This is a problem from the 2004 qualifying quiz for the Mathpath camp.

For positive integers m, n define m ↑ n as follows:
m ↑ 1 = m
m ↑ (n+1) = m^(m↑n).

a) Find the smallest integer n so that 2 ↑ n is greater than a googol (1 followed by 100 zeros).

b) Find the smallest integer n so that 2 ↑ n is greater than a googolplex (10^googol).

I got the first part, but I just can't think about numbers as big as a googolplex. Can someone help me with this problem?

[quote="nr1337"]This question was on the MathPath Qualifying Quiz, wasn't it?[/quote]
It says so at the very beginning of the question...
maybe you could use the fact that 1024 is close to 1000?
and i have no idea how mathpath works so are you allowed to use a calculator?

OK, so for part 2:

Our 2 ↑ n go 2, 2^2, 2^4, 2^16, 2^65536, 2^(2^65536)...

We know 2^65536 is the least one greater than a googol, so we do not need to check the cases below it for the googolplex. So, checking for this one, we get 2^65536 ?> 10^(10^100). But, the base on the left is less than the base on the right, and the exponent on the left is less than the exponent on the right, so this clearly is not true.

We check the next case: 2^(2^65536) ?> 10^(10^100).
Taking log of both sides,
we get 2^65536 * log(2) ?> 10^100.
We know 2^7 > 10^2, so we may substitute 2^350 for 10^100 on the right side -->
2^65536 * log(2) ?> 2^350 -->
log(2) ?> 2^(350-65536).
log(2) > 0.25,
which is clearly greater than 2^(-65186).

Thus, n = 6.

I hope my explanation is OK.

[quote="Scrambled"][quote="nr1337"]This question was on the MathPath Qualifying Quiz, wasn't it?[/quote]
It says so at the very beginning of the question...
[/quote]

Oops... :oops:


It struck me the other day that many introductory combinatorics and number theory topics show up in some introduction to discrete mathematics courses.

People who wrote the earliest math contests could not base their ideas and tests off previous math competitions because they didn't exist, right? Where do most adults who write non-calculus high school math contests learn their math to write the contests? Some topics like number theory and combinatorics are only lightly touched in some high school curricula so would they have had to either do one of the following?

1) be a math major and study outside high school topics like number theory and combinatorics in depth
2) somehow remember number theory or combinatorics touched on in college courses such as discrete mathematics?

I realize this is somewhat evasive, but in fact, the earliest math contest writers did not make as interesting problems as the contest writers do now. Try comparing the IMO's in the 70's to the IMO's now.


I have a doubt regarding sending private messages?

When do I know---- that the message I had 'typed' and sent, is actually sent to the desired person? I have tried sending two messages just now, but I still see them in the 'outbox' , and not in the 'sent box', does it take time to send these messages?

How should I know how to correct this problem.

They stay in the outbox until the addressee reads them. This is the way for you to know when your message has been actually read, not just delivered to the other person's mailbox :).


How can we prove that, for any n >= 2,

 (sin(pi/n))*(sin(2pi/n))*...*(sin((n-1)pi/n)) = n/(2^(n-1))?

See here.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=819011#819011[/url]


limit inequalities logarithms factorial calculus integration geometry

$ \lim_{n\to\infty} \sqrt[n\plus{}1] { \prod_{k\equal{}0}^{n} {n \choose k} }$

The limit is $ \infty.$

Tempting as it is to get something out of the AM-GM inequality, the inequality sign has the wrong sense for that.

Let $ A_n\equal{}\prod_{k\equal{}0}^n\binom{n}{k}.$ We're trying to find $ \lim_{n\to\infty}\sqrt[n\plus{}1]{A_n}.$

If I can show that $ \lim_{n\to\infty}\frac{A_{n\plus{}1}}{A_n}\equal{}\infty,$ it will follow that $ \lim_{n\to\infty}\sqrt[n\plus{}1]{A_n}\equal{}\infty.$

$ \frac{A_{n\plus{}1}}{A_n}\equal{}\prod_{k\equal{}0}^n\frac{\binom{n\plus{}1}{k}} {\binom{n}{k}}\equal{} \prod_{k\equal{}0}^n\frac{n\plus{}1}{n\plus{}1\minus{}k}\equal{} \prod_{k\equal{}0}^n\left(1\minus{}\frac{k}{n\plus{}1}\right)^{\minus{}1}.$

Take the logarithm of that:

$ \ln\left(\prod_{k\equal{}0}^n\left(1\minus{}\frac{k}{n\plus{}1}\right)^{\minus{}1}\right)\equal{} \minus{}\sum_{k\equal{}0}^n\ln\left(1\minus{}\frac{k}{n\plus{}1}\right)$

$ >\sum_{k\equal{}0}^n\frac{k}{n\plus{}1}\equal{}\frac{n}2$

Since this tends to infinity, we are done. We have that $ \frac{A_{n\plus{}1}}{A_n}>e^{n/2}.$

So, a stronger question: what does $ \ln \prod_{k\equal{}0}^n \binom{n}{k}$ look like asymptotically?
Using the factorial representation of the binomial coefficients, it's $ \ln\frac{(n!)^{n\plus{}1}}{\prod_{k\equal{}0}^n (k!)^2}\equal{}\sum_{j\equal{}1}^n (2j\minus{}n\minus{}1)\ln j$

We estimate that sum with an integral: $ \int_1^n (2x\minus{}n\minus{}1)\ln x\,dx\approx \sum_{j\equal{}1}^n (2j\minus{}n\minus{}1)\ln j\minus{}\frac12\ln 1\minus{}\frac12(n\minus{}1)\ln n$ by the trapezoid rule. The error is a sum of terms involving the second derivative $ \frac2x\plus{}\frac{n\plus{}1}{x^2}$ evaluated on each interval, which is $ O(n)$ as $ n\to\infty$. Most of that error comes from the first few terms of the sum, and from the $ \frac{n\plus{}1}{x^2}$ term of the second derivative.

The integral $ \int_1^n (2x\minus{}n\minus{}1)\ln x\,dx$ evaluates to
$ \left[x^2\ln x\minus{}\frac12x^2\minus{}(n\plus{}1)(x\ln x\minus{}x)\right]_1^n$
$ \equal{} n^2\ln n\minus{}\frac12n^2\minus{}(n^2\plus{}n)\ln n\plus{}(n^2\plus{}n)\plus{}(n\plus{}1)$
$ \equal{}\frac12n^2\plus{}n\ln n\plus{}2n\plus{}1$

Combining with the error estimate and the additional $ \frac12(n\minus{}1)\ln n$ term, $ \ln\left( \prod_{k\equal{}0}^n \binom{n}{k}\right)\equal{}\frac12n^2\plus{}\frac32n\ln n\plus{}O(n)$ as $ n\to\infty$. It shouldn't be hard to get a decent bound on the $ O(n)$ term as well.
Kent's method essentially found the dominant $ \frac12n^2$ term of the logarithm.

$ \lim_{n\to\infty} \sqrt[n\plus{}1] { \prod_{k\equal{}0}^{n} {n \choose k} }\equal{}\sqrt{e}$
why?

That's false. You'll only get that if you take an $ n^2$ root.

Please make better titles for your posts. "Limit" says nothing about the content, and multiples of the same useless title are even worse.

Hey int, ZetaX posted some good guidelines for making posts.  They're here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135914&sid=b3cd8382b67f0eb870ad0db0f35691e8

Maybe we could make a sticky in each forum about such guidelines.  How about this, let me try:  "Limit of n'th root of combinatorial products"; it'll help with searching if the title is more meaningful.


$0<a<1$ an integer. Prove that there is no bounded sequence $(x_n)$ of real numbers
such that for any $i<j$, $|x_i-x_j|>\frac{1}{|i-j|}$. And for $a=1$ ?

Where does $a$ (which, by the way, cannot be an integer, as you wrote, if $0<a<1$ :)) enter the picture?

Using Dirichlet's principle, one can easily conclude that you and killer have at least a friend with a very large number of nice problems.  :D . About this one, one can easily see that for a suitable constant $C$ the sequence $C{n\cdot\sqrt{2}}$ is good. For the other part, for any $n$ choose a permutation $f$ of $1,2,...,n$ such that $x_{f(1)}\geq ...\geq x_{f(n)}$ then write that the sequence ${\sum_{i=1}^{n-1}{\frac{1}{|f(i+1)-f(i)|^a}}}$ is bounded. But the last value is at least $\frac{n-1}{n^a}$. Conclusion.

IMO 1991
B3

http://www.kalva.demon.co.uk/imo/imo91.html


Do we have to prove things that are inherently obvious?

When in doubt, prove it, if it is "inherently obvious," it shouldn't take too much time to do so. You will never lose points for proving more than you have to.

If it's, say, a theorem or something, you can simply cite it; otherwise, it's probably good to give some justification.

What would not be good, however, would be too assume something or claim something is "obvious" that the grader thinks is not obvious.


Find all sets of three distinct prime numbers such that the sum of any two of them is
a square.

I like it.

[hide="This hint is all you need for a solution"][hide="so you should try the question out before looking"][hide="I am really serious about this"]mod 4[/hide][/hide][/hide]

So, there is no such set.

So much effort by nothing  :rotfl:  . But it's a nice problem :D

[quote="demji"]how do you prove that though.[/quote]

Just use the hint. BTW thanks JBL. :P

[quote="demji"]oh so the squares alternate 1,0 mod 4 and all primes can be written as 4k+1 or 4k+3 so that they equal 2 mod 4 when summed.  Is that right?[/quote]

Yeah. You got it. :coolspeak:

Actually, that's not [i]quite[/i] it: there are primes congruent to 1, 3 and also 2 mod 4.  However, because there's only one such prime, it doesn't work out.  There are solutions if you relax the distinctness condition, though:
2, 2, 2
2, 2, 7
2, 2, 23
and so on.


Is it possible to draw in Asymptote a surface $ \{(x,y,z): f(x,y,z)\equal{}0\}$ where $ f$ is a polynomial in three real variables $ x, y$ and $ z$?

Try this code with the latest version of Asymptote.
[code]import graph3;
import contour3;

real f(real x, real y, real z)
{
  return x^4+y^4+z^4-(x^2+y^2+z^2);
}

draw(surface(contour3(f,(-1.2,-1.2,-1.2),(1.2,1.2,1.2),15)),blue);[/code]

Trying that gets:

[asy]import graph3;
import contour3;

real f(real x, real y, real z)
{
  return x^4+y^4+z^4-(x^2+y^2+z^2);
}

draw(surface(contour3(f,(-1.2,-1.2,-1.2),(1.2,1.2,1.2),15)),blue);[/asy]

Strange... why would import be blacklisted? I can usually use "import three;" in diagrams.

[quote="mewto55555"]Trying that gets:
[Unparseable or potentially dangerous asy code. This code contains one or more blacklisted tags: import][/quote]
Please, compile the example with your own computer; you could manipulate the picture in an OpenGL window, export to eps, pdf, png, gif etc.

It seems like "contour3" is still blacklisted; is this by intent?


Can any string of consecutive terms in the harmonic sequence add up to 1?

in general, the sum of k terms of the harmonic sequence can never be an integer, except for the first term which is 1...
but i don't think that kind of proofs belongs in pre-olympiad  :?  maybe if someone knows a more easy proof than the one i have read...

Why not belonging here¿
The proof is easy I think:
[hide]
The denominator of $\sum_{i=n}^{n+k} \frac{1}{i} $ is divisible by two when $k>0$.
[/hide]

Yes, and why can't the nominator (edit: I mean numberator, sorry) be divisible by 2 too?

1/2 + 1/3 + 1/4 = (12+8+6)/24

if $h_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
Let $k$ be the largest power or $2$ less than or equal to $n$

Now if sum is $h_n=p$, multiply both side by $\frac{k}{2}*$(product of all odd numbers up to $n$ ) then except for the term corresponding to $\frac{1}{k}$ every other term would be an integer so: contradiction.

besides if harmanic sequence didn't start with $1$, then the sum would always be a odd number over an even number

[quote="Stijn"]Yes, and why can't the nominator be divisible by 2 too?

1/2 + 1/3 + 1/4 = (12+8+6)/24[/quote]

I didn't tell that it isn't so, but the denominator is divisible by two everytime! (and so different from one)

I don't really get the point ZetaX is trying to make, but amirhtlusa's proof makes perfectly sense, well done!

[quote="ZetaX"]I am telling exactly what amirhtlusa did explicitly!
If the denominator of a fraction contains a factor $2$ in it's prime factorisation ( independent from having coprime numerator or not!, or in other words: even when they are coprime ) it can't be an integer.[/quote]

I'm kind of not getting this.  8/4 is an integer...

[quote="ZetaX"]sorry when my explaination is bad: I mean: when independently how you write that rational number as fraction a factor $2$ occurs in the denominator, then it can't be an integer. Equivalent to that is the statement that when $2$ divides the denominator of the reduced fraction it can't be an integer.
Your example doesn't work, since $\frac{8}{4} = \frac{2}{1}$ and $1$ is not even.[/quote]

Ah, you mean a fully reduced/simplified fraction.

[quote="paladin8"][quote="amirhtlusa"]
besides if harmanic sequence didn't start with $1$, then the sum would always be a odd number over an even number[/quote]

Well ZetaX pointed out that even if it does start with $1$, the sum is odd over even. It's basically the same as what you said, by considering the highest power of $2$.[/quote]
 :yup:
(i never disagreed :) )

[quote="seamusoboyle"]Guys, read the question. It dosnt nescilarrly start at 1. It "consuctuive terms" you can prove that starting at one, consider the higher power of 2 under n. I think the same approach might work for the actual question[/quote]

Yeah. Check ZetaX's first post. It's a summation from $n$ to $n+k$ and we can show it cannot be an integer the same way as amirhtlusa showed the harmonic series from $1$ cannot be an integer.


function
logarithms
calculus
calculus computations

How to solve this function in respect to x: [b]c*x^d = a*b^x[/b], where a, b, c and d are constants.

Thanks in advance!

That is $ \ln c\plus{}d\ln x\equal{}\ln a\plus{}x\ln b$.  :maybe:

There is no general solution in terms of elementary functions.

I think should be $ b>0$
I solve the case $ x>0$ (the case $ x<0$ have same method.
Because $ ac>0$ so enogh to solve
$ \Longleftrightarrow \ln{c}+d\ln{x}=\ln{a}+x\ln{b}$
$ f(x)=\ln{b}x+\ln{a}-\ln{c}-d\ln{x}$
$ f'(x)=\ln{b}-\frac{d}{x}$
Use this to find out the inteval of root.
The case $ x<0$ change sign of $ a,c$ and dolve with the same method.

Sorry but i don't understand this point:

[quote="TTsphn"]
$ \Longleftrightarrow \ln{c} + d\ln{x} = \ln{a} + x\ln{b}$
$ f(x) = \ln{b}x + \ln{a} - \ln{c} - d\ln{x}$
$ f'(x) = \ln{b} - \frac {d}{x}$
[/quote]

I should prove this but i don't see how is it possible:

[b]lnb = x(lnb - (lnx)/x)[/b]


[quote="Lovasz"]Find all integer triples $(a,b,c)$ satisfying: $2^a + 3^b = 5^c$ .[/quote]
$3^b \ and \ 5^c \ are \ olds \Rightarrow 2^a \ is \ even$
If a=1: $c\neq0$ if c=1 $\Rightarrow$ (1,1,1) is a solution
if  $c\ge2$ In mod 25 result: $b\equiv13(mod \ 25)$
$\Rightarrow b\equiv3(mod \ 5)$
$\Rightarrow 3^b\equiv5(mod \ 11)$
$\Rightarrow 5^c\equiv7(mod \ 11)$(impossible)
If $a\ge2$ $\Rightarrow 5^c\equiv1(mod \ 4)$
$\Rightarrow 3^b\equiv1(mod \ 4)$
$\Rightarrow \ b \ is \ even$
If c is old $\Rightarrow 5^c\equiv5(mod \ 8)$
since b is even $\Rightarrow 3^b\equiv1(mod \ 8)$
$\Rightarrow 2^a\equiv4(mod \ 8)$
$\Rightarrow$ a=2
$\Rightarrow$ $4+3^b\equiv1(mod \ 3)$
$\Rightarrow 5^c\equiv1(mod \ 3)$
$\Rightarrow$ c is even(contradiction)
$\Rightarrow$ c is even
$\Rightarrow$ $b=2x \ and c=2y$
$\Rightarrow 2^a=(5^x+3^y)(5^x-3^y)$
$\Rightarrow 5^x+3^y=2^m \ and \ 5^x-3^y=2^n$
$\Rightarrow 2. 3^y=2^m-2^n; \ m>n$
$\Rightarrow 2^n \ is \ even$
$\Rightarrow 3^y=2^{m-1}-2^{n-1}$ m-1>n-1
$\Rightarrow 2^{n-1} \ is \ old$
$\Rightarrow n=1$
$\Rightarrow 5^x-3^y=2$
By the proof above x=y=1
$\Rightarrow b=c=2$ $\Rightarrow a=4$
(4,2,2) is other solution
Hence the solutions are (1,1,1) and (4,2,2)


Σιλουαν πολυ ομορφη ασκηση αλλα μηπως ειναι λαθος? Αληθεια ειδε κανεις το Νικο Ραπανο στην τηλεοραση?Μπραβο Νικο παντα τετοια :)  :)  :)

Ναι, είναι λάθος. Το δεξιό μέρος πρέπει να είναι 1.

Ομως για ν=3 νομιζω δε βγαινει 1

Με συγχωρείτε  :oops:  Έκανα  τυπογραφικό λάθος. Τώρα την διόρθωσα ξανακοιτάξτε την...... :P

Αν και δε βλεπω την ασκηση συμφωνω ρε φιλε:πρεπει να ετοιμαζεσαι για το καλυτερο και να εισαι ετοιμος για το χειροτερο. σωστος :)   :)

Δείξτε ότι  $\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^{k}}=2^n$

. Την είχα διορθώσει πάνω   ;)

δεν πιστευω να λυνεται(αν λυνεται μονο με επαγωγη). το λεω γιατι την αποφευγω σετις περισοτερες ασκησεις :lol:  :lol:

όχι δεν λύνεται με επαγωγή (τουλάχιστον η λύση μου δεν είναι με επαγωγή  ;) ) και λύνεται σε λιγότερο από πέντε σειρές.....

[quote="silouan"]Απλά καταπληκτικό .  :lol:  . Πολύ ωραία......
Η λύση μου χρησιμοποιεί ότι
 $\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^{n+k}}=\sum_{k=0}^np_k=1$[/quote]

Στραβώθηκα για να το γράψω...

Πώς αποδεικνύεται ???  :?

Το  αριστερό μέλος είναι άθροισma πιθανοτήτων $p_k=\binom{n+k}{k}\frac{1}{2^{n+k}}=\frac{1}{2}\binom{n+k}{k}\frac{1}{2^{n+k}}+\frac{1}{2}\binom{n+k}{k}\frac{1}{2^{n+k}}=P(A_k)+P(B_k)$

Πιθανοτητες???!!!! ρε μεγαλε το εξηγεις λιγο καλυτερα?και γιατι ξαναγραφεις το n+k για εκθετη? :)


Let $ x,y,z$ be positive numbers satisfying $ x(x\plus{}y\plus{}z)\equal{}3yz$.Prove that
\[ (x\plus{}y)^3\plus{}(x\plus{}z)^3\plus{}3(x\plus{}y)(y\plus{}z)(z\plus{}x)\le 5(y\plus{}z)^3\]

[quote="Allnames"]Let $ x,y,z$ be positive numbers satisfying $ x(x \plus{} y \plus{} z) \equal{} 3yz$.Prove that
\[ (x \plus{} y)^3 \plus{} (x \plus{} z)^3 \plus{} 3(x \plus{} y)(y \plus{} z)(z \plus{} x)\le 5(y \plus{} z)^3\] (*)
[/quote]

$ 3(x\plus{}y)(x\plus{}z)(y\plus{}z)\equal{}12yz(y\plus{}z)$

We have :

$ (*) \Leftrightarrow 4(y^3\plus{}z^3)\plus{}3yz(y\plus{}z) \geq 2x^3\plus{}3x^2(y\plus{}z)\plus{}3x(y^2\plus{}z^2)$

By AM-GM , we have :

$ 3yz\equal{}x(x\plus{}y\plus{}z) \geq 3x\sqrt[3]{xyz} \Leftrightarrow yz \geq x^2$

We have :

* $ 3yz(y\plus{}z) \geq 3x^2(y\plus{}z)$

* $ y^3\plus{}z^3 \geq 2\sqrt{(yz)^3} \geq 2x^3$

* $ (y^3\plus{}z^3)^2(2) \geq (y^2\plus{}z^2)^2(y^2\plus{}z^2) \geq (y^2\plus{}z^2)^{2}2yz \geq2x^2(y^2\plus{}z^2)^2$

$ \Leftrightarrow 3(y^3\plus{}z^3) \geq 3x^(y^2\plus{}z^2)$

Done !

:)

If $ x\plus{}y\equal{}2a , x\plus{}z\equal{}2b , y\plus{}z\equal{}2c$ $ \Longrightarrow$ $ a^2\minus{}ab\plus{}b^2\equal{}c^2$

Inequality equivalents to $ (a\plus{}b)c\plus{}3ab \leq 5c^2 \Longleftrightarrow \sqrt{a^2\minus{}ab\plus{}b^2}.(a\plus{}b) \leq 5a^2\minus{}8ab\plus{}5b^2 \Longleftrightarrow$

$ [\frac{(a\plus{}b)^2}{ab}\minus{}4][24. \frac{(a\plus{}b)^2}{ab}\minus{}81]\geq0$ which is obviously true. :wink:

[quote="Materazzi"][quote="Allnames"]Let $ x,y,z$ be positive numbers satisfying $ x(x \plus{} y \plus{} z) \equal{} 3yz$.Prove that
\[ (x \plus{} y)^3 \plus{} (x \plus{} z)^3 \plus{} 3(x \plus{} y)(y \plus{} z)(z \plus{} x)\le 5(y \plus{} z)^3\]
(*)[/quote]

By AM-GM , we have :

$ 3yz \equal{} x(x \plus{} y \plus{} z) \geq 3x\sqrt [3]{xyz} \Leftrightarrow yz \geq x^2$

:)[/quote]
Can you explain this step? It means that you assume $ x\equal{}min\{x,y,z\}$ so
$ 3yz \equal{} x(x \plus{} y \plus{} z) \geq 3x\sqrt [3]{xyz} \ge 3x^2 \Leftrightarrow yz \geq x^2$

[quote="zaizai-hoang"][quote="Materazzi"][quote="Allnames"]Let $ x,y,z$ be positive numbers satisfying $ x(x \plus{} y \plus{} z) \equal{} 3yz$.Prove that
\[ (x \plus{} y)^3 \plus{} (x \plus{} z)^3 \plus{} 3(x \plus{} y)(y \plus{} z)(z \plus{} x)\le 5(y \plus{} z)^3\]
(*)[/quote]

By AM-GM , we have :

$ 3yz \equal{} x(x \plus{} y \plus{} z) \geq 3x\sqrt [3]{xyz} \Leftrightarrow yz \geq x^2$

:)[/quote]
Can you explain this step? It means that you assume $ x \equal{} min\{x,y,z\}$ so
$ 3yz \equal{} x(x \plus{} y \plus{} z) \geq 3x\sqrt [3]{xyz} \ge 3x^2 \Leftrightarrow yz \geq x^2$[/quote]

No ! it is just a simplified step and you do not need to assume x min

Let $ x,y,z$ be positive numbers satisfying $ x(x\plus{}y\plus{}z) \equal{} 3yz$ .Prove that
\[ (x\plus{}y)^3\plus{}(x\plus{}z)^3\le 2(z\plus{}y)^3\]

[quote="zaizai-hoang"]I understand this one, everything comes from the condition :maybe:  I have another question. Can we assume that without loss of generally $ x \equal{} min\{x,y,z\}$ ? I think we can not  :maybe: Right or wrong? Who can answers my question? Thank u so much  :)[/quote]
Of course the answer is No, dear zaizai, Because The triple $ (2;6;1)$ satisfies the condition and neither of these  triples $ (1,6,2)$,$ (6,2,1)$ does.But $ x$ is not maximal number.

[quote="zaizai-hoang"]Let $ x,y,z$ be positive numbers satisfying $ x(x \plus{} y \plus{} z) \equal{} 3yz$ .Prove that
\[ (x \plus{} y)^3 \plus{} (x \plus{} z)^3\le 2(z \plus{} y)^3\]
[/quote]
The inequality is homogeneous, thus we can assume that $ x\plus{}y\plus{}z\equal{}1$. Hence $ x\equal{}3yz\equal{}1\minus{}y\minus{}z$.
Your inequality is equivalent to
\[ (y\plus{}z)^2(3(y\plus{}z)\minus{}2)\ge 0\]
Which is obvious since $ 1\minus{}y\minus{}z\le \frac{3(y\plus{}z)^2}{4} \leftrightarrow y\plus{}z\ge \frac{2}{3}$.
It is also the method I used to prove the first problem!


Hi guys!
I am looking for a Physics book in the net!
I want something like "Introduction to Physics" :) .
I hope you can help me!
Thnks beforhand!! :wink:

Physics for scientists and engineers, by Tipler and Mosca:

http://lib.org.by/info/P_Physics/PG_General%20courses/Tipler%20P.A.,%20Mosca%20G.%20Physics%20for%20scientists%20and%20engineers%20(5ed.,%20extended%20version)(no%20p.399)(600dpi)(T)(C)(1395s).djvu

Click http://windjview.sourceforge.net/ to get a djvu reader to read this scanned book with.


Well, everyone knows that complex numbers can be expressed in exponential form with modulus $ r$ and argument $ \theta$, written as $ e^{i\theta}$.  But $ e\equal{}e(1)\equal{}e(e^{2i\pi})\equal{}e^{2i\pi\plus{}1}$, but that seems to imply that $ e^{i\theta}\equal{}e^{(2i\pi\plus{}1)(i\theta)}\equal{}e^{\minus{}2\pi\theta\plus{}i\theta}\equal{}ke^{i\theta}$ where $ k\equal{}e^{\minus{}2\pi\theta}$ is generally not 1, and thus the equality doesn't usually establish.  Did I have a mistake in the above equations (or inequations) or is this actually a loophole of the set of complex numbers?  If so, are there tacit assumptions that decrease ambiguity?

I think your mistake is assuming that $ 1 \equal{} 2i\pi \plus{} 1$ because $ e \equal{} e^{2i\pi \plus{} 1}$. This assumption gives $ i \equal{} 0$ which is not true.

Edit: Somebody correct me if I'm wrong but here's why I think the assumption doesn't work:
$ e$ is a purely real number and when you multiply it by a complex number, $ z\equal{}e^{i\theta}$, only $ |z|$ is affected. In other words, the multiplication has no effect on $ \arg(z) \equal{} \theta$.

[quote]I think your mistake is assuming that $ 1 \equal{} 2i\pi \plus{} 1$ because $ e \equal{} e^{2i\pi \plus{} 1}$. This assumption gives $ i \equal{} 0$ which is not true.[/quote]

Not quite.  It tells you that $ 2\pi i \equal{} 0$, which can also imply, with equal incorrectness, that $ \pi \equal{} 0$.

$ e$, as a complex number, can be written as $ e \bullet e^{i\pi}$, which basically brings about the same problem that Sorcerer presents since expanding would result in a totally different (and ugly) complex number, namely $ e^{i(2\pi \plus{} \frac {1}{i})} \equal{} \cos(2\pi \plus{} \frac {1}{i}) \plus{} i\sin(2\pi \plus{} \frac {1}{i})$.  As kcn2rivers points out, this cannot be simplified, for reasons that probably come down to the inability to use regular exponentiation laws when working with such numbers.

The complex exponential doesn't satisfy $ (e^a)^b \equal{} e^{ab}$ for complex $ b$.  This is because there is no unambiguous way to define $ x^y$ for complex $ x, y$; the problem boils down to describing the [url=http://en.wikipedia.org/wiki/Complex_logarithm]complex logarithm[/url].


inequalities
function
logarithms
inequalities proposed

Let $k > 1$ and $n > 1$ be natural numbers. Let $x_1, x_2, ..., x_n$ be $n$ positive real numbers whose product is 1.

Show that $\sum {x_i}^{k+1} \ge \sum {x_i}^k$.

[i]Edited by Myth[/i]

Sorry for the ignorant question: What is 'Chebyshev' and how would you apply it here?

http://planetmath.org/encyclopedia/ChebyshevsInequality.html
Chebyshev inequality is here.
Now, as $x_i^k$ and $x_i$ are similarly sorted, you may apply chebyshev inequality and then amgm to get the result.

Ahh! It makes it so simple!

For a challenge, try not using the Chebyshev inequality... only AM-GM suffices for this problem.

[quote="Aryabhatta"]Let $k > 1$ and $n > 1$ be natural numbers. Let $x_1, x_2, ..., x_n$ be $n$ positive real numbers whose product is 1.

Show that $\sum {x_i}^{k+1} \ge \sum {x_i}^k$.[/quote]

This can be proven without the use of Chebyshev. We can prove, more generally:

[color=blue][b]Theorem.[/b] If $x_1$, $x_2$, ..., $x_n$ are n positive real numbers such that $x_1x_2...x_n\geq 1$, then the function

$f\left(t\right)=x_1^t+x_2^t+...+x_n^t$

is monotonically increasing for $t\geq 0$.[/color]

[i]Proof.[/i] We have to prove that if u and v are two nonnegative reals such that $u\geq v$, then $f\left(u\right)\geq f\left(v\right)$.

Well, let $q=\frac{u-v}{n}$. Since $u\geq v$, we have $u-v\geq 0$ and thus $q=\frac{u-v}{n}\geq 0$.

Now,

$\frac{v+q}{u}+\underbrace{\frac{q}{u}+\frac{q}{u}+...+\frac{q}{u}}_{n-1\text{ times}}=\frac{\left(v+q\right)+\left(n-1\right)q}{u}$
$=\frac{v+nq}{u}=\frac{v+n\frac{u-v}{n}}{u}=\frac{v+\left(u-v\right)}{u}=1$;

thus, we can apply the weighted AM-GM inequality to get

$\frac{v+q}{u}x_1^u+\frac{q}{u}x_2^u+\frac{q}{u}x_3^u+...+\frac{q}{u}x_n^u\geq \left(x_1^u\right)^\frac{v+q}{u}\left(x_2^u\right)^\frac{q}{u}\left(x_3^u\right)^\frac{q}{u}...\left(x_n^u\right)^\frac{q}{u}$
$=x_1^{v+q}x_2^qx_3^q...x_n^q=x_1^v\cdot\left(x_1^qx_2^qx_3^q...x_n^q\right)=x_1^v\cdot\left(x_1x_2...x_n\right)^q$.

Since $x_1x_2...x_n\geq 1$, this becomes

$\frac{v+q}{u}x_1^u+\frac{q}{u}x_2^u+\frac{q}{u}x_3^u+...+\frac{q}{u}x_n^u\geq x_1^v$.

By cyclic permutation, this inequality yields

$\frac{v+q}{u}x_2^u+\frac{q}{u}x_3^u+\frac{q}{u}x_4^u+...+\frac{q}{u}x_1^u\geq x_2^v$;
$\frac{v+q}{u}x_3^u+\frac{q}{u}x_4^u+\frac{q}{u}x_5^u+...+\frac{q}{u}x_2^u\geq x_3^v$;
...;
$\frac{v+q}{u}x_n^u+\frac{q}{u}x_1^u+\frac{q}{u}x_2^u+...+\frac{q}{u}x_{n-1}^u\geq x_n^v$.

After summing up all these inequalities, we arrive at

$\left(\frac{v+q}{u}+\underbrace{\frac{q}{u}+\frac{q}{u}+...+\frac{q}{u}}_{n-1\text{ times}}\right)\left(x_1^u+x_2^u+...+x_n^u\right)\geq x_1^v+x_2^v+...+x_n^v$.

But since $\frac{v+q}{u}+\underbrace{\frac{q}{u}+\frac{q}{u}+...+\frac{q}{u}}_{n-1\text{ times}}=1$, $x_1^u+x_2^u+...+x_n^u=f\left(u\right)$ and $x_1^v+x_2^v+...+x_n^v=f\left(v\right)$, this becomes $1\cdot f\left(u\right)\geq f\left(v\right)$, so that $f\left(u\right)\geq f\left(v\right)$, and we are done.

  Darij

[quote="Aryabhatta"]Let $k > 1$ and $n > 1$ be natural numbers. Let $x_1, x_2, ..., x_n$ be $n$ positive real numbers whose product is 1.

Show that $\sum {x_i}^{u} \ge \sum {x_i}^v$ for $u>v$.[/quote]
To present a short way...
We have $\sum {x_i}^{u} \geq \sum {x_i}^v(x_1...x_n)^{\frac{u-v}{n}}$ by Murheid. Finally, $ \sum {x_i}^v(x_1...x_n)^{\frac{u-v}{n}}= \sum {x_i}^v$.

You guys are good!

Anyway, an 'elementary' solution, if you are willing to call it that  :-)

We show that for _any_ positive reals $x_1, x_2, ..., x_n$

$(\frac{\sum x_i^{k+1}}{\sum x_i^k})^{\sum x_i ^k} \ge   x_1^{x_1^k}x_2^{x_2^k}...x_n^{x_n^k} $

First assume each $x_i$ is a natural number.

Writing $x_i^{k+1}$ as $x_i + x_i + ...+ x_i, x_i^k$ times and applying AM-GM we get the result. Now it is easy to move to rationals and hence to reals.

Now given that  $x_1x_2...x_n = 1$, all we need to show is $x_1^{x_1^k}x_2^{x_2^k}...x_n^{x_n^k} \ge 1$

Wlog assume $x_1 \ge x_2 \ge ... \ge x_n$.
Let $r$ be the smallest such that $x_r < 1$.
Rewriting using $x_r = {\frac {1} {x_1...x_{r-1}x_{r+1}...x_n}}$ we have

$x_1^{x_1^k}x_2^{x_2^k}...x_n^{x_n^k} = x_1^{x_1^k - x_r^k}...x_n^{x_n^k - x_r ^k} \ge 1$ as each term $x_i^{x_i^k - x_r^k} \ge 1$

the same as darij said put $f(y)=x_{1}^{y}+...+x_{n}^{y}$ we want to show that for every $u,v$ such that $u\geq v$ we have $f(u)\geq f(v)$ so we have to show that $f$ is increasing,so its sufficient to show that $f'\geq 0$ (note that $f$ is the sum of $n$ derivable functions so $f$ it self is derivable too)
we have $f'(y)=(x_{1}^{y})'+...+(x_{n}^{y})'=(x_{1}^{y}.\ln x_{1})+...+(x_{n}^{y}.\ln x_{n})=\ln (x_{1}^{x_{1}^{y}}...x_{n}^{x_{n}^{y}})$
in the last equation we used the facts that $\log_{c}^{a}+\log_{c}^{b}=\log_{c}^{ab}$ and $b\log_{c}^{a}=\log_{c}^{a^{b}}$
now we know that the $\log$ function is increasing in the interval $[1,+\infty)$ and we know that $\ln 1=0$ so the only thing that remains is to show that $x_{1}^{x_{1}^{y}}...x_{n}^{x_{n}^{y}}\geq 1$...
[hide][b]lemma:for every positive numbers $a_{1},...,a_{n}$ we have $a_{1}^{a_{1}}...a_{n}^{a_{n}}\geq (\frac{a_{1}+...+a_{n}}{n})^{a_{1}+...+a_{n}}$
proof:WLOG assume that $a_{1}+...+a_{n}=1$ so we have to show that $a_{1}^{a_{1}}...a_{n}^{a_{n}}\geq \frac{1}{n}$ but from weighted AM-GM we know that $a_{1}^{\alpha_{1}}...a_{n}^{\alpha_{n}}\geq \frac{1}{\frac{\alpha_{1}}{a_{1}}+...+\frac{\alpha_{n}}{a_{n}}}$ if we put $\alpha_{i}=a_{i}$;for every $i=1,...,n$;we get that $\frac{1}{n}\leq a_{1}^{a_{1}}...a_{n}^{a_{n}}$...[/b]
now in the lemma above put $a_{i}=x_{i}^{y}$;$i=1,...,n$; so we get that:
$x_{1}^{yx_{1}^{y}}...x_{n}^{yx_{n}^{y}}\geq (\frac{x_{1}^{y}+...+x_{n}^{y}}{n})^{x_{1}^{y}+...+x_{n}^{y}}\geq (\sqrt[n]{x_{1}^{y}...x_{n}^{y}})^{x_{1}^{y}+...+x_{n}^{y}}\geq 1$
[/hide]


algebra
polynomial
number theory unsolved
number theory

Let $k$ be even number; let $p>2$ be prime number. Prove that iff $p-1$ doesnot divide $k$ then $p^{k+1}$ divides $\displaystyle \sum_{i=1}^{p-1}(\binom {p}{i})^k$.

A similar problem has been solved on the forum recently. In fact, this is a very natural generalization of that problem. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15433&highlight=]Here[/url]'s the link.

We need to show that, in $\mathbb Z_p$, $S=\displaystyle \sum_{i=1}^{p-1}(\frac{\binom pi}p)^k=0$. Because $k$ is even, we can show fairly easily that $\displaystyle(\frac{\binom pi}p)^k=i^{-k}$ (for a proof check out that link), so $S=\displaystyle\sum_{i=1}^{p-1}i^{-k}=\sum_{i=1}^{p-1}i^k$. This is $0$. We prove it in the same way the following well-known theorem about the roots of unity is proven: if $z_i,\ i\in\overline{0,n-1}$ are the roots of the polynomial $x^n-1$, then $\displaystyle\sum_{i=0}^{n-1} z_i^k= \{\begin{array}{c}0,\ n\not |\ k\\n,\ n\ |\ k\end{array}$. We just work with the polynomial $x^{p-1}-1$ in $\mathbb Z_p$ instead of $x^n-1$ in $\mathbb C$.

For a quick proof, we can use the [url=http://mathworld.wolfram.com/Newton-GirardFormulas.html]Newton-Girard formulas[/url].

since $S=\sum_{i=1}^{p-1}i^k$ taking a primitive root of $a\in\mathbb{Z}_p$ ($a^k\neq 1$ for $k<p-1$) $\{1,2,\ldots,p-1\}=\{a,2a,\ldots,(p-1)a\}$ then
$S=a^kS$ then we can conclude if $p-1$ doesnt divides $ k$ then $S=0$


function
logarithms
limit
inequalities
trigonometry
real analysis
real analysis unsolved

If $xf(x) \to \infty, xf''(x)\to \infty$ when $x\to \infty$ and $f\in C^2(R)$ then prove that $xf'(x) \to \infty.$
Sorry, if this problem already was discused on the forum :?

$f'(x) - f'(x-1) = f''(\xi)$ for some $\xi \in (x-1,x)$. From this we  get that $x(f'(x) - f'(x-1)) \to \infty$ as $x \to \infty$. I am tired right now, so I am probably entirely incorrect, but what other possibilites are there?

The first condition is superfluous. Since $xf''(x)\to\infty$, we can choose some $a,M>0$ such that $f''(x)>\frac{M}x$ for all $x>a$. Integrating, $f'(x)>M\ln\left(\frac xa\right)+f'(a)\to\infty.$ (That's only true for $x>a$, but it's all we need)

I'm sorry :blush:
I wanted $xf(x) \to 0, xf''(x)\to 0$ when $x\to \infty$ and $f\in C^2(R)$ and we need to prove that $xf'(x)\to 0$ when $x\to \infty$
Sorry one more  :blush:

[b]This solution is entirely wrong[/b]
[hide]
$f'(x) - f'(x-1) = f''(\xi)$ for some $\xi \in (x-1,x)$. We have then that $\xi (f'(x) - f'(x-1)) \to 0$ as $x \to \infty$. Hence, $f'(x) - f'(x-1) = O(\xi^{-p}) = O\left(\frac{1}{(x-1)^p}\right)$ for some p > 1.

We also have $f(x) - f(x-1) = f'(\eta) = O\left(\frac{1}{(\eta-1)^p}\right) + f'(\eta - 1)$,  where  $\eta \in (x-1,x)$. Multiplying by x, we have that $xf(x) \to 0$ and $xO\left(\frac{1}{(\eta-1)^p}\right) = xO\left(\frac{1}{(x-2)^p}\right) \to 0$, so we get that the limits of $-xf(x-1)$ and $xf'(\eta - 1)$ are the same.

We have that $xf(x-1) = (x-1)f(x-1) + f(x-1) \to 0$ as $x\to \infty$, where the limit of $f(x-1)$ is obvious (it must be 0). So we have that $\lim_{x\to\infty}xf'(\eta-1) = 0$, which must mean that $\lim_{x\to \infty}xf'(x) = 0$ (I think you could make a formal argument, but I reason that $f'(x)$ goes to zero more quickly than $f'(\eta-1)$.[/hide]

[quote="Kalle"] We have then that $\xi (f'(x) - f'(x-1)) \to 0$ as $x \to \infty$. Hence, $f'(x) - f'(x-1) = O(\xi^{-p}) = O\left(\frac{1}{(x-1)^p}\right)$ for some p > 1.
[/quote]

You have $\xi f \to 0.$ when $\xi \to \infty$And you make a conclusion that $\exists p>1$ such that $\frac{f}{\xi ^{-p}}$ is bounded?
But if $f(\xi)=\frac{1}{\xi\ln \xi}$ than  $\frac{f}{\xi ^{-p}}=\frac{\xi^p}{\xi\ln \xi} = \frac{\xi^{p-1}}{\ln\xi}$ and it not bounded for any $p-1>0$.
Or, maybe i not undestand you :?:

Hmm. What you say is indeed a counterexample. What I wanted to get at is that when you multiply $f'(x) - f'(x-1)$ by $x$, then the limit is 0. Maybe that is not correct either? Then it doesn't work.

This proof is now in a more clear version a couple of posts down :P

[hide]$f(x) - f(x-1) = f'(\eta(x))$ where $\eta(x) \in (x-1,x)$ by the mean value theorem. From the given conditions we have that $(x-1)(f(x) - f(x-1)) = xf(x) - (x-1)f(x-1) - f(x) \to 0$ when $x \to \infty$ since the limit of $f(x)$ has to be 0. This means that $(x-1)f'(\eta(x)) \to 0$ when $x \to \infty$. Since $x$ goes to infinity more quickly than $\eta(x)$, this bears the implication that $(x-1)f'(x) \to 0$ when $x \to \infty$*.

Now, $f'(x) - f'(x-1) = f''(\xi(x))$ where $\xi(x) \in (x-1,x)$. From the given conditions we have that $|(x-1)f''(\xi)| < |\xi(x)f''(\xi(x))| \to 0$ as $x\to \infty$. So, rearranging and multiplying by $(x-1)$ we have that

$\lim_{x\to\infty}(x-1)f'(x-1) = \lim_{x\to\infty}\left((x-1)f'(x) - (x-1)f''(\xi(x))\right) = 0$

*Proof: We have that $\lim_{x\to\infty}(\eta(x) - 1)f(\eta(x)) = 0$ since $0 < |(\eta(x) - 1)f(eta(x))| < |(x - 1)f(eta(x))| \to 0$. Since $(y-1)f(y)$ is a continous function the statement follows.

edit: FINALLY. This has to be right. Only took the entire day for me to come up with a full solution  :D :P [/hide]

I don't understad your argument. can you be more detalied? Write it in the logic order. Thank you.

Hmm. Ok. I've spent a lot of time on it anyway, so why not.

For a constant $b$, the mean value theorem states that there exists a $\eta \in (b-1,b)$ such that $f(b) - f(b-1) = f'(\eta)$. Therefore, there exists a function $\eta(x)$ defined by the equality $f(x) - f(x-1) = f'(\eta(x))$. This function does not have to be continous or anything, but it exists according to the mean value theorem. For this function, it must also be true that $x-1 < \eta(x) < x$ for every x.

We must have that $\lim_{x\to\infty}f(x) = 0$, or it would not be possible that $\lim_{x\to\infty}xf(x) = 0$. From this we can conclude that:
$(x-1)f'(\eta(x)) = (x-1)(f(x) - f(x-1)) = xf(x) - f(x) - (x-1)f(x-1) \to 0$ when $x\to\infty$, since every term on the right hand side goes to zero. From this, it is quite trivial to prove that $\lim_{x\to\infty}(x-1)f'(x) = 0$, which is a result from the fact that $x-1 < \eta(x) < x$. The proof is marked under * at the end of this message.

Note how we've now proven a very similar thing to what's asked for, but it is slightly weaker. $(x-1)$ grows slower than $x$. So we need to strengthen the conditions a little.

With similar reasoning to in the beginning of the proof, there exists a function $\xi(x)$ which fulfills the equality $f'(x) - f'(x-1) = f''(\xi(x))$ and the inequality $x - 1 < \xi(x) < x$. This inequality shows that $\lim_{x\to\infty}\xi(x) = \infty$. Since $f''(x)$ is a continous function, we have that $\lim_{x\to\infty}\xi(x)f''(\xi(x)) = 0$ by the initial conditions. From this we can prove that $\lim_{x\to\infty}(x-1)f''(\xi(x)) = 0$, since we have the inequality $|(x-1)f''(\xi(x))| < |\xi(x)f''(\xi(x))|$ at least for large x.

Now we have enough information to prove what is required. From multiplying the equality $f'(x) - f'(x-1) = f''(\xi(x))$ by $(x-1)$ we find that
$\lim_{x\to\infty}(x-1)f'(x-1) = \lim_{x\to\infty}\left((x-1)f'(x) - (x-1)f''(\xi(x))\right) = 0$.

Which is an equivalent statement to what's asked for. Is it clear now? I think it is! I realize that it's a very long and messy solution, but I am fairly confident that it is right.

*We have that $\lim_{x\to\infty}(\eta(x) - 1)f'(\eta(x)) = 0$ since $0 < |(\eta(x) - 1)f'(\eta(x))| < |(x - 1)f'(\eta(x))|$ at least for large x, and we've shown that $\lim_{x\to\infty}(x-1)f'(\eta(x)) = 0$. The inequality $x-1 < \eta(x)$ shows that $\lim_{x\to\infty}\eta(x) = \infty$. Since $f'(y)$ is a continous function we can make the variable substiton $y = \eta(x)$ to get that $\lim_{y\to\infty}(y - 1)f'(y) = 0$

[quote="Kalle"]
*We have that $\lim_{x\to\infty}(\eta(x) - 1)f'(\eta(x)) = 0$ since $0 < |(\eta(x) - 1)f'(\eta(x))| < |(x - 1)f'(\eta(x))|$ at least for large x, and we've shown that $\lim_{x\to\infty}(x-1)f'(\eta(x)) = 0$. The inequality $x-1 < \eta(x)$ shows that $\lim_{x\to\infty}\eta(x) = \infty$. Since $f'(y)$ is a continous function we can make the variable substiton $y = \eta(x)$ to get that $\lim_{y\to\infty}(y - 1)f'(y) = 0$[/quote]
Hmm...
Denote $E=\{\eta(x)|x>0\}$
we have $\lim_{x\to \infty}(\eta(x) - 1)f'(\eta(x)) =0$
If $\eta(x)$ is not continuous then it is possible case when $\forall N\ \; (N;\infty)\setminus E \ne \emptyset$.
then we can prove only that \[ \forall \varepsilon \exists N \; (x>N)\bigwedge(x \in E) \Rightarrow |(x-1)f'(x)|<\varepsilon \] but we need \[ \forall \varepsilon \exists N \; (x>N) \Rightarrow |(x-1)f'(x)|<\varepsilon \]
 :?:  :?:

:(

If you don't see a way to save it, I probably don't either. There is no way to get around that problem with some continuity arguments? It is possible that the solution could be saved, since the "1" was chosen arbitrarily and we could repeat the argument with $f(x) - f(x-\varepsilon)$, but I don't know. I retire. I've spent a lot of time on this problem, probably like 5+ hours, so that's it.

Post the real solution.

I'm don't understand why it's wrong. If we have that $f$ is continous and that $\lim_{x\to\infty}g(x) = \infty$. Do we not have that
$\lim_{x\to\infty}f(g(x)) = \lim_{y\to\infty}f(y)$ ?
I thought it was so.

Would it be true if $\lim_{x\to\infty}g(x) = A$ so that $\lim_{x\to\infty}f(g(x)) = \lim_{y\to A}f(y)$ ?

[quote="Kalle"]I'm confused as to why it doesn't work. If we have that $f$ is continous and that $\lim_{x\to\infty}g(x) = \infty$. Do we not have that
$\lim_{x\to\infty}f(g(x)) = \lim_{y\to\infty}f(y)$ ?
I thought it was so.
[/quote]
i try to construct counterexample...
let $g(x)=[x] \to \infty,$ when $x\to \infty$ and let for example $f(x)=\sin(\pi x)$.
then we have $f(g(x))=\sin(\pi [x]) = 0 \to 0$,  but $\sin (x) \nrightarrow 0$ when $x\to \infty$...

[quote="Kalle"]Would it be true if $\lim_{x\to\infty}g(x) = A$ so that $\lim_{x\to\infty}f(g(x)) = \lim_{y\to A}f(y)$ ?[/quote]

let $g(x)=\frac{1}{[x]\pi}$ and $f(x)=\sin \frac{1}{x}$. Then $g(x)\to 0$,$f(g(x))=\sin([x]\pi)=0 \to 0$ and $f(x)\nrightarrow 0$ when $x\to 0$... :(

Yes, I see you are right. My book actually does write $\lim_{x\to\infty}f(g(x)) = \lim_{y\to\infty}f(y)$ at places. Do they actually mean that if $\lim_{y\to\infty}f(y) = A$ where $A$ is a constant, "$\infty$" or "$-\infty$" then we know that $\lim_{x\to\infty}f(g(x)) = A$? But if the first limit diverges in some other way, we don't know anything about the second. And if we know something about the second we don't know anything about the first limit?

I thought I understood limits, but now I don't understand simple calculations anymore :(

if there $\exists \lim_{x\to \infty} f(x)$ and $\lim_{x\to \infty} g(x) =\infty$ and  $\lim_{x\to \infty} f(g(x))=A$
then $\lim_{x\to \infty} f(x)=A.$ but when we not knew $\exists \lim_{x\to \infty}f(x)$ or not, then existence $\lim f(g(x))$ can't guarantee existence $\lim f(x)$...  ;)
and when $\exists \lim f(x)$ then $\forall g$ such that $g \to\infty$ when $x\to \infty$ we have \[ \lim f(g(x)) = \lim f(x) \]

Ok, thanks for the help! I am going to think through limits and how they work tomorrow. I am sorry if you feel that I have wasted your time. I thought my solution was correct.

Do you know a real solution to this problem? What "level" is this problem meant for? All I know is the mean value theorem ;) Would it be possible to show that the limit actually exists in some way, so that my solution could work?

[quote="Kalle"]Ok, thanks for the help! I am going to think through limits and how they work tomorrow. I am sorry if you feel that I have wasted your time. I thought my solution was correct.

Do you know a real solution to this problem? What "level" is this problem meant for? [/quote]

Quite the contrary Kalle. Thank you because you think about my problem :)
Reason is that i don't knew solution :(  This problem is give me my friend who is a student of first year of study in our  university...

Two related problems
1) Let $A=\sup |f(x)|$ $B=\sup|f''(x)|$, then $\sup |f'(x)|\le 2\sqrt{AB}$ (Landau ineq).
2) If $f\to 0$ as $x\to \infty$ and $f''$ is bounded then $f'(x)\to 0$ as $x\to \infty$.(Hardy-Littlewood th)

I know the proof for the first, and I think that the first serve in the proof of the second problem. The second problem is quite similar to our problem. I think that those two problems will help you to find a solution of the problem.


[quote="brokenbone"]Hi!

I have a \newcounter command for definitions in my paper. I use newcounter, because i don't like to have italic text in my definitions, as it would be if i used \newtheorem command. The problem is that i can't use \label and \ref commands on my definitions. OK, i can use them, but it doesn't work properly  :lol: .
Can somebody please help me with this. Or is there a way to define newtheorem command that it wouldn't have italic text. (But just for definitions, not for theorems, lemmas, ...)

Thank you for your answer(s)![/quote]

Usually what I did is to put
\theoremstyle{definition}

So for example
\newtheorem{theorem}{Theorem}
\theoremstyle{definition}
\newtheorem{remark}{Remark}
\newtheorem{definition}{Definition}
above all \newtheorem that you want to write in definition style (not italic text).
But I forgot whether I should put \usepackage{amsthm} or \usepackage{theorem}

Thank you!

That works (with amsthm package)! The thing that bothers me now (never satisfied ) :D is that definitions and theorems use the same counter.

For example:
[b]Theorem 1.1[/b]
...
[b]Theorem 1.2[/b]
...
[b]Definition 1.3[/b]

I'd like to have [b]Definition 1.1[/b] in this example.

I'm sorry, it doesn't count them like that, it works fine.
Thanks again for your answer


algebra
polynomial
logarithms
Gauss
algebra proposed

let $k,n\in \mathbb{N}$ and $P(x)$ apolynomial of degree $n$ whose coefficients are all in the set $\{-1,0,1\}$. We are given that $(x-1)^{k}|P(x)$ and that there exists a $q$ prime such that $\frac{q}{\ln q}<\frac{k}{\ln (n+1)}$. Show that at least one root of unity of index $q$ will be a root of $P(x)$.

As Arqady would say: Unsolved in 2006!
anyway this problem is not so difficult, I'm surprised it recieved no solution (or attention :P )

I guess it's the presence of those $ \ln$ that made the problem sound difficult :D . Anyway, it is more or less trivial: assuming the contrary and denoting $ x_{1},x_{2},...,x_{q}$ the roots of order $ q$ of unity, we have on one hand $ |P(x_{1})...P(x_{q})|\leq n^{q}$ because of the coefficients and degree conditions. On the other hand, write $ P\equal{}(x\minus{}1)^{k}Q$ then observe that $ |P(x_{1})...P(x_{k})|\equal{}q^{k}|Q(x_{1})...Q(x_{k})|\geq q^{k}$, because by fundamental theorem of symmetric polynomials $ Q(x_{1})...Q(x_{k})$ is an integer (ok, Gauss lemma too).

That's OK,  :lol:
And you meant $ |P(x_{1})\cdots P(x_{q})|\le (n\plus{}1)^{q}$ of course  :)


AMC USA(J)MO USAMO AIME trigonometry vector geometry

Great Idea Rep!

That'll give someone time to prepare the Mock AIME also. Really appreciate it! Will this test be in the same format as the previous one by JSRosen?

Very good idea!

(thinks to self: The more people we have, the more likely we are to have a Mock USAMO! muahahaha....)

Edit: Actually, assuming that I don't end up writing the Mock AIME, I'd like to co-write this test. It'd probably help me more than taking it...

I MUST participate!  I mad SOO many stupid mistakes on this one that I need to make a better score.

[quote="Rep123max"]Ya, it would be the same way (since that way worked really well). I'll do it if we have a minimum of 15 people. And is a .pdf of the test okay?[/quote]

Rep: Honestly, I want to know if I could co-write the test. I may have to captain (and possibly train) my math team this year, so I need experience in problem-writing.

I'm in! It'll be my chance to redeem myself after a very VERY embarrassing 53.5  :)

I'd like to take it...

even though I made the Mock AIME already (barely... i was 14th/29 I think)

I need all the practice I can get though... so bring it on!

Will this one be a bit easier than the first mock AMC contest?

I'm in.  Lol.  I messed up on the first one. Trigs and number theory beat me on the face good. :(

P.S.  Is yours gonna be easier than JSRosens'? I hope it is..

I suppose I already made mock AIME, but I'd like to take it anyway.

If your going to make another AMC, please make it easier.  Try make the first half of the test the same difficulty as about the first 5 questions in JRosen's test.

Mabye then I can get at least a 100 on something besides math class.

[quote="kool_dudy"]Yeh...Rep...you might want to think about making this AMC contest lean a bit more to AMC 10 than AMC 12 since this is after all a bonus qualifier for the AIME.[/quote]

It may turn out that way, as I can't write trig problems.  :D

We'll work on it.

I'm definitiely up for it...I have to beat yif man12 for lowest score!

Rep123max wrote:I wrote their scores on a piece of scratch paper, but since I am 99% positive I will lose it, I thought to put it on Excel.

This can lead rise to a math problem: if there is a 99% chance that Rep123max will lose a piece of scratch paper with scores on it, what is the fewest number of copies Rep123max must make in order to be at least 50% sure he will still be able to find at least one of them (the original sheet doesn't count as a copy)?

Answer (in spoler):[hide]68.[/hide]

I somehow doubt it took him very long.

Plus, I found it quite amusing.

(Yay, AMC10B is tomorrow!)

[quote="genius1601"]somebody has too much free time on their hands.  :lol:[/quote]

Probably you, David.

I should be fine with Wed. 6PM EST.  If not the hopefully I'll be able to PM you before or during to let you know I can't do it.

I can't find the the topic Problems of Mock AMC B, can someone post the link to me?

Yeah I can't find it either.
Where'd it go?

[quote="JSRosen3"]I'd advise against that, simply because it's generally not good to have the person who got the lowest score know they got the lowest score...it can be very demoralizing  :([/quote]

hardy har har.  :)


probability
function
probability and stats

I have come across a problem in A N Shiryaev, Probability page no. 387 . There he is saying the following obvious.
$ \sum \mathbb{E}\left(\frac{X_n ^2}{1\plus{}|X_n|}\right) < \infty  \Rightarrow \sum \mathbb{E}(X_n ^2 I_{\{|X_n| \leq 1\}} \plus{} |X_n| I_{\{|X_n| > 1\}}) < \infty$
where $ X_n$ are independent random variables with $ \mathbb{E}(X_n) \equal{}0$
Can anyone justify this? Please...

Thank you.

Let $ f$ be the function $ f(x) \equal{} \frac{x^2}{1 \plus{} |x|}$.  Let $ g$ be the function such that $ g(x) \equal{} x^2$ for $ |x| \le 1$ and $ g(x) \equal{} |x|$ for $ |x| > 1$.  Then $ g(x) \le 2f(x)$.  That should justify the implication.


geometry
circumcircle
Euler
incenter
geometric transformation
homothety

I don't have time to finish it right now, but I guess the following approach works.
For part a, just note that DK bisects the arc BC than containing A.
For part b, use the well known fact that O lies on the Euler line of triangle DEF.

$D$ - point which is center of omothety of one $\omega_a$ and incircle,
$K$ - point which is center of omothety of one $\omega_a$ and circumcircle.
It rezults that point which is center of omothety of circumcircle and incircle lies on $DK$.
Hence $P$ is exactly point of similitude of incircle and circumcircle (I think so called).
So it lies on $OI$ and from Iran 1995 it contains ortocenter of $DEF$.

Can anyone give a clear solution please ??  :?

For Erdos, I am rewriting Prowler's solution:

[quote="grobber"]Let $(I),(O)$ be the incircle, and, respectiely, circumcircle of $ABC$. $(I)$ touches $BC,CA,AB$ in $D,E,F$ respectively. We are also given three circles $\omega_a,\omega_b,\omega_c$, tangent to $(I),(O)$ in $D,K$ (for $\omega_a$), $E,M$ (for $\omega_b$), and $F,N$ (for $\omega_c$).

[b]a)[/b] Show that $DK,EM,FN$ are concurrent in a point $P$;

[b]b)[/b] Show that the orthocenter of $DEF$ lies on $OP$.[/quote]

[i]Solution.[/i] Let Q be the internal center of similitude of the circumcircle (O) and the incircle (I) of triangle ABC. Then, of course, this point Q lies on the line OI which joins the centers O and I of these two circles.

By the Monge theorem, the internal center of similitude of the circles (O) and (I), the internal center of similitude of the circles (I) and $\omega_a$, and the external center of similitude of the circles $\omega_a$ and (O) are collinear. But the internal center of similitude of the circles (O) and (I) is the point Q; the internal center of similitude of the circles (I) and $\omega_a$ is the point D (since these two circles externally touch at D), and the external center of similitude of the circles $\omega_a$ and (O) is the point K (since these two circles internally touch at K). Hence, the points Q, D and K are collinear; in other words, the point Q lies on the line DK. Similarly, the point Q lies on the lines EM and FN. Thus, the lines DK, EM and FN are concurrent at the point Q. This solves part [b](a)[/b] of the problem, and shows that the point of concurrence P of the lines DK, EM and FN is the point Q.

Now, in order to prove part [b](b)[/b] of the problem, it is enough to show that the orthocenter of triangle DEF lies on the line OQ. Since the point Q lies on the line OI, the line OQ is the line OI; hence, it remains to show that the orthocenter of triangle DEF lies on the line OI. In other words, it remains to show that the orthocenter of triangle DEF, the incenter I of triangle ABC and the circumcenter O of triangle ABC are collinear. But this was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 .

Problem solved.

  Darij

Dear Mathlinkers,
an article concerning the “Ayme’s theorem’’ and this question have been put on my website.
http://perso.orange.fr/jl.ayme    vol. 20 p. 17, 18
You can use Google translator
Sincerely
Jean-Louis

[hide="a"]By Monge d'Alembert, they concur at the insimilicenter of $(I), (O)$.[/hide]

[hide="b"]By a, $P\in OI$, so it suffices to show that $H\in OI$, where $H$ is the orthocenter of triangle $DEF$. Let $XYZ$ be the orthic triangle of triangle $DEF$. Angle chasing yields triangles $XYZ$, $ABC$ are homothetic, so let $\mathcal{H}$ be the homothety that sends them to one another. $\mathcal{H}$ sends their incenters $H$, $I$ and their circumcenters $N$, $O$ to one another, where $N$ is the nine-point center of triangle $DEF$; hence $H\in OI$ as desired. $\blacksquare$[/hide]

Notice the similarity to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=268390]2003 Vietnam TST #2[/url], which concerns the incenter of the orthic triangle rather than the orthocenter of the intouch triangle.

Here is my solution for this problem
[b]Solution[/b]
a) We define again point $K$ is second intersection of $\bigodot(AI)$ with ($O$)
Let $X$, $Y$, $Z$ be midpoint of $\stackrel\frown{BC}$ which not containing $A$, $\stackrel\frown{CA}$ which not containing $B$, $\stackrel\frown{AB}$ which not containing $C$
Then: $K$ is Miquel point of quadrilateral $BCEF$
So: $\triangle$ $KBF$ $\stackrel{+}{\sim}$ $\triangle$ $KCE$
Hence: $\dfrac{KB}{KC}$ = $\dfrac{BF}{CE}$ = $\dfrac{BD}{CD}$ or $KD$ is internal bisector of $\widehat{BKC}$
Let $J$ $\equiv$ $OK$ $\cap$ $ID$
We have: ($KD$; $KJ$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($CK$; $CX$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($DK$; $DB$) $\equiv$ ($DJ$; $DK$) (mod $\pi$)
So: ($J$; $JK$) $\equiv$ $\omega_a$
Then: $DK$ passes through $X$
Similarly: $Y$ $\in$ $EM$, $Z$ $\in$ $FN$
But: $EF$ $\parallel$ $YZ$, $FD$ $\parallel$ $ZX$, $DE$ $\parallel$ $XY$ then: $DX$, $EY$, $FZ$ concurrent at point $P$ is external homothetic center of ($I$) and ($O$) or $DK$, $EM$, $FN$ concurrent at $P$
b) Let $H$ be orthocenter of $\triangle$ $DEF$
Since: $I$ is orthocenter of $\triangle$ $XYZ$, we have: $I$, $H$, $P$ are collinear
But: $P$, $I$, $O$ are collinear then $I$, $H$, $P$, $O$ are collinear of $H$ $\in$ $PO$


[quote="sharkman"]six-sevenths in a fraction of the party delegates were proclaiming
their positions. If 1100 were not involved in his behavior, how many delegates were there in all?[/quote]

[hide]
1100 is $\frac{1}{7}$ th of the total. so 7700 is the total number of delagates.
[/hide]

[quote="rd5493"][quote="diophantient"][quote="sharkman"]six-sevenths in a fraction of the party delegates were proclaiming
their positions. If 1100 were not involved in his behavior, how many delegates were there in all?[/quote]
Note that it doesn't specify behavior.
Using Izebiglips's Lemma, we see that the result is 7700.[/quote]

what exactly is Izebiglips's Lemma?  Is it a type of formula of some kind?

From what I can tell, the answer is 7700, which makes this problem way too easy for this forum if I can figure it out that quickly[/quote]

From my sources, there is no Izebiglips's Lemma.


How far is the line x+y=0 from the circle centered at (7,4) having a radius of 3 times the square root of 2?

[quote="i_like_pie"][hide]In slope-intercept form, the first line (line $m$), is $y=-x.$ The line perpendicular to line $m$ that goes through $(7,\,4)$ is $y=x-3.$ The intersection of these two lines is at $\left(\frac{3}{2},\,-\frac{3}{2}\right).$ The distance from this point to the center of the circle is $D=\sqrt{\left(4+\frac{3}{2}\right)^{2}+\left(7-\frac{3}{2}\right)^{2}}=\sqrt{\frac{242}{4}}=\frac{11\sqrt{2}}{2}.$ The distance from the circle is $\frac{11\sqrt{2}}{2}-3\sqrt{2},$ or $\boxed{\frac{5\sqrt{2}}{2}}.$[/hide][/quote]

it didn't show up right the first time, but now i see it. thanks


prove that the square of any prime, $p>3$, is one more than a multiple of 24.
have fun. i think this is really easy, just want to see how different people solve it.

[hide]Any prime greater than 3 is $1\mod{3}$.  Hence, the square of a prime is $1\mod{3}$.  Thus, any square of a prime is already one more than a multiple of 3.
Now take primes mod 8.  Clearly, a prime has to be one of $1\mod{8}$, $3\mod{8}$, $5\mod{8}$ and $7\mod{8}$.  Any of these yields that the square of a prime is $1\mod{8}$.  Thus, any square of a prime is one more than a multiple of 8.
So, being one more than than a multiple of 8 and one more than a multiple of 3, the square of a prime must be one more than a multiple of 3*8=24, because 3 and 8 are relatively prime.  QED[/hide]

p^2 = 24n + 1
p^2 - 1 = (p+1)(p-1) = 24n
24|(p-1)(p+1)

p-1 and p+1 are both even numbers because p must be odd, and one of them will also be a multiple of 4, so we know 8|(p-1)(p+1).

Along with that, because p cannot be a multiple of 3, p-1 or p+1 has to be divisible.  We could use casework with mods to see this.

Since 8 and 3 are both divisors of (p-1)(p+1), 24 is a factor of p^2 - 1.  Therefore, p^2 is one greater than a multiple of 24 for all primes p.


1. Show that $ 1987$ divides $ 2903^n\minus{}803^n\minus{}464^n\plus{}261^n$ for all positive integers $ n$.

2. Show that $ S\equal{}5^n\plus{}2*3^(n\minus{}1)\plus{}1$ is divisible by $ 8$ for all integers $ >\equal{}1$.

3. Let $ n$ be a positive integer. Show that any number greater than $ frac{n^4}{16}$ can be written in at most one way as the product of two of its divisors having difference not exceeding $ n$.

4. Let $ k$ and $ n$ be positive integers. Show that the only solutions $ (k,n)$ of the equation $ (n\minus{}1)!\equal{}n^k\minus{}1$ are $ (1,2), (1,3)$ and $ (2,5)$

[quote="kl2836"]1. Show that $ 1987$ divides $ 2903^n \minus{} 803^n \minus{} 464^n \plus{} 261^n$ for all positive integers $ n$.

2. Show that $ S \equal{} 5^n \plus{} 2*3^(n \minus{} 1) \plus{} 1$ is divisible by $ 8$ for all integers $ > \equal{} 1$.

3. Let $ n$ be a positive integer. Show that any number greater than $ frac{n^4}{16}$ can be written in at most one way as the product of two of its divisors having difference not exceeding $ n$.

4. Let $ k$ and $ n$ be positive integers. Show that the only solutions $ (k,n)$ of the equation $ (n \minus{} 1)! \equal{} n^k \minus{} 1$ are $ (1,2), (1,3)$ and $ (2,5)$[/quote]

2. $ n\equal{}2k$ ----> $ 5^{2k}\plus{}2*3^{2k\minus{}1}\plus{}1\equiv25^{k}\plus{}6*9^{k\minus{}1}\plus{}1\equiv1^{k}\plus{}6*1^{k\minus{}1}\plus{}1\equiv 0 (mod8)$
    $ n\equal{}2k\plus{}1$ ----> $ 5^{2k\plus{}}\plus{}2*3^{2k}\plus{}1\equiv5*25^{k}\plus{}2*9^{k}\plus{}1\equiv5*1^{k}\plus{}2*1^{k\minus{}1}\plus{}1\equiv 0 (mod8)$


[quote="mdk"]Find the sum of the digits of (100000+10000+1000+100+10+1)^2[/quote]

$ (10^{5}+10^{4}+10^{3}+10^{2}+10^{1}+10^{0})^{2}$ has 36 terms, and each of them are powers of 10. Since no single power of 10 is counted more than nine times, the answer is 36.


The only known virginia ARML team is TJ's.  But this blocks many good math students from competing.  Is there any way to go without going to TJ?

This questions has also been posted in the ARML forum.

last years arml sucked(for both of us but for me more than you)

last year's ARML owned. XD Can't believe I lost to allen and Kevin by 1 T.T

it was really hard (for me at least).

was it just me or was there an abnormally large number of geometry problems?

anybody interested in trying to make a VA ARML team not from TJ?


For which real numbers $ x>1$ is there a triangle with side lengths $ x^4\plus{}x^3\plus{}2x^2\plus{}x\plus{}1, 2x^3\plus{}x^2\plus{}2x\plus{}1,$ and $ x^4\minus{}1$?

$ (x^4 \plus{} x^3 \plus{} 2x^2 \plus{} x \plus{} 1) \plus{} (x^4 \minus{} 1) > (2x^3 \plus{} x^2 \plus{} 2x \plus{} 1)\Rightarrow 2x^4 \minus{} x^3 \plus{} x^2 \minus{} x \minus{} 1 > 0$
$ (2x^3 \plus{} x^2 \plus{} 2x \plus{} 1) \plus{} (x^4 \minus{} 1) > (x^4 \plus{} x^3 \plus{} 2x^2 \plus{} x \plus{} 1)\Rightarrow x^3 \minus{} x^2 \plus{} x \minus{} 1 > 0$
$ (2x^3 \plus{} x^2 \plus{} 2x \plus{} 1) \plus{} (x^4 \plus{} x^3 \plus{} 2x^2 \plus{} x \plus{} 1) > (x^4 \minus{} 1)\Rightarrow 3x^3 \plus{} 3x^2 \plus{} 3x \plus{} 3 > 0$

These are all true for $ x > 1$

Or, one can realize that $ x^4\plus{}x^3\plus{}2x^2\plus{}x\plus{}1$ is the largest side, so only the second line is actually necessary.


function
geometry
pigeonhole principle
ARML

I thought it was significantly easier this year.

1: almost trivial, just parity analysis.
2: also very easy, definition of the functions
3: pretty simple, I expressed areas in terms of side lengths and solved for AB*BC. My friend used guess and check to solve it, which I thought was unfortunate...
4: I thought this was the hardest one, I think i got it though
5: (a) and (b) were easy, (c) was tough, I don't think I got (c).

#4 was really complicated, and I just gave up. I got everything else though.

How did you do 5(c)?

For 4, I showed that there existed a one-to-one correspondence between any arbitrary 7x7 board and 8x8 boards that satisfy the conditions. So I got an answer of $ 2^{49}$?

That seems right, now that I think about it.  :wallbash:

For 5c, I basically considered the strategies of the $ 2009$ people when they had to consider the cases when they saw $ 1004$ red hats and $ 1004$ blue hats. By pigeonhole, at least $ 1005$ people make the same choice. Give $ 1005$ people that do this the hat that they would not expect in such a situation, and give the other $ 1004$ people the other color hat. Since $ 1005$ people are guaranteed to be wrong, there is no such strategy for $ 2009$ people.

I was confused as to how we know for sure that for 2008, 1004 is the optimum number of correct hats the people can get. I might just be being dumb right now. :( And why are 1005 people "guaranteed to be wrong"?

It's optimum in that $ 2009 \minus{} 1005 \equal{} 1004$.

By pigeonhole, at least $ 1005$ people will decide on, assuming WLOG, choosing a red hat in such a case. Give an arbitrary subset of $ 1005$ of those people a blue hat, and the other $ 1004$ people red hats. Those $ 1005$ people with blue hats will see $ 1004$ red hats and $ 1004$ blue hats, and deduce they are wearing red hats. Therefore, they are all wrong.

Why will the 1005 people deduce they are wearing red hats, from the fact that they see 1004 red and 1004 blue hats? Sorry, I don't think I'm following you...  :maybe:

Consider the strategies of the $ 2009$ people when they have to decide on the color hat they're wearing when they see $ 1004$ people wearing red hats and $ 1004$ people wearing blue hats. By the pigeonhole principle, at least $ 1005$ people must share the same choice in the color hat they choose. Assume without loss of generality that these people guess that they're wearing a red hat when confronted with this situation. Take $ 1005$ people that make this decision, and give them blue hats. The other $ 1004$ people are given red hats.

When it is time to guess, the $ 1005$ blue-hat-wearers will observe that $ 1004$ people are wearing red hats, and $ 1004$ people are wearing blue hats. Therefore, they will all guess they are wearing red hats. Since $ 1005$ people are already wrong, a maximum of $ 1004$ people can be right, so no set of strategies exists that consistently yields $ 1005$ or more people guessing correctly.

Oh, so the people are using Pigeonhole. I see what you mean. That was clever, to take a specific case of hat distribution, because that's all that was really needed....

Can anyone post the problems for part2, please?

Also, solutions are up! Looks like $ 2^{49}$ is right for number 4. :)

Congratulations! Maybe the first place winner won't come from Montgomery County this year. :P

Haha no, I definitely won't be first place. I'm sure there's someone out there who will ace part 2. And do better on part 1.

He's a 10th grader at Eleanor Roosevelt. We were on the same ARML team.

(I'm just hoping we can continue this trend of not having a Blair person win. Not that I have anything against Blair, but they did keep us from a first place finish at UMD HSPC. :P)

I know what you mean... do you go to Roosevelt?

Nope. I left a hint in my previous post as to what school I go to. :oops:


function
vector
limit
abstract algebra
real analysis
real analysis unsolved

Let $ a$ be a stationary point of a twice continuously differentiable function $ f, f: \mathbb{R}^{n}\rightarrow \mathbb{R}$. Show, that if $ \det{Hf(a)}\neq 0$, then $ a$ is an isolated stationary point, that means there does not exist any other stationary point in the small neighbourghood of $ a$.

More general statement: Suppose $ g_1,\dots,g_n$ are continuously differentiable functions such that
(I) $ g_i(a)\equal{}0$ for all $ i$;
(II) the vectors $ \nabla g_i(a)$ are linearly independent.
Then $ a$ is an isolated point of the set $ Z\equal{}\{x\colon g_i(x)\equal{}0\ \forall i\}$.

Idea of proof: suppose there is a sequence $ x_i\to a$ of other elements of $ Z$. Passing to a subsequence, we may assume that $ \frac{x_i\minus{}a}{\parallel{}x_i\minus{}a\parallel{}}$ converges to some unit vector $ v$. Prove that $ v$ is orthogonal to $ \nabla g_i$ for every $ i$.

Vectors $ \nabla g_{i}(a)$ $ \forall i$ are linearly independent. Therefore $ \det(Df(a))\neq 0$ ($ f\equal{}(g_1,g_2,\ldots,g_n)$). Hence $ Z$ is a manifold.

But then $ v: \equal{}\lim_{i\to \infty} \frac{x_{i}\minus{}a}{\parallel x_{i}\minus{}a\parallel }$ lies in the tangent space of $ Z$ for the point $ a$. Indeed, we can make a smooth path $ \xi: (\minus{}\delta,\delta)\rightarrow M$, which goes through $ a$ and $ x_i$. Since $ x_i$ can come arbitrarily close to $ x$, $ x\minus{}a_i$ will be a tangent vector on $ Z$ in $ a$ for $ i  \to \infty$. We know, that in the tangent space all vectors are orthogonal to $ \nabla g_{i}(a)$ $ \forall i$. But since $ v\neq 0$, we get a contradiction with nonsingularity of $ Df(a)v$, $ Df(a)v\equal{}0$.

There shall also exist a solution to my problem using Taylor series for $ \frac {\partial f}{\partial x_i} \forall i$ in the point $ (a \plus{} h)$.

I get $ \frac {\partial f}{\partial x_i}(a \plus{} h) \equal{} \sum^{n}_{j \equal{} 1}\frac {\partial^2 f}{\partial x_j\partial x_i}(a \plus{} th) h_j$, where $ t\in (0,1)$. Using continuity of $ \frac {\partial^2 f}{\partial x_j\partial x_i}$ in $ a$, I get
$ \frac {\partial f}{\partial x_i} (a \plus{} h) \equal{} \sum^{n}_{j \equal{} 1}\left(\frac {\partial^2 f}{\partial x_j\partial x_i}(a) \plus{} \epsilon_{ji}\right)h_j.$

Now if $ \frac {\partial f}{\partial x_i} (a \plus{} h) \equal{} 0 \forall i$ and some arbitrarily small $ h$, then $ \sum^{n}_{i \equal{} 1}\frac {\partial f}{\partial x_i} (a \plus{} h) \equal{} 0$. But then

$ \sum^{n}_{i \equal{} 1}\frac {\partial f}{\partial x_i} (a \plus{} h) \equal{} \sum^{n}_{i \equal{} 1}\sum^{n}_{j \equal{} 1}\left(\frac {\partial^2 f}{\partial x_j\partial x_i}(a) \plus{} \epsilon_{ji}\right)h_j \equal{} \\
\sum^{n}_{i \equal{} 1}\sum^{n}_{j \equal{} 1}\frac {\partial^2 f}{\partial x_j\partial x_i}(a)h_j \plus{} \sum^{n}_{i \equal{} 1}\sum^{n}_{j \equal{} 1}\epsilon_{ji}h_j \equal{} 0.$

Now I do not see , why this is necessarily a contradiction. Starting a little bit different, we can have a control over  $ \sum^{n}_{i \equal{} 1}\sum^{n}_{j \equal{} 1}\epsilon_{ji}h_j$, which means, that we can make it arbitrarily small. However, I do not see, why $ \sum^{n}_{i \equal{} 1}\sum^{n}_{j \equal{} 1}\frac {\partial^2 f}{\partial x_j\partial x_i}(a)h_j$ is not $ 0$. How is this connected with a nonzero determinant?

You can't arrive at a contradiction starting from $ \sum^{n}_{i \equal{} 1}\frac {\partial f}{\partial x_i} (a \plus{} h) \equal{} 0$, because this equality may well occur in a lot of places.

If you don't sum over $ i$, your argument should yield $ \sum^{n}_{j \equal{} 1}\frac {\partial^2 f}{\partial x_j\partial x_i}(a)h_j\equal{}0$ for every $ i$. And this is a contradiction: $ h$ is in the kernel of the Hessian matrix.


Each day, two people go to a 2 mile circular track. One walks around it once and the other jogs around it once. They each go a constant speed.
One day, they went in the same direction. They both started in the same place but at different times. The walker had gone 3520 ft. when the jogger started. The jogger passed the walker 10 minutes later. After the jogger had passed him, the walker took 30 minutes to finish.
How fast is the jogger in mph?

[hide]3520/5280 = 2/3
anything sub 1 = walker
anuthing sub 2 = jogger
Rate = miles/hour

D=RT
D = RT
R_2 = R_1+(2/3)/(1/6) = R_1+4

2 miles for walker = 2/3 mile + 10 min + 30 min
4/3 mile for walker = 40 min
1 mile for walker  = 30 min
2 = R_1*1/2
R_1 =4

R_2 = 4+4
R_2 = 8
>>8<<[/hide]

sorry, that's wrong :( ... the rate of the walking person is wrong in your answer, too.


Find all $ f: R_{\plus{}}\rightarrow R_{\plus{}}$ that satisfy two conditions:
I)$ \lim_{x\to \infty}f(x)\equal{}0$
II)$ \forall_{x,y, \in R_{\plus{}}} f(xf(y))\equal{}yf(x)$
Unfortunately we don't have the continuity.
Well, I proved few things:
$ f(1)\equal{}1$
$ f(x)\equal{}f^{\minus{}1}(x)$
f is an injection
$ \lim_{x\to 0}f(x)\equal{}\infty$
Unfortunately I couldn't think of anything more  :blush:
The quite obvious function would be $ f(x)\equal{}\frac{1}{x}$, but are there any more? If the answer is positive would the continuity be enough to prove $ f(x)\equal{}\frac{1}{x}$ is the only function?
Many questions, not many answers...  :wink:

[quote="polskimisiek"]Find all $ f: R_{ \plus{} }\rightarrow R_{ \plus{} }$ that satisfy two conditions:
I)$ \lim_{x\to \infty}f(x) \equal{} 0$
II)$ \forall_{x,y, \in R_{ \plus{} }} f(xf(y)) \equal{} yf(x)$
Unfortunately we don't have the continuity.
Well, I proved few things:
$ f(1) \equal{} 1$
$ f(x) \equal{} f^{ \minus{} 1}(x)$
f is an injection
$ \lim_{x\to 0}f(x) \equal{} \infty$
Unfortunately I couldn't think of anything more  :blush:
The quite obvious function would be $ f(x) \equal{} \frac {1}{x}$, but are there any more? If the answer is positive would the continuity be enough to prove $ f(x) \equal{} \frac {1}{x}$ is the only function?
Many questions, not many answers...  :wink:[/quote]

Let $ P(x,y)$ be the property $ f(xf(y)) \equal{} yf(x)$

$ P(1,x)$ implies $ f(f(x))\equal{}f(1)x$ and so $ f(x)$ is bijective (since $ f(1)\neq 0$)
Then $ P(1,1)$ gives $ f(f(1))\equal{}f(1)$ and so $ f(1)\equal{}1$ since $ f(x)$ is injective. As a consequence, we have $ ff(x))\equal{}x$
Then, $ P(x,f(y))$ implies $ f(xy)\equal{}f(x)f(y)$ and $ f(x)$ is multiplicative.
Then, Let $ x>1$ : $ f(x^n)\equal{}f(x)^n$ and so $ f(x)<1$, else $ \lim_{x\to \plus{}\infty}f(x) \equal{} 0$ would be false (take $ n\rightarrow \plus{}\infty$)
So $ x>y$ implies $ \frac{x}{y}>1$ and so $ f(\frac{x}{y})<1$ and so $ f(y)\equal{}f(x\frac{y}{x})\equal{}f(x)f(\frac{x}{y})<f(x)$. So $ f(x)$ is strictly decreasing.
Then, $ f(x)$ is a bijection strictly decreasing and so is continuous.

It is very well known that the only multiplicative continuous functions $ R_{ \plus{} }\rightarrow R_{ \plus{} }$ are $ f(x)\equal{}x^c$ for some $ c\in\mathbb{R}$

Then $ f(f(x))$ implies $ c\equal{}1$ or $ c\equal{}\minus{}1$ and so $ f(x)\equal{}x$ or $ f(x)\equal{}\frac{1}{x}$

And, since $ f(x)$ is strictly decreasing, $ f(x)\equal{}\frac{1}{x}$ and it is easy to check that this necessary condition is sufficient.

another approach..will use some properties pco proved in the above post..

Let $ u$ be a fixed point of $ f$, suppose $ u \not \equal{} 1$. Since $ f$ is multiplicative, also $ u^n$ is a fixed point of $ f$, $ \forall n \in \mathbb{N}$.
But since $ \displaystyle lim_{x \to \infty} f(x)\equal{}0$, we conclude $ u < 1$. Now plugging $ x\equal{}u$, $ y\equal{} \frac {1}{u}$, and using $ f(1)\equal{}1$ and $ f(f(x))\equal{}x$, gives that $ \frac{1}{u}$ is also a fixed point of $ f$ and $ \frac{1}{u} > 1$. But this contradicts that every fixed point $ t$ must satisfy $ t < 1$. So only fixed point of $ f$ is $ 1$. Plugging $ x\equal{}y$ to the initial equation gives $ f(x)\equal{}\frac {1}{x}$.

Thank you very much :wink: Nice proof, well done :wink: That problem was not done by a doctor of mathematics, but it seems not so terrible after all :wink:


http://www.mathlinks.ro/Forum/topic-66332-20.html
bai 6 nam 1993 cua trung quoc


Can anyone solve this question??

(x+2)(x+3)(x+5)(x+6)=10

[hide="Solution"]$ (x\plus{}2)(x\plus{}6) \equal{} (x\plus{}4\plus{}2)(x\plus{}4\minus{}2) \equal{} (x\plus{}4)^2\minus{}4$
$ (x\plus{}3)(x\plus{}5) \equal{} (x\plus{}4\plus{}1)(x\plus{}4\minus{}1) \equal{} (x\plus{}4)^2 \minus{} 1$

Let $ y \equal{} (x\plus{}4)^2$. Then:

$ (y\minus{}4)(y\minus{}1) \equal{} 10$. This is a quadratic equation, and we can solve it (or inspect) to determine that we must have $ y\equal{}6$ or $ y\equal{}\minus{}2$.

In the former case, $ (x\plus{}4)^2 \equal{} 6 \implies x\plus{}4 \equal{} \pm \sqrt{6} \implies x \equal{} \minus{}4 \pm \sqrt{6}$
In the latter case, $ (x\plus{}4)^2 \equal{} \minus{}3 \implies x\plus{}4 \equal{} \pm i \sqrt{2} \implies x \equal{} \minus{} 4 \pm i \sqrt{2}$

Thus, those are the four solutions to your first equation.[/hide]

There is a mistake in the second pair of solutions. It should be

[hide] $ y \equal{} \minus{} 1$, so the solutions are $ x \equal{} \minus{} 4 \pm \sqrt {6}$ and $ x \equal{} \pm i \minus{} 4$ [/hide]

This type of problem is pretty popular in math competitions. Here is a slightly different method

[hide="Method 2"]

"Pair up" the equations as $ (x\plus{}2)(x\plus{}6)$ and $ (x\plus{}3)(x\plus{}5)$. We like to do this because...

$ (x^2\plus{}8x\plus{}12)$ and $ (x^2\plus{}8x\plus{}15)$ is what we get.

Let $ x^2 \plus{} 8x \plus{} 12 \equal{} y$ then:

$ y(y\plus{}3) \equal{} 10$

From here, it's relatively simple quadratic. [/hide]

[hide="I propose this way"]
1st lets make substitution $ k\equal{}x\plus{}4$
then:
$ (k\minus{}2)(k\minus{}1)(k\plus{}1)(k\plus{}2)\equal{}10$

$ (k^2\minus{}4)(k^2\minus{}1)\equal{}10$

$ k^4\minus{}5k^2\minus{}6\equal{}0$

again lets substitute: $ k^2\equal{}t$

$ t^2\minus{}5t\minus{}6\equal{}0$

and when we solve t, we substitute until we get x, so

$ x\equal{}\minus{}4\pm\sqrt{6}$ and $ x\equal{}\minus{}4\pm i$[/hide]

[quote="tony77"][quote="Silverfalcon"]This type of problem is pretty popular in math competitions. Here is a slightly different method

[hide="Method 2"]

"Pair up" the equations as $ (x \plus{} 2)(x \plus{} 6)$ and $ (x \plus{} 3)(x \plus{} 5)$. We like to do this because...

$ (x^2 \plus{} 8x \plus{} 12)$ and $ (x^2 \plus{} 8x \plus{} 15)$ is what we get.

Let $ x^2 \plus{} 8x \plus{} 12 \equal{} y$ then:

$ y(y \plus{} 3) \equal{} 10$

From here, it's relatively simple quadratic. [/hide][/quote]

can you complete it all the way. I still get only 2 answers. Thanks[/quote]

[hide="Rest"]

$ y^2\plus{}3y\minus{}10 \equal{} (y\plus{}5)(y\minus{}2) \equal{} 0$

So, $ y \equal{} \minus{}5, 2$. Now, substituting it back:

$ x^2\plus{}8x\plus{}12 \equal{} \minus{}5 \rightarrow x^2\plus{}8x\plus{}17 \equal{} 0$
$ x^2\plus{}8x\plus{}12 \equal{} 2 \rightarrow x^2\plus{}8x\plus{}10 \equal{} 0$

From here, it's just quadratic formula. [/hide]


Let $ a,b,c$ be a real numbers such that $ 0<a<1,0<b<1,0<c<1$.Prove that these three products,
$ (1\minus{}a)b,(1\minus{}b)c,(1\minus{}c)a$ cannot be bigger than $ \frac{1}{4}$ at the same time?

Suppose for the sake of contradiction that they all are.

Then $ a \plus{} b \plus{} c \minus{} ab \minus{} ac \minus{} bc > 3/4$, that is, $ 1 \minus{} abc \minus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) > 3/4$. Rearranging, $ abc \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) < 1/4$.

I'm a total failure at inequalities, so I'll let someone else find a contradiction.

[hide="solution"]We have $ \left(\frac {1}{2} \minus{} a\right)^2\ge 0, \left(\frac {1}{2} \minus{} b\right)^2\ge 0,$ and $ \left(\frac {1}{2} \minus{} c\right)^2\ge 0$.

We expand each of these: $ a^2 \minus{} a\ge \minus{} \frac {1}{4}$ and symmetric.

We reverse the sign and inequality symbol: $ a(1 \minus{} a)\le \frac {1}{4}$ and symmetric.

We then take the product of all these: $ abc(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\le \left(\frac {1}{4}\right)^3$

This is equal to $ b(1 \minus{} a)\cdot c(1 \minus{} b)\cdot a(1 \minus{} c)$.  Now none of these multiplicands are negative, so there must be one that is less than $ \frac {1}{4}$.  (For if they were all greater than 1/4, then their product would be greater than 1/64.) $ \boxed{\text{QED}}$.[/hide]

Edit: Thanks, Farenhajt.

Just a typo: you should have $ \left({1\over 4}\right)^3$


function
trigonometry
limit
calculus
calculus computations

Determine all real numbers $ a$, for which the function $ \tan(x) \plus{} a\tan(3x)$ has a finite limit at $ \frac {\pi}{2}$ and calculate the limit for those values of $ a$.

Now I got it.

We know that $ \cos(3x)=\cos^{3}(x)-3\sin^{2}(x)\cos(x)$ and $ \sin(3x)=-\sin^{3}(x)+3\cos^{2}(x)\sin(x)$.

So we have

\begin{eqnarray*}
   \tan(x)-a\tan(3x) &=& \frac{\sin(x)\cos(3x)+a\sin(3x)\cos(x)}{\cos(3x)\cos(x)} = \\
   &=& \frac{\sin(x)\cos^{3}(x)-3\sin^{3}(x)\cos(x)-a\sin^{3}(x)\cos(x)+a3\cos^{3}(x)\sin(x)}{\cos^{2}(x)(\cos^{2}(x)-3\sin^{2}(x))}= \\
   &=& \frac{(1+3a)\sin(x)\cos^{2}(x)-(3+a)\sin^{3}(x)}{\cos(x)(\cos^{2}(x)-3\sin^{2}(x))}
\end{eqnarray*}

Now since the denominator goes toward $ 0$, also the numerator must go toward $ 0$ in order for the finite limit. Since $ (1+3a)\sin(x)\cos^{2}(x)$ in the numerator goes toward $ 0$, also $ (3+a)\sin^{3}(x)$ must go toward $ 0$. Hence $ a$ must equal $ -3$.

$ a=-3$:

$ \lim_{x \to \frac{\pi}{2}} \frac{-8\sin(x)\cos^{2}(x)}{\cos(x)(\cos^{2}(x)-3\sin^{2}(x))}=\lim_{x \to \frac{\pi}{2}} \sin(x)  \lim_{x \to \frac{\pi}{2}} \frac{-8\cos(x)}{\cos^{2}(x)-3\sin^{2}(x)}= 0$


How do you make the conjugate line longer over longer expressions like a-bi? When I use the bar command I get a shorter line:
$\bar{a-bi}$


We say that a natural number n is [i]nice[/i] if there exist integers a1, a2, . . . , an such that
a1 + a2 + ... + an = a1a2...an = n.
Find all nice numbers.

[quote="pbornsztein"]Is it allowed to use negative integers a_i ?

Pierre.[/quote]

I think the answer is yes, because if a_i > 0 then only n=1 have this property.

I didnt solve this prb, but I found that if  n=4k+1 then n is nice.

For example  5 = 5.1.1.(-1).(-1) = 5 + 1 + 1 + (-1) + (-1).

Also, we can see that if  n is prime then n is nice <=> n = 4k+1.

Try to continue, friends ...


Suppose we have a number in base 3 between 0 and 1.  Suppose this number has the form
$ 0.a_{1}a_{2}\cdots022\cdots\equal{}0.a_{1}a_{2}\cdots100\cdots$
Naturally, these two numbers are the same.  I need to prove that such a number has two representations in binary.  Any help?  Thank you.

I hope you don't need to prove this, because it's not true: $ \frac {1}{3} \equal{} 0.1_3 \equal{} 0.0222\ldots_3$, but $ \frac {1}{3} \equal{} 0.01010101\ldots_2$ has a unique binary representation.  I think you are misinterpreting something in the problem that this came from.

Okay, that is what I want to hear.  In my problem, I am using the Cantor set and the Cantor function. Obviously, $ f(1/3)\equal{}f(2/3)$.  The problem asks to prove that if $ x<y$, $ x,y\in\Delta$ and $ f(x)\equal{}f(y)$, then $ x$ has two distinct binary representations.  But as you pointed out, $ 1/3$ has only one binary representation, which would contradict the book.  Any thoughts?

See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227672]other thread[/url].


Show that there is no arithmetic progression of 1999 distinct positive primes all less than 12345.

[hide]Assume there is one. Since there is only 1 even prime, the common difference for the 1999 primes must be even. Since 12345/1999  :approx:  6.2, the common difference can be at most 6. Let p be the fourth prime in this arithmetic sequence, and note that p>5 since 5 is the third prime. Also let the common difference be d with d = 2, 4, or 6. Then, since p, p+d, p+2d, p+3d, p+4d are all different modulo 5, one of them is divisible by 5. Since p>5, we have a composite number among these 5 terms. contradiction[/hide]

zabelman, I don't think the problem precludes having composite numbers in the list. It's just that it is necessary for there to be 1999 primes. But[hide]there aren't 1999 positive primes under 12345.[/hide]

I'm not entirely sure what the question is asking.  Although I'd be inclined to agree that complexzeta's objection is the stronger of the two.


calculus integration trigonometry logarithms parameterization function limit

$\frac{1-\cos x}{x}=\int_{0}^{1}\sin(xt)\,dt$.

The integral becomes $2\int_{0}^\infty\int_{0}^{1}\sin(xt)e^{-x}\,dx\,dt$.
Apply Fubini and switch the order of integration. $\int_{0}^{\infty}\sin(tx)e^{-x}\,dx=\text{Im}\int_{0}^{\infty}e^{-(1-it)x}\,dx =\text{Im}\frac1{1-it}=\frac{t}{1+t^{2}}$

Now $\int_{0}^{1}\frac{2t}{1+t^{2}}\,dt=\left.\ln(1+t^{2})\right|_{0}^{1}=\ln 2$.

Introduce a parameter $ t > 0$:
\[ I(t) = \int_{0}^\infty \frac{2-2 \cos x}{x}\cdot e^{-tx}\, dx . \]
We have that
\begin{eqnarray*}I^\prime (t) &=&-\int_{0}^\infty \left( 2-2 \cos x \right) e^{-tx}\, dx \\ \ &=&-2 \cdot \text{Re}\int_{0}^\infty \left( e^{-tx}-e^{-(t-i)x}\right) \, dx \\ \ &=&-2 \cdot \frac1{t}+\frac{2t}{t^{2}+1}.\end{eqnarray*}
Hence, $ I(t) = \log \left( 1+\frac1{t^{2}}\right)+\mathcal C$, where $ \mathcal C$ is some constant.
Now notice that $ x \mapsto \frac{2-2 \cos x}{x}$ is a continuous, bounded function on $ \left[ 0, \infty \right)$ and that $ \lim_{t \to \infty}\int_{0}^\infty e^{-tx}\, dx = 0$. Hence, $ \lim_{t \to \infty}I(t) = 0$. From this we infer that $ \mathcal C = 0$, so $ I(1) = \log 2$.

PS: As always, I don't know how to justify the differentiation under the integral sign.

[quote="perfect_radio"]As always, I don't know how to justify the differentiation under the integral sign.[/quote]
That's one of the big reasons why I like writing this sort of thing in terms of iterated integrals- it's much easier to see when you can apply a Fubini-type theorem.

With that idea, we write $ I(t)=\int_{0}^{\infty}\int_{0}^{t}(2-2\cos x)e^{-sx}\,ds\,dx$
The interchange is easy to justify, and leads to your computation.

[quote="perfect_radio"]PS: As always, I don't know how to justify the differentiation under the integral sign.[/quote]

i personally think leibniz's integral rule i.e. "differentiation under the integral sign" kicks
some major butt !!  :coolspeak:  :yup:


conics
parabola
analytic geometry
graphing lines
slope
algebra unsolved
algebra

Find  value $ m \in R$ for the equation later existence of root $ x \in R$

$ \left( {\sqrt {1 \plus{} x^2 } \plus{} \sqrt {1 \minus{} x^2 } \plus{} 1} \right).m \equal{} 2\sqrt {1 \minus{} x^4 } \plus{} \sqrt {1 \plus{} x^2 } \plus{} \sqrt {1 \minus{} x^2 } \plus{} 3$

[quote="kunny"]Answer: $ 2\sqrt {2} \minus{} 1\leq m\leq \frac {7}{3}$.

Let $ t \equal{} \sqrt {1 \plus{} x^2} \plus{} \sqrt {1 \minus{} x^2}\Longrightarrow 2\sqrt {1 \minus{} x^4} \equal{} t^2 \minus{} 2$, thus the given equation become $ m(t \plus{} 1) \equal{} t^2 \plus{} t \plus{} 1$ where $ 2\leq t^2\leq 4\Longleftrightarrow \sqrt {2}\leq t\leq 2$. We are to find the condition that the parabola $ y \equal{} t^2 \plus{} t \plus{} 1\ (\sqrt {2}\leq t\leq 2)$ has one more intersection points with the line $ y \equal{} m(t \plus{} 1)$ which passes through the point $ ( \minus{} 1,\ 0)$ regardless any slope $ m$. Let $ f(t) \equal{} t^2 \plus{} t \plus{} 1$, we have $ f(\sqrt {2})\leq m\leq f(2)$, yielding the answer $ 2\sqrt {2} \minus{} 1\leq m\leq \frac {7}{3}$.[/quote]

Thank kunny very much. It's nice. :P


O, A, B, C, D  are five distinct points on a circle.

$L_1$, $L_2$, $L_3$, $L_4$, are the Simson lines of O with respect to the triangles  BCD, CDA, DAB, ABC respectively.

Prove that OD is a diameter of the circumcircle of the triangle formed by the lines $L_1$, $L_2$, $L_3$.

Prove also that the feet of the perpendiculars from O to $L_1$, $L_2$, $L_3$, $L_4$ are collinear.

Let $N_{12}, N_{13}, N_{14}, N_{23}, N_{24}, N_{34}$ be the feet of normals from the point $O$ to the lines $AB, AC, AD, BC, BD, CD$, respectively.

The Simson lines $l_2, l_3$ of the triangles $\triangle CDA, \triangle DAB$ meet at the point $N_{14} \equiv l_2 \cap l_3$ on their common side $DA$. Since the angle $\measuredangle ON_{14}D = 90^o$ is right, the locus of $N_{14}$ is a circle with diameter $OD$.

The Simson lines $l_3, l_1$ of the triangles  $\triangle DAB, \triangle CBD$ meet at the point $N_{24} \equiv l_3 \cap l_1$ on their common side $BD$. Since the angle $\measuredangle ON_{24}D = 90^o$ is right, the locus of $N_{24}$ is a circle with diameter $OD$.

The Simson lines $l_1, l_2$ of the triangles  $\triangle BCD, \triangle CDA$ meet at the point $N_{34} \equiv l_1 \cap l_2$ on their common side $CD$. Since the angle $\measuredangle ON_{34}D = 90^o$ is right, the locus of $N_{34}$ is a circle with diameter $OD$.

Thus the vertices of the triangle $\triangle N_{14}N_{24}N_{34}$ formed by the Simson lines $l_1, l_2, l_3$ all lie on the circle with diameter $OD$, the circumcircle of this triangle.

In an entirely similar way, it can be shown that the circumcircle of the triangle $\triangle N_{13}N_{23}N_{34}$ formed by the Simson lines $l_1, l_2, l_4$ is a circle with diameter $OC$, the circumcircle of the triangle $\triangle N_{12}N_{23}N_{24}$ formed by the Simson lines $l_1, l_3, l_4$ is a circle with diameter $OB$, and the circumcircle of the triangle $\triangle N_{12}N_{13}N_{14}$ formed by the Simson lines $l_2, l_3, l_4$ is a circle with diameter $OA$.

Let $M_1, M_2, M_3, M_4$ be the feet of normals from the point $O$ to the Simson lines $l_1, l_2, l_3, l_4$. Since the point $O$ is on the circumcircle with diameter $OD$ of the $\triangle N_{14}N_{24}N_{34}$, $M_1M_2M_3$ is a Simson line of this triangle. Since the point $O$ is on the circumcircle with diameter $OC$ of the triangle $\triangle N_{13}N_{23}N_{34}$, $M_1M_2M_4$ is a Simson line of this triangle. Since the point $O$ is on the circumcircle with diameter $OB$ of the triangle $\triangle N_{12}N_{23}N_{24}$, $M_1M_3M_4$ is a Simson line of this triangle. Finally, since the point $O$ is on the circumcircle with diameter $OA$ of the $\triangle N_{12}N_{13}N_{14}$, $M_2M_3M_4$ is a Simson line of this triangle. As a result, the point $M_1, M_2, M_3, M_4$ are all collinear.

This is one of the possible proofs of the following theorem:

The 4 circumcircles of 4 triangles formed by any 4 lines of general placement are concurrent.


Depends how good you are at guessing then..
Really -theres the basic cases where it looks like you have to guess but can actually logically work out which one must be a mine etc from other squares, but even if there is a guess involved usually one of the squares can be figured out to be a lot better to guess than the other.

you gan get $1000000 if you crack the code

[quote="Valentin Vornicu"][hide="I was"]back in 1995 when I saw Win95 for the first time :D[/hide][/quote]

i thought 95 still had chip's challenge on it? now THAT was an addicting game.

[quote="hello"]I like minesweeper but almost always, especially in expert you have to guess (at least that is what it seems to me) and so its hard to beat it. Its doesnt seem to be too mathematically challenging.[/quote]

As the $1M comment suggests, it is NP-complete (in some deterministic version of the problem that can be formulated as a Yes/No decision problem).

Yes! I am sooo addicted to that game. I play it way to much its just addicting I guess.

Wow... I was so addicted back when I was 6 or 7 - my parents yelled at me cuz I was hogging the computer so much.

Look below to see how good I am at Minesweeper  :D

so im not the only one :P just out of curiosity, what are your best times?

Well done h_s_potter2002!

I never played that game, but the MathChauffeuress had a 95 and a bundle of 97-99s at the expert level.

Back then I did not know whether to sue the authors for taking so much of her from me, or to write them a thank you note for keeping her away from interfering with my own addictions  :D

When I was an actuarial technician 8 or 9 years ago, I would play for 10 to 20 minutes every day -- until I won once going at a fairly fast pace.  I can't recall for sure, but my best time was somewhere in the high double digits, but that involved hitting around 10 or 12 random squares to start each game.

At some point I got very bored of it as I eventually have all video games.

I was about a year ago.  My best score was 126 :(.

[quote="MathChauffeur"][quote="furious"]on the mindsweeper that came w/ windows 98 and earlier, the minesweeper clock could be stopped by clicking both mouse buttons and escape all at the same time. i used that trick and beat expert in 2 seconds. too bad it doesn't work on the newer windows. it's kinda impressive to show ppl that. anywho, i also don't like the fact that you do have to usually guess on expert - that takes the fun out of it.[/quote]It's good that you showed people 2 seconds. If you showed them 67, they might believe the score was real and would be scared away from you till eternity.  :D

We found the 67 score posted on the net by somebody a few years back. Judging the way the story was told, we believed the score was real. We wished we had kept the story and his name, just to see what he’s up to these days. Math may be too easy for him  :)[/quote]

the current highscores are somewhere in the low 40s iirc although i did hear of someone breaking the 40 barrier for expert (may have been a fake)

my personal best is 100, which i got several days ago :)

[quote="alan"]the current highscores are somewhere in the low 40s...[/quote] WOW! It would be something to see those minds and fingers in action. I don't even know how to imagine somebody doing it that fast  :D

You should see the masters of tetris and speed rubix cubing. Gods of players.

[quote="sunnyray"]um.... whats minesweeper...

i've never heard of it, never played it, never got addicted to it, etc, etc

the only thing im addicted to now is spamming :D (j/k) and neopets[/quote]

go to start and games and wala! there your go. Minesweeper is there.

BTW, i played neotpets for a while

[quote="MathChauffeur"][quote="jli"]i finished in 999 seconds but for somereason, it woulnd't let me put my high score up :([/quote] According to "reliable source", high score is defined as 998 or less.[/quote]

ohh... :blush:  :blush:

[quote="alan"]is anyone else addicted to this game? :lol:[/quote]

Well, I like to play once in a while, and is a good time killer, but it is not like an addiction, I really don't play it that much.

The games that I find addictive are those that include some "strategy" issue, for example, FREECIV, I have spent 26 hours in a row playing it, that is why I forbided to my self to play it ever if I am not in vacations (or spring break or something like that).

Best regards,

i was an addicted to...
i was quite good in the intermediate, with a best time of 31 seconds... and quite good also at the expert: 93 seconds (if i recall correctly)...
but it's not a silly game... it's just recognizing some recurring paths, and be able to see much more than it's shown...
 ;)

I like minesweeper but i'm bad it it...  :blush:
I don't think i've ever beat expert but i don't try to often. ;)

I love it but not very good at it..........
My best score is only 150.......... :(

Can anyone say ........... 4 in 5 years?

[quote="alan"][quote="RC-7th"][quote="jli"]it is impossible to get a mine on your first try[/quote]

Why is that?[/quote]

thats how the program works.  i think they want to give you a fighting chance :)[/quote]

Well i always fail at the second......


These are all essentially the same problem  :P

[hide="Solution for 1"]$ 2^{5n\plus{}1} \equal{} 2 \cdot (2^5)^n \equiv 2 \cdot 5^n \pmod{27}$
$ 5^{n\plus{}2} \equal{} 25 \cdot 5^n$
$ \therefore 2^{5n\plus{}1} \plus{} 5^{n\plus{}2} \equiv 27 \cdot 5^n \equiv 0 \pmod{27}$[/hide]
[hide="Solution for 2"]$ 4^{2n\plus{}1} \equal{} 4 \cdot (4^2)^n \equiv 4 \cdot 3^n \pmod{13}$
$ 3^{n\plus{}2} \equal{} 9 \cdot 3^n$
$ \therefore3^{n\plus{}2} \plus{} 4^{2n\plus{}1} \equiv 13 \cdot 3^n \equiv 0 \pmod{13}$[/hide]
[hide="Solution for 3"]$ 3^{2n\plus{}1} \equal{} 3 \cdot (3^2)^n \equiv 3 \cdot 2^n \pmod{7}$
$ 2^{n\plus{}2} \equal{} 4 \cdot 2^n$
$ \therefore 2^{n\plus{}2} \plus{} 3^{2n\plus{}1} \equiv 7 \cdot 2^n \equiv 0 \pmod{7}$[/hide]


geometry
3D geometry
icosahedron
USAMTS
pyramid
trigonometry
AMC

I'm wonding what scores you think you got in round 2.
I think i had 21,  52554

I'm guessing, probably on the better side though.

1. 3 or 4
2. 5
3. 5
4. 3
5. 5

Total: 21-22

Hmm...

4/5
5
5
5
1/2/3 (hope I get a nice grader!)

At worst, 20. At best, 23.

Hopefully, something like 52555

I got all the answers right but basically guessed on problem 2. I'm confident that i nailed the other problems, even though i actually had to make an icosahedron to visualize number 5. I'll be happy with anything that beats the 20 I got on round 1.

I'm thinking 55555, but I'm not sure if my number 3 was rigorous enough...so possibly 55455. I'll be angry if I get anything below 24 though, because I'm in 10th grade, and I really want to use USAMTS as a safety net.

Did anybody else write a program to brute-force test all 512 combinations of N=9?

Which problem did you guys think was hardest on round 2?

Without doubt #5 was the hardest because unless you already knew what an icosahedron looked like, you had to look it up in a book or the internet.  However, you could also say #1 was the hardest to find an elegant solution to without tons of casework.

Hopefully 5 5 5 5 4, depends how nice my grader is for number 5.

Hm.. last year round 2 I had $ 0$ commends, round 3, $ 1$ commend, round 4, $ 2$ commends, this year round 1, $ 3$ commends. Therefore my commends go in an arithmetic progression, and I will have four commends this round, and round 4 this year I will be getting 6 commendeds. $ \square$

Oh, and about which problem was the hardest, I go with #2. I finished #4 before the other problems somehow .. and #1 (just a couple of cases), 5 (pyramids) weren't too bad.

Compared to previous years, are this year's problems harder or easier?

I guess thats based upon opinion, here's my feeling for rounds 2:

1: Both are a lot of casework, but I think this years was slightly easier.
2: I found this years rather difficult, but last year's wasn't very hard
3: Neither was very difficult, I suppose last year's was easier proof-wise
4: Both were fairly brutal (another coloring problem w/ contradictions!). I think last year's was harder.
5: Last year's geometry problem was pretty easy, so I go with this year's being harder.

Adding them up, that makes this years harder.

As for round 1, I did terrible last yr's round 1, this yr I finished in two days, but that's not a very accurate representation of the difficulty of the problems.

Hmm, I made two typos on #5... wrote $ \sin$ as $ \cos$, and forgot to -not- type a square root. :huh:  Perhaps 55554, if I'm not lucky...

#1, I was like, dude! casework! and there were only 3 cases... WAIT #$ *$(#^@# I DID NOT SHOW FOR N<8... w/e... argh
#2, Newton's Sums.
#3, I was lucky that I found one where 120 works... I thought it was 120, and then I was like "oh no I keep getting 240"... but yay, 120 works.
#4, uh? what?
#5, made two typos.

bleh.

[quote="123456789"]Are problems randomized in rounds, or are they categorized per round? Like, for Round 3s of different years, are the problem types similar?[/quote]

I'm fairly certain that problem types being similar would just be coincidences. I imagine that the writers would try to balance the types of questions on each round, which would probably cause the coincidences.

[quote="123456789"][quote="diophantient"]Initially USAMTS was sorted by subject, but in recent years this hasn't been true (now it's more sorted in terms of difficulty).[/quote]

When you say "difficulty", do you mean that Round 1 is easier than Round 2, which is easier than Round 3? Or do you mean the problems on each round (#1 is easier than #2, is easier than #3, etc.)?[/quote]
Both, I think.

#1 wasnt that hard i thought. heres what i did: (see if you can decipher my diagram :) )

ATTACHED

That's essentially the same as the brute force argument.

Well it took all of five minutes, which is not, in my opinion, long enough to be awarded the title of "brute forced".

It is just listing out all of the cases in a shorthand way. Yes, it does not take as long as writing them out one at a time, but how long a solution takes to find does not determine whether or not a solution is brute force, the actual solution in comparison to what is percieved as an elegant and concise solution determines whether or not it is brute force.

Um, listing them all out is "brute force." Definitely better brute force than programming though, but nevertheless brute force.

Enough arguing.

I'm thinking 54421 = 16 at worst, and 55543 = 22 at best.

That means I'll need a 21-27 next round to make AIME :( .

I thought you said you had handwaved #3?

i think I did but i'm not sure, hence my predictions


Let $ n$ be a natural number. Prove that there exists integers $ a_1, a_2, a_3,...,a_k > 1$, such that $ a_1\plus{}a_2\plus{}...a_k \equal{} n(\frac{1}{a_1}\plus{} \frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_k})$.

My attempt:

[hide]By AM-HM,
 $ \frac{a_1\plus{}a_2\plus{}...a_k}{k} \geq \frac{k}{\frac{1}{a_1}\plus{} \frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_k}}$

Since we're exists a set of numbers, there is the case when $ n\equal{}k$, since n is just any natural number. So,

$ \frac{a_1\plus{}a_2\plus{}...a_k}{n} \geq \frac{n}{\frac{1}{a_1}\plus{} \frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_k}}$

I don't know what happens after this though..[/hide]

Inequalities tell you that things can't happen; generally, they don't say much about whether things [b]can[/b] happen.  You should try a different approach entirely.

[hide="Big hint"] What happens when you try to pair the terms on the LHS with the terms on the RHS? [/hide]

What if $ n \equal{} 1$? Then we have $ a_1 \plus{} \dots \plus{} a_k \equal{} \frac 1{a_1} \plus{} \dots \plus{} \frac 1{a_k}$, but since $ a_j > 1$, $ a_j^2>1$ as well, so it follows that $ a_j > \frac 1{a_j}$ for all $ j$, so equality can't be achieved. :maybe:

hmm, the solution from the book i got it from is stupid. It says "since n is a composite number, it can be written as $ n\equal{}a*b$, where a and b are integers greater than 1. Consider $ a*b(\frac{1}{a}\plus{}\frac{1}{b})$. Notice how it's equal to $ (a\plus{}b)$, which satisfies the conditions of the problem"... I forgot to put that n is composite but I didn't know it'd be that crucial for the solution  :blush:

[quote="modularmarc101"]I forgot to put that n is composite but I didn't know it'd be that crucial for the solution  :blush:[/quote]
I was about to say... anyway, why is that a "stupid" solution?

Well it was the last problem of an Olimpic test and the solution is like 3 sentences when the problem looks so hard (at least for me)

OK, so the solution given in the book shows that just 2 nos. are enough for any $ n$.
How does one think of something like that? I was trying for a generic proof (any no. of nos.)
The solution given in the book certainly didn't strike me.
How does one avoid traps like that, and think of the simplest way possible? Is experience the key?

Edit: Oh yeah, I didn't know that $ n$ was composite, but anyway, could someone answer my question with a general perspective (not for this particular prob.)

The problem itself is vague; do you mean "there exists a $ k$" or "for any $ k$"? I'm pretty sure it's "there exists a $ k$", seeing as the "any" statement is not true.


function
combinatorics theorems
combinatorics

I want to find a closed expression for $ \sum_{n \equal{} 0}^\infty B_n \frac {x^n}{n!}$, with $ B_n$ being the $ n^{th}$ Bell number. How do I go about that?

Thanks in advance,

Los

The following is a very good reference on generating functions:
[url]http://www.math.upenn.edu/~wilf/DownldGF.html[/url]

I already heard that generating function is the most powerful tools to solve combinatorics, is this right? Who can give me some example for this

See for example: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141592