limit
videos
complex numbers
number theory

If anyone needs, I've created a good list of random numbers to use in our daily math life. I would appreciate if anyone could post more random numbers here so that the list could get complete.

Thanks, here it goes:

[u]Random Numbers List[/u]

$2476$
$26597$
$25678$
$378$
$4837894$
$657983645$
$57847$
$501728$
$70988$
$818744$
$75473$
$928274$
$87483$
$0928$

It's kind of small, but I'm still working on it. Please send suggestions.

cheers

$92343$.

 a friend of mine would probably suggest

 $33554433$ inspite of my insistence that it is not random, hru being $2^{2}5+1$ :P  :oops:

sugoi!!

I made a conjecture. called "Yagaron's Conjecture of Random Numbers"

it is the following.

"All Random number can be expressed as a sum of primes numbers"

because the distribuition of the primes numbers is RANDOMIC! subarashi

we can express a certain random number as a sum of another randomic number.

making the limit of $n$ tendind to infinity we get that random numbers is dependent of itself.

so we can make a theorem "we will never know that a number is randomic or not, because it change very very fast, with $v \to  c$"

hey yagaron

good conjecture, but it seems like you are somewhat joking   :(

this is a serious topic, because that random numbers always happen to appear around.

by the way, I took a couple of hours today searching for more random numbers:

$647782$
$65626$
$897070$
$157675$
$29492$
$54874$
$9468$
$3112273$

cheers

The most random number, of course, is 17.

What's the point of this thread? It's so... random. LOL

the random number is the most important type of number of all infinity (positive and negative too, i'm working to expand it to complex numbers ;) )

dude, let's show a secret.
go to
http://www.google.com/intl/en/

type some text and push "I'm Feeling lucky"

it will redirect yourself directly to a Random Page, or RP (for more information read about Google on Wikipedia, for that just find Wikipedia on google)

The Corporation Google has a entire html (jv, cgi, py ,c) made by random numbers and random characters.

Some persons said that this is the complete proof that God or someting similiar does not exist!

Others says that Google is a Evil corporation made to destroy earth.

the software Google Earth has this purpose, made, of course, with random numbers!

dude. Random Numbers is some kind of Demoniac Weapon.

stop this topic now, save the little rabbits

[quote="yagaron"]dude. Random Numbers is some kind of Demoniac Weapon.[/quote]

wow!

err.... being demoniac or not, I'm working hard to find more of them.

here are some that I've just calculated:

$71443$
$092754658$
$44556$
$873216$
$2561$
$156$

the number 9875648736543278674983030  is too big to be a random number.

the previous post is just a spam, oh man!

if we split the number in 4 we have 4 random numbers

987564
873654
327867
498303

suppose we have a random number with $n$ digits, so it's possible to split this number infinitely many times, and each time we will get a random number.

so. we can think that exist a master random number, such as all others random numbers is derived from or integraded etc

this is number theory or random theory?

Those numbers are much too small.
Since we appear to be working in the positives only,

$0\le r \le \infty$

Thus, the average r would be:

$\frac{\infty-0}{2}=\infty$

The average random number is infinity.

Let the set of random numbers be $\text{R}$ where $\text{R}_{n}$ is the $n^{\text{th}}$  random number.
Then the set $\text{R}$ is $\{ \text{R}_{1}\ , \ \text{R}_{2}\ , \ \text{R}_{3}\ ... \ , \ \text{R}_{n}\}$.

And if we additionally restrict $n$ to the set of random numbers $\text{R}$, we get the space $\mathbb{R}^{2}$

:roll: :spam:

I made some useful theorem's about Random Number Theory

1 - The sum of small random numbers is a big random number
${R=\sum_{i=1}^{\infty }r_{i}}$

2 - The last Small Random number is equal to a fraction of the Big Random number
${\lim_{i\to \infty }\, r_{i}=\lim_{i\to \infty }\, \frac{R}{i}=357}$

3 - Laplace and Fourier are friends
${\mathcal{L}_{i}\left[r_{i}\right](s)=\mathcal{F}_{i}\left[r_{i}\right](\omega )}$
${\mathcal{L}_{i}^{-1}\left[r_{i}\right](s)=\mathcal{F}_{i}^{-1}\left[r_{i}\right](\omega )}$

  A further generalization
${\mathcal{L}_{i}^{n}\left[r_{i}\right]( s)=\mathcal{F}_{i}^{n}\left[r_{i}\right]( \omega )}$

the last theorem is so easy to demonstrate that i will hide the solution

[hide]
Solution
[/hide]

Do you people have any friends?

[size=150][color=RED][u][b]TREETHING HAS BEEN MOD-BANNED[/b][/u][/color][/size]

[size=25][color=white]kidding[/color][/size]

Start small dude

$0$
$1$
$2$
$3$
$4$
$5$
$6$
$7$
$8$
$9$

see? Its so easy. And its completely random (you check that the digit frequencise are roughly equal)

[quote="Temperal"]Uh, why'd you post that twice?[/quote]

It's the glitch that happens with my computer  :mad:

Guys, extremely good news!

After 1 year of researches, I've just discovered the Grand-Master-Random-Number.

It's in the Magic Link Below. Keep derivating it to the answer.

[hide="Magic Link Below"]
[hide="Magic Link Below"]
[hide="Magic Link Below"]

Oops, I was so busy hiding the magic link that I forgot the number   :(

[/hide]
[/hide]
[/hide]

This thread reminds me of this video:
[youtube]qjOZtWZ56lc[/youtube]

~lingomaniac-what the heck is taht supposed to be?

my brother says 64. a lot. and 24.

OMG WHAT A REVIVE!!! :clap:  :clap:  :clap:  :clap:  :clap:

[quote=GCampos]10 YEARS!!! :wacko:  :wacko:  :wacko:  :wacko:  :wacko:[/quote]

5 years! :wacko:


limit
trigonometry
calculus
calculus computations

[b](1) limit x-->infinity (x-sinx)/(x+cos^2x) =
here cos^2x means cos square x.[/b]

This is all about size. Focus on the big terms in the numerator and denominator; can we get rid of the other stuff?

$ L \equal{} \lim_{x \to \infty} \frac{x \minus{} \sin x}{x \plus{} \cos ^2 x} \\ \\ \\ \equal{} \lim_{x \to \infty} \frac{1 \minus{} \frac{\sin x}{x}}{1 \plus{} \frac{\cos ^2 x}{x}}$

As $ x \to \infty$, sin (x) and cos (x) take a value between -1 and +1, the graph it oscillates infinitely between -1 and +1 ...

so, $ \lim_{x \to \infty} \frac{\sin x}{x} \equal{} \frac{finite \ value}{infinity} \equal{} 0$

similarly, $ \lim_{x \to \infty} \frac{\cos ^2 x}{x} \equal{} 0$

This gives

$ L \equal{} \frac{1\minus{}0}{1\plus{}0} \equal{} \boxed{1}$


I have been working on this problem for about an hour now and don't know where to go...any help would be appreciate.  I know that one is real and the other two are conjugates.  I think that one might be an "i."

x+y+z=3
x^2+y^2+z^2=3
x^3+y^3+z^3=6

Any help is appreciated.

A cubic equation with roots $x,\ y,\ z$ with respect to $t$ can be obtained $(t-x)(t-y)(t-z)=0\Longleftrightarrow t^{3}-(x+y+z)t^{2}+(xy+yz+zx)t-xyz=0.$
Now find the values of $xy+yz+zx$ and $xyz.$

[quote="Phelpedo"]Are you sure that this system has solutions? Somehow I'm getting that it doesn't:

$(x+y+z)^{2}= x^{2}+y^{2}+z^{2}+2(xy+yz+xz) = 9$

Subtract $x^{2}+y^{2}+z^{2}= 3$ from this; we get

$2xy+2yz+2xz = 6$.

Thus, $2x^{2}+2y^{2}+2z^{2}= 2xy+2yz+2xz$. However, by rearrangement or trivial inequality, $2x^{2}+2y^{2}+2z^{2}\geq 2xy+2yz+2xz$, which equality at $x=y=z$. Thus, it must be the case that $x=y=z=1$, which does not satisfy the third equation.[/quote]

The inequalities only hold for real numbers, complex solutions may not satisfy the trivial inequality.

We can use

$(x+y+z)^{3}=-2(x^{3}+y^{3}+z^{3})+3(x+y+z)(x^{2}+y^{2}+z^{2})+6xyz$

to get the last term in the polynomial.

I thought it was implied that x,y,z were real.

[quote="deakstr"] I know that one is real and the other two are conjugates. [/quote]

Why would this statement be there then... but anyway, someone should solve this.


function
absolute value
irrational number
calculus
calculus computations

Let $ f_{n}: \mathbb{R}\rightarrow\mathbb{R}$ be continuous, and suppose that $ f_{n}$ is uniformly convergent on $ \mathbb{Q}$.  Prove that it is in fact uniformly convergent on $ \mathbb{R}$.

Suppose that $ (x_{n})$ is a sequence such that $ x_{n}\rightarrow x$, with $ x_{n}\in\mathbb{Q}$, $ x\in\mathbb{R}$.  Then $ \forall\varepsilon>0$, $ \exists\delta>0$ such that if $ |x_{\ell}\minus{}x|<\delta$, then $ |f_{n}(x_{\ell})\minus{}f_{n}(x)|<\varepsilon$.  Also, $ \forall\varepsilon>0$, $ \exists N\geq 1$ such that if $ n\geq N$, then $ |f_{n}(x)\minus{}f(x)|<\varepsilon$ $ \forall x\in\mathbb{Q}$.  But $ \exists y\notin\mathbb{Q}$ such that $ x_{i}<y<x_{i\plus{}1}$ $ \forall x_{i}\in (x_{n})$.  Then
$ |f_{m}(x_{\ell})\minus{}f_{n}(y)|\leq |f_{m}(x_{\ell})\minus{}f_{n}(x_{\ell})|\plus{}|f_{n}(x_{\ell})\minus{}f_{n}(y)|<|f_{m}(x_{\ell})\minus{}f_{n}(x_{\ell})|\plus{}|f_{n}(x_{\ell})\minus{}f_{n}(x_{k})|$.

My question arises in this last step.  Am I justified in saying that this is less than $ \varepsilon$?  The first absolute value is just a statement of uniform convergence so I know that's fine.  The second is a statement of continuity.  But must it be uniform continuity?  If so, how do I prove that I have it?  Or is my proof completely wrong anyway?

I'm not sure what you're defining $ y$ for.  It seems like you want to, but do not actually use, the Intermediate Value Theorem.

In any case, it looks like you're trying to apply the Cauchy criterion, but applying the Cauchy criterion requires expressions of the form $ |f_n(x) \minus{} f_m(x)|$, which you don't have - you've changed both the argument and the index.

Try instead the following argument:  consider the sequence $ |f_n(x_k) \minus{} f_m(x_k)|$ in $ k$.  You can get both terms within $ \frac {\epsilon}{3}$ of $ f_n(x), f_m(x)$ respectively, and you can get $ |f_n(x_k) \minus{} f_m(x_k)| < \frac {\epsilon}{3}$ - therefore...?

I assume, in the last part, you're using the same $ (x_{n})$ and $ x$ that I defined?  So $ x$ is an arbtrary real number.

$ \left|f_{n}(x)\minus{}f_{m}(x)\right|\leq\left|f_{n}(x)\minus{}f_{n}(x_{k})\right|\plus{}\left|f_{n}(x_{k})\minus{}f_{m}(x_{k})\right|\plus{}\left|f_{m}(x_{k})\minus{}f_{m}(x)\right|<\frac{\varepsilon}{3}\plus{}\frac{\varepsilon}{3}\plus{}\frac{\varepsilon}{3}\equal{}\varepsilon$
The first part follows from continuity of $ f_{n}$, the second from uniform convergence, and the third again from continuity, yes?

[quote="t0rajir0u"]I'm not sure what you're defining  for. It seems like you want to, but do not actually use, the Intermediate Value Theorem.
In any case, it looks like you're trying to apply the Cauchy criterion, but applying the Cauchy criterion requires expressions of the form $ |f_{n}(x)\minus{}f_{m}(x)|$, which you don't have - you've changed both the argument and the index.[/quote]

I was trying to show that if $ f_{n}(x_{k})$ is getting closer to $ f_{n}(x_{k})$, well, it gets even closer to some irrational number because there's always an irrational in between.  So it's Cauchy on all of $ \mathbb{R}$.  But I guess your method is better.

There is a problem with similar conditions where the family of functions is defined on a bounded open interval and one must infer the uniform continuity on the closure of the interval. The ideas appealed to are the same.


algebra
polynomial
induction
arithmetic sequence
algebra unsolved

Find polynomial $ P \in R[x]$ such that : $ P(x)\plus{}P(1)\equal{}\frac{P(x\plus{}1)\plus{}P(x\minus{}1)}{2}$  for all $ x \in R$

we have for every positive integer $ n$ ,$ P(n)\plus{}P(1)\equal{}\frac{P(n\plus{}1)\plus{}P(n\minus{}1)}{2}$, by induction it 's easily to prove that $ p(n)\equal{}p(1)n^2.$
now cosider the polynominal $ F(x)\equal{}P(x)\minus{}P(1)x^2$ ,it has infinitely many root (every positive integer is a root) , so we must have $ F\equal{}0$, so $ P(x)\equal{}P(1)x^2$ , is the solution to your problem .

and what about P(x)=x-1 ( for to prove P(n)=n²P(1) you must to "assure" P(2)=2²P(1)  :)  )

P'(x)=H(x) ==> H(x)=(H(x+1)+H(x-1))/2

suppose degH=2k#0  so ; it existe a : H(a)=H(a-1) ==> H(a)=H(a+1)

conclusion H(a)=H(a+1)=..H(a+2k) so degH>2k contradiction ==> 2k=0

degP=1;0

if degH=2k+1 ==>degH'=2k and you  can conclur k=0 ==>degP=2

P(x)=0 or P(x)=a(x-1) or P(x)=ax²+b(x-1)

ok , $ P(n) \equal{} P(1)n^2$ is just a specail case ,

now I give  the  full solution to find general term , consider two sequence $ (u_n)$ and $ (v_n)$ defined by $ u_n \equal{} P(n)$ and

$ v_n \equal{} u_n \minus{} u_{n \minus{} 1}$ , we have $ v_{n \plus{} 1} \equal{} 2v_n \minus{} v{n \minus{} 1}$, so $ (v_n)$ is an arithmetic sequence , there exist $ a$ and $ b$ such

that $ v_n \equal{} an \plus{} b$ therefore $ u_n \minus{} u_1 \equal{} v_n \plus{} v_{n \minus{} 1} \plus{} ...v_2 \equal{} a(\frac {n(n \plus{} 1)}{2} \minus{} 1) \plus{} (n \minus{} 1)b$ ,so $ P(n) \equal{} a(\frac {n(n \plus{} 1)}{2} \minus{} 1) \plus{} (n \minus{} 1)b \plus{} P(1)$ compare this with the original equation to find relation between $ a$ and $ b$ and

with the same method (that I wroted in my previous post )  we find all possible solutions.
[quote="mohamed_3ab9or_maths"]
suppose degH=2k#0  so ; it existe a : H(a)=H(a-1) ==> H(a)=H(a+1)
[/quote] can you explain more the existence of $ a$.

[quote="conan_naruto236"]Find polynomial $ P \in R[x]$ such that : $ P(x) \plus{} P(1) \equal{} \frac {P(x \plus{} 1) \plus{} P(x \minus{} 1)}{2}$  for all $ x \in R$[/quote]
An easy problem Let $ q(x) \equal{} P(x) \minus{} P(x \minus{} 1)$ Then we have $ 2P(1) \equal{} q(x \plus{} 1) \minus{} q(x )$ We only find it  in set $ N$
The last is easy for you conan_naruto236.Ok

[quote="tchebytchev"] [quote="mohamed_3ab9or_maths"]
suppose degH=2k#0  so ; it existe a : H(a)=H(a-1) ==> H(a)=H(a+1)
[/quote] can you explain more the existence of $ a$.[/quote]

for all polynomial $ H$ $ degH \equal{} 2k$ (for all $ d$ ) (existe x) $ P(x) \minus{} P(x \plus{} d) \equal{} 0$

voila la demonstration :

$ P(x) \equal{} a_nx^n$ +$ a_{n \minus{} 1}*x^{n \minus{} 1}$..... + $ a0$

$ P(x \plus{} d)$ = $ a_n*(x \plus{} d)^n$ + $ a_{n \minus{} 1}$*$ (x \plus{} d)^{n \minus{} 1}$.........   (for d#0)

          =$ a_nx^n$ + $ a_n*n*d*x^{n \minus{} 1}$ +$ a_{n \minus{} 1}x^{n \minus{} 1}$ + ......... (use the binome the newton )

P(x)-P(x+d)=$ n*d*a_nx^{n \minus{} 1} \plus{} .............$

so $ deg(P(x) \minus{} P(x \plus{} d))$=$ 2k \minus{} 1$ so he has the root


rotation
geometry
geometric transformation
reflection

Prove that for any triangle ABC, if you draw external equilateral triangles (or for that matter, internal) and join the centers of the equilateral triangles, you will find yet another equilateral triangle.

[hide="Solution (sort of hard to follow)"]Let $ \triangle XAB$, $ \triangle ZAC$, and $ \triangle YBC$ be equilateral. Let $ \triangle PQR$ be the outer Napoleonic triangle of $ \triangle ABC$, where $ P$ is the center of $ \triangle XAB$, $ Q$ is the center of $ \triangle YBC$, and $ R$ is the center of $ \triangle ZAC$.

For the sake of simplicity, define $ \text{rot}(K,\theta)$ as a transformation that rotates around $ K$ by $ \theta$, and define $ \text{ref}(l)$ as the transformation that reflects over line $ l$.

We see that $ \text{rot}(P,120)\cdot \text{rot}(Q,120)\cdot \text{rot}(R,120)=I$, where $ I$ is the identity transformation. We also notice that $ \text{rot}^{-1}(P,120)=\text{rot}(P,-120)$. Suppose $ P'$ is the point on the appropriate side of $ \triangle ABC$ such that $ \triangle RQP'$ is equilateral.We have the following: \begin{eqnarray}
\text{rot}(Q,120)\cdot \text{rot}(R,120)&=&\text{rot}(P,-120)\\
\text{rot}(R,120)&=&\text{ref}(RP')\cdot\text{ref}(RQ)
\end{eqnarray}

 Applying $ (2)$ to $ (1)$ gives $ \text{rot}(Q,120)\cdot\text{rot}(R,120)=\text{ref}(P'Q)\cdot\text{ref}(QR)\cdot \text{ref}(QR)\cdot \text{ref}(P'R)=\text{ref}(P'Q)\cdot \text{ref}(P'R)=\text{rot}(P',-120)$.

Therefore, from $ (1)$, $ P'=P$. $ \Box$[/hide]

Here's a property of the outer Napoleonic triangle, if anyone wants to try: Prove that $ AY=BZ=CX$.
[hide="Hint"]Consider rotations by $ 60$.[/hide]

It's also possible to prove them concurrent...


trigonometry
real analysis
real analysis unsolved

Derive the Taylor series expansion of exp(x)/cos(x) and exp(2x)/cos(x) please.

The answer is that there's no nice formula. Dividing by $ \cos$ is bad; even $ \frac1{\cos x}$ has an ugly power series.

I don't care if the formula is "nice" or not... please just show me  :)

This is straight forward if you just need a few terms of the taylor series: perform a longdivision $ (1\plus{}x\plus{}x^2/2!\plus{}\ldots) : (1\plus{}x^2/2!\plus{}\ldots)$


Several positive reals are written on paper. The sum of their pairwise product is 1. Prove that you can cross out one number so that the sum of the remaining numbers is less than $ \sqrt2$

Where are you getting all of these problems that you are posting?

Well, several of them look like from Engel.  I'd imagine they're from books or he wouldn't have so many.

mdk, you should probably realize that many of these problems have been posted before.

[quote="mdk"]few of them are from Engel. Others are from questions I remembered from a while back and others are from a source that I cannot disclose or else I will lose my NBA range for a year.

I don't even have that many problems. I could post a lot more, but then I wouldn't really learn other people's solutions[/quote]

NBA? As in basketball? :blush: Sorry, I'm naive.

And what's pairwise product?

The sum of every possible product of a pair of them.  In other words, $ \sum_{i \neq j}a_{i}a_{j}$.

[quote="mathnerd314"][quote="mdk"]few of them are from Engel. Others are from questions I remembered from a while back and others are from a source that I cannot disclose or else I will lose my NBA range for a year.

I don't even have that many problems. I could post a lot more, but then I wouldn't really learn other people's solutions[/quote]

NBA? As in basketball? :blush: Sorry, I'm naive.
[/quote]

yes, as in basketball


[quote="Fermat -Euler"]$x , y$ are two réels $(x\neq 0 , y\neq 0)$ satisfaies $(x-y)(3x-2y)=2xy$
find   $\frac{x+y}{x-y}$[/quote]

[hide]

$(x-y)(3x-2y)= 3x^2-2xy-3xy+2y^2=2xy$

$3x^2 + 2y= 7xy$

$3x^2 - 7xy + 2y = 0$

$(3x-y)(x-2y)=0$

  Set each as zero, and solve:

$3x-y=0$

$3x=y$

or

$x-2y=0$

$x=2y$

  Plugging in each case for $\frac{x+y}{x-y}$ , yields:

$\frac{x+3x}{-2x}=\frac{4x}{-2x}=-2$

or

$\frac{2y+y}{2y-y}=\frac{3y}{y}=3$

  So our solutions are: $(-2, 3)$

We know that we have fufilled the requirement: $(x\neq 0 , y\neq 0)$ because if either $x$ or $y$ were 0, both would have to be 0.  And $\frac{x+y}{x-y}=\frac{0}{0}=0$  An answer we have not come across...
[/hide]

Just for the record, $\frac{0}{0} \neq 0$ !!!


[quote="nguoivn"]Given $ a, b, c$ are the [b]reals[/b]. Prove that: $ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}$
 :)[/quote]
$ \iff \sum_{cyc}{\frac{(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b)^2}} \leq 9\sum_{cyc}a^2$
$ \iff\sum_{cyc}{\frac{8abc}{a\plus{}b}}\plus{}\sum_{cyc}\frac{4a^2b^2}{(a\plus{}b)^2} \leq 5\sum_{cyc}a^2$
$ \sum_{cyc}{\frac{8abc}{a\plus{}b}} \leq 4\sum_{cyc}a^2$and$ \sum_{cyc}\frac{4a^2b^2}{(a\plus{}b)^2}\leq \sum_{cyc}a^2$

[quote="ye109"]
$ \sum_{cyc}{\frac {8abc}{a \plus{} b}} \leq 4\sum_{cyc}a^2$[/quote]
How did you prove this ineq with $ a, b, c$ are the [b]reals[/b] ?  :maybe:

Your proof was wrong, ye109  :maybe:
[quote="ye109"]
$ \sum_{cyc}{\frac {8abc}{a \plus{} b}} \leq 4\sum_{cyc}a^2$[/quote]
This ineq is not trues with $ a, b, c$ are the [b]reals[/b]. Try $ (a, b, c) \equal{} (1, 1, \minus{}1,000001)$
Wait for another proof  :)

[quote="nguoivn"][quote="nguoivn"]Given $ a, b, c$ are the [b]reals[/b]. Prove that: $ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}$
 :)[/quote]
I just have a proof by classic ineq. Can somebody give me another hint ?  :maybe:[/quote]

We have:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{1}{2} \ge \sum \frac{a}{b\plus{}c}$  :)

[quote="caubetoanhoc94"]

We have:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {1}{2} \ge \sum \frac {a}{b \plus{} c}$  :)[/quote]
This ineq is only trues with $ a, b, c \ge 0$, Dang.
Besides, we also can't multiply both of sides with $ \frac {1}{ab \plus{} bc \plus{} ca}$ because $ ab\plus{}bc\plus{}ca$ isn't always nonegative  :wink:

[quote="nguoivn"]Given $ a, b, c$ are the [b]reals[/b]. Prove that: $ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)^2} \geq\ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}$
 :)[/quote]
The inequality does not hold, nguoivn. For example, you can check the case $ (a,b,c) \equal{} ( \minus{} 2,1,1).$

Indeed, I found my mistake when created it.
Very sorry, it's only trues with $ a, b, c \ge 0$  :(


function
calculus
derivative
real analysis
real analysis unsolved

[quote="dickchimney"][quote="harazi"]Let g be a real function from [0,1] a function with the intermediate value property which is the punctual limit of a sequence of continuous functions. Then there exists a strictly increasing continue function f such that g(f) is the derivative of a function.[/quote]

 Again, does "punctual limit" means "pointwise limit" :D . Sorry because I'm not familiar with that word. :blush: . This must be an interesting problem[/quote]
Yes, you are right! It is "pointwise limit".

Do you have the solution now, harazi?? :D

Well, I thought it was in the unsolved section. Am I blind?  :D


Let be $P$ any (non-self-intersecting) $n$-gon and define the set $T$ as the set of all points in the interior of $P$ from where you can see all vertices of $P$. Prove that $T$ is a convex polygon with less than or equal to $n$ sides.

Looks like the set $T$ is the intersection of $n$ half-spaces.

For each side $s$ of $P$ consider the half-plane determined by the supporting line $\ell_s$ of $s$ which contains points belonging to the interior of $P$ and arbitrarily close to the points of $s$. If I am not mistaken, $T$ is precisely the intersection of these $n$ half-spaces. If this is right, the result follows, since it's easy to prove that any bounded intersection of $n$ half-spaces is a convex polygon with at most $n$ sides.

Well, Darij argued that way, but the jury didn't accept his solution totally,giving 8/12 points (I didn't really understand what their problem was...).
Peter says, that it can be completed by the way.


Once again, thanks again for all for your help in future advance.

I have one troublesome problem - The question reads [b]"Derive an expression for the potential energy V between two parallel dipoles seperated by distance r." [/b]There is a picture beneath the problem which simply shows the two parallel dipoles, one with charge +q1 and -q1 on the sides, and the other with +q2 and -q2 on the sides. The length of the dipole is l, and the distance between the two dipoles, as described in the problem, is r. (the picture sort of looks like parallel plate capacitors, if that helps - eventhough capacitance has nothing to do with the problem.)

I attempted solving the problem, and I believe that the pythagorean theorem is necessary to be used.

If you all could help me out I once again would really appreciate it. Thanks!


geometry
perimeter
calculus
integration
cyclic quadrilateral
area of a triangle
Heron\u0027s formula

Suppose that length of each side of a convex polygonal is a natural number and that its perimeter is odd. Prove that its area is at least $\frac{\sqrt{3}}{4}$.

Suppose the polygon has at least four sides.  Choose four vertices $A, B, C,$ and $D$.  We will deform the quadrilateral $ABCD$, while keeping the rest of the polygon outside of $ABCD$ fixed.  Since the area of the polygon outside of $ABCD$ remains fixed, we seek to minimize the area of $ABCD$.  Note that the area of $ABCD$ is maximized when the vertices form a cyclic quadrilateral, and it gets smaller the further the quadrilateral is from cyclic.  We can stretch or shrink the diagonal $AD$ until stretching or shrinking any further would cause the original polygon to no longer be convex;  the minimum area occurs at one of these extremes.  The boundary occurs when two adjacent sides are exactly parallel, i.e. they form a single side.  So we have transformed a polygon of $n$ sides into one of $n-1$ sides, while decreasing the area.  Continuing in this fashion, we eventually reach a triangle.  But the area of a triangle is given by Heron's formula:

$K = \sqrt{s (s-a)(s-b)(s-c)}$,

and for $a,b,c$ integral with $a+b+c$ odd, we have $s \geq 3/2$ and $s-a,s-b,s-c \geq 1/2$.  So the area is at least $\frac{\sqrt{3}}{4}$.


Im a student 16 years old from greece. I hve some questions about colleges. I havent participate in an I.M.O but still want to go to a very good college. Can i do that?

Most people haven't participated in an IMO, or any other international olympiad for that matter.  I didn't participate in any of these and I think my university (Johns Hopkins) is pretty good (top 15 in the world actually).

So all that it just want is to study hard maths and take a good grade in the uinversity. Oh! something more. Can i participate a college from my early 19s withouht going to a university here in Greece?

I said if i can go to a college without visiting a university in my country.

I'm still not entirely sure what you're asking.  If you want to know if you can go to a university in the US without going to one in Greece first, yes.

If you are lucky you will get some financial aid. But unless you get into certain VERY selective schools, it's going to take a LOT of luck.

[quote="JRav"]College is really expensive.  A top private school in the US costs around 45,000-50,000 a year (US).[/quote]
 :( that is very expensive.....

[quote="Cyrazeno"]Private top colleges... of course expensive

Good luck Γιώργος[/quote]

Thank you Cyrazeno :)

You're welcome  :lol:

If you need more help, go to: [url]http://talk.collegeconfidential.com/[/url].
Should help you out a bit.

Actually, be careful about collegeconfidential.  Yes it has a lot of information, but it could make you even more depressed.

It will depress him only if he allows it to.
Otherwise, it will perhaps motivate him to do more. Give him more of an idea of what these top colleges look for.

[quote="Cyrazeno"]Otherwise, it will perhaps motivate him to do more. [/quote]

But make sure you understand what you're doing. Taking full APs during senior year is [i]not[/i] a good idea, even though top colleges say they're prefer a lot of AP courses.


function
trigonometry
real analysis
real analysis unsolved

Let $f: [0,1]\rightarrow \mathbb{R}$ continuous. We say that $f$ crosses the axis at $x$ if $f(x)=0$ but $\exists y,z \in [x-\epsilon,x+\epsilon]: f(y)<0<f(z)$ for any $\epsilon$.

(a) Give an example of a function that crosses the axis infinitely often.

(b) Can a continuous function cross the axis uncountably often?

For the first part - I guess $f(x) = \sin{\frac{1}{x}}$ works?

It's not continuous in $0$, Arne, but $x\cdot\sin\frac 1x$ does work (if you define it to be $0$ in $0$).

For the second part, the answer is positive. You can find a continuous function which crosses the axis in all the points of the Cantor set, for example. Just define it to be zero in all those points, and something similar to the example above in the open intervals between the points of the Cantor set. I don't like numerical examples :).

Edit:

I sort of remember posting the following problem, inspired by this one:

Characterize those sets $A\subset R$ for which there exists a continuous function $f: \mathbb R\to\mathbb R$ which crosses the axis precisely in the points of $A$.

Hmm- \exist doesn't work, so it looks like a quantifier is missing.

For grobber's variant, closed sets with empty interior should do the trick.

[quote="jmerry"]Hmm- \exist doesn't work, so it looks like a quantifier is missing.[/quote]Has been corrected. :)


Hey guys, I would like some help in this problem:

Simplify the expression:

$ \frac{x^7\plus{}y^7\plus{}z^7}{xyz(x^4\plus{}y^4\plus{}z^4)}$

Knowing that $ x\plus{}y\plus{}z\equal{}0$

[quote="triplebig"]Hey guys, I would like some help in this problem:

Simplify the expression:

$ \frac {x^7 \plus{} y^7 \plus{} z^7}{xyz(x^4 \plus{} y^4 \plus{} z^4)}$

Knowing that $ x \plus{} y \plus{} z \equal{} 0$[/quote]

$ \frac {x^7 \plus{} y^7 \plus{} z^7}{xyz(x^4 \plus{} y^4 \plus{} z^4)} \equal{} \frac {(x^3 \plus{} y^3 \plus{} z^3)(x^4 \plus{} y^4 \plus{} z^4)}{xyz(x^4 \plus{} y^4 \plus{} z^4)} \plus{} \frac{1}{2} \equal{} \frac {(x^3 \plus{} y^3 \plus{} z^3)}{xyz} \plus{} \frac{1}{2} \equal{} \frac {3xyz}{xyz} \plus{} \frac{1}{2} \equal{} \frac{7}{2}$

Hi!

Let $ p(m) \equal{} (m \minus{} x)(m \minus{} y)(m \minus{} z) \equal{} m^3 \minus{} T_1m^2 \plus{} T_2m \minus{} T_3$, where:

i) $ T_1 \equal{} x \plus{} y \plus{} z \equal{} 0$
ii) $ T_2 \equal{} xy \plus{} xz\plus{} yz \equal{} p$
iii) $ T_3 \equal{} xyz \equal{} r$

By Newton's Formulae, we have: $ S_n \equal{} \minus{} pS_{n \minus{} 2} \plus{} rS_{n \minus{} 3}$ (*). In This Formulae, $ S_n \equal{} x^n \plus{} y^n \plus{} z^n$

We now must put some values to n in (*):

. $ n \equal{} 3$: $ S_3 \equal{} \minus{} pT_1 \plus{} rS_0 \equal{} 3r$
. $ n \equal{} 4$: $ S_4 \equal{} \minus{} pS_2 \plus{} rT_1 \equal{} \minus{} p(\minus{}2p) \equal{} 2p^2$, because $ S_2 \equal{} T_1^2 \minus{} 2T_2$
. $ n \equal{} 5$: $ S_5 \equal{} \minus{} pS_3 \plus{} rS_2 \equal{} \minus{} p(3r) \plus{} r(\minus{}2p) \equal{} \minus{}5pr$
. $ n \equal{} 6$: $ S_6 \equal{} \minus{} pS_4 \plus{} rS_3 \equal{} \minus{}p(2p^2) \plus{} r(3r) \equal{} 3r^2 \minus{} 2p^3$
. $ n \equal{} 7$: $ S_7 \equal{} \minus{} pS_5 \plus{} rS_4 \equal{} \minus{}p(\minus{}5pr) \plus{} r(2p^2) \equal{} 7p^2r$

So: $ \frac{x^7\plus{}y^7\plus{}z^7}{xyz(x^4\plus{}y^4\plus{}z^4)} \equal{} \dfrac{S_7}{rS_4} \equal{} \dfrac{7p^2r}{r(2p^2)} \equal{} \dfrac{7}{2}$.

Well, to simplify the problem you want to get rid of the annoying $ x^{4}\plus{}y^{4}\plus{}z^{4}$ in the denominator, so you think of dividing
$ x^{7}\plus{}y^{7}\plus{}z^{7}$ by $ x^{4}\plus{}y^{4}\plus{}z^{4}$ so that you can
express $ x^{7}\plus{}y^{7}\plus{}z^{7}$ as two factors and cancel out
the $ x^{4}\plus{}y^{4}\plus{}z^{4}$ in the denominator.

Well yes, but that transformation seems too simple and it's hard to rearrange it back so easily. I'm thinking that there is an easier way in which he found that out.

[quote="triplebig"]First off, thank you SaYaT and mffmartinelli for your solutions. I understood mffmartinelli's solution, but there is one thing in SaYaT's solution which I don't understand.

[quote="SaYaT"]
$ \frac {x^7 \plus{} y^7 \plus{} z^7}{xyz(x^4 \plus{} y^4 \plus{} z^4)} \equal{} \frac {(x^3 \plus{} y^3 \plus{} z^3)(x^4 \plus{} y^4 \plus{} z^4)}{xyz(x^4 \plus{} y^4 \plus{} z^4)} \plus{} \frac {1}{2}$[/quote]

It is equal and can be verified, but how did you find that equality in the first place? Seems kind of a long shot, is there something to look for?[/quote]

[hide="Index"] At first, i found out expreion $ x^4\plus{}y^4\plus{}z^4$ by squaring $ 2$ times $ x\plus{}y\plus{}z$.
Then i multiplied $ x^3\plus{}y^3\plus{}z^3$ to $ x^4\plus{}y^4\plus{}z^4$ to have $ 7^{th}$ degree and then through this equation i found  $ x^7\plus{}y^7\plus{}z^7$. Then i think everything else is clear.[/hide]

Uhh..How were you able to find $ x^{4}+y^{4}+z^{4}$ by just squaring $ x^{1}\+y^{1}+z^{1}$ twice?  There would be a lot of left over terms...

[quote="Z-coli"]Uhh..How were you able to find $ x^{4} \plus{} y^{4} \plus{} z^{4}$ by just squaring $ x^{1}\ \plus{} y^{1} \plus{} z^{1}$ twice?  There would be a lot of left over terms...[/quote]

 :lol: Not. i'm not "just" squaring(yeah, i didn't mentioned, but i thought it's clear):
first squaring:
$ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2xz\plus{}2yz\equal{}0$
or
$ x^2\plus{}y^2\plus{}z^2\equal{}\minus{}(2xy\plus{}2xz\plus{}2yz)$
Then we do second squaring=)):
we have:
$ x^4\plus{}y^4\plus{}z^4\plus{}2x^2y^2\plus{}2x^2z^2\plus{}2y^2z^2\equal{}4x^2y^2\plus{}4x^2z^2\plus{}4y^2z^2\plus{}8xyz(x\plus{}y\plus{}z)\equal{}4x^2y^2\plus{}4x^2z^2\plus{}4y^2z^2$
or
$ x^4\plus{}y^4\plus{}z^4\equal{}2x^2y^2\plus{}2x^2z^2\plus{}2y^2z^2$
so we able to find this expression=))


We condole with victims on [url=http://news.yahoo.com/s/ap/20080512/ap_on_re_as/china_earthquake]China's Earthquake.[/url]

In 1995, there was a big earthquake; Great Hanshin-Awaji earthquake in Japan.
I remember and was shocked at seeing that the bridges of the expressway had broken down.

Japanese Communities Moderator

kunny

:O My grandparents and a lot of other relatives are in Chengdu...
EDIT: I called my mom, it appears they're all fine, except my grandma's LCD TV broke :( they're pretty expensive in China too :(

Apparently the building structures are not that sound, from what I've heard. Or is that in other places? I was told that one collapse can make huge differences...
And so many students.
I'm glad your family members are safe.  :)

Is it seriously the worst Chinese earthquake in 70 yrs?

Thank you kunny.

The latest statistics in China is that the death number has reached 9219, but the horrible thing is, people still cannot get in touch with 3 towns nearest to the location of the earthquake.

Bless them.

[quote="academicsrule"]
Is it seriously the worst Chinese earthquake in 70 yrs?[/quote]

See here: http://earthquake.usgs.gov/regional/world/most_destructive.php

The Tangshan earthquake is 7.5 magnitude, but it caused a death of 255,000. It is a nightmare for all Chinses. This should be the worst in China.

lol I'm Chinese too, I know about that...
I just knew less about this one, especially since the last time I looked the death toll wasn't exact, right???? But the 70 years thing I read about an hour ago.
But yea, I love my country and I feel really bad for those whose lives were impacted. And I think that what matters is that lives were impacted, whether it's one person or 5 billion, because they're all people and they're all important.

7.9 will knock down even some stronger buildings...I heard (after making the previous post) that my grandma's house had a crack in the wall (and this building was built last year, pretty decent I've been there).

[quote="Kent Merryfield"]On a local television station in Los Angeles, I just saw an interview with Lucy Jones, who's been a spokesperson for USGS in Pasadena for a long time. She thinks the death toll will get much larger than has been reported so far, just because of the size of the population that was subjected to very strong shaking intensity. This is a very large disaster.[/quote]
Indeed. Just as I said, the Death toll 9219 is just a statistic around the epicenter. People have not been able to get in touch with 3 towns of more than 20000 residents (by the end of 2007). The PLA are fighting their way through the dangerous mountains and forrests. Hopefully they will get there before midnight, that is, in 10 hours, to start the rescue operation.

[quote="7h3.D3m0n.117"]I just saw it on the evening news. That's terrible! According to the reports, only one student at one of the high schools were recovered and if anymore aren't found alive, then the entire graduating class is more or less wiped out  :o .

At least the government is willing to accept foreign aid (oops was that outloud?  :D ).[/quote]
Yep...People are saying badly constructed buildings, but being basically right next to the epicenter of a 7.9 doesn't exactly help, even if you have a pretty strong building...
I don't think China will accept foreign aid, as it's a much much stronger country now with a lot more money. That's my guess at least...

Apparently my grandpa in Shanghai felt it...wow that's scary.

For you guys who didn't know btw, Sichuan is China's largest and most populous province. (including Chongqing)

What is on earth, why sacred torch relay was held in such urgent situation? :mad:

It isn't now when  a celeblation event is held on.

I hear that about 1000 pupils were buried alive, it is very sorrowful.

Japanese Government desided to send rescue troop, but Chinese Government don't accept it.

I can't understand the reason at all.

[quote="kunny"]
Japanese Government decided to send rescue troop, but Chinese Government don't accept it. I can't understand the reason at all.[/quote]
Well, China has no shortage in rescue troops, helicopters, medications, etc. of its own. Efficient organization of the operation may be a problem though, especially under the current weather conditions and with roads almost entirely destroyed all around the place. Some news tell that the rescue troops come to the destination and can do virtually nothing because it would take quite a lot of heavy equipment to clean the rubble and there is absolutely no way to deliver it, so they have to resort to digging with spades and bare hands through concrete and stone. I'm also afraid that the politicians may be concerned more about concealing the information than about the fate of people...

As to the torch relay, it is a separate thing altogether. We do not stop power plants, still maintain our regular work schedule, etc. no matter how big a tragedy occurs. The life is going on and there should be something in it to lift the human spirit. So, I do not think that the torch relay (or Olympic games) should be cancelled. The current decision to start it with a minute of silence and to cut back the purely celebrative part seems quite right to me. The alternative is to start an endless period of grief. We care about the earthquake human victims, of course, but if you think a bit about how many people die in senseless local wars, starve, etc. or, if you identify yourself with a bit wider entity than just human race, how many species become extinct every year (and here there is no way back), you'll realize that, following your logic, we shouldn't celebrate anything at any time but start every day with a minute of silence and put all our efforts into the rescue operations. Maybe that would be right, but people won't tolerate such life style for long. So they need this relay, birthday parties and whatever else.

[quote="fedja"]I'm also afraid that the politicians may be concerned more about concealing the information than about the fate of people...
[/quote]
Don't be afraid. This time the information is not concealed as you worried.

Maybe China Government doesn't accept other country's human support partly because there is a place in which foreigners are not allowed to enter for some reaon. When I see the image of collapsed school, we can see holes which should essentially contain ferro, or there are no reinforced concretes.  That would be called ''corner-cutting"". "bean curd lees costruction" in Japanese. this would be man-made disaster.

At any rate, It has been paseed 3 days (72 hours) , I do hope there would be survivor as many as possible. We condole with victims on the disaster and pray for the souls of the dead afresh.

Fortunately Myanmar accepted a human support by some countries, say, Thailand, which makes me a little relief.

kunny why would not accepting foreign aid when you don't need it be bad? It seems to be more of an act of politeness.

Also the concrete blocks...I don't think they cut corners. The pieces of mild steel probably were pulled out when the building collapsed.

[quote="serialk11r"]kunny why would not accepting foreign aid when you don't need it be bad? It seems to be more of an act of politeness.[/quote]

There is no choice to save life more than one as posiible except ''Human-wave sweep'', China should ask for help to U.N. , much less it is a main country of U.N.

[quote="serialk11r"]

Also the concrete blocks...I don't think they cut corners. The pieces of mild steel probably were pulled out when the building collapsed.[/quote]

It is pointed out by a Chinese Architect. It haven't passed 10 years since the school building was built.

In Japanese earthquake's case in Kobe, there was partly cause of the decrepit of the buildings.

By the way, how do you think about the torch relay, as for yourself, serialk11r?

I still don't see how having foreign aid would help much at all. China has a lot of resources, a very well trained military and other forces already on the job, unlike Myanmar.

And about the school, I think it was VERY close to the epicenter, and anything would be damaged pretty badly in that range.

As for the torch relay, I agree with the post up there.

At last, Chinese Government seemed to accept the offer to the rescue man of Japanese Government.
It's too late. Sorry, politeness aside, I dare to say, Which is vital, the face of the country or lives?

To be fair, this was hundreds of times worse than Hurricane Katrina, and we consider THAT disastrous. No offense to China, but I wouldn't be surprised if officials in that part of China (which is supposed to be a lot poorer than Shanghai, Beijing, etc.) would be even less competent than Ray Nagin. This is terrible that so many people have died :(

This thread was originally started to give sympathy to the hundreds of thousands of chinese whose lives have been horribly affected by this disaster.  As a sign of respect to those who have lost loved ones and friends, let's not let this turn into a talk about politics or the chinese government's policies please.

Yes, I agree with Quickster.

I had to listen to a sad news story about the disaster on NPR just this morning, talking about the schoolchildren. It was hard to listen to.

I mourn for the lives who have been affected by this horrendous disaster. I pray for all of them.
I should hope the world would realize this.

Unfortunately, many of my peers are not aware of the disaster. And they are typically the informed ones. I have only come across Chinese individuals who know about this. Is this the case elsewhere? (Sorry is this is a bit off topic--feel free to ignore it if it is.)

Yea at school a lot of people didn't even know about it, including my [b]chinese[/b] bio teacher.

In Japan, the disaster's news is on air every day so that Japanese people are very concerned about it, especially Japan had experienced several big earthquakes , in Kobe and Niigata etc. within 10 years.

All I have heard about it is my Chinese friends asking each other if their families are ok, and my Chinese friends joining some Facebook group with a theme similar to this thread. But I don't know if this is because the Western media is ignoring it, as the media doesn't really penetrate into Princeton campus much during exams. Basically nobody knows what's going on outside unless they hear from friends.

But there was some event here recently in support of relief efforts in Myanmar so maybe soon someone will organize one for this disaster as well. I hope so....

edit: http://news.bbc.co.uk/2/hi/asia-pacific/7399004.stm

[quote="Xevarion"]I don't know if this is because the Western media is ignoring it[/quote]

Ignorance is curable :P.  The New York Times has had at least one article about it on the front page of their website for all of this week.

wow... going near to 50000 died people :(

And, I have read something about a disaster at 1910 or something, at China. There was a really strong earthquake. A side of a mountain collapsed and destroyed a whole city... there were about 250000+ dead people.

There was also a horrible disaster at Colombia. At 1985, the volcano Nevado del Ruiz made eruption, and melt down the snow of the top. Snow, coal and rocks rolled down the mountain, and stuck at the little village of Armero. A lot of people died  :(

And, the famous Tsunami of Java, at December 26, 2006... up to 200000 people died  :(

poor people

my parents (we live in U.S.) and me were so worried about my grandparents in china! we spent like an entire noght just making phone calls. I felt so lucky that all my relatives in china were faaar away from the earthquake, though. I wonder how many people have already died in the most devasted city...?  :(

pacman, I think you might be talking about the Tangshan earthquake of 1976

In times of disaster like this, the world really bands together to help, it's great to see the outpouring of humanity, no matter wat race, religon, nationality, etc.

Chinese Official reports:

The Deth Toll 32476
The Injured   220,000
The buried alive 9500

In 北川(Hoku-Seng), there seems to be anxious about washouting of 21 seismo-lake.

Chinese Governmemt decides to issue food in 500g and 1.5dollars to a person.

are the buried ailve included in that death toll?  I hope yes, but my mind tells me no...

i've seen this picture of a girl who dies while she was crawling out of a building!  :( now i pray for the survivors every day.

A man was rescued 11 hours ago after buried for 179 hours. What a strong life!

Now the death toll is 34073.

I heard that one of the buildings owned by the company my uncle works for collapsed and everyone in it died :(

That's too bad. :(  God bless everyone, victim to the  disaster.


calculus
integration
modular arithmetic
algebra
function
domain
Ring Theory

[quote="fakie6623"]We know in general that a product bc in in an ideal if one of the two factors is in it. We call an ideal I prime if having bc in I forces one of the factors to be in I. Prove that an ideal I is prime precisely when R/I is an integral domain.[/quote]

(I suppose thar $ R$ is a commutative ring.)
Assume $ I$ is prime and let us prove that $ R/I$ is an integral domain.
Let $ A$, $ B$ elements of $ R/I$ such that $ AB \equal{} 0$; we claim that $ A \equal{} 0$ or $ B \equal{} 0$ in $ R/I$. Choose $ a$ in $ A$ and $ b$in $ B$; thus $ A\equal{}a\plus{}I$ and $ B\equal{}b\plus{}I$. By hypothesis, $ (a \plus{} I)(b \plus{} I) \equal{} 0$ in $ R/I$, i.e. $ ab \plus{} I \equal{} 0$ in $ R/I$, i.e. $ ab \equiv 0 \pmod{I}$, i.e. $ ab$ is in $ I$; since $ I$ is prime, $ a$ or $ b$ is in $ I$, i.e. $ a \plus{} I$ or $ b \plus{} I$ is null in $ R/I$, i.e. $ A \equal{} 0$ or $ B \equal{} 0$ in $ R/I$, which is our thesis.
Reciprocally, assume that $ R/I$ is an integral domain and let us prove that $ I$ is prime. Let $ a$ and $ b$ be elements of $ R$ such that $ ab$ is in $ I$, we claim that $ a$ or $ b$ is in $ I$.
From $ ab \in I$ results $ ab \plus{} I \equal{} I$, i.e. $ (a \plus{} I)(b \plus{} I) \equal{} I$, i.e. $ (a \plus{} I)(b \plus{} I)$ is the null element of $ R/I$. Since $ R/I$ is an integral domain, $ a \plus{} I$ or $ b \plus{} I$ is the null element of $ R/I$, i.e. $ a$ or $ b$ is in $ I$, which is our thesis.

More briefly : "$ I$ is prime" means "$ \forall a,b \in R, (ab \in I \Rightarrow (a \in I$ or $ b \in I))$".
 On the other hand, "$ R/I$ is an integral domain means "$ \forall a,b \in R, ((a \plus{} I)(b \plus{} I) \equal{} 0$ in $ R/I \Rightarrow (aI \equal{} 0$ or $ bI \equal{} 0$ in $ R/I))$.
The left members of these two implications are equivalent, and so are the right members, thus the two implications are equivalent.

In fact, this is very very standard.

Friendly,
P.


to tanpham90
 please more clearer I can't understand it completely


geometry
circumcircle
trigonometry
geometry proposed

Take $P$ between $A$ and $M$.
Taking$\angle ADB=\angle BAC=a;\angle ACB=\angle BAD=b;\angle MCA=x;$
$\angle ACP=\angle ADP=y;\angle ADN=m$ and $\angle DCB=z$
We have $\angle ACD=\angle APD=b-z;\angle AND=\angle AMC=a+b$
$\angle NDB=\angle MAB\rightarrow \angle MAC=\angle NDA=m$
$\angle NAB=\angle MCB\rightarrow \angle NAD=\angle MCA=x+y$
and $\angle NAB=\angle PCD\rightarrow \ b-z=x$  $(1)$
So $\frac{AN}{AP}=\frac{AN}{AD}\frac{AD}{AP}=\frac{sin(m)}{sin(a+b)}\frac{sin(b-z)}{sin(y)}$   $(2)$
and $\frac{PM}{AP}=\frac{PM}{PC}\frac{PC}{AP}=\frac{sin(x)}{sin(a+b)}\frac{sin(m)}{sin(y)}$   $(3)$
From $(1);(2)$ and $(3)$ we have $AN=PM$
If $P$ is between $A$ and$N$ we have $PM=AN\rightarrow\ PA+AM=AP+PN \rightarrow\ AM=PN$ and it is similar to when $P$ is between $A$ and $M$

Yeah cefer, this is also my solution, but I send it to Virgil to find a better no trig solution, and I think he has it. Do you Virgil ?

[quote="Virgil Nicula"][quote="cefer"]Take $P$ between $A$ and $M$.
Taking$\ldots\ldots\angle MCA=x\ldots\ldots$
We have $\ldots\ldots\angle MCA=x+y\ldots\ldots$
[/quote]Cefer, I think and I hope is a mistake in at least a letter and you didn't use in proof $B\in CD$ ! If yes, please correct now for after I delete this reply. You write nothing now after me. After I deleted my reply, you can write ![/quote]I think $z=x-y-b$ (on my figure, where the line $CD$ don't separares the points $A$ and $B$).

[hide="Here is two methods (the first is somehow similarly with the Cefer-s proof)."]
Prove rather easily the following equivalencies between angles :
$(0)\ \blacksquare\widehat {ADP}\equiv\widehat {ACP}\blacksquare\widehat {AND}\equiv\widehat {CMP}$$\blacksquare\widehat {CAP}\equiv\widehat {ADN}\blacksquare\boxed {\widehat {MCP}\equiv\widehat {APD}}\equiv\widehat {ACD}\blacksquare\ .$

[b]The first (metrical) method.[/b]

Apply the [u]Sinus' theorem[/u] in the following triangles :

$(1)\ \blacktriangleright\triangle\ CMP\ : \ \frac{MP}{\sin \widehat {MCP}}=\frac{PC}{\sin \widehat {CMP}}\ \blacktriangleleft\ .$

$(2)\ \blacktriangleright\triangle\ ADN\ : \ \frac{AD}{\sin \widehat {AND}}=\frac{AN}{\sin \widehat {ADN}}\ \blacktriangleleft\ .$

$(3)\ \blacktriangleright\triangle\ ACP\ : \ \frac{PC}{\sin \widehat {CAP}}=\frac{PA}{\sin \widehat {ACP}}\ \blacktriangleleft\ .$

$(4)\ \blacktriangleright\triangle\ ADP\ : \ \frac{PA}{\sin \widehat {ADP}}=\frac{AD}{\sin \widehat {APD}}\ \blacktriangleleft\ .$

$(0)$ and the product of the relations $(1)\wedge (2)\wedge (3)\wedge (4)\Longrightarrow \boxed {\ AN=MP\ }\ .$

[u]Remark.[/u] The $\mathrm {\ clue\ }$ of this problem is the four equivalence $\widehat {MCP}\equiv\widehat {APD}$. Indeed,
$\widehat {MCP}+\widehat {ACP}\equiv\widehat {DAN}\equiv\widehat {DCP}\equiv\widehat {ACD}+\widehat {ACP}\Longrightarrow$ $\widehat {MCP}\equiv \widehat {ACD}\equiv\widehat {APD}\ .$

[b]The second (syntetical) method.[/b]

Denote $MP=x$, $PA=y$, $AN=z$. From the equivalencies $(0)$ results :
$\blacktriangleright\ \triangle AND\sim \triangle CMA\Longleftrightarrow \frac{AN}{CM}=\frac{ND}{MA}=\frac{AD}{AC}\Longrightarrow AM\cdot AN=CM\cdot DN\ .$
$\blacktriangleright\ \triangle CMP\sim \triangle PND\Longleftrightarrow \frac{CM}{PN}=\frac{MP}{ND}=\frac{CP}{DP}\Longrightarrow PM\cdot PN=CM\cdot DN\ .$
Therefore, $AM\cdot AN=PM\cdot PN\Longleftrightarrow z(x+y)=x(y+z)\Longleftrightarrow x=z\Longleftrightarrow \boxed {\ AN=MP\ }\ .$

[u]Remark.[/u] $\boxed {\mathrm {\ The\ second\ proof\ made\ me\ happy\ }}$
[/hide]

In the last reply I added the second (syntetical) method.

Dear VIRGIL I made a letter mistake in my first prove :blush: .I mean to take $\angle MCP=x$
And when $AB$ don't intersect $CD$ we have $\angle APD=b+z$
and $\angle NAD=\angle PCD\rightarrow\ b+z+y=x+y$ so $sin(b+z)=sin(x)$
$\frac{AN}{AP}=\frac{AN}{AD}\frac{AD}{AP}=\frac{sin(m)}{sin(a+b)}\frac{sin(b+z)}{sin(y)}$
and $\frac{PM}{AP}=\frac{PM}{PC}\frac{PC}{AP}=\frac{sin(x)}{sin(a+b)}\frac{sin(m)}{sin(y)}$
And it is similar when $B\in C$  :oops:

Extremely nice synthentic proof Virgil. Congratulations :D

[quote="Virgil Nicula"]Given are two triangles $ABC$, $ABD$ so that: the line $AB$ separates the points $C$, $D$; $\widehat {BAC}\equiv \widehat {BDA}$; $\widehat {BCA}\equiv \widehat {BAD}$. Denote the circumcircles $w_1$, $w_2$, $w_3$ of the triangles $ABC$, $ABD$, $ACD$ respectively. For a $\boxed {\mathrm {mobile\ circle\ w}\ ,\ A\in w}$ which don't separate the points $C$, $D$ denote the second intersections $M$, $N$, $P$ of the $\boxed {\mathrm {circle}\ w}$ with the circles $w_1$, $w_2$, $w_3$ respectively. Prove that $PM=AN$.[/quote][color=darkred]I suggest you (Silouan, Cefer a.s.o.) a research them : if instead of a mobile line $d,\ A\in d$ we will consider a mobile circle $w,\ A\in w\ ?$ See about (in "quote") this new hypothetical proposed problem. What are the your opinions ? In my opinion, it is worth looking, even if we add something to the hypothesis !.
[hide="I wish you every success !"][/color]I hadn't the time and the your young mind (or imagination). I am a retired teacher who loves and respect the young mathmen and their universal language Math (U.L.M.). I wish you every success ![/hide]

I think that the problem you proposed is much more difficult than the initial one.

BTW, I find a nice solution for the initial problem
Extend CM to meet $C_2$ at E. Then it's easy to show (i)MNDE is an isosceles trapeizum, (ii) $\triangle ACM\sim\triangle DAN$.
From this, we obtain $AM\cdot AN=DN\cdot CM=ME\cdot CM=AM\cdot MP$. The conclusion follows

Silouan,
 which do you mean as <<  initial proplem >> ?. I just want to study your solution and I want to make the figure.

  [u]Babis[/u]

This problem i meant mr Babis .

[quote="Virgil Nicula"]Given are two triangles $ABC$, $ABD$ so that: the line $AB$ separates the points $C$, $D$; $\widehat {BAC}\equiv \widehat {BDA}$; $\widehat {BCA}\equiv \widehat {BAD}$. Denote the circumcircles $w_1$, $w_2$, $w_3$ of the triangles $ABC$, $ABD$, $ACD$ respectively. For a mobile line $d,\ A\in d$ which don't separate the points $C$, $D$ denote the second intersections $M$, $N$, $P$ of the line $d$ with the circles $w_1$, $w_2$, $w_3$ respectively. Prove that $PM=AN$.[/quote]

Also, for the initial problem:
$\triangle PND \sim \triangle PCM$: $\frac{PN}{PD}=\frac{CM}{PC}$
$\triangle ACM \sim \triangle CPD$:$\frac{AM}{PD}=\frac{CM}{PC}$
So: $AM=PN$ i.e. $AN=PM$

This beautifull problem was given in a  recent team sellection test in Romany , but formulated in a little different way. If you are looking for << Who has it reformulated ?>> , his name is VIRNIC !

 babis


Let a,b,c >0. Prove that
$\sum_{cyc}\sqrt{\frac{a}{a+8b}}\ge 1$

I think this problem is very simple.
It 's a problem of IMO 2001 in Korea.
 $\sum_{cyc}$  ${ \frac{x}{ \sqrt{x^{2}+8yz}}\geq 0}$.

I beleive IMO 2001 was not in Korea But in USA


limit
logarithms
Euler
number theory open
number theory

Have we already got a proof for ${\gamma = \lim_{n\to+\infty}\left\lgroup 1+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{n}-\ln n \right\rgroup }$ is a transendental number or not?

The Euler-Mascheroni constant $\gamma$ is not known to be irrational/transcendental.

Right. But I would be  :o if $\gamma$ turned out to be a rational number.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Hi Virgil Nicula.
Please explain me why $ \int_{0}^\infty{f(u)}du$ would be easier to calculate than $ \int_{0}^{1}f(u)du$ (with the notation of my post).


calculus
integration
logarithms
trigonometry
calculus computations

Express $\frac{x^{2}}{x^{2}+2}$ in the form $a-\frac{b}{x^{2}+2}$ where $a,b\in\Re$

By considering this result, but without performing any further division, find

$\int{\frac{x^{3}}{x^{2}+2}}\,dx$

$\int\frac{x^{3}}{x^{2}+2}\,dx = \int x-\frac{2x}{x^{2}+2}\,dx = \frac{x^{2}}{2}-\ln (x^{2}+2)$

This is a pretty nice problem.  Is there any other way other than factoring out an $x$ or $x^{2}$ to begin with?

To be honest i just dreamt up this idea whilst doodling, and toying with $\int{\frac{x^{2}}{x^{2}+2}}\,dx$, which I was unable to progress greatly on...and yes, your answer is correct  :blush:

EDIT: Doh! Of course: inverse trig comes into that integral  :blush:  :blush:


Each of the first eight prime numbers is placed in a bowl. Two primes are drawn without replacement. What is the probability, expressed as a common fraction, that the sum of the two numbers drawn is a prime number?

The only way we can get a prime number here is $ 2\plus{}p$ for some odd prime $ p$. It turns out that only $ p\equal{}\{3,5,11,17\}$ work, for a probability of $ \frac4{\binom82}\equal{}\frac4{28}\equal{}\boxed{\frac17}$.


For every $\Delta ABC$.Prove that:\[ \frac{h_b}{m_c}+\frac{h_c}{m_b}\le\frac{4}{\sqrt{3}} \]

I remember that this problem was post in the forum


function
vector
LaTeX
trigonometry
geometry
geometric transformation
rotation

Hi

I wasnt sure wheres that topic belong but I hope here is appropriate.

How to prove if the following function is convex (strict or self concordant) ?

The Ai's are input vectors but that shouldnt matter I believe.

[url=http://img177.imageshack.us/my.php?image=helpkk7.jpg][img]http://img177.imageshack.us/img177/2345/helpkk7.th.jpg[/img][/url]

Sorry for the image, but im having hard time with latex :).

Thanks!

Show that each term in the sum is convex by showing that $\frac12t^{2}+\sin t$ is a convex function of one variable.

Are you sure? because it looks pretty much dependant on the coefficients (a_i's).

So basicly if some of them are negative, we will get after deriving twice a function that is < 0 for all x (or t).

Thanks.

The $a_{i}$ don't work that way; they only create rotations in the domain, since we're looking at functions of $\langle a_{i},x\rangle$. A convex function is still convex after we apply a linear transformation to the domain.
Each term is strictly convex in one direction and flat in all orthogonal directions.

One small thing though.

Did you mean to show $\frac{1}{2}t^{2}-\sin{t}$ is convex?

(notice its $-\sin{t}$ and not $+$).

Yeah, that's it. It was just a typo on my part.


geometry
rectangle
rotation
trigonometry
analytic geometry
geometric transformation

[quote="lingomaniac88"]If you're given a point $P$ in rectangle $ABCD$ along with the measures of $\overline{PA}$, $\overline{PB}$, $\overline{PC}$, and $\overline{PD}$, how do you find the area of the rectangle?[/quote]

I know it is possible in a square, but I don't know for a rectangle.

Wow, but we don't uniquely know the coordinates of a, b, c, and d, do we?

I'm pretty sure I supplied them above...

wlog assume the sides of the rectangle are parallel to the axes.

When you solve it out, you'll get a few distinct but equivalent results; the distinct results come from all the squaring you do with the distance formula.

[quote="The Zuton Force"]Let P be at the origin. Let point A be $(a+c,b+d)$, point B be $(a+c,-b)$, point C be $(-a,-b)$, and point D be $(-a, b+d)$.

You can now use the distance formula for each of these cases and set them equal to your known values of $PA,PB,PC,PD$ and solve the system of 4 equations in 4 variables for $a,b,c,d$.

Now your area is just $(2a+c)(2b+d)$.

 :first:

(you're welcome :rotfl: )[/quote]

But the problem is that any three of the equations imply the fourth equation.  So, you technically only have 3 equations to do and hence you can't assume you know the position.

Well, true, but no one likes taking a long time.

[quote="Altheman"]you rotate 90 degrees about B...and then you are like omg there is a triangle with sides PA, PB\sqrt{2} and PC, and you get the cosine of the angle opposite PA, cos x, then you get cos (x+45), and use that for triangle ABP

[also note for any rectangle, AP^2+CP^2=BP^2+DP^2, this is the british flag theorem, proven by coordinates][/quote]

Can you explain this a bit cleaner.....

I can't see the triangle ABP.  :maybe:

[quote="mathgeniuse^ln(x)"]First, in a square draw the diagonal.

Then whichever right triangle has three segments within it, take that triangle and rotate it 90 degrees.

Then, by rotation the angle formed by P'BP, where P is the image, is 90 degrees.

Then, this also means that P'B=PB, so this means that the angle is 45 degrees.

Then Altheman continues.[/quote]

But does Altheman's method also work in a rectangle, too?

I don't think so, but that is me saying it.

Right, there's not enough information to find the area of the rectangle.

[quote="1234567890"]OK, thanks. I was trying use Altheman's method for any rectangle, so that's why I messed up.[/quote]

my method cannot work for an arbitrary rectangle because the rotation does not map one side to another...


Hi all!

Find all the integers n such that $2^{n} + 1$ can be divisible by 3.

:)

Yeah, if you say so.. :P
But, how to prove it? :)

We want $2^n + 1 \equiv 0 \text{ mod } 3 \Rightarrow 2^n \equiv 2 \text{ mod } 3$.

This works for any odd $n$ and can be proven inductively:

$2^1 = 2 \equiv 2 \text{ mod } 3$

$2^{n+2} = 2^2 \cdot 2^n \equiv 1\cdot 2 = 2 \text{ mod } 3$

[hide]We will make use of the binomial theorem:

$2^n + 1 = (2 \mod 3)^n -1 \equiv$
$2 \mod 3 - 1 \equiv 0 \mod 3$ for odd n
$1 \mod 3 + 1 \equiv 1 mod 3$ for even n

So all odd $n, 2^n + 1$ is divisible by $3$[/hide]

Okay, I didn't feel like writing out the solution nicely, but you'll get the gist of it.

Yeah, that works :)
I particularly like Jimmy's one

I had it done the same way as you 1234567890.


x,y,z >0, xy+yz+xz=1.Prove that
$ \sum\limits_{cyc}^{x,y,z} {\frac{x}{{1 \plus{} x^2 }} \ge \sum\limits_{cyc}^{x,y,z} {\frac{{2x(1 \minus{} x^2 )}}{{(1 \plus{} x^2 )^2 }}} }$ :roll:

[quote="TRAN THAI HUNG"]x,y,z >0, xy+yz+xz=1.Prove that
$ \sum\limits_{cyc}^{x,y,z} {\frac {x}{{1 \plus{} x^2 }} \ge \sum\limits_{cyc}^{x,y,z} {\frac {{2x(1 \minus{} x^2 )}}{{(1 \plus{} x^2 )^2 }}} }$ :roll:[/quote]
See here:

http://www.mathlinks.ro/viewtopic.php?t=222187


from:
http://www.mathlinks.ro/viewtopic.php?t=305793&sid=c9b873f788d8c6e189588dba802313fb

Let $ {x_1,x_2,x_3,x_4,x_5}\in {R},x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \equal{} 0,$ prove that

     $ {\frac { \minus{} \sqrt {5} \minus{} 1}{4}}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2 \plus{} x_5^2}}\leq{\frac {\sqrt {5} \minus{} 1}{4}},$

in which  $ {\frac { \minus{} \sqrt {5} \minus{} 1}{4}},{\frac { \sqrt {5} \minus{} 1}{4}}$ are the best coefficient .

[quote="can_hang2007"][quote="fjwxcsl"]from:
http://www.mathlinks.ro/viewtopic.php?t=305793&sid=c9b873f788d8c6e189588dba802313fb

Let $ {x_1,x_2,x_3,x_4,x_5}\in {R},x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \equal{} 0,$ prove that

     $ {\frac { \minus{} \sqrt {5} \minus{} 1}{4}}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2 \plus{} x_5^2}}\leq{\frac {\sqrt {5} \minus{} 1}{4}},$

in which  $ {\frac { \minus{} \sqrt {5} \minus{} 1}{4}},{\frac {\sqrt {5} \minus{} 1}{4}}$ are the best coefficient .[/quote]
Very nice your inequality, fjwxcsl. I think we can prove this inequality by AM-GM, can someone figure it out?[/quote]

We have also the following best coefficient inequality:

 $ { \minus{} \frac {2}{3}}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2}}\leq{0}(x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}0),$

  $ { \minus{} 1}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_6 \plus{} x_6x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2 \plus{} x_5^2 \plus{} x_6^2}}\leq{\frac { 1}{2}}(x_1\plus{}...\plus{}x_6\equal{}0),$

  $ .........$

Looking forward to your wonderful proof !

[quote="fjwxcsl"][quote="can_hang2007"][quote="fjwxcsl"]from:
http://www.mathlinks.ro/viewtopic.php?t=305793&sid=c9b873f788d8c6e189588dba802313fb

Let $ {x_1,x_2,x_3,x_4,x_5}\in {R},x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \equal{} 0,$ prove that

     $ {\frac { \minus{} \sqrt {5} \minus{} 1}{4}}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2 \plus{} x_5^2}}\leq{\frac {\sqrt {5} \minus{} 1}{4}},$

in which  $ {\frac { \minus{} \sqrt {5} \minus{} 1}{4}},{\frac {\sqrt {5} \minus{} 1}{4}}$ are the best coefficient .[/quote]
Very nice your inequality, fjwxcsl. I think we can prove this inequality by AM-GM, can someone figure it out?[/quote]

We have also the following best coefficient inequality:

 $ { \minus{} \frac {2}{3}}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2}}\leq{0}(x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 0),$

  $ { \minus{} 1}\leq{\frac {x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_6 \plus{} x_6x_1}{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} x_4^2 \plus{} x_5^2 \plus{} x_6^2}}\leq{\frac { 1}{2}}(x_1 \plus{} ... \plus{} x_6 \equal{} 0),$

  $ .........$

Looking forward to your wonderful proof ![/quote]
The case $ n\equal{}6$, I have found a proof by AM-GM, too. I will post it in some days. I also found that the general ineuality is known as Ky Fan's Inequality, but in this case, I only know a proof by using Lagrange Multiplier.

[quote="gb2124"]Dear fjwxcsl
Can you show us your nice proof?
It seems that your method can generalize the problem to n varibles.[/quote]

Sorry, we are trying to build a new  algebra inequality mechanical proving algorithm, which can be read  , but  still not perfect...

[quote="Vasc"]I found in my math. research material the following.

If
$ E_n(k) \equal{} \sum x_1^2 \plus{} k\sum x_1x_2 \minus{} \frac 1{n}(1 \plus{} k)(\sum x_1)^2$,
then

$ 20E_5(k) \equal{} \sum [2(x_1 \minus{} x_4) \plus{} k(x_2 \minus{} x_3)]^2\ge 0, \ for \ \ k \equal{} \minus{} 1\pm \sqrt 5\ ;$

$ 3E_6( \minus{} 2) \equal{} \frac 3{2}(x_1 \minus{} x_2 \plus{} x_3 \minus{} x_4 \plus{} x_5 \minus{} x_6)^2 \plus{} (x_1 \minus{} x_3 \plus{} x_4 \minus{} x_6)^2 \plus{} (x_2 \minus{} x_4 \plus{} x_5 \minus{} x_1)^2 \plus{} (x_3 \minus{} x_5 \plus{} x_6 \minus{} x_2)^2 \ge 0\ ;$

$ 8E_8( \minus{} \sqrt 2) \equal{} (1 \plus{} \sqrt 2)(x_1 \minus{} x_2 \plus{} x_3 \minus{} x_4 \plus{} x_5 \minus{} x_6 \plus{} x_7 \minus{} x_8)^2 \plus{} \sum [x_1 \minus{} x_5 \minus{} \sqrt 2(x_2 \minus{} x_4)]^2\ge 0\ .$[/quote]
Very nice your identities, Mr Vasile. I really like them, I will try to figure out the way you established them.


It's new only in the sense that it hasn't been mentioned on this
list before, but the 10th annual internet high school math
competition will begin October 31st.

The competition is open to students in Martin, Palm Beach
and Broward Counties in Florida.

More details are available at
http://www.geocities.com/stuyfla/mathcontest.htm

Couldn't you open it up to Seminole and Orange counties (in Orlando) too... :roll: ?

I'm j/k, I wonder if there's anything like that in Orlando.

Darn. South Florida gets to have all the fun.

Miami-Dade and Broward counties should merge...

Anyways, isn't registration closed, as of October 4th?

Jason, your county has the Essner though and besides Miami-Dade is huge.

Inteluser asks:  Anyways, isn't registration closed, as of October 4th?

Registration is now open and will remain open until the end of the
contest, on December 4.

But the questions won't be available until October 31

And anyone who registers on December 4 had better be able
to solve those problems quickly.


can someone come up with a riddle?

u cant think of any...

how about... oh well...

OUR school ends this friday, 6-11.  we get to go to great a. mon.!

when's ur school ending?

some people's school, such as mine, have ended 3-4 weeks ago.

Mine ended on the 28th of May I think.  Though I start summer school tomarrow and that will end on July 2nd.

BHorseMath, ur school was out on the May 27th, mine was out on the May 28th

ehh I knew it was one of the 2 wishful thinking I guess.

[quote="beta"]Mine ends in June 18th, still 2 more weeks of school[/quote]
lol, ur school ends on my birthday

eh...how bout options like "the month before" or "2 weeks after"?...mine ended mid-may...

Mine ended the 27th of May, so I've been on break for a bit over a week now.

Mine ends on June 11th but my last day will be June 10th due to being exempt from a couple of exams...

For all these schools that end so early...when did they start? I've never heard of any school ending as early as mid-May; most schools around here end in early/mid-June. This year my school went from September 10 to June 3. Is September 10 relatively late for starting the year?

Mine started August 25 and went to June 4th.

i find it hard to be bored in summer. you could go to the beach, go camping, play a ton of sports, swim a bit, go hiking or running, and so much other cool stuff! of course, i'm a really outdoors type of person, but there should be a ton of stuff for everyone! The only thing i regret is not being able to go hiking in new hampshire this year...

[quote="bleumoose"]is there not a scout camp that you can go to? in middle school, i would go to girl scout camp for three weeks +.[/quote]
the problem is, and please correct me if i'm wrong, most camps have already closed registration now. in my town, you've got to sign up for camps by the end of april or something like that. it varies from camp to camp though

thats true. but there are still many volunteer ops EVERYWHERE, and i can think of very few places that will turn you away from volunteering because you registered too late. also, there are a lot of places that will take <14 kids.

I have gone to scout camps.  Unfortunatly the ones near here have horses and I can't stand to see them being mistreated like they are.  I complained and they just said don't come back.  I work with a lot of places during the summer though.

if you like a sport, something you could do is referee or umpire it! i've been umpiring baseball during the spring. getting a job is usually pretty interesting. too bad the ice cream parlor half a mile from my house isn't hiring...

[quote="BHorseMath"]Mathcamp....that would be a dream come true.  But my parents won't let me go to them or even any camp really.  So schools the best thing I love it its what I'm good at.[/quote]
[quote="BHorseMath"]I'm an outdoors person too, but I can't go anywhere because my parents won't take me anywhere. I can't even go in our front yard to play alone. I am an equestrian but we have to drice 45 minutes to get to the stable where my horse is so I only get out there once or twice a week.[/quote]
It's sad how so many people's parents restrict their children's options instead of broadening them.

[quote="nr1337"]I want to go to MathPath, but it is WAY to expensive.[/quote]
I know that Mathcamp offers discounts, scholarships, and fund raising opportunities. You should check whether MathPath does the same?

I'll try next year.  I think I'm one month too young to go this year anyway.

[quote="Yasha"][quote="BHorseMath"]Mathcamp....that would be a dream come true.  But my parents won't let me go to them or even any camp really.  So schools the best thing I love it its what I'm good at.[/quote]
[quote="BHorseMath"]I'm an outdoors person too, but I can't go anywhere because my parents won't take me anywhere. I can't even go in our front yard to play alone. I am an equestrian but we have to drice 45 minutes to get to the stable where my horse is so I only get out there once or twice a week.[/quote]
It's sad how so many people's parents restrict their children's options instead of broadening them.[/quote]
yeah, like sending you to NC! :evil:

[quote="Yasha"][quote="yif man12"]yeah, like sending you to NC! :evil:[/quote]

Well, staying in CT would have restricted my mom's job options, so you have to weigh in everything, not just yourself. :-)

But saying you can't go to a summer camp for a reason other than money is just bad. And Mathcamp gives enough scholarships and fund raising that even money wouldn't be a good excuse.[/quote]
lol, i know. i was just kidding. anyways, it looks like our team's not even going hiking this year...
at least we'll remember that "no talking challenge". that was epic

This is not really related to the current discussion, but:

ONE MORE DAY OF SCHOOL FOR ME!

ONE MORE ESSAY TO WRITE!

ONE MORE STRESSFUL MORNING!

The second I come home I will celebrate. I am so happy that school is finally over-- junior year was quite a nightmare, and I have a feeling senior year won't be fun either. This is why I am looking forward to the summer.

Even though in summer there is less work and more time to relax, I'm looking foward to school next year because I'm to young for summer classes or math camps this year, and I'm not an outside person.

Mine was the same as bhorsemaths the 28th of may. But we start on the 13th of august so I guess uve all got the better deal.

I already had my vacation. We went floating a river in south texas. I passed out because of severe sunburn (my legs and arms were blistered and I was severely dehydrated so I had what is called a sun stroke) Then we went up into east texas up into the mountains stayed for awhile. Then we went to the arkansas/missouri border line and stayed in the ozark mountains. It was fun. We were camping for two weeks and got rained on every night ooogggers... That was the only not fun part of it.


geometry inequalities algebra polynomial function trigonometry Vieta

Hi,

IMO2007 is coming soon. I am asking for your opinion regarding certain issues which might occur ?

My prediction (also this is the list I want  :wink: )
1.Geometry
2.Inequality
3.Combinatorics (maybe not geometric this time)
4.Number theory
5.Geometry
6.Functional equation (i want) , Combinatorial geometry (I predict )  :D

Tell your opinion :)

Ok!  :D  This is what I want:

1 - Combinatorics
2 - Algebra (absolutely not an inequality)
3 - Geometry

4 - Algebra (functional equation)
5 - Number Theory
6 - Combinatorial geometry

My prediction:

1.Number theory
2.Geometry
3.Combinatorics
4.Geometry
5.Inequality
6.Number theory

the first question has been geometry for the last three years... this is gettin a bit too previsible, so that will probably change. Furthermore, there hasn't been a functional equation for some years now. So I are rather inclining to a number theory or algebra problem as question 1. But, why not geometry as question 4?

[b]
1) Geometry
2) Inequality
3) Combinatorics

4) Number theory
5) Geometry
6) Inequality

 :thumbup: [/b]

I think
1) Geometry
2) Inequality
3) Combinatorics
4) Number Theory
5) Geometry
6) Combinatorics

I hope:
1) Inequality
2) Geometry
3) Number Theory
4) Algebra
5) Geometry
6) Combinatorics

In my opinion, it will be the following:

1. Geometry
2. Inequality
3. Number Theory
4. Algebra
5. Functional equation
6. Combinatorics

For me:

1. Geometry
2. Number Theory
3. Combinatorics
4. Inequality
5. Functional equation
6. Combinatorics

My prediction:
1 geometry
2 Inequality
3 Number Theory
4 Functional equation
5 Geometry
6 Combinatorics

1. Number theory
2. Geometry
3. Combinatorics
4. Geometry
5. Inequality
6. Functional equation

Ok, let me try :lol:

1. Number Theory
2. Geometry
3. Algebra(not an inequality)

4. Combinatorics
5. Number theory
6. Geometry

Good luck! :wink:

Here goes:

1) Number Theory
2) Geometry
3) Combinatorics
4) Combinatorics
5) Geometry
6) Inequality

1) combinatorics
2) geometry
3) number theory

4) number theory
5) algebra
6) geometry

^^

Naphthalin

1) Geometry
2) Inequality
3) Combinatorics

4) Number Theory
5) Functionnal equation
6) Geometry

But we can change the functionnal equation by a polynomial problem.

1. Geometry
2. Number Theory
3. Combinatorics
4. Inequality
5. Geometry
6. Number theory

My opinion like my interest :D
1. Number Theory
2. Combinatorics
3. Geometric inequality
5. Geometry
4. Inequality
6. Combinatorics

own prediction:
1.Geometry
2.Algebra
3.NT
4.Combinatorics
5.Geometry
6.Algebra

My prediction and my hope

1. Geometry
2. Inequality
3. Functional Equation
4. Combinatorics
5. Geometry
6. Number Theory

I think it will be :
1. Geometry (Easy)
2. Combinatorics (Fairly hard)
3. Inequality (Hard)
4. Functional equation (Very easy)
5. Geometry (Hard)
6. Number Theory (Killer)

Furthermore, I think the combinatorics one will not be combinatoral geometry again. :wink:

[quote="ychjae"]Furthermore, I think the combinatorics one will not be combinatoral geometry again. :wink:[/quote]

This is, IM(H)O, the only wellfounded prognosis one can make about this year's IMO.
[size=59]Too bad I can't participate anymore...[/size]

  Darij

my guess is
1. algebra (some sequence maybe)
2. geometry
3. combinatorics

4. geometry
5. number theory
6. a hard algebra problem (maybe mixed with NT)

Let me think... hope it right

1. Geometry
2. Algebra (maybe Function Equation).
3. Number Theory

4. Algebra (maybe Inequality).
5. Geometry.
6. Combinatoric (mix with Geometry).

1.Number theory (maybe related to algebra, some sequence or sth)
2.Algebra
3.Geometry
4.Geometry
5.Combinatorics
6.Number theory

[quote="Albanian Eagle"]Guys I predicted correctly the problems of the first day !  :lol:[/quote]

Great!!! So how is your test, Albanian Eagle?

[quote="Albanian Eagle"]Guys I predicted correctly the problems of the first day !  :lol:[/quote]
okie  :lol: we will wait for the second day, hope you are right, too. Goodluck!

So do I win the prediction contest  :lol:

Congratulation Albanian Eagle  :lol: , I think all members of your team think as you and they won't be suddenly by the choosen of Vietnamese   :D

You are an expert Eagle!!! :wink: Congrats! :wink:

Here is what I did at home sitting before the computer:
1: Solved quickly, albeit not necessarily correctly, at least not part (b). Dunno how long it would taken me to fix the mistake (I was told the right answer too soon, so no possibility to try out, but it isn't quite the hardest problem).
2: Solved in about 20 minutes, but Leepakhin had already submitted his post when I was 1/3 done with writing it up, and it was pretty much the same solution as mine (modulo unnecessary steps in mine), so I didn't bother to complete my post.
3: Nothing valuable.
4: Solved by trigonometry and didn't try any other approach since the problem is not particularly exciting. Actually the same holds for number 2, but trigbashing 2 is somewhat more difficult.
5: Got the equivalence to $ 4ab-1\mid\left(a-b\right)^{2}$ within 2 minutes, then it took me half an hour to figure out how to solve such equations once again. One year without olympiads leaves its traces... But come on, an IMO problem should require maybe 75% brains and 25% preparation and not the other way round. I really wonder why we have the same style of Vieta jumping over and over on IMO's since the 1988 problem by Arthur Engel. But then again, the quality of number theory problems in the IMO is suffering an infinite descent...
6: Came up with the generalization ($ 0\leq x_{i}\leq n_{i}$ for every $ i\in\left\{1,2,...,k\right\}$, in $ \mathbb{R}^{k}$). Proved it for $ k=2$, what is one dimension less than required. Figured out by a strange kind of intuition that the problem reminded me of the Cauchy-Davenport theorem (an interesting foreboding of the fact that it follows from the Combinatorial Nullstellensatz). I really wonder how many marks this would yield on the IMO, probably 1/7... :D

  dg

I also came up with the general version of #6 and "found" a proof based on strong induction and recursive ideas, but I'm going to check it once again.
as for the two geometry problems: better than last year's  :D  but boring.
The third was nice and required much thought (almost a perfect olympiad problem don't you think ?) Is there a solution based on known ideas?

Ps. I was really surprised to have such strong guessing capabilities  :D

[quote="Davron"]Ok, guys whom I know... tell me how was your contest?? Eagle, Tiks :?

D.[/quote]

You maybe don't know Davron but Eagle didn't come to Vietnam. As for me, I am not doing very well here. :(  I solved the second problem yesterday and part (a) of the first problem. Today I solved first twoproblems and wrote something from the last one but I don't think I will get a point for it. So, probably, I will end up with a silver medal again. I don't mind it :blush: .

[quote="darij grinberg"]6: Proved it for $ k=2$, what is one dimension less than required. [...] I really wonder how many marks this would yield on the IMO, probably 1/7... :D[/quote]
I'll tell you how much you would have gotten: 0 points :)

[quote="Peter"]Does anyone know who are the authors of the problems? Can we congratulate Hojoo Lee again for the n-th year in a row? :P[/quote]

1: New Zealand
2: Luxembourg
3: Russia
4: Czech Republic
5: United Kingdom
6: Netherlands

Problem 6 is an algebra problem on the shortlist, so in this manner Albanian_Eagle's guess is perfectly correct :).


Let's discuss three cases:

1)  If one of the numbers is 0, then it is easy to prove that all the numbers will be 0.
2) $a,b,c,d,e>0$. Simply using the AM-GM equation we will get that $a^4\ge 4(bc)^3$, using this for other equations of the system we will get $(abcdef)^4\ge 4^5(abcde)^6 \Longrightarrow (abcdef)^2\le \frac{1}{2^5}$. There will always exist two neighboring numbers (one of the pairs $(a,b);(b,c);(c,d);(d,e);(e,a)$) that are either both greater or equal to $\frac{1}{2}$ or both less than $\frac{1}{2}$, by using the equations of the system it is clear that there will be two possible choices: 1) all the numbers are greater or equal to $\frac{1}{2}$ 2) all the numbers are less than $\frac{1}{2}$. The only choice that won’t lead to a disaster is when all the numbers are $\frac{1}{2}$, which is actually a solution.
3)Some of the $a,b,c,d,e$ are negative. We can just pick up $a$ as a negative. $d^2> 0 \Longrightarrow |e|> |a| \Longrightarrow e^2> a^2$ also we can understand that $b^3>e^2$, from these last two inequations we can understand that $b^3>a^2 \Longrightarrow c<0$. If we continue the result will be that $a,b,c,d,e<0$, which is impossible.

Can the problem be solved by extreme principle??
Like monotonizing the variables, $a\geq b\geq c\geq d\geq e$?

[quote="TomciO"]For the first question answer is yes, this is the most natural way to solve it.
For the second obviously not, it's not allowed here.[/quote]
What do mean by "It's not allowed here"?

You mean it doesn't work?

Then how could u apply extreme principle? :?

[quote="Cossack"] 2) all the numbers are less than $\frac{1}{2}$. The only choice that won’t lead to a disaster is when all the numbers are $\frac{1}{2}$, which is actually a solution.[/quote]

How does this case lead to a disaster?  :?

[quote="ZetaX"]He means that you can't assume $a \geq b \geq c \geq d \geq e$ because it's not symmetric.[/quote]
Adding the system together and moving everything to the $RHS$ we'd get
$0=a^2(2a-1)+b^2(2b-1)+c^2(2c-1)+d^2(2d-1)+e^2(2e-1)$
How is it not symmetric?
Case 1a (easy zeroes): if $a=b=c=d=e=\frac{1}{2}$, then the equations would all work    because $\frac{1}{2^3}+\frac{1}{2^3}=\frac{1}{4}$ $(=\frac{1}{2^2})$
Case 1b (trivial solution): everything's just equal to $0$! :D
Case 2: There are other, more difficult solutions to find :o , but i would just accept Cossack's logic

Did the problem ask how many (nontrivial? ;) ) solutions there were?  Otherwise, i'd just have said everything equals $\frac{1}{2}$ and tried the next problem ifi took that test.

[quote="bui_haiha2000"]Just all of idea , no one solve completely. Anyone has the solution , please post completely[/quote]
[quote="cossack"]Some of the $a,b,c,d,e$ are negative. We can just pick up $a$ as a negative. $d^{2}> 0 \Longrightarrow |e|> |a| \Longrightarrow e^{2}> a^{2}$ also we can understand that $b^{3}>e^{2}$, from these last two inequations we can understand that $b^{3}>a^{2}\Longrightarrow c<0$. If we continue the result will be that $a,b,c,d,e<0$, which is impossible[/quote]
for $a,b,c,d,e\geq 0$suppose $a\ge b,c,d,e$
then $e^{2}=a^{3}+b^{3}\ge c^{3}+b^{3}=a^{2}$
then $e\ge a$ simialarly ,we can get $b\ge \ge c\ge d \ge e \ge a$
so $a=b=c=d=e$


Let $A$ be a $n \times n$ matrix and $B$ be a $n \times 1$ matrix.

I want to find $A*(A*(\cdots (A*(A*B))\cdots))$,

I know how to find eigenvalues and eigenvectors of $A$ but I am quite unsure as to how to use them to compute the above.

What happens if $B$ is an eigenvector?

Now, write a general $B$ as a linear combination of eigenvectors.

Can you explain what to do in each of these instances?  :huh:

If $b$ is an eigenvector of $A,$ then $Ab=\lambda b.$

Then $A(Ab)=A(\lambda b)=\lambda Ab=\lambda^{2}b.$

You can continue this by induction, getting $A^{n}b=\lambda^{n}b.$

Here's what jmerry was asking you to consider: what if $b=c_{1}v_{1}+c_{2}v_{2}+\cdots+c_{k}v_{k}$ where $Av_{1}=\lambda_{1}v_{1},\dots,Av_{k}=\lambda_{k}v_{k}$ ?


I'm thinking of an integer,
let's call this number P.
P contains four digits,
to find it you need x, y and z.

x is a 2 digit number,
one that is less than N
where N is the largest prime factor
of 667.

y is a prime number,
one of the first few.
But it isn't equal to x or z,
nor 23, 5, or 2.

If you simplify the square root of
the number of possibilities for P,
the number under the radical
plus 3, will equal z.

Two last clues I'll give you,
P is a power plus 1,
and also the product of x, y and z.
Find P. Have fun!

(This has one solution in mind. extra credit if you can find any others...)

[hide]
i wasn't sure about what the "power" meant, but i assumed $P=k^n+1$

i don't know how to do this without a calculator--i wrote a program.
P=2210, 5330, 6890, 4901, 9802

i think it turns out that $n=2$ for all of them.
[/hide]

[hide="solution"]

so P = xyz.

since the largest prime factor of 667 is 29, x is

$10 \le x \le 28$

$y = 7, 11, 13, 17, 19, ...$

since there are 9*10*10*10 ways of P (a four-digit #),
squarerooting that equals $30 \sqrt{10}$, and add 3 to 10
equals our z, 13.

so P is in form of $n^k + 1$, and since P is also equal to
xyz, there is only one way: x=11, y=7, z=13

so $P = 11 * 7 * 13 = 10^3 +1 = 1001$
[/hide]

oops there was something wrong with what i did.  i think i fixed it.
this is what i got:
[hide]
$P$	2210	5330	6890	1001	4901	1729	9802
$x$	10	10	10	11	13	19	26
$y$	17	41	53	7	29	7	29
$z$	13	13	13	13	13	13	13
$k$	47	73	83	10	70	12	99
$n$	2	2	2	3	2	3	2

the numbers correspond down the column
with $P=k^n+1$ and $xyz=P$

[/hide]


Please continuous here http://www.mathlinks.ro/Forum/viewtopic.php?t=129690 .


logarithms
real analysis
real analysis unsolved

$ P_{n}Q_{n} \equal{} 1, Q_{n} \equal{} Q_{0} \plus{} \sum _{k \equal{} 0} ^{n \minus{} 1} P_{k}$ for all $ n \in N_{0}, Q_{0} \equal{} \alpha$
Following properties are given.
1. $ P_{n}$ is decreasing and positive sequence
2. $ Q_{n} \succ \sqrt {\alpha^2 \plus{} 2n}$
3. $ Q_{n} \prec \sqrt {\alpha^2 \plus{} 2n \plus{} \alpha^{ \minus{} 2} \plus{} \frac {1}{2}ln(\frac {2n}{\alpha^2} \plus{} 1)}$
Then, find minimum value of n that
$ P_{n} \prec \frac {1}{2e}$

If I only had the slightest idea of what $ \succ$ and $ \prec$ mean here...  :roll:

sorry?

It just mean 3<5 and 5<7 like...
$ 6 \prec 7$
can i understand what you mean right?

\prec, $ \prec$ is not the same as <, $ <$.

I assume you mean $ <$ when you write $ \prec$.

Also, use \ln to denote $ \ln$.

ah.... I didn't know before.
Thank you I will fix it.
but.. i want to know solution ...

$ P_{n}Q_{n} \equal{} 1, Q_{n} \equal{} Q_{0} \plus{} \sum_{k \equal{} 0}^{n \minus{} 1}P_{k}$$ n\in N_{0}, Q_{0} \equal{} \alpha$
Following properties are given.
1. $ P_{n}$ is decreasing and positive sequence
2. $ Q_{n} > \sqrt {\alpha^{2} \plus{} 2n}$
3. $ Q_{n} < \sqrt{\alpha^{2} \plus{} 2n \plus{} \alpha^{ \minus{} 2} \plus{} \frac {1}{2}\ln(\frac {2n}{\alpha^{2}} \plus{} 1)}$
Then, find minimum value of n that

$ P_{n} < \frac {1}{2e}$

In other words, $ Q_n$ are defined recursively by $ Q_0\equal{}\alpha$ and $ Q_n\equal{}Q_{n\minus{}1}\plus{}\frac{1}{Q_{n\minus{}1}}$ and you want to know the least $ n\equal{}n(\alpha)$ such that $ Q_n>2e$. This can be done by the reverse iteration. Note that $ Q_1\ge 2>1$. Now, if $ n\ge 2$, $ y>2$, then $ Q_n>y$ if and only if $ Q_{n\minus{}1}>F(y)\equal{}\frac y2\plus{}\sqrt{\frac{y^2}{4}\minus{}1}$. So, if you want to find $ \alpha$ for which $ n(\alpha)\equal{}5$, you just say that we mush have $ Q_5>2e$ but not $ Q_4>2e$, which translates into $ F^{4}(2e)<Q_1\le F^{3}(2e)$ (the powers on $ F$ mean the number of iterations) and, finally, $ \alpha\in[\tfrac 1b,\tfrac 1a)\cup(a,b]$ with $ a\equal{}F^5(2e), b\equal{}F^4(2e)$.

There is a small caveat here, which is that some reverse iteration may become less than $ 2$, so the next iteration will be impossible (it is not the case for $ n(\alpha)\equal{}5$, but when you get, roughly speaking, to $ 20$, you face this problem). I leave it to you to figure out what to do then. ;)

All the properties you listed are not so relevant here and the only difficulty I see in this problem is writing the answer explicitly in terms of square roots.


calculus
linear algebra
real analysis
real analysis theorems

I want to study PDEs for the first time and my school uses the book by Strauss, which I have heard is a nightmare. I have calculus, ODE, and linear algebra backgrounds. I'm studying complex analysis. Do you have suggestions for a rigorous but introductory (English) text on the subject?

I'll just add. Didnt want to create new post. Maybe someone know where to find Spivak Calculus.pdf or smth like that around here, I mean internet  :?:


[hide="hint"]
$ 7\cdot 11\cdot 13\equal{}1001$
[/hide]

[quote="mathemagician1729"]The sum of the digits of a three-digit number is 26. The number is multiplied by 7 then by 11, and finally by 13. How many 9's are in the final product?

How to do without multipliying.[/quote]

observe that 7*11*13 is 1001.so if the original 3 digit number N has digits xyz then the final product is of form
xyzxyz. As the sum of digits is 26,they must be 8,9,9 (in any order) so in final product you have 4 9's.

[quote="mathemagician1729"]The sum of the digits of a three-digit number is 26. The number is multiplied by 7 then by 11, and finally by 13. How many 9's are in the final product?

How to do without multipliying.[/quote]

[hide]

You have to multiply $ 7 \times 11 \times 13 \equal{} 1001$.  The highest sum of numbers you can get which are 3 digit numbers is 27 (999), one less would mean 899 in any order.  $ 1001$ has two ones, so for each $ 9$, you have $ 2$ nines.  There are two, so $ 2 \times 2 \equal{} \boxed{4}$[/hide]


Prove that $\sum_{k=1}^{\infty}p_{k}\sim \frac{1}{2}np_{n}$

$p_{k}$ is the $k$-th prime.  Also, the question is meant to read, $\sum_{k=1}^{n}p_{k}\sim \frac{1}{2}np_{n}$.

The problem is easy to solve if we can use the prime number theorem ($p_{k}\sim k\log k$), but it's probably hard without the PNT.


\sum x 2 /(x+y) \geq (x+y+z)/2
So we have to prove that x+y+z \geq  \sqrt xy+ \sqrt yz+ \sqrt zx which is obvious


Can someone tell me how to approach problem 5 of 2007 paper? Just a hint would be enough. no solutions. I tried a trigonometric approach but was going nowhere. any help wold be appreciated.

Can you post problem 5.
Too lazy  :P   :D  :D

I was actually searching for a PDF copy of the question paper when I found the solns as well. :D

Here you go http://www.isid.ac.in/~rbb/inmosol_07.pdf

i didnot even know generalized appolonius theorem..... :(

actually last year both the geometry were brute force...there was nothing actually really beautiful or interesting in them....i had solved this using the method i posted there seems they didn't understand that :D

too bad.....but 2006 paper was even easier....both the geometry were elementary....

yes 2006 paper was good from all perspectives....
i beleive except the 6th inequality and probably the combi+quadratic nothing deserved to be there in that paper


geometry quadratics algebra polynomial trapezoid probability MATHCOUNTS

You can add your questions and use this like a marathon. You can only post a question if you answered the previous question.

What do you mean you already have one going on? Well, there isno harm to start another one.   :)  :D

the other one is stickied.

All right then. Here goes nothing.

[size=200]Problem 1[/size]

For what value of x does $216\times6^6=6^x+6^x+6^x+6^x+6^x+6^x$?

Who said you get the first question? :?: And why did you say here goes nothing. please change it.

[hide="Question 1"]Well, anyway,

216=$6^2$

$6^8$=$6(6^x)$

$6^8=6^{x+1}$

$x=7$[/hide]

[size=200]Question 2:[/size]

What is the area of a right triangle in which one leg is 33 and 44?

Question 1 is wrong  :D  What is 216 again?

[hide="2"]33*44/2=726[/hide]

[size=200]Problem 3[/size]

What is the 80th term to the following sequence:

0, 3, 8, 15, 24, 35, 48, 63, ....

[quote="236factorial"]Question 1 is wrong  :D  What is 216 again?

[hide="2"]33*44/2=726[/hide]

[size=200]Problem 3[/size]

What is the 80th term to the following sequence:

0, 3, 7, 15, 24, 35, 48, 63, ....[/quote]

Question 1: I meant x to be 8, because 21 is $6^3$. :starwars:

[hide]Question 3:

Is there a pattern for this?
[/hide]

[quote="anirudh"] 21 is $6^3$

[/quote]

 :huh:   :?

[quote="236factorial"]He meant 216.[/quote]

Yeah. That's what i meant. Anyway is prroblem 3 solvable?

[hide][quote="236factorial"]What is the 80th term to the following sequence:

0, 3, 7, 15, 24, 35, 48, 63, ....[/quote]

The differences are

[code]
0, 3, 7, 15, 24, 35, 48, 63, ....
  3  4   8   9    11  13  15
[/code]

Don't know what went on in the beginning, but...
Take out the first 4 terms, then we get 24, 35, 48, 63, ...
Recognize that this can be written as a quadratic polynomial, so think of the differences between the numbers and perfect squares. Each term is one less than a perfect square, so the equation is $f(x)=(x+1)^2-1 \Rightarrow 81^2-1=82 \cdot 80=\fbox {6560}$
[/hide]

[size=150]Problem 4[/size]

Some random person spammed 97 times in one week. The second week, he spammed 128 times. If his spam rate increases linearly, how many of his posts will be spammed by the 10th week?

Oh sorry, for problem 3, I fixed the sequence up  :blush:
Now try it.

[hide="problem 4"]Common difference is 31.
91+31*9 = 376. [/hide]

[size=200]Problem 5[/size]

What is the greatest value of $x+y$ such that $x^2+y^2=90,\&xy=27$?

[hide]$x^2+y^2=90=(x+y)^2-2xy=(x+y)^2-54 \Rightarrow (x+y)^2=144 \Rightarrow x+y=\pm \boxed {12}$[/hide]

[size=150]Problem 6[/size]

Today is August 2, 2005. We've solved 5 problems in one day, so far. If we solve one more problem each day, on what day will we have solved $1,000$ problems?

[hide]After 45 days, we would solve 1000 problems. That's September 16? [/hide]

[size=200]Problem 7[/size]

What is the mean of all the elements in the sequence -1, 3, -5, 7, -9, 11, ...., -201?

[quote="236factorial"][hide]After 45 days, we would solve 1000 problems. That's September 16? [/hide]

[hide="problem7"][size=200]Problem 7[/size]

What is the mean of all the elements in the sequence -1, 3, -5, 7, -9, 11, ...., -201?[/quote]

aka -(1, 3, 5...201)

202*50.5=10201

10201/110=92.7363636...

ans:-92.7363636...

[/hide]

[size=200]Problem 8[/size]

What will be the the area of the trapezoid if the measurements are in the image.

[quote="PenguinIntegral"][quote="236factorial"]That's what I was thinking through. Have you tried it yourself?[/quote]

Case Middle=1/3 chance of choosing a different one within 2.5

Case Side=7/15 chance of choosing a different one within 2.5

Case Corner=1/3 chance of choosing a different one within 2.5

I will let you figure out how to combine them.
 :)[/quote]

what cases rey you guys talking about?

[quote="236factorial"]We're talking about problem 11, a while ago.

If I add the fractions up, I get a number over 1.... that can't be correct.[/quote]

you were supposed to multiply them :)

[quote="236factorial"]I get 7/108 that way, and that looks incorrect to me.[/quote]

But you had to have an x. ;) How did you cancel out the x's. You were just supposed to simpliy

Wait, are you still talking about the probability one or your problem?

Here are questoins 1-13 in a pdf format for people who haven't soled all the maththon questions.

I get 12/30 on my problem.

Think of it this way:
1/2 of the time you have a 7/15 chance
1/4 of the time you have a 1/3 chance
1/4 of the time you have a 1/3 chance

1/2*1/3+1/2*7/15=12/30

If found all the line segments possible and then found the ones under 2.5. Count all the segments. How many did you find?

I didn't coun't line segs, rather points within 2.5 units.

okay... let's work on anirudh's problem now.

I believe that tis is becoming rather inactive. well, anyway, a hint is that you first square everything. :)

[quote="236factorial"]Are these numbers out of your head?  ;) The numerator of the second one isn't really a quadratic.[/quote]

Yes they are out of my head. Look more carefully at the second.  :pilot:  :jump:  :starwars:


I never thought I'd have to do one of these again after Part XI.

Is there a faster way rather than just listing all of the possibilities out?

1. How many ways can you have pennies, nickels and/or dimes with a total value of 25 cents?

2. How many ways can you have pennies, nickels, dimes, and/or quarters with a total of 40 cents?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1593204#1593204]this[/url].
[hide="NOTE for 1"]
For number 1, you cannot use quarters.  If you could use quarters, there is one way in which you can make 25 cents with at least one quarter (use one quarter!).  So you must subtract $ 1$ from the number you get when using Mewto's formula for number 1.[/hide]