calculus
quadratics
analytic geometry
algorithm
number theory unsolved
number theory

[quote="cosinerburc"]Find in explicit form all ordered pairs of positive integers $(m, n)$ such that $mn-1$ divides $m^2 + n^2$.[/quote]

This problem is es exactly to another one, but I could not find it, but I remember tha if $a,\, b$ and $\frac{a^2+b^2}{ab-1}=q$ are integers then $q=5$ so, in this case $m^2+n^2=5mn-5$ and how you can see in the other Topic the solution is $(2,1)$.

Best Regards.

RX the Crazy Math

[quote="RobertuX"]if $a,\, b$ and $\frac{a^2+b^2}{ab-1}=q$ are integers then $q=5$[/quote]
I tried to prove it but I failed :?
Can you give your proof?

I think stuff like this has been proven before on the forum. In fact, I think I can recall Myth solving this very problem. Try an infinite descent approach. I don't have the time to work out the details, but try to find, from a solution $(a_n,b_n)$ with $a\le b$, a solution $(a_{n+1},b_{n+1})$ with $a_{n+1}\le b_{n+1}$ and $b_{n+1}=a_n$. You have to solve a second degree equation in order to do this.

[quote="cosinerburc"]i know that q is 5,i mean i read it in a book,also the solution after realising that q=5; but do u know how i can find that q=5[/quote]

I think will be a good idea to find the another TOPIC, I couldn't find, so maybe Groober knows, ask him and in the topic you can find the solution of the problem.

Best Regards
RX  The Crazy Math

Here goes the solution.
The infinit descent grobber was talking about is to show that if $(a,b)$ is a solution with $a\leq b$ then $a,ka-b$ is also a solution and is a smaller one (that is, $(x,y)$ is smaller than $(u,v)$ iff $x<y$ or $x=y$ and $u<v$).

Just check to see that indeed $(a,ka-b)$ is a solution. We still have to prove that $ka-b>0$ and to prove that this solution is smaller, we have to show $ka-b<b$. So, let's write the quadratic.

$a^2-a(kb)+b^2+k=0$. Let $c$ be the other solution, different from $a$ of this quadratic. Then we have $a+c=kb, ac=b^2+k$. Now, if $k\leq\frac ba\Rightarrow a+c\leq \frac {b^2}a\Rightarrow a^2+b^2+k\leq b^2$, false. $\frac {a^2+b^2}{ab-1}>\frac {2b}a\Leftrightarrow 2b>a(b^2-a^2) (1)$. But $a=b$ gives no solution, so $a^2\leq (b-1)^2$. So putting this upper, we have $a(b^2-a^2)\geq a(2b-1)\geq 2(2b-1)>2b$, if $a\geq 2$, so we can't have $(1)$. If $a=1$, we must have $b=2$ to have (1) true.

So, unless we have reached the minimal solution $(1,2)$, we can find a smaller solution from $(a,b)$, so go lower 'till we must reach the $(a,2)$ pair at some point. Because $k$ remains unchanged, it follows that $k=5$ for every such a pair.

one more thing
I know I left the problem unfinished, but it's really hard to determine all the solutions and probably it would cost a lot of extrawork because, when trying to go reverse you can obtain 2 different bigger solutions from a given solution.
For example, from $(1,2)$ you can obtain $(1,5\cdot 1-3)$ and $(2,5\cdot 2-1)$. i don't know how all these solutions can be generated.

Something wrong in ikap's last statement. Solutions can be generated in the unique way. :?

oops, right  :blush:

so $5a_n-b_n=a_{n-1} \&\& a_n=b_{n-1}$. This is a recurence formula only starting from pairs $(2,9)$ and $3,14$. From here we can prove that $4a_n\leq b_n\Leftrightarrow 4b_{n-1}\leq 5b_{n-1}-a_{n-1}$, and from this it follows that the other pair $(a_{n+1},b_{n+1})$ than the one described upper is in fact smaller than $(a_n,b_n)$ so it can't be.

Solving the above recurence and adding tot those solutions $(1,2), (1,3)$ gives us all the solutions.

[quote="ikap"] $\frac {a^2+b^2}{ab-1}<\frac {2b}a\Leftrightarrow 2b>a(b^2-a^2) (1)$[/quote]
Well this is NOT ok, you need to do better your calculus because $> \to <$ in the 2nd expression.

Sorry and best regards.

RX the Crazy Math

Yes, Robertux, but if u would have followed the line of my proof you would have noticed that what I needed to show was that $k\leq \frac {2b}a$, so I supposed $k> \frac{2b}a$ so only the first sign should be inverted in what you have quoted.
If you don't understand the solution don't hesitate to ask.
And, something more. Calculus is something completely different from computation.

[quote="ikap"][quote="Myth"][quote="ikap"]And, something more. Calculus is something completely different from computation.[/quote]
WoW, I see a philosophical underlying theme here. :)[/quote]

It goes far beyond my power of understanding :blush:. Maybe a hint to make me understand where the philosophy is :)[/quote]

mmmm, anyway if you don't understand your mistakes, is not my problem  :D   :P  (I assume that $a b-1 \ge  0$).

[quote="ikap"]I've already told you which was the one and only mistake, and now it's fixed.
It would be another mistake there $5\cdot 1-3\neq 2$  :lol: but that one remains for posterity. :D[/quote]

Well. move this TOPIC to solved topics, I only know that was a simple problem and I am ready to "try" new problems, I suppose that the idea in the forum is to know new ways to solve problems, and not only solve the problems. So, I hope people learn math and technics to solve it. As you know, I am teacher in the university and this is one of my interest in science.

Best regards.
RX the Crazy Math

BTW, as grobber said, this problem really appeared three times on forum. And I wrote a solution for it some weeks ago... ;)

[quote="ikap"]ok, ok, sorry, be sure to have all my respect, i didn't wanted to irritate you, just wanted to make it clear.[/quote]
I never was irritate, don't worry about that, I only want to help another guys learn.

About the problem, see also http://www.mathlinks.ro/Forum/viewtopic.php?t=22810 . Some other topics should exist, too.

  Darij

Let $ \frac{m^2 + n^2}{mn - 1} = k $ for some $ k \in \mathbb{Z}^+ $.  We can write this as $ m^2 - knm + (n^2 + k) = 0 $ and so $ m $ is a solution to the quadratic in $ x $, $ x^2 - knx + (n^2 + k) = 0 $.  The other solution is $ m' = kn - m = \frac{n^2 + k}{m} $.  Now clearly $ m \ne n $, so assume WLOG that $ m > n $.  For now, assume that $ n > 1 $.

These assumptions imply that $ n(m - n) \ge 2 \Longrightarrow m(mn - 2) > n^2 $.  Rearranging this means that $ n(m^2 + n^2) < 2m(mn - 1) \Longrightarrow m' = kn - m < m $.  So if we consider the process of changing pairs $ (m, n) $ into a new pair $ (km - n, n) $ (rearranged so that the first coordinate is the larger of the two) until one of the elements of the pair is $ 1 $, this process clearly terminates at either $ (2, 1) $ or $ (3, 1) $.  Since $ k $ has stayed the same throughout this process, this argument implies that $ k = 5 $.

This implies that all solutions are given by taking the sequences $ x_1 = 1, x_2 = 2, x_n = 5x_{n - 1} - x_{n - 2} $ and $ y_1 = 1, y_2 = 3, y_n = 5y_{n - 1} - y_{n - 2} $ and taking pairs $ (x_m, x_{m + 1}) $ and $ (y_m, y_{m + 1}) $ for any $ m \in \mathbb{N}. $  The explicit forms of these pairs can then be found in a standard fashion with generating functions or just taking the characteristic polynomial of the linear recurrences.

[quote="Wolstenholme"]Let $ \frac{m^2 + n^2}{mn - 1} = k $ for some $ k \in \mathbb{Z}^+ $.  We can write this as $ m^2 + knm + (n^2 + k) = 0 $ and so $ m $ is a solution to the quadratic in $ x $, $ x^2 + knx + (n^2 + k) = 0 $.  The other solution is $ m' = kn - m = \frac{n^2 + k}{m} $.  Now clearly $ m \ne n $, so assume WLOG that $ m > n $.  For now, assume that $ n > 1 $.

.[/quote]

The correct is $m^2-kmn+n^2+k-0$

$x^2+y^2=k(xy-1)$

Set

$xy-1=a$

$x^2+y^2=ka$

We want to solve this system...

$x=\dfrac{a+1}{y}$

So put $x$ at the equation  and will get..

$y^4-kay^2+(a+1)^2=0$

$D=k^2a^2-4(a+1)^2=w^2$   ,   (then $y^2=\dfrac{ka+-w}{2}$)

$(ka)^2=(2a+2)^2+w^2$  pythagorian triples and the rest is easy...

Δημήτρης

[quote=realquarterb]I claim the only pairs $(m,n)$ for which $mn-1 \mid m^2+n^2$ are $\boxed{(m,n)=(2,1),(3,1),(1,2),(1,3)}$.

If $mn-1 \mid m^2+n^2$ we must surely have $$\frac{m^2+n^2}{mn-1}=c~~~c\in\mathbb{Z}.$$ Now this implies that we have $m^2+n^2=c(mn-1)$ or $m^2-cnm+n^2+c^2=0$.

Now let $(m,n)$ be a solution to this diophantine, where $m+n$ is minimized. If $(m,n)$ is a solution to the diophantine, it must surely be true that $(m',n)$ is a solution to the equation, where $m'=cn-m=\frac{n^2+c}{m}$. Clearly, $m'$ is positive, since $c$ is positive, so $cn-m>0$. Since $m'\in\mathbb{Z}, cn-m \ge 1$.

Now clearly $m \ne n$, so without loss of generality, $m>n$. Now I claim that $n=1$.
For sake of contradiction, assume that $n \ne 1$. Since $n$ is a positive integer, we must have $n>1 \implies m>n>1$. Now by minimality of $m+n$, we must have $$m+n \le m'+n \implies m \le m'.$$ However, observe that since $m>n>1$, $$n(m-n) \ge 2 \implies mn-2 \ge n^2 \implies m(mn-2) \ge mn^2 >n^3 \implies n(m^2-n^2) >2m$$ $$\implies m^2n >2m+n^3 \implies 2m^2n-2m >m^2n+n^3 \implies 2m(mn-1) > n(m^2+n^2) \implies m>kn-m=m'$$ a contradiction. Thus, $n=1$.

Now since $n=1$, we must have $$\frac{m^2+n^2}{mn-1}=\frac{m^2+1}{m-1}=(m+1)+\frac{2}{m-1}$$ Clearly this an integer if and only if $m-1 \mid 2$, and since $m \ge 0$ we must clearly have $m=2$ or $m=3$. Thus, the only solutions for which $m>n$ is $(2,1)$ and $(3,1)$.

Now since the expression is symmetric about $m$ and $n$ the only pairs $(m,n)$ for which $mn-1 \mid m^2+n^2$ is $\boxed{(2,1),(3,1),(1,2),(1,3)}. \blacksquare$

P.S: 1000th post![/quote]

This is wrong of course you've shown using vieta jumping that the smallest solution should have $n=1$ but there is nothing about the bigger once however applying vieta jumping conversly generates the other solutions(There are infinity of them for example $(9,2)$ works too.)

Here's my solution:
${\color{red} \rule{\linewidth}{0.5mm} }$
So first we will claim that if [b]Claim:[/b] $\frac{a^2+b^2}{ab-1}=k$ then $k=5$
[b]Proof:[/b] First we use a [b]Lemma 1:[/b] If $\frac{A^2+B^2}{AB-1}=n$ has $(A,B)$ as a solution then $(A,\frac{B^2+n}{A})$ is also an integer solution.
[b][color=#f00]Proof:[/color][/b] The solution to $\frac{A^2+B^2}{AB-1}=n$  will also be a solution to $x^2-nBx+(B^2+n)=0$.By our initial assumption $A$ is obviously a root,and by vieta's relations $\frac{B^2+n}{A}$ is also a root.Putting this in the equation gives the result as $n$ so this is another root.
[b]Lemma 2:- [/b] If $A>B \ge 3$ then $\frac{b^2+n}{a}<b$ where $min(a,b)=(A,B)$
[b][color=#ff9a00]Proof:[/color][/b] Since $b \ge 3$ we must have $b^2-1>2b$.
Using this:
$\frac{b^3+b}{b^2-1}=b+\frac{2b}{b^2-1}<a->b^3+b<ab^2-a->a+b^3<ab^2-b->\frac{a+b^3}{ab-1}<b->\frac{b^2+n}{a}<b$ as required.
So the only cases we need to check are $b=1,2$
Now we check case by case:-
1)$b=1$,then we must have $\frac{2}{a-1} \in N$ so the only possible value of $n=5$
2)$b=2$,then $\frac{17}{a-1} \in N$ and checking yields $n=5$
Hence proved.
${\color{red} \rule{\linewidth}{0.5mm} }$
Now we want to solve the equation $\frac{m^2+n^2}{mn-1}=5$ so we get $m^2+m^2=5(mn+1)$ and then suppose that $(m,n)$ is the smallest root.
Now by vieta's relations the other root will be $(m,n)->(5m-n,m)$ yielding infinite solutions to the equation.
So to find its explicit formula we need to calculate ${a_1},{b_1}$
Now as @realquaterb showed,we will show that the smallest solutions must have $n=1$.
So,If $mn-1 \mid m^2+n^2$ we must have$$\frac{m^2+n^2}{mn-1}=c$$,And by this we get the quadratic:$m^2+n^2=c(mn-1)$ or $m^2-cnm+n^2+c^2=0$.

Now let $(m,n)$ be a solution to this diophantine, where $m+n$ is minimal.And also assume that the other root of this quadratic is $M$

Now since  $m \ne n$, so WLOG, $m>n$. Now we show that $n=1$.
FTSOC, assume that $n \ne 1$. Since $n$ is a positive integer, we must have $n>1 \implies m>n>1$. Now  since  $m+n$ is minimum, we must have$$m+n \le M+n \implies m \le M.$$However, observe that since $m>n>1$,$$n(m-n) \ge 2 \implies mn-2 \ge n^2 \implies m(mn-2) \ge mn^2 >n^3 \implies n(m^2-n^2) >2m$$$$\implies m^2n >2m+n^3 \implies 2m^2n-2m >m^2n+n^3 \implies 2m(mn-1) > n(m^2+n^2) \implies m>kn-m=M$$a contradiction. Hence, $n=1$.
So we obtain $(1,2),(1,3)$ and it's permutations to be ${a_1},{b_1}$ respectively.
Now all the solutions generated by this sequence are unique since if say $(m,n)$ was a solution then $(5m-n,m)$ is greater than our previous solution since $5m-n>5m-m>4m>m,m>n$
Therefore, $\boxed{({a_1},{b_1})=(1,2),(1,3),(2,1),(3,1) \text{ and the explicit formula is } 5a_n-b_n=a_{n+1},b_n=a_{n-1}}$
${\color{black} \rule{\linewidth}{0.5mm} }$

The answer is the following two classes of solutions, up to symmetry:
[list]
[*]$(a_k, a_{k+1})$ where $a_1 = 1, a_2 = 2, a_n = 5a_{n-1} - a_{n-2}$.
[*]$(b_k, b_{k+1})$ where $b_1 = 1, b_2 = 3, b_n = 5b_{n-1} - b_{n-2}$.
[/list]
Assume that $(m, n)$ is a solution to the equation $$\frac{m^2+n^2}{mn-1} = c$$ for some fixed positive integer $c$, which we will call $\dagger$. Then, $$m^2+n^2=c(mn-1),$$ which expands to $$m^2+n^2-cmn+c=0.$$ As a result, if $(m, n)$ is a solution to $\dagger$, so is $(m', n) = (cn-m, n)$, a transformation which we will call $T$.

Now, suppose that $m+n$ is minimal across all solutions $(m, n)$. Then $$m' = \frac{n^2+c}m \geq m \iff m^2-n^2 \leq \frac{m^2+n^2}{mn-1}.$$ With some algebra, this expands to $n(m^2-n^2) \leq 2m$. However, consider the chain of inequalities $$mn(m-n) < n(m+n)(m-n) \leq 2m,$$ which implies that $n(m-n)\leq 1$ for this minimal pair. It follows that $(m, n) = (2, 1)$ is the only solution to $\dagger$, which yields the constant $c=5$.

It remains to classify all solutions. All solutions can be obtained by alternating applying $T^{-1}$ and inverting $(m, n) \to (n, m)$ (as applying any two of these operations consecutively returns to the identity.) This yields two classes of solutions, up to symmetry:
[list]
[*]$(a_k, a_{k+1})$ where $a_1 = 1, a_2 = 2, a_n = 5a_{n-1} - a_{n-2}$.
[*]$(b_k, b_{k+1})$ where $b_1 = 1, b_2 = 3, b_n = 5b_{n-1} - b_{n-2}$.
[/list]
These are all the solutions, the first class obtained by inverting the base solution $(2, 1)$ to $(1, 2)$ and applying $T^{-1}$ and performing an inversion alternately, and the second class obtained by performing $T^{-1}$ on the base solution and inverting it, yielding $(1, 3)$, before we apply the alternating series of operations. Notice that because some series of operations $T$ and inversions will transform any solution into $(2, 1)$ and thus have $c=5$, this classifies all solutions.

Essentially identical to 88IMO6.

Suppose $(m,n)$ is a solution to $\tfrac{m^2+n^2}{mn-1}=k$ for some positive integer $k.$ We have \[m^2+n^2=k(mn-1) \implies m^2-m(kn)+(n^2+k).\] It follows that $(m',n)$ is also a solution, where $m'=\tfrac{n^2+k}{m} = kn-m.$ Suppose that $m+n$ is minimal across all solutions $(m,n).$ Then \[m'=\frac{n^2+k}{m} \geq m \Longleftrightarrow n^2+k \geq m^2 \Longleftrightarrow m^2-n^2 \leq \frac{m^2+n^2}{mn-1}.\] A bit of algebra reduces this to \[(m^2-n^2)(mn-1) \leq m^2+n^2 \implies n(m^2-n^2)=n(m-n)(m+n) \leq 2m.\] Observe \[mn(m-n) < n(m-n)(m+n) \leq 2m,\] So $n(m-n) \leq 1.$ This forces $(m,n)=(2,1),$ which gives $k=5.$ From here, classifying all solutions is not hard. We can generate solutions with our transformation $(kn-m, n) \to (m,n),$ and inverting the pair $(m,n) \to (n,m).$ It follows we can obtain all solutions with:

1. $(a_k, a_{k+1}),$ where $a_1=1,$  $a_2=2,$ $a_n=5a_{n-1}-a_{n-2}.$
2. $(b_k, b_{k+1}),$ where $b_1=1,$ $b_2=3,$ $b_n=5b_{n-1}-b_{n-2}.$

The first solution is attained by alternating between a transformation and inversion from the base pair $(1,2),$ and the second solution is attained by applying a transformation and then an inversion from the base pair $(2,1),$ and then continuing by alternating between transformations and inversions. Observe this classifies all solutions, because a sequence of inversions and transformations will take any solution to our original base pair $(2,1),$ as desired.


Let $n$ be a positive integer such that $2+2\sqrt{28n^2+1}$ is an integer. Show that $2+2\sqrt{28n^2+1}$ is the square of an integer.

Masoud Zargar

Appeared recently in http://www.mathlinks.ro/Forum/viewtopic.php?t=86450.
Other ones:
http://www.mathlinks.ro/Forum/viewtopic.php?t=23315
http://www.mathlinks.ro/Forum/viewtopic.php?t=43334


hello

can you help me please in this question

thanku

[img]http://img105.imageshack.us/img105/7439/5aayv9.jpg[/img]

[hide="Quickly"] $e \to 2$
$b \to 5$
$d \to 3$
$a \to 1$
$c \to 4$ [/hide]

hello
thanx

i am trying to understand your solution i am sure its easy but some how i can't get it...

why you did not assume ?
b-->3
d-->5

can we do that please explain it in details i do appreciate your time thanx much

There's more than one isomorphism. There are, I think, four.

woah that reminds me: [url=http://www.addictinggames.com/planarity.html]Planarity[/url]
but sorry for spam  :D


Prove that

\[ \displaystyle{\sum_{i_1\equal{}0}^n \sum_{i_2\equal{}0}^{i_1} ... \sum_{i_m\equal{}0}^{i_{m\minus{}1}} 1 } \equal{} \displaystyle{\frac{(m\plus{}n)!}{m! n!}}\]

thx

[hide="Combinatorial Solution"]The summation counts $ 1$ for each $ m$-tuple of integers $ (i_1,i_2,\ldots,i_{m\minus{}1},i_m)$ such that:
$ n \ge i_1 \ge i_2 \ge \ldots \ge i_{m\minus{}1} \ge i_m \ge 0$

So, the value of the summation is equal to the number of such $ m$-tuples.

Let $ x_j \equal{} i_j \plus{} (m\minus{}j)$.
Clearly, there is a one-to-one correspondence between $ m$-tuples $ (x_1,x_2,\ldots,x_{m\minus{}1},x_m)$ and $ m$-tuples $ (i_1,i_2,\ldots,i_{m\minus{}1},i_m)$.

Note that:
$ n \ge i_1 \ge i_2 \ge \ldots \ge i_{m\minus{}1} \ge i_m \ge 0$
$ \Updownarrow$
$ n\plus{}m\minus{}1 \ge i_1\plus{}(m\minus{}1) > i_2\plus{}(m\minus{}2) > \ldots > i_{m\minus{}1}\plus{}(1) > i_m\plus{}(0) \ge 0$
$ \Updownarrow$
$ n\plus{}m\minus{}1 \ge x_1 > x_2 > \ldots > x_{m\minus{}1} > x_m \ge 0$

So, the number of $ m$-tuples of integers $ (i_1,i_2,\ldots,i_{m\minus{}1},i_m)$ satisfying the conditions is equal to the number of $ m$-tuples of integers $ (x_1,x_2,\ldots,x_{m\minus{}1},x_m)$ satisfying:
$ n\plus{}m\minus{}1 \ge x_1 > x_2 > \ldots > x_{m\minus{}1} > x_m \ge 0$

This is equal to the number of ways to choose $ m$ distinct integers from a set of $ n\plus{}m$ integers which is $ \dbinom{n\plus{}m}{m} \equal{} \dfrac{(m\plus{}n)!}{m!n!}$ as desired.[/hide]


I'm codin a programm that resolves linear problems by the simplex method well i divided the programm into elementary functions and made sure that they all give the right results but when i tried to gather them in an iteration tentative in the main i had a segmentation fault ,i've changed the fgets into a scanf but it still doesn't work moreover i've heard about a conflict problem while usin scanf and fgets in the same code and that we av to use fflush but franckly i don't know exactly what this is conflict about here's my main function ,with the definition of ab ,co and aff :
int aff(int *a,int x):it prints a table on the screen
int co(char* tab,int n,int **a,int z):for example for tab="2x3+7x4",n=5,and z=0 it defines a[0]={0,0,2,7,0}
and here's the main function:
int main()
{
int **a,i,j,k,l;
char**b;
a=(int**)malloc(sizeof(int*)*4);
for (i=0;i<4;i++)
   {
    a[i]=(int*)malloc(sizeof(int)*5);
   }
b=(char**)malloc(sizeof(char*)*4);
for (i=0;i<4;i++)
   {
    b[i]=(char*)malloc(sizeof(char)*1000);
   }
for (i=0;i<4;i++)
{
scanf(" %100[1234567890x><=]", b[i]);
l=co(*(b+i),5,a,i);
k=aff(a[i],6);
}
free(b);
b=NULL;
return 0;
}

might someone detect where the error is plz ??
Thx in advance

Hm, I don't know, it works fine on my computer..... Well, there's one thing - you should add '+' to the line with

scanf(" %100[1234567890x><=]", b[i]);

so that it reads

scanf(" %100[1234567890x><=+]", b[i]);


Here are the top 7 at comprehensive, plus a few more tentative...

1. Misha Lavrov                  Enloe
2. John Berman                   Hoggard
3. Arnav Tripathy                East Chapel Hill
4. Jeremy Hahn                   East Chapel Hill
5. John Pardon                    Durham Academy
6. Daniel Vitek                     Enloe
7. Vivek Bhattacharya          Enloe
8. Logan Bailey???
9. Joe Lozier???                   NCSSM
10. Jenny Chen???               Arendell Parrott Academy
11. Ray Wang???                 NCSSM
12. ???
13. ???
14. Yakov Berchenko-Kogan   Broughton
15. Brendan Fletcher            Charlotte Homeschool

I think that Drew Boyuka may have made it in 12th or 13th, and I'm not sure about the placement of Ray and Jenny.  Maybe Yasha got 12th, I can't remember.  Does anybody know who Logan Bailey is?

EDIT: I posted the first vote for the 6th place option.  The 'liar' description applies to anybody else who votes this.

okay, here's the official top 20 list with rank, name, score, school, teacher thanks to Damien (Tian-yi) Jiang:

1 Mikhail Lavrov 118 Enloe High School James Cryer
2 John Berman 113 J. T. Hoggard High School Colleen St. Ledger
3 Arnav Tripathy 111 East Chapel Hill High School Michael Tylor
4 Jeremy Hahn 107 East Chapel Hill High School Shannon McGinnis
5 John Pardon 106 Durham Academy Upper School Andrew Ferrari
6 Daniel Vitek 100 Enloe High School James Cryer
7 Vivek Bhattacharya 98 Enloe High School John Noland
8 Logan Bailey 90 TC Roberson High School Beverly Guthrie
9 Joseph Lozier 89 NCSSM Philip Rash
10 Jenny Chen 83 Arendell Parrot Academy Mary Leighton
11 Bangchen Wang 82 AC Reynolds High School Kristen Kutob
12 Ray Wang 82 NCSSM Peggy Craft
13 Drew Boyuka 79 Charlotte Home School Harold Reiter
14 Yakov Berchenko-Kogan 78 Broughton High School Kevin Ledger
15 Brendan Fletcher 78 Charlotte Home Educators Susan Schaeffer
16 Yutong Chen 77 NCSSM Philip Rash, Dan Tea
17 Hyun Ji Bae 75 East Chapel Hill High School John Wilson III
18 Pom McCabe 75 Pinecrest High School Mike Apple
19 Steven Ji 75 NCSSM Dan Teague
20 Kevin Lang 74 Myers Park High School Adele Fielding


Let $ ABC $ be a triangle . The incircle of the triangle ABC tangents BC at D . G is the centroid of the triangle ABC . Prove that : DG passes through the midpoint of the arc BC ( contains A )

This is certainly not true. If $M_a,M_b,M_c$ are the midpoints of teh large arcs $BC,CA,AB$ (I mean the arc $BC$ which contains $A$, and so on), and $D,E,F$ are the points where the incircle touches $BC,CA,AB$ respectively, then $DM_a,EM_b,FM_c$ are concurrent in the internal similarity center of the incircle and circumcircle, which is different from the centroid, in general.


Let A be a set of number in [0,1] whose tenary expansion have only finitely many 1's. Prove that $\mu(A) = 0$. ($\mu$ means Lebesgue measure)

Thanks!

It's just countably many copies of something which looks like the Cantor set.


In case someone doesn't know this yet, TOH is this one legend, but the problem goes like this:
There are 5 rings stacked up, biggest on the bottom and smallest on top. They are stacked on a pole and there are two other poles. You have to transfer the 5 rings to the third pole, but you can only place a ring on an empty pole or a larger ring.

That is fairly easy to do. Here's a few interesting questions:

Is there a general way to solve this, if there were not 5 but n rings?

What is the formula for the minimum amount of moves?

What is the minimum amount of moves needed to move 52 rings(say you used a deck of cards instead of rings)? Do this efficiently.

If you make one move per second how long would it take to move that whole deck of cards? What about if you got a computer to do it for you?

Have fun! :)

I don't know if   the problem here for a chalange ( otheres to solve) , or you really want to know the answer.

In either case the answer is fairly well known (and easy). All you have  to do for n is to move top  (n-1) rings  to  to B; move bottom ring  to A and then agaim move (n-1) rings to A. Total number of moves  thus would be [b](2^n -1)[/b] .
(If you know how to do this for  n-1 rings, you can do n rings, and you do know how to move 1 ring)

Legend has it that (Well, the 'movers of the rings' in  this  legend  are some   priests in the temple  in Varanasi)  this move (for 64   rings, in this case) started  at the same  time as  the 'big bang'  and the universe will end (at least this cycle) when all these moves are  finished.  For reference universe, at present, is about 2^58 seconds old.

Did this get moved or something? 'Cause I don't think I posted this in this forum...

Yes, Gyan, this is for others to solve because I wanted to see what others think. The legend is fasinating, isn't it? It seems like they would be done moving the rings, but actually they have $2^{12}$ as much time left as they already had. That means the world won't end for a very long time. :D  :D

On a tiny-off-topic note, if anyone actually wanted to solve this problem by hand, theres a much easier method to remember other than the recursive way, in which you may get confused part way through.
It is: move the smallest ring left. Move another ring. Repeat.

Change left to right and the pile ends up on the other empty ring.

[quote]move the smallest ring left. Move another ring. Repeat.
[/quote]
Very nice. (After all half (actually more than half) the moves  move the tiny ring - every other move - so that should work very well)

The funny thing is that as I was literally doing it for real, and right before I got bored, I noticed the same thing too. I was just holding on to the smallest one because I knew I'd have to move it again. Then I gave up! :D


i have auto login.  i logged in, and on the side, it still says log in.  I can still change my profile.  However, it doesnt say "my classes" on the side, so it doesnt allow me to view transcripts or challenges.

great.  I opened artofproblemsolving on another window and clikced login, to see if that might help.  Now it says "please enter your username and password to log in"  and on the other window, i can still go to my onlline class forums/.


Ο υπερσιβηρικός σιδηρόδρομος, ως γνωστόν, ξεκινάει από τη Μόσχα και καταλήγει στο Βλαδιβοστόκ. Η απόσταση που διανύει είναι πολύ μεγάλη. Μέχρι να φτάσει, κάνει ένα μεγάλο αριθμό από στάσεις.
Σε ένα ταξίδι του υπερσιβηρικού, στο τραίνο είχε επιβιβαστεί ένα πλήθος ποινικών κρατουμένων οι οποίοι θα μετέβαιναν από τη Μόσχα, σε μια φυλακή που βρίσκεται κοντά στο Βλαδιβοστόκ.
Οι κρατούμενοι επιβιβάστηκαν όλοι στο ίδιο βαγόνι συνοδευόμενοι από ένοπλους φρουρούς.
Πριν ξεκινήσει το τραίνο, ο επικεφαλής της φρουράς ζήτησε να γίνει ησυχία και απευθύνθηκε προς τους κρατούμενους με τούτα τα λόγια:

«Στην πλάτη καθενός από εσάς υπάρχει μια αυτοκόλλητη πινακίδα. Το χρώμα κάθε πινακίδας μπορεί να είναι είτε κόκκινο είτε μαύρο. Αν, χρησιμοποιώντας μόνο το μυαλό σας, καταφέρετε σε κάποια από τις στάσεις του τραίνου να σηκωθείτε και να κατεβείτε όλοι όσοι έχετε κόκκινες πινακίδες, τότε θα αποκτήσετε την ελευθερία σας. Ο καθένας από εσάς, χρησιμοποιώντας τα μάτια του μπορεί να ξέρει τα χρώματα των πινακίδων όλων των υπολοίπων. Δεν μπορεί όμως να ξέρει το χρώμα της δικής του πινακίδας. Οποιαδήποτε απόπειρα να συνεννοηθείτε μεταξύ σας ανταλλάσσοντας πληροφορίες, τιμωρείται με θάνατο».

Οι κρατούμενοι έμειναν σιωπηλοί. Το τραίνο ξεκίνησε. Σε κάποια από τις στάσεις, σηκώθηκαν όλοι όσοι είχαν κόκκινες πινακίδες και κατέβηκαν. Η ζωή ήταν δική τους χάρη στην ικανότητά τους να χρησιμοποιούν το μυαλό τους.

Πώς τα κατάφεραν;

To πρόβλημα αυτό είναι αρκετά έξυπνο...
Εγώ το ξέρω και με μια άλλη διατύπωσει:
<<Μέσα σένα σκοτεινό δωμάτιο βρίσκονται 100 κρατούμενοι.Καθένας φοράει από ένα σκουφάκι μαύρο ή άσπρο.Οι δεσμοφύλακες επιτρέπουν να βγαίνει κάθε φορά και από ένας μόνο κρατούμενος.Πως γίνεται οι κρατούμενοι να φτιάξουν δύο ομάδες που οι καθεμία να έχει αυτούς που το σκουφάκι τους έχει το ίδιο χρώμα χωρίς να συνεννοηθούν με κανέναν τρόπο;;;;;>>

Πάντως εγώ το έλυσα όταν ήμουν για καφέ και με βοήθησαν τρία ποτήρια.....
Μπορει να βοηθήσει κι αυτή η παρατήρηση...... :lol:

prokeitai gia poly omrfo provlima...ki ego ksero mia alli ekdoxi kata tin opoia oi kratoumenoi kathontai se mia skala kai prepei na mantepsoun o kathenas ti kapelo foraei o idios-allios....
ekei vevaia epeidi kathontan se skalopatia o kathenas evlepe mono ta kapela ton brostinon tou kai oxi olon ton ypoloipon,opos edo.alla ki ekei douleuei to kolpo mia xara mono pou autos pou kathetai sto kefaloskalo exei 50 tois ekato pithanotites na  petyxei to xroma.

[hide]me artious-perittous den to kaneis??[/hide]

Τα παραπάνω είναι τρία διαφορετικά προβλήματα
Τα ξέρω και τα 3 (σε άλλες παραλλαγές) γι αυτό δε θα δώσω απάντηση.
[color=olive]Το πρώτο είναι [u]καθαρά μαθηματικό[/u], και (συνεπώς) είναι από τους καλύτερους γρίφους που έχω ακούσει[/color]

[color=brown][b]1ο (fotis)[/b]
Η ομοιότητα με το 2ο γρίφο είναι ότι θα χωριστούν κι εδώ σε δύο ομάδες, αλλά καθοριστικό ρόλο παίζει το γεγονός ότι έχουμε διαδοχικές στάσεις. Δηλαδή μπορεί να κατέβει κάποιος από το τραίνο αλλά σε διακριτό χρόνο.
Πιστεύω πάντως ότι μπορεί να λυθεί και με τον τρόπο που λύνεται το (2), αλλά τότε χάνει την ομορφιά του.
Η παραλλαγή που ξέρω εγώ είναι με μοναχούς που έχουν βγάλει ένα "διαβολικό σημάδι" στο μέτωπο.[/color]

[color=blue][b]2ο (bouzoukman)[/b]
Με μπερδεύει λίγο το[/color] [quote]Οι δεσμοφύλακες επιτρέπουν να βγαίνει κάθε φορά και από ένας μόνο κρατούμενος[/quote]
[color=blue]Βγαίνει και ξαναμπαίνει;

Αν εννοείς ότι βγαίνει στο προαύλιο και μένει εκεί τότε είναι ακριβώς αυτό που ξέρω κι εγώ.

Σ' αυτό συννενοούνται από την προηγούμενη μέρα οι κρατούμενοι για τον αλγόριθμο που θα ακολουθήσουν

[hide="@bouzoukman"][quote]το έλυσα όταν ήμουν για καφέ και με βοήθησαν τρία ποτήρια..... [/quote]Τρία ποτήρια  καφέ???   :o :D [/hide]
[/color]

[color=green][b]3ο (Nick Rapanos)[/b]
Ένα παρόμοιο ξέρω με τρία άτομα στη σειρά (ο ένας πίσω από τον άλλο) στους οποίους έχουν φορέσει καπέλα από ένα πλήθος 2 μαύρων και 3 άσπρων. Ο  πρώτος λέει "δεν ξέρω", ο δεύτερος "δεν ξέρω" και ο τελευταίος στη σειρά βρίσκει το χρώμα του.[/color]
[color=red]EDIT: Α ναι, σωστά: "Ρωτάν τον καθένα ξεχωριστά,αρχίζοντας από πίσω"[/color]

[color=green]Αυτό με τους περιττούς που γράφεις μου θυμίζει ένα πολύ καλό που έγραψαν στο αγγλικό φόρουμ και το έλυσαν με ισοδυναμίες - πολύ καλό. 7 άτομα και 7 διαφορετικά χρώματα (γνωστά στους κρατουμενους) Αν θέλετε δίνω το link.

Και στα δυο αυτά αρκεί τουλάχιστον ένας να βρει το χρώμα που φοράει[/color]

Πράγματι τα τρία προβλήματα είναι διαφορετικά μεταξύ τους, όπως επίσης πολύ διαφορετικές είναι και οι λύσεις τους.

Να κάνω μόνο κάποιες διευκρινήσεις:
Στο πρώτο πρόβλημα δεν απαιτείται καμιά συννενόηση μεταξύ των κρατουμένων, ούτε πριν μπουν στο τρένο, ούτε κατά τη διάρκεια της διαδρομής.

Στο δεύτερο πρόβλημα χρειάζεται μια συννενόηση μεταξύ των κρατουμένων πριν τους βάλλουν τα σκουφιά ώστε να σχεδιάσουν ποια στρατηγική θα ακολουθήσουν. Συνεπώς το πρώτο πρόβλημα δε μπορεί να λυθεί όμοια με το δεύτερο.

Το τρίτο πρόβλημα γενικεύεται με $n$ άτομα και $k$ χρώματα καπέλων. Τα άτομα μπαίνουν ο ένας πίσω από τον άλλο και ο καθένας βλέπει μόνο τα καπέλα των μπροστινών του. Ρωτάν τον καθένα ξεχωριστά,αρχίζοντας από πίσω, τί καπέλο φοράει. Ο τελευταίος έχει $\frac{1}{k}$ πιθανότητες να το βρει και οι άλλοι μπορούν να το βρουν σίγουρα. Και αυτό το πρόβλημα απαιτεί συννενόηση των ατόμων πριν τη διαδικασία αυτή για να αποφασίσουν τι πρέπει να κάνουν.

Επειδή έτυχε κι εγώ να γνωρίζω και τους τρεις γρίφους από πριν, δε βάζω λύση. Προτρέπω όμως όποιον δε τους ξέρει να προσπαθήσει να τους λύσει, γιατί είναι αρκετά ενδιαφέροντες και έχουν έξυπνες λύσεις. :wink:

Μπορεί να μην το διατύπωσα σωστά το πρόβλημα γιατί θυμόμουν περισσότερο την λύση παρά την εκφώνηση..
Σωστά pontios με τις προσθέσεις που έκανες είναι αυτός που μου είπανε..
Κάθε φορά βγαίνει και από ένας κρατούμενος στο προαύλιο..


Show that $[\sqrt[n]{x}]=[\sqrt[n]{[x]}]$

 $\forall n \in \mathbb{N}$, $\forall x \geq 0$

 :)

The function $x\mapsto[\sqrt[n]{x}]$ is increasing, therefore
\[ [\sqrt[n]{x}]\geq[\sqrt[n]{[x]}]. \]

On the other hand $[\sqrt[n]{x}]^n$ is an integer, and $[\sqrt[n]{x}]^n \leq x$. So $[\sqrt[n]{x}]^n \leq [x]$. Hence
\[ [\sqrt[n]{x}] \leq \sqrt[n]{[x]}. \]
Finally,
\[ [\sqrt[n]{x}] \leq [\sqrt[n]{[x]}]. \]

i don't think that $x$ must be an integer...

here is my solution:
as Yury said, we have
$[\sqrt[n]{x}]\geq[\sqrt[n]{[x]}]$

if i show that it cannot be
$\sqrt[n]{x}\geq a>\sqrt[n]{[x]}$
where a is an integer, i end my proof.

$\sqrt[n]{x}\geq a>\sqrt[n]{[x]}$
$(\sqrt[n]{x})^n\geq a^n>(\sqrt[n]{[x]})^n$
$x\geq a^n>[x]$
$x\geq a^n\geq[x]+1$
because $a^n$ and $[x]$ are integer, and so
$x\geq[x]+1$
that's impossible.

bye

[quote="frengo"]i don't think that $x$ must be an integer...[/quote]
Not sure whether you refer to my solution :? I don't assume that $x$ is integer  ;)

Yes, Yury's solution does not assume $x$ is an integer.
(Good solution, too.  Mine was exactly like frengo's, but I came too late.)

The problem is meaningless if x must be an integer


Let the positive real number $a_{1},a_{2},....a_{n}$ and $k$ a positive integer, such that $\sum_{j=1}^{n}a_{j}=1$ prove that:
$\prod_{i=1}^{n}(\frac{1-a_{i}^{k}}{a_{i}^{k}})\geq (n^{k}-1)^{n}$

I think you have typos, it is not even close to true

(just an example: $k=1, n=2, a_{1}=2,a_{2}=.5$) actually, in fact ($k=n=1, a_{1}<1$ is also a counter example)
maybe you misplaced n with k?

See here for a solution
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mop&t=84502


abstract algebra
superior algebra
superior algebra unsolved

A module P is projective if for every surjective module homomorphism $ f : N \twoheadrightarrow M$ and every module homomorphism $ g : P \rightarrow M$, there exists a homomorphism $ h : P \rightarrow N$ such that $ f \circ h \equal{} g$.

Prove that if $ P$ is projective and $ h: P \rightarrow M$ is a surjective $ R\minus{}$module homomorphism, then there exists some $ R\minus{}$module homomorphism $ s: M \rightarrow P$ such that $ h \circ s\equal{}1_M$.

This is a section map. I don't know how to prove this. I was thinking to let $ N\equal{}M$, but it doesn't seem that easy.

I believe that in what you're trying to prove, you mixed up $ M$ and $ P$  :wink: . Specifically, you should be given a surjective homomorphism $ h: M \twoheadrightarrow P$, and you're trying to find $ s: P \to M$.


integration
logarithms
calculus
calculus computations

Let $(u_n)$ be the sequence defined by : $ \forall n \in \mathbb N : u_n=\int_{1}^{e}\left( \ln t\right) ^{n}dt$
Find the explicit term of $u_n$ in terms of $n$.

I'd rather do a change of variables and write it as

$u_n=\int_0^1x^ne^x\,dx.$

Integration by parts leads to the recursion

$u_n=e-nu_{n-1}.$

Following this along then leads to the intermediate formula:

$u_n=(-1)^n(a_ne-n!)$ where the integer $a_n=na_{n-1}+(-1)^n.$

A more refined form of the answer looks like this:

$u_n=(-1)^nn!\left(\left(1-1+\frac12-\frac1{3!}+\cdots+\frac{(-1)^n}{n!}\right)e-1\right).$

Asymptotically, $u_n$ behaves like $\frac{e}{n+1},$ which is exactly what you would anticipate from the integral.


Just add  1 to all fractions and then apply AM-GM.

Namdung


geometry
circumcircle
geometric transformation
reflection
homothety
ratio
perpendicular bisector

Let  $ AD  $  the altitude of the triangle  $  ABC  $ and   $  G  $ the point of intersection of its medians. The ray  $  DG   $ meets the circumcircle of triangle  $  ABC  $ at  point   $ E $ (  G is between D and E  )  . Prove that    $  GE = 2DG $.

What a sweet little problem! :)

Let $M$ be the midpoint of $BC$, $D'$-the reflection of $D$ in $M$, and $G_a,G_d$-the projections of $G$ on $BC,AD$ respectively. We have $DG_a=\frac 13\cdot DD'$ and $DG_d=\frac 13\cdot DA$. This means that the homothecy centered at $D$ of ratio $3$ takes $G$ onto a point $E'$ which satisfies $DE'=3DG$, and which is, at the same time, the intersection between the perpendiculars in $A,D'$ to $AD,BC$ respectively. This intersection lies on the circumcircle $(ABC)$ (because it is the reflection of $A$ in the perpendicular bisector of $BC$), so $E'=E$, and we get $DE=3DG$, as desired.

I like this easy and nice problem !

I note $w=C(O,R),\ \{A,F\} =AD\cap w,\ E'\in w,\ O\in (FE')$. From the relations
$DH=DF,\ OF=OE',\ HG=2\cdot GO$, results that the point $G\in DE'\ (E'\equiv E)$,
the point $G$ is the centroid of the triangle $HFE'$ and $GE=2\cdot DG.$

[u]Remark[/u]$.\ DO\parallel HE$.

First let's slightly rewrite the problem:

[color=blue]Given a triangle ABC, let D be the foot of its altitude from the vertex A. Also, let G be the centroid of triangle ABC, and let E be the point on the line DG such that $GE=2\cdot DG$, where we use directed segments. Prove that this point E lies on the circumcircle of triangle ABC.[/color]

Now, the [i]solution[/i] becomes very easy: If X, Y, Z are the midpoints of the sides BC, CA, AB of triangle ABC, then the segments AX, BY and CZ are the medians of triangle ABC and intersect at its centroid G. Since the centroid of a triangle divides every median in the ratio 2 : 1, we have $\frac{AG}{GX}=\frac{BG}{GY}=\frac{CG}{GZ}=2$. On the other hand, $\frac{EG}{GD}=\frac{GE}{DG}=\frac{2\cdot DG}{DG}=2$. Thus, $\frac{AG}{GX}=\frac{BG}{GY}=\frac{CG}{GZ}=\frac{EG}{GD}=2$, so that $\frac{GA}{GX}=\frac{GB}{GY}=\frac{GC}{GZ}=\frac{GE}{GD}=-2$. Thus, the points A, B, C, E are the images of the points X, Y, Z, D under the homothety with center G and factor -2. Now, the points X, Y, Z, D lie on one circle - namely, on the nine-point circle of triangle ABC. Thus, their images A, B, C, E also lie on one circle, i. e. the point E lies on the circumcircle of triangle ABC.

A similar argument yields OD || HE.

  darij

More or less the same:
Reflect the points A and D about the perp. bisector of BC to obtain A' and D'. Let M be the midpoint of BC. $AM\cap A'D=G'$. Then $\frac{AG}{AH}=\frac{AA'}{DM}=2\Rightarrow G=G'$. The conclusion follows since B,C,A,A' are concyclic.

When I posted the problem , I could not imagine that it is so interesting.As I have said , Mr Kontogiannis , the Autor of the book (in greek) ,  was (and is) IMO team leader(with Bolis or Fellouris) and trainer for the greek team. He has the best bibliografy in Greece and the best books from all countries , especially from <<  East Europe  and the countries from ... Soviet Union >>.I' ll try to translate some more of the unsolved problems from this book , for all friends in this forum .

Thanks to all of you , solvers or visitors  , on this topic.

  Babis


What are your favorite games on any system or device?

My favorite game is called 'Settlers of Catan' and it's played on a nifty little device called a Table.

Hehe, Settlers of Catan is pretty fun, although if you play with a random board it sometimes comes down to a diceroll to see who gets the best starting position. Also, it's usually more fun with more people.

Anyway, some awesome games I have played:

Chrono Trigger (DS)
Paper Mario: Thousand Year Door (GC)
Twilight Princess (Wii)
Okami (Wii) - Not done yet
SSBM (GC)
Super Mario Galaxy (Wii)

Although SSBM gets boring really quickly if you have nobody about your level to play with. Most of the awesome games I have played are RPGs. I decided to throw in a platformer in there for the heck of it. That, and Galaxy was pretty fun.

I'm probably forgetting something. Ah well.

WoW. GTA4, (PC), Texas Hold'em, Touchgrind, Klaus :D (iPhone).

My favorite game is called kemps and it's played with nifty little devices called cards. Cards are really a revolutionary gaming system, with a countless amount of different possible games and the ability to make your own games without writing a piece of code or software. It is probably the only one with major tournaments held throughout the world, including a world series for one of the more major games, and different types of games can be played from 1 to over 10 people. However, for some reason, many people think that pressing random buttons on a piece of plastic and watching random images on a screen move around is more fun than using these cards. Take a look at the price range: less than one dollar to buy your own cards, while it costs more than 100 dollars for most of these so called "video games". Personally, I think a game where you talk with your friends in real life is better than a game where you sit staring at a screen with your mouth open, saying things like "HAHA U GOT PWND" every so often.

(note: the above is written as a complete joke and is not meant to offend anyone who plays video games, including me)

Although your pricings are somewhat skewed:

about 25-50 dollars for a decent priced video game
vs.
3 dollars for a passable deck of cards.

Cards deteriorate over time.

I like mario party 8 for wii. although its not fun when you're playing alone :(

FTW - not fun when no one is playing, [size=9]or when you are muted for breaking a rule that a certain mod *cough* breaks almost everyday.[/size]

[quote="aleph0"][quote="jjx1"]no... POKEMON games are fun even though I don't like pokemon.[/quote]
 :rotfl: Isn't that a contradiction?[/quote]
Actually, it isn't. You assumed that jjx1 likes fun. Or, he could mean that he doesn't want to like pokemon but still thinks it is fun.

[quote="coolts"]Anyone here a fan of hardcore shooters like Gears 2 or COD?[/quote]
I do, but I like Halo and Unreal more then Gears of war 2. But i do like Call of duty.

There's another nifty game that I enjoy; it is called 'Life'. It is played in a nifty little thing called a 'Universe'.

How about "Conway's Game of Life"... there are some really cool patterns out there!

Most games though tend to get boring after a while. i think pokemon lasts a long time of fun. But once you win its pretty boring.

lol runescape was pretty addicting but i quit like 5 years ago. I think my account is still there if I didn't get hacked.  Back then people loved that game but now people laugh at it.

yeah. it used to be number one on miniclips. everyone was shocked when it dropped to 2nd with club penguin taking over. after that, the game went downhill.

1. Halo (XBOX)

2. Call of Duty 4 (XBOX)

3. Egyption Ratscrew (card game)

[quote="moldovan"][size=150][b]MONOPOLY [/b][/size][/quote]

Go Monopoly! It rocks!

Yeah, Conway's Game of Life is pretty cool. You could spend hours playing with one of those applets out there.


calculus
integration
trigonometry
logarithms
calculus computations

For the fourth one, it should perhaps be easier to make the substitution $\tan x = t$. Then,

$dx = \frac{dt}{1+t^{2}}, \cos x = \frac{1}{\sqrt{1+t^{2}}}, \sin x = \frac{t}{\sqrt{1+t^{2}}}$

and so

$\int{\frac{1}{(3\sin x-4\cos x)^{2}}}dx = \int{\frac{1}{\left(\frac{3t}{\sqrt{1+t^{2}}}-\frac{4}{\sqrt{1+t^{2}}}\right)^{2}}\cdot\frac{1}{1+t^{2}}}dx$

$= \frac{1}{3}\int{3\cdot(3t-4)^{-2}}dx =-\frac{1}{3(3t-4)}$

$=-\frac{1}{3(3\tan x-4)}+C$

Carcul

\[\begin{array}{l}\int{\frac{{dx}}{{1+\sin x+\cos x}}}\\ \\ = \int{\frac{{dx}}{{1+\sin x+\cos x}}}.\frac{{1-\left({\sin x+\cos x}\right)}}{{1-\left({\sin x+\cos x}\right)}}dx \\ \\ = \int{\frac{{1-\sin x-\cos x}}{{1-\left({\sin x+\cos x}\right)^{2}}}dx = \int{\frac{{1-\sin x-\cos x}}{{1-\left({\sin x^{2}+\cos x^{2}+2\sin x\cos x}\right)}}}}dx \\ \\ = \int{\frac{{1-\sin x-\cos x}}{{1-\left({1+2\sin x\cos x}\right)}}}dx = \int{\frac{{1-\sin x-\cos x}}{{1-1-2\sin x\cos x}}}dx \\ \\ = \int{\frac{{1-\sin x-\cos x}}{{-2\sin x\cos x}}}dx \\ \\ =-\int{\frac{1}{{2\sin x\cos x}}}dx+\int{\frac{{\sin x}}{{2\sin x\cos x}}}dx+\int{\frac{{\cos x}}{{2\sin x\cos x}}}dx \\ \\ =-\int{\frac{{dx}}{{\sin 2x}}}+\frac{1}{2}\int{\sec x.dx+\frac{1}{2}\int{\csc x.dx}}\\ \\ =-\int{\csc 2x.dx}+\frac{1}{2}\int{secx.dx}+\frac{1}{2}\int{\csc x.dx}\\ \\ = \frac{1}{2}\ln \left|{\csc 2x+\cot 2x}\right|+\frac{1}{2}\ln \left|{\sec x+\tan x}\right|-\frac{1}{2}\ln \left|{\csc x+\cot x}\right|+c \\ \end{array}\]

is that correct?

That is correct, but you could simplify considerably your answer.


farz konid $s=\{n_{1},n_{2},....\}$ majmoe tamam adadhaye tabeei bashad ke dar namayesh ashary anha sefr vojod nadarad(for example $7 \in s$,$101 \notin s$) sabet konid$\sum \frac{1}{n_{i}}$ hamgera ast.

There are $9^{d}$ memebers of $s$ with $d$ digits. Replacing all of them with $\frac 1{10^{d-1}}$, $\sum \frac 1{n_{i}}$ does not decrease. Since $\sum \frac{9^{d}}{10^{d-1}}$ is convergent, we are done!


[quote="Magnara"]Is there any mathematical shorthand version of writing "such that"?  I really feel that there should be, but I can't find one.[/quote]

I don't know of any symbol (though there very well may be one), but I think people often just write "s.t."

There is a such that shorthand that I really hate. It looks like a cross between [tex]\exists[/tex] and [tex]\ni[/tex], so if you write it by hand, it's really hard to tell what it is.

I think it is widely accepted that "s.t." = such that

[quote="tokenadult"]In set builder notation "such that" is indicated by a vertical bar character

|

but that would look ghastly in a sentence of connected prose.

I would write out that kind of phrase in full, living as I do in a world in which most educated people can't reliably distinguish between "e.g." ("for example") and "i.e." ("that is"). Sometimes a few more seconds of writing results in a great gain in clarity.[/quote]

It's just for my own uses.


got a one-liner....it's Minkowski's Inequality!  :D

http://mathworld.wolfram.com/MinkowskisInequalities.html

I borrowed this proof from a Polish book : "Macierze i operatory" (Polish people : don't buy it :)) :

Let [tex]p > 1[/tex] and [tex]q[/tex] s.t [tex]\frac{1}{p} + \frac{1}{q} = 1[/tex] or equivalently we have [tex]( p - 1 ) q = p[/tex]. We will make use of the Holder Inequality:
[tex]\sum_{i = 1}^n |x_i y_i | \leqslant \left( \sum_{i = 1}^n |x_i |^p \right)^{\frac{1}{p}} \left( \sum_{i = 1}^n |y_i |^q \right)^{\frac{1}{q}}[/tex]
 holding for any [tex]p > 1[/tex] and [tex]q[/tex] s.t [tex]\frac{1}{p} + \frac{1}{q} = 1[/tex].
[list]
[tex]\sum_{i = 1}^n |x_i + y_i |^p = \sum_{i = 1}^n |x_i + y_i ||x_i + y_i |^{p -1}[/tex]
[tex]\leqslant \sum_{i = 1}^n x_i |x_i + y_i |^{p - 1} + \sum_{i = 1}^n y_i |x_i + y_i |^{p - 1}[/tex]
[tex]\leqslant \left( \sum_{i = 1}^n |x_i |^p  \right)^{\frac{1}{p}} \left( \sum_{i = 1}^n |x_i + y_i |^{( p - 1 ) q} \right)^{\frac{1}{q}} + \left( \sum_{i =
1}^n |y_i |^p  \right)^{\frac{1}{p}} \left( \sum_{i = 1}^n |x_i + y_i |^{( p - 1) q} \right)^{\frac{1}{q}}[/tex]
[tex]= \left[ \left( \sum_{i = 1}^n |x_i |^p \right)^{\frac{1}{p}} + \left( \sum_{i = 1}^n |y_i |^p \right)^{\frac{1}{p}}  \right] \left( \sum_{i = 1}^n |x_i + y_i |^p
\right)^{\frac{1}{q}}[/tex]
[/list]
Now divide both sides by [tex]( \sum_{i = 1}^n |x_i + y_i |^p)^{\frac{1}{q}}[/tex], to get
[tex]\left( \sum_{i = 1}^n |x_i + y_i |^p \right)^{1 - \frac{1}{q}} = \left( \sum_{i = 1}^n |x_i + y_i |^p \right)^{\frac{1}{p}} \leqslant \left( \sum_{i = 1}^n |x_i |^p \right)^{\frac{1}{p}} +\left ( \sum_{i = 1}^n |y_i |^p \right)^{\frac{1}{p}} [/tex]
which is the Minkowski inequality.


If the grocery store has 15 different kinds of fruit and 10 different kinds of canned soup, in how many ways can Michael choose 2 different kinds of fruit and 3 different kinds of soup to buy?

[quote="akhil0422"]$ \dbinom{15}{2}\dbinom{10}{3} \equal{} 105\cdot105 \equal{} 11025$.[/quote]

$ \dbinom{10}{3}$ actually equals $ 120$.  Therefore, we have $ 105\cdot120\equal{}\boxed{12600}$

How can a product of a number that is a multiple of 10 and any other integer not be a multiple of 10?

[quote=wuwang2002][quote=Harry10]$\binom{15}{2}=105$ Those are the ways to pick the fruit.
$\binom{10}{3}=120$ Those are the ways to pick the soup
$105\cdot 120=11025$[/quote]

You multiplied wrong[/quote]

I think they multiplied wrong.

$105*120$ is $ 12600 $.
I used a calculator.


floor function
limit
ratio
number theory proposed
number theory

Let a non-empty set A whose elements are integers. Know that if $ a\in A$, then $ 4a\in A$ and $ [\sqrt{a}]\in A$, too. Prove that $ A\equal{}\mathbb N$

This problem has been posted and solved, although the last time it was posted the deadline was not yet over and I believe megus moved the thread.  A sketch:  if $ A$ is non-empty, use the second condition until we can conclude that $ 1 \in A$, so every power of $ 4$ is in $ A$.  Then let $ S(k)$ denote the set of positive integers $ a$ such that $ \lfloor \sqrt{a} \rfloor \equal{} k$.  Then

$ S(k) \equal{} \{ a : k^2 \le a \le (k \plus{} 1)^2 \minus{} 1 \}$

so $ S^r(k) \equal{} \{ a : k^{2^r} \le a \le (k \plus{} 1)^{2^r} \minus{} 1 \}$.  Clearly $ \lim_{r \to \infty} \frac{ (k\plus{}1)^{2^r} \minus{} 1}{k^{2^r}} \equal{} \infty$, so in particular there exists $ r$ such that this ratio is greater than $ 4$ and $ S^r(k)$ contains a power of $ 4$.


Let ABC be acute angled triangle and P point in the plane of triangle.  Let x,y z be distances from the sides, and u,v,w distances from the vertices of triangle. Let a, b, c be sides of triangle. Prove or disprove the following inequality

4(x+y+z)^2+(a+b+c)^2 <= 4(u+v+w)^2.

It's here: http://www.mathlinks.ro/Forum/viewtopic.php?t=5428

  darij

Thank you very much on help. It is very nice solution


Three circles having each a radius of 7 cm are tangent to each other. Consider the triangle formed by joining their three centers. What is the number of square centimeters inside this triangle but outside the three circles. Give your answer to the nearest hunderdths and use $\pi=3.14$ .

too lazy to simplify

[hide="Answer"]
49sqrt(3) - 49pi/2
[/hide]

To Getting Started, as this is a challenging high-school problem...


calculus
derivative
function
advanced fields
advanced fields unsolved

This is really bad...I have an exam coming up and my professor is out-of-town for a few days and I need help with fundamentals (nevermind everything else).

I have absolutely no idea how to solve equations like:

au_x+bu_y+cu=0

or:

u_x+u_y=1

or:

u_x+2u_xy+u=1 given a condition, say, (x,1)=0

Basically, I have no idea how to find general solutions of basic partial derivatives if they're non-homogeneous or don't have a form like u_xy=0 or u_x+u_y=0.

I really, really need help. This is the first time I've struggled completely conceptually in a math course. I know this is beneath a lot of you, but I need help.

If you know how to solve something like $ u_x \plus{} u_y \equal{} 0$ then you (sort-of) know how to solve everything like $ u_x \plus{} u_y \equal{} f(x, y)$: any particular solution added to the general solution of the homogeneous equation gives you the general solution of the inhomogeneous equation.

There's a nice trick (the "method of characteristics") for converting PDEs like these into ODEs.

Let $ u$ be a function satisfying $ u_x \plus{} bu_y \plus{} cu \equal{} 0$.
Suppose $ g(t) \equal{} u(t,x(t))$ for some function $ x$.  Then $ g'(t) \equal{} u_x(t,x(t))\cdot 1 \plus{} u_y(t,x(t))x'(t)$.  If $ x$ happens to have the property that $ x'(t) \equal{} b$, then we have discovered that $ g'(t) \equal{} \minus{} cu(t,x(t)) \equal{} \minus{} cg(t)$.  This ODE, together with the initial condition $ g(0) \equal{} u(0,x(0))$, can be solved.  So what you should do is, pick an $ x$ that happens to have this nice property, then solve for $ g$, and then you will know the value of $ u$ at every point of the form $ (t,x(t))$.  By using different choices for $ x$, you can figure out the value of $ u$ anywhere you like.

Strauss is a good intro to PDEs book.

Thanks for the help...though I'm one of those people who likes to see examples. Can anyone show me how (or at least work me through) to solve the PDE:

u_x-yu_y+2u=1, subject to (x,1)=0?

I'll give you some help, and I'll solve it using characteristic method ( since I like it so much, and it isn't that hard to understand  :lol: ), also I will assume you know something about it ;

denote $ a\equal{}(1,\minus{}y),~b\equal{}1\minus{}2u$, and I believe the boundary value is $ u(x,1)\equal{}0$. I will change the notation to make it simpler $ x\rightarrow x_1,~y\rightarrow x_2,~u\rightarrow y$. Solve ODE;

$ \frac{dx_1}{d\tau}\equal{}1,~\frac{dx_2}{d\tau}\equal{}\minus{}x_2,~\frac{dy}{d\tau}\equal{}1\minus{}2y$, in the first ODE you denote boundary value $ x_1^0$ and in the second $ x_2^0$, $ y^0$ in the third( you will get rid of them later ). You should get $ x_1(\tau)\equal{}x_1^0\plus{}\tau,~x_2(\tau)\equal{}x_2^0 e^{\minus{}\tau}$, and $ y(\tau)\equal{}\frac{(2y^0\minus{}1)e^{\minus{}2\tau}\plus{}1}{2}$. You get (from boundary values) that $ x_2^0\equal{}1,~y_0\equal{}0$. For the first two equation you get $ \tau$ and $ x_1^0$ ( but you wont really need $ x_1^0$ ). Putting it into $ y$, denote $ y$ back to $ u$ (as a function of $ x$ and $ y$) you get (if I haven't made computational mistakes)

\[ u(x,y)\equal{}\frac{1}{2}\minus{}\frac{y^2}{2}.\]

Now check that this is the solution.

How do we get those numbers for the constants from the boundary values?

Sorry for the late reply. From $ u(x,1)\equal{}0$.


induction
inequalities solved
inequalities

Show that
0<a_i<1 for all i in {1,2,..,n} iff x_i>0, x1+..+xi<1 for all i in {1,2,..,n}
where a1*a2*..*ai=xi+x_(i+1)+..+x_n for all i in {1,2,..,n}, n>=2

[i]D.S. Mitrinovic[/i]

This one is very easy and very nice:
   Suppose that 0<a_i<1. Then 0<a_1...a_n<1 and so 0<x_n<1.Then,
 x_n=a_1*...*a_n<a_1*...*a_n-1=x_n+x_n-1 and so x_n-1>0. Then x_n+x_n-1+x_n-2=a_1*...*a_n-2>a_1*...*a_n-1=x_n+x_n-1 and so x_n-2>0. Similarly, one can prove that x_i>0. But for all i we have x_1+...+x_i<=x_1+...+x_n=a_1<1.
   Suppose now that x_i>0 and x_1+...+x_i<1. So, a_1=x_1+...+x_n and 0<a_1<1. Since a_1*a_2,....,a_1*...*a_n>0 and a_1>0, an easy induction shows that a_i>0. And since a_1>a_1*a_2>a_1*a_2*a_3>...>a_1*a_2*...*a_n, we also have a_i<1.


algebra
polynomial
ratio
binomial theorem
AIME

Ok, this was my favorite problem of this years Aime. Therefore, I'm posting my solution. :P

[hide="Click to Reveal Hidden Awesomeness"]First we turn $y=x+1$ into $y-1=x$, so that we can include $y$ in our equation. Now, it wouldn't be extremely difficult to plug $y$ in and start pounding out the coefficients of all the $y^2$ terms. However, this is a very long process and it's easy to make errors. Therefore, we look for a cleaner approach.

Recalling expressions such as the difference of cubes, we see that $1-x^{18}=(1-x+x^2-x^3+\cdots+x^{16}-x^{17})(1+x)$ and $1-x+x^2-x^3+\cdots+x^{16}-x^{17}=\frac{(1-x^{18})}{(1+x)}$. Now this is Exactly the type of equation we like.

Plugging in $x=y-1$, our expression becomes $\frac{(1-(y-1)^{18})}{y}$, so we need only find the coefficient of $y^3$ in the numerator to get the answer. Applying the Binomial Theorem, our answer is ${18 \choose 3}=17 \cdot 48= \boxed{816}$. [/hide]

[quote="4everwise"] $\boxed{816}$[/quote]
Nice, once more, they managed to involve fibonacci... :)

Could someone explain how that involved Fibonacci?

816 -> reversed : 618, now, is it more striking? ;)

[quote="Hokkage"]Again by use of almighty Google, 618 are the first three digits after the decimal place in $\phi$ and appear in the Fibonacci sequence because the ratio between consecutive Fibonacci numbers (as they get bigger) gets closer and closre to $\phi$. I still don't know how this relates spectacularly to 816... the digits are just reversed.[/quote] :rotfl:

Googling skills!!  :rotfl:

There are 10 types of people in this world, those who can Google and those who can't.

[quote="Hokkage"]I still don't know how this relates spectacularly to 816... the digits are just reversed.[/quote]

Yeah mathmanman, I don't think it really involves Fibonacci... :huh:

Well, what I meant is that it was related with him ;)

[hide="Solution"]The coefficient of $ y^{2}$ in $ ( \minus{} 1)^{k}(y \minus{} 1)^{k} \equal{} (1 \minus{} y)^{k}$ is $ \binom{k}{2}$ by the Binomial Theorem.  Therefore, $ a_{2} \equal{} \binom{2}{2} \plus{} \binom{3}{2} \plus{} \cdots \plus{} \binom{17}{2} \equal{} \binom{18}{3} \equal{} 816$ by Hockey-Stick.[/hide]

[quote="matt276eagles"][quote="4everwise"]The polynomial $1-x+x^{2}-x^{3}+\cdots+x^{16}-x^{17}$ may be written in the form $a_{0}+a_{1}y+a_{2}y^{2}+\cdots+a_{16}y^{16}+a_{17}y^{17}$, where $y=x+1$ and thet $a_{i}$'s are constants. Find the value of $a_{2}$.[/quote]

[hide="Solution"]The coefficient of $y^{2}$ in $(-1)^{k}(y-1)^{k}=(1-y)^{k}$ is $\binom{k}{2}$ by the Binomial Theorem.  Therefore, $a_{2}=\binom{2}{2}+\binom{3}{2}+\cdots+\binom{17}{2}=\binom{18}{3}=816$ by Hockey-Stick.[/hide][/quote]

What is Hockey-Stick (and how is it used)?

[quote="matt276eagles"][quote="4everwise"]The polynomial $1-x+x^{2}-x^{3}+\cdots+x^{16}-x^{17}$ may be written in the form $a_{0}+a_{1}y+a_{2}y^{2}+\cdots+a_{16}y^{16}+a_{17}y^{17}$, where $y=x+1$ and thet $a_{i}$'s are constants. Find the value of $a_{2}$.[/quote]

[hide="Solution"]The coefficient of $y^{2}$ in $(-1)^{k}(y-1)^{k}=(1-y)^{k}$ is $\binom{k}{2}$ by the Binomial Theorem.  Therefore, $a_{2}=\binom{2}{2}+\binom{3}{2}+\cdots+\binom{17}{2}=\binom{18}{3}=816$ by Hockey-Stick.[/hide][/quote]

Nice.  Though more people would probably do it the other way.

I wouldn't have originally thought of that, but its a much slicker solution.

Wow how did this topic get revived after 8 months.

[quote="cincodemayo5590"]

What is Hockey-Stick (and how is it used)?[/quote]

[url=http://mathworld.wolfram.com/ChristmasStockingTheorem.html]Mathworld[/url] will answer your first question.  As for how it is used, you saw one example in my solution above.

[hide="Solution"]

Substitute $x=y-1$ into the equation.  In $x^2$, the coefficient of $y^2$ is $\binom{2}{2}$.  In $x^3$, the coefficient of $y^2$ is $-(-)\binom{3}{2}$.  This pattern repeats, and we get $\binom{2}{2}+\binom{3}{2}+\cdots \binom{17}{2}$.  When I did this problem as part of the AoPS book without any resources, I derived the sum of this using method of finite differences, to get that the sum is $\frac{1}{6}(16^3)+\frac{1}{2}(16^2)+\frac{1}{3}(16)=816$.  With resources, it's the hockey stick identity which gives us $\binom{18}{3}=816$.  Both ways, the answer is $\boxed{816}$.  [/hide]

[quote=4everwise]The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$, where $y=x+1$ and thet $a_i$'s are constants. Find the value of $a_2$.[/quote]

[hide="My solution"]let $f(x)$ denote the given expression. Then it is easy to see that,
 $$f(x)=\frac{1-x^{18}}{1+x}=\frac{1-(y-1)^{18}}{y}=f(y-1)$$
Therefore, $a_2$ is equal to the coefficient of $y^3$ in the expansion of, $1-(y-1)^{18}.$ Or,
$$a_2=\binom{18}{3}=816.$$ [/hide]

[hide=Linear Algebra Solution]
Let $V$ be the vector space of polynomials of degree $\leq 17, $ and let $B = \{1, x, x^2, ..., x^{17} \}$ and $C = \{1, (x+1), (x+1)^2, ..., (x+1)^{17} \}$ be two bases for $V$.
Let $\vec{v} \in V$ be the polynomial given in the problem, and it is easy to see that $[ \vec{v} ]_B = \langle 1, -1, 1, -1, ... , 1, -1 \rangle.$

Note that the transformation matrix from $C$ to $B$ can be easily found to be $P_{C \to B} = [ [\vec{c_1}]_B [\vec{c_2}]_B ... [\vec{c_3}]_B ] = \begin{bmatrix} \tbinom{0}{0} & \tbinom{1}{0} & \tbinom{2}{0} & \cdots & \tbinom{17}{0} \\ 0 & \tbinom{1}{1} & \tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & \tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$

I claim that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} ,$ where the term $\dbinom{n}{k}$ is negated if $n+k$ is odd.

One can prove that the $i$th row of $P_{C \to B}$ dotted with the $j$th column of $P_{B \to C}$ is $\delta_{i, j}$ by using combinatorial identities, which is left as an exercise for the reader. Thus, since the two matrices multiply to form $\mathbb{I}_{18},$ we have proved that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$

To find the coordinates of $\vec{v}$ under basis $C,$ we compute the product $[ \vec{v} ]_C = P_{B \to C} [\vec{v} ]_B = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 1 \\ \vdots \\ -1 \end{bmatrix} = \begin{bmatrix} \sum_{n=0}^{17} \tbinom{n}{0} \\ -\sum_{n=1}^{17} \tbinom{n}{1} \\  \sum_{n=2}^{17} \tbinom{n}{2} \\ \vdots \\  -\sum_{n=17}^{17} \tbinom{n}{17} \end{bmatrix} = \begin{bmatrix} \tbinom{18}{1} \\ - \tbinom{18}{2} \\ \tbinom{18}{3} \\ \vdots \\ -\tbinom{18}{18} \end{bmatrix},$ where the last equality was obtained via Hockey Stick Identity.

Thus, our answer is $a_2 =  [ [ \vec{v} ]_C ]_3 = \dbinom{18}{3} = \boxed{816}.$


For 4 noncoplanar points $A, B, C, O$, $\angle AOB=\alpha$ and $\angle BOC=\beta$. What is the measure of $\angle AOC$?

consider a tetrahedron...this is obviously not uniquely defined

sorry, I forgot plane AOB is perpendicular to plane BOC


let a,b,c>0,prove
(1)$ \prod \left (\frac {a}{b} + \frac {a} {a+b+c}\right)  \leq \frac {64} {27} $
(2)$ \prod \left (\frac {a}{a+b} + \frac {c+a} {a+b+c}\right)  \leq \frac {343} {216} $

One (somewhat) tricky application of AM-GM seems to do it...

$\prod_{\text{cyc}}\left(\frac{a}{a+b} + \frac{a+c}{a+b+c}\right)$

$= \prod_{\text{cyc}}\left( \frac{2a(a+b+c)+bc}{(a+b)(a+b+c)}\right)$

$= \sqrt{\prod_{\text{cyc}} \left( \frac{2a(a+b+c)+bc}{(a+b)(a+b+c)}\right)\prod_{\text{cyc}} \left( \frac{2a(a+b+c)+bc}{(a+c)(a+b+c)}\right) }$

$\leq \frac{1}{6^3} \left( \sum_{\text{sym}}\frac{2a(a+b+c)+bc}{(a+b)(a+b+c)}\right)^3$

$= \frac{1}{6^3} \left( \sum_{\text{sym}} \left(\frac{2a}{a+b}\right) + \sum_{\text{sym}} \left(\frac{bc}{(a+b)(a+b+c)}\right) \right)^3 $

$= \frac{1}{6^3} \left( \sum_{\text{cyc}} \left(\frac{2a+2b}{a+b}\right) + \sum_{\text{cyc}} \left(\frac{bc+ca}{(a+b)(a+b+c)}\right) \right)^3$

$= \frac{1}{6^3} \left( 2\cdot 3+1 \right)^3 = \frac{343}{216}.$


probability
analytic geometry
graphing lines
slope

[asy]size(100);
for(int i = 0; i<6; ++i)
for(int j = 0; j<6; ++j)
dot((i,j),linewidth(7));[/asy]

The following represents a square array of points in the plane.  If three distinct
points are chosen at random, what is the probability that they are collinear
(all lie on the same line)?  You can round your answer to the nearest
whole number of a percent.

[hide="solution"]This is what I got.  There are 36 points and $ \binom{36}{3} \equal{} 7140$ ways to choose any 3 points out of those 36.  Now we have to find the amount of ways three points can be collinear.  First we think about the ways 3 points can be collinear.  They could be collinear horizontally, vertically, or diagonally.

Case 1 Horizontal : There are 6 rows and in each one you can choose 3 out of the 6.  So there are $ 6\binom{6}{3} \equal{} 120$ ways.

Case 2 Vertical:  The same thing as case 1.  There are 120 ways here as well.

Case 3 Diagonal: This case is a bit more complicated because there are different lengths of diagonals.  We can use the diagonals with 6 points, 5 points, 4 points, or 3 points.  Any lower than three and we wouldn't be able to choose 2 points.  The number of ways is $ \binom{6}{3} \plus{} 2\binom{5}{3} \plus{} 2\binom{4}{3} \plus{} 2\binom{3}{3} \equal{} 20 \plus{} 20 \plus{} 8 \plus{} 2 \equal{} 50$

That's a total of 290 possibilities for 3 collinear points.  The probability is about 4%.[/hide]

Sorry if I'm not understanding this, but for case 3 should you multiply everything by two because of how you can 'flip' everything?

It's because there are two diagonals to choose from.  Ex:  On both the left and right side of the diagonal with 6 points are diagonals with 5 points.

Right, but shouldn't you also multiply your sum of 50 by two because you have two 6 dot diagonals from reflecting all the points.

What about diagonals which don't have a slope of $ \pm 1$?

Oh that's true,  I didn't think about that.  Now I'm too lazy to take that into account so anyone who wants to can solve the rest of the problem.  :P


[hide="Welcome!"]Merhaba! Ongi etorriak! Shagatom! Vítejte! Tervetuloa! Bienvenue! Xαιρε! Καλως ορισατε! Aloha mai!
Swaagatam! Benvenuto! Irashaimasu! Salvete! Sveiki! Ximocehuitzino! Velkomen! Bun venit!
добро пожаловать! Fàilte! Bienvenidos! Välkommen! Croeso!
!זאית וילכום! ברוחים הביים[/hide]

Sorry if I got anything wrong up there...

Post here and say hi, so we can see how much support this forum has. And make sure to tell your friends about it!

Hi.  I'm interested in linguistics; perhaps "interested amateur" would be the best description.  And I happen to know both of the other current users of the forum personally, funnily enough.

And to add Gaelic :  Fàilte !

I am glad we have this forum. I just stumbled upon it. :ninja:

Same here funcia.  I just found this too... This seems like a great idea... We'll have to see how this pans out...  :lol:

Change "Merhaba!" to "Hoş geldiniz!"

"Merhaba" means "Hello" in Turkish.

I want to be an archaeologist specializing in ancient languages.

I'm a conlanger too! I specialize in computational linguistics.

It has a heavy emphasis on logic. Look up the North American Computational Linguistics Olympiad (NACLO) and try some of their problems.

Just to clarify... I think what QuantumTiger meant by "logic" in the above post was the general process of deducing one thing from another, as in when solving crossword or sudoku puzzles, rather than the specific discipline that overlaps with mathematics, philosophy, and linguistics and deals with fancy words and intimidating strings of symbols. If I'm not mistaken, real computational linguists have computers work out the "logic" that linguistics olympiad contestants have to work out by hand.

Yes, that's what I meant. Thank you for clarifying.

Real computational linguists do utilize computers to cover this process, but I personally enjoy the process of working the logic by hand.

[url]http://www.naclo.cs.cmu.edu[/url] has old NACLO problems.

Yes, the problems in the link not_trig provided are a perfect example.


I have to graph equations in Cartesian Plane on Texnic. Apparently with me the Asymptote codes are not working with Texnic

Impossible to answer your question without more detail of what you've done and what goes wrong. I assume you have followed the Asymptote installation instructions starting at [[Asymptote:_Getting_Started]] and read [[Asymptote:_Help]] but need more help.


1. If $ f(x) \equal{} 2 \sin(2\sqrt2x/3) \plus{} 3\cos(3x/2)$, what is the number of values of p
such that $ 0 < p < 40$ and $ f(p \plus{} x) \equal{} f(x)$ for all $ x\in\mathbb{R}$? Justify your answer.
2. Find the period of $ y \equal{} 12\sin(4\pi x/7)\plus{}13\cos(2\pi x/5\plus{}\pi/2)$

3. Find the minimum value of $ f(x)\equal{}\sqrt{x^4\minus{}x^2\plus{}1}\plus{}\sqrt{x^4\minus{}5x^2\minus{}4x\plus{}13}$.


Republicans are just a disaster. THe only thing they are good at is wasting goverment monye (tax cuts, war, and MANY MORE). I bet you anything that no body has an argument the fact that tax cuts for those millionares are necessary.
I'm not saying that Republicans are racist, but the racist are in the republican party (Rush Limbaugh, Shawn Hannity, Newt Gringrich).
Republicans use money (they get tons from big businesses to trick the people). They create a huge scandal for Bill Clinton, and practically nothing for those CEOS (bush had enron stocks and cheny had halliburton). If it was a democratic president I ll bet you anything that there would be a huge case (the reason is because republicans are in control of legislature, judicial, and executive).
Republicans lie and cheat. THey had a humongous case for Bill, and now say that Arnio Scharneger was just doing stuff.
Republicans say that Democrats aren't patriot, when they themseveles never fought in war (George Bush, Dick Cheny).
Republicans don't actually provide jobs with tax cuts because a greater percentage of the money just gives the rich more money.
One' of the worst cases of Republican cheating was the Mcarthy who accused all democrats as being commie-symps.
Democrats have always improved the economy because they take the money from the very rich into the government. THe government in return employs people who serve the community. THis provides jobs and a better community. More jobs means more consumers so businesses also prosper thanks to the taxes. Republicans on the other hand say that businesses are getting tax cuts, how3ever they in reutrn get a bad economy and are even loosing customers. Also because of the tax cuts, the mmoney that the rich have aren't just stored in bank accounts they are passed around the economy.

No offense to thre Rebulicans
I like the immigration thing that Bush did and the cooperation with China and urging Taiwan not to be aggresive.

I'll allow the thread, but not with that topic. Others may feel free to post rants about how horrible the democrats are.

my subject is consummate, perfect, flawless, irreproachable, impeccable, indefectible, falutless, letter-perfect facts about the Republican party That is why it is not a rant, but however a FAIR and Balanced Manifesto of the right.

what was the original name of the thread?

Democracy makes these things confusing - yes I said Democracy. In Communism - everybody is happy(unless they are reallllllly egocentric). There are no major problems concerning taxes, retirement policies, political lawsuits, constitutional ammendments, and whatnot. Life is simple. ANd that is right... I voted for Communism. The Kremlin, and some Chinese Politicians have given it a bad name, but you don't wee China struggling under Communism today do you? That is because life isn't much different in CHina than in the US - besides social issues.

What about the Democrats' shameless supporting of horrible crimes like abortion?

well a republican named Arnold Scharwarnegger supported women's rights including abortion. (Gray Davis and Bustamante would'nt because they are Catholics)
so topper so you should change your party affiliation.

Oh dear.

Abortion is not a political issue -- the fundamental disagreement over religion is religious, not political.  Please don't bring it up here.

well bush tries to win votes by calling democrats abortionist which is not all true and bush also try to win reliogious support by going against homosexual marriages

i like talking about this because I won the state geography bee last year.

[quote="white_horse_king88"]In Communism - everybody is happy(unless they are reallllllly egocentric). There are no major problems concerning taxes, retirement policies, political lawsuits, constitutional ammendments, and whatnot. Life is simple.[/quote]

Can't resist saying it: It isn't that everybody's full. It's just that everybody's equally hungry.

And life is simple because the Communist gov't makes sure you don't know what's going on.

[quote]That is because life isn't much different in CHina than in the US - besides social issues.[/quote]

You're right -- but China isn't entirely Communist anymore. They saw what capitalism did to the US economy, so they're trying it in China. Slowly, gradually, with constraints. They're being smart about it for once.. hence, the sudden boom in the Chinese economy.

It definitely takes guts to post support for Communism, though. I would... if only Communism worked. There are no true Communist states ("utopias"). But who knows? Didn't Marx say something like democracy will evolve into communism anyway? Hmm. Or maybe I botched that up.

By the way, China also supresses non-state religion.

Last I checked, China didn't have a state religion? I know it suppresses Christianity. Bush, on the other hand, promotes the Christian state unabashedly.. I think this man should reread the Constitution <i>very</i> carefully. Especially that bit about separation of Church and State.

The most ridiculous position he's taken (well, one of the most) is to advocate an "abstinence program" instead of "safer sex program." Yes, yes, that is all very fine and good but who is he kidding? He expects teenagers to listen to "Premarital sex is immoral" instead of "Unsafe sex will kill you?"

[quote="churchilljrhigh"]i like talking about this because I won the state geography bee last year.[/quote]

you lost me... geography=/politics

china allows the old religions (chrsitianity, judaism, etc). I have a relative in China who is a christian. It just doesn't allow new religions that are bit twisted. Recently there was this thing called falongong and because of it people who got sick don't go see doctors (it says the cure is to read the bible or something) and all sorts of weird cult stuff. THe chinese government noticed and tried to stop it. and now these people are making up all sorts of stuff about the chinese government and say they were politcally prosecuted so they can come to the U.S.

[quote="JBL"]
By the way, I voted Democrat (being a registered member of the party) but my true political leanings are far to the left of the mainstream of the Democratic party.
[/quote]

Just curious: Would you vote for Howard Dean in a primary?

Okay, I am absolutely 100% serious now -- do not talk about abortion here.  Go start a seperate thread if you want to talk about abortion.  Whether abortion is bad or not is a fundamentally non-political issue that has been politicized.  It shouldn't be part of this discussion if not for any other reason than it will completely dominate it and drive all other issues to the side of this post.  Don't talk about it.

I don't know how I'm going to vote for in the primary yet.  In all probability, by the time NY (where I'm registered) gets around to voting, all 9 candidates won't be around any more.  Lieberman is way, way too conservative for me.  On the widest number of issues, I think I share my viewpoints with Kucinich (whose name I can't spell worth a darn).  Dean is pretty moderate on quite a large number of issues, although he is liberal on others.  I am well beyond liberal most of the time.

I will be voting for ABB in the general election.  (ABB = anyone but Bush.)

wyra -- Republicans are for small government when they aren't in power.  When they are in power, they're as much for big spending as anyone else.  It's just that they like to shift the spending to benefit different groups of constituents.  The only people who consistently advocate for small government are libertarians, and Republicans borrow their philosophy as long as it lets them dismantle social welfare programs.

aalindsey, with respect to your comments on president Bush and Johnson:  Johnson did some good things and some bad things.  His work on civil rights was incredible, especially as a southerner.  However, he obviously made a major mistake with respect to Vietnam.  Realizing his mistake, he decided not to run for another term.  That's more than we can hope for with Bush.
If you want to know what Bush has done wrong, look at the changes to environmental policy, consumer protections and nearly every program benefiting people under the poverty line.  In fact, look at anything he's done on the domestic front.  It's all atrocious.

One question before i start ranting...

Have we stopped discussing about China???

no martin we haven't
republicans get the credit for our good relatioonship with CHina. The commencement was from Nixon who secretly sent kissinger to China to talk and today bush has urged Tawian from not being to aggresive. the u.s. keeps on blabbering about China's communist stuff. However most of it untrue. CHina only goes after those who go agianst and criticize the government and those terroists in Tibet (from something I read about Tibet in ENcarta). China has the full reason to this just as bush does with his patriot acts. Every country needs to pereserve the security of the nation. Look at Russia, whose lack of security by Boris Yelstin resulted in an a powerful CHechen suicide bombers everywhere.

Ok I said nothing about JFK if I was alive back in that time I probably wouldve been to assassinate Oswald. I liked JFK but I didnt like Johnson I thought it was funky that he was sworn in right on the plane it made it seem almost if he had planned it that way. I liked JFK because he did wonders for intigration and civil rights for African American peoples in America. I like the way he handled the Cuban Missle Crisis too I dont know I just liked him he was an interesting president as well as an interesting person.

Hmm... interesting, but i have to do an essay for some stuff... and who might've thought, the essay's on the Boxer Rebellion... HAHAHAHA...lol. and im writing it from the Chinese point of view.

I was being sarcastic sorry I didnt say so in the first place and Oswald was one of the assassins at least thats what I believe and LBJ was also in my theory but Im not going to get started on that the assassination theory anyways (we just got done in english with unit over JFK.)

aalindsey you are a texan and you are acusing an innocnet texan. I think the case pretty much solved Oswald was a commy symp. and jack ruby was a patriot who got mad at oswald. LBJ is totally inoncent. the only assasination that i think is very confusing is the lincoln visit [url]http://members.aol.com/RVSNorton/Lincoln.html[/url] to learn about the different theories.

I know what Im saying but I have a grudge agianst LBJ and I only think he was part of it not all of it. There was an inside track somewhere but thats the info we are missing. I have a weird theory cause if u think about it is there even a such thing as visual evidence or do human emotions get in the way to much Ill leave u pondering with that question and discuss it when I get an answer.

One comment then I'm outta here: Woot! Can't believe commusism's not the one with the least votes, I'm not in the minority, ya-ha, ya-ha :lol:

[quote="Tare"]One comment then I;m outta here: Woot! Can't believe commusism's not the one with the least votes, I'm not in the minority, ya-ha, ya-ha :lol:[/quote]
If only you truly knew what communism is, was, and what it has done... if only...

confuted, you are confusing communism with totalitarianism.

And by the way, capitalism has done some pretty nasty things, too.

[quote="MysticTerminator"]Perhaps Matt just did that because of his personal grudge against Tare (:)) Anyway, this is a very interesting thread, although I won't start spouting my (probably vastly misinformed) views for a while. It does interest me that confuted allowed this post in the first place, what with the fallacies in the opening page.[/quote]
I probably only allowed it because I didn't read it.

Whoops, sorry guys. After several requests, I was attempting to move the drastically off topic / slanderous / nonsense comments from this thread, but accidentally clicked the wrong button and ended up deleting many posts that I did not intend to delete. So if you had a constructive post and it got deleted - sorry.


Let ABC be a triangle that  the length of any side is  4.The other sides of triangle are roots(zeros)of  the equation  x^2-m.x+1=0.Find the smallest interval for m.

Hello, ssupermath!

Quote:Let ABC be a triangle that  the length of any side is  4.
The other sides of triangle are roots(zeros)of  the equation  x^2 - mx + 1 = 0
Find the smallest interval for m.

[hide]

Let side a = 4.  The other two sides are b and c, which are roots of the quadratic:

b =  [m + sqrt{m^2 - 4}]/2 and c = [m - sqrt{m^2-4}]/2

These sides must satisfy the Triangle Inequality.

(1) a + b > c

Then: 4 + [m + sqrt{m^2 - 4}]/2 > [m - sqrt{m^2 - 4}]/2

which simplifies to: sqrt{m^2-4} > -4 ... which is always true.

(2) a + c > b

Then: 4 + [m - sqrt{m^2 - 4}]/2 > [m + sqrt{m^2 - 4}]/2

which simplifies to: m^2 < 20 . . . |m| < 2sqrt{5}

(3) b + c > a

Then: [m + sqrt(m^2 -  4}]/2 + [m - sqrt{m^2 - 4}]/2 > 4

which simplifies to: m > 4

Therefore, the interval is: (4, 2sqrt{5})

[/hide]

Its in spoiler text - just highlight it with your mouse to read it (so that other people can try the problem without seeing answers).


If $a=-7$ and $b=a+3$, what is the value of $\frac{b^3(a-b)}{b-a}$?

[hide]$a=-7$ and $b=-4$.  The expression becomes $\frac{-64(-3)}{3} \Rightarrow 64$.[/hide]

[hide]As b=-7+3, all b can be substituted as -4. All a can be substituted as -7. -4^3(-7-(-4)/-4-(-7)=-64(-3)/3=[b]64.[/b][/hide]

find the a and the b, which are -7 and -4 respectively. Plug them into the equation, simplify, and you have the answer.

Plug in the numbers for the respected variables and you will get 64


Solve: Antiderivative-Sin^2(x)cos(x)dx

I know that i should use substitution. But still no clue...

For $ n\neq-1$, try to differentiate $ \frac{1}{n+1}\{f(x)\}^{n+1}$.

thanks a lot

I am not sure.  i remember that concept tho.


inequalities
trigonometry
inequalities unsolved

Let $ 0 \leq a,b,c,d \leq 1$. Prove that

\[ \sqrt{(1\minus{}a b)(1\minus{}c d)}\minus{}\sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)}\minus{}\sqrt{a c (1\minus{}b)(1\minus{}d)}\minus{}\sqrt{b d (1\minus{}a)(1\minus{}c)}\geq 0\]

Define the angles $ 0 \leq A,B,C,D \leq \pi/2$ and use the substitution of variables $ a\equal{}\cos^2 A$, $ b\equal{}\cos^2 B$, $ c\equal{}\cos^2 C$ and $ d\equal{}\cos^2 D$, to obtain the equivalent inequality

\[ \sqrt{(1\minus{}\cos^2 A \cos^2 B)(1\minus{}\cos^2 C \cos^2 D)}\geq \sin A \sin B \sin C \sin D \plus{} \cos A \sin B \cos C \sin D \plus{} \sin A \cos B \sin C \cos D\]

which can obviously be square on both sides if you don't like square-roots. Can anyone prove the inequality above holds? Or does that remind you of any "classical inequality"? (please?!  :roll: )

Substitute $ a' \equal{} 1\minus{}a, \ldots$, divide the inequality by $ \sqrt{a'b'c'd'}$ and
substitute $ a'' \equal{} (1/a')\minus{}1$. Then $ a'',\ldots \in [0,\infty]$ and the inequality
becomes
\[ \sqrt{a'' \plus{} b''\plus{}1} \sqrt{c''\plus{}d''\plus{}1} \ge 1 \plus{} \sqrt{a''c''} \plus{} \sqrt{b''d''}.
\]
In order to see that this holds, e.g. you can use brute force: Square it, simplify and
use AM-GM twice.

Thank you for you proof Solyaris, it's a nice one!

Just for record I'd like to add that it's fine to divide by $ \sqrt{a^\prime b^\prime c^\prime d^\prime}$.

The reason is this: if any of the variables is 0, then the inequality can be proved in a trivial way --- we are thus interested in proving the general case, i.e., $ a^\prime, b^\prime, c^\prime, d^\prime \neq 0$. Here, Solyaris' procedure is well-defined.


analytic geometry
advanced fields
advanced fields unsolved

Prove that the mapping $ f : S^{n\minus{}1}\to \mathbb{R}P^{n\minus{}1}$,defined by $ f(x_{1},x_{2},...,x_{n}) \equal{} (x_{1}: x_{2}: ... : x_{n})$ has a rank $ n\minus{}1$ and, thus, an immersion.

The restriction of $ f$ to $ S^{n\minus{}1}\cap\{x_{1}> 0\}$ is equivalent, up to composition with diffeomorphisms, to the identity map on $ \mathbb R^{n\minus{}1}$. Same for other coordinate patches.


For $ x,y,z$ positive real numbers, prove that

$ (2x\minus{}y\minus{}z)\cdot\frac{x\plus{}y}{y\plus{}z}\plus{}(2y\minus{}z\minus{}x)\cdot\frac{y\plus{}z}{z\plus{}x}\plus{}(2z\minus{}x\minus{}y)\cdot\frac{z\plus{}x}{x\plus{}y}\geq0$

P.S.
I hope that this inequality is a new one! I haven't seen it before.

[quote="marin.bancos"]For $ x,y,z$ positive real numbers, prove that

$ (2x \minus{} y \minus{} z)\cdot\frac {x \plus{} y}{y \plus{} z} \plus{} (2y \minus{} z \minus{} x)\cdot\frac {y \plus{} z}{z \plus{} x} \plus{} (2z \minus{} x \minus{} y)\cdot\frac {z \plus{} x}{x \plus{} y}\geq0$

P.S.
I hope that this inequality is a new one! I haven't seen it before.[/quote]

$ \Longleftrightarrow {\frac { 2\left( x\plus{}y\plus{}z \right)  \left(x^3\plus{}y^3\plus{}z^3\minus{}xy^2\minus{}yz^2\minus{}zx^2\right) }{ \left( y\plus{}z \right)  \left( z\plus{}x\right)  \left( x\plus{}y \right) }} \ge 0$

[quote="FelixD"]You can rewrite the inequality as $ \sum \frac {(x \minus{} y)^2}{x \plus{} z} \plus{} \sum \frac {(x \minus{} y)(x \minus{} z)}{y \plus{} z} \ge 0$. The first sum is clearly greather than or equal to $ 0$. Hence we have to prove $ \sum (x^2 \minus{} y^2)(x^2 \minus{} z^2)\ge 0$, which is obvious after multiplying out.  :)[/quote]

I don't think that the initial inequality is equivalent to your inequality. In fact, your inequality is equivalent to
$ (2x\minus{}y\minus{}z)\cdot\frac{x\minus{}z}{y\plus{}z}\plus{}(2y\minus{}z\minus{}x)\cdot\frac{y\minus{}x}{z\plus{}x}\plus{}(2z\minus{}x\minus{}y)\cdot\frac{z\minus{}y}{x\plus{}y}\geq0$
Am I right?

[quote="rachid"][quote="marin.bancos"]For $ x,y,z$ positive real numbers, prove that

$ (2x \minus{} y \minus{} z)\cdot\frac {x \plus{} y}{y \plus{} z} \plus{} (2y \minus{} z \minus{} x)\cdot\frac {y \plus{} z}{z \plus{} x} \plus{} (2z \minus{} x \minus{} y)\cdot\frac {z \plus{} x}{x \plus{} y}\geq0$

P.S.
I hope that this inequality is a new one! I haven't seen it before.[/quote]
We have to prove that :

$ \frac {x(x \plus{} y)}{y \plus{} z} \plus{} \frac {y(y \plus{} z)}{z \plus{} x} \plus{} \frac {z(z \plus{} x)}{x \plus{} y} \geq x \plus{} y \plus{} z$

Or,equivalently :

$ x \left ( \frac {1}{y \plus{} z} \plus{} \frac {1}{x \plus{} y} \right ) \plus{} y \left ( \frac {1}{z \plus{} x} \plus{} \frac {1}{y \plus{} z} \right ) \plus{} z \left ( \frac {1}{x \plus{} y} \plus{} \frac {1}{z \plus{} x} \right ) \geq 3$

Which can be reduced to :

$ \frac {x \plus{} y}{y \plus{} z} \plus{} \frac {y \plus{} z}{z \plus{} x} \plus{} \frac {z \plus{} x}{x \plus{} y} \geq 3$

Which is true according to AM-GM's inequality.Equality holds iff : $ x \equal{} y \equal{} z$[/quote]

Why your first inequality is equivalent to the second? Could you explain?

[quote="marin.bancos"]Why your first inequality is equivalent to the second? Could you explain?[/quote]
Of course,I can  :)

The first inequality is equivalent to :

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

Or :

$ x\left (\frac {x}{y \plus{} z} \plus{} 1 \plus{} \frac {z}{x \plus{} y} \plus{} 1\right ) \plus{} y\left (\frac {y}{z \plus{} x} \plus{} 1 \plus{} \frac {x}{y \plus{} z} \plus{} 1\right )$$ \plus{} z\left (\frac {z}{x \plus{} y} \plus{} 1 \plus{} \frac {y}{z \plus{} x} \plus{} 1\right )\geq 3(x \plus{} y \plus{} z)$

Then,we eliminate the term $ x \plus{} y \plus{} z$ from both sides and we obtain the second inequality.

[quote="marin.bancos"][quote="FelixD"]You can rewrite the inequality as $ \sum \frac {(x \minus{} y)^2}{x \plus{} z} \plus{} \sum \frac {(x \minus{} y)(x \minus{} z)}{y \plus{} z} \ge 0$. The first sum is clearly greather than or equal to $ 0$. Hence we have to prove $ \sum (x^2 \minus{} y^2)(x^2 \minus{} z^2)\ge 0$, which is obvious after multiplying out.  :)[/quote]

I don't think that the initial inequality is equivalent to your inequality. In fact, your inequality is equivalent to
$ (2x \minus{} y \minus{} z)\cdot\frac {x \minus{} z}{y \plus{} z} \plus{} (2y \minus{} z \minus{} x)\cdot\frac {y \minus{} x}{z \plus{} x} \plus{} (2z \minus{} x \minus{} y)\cdot\frac {z \minus{} y}{x \plus{} y}\geq0$
Am I right?[/quote]
My calculations are definitely correct.

Sorry, Felix! You are right! In fact
$ (2x \minus{} y \minus{} z) \cdot \frac {x \minus{} z}{y \plus{} z} \plus{} (2y \minus{} z \minus{} x) \cdot \frac {y \minus{} x}{z \plus{} x} \plus{} (2z \minus{} x \minus{} y)\cdot\frac {z \minus{} y}{x \plus{} y}\geq0 \ \Leftrightarrow$
$ \Leftrightarrow \ (2x \minus{} y \minus{} z)\cdot\frac {x \minus{} z}{y \plus{} z} \plus{} (2x \minus{} y \minus{} z) \plus{} (2y \minus{} z \minus{} x)\cdot\frac {y \minus{} x}{z \plus{} x} \plus{} (2y \minus{} z \minus{} x) \plus{}$
$ \plus{}(2z \minus{} x \minus{} y)\cdot\frac {z \minus{} y}{x \plus{} y} \plus{} (2z \minus{} x \minus{} y) \geq0\ \Leftrightarrow$
$ \Leftrightarrow (2x\minus{}y\minus{}z)\cdot\frac{x\plus{}y}{y\plus{}z}\plus{}(2y\minus{}z\minus{}x)\cdot\frac{y\plus{}z}{z\plus{}x}\plus{}(2z\minus{}x\minus{}y)\cdot\frac{z\plus{}x}{x\plus{}y}\geq0$

Thank you!

Taking into account the previous remark, it's possible to give a new shape to this inequality:

For $ x,y,z>0$ and $ k\in R$, prove that

$ \frac{(2x\minus{}y\minus{}z)\cdot[x\plus{}(k\plus{}1)y\plus{}kz]}{y\plus{}z}\plus{} \frac{(2y\minus{}z\minus{}x)\cdot[y\plus{}(k\plus{}1)z\plus{}kx]}{z\plus{}x}\plus{}$
$ \plus{} \frac{(2z\minus{}x\minus{}y)\cdot[z\plus{}(k\plus{}1)x\plus{}ky]}{x\plus{}y}\geq0$

REMARK
For example, we can be "in fashion" taking k=-2010.

For $ x,y,z>0$, prove that

$ \frac{(2x\minus{}y\minus{}z)\cdot(x\minus{}2009y\minus{}2010z)}{y\plus{}z}\plus{} \frac{(2y\minus{}z\minus{}x)\cdot(y\minus{}2009z\minus{}2010x)}{z\plus{}x}\plus{}$
$ \plus{} \frac{(2z\minus{}x\minus{}y)\cdot(z\minus{}2009x\minus{}2010y)}{x\plus{}y}\geq0$

[quote="marin.bancos"]Taking into account the previous remark, it's possible to give a new shape to this inequality:

For $ x,y,z > 0$ and $ k\in R$, prove that

$ \frac {(2x \minus{} y \minus{} z)\cdot[x \plus{} (k \plus{} 1)y \plus{} kz]}{y \plus{} z} \plus{} \frac {(2y \minus{} z \minus{} x)\cdot[y \plus{} (k \plus{} 1)z \plus{} kx]}{z \plus{} x} \plus{}$
$ \plus{} \frac {(2z \minus{} x \minus{} y)\cdot[z \plus{} (k \plus{} 1)x \plus{} ky]}{x \plus{} y}\geq0$

REMARK
For example, we can be "in fashion" taking k=-2010.

For $ x,y,z > 0$, prove that

$ \frac {(2x \minus{} y \minus{} z)\cdot(x \minus{} 2009y \minus{} 2010z)}{y \plus{} z} \plus{} \frac {(2y \minus{} z \minus{} x)\cdot(y \minus{} 2009z \minus{} 2010x)}{z \plus{} x} \plus{}$
$ \plus{} \frac {(2z \minus{} x \minus{} y)\cdot(z \minus{} 2009x \minus{} 2010y)}{x \plus{} y}\geq0$[/quote]

yeah...

$ \frac {(2x \minus{} y \minus{} z)\cdot[x \plus{} (k \plus{} 1)y \plus{} kz]}{y \plus{} z} \plus{} \frac {(2y \minus{} z \minus{} x)\cdot[y \plus{} (k \plus{} 1)z \plus{} kx]}{z \plus{} x} \plus{}$  $ \frac {(2z \minus{} x \minus{} y)\cdot[z \plus{} (k \plus{} 1)x \plus{} ky]}{x \plus{} y}$

$ \equal{} (2x \minus{} y \minus{} z)\cdot\frac {x \plus{} y}{y \plus{} z} \plus{} (2y \minus{} z \minus{} x)\cdot\frac {y \plus{} z}{z \plus{} x} \plus{} (2z \minus{} x \minus{} y)\cdot\frac {z \plus{} x}{x \plus{} y}$

$ \equal{} {\frac { 2\left( x \plus{} y \plus{} z \right) \left(x^3 \plus{} y^3 \plus{} z^3 \minus{} xy^2 \minus{} yz^2 \minus{} zx^2\right) }{\left( y \plus{} z \right) \left( z \plus{} x\right) \left( x \plus{} y \right) }}$

Dear [b]kuing[/b],

This is what I wanted to say, [b]kuing![/b] Both inequalities are equivalent.
Thank you for your solution!

[quote]$ \frac {x(x \plus{} y)}{y \plus{} z} \plus{} \frac {y(y \plus{} z)}{z \plus{} x} \plus{} \frac {z(z \plus{} x)}{x \plus{} y} \geq x \plus{} y \plus{} z$
[/quote]also weighted Jensen for $ f(x)\equal{} \frac{x}{S\minus{}x}$ where S=x+y+z works here  :)


geometry
trigonometry
conics
perpendicular bisector

Here are the problems: http://www.math.sc.edu/contest/1990/exam/PDF/exam1990.pdf

If anyone wants to do number 18, be my guest...

[hide="16"]Let $x$ be the radius of the small circle and $y$ be the radius of the big circle. Extend $a$ such that it is the perpendicular bisector of the chord. We can construct a right triangle with hypotenuse $y$ and legs $5$ and $x$. Thus $x^{2}+5^{2}=y^{2}$. But the area of the region we want is $(y^{2}-x^{2})\pi$. The area is thus $25\pi$.[/hide]

 [hide="17"]Letting $x=1$, we are left only with the coefficients. The sum of half of the coefficients is thus $\frac{4^{5}}{2}=528$.[/hide]

[hide="19"]Converting everything in terms of $\sin$ and simplifying, we have $\frac{2 \sin{45}\cos{30}}{2 \cos{45}\sin{30}}=\sqrt{3}$[/hide]

[hide="20"]There are more ways for the numbers to sum up to $2$ or $-2$ than any other integer listed (or there negative values).[/hide]

[hide="21"]Let $n$ be the number of tests John has already taken and $x$ be the average score of each of those $n$ tests. We have:

$\frac{nx+97}{n+1}=90$
$\frac{nx+73}{n+1}=87$

From this, we have that $90(n+1)-97=87(n+1)-73$ and $n=7$.[/hide]

[hide="22"]The last digit of $17^{9}$ is $7$ and the last digit of $9^{9}$ is $9$, so therefore the last digit of the expression is $8$. The power of two therefore must be $3 \mod{4}$, so only choices $a)$ and $b)$ are possibilities. Checking again, we also see that $17^{9}-9^{9}$ must be divisible by $17-9=8$ and therefore $2, 2^{2}, 2^{3}$ are the only even factors of the expression and $n=3$ is the largest solution.
[/hide]
[hide="23"]Let $\angle{POQ}=a$, then $\angle{PQO}=a$, and $\angle{P}=180-2a$. Also, this means that $\angle{QOR}=\angle{QRO}=60-a$. But $\angle{P}+\angle{R}=120^\circ$. Therefore, we have that $180-2a+60-a=120$ and $a=40^\circ$. Therefore $\angle{QRO}=20^\circ$.[/hide]

[hide="24"]After rationalizing the denominator, we get $\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}=-1+5=4$[/hide]

[hide="25"]Using $R=\frac{abc}{4A}$ and Heron’s formula, we have that $R=\frac{8 \sqrt{15}}{15}$.[/hide]

[hide="26"]We use combinations since then we can arrange the four numbers any which way we like. We choose $4$ out of the $7$ spots to place our numbers and the remaining three numbers left are arranged in different ways also. Therefore our answer is $\binom{7}{4}\cdot 3!=210$.[/hide]

[hide="27"]$r^{2}+s^{2}+t^{2}=(r+s+t)^{2}-2(rs+st+rt)=4-2(5)=-6$.[/hide]

[hide="28"]Notice that $f(\frac{1}{2})=\frac{5}{2}$ so choice $a)$ is automatically eliminated. Now using the answer choices to our advantage, we can try to solve the problem based on the choices themselves.

Starting with choice $c)$, we take the value of $x$ for the minimum:
$4x+1=8/3 \Longrightarrow x=1/6$
$x+2=8/3 \Longrightarrow x=2/3$
$-2x+4=8/3 \Longrightarrow x=2/3$

Continuing this way with all the answer choices, we have that the values of $x$ are $1/6,2/3,11/24,5/6,7/12,1/2,1$. Trying $f(x)$ for all these values gives the maximum value of $f(x)$ as $f(2/3)=8/3$[/hide]
[hide="29"]Seeing the multiples of $20$, we can think of double angle formula.

Let $S$ be the desired product. Now notice that:

$\sin{20}\cdot S=\sin{20}\cos{20}\cos{40}\cos{80}=\frac{\sin{40}\cos{40}\cos{80}}{2}=\frac{\sin{80}\cos{80}}{4}=\frac{\sin{160}}{8}$.

Thus $S=\frac{\sin{160}}{8 \cdot \sin{20}}=\frac{1}{8}$.[/hide]

[hide="30"]Using the divisibility test for $13$ or simply trial and error, we find that $111111$ is divisible by $13$ yet is not divisible by $17$.[/hide]

Shobhit needs to work on his conic sections!

[hide="18"](a) We want the limit as x goes to infinity of y/x to be 2. In the first equation, we have $10-(y/x)^{2}-3(y/x)=\frac{10}{x^{2}}$, so $y/x$ can go to two since $10-2^{2}-3*2=0$. The second equation is a parabola. The third equation is $1-2(y/x)^{2}+\frac 5{x^{2}}=0$ which doesn't allow 2. The fourth equation is an ellipse. The fifth equation has $1-(y/x)^{2}+\frac{3}{x}-2(y/x)=\frac 2{x^{2}}$, which doesn't allow 2. Hence only the first equation works.[/hide]


I really have no idea how to do this, any help would be appriciated.

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

If $\{ V, (.|.)\}$ is a prehilbertarian space and if H is the completion of V with respect to the metric induced by the norm on V, then there is an inner product $(.|.)_{H}$ on H such that $(x|y)_{H}$ = $(x|y)$ $\forall x,y \in V$ and the metric on H is induced by this inner product.

Every element of $H$ can be written as a limit of elements of $V$. Then you can define $<x,y>_{H}=lim <x_{n},y_{n}>$ where $(x_{n})_{n} \rightarrow x$ and $(y_{n})_{n} \rightarrow y$. The properties of the inner product $<.,.>$ will be preserved for $<.,.>_{H}$ because of the limit process. The existence of the limit itself is then guaranteed by the continuity of $<.,.>_{H}$ (continuity proved by Cauchy-Schwarz inequality).


Find the product of the square root of 1764 and the largest prime factor of 1764.

$ \sqrt{1764}\equal{}42$ (OMG its 42!!!) Largest prime factor of 42 = 7. $ 42*7\equal{}294$

Because it's a perfect square, all prime factors of the square are also factors of the square root.  The perfect square can be written as:

\[ n\equal{}{p_1}^{2a_1}{p_2}^{2a_2}{p_3}^{2a_3}\dots{p_k}^{2a_k}
\]

so taking the square root yields all the same prime factors.

Do you just factor 1764, or is there another quicker method?


number theory
least common multiple
superior algebra
superior algebra solved

Let $(G, \cdot)$ be a group and $e$ the neutral element. For all $a \in G$ and all $n \in \mathbb{Z^{+}}$ define $a^{n}: =a\cdot a\cdot \ldots \cdot a$ with $n$ factors. Show that if $G$ is finite, then for all $a \in G$ exists a $n \in \mathbb{Z^{+}}$ that $a^{n}=e$. There exists a unique $n$ which work for all elements in $G$?  [hide=" Hint"]Consider $e, a, a^{2}, \ldots ,a^{m}$ where $m$ is the number of elements of the group and use the cancellation law.[/hide]
I don't know how to use the hint... sorry for the bad english.
Mathlinks is awesome

Isn't it really easy?

The sequence $a^{0}, a^{1}, a^{2}, \ldots$ is infinite and takes only finitely many values, so there are $u,v \in \mathbb N_{0}$ such that $a^{u}= a^{v}$. Thus, $a^{u-v}= e$ (WLOG assuming that $u \geq v$). [EDIT]WLOG assuming that $u > v$[/EDIT]

And of course there is a number for which we have $a^{n}= e, \, \forall a \in G$. Just take $n = \textrm{lcm}\left\{ n_{a}| a \in G \right\}$.

Of course $n$ is not unique ... every multiple of it also works.

@perfect_radio: $u > v$ ;)