Let $n$ be a natural number.  Define $\sigma(n)$ and $d(n)$, respectively, to be the sum of the divisors of $n$ and the number of divisors of $n$.  Show that:

$\frac {\sigma (n)}{d(n)}\geq \sqrt {n}$

When does equality hold?

If you use any formulae for $\sigma (n)$ or $d(n)$, I expect you to prove them.

Cute!!
[hide][hide] Don't Peek  [hide] Really  Don't peek  [/hide]
[hide](BIG Hint: Isn't the left hand side just the average([b]AM,[/b] that is)  value of sigma(n))[/hide][/hide][/hide]

solution below:

[hide]

lemma 1:  the product of the divisors of n is $n^{\frac{d(n)}{2}}$

pf:  we separate the cases of n a square, and n a nonsquare

1) if n is a non square, then we can pair up each of the divisors of the form d and n/d, which has product n.  Therefore in the product of all the divisors, we have d(n)/2 pairs, each with product n, so the overall product is $n^{\frac{d(n)}{2}}$

2) if n is a square, we can do the same as above, but we cannot pair $\sqrt{n}$ with anything so our product is $n^{\frac{1}{2}}*n^{\frac{d(n)-1}{2}} = n^{\frac{d(n)}{2}}$

now, s(n)/d(n) is just the arithmetic mean of all the divisors of n, so by the AM-GM inequality we have

$\frac{\sigma (n)}{d(n)} \ge (n^{\frac{d(n)}{2}})^{\frac{1}{d(n)}} = \sqrt{n}$

equality occurs when all the divisors of n are equal, but this cannot happen for n > 1 (since 1 and n are divisors of n) so equality occurs for n = 1.[/hide]

Nicely done.  Here was my solution (gyan found the same)

[hide]$\sigma (n)=\prod_{p\vert n}(1+p_i+p_i^2+\cdots p_i^r_i)$ where ${p_i}^{r_i}\vert\vert n$

and $d(n)=\prod (r_i+1)$

and, by the FTA, $\sqrt n=\prod p_i^\frac {r_i}{2}$

However, $\frac {(1+p_i+p_i^2+\cdots p_i^r_i)}{(r_i+1)}\geq p_i^\frac {r_i}{2}$ by AM-GM.  After we run through all the $p_i$'s and multiply them together, we're done.[/hide]

The following problem appeared in Belarus Olympiad 1999:

For any positive integer $n>1$ show that $d(n)\sqrt{n} < \sigma(n) < \sqrt{2d(n)}n$.

I thought as zscool, but I think it is easier than all that :
For each positive divisor $d$ of $n$, the number $\frac n d $ is also a divisor of $n$.
Then, we may pair the positive divisors $(d, \frac n d)$ for $d < \sqrt n$. Note that the only divisor left is $\sqrt n$ in case where $n$ is a square.
Moreover $d + \frac n d \geq 2 \sqrt n$.
Summing over all the pairs (and including $d= \sqrt n$ if necessary), we obtain the desired result.

Pierre.

[quote="blahblahblah"]Are you sure the problem is correct?  All that is left is to prove $1<\sqrt {2d(n)}$, which is obvious.[/quote]

Sorry about that. I corrected the problem.


What angle does the minute hand and hour hand of an analog clock make at 4:25?

[hide="does"]
It's basically the same, but for each forth of the space between the two numbers, there is 7.5 degrees.
So, the answer would be $\boxed{17.5}$[/hide]
EDIT: Ok, it's the second one.

I'm guessing in your question, the hour hand moves as the min hand moves since they do in most questions.
[hide]

at 4:00, the hands forme a $4 \cdot 30=120 \text{ degrees}$

The speed of the MIN hand is: $\frac{360}{60}=6 \text{ degrees/minutes}$
The speed of the HOUR hand is: $\frac{30}{60}=0.5 \text{ degrees/minutes}$

Therefore they close in by $6-0.5=5.5 \text{ degrees/minutes}$

Since it has been closing in 5.5 degrees ever minute since 4:00, it has closed in $25\cdot5.5=137.5 \text{ degrees}$

Distance between the hands equal $\mid120-137.5=-17.5\mid=17.5 \text{ degrees}$

[/hide]

[quote="pascal_1623"]I think there are 2 solutions.
One with the little hand moved because the minutes is 25, or it's just 4.
I don't know which one the problem is asking for :?[/quote]

A clock always has its hour hand moving [i]with[/i] its minute hand.  ;)  The hour hand never stays still during the hour.

akimatsu's answer is correct.


Consider the parabola $y=a(x-h)^2+k$.  What is its focal length?

I've been looking over some old Minnesota competitions, and the official solutions appear to be using $\frac{a}{4}$ for the focal length, but my own calculations show over and over that it is $\frac{1}{4a}$.  Which is correct and why?

$\frac{1}{4a}$ is correct since the focus of that parabola is $(h, k+a)$.


#2

We will say that the integer $n$ is fortunate if it can be expressed in form of $3x^2 + 32y^2$, where $x$ and $y$ are integers. Prove that if $n$ is fortunate, then so is $97n$.

I have no idea on how to do this problem. Sorry if it was posted before, and if it did, somebody can direct me to that link.

It seems reasonable to guess that $97(3x^2+32y^2) = 3(ax\pm by)^2 + 32(cx\mp dy)^2$, where $a,b,c,d \in \mathbb{Z}$.  This is the main idea, since if we can find those coefficients, the result follows.  Comparing coefficients, we have

$3a^2+32c^2=3\cdot 97$
$3b^2+32d^2=32\cdot 97$
$6ab = 64cd \iff 3ab = 32cd$

From this first equation, we note that $3|c$ so we try $c=3$ (because we cannot have $c=0 \iff a^2=97$), and $a=1$ works.  Now, from the second equation, we note that $8|b$, so we do the same thing as before.

$b=0 \iff d^2=97$
$b=8 \iff 6+d^2=97$
$b=16 \iff 24+d^2=97$
$b=24 \iff 72+d^2=97$
$b=32 \iff 96+d^2=97$

Seeing that the last two have a solution, we have $b=32$, $d=1$, since this satisfies $3ab = 32cd$ and the other solution does not.  So finally, we arrive at the identity

$97(3x^2+32y^2) = 3(x\pm 32y)^2 + 32(3x\mp y)^2,$

and we are done.

Here's a related problem (probably posted on the forum somewhere):

Prove that for all integers $n\geq 3$, there exists integers $x,y,$ both odd, such that $7x^2+y^2 = 2^n$.


Euler
function
calculus
calculus computations

Ok so I dug through the internet & somewhat understand the Euler method theory but I'm a practical person.
I have y'= -2-3x-2y with initial values of y(0)=2, step size of 0.5
I need the y values of y1, y2, ...y4

All i need is to see it done once. Could someone run me through the steps. I assume you need to solve for Y(x). If that's so should I use undertermined coefficient technique to get exact solution? Then how do I plug into Yn+1=Yn+Y'(delta)x

Thanks

In this problem, we have a function $f$ such that $f(x,y)= -2-3x-2y$ for all real numbers $x,y$.  We are trying to approximate the solution of the ODE $y'(x)=f(x,y(x)), y(0)=2$.

$y_0=2$, $x_0 = 0$, and $\Delta x = .5$.  $y_{k+1}=y_k + \Delta x f(x_k,y_k)$ for all non-negative integers $k$.

$y_1=y_0 + \Delta x f(x_0,y_0) = y_0+\Delta x(-2-3x_0-2y_0)=2+.5(-2-3*0 -2*2)=0$.

$x_1=x_0 + \Delta x = 0 + .5 = .5$.

$y_2=y_1 + \Delta x f(x_1,y_1)=0 + .5*(-2-3*.5-2*0)=-\frac{7}{4}$.

$x_2=x_1+\Delta x = .5 + .5 = 1$.

And so on.

You don't need to solve for $y(x)$ in order to use Euler's method.  In fact, the reason Euler's method is useful is that it allows us to approximate the values of $y$ without having a nice, explicit formula for $y$.  Most differential equations are too difficult to solve exactly; that is why we have to resort to numerical methods like Euler's method.

Hope that helps.

Out of curiosity, what are you learning this for?

:idea: That was the best way anyone has explained it to me (& the 1st time I've finally understood it). Thanks! I'm a conditional grad student for Fire Protection Engineering. One of the prerequisites was to take some undergrad engineering courses that I didn't take during my undergrad. Diffy's being one of them :-) yea!

OK so normal equations, I'm all over now thanks. But now I'm thrown a curve ball. I did a Newton's Law of Cooling problem which I solved for everything with

DT/dt=k(T-T(of surroundings)).

My constant (k) ended up being -2/50 with initial temp of 90 and surrounding temp of 25. Then I'm asked to use Euler with step size of 2min to estimate what temp will be after 10min [T(10)].

I assume I need to set this up in the Yn+1=Yn+(f(y,x))(step) format.

Using like symbols I set Y=T [temp of object] and X=t [time]
So I did T1=90+(-2/50(90-25))(2).

I know this is wrong so I'll use this funny little emoticon to show what I think you are all doing after reading this  :rotfl:

I'm not asking anyone to solve this for me (I've been recently blamed of trying to get easy homework solutions), but if you could steer me in the right direction on how to set the right Euler equation up that would be swell.

What you did looks correct to me.  What makes you think it's wrong?

[quote]So I did T1=90+(-2/50(90-25))(2). [/quote]

I would write that as $T_1=90+(-2/50)(90-25)(2)$, just to make sure the order of operations is clear.  Since M comes before D in "Please excuse my dear Aunt Sally", an expression like 8/2*4 for example could be interpreted as 1 instead of 16.

[quote]I've been recently blamed of trying to get easy homework solutions.[/quote]

I don't have that impression.  Your questions seem reasonable to me.  The goal is to understand the material as well as possible, and this website seems to be helping you with that.

WOO HOO. That was right I guess. My algebra was crapola so I kept coming up with a higher temp than the object normally was that's why I thought it was wrong. So I have to say these forums rock. I went with not getting Euler at all to being able to apply it to (I think) a reasonably complicated application in a short amount of time. Thanks of course to people like yourself who are willing to help.

I posted this same thing in another forum and some guy was basically attacking me. First he thought it impossible that i "searched the internet" and couldn't find one single solution. That's because all of those sites run you through the thoery which sometimes is hard to grasp unless you see an actual example with numbers like you showed me. Then that's where he said that no one wanted to help because they thought I was just trying to get answers to a specific homework problem. Anyway sorry to ramble in this post. Just feels good to have my math confidence back.

Cheers.

Cool.

So, just to make sure it's clear... You calculated $T_1$ correctly.  $T_1$ is an Euler's Method approximation to $T(t_1)=T(2)$.  Since you're really interested in $T(10)$, you'll need to keep going with Euler's Method, generating $T_2, T_3, T_4,$ and finally $T_5$, which is an approximation of $T(t_5)=T(10)$.

I think this is a great website too.   :)

Oh yeah about that. I calculated out to y10,x10 & it wasn't coming out right then I realized ohhh since the step size is 2 minutes T(10) is only at (y5,x5). That's how we learn i guess.


Write a program that, given two numbers, $a$ and $b$, determines the smallest number in $\overline{a,b}$, which has the maximum number of divisors (call this number $t$). If there are other numbers in the interval that have $t$ divisors, write them also.

Restrictions: $1 \leq a \leq b \leq 10^9$ and $0 \leq b-a \leq 10^4$.

Maximum execution time: $1 ~ \frac{\textrm{second}}{\textrm{test}}$.

I originally thought of a brute-force algorithm, but I realised that would too slow. When I looked at the official "hints", I saw that they practically encouraged that (i.e. factorizing the numbers).

The program written according to these hints runs in $\approx 7 \textrm{ seconds}$ for the worst-case scenario. I think I'm doing smth wrong for sure. Can you help me improve my program?

Another strange thing is that the evaluator they used gives me the maximum $100$ for my program (They don't actually consider the worst-case scenario in their tests).

I have solved this problem also, and my program runs in 0.1s for worst-case from test data if compiled with GCC. Since b-a is pretty small (only 10.000) , you just have to optimize the way you count how many divisors does a number have. A classic way is to use a O(sqrt(n)) factorizing algorithm (try each divisor <= sqrt(n) , if there is a divisor > sqrt(n) there is only one and it is easy to find it out). To optimize this, precompute first a list of prime number up to sqrt(10^9) using Eratostene's sieve (which is O(n*lg n) for primes <= n), and when finding out divisors divide with prime numbers from that list and find it's power. I have also attached code if my explanation is not clear enough.

[code]#include <stdio.h>

#define FIN "maxd.in"
#define FOUT "maxd.out"

long A, B, P[4000], np, n, max_d, cnt;
char sieve[32000];

long divisors(long n)
{
    long i, j, d = 1;

    for (i = 0; i < np && P[i] * P[i] <= n; i++)
    {
        for (j = 0; n % P[i] == 0; j++, n /= P[i]);
        d *= j + 1;
    }
    if (n > 1) d *= 2;
    return d;
}

int main(void)
{
    long i, j, d;

    freopen(FIN, "r", stdin);
    freopen(FOUT, "w", stdout);

    scanf("%ld %ld", &A, &B);

    P[np++] = 2;
    for (i = 3; i <= 32000; i += 2)
    {
        if (sieve[i]) continue;
        P[np++] = i;
        for (j = i * i; j <= 32000; j += 2 * i)
            sieve[j] = 1;
    }

    for (i = A; i <= B; i++)
    {
        d = divisors(i);
        if (max_d < d)
           n = i, max_d = d, cnt = 1;
        else
        if (max_d == d)
           cnt++;
    }

    printf("%ld %ld %ld\n", n, max_d, cnt);

    return 0;
}[/code]

I assume that $\overline{a,b}$ means an interval between a and b.

In that case, I think it suffices to check all of the numbers in the interval - the question is how.

I'd do it this way: for a number X, decompose it into primes (it is enough to check the numbers from 1 to sqrt(X))(you can pre-compute the primes to make things faster, there is .. I guess no more that 1000 primes less than sqrt(10^9)). With basic combinatorial ideas you can compute the number of divisors.
I think this should work under a second.

I hope it helps ...

domino, on my borland c++ 3.1 your program runs in $\approx 1.5 \textrm{ seconds}$. way lot better than mine.

how is the big time difference explained? ( i tried to modify my program to look more like yours, but i got about the same time; below is the first version of the program )

[code]#include <fstream.h>
#include <time.h>
#include <math.h>
#include <conio.h>

unsigned int prime[10000],pt;
fstream f;

int prim(unsigned int n)
 {
  if(n<=1)
   return 0;
  if(n==2)
   return 1;
  if(n%2==0)
   return 0;
  unsigned i;
  for(i=3;i<=sqrt(n);i+=2)
   if(n%i==0)
	return 0;
  return 1;
 }

void gen(void)
 {
  unsigned int j,i;
  prime[0]=2;
  i=1;
  for(j=3;j!=1;j+=2)
   {
	if(prim(j))
	 {
	  prime[i]=j;
	  i++;
	 }
   }
  pt=i-1;
 }

void main(void)
{
long ce,cs=clock();
clrscr();
long a,b,t,nm,y,p,g,max=0,contor=0,z;
gen();
f.open("maxd.in",ios::in);
f>>a>>b;
f.close();
for(t=a;t<=b;t++)
 {
  y=t;
  p=1;
  for(g=0;g<=pt && prime[g]<=sqrt(t);g++)
   {
	 z=0;
	 while(y%prime[g]==0)
	  {
	   z++;
	   y/=prime[g];
	  }
	 p*=(z+1);
   }
  if(p>max)
   {
	contor=1;
	max=p;
	nm=t;
   }
   else
  if(p==max)
   contor++;
 }
f.open("maxd.out",ios::out);
f<<nm<<' '<<max<<' '<<contor<<endl;
f.close();
ce=clock();
cout<<(ce-cs)/CLK_TCK<<" secunde"<<endl;
}[/code]

btw, i don't think you do the prime test in $O(\sqrt{n})$, but rather in exponential time. the same for factorization

The time difference is because of efficient coding. As a general advice, you should quit using "fstream" because it is very slow for big files (this is not the case now). Code like for (i = 3; i <= sqrt(n); i += 2) is really bad, it's much better to do for (i = 3; i * i <= n; i += 2). Also, for finding primes you use an algorithm which is worse that Eratostene's sieve (which is O(n*lg n) like i said before), so try using the sieve. Also, i don't understand why you say that the the prime test and factorization are exponential!  :huh:

These slow algorithms do not cause such a big difference (about $0.07$ seconds), since here we are dealing with relatively small numbers

anyway, i realised that my mistake was writing

[code]for(t=a;t<=b;t++)
 {
  y=t;
  p=1;
FOR(G=0;G<=PT && PRIME[G]<=SQRT(T);G++)
   {
    z=0;
    while(y%prime[g]==0)
     {
      z++;
      y/=prime[g];
     }
    p*=(z+1);
   } [/code]

instead of

[code]for(t=a;t<=b;t++)
 {
  y=t;
  p=1;
FOR(G=0;G<=PT && PRIME[G]<=SQRT(Y);G++)
   {
    z=0;
    while(y%prime[g]==0)
     {
      z++;
      y/=prime[g];
     }
    p*=(z+1);
   } [/code]

my program, with this tiny modification, runs the worst-case in $\approx 1.6 \textrm{ seconds}$


trigonometry
function
complex numbers
calculus
calculus computations

If arcsin z is the inverse of sin z, the sine function defined in the usual way on the complex numbers, find an expression for arcsin z.

Just use the identity

$ \sin z \equal{} \frac{e^{iz} \minus{} e^{\minus{}iz}}{2i}$

and then replace $ z$ by $ \arcsin z.$


limit
probability
expected value
probability and stats

Suppose we have a  n X n grid.  The computer generates a random ordering on the set {1,2,...,n^2} and starts to occupy the girds on the chessboard at the rate of 1 position per second. We assume that a number 1<= m <= n^2 is corresponding to (i,j) where m=i+(j-1)n.

The question is: If 1<=k<=n, what is the expectation of the seconds E(k) when we observe the FIRST kXk occupied grid?

For instance, trivial answers: E(1)=1, E(n)=n^2.

This question can generalize to higher dimensions, then ask the first time to find a k-cube.

let's be less ambitious and find $ \beta$, if it exists, such that
\[ \lim_n \frac{E(n,k)}{n^{\beta}}\]
is non-trivial, where $ E(n,k)$ is the expected first time that a square of side $ k$ appears inside a big square of length $ n$. (Poissonization?)

@hankbai: do you have an answer to this question ?

I make a bet that for every $ \epsilon > 0$ and $ k \geq 2$ we have $ \lim_{n \to \infty} \frac {E(n,k)}{n^{\beta \plus{} \epsilon}} \equal{} 0$ while $ \lim_{n \to \infty} \frac {E(n,k)}{n^{\beta \minus{} \epsilon}} \equal{} \plus{} \infty$ where $ \beta \equal{} 2 \minus{} \frac {2}{k^2}$, even though I have absolutely no proof  :lol:

ps: how would you compute a reasonable approximation of such a quantity, say for $ k\equal{}10$ ?

I guess the answer depends on whether you consider a subgrid as contiguous or just a product $ A \times B$ where $ A$ and $ B$ are $ k$ element subsets.  Either way, it seems like you may be able to (for fixed $ k$)  get good estimates in the limit as $ n$ tends to infinity by looking at the first and second moments.

There's a general "coupling" rule here that may make things easier (again, assuming you're talking about limits instead of fixed $ n$)  Instead of considering a model where we choose $ m$ points, consider one where each point is [b]independently[/b] selected with probability $ \frac {m}{n^2}$.  Assuming both $ m$ and $ n^2 \minus{} m$ tend to infinity, this will give you the right threshold up to a $ 1 \plus{} o(1)$ multiplicative factor (essentially this is because the containing a grid is a monotone property -- if it holds almost surely in our new model and our new model almost surely has fewer than $ m_0$ selected squares, it also holds almost surely for a random set of $ m_0$ selected squares).

that would be very interesting if one could make this a little more quantitative  :)  As you, to get this $ 2 \minus{} \frac {2}{n^2}$, I assumed that each point was independently selected with a fixed probability. Nevertheless, I cannot quite understand what you really meant in your last post ?

What I meant by my post was this:

Suppose we knew that a set of points independently selected with probability $ p$ almost surely contained a grid.  I claim that a randomly chosen selection of exactly $ (1 \plus{} \epsilon)pn^2$ points also contains such a grid, for any positive $ \epsilon$.  To see this, consider the following process.

I independently select each point from the grid with probability $ p$.  If I ended up selecting more than $ (1 \plus{} \epsilon)pn^2$ points, I call my attempt a failure and quit.  Otherwise, I randomly choose additional points from the grid until I have $ (1 \plus{} \epsilon)pn^2$ points in total.

If I condition on my not failing, I achieve a uniform distribution on the sets of $ (1 \plus{} \epsilon)pn^2$ points.  But I know that my failure probability is $ o(1)$ (law of large numbers), and I know that I almost surely had a grid even before adding the additional points in.  This means the uniform distribution on $ (1 \plus{} \epsilon)pn^2$ points also contains such a grid.  The same argument would show that if selecting independently with probability $ p$ almost surely does not contain a grid, then a random selection of $ (1 \minus{} \epsilon)pn^2$ points also almost surely does not.

If you've worked with random graphs at all, this sort of coupling argument is used to show the equivalence of the two different models of Erdos-Renyi graphs (selecting edges independently, or selecting $ M$ edges uniformly) for many purposes.

yes thanks. And in the original question, one easily gets the right threshold $ \beta \equal{} 2\minus{}\frac{2}{k^2}$ by computing the expected number of $ k \times k$ squares, which is $ E_p[N] \equal{} (n\minus{}k)^2 \cdot p^{k^2}$ where $ p\equal{}\frac{n^{\beta}}{n^2}$. Easy arguments then show that if one put less that $ n^{\beta\minus{}\epsilon}$ boxes, one will almost never get a $ k \times k$ square, while if one puts more that $ n^{\beta\plus{}\epsilon}$ boxes, this will almost surely happen (as $ n \to \infty$).


[quote="2006"]$\ a,b,c\geq0$ and $\ a+b+c=3$ ,Prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3 \geq a^{2}+b^{2}+c^{2}-3$
    Good luck  :D[/quote]

i think you tryed to change a bit Vasc`s inequality from the Romanian TST 2006, but you didn`t have many success, and i don`t understand what`s with the $-3$ in each side, and a friendly advice, stop spamming. :blush:

[quote="pohoatza"]i don`t understand what`s with the $-3$ in each side[/quote]It's a test problem of fills up measures.

For positive real numbers  $x,y,z$, the following inequalities

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3xyz \geq x^{2}+y^{2}+z^{2}-3$,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3\min\{x,y,z\}\geq x^{2}+y^{2}+z^{2}-3$,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3 \geq x^{2}+y^{2}+z^{2}-3\max\{x,y,z\}$,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3 \geq x^{2}+y^{2}+z^{2}-3\max\{x^{2}y,y^{2}z,z^{2}x,xy^{2},yz^{2},zx^{2}\}$

hold with $x+y+z=3$.

[quote="Ji Chen"]For positive real numbers  $ x,y,z$, the following inequalitiy

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 3xyz \geq x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} 3$

holds with $ x \plus{} y \plus{} z \equal{} 3$.[/quote]$ \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)\left(\frac {x \plus{} y \plus{} z}{3}\right)^4 \minus{} 3xyz \minus{} \left(x^{2} \plus{} y^{2} \plus{} z^{2}\right)\left(\frac {x \plus{} y \plus{} z}{3}\right)$

$ \plus{} 3\left(\frac {x \plus{} y \plus{} z}{3}\right)^3$

$ \equal{} \sum{\frac {(y \minus{} z)^2}{2yz}\left[x\left(8x^2 \plus{} 51yz\right) \plus{} (y \plus{} z)\left(6x^2 \plus{} 2y^2 \plus{} yz\plus{} 2z^2 \right)\right]}\geq 0.$

This problem is record as Exercise 2.24 on page 87 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.china-pub.com/1786406]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

Another proof due to Yi-Chao Ye on pp. 210~211.

[quote="Ji Chen"]For positive real numbers  $ x,y,z$, the following inequalitiy

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 3\min\{x,y,z\}\geq x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} 3$

holds with $ x \plus{} y \plus{} z \equal{} 3$.[/quote]

assume $ x \geq y \geq z$,

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} 3 \minus{} 3z \minus{} x^{2} \minus{} y^{2} \minus{} z^{2}$

$ \equal{} \frac { x \plus{} y}{xy} \plus{} 2xy \minus{} (x \plus{} y)^{2} \plus{} \frac {1}{z} \plus{} 3 \minus{} z^{2} \minus{} 3z$

 $ \geq 2\sqrt {2(3 \minus{} z)} \minus{} (3 \minus{} z)^{2} \plus{} \frac {1}{z} \plus{} 3 \minus{} z^2 \minus{} 3z$

$ \equal{} \frac {4 \minus{} 4z}{\sqrt {2(3 \minus{} z)} \plus{} 2} \plus{} \frac {(1 \plus{} 2z)(1 \minus{} z)^{2}}{z} \geq 0 .$

[quote="Ji Chen"]
For positive real numbers  $ x,y,z$, the following inequality

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 3 \geq x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} 3\max\{x,y,z\}$

holds with $ x \plus{} y \plus{} z \equal{} 3 .$[/quote]

assume $ x\geq y\geq z$.

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 3 \plus{} 3x \minus{} x^{2} \minus{} y^{2} \minus{} z^{2}$

$ \equal{} \frac {3 \minus{} x}{yz} \plus{} 2yz \minus{} (3 \minus{} x)^{2} \plus{} \frac {1}{x} \minus{} 3 \minus{} x^{2} \plus{} 3x$

$ \equal{} \frac {(2(1 \minus{} yz) \plus{} yz(x \minus{} 1))( y \minus{} z)^{2}}{2yz(3 \minus{} x)} \plus{} \frac {3(x \minus{} 1)(x(3 \minus{} x)^{2} \plus{} 2(x \minus{} 1))}{x(3 \minus{} x)} \geq 0.$

[quote="Ji Chen"]

For positive real numbers  $ x,y,z$, the following inequality

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 3 \geq x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} 3\max\{x^{2}y,y^{2}z,z^{2}x,xy^{2},yz^{2},zx^{2}\}$

holds with $ x \plus{} y \plus{} z \equal{} 3$.[/quote]

assume $ x\geq y\geq z$.

$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \minus{} 3 \plus{} 3x^{2}y \minus{} x^{2} \minus{} y^{2} \minus{} z^{2}$

$ \equal{} \frac {3 \minus{} x}{yz} \plus{} 2yz \minus{} (3 \minus{} x)^{2} \plus{} \frac {1}{x} \minus{} 3 \minus{} x^{2} \plus{} 3x^{2}y$

$ \equal{} \frac {(2(1 \minus{} yz) \plus{} yz(x \minus{} 1))( y \minus{} z)^{2}}{2yz(3 \minus{} x)}\plus{}\frac {3(x \minus{} 1)(x^{2}(2x \minus{} 5)^{2} \plus{} (4x \minus{} 1)(3 \minus{} x)^{2} \plus{} (2x \plus{} 1))}{8x(3 \minus{} x)}$

$ \plus{}\frac {3x^{2} (y \minus{} z)}{2}\geq 0.$


Source: Me

Give an explicit formula $f(n)$ that when taken at all nonzero integers (in order) gives the sequence $\ldots,\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{1}{1},1,2,3,4,\ldots$

What do you mean by "in order" for the non-zero integers? [hide] Do you mean that you want $f(n) = n, f(-n) = \frac{1}{n}$ for $n >0$?[/hide]

[quote="JBL"]What do you mean by "in order" for the non-zero integers?[/quote]I thought that might be a bit ambiguous. Basically I'm looking for $f(n)$ such that $f(1)=1,f(2)=2,f(3)=3,\ldots$ and $f(-1)=\frac{1}{1},f(-2)=\frac{1}{2},f(-3)=\frac{1}{3},\ldots$. It should be pretty simple now.[quote][hide] Do you mean that you want $f(n) = n, f(-n) = \frac{1}{n}$ for $n >0$?[/hide][/quote]That would work but I'm looking for a single explicit expression for $f$.

I didn't mean that as an answer, really -- I was just trying to express what you were asking for.  It's a little hard to do without making something that begins to look like the answer.

Oh I see now. In that case you're exactly right.

[quote="Treething"]Hmm...

f(n) = n^(n/|n|)

I feel special :D ! I beat the mathematical masterminds of the AoPS :D ! j/k[/quote]

Close.. theres one thing different from blahblahblahs answer that makes yours a tiny bit wrong :)

[quote="santosh_blr"][quote="blahblahblah"]Define $f:\mathbb{N}^{*}\rightarrow \mathbb{Q}$

Let $\text{sgn }{n}=\frac {\vert n\vert}{n}$

Then $f(n)={\vert n\vert}^{\text{sgn }{n}}$[/quote]

Just because sgn(x) can't be explicitly given by a single formula,

we can have f(n) = |n|^(|n|/n) for all integers n != 0[/quote]

hi blahblahblah..please ignore my last post..
actually I didn't notice that u have defined sgn(n) as |n|/n..

my solution is then exactly the same as urs...my apologies


Find the resultant pH when 10.0 mL of 0.950 M NaHO is added to 100 mL of a buffer solution that consists of 0.10 M CH3COOH and 0.10 M NaCH3COO. Ka for CH3COOH is 1.75x10^-5.

Is it $ 4$?  :maybe: Maybe it's wrong since I don't see the role of $ NaCH_{3}COO$

Yes, that's wrong. We have a buffer solution of a weak monoprotic acid and its conjugate base, both present with equal concentrations: so, the initial pH of the buffer is pH = pKa = 4.76. As we are adding a strong base, the final pH must be higher than 4.76.

$ pH\equal{} 6.43$ is around what I got?

Incorrect BanishedTraitor, but very close.

[hide="Correct answer"]pH = 6.35[/hide]


[quote="maggot"]Let $a,b,c\geq 0$. Show that $a^{4}+b^{4}+c^{4}\geq 2bc(a-b)(a-c)$[/quote]
$\frac{2}{27}\cdot a^{4}+b^{4}+c^{4}\geq 2bc(a-b)(a-c)$ is true too. :wink:

[hide="the stronger one"]
Let $g=\sqrt{bc}$. By AM-GM, $b^{4}+c^{4}\geq 2g^{4}$, and $(a-b)(a-c)\leq (a-g)^{2}$ (which reduces to AM-GM after expanding). So it remains to prove
\[a^{4}+27g^{4}\geq 27g^{2}(a-g)^{2}\]
But the difference is $a(a-3g)^{2}(a+6g)$, so it's true.
[/hide]

[quote="maggot"][quote="scorpius119"]
$a^{4}+27g^{4}$[/quote]

Why is this not $2$ instead of $27$, then?[/quote]

Well, after using the AM-GM's and multiplying by 27, it becomes $2a^{4}+54g^{4}\geq 54g^{2}(a-g)^{2}$, and we can just divide by 2 for simplicity. (the coefficient of $g^{4}$ is 54 and not 27 because it's $b^{4}+c^{4}\geq 2g^{4}$)


Please do not double post. :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199410

sry, delete the other one..
Can someone solve this problem?

[hide="sketch"]
$ \gcd (n,k)$ is a linear combination of $ n$ and $ k$. Of course $ n$ divides $ n\binom{n}{k}$, and some algebraic manipulation can be used to express $ k\binom{n}{k}$ as $ n$ times a binomial coefficient.
[/hide]

Wow thats cool. ( I think we'd want to handle k=0 carefully).

We know that there are integers $ a,b$ such that $ \gcd(n,k)\equal{}an\plus{}bk$. So it suffices to show that $ n$ divides \[ an\cdot\binom{n}{k}\plus{}bk\cdot\binom{n}{k}\]

Obviously, $ n$ divides the first fraction. If $ k\equal{}0$, the second fraction is zero, and we are done. Otherwise, $ k\cdot\binom{n}{k}\equal{}k\cdot\frac{n}{k}\cdot \binom{n\minus{}1}{k\minus{}1}$ which is divisible by $ n$, and we are done.


I was the school winner last year for the AMC10 and this year for the AMC12. So I should get a bronze medal, right? When are they supposed to be mailed out? I know that the school winner pins were mailed along with the AIME papers, but my teacher received no medals.

Sorry, switching from AMC 10 to AMC 12 means you start over with respect to these medals.  Those are AMC's rules, not mine...

What exactly are the rules for amc medals?


Roman numerals are letters that represent numbers. Some of the most common ones are $ I\equal{}1$, $ V\equal{}5$, $ X\equal{}10$, $ L\equal{}50$ and $ C\equal{}100$. The Roman system represents the number 3 as $ III$ and 30 as $ XXX$. If a symbol with a lower value is placed in front of a symbol with a greater value, the lesser number is subtracted from the greater number. For example, $ IV$ represents $ 5 \minus{} 1 \equal{} 4$. What is the three-digit value of the following subtraction problem: $ CCCXLVII \minus{} CLXVIII$ ?


invariant
superior algebra
superior algebra unsolved

Let $V$ be a vecto space over field $k$, and  $V^{T}$ be the $k[x]$-module. Prove that $W$ is a submodule iff $W$ is a subspace of $V$ for which $T(W)\leq W$ ($W$ is often called a $T-invariant subspace$ of $V$).

$(\Leftarrow)$ is easy.
$(\Rightarrow)$ If $W$ is a submodule, so also a subspace (by definition). Help me prove that $T(W)\leq W$

That's easy. As W is a submodule, it is closed under multiplication with scalars. Since $x\cdot v: =T(v)$ we get $T(W)=x\cdot W\subseteq\ W$.


[quote="shinwoo"][quote]LuCy4EvA wrote:
I agree with Shinwoo, if ppl doesn't like math, there's no point in making a "fun" forum for them because it doesn't help them in math at all. But if you truly like math, you'll come no matter how not "fun" the site is. The only suggestion is to tell more people about it, I know there must be lots of people who likes math out there.

But you have to realize, many people living in Wyoming, for example, might not be able to find any friends that like maths as much as they do. This is their place to socialize. I believe it actually helps peoples maths since it encourages them and makes them feel less lonely while doing maths[/quote]

this is funny story

There is a korean professor who is in the head of math department in MIT , and he said, people are supposed to learn math privately and very secretly so noone would notice the actually learn their abilities.

he also said, reveiling the math secret is like telling the secret in magic[/quote]

Huh? I don't get what you're trying to say here  :? .

[quote="shinwoo"][quote]LuCy4EvA wrote:
I agree with Shinwoo, if ppl doesn't like math, there's no point in making a "fun" forum for them because it doesn't help them in math at all. But if you truly like math, you'll come no matter how not "fun" the site is. The only suggestion is to tell more people about it, I know there must be lots of people who likes math out there.

But you have to realize, many people living in Wyoming, for example, might not be able to find any friends that like maths as much as they do. This is their place to socialize. I believe it actually helps peoples maths since it encourages them and makes them feel less lonely while doing maths[/quote]

this is funny story

There is a korean professor who is in the head of math department in MIT , and he said, people are supposed to learn math privately and very secretly so noone would notice the actually learn their abilities.

he also said, reveiling the math secret is like telling the secret in magic[/quote]

if he's a math professor, he should know that if no one shares how they figure out a problem, then he wouldn't be a math professor, because I bet that he didn't have to figure out everything he knows about math. and everyone would know so little about math, just the part that they figure out themselves, then math wouldn't really be helpful.

[quote]if he's a math professor, he should know that if no one shares how they figure out a problem, then he wouldn't be a math professor, because I bet that he didn't have to figure out everything he knows about math. and everyone would know so little about math, just the part that they figure out themselves, then math wouldn't really be helpful.[/quote]

that makes sense...

well, he is not just a normal math professor..

he is the first guy who has announced about image math

using the rigth brain and the left at the same time

The head of the math department at MIT is not Korean.

really?

but I'm pretty sure that he is related to MIT math department

I just found out the exact stuff

He is the head of MIT Mathmatical Acadamey

what ever it is

shinwoo,

As far as I know, there are not many Koreans in the
upper echelons of pure mathematics academia, outside of
Korea, and few or none who are full professors at top schools.
There are some younger Koreans with high-level
postdocs or tenure track jobs (Princeton, UCLA, etc) who are coming
up through the system after getting PhD's in the United States
and will probably hold such jobs in the future.
But no current tenured professors at MIT, Harvard, and the like,
as far as I know.

I would expect representation to be greater in applied
mathematics, though not enormously so, but
am not in that field so can't say for sure.
But in pure mathematics I'm not aware of many Korean
professors at the top institutions.  I think the reputation
of the institutes in Korea is growing, but they are not major
players right now.

Also, it's unclear whether "MIT Mathematical Academy", whatever
that is, has anything to do with MIT or any personnel in common.

[quote="fleeting_guest"]Also, it's unclear whether "MIT Mathematical Academy", whatever
that is, has anything to do with MIT or any personnel in common[/quote]

I took few hours of research yesterday, but I couldn't find his name in MIT mathmatic Proffessor list

so,

I kinda doubt that he works in MIT,  but he is works in Korea Mathatic Education Department

The other stuff that's true is that he is the first guy to announce about using right brain when you do math

What is the man's theory about left + right brain mathematics?
Any URL?
I hope it is not something obvious such as "visualization is good".

It's like using your your brains at the same time

He calls it image math

He describes it as finding a way to solve problems in the easiest way as possible

since left brain mostly does math and right brains shows visulization, you use them together
(this info is found by Dr.Sperry who got the Nobel Prize for it)

check this out..

label 16 dots

4 vertically and 4 horizontly which makes a square

if your done labeling, count them diagonaly..

it would be like 1+2+3+4+3+2+1 and it's middle number squared

same with other labeling

it would work the same with 25 dots or 36 dots.. That 's where you get the concept of Squre root and powers...

now do this

how do you find the sum of these odd numbers?

1+3+5+7+...+n

you could use the same way except that you have to use the dots differently

label them and count them in 90angle lines

AHHH!!!
this is hard to explain..

I will send you the picture..

he wrote a book about it

Do you mean something like what in English is called
"proof without words"?   (Like the animated cartoons at
AoPS website illustrating various formulas).  Visual
proofs of various algebraic/numerica/combinatorial formulas?

Or is he saying something more specific about ways
of doing things that employ both brain hemispheres?

however I love the idea that reveiling the math secret is like telling the secret in MAGIC
I wrote some fiction stories involving that concept.

[quote]Do you mean something like what in English is called
"proof without words"? (Like the animated cartoons at
AoPS website illustrating various formulas). Visual
proofs of various algebraic/numerica/combinatorial formulas?

Or is he saying something more specific about ways
of doing things that employ both brain hemispheres[/quote]

That's not what he means....

It's more like using creativity while you do math....

I'll try to send you a picture someday..

He wrote a book( in korean ) about it


What is the fewest number of colors needed to paint the faces of a regular octahedron so that no two adjacent faces are the same color?

In an octahedron, there are $ 8$ faces.  Let the top faces be numbered $ 1, 2, 3,$ and $ 4$, and let the bottom faces be numbered $ 5, 6, 7,$ and $ 8$ (each below the top faces, respectively).  We could paint $ 1, 6, 3, 8$ one color and $ 5, 2, 7, 4$ the other.  Thus, the minimum number of colors is $ \boxed{2}$.


For some reason, I would be led to believe that Las Vegas bookmakers accepted wagers on Mathcounts at a national level. Or wagers on IMO rankings.

Were there any instances in which there was gambling on any aspects of math contests?

:rotfl:

Does betting $ \$2$ on an ARML relay contest count?

[quote="blahblahblah"]do you have the slightest clue what 'oddsmaking' entails?[/quote]

I know it's something quite complex that require some knowledge of the playing field to be done properly.

[quote]Why are you led to believe this?[/quote]

Because these competitions are high-profile competitions that would be sufficiently popular to have a little gambling on it.

Vegas doesn't take wagers on stuff like this.  For example, I'm not even sure if you can bet on the Oscars in Vegas.  You certainly can't do it at the major sportsbooks, although they may make odds on the Oscars for fun.

On the other hand, they bet on all kinds of crazy stuff in London, but I doubt this includes the IMO or anything like it.  Presumably there isn't enough interest.

However, there's always the wild world of the internet.  If I'm not mistaken, there are websites where you can essentially set up wagers on just about anything.  (But you'd still have to find others to wager against you.)

I imagine it would be a lot like a horse race, but with more people; there would be favorites, but odds are the favorites wouldn't win.

Maybe Dutch bookmakers take wagers on IMO.

Really, Dutch bookmakers? Why [i]Dutch[/i]? I'm curious.

I tried searching for stuff on betting on math contests, and the only on-topic results I got were this thread, and some other thread on this forum in which you were trying to write a story. So, I'm going to guess that it doesn't happen in general, apart from perhaps some small-scale wagering (like [b]cosinator's[/b] 2 dollar bet). Also, such wagering is against the spirit of math contests, so most of the people that have any knowledge that might remotely help them make a bet is probably not going to actually make a bet.

Dutch bookmakers could probably be more open to allow betting on high-profile math contests because I heard Dutch bookmakers operate within a [b]state-owned frame[/b] that encourages them to wager on more "outlandish" things. Again, it is Netherlands that we are speaking of and all legal gambling is state-owned, including bookmaking operations.

Maybe the Dutch government wants the Netherlands to be the gambling hub of Europe and that they desire to show that they are open to all kinds of gambling. If they take IMO wagers it's probably a state-inspired effort of openness.

you sure are one weird guy (no one is going to read your story and it's completely implausible, so why do you care about factual accuracy?)

Just to scale how much of a goof (or spoof) my story is.


calculus
integration
calculus computations

[quote="Extremal"]find $\int\frac{dx}{(x^{2}-x+1)(\sqrt{x^{2}+x+1})}$[/quote]

substitue $x=\frac{1-t}{1+t}$  and then after few calculus we use integral #3, I mean this $\int\frac{\sqrt{x^{2}+A}}{x^{2}+B}dx$
The idea was next: we substitute $x=\frac{a+bt}{1+t}$ and then after few calculation we put needed $a,b$. done.


Given are $\triangle ABC$ inscribed $(O)$ with two medians $AA',BB'(A'\in BC,B'\in CA)$ the ray $B'A'\cap (O)=\{M\}$ prove that
$\frac{b}{MB}=\frac{c}{MC}+\frac{a}{MA}$  :idea:  :?:

$\\ \frac{b}{MB}=\frac{c}{MC}+\frac{a}{MA}\Leftrightarrow b \cdot MC \cdot MA = c \cdot MA \cdot MB+a \cdot MB \cdot MC \Leftrightarrow \\ \\ \\ \frac{b \cdot MC \cdot MA}{4R}= \frac{c \cdot MA \cdot MB }{4R}+\frac{a \cdot MB \cdot MC}{4R}\Leftrightarrow E_{MAC}=E_{MBA}+E_{MBC}\Leftrightarrow \\ \\ E_{MA'C}+E_{AA'C}+E_{MA'A}= E_{MBA'}+E_{ABA'}-E_{MA'A}+E_{MBA'}+E_{MCA'}\Leftrightarrow \\ \\ E_{AA'C}+2 \cdot E_{MA'A}= 2 \cdot E_{MBA'}+E_{ABA'}\Leftrightarrow$

But $E_{AA'C}= E_{ABA'}$ therefore we must show $E_{MA'A}= E_{MBA'}$

which is true because they have the same base and $BA \parallel MA'$ .

 :D

yes that's a good solution Ramanujan :) and I have two same problems :D
$+)$change $AA',BB'$ by internal bisector prove that $\frac{1}{MB}=\frac{1}{MC}+\frac{1}{MA}$
$+)$ change $AA',BB'$ by altitude (of acute tringle )prove that
$\frac{\cos B}{MB}=\frac{\cos C}{MC}+\frac{\cos A}{MA}$
 hope you are interested in !  :D


vector
calculus
real analysis
real analysis theorems

I was wondering, is there really a significant difference between the number $0$ and the vector $\vec{0}$? I mean, when working with vectors is it really "worth it" to put a vector sign over the zero?

Depends on your target audience. When writing for mathematicians, don't worry about the arrow. When teaching multivariable calculus, worry about everything.  :roll:

I'm used to seeing the bold zero vector $\mathbf{0}$, but in context they're all the same.


Prove that there is no triangle whose altitudes are of length 4, 7, and 10 units.

[hide]computing the area of the triangle with each height makes the sides a/10, a/7, a/4, but 1/10+1/7<1/4.[/hide]

[hide]If a=area, then technically it's 2a/10, 2a/7, 2a/4 (:P), but yeah, that's the solution.[/hide]


[quote="Virgil Nicula"]Prove that in an acute triangle $ABC$ exists the following inequality : $2r\cdot\sum a^{2}+R\cdot \sum bc\le R\cdot\sum \left[a^{2}\left(\frac{b}{c}+\frac{c}{b}\right)\right]$.[/quote]

Hm, it's easy, my friend Virgil Nicula!

$2r \leq R$ and $\frac{b}{c}+\frac{c}{b}\geq 2$, so, we only have to prove that $(a^{2}+b^{2}+c^{2}) \geq (ab+bc+ca)$.


geometry
3D geometry
number theory unsolved
number theory

find all prime numbers $ p$, $ q$ and $ r$ such that $ p^3\equal{}p^2\plus{}q^2\plus{}r^2$

$ p^2(p-1)=q^2+r^2\Rightarrow q^2\equiv -r^2[p]$

if $ r=p$ then $ q^2\equiv 0[p]$ gives $ q=p$ and we get $ p^3=3p^2%Error. "Ritharrow" is a bad command.
p=3$

if $ r\neq p,\exists a: \ ra\equiv 1[p]$ then $ (qa)^2\equiv -1[p]\Rightarrow p\equiv 1[4]$
then $ 4|p^2(p-1)=q^2+r^2\Rightarrow 2|r,2|q\Rightarrow r=q=2;\ p^2(p-1)=8(impossibe)$

$ (p,q,r)=(3,2,2)$

[quote="FOURRIER"][quote="aviateurpilot"]
if $ r\neq p, \exists a: \ ra\equiv 1[p]$ [/quote]

why ??[/quote]

$ r\neq p\Rightarrow gcd(p,r)\equal{}1\Rightarrow \exists (a,b): \ bp\plus{}ar\equal{}1$


[quote="pvthuan"]Let $a,b,c$ be positive real numbers such that $a+b+c=1$, prove that
\[ 8^a+8^b+8^c+8^{(abc)^{1/3}}\geq8.  \][/quote]
This is wrong. Try $a=b\rightarrow0.5^{-},$ $c\rightarrow0^+.$ ;)

[quote="pvthuan"]Arqady, but I think $4\sqrt2+2\geq8$. ;)[/quote]
$4\sqrt2+2\geq8\Leftrightarrow2\sqrt2\geq3\Leftrightarrow8\geq9$ :?

So it should be as follows: Let $a,b,c$ be non-negative real numbers such that $a+b+c=1$, prove that
\[ 8^a+8^b+8^c+8^{(abc)^{1/3}}\geq4\sqrt2+2.  \]

:D But if $x=m,y=n,z=p$ the Lefthand's smaller than the Righthand, so the problem should be as follow
\[ 8^a+8^b+8^c+8^{abc/3} \ge 8^m+8^n+8^p+8^{mnp/3} \]
 :P


function
number theory open
number theory

what do you think / what do you know about the problem of estimation of the next prime for a power of 10 ?

Chebychev theorem (Bertrand's postulate) imply that every interval $ (10^n, 2\cdot 10^n)$ contains a prime (at least one).

I claim that for every positive integer $ n$ there exists prime $ p$ in $ \left(10^n, \frac{3}{2}\cdot 10^n\right)$.

I have checked "not so big" cases using Derive's NEXT_PRIME built-in function, and it looks optymistic :) Even stronger estimation for NEXT_PRIME(10^n) should be true....

If I have remembered the result right, the statement is correct and the proof follows along roughly the same lines.

Edit:  The paper is called "Primes in the Interval [2n, 3n]" by M. El Bachraoui, Int. J. Contemp. Math. Sci., Vol. 1, 2006, no. 13, 617-621.  I can't find my original link, so [url=http://qc.fengyuan.com/random/elbachraouiIJCMS13-16-2006.pdf]here it is[/url] :)

strongest result (we want to catch all positive integers $ n$ in it, not onlt sufficiently large) will be:

Let P(n) = NEXT_PRIME(10^n) (the smallest prime greater then $ 10^n$).  For all $ n$ holds
\[ P(n)\le \frac{11}{10}10^n\equal{}10^n\plus{}10^{n\minus{}1}\]

It's known that $ P(n) \le 10^n \plus{} O(10^{\left( \frac {6}{11} \plus{} \epsilon \right) n })$ (see [url=http://mathworld.wolfram.com/BertrandsPostulate.html]the Mathworld article[/url]), so the result is clearly true for $ n$ sufficiently large.  It remains to bound the corresponding constant with some calculations, I think.

Thanks very much for this very interesting text and informations :) :)


function
integration
limit
real analysis
real analysis unsolved

Suppose that $f$ is a continuous function, whose period is $2\cdot\pi$ with complex values. Let $H$ the Poincaré demi-plan ( set of points $(x,y)$ with $y>0$). Prove that there exists a unique harmonic bounded, $C^2$ class function $u$ defined on $H$ with complex values such that for all $y>0$ the function $u_y(x)=u(x,y)$ is of period $2\cdot\pi$ and $u(x,y)$ has limit $f(x)$ when $y$ is close to 0, for all $x$. I couldn't solve it and the solution I saw spanned over 7 pages in RMS. Does anyone have normal approach?

$u(x,y) = \frac{1}{\pi} \int_{\mathbb{R}} f(t) \frac{y}{(x-t)^2 + y^2} dt$ should work. It is a classical solution to the Dirichlet problem with Poisson integral.

Thanks a lot, I will try to see if I can prove it verifies all conditions. Thanks again.

Seems like the main issue here is not existence but uniqueness. Note that harazi's boundary condition ($\lim_{y\to 0+}u(x,y)=f(x)$ for all $x$) is much weaker than the standard boundary condition of continuity up to the boundary and the maximum principle does not directly apply here. Also, some conditions are missing because, as it is, we can add $Cy$ to any solution and get another solution. So, I would like harazi to double check the statement of the problem. By the way, "demi-plan" is "half-plane" in English  :P .

Thanks for "half-plane". I corrected the statement, the condition of boundedness was missing. Instead, all the other hypotheses were correct.


function
trigonometry
limit
real analysis
real analysis unsolved

Let $f$ be a function defined: $f(x) = x^\alpha \sin\frac{1}{x}$ for $x>0$ , and $f(0)=0$.
Proof that $f$ is continue $\Longleftrightarrow$ $\alpha > 0$.

Thanks!

We want to prove \[ x^{\alpha}\sin\frac{1}{x} \to 0 \Leftrightarrow \alpha>0. \]
$\Leftarrow$
$x^{\alpha}\to 0, \left|\sin\frac{1}{x}\right| <2$ so $x^{\alpha}\sin\frac{1}{x} \to 0.$
$\Rightarrow$
Suppose $\alpha=0$ then $\nexists \lim_{x\to 0} \sin\frac{1}{x}$.
Suppose $\alpha<0$ then also $\nexists \lim_{x\to 0} x^{\alpha} \sin\frac{1}{x}$ because $\exists \{x_n\}_{n=1}^{\infty},\{y_n\}_{n=1}^{\infty}$ such that $\lim_{n\to \infty} x_n =\lim_{n \to \infty}y_n = 0$ and $\lim_{n \to \infty}f(y_n) = +\infty$,
$\lim_{n \to \infty}f(x_n) = -\infty$. So $\alpha >0.$

Thanks  :)
But what are the sequences $y_n$ and $x_n$?  :?

$\frac{1}{2n\pi+\frac{\pi}{2}}$ and $\frac{1}{2n\pi+\pi}$

$x_n=\frac{1}{2\pi n +\frac{\pi}{2}}, y_n=\frac{1}{2\pi n +\pi}$ - good.
You have
\[ \lim_{n\to \infty} f(x_n) =\infty, \lim_{n\to \infty} f(y_n) = 0 \]


Let $0 = a_1 < a_2 < a_3 < \cdots < a_n < 1$ and $0 = b_1 < b_2 < b_3 \cdots < b_m < 1$ be real numbers such that for no $a_j$ and $b_k$ the relation $a_j + b_k = 1$ is satisfied. Prove that if the $mn$ numbers ${\ a_j + b_k : 1 \leq j \leq n , 1 \leq k \leq m \}}$ are reduced modulo $1$, then at least $m+n -1$ residues will be distinct.

[quote="Rushil"]Let $ 0 = a_1 < a_2 < a_3 < \cdots < a_n < 1 $ and $ 0 = b_1 < b_2 < b_3 \cdots < b_n < 1 $ be real numbers such that for no $a_j $ and $b_k$ the relation $a_j + b_k $ is satisfied. Prove that if the $mn$ numbers ${\ a_j + b_k : 1 \leq j \leq n , 1 \leq k \leq m }\ $ are reduced modulo 1, then at least $m+n -1 $ residues will be distinct.[/quote]

A few question.

(1) How to understand ''for no $a_j $ and $b_k$ the relation $a_j + b_k $ is satisfied.''? What relation?

(2) Does the sentence ''the $mn$ numbers ${\ a_j + b_k : 1 \leq j \leq n , 1 \leq k \leq m }\ $'' coincide with ''$ 0 = a_1 < a_2 < a_3 < \cdots < a_n < 1 $ and $ 0 = b_1 < b_2 < b_3 \cdots < b_n < 1 $''? In one sentence, the indice of $b$ is $m$, while another is $n$?

(3) How to understand ''modulo 1''? If then, it will all be 0.

Sorry!! (i) and (ii) were typos...

Regarding (iii), I've checked that it is not a typo. I considered it referring to the fractioanl part. The peoposers didn't clarify!!!


Today we went over a test on probability and couldn't agree on an answer for one of the problems.

(paraphased)
"Players A, B, C, and D have each been dealt 13 cards from a standard 52-card deck. If players A and B are dealt 9 hearts, what is the probability that players C and D have two hearts each?"

The answer given is $\frac{1}{5}$. Mrs. Tipton argued that the possibilities for C and D's number of hearts are 4-0, 3-1, 2-2, 1-3, and 0-4, and 2-2 is one of five possibilities. She says that the arrangement of the cards is irrelevant because they have already been dealt.

A few others came along and argued $\frac{3}{8}$. They stated that the ways of being dealt 4 and 0, 3 and 1, or 2 and 2 are not the same, so the probability isn't equal. I think their answer comes from
\[\frac{\binom{4}{2}}{\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}} = \frac{3}{8}\]

My answer was $\frac{234}{575}$. Not only do the ways to choose n hearts from the 4 hearts change, but also the ways to choose the remaining 13-n cards. From this I get for person C: the way to choose 2 from the 4 hearts times the remaining 11 cards from 22 that are not hearts divided by the way to choose any 13 cards from the 26. Then you would multiply by the way person D chooses the remaining cards, which is just 1.
\[\frac{\binom{4}{2} \binom{22}{11}}{\binom{26}{13}} = \frac{234}{575}\]

Only one person will agree with me so far, but I haven't heard anything to convince me that my answer is wrong. We're curious as to what Mathew Crawford's opinion is. :)

Thanks!

You are correct on this one.  Very nice work.

As far as convincing them goes, try working simpler examples where there are 8 cards dealt between 2 players instead of 26.  4 are hearts.  What is the probability that a play gets exactly 2 hearts.  Then build up to the more general problem.

This is ultimately a counting problem.  There are clearly $\binom{26}{13}$ ways a player could be dealt 13 of the remaining 26 cards, permutations aside.

Now count the number of ways to get dealt 2 hearts out of the 4.

Now multiply that by the number of ways to get 11 non-hearts out of the 22 non-hearts.

And you have the answer you proposed.


modular arithmetic
algorithm
number theory
least common multiple
greatest common divisor
Euclidean algorithm

Skimming through AoPS V1 divisibility section I couldn't find anything on this type of problem. It's from a calculator Algebra contest.

Find the smallest positive integer which, when divided in turn by 21, 23, and 25, leaves respective remainders of 8, 1, and 11.

It's called solving a system of congruences.

You have
\begin{align*}
n & \equiv 8 \pmod{21} \\
n & \equiv 1 \pmod{23} \\
n & \equiv 11 \pmod{25}
\end{align*}

You can turn these into algebraic equations and solve, or you can use the Chinese Remainder Theorem.

I can get this far, but I'm not very familiar with modular arithmetic, so could you show me how to solve a system of congruences when there are three of them?

Transform them into algebraic equations ($ a\equiv b\pmod m\implies a\equal{}mk\plus{}b$) and then substitute into other congruences to find a closed form that satisfies all three.

That's not saying much; what you end up having to do is solve linear Diophantine equations, so the actual content of the solution is the [url=http://en.wikipedia.org/wiki/Euclidean_algorithm]Euclidean algorithm[/url].

(The Chinese Remainder Theorem tells you what the solution should look like, but actually computing it still requires modular division, which is what the Euclidean algorithm is for.)

The number we are looking for should be Greater than the LCM of the 3 numbers..right?

If you mean the divisors, then I don't believe it has to be greater, because our number "n" isn't perfectly divisible, so it sort of "offsets" the pattern of the multiples. So I think it could be smaller than their LCM. I read the Euclidean algorithm wiki that was posted, and I remembering learning that in computer science, but I thought that was only to find LCM's or GCF's.

The number can be represented as 21x+8, 23y+1, 25z+11. So
$ 23y \plus{} 1 \equiv 25z \plus{} 11 \pmod {23}$
$ 0 \equiv 2z \plus{} 10 \pmod {23}$
$ z \equiv \minus{} 5 \equiv 18 \pmod {23}$
So we find $ z \equal{} 18$, $ y \equal{} 20$ works for these two equations. But then, the value 27 for x makes all three work. This gives a specific solution:$ (27,20,18)$. Then general solution is $ (27 \plus{} 23 \cdot 25k, 20 \plus{} 21\cdot 25k, 18 \plus{} 21\cdot23k)$. Subbing any of these in gives $ 21 \cdot 23 \cdot 25k \plus{} 575$ as the answer.

MathleteWannabe:

The euclidean algorithm gives a way to find $ (x,y)$ such that
$ ax \plus{} by \equal{} c$, provided that $ c$ is a multiple of $ gcd(a,b)$.  This method is the extended Euclidean algorithm.

for example:

$ 7x \plus{} 5y \equal{} 1$

Lets find the gcd of $ 7$ and $ 5$, via the euclidean algorithm:
$ 7 \equal{} 5*1 \plus{} 2$ dividing 7 by 5
$ 5 \equal{} 2*2 \plus{} 1$ dividing 5 by 2

Hence $ 1 \equal{} 5 \minus{} 2*2 \equal{} 5 \minus{} 2*(7 \minus{} 5*1) \equal{} 3*5 \minus{} 2*7$ so $ (x,y) \equal{} ( \minus{} 2, 3)$ is one solution.

The problem of finding an inverse is equivalent:
$ aa^{ \minus{} 1} \equiv 1 \pmod{p}$ means $ aa^{ \minus{} 1} \equal{} 1 \plus{} kp$ implies $ aa^{ \minus{} 1} \minus{} kp \equal{} 1$.  $ a$ and $ p$ are known, so you can find integer solutions for $ a^{ \minus{} 1}$ and $ k$ by the euclidean algorithm.

btw, dgreenb801, your solution doesn't work.  $ x$ is not 27...

Anyway, you have found your number is $ 461 \pmod{575}$.  Then $ 461 \equiv 21x \plus{} 8 \pmod{575}$ implies $ 453 \equiv 21x \pmod{575}$.  the inverse of this is $ \minus{} 219 \equiv 356$ by euclidean algorithm, so $ x \equiv 453*356 \equiv 268 \pmod{575}$, so your number is $ 21(268) \plus{} 8 \equal{} 5636$ which works.

[quote="facis"]
Anyway, you have found your number is $ 461 \pmod{575}$.  Then $ 461 \equiv 21x \plus{} 8 \pmod{575}$ implies $ 453 \equiv 21x \pmod{575}$.  [/quote]

^^ I got that far but didn't know where to go from there. Thanks for the explanation! That's the right answer!

no problem.  I've actually seen some problems like this before, but haven't really understood how to do them until now.

^^ I runned it in C++.
Worked like a charm.


Express $ (\minus{}8)^{\minus{}\frac{1}{3}}$ as a common fraction.


A broken circuit set has wires of length 3 in that can connect with a) other wires or b) battery packs.  The circuit set is broken because electricity can only be carried 4 in before dying out.  A short circuit is formed when electricity from one battery pack reaches another battery pack (remeber - a six inch wire between battery packs is SAFE).  If the battery packs are also 3 in wide, compute the number of linear 39 in circuits with no short circuits.  Remember: The only restriction is no short circuits.  Three battery packs in a row is FINE.  A wire leading nowhere is FINE.

[hide="Answer to check your answer with"]
1897
[/hide]

[hide="Hint"]
Call battery packs 0 and wires 1 and find the number of 13 bit strings with no occurances of 010.
[/hide]

This is a very good example of a bit-string problem.  Will somebody do it? :maybe:


linear algebra
matrix
linear algebra solved

Is it true that if $A$ is an $n\times n$ matrix with integer entries and $det(A) = \pm 1$ then $A^{-1}$ also has integer entries? How to prove it?

This must be ancient but I searched diligently and I couldn't find a thread about it... if someone remembers one, please refer me to it. Thanks.

More generally, a square matrix $A$ with elements from a commutative ring $R$ is invertible in $M_{n}(R)$ if and only if $\det A$ is a unit (aka an invertible element) of $R$.
Proof: recall how to compute the inverse matrix using $(n-1)\times(n-1)$ minors. This does not involve any division until the very last step, when you divide all elements by $\det A$.

In other words, if $A$ is invertible, then $A^{-1}=\frac{1}{det(A)}com(A)^{t}$. I must admit I didn't think about it, I need to revise my linear algebra ...

if it helps any more than what julien and mlok have already said.......

if $A$ has integer entries then clearly its adjoint  $A_{adj}$, also has integer entries. so now use that $A^{-1}=\frac{1}{detA}A_{adj}$ and that $detA=\pm1$

Well, yes, it's true, since, as [b]mlok[/b] said,
[quote="mlok"]More generally, a square matrix $A$ with elements from a commutative ring $R$ is invertible in $M_{n}(R)$ if and only if $\det A$ is a unit (aka an invertible element) of $R$.[/quote]

Now, as for this particular case :
Let $A$ be invertible with integer entries, then $A^{-1}$ has integer entries.
But we know that $\det A \det A^{-1}= \det I_{n}= 1.$ But then (since the two elements of that product are integers) it follows that $\det A = \pm 1.$

Not necessarily.
$ A: \equal{}\begin{bmatrix}
2& 0& 0\\
0& 2& 0\\
0& 0& 2
\end{bmatrix}$

A is invertable with integer entries, but $ A^{\minus{}1}$ has rational entries.
You can not drop the condition $ det(A)=\pm 1$.

[quote="mathmanman"]Let $ A$ be invertible with integer entries, then $ A^{ \minus{} 1}$ has integer entries.[/quote]
I think there's a poor choice of word in this sentence. I suspect that if you replaced "then" with "and" you would come closer to mathmanman's intent - and vesszabo wouldn't have felt the need to reply.


Given are $\triangle ABC$ area $S$ with any point $M$ on plane prove that
$\frac{1}{2}(MA^{2}+MB^{2}+MC^{2})+MB.MC+MC.MA+MA.MB\ge 2\sqrt{3}S$
equal when $ABC$ is equaliteral triangle and $M$ is center!
$\frac{1}{2}$ is the best const that I can find and there is a hard question:
is there a inequality form $MB.MC+MC.MA+MA.MB\ge const$ I belive there is it! :)

We have to prove the inequality $(MA+MB+MC)^{2}\geq 4\sqrt{3}S$.As we know $MA+MB+MC$ is minimum for the Toricelli point.For that point we have $(MA+MB+MC)^{2}=\frac{a^{2}+b^{2}+c^{2}+4S\sqrt{3}}{2}$.So it is simple now to prove $a^{2}+b^{2}+c^{2}\geq 4S\sqrt{3}$. :lol:

yes yes I see thank mateivld  :lol: sorry for the title "a hard problem"  :)
but there is a similar problem for polygon that I can't solve too
Given regular polygon $A_{1}..A_{n}$ with any $M$ on plane find min of
$\sum_{1\le i,j\le n}MA_{i}.MA_{j}$
for equaliteral triangle we have $MBMC+MCMA+MAMB$
for square we have $MAMB+MAMC+MAMD+MBMC+MCMD+MDMB$
It is a problem in M&Y of Viet Nam but the solution false!


vector
geometry
3D geometry
sphere
real analysis
real analysis solved

Given a simple region in space bounded by a simple closed surface, and a point inside the region. Must there exist a plane through this point such the point is the centroid of the cross-section with the region?

(I thought of this problem from the 2-dimension version)

I remember giving an answer (which I am not sure about) before the site crashed.

I think the answer is positive, and one way to prove it would be employing [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=hairy+ball&t=19830]this[/url] theorem (the Hairy Ball Theorem).

For each plane $\mathcal P$ through our point $P$ and each half-space $\mathcal H$ determined by the plane, consider the point $X=X(\mathcal H)$ s.t. $PX\perp\mathcal P$, and $PX=1$. This point describes a unit sphere. Now, for each pair $(\mathcal P,\mathcal H)$, consider the centroid $G$ of the cross-section determined by $\mathcal P$, and construct the vector $\overrightarrow{XT}=\overrightarrow{PG}$.

The vectors $\overrightarrow{XT}$ form a field of vectors tangent to the sphere, and it's clear that this field should be continuous, so there is a null vector among them, meaning that for that particular centroid $G$, we must have $\overrightarrow{PG}=\vec 0$.

[quote="grobber"]The vectors $\overrightarrow{XT}$ form a field of vectors tangent to the sphere, and it's clear that this field should be continuous...[/quote]
It is a wrong statement ;)

Hmm.. Why? Doesn't everything depend continuously on the position of the half-space $\mathcal H$? (don't think about the plane: the half-space determines everything)

You can construct simple counter-example. Consider polygon (0,0); (0,2); (3,2); (3,1); (2,1); (2,0) and point (1,1).

:blush:  :blush: I see what you mean... Well, at least it works for convex bodies, doesn't it?

But if you also consider concave figures, I think not even the 2 dimensional problem holds: is there always a chord s.t. the chosen point is its midpoint? I think your example can be modified to give a counterexample to this. If this is true (I'm sure you'll disprove it, since I seem to be making lots of stupid mistakes lately :D), then maybe we had nothing to prove in the first place.

Myth's example (  or his method of constructing examples ) shows that this problem has nagative answer even in the 2-dimensional case, isn't it?
( the region is not convex )
But Grobber's solution can deal well with the case where the region is convex.

I edited my post before seeing the last message :D.

[quote="grobber"]:blush:  :blush: I see what you mean... [/quote]

 And that is what your message looked like when I posted mine :lol: :D


Let $ f$ be a function whose domain is $ S \equal{} \{1, 2, 3, 4, 5, 6 \}$ and whose range is contained in $ S$. Compute the number of different functions $ f$ which have the following property: no range value $ y$ comes from more than three arguments $ x$ in the domain. For example, $ f \equal{} \{ (1,1), (2,1), (3,1), (4,4), (5,4), (6,6) \}$ has the property, but $ g \equal{} \{ (1,1), (2,1), (3,1), (4,1), (5,3), (6,6) \}$ does not.

There are $ 6^{6}$ functions

Of those functions, $ 6 \cdot 1 \cdot 1 \cdot 1 \cdot 6 \cdot 6$ will contain a group of at least $ 4$ values the same
That will count $ 6 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 6$ this many twice
So there are

Hence there is $ 6^6 \minus{} 6^3 \plus{} 6^2 \minus{} 6$ possible functions.

Bear in mind my combinatorics is very poor

[hide="Solution"]You start with 6^6 functions that map S into S.

We must discount the cases in which we have four of a number, five of a number, and six of a number.

There are 6 ways to have 6 of one number (pick the number).

For 5 of one number, there are 6 ways to pick the number of which there are 5, 5 ways to pick the second number, and 6 x values to assign the second number, for a total of 6*5*6 = 180 functions.

For 4 of a number, we can either have one other number or two other numbers.

If there is one other number, then there are 6 ways to pick the number that occurs 4 times, 5 ways to pick the number which occurs twice, and C(6,2) = 15 ways to position the numbers of which there are two, for a total of 6*5*15 = 450 functions.

If there are two other numbers, then there are 6 ways to pick the number of which there are 4 and C(5,2)=10 ways to pick the other two numbers. Of these, there are 6 positions for the larger of these two and 5 positions for the smaller, for a total of 6*10*6*5 = 1800 functions.

Hence, there are 6^6 - 6 - 180 - 450 - 1800 = [b]44220 [/b] functions that satisfy the given conditions.[/hide]


Hi, I have the following problem:
I want to write an article (documentclass) and I'm adding images. When I write the \caption command inside a figure environment I obtained a figure with a simple numeration as result, i.e., 1,2,3, etc. The problema is that I need a numeration that has a relation with the current section or subsection, etc.
For example, if I put a figure environment within the section 2, I want to see in the output the FIGURE 2.1

How can I do this ???

[code]\renewcommand{\figurename}{FIGURE}[/code]changes Figure to FIGURE
[code]\renewcommand\thefigure{\arabic{section}.\arabic{figure}}[/code]gives 2.1 as you want.
You may also want to have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=caption]caption package[/url] which allows you to customise further.

I think that these commands don't work with the packages:
\usepackage[spanish]{babel}

And I need to use it, due to my language.

Why won't it work with spanish babel? The renewcommands are not affected by the language so they work fine.

Look at the code.
1) It doesn't change the name figure. It appears "Figura" instead of "FIGURE" as I defined in the preamble.
2) Now I have other doubt. If you look the pdf file you'll see three images. The first two of them are well numerated: 2.1 and 2.2. However in section 3 (Otros) it shows 3.3 and I want to see 3.1 as the first figure of the section 3. How can I improve that?

Thanks for your help.

1. Babel redefines many names at \begin{document} so you need to add your definition afterwards. So put \renewcommand{\figurename}{FIGURE} after \begin{document}.

2. The figures are being numbered consecutively so, ignoring sections, they are numbered 1,2,3 correctly. To change this you need the amsmath package (which you should be using in all documents along with amssymb) and then add the line \numberwithin{figure}{section} after the \renewcomands for the figures and tables.


yo just wondering if someone can solve this for me, as iv forgot prety much everything i knew about maths scince college

heres a diogram i made
http://img9.imageshack.us/my.php?image=radiusb.jpg

cheers

First find the difference of the radius and the given height, which is $ 1.6\minus{}1\equal{}0.6$.

Now draw the related radius and apply Pythagorean theorem to get $ a\equal{}\sqrt{1.6^2\minus{}0.6^2}$.


What is the minimum number of guests that must be invited to a par so that at least 3 wll know each other OR at least 3 will not know each other

[hide="Solution"]
It can be shown that 5 doesn't fulfill the terms, and 6 does (Pigeonhole). So the answer is 6. [/hide]

Hm, isn't it 3? Just let them all know each other or all not know each other.

But I'm guessing that's not the correct answer, or the problem is not the correct problem, since it's so simple...

What we want is the minimum number of people necessary so that the result is true no matter what the relationships between the people are.  In other words, the problem asks to compute the [url=http://en.wikipedia.org/wiki/Ramsey%27s_theorem]Ramsey number[/url] $ R(3, 3)$.

The answer is $6$.
Proof:
Try to engineer a counterexample, i.e, a case where there are $6$ people at a party but no three people who know each other and no three people who do not know each other: Let the people at the party be represented by the vertices $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, $p_{5}$ and $p_{6}$ of a regular hexagon. Let acquaintances be connected by a line segment and strangers by a dashed line. The person represented by $p_{1}$ must be acquainted with or unknown to at least three people, and therefore connect $p_{1}$ to  $p_{2}$, $p_{3}$ and  $p_{4}$ with a dashed line. Now, since a counterexample is the desired result,  $p_{2}$,  $p_{3}$ and  $p_{4}$ have to be connected to each other with a full line, as otherwise a triangle whose edges are all dashed will be created, and that is not the desired result. However if $p_{2}$,  $p_{3}$ and  $p_{4}$ are connected to each other with full lines, a triangle whose edges are all full will be created, which is also not the desired result! Therefore, it is impossible to connect the vertices of a hexagon with dashed and dotted lines in such a way that there are no triangles created, i.e. in a party with six people, there will always be three people who are acquaintances or three people who do not know each other. $Q.E.D$


There is a block and tackle system composed of two pulleys and a rope. The distance between the centers of the pulleys is 113 centimeters and the pulleys each have a radius of 15 cm. What percent of the circumference of the bottom pulley is not touching the rope?  Diagram [img]http://www.artofproblemsolving.com/Forum/album.php?mode=attach&id=8522[/img]

[hide]Um, wouldn't it just be 50%? The two pulleys are the the same size, so when you draw tangents it becomes one line going through the center, aka the diameter, which splits the circumfrence in 1/2.[/hide]

Well, I'm not sure exactly what you're doing to get that answer but its not exactly right, but its close.

umm...i think superwizard is right.  maybe you should explain what you mean by 'block and tackle' more clearly or draw a diagram.

By block and tackle, I mean the following: two pulleys, with one rope coming down from the middle of the first, winding around the second and then going back all the way around the top pulley until it reaches the middle again.  Is there a way I can give a diagram if the picture is not on a website?

[quote="sharkboy"][quote="kohjhsd"][quote="sharkboy"]Is there a way I can give a diagram if the picture is not on a website?[/quote]

1) Upload it on the gallery
2) Upload it on other image sites and link to the image with [img] tags.[/quote]

Ok. I cut the problem out of a PDF file. Then I copy and pasted it into word.  I'm not sure how to save a picture from word. Any help?[/quote]

If you are using MS Windows, the fastest (and cheapest) way is to paste it into mspaint and save it as a .png or .gif file.

Ok, so the diagram is in the link in the problem. It's just a link to the galley in AoPS.

Please do not leave spaces between the tags. This is modification to make it easier to see...
[img]http://img237.imageshack.us/img237/7195/chapter1rp5.png[/img]
This looks like physics.

[hide="Using calculator"]
In radian mode,
$\frac{\arccos \left( \frac{15}{113}\right)}{2\pi}\cdot 100 \approx \boxed{47.88 \%}.$
[/hide]

Yeah, that's about right, since there is just a bit more than 50% covered, if you just look at it the right way.


Hello. I've been looking for some interesting mathematic scholarly papers contests on Internet but i couldn't find any. I prefer economic theory and it would be better if the contest is for economic matemathic scholarly paper. Can you tell me some contests?


How many ways can you color a cube with 3 colors?

I don't know the answer, but I think I got close to it.
Is there an easier way than brute forcing it?

I'm confused myself but i'm guessing that 3 colors can go on each face so could it be 3^6 which is 729?

Wait, does it mean that you are allowed to color any three sides with different colors or that you are allowed to color the entire cube?  And neurosurgeon1118, I don't think you're right because you can rotate the cube.  Then again, I'm not sure about the answer either. :maybe:

Well according to what you said, there are no restrictions to how you color the cube, so for each face, there are 3 options, and there are 6 faces in total so it is:

$ 3^6$

But happy me, you can still rotate the cube to make the same arrangement of colors.  Otherwise, the question would be too easy.  It's kinda like the number of arrangements of things around a circle.  So I guess you need to "anchor" or set one face as a certain color and then move on from there.  :stink:

Well, this all depends on if the faces are distinct.  If they are, which I think it is (or else...), then the answer is easily 3^6; however, if they are not, then we have an exponentially harder problem.  So, basically, the argument you guys are having is if the faces are distinct, or if they are indistinguishable.

wasn't this a mc poster problem like 2 yrs ago?

That poster had only white and red and the faces were distinct (so no rotating).

I think the answer was 10, but don't ask me, I was pissed off the day I saw that poster.

Sorry if I didn't make it clear..

The restriction is that you can NOT have the same colored figure. Like you paint it one way and you can NOT have a figure that is derived by flipping, rotating or transforing.

[quote="happyme"]Is the answer like exponentially huge?

My brain is dead, not willing to do work.[/quote]
NO, I don't think it's that big.

Start off with this. How many ways can you paint a cube with 2 colors?
[hide="Case work for 2 colors"]
Let's say we have yellow and blue as our colors.

[b]Case 1[/b]- No Yellow
You only have 1 way to paint it: All blue.
$ \boxed 1$

[b]Case 2[/b]- 1 Yellow, 5 blue
You can paint it 1 way, because as side you pick, it's going to look the same.
$ \boxed 1$

[b]Case 3[/b]- 2 Yellow, 4 blue
You can paint it 2 ways: one facing the other and one next to other.
$ \boxed 2$

[b]Case 4[/b]- 3 yellow, 3 blue
You can paint it 2 ways: one facing the other with one next to it, and all of them next to each other.
$ \boxed 2$

[b]Case 5[/b]- 4 yellow, 2 blue.
Symmetric to [b]Case 3[/b].
$ \boxed 2$

[b]Case 6[/b]- 5 yellow, 1 blue
Symmetric to [b]Case 2[/b].
$ \boxed 1$

[b]Case 7[/b]- All yellow
Symmetric to [b]Case 1[/b].
$ \boxed 1$

Add those to get $ 10$ possible ways.[/hide]


limit
logarithms
inequalities
calculus
calculus computations

Prove that $ \lim_{n\to\infty}\sqrt[n]n\equal{}1.$

Of course, without L'Hôpital's Rule.

The "elementary" proof:

If we take $ \sqrt[n]{n} : \equal{} 1\plus{}\delta_n$ we get

$ \displaystyle n \equal{} (1\plus{}\delta_n)^n \equal{} 1\plus{}{n\choose1}\delta_n\plus{}{n\choose2}\delta_n^2\plus{}\ldots\plus{}{n\choose n}\delta_n^n$

For $ n>1$ following holds:

$ \displaystyle n > {n\choose2}\delta_n^2 \equal{} \frac{n(n\!\minus{}\!1)}{2!}\delta_n^2,\quad \delta_n^2 < \frac{2}{n\minus{}1},\quad 0 < \delta_n < \sqrt{\frac{2}{n\minus{}1}},$

So $ \displaystyle\lim_{n\to\infty}\delta_n \equal{} 0$

Hence: $ \displaystyle \lim_{n\to\infty}\sqrt[n]{n} \equal{} \lim_{n\to\infty}(1\plus{}\delta_n) \equal{} 1$

(Proof taken from PlanetMath).

Another approach would use Stolz-Cesaro ,as following :

$ \ln \lim_{n \to \infty} \sqrt[n]{n}\equal{} \lim_{n \to \infty} \ln \sqrt[n]{n}\equal{}\lim_{n\to \infty} \frac{\ln n}{n}\equal{}\lim_{n\to \infty} (\ln(n\plus{}1)\minus{}\ln(n))\equal{}\lim_{n \to \infty} \ln \left(\frac{n\plus{}1}{n} \right)\equal{}0$
So $ \lim_{n \to\infty} \sqrt[n]{n}\equal{}e^0\equal{}1$

By Bernoulli's inequality, $ \left(1\plus{}\frac{1}{\sqrt{n}}\right)^n\geq 1\plus{}\sqrt{n}>\sqrt{n}\ \therefore 1\leq \sqrt[n]{n}<\left(1\plus{}\frac{1}{\sqrt{n}}\right)^2$,

yielding $ \lim_{n\to\infty} \sqrt[n]{n}\equal{}1$.

yes, bernomlli inequality is a nice approach


Find all polynomials P(x) which satisfy the equation

$P(x^{2}-2x) =[P(x-2)]^{2}$

$\ P((x+1)(x-1))=[P(x-1)]^{2}$

So for the coefficient of the biggest power of $\ P(x)$

we have: $\ a_{n}(x+1)^{n}(x-1)^{n}=a_{n}^{2}(x-1)^{2n}$

this gives $\ a_{n}= (\frac{x+1}{x-1})^{n}$

But by definition $\ a_{n}$ is a constant value, so we conclude that $\ P(x)$
most be constant $\ (n=0)$.

So $\ P(x)=1$ or $\ 0$.

[quote="maokid7"]how do u know that $a_{n}$ must be constant.[/quote]
Because any polynomial of degre $\ n$  can be written in this way:

$\ P(x)=\sum_{i=0}^{n}a_{i}x_{i}^{i}$ and that I dont need to prove.

with the $\ a_{i}$ real constants. :) there cannot be a polynomial  $\ P(x)\in R$without real constant coefficients.

[quote="El fisico"]So for the coefficient of the biggest power of $\ P(x)$

we have: $\ a_{n}(x+1)^{n}(x-1)^{n}=a_{n}^{2}(x-1)^{2n}$[/quote]

This first step doesn't make any sense -- you've included multiple powers of $x$ on both sides.  In fact, the coefficient of the largest power on the left is $a_{n}$ and the coefficient of the largest power on the right is $a_{n}^{2}$.

oups! sorry I make a mistake.  :ewpu: :blush: You are right JBL. In fact $\ a_{n}=1$ Next time I give a solution. :)


(A) I  have 2 children. They aren't both boys. What is the probability that both children are girls ?
(B) An artist has 2 children. The older is a boy. What is the probability that both children are boys ?

[i]from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/i]
 :ninja:  :rotfl:

A. There are 4 equally likely events, boy-boy, girl-girl, boy-girl, and girl-boy. Thus the probably of two girls is $ \boxed{\frac{1}{3}}$.

B. Going through what we did for A, gives $ \boxed{\frac{1}{2}}$?

I think this is a variant of the boy-girl paradox, right? http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

Isn't the answer to (a) 1/4 because there's a 1/2 chance the first child is a girl and a 1/2 chance the 2nd child is a girl, and 1/2*1/2 is 1/4.

No, not really, that's what it would seem like, but remember that these are puzzles.

Read my first post in this topic. (Order does matter)

... for the first one it would be 100% because theyre both not boys then for the second it would be 50% because its either going to be a boy or a girl.

For number 1, it isn't 100%. He could have a boy and a girl... It is $ \frac{1}{3}$ as previously explained.

Wow, I'm so stupid. I didn't read that both children couldn't be boys. SO STUPID!

my answer for b) is 1/2 because the artist has two children and one has to be a

boy so the two possibilities are boy,boy or girl,boy


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

I'll use that $ \int \cos ^2(\theta ) \, d\theta \equal{} \frac {\theta }{2} \plus{} \frac {1}{4} \sin (2 \theta )$

assuming:

$ x \equal{} e^{ \minus{} \phi }$ $ y \equal{} e^{\phi } \tan (\theta )$

$ 0 < \theta < \frac {\pi }{4}\land 0 < \phi < \minus{} \log (\tan (\theta ))$

$ \left|\left( \begin{array}{ll} \frac {\partial x}{\partial \theta } & \frac {\partial x}{\partial \phi } \\
\frac {\partial y}{\partial \theta } & \frac {\partial y}{\partial \phi } \end{array} \right)\right| \equal{} \sec ^2(\theta )$

The integral becomes

$ P \equal{} \int _0^{\frac {\pi }{4}}\int _0^{ \minus{} \log (\tan (\theta ))}\frac {\cos ^2(\theta )}{\log (\tan (\theta ))}d\phi d\theta \equal{} \minus{} \int_0^{\frac {\pi }{4}} \cos ^2(\theta ) \, d\theta \equal{} \minus{} \frac {1}{8} (2 \plus{} \pi )$

I used substitution $ u\equal{}xy, v\equal{}\frac{x}{y}$.Then we have Jacobian $ J\equal{}\minus{}\frac{1}{2v}$ and :
$ I\equal{}\minus{}\int_0^1\;du \int_u^\frac{1}{u}\;\frac{dv}{2v} \frac{1}{\ln\,u \cdot (1\plus{}u^2)^2}\equal{}\minus{}\int_0^1\; \frac{du}{\ln\,u \cdot (1\plus{}u^2)^2} \int_u^\frac{1}{u}\;\frac{dv}{2v}\equal{}\minus{}\int_0^1\; \frac{du}{\ln\,u \cdot (1\plus{}u^2)^2} \ln\,u\equal{}\minus{}\int_0^1\; \frac{du}{(1\plus{}u^2)^2}$
Then we use substitution $ u\equal{}\tan\,t$ and:
$ \int_0^1\; \frac{du}{(1\plus{}u^2)^2}\equal{}\int_0^\frac{\pi}{4}\; \frac{\frac{dt}{\cos^2\,t}}{(\frac{1}{\cos^2\,t})^2}\equal{}\int_0^\frac{\pi}{4}\; \cos^2\,t dt\equal{}\frac{2\plus{}\pi}{8}$

[quote="misan"]
$ \int_0^1\;\int_0^1\;\;\dfrac{\textbf dx\;\textbf dy}{(\ln\,x \plus{} \ln\,y)\cdot(1 \plus{} x^2\,y^2)^2}\;\; \equal{} \;\;\boxed{ \minus{} \frac {1}{8}\,(2 \plus{} \pi)}$[/quote]
Similar:

Define $ u \equal{} xy$ and reverse the integration order

$ \int_0^1 {\int_0^1 {\frac {1}{{\ln (xy)(1 \plus{} x^2 y^2 )^2 }}\,dx} \,dy} \equal{} \int_0^1 {\int_0^y {\frac {1}{{y(1 \plus{} u^2 )^2 \ln u}}\,du} \,dy} \equal{} \int_0^1 {\underbrace {\int_u^1 {\frac {1}{y}\,dy} }_{ \minus{} \ln u} \cdot \frac {1}{{(1 \plus{} u^2 )^2 \ln u}}\,du}.$

The conclusion follows from greenvert's computation for the last integral.


Two scientists are collecting rock samples when suddenly they encounter a threatening animal. The first scientist, remembering his training, leaves slowly and lives. The second scientist panics and flees quickly; he dies.

- Is the animal common enough that a layperson would be familiar with it?
- Does the second scientist die as a result of actions on the part of the animal?

- Is the animal common enough that a layperson would be familiar with it? [b]yes[/b]
- Does the second scientist die as a result of actions on the part of the animal? [b]No[/b]

[quote="t0rajir0u"][hide="Hypothesis"] The two scientists are divers in the deep ocean collecting rock samples near heat vents or something of the like.  The animal is a shark or some other dangerous deep-sea creature.  The second scientist dies because he surfaces too quickly, getting the bends. [/hide][/quote]
correct!


Solve

1.In the island of Camelot 13 purple camels live, 15 greens and 17 yellows. When two of distinct colors if find, move simultaneously for the third color. Could be given to it situation in which all has the same color?

Good bye.........

[hide]  no because the difference between the number of each color is invariant  mod 3 (i.e g-p is always 2 mod 3, etc.)  If it can happen, then at some point they will be 43, 1, 1.  But this means that at the start there is a pair of colors whose difference is 1-1 ==0(mod 3), which is false.[/hide]


A cyclic quadriteral $ ABCD$ is given. If $ M$ and $ N$ are midpoints of $ AC$ and $ BD$ and if $ BD$ is bisector of $ \angle ANC$ prove that $ AC$ is a bisector of angle $ \angle BMD$

See here:
http://www.mathlinks.ro/viewtopic.php?search_id=2088370956&t=6557

Best regards,
sunken rock


today a friend told me that : for any compact subset $K$ of $[0;1]$, there exists a real number $x$ such that the accumulation points of $u_{n}=cos(n! x)$ is precisely $K$. Is it true ?

There is an article by Richard Antetomaso in the last RMS about that.

really!  I thought it would be easy by looking for an $x$ of the form $x = \sum \frac{a_{k}}{k!}$ with $|a_{k}| \leq n$ for example. My proof must be wrong.


The middle third of a line 1m long is erased. From each of the two remaining thirds, their middle thirds are erased. From each of the four remaining ninths, their middle thirds are erased. If this process is continued indefinitely, find the sum of the remaining segments.[/quote]

You mean the sum of their [i]lengths.[/i]  This describes (up to fiddling with endpoints) the construction of the [url=http://en.wikipedia.org/wiki/Cantor_set]Cantor set[/url].


function
group theory
abstract algebra
number theory open
number theory

Let $ a$ and $ n$ be positive integers with $ a > 1$. Prove that $ n|\phi(a^n \minus{} 1)$.

$ \mathbf{M}_{a^{n} \minus{} 1} \equal{} \{m \in \mathbb{N} : 1 \leq m \leq a^{n} \minus{} 1 \mbox{ and }(m,a^{n} \minus{} 1) \equal{} 1 \}$ is a group under multiplication modulo $ a^{n} \minus{} 1$ and $ \mathbf{S} \equal{} \{1,a, a^{2}, \ldots, a^{n \minus{} 1}\}$ is a [i]distinguished[/i] subgroup of $ \mathbf{M}_{a^{n} \minus{} 1}$.

The purported result is an immediate consequence of [url=http://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)]Lagrange's theorem[/url] now.


On problem 184 of AoPS book 1, it says this:

$\angle B$ of $\triangle ABC$, is trisected by $BD$ and $BE$ which meet $AC$ at $D$ and $E$ respectively.

But how are you supposed to know if it is $D$ or $E$ that is closer to point $A$? It makes a big difference in the problem.

I think the use of the word "respectively" in that problem, and the order in which the line segments and points are named, makes it clear enough that D is closer to A, and E is closer to C.

After edit: I checked the solution volume for that problem, and what is said is correct about the order of the points. The authors' solution is interesting.

I agree that it isn't 100% conclusive though. If the problem appeared in an olympiad, they would state explicitly which point was closer to A.

But just as a regular problem in a book or something, its reasonably guessable.


If a scientist is sitting inside the bob of a huge pendulum will he feel any gravitational acceleration? and will it be constant?

Yes, he would, and it would be far from constant. You can actually calculate it as a sum of gravitational acceleration, centrifugal acceleration and acceleration of the pendulum itself. It should look something like this:

$\vec{g'}=\vec{g}+\frac{v^2}{r^2}\vec{r}-\vec{a}$

where

$\vec{v}=\dot{\vec{r}}$ (velocity)

$\vec{a}=\dot{\vec{v}}$ (acceleration)

and $\vec{r}$ is position of the pendulum bob.


videos
MIT
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linear algebra
Functional Analysis

Hi, I am starting to learn Linear Algebra now, can anyone please suggest me some books?

I am reading two books, one by Hoffman and Kuzne and the other by Serge Lang.

[u][i]Linear Algebra Done Right[/i][/u], by Sheldon Axler
ISBN 978-0-387-98258-8

I always suggest this book to people who are starting linear algebra. I'm not familiar with the one you're using, but this book really develops and motivates the concepts in a way that actually gave me a sense of how linear algebra works. Before I read it, I didn't get linear algebra, and after I read it, I understood it. I recommend you at least take a look  :)

http://joshua.smcvt.edu/linearalgebra/

You can download this pdf book there for free. It's the one I'm reading and it seems pretty good to me.

[b]darij[/b], thanks for the link! Some great problems there.

I still think Axler's book is more appropriate for someone who's starting linear algebra -- throwing abstract algebra into the picture is a (much) later step.

Thanks very much, specially Darij I liked it.....

Glad that you liked the Prasolov book. The main downside (which it has in common with 90% of Russian literature) that it never explicitely says which field (or even ring) one is working over, so it can be considered a part of the problem to find out whether it needs $ k$ algebraically closed or $ \mathrm{char} k\equal{}0$ or not...

  darij

I see Halmos' Finite Dimensional Vector Spaces hasn't been mentioned yet.

I have a question myself: I am working through Axler at the moment (mainly as as preparation for functional analysis and quantum mechanics, but also just because I want to know this stuff), and it is going very well. What comes next?
Roman's [url=http://www.amazon.com/Advanced-Linear-Algebra-Graduate-Mathematics/dp/0387247661]Advanced Linear Algebra[/url] and Peter Lax' [url=http://www.amazon.com/Linear-Algebra-Applications-Applied-Mathematics/dp/0471751561/ref=sr_1_1?ie=UTF8&qid=1249142564&sr=8-1]Linear Algebra[/url] seem pretty good and doable as a follow up for Axler. What do you guys think?
Or is it wise to do some abstract algebra first? I have a copy of Dummit & Foote, but have only done group and ring theory. Should I do vector spaces and modules from Dummit and Foote? Thanks.

[quote="darij grinberg"]The book by Axler should rather be called "Linear algebra without the algebra"...

My favorite workbook for linear algebra: [url=http://www2.math.su.se/~mleites/Prasolov/prasLinAlg]Victor Prasolov, [i]Problems and theorems in linear algebra[/i], translated by Dimitry Leites[/url]. (The big file, pr-linAlg-main.dvi, is the important one. You need some DVI previewer to read it.) This is a Russian-style problem collection, but rather challenging even for the Russian level. The explanations at the beginning of the chapters could be used as a rather compressed textbook, but probably it is best used along with a real textbook - Prasolov is a source of beautiful and hard problems rather than something to read from the beginning to the end.

Also, Georgi E. Shilov, [i]Linear Algebra[/i] looks nice to me, but I haven't read enough of it to judge its style - I just like the selection of material. It's a book that (unlike Axler ;) ) does at least some of the hard stuff (though still no tensor products).

  darij[/quote]

What DVI previewer do I get? I have got two : LEd and DVIViewer and both don't seem to work. Are you sure I have to download only that file?

[quote="orl"]I can recommend the text book on the topic by Gilbert Strang. You may also watch the video lectures on [url=http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/index.htm]MIT OCW (Youtube).[/url][/quote]

I have his work:  [url=http://www.amazon.co.jp/Introduction-Linear-Algebra-Gilbert-Strang/dp/0961408898/ref=sr_1_7?ie=UTF8&s=english-books&qid=1250436766&sr=1-7]Introduiction to Linear Algebra[/url]. MIT OCW by Gilbert Strang is good.

Personally Linear Algebra by Serge Lang is always at hand.


logarithms
trigonometry
probability
percent
geometry
analytic geometry
graphing lines

I decided to post furman solutions for 2002 in order to revive the SC forum :wink:

http://math.furman.edu/tournament/tournament.html

Need [b]#26, 30, 32[/b]

1. Because a leap year occurs every $4$ years, it will take $7 \cdot 4=28$ years for a leap year starting with a Saturday. Thus, including $1900$, there are $4$ leap years fulfilling those conditions.

2. Let the base be $b$. We have that $b+3, 4b+2, b^{2}+1$ are in an arithmetic progression. Thus, $b^{2}+1-(4b+2)=4b+2-(b+3) \Longrightarrow b=7$.

3. The sum is $\log_{6}{4}+\log_{6}{9}=\log_{6}{36}=2$.

4. Let $\angle{A}=\angle{C}=\theta$. We have that:

$2^{2}+5^{2}-2(2)(5)(\cos{\theta})=4^{2}+4^{2}-2(4)(4)(\cos{\theta})$. Solving for $\cos{\theta}$ and plugging back in gives us $BD=2\sqrt{6}$.

5. We have $(1+2)(1-2)+(3+4)(3-4)+...+(99+100)(99-100)=-1(1+2+3+...+100)=-5050$.

6. We keep squaring the expression to get: $1+x+\sqrt{x}=x+\sqrt{x+5}\Longrightarrow 1+2\sqrt{x}+x=x+5 \Longrightarrow x=4$ (since $x>0$).

7. Let the ages of the kids be $x,x,y$. We have $2x+y=13$ and we want to know $x^{2}y$. We deduce $x=6$ and $y=1$ (since the professor cannot be in the 80s and have such young kids). The answer is $36$.

8. We have:

$2^{k\mod{4}}\equiv 1 \mod{5}$
$2^{k\mod{6}}\equiv 1 \mod{7}$
$2^{k\mod{30}}\equiv 1 \mod{30}$.

Letting $k \equiv 0 \mod{4}, \mod{6}\mod{30}$ ensures the smallest $k$. Thus, $k=60$.

9. Let the roots be $a,b,c,d$.

We have:

$a+b+c+d=9$
$abcd=-11$.

Because $a,b,c,d$ are all integers, and $11$ is prime, and looking to see how the sum of the roots is positive, we deduce one of the roots is $11$. Thus, there are three $-1$s as roots, and one $1$.

10. The answer is $\binom{6}{2}+\binom{5}{2}-\binom{2}{2}=24$.

11. We have that $\frac{1}{b-1}=r$ and $.2b=r$ in base $2b$. $b=5$ satisfies our conditions with $r=\frac{1}{4}$.

12. The equation can be simplified to $(\log{2})^{2}-(\log{3})^{2}=(\log{3}-\log{2})\log{x}$ and thus $\log{x}=\log{1/6}\Longrightarrow x=1/6$.

13. The equation simplifies to:

$4^{t}(4+16+64+256)=170 \Longrightarrow 4^{t}=1/2 \Longrightarrow-1/2$.

14. $9241$ satisfies the conditions through simple long division.

15. The distance between the poles is the sum of their heights.

16. There are $10$ digits and since no two digits can be the same (or it would violate $a>b>c$), we have that the probability is $\frac{\binom{10}{3}}{10^{3}}=\frac{3}{25}$.

17. Factoring, we have $25!(27)$. $27=3^{3}$ and $25!$ contains $[\frac{25}{3}]=8, [\frac{8}{3}]=2 \Longrightarrow 10$ threes. Therefore, the answer is $10+3=13$.

18. Let the original gasohol be $\$100$ ($\$90$ gasoline and $\$10$ alcohol). Then, gasoline costs $\$\frac{100}{1.05}$. After the doubling of the price, gasohol will be $\$190$ and gasoline will be $\$\frac{200}{1.05}$. The percent difference is $\frac{.5}{200}=2.5 \%$.

19. We have that $198 \cdot a=x92y$, where $x$ and $y$ represent two digits of the total sum. Notice also that $a>5$. Trying out $a$ as multiples of $10$, we realize that $a=40$.

20. Let the radius of the cylinder be $x$. We have that $2x^{2}\pi+8=3x^{2}\pi \Longrightarrow x^{2}\pi=8$. Thus, the original volume is $8 \cdot 2=16$.

21. Notice that $\tan{x}=\cot{(90-x)}$ and $\tan{x}\cdot \cot{x}=1$. It follows that the product evaluates to $1$.

22. We continuously perform long division on the two given terms and then their remainders. Eventually, we see that $\frac{17}{n+10}$ must be an integer, giving us that $17$ is our answer when $n=7$.

23. Let $n=x^$. We have that $x^{2}+1001=k^{2}$ for some positive integer $k$. Thus $(k+x)(k-x)=1001$. $1001=7 \cdot 11 \cdot 13$. Using these values (along with $1001 \cdot 1$), we see that there are four values of $n$.

24. Notice that $\alpha=i$ and $\beta=cis60$. There are 4 values that powers of $i$ form, and 3 values that powers of $cis60$ can form. Thus the answer is $3 \cdot 4=12$.

25. Notice that if the largest number of the $n$th group is $\frac{n(n+1)}{2}$. Thus, the largest number of the $100$th group is $5050$. The sum of the $100$ numbers in that group is $500050$.

27. The figure is symmetric, so let's find the area of the region we want for the top part, and then multiply by $2$ at the end. Call the region we are looking for $X$, and the other two regions $Y, Z$, with $Y>Z$. Using the fact that we have semicircles to work with, we find that the area of $Z=\frac{1}{18}$ and the area of $Z+X=\frac{4}{18}$. Thus, $2X=\frac{1}{3}$.

28. $A$ has $18$ divisors and the product of it's divisors is $A^{\frac{18}{2}}$. Thus, the answer is $9$.

29. In order to divide the square into parts of equal area, the line must go through the midpoint of $(-2,-1)$ and $(4,5)$ which is $(1,2)$. The slope is thus $2$.

31. Label the squares from left to right $1,2,3,4,5$. The striped bug starts off at $2$ and can't go to $1$, and the spotted bug starts off at $4$ and can't go to $5$. Because the two bugs move simultaneously one space, it follows that every time the striped bug is on an odd number, the spotted bug is on an even number and vice versa. Thus, the two bugs are never in the same square.

[hide]

26. . Let . Then we want . This is clearly .

32. Um, cube , match coefficients, you get , so the answer is .

30. Ok, let  be distance,  = sara bike, and so on. Then we are given (by d=rt)

Substituting first and third equations, we get . Substituting second and third, we get , or equivalent, . Adding the two equations we got, we get . Substituting  and isolating , we get , so it takes them 8 minutes, and they meet at 3:38.

[/hide]


All faces of a convex polyhedron are parallelograms. Can the polyhedron have exactly 1992 faces?

the official solution, below quoted, is not correct, so it remains unsolved

[quote]Solution. No, it cannot. Let us call a series of faces $F_1, F_2, .. , F_k$ a ring if the pairs $(F_1, F_2), (F_2, F_3),.. ,
(F_{k-1}, F_k), (F_k, F_1)$ each have a common edge and all these common edges are parallel. It is not difficult to see that any two rings have exactly two common faces and, conversely, each face belongs to exactly two rings. Therefore, if there are n rings then the total number of faces must be  $2 \binom{n}{2} = n(n - 1)$. But there is no positive integer n such that $n(n - 1) = 1992$.

Remark. The above solution, which is the only one proposed that is known to us, is not correct. For a counterexample, consider a cube with side $2$ built up of four unit cubes, and take the polyhedron with $24$ faces built up of the faces of the unit cubes that face the outside. This polyhedron has rings that do not have any faces in common. Moreover, by subdividing faces into rectangles sufficiently many times, we can obtain a polyhedron with $1992$ faces.[/quote]

[quote=parmenides51]so it remains unsolved[/quote]
Are you sure? Isn't the last sentence of the remark saying that the answer actually is YES with a very simple construction:
Similar to the counterexample, we divide the cube into $2 \times 18 \times 48$ rectangular boxes and obtain a polyhedron with $2(2 \cdot 18+2 \cdot 48+18 \cdot 48)=1992$ rectangular faces (in particular parallelograms).

I do not know anything about Solid Geometry (shame on me, I know). I just quoted everything written in the official solutions. So I cannot understand any of these. I just though it was worth mentioning the comment found there.

[url=http://www.math.olympiaadid.ut.ee/eng/archive/bw/bw92sol.pdf]here [/url]  is the solutions' pdf


Hey guys!
Okay, I have a question regarding the phi function and the generalized form of fermat's little theorem (which is related to the phi function). I've seen many number theory problems (some even olympiad level), but I have yet to see a solution or step that requires or came as a result of the phi function or its properties. In other words, what's it good for? :lol:

It is very useful and there are quite a few problems which require it. For example this Indian RMO problem-1990 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=347213#347213

Generally, problems like "find all integers $ x,y,z,w$ with $ x^4\plus{}y^4\plus{}z^4\equal{}4w^4$" can be bashed easily with Euler-Fermat. The problem I posted becomes close to trivial; others can be reduced in complexity.

It can also be used to trivialize different types of diophantine equations.

Most Diophantine proofs using mods use prime mods, since if an equation has no solution $ \bmod m$ then it must have no solutions $ \bmod p^k$ for some prime $ p$.  And I have very rarely seen a problem that required $ k > 1$.

One example of a class of problems that requires the phi function is the "last two digits" or "last three digits" kind of problems, since $ 100, 1000$ (respectively) aren't prime (for example, we might want to find the last three digits of $ 2004^{2003^{2002}}$).  Of course, it's more efficient to use [url=http://mathworld.wolfram.com/CarmichaelFunction.html]the Carmichael function[/url] instead.


algebra
polynomial
inequalities
real analysis
real analysis unsolved

Given any monic polynomial of degree $ d$ it holds $ \max_{\minus{}1 \leq x \leq 1}|P(x)|  \geq 2^{1\minus{}d}$.

Yes, this is well-known fact about Chebyshev polynomials http://en.wikipedia.org/wiki/Chebyshev_polynomials. To prove this theorem one can use Alternate Chebyshev theorem (try to google it).