The integers 1,...,2n are arranged in any order on 2n places numbered 1,...,2n. Now we add its place number to each integer. Prove that there are two among the sums which have the same remainder mod 2n

[hide]Let the permutation be $ a_{1}, a_{2}, ...a_{2n}$. Assume to the contrary; then the new sums form a complete residue class mod $ 2n$, and we have $ (a_{1}+1)+(a_{2}+2)+...+(a_{2n}+2n) \equiv 1+2+...+2n = n(n+1) \neq 0 \pmod{2n}$, but $ a_{1}+a_{2}+...+a_{2n}+1+2+...+2n \equiv 2n(n+1) \equiv 0 \pmod{2n}$. [/hide]

Minor correction: $ 1+2+\ldots+2n = n(2n+1)$; you need the fact that $ 2n+1$ is odd to make that work.


1 should be $ 7 \mid (5^{2n} \plus{} 3)(2^{5n}\minus{}2)$
[hide="1"]Mod 7, $ 5^{2n}\plus{}3 \equal{} 25^n \plus{} 3 \equiv 4^n \plus{} 3$, which is equivalent to $ 0$ for odd $ n$. $ 2^{5n} \minus{} 2 \equal{} 32^n \minus{} 2 \equal{} 5^n \minus{} 2$, which is equivalent to $ 0$ for even $ n$.[/hide]

[hide="2"]Mod 27, $ 2^{5n\plus{}1} \plus{} 5^{n\plus{}2} \equal{} 2 \cdot 32^n \plus{} 25 \cdot 5^n \equiv 2 \cdot 5^n \plus{} (\minus{}2) \cdot 5^n \equal{} 0$[/hide]

[quote="MellowMelon"]1 should be $ 7 \mid (5^{2n} \plus{} 3)(2^{5n} \minus{} 2)$
[hide="1"]Mod 7, $ 5^{2n} \plus{} 3 \equal{} 25^n \plus{} 3 \equiv 4^n \plus{} 3$, which is equivalent to $ 0$ for odd $ n$. $ 2^{5n} \minus{} 2 \equal{} 32^n \minus{} 2 \equal{} 5^n \minus{} 2$, which is equivalent to $ 0$ for even $ n$.[/hide]

 I don't know if i understand you correctly, but how come that $ 4^n \plus{} 3$, is 0 mod 7 if n = 3?

[quote="MellowMelon"]Okay, so my correction is wrong, not to mention both lines of my solution. It's true only when $ n \equiv 1,4 \pmod{6}$. Do you have the right version of the problem?[/quote]

The $ \minus{}2$ in the second binomial should be at the exponent,  you have misread the problem. :D


search
real analysis
real analysis theorems

please, send me any e-book you have related to Limits. [looking for a link or a book]

thanks anyway
dl


[quote="Inspired By Nature"]Find a polynomial such that $xP(x-1)\equiv (x-3)P(x)$.[/quote][hide] \[x=3\qquad\Rightarrow\qquad 3P(2)=0P(3)\qquad\Rightarrow\qquad P(2)=0.\]  \[x=0\qquad\Rightarrow\qquad 0P(-1)=-3P(0)\qquad\Rightarrow\qquad P(0)=0.\] Let $P(x)=(x-2)(x)Q(x)$. Then \[x(x-3)(x-1)Q(x-1)=(x-3)(x-2)xQ(x),\]  \[(x-1)Q(x-1)=(x-2)Q(x).\]  \[x=1\qquad\Rightarrow\qquad 0Q(0)=-1Q(1)\qquad\Rightarrow\qquad Q(1)=0.\] Let $Q(x)=(x-1)R(x)$ or $P(x)=(x)(x-1)(x-2)R(x)$. Then \[x(x-1)(x-2)(x-3)R(x)=(x-3)(x)(x-1)(x-2)R(x),\] which is always true. Hence, \[P(x)=x(x-1)(x-2)R(x),\] where $R(x)$ is some polynomial.[/hide]

[quote="rzsolt"]Does the $\equiv$ means something special here?[/quote]

it means congruent

Yeah, like in modular arithmetic, but here we treated it just like an equal sighn, right? So why was it used here?

Does this imply that $P(x)$ has to be a polynomial of degree higher than $3$?

By the way, the $\equiv$ sign was used to mean that the two polynomials are congruent.

[quote="JBL"]Saying "the $\equiv$ sign means congruent" is not helpful unless you explain what you think it means for two polynomials to be congruent.
Most likely, in this case it meant to emphasize that they are equal as polynomials, i.e. not that they are equal for some value of $x$ but rather that they are equal for all values of $x$ (that is, they are the same polynomial).[/quote]
Thats exactly what the question means, thank you JBL. :)
[hide="hint"]Just find the roots of the given polynomial and you will find $p(x)$.[/hide]
[quote="towersfreak2006"]
P(x)=x(x-1)(x-2)R(x),
where $R(x)$ is some polynomial.[/quote]
hmm, you might want to check that over. :read:

[quote="Inspired By Nature"][quote="JBL"]Saying "the $\equiv$ sign means congruent" is not helpful unless you explain what you think it means for two polynomials to be congruent.
Most likely, in this case it meant to emphasize that they are equal as polynomials, i.e. not that they are equal for some value of $x$ but rather that they are equal for all values of $x$ (that is, they are the same polynomial).[/quote]
Thats exactly what the question means, thank you JBL. :)
[/quote]

Or put in the simplest manner: $=$ denotes an equation, $\equiv$ denotes an identity.

[quote="towersfreak2006"] Hence, \[P(x)=x(x-1)(x-2)R(x),\] where $R(x)$ is some polynomial.[/quote]

This works when $R(x)=1$, but if $R(x)=x$, we get $x-1=x$.  :maybe:

yeah, thats why I asked towersfreak2006 to check it over.

[quote="calculuslover800"][hide="solution, and some other stuff"]
$P(x) = C*x(x-1)(x-2)$
where $C$ is a constant,....  which is what towersfreak put, without the extra polynomial.
If could multiply $P(x)$ by a polynomial $R(x)$, then $R(x-1)=R(x)$ for all $x$, but then if $R(x)$ had any roots, it must have an infinite number of roots, which is a contradiction.   Therefore the only polynomail you could multiply $P(x)$ by is just a constant term.
[/hide][/quote]
I got $P(x)=x(x-1)(x-2)$. I didn't find the need to add a constant. All I did was simply find all the [hide="roots"]0, 1, and 2[/hide] and multiply them together to get the value of $P(x)$. I don't think the $C$ is needed.

[quote="Farenhajt"][quote="matt276eagles"]$Q(x)$ is constant $\implies Q(x)=0$[/quote]

Please explain your reasoning here. Is zero only existing constant?  :roll:[/quote]

Oops... I forgot that it was $R$ that had roots at $0,1,2$, not $Q$.  I'll edit.  :blush:

[quote="matt276eagles"]I'm not sure about this, but here's my solution:
[hide]For $x=0$, we have $0=-3P(0)$, so $P(0)=0$.
For $x=1$, we have $P(0)=-2P(1)$, so $P(1)=0$.
For $x=2$, we have $2P(1)=-P(2)$, so $P(2)=0$.
Let $P(x)=x(x-1)(x-2)Q(x)$, where $Q$ is a polynomial.  Substituting, we obtain $x(x-1)(x-2)(x-3)Q(x-1)=(x-3)x(x-1)(x-2)Q(x)$.  Therefore, for all $x\ne 0,1,2,3$, we have $Q(x-1)=Q(x)$.  Since $Q$ is a polynomial, $Q(x)$ is constant.  Let $Q(x)=c$, so that $P(x)=cx(x-1)(x-2)$.
Plugging this back into the original equation, we get $xc(x-1)(x-2)(x-3)=(x-3)cx(x-1)(x-2)$, which is true for all $x$, as desired.[/hide][/quote]
hmm, makes sense. Its a more specific solution than mine. Thanks everybody. :)


A nuclide $X$ of decay constant $\lambda_{1}$ decays to a nuclide $Y$ of decay constant $\lambda_{2}$.Show that the instantaneous rate of decay of nuclide $Y$ in a sample is given by $\frac{dN}{dt}=\lambda_{1}N_{X}-\lambda_{2}N_{Y}$ where $N_{X}$ and $N_{Y}$ are the instantaneous numbers of nuclei of $X$ and $Y$ respectively in the sample .

I dont quite know the way to show .Anyone help?

[quote="Dr. No"][quote="shyong"][quote="Dr. No"]First, define $N = N_{Y}-N_{X}$. (The net increase of $Y$ particles.)

Second, take the derivative, $\frac{d N}{dt}= \frac{dN_{Y}}{dt}-\frac{dN_{X}}{dt}$

Then it's simply a matter of substitution as:

$\frac{dN_{Y}}{dt}\equiv-\lambda_{2}N_{Y}\\ \frac{dN_{X}}{dt}\equiv-\lambda_{1}N_{X}$

$\frac{dN}{dt}= \lambda_{1}N_{X}-\lambda_{2}N_{Y}\,\,\, QED$[/quote]

isnt that $N_{Y}<N_{X}$ ? so it is net decrease of $Y$ particles instead ? or my concept is wrong ?[/quote]

I don't know if I said it exactly the right way, but the idea is that you have particles decaying from state $Y$ while other particles are entering state $Y$ from state $X$; it is that net rate of decay that you're trying to show. Besides if there isn't an increase in $Y$ particles that will show in the sign of your final answer.[/quote]

ok thanks , i get what you mean . :)


For every positive $a$, $b$, and $c$ such that $a^2+b^2+c^2=1$, prove that

\[ {a\over a^3+bc} + {b\over b^3+ca} + {c\over c^3+ab} > 3 . \]

[quote="hungkhtn"]I don' think so.
If we take $b=c\to 0,a\to 1$? The RHS is $1$.[/quote]
no, I think three is fine. Anyway, this has been posted already.

[quote="indybar"]Hmm... interesting. I took it from tuymaada post http://www.mathlinks.ro/Forum/viewtopic.php?t=44430, which seems all solutions are wrong..[/quote]
  I posted a solution on ML and I think Iura's solution is right .


Let $ Y$ be the complement of the following subset of the plane $ \mathbb{R}^2:$
$ \{(x,0) \in \mathbb{R}^2: x \in \mathbb{Z} \}$
Prove that $ \pi_1(Y)$ is a free group on a countable set of generators.

I don't know how to start this problem.  I just need a few helpful hints.  Thanks in advance.

The generators can, of course, be identified with loops around each missing point.

One standard tool for identifying a fundamental group is the Seifert-Van Kampen theorem. Can you use this to show that the fundamental group of the slice $ \minus{}n\minus{}\frac12< x< n\plus{}\frac12$ is free on $ 2n\plus{}1$ generators? Then, can you show that the fundamental group of $ Y$ is the limit of these?

The Van Kampen theorem (as stated in May's "A Concise Course in Algebraic Topology", p. 17) works on covers by infinitely many open subsets; you don't need to make
a limiting argument after using it. Either way, there's some machinery needed (either in the proof of Van Kampen or on how $ \pi_1$ behaves with
respect to limits and colimits.)

EDIT: May have misread May a little, but the version in his text definitely at least shows that $ \pi_1$ takes colimits of open sets
to the colimit of the $ \pi_1$s.

Actually, I take that back; $ Y$ deformation retracts to an infinite sequence of adjacent circles
(I will call this space $ Z$). Try to find a covering space $ Z$. (Hint: the real line is a universal covering
space for a circle; the cayley graph for $ F_2$ is a covering for two adjacent circles, etc.
Try looking at the cayley graph for $ F_\infty$ ... and show this is a universal covering space for $ Z$.
I haven't worked out all the details but this doesn't look too hard.)


Compute the sum of the frist 2006 numbers of the sequence 1,2,2,3,3,3,4,4,4,4.....


LaTeX
inequalities
inequalities unsolved

Sorry Siuhochung If I say your ineq not true.
You can Take
$a=k^5,b=k,c=k^{-6}$
and You have:
$\frac{1}{1+b^3+c^3} \ge \frac{ab+bc+ca}{a^2+b^2+c^2}$
Because:
$\frac{1}{1+k^3+k^{-18}} \ge \frac{k^6+k^{-1}+k^{-5}}{k^{10}+k^2+k^{-12}}$
It will satisfy if $k \rightarrow +infinitive$

I think I have a nice solution for the problem.
we can prove 1/(1+a+b) <= a k/(a k+b k +c k )
k=1/3

In addition,I have downloaded Latex installated it.But I can't find the mian pile.Who can help me? :)

Let a,b,c>0, abc=1.We have
1/(1+a+b)+1/(1+b+c)+1/(1+a+c)1/(a+2)+1/(b+2)+1/(c+2)
the ineq should come from 02-03 china team training.

[quote="zhaobin"]In addition,I have downloaded Latex installated it.But I can't find the mian pile.Who can help me? :)[/quote]
What do you mean "main file"? There are many files and all they are main :)
You need a LaTex Editor, like Texniccenter, to write and compile latex documents.

Yes!
Actually yesterday we had an exam in a math training camp and that was the hardest problem.

[quote="hardsoul"]Can anyone prove 1/(1+a+b)+1/(1+b+c)+1/(1+a+c) <= 1/(a+2)+1/(b+2)+1/(c+2).[/quote]

This inequality is generally wrong (e. g. take a = b = c = 0.5). But if you assume that abc = 1, then this is the problem from http://www.mathlinks.ro/Forum/viewtopic.php?p=115579 . Maybe it is also true for all positive a, b, c such that $abc \geq 1$ ?

  darij

what about this problem: abc=1 (a,b,c>0), prove that
1/(a+b+1) + 1/(b+c+1) + 1/(c+a+1) > 1/2

sory, Myth!!
i was stupid when let c=1 then take b close to 0 :blush:  :blush:  :blush:

[quote=Soarer]In a very old post it is proven that if abc=1, then
1/(1+a+b)+1/(1+b+c)+1/(1+c+a)<=1
Auh Cuong had a nice proof by substituting a=x^3,b=y^3,c=z^3
What about this, if xyz=1 ,
1/(1+x^3+y^3)+1/(1+y^3+z^3)+1/(1+z^3+x^3)<=(xy+yz+zx)/(x^2+y^2+z^2)
is this true?[/quote]
In a very old post it is proven that if $abc=1$, then
$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq1$$
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=409496&hilit=Moscow+1997]Moscow Olympiads 1997[/url]


algebra
function
domain
integration
limit
logarithms
complex analysis

[quote="yanglunj"]Find the domain and range of the two functions f(x) = x^x and g(x) = x^-x. Justify your answer.

I could not find any way to solve this problem completely, given that I was unable to evaluate this function for irrational x with the method below as a definition for irrational exponents, as I am unfamiliar with integration.

$ \int_{1}^{y} \frac {dt}{t} \equal{} p\int_{1}^{a} \frac {dt}{t}$

This is how far I got for the domain of f(x) and g(x)

Subsets of its domain that I found with overlaps:
1) All real numbers greater than zero
2) All x = a/b, a does not equal zero, where a and b have no common factors except 1, and b is odd. *

*This is what throws me off.
3) All x = a/b, where a and b have no common factors except 1, b does not equal zero, and b is even is undefined.

2) and 3) are both infinite sets. Given two rational numbers on a number line, there is an infinite number of rational numbers in between those two numbers. So, f(x) is defined at an infinite number of points, and undefined at another infinite number of points...

For the range, f(x) appears to cover (-inf,inf), however, there appears to be holes in the range, but I am unable to determine their locations.

g(x) has one noticeable gap in the range at [0, -0.698...), but I could not find others.

If anyone here is able to help, I would deeply appreciate it.  :)[/quote]

The domain of $ f(x)$ and $ g(x)$ is $ \mathbb{R}^ \plus{} .$ (Otherwise, the definitions don't really make sense...)

First, clearly $ f(x) > 0$ and $ \lim_{x\to\infty} f(x) \equal{} \infty.$ Also $ f'(x) \equal{} \left(e^{x\ln x}\right)' \equal{} \left(x\cdot\frac {1}{x} \plus{} 1\cdot\ln x\right)x^x \equal{} \left(1 \plus{} \ln x\right)x^x$ and $ f''(x) \equal{} \left(\left(1 \plus{} \ln x\right)^2 \plus{} \frac {1}{x}\right)x^x > 0$ so $ f$ has a minimum if $ 1 \plus{} \ln x \equal{} 0$ or $ x \equal{} e^{ \minus{} 1}$ which gives the range $ \left[e^{ \minus{} \frac {1}{e}},\infty\right).$

We also have $ g(x) \equal{} \frac {1}{f(x)}$ which gives a range of $ \left(0, e^{\frac {1}{e}}\right].$

[b]bertram[/b]'s solution is correct, but I would not want to take the second derivative. The formula $ f'(x)\equal{}(1\plus{}\ln x)x^x$ already shows that $ f$ is decreasing for $ x<e^{\minus{}1}$ and increasing afterwards.

[quote="yanglunj"][quote="bertram"]
The domain of $ f(x)$ and $ g(x)$ is $ \mathbb{R}^ \plus{} .$ (Otherwise, the definitions don't really make sense...)

First, clearly $ f(x) > 0$ and $ \lim_{x\to\infty} f(x) \equal{} \infty.$ Also $ f'(x) \equal{} \left(e^{x\ln x}\right)' \equal{} \left(x\cdot\frac {1}{x} \plus{} 1\cdot\ln x\right)x^x \equal{} \left(1 \plus{} \ln x\right)x^x$ and $ f''(x) \equal{} \left(\left(1 \plus{} \ln x\right)^2 \plus{} \frac {1}{x}\right)x^x > 0$ so $ f$ has a minimum if $ 1 \plus{} \ln x \equal{} 0$ or $ x \equal{} e^{ \minus{} 1}$ which gives the range $ \left[e^{ \minus{} \frac {1}{e}},\infty\right).$

We also have $ g(x) \equal{} \frac {1}{f(x)}$ which gives a range of $ \left(0, e^{\frac {1}{e}}\right].$[/quote]

Those ranges seem to only apply to the positive portion of the graph, as f(-1) = -1 which lies below e^(-1/e), although I have not thought about using natural logarithms to solve it. Thank you =)

Something I found out while playing around with my calculator:
The graph of f(x) for x < 0 seems to be the union of g(-x)and -g(-x)

The graph of g(x) for x < 0 seems to be the union of f(-x)and -f(-x)

For the TI-83+: Put the viewing window as
x: -3.76,0
y: -2, 2

and plot f(x) , g(-x), and -g(-x) on sequential. (set f(x) as dot)

You can repeat for g(x), f(-x), and -f(-x)[/quote]

The real pain with negative numbers is that you can't really expand it there, since it won't be defined for any irrational number and it will be discontinouus about everywhere. So it really makes sense to restrict it to $ \mathbb{R}^\plus{}$ (One could use De l'Hospital to include 0 with $ f(0)\equal{}g(0) \equal{} 1$).

Conclusion: Negative bases are baaaaaaaaad.

You can leave the negative integers in the domain

You can, but do you want to? Do you also want to say that $ \minus{}\frac13$ is in the domain, because $ (\minus{}1/3)^{\minus{}1/3}\equal{}\minus{}3^{1/3}$? I'd rather not to. For me, the natural domain of $ f(x)\equal{}x^x$ is $ (0,\infty)$.

Well its not an interesting problem if we dont try find these points!

[quote="mlok"]You can, but do you want to? Do you also want to say that $ \minus{} \frac13$ is in the domain, because $ ( \minus{} 1/3)^{ \minus{} 1/3} \equal{} \minus{} 3^{1/3}$? I'd rather not to. For me, the natural domain of $ f(x) \equal{} x^x$ is $ (0,\infty)$.[/quote]

That's common sense =)

They are apparently discontinuous everywhere to the left of 0, as bertram pointed out. I went into it a bit today and thought

For all a and b satisfying above conditions
f(a/b) is positive for all even a, and negative for all odd a

[quote="bertram"]

The real pain with negative numbers is that you can't really expand it there, since it won't be defined for any irrational number and it will be discontinouus about everywhere. So it really makes sense to restrict it to $ \mathbb{R}^ \plus{}$ (One could use De l'Hospital to include 0 with $ f(0) \equal{} g(0) \equal{} 1$).

Conclusion: Negative bases are baaaaaaaaad.[/quote]

It's not defined at irrationals? Thanks, I guess that sums up all numbers in R now and whether it's defined or not =)

$ x^x\equal{}e^{x \ln{x}}$ and thus is holomorphic except at zero, and probably along, say, the negative real axis. Basically, in Complex Numbers its continuous. So we can put any number in and get out a complex answer. Even for irrationals, you can take a limit of rational points that converge to it

[quote="seamusoboyle"]$ x^x \equal{} e^{x \ln{x}}$ and thus is holomorphic except at zero, and probably along, say, the negative real axis. Basically, in Complex Numbers its continuous. So we can put any number in and get out a complex answer. Even for irrationals, you can take a limit of rational points that converge to it[/quote]

OK, but if we use complex numbers we might just as well say that the domain is $ \mathbb{C}.$ And then there's also the problem of multi-valued logarithm etc...

thats why we take out the negative real axis  :lol: so were only working on one branch. I was just pointing out that it can still be continous, and defined for irrational values.

Bertram: when you're responding to the post directly preceding yours, it's bad form to quote it.  That's what you are responding to by default, so all it does is makes the thread more difficult to read.

It's continuous in the complex numbers only as a multivalued function.  I've discussed related questions several times before on the forum, so you might try a search.*  The only negative points where it's defined as a real function are rational numbers with odd denominators (in particular, not for any irrational number).  However, if we use the given definition of exponentiation (via integration), it's not defined anywhere left of the origin.

In somewhat related topics, I found http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=983130192&t=149758
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=983130192&t=77380
etc.  (I was searching for my posts with the word "complex" in them.)


calculus
integration
inequalities
function
calculus computations

the function f(x) is continuos on $ [0,\pi]$ satifying $ xf(y) \plus{} yf(x) \leq 1$ for all $ x \in [0,1]$. prove that:
$ \int_0^1 f(x)dx \leq \frac {\pi}4$
i'm sorry if it was posted :blush:

i am sorry if  i am wrong but donet we need any limits  on  `y` :(  ?

If we suppose inequality holds for each x, y in [0,1] then put $ x \equal{} sin(t), y \equal{} cos(t)$, t in $ [0,\frac{\pi}{2}]$ and function $ G(t) \equal{} F(sin(t)) \minus{} F(cos(t)) \minus{} t$ is decreasing on $ [0,\frac{\pi}{2}]$, having $ G'(t) \leq 0$, where $ F(x) \equal{} \int_0^x f(u) du$, so $ G(0) \geq G(\frac{\pi}{2})$. From here, $ F(1) \leq \frac{\pi}{4}$

One interesting question is: can we find a function f with initial properties, such that $ F(1) \equal{} \frac{\pi}{4}$ ?


I am interested in buying these books/textbooks due to the fact that I feel like I am missing bits and pieces of information from algebra/geometry/problem solving in general. My friend says that they are a waste due to the fact that he does not understand any of the material inside the book (vol. 1). I've finished precalc with a little knowledge in calc, what books do you suggest I should purchase? I don't want to buy the expensive ($ \$$50+) textbooks that I have easily mastered 90%+ of the material. But I do not want to purchase (vol 1.) and then not have a clue what they are talking about... So I'm just wondering what do I need to know in order to successfully understand (vol 1.)? What should I buy?

Thanks

If you do not understand something, how is it a waste of money? I believe you should buy volume 2 as volume is basic material (in fact, you probably learned the majority in middle school). Volume 2 covers introductory calculus, as well as vectors and advanced trig.

Take a look at the table of contents for [url=http://www.artofproblemsolving.com/Books/Vol1/toc.pdf]volume 1[/url] and [url=http://www.artofproblemsolving.com/Books/Vol2/toc.pdf]volume 2[/url] to decide  :) .

Are you getting these books to prepare for the AMC?  If so, how do you do on the AMC?

I got Volume 1 over this past school year, when I was taking Algebra, and some parts were a bit difficult, all you have to do is reread and reread and try all the exercises. I'm going into 7th grade this year, and it's a challenge for me, but rewarding. I really recommend it.

Also, if you're still not sure about it, ask your friend if you can borrow it and take a look.


geometry
circumcircle
conics
projective geometry
geometry unsolved

One circle meets BC at $ D_1, D_2$, meets CA at $ E_1, E_2$, meets AB at $ F_1, F_2$.

Also, $ D_1 E_1$ and $ D_2 F_2$ meets at L

        $ E_1 F_1$ and $ E_2 D_2$ meets at M

        $ F_1 D_1$ and $ F_2 E_2$ meets at N

Show that AL, BM, CN meets at one point.

By Pascal theorem, $ A \equiv E_1E_2 \cap F_1F_2,$ $ L \equiv D_1E_1 \cap F_2D_2,$ $ X \equiv D_2E_2 \cap F_1D_1$ are collinear, let $ p_a \equiv XAL.$ Similarly, if $ Y \equiv E_2F_2 \cap D_1E_1,$ $ Z \equiv F_2D_2 \cap E_1F_1,$ then $ B, M, Y$ are collinear and $ C, N, Z$ are collinear. Let $ p_b \equiv YBM, p_c \equiv ZCN.$ By Steiner theorem, the 3 Pascal lines $ p_a, p_b, p_c$ are concurrent.

Proof of Steiner theorem: Consider two arbitrary triangles $ \triangle D_1E_1F_1, \triangle D_2E_2F_2$ with a common circumcircle, their six sidelines  $ F_1D_1 \equiv XN,$  $ E_2F_2 \equiv NY,$ $ D_1E_1 \equiv YL,$ $ F_2D_2 \equiv LZ,$ $ E_1F_1 \equiv ZM,$  $ D_2E_2 \equiv MX$ are tangent to a common conic (see [url]http://www.mathlinks.ro/viewtopic.php?t=233430[/url]). Therefore, the main diagonals $ XL, YM, ZN$ of the (non-convex) hexagon $ XNYLZM$ are concurrent by Brianchon theorem.

Dear Yetti and Mathlinkers,
thank you for your nice proof inolving three Pascal's lines. I think that the point of concurs is a Steiner's points...
Do you have a technik for chasing all the Steiner's points from starting with a cyclic hexagon?
Sincerely
Jean-Louis

Dear Jayme,
sorry to disappoint you. I have never heard of this Steiner theorem about concurrent Pascal lines, until I tried this problem. After finding the 3 Pascal lines, I looked it up on [url]http://www.cut-the-knot.org/Curriculum/Geometry/PascalLines.shtml[/url], but I did not like the proof there. I think that given a cyclic hexagon, there are $ \frac {1}{2} \cdot \left(\begin{array}{c}6 \\
3\end{array}\right) \equal{} 10$ different combinations to pick the 2 triangles with common circumcircle, which would result in 10 Steiner points. But I know that there are 20 Steiner points. So, there should be 2 Steiner points per triangle pair.

Let $ P \equiv E_1F_1 \cap E_2F_2,\ \ \ Q \equiv F_1D_1 \cap F_2D_2,\ \ \ R \equiv D_1E_1 \cap D_2E_1$
and $ P^* \equiv D_1E_2 \cap E_1F_2,\ Q^* \equiv E_1F_2 \cap F_1D_2,\ R^* \equiv D_1E_2 \cap E_1F_2$
(running out of alphabet).

$ PP^*L,\ QQ^*M,\ RR^*N$ would be the other concurrent triple of Pascal lines per triangle pair ?

Dear Mathlinkers,
for an answer to my question, see
http://perso.orange.fr/jl.ayme vol. 12 Hexagramma mysticum p. 54
Sincerely
Jean-Louis

This problem from China-2005 TST or NMO.
Easy by cheva's theorem(trigonametric).

Dear Mathlinkers,
now xe can say that the point of concur is a Kirkman's point.
Sincerely
Jean-Louis

We have that by Pascal's theorem  $ ABC $ and $ LMN $ are perspective triangles, so $ AL,BM,CL $ are concurrent by Desargues's theorem.


So basically, we just have to consider 2,3,7. Then we can make a 3-dimensional mod chart.

[quote="i_like_pie"][quote="cincodemayo5590"]What is the largest score that is [u]not[/u] attainable in an (American) football game? (The different types of scoring are, in points, $2, \ 3, \ 6, \ 7, \ 8$)[/quote]
How do you get 2 and 8?[/quote]

2 is from a safety (the opposing team gets pushed back into their own endzone) and 8 is from when you get a touchdown, and then get the ball into the endzone instead of kicking it.

How do you do a 3-dimensional mod chart?

Isn't it 1? anything above 8 you get by adding 2s to 7 or 8, 5=2+3 and 4=2+2

Anything even can obviously be made. Anything odd, except 1 can too. Since anything odd can be expressed as 2k + 1 = 2(k-1) + 3. Except when k = 0.

Oh oops, I didnt see how trivial this problem was when i was writing it  :oops:

Here's a better problem: Get rid of the $2$-point scores, so you've got points in $3, \ 6, \ 7, \ \text{and }8$ points. What's the highest score not attainable then?

The point is to find a general formula like the Chicken McNugget formula for these sorts of problems.

Anything other than $1,2,4,5$ can be made be adding the correct number of $3$'s to one of $6,7,8$- which cover between them all residues $\mod 3$.

Tim, you're right.
Phelpedo, you're right also.

(LOL i'm sorry, this problem has been pretty trivial.)

[u]Ok, here's the actual problem:[/u] :D

What is the largest number that cannot be expressed as a linear combination of relatively prime positive integers x, y, and z?

[quote="cincodemayo5590"]Tim, you're right.
Phelpedo, you're right also.

(LOL i'm sorry, this problem has been pretty trivial.)

[u]Ok, here's the actual problem:[/u] :D

$\text{What is the largest number that cannot be expressed as a linear combination of positive integers x, y, and z?}$[/quote]

well, if x,y,z are not relatively prime, then the answer is infinitely large

[quote="junggi"][quote="cincodemayo5590"]Tim, you're right.
Phelpedo, you're right also.

(LOL i'm sorry, this problem has been pretty trivial.)

[u]Ok, here's the actual problem:[/u] :D

$\text{What is the largest number that cannot be expressed as a linear combination of positive integers x, y, and z?}$[/quote]
i think I heard somewhere(stanford math circle, i think) that this problem was unsolved(meaning there's no nice closed form or w/e) when x,y,z are pairwise coprime(or something like that)[/quote]

ok. thanks!


Does there exist $ f: N\rightarrow N$ such that
$ f(f(n))\equal{}n\plus{}2[\frac{n}{2}]\plus{}1$ for all $ n\leqslant 2006$


function
limit
algebra
domain
real analysis
real analysis unsolved

[quote="perfect_radio"]Find all $f : ] 0, \infty [ \mapsto \mathbb{R}$ s.t. $f(x) - x \to 0$ as $x \to \infty$ and $f(f(f(x)))+2x=f(3x)$.[/quote]

 :blush:

It's a nice one ;). First considr $\alpha(x)=f(x)-x$ so your condition can be reformulated as:
\[ \lim_{x\longrightarrow \infty}\alpha(x)=0\quad and\quad \alpha(3x)=3\alpha(x)  \]
Now for $x\in]0,\infty[$, we have $\alpha(3^nx)=3^n\alpha(x)\longrightarrow 0$ as $n\longrightarrow\infty$,  that means that $\alpha(x)=0$, then $f$ must be the identity.

A problem which is solved in the same way:
   Let $f: (0,\infty)\to\Re$ such that :
 $1)f(x)+2x=f(3x),\forall x>0$
 $2)\lim_{x\to\infty}(f(x)-x)=0$
   Determine $f$.

No, you solved only that one from the shortlist, which unfortunately had a typo. The original problem (that I submitted and which is posted by Perfect Radio) is trickier.  ;)  Needless to say, the solution is wrong (for the original problem, the other one being of course trivial).

You are right, I couldn't sleep until finished it  :P . The correct reformulation is:
\[ \lim_{x\longrightarrow \infty}\alpha(x)=0\quad and\quad \alpha(3x)=\alpha(x)+\alpha(f(x))+\alpha(f(f(x)))  \]
Since $f$ takes positives values, $f(3x)=f(f(f(x)))+2x>2x$ or $f(x)>(2/3)x$.
Now if $f(x)>\beta x$, $f(f(f(x)))>\beta^3x$ then $f(3x)>\beta^3x+2x$ or \[ f(x)>\beta x\Longrightarrow f(x)>\frac{\beta^3+2}{3}x  \] Consider $h(\beta)=\frac{\beta^3+2}{3}$, then $f(x)>(2/3)x\Longrightarrow f(x)>h(2/3)x\Longrightarrow f(x)>h(h(2/3))x\ldots$ that is \[ \forall\,n\in\mathbb{N},\,f(x)>h^n(2/3)x \]
Since $\{h^n(2/3)\}\subseteq [0,1]$ is monotone it must converges to a fix point of $h$ so it must be 1, then $f(x)\geq x$ or $\alpha(x)\geq 0$. Now the reformulation of the second condition implies
\[ \alpha(3x)\geq \alpha(x)  \]
So the same argument will work :lol: (for each $x$ we have $0\leq\alpha(x)\leq \alpha(3^nx)\longrightarrow 0$ as $n\longrightarrow\infty$)

[quote="hcast"]Since $f$ takes positives values, $f(3x)=f(f(f(x)))+2x>2x$ or $f(x)>(2/3)x$.[/quote]

The problem doesn't state this.

mmm.. it must, if not how can you iterate it?, in other words if $f(f(f(x)))+2x=f(3x)$ have sence for all $x>0$ as I tought, $f(x)$ must be in the domain of $f$, $]0,\infty[$, in other case there must be a condition about the $x$ st $f(f(f(x)))+2x=f(3x)$.


List of threads pertaining to High School Chemistry or General Chemistry

Discussion of chemistry books that are good for independent study: click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=254832]here[/url].

General Discussion of Unanswered Problems in Chemistry.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181194


Για σας παιδια είμαι ο Γιώργος

Γεια σου γιώργο πολύ σωστή η πρόταση αυτή
 :wink:

Καλώς ήρθες στον ''τεκέ'' μας Γιώργο. Εδώ όλοι τη ''βρίσκουμε'' με τα μαθηματικά  :)

:welcomeani: Γιώργο και από εμένα.

Και μιας και το topic ονομάζεται έτσι, βρήκα ένα ανέκδοτο το οποίο μου άρεσε παρα πολύ, γιατί νομίζω ότι δεν απέχει από την πραγματικότητα

"So how's your boyfriend doing, the math student?"
"Don't mention that crazy pervert to me anymore! We broke up."
"How can you say such a nasty thing about him? He seemed to be such a nice boy."
"Imagine! He was restless during the days and couldn't sleep at night - always trying to solve his math problems. When he had finally done it, he wasn't happy: he would call himself a complete idiot and throw all his notes into the garbage. One day, I couldn't take it anymore, and I told him to drop math. You know what he told me?"
"No."
"He said, he enjoyed it!!!"

Παρεμπιπτόντως, υπάρχει ένα άλλο ανέκδοτο, το οποίο αρχίζει

Q: How can you tell that a mathematician is extroverted?

Δεν θα πω τώρα την απάντηση (μπορεί να βρεθεί εύκολα με λίγο ψάξιμο στο Internet), αλλά ας προσπαθήσει να δώσει ο καθένας μια πρωτότυπη δική του απάντηση, η οποία να είναι χιουμοριστική.


Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3,4,12, and 13, respectively, and $\measuredangle CBA$ is a right angle. The area of the quadrilateral is

[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
real r=degrees((12,5)), s=degrees((3,4));
pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s));
draw(A--B--C--D--cycle);
markscalefactor=0.05;
draw(rightanglemark(A,B,C));
pair point=incenter(A,C,D);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$3$", A--B, dir(A--B)*dir(-90));
label("$4$", B--C, dir(B--C)*dir(-90));
label("$12$", C--D, dir(C--D)*dir(-90));
label("$13$", D--A, dir(D--A)*dir(-90));[/asy]

$\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$

[hide]Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=36\Rightarrow\boxed{B}$.[/hide]

That, and none of the other answers seem remotely possible. :P

EDIT: Looked at the LaTeX. The answer choices should look like this:

$\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$

-BLt

[quote="breez"]How do you know that the larger triangle is a right triangle?[/quote]

When you join C and A, you know that that length is 5, because triangle ABC is a right triangle. Now the larger triangle has side lengths 5-12-13, which is a Pythagorean triple (or, they satisfy $5^2+12^2=13^2$, making it a right triangle).

Ahh, ok so thats just like how 3 sides can only make 1 unique triangle, thus if it is pythagorean triplet, the resulting triangle must be a right triangle.

[hide="Solution"]$\text{Answer: (B)}$

First, we use the right angle. Applying the Pythagorean theorem, $AC=5$. Next, we recognize the lengths 5-12-13 to be in the form of a Pythagorean Triple, which means that $\triangle ACD$ is right.

Our answer is simply the sum of the areas of these two right triangles, which is $\frac{5\cdot 12}{2}+\frac{3\cdot 4}{2}=36$[/hide]


Consider the real numbers $a,b,c$ such that the  equation $ax^2+bx+c=0$ has  two roots in the segment [0,1]. Find the greatest value of the expression $P=\frac{(a-b)(2a-b)}{a(a-b+c)}.$

Hint: using Viete's formula and...
Case 1: $a=0$
Case 2: $a\neq 0$
$P=\frac{(a-b)(2a-b)}{a(a-b+c)}= \frac{(1-\frac{b}{a})(2-\frac{b}{a})}{1-\frac{b}{a}+\frac{c}{a}}$
Good luck. :D

Lovasz,why don't you just post the entire solution?

[hide]Considering that $m$ and $n$ are the roots of the equation,by Viete's formula: $-a(m+n)=b$ and $a(mn)=c$ substituing in $P$ we get:

$P=\frac{(m+n+1)(m+n+2)}{m+n+1+mn}=\frac{m}{n+1}+\frac{n}{m+1}+2$

notice that when $m$ and $n$ tends to zero the two fractions tends to zero too,consequently the maximum value for $P$ will occur when $m,n=1$:

$P_{max}=\frac{1}{2}.2+2=3$[/hide]


Solve for real numbers $ x$:

a) $ \frac{x\minus{}2}{x^2\minus{}3x\minus{}4} \ge 0$.

b) $ \frac{5}{x\plus{}4} \le 1$.

[hide="a."]
First start by factoring the denominator and finding the restrictions.
$ x^2-3x-4 = (x-4)(x+1)$. $ x$ cannot be 4 or -1 because that would make the denominator equal 0.
Multiplying both sides by (x-4)(x+1),
$ x-2\ge 0\\
x\ge 2$
If you plot the restrictions and solution on a number line, you can test all the intervals to see which make the original equation true. The intervals are $ (-\infty,-1), (-1,2], [2,4), (4,\infty)$.
Of these, the intervals $ (-1,2]$ and $ (4,\infty)$ satisfy the original equation.
So the values of x that work are $ \boxed{(-1,2]\cup (4,\infty)}$.
[/hide]
[hide="b."]
Again start by finding the restriction.
x cannot be -4 because that would make the denominator equal 0.
Multiplying both sides by (x+4),
$ 5\le x+4\\
1\le x\\
x\ge 1$
Plot the restriction and solution and then test all the intervals, which are: $ (-\infty,-4), (-4,1], [1,\infty)$
Of these, $ (-\infty,-4)$ and $ [1,\infty)$ satisfy the original equation.
So the values of x that work are $ \boxed{(-\infty,-4)\cup [1,\infty)}$.
[/hide]

You cannot just multiply both sides by a quantity, you must make sure that a) it's nonnegative or b) you flip the sign when it's negative.

Hint:

Break into cases: when the denominator is nonnegative and when it is negative.

okay...I think this would be how you would do the cases (but I'm not completely sure):
[hide="a."]
The denominator is positive when x is in $ (-\infty,-1)\cup (4,\infty)$.
It is negative when x is in $ (-1,4)$, and undefined when $ x=-1$ or $ x=4$.

Starting with the case when the denominator is positive, multiplying both sides by the denominator (x-4)(x+1) will give:
$ x-2\ge 0 \\
x \ge 2$
Then the only solutions when the denominator is positive are $ (4,\infty)$, since x must be in $ (-\infty,-1)\cup (4,\infty)$ but also must be $ \ge 2$.

When the denominator is negative, multiplying both sides by the denominator (x-4)(x+1) will switch the inequality sign and give:
$ x-2\le 0 \\
x \le 2$
Then the only solutions when the denominator is negative are $ (-1,2]$, because x must be in $ (-1,4)$ but also must be $ \le 2$.

Combining these, the solutions are $ \boxed{(-1,2]\cup (4,\infty)}$.
[/hide]
[hide="b."]
The denominator is positive when x is in $ (-4,\infty)$.
It is negative when x is in $ (-\infty,-4)$, and undefined when $ x=-4$.

When the denominator is positive, multiplying both sides by the denominator (x+4) will give:
$ 5\le x+4\\
x\ge 1$
Then the solutions when the denominator is positive are $ [1,\infty)$, because x must be in $ (-4,\infty)$, but also $ \ge 1$.

When the denominator is negative, multiplying both sides by the denominator (x+4) will switch the inequality sign and give:
$ 5\ge x+4\\
x\le 1$
Then the solutions when the denominator is negative are $ (-\infty,-4)$, because x must be in $ (-\infty,-4)$ and also $ \le 1$, and all values of x in this range are less than 1.

Combining these, the solutions are $ \boxed{(-\infty,4)\cup [1,\infty)}$.
[/hide]

I came up with the same answers, though making the cases is probably a better solution.
I think the first time I should have said to solve the inequality with equal signs instead, and then test the intervals (especially because I didn't use the inequality at the end anyway).

a.

$ \frac{x\minus{}2}{x^{2}\minus{}3x\minus{}4}\ge 0$

$ \frac{x\minus{}2}{(x\minus{}4)(x\plus{}1)} \ge 0$

By looking at either side of the critical values $ \minus{}1, 2$ and $ 4$, we see that the required range is

$ \minus{}1\le x \le 2$ and $ x\ge 4$.

b.

$ \frac{5}{x\plus{}4} \minus{} 1 \le 0$

$ \frac{5\minus{}(x\plus{}4)}{x\plus{}4} \le 0$

$ \frac{1\minus{}x}{4\plus{}x} \le 0$

$ x\le \minus{}4$ or $ x\ge1$.


geometry
3D geometry
analytic geometry
vector

How can it be proven that the line [b]A`C [/b]is perpendicularly on the plane [b] AB`D` [/b] in the cube[b] ABCDA`B`C`D`[/b]?
Has anyone an idea for the above problem?

Please pay attention to the posting guidelines of each forum where available.  This problem is far too difficult for the forum in which you posted it.  Thanks.

3D coordinate system and vectors would help.

It is a problem from the Romanian 8th class book :)


Specifically,what kind of algorithm[b] is use in Robotics[/b] artificial[b] intelligence[/b] programming? Is there any available free[b] information that deals specifically with Artificial Intelligence in the internet?[/b]

All I know is that you need QBASIC and it's very simple to program them...don't know how...

The simplest algorithm would be to list all rational available choices to the program powering the AI, and after what happens, if it was successful, to put more *weight* on the option (i.e. to make it more likely to choose that option once the original branching is encountered again), or if it was a failure, to remove weight.

Your use of bold is very distracting, please stop.


Any AoPSer do Myspace?

mine is myspace.com/blysh3r

[quote="Valentin Vornicu"]I think most AoPS-MathLink-ers have their own blog right here on the site :P http://www.mathlinks.ro/Forum/apsjour.php ;)[/quote]

Yes, you are right.  But there are probably many AoPSers with myspace as well.

You can find mine [url=http://www.myspace.com/index.cfm?fuseaction=user.viewProfile&friendID=7810069&Mytoken=20050530061907]here[/url].

I started my account not too long ago...so I don't have all that many friends compared to a lot of ppl I know :(


What would be some good classes to take in high school if I am planning to have a career in politics?  I know history would be good, but what else?  Psychology?

I'm not quite sure there is really good class you can prepare for politics in high school. Just get good grades on main subjects..

Why don't you ask this question on College Forum for the detailsed courses in college?

It sure would be better than asking about high school.

[quote="peter"]just wondering...do you want to know for curiosity, or are you actually considering a career in politics?[/quote]

Hehe, good question.

Well, with everything politics have become today, you might want to take Public Speaking. But if they're offered at your high school, try Bending the Truth, Digging up Dirt, and Criticism 101.

[quote="tetrahedr0n"][quote]But if they're offered at your high school, try Bending the Truth, Digging up Dirt, and Criticism 101.[/quote]

Debate.[/quote]

 :blush:  :blush:

Kind of forgot about that...

Statistics, so you can manipulate the truth while claiming that you're doing it in a mathematically rigorous way.

This is not really a class but if your school has it, join Model UN. Also, take classes like Constitutional Law (if your school has it of course)

dont concentrate too much on classes. i dont believe there are any good politics-related high school classes. instead id suggest reading books on it.

most of your political knowledge will be gained in college. for now, just take as many advanced courses as possible (AP/IB) and maintain good grades. also, join forensics (speech and debate) or Model United Nations (it rules all) to develop your public speaking skills. if your school has a JSA chapter (jsa.org), join it and consider attending some of the conferences. if not, hey, start your own chapter! that's what i did, and the organization makes it very easy for you. oh yes, and AP Government, but most take that course anyway. perhaps look into some summer programs or evening classes at a local university? volunteer for a campaign? intern for a law firm? politics is an insanely competitive field; let me just tell you that.

i suppose, for now, just focus on satisfying and developing your interest in politics, as well as gaining admission to a good college. colleges love the passionate type, so if politics is really your passion, pursue it! taking initiative is key.

Okay...the most obvious one here is probably Speech and Debate.

Many public schools run Model UN programs but let me tell you right now that such programs don't come close to portraying how politics realli is, unless you become an ambassador to the UN...

YMCA runs a program known as Youth and Government in 36 states. Check to see if your state has it. Each state has some freedom with their own program, but generally this is actually an extremely good simulation of real politics. In Texas, we have all three branches of government and extensive offices in each plus some groups that are not actually part of the government (media, executive government planners, lobbyists, etc). The organization is almost completely identical to that of the actual state government. (In fact we actually use the Capitol chambers and other government buildings to compete in.) There's officers, elections, campaigns, political parties, etc. You use parliamentary and/or courtroom procedures depending on where you are. But all-in-all it's just an awesome thing to pursue. Each state also sends representatives to the Conference on National Affairs every July, which is a huge honor. This program is actually not all that advertised because you participate through your YMCA instead of your school.

If you need more information, let me know. But I'll tell you straight out that this is the single best way you can actually experience what it's like to be in public service. No other program can come close to giving you the experiences and memories that this does. And in case you're wondering, I am not unfamiliar with the "normal" speech and debate programs.

psycology would be one to consider. Also speech and debate. :)
There is also political science.

what???????? jk

I recomment AP Government and Politics.

It has Politics in the title


calculus
integration
function
calculus computations

If the function $ f(x) \equal{} 3x^2 \plus{} 2ax \plus{} b$ satisfies $ \int_{ \minus{} 1}^1 |f(x)|\ dx < 2,$ then show that $ f(x) \equal{} 0$ has distinct real roots.

$ f(x) \equal{} 3x^2 \plus{} 2ax \plus{} b\equal{}3(x\plus{}\frac{a}{3})^2\plus{}b\minus{}\frac{a^2}{3}$

Suppose $ f$ has not distinct roots then $ b\minus{}\frac{a^2}{3}>\equal{}0$ and $ |f(x)|\equal{}f(x)$

We have  $ \int_{ \minus{} 1}^1 |f(x)|\ dx \equal{} \int_{ \minus{} 1}^1 f(x)\ dx \equal{}2\plus{}2b< 2$ then $ b<0$ but $ b\minus{}\frac{a^2}{3}>\equal{}0$=>$ 0>a^2$ contradiction.


How many ways are there to choose four cards of different suits and different values from a deck of 52 cards?

I have three different ways of reasoning this:

1. For suit 1, we have 13 choices for values.

For suit 2, we have 12 choices for values.

For suit 3, we have 11 choices for values.

For suit 4, we have 10 choices for values.

Thus, we have 13 x 12 x 11 x 10 choices.

2. For value 1, we have 4 choices for suits.

For value 2, we have 3 choices for suits.

For value 3, we have 2 choices for suits.

For value 4, we have 1 choices for suits.

Thus, we have 4 x 3 x 2 x 1 choices.

3. For card 1, we have 52 choices for cards.

For card 2, we have (52 - 13 - 1) choices for cards.

For card 3, we have (52 - 13 - 1) - 13 - 1 choices for cards.

For card 4, we have (52 - 13 - 1) - 13 - 1) - 13 - 1 choices for cards.

Thus, we have 52 x (52 - 13 - 1)  x ((52 - 13 - 1) - 13 - 1 ) x ( (52 - 13 - 1) - 13 - 1) - 13 - 1 ) choices for cards.

Each of these reasonings seem correct to me.

What is wrong with the reasoning in 2 and 3?

Please help because I am very confused.

is there supposed to be something wrong? because it seems correct, at least at first glance.

My solution is none of those three.

[hide]Consider the values first. There are 13C4 ways of choosing 4 cards.

Then consider the suits. They have to appear in some particular order to match up with the values; then there are 4P4 = 4! ways of doing that.

So, there are 13C4 * 4! = 17 160 ways of choosing 4 cards of different suits and different values from a deck of 52 cards.[/hide]

Edit: Sorry, this is really identical to solution #1.

first you can choose one of the 52 cards, then 52-13-3 cards, then 52-13-3-12-2, and finally you can choose from 10 cards, so it 52*36*22*10=411840,

additionally your no. 3 is wrong because after taking away 13 and 1 (you should have taken away 13 and 3 (thirteen for suit three for number, then next time you only take away 12 for suit since 1 number is gone.... etc.)

Order doesn't matter though, so you should divide 411 840 by 24 (4!) to get 17 160.

your number 2 reasoning is mistaken because you have not chosen the actual values. when you do, you get osiris's answer (you will have to multiply by 13C4). also, osiris got the same thing as your reasoning in number 1. (13C4*4!=13P4=what you got)


If your inbox is full and someone sends you messages, what will happen? Would you still be able to receive them (That is, if you delete the older ones) or they'll go back to the sender?

Thanks.

sadly, it just deletes the old ones without confirmation ... [here stood some very evil trick :censored:]

[quote="Peter VDD"]sadly, it just deletes the old ones without confirmation. ^^[/quote]

That's harsh but hilarious.  :rotfl:  :rotfl:  Don't do that to me though Peter.  :lol:

Yeah I'll do something about that soon ...

argh, censor! censor! :P:P j/k

Anyway Valentin, I think there's something more annoying, but caused by ourselves... we have 120 inbox slots now (more than one can use actually) but the outbox is very very limited... maybe you could increase the outbox size? It is annoying when sent messages get gone because you send some other pm's, and it would not take any new disk space as the message is already stored for the recipient's inbox. :)

[quote="Peter VDD"]argh, censor! censor! :P:P j/k
[/quote]Yes, let's not give people ideas  :diablo:

I will see what I can do about the outbox thing.

[quote="Peter VDD"]*just for the purpose of details: I meant sentbox, which is different from outbox*[/quote]

Hmm.. I meant "inbox" though.. Outbox usually doesn't matter to me though. The best way is to keep quoting the text.  :D That way, you can see what you said first to what other person said last (assuming both of you are keep quoting).


[quote="0mega"]Hey i need help making a program that can be interjected into other programs easily without going through the necessarily implications of assigning variables.... for example... is there a way to just insert a program into the calculator w/o needing a prompt or input maybe just assign the variable in a pair of parenthesis following the program name?  Trying to make a program that basically goes x(abs(x)), very simple.  I have a TI-84 SE. Help![/quote]

you can't do that on a TI-84 SE, only on a TI-89


number theory
greatest common divisor
Divisibility Theory

Determine the greatest common divisor of the elements of the set \[\{n^{13}-n \; \vert \; n \in \mathbb{Z}\}.\]

Since this kind or problem occurs so often, I think we should directly solve the following:

Determine the greatest common divisior $d_k$ of all the elements of $S_k = \{n^k-n |n \in \mathbb{Z} \}$.

The result is $d_k=\prod_{\substack{p \in \mathbb P \\ (p-1)|(k-1)}} p$ (being the denominator of the $(k-1)$-th Bernoulli number if $2|(k-1)$) and follows directly if using primitive roots.

* note to self: work out ZeTaX's solution with primitive roots and move to that chapter if appropriorate *

For the initial one:
2^13-2 factors as 2*3^2*5*7*13 so the only possible factors of the gcd are 2,3(twice),5,7,13
But using Fermat's little theorem we can prove that 2,3,5,7,13|(n^13-n)
We now need to consider whether 9|n^13-n for every n, but this does not happen for n=3 so the gcd is 2*3*5*7*13=2730

PostScript
In Fermat's little theorem we can only assume that n is not congruent to zero modp(where p prime).
But obviously n^13 is congruent to n for every n congruent to zero modp

 there will not exist a prime with its exponential more than 1 as when we let $n=p$ then it is clear that $V_{p}p(p^{12}-1)=1$
$ord_p(n)| p-1$ and also $ord_p(n)|12$
so the possible values for $ord_p(n)= 1,2,3,4,6,12$
We must note that p cant be bigger than 12 as there must exist a primitive root so any prime bigger than 12 will not be a common divisor
so the only possible primes are $3,5,7,13$
greatest common divisor: $3\times5\times7\times13$


inequalities
algebra
function
domain
inequalities unsolved

hi guys
i was scrolling through one math jam transscript and discovered the following problem.
i tried to solve but i couldnt, so im asking u to do so ...
[hide]without wanting to spoil anything i tell u it was asked in the context of cauchy-schwarz[/hide]
[img]http://www.artofproblemsolving.com/Images/ClassLaTeX/lx-15398132.gif[/img]

thx 4 ur help
peeta

Right now I have only a solution using AM-GM.

We will actually prove that if x, y, z are three positive reals satisfying $x^4+y^4+z^4=1$, then

$\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}\geq\frac{9\sqrt[4]{3}}{8}$, with equality if and only if $x=y=z=\frac{1}{\sqrt[4]{3}}$.

In fact, from the AM-GM inequality,

$\frac13+\frac13+\frac13+\frac13+\frac13+\frac13+\frac13+\frac13+3\sqrt[4]{3}x^9\geq 9\sqrt[9]{\frac13\cdot\frac13\cdot\frac13\cdot\frac13\cdot\frac13\cdot\frac13\cdot\frac13\cdot\frac13\cdot 3\sqrt[4]{3}x^9}$.

In other words,

$\frac83+3\sqrt[4]{3}x^9\geq 9\sqrt[9]{\frac{1}{3^8}\cdot 3\sqrt[4]{3}x^9}=9\sqrt[9]{\frac{1}{3^7}\cdot\sqrt[4]{3}}x=3\sqrt[4]{3}x$

(since $9\sqrt[9]{\frac{1}{3^7}\cdot\sqrt[4]{3}}=3\sqrt[4]{3}$). Hence,

$\frac83\geq 3\sqrt[4]{3}x-3\sqrt[4]{3}x^9=3\sqrt[4]{3}x\left(1-x^8\right)$.

Multiplying this with $\frac38x^3$, we get

$x^3\geq\frac{9\sqrt[4]{3}}{8}x^4\left(1-x^8\right)$.

Since $x^4+y^4+z^4=1$, while the numbers x, y, z are positive, we conclude that $x^4<1$, so that x < 1, and thus $1-x^8>0$. Hence, we can divide this inequality by $1-x^8$ and obtain

$\frac{x^3}{1-x^8}\geq\frac{9\sqrt[4]{3}}{8}x^4$.

Similarly,

$\frac{y^3}{1-y^8}\geq\frac{9\sqrt[4]{3}}{8}y^4$;
$\frac{z^3}{1-z^8}\geq\frac{9\sqrt[4]{3}}{8}z^4$.

Adding up these three inequalities, we get

$\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}\geq\frac{9\sqrt[4]{3}}{8}x^4+\frac{9\sqrt[4]{3}}{8}y^4+\frac{9\sqrt[4]{3}}{8}z^4$
$=\frac{9\sqrt[4]{3}}{8}\left(x^4+y^4+z^4\right)=\frac{9\sqrt[4]{3}}{8}$

(since $x^4+y^4+z^4=1$). This completes the solution. The equality case is an exercise for the reader. ;)

Probably the Cauchy-Schwarz solution is trivial, but I don't see it (what could partly be due to my general blindness with respect to Cauchy-Schwarz solutions, but also please take into account that I'm very tired now).

  darij

thanx darij
but yet i dont understand, why this problem was asked in context with cauchy-schwarz.

thx to all who might post a possible solution involving cauchy-schwarz

CS soln:

$\sum \frac{x^3}{1-x^8} = \sum \frac{x^3}{2-2x^4} + \frac{x^3}{2+2x^4} \geq \frac{(\sum x^{\frac{3}{2}})^2}{4} + \frac{(\sum x^{\frac{3}{2}})^2}{8} = \Bigg( \frac{3}{8} \Bigg) \Bigg(\sum x^{\frac{3}{2}} \Bigg)^2$

It remains to minimize $\sum x^{\frac{3}{2}}$.  Obviously it occurs when $x=y=z$.  Then $x = y = z = \frac{1}{\sqrt[4]{3}}$ and the minimum is $\frac{9\sqrt[4]{3}}{8}$ as darij said.

[quote="peeta"]thx singular
although i wouldnt call $\frac{x_1^2}{y_1}+\ldots+\frac{x_n^2}{y_n}\ge\frac{(x_1+\ldots+x_n)^2}
{y_1+\ldots+y_n}$ exactly cauchy-schwarz, i guess it was reminding me of this useful inequality, which is good.

sorry for being so picky, but is there anyone out there in this [size=59]web[/size]space that can find a solution with the $\sum x_i^2\cdot \sum y_i^2\ge \left(\sum x_iy_i\right)^2$-inequality. u can check [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=74]the transcript[/url], but i just think the is a simple solution with cs.[/quote]

I don't think you're going to get closer than that to Cauchy... in fact that IS Cauchy, just written differently.

$(\sum a_i^2)(\sum b_i^2) \ge (\sum a_ib_i)^2$

Let $a_i = \frac{x_i}{\sqrt{y_i}}$ and $b_i = \sqrt{y_i}$

Then $(\sum \frac{x_i^2}{y_i})(\sum y_i) \ge (\sum x_i)^2 \Rightarrow \sum \frac{x_i^2}{y_i} \ge \frac{(\sum x_i)^2}{\sum y_i}$.

[quote="siuhochung"][quote="Singular"]
It remains to minimize $\sum x^{\frac{3}{2}}$.  Obviously it occurs when $x=y=z$.  [/quote]

I'm sorry, I think the MAXIMUM of this is achieved when the variables are equal. (Think about power mean)[/quote]

If what you said is true, then let $r = 3^{-4}$, and $f(x,y,z) =  \sum x^{\frac{3}{2}}$.  Then you claim $f(r,r,r) \geq f(x,y,z)$ for all $x,y,z$ such that $\sum x^4 = 1$.  But clearly $f(r,r,r) \leq f(1,0,0)$ because $3r^{3/2} \leq 1$, contradicting your statement.

then can you prove your assertion? It's not obvious to me.

[quote="Singular"]because $3r^{3/2} \leq 1$[/quote]I'm afraid you made a mistake here : with $r=3^{-1/4}$, we have

$3r^{3/2}=3 \cdot 3^{-(1/4\times 3/2)}=3\cdot 3^{-3/8}= 3^{5/8} > 3^{4/8}=\sqrt{3} > 1$.

I suggest the following example to illustrate that [b]siuhochung[/b]'s statement is correct : let $f(x,y)=x+y$ and $g(x,y)=x^2+y^2$. Consider the problem of searching

$\max f()$ and $\min f()$ under the constraint $g(x,y)=1$.

If we draw the (quarter) circle $x^2+y^2=1$ in the $(x,y)$-plane and play with the line $x+y=C$ for different values of $C$, we see that the maximum is achieved for $C=\sqrt{2}$, at $x=y=1/\sqrt{2}$ (inside the domain), where the line is tangent to the circle, while the minimum is attained for $C=1$, at $(x,y)=(0,1)$ or (1,0) (on the boundary).

If we consider the inverse problem, that is, to seek

$\max g()$ and $\min g()$ under the constraint $f(x,y)=1$,

we now have to draw with the line $x+y=1$ and play with the circles $x^2+y^2=C$. We now see that, now, the minimum is achieved for $C=1/2$ at $x=y=1/2$ (inside), where the circle is tangent to the line, while the maximum is attained for $C=1$, at $(x,y)=(0,1)$ or (1,0) (boundary).

Conclusion (based on this and other experiments) : when the exponent of the objective function is higher than the exponent of the constraint, the maximum is inside and the minimum is on the boundary ; when the exponent of the objective function is lower than that of the constraint, the minimum is inside and the maximum is on the boundary.


Find all functions $f(x)$ such that $f^{-1}(x)=\frac{1}{f(x)}$.

[hide="Discussion"] In other words, $x = f \left( \frac{1}{ f(x) }\right)$.

Suppose we have some point $f(a) = b$ on this graph.  Repeatedly plugging in we have

$a = f \left( \frac{1}{b}\right)$
$\frac{1}{b}= f \left( \frac{1}{a}\right)$
$\frac{1}{a}= f \left( b \right)$

In other words, the value of $f$ at $a$ uniquely determines the value of $f$ at three other points [b]only[/b], so we can partition the reals (or the rationals, for that matter) nearly arbitrarily for a number of really strange functions that can't be succinctly described, so we should probably restrict $f$ continuous.

Although, compounding two appropriate pairs of these, we conclude that

$f(f(x) = \frac{1}{x}$

This might restrict solutions a little.  We do have the strange complex solution $f(x) = x^{i}$, which gives $f^{-1}(x) = f(x)^{-1}= x^{-i}$.  [/hide]


Five real numbers are chosen and put in order from smallest to largest. The average of all five is 14 . The average of the three middle numbers is only 13 . What is the average of the largest and smallest ?

The sum of the five numbers is 70, the sum of the middle three is 39.  Hence the remaining two have sum 70-39=31, with average 15.5.


Let $A$ be a $2\times 2$ matrix over the real field, which is similar exactly to one matrix (i.e. itself). Prove that $A=\alpha I, \alpha \in R$.

This can be generalized and is probably better known in the following form:
\[ Z(M_n(K)) = K I_n  \]
This means: Quadratic matrices, commuting with all other matrices of the same size (hence only similar to itsself), are exactly the scalar matrices, i.a. matrices of the form
\[ \begin{pmatrix}k & 0 & \cdots & 0 \\ 0 & k & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & k \end{pmatrix}  \]

See attachment for a proof.

Thanks for your reply. But would you please clarify this part? "commuting with all other matrices of the same size "
I suppose we are required to check this condition: "commuting with all nonsingular matrices of the same size"
rather than that.
Does it change anything?

You're right. "similar only to ittself" is weaker then " commuting with all other matrices". So let's enjoy $n=2$ ;-)

Write
\[ A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
From
\[ A \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} A \]
we get $b = c$ and $a = b + d$. And from
\[ A \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} A \]
we get $a+b = a$, hence $b = 0$. So, we have
\[ A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} \]


algebra
polynomial
complex numbers
imaginary numbers

$\sqrt{x} + y = 11$

$\sqrt{y} + x = 7$

proper method is required ,please.I know you can do it by hit and trial, but i would appreciate a proper soln  :D

P.S. This question is NOT for shadysaysurspammed as he knows the soln and answer too!

move over the y and square for the first, and mover over the x and square for the second, and any answer that is negative is bad, since it makes an imaginary number

I got a sol but it is kinda factors and all.
and then you have to assume.
[hide="Solution"]
substracting the 2 equations,
$\sqrt{x}-\sqrt{y}+y-x=4$
$\sqrt{x}-\sqrt{y}-(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=4$
$(\sqrt{x}-\sqrt{y})(1-\sqrt{x}+\sqrt{y})=4$
Let $\sqrt x=a$
and$\sqrt y=b$
$(a-b)(1-a-b)=2*2*1$
Im taking into consideration that a and b are integers but they also have to be positve because root of something is always positive except $i$
Test them out and then find the possible value
[/hide]
Another try,
[hide]
Just substitute x in both equations and you will get a 4 degree polynomial and then you have to factorise it
[/hide]


There is no way to predict anything.  No one knows what seniors in 2011 will score, but you can assume the percentiles will be similar to every other year.  If you got a 173 on the PSAT as a sophomore, that's okay but nothing amazing.

[quote="mj93"]Say I recieved a 1730 (75th percentile of college-bound seniors) on the SAT as a freshman.

What percentile of college-bound seniors of 2011 will I be in?

What score will I get on the SAT as a sophomore, junior, senior, assuming that I stay in the same percentile?

You might have to find the data .. I'm not sure.

But if you could find a way to figure this out, it would be great!

Thanks.[/quote]
If all freshman score like you and do everything like you, and all freshman this year become seniors in 2011, and only seniors of 2011 are part of college-bound seniors of 2011, then on average you will be right in the middle.
That's the only thing I can predict.
However, since you're a freshman and you were ranked among college-bound seniors, then i doubt that the college-bound seniors of 2011 will only have seniors.


[quote="Klebian"]Well, I could've made the title a bit less revealing, but I decided not to.

My statement here is that new moderators need to be assigned to Games and Fun Factory.
No, I don't mean adding moderators.
I mean revoking moderatorship of the two mods there now, and adding 2-3 competent members to mod the forum.

Mathfiend has not posted this year, in fact it has been almost 6 full months since her last post.
MithsApprentice has 14 posts in the last 2 months+. That's 7 posts a month, which means not even 1 post every 4 days.

Also, not only is G&FF the most likely forum for people to spam uselessly, it IS the forum where people spam most uselessly.
On top of that, every once in a while we have a user like kotsba advertising for gambling sites, and I recall only 1 or 2 of his topics being locked.

I can see someone arguing along the lines of moderators not really necessary, or something of the sort; however, what is the point of moderators then?

As I recall, the point of moderatorship of a forum is so that there is not too much spam, and so the users of the forum can feel comfortable posting in the forum.

I don't believe this is accomplished by either, nor both, of MathFiend and MithsApprentice.[/quote]
I agree. There was a discussion about this a while ago, and someone said that there would be no mod changes for a while...but the G&FF forum is going to get out of control soon (If it already hasn't)

Actually,FF does not really need constant monitering but lately people have started to psam a lot,so yes,amybe keeping a mod who is regular is fine but the mod should not be "not fun"

[quote="Centy"]How about just removing G&FF posst from people's post count like in the Test Forum. That might curb some of the spamming.[/quote]
Just my point :)
There is only one hand of pure spammers there, all others behave like they should.
(and doing the moderation of it should be hard and a lot of work...)

We'll take this under consideration.

Also, feel free to report commercial spamming to an AoPS admin like me.  I'll happily ban gambling site spammers.

If you still are looking for a moderator of any forum, I would like to be a candidate because i have good leadership qualities, i haven't spammed in a while, and i can control levels of spam. :)

[quote="MithsApprentice"]To be clear, the point of having mods around is not for us to lead forum discussions...at least, not in F&G. On spam-acceptable boards, we just make sure the content is appropriate and productive, sort of. Of course, any commercial advertising should've been deleted. If anything like this occurs, please bring it to our attention via PM. Thanks.[/quote]

However, if we're going to have mods, why not choose a more active user over an inactive user?
I mean, why NOT have a mod lead forum discussions?
And if a moderator isn't active enough to see a user posting gambling advertisements, should he really be a moderator?

Klebian,

You're a moderator now.  Those questions are better raised at this point in the moderators community.  Miths does a good job moderating.  If a commercial poster goes unnoticed, it's in G&FF -- there's an enormous amount of posting to keep up with and not every link is going to be clicked through.  Now that you've noted the problem, the user and IP is banned and everyone will know better what to look for in the future.  Remember, this whole message board is an experiment in human resource management -- and overall the community does a remarkable job at fixing problems as they arise.  Thanks for helping out.

[quote="robinhe"][quote="Peter VDD"][quote="math92"]i haven't spammed in a while[/quote] :D[/quote]

What about the "bump the thread" topic? That's pure spam :)[/quote]
I thought i posted that in the test forum, where it eventually gets deleted.

Depends what you call spam. I tend to live the Games and Fun Factory forum more liberal, a more friendly place for users of young age. We will soon change part of the forum's software, so posts in that forum will not be counted anymore. Other than that I see nothing wrong in having some spam in Games and Fun (or course some limitations will apply, like no 2 consecutive posts, no more than 2 posts per minute in that forum, etc.).

And of course all "real spam" (like commercial, or xxx ads) should be deleted and admins notified for user/IP banning (please send links with the posts before you delete them as well as links to the users profiles to make the job easier).


let $ a,b,c>0$ prove that $ \sum(\frac{a^2}{b}\plus{}c)^2 \ge 12\frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c}$

[quote="dhvdochilinh"]let $ a,b,c > 0$ prove that $ \sum(\frac {a^2}{b} \plus{} c)^2 \ge 12\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}$[/quote]
[hide="Hint(MAYBE)"]
I am new to the idea of homogenisations so I apologise if this idea is not correct.... :(

Replace $ a, b, c$ with $ ka, kb, kc$ in your inequality to notice that for $ k\neq0$, your inequality does not change at all.
So we can (perhaps) assume $ a\plus{}b\plus{}c\equal{}1$... :|
Sorry if I am incorrect :oops: .[/hide]


algebra
polynomial
Ring Theory
superior algebra
superior algebra theorems

Hi, I'm getting stuck trying to solve this question: what are the maximal ideals in $ Z[x]$?

Thanks

Grammar note: ideals are not people, so "who" is inappropriate. Also, there are many maximal ideals, and thus "is" is the wrong form. That should be "what are the maximal ideals in $ \mathbb{Z}[x]$".

Each maximal ideal is generated by a prime $ p$ and a polynomial $ f$ which is irreducible mod $ p$.

Thank you very much for your answer, and sorry for my english..........


Im studying for a prelim for January and need help with the following:

Suppose $ X \subset %Error. "field" is a bad command.
{R}^n$ is not homeomorphic to any proper subset of itself.  Show that $ X$ has empty interior.  Give an example of an infinite space $ X$ satisfying the above.

Any hints would be greatly appreciated.

For the first part, just note that any open ball is homeomorphic to the entire $ \mathbb R^n$, so you can reproduce a topological copy of $ X$ in it.

The second part is easy if $ n\ge 2$: just take a circle. I still have to think what to do when $ n\equal{}1$ though...

Thanks for the response fedja.  I think I understand where you want to go with this-am I supposed to
think about what this homeomorphic copy of X looks like (no segments of the ball?)?


Show that there is an infinite number of pairs $(a,b)\in N$x$N$ such that $a$ divides $b^2+b+1$ and $b$ divides $a^2+a+1$

After some (rather hard :mad:)  tries the  following works :

Let's define the following sequence :

$u_0 = 1, u_1 = 1, u_{n+1} = ({u_i}^2+u_i+1)/u_{i-1}$

Then each $(u_n, u_{n+1})$ works.

Just wondering - is it trivial that all terms of that sequence are integers? I can prove it, but it is not that trivial that I would omit it. So, am I missing something?

Hi,

I think the following can be proved by induction :

u_{n+3} = 6 u_{n+2} - 6 u_{n+1} + u_{n}

[quote="tµtµ"]

$u_0 = 1, u_1 = 1, u_{n+1} = ({u_i}^2+u_i+1)/u_{i-1}$

[/quote]

Well, how did you come up with this?


induction
ratio
graph theory
combinatorics proposed
combinatorics

Prove that one can colour the $n$ vertices of a graph, each having degree at most 5, with 2 colours such that at least $\frac 35$ of it's edges have different coloured endpoints.

The uckrainean problem was proposed for a regular $n$-gon with 2004 vertices.

We can use induction: I’ll always assume that whenever we remove some vertices from the graph, a coloring with the required proeprty (call it $P$) is possible, and whenever I say that we color the vertices of a graph, I’m referring to a coloring with $P$.

First of all, we may assume that all vertices have degree $2$ or $4$: if a vertex has odd degree (we disregard isolated vertices, so no vertex has degree $0$), then it’s either $3$ or $5$. Assume it’s $5$. Remove the vertex, color the remaining vertices, and add the vertex we have removed. It’s connected to at least $3$ vertices of one color (red) and at most $2$ vertices of the other color (blue). If we color it blue, $P$ is preserved. We do something similar if the vertex has degree $3$.

Assume now we have a vertex $x$ of degree $2$, and let $y$ be one of its neighbours, which we assume to have degree $4$ (if it has degree $2$, the ratios are even better, as can easily be checked, so assuming it has degree $4$ will suffice). Remove $x$, then remove $y$, and then color the rest of the graph. When we add $y$ again, we get at least $2$ bicolor edges out of $3$, and when we add $x$ again we get at least $1$ bicolor edge out of $2$, so, in total, we have added at least $3$ bicolor edges out of $5$, so $P$ is preserved. This means that we can now assume that all the vertices have degree $4$.

Assume we have a vertex $a$ connected to $a,b,c,d$, and that $a,b$ are not connected to each other. We do something very similar to the above, removing $x$, then $a$ and then $b$, coloring the rest, and, after we add the three vertices, we will have added $10$ edges ($4$ for $x$, $3$ for $a$, and $3$ for $b$), out of which at least $6$ will be bicolor ($2$ for $x$, $2$ for $a$, and $2$ for $b$), and we’re done.

This shows that we may assume that whenever $x$ is connected to $a,b,c,d$, these $4$ vertices form a complete graph $K_4$. This means that there are no connecttions between the set $x,a,b,c,d$ and the rest of the vertices, so our graph consists of several connected components, each one being a $K_5$, and it’s easy to color each $K_5$ so as to obtain $P$.


Suppose that d is a factor of $ n^4 \plus{} 2n^2 \plus{} 2$ such that $ d>n^2 \plus{} 1$,where n is some natural number n>1. Prove that $ d > n^2 \plus{} 1 \plus{} \sqrt {n^2 \plus{} 1}$

$ n^{2}\plus{}1\equal{}x$  $ \implies$  $ d|x^{2}\plus{}1$ and $ d>x$
$ \implies$  $ d\equal{}x\plus{}a$
$ \frac{x^{2}\plus{}1}{x\plus{}a}\equal{}x\minus{}a\plus{}\frac{a^{2}\plus{}1}{x\plus{}a}$  $ \implies$  $ x\plus{}a|a^{2}\plus{}1$
$ a^{2}\minus{}a\plus{}1>x$  $ \implies$  $ a>\sqrt{x}$  QED :wink:

Why do you have this?
  $ x \plus{} a|a^{2} \plus{} 1$
$ \equal{}> a^{2} \minus{} a \plus{} 1 > x$

$ x \plus{} a|a^2 \plus{} 1\implies a^2 \plus{} 1 \geq x \plus{} a$, or $ a^2 \minus{} a \plus{} 1 \geq x$. But $ x\equal{}n^2\plus{}1$, so $ a^2\minus{}a\plus{}1\geq n^2\plus{}1\implies a^2\minus{}a\geq n^2$. Clearly equality cannot occur so we have $ a^2\minus{}a>n^2$, and $ a^2\minus{}a\plus{}1>x$.


probability calculus articles integration Putnam

An infinite number of lines, all parallel to each other and spaced one metre apart, cover an infinite floor. If I stand on this floor and throw a rod of negligible thickness and 2 metres long up in the air, what is the probability that the rod lands and touches a line?

This looks very, very much like Buffon's Needle.

Yup i just read it on Wikipedia.. is there a way to solve this without calculus though?

[quote="CircleSquared"]It is a problem exactly like this one, except the needle is shorter than the distance between the lines.[/quote]
The article on Wikipedia has a formula for this case also, but it is simply a variation of the original solution and uses calculus.

I dont think this is that difficult a problem at all! It is a simple case of buffon's needle.

First note that the only thing that makes a difference is that the length. the thing about whether its curved or straight will not make a difference logically.

Now, consider the circle with radius 0.5m.

Obviously, it will land such that it cuts the line twice (or cuts two lines once)
so each throw of the "needle" will result in 2 cuts of the line.

now, we know that the length of the needle is 2*Pi*0.5=Pi
so we know that for a needle Pi m long, it hit 2 lines(on average). For a needle 2 meters long, It will hit 4/Pi lines(on average).

However note that the question here is ambiguous as the question asks for the probability that it hits a line and not on average how many lines does it hit. (note the difference between these two statements I solved the question for my statement and not Aidan's case)

Hi aidan, its Chai here! :)

just treat the angle the rod makes to the lines and distance to the nearest line from the center of the rod as two uniform random variables and solve it

Can you elaborate? maybe offer a solution?

Hi Chai, didn't see that u'd replied =P Usually AoPS informs me when someone replies to a thread i start. Why does the curvature of the needle not matter? Actually, i asked Mr Lum (my math tutor in NJC) and he had no idea how to start. That's a good common test problem =P

that would be hard to explain. But let me try.
consider a bend as two disjoint needles.

so a needle which looks like |_  has a bend. Now we consider the probability that it hits a line.

notice that we can just look at one part of the needle. i.e. we can consider it as just |+_. then we just sum the number of times each of these smaller needles will cross a line and we are done.
using that as an example for your problem, you can consider your 2m long needle to be two individual 1m long needles. On average, a buffon's needle hits 2/Pi times. So yours will hit 2(2/Pi) times.

Hope that helps!

Ok i get it now, thanks.  :)  However, how would that lead to an answer to my original problem? I suspect some integration would have to be done as the angle at which the needle lands relative to the lines has to be considered. 30 degrees relative to the lines would be the limit, but smaller the angle than 30 degrees, the more leeway the needle has to settle in between any two lines. That looks hard!

And what if we throw the needle on a grid? Haha.. Say if the grid consisted of 2m by 2m squares

The problem can be solved by calculus too... you basically treat the angle and the position of the middle of the rod as two random variables and integrate over the possibility with all angles. :)

Ok thanks.. that implies it will never appear in any math contest (save Putnam maybe).  :lol:


limit
pigeonhole principle
irrational number
calculus
calculus computations

Let $ f: [0,1]\rightarrow\mathbb{R}$ for which:
if $ x\equal{}\frac{p}{q}$; $ (p,q)\equal{}1$; $ \{p,q\}\in\mathbb{Z}^{\plus{}}$, then $ f(x)\equal{}\frac{1}{q}$.
else, $ f(x)\equal{}0$ [i.e. when $ x$ is irrational]

i) Determine $ \lim_{x\rightarrow a}f(x)$ where $ a\in[0,1]$ is irrational.

Proof:
I will use two 'basic' facts:
1) The statement: "two nonequal real numbers are arbitrarily close" is nonsense.
2) For any irrational number, $ \alpha$, there are other irrational numbers that are arbitrarily close to $ \alpha$.

The limit either exists or it doesn't. If it exists, it must be $ 0$ by 2). Lets assume that the limit does not exist.

Then for some $ \epsilon_{0}>0$, there does not exist some $ \delta$ such that
$ 0<|x\minus{}\alpha|<\delta\implies |f(x)\minus{}0|<\epsilon_{0}$.

In other words, we can always find an $ x_{i}$ that is arbitrarily close to $ \alpha$ for which |$ f(x_{i})|\ge\epsilon_{0}$. But by assumption 1), $ x_{i}$ must be a sequence because it is nonsense to say that one number is 'arbitrarily close' to another number.

Noting that $ \epsilon_{0}>0\implies f(x_{i})>0$ which implies that $ x_{i}$ is rational. Define $ x_{i}\equal{}\frac{p_{i}}{q_{i}}$. If we have that $ \lim_{n\rightarrow\infty}\frac{1}{q_{n}}\equal{}0$, then we our limit is $ 0$. Assume this is not true. Then there exists some $ M\in\mathbb{Z}^{\plus{}}$ such that $ \frac{1}{q_{i}}\ge\frac{1}{M}$ for all $ i$. But then we are saying that $ q_{n}\le M$. But as $ q_{n}$ has to be an infinite sequence of positive integers, it follows by the pidgeonhole principle, that there is an infinite sequence, $ y_{i}\equal{}\frac{a_{i}}{q}$ that gets arbitrarily close to $ \alpha$. We have:
$ \left|\alpha\minus{}\frac{a_{i}}{q}\right|$ or $ |q\alpha\minus{}a_{i}|$ gets arbitrarily small [but not $ 0$]. However, this is nonsense since $ q\alpha$ is constant and $ a_{i}$ is an integer. Contradiction. Hence the limit is equal to $ 0$.

Comment: the end of the argument can be tightened up with the following note: We define our sequence $ x_{i}$ to get closer to $ \alpha$. Furthermore, we can take this sequence to be the one where we have all possible $ q_{i}$. [In my proof, I took $ x_{i}$ to be an arbitrary sequence. But by the constructive nature of this proof, we can add in extra terms 'in between' so it satisfies the properties that we want it to.]

Hence $ |q\alpha\minus{}a_{i}|$ is strictly decreasing. But if we consider $ \parallel q\alpha\minus{}a_{2i}|\minus{}\{q\alpha\}|$, we have an infinite decreasing sequence of positive integers that is bounded below, which is a contradiction. [the $ 2i$ accounts for the fact that we can approximate a noninteger on the left and righthand sides.]

ii) Determine $ \lim_{x\rightarrow a}f(x)$ were $ a\in [0,1]$ is rational.

Proof: The statement is the same as the previous where we just need to pay attention to $ 0 < |x\minus{}a|$.

---

It trivially follows by definitions of continuity that $ f(x)$ is continuous at irrational inputs and discontinuous at rational inputs.


The question:
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2008&p=1088465
I found the quotients of the two sequences, then found that the quotient is increasing in a series of triangular numbers, so I used the (n)(n+1)/2 equation to find the number. I assume everything right up until I got 33*16 instead of 33*17, which is the right answer. Was n supposed to be 32 or 33? I'm guessing it's 33 because the original series start with n=0. If this is wrong, please correct me.

Please keep all the discussion confined to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088465#p1088465]original thread[/url].

The posted solution seems fine to me.


Daca $ x,y$ sunt numere pozitive, atunci $ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^{2}} \geq
\frac{1}{1\plus{}xy}.$

Solutia mea este una urata , o sa ma gandesc si la alta metoda , dar pana atunci :

[hide="Click Aici"]\[ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^{2}} \geq \frac{1}{1\plus{}xy} \Rightarrow \frac{(1\plus{}x)^2\plus{}(1\plus{}y)^2}{(1\plus{}x)^2(1\plus{}y)^2} \geq    \frac{1}{1\plus{}xy}\]

Mergem inainte , calculand :

\[ \Rightarrow (2\plus{}2x \plus{} 2y \plus{} x^2 \plus{} y^2)(1\plus{}xy) \geq (x^2\plus{}2x\plus{}1)(y^2\plus{}2y\plus{}1)\]

Desfacem parantezele , facem simplificarile de rigoare si ajungem la :

\[ x^3y\plus{}xy^3 \minus{}x^2y^2 \minus{}2xy \plus{}1 \geq 0\] Dar stim ca din $ AM\minus{}GM \Rightarrow$ ca $ x^3y \plus{} xy^3 \geq 2x^2y^2$ , deci ramane sa demonstram ca:

\[ x^2y^2 \minus{}2xy \plus{} 1 \geq 0 \Leftrightarrow (xy\minus{}1)^2 \geq 0\] . Care este evident adevarat . Deci $ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^{2}} \geq \frac{1}{1\plus{}xy}$ este adevarata .[/hide]

$ {\frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2}}\ge{\frac {1}{1 \plus{} xy}}$ $ \Longleftrightarrow$ $ {\frac {2 \plus{} 2(x \plus{} y) \plus{} (x^2 \plus{} y^2)}{[(1 \plus{} x)(1 \plus{} y)]^2}}\ge{\frac {1}{1 \plus{} xy}}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {(1 \plus{} xy)[2 \plus{} 2(x \plus{} y) \plus{} (x^2 \plus{} y^2)]}\ge{[(1 \plus{} xy) \plus{} (x \plus{} y)]^2}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {(1 \plus{} xy)[2 \plus{} 2(x \plus{} y) \plus{} (x^2 \plus{} y^2)]}\ge{(1 \plus{} xy)^2 \plus{} 2(1 \plus{} xy)(x \plus{} y) \plus{} (x \plus{} y)^2}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {(1 \plus{} xy)[2 \plus{} (x^2 \plus{} y^2]}\ge{(1 \plus{} xy)^2 \plus{} (x^2 \plus{} y^2 \plus{} 2xy)}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {2(1 \plus{} xy) \plus{} (1 \plus{} xy)(x^2 \plus{} y^2)}\ge{(1 \plus{} xy)^2 \plus{} (x^2 \plus{} y^2) \plus{} 2xy}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {xy(x^2 \plus{} y^2) \minus{} 2xy}\ge{(1 \plus{} xy)^2 \minus{} 2(1 \plus{} xy)}$ $ \Longleftrightarrow$ $ {xy(x^2 \plus{} y^2) \minus{} 2xy \plus{} 1}\ge{(1 \plus{} xy)^2 \minus{} 2(1 \plus{} xy) \plus{} 1}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {xy(x^2 \plus{} y^2) \minus{} 2xy \plus{} 1}\ge{(xy)^2}$ $ \Longleftrightarrow$ $ {x^3y \minus{} x^2y^2 \plus{} xy^3 \minus{} 2xy \plus{} 1}\ge{0}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ {xy(x \minus{} y)^2 \plus{} (xy \minus{} 1)^2}\ge{0}$.  $ {OK!}$.

[quote="Virgil Nicula"][color=darkred]Daca $ x,y$ sunt numere pozitive, atunci $ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^{2}} \geq \frac {1}{1 \plus{} xy}.$[/color][/quote]
[b]Metoda I[/b] ([u]Darij Grinberg[/u]). Vom aplica inegalitatea [b]Cauchy-Buniakovschi-Schwartz[/b] (C.B.S.) : $ x(1 \plus{} y)^2 \equal{} \left(\sqrt x\cdot 1 \plus{} \sqrt y\cdot \sqrt {xy}\right)^2\le$ $ (x \plus{} y)(1 \plus{} xy)$ $ \Longleftrightarrow$ $ \frac {1}{(y \plus{} 1)^2}\ge$ $ \frac {x}{(x \plus{} y)(1 \plus{} xy)}\ .$ Analog se arata $ \frac {1}{(x \plus{} 1)^2}\ge$ $ \frac {y}{(x \plus{} y)(1 \plus{} xy)}\ .$ Suma ultimelor doua relatii conduce la inegalitatea propusa. Avem egalitate daca si numai daca $ \frac {\sqrt x}{1} \equal{} \frac {\sqrt y}{\sqrt {xy}}$ si $ \frac {\sqrt y}{1} \equal{} \frac {\sqrt x}{\sqrt {xy}}$ , adica $ x \equal{} y \equal{} 1$ .

Va ofer trei demonstratii la nivelul clasei a VII - a, fara utilizarea inegalitatii celebre C.B.S.

[b]Metoda II.[/b] Notam $ x \plus{} y \equal{} s$, $ xy \equal{} p$ . Deoarece $ x > 0$ si $ y > 0$ avem $ s > 0$ , $ p > 0$ si $ (x \plus{} y)^2\ge 4xy$ , adica $ s^2\ge 4p$ . Inegalitatea devine : $ (p \plus{} 1)(s^2 \plus{} 2s \minus{} 2p \plus{} 2)\ge (s \plus{} p \plus{} 1)^2$  $ \Longleftrightarrow$   $ ps^2\ge 3p^2 \plus{} 2p \minus{} 1$ . Insa $ s^2\ge 4p$  $ \Longrightarrow$ $ ps^2\ge 4p^2\ge 3p^2 \plus{} 2p \minus{} 1$ deoarece $ (p \minus{} 1)^2\ge 0$ . Avem egalitate daca si numai daca $ p \equal{} 1$ si $ x \equal{} y$ , adica $ x \equal{} y \equal{} 1$ .

[b]Metoda III.[/b] $ x^2 \plus{} 1\ge 2x$ $ \Longrightarrow$ $ y(x^2 \plus{} 1)\ge 2xy$ $ \Longrightarrow$ $ y(x^2 \plus{} 1) \plus{} x(y^2 \plus{} 1)\ge 2xy \plus{} x(y^2 \plus{} 1)$ $ \Longrightarrow$ $ (x \plus{} y)(1 \plus{} xy)\ge x(y \plus{} 1)^2$ $ \Longrightarrow$ $ \frac {1}{(y \plus{} 1)^2}\ge \frac {x}{(x \plus{} y)(1 \plus{} xy)}$ . Analog se arata ca $ \frac {1}{(x \plus{} 1)^2}\ge \frac {y}{(x \plus{} y)(1 \plus{} xy)}$ . Din suma ultimelor doua relatii se obtine inegalitatea propusa. Avem egalitate daca si numai daca $ x \equal{} y \equal{} 1$ .

[b]Metoda IV.[/b] Vom folosi o identitate interesanta : $ y(x \plus{} 1)^2 \plus{} x(y \minus{} 1)^2 \equal{} (x \plus{} y)(1 \plus{} xy)$. Deoarece si $ x(y \plus{} 1)^2 \plus{} y(x \minus{} 1)^2 \equal{} (x \plus{} y)(1 \plus{} xy)$ vom obtine lantul $ 0\le \frac {x(y \minus{} 1)^2}{(x \plus{} 1)^2} \plus{} \frac {y(x \minus{} 1)^2}{(y \plus{} 1)^2} \equal{}$ $ \frac {(x \plus{} y)(1 \plus{} xy) \minus{} y(x \plus{} 1)^2}{(x \plus{} 1)^2} \plus{} \frac {(x \plus{} y)(1 \plus{} xy) \minus{} x(y \plus{} 1)^2}{(y \plus{} 1)^2} \equal{}$ $ (x \plus{} y)(1 \plus{} xy)\cdot\left[\frac {1}{(x \plus{} 1)^2} \plus{} \frac {1}{(y \plus{} 1)^2} \minus{} \frac {1}{1 \plus{} xy}\right]\ .$ Cu alte cuvinte, inegalitatea initiala este echivalenta cu inegalitatea evidenta $ 0\le \frac {x(y \minus{} 1)^2}{(x \plus{} 1)^2} \plus{} \frac {y(x \minus{} 1)^2}{(y \plus{} 1)^2}\ .$

[b]Remarca.[/b] Omogenizare : $ \{x,y,z\}\subset\mathcal R^*_ \plus{} \implies$ $ \frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(x \plus{} z)^{2}} \geq \frac {1}{x^2 \plus{} yz}\ .$

O interpretare geometrica a acestei inegalitati gasiti la

http://www.mathlinks.ro/Forum/viewtopic.php?t=172083

[quote=Virgil Nicula]Daca $ x,y$ sunt numere pozitive, atunci $$ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^{2}} \geq
\frac{1}{1\plus{}xy}.$$[/quote]
Let $a,b $ be positive real number such that $ab=1.$ Prove that $$2\left(\frac{1}{1+7a}+\frac{1}{1+7b}\right)\geq \frac{3}{2}\left(\frac{1}{1+5a}+\frac{1}{1+5b}\right)\geq \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{a+b}{a+b+2}\geq \frac{1}{1+3a}+\frac{1}{1+3b}\geq \frac{1}{2}$$


Give a proof,

suppose that one function is integrable and another is increasing,

prove Second Mean Value theorem for integrals, derived by Riemann-Lebesque Lemma

http://en.wikipedia.org/wiki/Mean_value_theorem

Could you state this theorem? There seems to be several variants of it on the internet, so I'm not sure which you meant.


$ a^{\frac{b}{c}}$ is defined as $ \sqrt[c]{a}^{b}$.  Or something like that...

Well, they really are the same thing, just $ b$ has been shifted a quarter of an inch.

$ \sqrt[16]{243}$.  Can't be simplified.

[quote="IIL"]so what would be the answer to 3^(5/16)  ?

Thanks again.[/quote]

Well, I think you use calculator to figure that one..

In radical form, it's $ {}^{16}\sqrt{3}^5$

I agree with DonkeyKong and alanchou.
This is in AoPS Volume 1.

For the first one, you would need a calculator to figure it out. But in general, you can write fractional exponents in radical form, so x^(a/b)=bth root of x^a

[quote="IIL"]Hi All,

3^(5/16)
x^(a/b)

What is the best way of working these out ?
What is the best way to deal with exponents containing fractions ?

Thanks.[/quote]

Hello there,

I'll answer your second question because the first has been well answered already.

For algebra, I recommend writing exponents as fractions because then you can work with them easily. First of all, don't be frightened by them. Just treat them like numbers. Secondly, try to express square roots as fractions too.


[quote="arqady"]Let $ a,$ $ b$ and $ c$ are different non-negative numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {c(a \minus{} b)^5 \plus{} a(b \minus{} c)^5 \plus{} b(c \minus{} a)^5}{\sqrt [3]{(a \minus{} b)^5(b \minus{} c)^5(c \minus{} a)^5}}\geq7.5
\]
[/quote]

We have

$ \sum c(a \minus{} b)^5\equal{} (a\minus{}b)(b\minus{}c)(c\minus{}a)[\sum a^3\plus{}(a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}11abc]$

so after cancellation it remains to prove

$ \sum (3c\plus{}a\plus{}b)(a\minus{}b)^2 \geq 15(\sqrt [3]{(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2})$

true by AM-GM.

Yes! :lol:
But what is a minimal value of the following expression?
\[ \frac{c(a\minus{}b)^5\plus{}a(b\minus{}c)^5\plus{}b(c\minus{}a)^5}{\sqrt[3]{(a\minus{}b)^5(b\minus{}c)^5(c\minus{}a)^5}}\]
It's not $ 7.5$  :wink:


Is there any way we can find a list of schools participating in the AMCs?

I want to take the AMC8 and both days for the AMC10/12. If my school doesn't do that, and I can't convince them, can I take the test in a city that is about 70 miles away?

1st:  Read the AMC FAQ #4, follow the advice there.

2nd:  Pay attention to the listing at:
http://unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="PowerOfPi"]I did that. The closest place on the list was about 5 hours away. Thanks anyway.[/quote]

Wait until about a month before the AMC 10/12 A. It's possible some schools just haven't registered yet.

If not and you're really desperate you COULD try to take it at that place 5 hours away.

[quote="Paiev"]Out of curiosity... my school offers only the B date. What should I do if I want to take the A test too?[/quote]

There's not much you can do besides take the A at another school.

Don't worry, I have a similar problem. I'm taking A at my school, B at another.

[quote="Paiev"]Out of curiosity... my school offers only the B date. What should I do if I want to take the A test too?[/quote]

Someone correct me if I'm wrong, but I don't think you can do much to take the A test. The colleges that offer the AMC only offer it on the B dates, so unless if you actually go to another high school to take the A test (which would probably be inconvenient for you, the high school, and the MAA), I don't think you will be able to.

I was in the same position as you the last 2 years, and in the end, I only was able to take the B test both years.

The ideal solution is of course to convince your school to offer the AMCs, since that would allow some other students to take it as well, and hopefully continue even after you're gone. More participation is always good.
Students taking multiple tests is a loophole created in recent years. Don't expect support for it, since the two dates are really there to allow schools to pick one that is more convenient.

Worst case, just don't take both if the school won't let you. It won't kill you. Why I'm saying this is it's harder than you think sometimes to change the school's mind, remember they pay money for this.

[quote="pascal12"]Wow, in my school, it's all arranged by a student-founded club, and the members have to contribute to the costs of the tests. We're not even sure some years if we can even take one test, let alone 3, due to lack of school funding.[/quote]

You could ask for donations, although i'm not sure that will help much. Too many jocks at my school :P

[quote="jmerry"]My school always ran it with the sponsoring teacher paying the registration fee, and then having students contribute a dollar or two each for the cost of the tests. It's currently \$40-\$60 (per date) for AMC 10/12 registration, plus \$16 per ten tests. If you're worried about funding, try that system- and a teacher who's invested in it that way is likely to keep it going longer than the club.[/quote]

There are very few teachers that will pay from their own bank account to some competition that only 3 people (at max) participate in...


Let $ {a},{b},{c}$ are positive reals, such that $ {a}\plus{}{b}\plus{}{c}\equal{}1$. Find the maximum value of $ \frac{a^{2}}{b\plus{}1}\plus{}\frac{b^{2}}{c\plus{}1}\plus{}\frac{c^{2}}{a\plus{}1}$

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=278128[/url]