$\lim_{x\to\infty}\sqrt[3]{x^3-6x^2+5x+4} - \sqrt[3]{x^3+3x^2+3x+2}$

do you multiply it by its conjugate or something? i dont really know.. haha

Well, I can tell you a different problem that if you can solve will answer your question.  What is ${\lim_{x\to\infty}\sqrt[3]{x^3+ax^2+bx+c} - x}$?

To answer that, you could try multiplying by the conjugate (which in this case is of the form z^2 + zx + x^2 where z is ugly) and then applying L'Hopital's rule, but that isn't very pretty ... the answer is quite nice, though.  I'm continuing to think about it.

Okay, got it.  Do what I said.  Solve the question I asked by multiplying by the conjugate.  It's not messy at all.

[hide="Messy approach using conjugates"]
Let [tex]\displaystyle P(x)=\sqrt[3]{x^3+ax^2+bx+c}[/tex] and [tex]\displaystyle Q(x)=\sqrt[3]{x^3+dx^2+ex+f}[/tex]
[tex]\displaystyle\lim_{x\to\infty}P(x)-Q(x)=\lim_{x\to\infty}\frac{(P(x))^3-(Q(x))^3}{(P(x))^2+P(x)Q(x)+(Q(x))^2}[/tex]
[tex]\displaystyle\lim_{x\to\infty}P(x)-Q(x)=\lim_{x\to\infty}\frac{x^3+ax^2+bx+c-x^3-dx^2-ex-f}{(P(x))^2+P(x)Q(x)+(Q(x))^2}[/tex]

Define [tex]\displaystyle (P(x))^2=\sqrt[3]{x^6+gx^5+hx^4+ix^3+jx^2+kx+l}[/tex]
[tex]\displaystyle P(x)Q(x)=\sqrt[3]{x^6+mx^5+nx^4+ox^3+px^2+qx+r}[/tex]
[tex]\displaystyle Q(x))^2=\sqrt[3]{x^6+sx^5+tx^4+ux^3+vx^2+wx+y}[/tex]

For more simplification, let [tex]\displaystyle A(x)=gx^5+hx^4+ix^3+jx^2+kx+l[/tex]
[tex]\displaystyle B(x)=mx^5+nx^4+ox^3+px^2+qx+r[/tex]
[tex]\displaystyle C(x)=sx^5+tx^4+ux^3+vx^2+wx+y[/tex]

Now that we have traversed the alphabet, we may continue manipulating the given limit,
[tex]\displaystyle\lim_{x\to\infty}P(x)-Q(x)=\lim_{x\to\infty}\frac
{(a-d)x^2+(b-e)x+(c-f)}{\sqrt[3]{x^6+A(x)}+\sqrt[3]{x^6+B(x)}+\sqrt[3]{x^6+C(x)}}[/tex]
[tex]\displaystyle\lim_{x\to\infty}P(x)-Q(x)=\lim_{x\to\infty}\frac{(a-d)x^2+(b-e)x+(c-f)}{x^2\left(\sqrt[3]{1+\frac{A(x)}{x^6}}+\sqrt[3]{1+\frac{B(x)}{x^6}}+\sqrt[3]{1+\frac{C(x)}{x^6}}\right)}[/tex]

Since the degree of [tex]\displaystyle A(x), B(x), C(x) < 6 [/tex], then we have:
[tex]\displaystyle\lim_{x\to\infty}1+\frac{A(x)}{x^6}}+\sqrt[3]{1+\frac{B(x)}{x^6}}+\sqrt[3]{1+\frac{C(x)}{x^6}}\right)=3[/tex]
[tex]\displaystyle\lim_{x\to\infty}P(x)-Q(x)=\lim_{x\to\infty}\frac{(a-d)x^2+(b-e)x+(c-f)}{3x^2}[/tex]

Thus, the general form (recall that the first terms are [tex]\displaystyle x^3[/tex]):
[tex]\displaystyle\lim_{x\to\infty}P(x)-Q(x)=\lim_{x\to\infty}=\frac{a-d}{3}[/tex].

In this case, [tex]\displaystyle a=-6, d=3[/tex], so [tex]\displaystyle \lim_{x\to\infty}\sqrt[3]{x^3-6x^2+5x+4} - \sqrt[3]{x^3+3x^2+3x+2}=\fbox{-3}[/tex]

Whew!
[/hide]

Good, but you should have followed my suggestion!  Because we may write your initial expression as the difference of two of my (much nicer!) expressions, requiring only half the alphabetics.  :)  Your "make-the-denominator-disappear" step seems like it might be a little questionable, as well, but just a tiny bit -- I don't think it's generally good form to get rid of one portion of a limit without touching the rest of it.

Ah. That seems so much simpler. Can this concept be extended?

For example,

[tex]\displaystyle\lim_{x\to\infty}\sqrt[5]{(32x^5-240x^4+1000x^3-23x^2+42x-\pi)}-\sqrt[3]{(8x^3+60x^2-100x+42)}=\lim_{x\to\infty}\sqrt[5]{(2x-3)^5+P(x)}-\sqrt[3]{(2x+5)^3+Q(x)}=-3-5=-8[/tex]

I've tried using that method with many different radicals, and it's always seemed to work (and it actually makes sense when I think about it).

kunny's method works, but it needs justification.  As Kent Merryfield has been explaining in some of his recent posts, Big O notation is one nice method.  As $\epsilon$ approaches 0, we have
\[\sqrt[3]{1 + \epsilon} = 1 + O(\epsilon).\]
Thus
[tex]\begin{eqnarray*}
\sqrt[3]{(x+1)^3+1}
  & = & (x+1) \sqrt[3]{1 + \frac{1}{(x+1)^3}} \\
  & = & (x+1)(1+O(\frac{1}{(x+1)^3})) \\
  & = & x + 1 + O(\frac{1}{(x+1)^2}).
\end{eqnarray*}
[/tex]
In other words, as kunny suggests, that term is $x+1$ plus a negligible amount.


How many questions do I need right on AIME to get to USAMO if I got a 100.5 on the AMC 12A, but a 110 on the AMC 12B? Oh, and I am a junior, if that makes any difference.

doesn't make a difference what grade ur in, but i think u need a 9-10 on AIME to make it (with ur relatively low AMC 12 score)

Look on the [url=http://www.unl.edu/amc]AMC website[/url] for qualification process details.

I've never taken AIME before, as this is the first year I even heard of it. Is it generally hard to get 10 questions right?

Yes, very difficult. You can look at some of the old tests in the Contests section; look at the top of your screen.

[quote="ZzZzZzZzZzZz"]Actually, I would disagree.  AMC and AIME are such different tests.  It is quite easy to mess up on the AMC because of the time constraints.  The AIME has harder problems but could be easier for people who like to take their time.  It all depends on whether you're a deep thinker or a speed tester.[/quote]If it's the first time he's heard of it, it is going to be very difficult for him to score a 10. I don't think that's arguable.


calculus
integration
limit
probability
logarithms
geometry
expected value

Real numbers are chosen at random from the interval [0,1]. If after choosing the nth number the sum of the numbers so chosen first exceeds 1, show that the expected or average value for n is e.

It has appeared before, and it's clearly not Intermediate level.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22813]This thread[/url] contains links to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20753]this thread[/url] and a follow-up question at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20848]this thread[/url].

JBL, do you know where we can find a solution on the forums?  it's a nice problem.

Thanks JBL.  Converting the problem equation to a differential/integral equation is a pretty robust technique.  For mdk (does he even read the threads he posts)?  I'll write a solution using this method and a couple of similar problems that can be approached analogously.

Let $ E(t)$ denote the expected number of turns for the sum to exceed $ 0<t\leq 1$, starting at $ 0$.  You reach $ t$ (or more) a proportion of $ 1-t$ times, and you reach a number in $ [0,t]$ a proportion of $ t$ times, each with equal probability.  We plug into the definition of expected value and note that the last term corresponds to an integral to get
\[ E(t) = (1-t)+\int^{t}_{0}(E(x)+1)dx \]
Turning this into a differential equation, we get $ E(t) = a\exp(t)$.  We need to determine $ a$.  Note that $ \lim_{t\searrow 0}E(t) = 1$ (as $ t$ gets smaller and smaller, the expected number of turns we need to get a number $ >t$, starting at $ 0$, is just $ 1$).  So $ E(1) = \exp(1)=e$

Now for a related problem.  This was the June 2007 IBM Ponder This problem.  This solution wasn't one of their official solutions but I think it's quicker and nicer.

[quote]
You are playing a game in which you can ask for numbers. Each such number is a random value uniformly (and independently) distributed between 0 and 1. After you receive each number you can decide whether to request an additional number or to stop. When you stop your score is the sum of all the numbers you have received. Let 0<x<1. We ask two questions.
Suppose you are trying to achieve a score between x and 1. What are your chances of success?
[/quote]

In an analogous manner, if $ E(t)$ denotes your chances of success starting at $ 0\leq t\leq x$ and trying to achieve a score between $ x$ and $ 1$, we get the integral equation
\[ E(t) = (1-x)+\int^{x}_{t}E(x)dx \]
Differentiating gives $ E'(t) =-E(t)$, so $ E(t) = a\exp(-t)$.  It remains to determine $ a$.  This time the boundary condition becomes $ \lim_{t\nearrow x}= 1-x$.  This means that $ a\exp(-x)=1-x$, or $ a=(1-x)\exp x$.  So $ E(t) = (1-x)\exp x\exp(-t)$, and the answer that we are looking for is precisely $ E(0) = (1-x)\exp x$.

One more harder problem, which I posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152328]here[/url].  There's a quicker solution (imo) than the one posted.

Let $ P(Y|x)$ denote the probability that you win with a remaining stick of length $ 1-x$ $ (0\leq x\leq 1/2)$.  The probability that you win on the next cut is $ (1/2-x)/(1-x)$, and the probability that you win on a later cut is given by $ 1/(1-x)\int^{1/2}_{x}P(Y|t)dt$, so that
\[ P(Y|x) = \frac{0.5-x}{1-x}+\frac 1{1-x}\int^{0.5}_{x}P(Y|t)dt \]
so converting to a differential equation and solving, you get $ P(Y|x) = \log(1-x)+C$ and, since we have the boundary condition $ P(Y|0.5) = 0$, you see that $ C =-\log(0.5) = \log 2$.

From there, the desired probability is $ P(Y|0) = \log(1)+\log(2) = \log 2$, and that's that :)

Maybe this entire thread should be moved to a different forum?  I don't want this post to be ignored :)

e: to fix silly typo

[quote="blahblahblah"]mdk (does he even read the threads he posts)?
...
Maybe this entire thread should be moved to a different forum?  I don't want this post to be ignored :)[/quote]

All evidence indicates that mdk doesn't read anyone else's posts, ever.

Would a moderator please move this to, say, Calculus Unsolved Problems?

[hide="Sketch of a Solution"]I have a feeling that this will lead to $ \sum_{n=0}^{\infty}\frac{1}{n!}$, which equals $ e$.

It would probably involve the fact that the volume (or area) in $ n$-dimensional space with axes $ x_{1},x_{2},\ldots x_{n}$ bounded by the hyperplanes $ x_{1}=0$, $ x_{2}=0$, $ \ldots$ $ x_{n}=0$, and $ x_{1}+x_{2}+\ldots+x_{n}=1$ is $ \frac{1}{n!}$.  That could probably be proven by integrating $ u$ repeatedly with respect to $ u$.  You'd get $ \frac{1}{2}u^{2}$, $ \frac{1}{6}u^{3}$, $ \frac{1}{24}u^{4}$, and so on; generally, $ \frac{1}{n!}u^{n}$.  Finally, setting $ u=1$ yields $ \frac{1}{n!}$.

Adding them up from $ n=0$ to $ \infty$ will yield $ \sum_{n=0}^{\infty}\frac{1}{n!}=e$.

I don't know, but this is probably what it'd look like.[/hide]

some detail on lingomaniac88's approach:

(Expectation of the number of events that happened)
Let $ (E_{n})$ be a sequence of probabilistic events. Let $ N$ be a random variable that counts the number of events (among $ (E_{n})$) that happens. Then the expectation of $ N$ is $ E(N) = \sum_{n}Pr(E_{n})$. This sum equality is true even if some of the events are dependent because this equality is a case of [url=http://planetmath.org/encyclopedia/MonotoneConvergenceTheorem.html]monotone convergence theorem[/url]. (is there a name for this?)

In our case, If $ N$ is the smallest $ n$ such that $ x_{1}+x_{2}+\ldots+x_{n}> 1$, then $ N$ is the number of $ n$ for which $ x_{1}+x_{2}+\ldots+x_{n}\le 1$ (including $ n=0$). So we can apply the sum equality.


function
algebra
domain
calculus
derivative
complex numbers
geometry

1. Prove that if $ f(z) \equal{} u(x,y) \plus{} iv(x,y)$ is analytic in some doamin, then $ g(z) \equal{} v \minus{} iu$ must also be analytic in this domain.

2.

a) Prove that if an analytic function is purely real throughout some domain, then it must be constant in that domain.

b) Repeat that question in part (a) with the word "imaginary" substituted for "real."

2a. Let $ f\equal{}u\plus{}iv$, then $ v\equal{}0$. Since $ f$ is analytic, by the Cauchy-Riemann equations: $ \frac{\partial u}{\partial x}\equal{}\frac{\partial v}{\partial y}\equal{}0$. This tells us that $ u$ is a constant in $ x$. Also, $ \frac{\partial u}{\partial y}\equal{}\minus{}\frac{\partial u}{\partial x}\equal{}0$ tells us that $ u$ is a constant in $ y$. Thus $ u$ is constant.

For b, this can be imitated the other way.

a geometric reason: at any nonsingular point of an analytic function, an infinitesimal shape will be transformed under the mapping by the derivative (by the definition of the derivative), so choosing an infinitesimal circle about $ z$ (for simplicity), we see that it gets mapped to an infinitesimal circle about $ f(z)$ acted upon by the spiral similarity represented by $ f'(z)$. However there is not actually any space to fit a circle in the real line so the derivative must vanish. Hence the derivative vanishes identically and the mapping is constant.


integration
calculus
derivative
function
calculus computations

prove that if f is continuous, then

$ \int^{x}_{0} f(u)(x\minus{}u) du \equal{} \int^{x}_{0} (\int^{u}_{0} f(t) dt) du$

Good day. Here is a solution
$ \begin{array}{l}
 \int\limits_0^x {\left( {\int\limits_0^u {f\left( t \right)dt} } \right)du}  \equal{} \int\limits_0^x {u'\left( {\int\limits_0^u {f\left( t \right)dt} } \right)du}  \equal{}  \\
  \equal{} x\left( {\int\limits_0^x {f\left( t \right)dt} } \right) \minus{} 0\left( {\int\limits_0^0 {f\left( t \right)dt} } \right) \minus{} \int\limits_0^x {uf\left( u \right)du}  \equal{}  \\
  \equal{} \underbrace {\int\limits_0^x {xf\left( u \right)du} }_{t \equal{} u} \minus{} \int\limits_0^x {uf\left( u \right)du}  \equal{} \int\limits_0^x {\left[ {xf\left( u \right) \minus{} uf\left( u \right)} \right]du \equal{} }  \\
  \equal{} \int\limits_0^x {\left[ {\left( {x \minus{} u} \right)f\left( u \right)} \right]du}  \\
 \end{array}$

thanks mathxl! is this method right:

L.H.S = $ x \int^{x}_{0} f(u)du \minus{} \int^{x}_{0} uf(u)du$

taking the derivative with respect to x,

$ xf(x) \plus{} \int^{x}_{0} f(u)du \minus{} xf(x)$

= $ \int^{x}_{0} f(u)du$

the derivative of the right hand side is

$ \int^{x}_{0} f(t)dt$.

the derivatives of the right hand side and left hand side w.r.t x are the same. also, the right hand side and left hand side have the same value at x=0. so the functions on the RHS and LHS are the same.

$ \int^{x}_{0} f(u)(x \minus{} u)du$ = $ \int^{x}_{0} (\int^{u}_{0} f(t)dt)$


I found this math question in the Harvard magazine. The source is from a Chinese entrance exam to foregin business...

Find the next term of this sequence:

$2, 3, 13, 175...$

[quote="borz"]$2 * 13 + 175^2$ ?[/quote]

What is the explanation for that?  Can you show the connection between these terms?


number theory
prime factorization
number theory unsolved

(a) Let $a_n$ be the last nonzero digit in the decimal representation of the number $n!$ . Does the sequence $a_1,a_2,a_3,...$ become periodic after a finite number of steps.

[quote="Andreas"]For $n > 4$ the exponent of the power of $2$ in the prime factorization of $n!$ is greater than that of the power of $5$. So $a_{n}$ is even for all $n > 1$, and the sequence doesn't become periodic.[/quote]
Sorry but I think that this is not enough... we can take the sequence 2,6,4,8,2,6,4,8,....and it is periodic...or I didn't understand what you want to say. ;)

Ohh sorry I didn't see that we have $a_1$...more interesting question is about the sequense $a_2,a_3,...$


Which of the following sets of whole numbers has the largest average?

\[ \textbf{(A)}\ \text{multiples of 2 between 1 and 101} \qquad
\textbf{(B)}\ \text{multiples of 3 between 1 and 101} \\
\textbf{(C)}\ \text{multiples of 4 between 1 and 101} \qquad
\textbf{(D)}\ \text{multiples of 5 between 1 and 101} \\
\textbf{(E)}\ \text{multiples of 6 between 1 and 101}
\]

[hide="Click for solution"]
In a set of whole numbers which are equally spaced, the average of the numbers in the set is the average of the smallest number and the largest number.  For example, the average of $ \{1,3,5,7,9\}\equal{}\frac{1\plus{}9}{2}\equal{}5$, and the average of $ \{2,5,8,11,14,17\}\equal{}\frac{2\plus{}17}{2}\equal{}9.5$.  In this problem, the averages are:

$ \textbf{A: }\frac{2\plus{}100}{2}\equal{}51\\
\\
\textbf{B: }\frac{3\plus{}99}{2}\equal{}51\\
\\
\textbf{C: }\frac{4\plus{}100}{2}\equal{}52\\
\\
\textbf{D: }\frac{5\plus{}100}{2}\equal{}52.5\\
\\
\textbf{E: }\frac{6\plus{}96}{2}\equal{}51$

so the answer is $ \boxed{\textbf{(D)}}$.
[/hide]

The answer is D which is the multiples of 6.

[quote="AIME15"][hide="Click for solution"]
the answer is $ \boxed{\textbf{(D)}}$.
[/hide][/quote]

ummm....

1) fail much?
2) post a solution so we know how you got the answer. an empty "E" is of no use to otehrs hoping to find out the answer

Adding to what mz94 wrote, don't post something if the same thing (or for this matter, something more useful) has been posted.

Would this be posting the same thing? It's basically a different solution.
The multiples of any number in any range form an arithmetic sequence. It can be proven that the average of the numbers in an arithmetic sequence is simply the average of their highest and lowest entries, so you can just add the first term and the last term, and see which one is the largest (since the sum of two numbers is twice their average). $2 + 100 = 102$, $3 + 99 = 102$, $4 + 100 = 104$, $5 + 100 = 105$, $6 + 96 = 102$. Therefore, the answer is ${\boxed{\textbf{(D)}}}$ because it has the largest number.

That's basically the same thing. BTW, this thread is over 3 years old.

[hide=My thoughts]An add on to AIME15, we are looking for the largest number where one of its multiples is 100. Therefore, $\boxed{D}$ is correct. [/hide]

These are just the threads for the AJHSME contest collections.

Perhaps not, but perhaps yes for some users.


[quote="mdk"]Prove that a solution to the equation $ x^{x}=\pi$ exists among the set of irrational real numbers[/quote]

Seems easy
[hide]Let $ f(x) = x^{x}-\pi$.

Then $ f(1) = 1-\pi<0$ and $ f(2) = 4-\pi>0$.

By the continuity of $ f$, we may apply the Intermediate Value theorem to deduce that there exists some $ x_{0}$ such that $ f(x_{0}) = 0$.

Suppose that $ x_{0}$ were rational.

Let $ x_{0}= p/q$ for some integers $ p$ and $ q$, where $ q$ is not $ 0$, and the fraction is in lowest terms.
Then $ (p/q)^{p/q}= \pi$. But this implies that $ \pi$ is not transcendental, since $ (p/q)^{p/q}$ is not. (It is a root of the polynomial equation $ q^{p}x^{q}-p^{p}= 0$).

Hence, we have achieved our contradiction that $ \pi$ is not transcendental.

Hence, there does exist some irrational $ x_{0}$ such that $ x_{0}^{x_{0}}= \pi$.[/hide]


inequalities
geometry
perimeter
inequalities proposed

First of all, your 3 in the problem should be replaced by 3/2.
Let D be the midpoint of AB. Translate BC to DE, one will then obtain a triangle with sides $m_{a},m_{b},m_{c}$ in length, medians $3a/4,3b/4,3c/4$ in length, and area 3S/4. Thus it suffices to prove \[\frac{3abc}{8S}\ge\frac{\sum am_{a}}{\sum a}\] Note that $am_{a}\le2S$, thus the inequality is reduced to \[abc(a+b+c)\ge16S^{2}\] which is equivalent to \[abc\ge(a+b-c)(b+c-a)(c+a-b)\] Now anyone familiar with inequality should know what it is, if not, click here[hide]Schur[/hide]

[quote="gemath"]Given are $\triangle ABC$ with three median $m_{a},m_{b},m_{c}$  and area $S$ prove that
$\frac{m_{a}m_{b}m_{c}}{3S}\ge\frac{am_{a}+bm_{b}+cm_{c}}{m_{a}+m_{b}+m_{c}}$
 :)[/quote]

this does not hold for $a=b=c$

the correct version is probably

[quote="gemath"]Given are $\triangle ABC$ with three median $m_{a},m_{b},m_{c}$  and area $S$ prove that
$\frac{2 \cdot m_{a}m_{b}m_{c}}{3S}\ge\frac{am_{a}+bm_{b}+cm_{c}}{m_{a}+m_{b}+m_{c}}$
 [/quote]

which is not very hard.

 :)

yes you are right thank you for your solution

I think mecrazywong made a mistake. $am_{a}\geq 2S$

In order to do this one, apply the following inequality

Given $ABC$. Then $PA+PB+PC \geq 2 \frac{\sum am_{a}}{a+b+c}$.

Let $P=O$. Then $3R = \frac{3abc}{4S}\geq \frac{\sum am_{a}}{a+b+c}$. Now change it into the triangle with sides equal to medians $m_{a}, m_{b}, m_{c}$, we obtain the result.

[quote="treegoner"]I think mecrazywong made a mistake. $am_{a}\geq 2S$
[/quote]
Oh...what a stupid mistake :blush:

okk treegoner I see it as a private case of Jack Garfunkel's ineq:
Given are $\triangle ABC, M$ is any point inside it $A_{1},B_{1},C_{1}$is orthor project of $M$ on lines $BC,CA,AB$ and $A_{2},B_{2},C_{2}$ is trace of $M$ then $p(A_{2}B_{2}C_{2})\ge p(A_{1}B_{1}C_{1})$ ($p$ is notation of perimeter)
It is a private case when $M=G$ and use $m_{a},m_{b}m_{c}$ is three side of triangle!
Is there a good solution for Jack's ineq


Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac{1}{a^2\plus{}b^2}\plus{}\frac{1}{a^2\plus{}c^2}\plus{}\frac{1}{a^2\plus{}d^2}\plus{}\frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{b^2\plus{}d^2}\plus{}\frac{1}{c^2\plus{}d^2} \ge \frac{81}{2(a\plus{}b\plus{}c\plus{}d)^2}\]

Ps: My first solution for it used Mixing variables and it is very ugly. But now I have found a complete solution with Cauchy Schwarz. :)

[quote=can_hang2007]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac{1}{a^2\plus{}b^2}\plus{}\frac{1}{a^2\plus{}c^2}\plus{}\frac{1}{a^2\plus{}d^2}\plus{}\frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{b^2\plus{}d^2}\plus{}\frac{1}{c^2\plus{}d^2} \ge \frac{81}{2(a\plus{}b\plus{}c\plus{}d)^2}\]

[/quote]
[b]Nice.[/b]

Yes. The $(a+b+c+d)^2$ kinda gives the C-S clue away.

Titu's lemma :P
@below, am thinking about it :)

[quote=JustinLee2017]Titu's lemma :P[/quote]

How?
Can you show that? Thanks

[quote=Akatsuki1010]
A very nice solution sir![/quote]

I found a mistake

Yes, in the second line of the hided text.

There must be $4bd-3d^2$ instead of $4bd+3d^2$

[quote=can_hang2007]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac{1}{a^2\plus{}b^2}\plus{}\frac{1}{a^2\plus{}c^2}\plus{}\frac{1}{a^2\plus{}d^2}\plus{}\frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{b^2\plus{}d^2}\plus{}\frac{1}{c^2\plus{}d^2} \ge \frac{81}{2(a\plus{}b\plus{}c\plus{}d)^2}\]
[/quote]

There is a very simple proof for this inequality. See [url=https://artofproblemsolving.com/community/c6h2421202_harder_than_cans_inequality]here[/url] 5-variable version.
I am sure my proof is simpler than Can's proof  :) .

Let $a, b, c>0, a+b+c=3.$ [url=https://artofproblemsolving.com/community/c6h543527p18851613]Prove that[/url]
$$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}>\frac{10}{9}$$
Let $a,b,c$ be nonnegative real numbers .[url=https://artofproblemsolving.com/community/c4h2298775p18233651]Prove that[/url]
$$\frac{1}{a^2+b^2} + \frac{1}{b^2+c^2} + \frac{1}{c^2+a^2}\geq \frac{5}{2\sqrt{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}}}$$
Let $a,b,c$ be nonnegative real numbers such that $ab+bc+ca>0.$ [url=https://artofproblemsolving.com/community/c6h381263p2111211]Prove that[/url]$$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}\ge\frac{4}{a^2+b^2+c^2}+\frac{1}{2(ab+bc+ca)}\ge\frac{9}{2(a^2+b^2+c^2)}.$$
Given three non-negative real numbers   $   a,b,c   $   satisfying:  $   ab + bc + ca> 0   $ . [url=https://artofproblemsolving.com/community/c6h2421201p19927427]Prove that :[/url]
$$ \frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\frac{18}{(a+b+c)^2} \ge \frac{7}{a^2+b^2+c^2}+\frac{7}{2(ab+bc+ca)}.$$

Let  $a, b, c, $ be non-negative real numbers . [url=https://artofproblemsolving.com/community/c6h543527p18900472]Prove that[/url]
$$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2} \ge \frac{10}{(a+b+c)^2}$$
[quote=can_hang2007]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac{1}{a^2\plus{}b^2}\plus{}\frac{1}{a^2\plus{}c^2}\plus{}\frac{1}{a^2\plus{}d^2}\plus{}\frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{b^2\plus{}d^2}\plus{}\frac{1}{c^2\plus{}d^2} \ge \frac{81}{2(a\plus{}b\plus{}c\plus{}d)^2}\][/quote]
Let  $a,b,c,d,e$ are nonnegative real numbers such that at most one of them can be zero. [url=https://artofproblemsolving.com/community/c6h2421202p19927450]Prove that:
[/url]
$$\frac{1}{a^2+b^2}+ \frac{1}{b^2+c^2}+ \frac{1}{c^2+d^2}+ \frac{1}{d^2+e^2}+ \frac{1}{e^2+a^2}+ \frac{1}{a^2+c^2}+ \frac{1}{c^2+e^2}+ \frac{1}{e^2+b^2}+ \frac{1}{b^2+d^2}+ \frac{1}{d^2+a^2} \ge \frac{112}{(a+b+c+d+e)^2}$$


Use the formula $ \sum_{n\leq x}\tau (n) \equal{} x\operatorname{ log }x \plus{} (2\gamma \minus{} 1)x \plus{} q(x)$, where $ x\geq 1$ and $ |q(x)|\leq 4x^{1/2}$, to show that $ \sum_{n\leq 100}\tau (n) \equal{} 482$. How can we use that formula as $ 4\sqrt {100} > 1$?

I don't know, but perhaps you can start by observing that $ \tau(n)$ is even except when $ n$ is a square, and since there are $ 10$ squares less than or equal to $ 100$, the sum must be even. This at least cuts down the possible options for $ \sum_{n \leq 100} \tau(n)$ by half... but still not quite enough to pinpoint the value. I'm not sure what Jameson has in mind here.


AMC
AIME
trigonometry
quadratics
trig identities
Law of Cosines

Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

[asy]
defaultpen(linewidth(0.8));
draw((100,0)--origin--60*dir(60), EndArrow(5));
label("$A$", origin, SW);
label("$B$", (100,0), SE);
label("$100$", (50,0), S);
label("$60^\circ$", (15,0), N);[/asy]

Kalva's solutions is weird, with the velocity triangle and all. I don't get that. Haven't taken physics before. :(

Anyways, here's my solution.

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=642&sid=1c57b2f2c0d40693e8d1ddd11101ac34[/img]

We first assume that the two skaters will meet, and let $AC=8t$ and $BC=7t$, where $t$ is the time it takes to meet. Now, we apply the Law of Cosines to get a quadratic in terms of $t$: $49t^2=64t^2+10000-800t\implies 5(3t-100)(t-20)=0$. We take the smaller value of $t$ so that $AC=\boxed{160}$.[/hide]

I'm not sure about the Kalva solution, but here's another way to do it if you don't know the Law of Cosines. :)
You take the problem by horizontal and vertical components. Since the speeds are constant and they start and end at the same vertical height, the line AllieBillie is always parallel to AB (their original location). The horizontal distance between A and B decreases at a rate of $8\cos 60 + 7\cos {ABC}$. So
$(8\cos 60 + 7\cos {ABC})t=100$ (rt=d)
Express their final height:
$(8\sin 60)t = (7\sin {ABC})t = h_{final}$
From the above equation, we cancel t's and find that $\sin{ABC}=\frac{4\sqrt{3}}{7}$
$\cos{ABC} = \sqrt{1-\sin{ABC}^2}$
$\cos{ABC} = \frac{1}{7}$
Substitute into the first equation:
$t(8(\frac{1}{2})+1)=100$
$t=20$
$8t=160$
So we get the same answer as 4everwise.

A little different from 4everwise.

[hide=Solution]Let the intersection be $P$. Denote $AP=x$, so LoC gives $BP=\sqrt{x^2-100x+10000}$ and hence $7x=8\sqrt{x^2-100x+10000}$. Solving, we get $x=160, \frac{800}{3}$, and out of these $\boxed{160}$ is the smallest, giving the earliest meeting point.[/hide]


Find the ordered pair (x,y), where x and y are integers and 1 < x < y., such that xy+y+x=50

Thank you.

[hide]We note that, $(x+1)(y+1) = xy + x + y +1$
We know that $xy + x + y = 50$

$xy + x + y +1 = 50 +1$

$xy + x + y + 1 =51$

$(x+1)(y+1) = 51$

Ways to make 51:
$17*3, 1 * 51$

1*51 is ruled out because if we use it, we would get x as 0 and y as 50, but x and y must both me greater than 1.

So we must use 3*17.

$x+1 = 3, x = 2.$

$y + 1 = 17, y = 16.$

Ordered pair is $\boxed{(2, 16)}$[/hide]

[hide]Yes, use Simon's Favorite Factoring Trick.  [/hide]

[quote="kyyuanmathcount"]Yes, use Simon's Favorite Factoring Trick. [/quote]

Who is Simon and what is "Simon's Favorite Factoring Trick"

Simon is ComplexZeta (forum member) and Simon's favorite factoring trick is $xy+x+y+1=(x+1)(y+1)$


calculus
derivative
algebra
polynomial
quadratics
geometry
perpendicular bisector

Let $f(z) \in \mathbb{C}[z]$ be a polynomial of degree $\geq 2$, and let $\alpha,\beta$ be any pair of distinct zeros of $f$. If $\ell$ is the perpendicular bisector of the segment joining $\alpha$ and $\beta$ in the complex plane, show that the polynomial $f'(z)$ has a zero in either of the two (closed) half-planes determined by $\ell$.

[quote="vess"]show that the polynomial $f'(z)$ has a zero in either of the two (closed) half-planes determined by $\ell$.[/quote]
Sorry, but i don't understand this text! I know that zeros of $f'$ lies convex-hull of zeros of $f$.

The line $\ell$ divides the plane into two half-planes. Prove that either both these half-planes contain a zero of $f'(z)$; or $\ell$ passes through a zero of $f'(z)$ (as is for instance the case for quadratic polynomials).

In other words, you may think of the problem as follows: prove that there exists a zero $\omega$ of $f'(z)$ such that $|\alpha-\omega| \leq |\beta-\omega|$.


function
invariant
SFFT
special factorizations

$A=\{1, 2, 3, 4... n\}$ and $f(x,y)=xy+x+y$, where $x$ and $y$ are chosen randomly from $A$.  the new number is put into $A$ and $x$ and $y$ are removed.  find all possible final values of $A$ once the function has been applied $n$ times

[hide="hint"]simon's favorite factoring trick[/hide]
[hide="another hint"]invariants[/hide]

@mods: if this is in the wrong place, please move it

What is Simon's favorite factoring trick?

could a mod please move this to intermediate?  I think this may be a bit difficult for basics

I think I know how to do this, but yes, farther up than intermediate too.

Pre-Olympiad.

Can you hint us what the invariant might be?

why is it called simons favorite factoring trick anyways?

b/c simon (a.k.a. Complex Zeta) invented it.  or maybe he just made it popular.  its something like that :D


having $z=\sqrt{6}+\sqrt{2}+i(\sqrt{6}-\sqrt{2})$
how do i put this number into trigonometric form
i found it difficult to find $\phi$ so that $tan(\phi)=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$
other than that there wouldn't be any more problems.
at the answer section in my book they say that $\phi=\frac{\pi}{12}$
i would like to know why this is ?

pi/12=pi/3-pi/4
you can put 2sqroot2 as factor


geometry
probability
articles
expected value

A square dartboard of side 6 contains 3 concentric circles or radii 1,2, and 3. The numbers in each region represent the score for landing in that region. Darts are thrown so that they hit the board at random. What is the expected score for one throw? Leave your answer in terms of $pi$.

The Dartbord:
Since I don't have a scanner here goes:

The outter-most circle gives one point, the middle circle gives 2 points, and the inner-most circle gives five points. The area of the square that isn't part of the circle (the area near the corners) gives 0 points.

I'm getting somewhere with this one, but I don't get how you calculate the "expected score."

Then why does it ask you to answer in terms of $pi$?

[quote="sonny"]I think what they mean is that which score is most likely to come up. That would be the region which covers up the most area.[/quote]

Expected score(value) is calculated by multiplying each score to the probability to get each score and adding all.

For example, if the square is divided into two parts and each have score of 5 or 1, then your expected score is
$\frac{1}{2}\cdot 5+\frac{1}{2}\cdot 1=3$.

[url=http://en.wikipedia.org/wiki/Expected_value]Wikipedia article on expected value[/url]

In this case,
[hide]$\frac{\pi}{36}\cdot 5+\frac{3 \pi}{36}\cdot 2+\frac{5 \pi}{36}\cdot 1+\frac{36-9 \pi}{36}\cdot 0=\boxed{\frac{4}{9}\pi}$.[/hide]
in terms of $\pi$.

Thats what I thought, and I just realized what I was doing wrong.

Thanks!


quadratics
algebra
difference of squares
special factorizations
Divisibility Theory

Let $m$ and $n$ be natural numbers and let $mn+1$ be divisible by $24$. Show that $m+n$ is divisible by $24$.

Seems a little straightforward.

In $\mathbb{Z}_8$, $mn=-1$ has solutions $\{m,n\}=\{1,7\},\{3,5\}$ and in both cases $m+n=0$.
In $\mathbb{Z}_3$, $mn=-1$ has solution $\{m,n\}=\{1,2\}$ and again $m+n=0$, so $24|m+n$.

Let

$ m \equal{} 24k_1 \plus{} r_1$ and
$ n \equal{} 24k_2 \plus{} r_2$

$ 0 \le r_1,r_2 <24$

Since $ 24|mn \plus{} 1 \rightarrow mn \equiv r_1r_2 \equiv 23$  ( mod $ 24$).

Since $ 23$ is prime and $ r_1$ and $ r_2$ are integers then $ r_1 \equal{} 1$ and $ r_2 \equal{} 23$ (or $ r_1 \equal{} 23$ and $ r_2 \equal{} 1$).

So $ m \plus{} n \equal{} 24k_1 \plus{} r_1 \plus{} 24k_2 \plus{} r_2 \equal{} 24(k_1 \plus{} k_2) \plus{} 1 \plus{} 23 \equal{} 24(k_1 \plus{} k_2 \plus{} 1)$

[quote]Seems a little straightforward. [/quote]

yeah it is  :P

[quote="Wichking"]
Since $ 23$ is prime and $ r_1$ and $ r_2$ are integers then $ r_1 \equal{} 1$ and $ r_2 \equal{} 23$ (or $ r_1 \equal{} 23$ and $ r_2 \equal{} 1$).
[/quote]
you are wrong...if $ r_1\equal{}7 , r_2\equal{}17$ then $ r_1r_2\equiv23 (\mod 24)$

you are correct  :(  in fact

there are 4 such pairs

$ (1,23)$
$ (5,19)$
$ (7,17)$
$ (11,13)$

(we can show that these are the only such pairs and their sum is always 24)... but that makes my solution even more ugly :(

A related problem:
If $mn+1$ is divided by k,then $m+n$ is divided by k.
Find the maximimu of $k$.($k_{max}=24$)

[hide="A18"]
Let $mn=24k-1$. Every factoring will have $m$ and $n$ both be odd, so their product can be expressed as a difference of squares, say $a^2-b^2$, where $a-b=m$, $a+b=n$. Now, the quadratic residues mod 24 are $0, 1, 4, 9, 16, 12$, so if $a^2-b^2\equiv-1\mod24$, then $a^2\equiv0\mod24, b^2\equiv1\mod24 \implies a \equiv 0\mod12 \implies m+n=(a-b)+(a+b)=2a\equiv0\mod24 \boxed{}$[/hide]


A dartboard of radius 4 inches is placed on top of a dartboard of radius 8 inches. If twelve darts land randomly on the dartboard, what is the most likely value for the number of darts landing on the smaller dartboard?

[hide]The ratio of the areas is 1/4. So 1/4 * 12 = 3. [/hide]

[quote="mathnerd314"]I don't think you should say "most likely value" but instead "expected value". Like if you flip 100 coins, how many will you expect to land heads? 50. But how likely is that? Not 100%, obviously. There's all this chi-square junk that nobody wants to get into (especially in the Mathcounts forum) but I just wanted to clarify terms. Please correct me if I'm wrong.[/quote]

You're right, but we know what math92 meant  :P

[quote="mysmartmouth"]Wait, I don't get the question.  Can someone please explain?

Thanks![/quote]
You are trying to find the most likely number of darts that will land on a 4 inch diameter dartboard on top of an 8 inch diameter dartboard, if the darts have to hit a dartboard.

Hope that helps!

[quote="nat mc"][quote="mathnerd314"]I don't think you should say "most likely value" but instead "expected value". Like if you flip 100 coins, how many will you expect to land heads? 50. But how likely is that? Not 100%, obviously. There's all this chi-square junk that nobody wants to get into (especially in the Mathcounts forum) but I just wanted to clarify terms. Please correct me if I'm wrong.[/quote]
But 50 heads is the most likely value.  Although it is not very likely, it is the most likely with a probability of $\frac{{{100}\choose{50}}}{2^{100}}$, all number of heads have a lower value.  It is Pascals triangle, where the middle value is greatest[/quote]

Hmmmm... Right. :oops_sign:


geometry
rectangle
geometric transformation
rotation
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14577[/img]

 :(

[url]http://gallica.bnf.fr/ark:/12148/bpt6k99640w/f88.table[/url]

In my very old spanish version of the Petersen's book, with outlined solutions of the famous spanish problemist Jose Gallego Diez, one can read the following:
Problem 367 (its need its solution to solve 368): Two circles  c_1, c_2 are given; point A belong to c_1 and B to c_2. Determine on both circles points X and Y in a such manner that arcs AX and BY are similar and XY must have a given lenght.
Outline of solution: In the direct similarity between c_1 and c_2, we determine the simultaneous center  O of the rotation and the homotecy from which the similarity is composed, considering A and B as homologous points and also X and Y;  the triangles ABO and XYO are similar.
Back to problem 368: If we draw the two circles which are the locuses of two oppsed vertices of the rectangle, we can apply the previous problem.


how would you prove this:

let T : V -> W be a linear transformation
if {v1, v2, v3} is linearly dependant, prove that {T(v1), T(v2), T(v3)} is linearly dependant.

kinda lost =/

Start with the definition: what does linear dependence mean? Write it out explicitly for this case.


For each $ n\in \mathbb{N}$, $ a(n)$ denotes the units digit of $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} \cdots \plus{} n^2$.
How many $ n$ with $ 1\le n\le 2009$ are there such that $ a(n) \equal{} 9$ ?

[hide]The answer is $ 200$.

Only the units digit of a number is relevant in considering the units digit of a sum.

Call $ b(n)$ the units digits of $ n^2 \: \rm{mod} \: 10$. We trivially get the following:
$ b(0) \equal{} 0$
$ b(1) \equal{} 1$
$ b(2) \equal{} 4$
$ b(3) \equal{} 9$
$ b(4) \equal{} 6$
$ b(5) \equal{} 5$
$ b(6) \equal{} 6$
$ b(7) \equal{} 9$
$ b(8) \equal{} 4$
$ b(9) \equal{} 1$

$ b(n)$ is periodic with a period of $ 10$ by definition. If we compute $ a(n)$ for all positive $ n \le 20$, we get the following:
$ a(0) \equal{} 0$
$ a(1) \equal{} 1$
$ a(2) \equal{} 5$
$ a(3) \equal{} 4$
$ a(4) \equal{} 0$
$ a(5) \equal{} 5$
$ a(6) \equal{} 1$
$ a(7) \equal{} 0$
$ a(8) \equal{} 4$
$ a(9) \equal{} 5$
$ a(10) \equal{} 5$
$ a(11) \equal{} 6$
$ a(12) \equal{} 0$
$ a(13) \equal{} 9$
$ a(14) \equal{} 5$
$ a(15) \equal{} 0$
$ a(16) \equal{} 6$
$ a(17) \equal{} 5$
$ a(18) \equal{} 9$
$ a(19) \equal{} 0$
$ a(20) \equal{} 0$

$ a(n)$ is periodic with a period of $ 20$. If $ n \: \rm{mod} \: 20$ is equivalent to either $ 13$ or $ 18$, $ a(n) \equal{} 9$. There are $ 200$ such values of $ n$ when $ n \le 2009$.
[/hide]


Girls are outnumbered 3:1.  Therefore, let's start 1/3 of the way to 50:

[b]16[/b]

Same rules as last time: For new players, if you're a girl, you subtract 1 from the previous number.  If you're a boy, you add 1 to the previous number.  Girls, to win you have to get to 0.  Boys, to win you have to get to 50.

well its 2 v 0 and that's very fair isn't it?

Well, Blazer.. remember when it wuz just me against you, dillinger, and PowerOfPi?? :P

[b]10[/b]

What about round 1?  It was like 5 boys to 2 girls...

Heh, but the game started at 25

might as well [b]11[/b]

But I gtg now cyas!

[b]9[/b]

I know!!!
Well, looks like girls are going to win this round!!!  :D

[b]7[/b] WE're getting there!!! GO  GIRLS!

[b]3[/b] haha so close!!!!

the boys are losing... which is strange bcuz they outnumber us a lot to three..

[b]2[/b] Almost there!  It's not surprising.  We're the only people in the FF....   :P

[b]0[/b]  Hah!  We win!   :D  :rotfl:  :)

YAYYYYYY GIRLS WIN!!!!!!!!!! GOOD TRY BOYS..........

lolz.. they didn't even try!!! *coughBlazer123cough*

Good job westiepaw!! :D

@spaceguy524: please do not post in games that are finished, thanks!


How do you put comments (words) during an equation array?

hmm havn'et tried this but u can use the
[code]\text{blah blah blah}[/code] thing while during eqnarray...like i said try it out in the test forum im not sure if it works or not

Yes, for example:

[code]\begin{eqnarray*}
f(x) &=& x^2 + 4x + 4 \\
&=& (x+2)^2 \qquad \text{(by factoring the expression)}
\end{eqnarray*}[/code]
\begin{eqnarray*}f(x) &=& x^{2}+4x+4 \\ &=& (x+2)^{2}\qquad \text{(by factoring the expression)}\end{eqnarray*}

what about only labeling selected equations with numbers?

Use eqnarray instead of eqnarray* so all equations are numbered. Then use \nonumber for those lines you don't want numbered.
\begin{eqnarray}f(x) &=& x^{2}+4x+4 \nonumber \\ &=& (x+2)^{2}\qquad \text{(by factoring the expression)}\end{eqnarray}
[code]\begin{eqnarray}
f(x) &=& x^{2}+4x+4 \nonumber \\
 &=& (x+2)^{2}\qquad \text{(by factoring the expression)}
\end{eqnarray}[/code]


In the geometric sequence with a first term of $ 6$ and a second term of $ \minus{}6$, what is the $ 205^{th}$ term?

[quote="GameBot"]In the geometric sequence with a first term of $ 6$ and a second term of $ \minus{} 6$, what is the $ 205^{th}$ term?[/quote]

6. :)

For every odd power, is is positive 6. For every even power, -6. This is because the first odd power (1) is positive, and each time you multiply by -1.

:lol:

Geometric sequence : $ a, ar, ar^2, ...$, where $ a$ is the first term and $ r$ is the common ratio.

$ a \equal{} 6$

$ ar \equal{} \minus{}6$

$ r \equal{} \minus{}1$

The 205th term is $ ar^{204} \equal{} 6 \cdot (\minus{}1)^{204} \equal{} \boxed{6}$

*Note: This is a CD problem, so you should be able to recognize instantly that the ratio is $ \minus{}1$. Minamoto's is an example of a good, fast approach to the problem.


1) Find the differential equation whose solution represents the family $ y^{2} \equal{} 2c( x\plus{} \sqrt{c} )$

2 ) Find the differential equation whose solution represents the family  $ y\equal{} k (x \minus{} k)^{2}$

Please help me..

\[ 2y\frac{dy}{dx} \equal{} 2c\]
\[ y\frac{dy}{dx} \equal{} c\]
\[ \left(\frac{dy}{dx}^2 \plus{} y\frac{d^2y}{dx^2}\right)\equal{}0\]Since \[ y \equal{} \frac{c}{\frac{dy}{dx}}\]
\[ \left(\frac{dy}{dx}^2 \plus{} \frac{c}{\frac{dy}{dx}}\frac{d^2y}{dx^2}\right)\equal{}0\]

[quote="mismatchtea"]
\[ 2y\frac {dy}{dx} \equal{} 2c\]

\[ y\frac {dy}{dx} \equal{} c\]

\[ \left(\frac {dy}{dx}^2 \plus{} y\frac {d^2y}{dx^2}\right) \equal{} 0\]
Since
\[ y \equal{} \frac {c}{\frac {dy}{dx}}\]

\[ \left(\frac {dy}{dx}^2 \plus{} \frac {c}{\frac {dy}{dx}}\frac {d^2y}{dx^2}\right) \equal{} 0\]
[/quote]
@mismatchtea
for finding DE of a family of curves we need to eliminate the arbitrary constants . But u did just the opposite work .And as there is just one arbitrary constant the DE would be of $ 1st$ order only.
@vijay
just put the value of  $ c$ i.e $ y\frac {dy}{dx}$ in the equation $ y^{2} \equal{} 2c( x \plus{} \sqrt {c})$.
Second one is to be done in similar way .
$ y' \equal{} 2k(x \minus{} k)$
$ y' \equal{} 2k\frac {(x \minus{} k)^2}{(x \minus{} k)}$
$ x \minus{} k \equal{} 2\frac {y}{y'}$
$ k \equal{} x \minus{} 2\frac {y}{y'}$
now put this value of k in original  equation .
thank u .

Yeah, I see my mistake...I only completed Calc I & II so far, so I haven't dealt with any dif. eqs. as of yet.


group theory
abstract algebra
advanced fields
advanced fields unsolved

If $G$ is a Lie group, prove that there is a neighborhood of the identity of $G$ which contains no non-trivial subgroups of $G$.

The exponential map $\exp\colon\mathfrak g\to G$ is a diffeomorphism onto some neighborhood of the identity of $G$. The preimage of any subgroup contained in this neighborhood is a bounded additive subgroup of $\mathfrak g$.

Grobber,the result you stated is somehow related to Cartan's Theorem.

mlok's proof is wrong: exp is not a homomorphism of groups.

I suggest this proof:

Let u be an open neighborhood around 0 in g and U around 1 in G such that exp is an analytic diffeomorphism from u onto U.

Suppose U contains a closed Lie subgroup H. Let h be the corresponding subalgebra of g. Let X != 0 be any element of h. Then by suitably picking t != 0 in R (or C) we have tX in u, and exp(tX) is in H.

Pull back H with exp to a subset K of u closed in g. We saw above that K ^ h (intersection) is not empty. Let X != 0 be any of its elements. Since K is closed (in g!) there is a t such that tX is in u but not in K. But then exp(tX) is not in H. A contradiction.

Finally, let U' < U be a smaller neighborhood with closure contained in U. Suppose H is a non-closed Lie subgroup in U'. Then its closure is a closed Lie subgroup in U. We're through.


function
limit
number theory theorems
number theory

Could you please tell me any properties of the function integer part  of a number $x$ ($[x]$) :?:

here it is
http://en.wikipedia.org/wiki/Greatest_integer

Thank you ery much but I think that some other basic and interesting properties are missing.
For example I think that the following equation is true(but I am not quite sure about it :oops: )
$[nx]=[x]+[x+\frac{1}{2}]+[x+\frac{2}{3}]+...+[x+\frac{n-1}{n}]$

Hi bilot4 . the correct is

$[x]+[x+\frac{1}{n}]+[x+\frac{2}{n}]+...+[x+\frac{n-1}{n}]=[nx]$

you propably want some properties like

$[\frac{x}{2}]+[\frac{x+1}{2}]=[x]$ which is true for all real x.

Exactly Kostas ! That's what I want :) Thank u very much!
Does anyone else know any other property  :?: or at least where can I find them...

Ok, take these properties as (simple) exercises :

$\forall x \in \mathbb{R}, [x+1] = [x]+1,$
$\forall (x, y) \in \mathbb{R}^{2}, [x]+[y] \leq [x+y],$
$\forall (x, y) \in \mathbb{R}^{2}, [x]+[y]+[x+y] \leq [2x]+[2y].$

$\forall x \in \mathbb{R}, \lim_{n \to+\infty}\frac{[x]+[2x]+\ldots+[nx]}{n^{2}}= \frac{x}{2}.$

$\forall n \in \mathbb{N}, \forall x \in \mathbb{R}, \sum_{k=0}^{n-1}\left[x+\frac{k}{n}\right] = [nx].$

$\forall n \in \mathbb{N}, \forall x \in \mathbb{R}, \left[\frac{[nx]}n \right] = [x].$


advanced fields
advanced fields unsolved

I have 2 questions:

1. the complement of a finite set of points in a surface is dense?

2. every connected component of a Riemann surface S is an open subset of S, and hence is a Riemann surface?

3ks!

1. Let $ M$ be a Riemann surface and $ S$ a finite set of points. Let $ p \in S$. We wish to show that $ p$ is a limit point of $ M \setminus S$. Pick any open disk $ U$ around $ p$, because $ U$ has infinitely many points, we know that $ U\cap (M\setminus S)$ is nonempty. This shows that $ p$ is a limit point of $ M\setminus S$ and therefore the compliment of a finite set is dense.

2. Pick some point $ p$ in the component $ M$ that you're interested in. Consider the set of all points which can be joined to $ p$ by a path. This set is open.


This is both a challenge problem and a question. I think I got it, but I'm looking for easier, alternate solutions, so I would appreciate it if you show your work.

EDIT: Latex used :) Thanks FMako.
[color=red]
Given:
$\frac{log_{y} x - log_{x} y} {log_{x} 10 + log_{y} 10} = 1 - log_{0.1} 2$

and x > 0, y > 0, x != 1, y != 1 [/color][color=blue]
Find y in terms of x.
[/color]

[quote="Ender01"]This is both a challenge problem and a question. I think I got it, but I'm looking for easier, alternate solutions, so I would appreciate it if you show your work.

I looked for a way to write bases on these forums and couldn't find one, so forgive me for the math being ugly and hard to read.

Key: log_y (x) = z  <=>  log base y of x  = z
[color=red]
Given:
log_y (x) - log_x (y)
----------------------------   = 1 - log_0.1 (2)
log_x (10) + log_y (10)

and x > 0, y > 0, x != 1, y != 1 [/color][color=blue]
Find y in terms of x.
[/color][/quote]

Read the LaTeX link on the left side of your screen if you want to learn how to write math functions in nicer ways on this site.
If you replace the quotation marks with dollar signs then "\log_{f(a)} g(b)" $\rightarrow \log_{f(a)} g(b)$.

[hide="solution"] RHS is $log_{\frac{1}{10}} \frac{1}{20}$, by log difference equation and fact that $\log_{a} a = 1$
LHS is $\frac{\frac{(\log{x})^2-(\log{y})^2}{\log{x} \cdot \log{y}}}{\frac{\log{10}(\log{x}+\log{y})}{\log{x} \cdot \log{y}}}$ by change of base and common denominator.
LHS is then $\frac{\log{x/y}}{10} = \log_{\frac{1}{10}} {y/x} \rightarrow y = \frac{1}{20} x$, by division, log difference equation, and difference of squares.

Hope I agree with you.
[/hide]

[hide]For the left hand side, use the change of base formula to base 10 and multiply by $\log x\log y$ to get $\frac{(\log x)^2-(\log y)^2}{\log x+\log y}=\log x-\log y=\log \frac{x}{y}$. From change of base, $\log_{b^{-1}} c=-\log_b c$, making the right hand side $1+\log 2=\log 20$. Then $\frac{x}{y}=20$, so $y=\frac{1}{20}x$.[/hide]

[hide="another solution"]
$\frac{\log_y{x} - \log_x{y}}{\log_x{10} + \log_y{10}} = 1 - \log_{.1}2$

Using the change of base formula to change everything to $\log_{10}$:

$\frac{\frac{\log{x}}{\log{y}} - \frac{\log{y}}{\log{x}}}{\frac1{\log{x}} + \frac1{\log{y}}} = 1 - \frac{\log2}{\log{.1}}$

Multiplying the top and bottom of the LHS by $\log{x}\log{y}$:

$\frac{\log^2x - \log^2y}{\log{x} + \log{y}} = 1 + \log2$

The numerator of the LHS factors into $(\log{x} + \log{y})(\log{x} - \log{y})$, and cancels with the denominator:

$\log{x} - \log{y} = 1 + \log2$

$\log{xy} = \log{20}$

$y = \frac{x}{20}$

[/hide]


[hide]Write as $ |x|\minus{}|x\minus{}1| \equal{} 2 (1\plus{}|x| \cdot |x\minus{}1| )$ The RHS has range $ [2, \infty)$ and the LHS has range $ [\minus{}1,1]$.

No solution.[/hide]

[color=darkblue]$ \bullet$ If $ x \le 0$ Equaton $ \Longleftrightarrow \frac {2x \minus{} 1}{x^2 \minus{} x \plus{} 1} \equal{} 2 \Longleftrightarrow 2x^2 \minus{} 4x \plus{} 3 \equal{} 0 \Longrightarrow x \in \emptyset$

$ \bullet$ If $ 0 < x \le 1$ Equaton $ \Longleftrightarrow \frac {2x \minus{} 1}{ \minus{} x^2 \plus{} x \plus{} 1} \equal{} 2 \Longleftrightarrow 2x^2 \minus{} 4x \minus{} 3 \equal{} 0 \Longrightarrow \emptyset$

$ \bullet$ If $ x > 1$ Equaton $ \Longleftrightarrow \frac {1}{x^2 \minus{} x \plus{} 1} \equal{} 2 \Longleftrightarrow 2x^2 \minus{} 2x \plus{} 1 \equal{} 0 \Longrightarrow x \in \emptyset$

Hence eqution have none root [/color]

Nice bullets :) I'm going to use them!

[u]Two things:[/u]

$ \bullet$ the numerator of the first equation in the first case should just be $ \minus{}1$ (ex. $ |\minus{}4|\minus{}|\minus{}5|\equal{}\minus{}1$)

$ \bullet$ The solution you get in the second case falls outside of the interval of $ x$ values for which it is valid -- $ 0 < \frac{2\plus{}\sqrt{10}}{2} \leq 1$ is not a true statement.

[quote="TZF"]Nice bullets :) I'm going to use them!

[u]Two things:[/u]

$ \bullet$ the numerator of the first equation in the first case should just be $ \minus{} 1$ (ex. $ | \minus{} 4| \minus{} | \minus{} 5| \equal{} \minus{} 1$)

$ \bullet$ The solution you get in the second case falls outside of the interval of $ x$ values for which it is valid -- $ 0 < \frac {2 \plus{} \sqrt {10}}{2} \leq 1$ is not a true statement.[/quote]

Yes. Thank you give me mistake. I have change it. :lol:


AMC
USA(J)MO
USAMO
AwesomeMath
summer program
geometry
circumcircle

Today I'm releasing a mock USAMO for your late summer enjoyment!  It is my sincere hope that these problems will pique and hold your curiosity--or at least occupy a very LARGE amount of your time ;)  Approach it as you like--take it as a mock test, browse through the problems that look interesting, or simply ignore it altogether.  If you wish to submit solutions, please PM them to me within the next week, and I'll send back some comments.  Perhaps you'll find the solutions I haven't thought of yet.

Please note: The preliminary editions of the following six problems first appeared on the AwesomeMath Team Contest as problems 34, 51, 52, 54, 55, and 56.  The full team contest compilation, absent the mock USAMO problems, of course, is available here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218962
I encourage all to take a look, especially those who are just beginning their work on olympiad-caliber problems; the team contest document is a collection of intriguing problems over a wide range of difficulty, many of which are original.

Now to the problems:

DAY I:

1. (Alex Anderson) If $ m$ and $ n$ are positive integers, determine a closed-form expression for
\[ \prod_{a \equal{} 1}^m\prod_{b \equal{} 1}^n (x \minus{} e^{\frac {2\pi i a}{m}}e^{\frac {2\pi i b}{n}})
\]
2. (David B Rush) The ordered quadruple of integers $ (w,x,y,z)$ is transformed into $ (w,x,y \minus{} w,z \minus{} x)$ under the operation $ \alpha$ and into $ (w \minus{} y,x \minus{} z,y,z)$ under the operation $ \beta$.

Prove that, for all positive integers $ p$ and $ q$, there exists a sequence of operations $ \alpha$ and $ \beta$ that transforms $ (1,0,0,1)$ into $ (a,b,c,d)$ such that $ \lbrace |p|,|q|\rbrace \equal{} \lbrace a,b \rbrace$ if and only if $ p$ and $ q$ are relatively prime.

3. (Alex Zhai) Let $ \triangle ABC$ be a triangle, and let the excircle opposite $ A$ be tangent to $ BC$, $ CA$, and $ AB$ at $ D$, $ E$, and $ F$, respectively.  Let $ AD$ meet $ EF$ at $ P$ and intersect the excircle again at $ G$.  Let $ M$ be the midpoint of $ DG$, and let $ Q$ be the second intersection point of the circumcircles of $ \triangle MFB$ and $ \triangle MEC$.  Prove that $ PQ$ is perpendicular to $ BC$.

DAY II:

4. (Alex Anderson) Let $ ABCD$ be a cyclic quadrilateral.  Let $ R$ and $ S$ be the midpoints of $ AC$ and $ BD$, respectively, and let $ E$ be the intersection of $ AC$ and $ BD$.  Prove that there exists a point P on the circumcircle of $ \triangle RES$ such that $ \triangle ABP$ is directly similar to $ \triangle CDP$.

5. (Alex Zhai) At Phillips Exeter Academy, there are $ n$ student clubs.  Students may only be in one club at a time, but some clubs are affiliated with one another.  Every day, in a simulataneous process, if a certain club has $ m$ members and $ k$ affiliated clubs, for each of the $ k$ affiliates, $ \lfloor \frac {m}{k} \rfloor$ members will leave the club and join that affiliate, and all those remaining will surrender their membership in all clubs indefinitely.

If, for every club $ c$ and day $ x$, there exists a day $ y > x$ such that the number of members of club $ c$ is different on days $ x$ and $ y$, prove that it is possible to split the clubs into two groups such that no club is affiliated with any other in its group.

6. (David B Rush) Let $ c$ be the greatest real number such that for any positive integer $ n$ and integer-coefficient polynomials $ P$ and $ Q$ such that $ P(x)Q(x) \equal{} (1 \plus{} x \plus{} ... \plus{} x^{n \minus{} 1})^2$ and $ P(1) \equal{} Q(1) \equal{} n$, $ \log \max \lbrace \min \lbrace \deg \lbrace P,Q \rbrace \rbrace \rbrace \geq c \log n$.

Prove that:
(a) If the coefficients of $ P$ and $ Q$ are required to be nonnegative, then $ c > 0$.
(b) If the coefficients of $ P$ and $ Q$ are permitted to be any integers, then $ c \equal{} 0$.

Good luck!

Haha DBR, some of these problems look familar.   :wink:   Let's see if I remember how to solve #1.  :ewpu:

hmm

is number 6 a disguised version of the dice problem?

[quote="tjhance"][quote="gighiuhui"]If, for every club $ c$ and day $ x$, there exists a day $ y > x$ such that the number of clubs on day $ c$ is different on days $ x$ and $ y$[/quote]

What is that suppose to mean? First you call $ c$ a club, and later you refer to "day $ c$".[/quote]

I'm sorry--you're right I meant

"If, for every club $ c$ and day $ x$, there exists a day $ y>x$ such that the number of members of club $ c$ is different on days $ x$ and $ y$."

Perhaps a mod could help out here since I can't edit the original post anymore.

Ironic.  I checked the LaTeX pretty thoroughly, but not the actual text.

Original post edited.

[quote="LawOfSigns"]Problem 4 Specifics- A Question

[hide]
For problem 4, is it supposed to be $ \triangle ABP$ is directly similar to $ \triangle CDP$ (as stated), or is it $ \triangle ABP$ is directly similar to $ \triangle DCP$?  I think I have found a point P such that $ \triangle ABP$ is directly similar to $ \triangle DCP$, but I am having trouble finding a point P such that $ \triangle ABP$ is directly similar to $ \triangle CDP$- is this an intended feature to make the problem more interesting and challenging?
[/hide]
[/quote]
I'm pretty sure the problem is written correctly. Your variant is very easy, unless I'm missing something.

Quickie PDF:

Hey guys,

it seems no one has bothered to submit any solutions.

So I'm going to cancel this mock USAMO, not post solutions, and simply leave these problems up here for all who may be interested.   :(:(

That's too bad, really. I solved number one but my solution is very long and thus I did not want to type it up. Sorry.

You also might meet more success in, say, early April.

hey DBR, I really like these problems, but I'm still working on them.
Could you still post solutions/hints for those interested or at least pm me them? Thanks! Great USAMO, btw.
:D

[quote="gighiuhui"]

DAY I:

1. (Alex Anderson) If $ m$ and $ n$ are positive integers, determine a closed-form expression for
\[ \prod_{a \equal{} 1}^m\prod_{b \equal{} 1}^n (x \minus{} e^{\frac {2\pi i a}{m}}e^{\frac {2\pi i b}{n}})
\]
2. (David B Rush) The ordered quadruple of integers $ (w,x,y,z)$ is transformed into $ (w,x,y \minus{} w,z \minus{} x)$ under the operation $ \alpha$ and into $ (w \minus{} y,x \minus{} z,y,z)$ under the operation $ \beta$.

Prove that, for all positive integers $ p$ and $ q$, there exists a sequence of operations $ \alpha$ and $ \beta$ that transforms $ (1,0,0,1)$ into $ (a,b,c,d)$ such that $ \lbrace |p|,|q|\rbrace \equal{} \lbrace a,b \rbrace$ if and only if $ p$ and $ q$ are relatively prime.

3. (Alex Zhai) Let $ \triangle ABC$ be a triangle, and let the excircle opposite $ A$ be tangent to $ BC$, $ CA$, and $ AB$ at $ D$, $ E$, and $ F$, respectively.  Let $ AD$ meet $ EF$ at $ P$ and intersect the excircle again at $ G$.  Let $ M$ be the midpoint of $ DG$, and let $ Q$ be the second intersection point of the circumcircles of $ \triangle MFB$ and $ \triangle MEC$.  Prove that $ PQ$ is perpendicular to $ BC$.

DAY II:

4. (Alex Anderson) Let $ ABCD$ be a cyclic quadrilateral.  Let $ R$ and $ S$ be the midpoints of $ AC$ and $ BD$, respectively, and let $ E$ be the intersection of $ AC$ and $ BD$.  Prove that there exists a point P on the circumcircle of $ \triangle RES$ such that $ \triangle ABP$ is directly similar to $ \triangle CDP$.

5. (Alex Zhai) At Phillips Exeter Academy, there are $ n$ student clubs.  Students may only be in one club at a time, but some clubs are affiliated with one another.  Every day, in a simulataneous process, if a certain club has $ m$ members and $ k$ affiliated clubs, for each of the $ k$ affiliates, $ \lfloor \frac {m}{k} \rfloor$ members will leave the club and join that affiliate, and all those remaining will surrender their membership in all clubs indefinitely.

If, for every club $ c$ and day $ x$, there exists a day $ y > x$ such that the number of members of club $ c$ is different on days $ x$ and $ y$, prove that it is possible to split the clubs into two groups such that no club is affiliated with any other in its group.

6. (David B Rush) Let $ c$ be the greatest real number such that for any positive integer $ n$ and integer-coefficient polynomials $ P$ and $ Q$ such that $ P(x)Q(x) \equal{} (1 \plus{} x \plus{} ... \plus{} x^{n \minus{} 1})^2$ and $ P(1) \equal{} Q(1) \equal{} n$, $ \log \max \lbrace \min \lbrace \deg \lbrace P,Q \rbrace \rbrace \rbrace \geq c \log n$.

Prove that:
(a) If the coefficients of $ P$ and $ Q$ are required to be nonnegative, then $ c > 0$.
(b) If the coefficients of $ P$ and $ Q$ are permitted to be any integers, then $ c \equal{} 0$.

[/quote]
SAAYYY WHAAATTT???!??!?!!?!!!?!?!?!?!!?!?!?!!?!!??!?!??!!!!?!??!?!?!?!!?!!??
I have no idea what that means at all.

hey DBR, I've been trying to make progress on one. could you tell me if im going on the right track?
[hide]Consider the $ m\cdot n$th roots of unity.[/hide]

Yes--but that is obvious.

I'll give another hint: understanding Bezout's identity and/or the Chinese Remainder Theorem is helpful.  (You won't want to cite these facts, necessarily, but if you know how they work, you'll see that the same kind of logic works here).

[hide]the answer is $ (x^\frac{mn}{g} \minus{} 1)^g$, where $ g\equal{}gcd(m,n)$.[/hide]

[quote="gighiuhui"]2. (David B Rush) The ordered quadruple of integers $ (w,x,y,z)$ is transformed into $ (w,x,y \minus{} w,z \minus{} x)$ under the operation $ \alpha$ and into $ (w \minus{} y,x \minus{} z,y,z)$ under the operation $ \beta$.

Prove that, for all positive integers $ p$ and $ q$, there exists a sequence of operations $ \alpha$ and $ \beta$ that transforms $ (1,0,0,1)$ into $ (a,b,c,d)$ such that $ \lbrace |p|,|q|\rbrace \equal{} \lbrace a,b \rbrace$ if and only if $ p$ and $ q$ are relatively prime.[/quote]
Should it be $ \lbrace p,q \rbrace \equal{} \lbrace |a|,|b| \rbrace$ instead?

[hide="A HUGE hint for #5"]
Graph theory bipartite graphs.
Let f(v) for a vertic V be (number of elements)/(number of affiliates)
Consider min{$ f(V_1),f(V_2)...f(V_n)$} if those are the vertices.
Prove that that is an MONOVARIANT.
It has to eventually stop increasing.
Now, what happens if you have an odd cycle?
One of the vertices has to get non-minimalilty.
Then two of the other vertices will get non-minimality...
eventually what?
[/hide]

[quote="gighiuhui"]Yes--but that is obvious.

I'll give another hint: understanding Bezout's identity and/or the Chinese Remainder Theorem is helpful.  (You won't want to cite these facts, necessarily, but if you know how they work, you'll see that the same kind of logic works here).

[hide]the answer is $ (x^\frac {mn}{g} \minus{} 1)^g$, where $ g \equal{} gcd(m,n)$.[/hide][/quote]

thanks for the help. i'll continue looking into that.

[quote="pythag011"][hide="A HUGE hint for #5"]
Let f(v) for a vertic V
[/hide][/quote]
hahahaha david youre amazing...

[hide="3"]Let $X'$ denote the inverse of point $X$ with respect to the $A$-excircle. We will use directed angles modulo $180^\circ$.

First notice that $ADM$ is the $D$-symmedian of $\triangle{DEF}$. Furthermore, $A,F,I_a,M,E$ lie on a circle, so $M'EF$ is the $E$-symmedian of $\triangle{EDG}$. It follows from $A=EE\cap FF$ and $M'=DD\cap GG$ that $\triangle{MFD}\sim\triangle{MDE}$ (just angle chase with symmedians or note that $\triangle{M'F'D'}\sim\triangle{M'D'E'}$). Thus
\begin{align*}
\measuredangle{MEC} = \measuredangle{MED}+\measuredangle{DEC} &= \measuredangle{MDF}+\measuredangle{DFE}\\
&= \measuredangle{PDF}+\measuredangle{DFP} = -\measuredangle{FPD}.
\end{align*}Similarly, $\measuredangle{MFB}=-\measuredangle{EPD}$, so
\[\measuredangle{MQC}=\measuredangle{MEC}=-\measuredangle{FPD}=-\measuredangle{EPD}=\measuredangle{MFB}=\measuredangle{MQB},\]whence $Q\in BC$.

Hence $PQ\perp BC$ iff $M'MPQ$ is cyclic, since $\measuredangle{PMM'}=90^\circ$. But
\[\measuredangle{M'QM}=\measuredangle{CQM}=\measuredangle{CEM}=\measuredangle{FPD}=\measuredangle{M'PM},\]so we're done.[/hide]

[hide="another #3"][b]Claim 1[/b]: $Q$ is on $BC$
Notice $\angle AMI_a=\angle AEI_a=\angle AFI_a, \Rightarrow AFI_aME$ is cyclic.
$\angle BQM+\angle CQM=(180-\angle AEM)+(180-\angle AFM)=180,$ so $Q,B,C$ are colinear.

Let $OM\cap BC=T$

[b]Claim 2: [/b]$T$ is on $EF$
$A$ is on the polar of $T$ with respect to the A-excircle(DG), so $T$ is on polar of $A$ wrt the A-excircle(EF)

[b]Claim 3:[/b] $TQPM$ is cyclic.
Let $N$ be midpoint of $EF$.
Notice $\angle I_aNP=\angle I_aMP=90,$ so $I_aMPN$ is cyclic. $AP*AM=AN*AI_a=AE^2 \Rightarrow AEP\sim AME$
Then, $\angle TQM=180-\angle CEM=180-\angle APE=\angle TPM$.

Then $\angle PQT=180-\angle PMT=90,$ so $PQ\perp BC$ as desired.[/hide]

[quote="SnowEverywhere"][quote="abacadaea"]Let $DM\cap BC=T$
[/quote]

I thought that $D$ is on $BC$.[/quote]

oops that should say OM, my bad

[hide="5"]Interpret the problem graph theoretically where clubs are vertices and affiliated clubs have an edge between them. Assume that our graph is connected (or else we could consider a connected component of the main graph) . Let $f(x, v)$ be the number of members assigned to vertex $v$ on the $x$-th day. Then
\[
\begin {align*}
f(x + 1, v) = \sum_{u \in N(v)} \frac {f (x, u)} {|N(u)|}
\end {align*}
\]
Furthermore, by AM-HM
\[\begin {align*}
\sum_{u \in N(v)} \frac {|N(u)|} {f(x, u)} \ge \frac {|N(v)|^2} {\sum_{u \in N(v)} \frac {f(x, u)} {|N(u)|}} = \frac {|N(v)|^2} {f(x + 1, v)}
\end {align*}\]
Summing over all vertices $v$, each vertex $u$ is counted $|N(u)|$ times on the RHS, so
\[\begin {align*}
\sum \frac {|N(v)|^2} {f(x, v)} \ge \sum \frac {|N(v)|^2} {f(x + 1, v)}
\end {align*} \]
From the problem statement, there exists a day $d$ such that for $x > d$, $\frac {f(x, v)} {|N(v)|}$ is an integer for all vertices $v$ (or else all our members are eventually suspended). Furthermore, $f(x, v)$ is bounded since the total number of members is nonincreasing. Therefore, $\sum \frac {|N(v)|^2} {f(x, v)}$ can only take on finitely many values; hence it eventually becomes constant.
The equality case of AM-HM gives us that for a vertex $v$, and all $u \in N(v)$, $\frac {f(x, u)} {|N(u)|}$ is constant. (1)
Note that if $\frac {f(x, v)} {|N(v)|}$ is constant for all $v$, then the clubs never change number of members. Therefore, if two neighboring vertices $w, u$ have $\frac {f(x, w)} {|N(w)|} = \frac {f(x, u)} {|N(u)|}$, we can repeatedly apply (1) to all vertices to get that $\frac {f(x, v)} {|N(v)|}$ is constant for all vertices $v$, a contradiction. Since there are at most two possible values of $\frac {f(x, v)} {|N(v)|}$ by applying (1) repeatedly, our graph is bipartite, as desired.
 [/hide]

Is the use of gcd, lcm considered closed form?

[hide="Solution 1"]
We have \[ \prod_{a = 1}^m\prod_{b = 1}^n (x - e^{\frac {2\pi i a}{m}}e^{\frac {2\pi i b}{n}}) = \prod_{a=1}^m\left(x^n-e^{2\pi ian/m}\right) = \prod_{a=1}^m\left(x^n-e^{2\pi ian'/m'}\right) =\] \[=\left(\prod_{a=1}^m\left(x^n-e^{2\pi ia/m'}\right)\right)^{\gcd(m,n)} = \left(x^{mn/\gcd(m,n)}-1\right)^{\gcd(m,n)}\] where $\frac{n}{n'} = \frac{m}{m'} = \gcd (m,n)$, and we have used the fact that for relatively prime integers $r$ and $s$, $\{1, 2, ..., s\} \equiv \{r, 2r, ..., rs\}\pmod s$.
[/hide]

On [b]Problem 6[/b], I'm not entire sure what is meant by $\max \{\min \{\deg P, \deg Q\}\}$. Normally, I would assume it is asking for the maximum of $\min \{\deg P, \deg Q\}$ as $P,Q$ range over all polynomials satisfying the conditions, but this would mean that part b) is false, since the conditions of a part a) are a subset of the constraints of part b).

[hide="Outline 6B?"]
Let $S$ be the set of prime power divisors of $n$. Then taking $P(x)=\prod_{s\in S}\Phi_s(x)$ works, and $\frac{\deg P}{n^c}$ gets arbitrarily close to $0$.
[/hide]

1. For a fixed $a$, $\prod_{b=1}^n\left(x-e^{\tfrac{2\pi a i}{m}}e^{\tfrac{2\pi b i}{n}}\right)=\prod_{b=1}^n\left(e^{\tfrac{2\pi a i}{m}}\left(\dfrac{x}{e^{\tfrac{2\pi a i}{m}}}-e^{\tfrac{2\pi b i}{n}}\right)\right)=e^{\tfrac{2\pi an i}{m}}\left(\left(\dfrac{x}{e^{\tfrac{2\pi a i}{m}}}\right)^n - 1\right)=x^n-e^{\tfrac{2\pi a n i}{m}}$. Let $d=\gcd(m, n)$, $m=m'd$, and $n=n'd$. As $a$ ranges from $1$ to $m$, $e^{\tfrac{2\pi a n i}{m}}=e^{\tfrac{2\pi a n' i}{m'}}$ ranges through the $m'$th roots of unity $d$ times. Thus $\prod_{a=1}^m\left(x^n-e^{\tfrac{2\pi a n i}{m}}\right)=(x^{nm'}-1)^d=\left(x^{\tfrac{mn}{\gcd(m, n)}}-1\right)^{\gcd(m, n)}$.

problems have been posted in separate threads inside aops and are collected [url=https://artofproblemsolving.com/community/c3624995_2008_mock_usamo]here[/url]
(among others [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url], among USA Mocks/ USAMO Mocks)

Related post [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url]


Ok, some of you here may be interested in a thing called 'Bulgarian Math Olympiad.' The 2005 edition of our National olympiad was held on March 14-15 in Sofia. You may find the problems below. IMHO, the contest was slightly better thought off than last year's (nothing could be worse than last year's  :) ), though the problems were far too trivial. Anyway  :) .

Besides, there is a Russian team, comprising of six contestants and two leaders (Nazar Agakhanov and Pasha  Kojevnikov) participating in the event.

I [i]had[/i] to ...  :lol: (I'm doing this for my school. )

OK, So can you post the links to the discussions or shall I do it myself ?  :lol:

OK, I did it, here are the links to the problems  :D  ;)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37234]Problem 1[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37236]Problem 2[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37239]Problem 3[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37238]Problem 4[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37235]Problem 5[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=37237]Problem 6[/url]

The olympiad was held on 14 and 15 of May.I participated too,and I solved only two problems and a half.

As I know, there are two first diplomas: Kralev (40 out of 42) and Russian guest Stoliarov (39), next result is about 31.

I have to recognise that problems are much more interesing than on All-Russian olympiad.

[quote="Fedor Petrov"]
I have to recognise that problems are much more interesing than on All-Russian olympiad.[/quote]

Are you joking?  :?  This honestly surprises me ... IMHO, your problems are always incomparably better than ours. At least, they are almost always original -- and much more difficult, of course  :). In my opinion, this year the Bulgarian olympiad was the easiest in many years; each of the 6 problems can be safely solved in <1 hour; that I messed it up (the 1st day), then, is [i]nobody's fault but mine[/i] (as Led Zeppelin used to say  :D ) ... (I got 710777). As far as the Bulgarian scores are concerned (I know nothing about the results of your team; do you know the other scores?), the winner of the olympiad is Rosen Kralev with 41 (767777), 2nd is Ivan Dimitrov with 31 (730777), and I come third. For those easy problems, our results are quite poor, I think ...

Anyway, I didn't like the problems ... Problem 5, for example, is an absolute joke -- it is a blunder to give such a problem on any olympiad ... Problem 1 is also trivial and absolutely standard; Problem 6 is arguably one of the easiest #6 problems for many years on this contest (although it was, sadly, solved by only 3 (!) of us; I'm curious about how many Russians solved it during the exam?) I think that the All-Russian olympiad was much more interesting ... But this, IMHO.

--Vesselin

Probelms 2,3 and 6 are much better than all All-Russian of 2005 except maybe 8-th problems in each klass. I spent a lot of time for no.6. It depends a lot of how tricky you are when chosing things to denote.

Unfortunately, ARO problems are not always original. So, 2004.11.3 is from some old shortlist. 1998.10.5 is from old All-Soviet (!) These are only examples which I may call immediately.

Problem 2 is a not original - it is trivial follow from known "segment lemma" for externally touching circles.

Of course. Though, for my shame, I noticed it only after I solved the problem (by different method, though using Cayley theorem).

[quote="Valentin Vornicu"]Vess congrats for the nice result. When are the TSTs? :) Maybe you will come to the IMO this year just to check it out, or maybe just to meet some feelow MathLinkers :P :)[/quote]

Thank you, though by no means would I call 710777 a 'nice' result on such trivial problems (I'm rather weak on geometry because I don't practice it at all, but I still can't believe that I was so stupid to mess up such an easy problem as #3  :blush:  :blush:  :blush:).

Anyway, as our TSTs are over now, it is now official that I'll be going to IMO 2005. I have finally voted for it, just for the experience: to see what it's like, to meet with others, and of course, to see Mexico  :lol:.

[quote="Fedor Petrov"]
Probelms 2,3 and 6 are much better than all All-Russian of 2005 except maybe 8-th problems in each klass.
[/quote]

Maybe yes, though I doubt even this, but they (2,3,6) are *nothing* compared to  9.8, 10.8, and 11.8 from ARO. In my opinion, the quality of an olympiad is principally determined by the quality of the best problem in it ... Anyway, I'm glad that you liked our olympiad!  :)

[quote="vess"]Thank you, though by no means would I call 710777 a 'nice' result on such trivial problems (I'm rather weak on geometry because I don't practice it at all, but I still can't believe that I was so stupid to mess up such an easy problem as #3  :blush:  :blush:  :blush:).
[/quote]Well the rank matters :P And you've done a perfect day also. You have to catch up on your geometry, cuz you want to make a good impression at the IMO :)  :winner_first:


Given $ a, b, c > 0$ and $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that: $ (a^2\plus{}2b^2\plus{}3)(b^2\plus{}2c^2\plus{}3)(c^2\plus{}2a^2\plus{}3) \geq\ 64$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ (a^2 \plus{} 2b^2 \plus{} 3)(b^2 \plus{} 2c^2 \plus{} 3)(c^2 \plus{} 2a^2 \plus{} 3) \geq\ 64$
 :)[/quote]

We have:

$ (a^2 \plus{} 2b^2 \plus{} 3)(b^2 \plus{} 2c^2 \plus{} 3)(c^2 \plus{} 2a^2 \plus{} 3) \equal{} ((a\plus{}b)(a\plus{}c)\plus{}2(a\plus{}b)(b\plus{}c))((a\plus{}b)(b\plus{}c)\plus{}2(a\plus{}c)(b\plus{}c))((a\plus{}c)(b\plus{}c)\plus{}2(a\plus{}b)(a\plus{}c)) \ge 27(a\plus{}b)^2(b\plus{}c)^2(a\plus{}c)^2$

So it's left to prove:

$ 27(a\plus{}b)^2(b\plus{}c)^2(a\plus{}c)^2 \ge 64(ab\plus{}bc\plus{}ac)^3$

which is equivolent to:

$ 27(\sum_{sym}{a^4b^2}\plus{}\sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^4bc}\plus{}6\sum_{sym}{a^3b^2c}\plus{}\sum_{sym}{a^2b^2c^2}\plus{}4a^2b^2c^2) \ge 64(3\sum_{sym}{a^3b^2c}\plus{}\frac{1}{2} \sum_{sym}{a^3b^3}\plus{}\sum_{sym}{a^2b^2c^2})$

or
$ 27\sum_{sym}{a^4b^2}\plus{}27\sum_{sym}{a^4bc} \ge 5\sum_{sym}{a^3b^3}\plus{}30\sum_{sym}{a^3b^2c}\plus{}19\sum_{sym}{a^2b^2c^2}$
which is right by Muirhead inequality. :)

[quote="SaYaT"][quote="nguoivn"]Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ (a^2 \plus{} 2b^2 \plus{} 3)(b^2 \plus{} 2c^2 \plus{} 3)(c^2 \plus{} 2a^2 \plus{} 3) \geq\ 64$
 :)[/quote]

We have:

$ (a^2 \plus{} 2b^2 \plus{} 3)(b^2 \plus{} 2c^2 \plus{} 3)(c^2 \plus{} 2a^2 \plus{} 3) \equal{} ((a \plus{} b)(a \plus{} c) \plus{} 2(a \plus{} b)(b \plus{} c))((a \plus{} b)(b \plus{} c) \plus{} 2(a \plus{} c)(b \plus{} c))((a \plus{} c)(b \plus{} c) \plus{} 2(a \plus{} b)(a \plus{} c)) \ge 27(a \plus{} b)^2(b \plus{} c)^2(a \plus{} c)^2$

So it's left to prove:

$ 27(a \plus{} b)^2(b \plus{} c)^2(a \plus{} c)^2 \ge 64(ab \plus{} bc \plus{} ac)^3$

which is equivolent to:

$ 27(\sum_{sym}{a^4b^2} \plus{} \sum_{sym}{a^3b^3} \plus{} \sum_{sym}{a^4bc} \plus{} 6\sum_{sym}{a^3b^2c} \plus{} \sum_{sym}{a^2b^2c^2} \plus{} 4a^2b^2c^2) \ge 64(3\sum_{sym}{a^3b^2c} \plus{} \frac {1}{2} \sum_{sym}{a^3b^3} \plus{} \sum_{sym}{a^2b^2c^2})$

or
$ 27\sum_{sym}{a^4b^2} \plus{} 27\sum_{sym}{a^4bc} \ge 5\sum_{sym}{a^3b^3} \plus{} 30\sum_{sym}{a^3b^2c} \plus{} 19\sum_{sym}{a^2b^2c^2}$
which is right by Muirhead inequality. :)[/quote]
by AM-GM ineq:
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc \ge \frac {8}{9}(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \ge \frac {8}{9}\sqrt {3(ab \plus{} bc \plus{} ca)} (ab \plus{} bc \plus{} ca)$
then we have:
$ 27(a \plus{} b)^{2}(b \plus{} c)^{2}(a \plus{} c)^{2}\ge 64(ab \plus{} bc \plus{} ac)^{3}$(Q.E.D)


A circle and a point $P$ higher than the circle lie in the same vertical plane. A particle moves along a straight line under gravity from $P$ to a point $Q$ on the circle. Given that the distance travelled from $P$ in time $t$ is equal to $\dfrac{1}{2}gt^2 \sin{\alpha}$, where $\alpha$ is the angle of inclination of the line $PQ$ to the horizontal, give a geometrical characterization of the point $Q$ for which the time taken from $P$ to $Q$ is a minimum.

Let O be the center and r the radius of the given circle (O) and let the line PQ intersect the circle (O) at another point R different from Q. Denote $d_1 = PQ, d_2 = PR$. The time t is given by $t = \sqrt{\frac{2d_1}{g \sin \alpha}}$ and it is minimum when $\sqrt{\frac{d_1}{\sin \alpha}}$ is minimum, i.e., when $\frac{d_1}{\sin \alpha}$ is minimum (because the square root is an increasing function). The power p of the point P to the circle (O) is constant for any angle $\alpha$ and equal to

$p = d_1d_2 = (OP - r)(OP + r) = OP^2 - r^2$

Hence, $\frac{d_1}{\sin \alpha} = \frac{p}{d_2 \sin \alpha}$ and this expression is minimum, when $\frac{1}{d_2 \sin \alpha}$ is minimum or when $d_2 \sin \alpha$ is maximum (because the reciprocal value is a decreasing function). Let the vertical line through P intersect the horizontal line through R at a point S. The distance PS is equal to $PS = d_2 \sin \alpha$. It goes without saying anything more that this distance is maximum when the point R is identical the most bottom point B of the circle (O), i.e., with the bottom intersection of the vertical line through the center O with the circle (O). The desired point Q is the other intersection of the circle (O) with the line PB.


algebra
polynomial
geometry
number theory open
number theory

If the number of polynomials is larger than the number of variables this will usually be
false; vanishing of the polynomials will force $n = 0$ for some integer $n>0$, so they can all vanish only modulo the primes that divide $n$.   In the opposite situation where there are as many variables as polynomials, there is a solution modulo any large enough prime, because any algebraic variety has points over large finite fields.

I think, you misunderstood the problem. The question is whether it is true that there exists arbitrary large $p$ such that for any our polynomial (of one variable) $f(x)$ there exists an integer $n$ such that $p$ divides $f(n)$. $n$ may depend on $f$.

This interpretation (separate solutions of each polynomial) can be reduced to the situation I described (simultaneous solution of all polynomials),
by adding new variables $a_2, ..., a_k$  (where $k > 1$ is the number of polynomials) and discussing solvability of the system

$P_1(x) = P_2(x + a_2) = \dots P_k(x+a_k) = 0 \mod p$.

For polynomials of one variable, this is $k$ polynomials in $k$ variables, so will always have solutions for infinitely many $p$.

[quote="fleeting_guest"]  In the opposite situation where there are as many variables as polynomials, there is a solution modulo any large enough prime, because any algebraic variety has points over large finite fields.[/quote]
This I don't understand: the polynomial $x^2+1$ doesn't have a root modulo $p$ if $p=4k+3$ regardless of how large $p$ is. And you can change it to $(x_1+\dots+x_n)^2+1$ and have the same effect, which shows, by the way, that the formal number of variables is not exactly the right thing to look at. Or am I missing something as usual?  :?

[quote] $x^2+1$ doesn't have a root modulo $p$ if $p=4k+3$ regardless of how large $p$ is. [/quote]
I think this is the only type of problem that can arise; the question can be reduced to the situation of one polynomial of one variable, and the set of primes for which $f$ has a root modulo $p$.

[quote] And you can change it to $(x_1+\dots+x_n)^2+1$ and have the same effect, which shows, by the way, that the formal number of variables is not exactly the right thing to look at. Or am I missing something as usual?  :?[/quote]

The variety should be projectivized and it may be necessary to make one, finite degree, field extension in order to be in the situation where estimates on the number of points apply.
I will look up the theorem and try to formulate a precise statement.

"Number of variables" is a surrogate for dimension and codimension of the variety.
If more variables are added, a system without solutions can acquire solutions, for example $(x_1 + x_2)^2 + y^2 = 0$ will have solutions such as $(1,-1,0)$.

I found this topic while browsing old posts. The answer to Harazi's question is "yes," and the explanation is much simpler than most people probably thought -- we don't need any theorems from arithmetic geometry (or any theorems whatsoever at all). In fact, it is the simplest possible explanation!

[hide] It reduces to the (obvious) $n=1$ case ($n$ -- number of polynomials in our system) by a substitution. Not convinced? Then try the following: if $f_{1},\ldots, f_{n}$ are our polynomials, there exist integer nonconstant polynomials $h_{1},\ldots,h_{n}$ such that $f_{1}\circ h_{1}, \ldots, f_{n}\circ h_{n}$ have a common factor. You may even construct these $h_{i}$ explicitly! The result follows. [/hide]

Sorry, it was indeed a bit unclear indeed which case the composition was meant; but looking at what of these two assertions solves the initial problem, you can see that I meant $f_{i}(h_{i}(x))$.

Now, prove the assertion that I made :) Do you see how the conclusion follows?

How incredibly, fantastically and super nice proof! Ok, let me give a statement of your proof. I have to admit that this is extraordinarily elegant and the result is absolutely nontrivial (actually, the proof too; I have no idea how on earth that came to you, but thanks). Ok, so we are given $f_{i}$ nonconstant polynomials with integer coefficients. Now, we need some complex number $z$ and some polynomials $h_{i}$ with rational coefficients such that $f_{i}(h_{i}(z))=0$ for all $i$. The obvious approach is to consider a root $x_{1},...,x_{n}$ for each $f_{i}$ and to try to choose some $h_{i}$ with $h_{i}(z)=x_{i}$. Therefore, the continuation is obvious: all $x_{i}$ are algebraic numbers, thus the field generated by them over the field of rational numbers is a finite extension of $Q$, so it is generated by some $z$. Now, $z$ is an algebraic number, so by multiplying it with some integer we may assume that $z$ is actually an algebraic integer. Because $Q(z)$ contains $x_{i}$, one can find polynomials with rational coefficients $g_{i}$ with $g_{i}(z)=x_{i}$. Now, choose $h_{i}$ to be $g_{i}$ and we have found for now just some polynomials $h_{i}$ with rational coefficients such that $f_{i}(h_{i}(x))$ is divisible by some polynomial $P$, nonconstant and having rational coefficients. Now, take some $N$ such that all $Nh_{i}=g_{i}$ have integer coefficients and such large $d$. Then the polynomials $N^{d}f_{i}(\frac{1}{N}g_{i})$ have integer coefficients and are all divisible by some polynomial $S$ with rational coefficients. A standard argument using Gauss's lemma shows that in this case all polynomials $MN^{d}f_{i}(\frac{1}{N}g_{i})$ are divisible by some nonconstant polynomial $V$ with integer coefficients for some $M$. From here, it is very easy to conclude, by applying Schur's theorem to $V$.

Sorry, Harazi, but Gauss's Lemma seems strange to me. Can you please show me what it is

[quote="harazi"]We all know that for any polynomial $f$ with integer coefficients there are infinitely many primes such that $f$ has a root  modulo those primes. Is it true that given any finite family of polynomials with integer coefficients, there are infinitely many primes $p$ for which any member of the family has a root modulo $p$?[/quote]

Here's a different proof (but this one does use quite a few strong results :)):

It can be shown that the set of such primes has positive density (in the set of all primes). In fact, we can prove this even for primes modulo which all polynomials split into linear factors. Let $f_{i},\ i=\overline{1,n}$ be a finite set of integer polynomials (which we may as well assume irreducible), and let $K$ be a Galois extension of $\mathbb Q$ in which all $f_{i}$ split. If $\theta_{i}$ is a root of $f_{i}$ then, by a result of Kummer, almost all primes in $\mathbb Q$ (that is, all but a finite number; those which do not divide the discriminant of $f_{i}$ will do, for example) split in $\mathbb Q(\theta_{i})$ in the same way that $f_{i}$ splits modulo $p$. In particular, for almost all $p$, it's true that $f_{i}$ splits into linear factors modulo $p$ iff $p$ is the product of $[\mathbb Q(\theta_{i}): \mathbb Q]$ primes in $\mathbb Q(\theta_{i})$ (one usually says that $p$ [i]splits completely[/i] in $\mathbb Q(\theta_{i})$). From now on, disregard the finite exceptional set of primes for which this does not hold.

By Chebotarev's Density Theorem, the set of primes in $\mathbb Q$ which split completely in $K$ has density $\frac 1{[K: \mathbb Q]}$. A prime which splits completely in $K$ splits completely in all $\mathbb Q(\theta_{i})$, so all such primes meet our requirements. This finishes the proof.


by AM-GM, $ 1\plus{}\frac{1}{x}\equal{}1\plus{}\frac{1}{4x}\plus{}\frac{1}{4x}\plus{}\frac{1}{4x}\plus{}\frac{1}{4x}\geqslant 5\sqrt[5]{\frac{1}{4^4x^4}}$
then $ S\equal{}\left(1\plus{}\frac{1}{a}\right)\left(1\plus{}\frac{1}{b}\right)\left(1\plus{}\frac{1}{c}\right)\left(1\plus{}\frac{1}{d}\right)\geqslant 5^{4}\sqrt[5]{\frac{1}{4^{16}\times (abcd)^4}}$
by AM-GM $ abcd\leq \left(\frac{a\plus{}b\plus{}c\plus{}d}{4}\right)^4\equal{}\frac{1}{4^4} \Rightarrow \frac{1}{(abcd)^4}\geq 4^{16}$
from where we deduct that  $ S\geq 5^4$
equality holds when $ a\equal{}b\equal{}c\equal{}d$

[quote="thanhnam2902"]Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that $ \left(1 \plus{} \frac {1}{a}\right) \left(1 \plus{} \frac {1}{b}\right) \left(1 \plus{} \frac {1}{c}\right) \left(1 \plus{} \frac {1}{d}\right)\ge 5^4$[/quote]
By [b]Holder[/b], we have:
$ LHS\ge (1\plus{}\sqrt [4]{\frac {1}{abcd}})^4\ge (1\plus{}4)^4\equal{}5^4$ (By [b]AM-GM[/b]).
Our proof is completed then :lol:.

@: This is my $ 200^{th}$ post :thumbup:

[quote="mathVNpro"][quote="thanhnam2902"]Let $ a,b,c,d > 0$ and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that $ \left(1 \plus{} \frac {1}{a}\right) \left(1 \plus{} \frac {1}{b}\right) \left(1 \plus{} \frac {1}{c}\right) \left(1 \plus{} \frac {1}{d}\right)\ge 5^4$[/quote]
By [b]Holder[/b], we have:
$ LHS\ge (1 \plus{} \sqrt [4]{\frac {1}{abcd}})^4\ge (1 \plus{} 4)^4 \equal{} 5^4$ (By [b]AM-GM[/b]).
Our proof is completed then :lol:.

@: This is my $ 200^{th}$ post :thumbup:[/quote]
nice solution and good post,


[quote="lotrgreengrapes7926"][hide] 1 chicken will lay 1 egg in $1\frac{1}{2}$ days. There are 20 $1\frac{1}{2}$ days in a month, so 1 chicken will lay 20 eggs in a month. Then, 30 chickens will lay $30\times20=600$ eggs in a month. $\boxed{600}$ [/hide][/quote]
Isn't it $1\frac{1}{2}$ chickens laying $1\frac{1}{2}$ eggs in $1\frac{1}{2}$ days?

[hide="My Solution"]$1\frac{1}{2}$ chickens will lay $1\frac{1}{2}$ eggs in $1\frac{1}{2}$ days, so $1\frac{1}{2}$ chickens will lay $30$ eggs in $30$ days. Then, to make it $1$ chicken, $30\div1\frac{1}{2}=20.$ 1 chicken lays 20 eggs in 30 days.[/hide]

[hide] $1 1/2$ chicken lays $1 1/2$ eggs in $1 1/2$ days, so 1 chicken lays $20$ eggs in $30$ days. The farm has $30$ chicken , so in 30 days, the farm producé $600$ eggs.[/hide]

I forgot there were 30 chickens.  :oops:

[hide="In that case..."]30 chickens lay $600$ eggs in 30 days.[/hide]

[quote="spy101"]my son got this word problem for 8th grade homework, and we disagree with the "right" answer.
please tell us what you think :)

For your next vacation, your family has signed up for a space cruise with a theme of Mathematics in the Far Reaches of the Galaxy. The first stop of your cruise rocket is the planet Sesquipedalia, where every Sesquipedalien has one and a half fingers per hand. Because of this, they tend to count by one-and-a-halfs instead of by ones or fives or tens, like we Earthlings do.

	 When you visit a Sesquipedalian chicken farm producing Sesquipedalian eggs, they tell you that a chicken and a half lays an egg and a half in a day and a half.  This is confusing to you as an Earthling, so you decide to figure out how many eggs the whole farm produces in a month. If the farm has 30 chickens and a Sesquipedalian month has 30 days, what do you get?

P.S. – Please provide your answer in Earth numbers[/quote]

[hide]
1 chicken lays 1 egg in 1.5 days. $\frac{30}{1.5}=20$. So then you multiply $30\times 20=\boxed{600\text{ eggs in a month.}}$[/hide]

[quote="lotrgreengrapes7926"][quote="i_like_pie"]Isn't it $1\frac{1}{2}$ chickens laying $1\frac{1}{2}$ eggs in $1\frac{1}{2}$ days?[/quote]
Yes, but that can be immediately simplified. :D[/quote]
Yeah, I realized that later. :oops:

[quote="Arvind_sn"]Hey lotrgreengrapes7926 how did you box your answer like that? :huh:[/quote]
\boxed{stuff}

$1+2=\boxed{3}$

[quote="Arvind_sn"]Hey lotrgreengrapes7926 how did you box your answer like that? :huh:[/quote]
To box an answer, use the code "\boxed{answer}."
For example, "\boxed{\pi}" produces $\boxed{\pi}$.


inequalities
trigonometry
function
inequalities proposed

Let be $ x,y,z\in (0,\infty )$ such that $ x\plus{}y\plus{}z\equal{}xyz$ . Show that :

$ xy\plus{}yz\plus{}zx\ge 3\plus{}\sqrt{x^2\plus{}1}\plus{}\sqrt{y^2\plus{}1}\plus{}\sqrt{z^2\plus{}1}$

Nice! :D
[hide]We have to show
$ xy \plus{} yz \minus{} 2 \ge 2\sqrt {y^2 \plus{} 1}$
Squaring and dividing by y,
$ x^2y \plus{} yz^2 \plus{} 2xyz \ge 4(x \plus{} y \plus{} z)$
Using our condition,
$ x^2y \plus{} yz^2 \ge 2xyz$
Which is true by AM-GM[/hide]
Can we also use a trig substitution? It looks good....
Here's what I have so far:
[hide]Let $ x\equal{}tan A$, $ y\equal{}tan B$, $ z\equal{}tan C$, where A,B,C are angles of a triangle. Then $ xy\equal{}\frac{tanAtanBtanC}{tanC}\equal{}\frac{tanA\plus{}tanB\plus{}tanC}{tan C}\equal{}\frac{tanA\plus{}tanB}{tan C}\plus{}1$. So we have to show
$ \frac{tanA\plus{}tanB}{tan C}\plus{}\frac{tanA\plus{}tanC}{tan B}\plus{}\frac{tanB\plus{}tanC}{tan A} \ge secA\plus{} sec B \plus{} sec C$[/hide]

[quote="alex2008"]Let be $ x,y,z\in (0,\infty )$ such that $ x \plus{} y \plus{} z \equal{} xyz$ . Show that :

$ xy \plus{} yz \plus{} zx\ge 3 \plus{} \sqrt {x^2 \plus{} 1} \plus{} \sqrt {y^2 \plus{} 1} \plus{} \sqrt {z^2 \plus{} 1}$[/quote]
Funny solution :lol: :
Let $ P \equal{} 3 \plus{} \sqrt {x^2 \plus{} 1} \plus{} \sqrt {y^2 \plus{} 1} \plus{} \sqrt {z^2 \plus{} 1}$, $ Q \equal{} xy \plus{} yz \plus{} zx$. We easily get $ P,Q \ge 3$, and the function $ x \to x^2 \minus{} 6x$ is increasing in $ (3;\plus{}\infty)$, so it suffices to prove $ Q^2\minus{}6Q \ge P^2 \minus{}6P$.
But $ P^2\minus{}6P \equal{} \minus{}6 \plus{} \sum_{cyc}x^2 \plus{} \sum_{cyc}2\sqrt{x^2\plus{}1}\sqrt{y^2\plus{}1} \le 3\sum_{cyc}x^2$
And $ Q^2\minus{}6Q \equal{} \sum_{cyc}x^2y^2\plus{}2x^2yz\minus{}6xy$.
Then we have to prove:
$ \sum_{cyc}x^2y^2\plus{}2x^2yz \ge 3\sum_{cyc}x^2 \plus{} 6\sum_{cyc}xy \equal{} 3(x\plus{}y\plus{}z)^2\equal{}3xyz(x\plus{}y\plus{}z) \iff$
$ x^2y^2\plus{}y^2z^2\plus{}z^2x^2 \ge xyz(x\plus{}y\plus{}z)$, which is obvious :)


Remember when writing problems that word problems are by their nature harder than the math behind them.  This is because students must learn the mechanics behind the problem [i]and[/i] learn to make the analogy between the words and the math.  This means that some of the word problems being posted are best suited for the MATHCOUNTS and Getting Started forums.

Also, please do not post intentionally confusing or obtuse problems.  And be careful with the wording of the problems you post.  A few recent problems have left out important parts of the problem, making those posts confusing or nonsensical.

You have a really really awesome number theory book, are you going to release a intermediate number theory soon?

are you going to make an intermediate geo book soon?

noooooooooo!
Mr.Crawford why don't you work for AoPs any more????????????  :o  :o


integration
trigonometry
real analysis
real analysis unsolved

I have noticed a great way to evaluate :
$\sum_{k=1}^{\infty}\frac{1}{k^{2}+\alpha^{2}}$ .
Is there a elegant way to calculate
$\sum_{k=1}^{\infty}\frac{1}{k^{n}+\alpha}$ (or you could consider $\alpha^{n}$ ,it doesnt really matter). ,where $n\geq 3~,~\alpha \in \mathbb{R}$(or $\mathbb{C}$) ?
Thanks for any help.

Regards

You are right, there are a couple of techniques for calculating that sum, one idea is to use this trick:

$\int_{0}^{\infty}\frac{\sin \alpha x}{e^{x}-1}dx=\frac{\pi}{2}\coth \pi \alpha-\frac{1}{2\alpha}$, $\alpha >0$.

Then notice that:$\int_{0}^{\infty}e^{-kx}\sin \alpha x dx=\frac{\alpha}{k^{2}+\alpha^{2}}$, hence

$\sum_{k=1}^{\infty}\int_{0}^{\infty}e^{-kx}\sin \alpha x dx=\alpha \sum_{k=1}^{\infty}\frac{1}{k^{2}+\alpha^{2}}$.

A calculation shows that the left hand side equals $\int_{0}^{\infty}\frac{\sin \alpha x}{e^{x}-1}dx$, and the result follows.
As for the second sum, I do not know whether there are some techniques for the calculation of it, I may be wrong.

For the case when $n=3$ and $\alpha =1$, your sum resembles the famous Apery`s constant and since no closed formula is known to hold for $\xi(3)$, I think that this sum is hard to calculate and if it can be calculated then it is probably expressed in terms of nonelementary functions.


[quote="RaduSorici"]I can't even prove the case when $x_1=x_2=...=x_n$ :([/quote]
Come on! This particular case is not that hard. It is equivalent to the inequality $x^n+\frac1{x^n}-2\geq n^2 \left(x+\frac1x-2\right)$. Now put $x=e^t$ and differentiate twice  :) . The hard part is to reduce the general case to this special one.


The area of rectangle $ ABCD$ with vertices $ A$(0, 0), $ B$(0, 4),
$ C$($ x$, 4) and $ D$($ x$, 0) is 28 square units. If $ x > 0$, what is the
value of $ x$?

Because $ \frac{28}{4}\equal{}7$, $ x\equal{}\pm 7$.  However, we want $ x>0$, so $ x\equal{}\boxed{7}$.


1) Given $ g(x) \equal{}\frac{1}{2\minus{}x}$ find a formula for $ g(g(g(g(g(\cdots (g(x) ) ) ) ) ) )$ and prove it using induction.

2) Find the domain of $ f(x) \equal{}\sqrt{\sqrt{x\plus{}2}\minus{}x }$

[hide="(1)"]
Let $ g^{h}(x)$ indicate $ g(g(\cdots g(x)\cdots))$ where the function $ g$ is performed $ h$ times. Then the formula is $ \frac{(h-1)x-h}{hx-(h+1)}$.

The induction is fairly simple; it forks for $ h = 1$ (base case). Then for a $ k$, we have

$ g^{k}(x) =\frac{(k-1)x-k}{kx-(k+1)}$
$ g^{k+1}(x) =\frac{1}{2-\frac{(k-1)x-k}{kx-(k+1)}}$
And then we cleanup denominator and so forth and we get
$ g^{k+1}(x) =\frac{kx-(k+1)}{(k+1)x-(k+2)}$.
[/hide]

Don't have time to do (2) right now

[hide="(2)"]The domain is the set of all $ x$ such that $ \sqrt{x\plus{}2}\minus{}x\geq 0$, i.e. $ \sqrt{x\plus{}2}\geq x$. We see that $ x\geq\minus{}2$ is necessary to not have imaginary roots and notice that $ \minus{}2\leq x\leq 0$ works. Now we consider $ x > 0$: this allows us to square both sides to get $ x\plus{}2\geq x^{2}$. Manipulating gives $ (x\minus{}2)(x\plus{}1)\leq 0$. This holds for $ \minus{}1\leq x\leq 2$, but since we assumed $ x > 0$ we can only grab $ 0 < x\leq\minus{}2$. Combining this with our original interval we get that the domain is $ \boxed{\minus{}2\leq x\leq 2}$.[/hide]

[quote="t0rajir0u"][b]Follow-up:[/b]  Repeat 1) for $ g(x) \equal{}\frac{x\plus{}1}{x}$.[/quote]

$ g(x)\equal{}1\plus{}\frac{1}{x}$, then $ h(x)\equal{}\frac{1}{x}$ is cyclic with period 2, it shouldn't be too hard from there.


Prove that :

[tex]\displaystyle \sum_{k=1}^{n} \frac{1}{n-1+y_{k}} \leq 1[/tex] if:

1)all y_{k} are positive (>0)  and 2) [tex] \displaystyle \sum_{k=1}^{n} \frac{1}{y_{k}}=\sum_{k=1}^{n}  y_{k}[/tex]

I believe this one was solved on

  http://www.mathlinks.ro/viewtopic.php?t=4744

  darij


Let $a, b, c, d$ be integers with $a>b>c>d>0$. Suppose that $ac+bd=(b+d+a-c)(b+d-a+c)$. Prove that $ab+cd$ is not prime.

More generalize problem:
Let a, b, c, d be integers with $a>b>c>d>0$. Suppose that $ac+bd =k(b+d+a-c)$ $k \in \mathbb{Z}$. Prove that $ab+cd$ is not prime.

Hmm, that is a nice result. Do you have a proof for it?

Yes, but this result isn't of me. It's form an article form an author in Vietnam (I can't remember his name).
This is the solution
Assume that $ab+cd$ is prime.
We have
$ab+cd=(a+d)c+(b-a)a=m(a+d,b-c)$ $m\in \mathbb{N}$
Case $m=1$ We have $ab+cd=(a+d,b-c) < a+d$ (contradict)
case $ (a+d,b-c)=1$
Because $b+d+a-c|ac+bd$ so exist $p$ such that $ac+bd=p(b+d+a-c)$
In the other hand, $ac+bd=(a+d)b-(b-c)a=p(b+d+a-c)=p(a+d)-(b-c)p$
So $(p-b)(a+d)=(a+p)(b-c)$
But$ (a+d,b-c)=1$ which follows that $a+p=(a+d)k \rightarrow b-p=k(b-c)$
$\rightarrow a+b=k(a+b+d-c)$
If $k=1$ then $c=d$ (contradict)
If $k\ge 2$ then $a+b\ge 2(a+b+d-c) \rightarrow 2c\ge a+b+2d $ (contradict)

[quote="Peter"]Let $a, b, c, d$ be integers with $a\geq b>c>d\geq 0$. Suppose that

$ac+bd=(b+d+a-c)(b+d-a+c)$.

Prove that $ab+cd$ is not prime.[/quote]$(ab+cd)(ab-cd-bc)$

$=\left(b^2-c^2\right)\left(c^2-ca+a^2\right)+c^2\left[ac+bd-(b+d+a-c)(b+d-a+c)\right]$

$=\left(b^2-c^2\right)\left(c^2-ca+a^2\right)$,

where $(2a+b-c+2d)(ab-cd-bc)$

$=(b-c)\left[a^2+b^2+c^2+d^2+bc+2ad+(b-c)(a+d)\right]$

$+(b+c)\left[ac+bd-(b+d+a-c)(b+d-a+c)\right]$

$=(b-c)\left[a^2+b^2+c^2+d^2+bc+2ad+(b-c)(a+d)\right]>0$

$\Longrightarrow ab-cd-bc>0$,

$(a+b-c+d)\left[\left(b^2-c^2\right)-(ab-cd-bc)\right]$

$=(b-c)(c-d)(d+a-c)-b\left[ac+bd-(b+d+a-c)(b+d-a+c)\right]$

$=(b-c)(c-d)(d+a-c)>0$

$\Longrightarrow b^2-c^2>ab-cd-bc$,

$\left(c^2-ca+a^2\right)-(ab-cd-bc)$

$=cd+c^2-c(a-b)+(a-b)^2+b(a-b)>0$

$\Longrightarrow c^2-ca+a^2>ab-cd-bc$.

up to my calculations b|a and c|a is that true?