How many positive integer divisors of $2004^{2004}$ are divisible by exactly $2004$ positive integers?

a good start i think
[hide]$2004^{2004}=2^{4008}*3^{2004}*167^{2004}$
a number in the form $a^m*b^n*c^o*...$ has $(m+1)(n+1)(o+1)...$ factors
$2004=2^2*3*167$
we can have $a^{2003}$ (2+1)(1+1)(1+1)=12 combinations since a can be any factor of 2004
...
now im too lazy to work all of this out, but im quite sure this is how youd work it out[/hide]

[hide="Answer"]$2004^{2004}=2^{4008}\cdot 3^{2004}\cdot 167^{2004}$

We need all the numbers $n=2^a\cdot 3^b\cdot 167^c$ where $(a+1)(b+1)(c+1)=2004$. $2004=2^2\cdot 3\cdot 167$, which has $12$ distinct integer factors. Now, we need to see how many ways we can split these to get $(a+1)(b+1)(c+1)=2004$. If $(a+1)=2$, we have $(b+1)(c+1)=1002$, and there are $8$ different ways we can do this, since order does matter. If $(a+1)=3$, we have $(b+1)(c+1)=668$, and there are $8$ different ways we can do this. If $(a+1)=167$, we have the same. If we have $(a+1)=4$, we have $(b+1)(c+1)=501$, there are $4$ different ways we can do this, and the same if $(a+1)=6$, $(a+1)=334$, or $(a+1)=501$. If $(a+1)=12$, we have $(b+1)(c+1)=167$, and there are $2$ ways to do this, and the same for the other $2$ ways to do this. If $(a+1)=1$, we have $(b+1)(c+1)=2004$, and there are $12$ different ways we can do this. If $(a+1)=2004$, we have $(b+1)(c+1)=1$, and there is $1$ way to do this. Adding them together, we have $\boxed{63}$ different values.[/hide]

[b]EDIT:[/b] Okay, I've changed my answer a couple times, and with my most recent one, I probably messed up somewhere in the counting. :?

I might've done it wrong:

YOu know that it equals $2^{4008}(3^{2004})(167^{2004})$ that will be equal to $2^{2^{3(3)(167)}(3^{2^2(3)(167)})(167^{2^2(3)(167)})}$. Grouping it all together there are $7$ exponents of $2$, $3$ exponents of $3$ and $3$ exponents of 167. So $(7+1)(3+1)(3+1)=128$.

EDIT: Wait, maybe you just need to do $(4008+1)(2004+1)(2004+1)$ since the number has already been prime factored...

No, that would just be all the distinct positive integer divisors of $2004^{2004}$, not only the ones that have 2004 distinct positive integer divisors. Remember, that's what you're trying to find.

This was on the AIME, I think...
So $2004^{2004}=2^{4008}3^{2004}167^{2004}$
We know that any of the factors of this will have at most 3 prime factors.  Thus, let one with 2004 factors be written as
$2^{a-1}3^{b-1}167^{c-1}$
Thus, $abc=2004$.  So, what are the possibilities for a, b and c?  We already know the factors of 2004. Therefore, we have to distribute the four numbers 2, 2, 3, 167 between a, b and c, and we have to count the number of ways of doing so.  So this way works, I suppose...
(4, 0, 0)
1 way (multiplied by 3, there are 3 permutations of (4, 0, 0))
(3, 1, 0)
4 ways(multiplied by 6)
(2, 2, 0)
6 ways (multiplied by 3)
(2, 1, 1)
12 ways (multiplied by 3)
Adding it up, we get
3+24+18+36=81 ways

Something I missed?  I'm pretty sure i got it on the aime...

$2004^{2004} = 2^{4008} 3^{2004} 167^{2004}$.

The factors are in the form $2^a 3^b 167^c$.

$(a+1)(b+1)(c+1) = 2004 = 2^2 \cdot 3 \cdot 167$.

The four factors are split among $a+1$, $b+1$, $c+1$.

The number of ways to split the twos: one factor gets both (3) + two factors each get one (3) = 6.

The number of ways to split the threes: 3

The number of ways to split the 167's: 3

So in total $6\cdot 3 \cdot 3 = 54$

My way:

[hide]We have $2^{2004}=2^{4008}*3^{2004}*167^{2004}$

So we want to find all numbers of the form:
$2^a*3^b*167^c, 0\le a \le 4008,0\le b,c \le 2004$ such that $a,b,c$ are integers and $(a+1)(b+1)(c+1)=2004$.

-Here's how we'll approach this problem. We set $a= \text{divisor of }2004,n$, and then find out how many integers $(b,c)$ satisfy $(b+1)(c+1)=\frac{2004}{n}$, and add them for our result. This will just be the number of divisors of $\frac{2004}{n}$.

$a+1=1\Rightarrow (b+1)(c+1)=2004 \Rightarrow 12 \text{ possibilities.}$
$a+1=2\Rightarrow (b+1)(c+1)=1002 \Rightarrow 8 \text{ possibilities.}$
$a+1=3\Rightarrow (b+1)(c+1)=668 \Rightarrow 6 \text{ possibilities.}$
$a+1=4\Rightarrow (b+1)(c+1)=501 \Rightarrow 4 \text{ possibilities.}$
$a+1=6\Rightarrow (b+1)(c+1)=334\Rightarrow 4 \text{ possibilities.}$
$a+1=12\Rightarrow (b+1)(c+1)=167\Rightarrow 2 \text{ possibilities.}$
$a+1=167\Rightarrow (b+1)(c+1)=12 \Rightarrow 6 \text{ possibilities.}$
$a+1=334\Rightarrow (b+1)(c+1)=6 \Rightarrow 4 \text{ possibilities.}$
$a+1=501\Rightarrow (b+1)(c+1)=4 \Rightarrow 3 \text{ possibilities.}$
$a+1=668\Rightarrow (b+1)(c+1)=3\Rightarrow 2 \text{ possibilities.}$
$a+1=1002\Rightarrow (b+1)(c+1)=2\Rightarrow 2 \text{ possibilities.}$
$a+1=2004\Rightarrow (b+1)(c+1)=1 \Rightarrow 1 \text{ possibilities.}$  (Look at it symmetrically).

So there are a total of $12+8+6+4+4+2+6+4+3+2+2+1=\boxed{54}$ integer divisors of $2004^{2004}$ that satisfy the given conditions. [/hide]

[hide]Since $2004=2^2\cdot 3\cdot 167$, $2004^{2004} = 2^{4008}\cdot 3^{2004}\cdot 167^{2004}$. Let one of its factors be $n$. We can write $n$ in the form \[ n=2^{e_1}3^{e_2}167^{e_3}, \] where $e_1\le 4008$ and $e_2,e_3\le 2004$. If $n$ has 2004 factors, then \[ (e_1+1)(e_2+1)(e_3+1)=2^2\cdot3\cdot167. \] Thus the possibilities for $e_1+1$ are $1,2,3,4,6,12,167,167\cdot 2,167\cdot 2^2,167\cdot 3,167\cdot 2 \cdot 3, 167\cdot 2^2\cdot 3$.

If $e_1+1=1$, $(e_2+1)(e_3+1)=2^2\cdot 3\cdot 167$. The possibilities for $(e_2+1)$ are any of the factors of $2^2\cdot 3\cdot 167$. There are $(3)(2)(2)=12$ of them.

If $e_1+1=2$, $(e_2+1)(e_3+1)=2\cdot3\cdot167$. Therefore there are $(2)(2)(2)=8$ possibilities for $n$ in this case.

In each of the other cases we proceed similarly to find that there are $12+8+6+4+4+2+6+4+2+3+2+1=\boxed {54}$ possible values of $n$.[/hide]

kinda along the lines of K81o7's solution, but a different answer,

Once you get to the fact that $abc=2^2\cdot3\cdot167$, to find the number of solutions, note that all you need to do is ${4\choose{2}} \cdot {3\choose2} \cdot {3\choose2}=\boxed{54}$

the basic idea is that $abc=2004$, and we want the number of solutions.

note that $2004 = 2^2\cdot3\cdot167$. Each $a,b,c$ will be of the form $2^{x}\cdot3^{y}\cdot167^{z}$.

multiplying together, we get $2^{x_1+x_2+x_3}\cdot3^{y_1+y_2+y_3}\cdot167^{z_1+z_2+z_3}=2^2\cdot3^1\cdot167^1$.

Then we must find the number of solutions to $x_1+x_2+x_3=2$ which would be ${4\choose2}$...etc, and you go from there to get the other terms of the product. Hope that makes sense?

I also think that 54 is the right answer.

Sorry for the bump but I have a solution that is not the same as the solutions in this thread.

[hide=Solution]We see that $2004^{2004}$ factors into $2^{4008}\cdot 3^{2004}\cdot 167^{2004}.$ So, a factor of that number will be of the form $k=2^a\cdot 3^b\cdot 167^c.$ The number of factors of $k$ are $(a+1)(b+1)(c+1)=2004.$

We prime factorize $2004$ into $2^2\cdot 3\cdot 167.$ We can consider the following cases:

[b]Case 1: [/b]The two powers of $2$ are distributed evenly among two of $a+1,b+1,c+1.$

There are $\dbinom{3}{2}$ ways to pick which two get the power of $2.$ Then since there is only $1$ power of $3$ and $167,$ There is $\dbinom{3}{1}$ ways to distribute each of those powers. So, our answer for this case is $\dbinom{3}{2}\cdot \dbinom{3}{1}\cdot \dbinom{3}{1}=27.$

[b]Case 2:[/b] One of $a+1,b+1,c+1$ gets both powers of $2$

There are $\dbinom{3}{1}$ ways to pick which one of the 3 gets both powers. Then similar to the last case, there are $\dbinom{3}{1}$ ways to distribute the $3$ and $167$ powers. Hence, our final answer here is also $\dbinom{3}{1}^3=27.$

This covers all the cases so our final answer is $27+27=\boxed{054}$[/hide]

Write $2004 = 2^2 \cdot 3 \cdot 167$, so $2004^{2004} = 2^{4008} \cdot 3^{2004} \cdot 167^{2004}$. Then a factor would be in the form $2^a \cdot 3^b \cdot 167^c$, with $(a+1)(b+1)(c+1) = 2004$ and $a \leq 4008, b \leq 2004, c \leq 2004$. To get the factor of $2$ there are $6$ ways, to get the factor of $3$ there are $3$ ways, and to get the factor of $167$ there are $3$ ways, all by Stars & Bars. All possibilities satisfy our restrictions, since if we're multiplying positive integers to get $2004$ of course all of them will be less than or equal to $2004$. The answer is hence $6 \cdot 3 \cdot 3 = \boxed{54}$.


inequalities
induction
inequalities unsolved

Prove that $\big(\frac{1}{2}\big)^2\big(\frac{3}{4}\big)^2\cdots \big(\frac{2n-1}{2n}\big)^2 < \frac{1}{3n}, \forall \ n\geq 1$

Alexandros

P.S. I have in my mind a solution using strong induction. Have you any ideas to prove this without induction?

[quote="Elemennop"]Replace $3n$ with $3n+1$, then induction, I believe.[/quote]

Ok that's true. Or to prove that the left side is less than $\frac{1}{3n} - \frac{1}{12n^2}$ using induction.
But I asked for a proof without induction (maybe using well known inequalities).

Alexandros

Well, when you give us a definition of $\big(\frac{1}{2}\big)^2\big(\frac{3}{4}\big)^2\cdots \big(\frac{2n-1}{2n}\big)^2$ without induction...
And how do you even define $1,2,3,...$¿

So the answer is: you need induction  :P

Well, this expression is equivalent to
$\frac{(2n)!}{2^{2n}(n!)^2} < \frac{1}{\sqrt{3n}}$
Or
$\binom{2n}{n} < \frac{4^n}{\sqrt{3n}}$
Can we prove it without induction? It can be proved if a simple combinatorial argument is sustentable... And I cannot believe in it...

Perhaps replacing $\binom{2n}{n}$ with $\sum_{i=0}^{n} \binom{n}{i}^2$ will help?

Eh? We could always use Stirling's approximation formula:

$n! \approx n^n e^{-n} \sqrt{2\pi n}$ and check a few small cases...

well, the Stirling idea iis valid but not so elemntary...


$x,y,z > 0$ i $x^{2}+y^{2}+z^{2}=1$. Udowodnij, że

\[ \frac{x}{\sqrt{y^{4}+z^{4}}}+\frac{y}{\sqrt{x^{4}+z^{4}}}+\frac{z}{\sqrt{x^{4}+y^{4}}}\geq\sqrt{\frac{3}{2}}\frac{1}{x^{4}+y^{4}+z^{4}} \]

Ktoś może z was był na tych warsztatch ??? (Ja nie jestem z warszawy)


Suppose I want to post a 225 Kb animation file.  Will that slow down anything?  I noticed in the test forum that the animation runs only if you click on the file.  Does that prevent slow downs or does the entire 225 Kb need to be uploaded by the browser even though the animation is not running?

How would such a large file affect access to the thread for all users even if they have slow speed dial up?

Thanks,
Shaw

I believe that an attached animation will only display a one-slide preview, and won't slow anything down. If a person wants to see the actual animation, clicking on the preview will lead to it.

Try it here. 200kB though is definitely really small even for dial-uppers.

Didn't realize that was small; problem has changed though and I won't be using that particular animation but will perhaps use another one.

Thanks for replying.


@_@..... I have no idea why my teacher makes us learn it this way, but some are going to be on the final. Well I of course forgot and can't seem to find anyone that knows it at this time heh.

Ron is on a Ferris wheel of radius 35 ft. that turns counterclockwise at the rate of on revolution
     every 12 seconds.  The lowest point of the Ferris wheel is 15 feet above ground level at the
     point, (0,15) on  a rectangular coordinate system.  Find parametric equations for the position of
     Ron as a function of time t (in seconds) if the Ferris wheel starts with Ron at the point (35,50).

and

A Ferris wheel with a 20 ft radius turns counterclockwise one revolution every 12 seconds.  Eric
     stands at point D, 75 feet from the base of the wheel.  At the instant Jane is at a point parallel to
     the ground (see picture), Eric throws a ball at the Ferris wheel, releasing it from the same height
     as the bottom of the wheel.  If the ball’s initial speed is 60 ft/sec and it is released as an angle
     of 120 with the horizontal, does Jane have a chance to catch the ball?

are two sample problems. Could yall help me figure out how to write the equation for the ferris wheels? Thanks in advance

The equations for height in a ferris wheel is a sinusoid.


Let $a, b, c>0$ such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$ Prove the following inequality:

 $abc(ab+bc+ca)+a+b+c\leq (a+b+c)(ab+bc+ca)-3abc.$ ;)

Homogenuosize by

abc = (ab+bc+ca)/(a+b+c) and

a+b+c = abc(a+b+c)^2/(ab+bc+ca)

and then use Muirhead

From the condition we have $abc(a+b+c)=ab+bc+ca$. So
$(ab+bc+ca)^2\ge 3abc(a+b+c)=3(ab+bc+ca) \Rightarrow ab+bc+ca\ge 3$ (i)
$(a+b+c)^2\ge 3(ab+bc+ca)=3abc(a+b+c)\Rightarrow a+b+c\ge 3abc$ (ii)
(ii) $\Rightarrow abc(ab+bc+ca)\le \frac{1}{3}(a+b+c)(ab+bc+ca)$
(i) $\Rightarrow a+b+c=\frac{1}{3}3(a+b+c)\le \frac{1}{3}(ab+bc+ca)(a+b+c)$
(i & ii) $\Rightarrow 3abc\le \frac{1}{3}3abc3\le \frac{1}{3}(ab+bc+ca)(a+b+c)$
Summing up we get the desired result.
In same manner, we can also deduce
$(1+a)(1+b)(1+c)\le (a+b)(b+c)(c+a)$


It's given a triangle ABC with altitude AD. Let P be a point on AD, and E the intersection between BP and AC and F the intersection between CP and AB. Show that AD is the bisector of EDF.   :D

It's very easy to prove it in an analitic way, but I'm looking for a nicer synthetic solution  :wink:

You could also use harmonic divisions, but here is another way (The solution is not mine):
Let $d$ be a line parallel to $BC$ and let $DF\cap d= C'$ and $DE\cap d =B'$ .
Aply Ceva`s theorem for the triangle $ABC$ to obtain \[\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1.\]Now use the fact that $AC'\parallel BD$ and $AB' \parallel DC$ to obtain that \[\frac{AF}{FB}\cdot\frac{CE}{EA}=\frac{AC'}{BD}\cdot\frac{DC}{AB'},\] and from the Ceva above you can see that $AC'=AB'$. But $DA\perp B'C'$, therefore the triangle $B'C'D$ is isoscelles, which means $\angle FDA \equiv \angle EDA$,
qeq.

Too nice not to be well-known :D

[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1322618035&t=115732]Link 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20809]Link 2[/url]


geometry
algebra
polynomial
inequalities
number theory

I'm starting to learn math...and i guess it's really essential to be more "knowledged" rather than only "skilled"
so could u guys tell me wut kind of math formulas or theories or laws are important or useful? cuz i dont actually know any... :rotfl:  :rotfl:  :rotfl:  :(

thank you!!!!!

ps...mrry xmas evry1!!!!

lol, go to some websites like wikipedia or mathworld. There are plenty of formulas/theories/laws.

Merry Christmas!

I recommend AoPS vol. 1 and 2.  They provide a good general overview of problem solving mathematics, and are available here:
[url]http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php[/url]

[quote="Phelpedo"]Gosh there are so many. AoPS 2 is good but it covers mostly pre-olympiad stuff.

Geometry: Ceva, Menelaus, Apollonius' circle, 9-point circle, Fermat  point, etc....you should be able to find this stuff in Geometry Revisited I guess

Algebra: a handful of identities, factorizations like Sophie Germaine, Fibonacci-Brahmagupta identity (also attributed to Lagrange); for polynomials, stuff like Eisenstein's criterion, Lagrange Interpolation, Newton's sums, etc...

Number Theory: On a basic level, Fermat's Little Theorem, Euler's Totient, Wilson's, Chinese Remainder Theorem...more advanced stuff, Legendre/Quadratic Reciprocity, Gaussian integers, etc. these aren't really necessary but they do show up

Combinatorics: Pascal's, Vandermonde's, etc.

Inequalities: lots of theorems- AM-GM, Cauchy, Rearrangement, (Weighted) Power Mean, Holder, Chebyshev, Minkowski, Karamata, Bernoulli, Maclaurin, Newton, Muirhead. Some of these are quite obscure though.[/quote]

kkk...thx...that is really a lot to learn...just learned the fermat's little theoem and discovered that the 4th problem of 2004 cmo was so easy...i guess this is goibg to help me a lot...thank you... and merry xmas!!!!

just a question... how is Lagrange Interpolation formula often used in competition....

could someone post e.g. some examples of its uses, thanks!!!

That's exactly what I thought when Gabriel showed it to me. And then he dug up about 20 IMO#6's that apparently could be solved by it.

[quote="Phelpedo"]Gosh there are so many. AoPS 2 is good but it covers mostly pre-olympiad stuff.

Geometry: Ceva, Menelaus, Apollonius' circle, 9-point circle, Fermat  point, etc....you should be able to find this stuff in Geometry Revisited I guess

Algebra: a handful of identities, factorizations like Sophie Germaine, Fibonacci-Brahmagupta identity (also attributed to Lagrange); for polynomials, stuff like Eisenstein's criterion, Lagrange Interpolation, Newton's sums, etc...

Number Theory: On a basic level, Fermat's Little Theorem, Euler's Totient, Wilson's, Chinese Remainder Theorem...more advanced stuff, Legendre/Quadratic Reciprocity, Gaussian integers, etc. these aren't really necessary but they do show up

Combinatorics: Pascal's, Vandermonde's, etc.

Inequalities: lots of theorems- AM-GM, Cauchy, Rearrangement, (Weighted) Power Mean, Holder, Chebyshev, Minkowski, Karamata, Bernoulli, Maclaurin, Newton, Muirhead. Some of these are quite obscure though.[/quote]

Which book do you think is suitable for olympiad preparation?


Does there exist for every finite set $ V$ of positive integers a polynomial $ f \in \mathbb{Z}[X]$ such that $ f(1/n)\equal{}n$ for every $ n \in V$?

Let $ f(x) \equal{} a_kx^k \plus{} \cdots \plus{} a_1x \plus{} a_0$ where $ a_i\in\mathbb{Z}$. Then for all $ n\in V$ we have

$ {a_k\over n^k} \plus{} \cdots \plus{} {a_1\over n} \plus{} a_0 \equal{} n\iff a_0n^k \plus{} a_1n^{k \minus{} 1} \plus{} \cdots \plus{} a_k \equal{} n^{k \plus{} 1}$

That gives

$ n^{k \plus{} 1} \minus{} a_0n^k \minus{} \cdots \minus{} a_{k \minus{} 1}n \minus{} a_k \equal{} 0$

Obviously, the roots of the last polynomial are all $ n\in V$, hence $ k \equal{} |V| \minus{} 1$ and the coefficients $ a_i$ are generated from $ Q(x) \equal{} \prod_{n\in V}(x \minus{} n)$. Therefore, such polynomial does exist. And it's also clear that $ V$ can contain negative integers as well.


probability
counting
derangement
floor function
calculus
function

Four chairs with respective colors: brown, black, green and white are placed in this order in a row. Mr. Brown, Mr. Black, Mr. Green, and Mrs. White are seated at random in the chairs -- one person to a chair. Find the probability that none of the four persons sit in the chair whose color matches the person's name. Express your answer as a common fraction.

Label the four people $A,B,C,D$, and you want them to be ordered such that none of them are in there respective seats. Arrange the seats in the order $A,B,C,D$; then you merely want a permutation of those four letters such that there are no fixed points; that's simply a derangement of the four points, which is given by $!n=n!\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}$.

here is a better formula!

$!n=\left\lfloor \frac{n!}{e}+\frac{1}{2}\right\rfloor$

I have two questions:

1. What is $!n$ called?

2. How do you show that $n!\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{k!}=\left\lfloor\frac{n!}{e}+\frac{1}{2}\right\rfloor$?

For your first question, I think it's called a subfactorial.
The second: I have no idea  :P

[hide="Click to reveal hidden content"]We use the [url=http://mathforum.org/library/drmath/view/56185.html]derangements formula[/url]/[url=http://en.wikipedia.org/wiki/Subfactorial]subfactorial[/url] to find the number of ways that none of them get their matching chair, which is $9$. There are a total of $4!=24$ ways to seat them, so $\frac{9} {24}=\boxed{\frac{3} {8}}$[/hide]

Here is a glimpse into my mind  :P :
Both expressions have have a n! in common. The summation has 1 divided by factorials. The other on has e. Can we write e as a sum of 1 divided by factorials.

[hide="..."]taylor series for e, i believe that the 1/n+1!, etc in the expansion of e get rounded when we do the floor function....not sure exactly how to show this though[/hide]

hmm... thats interesting... ill have to try and prove that... on a non school night ahah but yea the whole e in the denominator definitely suggests some kind of taylor series

its a calculus thing... basically its like...

$\sum_{n=0}^{\infty} \frac{f^{(n)}(c) \cdot x^{n}}{n!}$ for some function $f(x)$ centered at $c$

You mean $(x-c)^n$ instead of $x^n$ but yeah

[hide="Hint"]$\left\lfloor x + \frac 12 \right\rfloor$ is the integer closest to $x$[/hide]

Quick question:

I noticed that you said $\frac{!4}{4!}$.  What does $!4$ mean?  I've seen $4!$ before, but not $!4$.

Um, we've already answered that...

It's called a subfactorial, and it gives the number of derangements you can have for a problem like this...


Solved this system
$ \sqrt 2 (x\plus{}y)(1\minus{}4xy)\equal{}\sqrt 3$
$ x^2\plus{}y^2\equal{}1$

Let $ x\plus{}y\equal{}t$, then $ 2xy\equal{}t^2\minus{}x^2\minus{}y^2\equal{}t^2\minus{}1$.
It give $ \sqrt 2 t(3\minus{}2t^2)\equal{}\sqrt 3$. Let $ 2\sqrt{\frac{2}{3}}t\equal{}sinz$, then $ \sin 3z\equal{}3\sin z(1\minus{}4z^2)\equal{}\frac 12.$
Therefore $ t\equal{}\sqrt{\frac 38 }sin(\frac{\pi}{18}\plus{}(\minus{}1)^k\frac{k\pi}{3}),k\equal{}0,1,..,5$.

Nice.my proof is used $ x\equal{}sin a then y\equal{}cos a$ and we also have an easy proof.thanks


[color=cyan]Compute F1/3 + F3/9 + F5/27 + F7/81 + . . ., where F1 = F2 = 1 and Fn = Fn-1 + Fn-2, n>2[/color]

Wait, F1/3 means (F_1)/3???

O man, no wonder i thought it was impossible, i thought it meant
F_(1/3)

i thought so too, but in the orriginal questions, Joel only said n>2, so i wasn't sure whether i should assume n is an integer,
and F1/3 seemed like F_(1/3) to me....

It could be interesting to try with F_(1/3)+F_(3/9)+F_(5/27)+..., where we can define those with Binet's Formula. It will probably converge, but that's just a guess right now.

but sadly, that kind of subscripture is not defined by the problem:
n>2

shoulda caught on to that sooner...

[color=cyan]Well, the terms will definitely approach 0, but the way that they do it is very odd . . .

Well, Mathematica doesn't think it converges.  The partial sum to 100 is a bit less than 3.  And it doesn't really seem to increase as you take the sum to 10,000.  So I don't really know.[/color]

So you would define those types of quasi-Fibonacci numbers with Binet's?

So would Binet's also work for, say F_i?

[color=cyan]Yes, if you know how to raise things to complex powers.  Or, I think so.[/color]

hmm, two questions

1. How do you raise things to irrational powers?
2. How do you raise things to complex powers?

According to [url]http://functions.wolfram.com[/url], F_n={:phi:^n-cos(n:pi:):phi:^(-n)}/sqrt(5). That only matches Binet's Formula when n is an integer.

Raising things to irrational powers: take it as the limit of things to powers of rational approximation.

Raising things to complex powers: raise it to the real part, then use the identity a^b=e^(b*ln(a)) and the fact that e^(i*theta)=cos(theta)+i*sin(theta).

sigh
i really ought to learn calculus....


calculus integration algebra polynomial Vieta blogs

1.
Let $i = \sqrt{-1}$. If $7-4i$ is one of the roots for $x$ of $x^{3}-24x^{2}+kx+w = 0$ where $k$ and $w$ are integers, find the value of $k+w$.

2.
One of the roots of $x^{3}-22x^{2}+148x-288=0$ is $7+\sqrt{13}$. Find the smaller of the other two roots.

When I look up the solutions...
$7-4i$ is one root and so is $7+4i$ for 1.
$7+\sqrt{13}$ and so is $7-\sqrt{13}$ for problem 2.

Why is that?

When you have integral coefficients for a polynomial, and one solution is imaginary, the conjugate must also be a solution.

Otherwise, you will end up with imaginary coefficients.

oh, of course  :) Thanks!

for #1 continuing on: Let r be the lone real root. Then the sum of the roots is $14+r$, which equals $24$, making $r = 10$.

Why is the sum $14+r$?

[hide="#1"]Using Vieta's Theorm, I get k+w=-445

Since the 7-4i is a root, then 7+4i is also a root.  So Vieta's theorm states that the sum of the roots is the negative of the second coefficiant(over the x^2 them) over the first coefficiant (of the x^3 term)

so 7+4i+(7-4i)=14, 14+x(third root)=24 so x=10

now -w/1 is the product of the roots-multiply the roots and we get -w=650 so w=-650

k/1 is the alternating sum of the roots (so ab+ac+bc wehre they are the three roots)-so we find k=65+70+70=205
add them and we get the answer to be -445[/hide]

[quote]now -w/1 is the product of the roots.
k/1 is the alternating sum of the roots (so ab+ac+bc wehre they are the three roots)-so we find k=65+70+70=205
add them and we get the answer to be -445[/quote]

I got lost in the middle of your solution  :blush:
What is the alternating sum of the roots? and -w/1? could you give me the general formula?

Here's another ending to it:
B/c we got the roots, we could write the equation in factored form then multiply all out, match the k and w to their numbers and get the answer.

[quote="jeez123"]When I look up the solutions...
$7-4i$ is one root and so is $7+4i$ for 1.
$7+\sqrt{13}$ and so is $7-\sqrt{13}$ for problem 2.

Why is that?[/quote]

This is a consequence of the Conjugate Root Theorem, which comes in two flavors.

One says that any polynomial with real coefficients that has a root $a+bi$ must also have $a-bi$ as a root, where $a, b$ are real.

The other says that any polynomial with [i]rational[/i] coefficients that has a root $a+\sqrt{b}$ (where $a, b$ are [i]rational[/i], $b > 0$, and $\sqrt{b}$ is not rational) must also have $a-\sqrt{b}$ as a root.

I wrote a blog post about this here:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=134466


induction
number theory proposed
number theory

i like this problem for its simplicity in first glance :
assume that $d_1,\dots,d_n$ is n divisors of n prove that sum of some of them is $n$. :)

yes :)  ofcourse they are not distinct divisors

Gm. An approach below is bit ugly.

Induction by $n$, cases $n=1$, $n$ prime are clear.

Let $p$ be a prime divisor of $n$. Then, if at least $n/p$ divisors of $n$ are divisible by $p$, we consider these divisors, $px_1,\,px_2,\dots,px_k$ and use the inductive hypothesis for divisors $x_1,\,x_2,\,\dots,x_k$ of $n/p$.

So, less then $n/p$ our divisors are divisible by $p$.

If $n=2^k$, it means that at least $n/2$ divisors are equal to 1, and there are no problems to get $n$: just arrange our divisors $d_1\ge d_2\ge \ge d_n$ and take maximal $k$, for which $d_1+d_2+\dots+d_k\le n$. Then the difference $n-(d_1+d_2+\dots+d_k)$ is not greater then number of 1-s in the set $d_{k+1},d_{k+2},\,\dots,d_{n}$ (because the number of 1-s in this set is at least $\min(n/2,n-k)$). Case $n=2^k3^l$ is quite analogous.

So, we assume that $p\ge 5$. Take all the $k$ our divisors, not divisible by $p$. Denote them $x_1\ge x_2\ge\dots \ge x_{k}$. Note that $k>n-n/p$. In any set of $n/p$ divisors $x_1,\,x_2,\dots,x_k$ we may find a subset with sum $n/p$. Do it for $x_1,\,x_2,\dots,x_{n/p}$. Mark the subset with sum $n/p$. Take now $n/p$ maximal not-marked $x$-s, chose a subset with sum $n/p$, and so on. If we may take it $p$ times, we are done. If not, after $(p-1)$ time less then $n/p$ numbers are left. Since $k>n-n/p$, it means that at least $n-2n/p$ divisors have total sum $n-n/p$. Therefore, at least $n-3n/p$ of these divisors are equal to 1. Next is easy and analogous to the argument of a previous paragraph.


the equation is (pi) r^2
r^2 is radius squared

[quote="shybox"]What is the formula for the area of the circle?

is it $2d\pi$?[/quote]
Circumference: $2\pi r$ or $\pi d$

Area: $\pi r^{2}$

[quote="shybox"]What is the formula for the area of the circle?

is it $2d\pi$?[/quote]

Google is your friend :wink:
\[\text{Area of a }\setlength{\unitlength}{.5cm}\begin{picture}(1,1) \qbezier(1,0.5)(1,1)(0.5,1) \qbezier(0.5,1)(0,1)(0,0.5) \qbezier(0,0.5)(0,0)(0.5,0) \qbezier(0.5,0)(1,0)(1,0.5) \end{picture}=\pi r^{2}=\frac{\pi d^{2}}{4}\\ \\ \\ \\ \text{Circumference of a }\setlength{\unitlength}{.5cm}\begin{picture}(1,1) \qbezier(1,0.5)(1,1)(0.5,1) \qbezier(0.5,1)(0,1)(0,0.5) \qbezier(0,0.5)(0,0)(0.5,0) \qbezier(0.5,0)(1,0)(1,0.5) \end{picture}=2\pi r=\pi d \]

My first picture so lets hope it works

Haha Bishoy, very funny.

In case you all didn't know, he was just messing with ya

yea, bishoy, dont post crap.  its annoying, and i think some people didnt know you were messing around.

[quote="anirudh"][quote="ragnarok23"][quote="mathgeniuse^ln(x)"]Who's bishoy?[/quote]

shybox[/quote]

I believe that he lives a few blocks from Walk Around the River...[/quote]
No, he lives about a 10 minute drive from me. He's quite far from Eli. But anyways, we all know each other and this thread should be locked or something


Hi, I have a question about AM-GM (more generally, QM-AM-GM-HM)

It is known that certain expression's GM is its AM's minimum value.

Is it true that that expression's AM is its GM's maximum value?

You know $ AM \geq GM$ for any particular valid combination of values.
[u]
Direct Consequences:[/u]

If GM is invariant, then it is the lower bound for AM.
If AM is invariant, it is the upper bound for GM.

Um, what are all of these weird abbreviations? [hide=" "]202[/hide]

[quote="r15s11z55y89w21"]Um, what are all of these weird abbreviations?[/quote]
AM=Arithmetic Mean
GM=Geometric Mean

Oh! I guess therefore, HM=harmonic mean? Then what's QM? I haven't heard of a fourth type of mean before. And also, isn't it possible for $ GM>AM$?

[quote="r15s11z55y89w21"]what's QM? I haven't heard of a fourth type of mean before.[/quote]
[url=http://en.wikipedia.org/wiki/Root_mean_square]Quadratic mean[/url].  Generally it's possible to define an infinite family of means called power means.

[quote="r15s11z55y89w21"]And also, isn't it possible for $ GM > AM$?[/quote]
Not for positive reals.  This is the content of the [url=http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means]AM-GM inequality[/url].

Anyway, FantasyLover, you're going to have to be more precise.  What do you mean by "an expression," and what do you mean by taking the AM or the GM of an expression?  Minimum and maximum with respect to what?

i think by expression he means a certain set of reals.

[quote="t0rajir0u"]Anyway, FantasyLover, you're going to have to be more precise.  What do you mean by "an expression," and what do you mean by taking the AM or the GM of an expression?  Minimum and maximum with respect to what?[/quote]
Let's say that there are two real numbers $ x$ and $ y$.

$ AM\equal{}\frac{x\plus{}y}{2}$ and $ GM\equal{}\sqrt{xy}$.

We know that minimum value of $ \frac{x\plus{}y}{2}$ is $ \sqrt{xy}$.

Basically what I'm asking is that is it true that the maximum value of $ \sqrt{xy}$ is $ \frac{x\plus{}y}{2}$?

[quote="FantasyLover"][quote="t0rajir0u"]Anyway, FantasyLover, you're going to have to be more precise.  What do you mean by "an expression," and what do you mean by taking the AM or the GM of an expression?  Minimum and maximum with respect to what?[/quote]
Let's say that there are two real numbers $ x$ and $ y$.

$ AM \equal{} \frac {x \plus{} y}{2}$ and $ GM \equal{} \sqrt {xy}$.

We know that minimum value of $ \frac {x \plus{} y}{2}$ is $ \sqrt {xy}$.

Basically what I'm asking is that is it true that the maximum value of $ \sqrt {xy}$ is $ \frac {x \plus{} y}{2}$?[/quote]

 yes, it is true that $ \sqrt{xy}$ is bounded by $ \frac{x\plus{}y}{2}$.in other words, the maximun of
$ \sqrt{xy}$ is obtained when $ \sqrt{xy}\equal{}\frac{x\plus{}y}{2}$.(this happens when $ x\equal{}y$).


There is a reunion of 201 people from 5 different countries. In each group of 6 people, at least two have the same age. Show that there must be at least 5 people with the same country, age and sex.

I think you are missing information or something, since you said nothing about sex in the problem.  Anyway here is what I get disregarding sex.
[hide]
By pigeonhole, if any 6 people have two with the same age, the max number of unique ages is 5.  We are given the number of unique countries is 5.  So the number of unique people without the same age and country is 5*5=25.  It follows that in 201 people, there will be a minimum of 8 people with the same country and age.
[/hide]


pigeonhole principle
MATHCOUNTS
trigonometry
geometry
3D geometry
prism
analytic geometry

I thought that Round 2 was slightly harder than round 1 overall.

Round 2, but I didn't spend as much time so I can't really say...

Well, I thought that Round 2 was too difficult for a round 2. Maybe that's just because of number 5, but round 2 shouldn't be this hard. Round 1 was slightly easier than normal (or I'm just getting better or math...).

Last year, my round 1 score was the lowest, because I didn't know how to use pidgeonhole and I ended skipping that particular problem on round 1.

I think round 1 was easier, and had less inspired solutions. 1.1 was pretty standard, basically just turn English into algebra and rearrange, then you don't even have to look for bounds, you can just check alot of factors. 1.2a was pretty much a MathCounts problem, and for 1.2b, it's pretty natural to draw in BE, and it opens up from there. 1.3 was tricky if you've never seen arctangents being added, but most people have, and the Diophantine wasn't that hard, even if you didn't see that a<4. 1.4 was very easy if you just drew a good diagram, and you could even just measure, then use the measurement to see what conditions would imply that. So basically the first four problems on round 1 were easy and pretty much standard. The trig substitution in 1.5 wasn't obvious, but with some work, you can find a solution without the substitution.

Round 2 was harder, in my opinion. 2.1 is not a new problem, and can be easily brute forced. 2.2 was nice, it's not clear exactly how to start, especially if you aren't experienced with Newton's Sums. 2.3 was tricky, you could dive into a bunch of casework, but if you think for a while, you can realize that it's unneccessary. The bijection in 2.4 is really nice, and brute force is just ugly (see my solution). It takes some thinking with the pencil down to get the bijection, it seems. 2.5 was really hard to visualize, so I made a cardboard model  :P, and putting a hexagonal prism around it was the easiest way to make the model, so my solution just went from there. But it was quite easily researchable, and there are plenty of proofs on the internet. If you don't just cite the coordinate definitions and the inradius/circumradius from mathworld, it's pretty tough to come up with them, but doable.

So yeah, round 2 was harder, and took more thought.

I agree. In round 1, 2/1 and 4/1 were fairly easy geometry problems, while 1/1, 3/1 could be solved by straightforward algebra with casework (3/1 are a little trickier since it's written in sums of arctan); 5/1 was the hardest, you had to be very careful at doing bounding arguments. Solutions to problems in Round 2, at least to me, were less obvious. I stared at 1/2 and eventually decided to brute-force it out. 2/2 was nice. I was also one of those who were trying to figure 3/2 out by casework, but fortunately I went back to the right track. 4/2 was also very nice, but I just forgot the concepts of combinatorial numbers and did some strange low-level counting, which led to massive summations and blah blah blah...(I figured out the general formula, though). 5/3 was probably the hardest one in this round, in my opinion, mostly because icosahedron was not well known to most people, so to me it was unbelievablly complicated at first sight. But doing some research on the internet, you would probably got some ideas by looking at models. The rest of work was just to deal with cross-sections, as always.

I don't think that Round 2 problem 5 was very difficult. I didn't know what an icosahedron was, so like other people I made a cardboard model of the icosahedron and the prism. After that, you just have to find the volume of a pyramid, which was fairly easy.

I found round 1 a bit easy, but #5 was pretty good. Round 2 was (slightly) trickier because of #5, but was not much harder.

(BTW it's "which round was harder" not hardest... comparative, not superlative :P )

EDIT: hmm, on second thought, maybe round 1 was harder... so my vote is wrong

[quote="CatalystOfNostalgia"]But it was quite easily researchable, and there are plenty of proofs on the internet. If you don't just cite the coordinate definitions and the inradius/circumradius from mathworld, it's pretty tough to come up with them, but doable.[/quote]

On the USAMTS, could you get full credit if you just cited the formulae for inradius/circumradius of an icosahedron?


How would you get an exact polar representation $r,\theta$ for $4+2i\sqrt{3}$?

In form $a+bi$
$r=\sqrt{a^{2}+b^{2}}$
$\theta = \tan^{-1}\left(\frac{b}{a}\right)$
Then $r \cos \theta+i \sin \theta$

Not good enough. How do you get an EXACT solution, for $\theta$ too, which $\tan^{-1}$ won't give you?

Hmmm (edited)
[hide]$z=4+2i\sqrt{3}\iff a=4 \text{ and }b=2\sqrt{3}$

$r=\sqrt{a^{2}+b^{2}}$
$r=\sqrt{4^{2}+(2\sqrt{3})^{2}}$
$r=2 \sqrt{7}$

$\theta=\tan^{-1}\left( \frac{b}{a}\right)$
$\theta=\tan^{-1}\left( \frac{2\sqrt{3}}{4}\right)$
$\theta=\tan^{-1}\left( \frac{\sqrt{3}}{2}\right)$

Solution: $\boxed{ \left( 2 \sqrt{7}, \tan^{-1}\left( \frac{\sqrt{3}}{2}\right) \right) }\iff z= \boxed{2 \sqrt{7}\left[ \cos \left[ \tan^{-1}\left( \frac{\sqrt{3}}{2}\right) \right]+i \sin \left[ \tan^{-1}\left( \frac{\sqrt{3}}{2}\right) \right] \right]}$

I would say that is as simple as it gets.  You can't evaluate $\tan^{-1}\left( \frac{\sqrt{3}}{2}\right)$ any further without a calculator.  The point is in the first quadrant so that will produce what you need.[/hide]


A fair, 8-sided die has faces numbered 1, 2, 3, 4, 5, 6, 7, and 8.  If the die is rolled twice, what is the probability that the sum of the two numbers will be 9?  Express your answer as a common fraction.

[hide] 1/8  i cant explain rite now! i have to go somewhere! ;) [/hide]

[hide="Solution"]
There are eight ways to get 9, and 64 possibilities in total. 8/64 = [b]1/8[/b].
[/hide]

Treething, thanks for showing your work

and

I got the same answer

[hide]8 ways to get 9 so its 8/64 or 1/8[/hide]


On Thursday, Oct 4, Art of Problem Solving will be hosting a Math Jam information session to discuss our upcoming Introduction to Algebra class.  Introduction to Algebra starts Wednesday, October 17.

At the Math Jam, AoPS instructors will discuss the content of the course, and answer any questions you may have about the class.  We'll also answer more general questions about our online school.

The Math Jam is FREE and will start at 7:30 PM Eastern (4:30 PM Pacific).

To join the Math Jam, please log on to http://www.artofproblemsolving.com using your Art of Problem Solving username, then click on the "Classroom" link on the left side of the page, and follow the instructions.

We hope to see you there.

Is this course different from Prealgebra and Algebra 1?

This course doesn't exist anymore.  The post you are responding to is from four years ago.  That course was split into Algebra 1 and Algebra 2.

Oh. Sorry :) I was wondering why I hadn't heard of that before...


Let$ (O)$is a circle with $ XY$is a diameter. $ A,B$ is two points in a line $ d$(doesn't have the common point with $ (O)$).Let $ A_1$, $ A_2$, $ B_1$, $ B_2$ is the intersection point of  $ AX, AY, BX, BY$with $ (O)$.
Prove that
$ A_2B_1$, $ A_1B_2$ and $ d$is concurrent :)

[quote="TRAN THAI HUNG"]Let$ (O)$is a circle with $ XY$is a diameter. $ A,B$ is two points in a line $ d$(doesn't have the common point with $ (O)$).Let $ A_1$, $ A_2$, $ B_1$, $ B_2$ is the intersection point of  $ AX, AY, BX, BY$with $ (O)$.
Prove that
$ A_2B_1$, $ A_1B_2$ and $ d$is concurrent[/quote]

[b]Generalized problem.[/b] [b][color=darkblue]Let $ X,Y$ be arbitry points on $ (O)$. Let $ A,B$ be two points wherever. $ AX$, $ AY$, $ BX$, $ BY$, respectively intersect $ (O)$ at $ A_1$, $ A_2$, $ B_1$, $ B_2$. Prove that $ A_1B_2$, $ B_1A_2$, $ AB$ are concurrent.[/color][/b]

Apply [i]Pascal theorem[/i] for $ YA_2B_1XA_1B_2$, we have,
\[ A\equiv XA_1\cap YA_2\]

\[ B\equiv XB_1\cap YB_2\]

\[ C\equiv A_1B_2\cap A_2B_1\]
Therefore $ A,B,C$ are collinear. The result is lead as follow. $ \square$


function
limit
integration
real analysis
real analysis unsolved

Let $ f$ be an integrable function and $ E_n\equal{}\{x: |f(x)|>n\}$.
Prove that $ \lim_{n\to\infty} \int _{E_n} |f| d\mu\equal{}0$

Hint: use the Dominated Convergence Theorem. Of course, this requires that you produce a sequence of functions to which you can apply the theorem.

what is the limit of $ \mu(E_j)$ as $ j\to\infty$?  what is the measure of $ \bigcap^\infty_{j\equal{}1} E_j$?

what can you say about the sequence of functions $ 1_{E_n}$ (the characteristic functions of the sets $ E_n$)?

what can you say about the sequence of functions $ |f|1_{E_n}$?

what is another way of writing $ \int_X |f|1_{E_n}d\mu$?


can magnet be a conductor or is a conductor?
if it can be what will be the effect of electric field on magnet?

Electric fields and magnetic fields are very closely intertwined.  Faraday's Law and the Maxwell-Ampere law say that changing magnetic/electric fields can induce an electric/magnetic field.

well, i too know about electromagnetic induction. tats why i asked that if we passed current in magnet, then would the magnetic field , increase or decrease ....and questions like that.
but first of all [quote]can magnet be a conductor or is a conductor? [/quote]

Well, a conventional magnet is a conductor (not really a good conductor, though.  :lol:  After all, it's iron... But current can go through, like through every metal.).

And the other questions you mentioned:
I pretty much agree with the response to a similar question I've read on physicsforums

[quote]Any heat generated by the electricity through resistance of the iron would cause demagnetization. The current would also induce a magnetic field concentric to itself. Depending on the orientation of this induced magnetic field to the domains in the magnet (with respective components parallel or antiparallel), the current would conventionally have a slight magnetizing or demagnetizing effect.[/quote]

nice, if you can share nore info. it would be nice as i am too curious here. or links to physics webpages or forums
tx in advance

Ok, here's the link to the physicsforums.com, the thread where I've found this

http://www.physicsforums.com/showthread.php?t=184882

You can prove it yourself, imagining the magnet like a common conductor, calculating the induction caused by the current going through it, giving it a vector, and then adding the effect of permanent magnet.


function
real analysis
real analysis unsolved

Suppose A is uniformly bounded family of right continuous non-decreasing functions defined on the interval $ [a,b]$, then show A is sequential compact.
i,e. for any sequence $ \{f_n\}_{n\in\mathbb{N}}$ in A, then there exists a $ f\in A$, and a subsequence $ \{f{}_n{}_k\}_{k\in\mathbb{N}}$ s.t for any point $ x\in [a,b]$, $ f{}_n{}_k(x)$ converges to $ f(x)$.

It is not really true. Consider $ [a,b]\equal{}[0,2]$, $ f_n(x)\equal{}0$ if $ x<1\plus{}1/n$, and $ f_n(x)\equal{}1$ if $ x\geq 1\plus{}1/n$. This sequence converges to a function which is not in $ A$.


Let $ a,b,c>0$.
Prove that:
$ \frac{(a\plus{}b)^2}{(a\plus{}b)c\plus{}a^2\plus{}b^2}\plus{}\frac{(b\plus{}c)^2}{(b\plus{}c)a\plus{}b^2\plus{}c^2}\plus{}\frac{(c\plus{}a)^2}{(c\plus{}a)b\plus{}c^2\plus{}a^2}\leq 3$

Marin Bancoș - ROMANIA

[quote="marin.bancos"]Let $ a,b,c > 0$.
Prove that:
$ \frac {(a \plus{} b)^2}{(a \plus{} b)c \plus{} a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)^2}{(b \plus{} c)a \plus{} b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)^2}{(c \plus{} a)b \plus{} c^2 \plus{} a^2}\leq 3$

Marin Bancoș - ROMANIA[/quote]
See here: http://canhang2007.wordpress.com/2009/11/09/inequality-59-m-bancos/

[b]can_hang2007[/b], you are amazing! Maybe your solution to this problem (which is also my solution) is the only one.


Let $x$ and $y$ be real numbers such that $x^2+y^2=2(x-y+1)$
Determine the largest possible value of $x^2+y^2$

[hide="informal geometric solution"]complete the square on the original equation and get $(x-1)^2+(y+1)^2=2^2$ so we have a circle $C$ with radius 2 centered at $(1,-1)$. let $d=x^2+y^2$ and this represents a set of concentric circles with radius $d$ and centered at the origin. so we are trying to find the maximum radius $d$ that intersects at least one point from the original circle $C$. the farthest distance from $C$ to the origin is $2+\sqrt2$ and so $d_\text{max}=(2+\sqrt2)^2=6+4\sqrt2$[/hide]


1/Find the maximum value of the sum:
$ S_n \equal{} a_1(1 \minus{} a_2) \plus{} a_2(1 \minus{} a_3) \plus{} \cdots \plus{} a_n(1 \minus{} a_1)$
where $ \frac {1}{2}\le a_i\le 1 \forall i \equal{} 1,2,\ldots ,n.$
2/Let $ n\ge 2$ and $ 0\le x_i\le 1\forall i \equal{} 1,2,\ldots ,n.$ Show that
$ (x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n) \minus{} (x_1x_2 \plus{} x_2x_3 \plus{} \cdots \plus{} x_nx_1)\le \lfloor\frac {n}{2}\rfloor$
and determine when there is equality.
3/Let $ a_1,a_2,\ldots,a_{19}$ be real numbers from the interval $ [ \minus{} 98,98]$. Determine the minimum value of $ a_1a_2 \plus{} a_2a_3 \plus{} \cdots \plus{} a_{19}a_1$
4/Prove that if $ 1\le x_k\le 2, k \equal{} 1,2,\ldots,n,$then
$ (\sum_{k \equal{} 1}^{n}x_k)(\sum_{k \equal{} 1}^{n}\frac {1}{x_k})\le n^3$

[hide="number one"]
$ a_1 \plus{} a_2 \plus{} a_3...a_n \minus{} (a_1\cdot a_2 \plus{} a_2\cdot a_3...a_n\cdot a_1)$

$ \Rightarrow \sum \frac {1}{2}a_1 \minus{} a_1\cdot a_2 \plus{} \frac {1}{2}a_2$

So we need to find the max value of $ a_1 \minus{} 2a_1\cdot a_2 \plus{} a_2$ given that $ \frac {1}{2}\le a_i \le 1$

If we fix $ a_2 \equal{} k$ then we have $ a_1 \minus{} 2ka_1 \plus{} k$ since $ 2k \ge 1$ the answer is maximised when $ a_1 \equal{} \frac {1}{2}$

Then we are left with only $ a_1 \plus{} k \minus{} k \equal{} \frac {1}{2}$.

So maximum of $ \frac {1}{2}a_1 \minus{} a_1\cdot a_2 \plus{} \frac {1}{2}a_2 \equal{} \frac {1}{4}$

$ \Rightarrow Max: S \equal{} \left\lfloor\frac {n}{4}\right\rfloor$

[/hide]

[hide="number two"]
Same as number one but this time we rewrite $ a_1\minus{}2a_1a_2 \plus{} a_2$ as;

$ 1\minus{}a_1a_2 \minus{} (1\minus{}a_1)(1\minus{}a_2)$ and since $ 0\le a_1 \le 1$ the expression is maximised when $ a_1\equal{}1$ and $ a_2\equal{}0$ or vice versa. So the expression is equal to $ \frac{1}{2}$

From here we find max $ a_1\plus{}a_2...a_n \minus{} (a_1a_2\plus{}a_2a_3...a_na_1) \equal{} \left\lfloor\frac{n}{2}\right\rfloor$

(note that the floor funtion is needed if n is odd)

Equality holds when even elements equal $ 1$ and odd elements equal $ 0$ or vice versa
Equality holds when [/hide]


[hide="Hint"] If $a+bi$ is a root of a polynomial with real coefficients, then $a-bi$ must also be a root. [/hide]

[hide=answer]$(x-i)$ is obviously the fastest way to get that root, but we can't, so we instead use $i^{2}=-1$ to get $(x^{2}+1)$. $(x^{2}+1)(x-\sqrt{2})=x^{3}-\sqrt{2}x^{2}+x-\sqrt{2}$[\hide]

[quote="t0rajir0u"][b]Follow-up:[/b]  Give the monic polynomial of minimum possible degree if we add the additional constraint that the coefficients must be [i]integer.[/i][/quote]

couldnt you just multiply what bill the maniac got by $x+\sqrt{2}$?

[hide="Hint"]
Replace $x-\sqrt{2}$ with $x^{2}-2$.
[/hide]

[quote="Ignite168"]If i and $\sqrt{2}$ are two roots of a polynomial equation with real coefficients, give the equation of the minimum possible degree.

This problem seems like it could have infinite answers so assume that the coefficient of the first term is 1.[/quote]

well $i$ and $\sqrt{2}$ must be the only roots then the minimum degree of an equation with those roots will have a degree of 3.

[quote="t0rajir0u"][quote="7h3.D3m0n.117"]well $i$ and $\sqrt{2}$ must be the only roots[/quote]

Not if you want the coefficients to be real... and by this logic, your answer would be 2.[/quote]

Well if a degree is 2  then either way there will be at least 2 roots, so it's impossible to have an imaginary?

I don't know i'm having a math breakdown this week  :blush:


In$Rt\triangle ABC,\angle C=90^\circ$,$AD$ is the bisector of $\angle A$, $BE$ is the bisector of $\angle B$, $D,E \in AC,BC$.
$M$ is the median point of $DE$.$AD\cap BE=I$.
Prove that $MI\bot AB$.

Let P and Q the midpoints of AD and BE. Than Ep and DQ are perpendicular on AD and BE. Also let oO the intersection of EP and DQ. I is the ortocenter of ODE and M the midpoint of DE. Now it is easy to see that MI is perp on PQ . Note that PQ is parallel with AB and that EOD is 45 :)

[quote="ciprian"]Let P and Q the midpoints of AD and BE. Than Ep and DQ are perpendicular on AD and BE. [/quote]
thx,but I can't find it right. :?:

Let $X$ be the intersection of $MI$ and $AB$. Also let $P$ be the intersection of $EX$ and $AD$,  and $Q$ the intersection of $XD$ and $EB$.

Then as $M$ is the middle point of $ED$, by Ceva $PQ$ is parallel to $AB$. Then $\frac{EP}{PX}=\frac{DQ}{QX}$ and because $AP$ and $BQ$ are angle bisectors, then $\frac{AE}{AX}=\frac{DB}{XB}$ and $\frac{AX}{XB}=\frac{AE}{DB}$

Look that $IM$ is perpendicular to $AB$ iff $X$ is the tangent point of the incircle with $AB$.

We can transform the equiality in terms of the sides of the triangle. $EA=\frac{ca}{b+c}$, $DB=\frac{cb}{a+c}$, $AX=\frac{a+c-b}{2}$, $XB=\frac{b+c-a}{2}$

Then. $\frac{\frac{ca}{b+c}}{\frac{cb}{a+c}}=\frac{\frac{a+c-b}{2}}{\frac{b+c-a}{2}}$

Multiplying we get:

$a^2b+a^2c+abc+c^2a-a^3-a^2c=b^2a+b^2c+abc+c^2b-b^3-b^2c$. After eliminate equal things and substracting by $a-b$ we get $ab+c^2=a^2+b^2+ab$ and this is true iff $C=90$ degrees


If x>-c>-b>-a what is the minimum of (x+a)(x+b)/(x+c)

We can make it arbitarily large and negative by making $a$ large and negative but keeping $x,b,c$ positive and small.
If $a>0$, then we can get the function as near to $0$ as we like (from above) be making $a,b,c,x \rightarrow 0$

Probably $a, b, c$ are fixed and only $x$ is variable. Indeed it is very easy to solve this with calculus, but I guess we are to look for an elementary solution. ;-)

[hide="Elementary solution"]By synthetic division we find

\begin{eqnarray*}f(x)=\frac{(x+a)(x+b)}{x+c}=x+a+b-c+\frac{ab-ac-bc+c^{2}}{x+c}=x+a+b-c+\frac{(a-c)(b-c)}{x+c}\end{eqnarray*}

which can be rewritten as

$f(x)=\left[(x+c)+\frac{(a-c)(b-c)}{x+c}\right]+a+b-2c$

Since $x>-c\iff x+c>0$ and $-c>-b>-a\iff a>b>c$, both $x+c$ and $\frac{(a-c)(b-c)}{x+c}$ are positive, hence we can apply AM-GM:

$f(x)\geqslant a+b-2c+2\sqrt{(a-c)(b-c)}=\boxed{(\sqrt{a-c}+\sqrt{b-c})^{2}}$, with the minimum obtained when $x+c=\frac{(a-c)(b-c)}{x+c}\iff x=-c+\sqrt{(a-c)(b-c)}$ (only plus sign because of $x>-c$)[/hide]


here is a practice proof i cannot figure out????

let x,y>=0, the let m and n be positive intgers.
Prove:

(x^m)(y^n)+(x^n)(y^m) < x^(m+n) + y^(m+n)

thanks for helping understand this guy[/u]

Put everything on one side, and factor: we want to show that $x^{m}x^{n}+y^{m}y^{n}-x^{m}y^{n}-x^{n}y^{m}=(x^{m}-x^{n})(y^{m}-y^{n})>0$

This isn't true in general. Can you find a counterexample?

By the way, this doesn't belong in Superior Algebra.


[color=darkblue]Consider a line of characters $ a_1a_2a_3\ldots a_n$. In how many ways could we divide this line in words.

An example of division for $ n \equal{} 6$ is
$ a_1a_2a_3a_4a_5a_6 \equal{} a_1a_2a_3 \plus{} a_4 \plus{} a_5a_6$.

Note. $ 1\le$ length of word $ \le n$
[/color]

Picking a division is equivalent to picking a subset of the $ n\minus{}1$ spaces between the words to insert $ \plus{}$ signs in.  This can be done in $ 2^{n\minus{}1}$ ways.


[quote="236factorial"]A mouse lives at the base of a 24-meter high tree. A hawk, sitting on the top of the tree, notices that the mouse when the mouse is heading home and is 72 meters from the base of the tree. The hawk plans to dive in a straight line and catch the mouse. If they both travel the same distance, how far from the base of the tree will the hawk catch the mouse?[/quote]

[hide]Draw right triangle ABC with C being the base of the tree, B being where the mouse it, and A being where the hawk is.  Then make point D which is the point where the hawk catches the mouse. DB=DA. Set CB-DB=x, which is what we're trying to find. AC=24 and CB=72.

Then, right trangle CAD has side lengths of  x, 24, and 72-x, with the last term being the hypotenuse.

Now we solve for x:

$x^2+24^2=(72-x)^2$

$x^2+576=5184-144x+x^2$

$576=5184-144x$

$144x=4608$

$x=32$[/hide]

[quote="shadysaysurspammed"]40 Metres from tree[/quote]

Why?  :? Drunken math disagrees  :D (and so do I).

[quote="shadysaysurspammed"]40 Metres from tree[/quote]
SILLY mistake! :D  $40$m is the distance TRAVELLED by the mouse so the distance from the tree base is $32$

[quote="shadysaysurspammed"]Wont the dsistance from the tree base be 72-x as drunken math did in his solution????x is what the mouse travels before gettin caught,right?or am i just actin stupid :blush:[/quote]

$x$ is how far the mouse was from the tree when it got caught...

Don't worry. Everyone make their share of careless mistakes...  :)

[hide]
If we draw a picture, We get a triangle with sides 24, x, and 72-x, plugging into pythagorean theorem, we get x=40, thus, the distance from the tree is 72-40=3200 cm.= 32 m=0.032 km.
[/hide]

For the hawk and the mouse to travel the same distance, the hypotenuse of a right triagnle is 72-x, and the other 2 legs are x and 24.

$24^2+x^2=(72-x)^2$

$576+x^2=5184+x^2-144x$

$4608-144x=0$

$4608=144x$

$x=32$

Therefore, the distance the hawk travels is 32

shady , we get x =40 hence z-x= 72-40 =32m

whats wrong??

Oh damn,I thought i got 32 as the distance travellrd by mouse and hawk :mad:

Sorry for bein stupid :oops:


AMC AIME counting distinguishability function Asymptote probability

This is a transplant from the old forum.  I just transcribed the messages.

Joe:
1. 651
2. 125
3. 17
4. 42
5.
6. 57?
7.
8.

I actually didn't work on 5 or 7... but oh well. 8, I think you use the unit circle? (Divide into 1997 sectors)

---------------

Yizheng:
I got something like:

1. 651
2. 365
3. 17
4. 42
5. 384 (?)
6. 64
7. 58
8. 665

-------------------------

Kevin:
1. 651
2. 125
3. 17
4. 42
5. a divisor of 576? possibly 144? (wild guesses)
6. 117
7. 58
8. 582 (though i'm not sure of this one either...i just got 3 answers in the last 5 minutes, one of which was 665, though i think that that's wrong. no guarentee that 582 is right)

so far three different answers for 6. heres mine:
pick a first card. pick a different second card. there is one, and only one, card that will make a set with the first two. however, each set will be counted 3! times.
so the final answer is 27*26/6=117

-------------------------

me:
I got
1. 651
2. 125
3. 017
4. 042
5. 060
6. 117
7. 058
8. ---

7 was by far the most interesting problem, although I took way too long to do it (found the exact value for f(x); was this necessary?).

6 I got by casework. If all three qualities are different, there are 36 ways: the three cards are three different shapes (1 choice), then the three colors can be sorted into the now-distinct cards in 6 ways, and the same for the shades.

If two qualities are different and one is the same: 3 ways to choose which quality will be the same, times 3 ways to choose which shape/color/shade will be the one repeated, times 1 way to sort the first different quality (they're indistinguishable before), times 6 ways to sort the second different quality.

If one quality is different, then there are 3 ways to choose which quality is different, 3*3 ways to choose which shape/color/shade will be repeated, twice, and 1 way to sort the different quality.

So in all there are 36 + 54 + 27 = 117. But Kevin's way is much simpler.

------------------

Kevin

i dont think you need to find the exact function for 6 (because well...i didnt)

if you think about what this graph looks like, it's basically 1/x shifted and skewed.
from algebra 2 (dont know what that translates into in exeter math), the vertical asymptote is at -d/c and the horizontal one is at a/c (which will be the value that the function never takes)

setting f(x) and the inverse of f(x) equal yields d=-a

then plugging in f(19)=19 and f(97)=97 yields a=58*c, thus a/c=58

---------------------

Sara C:

numba 7... let u = the # not in the range of f. it's the only number without an inverse so f^-1(u) = (b-du)/(cu-a) has to be undefined, so the denominator = 0, so u = a/c. same as kevin's horizontal asymptote OH SNAP!

--------------------

Elizabeth:

I've been having a bit of trouble w/ 7 and 8 (I'm gonna try Kevin's hint for 7 but any hints for 8?)...but I agree w/ Kevin and Keone's answers for 1, 2, 3, 4, and 6... 5 I got 90 for. I might be wrong, though.
What I did:
There are three cases if you chose the top two rows first.
Case 1: The numbers in top halfs of all the columns are the same. Then the rest of the grid is decided. There are 4C2 = 6 ways to choose which of the top half columns are positive. 4C2 is just the number of ways to choose two to be positive out of 4. The second row is then decided in this case. Eg.
1 1 -1 -1
1 1 -1 -1
-1 -1 1 1
-1 -1 1 1
Case 2: All are opposites in the top two rows. There are 4C2 = 6 ways to choose the top two rows, then you can choose anything you want for the 3rd row = 4C2, and the 4th row is then decided. 6*6 = 36
Case 3: 2 of the top halfs of the columns are the same, two are different... eg.
1 -1 1 -1
-1 -1 1 1
That means that we have
1 -1 1 -1
-1 -1 1 1
_ 1 -1 _
_ 1 -1 _
For choosing the blanks, there are 2 ways to do this. For the top half, there are 24 ways to choose it... this number is obtained from (6C2)^2 (the total number of ways to choose the top two rows with only the condition that there are two 1's and two -1's) - 6 (the number of ways choosing the top from case 1) - 6 (the number of ways choosing the top from case 2) = 24. 2*24 = 48.
6+36+48 = 90

-------------------------

Kevin:

well, that solution for #5 was a lot better than mine. it seemed fine to me....

well, here's how i did number 8:
the roots of the equation are the roots of unity, and so are evenly spaced around the unit circle.
it doesnt really matter what the two chosen roots are, only the relationship between them. the probability for picking a random root for v is the same as picking a specific root, so i just set it to be 1.
doing some trig, you can see that the second root will have to be within 30 degrees of the first root in order to get the required distance away from the origin. so on each side of the root, there will be floor(1997/12)=166 roots that will work, for a total of 2*166=332 roots.
332/1996=83/499
83+499=582

yizheng's answer of 665 can be gotten doing the same exact method, but divide by 6 instead of dividing by 12.

Here are my answers

1. 651
2. 125
3. 017
4. 042
5. 090
6. ---
7. ---
8. ---

For number 5, I just ignored the -1's. We want two 1's in each row and column, and you use simple casework.  I'll work on 6, 7, and 8 and post the answers later.


How do you prove that the circumcenter is the isogonal conjugate of the orthocenter?

Well consider triangle ABC with O as the circumcentre and H as the orthocentre.

Now angle BOC=2*angle BAC       ...angle subtended at the centre of the circumcircle

But triangle BOC is isosceles thus angle OBC is the complement of half of angle BOC

Thus angle OBC is the complement of angle BAC

But angle HBA is also the complement of angle BAC

Thus angle OBC = angle HBA

Thus the altitude is the isogonal conjugate of the radius to the centre and hence O and H are isogonal conjugates.

EDIT: OOPS wasted time in writing this by which Jelly World found the answer

[Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 for a proof.]

[quote="Vihang"]Well consider triangle ABC with O as the circumcentre and H as the orthocentre.

Now angle BOC=2*angle BAC       ...angle subtended at the centre of the circumcircle

But triangle BOC is isosceles thus angle OBC is the complement of half of angle BOC

Thus angle OBC is the complement of angle BAC

But angle HBA is also the complement of angle BAC

Thus angle OBC = angle HBA

Thus the altitude is the isogonal conjugate of the radius to the centre and hence O and H are isogonal conjugates.

EDIT: OOPS wasted time in writing this by which Jelly World found the answer[/quote]

Ahhh thanks anyway, at least now I have reference to a formal proof.


I was wondering whether I should study Plane Euclidean Geometry: Theory and Problems by A D Gardiner and C J Bradley or Geometry Revisited (Mathematical Association of America) by Coxeter and Greitzer.

What do you guys recommend?

Both. Also Crossing the Bridge by Gerry Leversha, The United Kingdom Mathematics Trust, 2008, ISBN 9781906001063 and New Problems in Euclidean Geometry by David Monk, The United Kingdom Mathematics Trust, 2009, ISBN 9781906001094. Perhaps also, Geometry: Ancient and Modern by John R. Silvester, Oxford University Press, 2001, ISBN 9780198508250.

unfortunately, I only have time for one book :(

So... which book do you guys recommend?


let $ x$ , $ y$ , $ z$ $ \in ]0,\plus{}\infty[$ , such that $ xyz\equal{}1$ , prove that :
$ 2\plus{}3(x\plus{}y\plus{}z) \geq xy\plus{}yz\plus{}zx$

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ , $ z$ $ \in ]0, \plus{} \infty[$ , such that $ xyz \equal{} 1$ , prove that :
$ 2 \plus{} 3(x \plus{} y \plus{} z) \geq xy \plus{} yz \plus{} zx$[/quote]
Try
$ x \equal{} 100,y \equal{} 100$ and $ z \equal{} \frac {1}{10000} .$

[hide="Secret"]
But the following inequality is absolutely true.
Let $ a, b, c$ be positive real numbers satisfying $ abc\equal{}1$. We have the property $ ab^2\plus{}bc^2\plus{}ca^2 \geq ab\plus{}bc\plus{}ca$.
On the other hand, we have $ ab^2\plus{}bc^2\plus{}ca^2 \geq a\plus{}b\plus{}c$.[/hide]

[quote="great math"][hide="Secret"]
But the following inequality is absolutely true.
Let $ a, b, c$ be positive real numbers satisfying $ abc \equal{} 1$. We have the property $ ab^2 \plus{} bc^2 \plus{} ca^2 \geq ab \plus{} bc \plus{} ca$.
On the other hand, we have $ ab^2 \plus{} bc^2 \plus{} ca^2 \geq a \plus{} b \plus{} c$.[/hide][/quote]

  Let
\[ a \equal{} \frac {x}{y}, b \equal{} \frac {y}{z}, c \equal{} \frac {z}{x}
\]
and first inequality become
\[ \sum_{cyc}{x^3y^3}\ge \sum_{cyc}{x^3y^2z}
\]
wich is true from Rearrangement's Inequality

   Similarly for \[ ab^2\plus{}bc^2\plus{}ca^2\ge a\plus{}b\plus{}c\]

[quote="great math"][hide="Secret"]
But the following inequality is absolutely true.
Let $ a, b, c$ be positive real numbers satisfying $ abc \equal{} 1$. We have the property $ ab^2 \plus{} bc^2 \plus{} ca^2 \geq ab \plus{} bc \plus{} ca$.
On the other hand, we have $ ab^2 \plus{} bc^2 \plus{} ca^2 \geq a \plus{} b \plus{} c$.[/hide][/quote]

Put $ a\equal{}\frac xz,b\equal{}\frac yx,c\equal{}\frac zy$. Then our inequality is equivalent to $ \sum\frac{x^2}{yz}\ge\sum\frac xy$ which follows from $ \sum x^3\ge \sum xy^2$, true by rearrangement. For the latter inequality substitute $ (a,b,c)\mapsto (c,b,a)$ and $ a\mapsto \frac 1a$ etc in the first inequality and we immediately get the result.

If a,b,c>o such that abc=1  Prove that
$ (a\plus{}b\plus{}c)^2\geq 4(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b})\plus{}3$


for non-negative real numbers like $a,b,c,d$ prove that :

$(1+\frac{2a}{b+c})(1+\frac{2b}{c+d})(1+\frac{2c}{d+a})(1+\frac{2d}{a+b})\ge 9$

Isn't it rather $\geq 9$ ? ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=23543

  darij

gosh ,you didnt even let me correct it ,i clicked the edit putton to put $9$ beside of $1$ and then darij had already answered it ,you are fast man  :D  :D (or i shall say dial up connection is to slow :) )


Find all positive integer $n$ such that there exists a $P\in\mathbb{R}[x]$ satisfying $P(x^{1998}-x^{-1998})=x^{n}-x^{-n}\forall x\in\mathbb{R}-\{0\}$.

That would be $n=1998m$ for some $m\in\mathbb{N}$ odd.

$x=1$ gives $P(0)=0$.
Set $\zeta=e^{\pi i/ 1998}$ a primitive $2\cdot 1998$-th root of unity.
Then $x=\zeta$ gives $0=P(-1-(-1))=\zeta^{n}-\zeta^{-n}$ and $\zeta^{2n}=1$, and so $n=1998m$ for some $m\in\mathbb{N}$.

Replacing $x$ with $x^{-1}$ gives
$P(x^{-1998}-x^{1998})=x^{-n}-x^{n}=-P(x^{1998}-x^{-1998})$
and $P(-x)+P(x)$ has an infinite number of roots, and so $P$ is odd.
If $m$ is even, then replacing $x$ with $\zeta x$ gives
$P(-x^{1998}+x^{-1998})=(\zeta x)^{n}-(\zeta x)^{-n}=x^{n}-x^{-n}=P(x^{1998}-x^{-1998})$.
Then $P(x)-P(-x)$ has an infinite number of roots, and $P$ is even.
Together we get $P(x)=P(-x)=-P(x)$ and $P\equiv 0$. Contradiction!
So $m$ is odd.

Conversely if $n=1998m$ with $m\in\mathbb{N}$ odd, then
$y^{m}+z^{m}\in\mathbb{Z}[y,z]$ is a [url=http://en.wikipedia.org/wiki/Symmetric_polynomial]symmetric polynomial[/url]
and $y^{m}+z^{m}=P(y+z,yz)$ for some $P\in\mathbb{Z}[y,z]$.
And so $(y,z)=(x^{1998},-x^{-1998})$ gives $x^{n}-x^{-n}=P(x^{1998}-x^{-1998},-1)$.

[quote="olorin"] gives $x^{n}+x^{-n}=P(x^{1998}+x^{-1998},1)$.[/quote]

And therefore...? Are you sure?

[quote="N.T.TUAN"]And therefore...? Are you sure?[/quote]
Yeah, saw that. Sorry. Look again.

We assume that $P(x) =$ $\large\sum\ limits_{i=0}^{m}a_{i}.x^{i}$ satisfies the conditions.
Set $k: = 1998$. By the hyposthesis, we have :
           $\large\sum\ limits_{i=0}^{m}a_{i}.(x^{k}-x{}^{-}{}^{k}) = x^{n}-x{}^{-}{}^{n}$
implies
$\large\sum\ limits_{i=0}^{m}a_{i}.x^{n}.(x{}^{2}{}^{k}-1)^{i}.x^{k}^{(}{}^{m}{}^{-}{}^{i}^{)}= x{}^{k}{}^{m}(x{}^{2}{}^{n}-1)$       $(2)$
for all $x$ ≠ $0$.
But DegLhs$(2) = n+2km$
     DegRhs$(2) = km+2n$
so $n = km$.
We will prove that $m$ is odd.
Indeed, we assume that $m$ is even. Set $y = x^{k}$, then :
  $(2)$ ↔ $P(y-$ $\large\frac{1}{y})$ = $y^{m}-$ $\large\frac{1}{y^{m}}$ $(3)$
           ↔ $large\sum\ limits_{i=0}^{m}a_{i}.(y^{2}-1)^{i}.y^{(}m-i) = y{}^{2}{}^{m}-1$
In $(3)$ choose $y = 2$ and $y =$ $\large\frac{-1}{2}$, we give :
    $P($ $\large\frac{3}{2})$ > $0$
    $P($ $\large\frac{3}{2})$ < $0$
contradiction!!
So $m$ is odd.
Now, by introduction, we will prove that if $n = km$, where $m$ is odd, then  there are a polynomial $P_{m}(y)$ with $degP = m$ satisfies our conditions.
Indeed :
 * If $m = 1$, then we choose $P_{1}(y) = y$
 * If $m = 2$, then we choose $P_{2}(y) = y^{3}+3y$
 We assume that we had $P_{1}, P_{2},...,P_{m}$, then we set
       $P_{m}_{+}_{2}= (y^{2}+2).P_{m}(y)-P{}_{m}{}_{-}{}_{2}(y)$
 We prove that polynomial satisfying our conditions.

I'm sorry about my poor English!!!

[quote="N.T.TUAN"][quote="olorin"]and so $P$ is odd.
[/quote]
From here $m$ is odd. Now, you are right![/quote]
What do you mean? I think my proof is correct. :huh:

:D If $P(x)\equiv-P(-x)$ then $\deg{P}$ odd. It is shorts than your solution :P

[quote="N.T.TUAN"]It is shorts than your solution :P[/quote]
Well, not really.
You need to show $\deg P$ odd $\Rightarrow m$ odd,
and therefore make ugly and long calculations like Dtrung.
I tried to avoid that. :P

:oops: I am wrong because I think that $\deg{P}=n$  :(


Hi,

I will be applying for RSI for 2010 and have some questions.

I live in Australia and as such i understand that because i am an international applicant competition will be fierce. I currently have had no competition results besides 2 distinctions in the Australian mathematics contest. I have done no research before besides my own investigations (generalizations).

I have accelerated 2 years of high school mathematics, and will be starting University mathematics next year (while in high school through distance education). Would this aid my application for RSI?

I am considering doing the SAT (for both my application towards university for next year) and for RSI. Will doing the SAT be good for my RSI application?

Could someone tell me what SAT is? When it is? And if i can take it in Australia. Also, whats PSAT and whats the difference?

Would it be possible for me to take the AP Calculus BC exam in Australia? When is the exam?

It is possible for me to pursue research in most likely a discrete field of mathematics eg graph theory and produce some results (as I can consult frequently with a professor in my region) would this boost my chances at RSI?

I dont have any school math clubs or science clubs but their is a chess "club" in the major city in my area that i go to, is this something that would be of value towards my RSI application?

I really really really appreciate every ones time and would be truly great full to anyone whom could help me with any of my questions.

[quote="JRav"][quote="kurt.math"]I am considering doing the SAT (for both my application towards university for next year) and for RSI. Will doing the SAT be good for my RSI application?

Could someone tell me what SAT is? When it is? And if i can take it in Australia. Also, whats PSAT and whats the difference?

Would it be possible for me to take the AP Calculus BC exam in Australia? When is the exam?[/quote]
You want to take all of these exams that you know absolutely nothing about.  The [url=http://www.collegeboard.com/student/index.html?student]College Board[/url] website would be a good start for information on the PSAT, SAT, and AP exams.[/quote]

Thankyou

Does any one have any suggestions about my other questions?  =)

Thankyou

anyone have any ideas about my previous questions?

I can pursue research in most likely a discrete field of mathematics eg graph theory and produce some results (as I can consult frequently with a professor in my region) would this boost my chances at RSI?

I have accelerated 2 years of high school mathematics, and will be starting University mathematics next year (while in high school through distance education). Would this aid my application for RSI?

Whats generally the average SAT score of some of the RSI students?

Because RSI is a research program, doing your own research would definitely be good as it shows RSI that you have experience. Having a strong math and science background will definitely aid in your application, although it is more like a base needed instead of something that will boost your chances.
I would think that the average SAT score is at least above 2100.


algebra
polynomial
function
Vieta
inequalities proposed
inequalities

Let $ a, c, n$ be positive integer numbers, such that $ a + c < n$. Consider the following polynomial
\begin{eqnarray*}
P (x, a, b, n) &=&
n \\
&&- (4 n + 1 + (n - 2) (a + c) - a^2 - c^2)  x \\
&&+ (4 - a^2 (c + 2) - c^2 (a + 2) + n (a c + 2 a + 2 c + 6) - 4 ac - 4 a - 4 c) x^2 \\
&&- (4 n + 6 + (n - 2) (a + c) - a^2 - c^2 - ac) x^3 \\
&&+ (n + 4) x^4 \\
&&- x^5
\end{eqnarray*}

Let $ x_1, x_2, x_3, x_4, x_5$ be the real roots of this polynomial.
Prove that the maximum of $ f = e^{x_1} + e^{x_2} + e^{x_3} + e^{x_4} + e^{x_5}$ is achieved for $ a = c = 1$.

Maybe the right approach would be to consider functions
\[ f (k) \equal{} x_1^k \plus{} x_2^k \plus{} x_3^k \plus{} x_4^k \plus{} x_5^k,\]
for all $ k \geq 1$, and use Vieta's and Newton-Girard formulas.
Using Taylor expansion of $ e^x$ the result follows...

It can be easily shown that $ f (1)$, $ f (2)$ and $ f (3)$ are maximal exactly for $ a\equal{}c\equal{}1$.


Find all the value of x:
$ 3x^2 \plus{} 11x \minus{} 1 \equal{} \sqrt {2x^3 \plus{} 2x^2 \plus{} x \minus{} 1}$
@Hung: Solve it for fun.

This doesn't seem to have "nice" roots. Is there a typo or a point to the question?

The graph will convince you their are no real roots.

Then you just have to prove that:

$ 3x^{2}\plus{}11x\minus{}1 \minus{}(\sqrt{2x^{3}\plus{}2x^{2}\plus{}x\minus{}1} )$ is an increasing function to show that there aren't any real roots.

Using Newtons iteration to the complex plane, I came up with these:

.095842272 + .082196274 i

.095842272 + .082196274 i


For some reason, the PDFs on the AMC and AIME contests are not loading at all; when I click on the "download PDF version" to print it out, it simply reloads the page. Can someone fix this please? I'd like to have the PDFs so I can practice for the upcoming AIME.

Thanks!   :)

Its not been working for a loong time and its been reported multiple times. I'm working on a script to make it printable. :)

Ok, great. Sorry to bug you, then. Thanks!  :)

[quote="now a ranger"]where did you find those pdfs? in wiki?[/quote]

It's in the section linked as "Contests" near the top of every page.

for which contests? I thought they were just latex documents with no pics.  :huh:


integration
calculus
calculus computations

$ E[x] \equal{} \int{xf(x)} dx$.  Let $ f(x) \equal{} \frac {1}{4}e^{ \minus{} \frac {1}{4}x}$, for $ x$ in $ (0,\infty)$

a.  Compute the mean $ (\mu)$ where $ \mu \equal{} E[x]$
b. Find $ E[x^{2}]$
c. Compute the variance $ ({\sigma}^2)$ where $ {\sigma}^2 \equal{} E[(x \minus{} \mu)^2]$.  Expand $ (x \minus{} \mu)^{2}$ first.
d. Compute the skewness of the distribution, $ f(x)$, by computing $ \frac {E[x \minus{} \mu]^3}{\sigma^3}$

For these questions, like b, is it okay just to calculate $ E[x]$ and then substitute in $ x^2$ or do you have to integrate it by substituting in $ x^2$ first.  Also, can someone clarify what c. and d. mean?  Isnt it just plugging it into the equation?

The solution for part [i]a[/i] is a constant, a real number. There is nowhere to plug anything in. The integral is supposed to be a definite integral, with the given limits ("$ x$ in $ (0,\infty)$").

To solve the other parts, replace the first $ x$ in the integral with the thing inside the square brackets. For example, to solve part [i]b[/i], compute $ \int^\infty_0x^2f(x)\; dx$.

[quote="kenn4000"][hide="d"]

$ E[(X \minus{} \mu)^3] \equal{} E[X^3] \minus{} 3\mu E[X^2] \plus{} 3\mu^2E[X] \minus{} \mu^3 \equal{} 384 \minus{} 12(32) \plus{} 12(4) \minus{} 64 \equal{} \minus{} 16$
so the skewness will be $ \minus{} \frac {1}{4}$ [/hide]

integration by parts.. hopefully correctly[/quote]

small calculation error there - $ 3\mu^2E[X] \neq 48$
The skewness of the exponential distribution is always 2

The rest are correct =)


inequalities
limit
three variable inequality

[quote="anas"]hello , let $ (a,b,c) \geq 0$ Prove that : $ \sum{a^3} + 3abc \geq \sum{ ab\sqrt {2(a^2 + b^2)}}$[/quote]

Which obviously wrong. Try $ c=0$.

But we have two nice followed inequality
\[ a^3  + b^3  + c^3  + 3abc \ge a\sqrt {a^2  + 3bc}  + b\sqrt {b^2  + 3ca}  + c\sqrt {c^2  + 3ab}
\]
\[ a\sqrt {a^2  + 3bc}  + b\sqrt {b^2  + 3ca}  + c\sqrt {c^2  + 3ab}  \ge ab\left( {a + b} \right) + bc\left( {b + c} \right) + ca\left( {c + a} \right)
\]
The first one was post by can_hang2007 and the second one was post by Tran Quoc Anh in toanthpt.  :)

[quote="Honey_S"][quote="anas"]hello , let $ (a,b,c) \geq 0$ Prove that : $ \sum{a^3} + 3abc \geq \sum{ ab\sqrt {2(a^2 + b^2)}}$[/quote]

Which obviously wrong. Try $ c = 0$.

But we have two nice followed inequality
\[ a^3 + b^3 + c^3 + 3abc \ge a\sqrt {a^2 + 3bc} + b\sqrt {b^2 + 3ca} + c\sqrt {c^2 + 3ab}
\]

\[ a\sqrt {a^2 + 3bc} + b\sqrt {b^2 + 3ca} + c\sqrt {c^2 + 3ab} \ge ab\left( {a + b} \right) + bc\left( {b + c} \right) + ca\left( {c + a} \right)
\]
The first one was post by can_hang2007 and the second one was post by Tran Quoc Anh in toanthpt.  :)[/quote]
no the ineq is true , with c=0 , the ineq is equivalent to:
$ (a-b)^2(a^4+b^4+2ab(a^2+b^2)+a^2b^2) \geq 0$

[quote="anas"]hello , let $ (a,b,c) \geq 0$ Prove that : $ \sum{a^3} + 3abc \geq \sum{ ab\sqrt {2(a^2 + b^2)}}$[/quote]
apply Cauchy-Schwarz, we have

$ \sum{ ab\sqrt {2(a^2 + b^2)}\le \sqrt {4(\sum (ab)^2)(\sum a^2)}}$

assume $ p = a + b + c = 1$ and setting $ q = ab + bc + ca$ and $ r = abc$

we need prove that $ \sum{a^3} + 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$

$ \Leftrightarrow 1 - 6q + 20r + 5q^2 - 52qr + 36r^2 + 8q^3\ge 0$

by Schur'inequality and AM-GM, we have $ \frac {(4q - 1)(1 - q)}{6}\le r\le \frac {q^2}{3}$

put $ f(r) = 1 - 6q + 20r + 5q^2 - 52qr + 36r^2 + 8q^3$

The first case: $ 20 - 52q + 72r\ge 0$

$ f'(x) = 20 - 52q + 72r\ge 0$

$ \Rightarrow f(x)\ge f( \frac {(4q - 1)(1 - q)}{6}) = \frac {4}{3}(3q - 1)(2q - 1)(2q^2 + 2q - 1)\ge 0$

The second case: $ 20 - 52q + 72r\le 0$

$ f'(x) = 20 - 52q + 72r\le 0$

$ \Rightarrow f(x)\ge f(\frac {q^2}{3}) = \frac {1}{3}(3q - 1)(2q - 1)(2q^2 - 3q + 3)\ge 0$

we have done  :wink: ~> :wink:

[quote="bunhiacovski"]
we need prove that $ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$

[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="arqady"][quote="bunhiacovski"]
we need prove that $ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$

[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.$ :wink:[/quote]
$ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$ is true with $ a\plus{}b\plus{}c\equal{}1$ :wink:

[quote="bunhiacovski"][quote="arqady"][quote="bunhiacovski"]
we need prove that $ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$

[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.$ :wink:[/quote]
$ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$ is true with $ a \plus{} b \plus{} c \equal{} 1$ :wink:[/quote]
OK. Try $ a\equal{}b\equal{}\frac{1}{2}$ and $ c\equal{}0.$ :wink:

[quote="arqady"][quote="anas"]hello , let $ (a,b,c) \geq 0$ Prove that : $ \sum{a^3} + 3abc \geq \sum{ ab\sqrt {2(a^2 + b^2)}}$[/quote]
Here is my proof:
$ \sum_{cyc}(a^3 + abc)\geq\sum_{cyc}ab\sqrt {2(a^2 + b^2)}\Leftrightarrow$ after squaring
$ \Leftrightarrow\sum_{cyc}(a^6 + 2a^3b^3 + 6a^4bc + 3a^2b^2c^2 - 2a^4b^2 - 2a^4c^2)\geq$
$ \geq4\sum_{cyc}a^2bc\sqrt {(a^2 + b^2)(a^2 + c^2)}.$
But $ 2\sqrt {(a^2 + b^2)(a^2 + c^2)}\leq2a^2 + b^2 + c^2.$ Thus, it remains to prove that
$ \sum_{cyc}(a^6 - 2a^4b^2 - 2a^4c^2 + 2a^3b^3 + 2a^4bc - 2a^3b^2c - 2a^3c^2b + 3a^2b^2c^2)\geq0,$
which easy to prove by SOS:
$ \sum_{cyc}(a^6 - 2a^4b^2 - 2a^4c^2 + 2a^3b^3 + 2a^4bc - 2a^3b^2c - 2a^3c^2b + 3a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^6 - 4a^4b^2 - 4a^4c^2 + 4a^3b^3 + 4a^4bc - 4a^3b^2c - 4a^3c^2b + 6a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6 - a^4b^2 - a^2b^4 + b^6) - 3\sum_{cyc}(a^4b^2 - 2a^3b^3 + a^2b^4) -$
$ - \sum_{cyc}(c^3a^3 - c^3a^2b - c^3ab^2 + c^3b^3) + 2\sum_{cyc}(a^4bc - a^3b^2c - b^3a^2c + b^4ac) -$
$ - 3\sum_{cyc}(a^3c^2b - 2a^2b^2c^2 + b^3c^2a)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a - b)^2((a + b)^2(a^2 + b^2) - 3a^2b^2 - c^3(a + b) + 2abc(a + b) - 3abc^2)\geq0.$
Let $ a\geq b\geq c.$ Then $ S_c\geq0,$
$ S_b = (a + c)^2(a^2 + c^2) - 3a^2c^2 - b^3(a + c) + 2abc(a + c) - 3ab^2c\geq0$ and
$ S_b + S_a\geq(a - b)^2(a^2 + b^2 - c^2 + ab + ac + bc)\geq0.$
Id est, $ \sum_{cyc}(a - b)^2S_c\geq S_a(b - c)^2 + S_b(b - c)^2\geq0.$
Done.
We can prove it by Mildorf's inequality:
\[ \frac {(1 + k)(a - b)^2 + 8ab}{4(a + b)}\geq\left(\frac {a^k + b^k}{2}\right)^\frac {1}{k}
\]
for $ a > 0,b > 0$ and $ k\geq - 1,$ $ k\in\mathbb Z$  ( if $ k = 0$ we take $ \lim_{k\rightarrow0}\left(\frac {a^k + b^k}{2}\right)^\frac {1}{k}$ ).
If $ k = 2$ we can prove our ( I think, it's Anh Cuong's ) inequality.[/quote]

Thanks arqady for your solution.  :)

@anas: sorry for my stupid thinking.  :(

[quote="bunhiacovski"]
we need prove that $ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$
ok :blush:  but I don't fine mistake in my proof[/quote]
It is wrong because this ineq is stronger than first ineq -> wrong

[quote="onlylove_math"][quote="bunhiacovski"]
we need prove that $ \sum{a^3} \plus{} 3abc \geq \sqrt {4(\sum (ab)^2)(\sum a^2)}$
ok :blush:  but I don't find mistake in my proof[/quote]
It is wrong because this ineq is stronger than first ineq -> wrong[/quote]
I said that I did't find mistake in my proof. It's mean in my proof where is mistake? :P

[quote="arqady"][quote="anas"]hello , let $ (a,b,c) \geq 0$ Prove that : $ \sum{a^3} + 3abc \geq \sum{ ab\sqrt {2(a^2 + b^2)}}$[/quote]
Here is my proof:
$ \sum_{cyc}(a^3 + abc)\geq\sum_{cyc}ab\sqrt {2(a^2 + b^2)}\Leftrightarrow$ after squaring
$ \Leftrightarrow\sum_{cyc}(a^6 + 2a^3b^3 + 6a^4bc + 3a^2b^2c^2 - 2a^4b^2 - 2a^4c^2)\geq$
$ \geq4\sum_{cyc}a^2bc\sqrt {(a^2 + b^2)(a^2 + c^2)}.$
But $ 2\sqrt {(a^2 + b^2)(a^2 + c^2)}\leq2a^2 + b^2 + c^2.$ Thus, it remains to prove that
$ \sum_{cyc}(a^6 - 2a^4b^2 - 2a^4c^2 + 2a^3b^3 + 2a^4bc - 2a^3b^2c - 2a^3c^2b + 3a^2b^2c^2)\geq0,$
which easy to prove by SOS:
$ \sum_{cyc}(a^6 - 2a^4b^2 - 2a^4c^2 + 2a^3b^3 + 2a^4bc - 2a^3b^2c - 2a^3c^2b + 3a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^6 - 4a^4b^2 - 4a^4c^2 + 4a^3b^3 + 4a^4bc - 4a^3b^2c - 4a^3c^2b + 6a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6 - a^4b^2 - a^2b^4 + b^6) - 3\sum_{cyc}(a^4b^2 - 2a^3b^3 + a^2b^4) -$
$ - \sum_{cyc}(c^3a^3 - c^3a^2b - c^3ab^2 + c^3b^3) + 2\sum_{cyc}(a^4bc - a^3b^2c - b^3a^2c + b^4ac) -$
$ - 3\sum_{cyc}(a^3c^2b - 2a^2b^2c^2 + b^3c^2a)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a - b)^2((a + b)^2(a^2 + b^2) - 3a^2b^2 - c^3(a + b) + 2abc(a + b) - 3abc^2)\geq0.$
Let $ a\geq b\geq c.$ Then $ S_c\geq0,$
$ S_b = (a + c)^2(a^2 + c^2) - 3a^2c^2 - b^3(a + c) + 2abc(a + c) - 3ab^2c\geq0$ and
$ S_b + S_a\geq(a - b)^2(a^2 + b^2 - c^2 + ab + ac + bc)\geq0.$
Id est, $ \sum_{cyc}(a - b)^2S_c\geq S_a(b - c)^2 + S_b(b - c)^2\geq0.$
Done.
We can prove it by Mildorf's inequality:
\[ \frac {(1 + k)(a - b)^2 + 8ab}{4(a + b)}\geq\left(\frac {a^k + b^k}{2}\right)^\frac {1}{k}
\]
for $ a > 0,b > 0$ and $ k\geq - 1,$ $ k\in\mathbb Z$  ( if $ k = 0$ we take $ \lim_{k\rightarrow0}\left(\frac {a^k + b^k}{2}\right)^\frac {1}{k}$ ).
If $ k = 2$ we can prove our ( I think, it's Anh Cuong's ) inequality.[/quote]
nice  :)
here is my proof:
lemma: for any reals $ a ,b$ , we have  $ (\frac {(a + b)^3 - (a - b)^3}{ab})^2 \geq 32(a^2 + b^2)$

proof of our lemma : it's equivalent to $ 4\frac {(a - b)^2(a + b)^2}{a^2} \geq 0$
back to the original inequality:
we assume that : $ c = min(a,b,c)$ , then we need to prove that:
$ a^3+b^3+c^3+3abc \geq \frac{ (a+b)^3-(a-b)^3+(a+c)^3-(a-c)^3+(b+c)^3-(b-c)^3}{4}$
which is equivalent to : $ \frac {(a - b)^2(2a + b - 3c)}{2} \geq 0$ , which is true :)
P.S: what do you think about my proof ?

[quote="anas"]
we assume that : $ c \equal{} min(a,b,c)$ , then we need to prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq \sum{\frac {(a \plus{} b)^3 \minus{} (a \minus{} b)^3}{4}}$[/quote]
I think, it's wrong.
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

Really sorry, my friend. Your proof is trues and my eyes are so bad . Verry sorry  :)

[quote="nguoivn"]
And here is my new proof: PDF-file allegated[/quote]

Sorry, nguoivn I checked your proof and I think there is again a mistake. :maybe:

I think that $ Sc\equal{}a^4 \plus{} b^4 \plus{} 2ab(a^2 \plus{} b^2 \minus{} c^2)\minus{}c^2(a^2\plus{}b^2)$ so $ Sb\equal{}c^4 \plus{} a^4 \plus{} 2ac(a^2 \plus{} c^2 \minus{} b^2)\minus{}b^2(a^2\plus{}c^2)$ can be negative and your proof fails. Am I right?

I don't think so. I also checked my proof and I don't think it was wrong. I only forget to write: assume $ a \geq\ b \geq\ c$
Besides, there are many ways to analyse a expression to SOS form. If you want to check, you can try to expand.
And with your way to analyse:
[quote="manlio"]
I think that $ Sc$ should be $ a^4 \plus{} b^4 \plus{} 2ab(a^2 \plus{} b^2 \minus{} c^2)$ so $ Sb$ cannot be positive and your proof fails. Am I right?[/quote]
With $ a \geq\ b \geq\ c$, $ Sb \geq\ 0$  obviously  :)

Sorry, nguoivn, but for $ a\geq b \geq c$ can you please show me that $ Sb\geq 0$ with

$ Sb\equal{}a^4\plus{}c^4\plus{}2ac(a^2\plus{}c^2\minus{}b^2)\minus{}b^2(a^2\plus{}c^2)$

Thank you very much.

[quote="manlio"]Sorry, nguoivn, but for $ a\geq b \geq c$ can you please show me that $ Sb\geq 0$ with

$ Sb \equal{} a^4 \plus{} c^4 \plus{} 2ac(a^2 \plus{} c^2 \minus{} b^2) \minus{} b^2(a^2 \plus{} c^2)$

Thank you very much.[/quote]
I agree with Manlio that $ S_b$ cannot alway be nonnegative, for example $ a\equal{}b\equal{}1, c\equal{}0.001$.

Oh yes. I was wrong when analyse, although the hint is true:
$ 2(ab\plus{}bc\plus{}ca)\sum\ ab(a^2\plus{}b^2) \leq\ (a^3\plus{}b^3\plus{}c^3\plus{}3abc)^2$
But I miss an positive:$ abc(a^3\plus{}b^3\plus{}c^3\plus{}3abc\minus{}\sum\ ab(a\plus{}b)$.
So, we can consider:$ Sb\equal{}a^4\plus{}c^4\plus{}2ac(a^2\plus{}c^2\minus{}b^2)\minus{}b^2(a^2\plus{}c^2)\plus{}2abc(a\plus{}c\minus{}b)$
Now, $ Sb \geq\ 0$. But $ Sa\plus{}Sb \geq\ 0$ seem not trues.
And $ a^2Sb\plus{}b^2Sa \geq\ 0$ works here  :)

[quote="Honey_S"]But we have two nice followed inequality

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge a^2\sqrt {a^2 \plus{} 3bc} \plus{} b^2\sqrt {b^2 \plus{} 3ca} \plus{} c^2\sqrt {c^2 \plus{} 3ab}$

$ \ge bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right) \plus{} ab\left( {a \plus{} b} \right)$

The first one was post by can_hang2007 and the second one was post by Tran Quoc Anh in toanthpt.  :)[/quote]The first one also see : http://www.mathlinks.ro/viewtopic.php?t=287567
[quote="nguoivn"]Hi, both of them are created by me. See it here:

http://www.maths.vn/forums/showthread.php?t=9244&page=8

But the stronger is so hard  :wink:

$ \sum{a^2\sqrt {a^2 \plus{} 3bc}} \geq\sum\ bc\sqrt {2(b^2 \plus{} c^2)}.$[/quote]An easier approach is

$ a^2\sqrt {a^2 \plus{} 3bc} \plus{} b^2\sqrt {b^2 \plus{} 3ca} \plus{} c^2\sqrt {c^2 \plus{} 3ab}$

$ \geq\frac {5(a^3 \plus{} b^3 \plus{} c^3) \plus{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 15abc}{6}$

$ \geq bc\sqrt {2b^2 \plus{} 2c^2} \plus{} ca\sqrt {2c^2 \plus{} 2a^2} \plus{} ab\sqrt {2a^2 \plus{} 2b^2}.$

see also : http://www.mathlinks.ro/viewtopic.php?t=196374

Actually we have

1) Anas strange solution

2) Arqady SOS solution

3) Can_hang2007 solution

4) Ji Chen first solution on this post

5) Ji Chen second solution at  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1614159#1614159

Dear can-hang2007,

as you said to have 9 solutions for this inequality (it is really incredible so many solutions  :o ) and I cannot imagine nothing of them, can you please share with us, if possible, when you have any time, any other solution different from these I illustrate here (Also only the ideas that support them).

Thank you very much.  :)

[quote="Honey_S"]But we have two nice followed inequality
\[ a^3  \plus{} b^3  \plus{} c^3  \plus{} 3abc \ge a^2\sqrt {a^2  \plus{} 3bc}  \plus{} b^2\sqrt {b^2  \plus{} 3ca}  \plus{} c^2\sqrt {c^2  \plus{} 3ab} \]
\[ a\sqrt {a^2  \plus{} 3bc}  \plus{} b\sqrt {b^2  \plus{} 3ca}  \plus{} c\sqrt {c^2  \plus{} 3ab}  \ge ab\left( {a \plus{} b} \right) \plus{} bc\left( {b \plus{} c} \right) \plus{} ca\left( {c \plus{} a} \right)\]
The first one was post by can_hang2007 and the second one was post by Tran Quoc Anh in toanthpt.  :)[/quote]
The first inequality is equivalent to $\sum \frac a{a+\sqrt{a^2+3bc}}\le 1$.

Actually, we can extend this inequality to 5 variables, as follows.
[i]If $a,b,c,d,e$ are positive numbers such that $abcde=1$, then[/i]
\[\frac 1{2+\sqrt{4+5a}}+ \frac 1{2+\sqrt{4+5b}}+ \frac 1{2+\sqrt{4+5c}}+ \frac 1{2+\sqrt{4+5d}}+ \frac 1{2+\sqrt{4+5e}} \le 1.\]

[quote="Vasc"]Actually, we can extend this inequality to 5 variables, as follows.
[i]If $a,b,c,d,e$ are positive numbers such that $abcde=1$, then[/i]
\[\frac 1{2+\sqrt{4+5a}}+ \frac 1{2+\sqrt{4+5b}}+ \frac 1{2+\sqrt{4+5c}}+ \frac 1{2+\sqrt{4+5d}}+ \frac 1{2+\sqrt{4+5e}} \le 1.\][/quote]
There are nice solutions by AM-GM and induction for the more general results:

Let $n\ge 3$ be a given integer. If $a_1,\,a_2,\, \ldots,\, a_n$ are positive real numbers such that $a_1a_2\cdots a_n=1,$ then
[list=1][*]$\frac{1}{n-1+\sqrt{(n-1)^2+4na_1}}+\cdots +\frac{1}{n-1+\sqrt{(n-1)^2+4na_n}} \le \frac{1}{2}.$
[*] $\frac{1}{\sqrt{(n-1)^2+(2n-1)a_1}}+\cdots +\frac{1}{\sqrt{(n-1)^2+(2n-1)a_n}} \le 1.$[/list]

[quote="can_hang2007"]
Let $n\ge 3$ be a given integer. If $a_1,\,a_2,\, \ldots,\, a_n$ are positive real numbers such that $a_1a_2\cdots a_n=1,$ then
[list=1][*]$\frac{1}{n-1+\sqrt{(n-1)^2+4na_1}}+\cdots +\frac{1}{n-1+\sqrt{(n-1)^2+4na_n}} \le \frac{1}{2}.$
[*] $\frac{1}{\sqrt{(n-1)^2+(2n-1)a_1}}+\cdots +\frac{1}{\sqrt{(n-1)^2+(2n-1)a_n}} \le 1.$[/list][/quote]

You are right, Can. :wink:
The proof is easy by HCF-Theorem.

[quote=anas]hello , let $ (a,b,c) \geq 0$ Prove that : $ \sum{a^3} + 3abc \geq \sum{ ab\sqrt{2(a^2+b^2)}}$[/quote]

[img]http://s15.sinaimg.cn/middle/006ptkjAzy76y393DoW3e&690[/img]

[quote=luofangxiang]
[img]http://s15.sinaimg.cn/middle/006ptkjAzy76y393DoW3e&690[/img][/quote]
But easier.
We need to prove that $\sum_{sym}(2a^5b-a^4b^2-a^3b^3-a^4bc+a^2b^2c^2)\geq0$, which is $f(w^3)\geq0$, where $f$ is a concave function.
Thus, by [url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] it's enough to prove our inequality in the following cases.
1. $c=0$, $b=1$, which gives $a(2a^4-a^3-2a^2-a+2)\geq0$ or $a(a-1)^2(2a^2+3a+2)\geq0$;
2. $b=c=1$, which gives $a^2(a+1)(a-1)^2\geq0$.
Done!


Soon enough I'm going to have to lay down my university preferences. I'm interested in studying computer science, but I've heard that computer science jobs are going overseas (i.e. to India). I'm not sure how the situation is in other countries, but I'm from Australia and would be glad of any feedback.

Do you think CS still has great prospects? Thanks, I appreciate it.

Computer science majors are in very high demand.  You don't have to get a job researching theoretical computer science or anything...  Knowing how to program and being a good problem solver are two attributes that a great deal of jobs value a lot.  You can't go wrong with a compsci major...unless you don't like compsci and are struggling through it just to get the degree but you obviously like compsci so that's not a problem.

I am a computer science student.  I have no problem with the programming classes in high level languages like C++, Java, Visual Basic.net and I enjoy it.  I have concerns over outsourcing though, too.  I don't think it is "that" bad.  I am more concerned with working long shifts, being treated badly by an employer and getting past all the math(College Algebra, Trig, Calc 1, Calc 2, Linear Algebra and Discrete Math - haven't had math for awhile and I never liked doing straight problems not applied to something concrete) and then Assembly language will probably not be fun.  It's a challenging degree program.  I like programming and the pay is supposedly good, so that's why I am pursuing C.S.  It's not my first passion though.

I have no problems with what's being taught in the course, only job prospects once I acquire the degree. I too don't want to work phenomenal shifts either.

My contact at the local university has loaded me with CS propaganda, so I'm fairly convinced that CS job prospects are good.[/i]

Maybe [url=http://www.washingtonpost.com/wp-dyn/content/article/2007/02/23/AR2007022301697.html]this[/url] article will be useful.


Let $ a_n$ be the following sequence:
$ a_0\equal{}1, a_1\equal{}0. a_2\equal{}2$, and $ a_{n\plus{}2}\equal{}2a_{n}\plus{}a_{n\minus{}1}$ for $ n \in N$.

Prove that for every $ m \in N$, there exist 2 consecutive terms ($ a_n$ and $ a_{n\plus{}1}$)  in the sequence such that both of them are divisible by m.

(Source - An Israeli Olympiad from 2005)

$ a_{n\plus{}1}\plus{}a_n\equal{}F_{n\plus{}1}$ (fibonacci)
so $ a_{n\plus{}2}\equal{}a_n\plus{}F_n$ gives $ a_{2k}\equal{}2\plus{}\sum_{i\equal{}2}^{k\minus{}1}F_{2i},a_{2k\plus{}1}\equal{}\sum_{i\equal{}2}^{k\minus{}1}F_{2i\plus{}1}$

[quote="bambaman"]Let $ a_n$ be the following sequence:
$ a_0 = 1, a_1 = 0. a_2 = 2$, and $ a_{n + 2} = 2a_{n} + a_{n - 1}$ for $ n \in N$.

Prove that for every $ m \in N$, there exist 2 consecutive terms ($ a_n$ and $ a_{n + 1}$)  in the sequence such that both of them are divisible by m.

(Source - An Israeli Olympiad from 2005)[/quote]
Here is my solution for this problem .
The first we have $ a_{-1}=0$
Cal $ \{r_n\}_{n=-1 ^{+\infty }}$ is the sequence define by :
\[ i) r_n\in {0,..,m-1\}
}\]
\[ ii ) r_n\equiv a_n (\mod m)
\]
We will prove that $ r_n$ is period from the first term .
Because there exist only finite pairs of $ (r_n,r_{n+1},r_{n+2})$ so exist $ k>n$ such that :
\[ r_{k}=r_n,r_{k+1}=r_{n+1},r_{k+2}=r_{n+2}
\]
Therefore :
\[ r_{k-1}=r_{n-1}
\]
Then easy to prove that
\[ r_{t+(k-n)}=r_k ,\forall t>0
\]
From $ r_1=r_{-1}=0$ then we have exist p such that
\[ r_{p+2}=r_{p}=0
\]
It gives $ r_{p-1}=0$
Pairs $ (a_{p},a_{p-1})$ satisfy condition of  this problem .


algebra
polynomial
induction
function
domain
vector
complex analysis

Let $\{P_{n}\}$ be the polynomial on $[0,1]$ with the highest degree is not greater than $m$($m$ is fixed number),and $\{P_{n}\}$ converges to $f$.Prove that $\{P_{n}\}$ converges to $f$ uniformly.

PS:I am sorry if it has been posted before.

By induction, you can show that $f$ must be a polynomial. Then, using the uniform continuity of the $P_{n}$ and the pointwise convergence at a finite number of points well chosen, the uniform convergence follows ...

Can you show me the detail?How to prove $f$ must be a polynomial?Thanks.

Hi,julien_santini,can you post the full solution?I haven't seen anything so far :blush:

Without loss of generality, we can assume that all the polynomials $P_{n}$ have same degree $N$. We will show by induction on $N$ that whenever $(P_{n}(x))$ converges for at least $N+1$ points $x$, then $(P_{n})$ converges pointwise to a polynomial. This will of course solve the problem. The case $N \leq 0$ is trivial; now assume the result to be true up to the rank $N \geq 0$. Consider $(P_{n})$ a sequence of polynomials having the same degree $N+1$, and such that $(P_{n}(a))$ converges for at least $N+2$ points $a$. For one such $a$, $P_{n}(x)=P_{n}(a)+(x-a)Q_{n}(x)$, where $deg(Q_{n}) \leq N$. Then $(Q_{n}(x))$ converges for at least $N+1$ points $x$ (it possibly diverges at $x=a$), therefore by induction hypothesis $(Q_{n})$ converges pointwise to a
polynomial $Q$. It implies that $(P_{n})$ converges pointwise to a polynomial, which solves the problem.

Can this be solved by complex analysis (by treating $P_{n}$ as polynomial in a bounded domain in $\mathbb C$) using normal family?

That would be a very complicated and strange approach to take. It also doesn't work, since you need an open region to do anything interesting with complex analysis.
We have to start by looking at some finite collection of points to either identify a candidate limit or bound some aspect of our polynomial. My variant:
Choose $N+1$ distinct points, and look at the interpolating basis at those points (polynomials which are 1 at one point and zero elsewhere). By pointwise convergence at these points, the coefficients of $P_{n}$ in this basis are bounded. By the Bolzano-Weierstrass property of finite dimensional real vector spaces, any subsequence of the $P_{n}$ has a convergent subsubsequence (in any norm; use the norm induced by the uniform norm on $C[0,1]$). Since the $P_{n}$ converge pointwise to some $f$, each such uniformly convergent subsubsequence must converge to $f$, and $f$ is in the polynomial vector space. A simple topological lemma due to Urysohn (not Urysohn's lemma) gives that the original sequence $P_{n}$ converges uniformly to $f$, and we are done.

I did learn that lemma in a complex analysis course. The full statement (Urysohn's principle):
Suppose that $X$ is a topological space and $x_{n}\in X$. If every subsequence of the $x_{n}$ has a further subsubsequence convergent to $x$, the $x_{n}$ converge to $x$.
The proof is easy, by considering the contrapositive.


calculus
integration
trigonometry
parameterization
real analysis
real analysis solved

p933 #4
Evaluate the line integral where C is the given curve.

$
\int_C [sin(x)] dx
$
C is the arc of the curve x = y^4 from (1, -1) to (1, 1)

I parametrized C as such:
$x = t^4$
$y = t$

this gives me
$\int^1_{-1} [sin(t^4) \sqrt{16t^6 + 1}] dt$

I tried trig substitution and integration by parts and I'm not getting anywhere.. I know I'm going to kick myself later

[quote="techmonkey4u"]p933 #4
Evaluate the line integral where C is the given curve.

$
\int_C [sin(x)] dx
$
C is the arc of the curve x = y^4 from (1, -1) to (1, 1)
[/quote]
As written, it seems to be an integral of a differential form rather than an integral with respect to the length measure :)

[quote="techmonkey4u"]p933 #4
Evaluate the line integral where C is the given curve.

$
\int_C [sin(x)] dx
$
C is the arc of the curve x = y^4 from (1, -1) to (1, 1)
[/quote]
I think you mean $\int_C [sin(x)] ds$?

Actually, I believe fedja on this one; if a problem in a calculus textbook leads to an impossible integral, you've probably made a mistake. If that $dx$ is intended, the answer should be 0.

yes, of course;  I mean $ds$ NOT $dx$
sorry about the typo

No, you're wrong; the integral is $\int_C \sin x\, dx$. I have Stewart 3rd edition here, in which the problem is #6 for section 14.2, and it is quite clearly $dx$.

Then it is easy...Isn't it?
It is just $dx$.
How do you obtain the complicated integral?

oh I just looked at the book again and it IS $dx$
yikes, sorry
so... what does that mean?  I don't need the jacobian for line integrals?

No, we don't need.

Let $C=C_1+C_2$ where $C_1$ is the part of $C$ from $(1,-1)$ to $(0,0)$, and $C_2$ is from $(0,0)$ to $(1,1)$

The integral is
$\int_{C_1}\sin x dx + \int_{C_2} \sin xdx = \int_1^0 \sin xdx + \int_0^1 \sin xdx= 0$

In parametric form, you would have $\int \sin (x(t))\cdot\frac{dx}{dt}\, dt$. With your parametrization, $t$ would run from $-1$ to $1$, and the integrand would be an odd function.


Forty percent of all glibs are blue. Sixty percent of all blue glibs are smooth. What percent of all glibs are blue and smooth?


AMC
USA(J)MO
USAMO
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Well, I've just thought of a word game off the top of my head. The rules are simple, Somebody will start off with a four-letter word, and other people will either:

1. Change 1 and only 1 letter of the word to make a new word.
2. Add a letter to make it a new word.
3. Delete a letter to make it a new word.

If nobody posts a new word within 2 days, the winner gets n points, where n is the number of words including the starting word in that round. Then, the winner will start off with another four-letter word for everybody. The limits for the words is that it can NOT be an abbreviation for anything, like USA or USAMO, and ABSOLUTELY no cuss words. "Stupid" and "dumb" and that kind of stuff are ok. Also, the maximum length for a word is seven letters, and the minimum is three letters. If you have questions, PM me.

I'll start off with the first word:
[b]Duck[/b]

[b]duct[/b] do you get points for posting a word?

nope, only if you stump people so that nobody enters another word in 2 days. Oh yeah, and also, words in each round can't be repeated.

[b]Dust[/b]

[b]rust[/b]

I suppose we can't repeat words?

yup, no repeated words.
(why did you post 3 times, mewto55555?)
[b]dusty[/b]

ok i cant find any other words so

[b]aunts[/b]

Nice huge bump... Almost half a year. Mods, please lock.

Yes. please keep that in mind. do not revive old threads.

We understand that you're a new member, but revivals like this are extremely annoying. Mods, please lock, and JSGandora, keep in mind that you should only post in threads in which the last post was less than a month ago.