lodish, Raven, etc. And then, there's always MIT OCW. You probably don't have time to go through all of that now, but in the future, keep that in mind.

Good luck with your cramming.


A shirt regularly priced at 30 dollars is on sale for 20% off. Mary has a coupon that will take another 25% off of the sale price. What single percent discount would give the same final price as the two successive discounts?

[hide] since Mary has a $ 25\%$ coupon, she will pay $ 30(.75)\equal{}22.5$. Without the $ 20\%$ discount, the shirt would cost $ 30(1.2)\equal{}36$. Hence, $ 36(1\minus{}x)\equal{}22.5 \implies x\equal{}.375\equal{}\boxed{37.5\%}$.[/hide]

[hide="My answer is different  :| "]The discount in the problem gives the shirt as 18 dollars.

12/30 then 4/10 - 2/5 12 dollars off is [b]40% discount[/b][/hide]

You're right, I misread the problem.  :oops:


Αν $ a,b,c,d,e$ πραγματικοί αριθμοί τέτοιοι ώστε η εξίσωση $ ax^2\plus{}(b\plus{}c)x\plus{}(d\plus{}e)\equal{}0$ έχει πραγματικές ρίζες μεγαλύτερες του 1 να αποδείξετε ότι η εξίσωση $ ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e\equal{}0$ έχει τουλάχιστον μία πραγματική ρίζα.

νομίζω ότι με γενικευμένες vieta δν θα έχουμε πρόβλημα. Ή μήπως κάνω λάθος????

γράψε λάθος βγαίνει $ 100\%$ με απαγωγή σε άτοπο αλλά δν προλαβαίνω να βάλω τη λύση μου

Οπως υποσχέθηκα στον Ale3androsG βάζω τη λύση σε αυτήν την άσκηση η οποία παρεπιμπτόντως αν δεν κάνω λάθος είναι από την Ελλάδα το 1995.

Λοιπόν κατ' αρχάς υποθέτουμε, όπως είπε κι ο meinstein ότι το συμπέρασμα είναι λάθος (δηλαδή ότι το πολυώνυμο δεν έχει πραγματικές ρίζες. Τότε wlog $ Q(x)>0$ για κάθε χ, όπου $ Q(x)\equal{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}e$.

An $ P(x)\equal{}ax^2\plus{}(b\plus{}c)x\plus{}(d\plus{}e)$ tote

$ Q(x)\equal{}P(x^2)\plus{}(x\minus{}1)(bx^2\plus{}d)$  $ (1)$

Tώρα έστω $ y$ μια ρίζα του $ P$ και $ u\equal{}\sqrt{y}\iff u>1$ kai $ P(u^2)\equal{}0$.

An stin 1 valoume opou $ x$ to $ u$ kai to $ \minus{}u$ antistoixa exoume:

$ Q(u)\equal{}(u\minus{}1)(bu^2\plus{}d)>0$ (i anisotita apo ypo8esi) kai
$ Q(\minus{}u)\equal{}(\minus{}u\minus{}1)(bu^2\plus{}d)>0$ (i anisotita apo ypo8esi)

Ομως οι 2 τελευταίες δεν μπορεί να συμβαίνουν ταυτόχρονα αφού το $ u\minus{}1$, $ \minus{}u\minus{}1$ είναι αντίστοιχα θετικό και αρνητικό (αφού $ u>1$). Αρα φτάσαμε σε άτοπο! :)


A positive integer $k$ is given. The sequence $(a_{n})$ is defined by
         $a_{1}= k+1$
         $a_{n}_{+}_{1}= a_{n}^{2}-k.a_{n}+k$ for all $\large n \geq 1$.
Show that for $\large m \neq n$, the number $a_{m},a_{n}$ are relative prime.

Sorry about my poor English!

[hide="hint"]We have $\frac{a_{n+1}-k}{a_{n}-k}=a_{n}$, therefore $a_{n+1}-k=a_{1}a_{2}...a_{n}$. Now, problem is easy.[/hide]


Yes, I also think it's easy. ;)

By the well-known inequality $x^2+y^2\geq 2xy$, we have $a^2+b^2\geq 2ab$, so that $a\cdot\left(a^2+b^2\right)\geq a\cdot 2ab$. In other words, $a^3+ab^2\geq 2ba^2$, and similarly $b^3+bc^2\geq 2cb^2$ and $c^3+ca^2\geq 2ac^2$. Adding these three inequalities together, we arrive at

$a^3+b^3+c^3+ab^2+bc^2+ca^2\geq 2ba^2+2cb^2+2ac^2$.

On the other hand, by the Schur inequality,

$a^3+b^3+c^3+3abc\geq bc^2+cb^2+ca^2+ac^2+ab^2+ba^2$.

Adding both inequalities together, we get

$2a^3+2b^3+2c^3+ab^2+bc^2+ca^2+3abc\geq ab^2+bc^2+ca^2+3ba^2+3cb^2+3ac^2$.

Upon subtraction of $ab^2+bc^2+ca^2$, this simplifies to

$2a^3+2b^3+2c^3+3abc\geq 3ba^2+3cb^2+3ac^2$;

in other words,

$2\left(a^3+b^3+c^3\right)+3abc\geq 3\left(ba^2+cb^2+ac^2\right)$.

Divide this by abc to get

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

On the other hand, $3\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\geq\left(\sqrt{\frac{a}{c}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{c}{b}}\right)^2$, as an application of the well-known inequality $3\left(x^2+y^2+z^2\right)\geq\left(x+y+z\right)^2$ to the numbers $x=\sqrt{\frac{a}{c}}$, $y=\sqrt{\frac{b}{a}}$, $z=\sqrt{\frac{c}{b}}$. Thus,

$2\left(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\right)+3\geq 3\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\geq \left(\sqrt{\frac{a}{c}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{c}{b}}\right)^2$,

and we are done.

  darij

[quote="Maverick"]$a,b,c>0$
\[\displaystyle 2\left( \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\right) +3 \geq \left(\sqrt{\frac{a}{c}}+\sqrt{\frac{b}{a}}+\sqrt{\frac{c}{b}}\right)^2\][/quote]
Just my vision of the short solution... By weighted AM-GM we have $\frac{1}{3}\frac{a^2}{bc}+\frac{1}{6}\frac{b^2}{ac}+\frac{1}{2}\geq \sqrt{\frac{a}{c}}$ and $\frac{2}{3}\frac{a^2}{bc}+\frac{1}{3}\frac{b^2}{ac}\geq \frac{a}{c}$. Hence the result.


Give $ a,b,c$ be non-negative real numbers with sum 3.Prove that
$ \frac{ab\plus{}bc\plus{}ca}3\plus{}3\ge abc\plus{}a^2\frac{b\plus{}c}2\plus{}b^2\frac{c\plus{}a}2\plus{}c^2\frac{a\plus{}b}2$

[quote="dduclam"]Give $ a,b,c$ be non-negative real numbers with sum 3.Prove that
$ \frac {ab \plus{} bc \plus{} ca}3 \plus{} 3\ge abc \plus{} a^2\frac {b \plus{} c}2 \plus{} b^2\frac {c \plus{} a}2 \plus{} c^2\frac {a \plus{} b}2$[/quote]
I think, it's very easy.
$ \frac {ab \plus{} bc \plus{} ca}3 \plus{} 3\ge abc \plus{} a^2\frac {b \plus{} c}2 \plus{} b^2\frac {c \plus{} a}2 \plus{} c^2\frac {a \plus{} b}2\Leftrightarrow$
$ \Leftrightarrow2(ab\plus{}ac\plus{}bc)(a\plus{}b\plus{}c)\plus{}2(a\plus{}b\plus{}c)^3\geq\sum_{cyc}(6abc\plus{}9a^2b\plus{}9a^2c)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^3\minus{}a^2b\minus{}a^2c)\geq0\Leftrightarrow\sum_{cyc}(a\plus{}b)(a\minus{}b)^2\geq0.$


After quite a bit of waiting, I got my four AoPS books (volumes 1 and 2, with solutions). I read through Volume 1 yesterday, and I have to say that I like it a lot :) .

I like it very much...

I think it's a great text book....

but i don't think there are enough practice problems...... :(

Well, It's the greatest math text book I've seen so far...

How many practice problems do you want?  I don't have a copy of volume 1 with me but Volume 2 has a little 500 and I know that volume 1 has even more than that.  There aren't as many exercises as the normal textbooks but the problems that are in there are quality problems.  Ninety percent of a regular text is mindless exercises whereas ninety percent of AoPS is challenge, non-routine problems.

Personally I think the AOPS books don't have enough mindless practice problems, which is good because the reader can make up those problems themselves or use a regular textbook.  I bought the AOPS books not really to learn pre-cal (which is what I'm doing anyways with volume 2 since I'm only in Algebra Ii right now) but to learn harder concepts and apply those with problems.

Edit:
To shinwoo: Patriots>Steelers.

I really like the AoPS books because they are concise. In regular math textbooks, the answer would be just given. But with AoPS, they teach you how to obtain the solutions and why it works. The solutions are very detailed. Also, the problems given in the books are from actual competitions so it gives great practice problems when practicing for competitions such as AMCs, Mandelbrots, etc. With regular textbooks, they just teach you concepts that will rarely appear on math competitions.

i think the AOPS books are a great tool. With volumes I and II, you can cover a whole lot of material. They are also good to study off of to review concepts. The problems are also great practice. These books beat a math textbook any day.

[quote="Jhyon8149"]I really like the AoPS books because they are concise.[/quote]

[quote="Jhyon8149"]The solutions are very detailed.[/quote]

Isn't that a bit ... contradictory?


Let $S_{1}$ deonote the sequence of positive integers 1,2,3..., and define $S_{n+1}$ in terms of $S_{n}$ by adding 1 to ghose integers in $S_{n}$ which are divisible by $n$. Thus, for example, $S_{2}$ is 2,3,4,..., and $S_{3}$ is 3,3,5,5,7,7.... Determine those integers $n$ with the property that the first $n-1$ integers in $S_{n}$ are $n$.

Answer: prime numbers.
[hide]
First, let us prove that such $n$ must be prime. Observe that
\[S_{n}(n-1) = n,\]
and
\[S_{2}(n-1) = (n-1)+1 =n.\]
So we have not incremented the $n$-th number in the sequence on steps $2,\dots, n-1$. That is, no number between $1$ and $n-1$ divides $S_{n}(n-1) = n$. We conclude that $n$ is prime.

Let us now assume that $p$ is prime. We shall prove that
\[S_{p}(1) = S_{p}(2)=\dots=S_{p}(p-1) = p.\]
Consider $k \in \{1,\dots,p-1\}$. Note that $S_{1}(k) = k < p$. If for some $t < p$, $S_{t}(k) = p$, then $S_{t}(k)$ is not divisible by $k$. Hence $S_{t+1}(k) = p$. Therefore, the sequence $S_{1}(k), S_{2}(k), \dots, S_{p}(k)$ cannot "jump over" $p$:
\[S_{p}(k) \leq p.\]
On the other hand, $S_{t}(k) \geq t$. So
\[S_{p}(k) \geq p.\]
This finishes the proof.
[/hide]


Show that for $a,b,c>0$ we have $4\sum_{cyc}\frac{bc}{a^{2}+\sqrt{3 abc\sum a}}\geq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$.
I am sorry for this inequality
But this one is verified $ 2\sum_{cyc}\frac{bc}{a^{2}+bc}\leq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$ :)

[quote="mateivld"]This one is realy nice.Show that for $a,b,c>0$ we have $4\sum_{cyc}\frac{bc}{a^{2}+\sqrt{3 abc\sum a}}\geq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$[/quote]
Multiplication for LS and RS on $\sqrt{abc}$ and try $a\rightarrow0^{+}.$ :wink:

You're right :blush: And i don't think that the reversed inequality takes place???

I hope you will enjoy the new version. :lol:

[quote="mateivld"]Show that for $a,b,c>0$ we have  $2\sum_{cyc}\frac{bc}{a^{2}+bc}\leq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$ :)[/quote]
Vornicu-Schur kills it. :wink:

My solution uses SOS. :lol:
Can somebody post a solution with out Vornicu Schur or expansion??

[quote="mateivld"]Show that for $a,b,c>0$ we have  $2\sum_{cyc}\frac{bc}{a^{2}+bc}\leq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$ :)[/quote]
And I think we can also prove:
$2\sum_{cyc}\frac{bc}{a^{2}+bc}\le 4\sum_{cyc}\frac{bc}{(a+b)(a+c)}$
with  Vornicu-Schur.So my inequality is stronger than it :wink:

Yes Zhaobin you are right. :lol:
But this one I think is good and strong.$2\sum_{cyc}\frac{bc}{(a+\sqrt{bc})^{2}}+\sum_{cyc}\frac{bc}{a^{2}+bc}\leq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$

[quote="mateivld"]Show that for $a,b,c>0$ we have $4\sum_{cyc}\frac{bc}{a^{2}+\sqrt{3 abc\sum a}}\geq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$.
:)[/quote]
what do you mean by this one?Is it right?

Zhaobin ,arqady gave a counterexample.You can try to see if the sign is reversed. :lol:

[quote="mateivld"]$4\sum_{cyc}\frac{bc}{(a+\sqrt{bc})^{2}}\leq \frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}a}$[/quote]
Ok,But I think this one is too stronger :) Then it is wrong.
Notice the equality will hold when $b=c \to 0$ in my inequality.
Try it to yours,you will find it is wrong. :wink:

[quote="mateivld"]Both tend to 0 if you say that.[/quote]
The right tend to 4.
but the left tend to 8.....
If I didn't made any mistake....

[quote="mateivld"]You are right again, Zhaobin. I edited it to look better.I will try it. :lol:[/quote]
Still wrong...the same case...

I am sorry :blush: .Take it as $ 2\sum_{cyc}\frac{bc}{(a+\sqrt{bc})^{2}}+\frac{9 abc}{2\sum a^{3}}\leq\frac{(\sum_{cyc}ab)^{2}}{abc\sum_{cyc}{a}}$


How many 12 digit strings consisting of only 0s and 1s do not have any 3 consecutive 0s or 1s.

solution
the number of such sequences is 2^12.
Now using PIE to find the number that we need to subtract from this we get:
512 + 512 - 64 = 1024 - 64= 960
so  our answer is 4096 - 960= 3036

[quote="Art of Owna"][quote="paladin8"]I made a blog [url=http://wangsblog.com/jeffrey/?p=303]post[/url] on a similar problem a while ago.[/quote]

sorry I cant really understand that.  can you explain it in simpler terms?[/quote]

Hmm, I think since you are considering both digits the method cannot be applied as simply... I'll think about it.

probably fibonacci numbers with a three-term recursion - which is how i would approach the problem, although i would not have known of tribonacci numbers.

Kamil Witek, your solution is wrong; in particular, $\frac{1}{4}+\frac{1}{4}-\frac{1}{16}= \frac{7}{16}$ths of all strings are bad (those that begin or end with 000 or 111), so the answer certainly cannot be larger than $\frac{9}{16}\cdot 4096 = 2304$; I suspect, due to possibilities of overlap and other nastiness, that a PIE solution will be intractible.  (Treating the first through third, fourth through sixth, seventh through ninth and tenth through twelfth bits as independent blocks (that is, ignoring situations like 110001...) and applying PIE shows that the answer is at most 1296, but this is still quite a bit too large.)

[quote="1=2"]uuhh... that are Tribonacci numbers????[/quote]
[url=http://www.goodsearch.com/Search.aspx?Keywords=tribonacci]Search engines[/url] are wonderful things.  (In fact, though, this hint was wrong and you should ignore it.)

[hide="Most of proof 1 including all the interesting ideas"]
You want to build a recurrence by adding digits to the end, one at a time.  For $n \geq 2$, let $a_{n}$ be the number of binary strings of length $n$ which have no three consecutive, identical digits and whose last two digits are not equal, and let $b_{n}$ be the number of binary strings of length $n$ which have no three consecutive, identical digits and whose last two digits are equal.  Then it's easy to see that $a_{2}= b_{2}= 2$.  Convince yourself that $a_{n+1}= a_{n}+b_{n}$ and that $b_{n+1}= a_{n}$.  Then solve this system of two recursive equations (or just calculate $a_{12}$ and $b_{12}$ from the recursion).[/hide]

[hide="Most of proof 2 including all the interesting ideas"]
Any example string of length $n+2$, $n \geq 1$, either ends in two copies of the same digit or two different digits.  In the first case, we can remove the last two digits to get a valid string of length $n$; in the second case, we can remove the last digit to get a valid string of length $n+1$.  This is a bijection, so our sequence obeys the Fibonacci recursion, and one just needs initial conditions.[/hide]

Bonus question: the sequence begins 1, 2, 4, 6 ... for strings of length 0, 1, 2, 3, ....  That "1" at the beginning should seem anomolous -- it doesn't work with the recursion, which is unusual.  Can you give a good explanation why?


The cube of what integer is closest to 800?

$ \boxed{9}$

We know $ 10^3\equal{}1000$, so it must be the next smallest number.
(By the way... $ 9^3\equal{}729$)


Good Day to the residents of the Grove. This is your governor speaking. Recently some of my agents have disappeared in your town. I suspect the work of the mafia. As a result of this I propose you organize a lynch system to murder them. There is one slight problem; we are running low on money. As a result I have decided that the lynches will be based on bidding.
So,
jackdillion
erabel
Hamster1800
mnmath
lingomaniac
CheeseIsGood
zapi2007
andersonw
miyomiyo
BanishedTraitor
1=2
kimmystar94
ZhangPeijin
alanchou
xpmath
shaggy75
distracted523
Temperal
LevatorNasolabialis
Peter
woser456
CircleSquared
BerylliumNitride
perfect628
cf249
pianoforte
samath
monkeyman238
The_Scintillator
Brut3Forc3
mz94
Flame
genuine
emannes
pythag011

Time to lynch some people.

Rules:
Everyone starts with $100.
You may use this money in order to bid on who you want dead.
Any money you use to bid will be taken away from this total.
You may only bid money you already posses.
If you want me to count it you must post [b]Lynch: Person $X[/b]
The person with the most money bid on them at the end of the day will be killed.
Each day will last 2 weeks or until there is no change in the leader for 24 hours.

In addition there will be bidding for some night roles. In addition to your normal night action you may bid for control of an action or actions. This includes the hospital (a doctor protection), the police (a cop), and the hired killer, (a vigilante), the lawyer (I’ll explain later) and the university (also explained later). You may also vote for one of the politicians (also explained later). All of this will be in your pm to me at night.

In order to be able to do this you must have the highest bid.
All bidding is private and there is no rebidding. Otherwise the same rules of bidding apply.

The lawyer is needed in order to write a will. This will guarantees that any money that remains goes to the person you choose.

The university allows you to change your job. Once you win the bidding I will tell you what job’s are available. Note this will not necessarily improve your income.

The politicians are running for mayor. As mayor they will be able to pass legislation that all of you must follow. This legislation must be approved by yours truly.

Each night you will earn some money based on this role. There are also roles that may influence the amount of money if any you receive.

Rule # 1: In order to earn any money you must fulfill your post restriction.
Rule # 2: You may not quote your pm.
Rule # 3: You may not also reveal your job as no one wants to know that you earn more or less money than them.

These are some additional rules imposed by the mayors:
As of now –Nada.

Any questions just pm me.

Ahh.. *takes fresh breath of air*
A brand new mafia game that I've been waiting for SO LONG!!
Life is good. Very good.

so.. any random lynching today?

[quote="Ars"]Each day will last 2 weeks or [b]until there is no change in the leader for 24 hours.[/b][/quote]
This should keep the game moving, even with the large number of people here.

I rather like how the lynching system is structured. *reads up on game theory*

This game is tight.  Props to you, Ars.

Quick question: is it possible to unbid?

OMG YES FINALLY! I'm tired of studying for the USAMO! :P

Another question: can we bid for not lynch?

I am pretty excited for the new Mafia game, which is finally here after that long wait!
It looks and is a whole lot different than the other boring ... one that's going on :|
The layout seems like a good new idea, especially after I read it over a few times because I didn't get it  :mad: , no offense.
The bidding sounds new, at least this game will move much faster than those other games where we have to wait for some to be lynched by majority.
Well, good luck to all!
(I hope this game does not degenerate into lurking like the other game did momentarily  :lol: )

[quote="zapi2007"]I hope this game does not degenerate into lurking like the other game did momentarily :lol:[/quote]
Somehow, I think that there will be some lurking.  (There's always one!)  But thankfully, it won't be as difficult to get rid of them.  I think that people will want to get rid of lurkers, and it would take somewhere between 24-48 hours to do so.

:P
Maybe we can spend $ \$$1 to lynch a lurker. Or 1 cent. Oh, stupid question: Can we bid in cents? (Yeah, I know... really stupid. Almost as stupid as math...) Just don't want to make assumptions...

"We" do not have to bid 1 to lynch a lurker. If you bid 1, and no one else bids, that's the high bid. ;)

[quote="genuine"]Quick question: is it possible to unbid?[/quote]

If you read carefully, no.

This is an interesting game... :)

Since I do not have a lepton of a clue what to do, I will wait.

Looks like a very nice setup! I wonder what the supply and demand of each of the night actions will be. Hopefully, the relative price of those items won't be too high.
The information asymmetry in this game better not deter us from getting rid of the scum.

I don't think placing random or joke votes is good in this game, because then every 24 hours, we would have to vote someone else, bidding more money.

Somehow I think we will have Price-is-Right-style overbidding.  "I bid $ \$1$ more than what he has." :P

[rant]We need to find out who the mafia are so we can put them before a grand jury and prove that they are guilty![/rant]

OK. Calming down now.

[quote="andersonw"]I don't think placing random or joke votes is good in this game, because then every 24 hours, we would have to vote someone else, bidding more money.[/quote]

There are 35 people in this game.  I don't think we'll have a problem getting someone to overbid and keep a person around-if there's a good reason for them doing so. [b]Lynch: andersonw $ \$1$[/b].

Well, I think I've figured out some of the post restrictions. lingomaniac88 has to...Um, wait, what was I saying? That happens sometimes with me. You know, you were thinking about something, someone interrupts you, and you forget it? Anyways, I just remembered: lingomaniac88 has to quote, and I think zapi(insert numbers here that I forget) has to use 3 smilies in his posts.

Yay, this game isn't going to be as slow and boring as the first few days of the other one was.
I'm going to have to read the setup again...

Mine was that I had to post something very happy!
because i was a FORMER factory worker and i could steal from people. i was neutral. i had to end up with the most amount of money.

This looked like an interesting set-up.

Does Ars still have the set-up?

I was a doctor who protected hamster second day onward

My role restriction was to post that I was better than everyone else 5 times a post.

I was one of three candidates for mayor, and my goal was to get elected four consecutive times and for the town to win.
I had to always try to prove that I was better than the other candidates.

My post restriction was to insult math in every post. (If you seriously didn't see that...)
Also, even though I wanted revenge on Hamster for getting lingo, I figured that the doctors would try to protect him, so I went for others. :P

[quote="lingomaniac88"]What was everyone's salaries?  Mine was $ \$100$.  Brut3Forc3 had a feeling that this amount was among the lower ones (I was a factory worker).[/quote]

Same here. And I was a doctor...

I want to comment on this post of mine:
[quote="lingomaniac88"]Due to Hamster1800's breach of contract and suspicious behavior, I shall do this: [b]Lynch: Hamster1800, $ \$136$[/b][/quote]
By this point, I had given up defending myself.  I just threw all my money (well, almost all of it) at Hamster1800.  Sure enough, Temperal placed a higher bid on me, like I hoped would happen.

[quote="Temperal"]I think my post restriction was the worst, except maybe for jackdillon's - I had to post in iambic pentameter. First I had to look up what that meant, and then I had to spend about 10 minutes poring over each post.

Like perfect said, royal mafia scum, 100 dollars a night, factory worker.[/quote]

LOL

ME: CANDIDATE FOR MAYOR. SAME AS XPMATH

[quote="Temperal"]I had to post in iambic pentameter.[/quote]
I'm wondering if anyone else went back and read through Temperal's posts to see that they are all in iambic pentameter.

I noticed Temperal's post restriction before he said it! Waaaayyy before he said it. :P

My initial thoughts were that he had THE suckiest post restriction. :lol:

Most of his posts fit, I think, if I remember correctly.

My salary wasn't the lowest, but one of the most varied, it just depended on owever much money was the highest bid. I tink I ended up with $236 or so, after killing two people.

[quote="Ars"]Eh yeah this is mod-abandoned

I over complicated this game and most people got lost. Since the game was so large this added even more chaos.
 This lead to extreme inactivity and therefore the game failed.

Sorry this site is not ready for such a complicated game for so many people.
I think this would have worked had it been a smaller game.
But since I tried to add in the new affect of bidding to a large game this game became overly inactive.
The post restrictions were almost entirely meant to discourage people from posting a ton day 1 and then just lurking.

Yeah I'll post everything about the game later but sorry for over complicating it.[/quote]

I'd like to see the setup too, even though I wasn't in the game.


I just read [url=http://www.americanscientist.org/template/AssetDetail/assetid/54428?&print=yes]this[/url] article and found it enjoyable.

What do you think?

Interesting, but I think math will manage to survive.  Even if it is becoming harder to validate modern proofs, actually proving that somemthing is true is not necessarily the most exciting thing about a proof.  For example, nobody expected Fermat's Last Theorem to be false.  The more exciting thing is often what methods are employed in the proof and what new connections are discovered.

Any chance the "M" in M. L. Wantzel could be "monsieur"? :lol:

@joml: I think what renatzu was getting at was that everyone expected FLT to be true, but the task was to prove it (or something like that).

[quote="joml88"]Um?

[quote]For example, nobody expected Fermat's Last Theorem to be false.[/quote]

Um.  His theorem was NOT false.  Andrew Wiles proved the theorem to be true.  Furthermore, the result of what Fermat theorized did not change.  What Wiles proved went against no one's expectations (er, well, [i]most[/i] people's expectations...I'm sure there were probably some people who believed it was false).[/quote]You misread what I said, I was saying that nobody was was surprised when it was proven true.  People were  not waiting to find  out whether or not it was true (everybody instinctively expected it to be true), they were waiting to find out how it was proven true.

I'm saying that the process was more important than the result, and even in an exhaustive modern proof, if the result is not 100% verifiable (maybe because it involves a computer checking something), people will still be able to see what insights the prover made, and that's the important part.

I dunno, it's hard for me to get excited about non-mathematicians saying durrrrrrrr proofs are too hard.  And also once I read about the four color theorem for the millionth time in a non-technical mathematical article my head started to hurt and I closed the tab.

[quote="drunner2007"]it's kind of scary to think that proofs are becoming so much larger and difficult to understand.  :(
also, i have seen articles where people worry about the increasing specialization in mathematics (which is needed since some areas of mathematics are so deep). i wonder if mathematics will continue to be more specialized.[/quote]

Actually there's nothing wrong with specialization, which applies in any general sense. It is better to have a group of people who have unique specializations and can thus solve a deep problem spanning across many fields rather than to have a group of "nearly-identical" people who have shallow development in all fields and thus cannot solve such a problem, as they have common weaknesses.


just wondering if any of you are familiar with social network analysis?

Just asking because I'm working on various applications of mathematics to the realms of pro sports, and I can't find any book on social network analysis applied to sports. Any of you can give a good idea of what these analysis can do and so on?

anyone knows about social network analysis?

[quote="PowerOfPi"]I don't know what it means. Remember, [url=http://www.lmgtfy.com/?q=Social%20Network%20Analysis]Google is your friend[/url].[/quote]

Uhh, the OP almost certainly knows what social network analysis is.  Way to be unnecessarily condescending though!

I'm not sure if these would help?

http://books.google.com/books?q=social+network+analysis&btnG=Search+Books

It's Limited Preview for most books, but worth a try. Again, I do not have any information on the concept though.


I'm not sure, but I think it is best to wait until tomorrow (the 29th), because after that day all of the exams will be in.

lawl, I remember when I was so obsessed about qualifying for camp this time last year.

Now I couldn't care less. :|


I was wondering, if some sort of cation reacted with hydroxide ions in solution to form a precipitate, would this ion be considered an acid as the [OH-] would go down, increasing [H+].

Also would this be considered a strong base?

The cation would be an acid since $[H^{+}]$ increases.  (An anion can be a base since it can cause $[H^{+}]$ to decrease.)  It could be considered relatively strong.

Thanks.

Also, I know that Lewis Acid/Bases are inclusive of Bronsted-Lowry acid/bases, but is Bronsted-Lowry inclusive of Arhennius? I would think not correct?

To add to what Zakary said and answer your new question:

It is ambiguous to simply ask the question, is some species a base.  There are at least three systems of acid/base classification taught in most chemistry classes: namely, Arrhenius, Bronsted-Lowry, and Lewis.  The cation you are speaking of would NOT be classified as either an Arrhenius or Bronsted-Lowry base.  Why not?  For Arrhenius, an acid is a species that increased the hydrogen ion concentration WHILE IN SOLUTION.  Notice the problem here: as soon as we start talking about precipitation, we are no longer talking about the species being in solution, and Arrhenius no longer applies.

In Bronsted-Lowry, we can talk about reactions that occur outside of water.  However, your ion still would not be considered an acid.  In order to be an acid, it must donate a hydrogen ion--which it does not.  (Note that some small, highly charged ions--such as $Fe^{3+}$ form a complex with water, which is able to donate hydrogen ions, and hence such a complex can be considered a B-L acid.)

Finally, however, your ion IS a Lewis acid.  It accepts a lone pair from hydroxide to form the precipitate.

It would not necessarily be a strong acid (which I assume is what you meant to ask).  Acid strength depends on the equilibrium constant for the reaction under consideration (ion + OH- --> ionOH); some such reactions proceed nearly to completion while others barely occur.

Finally, Bronsted-Lowry acids/bases include all Arrhenius acids/bases.  B-L simply generalized the work of Arrhenius without scrapping any species that fall within Arrhenius's domain.

Okay, I see that the definition for Arhennius acids/bases are substances that increase [H+] or [OH-]. The way my book worded it was awkward, using produce H+ and OH- instead of increase the concentration. So I was unsure as to whether something that did not directly dissociate said ions would be considered Arhennius.

It is possible, however, to have an Arrhenius base that does not actually contain an OH.  A classic example of this is $NH_{3}$.  I don't think it is possible to have an Arrhenius acid without an H, but perhaps there is something I have forgotten.

What about a strong metal cation that bonds with OH- to form a precipitate? Wouldn't that increase [H+]?

Again, that's not an Arrhenius acid.  Its behavior IN SOLUTION is what is key.  It only increases [H+] when leaving solution.


Since no one else has created a thread indicating a modship, I will.

[b]Kill Code II[/b]

Instead of being in a castle, you are 16 students and an English teacher. The teacher was talking about correct use of apostrophes when suddenly, the door sealed shut.

"AHHH!" screamed the really tall guy.

The English teacher instructed the class not to panic, but immediately started hyperventilating and fainted.

Over the intercom came a voice, "This is who all your base is belong to. The door will not open until one of you tells me what the kill code is, correctly. After that, we will be able to dominate the world with apostrophes and -." The message was cut off by some maniac laughing.

---

Gameplay:

Each of you will receive a role. In your role PM, it will say what your action is. During each turn (probably separated by 1-2 day intervals), you post ONCE in the thread a user who is in this game.

Say you are a mafia (your action would then be to kill someone). In thread, you would put "A" if you want to kill player A. However, it is recommended you put "A or B (or C)". I will choose A or B (or C) to randomly kill. That way, if A dies from your kill, and someone else also put A in their action list, you will draw a bit of suspicion away from you.

Rules for Signups

1) Do NOT PM me for sign ups. Post in thread.
2) Do not remind someone who obviously has not read the rules to read the rules again. We can think of something eventually for you people who read the rules.
3) Place your grade and gender when signing up.

[list]
    [*] Actions must be in bold. If you do not bold your action(s), it will not be counted.
    [*] The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
    [*] Once your death scene has been posted, you’re dead. Stop typing. The only thing you can type is a "bah" post. You may also talk as a ghost, but you cannot reveal any of the game plan.
    [*] Don’t edit/delete previously submitted posts. If you must, post again with EBWOP (Edit by way of post).
    [*] Don’t quote any PMs from me. Neither true, nor false. Also, put three lines of text in your sign-ups to confirm rule reading.
    [*] If you do not submit an action, you will do nothing.
    [*] Treat this game as a commitment. Be considerate – don’t leave us hanging.
    [*] Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.[/list]

There are 17 players most of you who will have different roles. Note: Some roles are going to be different this time.

[hide="participants"]
1.monkeygirl13
2. 15!!!
[/hide]
Why is everyone getting dates wrong today.
Both Thanksgiving and October 9th passed.

oops, I forgot to put my grade
and gender.
I am 8th grade male.

[quote="monkeygirl13"][b]Me signs up. [/b]
I like butterflies and rainbows
EDIT: 7th grade female[/quote]

Join
8th grade
male

[color=green]Mod edit (permission granted from AIME15):[/color]

[color=#3666D2][u]You find a radio
[hide="Use the Radio"][hide="Operate the AM"]Too much static, you can't hear anything. Unfortunately, you can't see what station it is either.
[hide="Operate the FM"]This looks like the right place, but you can't do much here[/hide][/hide]
[hide="Operate the CD"]A few buttons; none of them seem to work...yet[/hide]
[hide="Operate the Cassette"]Aha!
[hide="Play"]Nothing appears to be on the tape[/hide]
[hide="Record"]You hit record.Yes! Something worked!
Hmm
Well well, the play button seems to be the only working button...
...but it's red. Press it?KaboomYou got rickrolled! This is an FM station (somehow), but this is the first station where I have ever been rickrolled.
1191293122 Oh, that was only background noise..Did anything just happen?[/hide]
[hide="Rewind"]The tape is unused.[/hide][/hide][/hide][/u][/color]

I join, again.

8th grade boy.

Hmm... It says that someone needs to mod over Thanksgiving Break...  :rotfl:

[hide="participants"]
1. AceOfDiamonds
2. AIME15
3. 15!!!
4. royalflower
5. Yoshi
6. monkeygirl13
[/hide]
8th grade, male
I'll be nice to everyone. Hope you'll return the favor, everyone.

[u]People joined[/u]
1. monkeygirl13(7th, female)
2. 15!!!(8th, male)
3. AIME15(8th,male)
4. Yoshi(8th,male)
5. AceOfDiamonds(8th, male)

8th grade male majority :rotfl:

You guys make me feel old, so I'll just say I'm in first grade and I'm male.  :)  (I'm being serious that I am male though)...

[u]People joined[/u]
1. monkeygirl13(7th, female)
2. 15!!!(8th, male)
3. AIME15(8th,male)
4. Yoshi(8th,male)
5. AceOfDiamonds(8th, male)
6. hyperddude (8th, male)
7. zapi2007 (1st-ish, male)
8. andrewjjiang (7th, male)
9. westiepaw (8th, female)

There is an extreme majority of 8th grade males....

[u]People joined [/u]
1. monkeygirl13(7th, female)
2. 15!!!(8th, male)
3. AIME15(8th,male)
4. Yoshi(8th,male)
5. AceOfDiamonds(8th, male)
6. hyperddude (8th, male)
7. zapi2007 (1st-ish, male)
8. andrewjjiang (7th, male)
9. westiepaw (8th, female)
10. tennis123 (8th, female)

I'm just messing around with the rules right now; I don't actually have modship. Lol. But since AoD isn't doing anything, I guess it's fine. It is also following the rules, sort of like a loophole.You don't actually have to listen.

It said "type in purple", not "type in purple in purple" so I just do in purple.

[quote="zapi2007"][color=orange]every word[/color]

in purple, or in a color other than black?
[quote="dragon96"]
The Tyrant must send a whole bunch of new roles of each person alive (me, civilians, and active but dead people (except rg64)) to randomguy64. Every role must have a direct connection with the person he/she targets. Therefore, you cannot make everyone have a role that no matter he/she chooses, you will get something. There must be at least 5 people who start off with the kill code (A), and 5 people trying to get rid of the five (P).
[/quote]
has this happened yet?[/quote]
no
I can send out roles to everyone alive once there is a list of players. I am also very inactive, so I suggest tyrant-replacing the mod.

[size=200]Day 7 ends[/size]
[size=200]Day 8 begins[/size]

Too lazy to wait

WAT nothing happened?
No annoncement?
WTH?

(still no list of roles from AoD)

I object.

tyrant transfer modship: dragon96

Too lazy.

Wait, so its day?

This game has died somewhat...for the last two days.

and it's all my fault :(
just kidding, it's mostly dragon's fault.

Yeah, this is dead. I won't continue in that case.

PLEASE. Don't let it die. This is fun!  :o

I'll attach the list of roles, sent to me by meewhee, sent to her by dragon.

Wait...I can't do that. So I guess I'll have to post a screenshot of it, since I don't want to just type everything.
[hide="roles"]
[img]http://img685.imageshack.us/img685/1893/killcode2roles.png[/img][/hide]
If the image disappears, ask me to re-upload it.
[hide="roles explanation"]
[quote="meewhee009"]Column 1: The Side. Red = Roleswitcher. Blue = Story Role. Green = Other Power Role

Column 2: This is part of diversion 2. The Color corresponds to the color of the font down below. The Computer Room has already been opened, and four other riddles must be solved. When this happens, the Dragon room is open. You can be creative on what is inside.

Column 3: The Roles. If you don't understand, then do something random or one of your choice.

Column 4: The Person's Game #. This is what goes in columns 11+

Column 5/6: Info. In C5, red means they have an option of which one they wish to perform. usually even/odd or mod 1/2/3. Blue means their actions do nothing for now.

Column 7: Whether they have the Kill Code. If the Landlord receives, then you must give the entire kill code.

Column 8: Whether the person is poisoned or not. The person loses rand(1,2) or rand(1,3) every night. (Suspended does not lose HP)

Rows 16-n is the list of dead people. May somehow help.
There is also a random number generator there. The attachment will be starting on Day 6.

Most of all: Be creative if something is not listed.

Example 1: AoD attacks mod/Co-mod (becomes mod with a variation)
Example 2: Yoshi guards self (impossible) (someone divided by 0)

Send me the info stuff on the 30th. Thanks.
-dragon96[/quote][/hide]


Are hash tables useful for problems in the IOI-style?
I've been told not to worry learning this, becase we shouldn't spend time implementing hash tables when we have restricted time like in a contest. But some people say hash tables can be very helpful in some problems.
What do you guys think?

It can't hurt to know them... besides, they're really easy to implement, so especially with something like the IOI you have plenty of time. Also, if you code in C++ just use hashset or hashmap from STL.

Hash tables are necessary to solve some previous IOI tasks, so the answer to your question is simply: Yes.

can you tell me how to use hash table at STL?

is it O(1) in STL?

You can just Google it... http://www.sgi.com/tech/stl/table_of_contents.html has stuff on all STL stuff. I'm not completely sure if it runs in O(1), but I would expect so (I could be wrong, I don't use it that much).

I've never had to use a hash in any competition problem before, but I dunno.


Let $q\in\mathbb Q\ \cap ]0,1[$. Show that there is a [b]finite[/b] $K\subset\mathbb N$ such that $\sum_{k\in K} \frac{1}{k(k+1)} = q$.

First, $q$ can't be negative if $K\subset\mathbb{N}$, since all the terms are positive.

Next, by the definition of a set, all its elements are different. Hence for any finite $K$ we have $\sum_{k\in K}{\frac{1}{k(k+1)}}<\sum_{k=1}^{\infty}{\frac{1}{k(k+1)}}=1.$

Therefore, the statement can hold only if $0<q<1$, and for it to hold on whole $\mathbb{Q}$, it's necessary that $K$ is a collection instead of a set, and its elements must be from $\mathbb{Z}$.

It is true if 0<q<1. Because $q=\frac{a}{b}=\sum_s \frac{(-1)^{s-1}}{r_s}.$

it is 0<q<1 indeed. :) my solution looks a bit like rust's idea. Not that there's much choice on the idea here, probably.


Maybe you didn't consider the restriction that you need to put socks on before shoes. (It's hard to determine where you went wrong if you don't show your work: I see that the answer to 1+1 is 2, but I don't see how 340905 is wrong.)

[quote="unimpossible"]A spider has one sock and one shoe, for each of its eight legs.  In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

Ok i see how the answer is $ \frac {16!}{2^8}$, but i do not see how $ 8!8!\times2^8$ is wrong.

can someone explain how i overcounted?[/quote]

Why did you multiply by 2^8. Just explain on what logic you got your answer.

oops, sorry guys it was like 12:00 last night and i wasn't thinking.


linear algebra
matrix
inequalities
inequalities open

Suppose that two $ m$x$ n$ matrices $ A$ and $ B$ with real entries and all row sums equal to $ 1$ satisfy
\[ \prod_{i\equal{}1}^m (a_{i1}x_1\plus{}\dots \plus{}a_{in}x_n)  \leqslant \prod_{i\equal{}1}^m (b_{i1}x_1\plus{}\dots \plus{}b_{in}x_n)\ \ \forall \ x_1,\dots ,x_n\geqslant0.\ (*)\]

Prove or disprove that for all $ c\geqslant 0$, we then also have
\[ \prod_{i\equal{}1}^m (a_{i1}x_1\plus{}\dots \plus{}a_{in}x_n\plus{}c)  \leqslant \prod_{i\equal{}1}^m (b_{i1}x_1\plus{}\dots \plus{}b_{in}x_n\plus{}c)\ \ \forall \ x_1,\dots ,x_n\geqslant0.\]

Equivalently, if we denote the condition $ (*)$ by $ A\succ B$ and the all-1-Matrix of size $ m$x$ n$ by $ \mathbf 1$, prove $ A\plus{}\lambda\mathbf {1}\succ B\plus{}\lambda\mathbf {1}$ for all $ \lambda\geqslant 0$.

:) :roll:


number theory
greatest common divisor
geometry proposed
geometry

Let $AB=c$, $BC=a$ and $AC=b$ be the sides of a triangle $ABC$.Let $G$ to be the centroid of $ABC$ and $M$ to be the medium point of $AB$.We know that $mdc(a,b)=mdc(a,c)=mdc(b,c)=1$ and that $GM= \frac{ \sqrt{a^{2}+ab+b^{2}}}{6}$
Show that $c \equiv 1(mod 6)$.

I proposed that for the next edition of a brazilian math magazine (Eureka!).

May I please know what $mdc$ stands for?  Thanks in advance. :)

Sorry, I wrote it in portuguese!! mcd means gcd !!!Sorry again!

[quote="BoesFX"][quote="gabriel ponce"]hint:Use Stewart Theorem.[/quote]

Does that hint really help? :maybe:

And I think this is more suitable to be called NT problem.
I still don't know how to prove that $c \not \equiv 5 (\mod 6)$[/quote]

Yes, that hint really help( at least by my way of solution).I agree with you that is more a NT problem, but when I ask the moderator to put it at NT proposed and own problem, he put it here! :|

..........
Gabriel

Actually the stewart theorem, is just used to prove a very well know fact!

[quote="gabriel ponce"]And why is that a counter example?? :huh:[/quote]
Verify it yourself. The length of GM in my counterexample fits well with $\frac{\sqrt{a^{2}+ab+b^{2}}}{6}$.

[quote="gabriel ponce"]Are you sure :!:  :?: .
So remember that $(3GM)^{2}=\frac{2a^{2}+2b^{2}-c^{2}}{4}$ because $3GM$ is the lenght of the median with respect to $c$.So we get that $\frac{a^{2}+ab+b^{2}}{4}=\frac{2a^{2}+2b^{2}-c^{2}}{4}$ and substituting $a=33$,$b=16$ and $c=35$ we have that the equality don't hold!!!Thus, that is not a counter example!! :P  :wink:[/quote]
Oh... :oops:
I was so dumb that I though M is the midpoint of AC or BC (forgot which one).

OK, maybe I will reduce it to NT problem, I don't think it would help anyway. :P

$a,b,c \in \mathbb{N}$
$\gcd(a,b)=1$, $\gcd(b,c)=1$, $\gcd(c,a)=1$
$c^{2}=a^{2}-ab+b^{2}$
Prove that $c \equiv 1\mod 6$

I'm stuck here. :(

Sorry, but i can't give you any hint because I proposed that for a Brazilian math magazine.

Try to use quadract reciprocity(I'm not sure if I wrote it in the right way)!!!

$ \Delta\equal{}\minus{}3b^{2}$
If $ 3|c$ then $ 3|a,3|b$
If $ (3,c)\equal{}1$ then $ (ab,3)\equal{}1$
For every $ p|a^{2}\minus{}ab\plus{}b^{2}$ (p>3) has form $ 6k\plus{}1$
Easy to check $ (c,2)\equal{}1$
So $ c$  has form $ 6k\plus{}1$


Ben cok az biliyorum Belki yardimci olabilirim :)

Burda dolar işaretleri arasına yazıyorsun sanırım yazmak istediğin latex kodunu
Örneğin,
dolar   \int\prod_{a}^{b}\prod\sum  dolar yazınca
$ \int\prod_{a}^{b}\prod\sum$

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands
bu komutların listesi
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols
bu da sembollerin listesi ve yazarken iki $ arasına yazıyosun sonra alışırsın zaten

latex bilmeden de buradan yazdıklarınızı latex koduna çevirip,çıkan kodu dolar işaretleri arasında yazarsanız da olur
bit.ly/d85RrZ


calculus
derivative
function
vector
real analysis
real analysis unsolved

[sorry, posted this in the wrong section before -- i apology for the double post]

Suppose $ g(x)$ is a twice differentiable function in the open neighborhood of the origin, and let $ \mathbf{x}$ and $ \mathbf{y}$ be two unit vectors in $ \mathbb{R}^n$, not necessarily orthogonal to each other. Show that in an open neighborhood of the origin, $ \partial_{\mathbf{x}} \partial_{\mathbf{y}} g(\mathbf{0}) \equal{} \partial_{\mathbf{y}} \partial_{\mathbf{x}} g(\mathbf{0})$. Why is it to be expected that if $ \mathbf{0}$ is a local maximum of the function $ g$, then usually we will have $ \partial_{\mathbf{x}} \partial_{\mathbf{x}} g(\mathbf{0}) < 0$? Demonstrate an instance where this is not so.

sorry, didn't mention explicitly - $ \partial_{\mathbf{x}} f(a)$ refers to the partial derivative of the function $ f$ with respect to the unit vector $ \mathbf{x}$ at the point $ a$. this is very confusing, especially because we don't know anything about the unit vectors $ \mathbf{x}$ and $ \mathbf{y}$ directions. so i don't see how any of the results follows.

You mean the directional derivative, right? Do you know the formula for the directional derivative?

yes, sorry, directional derivative. i know

$ \partial_{\mathbf{x}} g(\mathbf{a}) \equal{} \nabla g(\mathbf{a}) \cdot \mathbf{x}$.

how do the results follow?


for real numbers $x,y,z >0$and$x+y+z=1$.
prove that :
$\frac{x}{\sqrt{1-x}} +\frac{y}{\sqrt{1-y}}+ \frac{z}{\sqrt{1-z}} \geq \sqrt{3/2}.$

WLOG let $x\geq y\geq z$ then $1/\sqrt{1-x}\geq 1/\sqrt{1-y} \geq 1/\sqrt{1-z}$ .Hence apply chebyserv

$\sum\frac{x}{\sqrt{1-x}}\geq \frac{1}{3}(\sum x)(\sum\frac{1}{\sqrt{1-x}}) = \frac{1}{3}\sum\frac{1}{\sqrt{1-x}}$

Then use Jensen as suggested by arqady we have $\sum\frac{1}{\sqrt{1-x}}\geq \frac{3\sqrt{3}}{2}$

 :)

I meant to use Jensen directly:
$\frac{x}{\sqrt{1-x}} +\frac{y}{\sqrt{1-y}}+ \frac{z}{\sqrt{1-z}}\geq\frac{1}{\sqrt{x(1-x)+y(1-y)+z(1-z)}}\geq\sqrt{\frac{3}{2}}.$  ;)

[quote="shyong"]I cant see which function u use for jensen here arqady  :( ........[/quote]
$f(x)=\frac{1}{\sqrt{x}}.$ :)

From where did you get this problem? i remember that I proposed it for RMNO 2005 and it even appeared in the shortlist. :)

this is a special case of a Chinese or Indian problem, it's something like

Let $x_1, x_2, ..., x_n$ be positive reals such that $x_1+x_2+...+x_n=1$, prove that
$\frac{x_1}{\sqrt{1-x_1}}+\frac{x_2}{\sqrt{1-x_2}}+...+\frac{x_n}{\sqrt{1-x_n}}\geq \sqrt{\frac{n}{n-1}}$

Wow , thats kool  arqady :) . I didnt see that weight jensen can reduce it an easier version of inequality . Now I learn how to use weight Jensen  :D

In fact it is no need to use weighted Jensen since the plain Jensen works.
See for it here.
http://www.mathlinks.ro/Forum/viewtopic.php?t=68638

By the way , trantranham also posted this in Geometry section :huh:

$\frac{x}{\sqrt{1-x}} +\frac{y}{\sqrt{1-y}}+ \frac{z}{\sqrt{1-z}} \geq \sqrt{3/2}.$
  we have by cauchy ineq's
        $\sqrt{3(1-x)}.\sqrt{2} \leq \frac{5-3x}{2}$
   likewise , we have      $\sqrt{3(1-y)}.\sqrt{2} \leq \frac{5-3y}{2}$

   and $\sqrt{3(1-z)}.\sqrt{2} \leq \frac{5-3z}{2}$
  hence  , we want to prove that :
 $\frac{x}{5-3x} + \frac{x}{5-3x} + \frac{x}{5-3x} \geq \frac{1}{4}$
      it's very easy

Let $f(x)=\frac{x}{\sqrt{1-x}}$, then $f'(\frac13)=\frac45\sqrt{\frac23}$.
So $f(x)\geq \frac54\sqrt{\frac23}x+\frac{11}{12}\sqrt{\frac23}$.
Summing yields the result.


[quote="citistarz"]The sum of two 5-digit numbers AMC10 and AMC12 is 123422. What is the value of A+M+C?[/quote]

Lol.
[hide]10+12=22, so 2(AMC00)=123400. AMC00=61700, A=6, M=1, and C=7, so A+M+C=14. [/hide]

Nice problem.
[hide]Let's start with C. You know it's either 2 or 7, as the units digit of its sum is 4. For M, a number plus itself can't have a units digit of 13, so C must be 7 with 1 carried over to M. Consequently, M is either 1 or 6.No 1 is carried over to A+A (as the units digit is even) so M is 1. A is 6. A+M+C=14. [/hide]


Don't post problems as pictures!

What have you tried so far¿ Show what your problems with using the definition are here.

sorry,I rarely come here, I tried but I couldn't use latex.


superior algebra
superior algebra unsolved

This is just unbelievable: any group of squarefree order is the semi-direct product of two cyclic groups!!!


geometry
circumcircle
symmetry
trigonometry
perpendicular bisector
angle bisector
power of a point

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$.

[i]Bulgaria[/i]

[color=blue][b](1)[/b][/color] Let $O_A, O_B$ be the circumcenters and $R_A, R_B$ the circumradii of the triangles $\triangle XAC, \triangle XBC$ with the common side XC. It is sufficient to show that the circumradii $R_A \neq R_B$ are not equal. The perpendicular bisectors of the sides CA, CB meet at the circumcenter O of the right angle triangle $\triangle ABC$, i.e., at the midpoint of the hypotenuse AB. The angle $\angle OO_AO_B = 90^\circ$ is right, because $OO_A, OO_B$ are the perpendicular bisectors of the triangle sides CA, CB and $CA \perp CB$. The angle $\angle OO_AO_B = OO_BO_A = 45^\circ$, because the $O_AO_B$ is the perpendicular bisector of the angle bisector segment CX, which forms the angle $45^\circ$ with both CA, CB and $OO_A \perp CA, OO_B \perp CB$. Thus the triangle $\triangle OO_AO_B$ is isosceles and its symmetry axis, the perpendicular bisector of the segment $O_AO_B$ is parallel to the angle bisector CL, identical with the radical axis of the circles $(O_A), (O_B)$. Therefore, these 2 circles do not intersect on the perpendicular bisector of the segment $O_AO_B$, which implies that their radii are different, $R_A \neq R_B$.

[color=blue][b](2)[/b][/color] Let $O_1, O_2$ be the circumcenters and $R_1, R_2$ the circumradii of the triangles $\triangle YAC, \triangle YBC$ with the common side YC. It is sufficient to show that the circumradii $R_1 \neq R_2$ are not equal. The perpendicular bisectors of the sides CA, CB meet at the circumcenter O of the right angle triangle $\triangle ABC$, i.e., at the midpoint of the hypotenuse AB. The angle $\angle OO_1O_2 = 90^\circ$ is right, because $OO_1, OO_2$ are the perpendicular bisectors of the triangle sides CA, CB and $CA \perp CB$. The angles $\angle OO_1O_2 = \angle CBA$ are equal and the angles $\angle OO_2O_1 = \angle CAB$ are equal, because the the perpendicular bisector $O_1O_2$ of the altitude segment CY is parallel to the hypotenuse AB, both being perpendicular to the altitude CH. Thus the triangles $\triangle O_2O_1O \sim \triangle ABC$ are centrally similar, having their corresponding sides parallel.  The lines connecting the corresponding vertices C, O pass through their similarity center. Since O is also the midpoint od AB, the line CO intersects the segment $O_1O_2$ at its midpoint M. Since CO is not parallel to the altitude CH and these 2 lines meet at C, the midpoint M does not lie on the altitude CH. The perpendicular bisector of the segment $O_1O_2$ passing through M is parallel to the altitude CH, identical with the radical axis of the circles $(O_1), (O_2)$, both being perpendicular to $AB \parallel O_1O_2$. Therefore, these 2 circles do not intersect on the perpendicular bisector of the segment $O_1O_2$, which implies that their radii are different, $R_1 \neq R_2$.

Part 1: If $\angle XAC$ = $\angle XBC$, then since we are given that $\angle ACX$ = $\angle BCX$, we have $\angle BXC$ = $\angle AXC$, which implies congruence by ASA. Thus, $\frac{XC}{BC}=\frac{XC}{AC} \longrightarrow AC=BC$, a contradiction.

Part 2: Since  $\bigtriangleup$ CAH and $\bigtriangleup$ BCH are similar, we have $\frac{AH}{AC}=\frac{BH}{BC} \qquad a (1)$

By law of sines, we have $\frac{sin CAY}{CY} = \frac{sin AYC}{AC}$
But, $sin(AYC) = sin (AYH) = \frac{AH}{AY}$, so plugging in yields

$\frac{sin(CAY)}{CY}=\frac{AH}{(AY)(AC) } \qquad a (2)$.
Similarly, we have
$\frac{sin(CBY)}{CY}=\frac{BH}{(BY)(BC) } \qquad a(3)$.

But if $\angle CBY$ = $\angle CAY$, $sin CBY = sin CAY$, so from (2) and (3) we now have:
$\frac{AH}{(AY)(AC)} = \frac{BH}{(BY)(BC)}$

Plugging in (1) yields $\frac{1}{AY}=\frac{1}{BY} \longrightarrow AY=BY$ Thus $\bigtriangleup$ AYB is isosceles. This forces $\angle XAB = \angle XBA \longrightarrow \angle BAC = \angle ABC \longrightarrow BC=AC$, a contradiction.

Part 2: Since $\bigtriangleup$ CAH and $\bigtriangleup$ BCH are similar, we have $\frac{AH}{AC}=\frac{HB}{BC} \qquad a (1)$

Similarity you used is wrong..  it must be

$\frac{AH}{AC}=\frac{HC}{BC} \qquad a (1)$

[hide="Part 1"]
All we need to prove is the contrapositive. Proceed as NuncChaos did.
[/hide]
[hide="Part 2"]
All we need to prove is the contrapositive. Let $b=CY$ and $a=\angle{CAY}=\angle{YBC}$. We have $\frac{b}{\sin{a}}$ as twice the circumradius of $\triangle{ACH}$ and $\triangle{BCH}$. But twice the circumradius is $AC$ and $BC$! Contradiction.
[/hide]

I did part 1 the same as NuncChaos did

[hide="Part 2"]

Assume there did exist a point Y on CH such that $\angle{YAC} = \angle{YBC}$

Let $A'$ be on AC and $B'$ on BC such that $YA' \perp AC$ and $YB' \perp BC$. Clearly $HBB'Y$ and $HAA'Y$ are cyclic. This means that $\angle{YBC} = \angle{YHB'}$  and $\angle{YAC} = \angle{YHA'}$

Notice that $A'B'$ is bisected by $YC$ because they are 2 diagonals of the same rectangle. Thus $HY$, which is the angle bisector of $\triangle{HA'B'}$, also bisects $A'B'$. Thus by the angle bisector theorem we must have $HA' = HB'$ so $YHA' \equiv YHB'$ by SAS.

This means that $YA' = YB'$, thus implying that $Y$ lies on the angle bisector. But since $AB \neq AC$ the triangle is not isoseles and the height and angle bisector only meet at C. But since $Y \neq C$ we arrive at a contradiction. Thus no point Y may exist.

[/hide]


I was investigating the four color theorem at http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/The_four_colour_theorem.html
and was wondering if there was any such limit for 3 dimensional regions.  I quickly came to the conclusion that there was no such limit.  Is that true?

Yup, thats right. You can design a 3d map that requires as many colours as you want.

There is an interesting branch of mathematics called topological graph theory, which deals with graphs imbedded on three dimensional surfaces.  There are theorems about the minimum colors needed to color graphs on those surfaces.

Well, 4-color-graph theory hasn't been proved yet, I mean a formal proof. They used Computers and test many cases so, they assumed the truthful of the problem...

A proof is a proof, even if it means checking thousands of cases by computer. The mathematical community accepts the proof of the four-color theorem, even if some members don't really approve of the technique.

The reason that we can place a limit on the number of colors required to vertex-color a planar graph is that not all graphs can be drawn on a plane without having crossing edges.

Not only can we make a 3D graph that requires an arbitrarily high number of colors[1], but we can "draw" (embed) ANY graph in 3-space.

As for the 4-color theorem: They didn't just check a bunch of cases and assume that it was true based on that. They proved that all graphs of a certain type would HAVE to be within a certain (big) set of cases, and used a computer to check all those cases, which is why you might have that impression.

[1]  Just have n vertices, and an edge between every pair of vertices, and you will need n colors.


Does anybody know if parents are allowed to watch the countdown round?
Also, is the countdown round live on ESPN or not?

At my countdown round in Texas there were parents... But I don't think it would be on TV. Only the nationals Countdown is I think.

Do my parents have to register to watch the countdown, or not?

I am pretty sure there was a fee for watching the countdown round.  You need a badge to get in, and you need to register($35) to get a badge.  There is a fee for each meal as well, but in order to do any of it, you must register.  Good luck.

[quote="solafidefarms"]Yes. And I don't think its live.[/quote]

IT IS LIVE.  You have to register as a guest fo $35.00.  Regn. fee is waived if you volunteer.  There are plenty of volunteer opportunities.

What you watch on ESPN is not live.

Naga

so is it live on tv... cuz i've never watched it before (first year in mathcounts) so it would be really coool to see it, cuz there was this kid in our chapter that totally wiped everyone out at countdown, and i think he's going to nationals, so i'd love to see smart people battling it out :D

ESPN broadcasts of MATHCOUNTS countdown started in 2003.

They have not been live in the past., so I assume they won't be live this year either.

The spelling bee IS always live on ESPN, and in 2003 and 2004, ESPN ran the MATHCOUNTS countdown during the one-hour lunch break of the final day of the spelling bee.  (They also reran it several times after that, but the initial broadcast came on the speling bee day.)

So, if previous patterns hold, we can expect to see the national countdown on ESPN at noon eastern time on June 2 (the final day of the spelling bee, according to http://www.spellingbee.com.)

Who knows if they'll change the broadcast time.

Ya, they seem to be pretty inconsistent and uninformative about it...

To be honest, though, I think the countdown round needs to be longer and more extensive to be worth broadcasting.  It's great for promoting MathCounts and stuff, but I mean all I got to see was myself getting wiped out for being slow  :? , then explanations of the easy countdown round problems (although it was by Mr. Ruszyck one year I think, so I saw what he looked like for the first time :)).  Sorry for my inability to spell the name  :blush:

Lets have a bracket of 48 instead of 12, then I might make it in (Not Really) ;)


Solve the equation in positive integer$y^2=x^3+4$

No if $x\vdots 7$?,and what about the equation$y^3=x^2+4$in positive integer


vector
topology
real analysis
real analysis unsolved

Prove that a separable normed vector space is a countable union of compact subsets.

I don't think that's true :?. What if this space is complete? Then this says that a complete metric space is a countable union of compact subsets. By Baire's Theorem, one of these subsets has to have non-empty interior, which, in particular, means that in our normed vector space the closed unit ball must be compact. This only happens if the space is finite dimensional.

$L^{2}([0;1])$ is a separable normed vector space, but does not seem to be the union of compact subset ?

Oh, god, I have no idea what I had in mind when posting this. The real problem is to prove that a connected and locally compact metric space is a countable union of compact subsets. I really apologize for this!  :blush:  :oops:  :roll:

Cover the metric space $X$ with open sets having compact closures. Because every metric space is paracompact, we can find a refinement $(V_\alpha)_\alpha$ of this open cover which is also an open cover, and, in addition, locally finite. If we manage to prove that every set $V_\alpha$ intersects only finitely many other sets $V_\beta$, we can finish the proof as follows:

Consider the graph having vertices $v_\alpha$, with $v_\alpha$ and $v_\beta$ being connected iff $V_\alpha\cap V_\beta\ne\emptyset$. We assumed that we can prove that this graph is locally finite (i.e. every vertex has finite degree). The connectedness of $X$ also implies that it is connected (every two vertices can be connected by a finite path). It’s easy to see that such a graph is at most countably infinite: start with a vertex, then notice that there are only finitely many vertices with distance $\le 1$ from the chosen vertex, then that there are only finitely many with distance $\le 2$ from it, and so on. In other words, we’ve shown that the family $(V_\alpha)_\alpha$ is at most countably infinite. $X$ is the union of the sets $V_\alpha$ and hence of the compact sets $\overline{V_\alpha}$, so we’re done.

Now for the proof of the fact that every $V_\alpha$ intersects only finitely many $V_\beta$.

For each point $x\in\overline{V_\alpha}$ choose an open neighborhood $U_{x}$ which intersects only finitely many $V_\beta$. Since $\overline{V_\alpha}$ is compact, it’s covered by finitely many $U_{x}$; call them $U_{i},\ i=\overline{1,n}$. If $V_\alpha\cap V_\beta\ne\emptyset$, then $V_\beta$ intersects one of the sets $U_{i}$. By the construction of the $U_{i}$, there are only finitely many such $V_\beta$. This finishes the proof.

[quote="alekk"]is a separable and locally compact space the countable union of compact subsets ?[/quote]

No, not in general. The plan is this: I want to find a separable, compact Hausdorff space $X$ which is not first countable at some point $x$ (i.e. $x\in X$ does not have a countable local base of open neighborhoods). The counterexample we're looking for will be provided by $Y=X\setminus\{x\}$.

Indeed, being an open subset of a compact Hausdorff space, $Y$ is locally compact. Clearly, since $X$ is separable, so is $Y$. Finally, suppose $Y$ is a countable union of compact sets $F_{n}$. $x$ must then be the intersection of the open sets $V_{n}=X\setminus F_{n}$. Let $U_{n}$ be an open neighborhood of $x$ whose closure is contained in $V_{n}$, and let $V$ be any open neighborhood of $x$. Because the compact set $X\setminus V$ is covered by the open sets $X\setminus\overline{U_{n}}$, it must be covered by finitely many of them, say those with indices from $1$ to $n$. This, however, means that the open neighborhood $\bigcap_{i=1}^{n}U_{i}$ of $x$ is contained in $V$, and hence that the finite intersections of $U_{n}$'s form a countable local basis at $x$. This contradicts the choice of $x$.

All that's left now is to actually find such an $X$. One possible choice is $X=I^{I}$, where $I$ is a compact interval. In other words, $X$ is the topological product of continuum many compact intervals. Both the fact that this space is separable and the fact that it is not first countable (at any point) have been discussed before on the forum. Another choice would be $X=\beta\mathbb N$, the [url=http://planetmath.org/encyclopedia/StoneVCechCompactification.html]Stone-Cech compactification[/url] of a discrete countable space. The points of $X$ are simply the ultrafilters on $\mathbb N$. As for $x$, any free ultrafilter will do ($X$ is first countable at principal ultrafilters; in fact, principal ultrafilters are clopen subsets of $X$). Indeed, suppose such an ultrafilter $x$ has a countable local basis of open sets. Then, since $X$ is totally disconnected, $x$ has a countable local basis of clopen sets. The clopen subsets of $X$ are simply the subsets of $\mathbb N$, so basically this says that we can find countably many subsets $A_{n}$ of $\mathbb N$ such that any subset of $\mathbb N$ contained in the ultrafilter $x$ contains one of $A_{n}$. $x$ is a free ultrafilter, so all $A_{n}$ are infinite. It is then easy to construct a subset $A$ of $\mathbb N$ such that neither $A$ nor $B=\mathbb N\setminus A$ contain any of the $A_{n}$. $x$ being an ultrafilter, either $A$ or $B$ is a member of $x$, and we get a contradiction.


I am a new member, I hope you help me!

$x,y,z$ are real postive number. Prove that:

$(x^{2} + y^{2} + z^{2})(\frac{1}{(x-y)^2} + \frac{1}{(y-z)^2} + \frac{1}{(z-x)^2}) \geq \frac{9}{2}$
 :pilot:

[quote="socrates"]Probably one condition is $x\ne y\ne z$..........[/quote]

OK $(x-y)(y-z)(z-x) \neq 0$
I speak English very bad...

inequality is wrong...

$deg(x^2+y^2+z^2)=2, deg(\frac{1}{(x-y)^3} + \frac{1}{(y-z)^3} + \frac{1}{(z-x)^3})=-3$, hence their product wil have degree $-1$, while the $deg(RHS)=0$, so there might be some missing condition!

Since

$\sum \frac{1}{(x-y)^2} \geq \frac{9}{\sum (x-y)^2} = \frac{9}{2(x^2+y^2+z^2-xy-yz-zx)}$

it suffices to prove

$\frac{x^2+y^2+z^2}{x^2+y^2+z^2-xy-yz-zx} \geq 1$
$\iff xy+yz+zx \geq 0$

I hope I'm not missing anything.  :)

soo at least two of $x,y,z$ must be zero, which means the product $(x-y)(y-z)(z-x) = 0$ so equality can't hold...

Wrong,ThAzN1. This one is very easy. But the author found it by indentify, very long method.

I'm probably blind but could you tell me what's wrong?

I think ThAzN1's solution is correct......It is just by Cauchy-Schwarz.

hungkhtn want to say that that ineq is still true for a,b,c [b] are reals number [/b]. OK.

Yeah... But It's only from himseft and I'm sure, no one can be find out it!  :)

[quote="robinho"][quote="Nameless"]Dao Hai Long's solution:
We have $(\frac{x}{y-z}+1)(\frac{y}{z-x}+1)(\frac{z}{x-y}+1)=(\frac{x}{y-z}-1)(\frac{y}{z-x}-1)(\frac{z}{x-y}-1) \Rightarrow\sum(\frac{y}{z-x}\frac{z}{x-y})=-1$(1)
And $\sum(\frac{x}{y-z})^2\ge-2\sum(\frac{y}{z-x}\frac{z}{x-y})$  (2)
(1),(2)$\Rightarrow(\frac{x}{y-z})^2+(\frac{y}{z-x})^2+(\frac{z}{x-y})^2\ge2$(5)
But we have
$(\frac{x+y}{x-y}+1)(\frac{y+z}{y-z}+1)(\frac{z+x}{z-x}+1)=(\frac{x+y}{x-y}-1)(\frac{y+z}{y-z}-1)(\frac{z+x}{z-x}-1) \Rightarrow\sum(\frac{y+z}{y-z}\frac{z+x}{z-x})=-1$(3)
and
$\sum(\frac{x+y}{x-y})^2\ge-2\sum\frac{y+z}{y-z}\frac{z+x}{z-x}$(4)
(3),(4)$\Rightarrow\sum(\frac{x+y}{x-y})^2\ge2 \Rightarrow\sum(\frac{(x+y)^2}{(x-y)^2}+1)\ge5 \Rightarrow\sum\frac{x^2+y^2}{(x-y)^2})\ge\frac{5}{2}$(6)
(5)+(6)$\Rightarrow$ we solved this ineq.
 :lol:[/quote]

it very not good because I posted in [b]Diendantoanhoc.net[/b]

[/quote]
Dao Hai Long's ineq is very famous, its solution is in many books, I don't know you posted it and this solution isn't yours

[quote="hungkhtn"]Of Course, two solution isn't nice. The full solution is in 3- line.[/quote]

to Hungkhtn: em xin lỗi anh Hùng, nhưng có lẽ cách chứng minh của anh là tự nhiên nhất bởi vì nó xây dựng Bất đẳng thức trên nền bất đẳng thức. theo trích dẫn của anh Mrmath thì cách giải đó chỉ sử dụng 2 bất đẳng thức phụ thôi và em thấy nó rất hay !!!

Xét x,y,z cùng dấu, giả sử c là số nhỏ nhất, sau đó đổi c bởi -c, quy về trường hợp 3 số ko cùng dấu, có thể giả sử $\large a>b>0\ge{c}$

Ta có $\large \frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}\ge{ \frac{1}{2}(\frac{1}{a-b}+\frac{1}{b-c})^2}\ge{\frac{1}{2}(\frac{4}{a-b+b-c})^2}=\frac{8}{(a-c)^2}$

Và bây giờ thì $\large \frac{9}{(a-c)^2}\ge{\frac{9}{2(a^2+b^2+c^2)}}$

Và vậy là :pe

Thank Hung

Yes! First I created it and even I think that It's own mine. But.. after I see that It's belong Dao hai Long. Here Is the more difficult and trange I'd posted in this forum :

$\frac{1}{(2a-b)^2}+\frac{1}{(2b-c)^2}+\frac{1}{(2c-a)^2} \ge \frac{11}{7(a^2+b^2+c^2}$

Um.., about the problem that hungkhtn has posted, take $2a-b=x,2b-c=y,2c-a=z$ then the problem is equal to
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{11}{3(x^2+y^2+z^2)+4(xy+yz+zx)}$
A symmetric inequality and it is easy now, let have a try.  :)
In fact, the more general problem can be given, however the idea doesn't differ from above one:
Give $x,y,z$ real and $n \in [-1,2]$ . Find the minimum value of:
$(x^2+y^2+z^2+n(xy+yz+zx))(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})$  :)

[quote="Anh Cuong"]Um.., about the problem that hungkhtn has posted, take $2a-b=x,2b-c=y,2c-a=z$ then the problem is equal to
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{11}{3(x^2+y^2+z^2)+4(xy+yz+zx)}$
A symmetric inequality and it is easy now, let have a try.  :)
[/quote]
  At first,when I found the solution of this,I also using your substitutions,but the last inequality is not so easy (at least for me), ;) ,because x,y,z is [b]real[/b] numbers

I know they are real, but say ABC and it is an easy type in my view.  ;)
But sorry that I can't post solution now  :(
In fact Dao Hai Long's inequality bothers me more, since it can't be transfered in this inequality class.

[quote="hungkhtn"]Ask him when the equal taken  :P[/quote]
Who???Dao Hai Long or Anh Cuong?  ;)

The equality in Dao Hai Long's inequality occurs when $a=0,b=-c$.
The equality in hungkhtn doesn't occur at all, the strict appraisal is that:
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{11.607...}{3(x^2+y^2+z^2)+4(xy+yz+zx)}$, the equality occurs when $x=y=-0.658..., z=1$   :read:

It's the reason why I don't post the general, because We can't have the detail result ( it's because the result's depended on the root of equality degree 5 ). But the private case is enough to descire the general :)

Hi hi, I even haven't try it yet, i just know that it can be solved. However, I have just solve it recently, the equality occurs when $x=y=$ a root of a third degree equation . So it is not a good idea to general it indeed  :) .

Because that then you don't see it's quite difficult than you guess. Because you must divice problem in two case. I don't say more and please seach my problem. Soarer and Zhaobin had completed two cases. ! You 'll see the real idea in it!

No, my solution doesn't need to seperate two cases  ;) , if $n$ is a definite number then the problem is really not hard in my view. I said that I hadn't had a try that mean I hadn;t have a try to find that if the equation has nice roots ,since we just need to solve a third degree equation to find the minimum point.  :)
However it is really a problem to me if $n<-1$ or $n>2$  :?


we have $55$ numbers from set ${1,2,...,100}$.for $k \leq 9$ prove that defference of two of these numbers is $k$.


There are 2008 bags numbered from 1 to 2008, with 2008 frogs in each one of them. Two people play in turns. A play consists in selecting  a bag and taking out of it any number of frongs (at least one), leaving $ x$ frogs in it ($ x\geq 0$). After each play, from each bag with a number higher than the selected one and having more than $ x$ frogs, some frogs scape until there are $ x$ frogs in the bag. The player that takes out the last frog from bag number 1 looses. Find and explain a winning strategy.

By Zermelo's theorem there is a winning strategy, indeed this is a finite (with at most $ 2008^2$ plays), bipersonal game with no tie possiblity.

The first player has the winning strategy:

As a first move remove $ 2007$ frogs from bag number $ 2$, this will result on $ 2008$ frogs on bag number $ 1$ and $ 1$ frog in the rest of bags. Then there will be $ 2\cdot 2007 \plus{}1$ frogs in game.

Then continue playing "symmetrically" to the second player:
  If the second player takes out a frog of bag number $ a \plus{} 1$, first player must leave only $ a$ frogs on bag number $ 1$. This results on $ a$ frogs on bag number $ 1$ and $ 1$ frogs in bags $ 2,3,...,a$.
  If the second player leaves only $ a$ frogs on bag number $ 1$, first player must take out a frog from bag number $ a \plus{} 1$, this results the same as the previous consideration.
In this way, at every turn beginning with the first play of second player, an even number of frogs are taken out from the bags (or escape...)

Continue this process until only $ 1$ frog remais in bag number $ 1$, when this case came it will be the second player the one who moves, so second player loses.

 :)

This is an instance of the game [url=http://en.wikipedia.org/wiki/Chomp]Chomp[/url] -- specifically, it's Chomp for a $ 2008 \times 2008$ rectangle.


Pascal\u0027s Triangle
number theory solved
number theory

how to prove theorem about odd numbers in n-th row of pascal triangle
number of odd numbers in n-th row of pacal triangle =2^(number of 1 in "n" binary system )
for example in 7th row
7 in binary =111 so 2^3= 8
in 7 th row there is 8 odd numbers
how to prove this theorem?

I think you can do that with the following theorem taking $p=2$:

LUCAS' THEOREM :

Let $a=(a_na_{n-1}...a_1a_0)^p$ (in representation of base $p$)
Let $b=(b_nb_{n-1}...b_1b_0)^p$ (in representation of base $p$)

Then, $C_a^b=C_{a_n}^{b_n}C_{a_{n-1}}^{b_{n-1}}...C_{a_1}^{b_1}C_{a_0}^{b_0} (mod p)$


geometry
trigonometry
quadratics
geometry solved

I didn't participate at the test, but I heard it was a very difficult problem.
All scored 0 :)

[color=red][[b]Moderator edit:[/b] Here is this problem:

[b]First Romanian IMO TST 2003, problem 2:[/b] Let ABC be a triangle with < BAC = 60°. Let P be a point in the interior of this triangle so that PA = 1, PB = 2 and PC = 3. Find the maximum area of the triangle ABC.][/color]

But I wanted to say that I found two very easy solutions, using a little calculus.

[b]Solution     I[/b]
Consider the area maximized for an angle X  with P in interior of angle A
The other Angle will be 60-X. Now write Cosine theorem with this angles,
And u will get a quadratic equation. solve it for side B. Same way for finding side C. They will be only of cos(x) and sin(x) . Product BC must be maximized, derivates and that's it.

[b]The second solution is based on analytic geom.[/b]
Take A(0,0), C(0,cy) B(b,2*b).
The angle A will be 60 degs. Then let the maximized area be for a postition P(x,y) in the plane. Then x^2+y^2=1. Also, you can express
coords of B(b,2*b) with x&y, cause it will be the intersection of 2 circles.
Same thing for C.
But we must maximize the area , so maximize the product of sides b & c.
Then again Derivates and that's it :) :) :) :).

I checked everything with maple, so really, derivates have only one zero,
easy to find.

[color=red][[b]Moderator edit:[/b] Generally, Romanian TST problems can be found at http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26 ][/color]

What's the actual answer to the problem?

It's being discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5308]here[/url].


calculus
integration
logarithms
complex analysis
algebra
partial fractions
difference of squares

That depends on whether one is allowed to use complex analysis or not. If yes, just apply an appropriate version of the residue theorem. If not, you still can do a simple fractions decomposition (which is quite ugly but doable in this case).

[quote="maokid7"]how would you solve

$\displaystyle\int_{-\infty}^{\infty}\frac{1-x^2}{1+x^4}dx$[/quote]

Since the integrand being even, we can write

$\int\limits_{ - \infty }^\infty {\frac{{1 - x^2 }}{{1 + x^4 }}} dx$

$= - 2\int\limits_0^\infty {\frac{{1 - \frac{1}{{x^2 }}}}{{x^2 + \frac{1}{{x^2 }}}}} dx$

$= - 2\int\limits_0^\infty {\frac{{d(x + \frac{1}{x})}}{{(x + \frac{1}{x})^2 - 2}}}$

$= - \frac{1}{{\sqrt 2 }}\left[ {\log \left( {\frac{{(x + \frac{1}{x}) - \sqrt 2 }}{{(x + \frac{1}{x}) + \sqrt 2 }}} \right)} \right]_0^\infty = 0$

I was thinking long-division, but that works too :).

That method was what's known as "getting lucky".

The actual partial fractions idea: We want to stay real, and that means we need to factor $1+x^4$. With some experience, a sum of fourth powers isn't that hard to deal with: $1^4+x^4=(1^2+x^2)^2-2\cdot1\cdot x^2$. Treating that as a difference of squares, $1+x^4=(1+x^2-\sqrt{2}x)\cdot(1+x^2+\sqrt{2}x)$. With that, you can write $\frac{1-x^2}{1+x^4}=\frac{ax+b}{1+x^2-\sqrt{2}x}+\frac{cx+d}{1+x^2+\sqrt{2}x}$. Those pieces can be integrated (it works out nicely), and everything works well.

This was a proper fraction from the start; long division doesn't help in that case.

[quote="maokid7"]how would you solve

$\int_{-\infty}^{\infty}\frac{1-x^{2}}{1+x^{4}}dx$[/quote]

Say $I = \int_{0}^{\infty}\frac{1}{1+x^{4}}dx$
Substitution of y=1/x yields
$I = \int_{0}^{\infty}\frac{y^{2}}{1+y^{4}}dy$
Subtract these two integrals
$0 = I-I = \int_{0}^{\infty}\frac{1-x^{2}}{1+x^{4}}dx$
The integrand is even, so
$\int_{-\infty}^{\infty}\frac{1-x^{2}}{1+x^{4}}dx = 0$


1).  If $\vec{a}$ and $\vec{b}$ are not collinear, and for real numbers x and y, we have:
\[(2x-y) \vec{a} + 4\vec{b} = 5 \vec{a} + (x-2y) \vec{b}\]
Find $x+y$.

2).  If O is the intersection of the medians of a triangle ABC, find: $\displaystyle\vec{OA}+\vec{OB}+\vec{OC}$.

[quote="cezar lupu"]the second statement is trivial.That sum is null vector.[/quote]
yep
we can extend one of the three medians and construct a parallelogram.

What about the first one?

[quote="seamusoboyle"]$1$[/quote]
you got it.

forgot about this! Yeah you can just eqate them