Yes, I think the stronger inequality $ 3\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{a\plus{}b\plus{}c}\geq\sum\frac{a^{2}\plus{}b^{2}}{a\plus{}b}$.

[Moderator edit: This inequality is discussed at http://www.mathlinks.ro/viewtopic.php?t=140571 .]

[quote="harazi"]$3\frac{a^2+b^2+c^2}{a+b+c}\geq\sum\frac{a^2+b^2}{a+b}$.[/quote]

Nice inequality! And quite easy. In fact,

$3\frac{a^2+b^2+c^2}{a+b+c}-\sum\frac{a^2+b^2}{a+b}=\sum\left(\frac{a^2+b^2+c^2}{a+b+c}-\frac{a^2+b^2}{a+b}\right)$
$=\sum\left(\frac{1}{a+b+c}\cdot\left(\left(a-b\right)ab\frac{1}{b+c}-\left(c-a\right)ca\frac{1}{b+c}\right)\right)$
$=\frac{1}{a+b+c}\cdot\sum\left(\left(a-b\right)ab\frac{1}{b+c}-\left(c-a\right)ca\frac{1}{b+c}\right)$
$=\frac{1}{a+b+c}\cdot\left(\sum\left(\left(a-b\right)ab\frac{1}{b+c}\right)-\sum\left(\left(c-a\right)ca\frac{1}{b+c}\right)\right)$
$=\frac{1}{a+b+c}\cdot\left(\sum\left(\left(a-b\right)ab\frac{1}{b+c}\right)-\sum\left(\left(a-b\right)ab\frac{1}{c+a}\right)\right)$
$=\frac{1}{a+b+c}\cdot\sum\left(\left(a-b\right)ab\frac{1}{b+c}-\left(a-b\right)ab\frac{1}{c+a}\right)$
$=\frac{1}{a+b+c}\cdot\sum\frac{\left(a-b\right)^2ab}{\left(b+c\right)\left(c+a\right)}\geq 0$,

and thus $3\frac{a^2+b^2+c^2}{a+b+c}\geq\sum\frac{a^2+b^2}{a+b}$. Proof complete.

  Darij

Well a little over two years later...

[hide="why harazis stronger?"]
The first is \[\iff \frac{a^{2}+b^{2}+c^{2}}{a+b+c}\geq \sqrt[3]{\frac{\prod(a^{2}+b^{2})}{\prod(a+b)}}\]
So the second is stronger because:
\[\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\geq\frac{\sum\frac{a^{2}+b^{2}}{a+b}}{3}\geq \sqrt[3]{\frac{\prod(a^{2}+b^{2})}{\prod(a+b)}}\]
[/hide]
[hide="darij soluition"]
\[3\frac{a^{2}+b^{2}+c^{2}}{a+b+c}-\sum\frac{a^{2}+b^{2}}{a+b}=\sum\left(\frac{a^{2}+b^{2}+c^{2}}{a+b+c}-\frac{a^{2}+b^{2}}{a+b}\right)\]
\[=\sum\left(\frac{1}{a+b+c}\cdot\left(\left(a-b\right)ab\frac{1}{b+c}-\left(c-a\right)ca\frac{1}{b+c}\right)\right) \]
[/hide]


I was wondering if anybody knows if there is a cutoff score on the AMC 10 to be able to take the AIME. I think the AMC 12 cutoff score is 100 so I guess it might be higher. I got a 126.5 on an AMC 10 practice test but I don't know if it's enough... :?

It is if you can consistently get that. The cutoff is usually around 120.

[quote="h_s_potter2002"]The cutoff score for the AMC 10 is either 120, or the top 1 perscent of scores, whichever one is lower.[/quote]

"All students who took the AMC 10 and were in the top 1% also qualify for the AIME."

RC-7th, hs_potter is right. 120 is guaranteed qualification to AIME even if the top 1% are in a higher score range.

[quote]"All students who took the AMC 10 and were in the top 1% also qualify for the AIME."[/quote]

We need a moratorium to stop quoting from the AMC website. It is too out of date to provide concrete evidence.

120 will definitly be enough, Last year i screwed up AMC10 and got only 110.5 :blush:
But i was still invited. lol

That's beucase the qualifying score for your test was 110.  For the B test, the qualifying score was 120, so it all depends on the test.  As long as you get at least 120 though, you're safe.

From: http://www.unl.edu/amc/b-registration/b1-archive/2004-2005/05AMC1012/05-1012-html/05HSbro-text.html

Students who take the contests on the official dates are eligible for state, regional and national awards. AMC 12 students who rank in the top 5% nationally (or score at least 100) will qualify for the American Invitational Mathematics Exam (AIME). AMC 10 students who rank in the top 1% nationally (or score at least 120) will also qualify for the AIME.

From:  http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM1012/html/05tm-05-text.html

Students who are in the top 5% of all AMC 12 participants or score at least 100 points on the AMC 12, and those who are in the top 1% of all AMC 10 participants or score at least 120 on the AMC 10 are invited to participate in the American Invitational Mathematics Examination.

The same statement also appears on the cover page of the AMC 10 and the AMC 12. (which are reproduced in the Teachers' Manual)

We will change the stantement which appears on the AIME page onthe Web.

Steve Dunbar
AMC Director


inequalities
algebra
system of equations
algebra unsolved

[quote="keyree10"]Find all non-zero real numbers $ x, y, z$ which satisfy the system of equations:
\[ (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2) \equal{} xyz\]

\[ (x^4 \plus{} x^2y^2 \plus{} y^4)(y^4 \plus{} y^2z^2 \plus{} z^4)(z^4 \plus{} z^2x^2 \plus{} x^4) \equal{} x^3y^3z^3\]
[/quote]

Acording to Cauchy-Shwarz inequality we have:

$ \prod_{cyc}(x^4 \plus{} x^2y^2 \plus{} y^4)(1 \plus{} 1 \plus{} 1) \ge \prod_{cyc}(x^2 \plus{} xy \plus{} y^2)^2$

Hence : $ x^3y^3z^3 \ge \frac {x^2y^2z^2}{27}$ and since $ xyz > 0$ we get $ xyz \ge \frac {1}{27}$

On the other hand we have $ x^2 \plus{} y^2 \plus{} xy \ge 3xy$ Thus $ \prod_{cyc}(x^2 \plus{} y^2 \plus{} xy) \ge 27 x^2y^2z^2$ It follows that $ xyz \le \frac {1}{27}$

Hence $ xyz \equal{} \frac {1}{27}$ and Am-Gm give the equality  if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$

(i don't know where , but i think i saw this problem here....)
$ x^4 \plus{} y^4 \plus{} x^2y^2 \equal{} (x^2 \plus{} y^2 \minus{} xy)(x^2 \plus{} y^2 \plus{} xy)$
by replacing each of the expressions by the above equality we get
$ \prod (x^2 \plus{} y^2 \minus{} xy) \equal{} \prod x^2$ (1)
since $ x^2 \plus{} y^2 \minus{} xy \ge |xy|$
we get
$ LHS(1) \ge RHS(1)$
and the equality occurs iff $ x \equal{} y \equal{} z$
now by considering the first given equality we get $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$

[quote="Abdek"]
On the other hand we have $ x^2 \plus{} y^2 \plus{} xy \ge 3xy$ Thus $ \prod_{cyc}(x^2 \plus{} y^2 \plus{} xy) \ge 27 x^2y^2z^2$ It follows that $ xyz \le \frac {1}{27}$

Hence $ xyz \equal{} \frac {1}{27}$ and Am-Gm give the equality  if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$[/quote]

I Dont think thats right because xy may or may not be positive, so u cant apply Am-Gm there, but alternatively,
To prove we can do, by AM-Gm $ (x^4 \plus{} x^2y^2 \plus{} y^4) \ge 3x^2y^2$ So, $ (xyz)^3 \ge27(xyz)^4$ or $ xyz \le \frac{1}{27}$

and that proves $ xyz\equal{}\frac{1}{27}$. Now, Equality occurs when \[ x^4 \equal{} y^4\] or $ x\equal{} \pm y$ , but putting two of $ x, y,z \equal{} \minus{}\frac{1}{3}$ doesnt hold right in the equation. So, we have only positive solution $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$

By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?

dear viewers solutions to the given set are updated by HBCSE

http://olympiads.hbcse.tifr.res.in/uploads/inmo-2010

Let $(1)$ denote the first equation and $(2)$ the second.
$(1)\Rightarrow \frac{x^3-y^3}{x-y}\cdot\frac{y^3-z^3}{y-z}\cdot\frac{z^3-x^3}{z-x}=xyz$

And

$(2)\Rightarrow \frac{x^6-y^6}{x^2-y^2}\cdot\frac{y^6-z^6}{y^2-z^2}\cdot\frac{z^6-x^6}{z^2-x^2}=x^3y^3z^3$

$(2)/(1)\Rightarrow x^2y^2z^2=\frac{x^3+y^3}{x+y}\frac{y^3+z^3}{y+z}\frac{z^3+x^3}{z+x}\geq\frac{x^2+y^2}{2}\frac{y^2+z^2}{2}\frac{z^2+x^2}{2}$

The above inequality is Chebyshev's. Then using AM-GM we immediately get the result.

[quote="keyree10"]Find all non-zero real numbers $ x, y, z$ which satisfy the system of equations:

\[ (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2) \equal{} xyz\]

\[ (x^4 \plus{} x^2y^2 \plus{} y^4)(y^4 \plus{} y^2z^2 \plus{} z^4)(z^4 \plus{} z^2x^2 \plus{} x^4) \equal{} x^3y^3z^3\][/quote]
We have \[ (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2) \equal{} xyz\]   $(*)$
and also \[ (x^4 \plus{} x^2y^2 \plus{} y^4)(y^4 \plus{} y^2z^2 \plus{} z^4)(z^4 \plus{} z^2x^2 \plus{} x^4) \equal{} x^3y^3z^3\] $(**)$
Using the identity $(x^2 \plus{} xy \plus{} y^2)(x^2 \minus{} xy \plus{} y^2)=x^4 \plus{} x^2y^2 \plus{} y^4$ , $(**)$ transforms to \[ (x^2 \minus{} xy \plus{} y^2)(y^2 \minus{} yz \plus{} z^2)(z^2 \minus{} zx \plus{} x^2) \equal{} x^2y^2z^2 \] We see that $LHS$ in $(*)$ is positive so $RHS$ must be positive so either two or none of $x$,$y$, $z$ is negative.
[b]Case 1:[/b]
Two of $x$,$y$,$z$ are negative. $WLOG$ $x=-a$,$y=-b$,$z=c$ where $a,b,c>0$.
Now because of $(**)$ and $a^2 \plus{} ab \plus{} b^2 \geq 3ab$ and $a^2 \minus{} ab \plus{} b^2 \geq ab$ we have \[ a^2b^2c^2\equal{} (a^2 \minus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2) \geq ab \cdot 3bc \cdot 3ca =9a^2b^2c^2  \] impossible
[b]Case 2:[/b]
$x$, $y$ and $z$ are positive.
Now \[ x^2y^2z^2\equal{}  (x^2 \minus{} xy \plus{} y^2)(y^2 \minus{} yz \plus{} z^2)(z^2 \minus{} zx \plus{} x^2) \geq xy \cdot yz \ zx=x^2y^2z^2\] which yields $(x-y)^2=(y-z)^2=(z-x)^2=0$ so $x=y=z=\frac{1}{3}$
So only solution is $x=y=z=\frac{1}{3}$

Here's my solution to this super easy problem :P

[b]Solution[/b]:

I claim that the only possible non- negative real solution to the system of equation is $\boxed{(x,y,z) = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})}$.

[b]Proof:[/b]
$\longrightarrow$ We apply the Cauchy-Schwarz inequality on the second equation to obtain:
$(\prod_{cyc} (x^4 + x^2y^2 + y^4))(1+1+1)$ $\ge \prod_{cyc}( x^2+y^2+xy)^2$$ \iff x^3y^3z^3 \ge \dfrac{x^2y^2z^2}{27} \iff xyz \ge \dfrac{1}{27}......(i)$

Now we apply $Am \ge Gm$ on the second equation to obtain

$\prod_{cyc} (x^4+x^2y^2 + y^4) \ge 27(xyz)^4 \iff xyz \le \dfrac{1}{27}$

Combining $(i)$ and $(ii)$ yields $xyz=\dfrac{1}{27}$

Now equality occurs when $ x^4 = y^4 \iff$$x = y$ or $x = -y$.

Putting two of $x,y,z = -\dfrac{1}{3}$ does not satisfy the equation and hence we conclude that $ x=y=z=\dfrac{1}{3}$


How many numbers can be expressed as the sum of two or more distinct elements of the set $ \{ 0, 1, 2, 4, 8, 16 \}$?

Please tell me if i did this wrong...
The largest possible sum is 31, and the smallest possible sum is 1. So, our answer is (31-1)+1, or 31.

The set of $ {2^0, 2^1, 2^2, ... 2^n}$ will always have $ 2^n \minus{} 1$ possibilities with $ 2$ or more elements of the set.

[quote="ernie"]Because the highest sum would be $ 2^n \minus{} 1$, and the lowest sum would be $ 1$.[/quote]
Ummmm, I'm not sure how that is supposed to explain it.  Can you please explain it in more detail?  Thanks.

Exploring a little, you can see that every number between 1 and 31 can be obtained.  This is because you have an odd number (which is 1) and the even numbers in the set can get you all the even numbers from 2 to 30 and with the 1 you can get all the odd numbers.

I think I didn't explain it that well...anyone else want to give it a shot?   :lol:

We see that the given numbers, expressed in binary, are $$\{0_2, 1_2, 10_2, 100_2, 1000_2, 10000_2\}.$$ Thus we can generate any positive integer less than $100 000_2 = 32$ using two or more of these numbers. So our answer is $\boxed{31}$ numbers.

It is from for the win. At least that is where I found it.

[quote=DP2003]Is that from Alcumus...??? I think you have to say you got if from Alcumus.[/quote]

This is the [b]For the Win![/b] problem dicussion


define  the  constants   $ a, b,  c$ ,    such   that  the  following   for  $ n \equal{} 1, 2,  3,.......$

这个了蛋的问题又跑到这里来发了,还不会阿?呵呵


Hello, I have another problem (sorry  :( )
When I have:
[code]
\begin{itemize}
\item aaa
\subitem zefzf
\end{itemize}
[/code]
And; the text under the subitem is MORE then 1 rule:
it looks like this:
[code]
texttexttext
 * itemone sdfq
      start of subitem .......... this is the end of first line...
 this is the second line , which starts under the "mainitem" en not under the "subitem".

[/code]

si, I want to add an indent before ALL of the following lines after the first subitemline.

Does anyone knows how I can do this ??

ok, it s not exactly the layout you want, but it s nice.  Use an enumerate environnement to create your subitem in the first enumerate, a bit like this

\begin{enumerate}
\item ALLO
\begin{enumerate}
\item   L'analyse vibratoire d'un moteur à combustion interne est très complexe. Pour simplifier le problème, nous avons posé plusieurs hypothèses concernant le comportement thermodynamique du moteur, la cinétique des pièces mobiles et le type de mouvement du système.
\item  L'analyse vibratoire d'un moteur à combustion interne est très complexe. Pour simplifier le problème, nous avons posé plusieurs hypothèses concernant le comportement thermodynamique du moteur, la cinétique des pièces mobiles et le type de mouvement du système.
\end{enumerate}
\item toi

\end{enumerate}

and in bonus, you ll have  automatic (a) (b)....

ledurt

np

[quote]Didn't knew this is possible...[/quote]

before you asked, me neither, so we both learn something ;)

ledurt

well, only if you answer my question about centering table...:P

[quote="plinius"]thanks a lot.
Didn't knew this is possible...[/quote]

Ya, the itemize, enumerate, and description environments may be nested up to four levels. Sorry I didn't see the question sooner.  :blush:


[quote="#H34N1"]Suppose $A$, $B$, and $C$ are three numbers for which $1001C-2002A=4004$ and $1001B+3003A=5005$. The average of the three numbers $A$, $B$, and $C$ is:[/quote]
[hide]add equations:
1001C+1001A+1001B=9009
divide by 1001
c+b+a=9
divide by 3
therefore the arithmetic mean is 3
but i am not sure if you want the mean and mode too....[/hide]

[hide]Add the equations to get

$1001C-2002A+1001B+3003A=9009$
$A+B+C=9$

Thus, the average of $A$, $B$, and $C$ is $3$[/hide]


inequalities
vector
inequalities unsolved

If $x,y,z$  have the same sign then we have an equality.
Otherwise suppose wlog $x<y<0<z$. Let $t=\min(|y|,|z|)$. Consider the new set $(x,y+t,z-t)$, then LHS will decrease and one of the numbers in the new set is 0, it means that new LHS is 0.

We must choose WLOG |z| as the greatest among the numbers given. Dividing by $|z| \neq 0$ we get
\[
\left|\frac x z\right|+\left|\frac y z\right|+1-\left|\frac x z+\frac y z\right|-\left|\frac y z+1\right|-\left|1+\frac x z\right|+\left|\frac x z+\frac y z+1\right|\ge 0
\]
Since $\left|\frac x z\right|,\left|\frac y z\right |\leq 1$ then $  1+\frac x z\ge 0, 1+\frac y z\ge 0$ we can write the relation
\[
\left|\frac x z\right|+\left|\frac y z\right|-\left|\frac x z+\frac y z\right|-\left(\frac x z+\frac y z+1\right)+\left|\frac x z+\frac y z+1\right|\ge 0
\]
which is clearly $\ge 0$

[quote="payman_pm"]x,y,z are real numbers prove that
lxl+lyl+lzl-lx+yl-ly+zl-lz+xl+lx+y+zl >= 0[/quote]

This inequality rewrites as $\left|x\right|+\left|y\right|+\left|z\right|+\left|x+y+z\right|\geq\left|y+z\right|+\left|z+x\right|+\left|x+y\right|$, and holds not only for arbitrary reals x, y, z, but also for arbitrary vectors x, y, z in $\mathbb{R}^n$. It is called the Hlawka inequality. See http://www.mathlinks.ro/Forum/viewtopic.php?t=16600 .

  Darij


[quote="nonie"][quote="B4k4ka"]A plane from Michigan leaves at 6:00 AM Michigan time. It arrives at England at 4:00 PM. England time. If England is 5 hours ahead of Michigan time, how many hours did the plane travel?[/quote]

[hide]6 to 4 equals 10 hours
minus 5 hours equals [b]5 hours[/b][/hide][/quote]
yup you are correct!  :D

[quote="236factorial"][hide]it must've arrived at england at 11am. 11-6=5 hours. [/hide][/quote]
thats right... i always give questions taht are too easy, or too hard... i cant seem to get it just right...

[hide]
6 hours + 4 hours = 10 hours
10 hours - 5 hours = 5 hours[/hide]

6+4=10
10-5=5 :idea: our answer.  Don't feel bad about posting easy problems. That is what this forum is about :)


How many three-digit positive integers are multiples of 4?

$ \left\lfloor\frac {999}{4}\right\rfloor \minus{} \left\lfloor\frac {100}{4}\right\rfloor \plus{} 1 \equal{} 249 \minus{} 25 \plus{} 1 \equal{} \boxed{225}$.

Or an easier way is this: You know the divisibility rule of 4 is: Look at the last 2 digits and see if that is divisible by 4. Thus, the first digit doesn't matter. So for a 3 digit number there is 9 choices (1,2,3,4,5,6,7,8,9) for the first number. Then for the last 2 digits there are 100/4=25. 9x25=225!


Let $ a,b,c,d > 0$,prove that:
(1)
$ {\frac {\left( a \minus{} b \right) \left( a \minus{} c \right) }{\left( a \plus{} b \right) \left( a \plus{} c \right) }} \plus{} {\frac {\left( \minus{} c \plus{} b \right) \left( \minus{} d \plus{} b \right) }{\left( b \plus{} c \right) \left( b \plus{} d \right) }} \plus{} {\frac { \left( \minus{} d \plus{} c \right) \left( \minus{} a \plus{} c \right) }{\left( c \plus{} d \right) \left( a \plus{} c \right) }} \plus{} {\frac {\left( d \minus{} a \right) \left( \minus{} b \plus{} d \right) }{\left( a \plus{} d \right) \left( b \plus{} d \right) }}\geq 0.$

(2)

Let $ a,b,c,d>0$,prove that

$ {\frac { \left( a\minus{}c \right)  \left( a\minus{}d \right) }{ \left( a\plus{}c
 \right)  \left( a\plus{}d \right) }}\plus{}{\frac { \left( a\minus{}d \right)  \left( a\minus{}
b \right) }{ \left( a\plus{}d \right)  \left( a\plus{}b \right) }}\plus{}{\frac {
 \left( b\minus{}a \right)  \left( b\minus{}d \right) }{ \left( a\plus{}b \right)  \left(
b\plus{}d \right) }}\plus{}{\frac { \left( b\minus{}a \right)  \left( b\minus{}c \right) }{
 \left( a\plus{}b \right)  \left( b\plus{}c \right) }}\plus{}{\frac { \left( c\minus{}a
 \right)  \left( c\minus{}b \right) }{ \left( a\plus{}c \right)  \left( b\plus{}c
 \right) }}\plus{}{\frac { \left( c\minus{}b \right)  \left( c\minus{}d \right) }{ \left(
b\plus{}c \right)  \left( c\plus{}d \right) }}\plus{}{\frac { \left( d\minus{}b \right)
 \left( d\minus{}c \right) }{ \left( b\plus{}d \right)  \left( c\plus{}d \right) }}\plus{}{
\frac { \left( d\minus{}c \right)  \left( d\minus{}a \right) }{ \left( c\plus{}d \right)
 \left( a\plus{}d \right) }}\geq 0.$

$ x_1,x_2,x_3,x_4,x_5>0,$,
$ 0\leq {\frac { \left( {\it x_1}\minus{}{\it x_2} \right)  \left( {\it x_1}\minus{}{\it
x_3} \right) }{ \left( {\it x_1}\plus{}{\it x_2} \right)  \left( {\it x_1}\plus{}{\it
x_3} \right) }}\plus{}{\frac { \left( {\it x_2}\minus{}{\it x_3} \right)  \left( {\it
x_2}\minus{}{\it x_4} \right) }{ \left( {\it x_2}\plus{}{\it x_3} \right)  \left( {\it
x_2}\plus{}{\it x_4} \right) }}\plus{}{\frac { \left( {\it x_3}\minus{}{\it x_4} \right)
 \left( {\it x_3}\minus{}{\it x_5} \right) }{ \left( {\it x_3}\plus{}{\it x_4} \right)
 \left( {\it x_3}\plus{}{\it x_5} \right) }}\plus{}{\frac { \left( {\it x_4}\minus{}{\it x_5}
 \right)  \left( {\it x_4}\minus{}{\it x_1} \right) }{ \left( {\it x_4}\plus{}{\it x_5}
 \right)  \left( {\it x_4}\plus{}{\it x_1} \right) }}\plus{}{\frac { \left( {\it x_5}
\minus{}{\it x_1} \right)  \left( {\it x_5}\minus{}{\it x_2} \right) }{ \left( {\it x_5}
\plus{}{\it x_1} \right)  \left( {\it x_5}\plus{}{\it x_2} \right) }}
.$

[quote="xzlbq"]Let $ a,b,c,d > 0$,prove that:
$ {\frac {\left( a \minus{} b \right) \left( a \minus{} c \right) }{\left( a \plus{} b \right) \left( a \plus{} c \right) }} \plus{} {\frac {\left( \minus{} c \plus{} b \right) \left( \minus{} d \plus{} b \right) }{\left( b \plus{} c \right) \left( b \plus{} d \right) }} \plus{} {\frac {\left( \minus{} d \plus{} c \right) \left( \minus{} a \plus{} c \right) }{\left( c \plus{} d \right) \left( a \plus{} c \right) }} \plus{} {\frac {\left( d \minus{} a \right) \left( \minus{} b \plus{} d \right) }{\left( a \plus{} d \right) \left( b \plus{} d \right) }}\geq 0.$
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-20-b-q-liu.html


It should be, given that registration ended three days ago and [url=http://mathcounts.org/Page.aspx?pid=355&srcid=315]this link[/url] exists.

That's an interesting question. I was just wondering that......

i'm pretty sure u can, if they are giving a sample state problem on their site :wink:

Was anyone else told to open their books before starting the test because the administraters wanted you to write in the O was the center of the circle? It gave people a opputunity to go through the problems and cheat.  :o

naw they put on powerpoint when they started the test like once you open the book....

No, there was a stapled page that had that clarified.

Yes, in my competition, they did do this. They told us to open our booklets and write that in. Then we quickly had to close our books.

[quote="diwavijay"]http://mathcounts.org/Document.Doc?id=341[/quote]

I believe that this was discussed in at least 2 other topics...


probability
number theory
least common multiple

What is the probability that a 3-digit number less than 900 is divisible by $3!$ or $3^2$ ?  Express your answer as a common fraction.

[hide]100-899 is 800 numbers. Numbers divisible by 6 in that range is from 102 - 894. To find the amount that this counts for 6x + 96 = 894 so x = 133. There are 133 numbers divisible by 6 in that range. Do the same thing for 9 and it comes out that 88 are divisible by 9 in that range. A lot of numbers overlap but we only need to subtract the multiples of the LCM of 9 and 6 which is 18. The amount of numbers divisible by 18 in that range is 44. 88 + 133 - 44 is 177 divisible by 9 or 6. The probability of picking one of those numbers is 177/800.[/hide]

[quote="ln(dx/dy)"][hide="My Solution"]149 numbers divisible by 6. 99 numbers divisible by 9. 149+99=248.  16 numbers divisible by 6 and 9. 248-16=232.[hide="My Answer"][b]232/899[/b][/hide][/hide][/quote]

This would work if it was all numbers but the problem says only 3-digit numbers.

[hide]There are 111 multiples of 9 smaller than 1000, and 11 smaller than 100, for 100 multiples of 9

There are 166 multiples of 6 smaller than 1000, and 16 smaller than 100, for 150 multiples of 6

There are 55 multiples of both 6 and 9 smaller than 1000, and 6 smaller than 100, for 49 double-counted numbers

So there are 201 numbers that fit, and 900 possibilities, so the probability is 201/900.[/hide]

Billy

Has to be less than 900 solafide, not 1000

tarquin and braves have it

[hide]$\frac{177}{800}$[/hide]


So.... Anyone know the results (top 10)? I wasn't there for the awards...

Individual
1 - Santosh 231
2 - Henry 230
3 - Eric 225
4 - Edward Park 223
5 - Tim Wu 212
6 - Boru 211
7 - Allen Park 204
8 - Claus 198
9 - Oliver 195
10 - Robert Grosse 188

Team
1 - Walton 910
2 - Chamblee 724
3 - Parkview 658
4 - Brookwood 606
5 - Brookstone 600

Did everyone just love the last four ciphering questions? :P

Anyways, nice job Santhosh and Walton!

And nice job Alpharetta and Northview. The tournament ran on time. The test was good. And everything seemed to go the way it was supposed to. I haven't checked to see if Maja killed anyone yet.  :D

Can you guys post the problems of individual round? I googled it but didn't get any result. :(

I would like to see the problems at the tourney...especially since the Norcross Math team decided not to go...or rather we had no clue this was even happening...I guess we weren't invited since it was in Fulton.

[quote="Benjamin Hu"]Top 10 are somewhat as expected.
Too bad I couldn't come...SAT Prep ruins certain things (especially things like these).
I wonder how well those three power question graders graded the power questions I created...[/quote]

Your power questions made grading over 9000 times harder cuz, last time I checked, we were supposed to do single-answer problems with justification as part of the solution, not full-blown proof questions. :mad:

1. It's up to the tournament hosts to post the questions if they wish. No one else should post these. Discussion of a question or two should be fine.

2. All Georgia tournaments are posted on the GCTM web site. Take a look here to see what is coming up. The next tournament is the Cobb County Invitational on October 25th.
[url]http://www.gctm.org/Default.aspx?pageId=130271[/url]

I'll be at Kennesaw Mountain, together with at least two other Soli Deo Gloria folks! (none of whom are from out-of-state...) Will any Vestavians come and try to beat us this year?


Hi, i was just wondering if someone could help me with the following question. I'm  actually totally stuck on it and would really appreciate the help - thanks in advance.

I've posted the question on my website - thanks guys.

http://members.optusnet.com.au/gujju_style/maths3.JPG

Can someone help please? Or have i posted this in the wrong forum/room?

Equivalence relations are reflexive, symmetric and transitive. Show that R is not transitive by finding a,b,c such that aRb and bRc but not aRc.
[hide="from the given hint"]Note that 1R41 and 1R7. Since R is symmetric, 41R1. However, 41R7 is false.[/hide]


inequalities
linear algebra
matrix
inequalities proposed

Hello,

this may be somewhat of a weird question, but with the municipal elections going on right now in Flanders, I was wondering how I could prove this. :oops:   Do tell me if you don't consider this algebra.

Let $a_{1},\ldots,a_{n}$ be a set of integers $>0$, and let $a_{1}>a_{2}\ldots> a_{n}$

Assume that $a_{1}> \frac{a_{2}+\ldots+a_{n}}{2}$
Let $S>0$ be an integer

Now make an infinite matrix :

$\frac{a_{1}}{2}\ldots \frac{a_{n}}{2}$
$\frac{a_{1}}{3}\ldots \frac{a_{n}}{3}$
$\frac{a_{1}}{4}\ldots \frac{a_{n}}{4}$
......

Now find the $S$ biggest elements.  Give each "party" $a_{i}$ for every element in its own column.  In case there is a dispute about seats at the end, give it to the largest party.

Prove that the number of seats for party 1 is strictly more than $\frac{S}{2}$

Thank you very much!

Let $a_{1}= 1002, a_{2}= 1001, a_{0}=1000$. Then we have $1002 > \frac{1001+1000}{2}$. But taking $S=3$, we see that $a_{1}$ only gets 1 seat in the parliament.

So I think your conclusion is wrong in general. The more all parties have approximately equal scores, the more imperiali gets closer to the D'Hondt system or to straightforward division (is that the right word?).

It's (conceptually) like when working with QM and AM: when all variables approach eachother, QM approaches AM, while in general $QM>AM$.

I think for $n=2$, your claim is true (d'uh), and for all others where $a_{1}<a_{2}+...+a_{n}$, one can always find a sufficiently large $S$ such that it becomes false. :)


CM trong 2k-1 so nguyen bat ky luon chon duoc k so co tong chia het cho k

Hmm... I don't know Vietnamese, but this will probably help: [hide]Consider when k is prime first.[/hide]


function
real analysis
real analysis unsolved

Suppose that $ f$ is a function with the property that

lim $ f(x)$ exists for all real $ c$
$ x \to c$

Define a new function $ g$ this way:

$ g(x) \equal{} lim f(t)$
$ t\to x$

Prove that  $ g$ is continuous everywhere, (i.e. continuous at every real number).

it's just a joke. :lol:
Give a definition of continious function and everything will be clear.


inequalities
calculus
integration
complex analysis
complex analysis unsolved

Suppose f(z) is analytic in the unit open disk and f(z)< c/(1-|z|) for all z in the unit disk and C>0, then show that |f'(z)|<(4*c)/(1-|z|^2)

[hide="hint"]try to parametrize a circle $ \gamma$ at center $ z$ with a suitable radius [b]depending on[/b] $ z$, such that $ \gamma \subset \mathbb{D}$  and then work with Cauchy integral to obtain the result. [/hide]


calculus
integration
logarithms
trigonometry
calculus computations

I don't believe your first step (the substitution changes the denominator), and you haven't finished the problem anyway.


Given a cube. How many pairs of skew lines can be formed if we connect any two of the 8 vertices?

[quote="Allan Z"]i got 186, if this is the correct answer, i will post up how i get my answer[/quote]
A little more.

you mean less? 'cause I forget some of them intersect

I mean that your answer is a little more than the correct one.

There are 8 vertices on a cube, therefore, you can draw ${8\choose2}=28$ different lines. Therefore, there are ${28\choose2}=378$ pairs of lines.

The relationship between a pair of line is:
1. Parallel
2. Intersected
3. Skew Lines

Let's look at the first case: parallel lines.
It's easy to observed, parallel lines can only find on the surface.
If they are edges of a cube, there are $3*{4\choose2}=18$ pairs
If they are diagonals of a face, there are $3*2=6$ pairs.
Therefore, there are $18+6=24$ pairs of parallel lines.

Let's look at the second case: Intersected lines.
If they intersect at a vertex, there are $8*{7\choose2}=168$ pairs.
If they intersect on a face(not including the vertex), the two lines must be diagonals of a face, there are 6 pairs.
If they intersect inside the cube, the two lines are space diagonals. There are 4 space diagonals, therefore, there are ${4\choose2}=6$ pairs.
Therefore, there are $168+6+6=180$ pairs of intersected lines.

Numbers of pairs of Skew Lines= $378-24-180=\boxed{174}$ pairs.

[quote="SplashD"]isn't there 4 space diagonals?[/quote]
[quote="Allan Z"]There are 4 space diagonals, therefore, there are ${4\choose2}=6$ pairs.[/quote]
 :wink:  Were you confusing it with the 6 or something?

oh.. i edited it.. i wrote there are 3 space diagonals. :wallbash:

I don't really know, if you count one by one, it may be too much, i guess.

[quote="SplashD"]Is there another way to do this without complementary counting?[/quote]
There is. First calculat the sides, then diagonals of faces, then space diagonals.


conics
ellipse
ratio
geometry solved
geometry

Though measurement I have found by measuring that for an ellipse of semi-major axis (a) and an ellipse of semi-major axis (b) there is a constant that is, the circumference of the ellipse divided by the sum of the axis's, which is aproximately 3.19

I do not know if my hypothesis is correct or not so I am asking you all what you think of it.

Take a=b. You get a circle, so your 'constant' must be $\pi$.
I don't think it is a constant.

I thought about this aswell but a circle is not an ellipse and it does not have a semi-major axis or a semi-minor axis.  Furthermore what defines an ellipse is the sum of the distance of two fixed points is a constant.  A circle only has one point to where the sum of the distance is a constant(the center).
So I probably should've stated that (a) can never equal (b) which I thought was implied

Basicly, if the formula would hold for every ellipse, then it would hold for circles, too, since a circle is a [i]limiting case[/i] of an ellipse (imagine the two focii coming nearer and nearer to each other - the ellipse becomes more and more like a circle, and in the limit it is a circle). Besides, most of people use to understand the circle as an ellipse.

Anyway, there is no "elliptical constant" as you suggest it, since it would be equal to $\pi$ in the limiting case when the ellipse is a circle, and it would be equal to 4 in the limiting case when the ellipse is just the segment between its two foci (passed twice), so the "elliptical constant" is not constant. You could also see this without referring to limiting cases if you would make some more precise measurings.

The question of measuring the circumference of an ellipse is a quite difficult one. See the formulas (61)-(63) from http://mathworld.wolfram.com/Ellipse.html .

  darij

sory about this post, it seems I just spouted out crazy ideas  because I was unable to find a circumfrencial equation for an ellipse without using elliptical integrals.

[quote="darij grinberg"]…a circle is a [i]limiting case[/i] of an ellipse (imagine the two focii coming nearer and nearer to each other - the ellipse becomes more and more like a circle, and in the limit it is a circle).[/quote]An ellipse is also a straight line when the foci separate to the infinite; as, it is a circle when the foci converge to the infinitesimal.

[quote="darij grinberg"]Anyway, there is no "elliptical constant"…[/quote]There is an Elliptical Constant.  The Elliptical Constant is such that the salient, intrinsic structural parts for [color=blue][i][b]any[/b][/i][/color] ellipse have the same algebraic relationship to one another.

It is the Elliptical Constant that demonstrates the relationship of the ellipse to the Golden Ratio and the Fibonacci sequence.


topology
advanced fields
advanced fields unsolved

suppose (E,p) is a covering space of B, and B is simple connected, then p: E to B is homeomorphism.

the conclution is obviouse,but i donot know how to prove.

you have to assume that $ E$ is nonemtpy.

$ B$ is path-connected, thus $ p$ is surjective (otherwise, lift a path from $ x \in im(p)$ to $ y \in B \backslash im(p)$ to optain a contradiction). since $ p$ is a local homeomorphism, it is open and continuous. but for every covering $ p$ and for every $ e \in E$, the index of $ \pi_1(E,e)$ in $ \pi_1(B,p(e))$ coincides with $ |p^{ \minus{} 1}(\{b\})|$. thus if $ B$ is simply connected, this is $ 1$ and $ p$ is injective.


Question 1: We know that many people are IPhO winners, but now I'm wondering what their careers be after they got IPhO medals, did any of them became famous physicists? Did they get any famous physics prizes? Such as Nobel Prize.

Question 2: As I know, the IPhO committee has published a book about IPhO around 1989 or 1990, I only know its Chinese name, so I don't know its exact English name, I can give a name similar: "International Physics Competition of Olympiads(Volume 1)" or " International Physics Olympic(Volume 1)", But I only have seen its Volume 1, after that, it seems that China has never translated volume2, volume3......So I'm wondering that if books volume2, volume3...... has been published or not.

for question one: I think, for Chinese students, none of them has become famous yet. But they are still young since China did not participate the early contests.

[quote="Alexander Lenin"]
Question 2: As I know, the IPhO committee has published a book about IPhO around 1989 or 1990, I only know its Chinese name, so I don't know its exact English name, I can give a name similar: "International Physics Competition of Olympiads(Volume 1)" or " International Physics Olympic(Volume 1)", But I only have seen its Volume 1, after that, it seems that China has never translated volume2, volume3......So I'm wondering that if books volume2, volume3...... has been published or not.[/quote]

Well, all the problems can be found [i]online[/i]
http://www.jyu.fi/tdk/kastdk/olympiads/
and I have Indian books with the olympiads in them, but I don't know about [u]volumes[/u]
Its probably only the Chinese books that are compiled in Volumes


Solve this multiplication problem (each letter is a unique digit from 0-9):

     OIL
[u]* SUN[/u]
MIAMI

Note: I created this problem myself, so there may be more than one solution.  However, I do know that there is at least one solution.

After one boring afternoon of calculator clicks, I got the(an) answer. I have no idea how you can get it(one) without resorting to brute force, guesswork, or calculator/computer.  :surrender:

[hide="The answer"]
oil = 128
sun = 409
miami = 52352
[/hide]

I haven't finished checking any MIAMI's larger that that, so this answer may not be unique.

Your solution is definitely correct, though it is not the one I came up with.

[hide]I came up with this by using multiples of 1001 for "miami", meaning the product would always be in the form ab0ab.  My solution is 168*572=96096.[/hide]

Well, that's almost the basis for my method too, really, except I added $ c \times 100$. I programmed my TI to do half the work (iterate all miami's, remove the repeats, factor) and I did the rest (combine the factors/remove absurd ones). *whew*

Anybody know any other ways? Any other answers?


calculus derivative integration algebra function domain polynomial

Can someone show me how $zeta^'(0)$ gets the value
-1/2 ln 2pi?
The mathworld entry isn't clear, what's the formula?

You guys feel like discussing this?  I'll dig up Edward's book on the subject and post later (when I figure it out)  how it's commonly done but what about just differentiating the defining equation for Zeta?  Wouldn't that work:

$\zeta(s)=\Gamma(1-s)\frac{1}{2\pi i}\int_{\multimap}\frac{z^{s-1}}{e^{-z}-1}dz$

the contour being about the negative real axis (looks like the symbol I use right?)

so:

$\frac{d}{ds}\zeta(s)\&=\frac{1}{2\pi i}\left\{\int_{0}^{\infty}t^{-s}e^{-t}dt \int_{\multimap}\frac{Log(z)z^{s-1}}{e^{-z}-1}dz-\int_{\multimap}\frac{z^{s-1}}{e^{-z}-1}dz\int_{0}^{\infty}ln(t)t^{-s}e^{-t}dt\right\}$

where I just used the chain rule and Leibnitz rule.  However I'm not absolutely sure about applying Leibnitz rule to a contour integral.  I think it's ok since zeta is perfectly continuous and differentialbe at s=0.  I could plug it all into Mathematica and check it numerically.  Might take a while though.

I believe differentiation across a contour integral is justified as long as the partial and it's antiderivative are analytic throughout the domain of integration.  This is met I believe for the zeta function.  I've drawn the zeta contour below.  Letting:

$z=re^{-\pi i}\;\text{on}\;C_{1}$

$z=e^{i\theta}\;\text{on}\; C_{2}$

$z=re^{\pi i}\;\text{on}\; C_{3}$

I calculate:

$\int_{\multimap}\frac{\partial}{\partial s}\left(\frac{z^{s-1}}{e^{-z}-1}\right)dz=I1+I2+I3$

with:

$I1=\int_{1}^{\infty}\frac{(Log[r]-\pi i)Exp[(s-1)(Log[r]-\pi i)]}{e^{r}-1}dr$

$I2=\int_{-\pi}^{\pi}\frac{-\theta e^{si\theta}}{Exp[-e^{i\theta}]-1}d\theta$

$I3=-\int_{1}^{\infty}\frac{(Log[r]+\pi i)Exp[(s-1)(Log[r]+\pi i)]}{e^{r}-1}dr$

The other integral is done similiarly so that I can show:

$\frac{d}{ds}\zeta(s)=Z(s)=\frac{1}{2\pi i}\left\{\Gamma(1-s)\int_{\multimap}\frac{\partial}{\partial s}\left(\frac{z^{s-1}}{e^{-z}-1}\right)dz+\frac{\partial}{\partial s}\Gamma(1-s)\int_{\multimap}\frac{z^{s-1}}{e^{-z}-1}dz \right\}$

and that:

$\zeta'(0)=Z(0)=-\frac{1}{2}ln(2\pi)$

Here's a closed-form solution based on my interpretation of Edwards, et. al.

First $\zeta[0]$ is needed and illustrates an  interesting phenomenon:

$\int_\multimap \frac{z^{n-1}}{e^{-z}-1}dz= \oint \frac{z^{n-1}}{e^{-z}-1}dz$

That is, when s is an integer, the zeta contour reduces to a  closed contour about the origin because the integral becomes single-valued (just a rational expression of a polynomial and an
exponential function in that case).

So:
\begin{eqnarray*}\zeta[0]&=&\Gamma[1]\frac{1}{2\pi i}\oint\frac{dz}{z(e^{-z}-1)}\\ &=& \Gamma[1]\frac{1}{2\pi i}2\pi i\mathop\text{Res}_{z\;= \;0}\left\{\frac{1}{z(e^{-z}-1)}\right\}\\ &=&\mathop\text{Res}_{z\;=\;0}\left\{\frac{1}{z(e^{-z}-1)}\right\}\\ &=&-1/2 \end{eqnarray*}

I now use the functional equation (see Ingham for derivation):

$\zeta[s]=\Gamma[1-s]2(2\pi)^{s-1}Sin[\pi s/2]\zeta[1-s]$

and combine it with the integral equation to obtain:

$\frac{\zeta[s]}{\Gamma[1-s]}=\frac{1}{2\pi i}\int_{\multimap}\frac{z^{s-1}}{e^{-z}-1}dz= 2(2\pi)^{s-1}Sin[\pi s/2]\zeta[1-s]$

Differentiating with respect to s:

$\frac{1}{2\pi i}\int_{\multimap}\frac{Log[z]z^{s-1}}{e^{-z}-1}dz= \frac{d}{ds}\left\{2(2\pi)^{s-1}Sin[\pi s/2]\zeta[1-s]\right\}$

I now wish to show that this derivative is zero at s=1.  Substituting s=1 and integrating the left side as previously described:

$\frac{1}{2\pi i}\int_{\multimap}\frac{Log[z]}{e^{-z}-1}dz =\frac{1}{2\pi i}\left\{ \int_{\epsilon}^{\infty}\frac{ln(r)-\pi i}{e^{r}-1}dr+\int_{|z|=\epsilon}\frac{Log[z]}{e^{-z}-1}dz-\int_{\epsilon}^{\infty}\frac{ln(r)+\pi i}{e^{r}-1}dr\right\}$

where this time I'm using $\epsilon$ as the diameter of the central contour (the integral is independent of path so I can use any diameter less than the distance to the first pole).

Simplifying:

\begin{eqnarray*}\frac{d}{ds}\frac{\zeta[s]}{\Gamma[1-s]}&=&\frac{1}{2\pi i}\left\{\int_{\epsilon}^{\infty}\frac{-2\pi i}{e^{r}-1}dr+\int_{-\pi}^{\pi}\frac{(ln(\epsilon)+i\theta)i\epsilon e^{i\theta}}{e^{-z}-1}d\theta\right\}\\ &=&\frac{1}{2\pi i}\left\{-\int_{\epsilon}^{\infty}\frac{2 \pi i}{e^{r}-1}dr+iln(\epsilon)\int_{-\pi}^{\pi}\frac{\epsilon e^{i\theta}}{e^{-z}-1}d\theta-\epsilon \int_{-\pi}^{\pi}\frac{\theta e^{i\theta}}{e^{-z}-1}d\theta\right\}\end{eqnarray*}

I mixed $\theta$ and z in the same integrals to make it easier to manage.  That's Ok (Edwards does also) as long as I keep everything straight. I now wish to take the limit as $\epsilon\to 0$.  In that case, the last integral goes to zero.
I'm left with:

$\mathop\lim_{\epsilon\to 0}\left\{-\int_{\epsilon}^{\infty}\frac{dr}{e^{r}-1}dr+\frac{ln(\epsilon)}{2\pi}\int_{-\pi}^{\pi}\frac{\epsilon e^{i\theta}}{e^{-z}-1}d\theta\right\}$

I convert the second integral back to z and obtain:

$\mathop\lim_{\epsilon\to 0}\left\{-\int_{\epsilon}^{\infty}\frac{dr}{e^{r}-1}dr-\frac{ln(\epsilon)}{2\pi}\oint\frac{i}{e^{-z}-1}dz\right\}$

which I then evaluate via residues:

$-\frac{ln(\epsilon)}{2\pi}\oint\frac{i}{e^{-z}-1}dz= \frac{ln(\epsilon)}{2\pi}\mathop\text{Res}_{z\;=\;0}\left\{\frac{1}{e^{-z}-1}\right\}=-ln(\epsilon)$

I now have:

$\frac{d}{ds}\frac{\zeta[s]}{\Gamma[1-s]}= \mathop\lim_{\epsilon\to 0}\left\{-\int_{\epsilon}^{\infty}\frac{dr}{e^{r}-1}dr-ln(\epsilon)\right\}$

Expressing the first integrand as a power series and integrating term by term, I have:

$\frac{d}{ds}\frac{\zeta[s]}{\Gamma[1-s]}= \mathop\lim_{\epsilon\to 0}\left\{ln(\epsilon)+ln(1-\epsilon/2+\epsilon^{3}/6+\cdots)-ln(\epsilon)\right\}=0$

Therefore:

$\frac{d}{ds}\left\{2(2\pi)^{s-1}Sin[\pi s/2]\zeta[1-s]\right\}$

is zero at s=1.  But this means the logarithmic derivative:

$\frac{ld}{ds}\left\{2(2\pi)^{s-1}Sin[\pi s/2]\zeta[1-s]\right\}= ln(2\pi)-\frac{\zeta'[1-s]}{\zeta[1-s]}+\pi/2\frac{Cos[\pi s/2]}{Sin[\pi s/2]}$

is also zero at s=1 and therefore:

$\frac{\zeta'[0]}{\zeta[0]}=ln(2\pi)$

Now, using the value $\zeta[0]=-1/2$ I get finally:

$\zeta'[0]=-1/2\;ln(2\pi)$


Let:

$ A\equal{}\begin{pmatrix}
  a & b \\
  c & d \\
  \end{pmatrix}$

Compute $ A^{\minus{}1}$ assuming that $ ad\minus{}bc \neq 0$

$ A^{ \minus{} 1}$=$ 1/$(ad-bc)$ \cdot\begin{pmatrix}d & \minus{} b \\
\minus{} c & a \\
\end{pmatrix}$, where we replaced each entry by its determinant, changed its sign relative to its position, transposed it, and multiplied by the reciprocal of its determinant.

Edits for latex failures.

Assume this problem was in the section BEFORE determinants.

Let  [u]C[/u]=$ \begin{pmatrix}e & f \\
g & h \\
\end{pmatrix}$.  As an inverse of [u]A[/u], we know that their product is equal to the matrix equivalent of one: the identity matrix [u]I[/u]$ \equal{} \begin{pmatrix}1 & 0 \\
0 & 1 \\
\end{pmatrix}$.  We can expand this product and set entries equal to solve for the matrix I got in my previous post.

Its easy enough to see the concept, but I'm asking for a solution.

How would you go from:

$ \begin{cases} ae \plus{} bg \equal{} 1 \\
cd \plus{} dg \equal{} 0 \end{cases}$

and

$ \begin{cases} af \plus{} bh \equal{} 0 \\
cf \plus{} dh \equal{} 1 \end{cases}$

to $ e \equal{} d$, $ f \equal{} \minus{} b$, $ g \equal{} \minus{} c$, and $ h \equal{} a$

EDIT:

After solving I get:

$ e \equal{} \frac {d}{ad \minus{} bc} \\
f \equal{} \minus{}\frac {b}{ad \minus{} bc} \\
g \equal{} \minus{}\frac {c}{ad \minus{} bc} \\
h \equal{} \frac {a}{ad \minus{} bc}$

How would I get rid of the denominator? I see that this can be achieved by multiplying the posted answer above by the scalar $ \frac{1}{ad\minus{}bc}$, but then they won't be inverse anymore.

Look again, he included a factor of $ \frac{1}{ad\minus{}bc}$ outside the matrix.

Oh, my bad. I thought the period separated the first expression from the second.


The figure in the dark black cloak stood motionless under the dim torch. Peeking from behind the brick corner, the two guards stood asleep, protecting the latched door in between the two of them. He stuck over, carefully removed a set of keys on one of the guard's belts, and pushed the door open. [i]What great guards the king hires.[/i]

In there, he shuffled through the files on top of the king's desk. Frustrated, he swept his hand in anger. When all the papers moved aside, there was a gold bar, with kill codes engraved in it. Everyone of royalty had his or her name on it - and there he saw King Daemon's name. [i]I have power. I can get anything I need.[/i] he chuckled, as he dissolved into the shadows.

---

King Daemon tore his hair out in agony. His list was evidently seen by someone. "Guards! Call a meeting in 5 minutes. This is an urgent issue!"

Five minutes later, 16 people of highest rank were seated around the table. "One of us has intruded into my study. I don't care what happens. Find him and kill him."

---

Gameplay:

Each of you will receive a role. In your role PM, it will say what your action is. During each turn (probably separated by 1-2 day intervals), you post ONCE in the thread a user who is in this game.

Say you are a mafia (your action would then be to kill someone). In thread, you would put "A" if you want to kill player A. However, it is recommended you put "A or B (or C)". I will choose A or B (or C) to randomly kill. That way, if A dies from your kill, and someone else also put A in their action list, you will draw a bit of suspicion away from you.

Rules (Stolen from Mafia [hide=":D"]
Yay you read the rules. Rules for joining:

1) Do NOT PM me for sign ups. Post in thread.
2) Continue a quote stack of the cumulative players rather than simply saying "I join".
4) Do not remind someone who obviously has not read the rules to read the rules again. We can think of something eventually for you people who read the rules.
5) Place your grade and gender when signing up.
6) I will need a substitute mod for a week during Thanksgiving break. Comment in the link below to sign up.

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=291377
[/hide]):
[list]
    [*] Actions must be in bold. If you do not bold your action(s), it will not be counted.
    [*] The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
    [*] Once your death scene has been posted, you’re dead. Stop typing. The only thing you can type is a "bah" post.
    [*] Don’t edit/delete previously submitted posts. If you must, post again with EBWOP (Edit by way of post).
    [*] Don’t quote any PMs from me. Neither true, nor false. Also, click on the only smiley in this post to confirm rule reading.
    [*] If you do not submit an action, you will do nothing.
    [*] Treat this game as a commitment. Be considerate – don’t leave us hanging.
    [*] Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.[/list]

There are 17 players, with no more spots open, most of you who will have different roles.

1. cf249 (M)
2. zapi2007 (M)
3. westiepaw (F)
4. 15!!! (M)
5. jjfun1 (F)
6. spaceguy524 (M)
7. andrewjiang97 (M)
8. james4l (M)
9. meewhee009 (F)
10. tcjy (assumed male*)
11. Eggylv.999 (M)
12. tennis123 (F)
13. royalflower (F)
14. isabella2296 (F)
15. ernie (M)
16. Yoshi (M)
17. pi31415926 (M)
17. pi31415926 (M)

* due to a male majority on AoPS

[u]signup list[/u]
15!!!

Hate people who don't read rules!
I'm 9th grade male.

[quote="15!!!"][u]signup list[/u]
15!!!
jjfun1[/quote]
Join! [hide=" :D "]What's a quote stack?[/hide]
I'm a 7th grade female.

[quote="jjfun1"][quote="15!!!"][u]signup list[/u]
15!!!
jjfun1
james4l[/quote]
Join! [hide=" :D "]What's a quote stack?[/hide]
I'm a 7th grade female.[/quote]
8th/M, join

[u]signups[/u]
15!!!
jjfun1
james4l
spaceguy524

I'm 7th grade male.

Hey! You forgot me![quote="zapi2007"]quote stacks get ugly.
and not to mention redundant.

signups
15!!!
jjfun1
james4l
spaceguy524
zapi2007
andrewjjiang97

I choose conceal my identity.

If you really need it, pm me.
More people should sign up![/quote]

But I'm not there... :( [quote="jjfun1"][quote="andrewjjiang97"]Hey! You forgot me![quote="zapi2007"]quote stacks get ugly.
and not to mention redundant.

signups
15!!!
jjfun1
james4l
spaceguy524
zapi2007

I choose conceal my identity.

If you really need it, pm me.
More people should sign up![/quote][/quote]
No, he didn't. :wink:[/quote]

[quote="andrewjjiang97"]But I'm not there... :( [/quote][quote="jjfun1"][quote="andrewjjiang97"]Hey! You forgot me![quote="zapi2007"]quote stacks get ugly.
and not to mention redundant.

signups
15!!!
jjfun1
james4l
spaceguy524
zapi2007

I choose conceal my identity.

If you really need it, pm me.
More people should sign up![/quote][/quote]
No, he didn't. :wink:[/quote][quote="andrewjjiang97"][/quote]
That's the point...

[quote="westiepaw"]@15!!!: How is Yoshi's innocence proven?  And why are you targeting like all the other players?[/quote]
first, 'yesterday', yoshi targetted only zapi and and was the only one to target zapi.
Nothing happened to zapi, so yoshi doesn't have ability to kill others.

for second question, it's because i hate my role.
i used to be a guard, but i got switched to a normal citizen.

Except now james4l and spaceguy524 have been eliminated. Few options.

(a) Start over completely (or end this game and have someone create a new thread)
(b) Redo night actions by rearranging the roles and giving a different  kill code
(c) Proceed as it is

ill join? is that okay, wait, maybe its too late. srry.

I'm in favor for starting over, from scratch. That way we can accommodate new people.

I have an interest in modding this sort of game.

[quote="dragon96"]Wait... according to the roles list, james4l is not dead. Neither is spaceguy524.

Anyone mind if we start over?[/quote]
Oh, crap. I knew I messed something up.
Maybe you could give james4l and spaceguy524 different roles?
And I thought I PM'd you the people that were supposed to die.
Bleh.

I also think that starting over is probably a good idea.  It's a little messed up now...

I'll probably start one in a few days, but currently, I'm too lazy to make a thread. Anyone is free to mod. I'll go ahead and lock.


Find all integers $ n\geq 3$ such that for any two positive integers $ m,r<n\minus{}1$, there exist $ m$ distinct elements of the set $ \{1,2,...,n\minus{}1\}$ whose sum is congruent to $ r$ modulo $ n$.

Maybe it helps:
some small cases for $ n$
$ n \equal{} 3$
Only case: $ (n,m,r) \equal{} (3,1,1)$ Ok
$ n \equal{} 4$
no solution for $ (n,m,r) \equal{} (4,2,2)$
$ n \equal{} 5$
solution for all $ 9 \equal{} (n \minus{} 2)^2$ cases
$ n \equal{} 6$
no solution for $ (n,m,r) \equal{} (6,4,3)$

It is imposible for $ n\equal{}2k$ because there is no solution for $ (n,m,r) \equal{} (2k,2k\minus{}2,k)$

[quote="mszew"]Maybe it helps:
some small cases for $ n$
$ n \equal{} 3$
Only case: $ (n,m,r) \equal{} (3,1,1)$ Ok
$ n \equal{} 4$
no solution for $ (n,m,r) \equal{} (4,2,2)$
$ n \equal{} 5$
solution for all $ 9 \equal{} (n \minus{} 2)^2$ cases
$ n \equal{} 6$
no solution for $ (n,m,r) \equal{} (6,4,3)$

It is imposible for $ n \equal{} 2k$ because there is no solution for $ (n,m,r) \equal{} (2k,2k \minus{} 2,k)$[/quote]

Right. I think the answer should be $ n$ odd. But I can't prove it yet  :P


Maybe it's too easy for this forum. But a mod can move this if necessary.

Two circles with radius $ 8$ and $ 10$ are orthogonal. Find the lenght of the common chord.

Let $ O_1$ and $ O_2$ the centers of the circles with radis $ R_1$ and $ R_2$.
Let $ B$ and $ A$ be the intersection points.

$ O_1O_2\equal{}\sqrt{R_1^2\plus{}R_2^2}$

$ (AO_2BO_1)\equal{}\frac{AB*O_1O_2}{2}\equal{}R_1*R_2 \Rightarrow AB\equal{}\frac{2R_1*R_2}{\sqrt{R_1^2\plus{}R_2^2}}$

 :)


Apparently. I don't keep track of all this stuff so idk

Impossible, a Korean who doesn't keep track of Starcraft???

[quote="Ubemaya"]He's freaking Japanese...[/quote]
Who's Japanese? xscapezaer posted above me...
"Kim" is a pretty darn Korean name.

[quote="not_trig"][quote="dgs01"][quote="not_trig"]a) old
b) so what? he has 2 silvers already
c) it was expected after fantasy somehow pwned ggplay with mech, because Stork has very good PvT
d) stork is out of msl after losing to... kal. did you know one of kal's akas is GOOJILLA? how does that make stork feel?
e) good job to stork for not being the king of 2nd place anymore, though[/quote]

a) It never gets old... Every sc fan will agree no one deserved a title as much as Stork
b) Total Starleague + Korea Wcg silvers will add up to like... 7?
c) It was Stork vs Silver curse
d) Stork lost 2-3.. In starcraft where the skill gap between the top progamers is so small, lack of preparation can lose games.  Apparently, Stork put more priority on OSl finals than MSL ro8.
e) STOOOOOOORRRRRKKK[/quote]

a) still old :P (as in why didn't he post this earlier)
b) 2 silvers as in 2 osl silvers
c) sure, ok... but like in terms of actual skill Stork PvT > Fantasy TvP by quite a bit
d) dude stork should learn to position goons better?

and finally stork's PvZ is still unproven to be S class. we'll see...[/quote]

a) I was at a karate seminar attempting to get un-fat (and learn how to beat up meggy-meg's other suitors).
b) Stork has failed to keep the silver mining industry alive (rofl @ TL picture).
c) Also yeah probably, game 4 was pretty wtf though.
d) Dude Stork probably forgot what PvP was like and was trying to play PvT... like in his ro16 game against ForGG, he forgot that he couldn't attack-move through Terran rofl.
e) STTOOOOOOOOOOOOOOORRRRRRRRRRRRRRRRRRRRRRRRKKKKKKKKKKKKKKKKKKK

[quote="Scrambled"]GUUUESSSSS WHO LOST TO BACKHO IN THE GSL??????[/quote]
...
fail... utter bs

[quote="Ubemaya"]GUESS WHO WON OSL?![/quote]
GUESS WHO GOT DESTROYED BY MVP AND IS ON THE WAY TO LEADING KHAN TO ITS LOWEST PROLEAGUE POSITION IN YEARS?
Storkkkkkk (and Jangbi).

[quote="grn_trtle"]for no good reason[/quote]
???
it is pretty true
and also pretty hilariuos

[quote="grn_trtle"]Guess who, for no good reason, just resurrected a thread that died like 4 months ago?  :roll:[/quote]
Dude what? SC is awesome.
Also wow....

just dropping by to ask

What the devil's eggplant is this thread?

idk but this thread will be locked bcuz of random thread revival


http://mandelbrot.org/dialogue/prevyear/pastprobs2002.html

Can someone look at 2002-6 and tell me how to do this problem... this has been bugging me for like forever.

ooh, I liked this problem :).  Consider, on the t[size=75]th[/size] day, the probability distribution of any given politician.

Once you've figured that out, think about any two politicians.  Compute the probability that they happen to be at the same place, using the distributions for both of them.  Then you need only to think about the pairs of politicians, and you'll be done.

If you want any more help, just ask.

Where's Setofallsets when we need him? He wrote the problem, after all.

I'm here.... I knew I should have replied when I saw the post earlier this afternoon, but I had to leave and I didn't get back until now.

Joel has pretty much covered it.  The important part of this problem is to remember what you can and can't do with expected values.  As he said, you need to reduce it to considering pairs of politicians, and use [i]linearity of expectation[/i].   When the problem was active last year I recieved a couple of (incorrect) solutions attempting to solve the problem by weird methods that involved the gamma function (!).

There are two ways to find the probability that two given politicians are at the same place on day t.  One of them involves using the probability distributions, and is pretty straightforward if you've done some combinatorics before: the position of the politician is determined by t and the number of steps, say, 'right', allowing nice calculation of a given politician being at a given point on the tth day.  The other is a bit cleverer, and you won't need to do any summation (but you will still need to know how many ways there are to reach a given position in a given number of steps).

Ask if you still need more info.

PS: Thanks for the compliment, Joel.

Lol Setofallsets I was one of those Gamma function solutions, which was obviously wrong seeing as for large values, my expression became negative. And then you told me expected values don't usually work that way--and then I got boggled down in this huge multinomial counting expression sum and never got any further.


Prove that : If $ a,n\ge 2$ are natural number and $ a^n\minus{}1$ is a prime then $ a\equal{}2$ and $ n$ is a prime.
We can see easily $ a\equal{}2$ but how about $ n$ is a prime ? I can not prove it.Please help me.
Thank you very much !

If    a =b+1  ,a^n= (b+1)^n ,           b|((b+1)^n)-1   then  b|(a^n-1)
So   a =2 .
Expanding (1+1)^n may be useful for the rest but I couldn't do it yet .

If n is not prime then write n as a*b where a,b are not 1

Then write $ 2^n \minus{} 1$ as $ (2^a)^b$-$ (1^a)^b$

Which has a factor $ (2^a\minus{}1)*l$

But $ 2^a\minus{}1$ is greater than 1

and l is ($ (2^a)^{b\minus{}1}$+...............+1) which is obviously greater than 1

So n should be prime

Or note that, if $ n\equal{}xy$, then you can replace $ 2^{xy}\minus{}1$ by $ a^y\minus{}1$ where $ a\equal{}2^x> 2$. As you already know that $ a$ must necessarily be $ \equal{}2$, you get a contradiction.


The Fibonacci sequence is 1, 1, 2, 3, 5, 8, $ \ldots$.  Each number after the first two numbers is the sum of the preceding two numbers.  What is the first perfect square greater than 1 to occur in this sequence?

Continuing the sequence, we find the smallest perfect square other than 1 is $ \boxed{144}$.


function algebra polynomial conics parabola domain

Show whether the set of ordered pairs (x,y) below defined by each equation is a function.

(a) y = x^3 - 3x

(b) x + 2y^2 = 1

What does it mean that a set of points is a function? Perhaps you want to know whether the equations define a function or not...

Then $y=x^{3}-3x$ is surely a function: it's polynomial of degree 3...
$x=-2y^{2}+1$ represents a parabola whose axis is the x-axis... so it's not a function, 'cause infinitely many points of the domain (really, all of them, except x=1) have a double image in the codomain...

actually, for part B...
[hide]

$x =-2y^{2}+1$ is both a function and not a function! As Ponnamperuma (correctly) pointed out, that equation is not a function of x.  However, it is a function of y because no matter what y you plug in... x only has one value.[/hide]

Yes, you're right, I just considered x as the independent variable, as usual...  :wink:


WHO THE MAN NOW HUH WHO THE MAN

I BELIEVE IT IS EVERYONE ON THE UNC BASKETBALL TEAM THANK YOU VERY MUCH

I THINK THERE IS NO QUESTION AS TO HOW GOOD UNC IS

WE ARE REALLY - GOOD

THERE YOU HAVE IT

HOLY CRAP YOU KIDDING ME
THE GOOD FOOD TEAM LOST?
NO WAY YOU ALIEN LIARS

[quote="not_trig"]NO ONE CARES ABOUT, eh, CANADA!!!![/quote]

of course no one cares  :lol:

[quote="not_trig"]eh, nice job

Duke is still better[/quote]

duke barely held off an unranked indiana team

stfun00bkthxh4xbye

yeah, because duke is HORRIBLE!

beyond your imagination.

and they smell like poop

you guys are just jealous that duke is going to cream unc when they play near the end of the season  :P

wtf? your signature is pretty messed up man I THINK THAT SCIENCE TEACHER WAS HIGH

OH YEAH THAT SCIENCE TEACHEr

is BERSERK!

shes been doing something

[quote="MysticTerminator"]buffalae[/quote]

*Edit* buffalus  :wink:

[quote="junggi"][quote="Imperial Effect"][quote="MysticTerminator"]buffalae[/quote]

buffalaus  :wink:[/quote]
buffali![/quote]
Canada? :maybe:

[quote="junggi"][quote="Imperial Effect"][quote="MysticTerminator"]buffalae[/quote]

buffalaus  :wink:[/quote]
buffali![/quote]

buffalus $=$ Fourth Declension Nom. P.  :ninja:

quaggai

muahahhaha

goose plural should be goosies.

[quote="Imperial Effect"][/quote][quote="junggi"][quote="Imperial Effect"][quote="MysticTerminator"]buffalae[/quote]

buffalaus  :wink:[/quote]
buffali![/quote][quote="Imperial Effect"]

buffalus $=$ Fourth Declension Nom. P.  :ninja:[/quote]

eh 1st declension nom. p.

what are you talking about man

and plural of quagga = quaggalis

eh? i have a concise latin dictionary i'll go check :lol:
it is urus,uri
aka wild ox aka bufflo
so pl of ox is oxen right?:P

[quote="MysticTerminator"]all right guys c'mon now[/quote]
what are you saying eh? :huh:


m is a fixed positive integer not equal to the integer powers of 2. Sequence {a_n} is defined by:
(1) a_1 is positive integer and 1 < a_1 < 2m;
(2) a_(n+1) = 2a_(n)  1 if a_(n) <= m or a_(n+1) = 2(2m + 1  a_(n)) if a_(n) > m.
Prove or disprove that for every possible value of a_1 as shown in (1), there always exist a positive integer h such that a_h = 2m.

i think i've got a counterexample...take e.g. m=7 and start with a_1=10 (1<10<14). then a_1>7 thus a_2=2(15-10)=10 so we get a constant sequence with 10's, but 2m=14 and 7 is not a power of 2...there must be some condition missing i think...

Thanks, maybe I should strength some conditions, but at this time, I had no idea, can someone help me?


1. How many positive divisors does 756 have besides 1 and 756?

2. Sally likes to make figurines from modeling clay.  She uses exactly 2/3rds of a box to make each figurine.  It did not take her long to discover that she could blend the left-overs from 3 boxes to make a new box of clay.  If Sally initially purchses 27 boxes of clay, how many figurines can she make before she needs to buy a new supply of clay?

3. Write x, y, & z in order from the smallest to the largest if x = 2^100, y = 3^75, and z = 5^50.

4. If a + b = 1 and a^2 + b^2 = 2 then a^4 + b^4 = what?

5. How many planes of symmetry does a cube have?

Sorry, I forgot that I discounted 1 and 756,
now you get #1.

I forgot to not include 1 and 756, but i changed it.

3)[hide]x < z < y[/hide]

5)[hide]9, 3 through faces and 6 through diagonals[/hide]

4) [hide]3.5[/hide]

No one else is here! No pressure!

theone gets #4.
Clean sweep.
7 other users were on-line, so I thought there might be decent competition.

I'll post the next set of 5 tomorrow after school around 4:00pm cst, I have to write a diaog between a turkey and a fruitcake for English tomorrow.  How fun.  Feel sorry for me.

I'll write my paper, then see if people feel like doing a set of 5 in about 30 minutes.

Cats...do you feel like you're too old to be answering the questions?  I just wasn't sure what you meant by your last post.

[quote="cats..."]im not too old, i dont think.  How old are most of you?   im 16.[/quote]
Meh? Not unexpected but...

I'm 14, pero I take 10th so by age-standard-thingy maybe I'm 15 :lol:

[quote="notfunny"]sure, ill do a set.[/quote]
? what do you mean?:?
i'm 13:cry:

I'll type the next 5 questions now.  Expect them in about 3 minutes.


inequalities
search
function
inequalities proposed

Prove that for $a,b,c$ positive numbers then ${\frac{a^2+b^2+c^2}{ab+bc+ca}+(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{a+b+c})^2}\geq 2$. I would like a "proof without words" from the most experienced members of this forum. There is one. ;)

Very nice inequality! I remember I created this inequality a long time ago. What do you mean "proof without words"?
 A figure,something like that???

[quote="harazi"]Then please post your solution to this inequality, Cezar. And excuse me for posting as being my creation, I thought nobody is so crazy to invent such monster. When I said no words I meant try to deduce it using some "mathlinks inequalities culture".[/quote]
mathlinks inequalities culture?  :?

It was a joke. I meant that there are two well discussed inequalities on the forum that imply this inequality. Which ones?

[quote="harazi"]Prove that for $a,b,c$ positive numbers then ${\frac{a^2+b^2+c^2}{ab+bc+ca}+(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{a+b+c})^2}\geq 2$. I would like a "proof without words" from the most experienced members of this forum.[/quote]

I have to be brief now:

By http://www.mathlinks.ro/Forum/viewtopic.php?t=14569 , if x, y, z are three nonnegative numbers satisfying x + y + z = 1, then $\left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)\geq 8\left(y^2z^2+z^2x^2+x^2y^2\right)^2$. Upon dehomogenizing, we obtain the inequality $\left(y^2+z^2\right)\left(z^2+x^2\right)\left(x^2+y^2\right)\left(x+y+z\right)^2\geq 8\left(y^2z^2+z^2x^2+x^2y^2\right)^2$ for any three nonnegative numbers x, y, z. Applying this to $x=\sqrt{bc}$, $y=\sqrt{ca}$, $z=\sqrt{ab}$, we get $\left(ca+ab\right)\left(ab+bc\right)\left(bc+ca\right)\left(\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\right)^2\geq 8\left(caab+abbc+bcca\right)^2$. After straightforward transformations, this becomes $\left(\frac{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}{a+b+c}\right)^2\geq\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}$. On the other hand, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=22815 or to http://www.mathlinks.ro/Forum/viewtopic.php?t=23109 post #2, we have $\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\geq 2$. Thus,

$\frac{a^2+b^2+c^2}{bc+ca+ab}+\left(\frac{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}{a+b+c}\right)^2\geq\frac{a^2+b^2+c^2}{bc+ca+ab}+\frac{8abc}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}$
$\geq 2$,

and it's done. I hope that "proof without words" did not mean "proof without links"; anyway, the problem is more an exercise on using the search function rather than on proving inequalities...

  darij

Excuse me, Darij, for posting such stupid problems, where I do not test skills of problem solving. I promise I won't ever do that. Even though, to be sincere, I think you are completely wrong when you say that. But anyway, if one prefers to make thousands of computations and not to think a little bit about different connections between problems, then so be it. I'm really stunned to hear that, but don't worry, I will never bother the forum with such stupid questions. There are two kinds of problems I posted: those which are really nice and which interest nobody (as usual) since they have ideas and not computations and those which are straightforward or have links and relations and also which interest nobody. It seems it isn't the right place for me this forum.  :(

if you want my idea harazi , i think you are one of guys that is most needed here and your place is exactly this forum(and of course Algebra  and number theorem too :lol: ).
I by myself really enjoy this kind of innovative thinking of numbers and constructing
huge functions .
Sometimes new inequalities can be killed by using the some old inequalities  as lemmas for them if your goal is to just and just solved the problem no matter how (as darij had done to your problem by using those links).
but be sure that problem in not designed by that way of thinking .
Harazi, I am , by myself, really thankfull that you put your time and energy here and also
the other instructors of mathlinks, Thank you all very much .

P.S harazi , what do you mean by the source of this link?url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14569[/url]

you and [b]Titu Andreescu[/b] ,  :o  :o  :w00t:  :w00tb:


function
superior algebra
superior algebra unsolved

Provide examples of binary operation on a set $ S$.
(i) Commutative but not associative.
(ii) Neither associative nor commutative.

i)  The complement of the intersection of sets.  (I think.)  Actually, most functions of a commutative operation should work.
ii)  [url=http://en.wikipedia.org/wiki/Octonion]Octonion multiplication[/url].


quadratics
algebra
polynomial
function
real analysis
real analysis unsolved

if a function satisfies the equation$f(x+h)-f(x-h)=2hf'(x)$ for all real x and for two fixed values of h, then it must be a quadratic polynomial.

Sorry for the stupid question, but can one of this values of $h$ be 0? Because otherwise it can't be a quadratic polynomial.

This is true for all $h$ when $f$ is quadratic.


A student tosses a fair coin and scores one point for each head that turns up, and two points for each tail. Prove that the probability of the student scoring $n$ points at some time in a sequence of $n$ tosses is $\frac{1}{3}\left(2+\left(\frac{-1}{2}\right)^n\right)$.

Masoud Zargar

hi

i tried it for \[ n=odd=2m+1 \]

i got this expression

\[ \sum_{k=0}^m {\frac{\binom{m+k+1}{2k+1}} {2^{m+k+1}} }\]

i tried a lot but cud not evaluate the sum

but some initial values of m do work

reply if u r able to evaluate this
hopefully it should match with ur ans :|

Looking at that formula, it definitely seems like it's based on the explicit form for a two term recursion... in fact, the recursion should be (if I'm not wrong) \[ x_{n+2}= \frac{x_{n+1}}{2}+\frac{x_n}{2}  \], with initial values of $x_1 = \frac{1}{2}$, $x_2=\frac{3}{4}$...now it's just enough to relate this back to the original problem, and then connect it all together...

it pays to see where a formula comes from.  ;)

edit> This actually makes perfect sense; suppose we want a sequence of $n$ tosses to have $n$ points; we can either take a sequence of $n-1$ tosses of $n-1$ points and add a head to it with 1/2 chance, or a sequence of $n-2$ tosses  with $n-2$ points and add a tails to it, again with 1/2 chance. It's clear that recursively this is defined exactly as the formula above, and then we merely need to find our probability for $n=1,2$ tosses, and then just go from recursive to explicit and we get the same thing the OP posted.

[quote="arpit goel"]well one more thing

ur q says n pts in a sequence of n tosses :!:

i dont think that the number tosses should be fixed

if tosses = n then the answer wud simply be \[ \frac{1}{2^n}  \] ;)[/quote]

It says scoring n points at SOME TIME in a sequence of n tosses. So it doesn't have to be all heads. :)

[quote="arpit goel"]...for \[ n=odd=2m+1  \]
...
\[ \sum_{k=0}^m {\frac{\binom{m+k+1}{2k+1}} {2^{m+k+1}}  }\]
[/quote]
Well here is what I say:
We know $n = 2T + H$ where T is number of tails and H is number of heads (this also tells us n and H have the same parity.) Since flips must be on the way to n we know $T + H \le n$. From this we get $H = 2(T + H) - (2T + H) \le 2(n) - n = n$.

With the $n = 2m + 1$ we know $H = 1, 3, \ldots, 2m + 1$. With this we get the ordered pairs $(H, T) = (2i + 1, m - i)$ for $i \in (0, m)$. For every pair that works we know that every sequence of flips beginning with that sequence will work. So we have one sequence of length $T + H = m + i + 1$ and 2 sequences of length $m + i + 2$ and $2^{m - i}$ sequences of length $2m + 1 = n$. So we have $\displaystyle\binom{m + i + 1}{m - i}$ such orders at each order pair, and also we have $\displaystyle\sum_{j=0}^{m-i}2^j = 2 \cdot 2^{m - i} - 1$ subsequent sequences so we have a total of $\displaystyle\sum_{i=0}^{m}\binom{m + i + 1}{m - i}\cdot\left(2 \cdot 2^{m - i} - 1\right)$ sequences that reach a sum of $n$ at some point. This includes different lengthed sequences so when we do our total count we get 2^1 sequences of length 1 etc. for a total of $\displaystyle\sum_{i=1}^{n=2m+1}2^i = 2^{2m + 2} - 2 = 4^{m + 1} - 2$ This leaves us with probability $\displaystyle\frac{\sum_{i=0}^{m}\binom{m + i + 1}{m - i}\cdot\left(2 \cdot 2^{m - i} - 1\right)}{4^{m + 1} - 2}$.

That was my first argument but it is flawed. I think we need only consider sequences of length $n$. Thus from:
$(H, T) = (2i + 1, m - i) \rightarrow \text{ length: } H + T = m + i + 1 \rightarrow \displaystyle\binom{m + i + 1}{m - i} \text{ sequences}$
We know we need a length $n = 2m + 1$ sequence so we have the first $m + i + 1$ spots filled and the last $m - i$ spots empty. Each spot can be filled with each value so there are $2^{m - i}$ for the rest which results in $\displaystyle\binom{m + i + 1}{m - i} \cdot 2^{m - i}$ sequences that reach a sum of $n$ somewhere along the way. Since we know $i$ varies we have a total of $\sum_{i=0}^{m}\binom{m + i + 1}{m - i}\cdot 2^{m - i}$ sequences. This gives us probability:
$\displaystyle\frac{\sum_{i=0}^{m}\binom{m + i + 1}{m - i}\cdot 2^{m - i}}{2^{2m + 1}}$

When $n = 2m$ it follows much the same
$(H, T) = (2i, m - i) \rightarrow \text{ length: } H + T = m + i \rightarrow \displaystyle\binom{m + i}{m - i} \text{ sequences}$
$\rightarrow \displaystyle\binom{m + i}{m - i}\cdot 2^{2m - m - i} \text{ length n sequences} \rightarrow \displaystyle\sum_{i=0}^{m}\binom{m + i}{m - i}\cdot 2^{m - i} \text{ possivle sequences}$
$\text{Pr} = \displaystyle\frac{\sum_{i=0}^{m}\binom{m + i}{m - i}\cdot 2^{m - i}}{2^{2m}}$

How to solve?

Edit: Formula would indicate:
$\displaystyle\sum_{i=0}^{m}\binom{m + i + 1}{m - i}\cdot 2^{m - i} = \frac{2^{2m+2} - 1}{3}$
and
$\displaystyle\sum_{i=0}^{m}\binom{m + i}{m - i}\cdot 2^{m - i} = \frac{2^{2m+1} + 1}{3}$


Let f(1)=f(2)=1 and f(3)=2. Define f(n+3)=(f(n+2)f(n+1)+5)/f(n) for all n>=1. Prove that f(n) is an integer for all positive integers n.

If you have seen this problem before, please don't post a solution.

[hide] Could we use induction? I haven't tried this but I've seen similar problems worked this way.[/hide]

Try it. I don't want to give away the solution yet.

Hi,

By induction :

. if n if odd then f(n+2) = 3*f(n+1)-f(n)
. if n is even then f(n+2) = 3*f(n+1)+2*f(n)-f(n-1)

HTH


Calculate alpha:

[img]http://img101.imageshack.us/img101/9803/cals.jpg[/img]

Hi luis. I think there might be somthing missing here as we could move $ F$ left or right without changing anything.

Yes, I'm confused too  :huh:
But it's like that picture, maybe has a mistake.

another thing: did you get this problem from any specific place? Because i would think that there would be a more complete question along with the diagram.

[quote="fishythefish"]Incidentally, what are you using to display the image? Because in certain threads, I seem to be getting a piece of text reading "Image" rather than a graphic.

EDIT: Luis, I was able to see the image in your other triangle post. What kind of file are you embedding?[/quote]

Hello, I don't understand your question, but the writes on the image it's because I tried to resolve this problem  :D
I erase the writes please see the another photo:

[img]http://img522.imageshack.us/img522/9803/cals.jpg[/img]

Question items:

a) 10 degrees
b) 20 degrees
c) 30 degrees
d) 40 degrees
e) 45 degrees

I found this question from my book, but isn't a university question, can be wrong or correct.  :(

It doesn't matter where you got the question from. The problem is there is(are) missing condition(s) that would make the value of $ \alpha$ unique. Just posting an image of a problem is not enough information to solve it. Do you understand what I am talking about because your English is not that good?

Yes, I understand  :D
But the only question is: "Calculate alpha", in the real question don't have more information.

Then the question doesn't have a single answer. In fact, it has infinite answers if we place $ F$ anywhere on $ EG$. (I labeled the center point on the base $ G$.)

Is there perhaps another chunk of text before the image in your book (which I presume is th e source of this problem)?


omg my friend just told me about it

i cant believe it

tears are falling on my keyboard right now  :(  :(  :(

rip....

Wow. I can't believe that you would waste your time like that. However, I wouldn't be surprised if she did die soon; she's such a prissy idiot that it'll probably go to her head.

[quote="life=tennis+math"]Am I the only one who thinks this should go to the Round Table :?:
This is important stuff guys  :roll:[/quote]

uh...yea...too bad shes not dead.

she is so annoying.  i hate spoiled brats

[quote="bubka"]I watched a porno flak or two by her in college and they were nice. :rotfl:[/quote]

Err, I think you could say the same about a whole buttload of women. Which, you know, sounds pretty funny.

[quote="MithsApprentice"]1) No commercial advertising. No exceptions.[/quote]

Please don't spam, Ignite.


here are the two options on my mind now:
1. cs + computational math
2. cs + pure math

a little background on myself:
I'm new to cs and I'm not really good at it, but I found it more interesting than all the other subjects I have encountered.
the reason why I want to double major is because there seem to be a lot of people in cs and I think double majoring in another plan can enhance my competiveness. Also, math is my second favourite.

Thank you

Talk to an adviser at your university. The details depend strongly on the exact requirements of those degrees, which are not a universal standard.
If you're not at a university yet, you don't need to decide right away. The introductory classes are the same either way.

If you want to double-major because you are interested in mathematics, then decide for yourself whether you are more interested in pure mathematics or applied mathematics.

If you want to double-major to make yourself more marketable, then you should first decide to whom you want to market yourself and/or possible career paths. Mathematics, economics, electrical engineering, and many other majors work very well with computer science in terms of making you marketable.

Come up with that and post back, and we'll be able to give you better advice.

There's also a trade-off of breadth versus depth that you might want to consider.  Double majoring is not necessarily more marketable than doing one major but taking it to great depths.  It also depends on what market you're trying to sell yourself on...


geometry
cyclic quadrilateral
geometry unsolved

Three circles that passed through a point $X$ are cut again, two by two, in points $A,B,C$. The line $AX$ cut one circle again at $D$. The same way, lines $BX$ and $CY$ produce points $E$ and $F$. Prove that $\triangle {BCD}$, $\triangle {CAE}$ and $\triangle {ABF}$ are similar ("semejantes" is "similar" or "congruent").

[quote="shobber"]$\angle{CDB}=\angle{FXB}=\angle{FAB}$.
$\angle{BCD}=\angle{BXC}=\angle{EXA}=\angle{AFB}$.
Therefore $\triangle{AFB} \sim \triangle{DCB}$.[/quote]

Thanks for replying... can you explain it with more detail??? (for example, why $\angle{CDB}=\angle{FXB}$)

[quote="Arne"]$\angle CDB = 180^{\circ} - \angle CXB$ (opposite angles of a cyclic quadrilateral)
$180^{\circ} - \angle CXB = \angle FXB$ (angles along a straight line)[/quote]

Thanks Arne, I got it now.


I'm really curious to know what is your favorite music.  :D

1. Boston
2. Santana
3. AC DC
4. Peter Frampton
5. Lynyrd Skynyrd
6. Journey
7. Twisted Sister
8. Linkin Park
9. Eric Johnson
10. Jimi Hendrix

Clearly, I'm a rock fan.  :lol:

EDIT: Forgot Metallica.

1. Micheal Jackson
2. Katy Perry
3. Linkin Park
4. Flo Rida
5. White Tie Affair
6. Lady Gaga
7. Owl City
8. We the Kings
9. The Ting Tings
10. Toybox

I apparently am a techno/pop type of person :D

1. Atreyu
2. Metallica
3. Red Jumpsuit Apparatus (before they started sucking)
4. Avenged Sevenfold
5. Nightwish
6. Three Days Grace
7. I Am Ghost
8. Disturbed
9. Muse
10. Apocalyptica

Sort of a Metal/Hard/Goth Rock mix...

1. Kristinia Debarge
2. Avril Lavinge
3. Lady Gaga
4. Taylor Swift
5. Evanescence
6. L'Arc~en~Ciel
7. Asian Kung-fu Generation
8. Nana Kitade
9. Ali Project
10. Nana Mizuki

Yeah, I'm a pop type of person. Although I like some rock also.

(By the way, the last 5 are Japanese, so I doubt anyone knows them. I love Jpop/Jrock.)

1. Boston
2. The Who
3. Pink Floyd
4. Eagles
5. Green Day
6. Metallica
7. The Beatles
8. The Frames
9. Damien Rice
10. Glen Hansard

Meh.

xD

I can't put them in order, so I'll just list them:

Michael Jackson
Weird Al Yankovich
Celtic Woman
Yanni
Youssou N'Dour
Anoushka Shankar
David and Steve Gordon
Bebel Gilberto
Pia
Tya

Good luck even [i]finding[/i] some of these artists.  :diablo:

1.Bryan Adams
2.U2
3.Zubeen Garg (Indian)
4. Dr.Bhupen Hazarika(Indian)
5. Hilary Duff
6.Hannah Montana  :oops:
7. Michael Jackson
8. Backstreet Boys
9. Eagles
10. Shakira

1. P!nk
2. Coldplay
3. t.A.T.u.
4. Avril Lavigne
5. Muse
6. Linkin Park
7. Evanescence
8. Maroon 5
9. Amy Winehouse
10. Céline Dion
 :)

In no particular order:

1. Pearl Jam
2. The Who
3. Nirvana
4. Linkin Park
5. Disturbed
6. Alice in Chains
7. Steve Miller Band
8. EDIT: Queen.
9. System of a Down
10. Airborne Toxic Event

What I don't get is how Linkin Park is showing up on all of these pop-artist-filled lists.

1) Dimrain47
2) Shiki
3) Silvia
4) DM Ashura
5) DJ SymBiotiX
6) ParagonX9
7) Digital Explosion
8) ZiGZaG
9) Kid2Will
10) Nobuo Uematsu

Weird..not many people put Bon Jovi. One of their biggest hits wuz Livin on a Prayer..

I think this would be thee wrong forum... It should go in music and arts

[quote="inoit"]there's a music and arts forum?![/quote]

Surprisingly yes!

actually it belongs here cause its not real discussion about arts and music just opinions and free talking

although it should probly go in fun factory...
anyways heres mine

1. Big Bang
2. The Academy Is...
3. Ft Island
4. 王光良(uhh..i think thats right ^_^)
5. Wonder Girls (teehee)
6. Nana Mizuki (teehee x2 ^_^)
7. Epik High
8. Metallica
9. Linkin Park
10. The Who

Lets see... (restricting myself to internationally famous performers)

Nick Cave
Gipsy Kings
Carlos Santana
Guns'n'Roses
Pink Floyd
Iron Maiden
Leningrad Cowboys
Alexandrov Choir
Panic! at the Disco
U2

1.Fall out Boy
2. John Mayer
3. Panic! At The Disco
4.Of Montreal
5. Vampire Weekend
6. Sondre Lerche
7. Spoon
8. Ben Folds
9. Beck
10. Cobra Starship

My current 10, in alphabetical order (yes I did just go down my iTunes list):

Anberlin
Boys Like Girls
Death Cab for Cutie
Demon Hunter
The Killers
Linkin Park
Paramore
Red
Switchfoot
Thrice

Complex_Ninja, I've got two from your top 10 too :D

Maroon 5
Nickelback
Fall Out Boy
The All-American Rejects
Green Day
Kanye West
Disturbed
Akon
Panic! At the Disco
The Red Jumpsuit Apparatus

[in no particular order]

the killers, the strokes, the offspring, foo fighters, nirvana, metallica, green day, system of a down, pearl jam, rage against the machine
if i've heard the name of an artist said out loud more than 10 times, it's probably the pathetic pop/rap bull that they plaster all over the itunes top 10 (and that my friends listen to >=( ).

[quote="Complex_Ninja"]From #1 - #10
1 Three days Grace
2 Senses Fail
3 Radiohead
4 Muse
5 Modest Mouse
6 Editors
7 Anberlin
8 Mono
9 Death Cab for a Cutie
10 Asobi Seksu

Wow only 1 person had 2 from  my top 10 :) . Basically Hard Rock/Post Rock/ indie[/quote]

Yeah surprising considering how amazing Three Days Grace and Muse are!
On another note, Dream Theater definitely deserves a spot on the top ten... forgot them before.

1)Linkin Park
2)Coldplay
3)Simple plan
4)Fort Minor
5)eminem
6)Maroon5
7)Uverworld
8)Akon
9)Kayne West
10)SNoW and Nana kitade and yuna ito(Jpop)
Simply luv Jpop :D
11)GaGa
12)Demi Lovato(sum tracks)

[quote="ernie"]Foo Fighters
Nickelback
Fall Out Boy
Godsmack
Green Day
Kanye West
Disturbed
Akon
Panic! At the Disco
Korn

[in no particular order][/quote]

I have changed. :)

I like the on republic whole new album-wakening up. I also like onerepublic- stop and stare. we see Ryan walking through the desert toward an? open grave where a Preacher stands delivering a eulogy. As he walks, we see flashes of scenes with Ryan completely submerged in a bathtub while fully clothed, in motel room #7 staring at a TV displaying static, ringing the service bell at the motel desk, and sitting and waiting in the motel lobby.

Hello,

I like to listen music most of the time.I am a musician  too.Here are some artist i like to here songs.

Enrique
A R Rehman
Linkin park
Akon
Michael Jackson
Big bang
Metallica
shaggy
eminem
bryan adams

Thank you!!

Not in any Order,I just like them all.

Linkin park
Metallica
Green Day
Foo Fighters
Theory Of a Death Man
3Doors Down
The All-American Rejects
Three days Grace
Iron Maiden
Dragon Force

Music Is Über Awesome!

No particular order:

-Kanye West
-Eminem
-Akon
-Beyonce
-Jay Sean
-Rihanna
-Jason Derulo
-Jason Mraz
-Jay-Z
-Chris Brown

Hip-Hop freak. One of the few ones on this website.

1.Iyaz
2.Jason Derulo
3.Weird Al Yankovic
4.Ke$ \$$ha
5.Jay-Z
6.Taylor Swift
7.Rihanna

Some favorite songs: Replay, In my head, Whatcha Say, Tik Tok, Empire State of Mind, Two is Better Than One, I look so good, White and Nerdy.

I think I'm mainly pop music.

1.Green Day
2.The Fray
3.All American Rejects
4.Rise Against
5.Offspring
6.Kanye West
7.Boys Like Girls
8.Lincoln Park
9.Taylor Swift
10.T.I.

It changes position a lot and in some places I can't distinguish which is better, but generally my top ten looks like  this

As you can see, I'm and alternative, country, pop kind of guy

Hey guys
Here is mine..
1) Jack Johnson
2) Matt Nathanson
3) five for fighting
4) everclear
5) REM
6) ParagonX9
7) Digital Explosion
8) ZiGZaG
9) Kid2Will
10) Nobuo Uematsu

plz don't revive things over 1 month old.
it screws up everything.

MODS.PLZ.LOCK.THIS.THREAD.

That's what users are generally told.  But in this instance, I feel as if it was a perfectly fine post.


The important thing to know for this problem is that $ \frac{n!}{k!}=n(n-1)(n-2)\cdots (n-k+1)$.  Do not memorize this, as it follows easily from the definition of factorials.

Multiplying the numerator and denominator by $ 5!$, we get
\begin{align*}
\frac{5!\left(\frac{5}{3!}+\frac{5}{4!}\right)}{5!\left(\frac{5^2}{5!}\right)} &= \frac{5!\left(\frac{5}{3!}\right)+5!\left(\frac{5}{4!}\right)}{5^2} \\
&= \frac{5\cdot 5\cdot 4+5\cdot 5}{25} \\
&= \frac{125}{25} \\
&= \boxed{5} \\
\end{align*}

hello, we have
$ \left(\frac {5}{3!} \plus{} \frac {5}{4!}\right)\cdot\frac {5!}{5^2} \equal{} \frac {5\cdot5!}{3!\cdot5^2} \plus{} \frac {5\cdot5!}{4!\cdot5^2} \equal{} \frac {5!}{3!\cdot5} \plus{} \frac {5!}{4!\cdot5} \equal{} 4 \plus{} 1 \equal{} 5$
Sonnhard.


Any ideas on how to solve this one?

A "Fibonoid" chain of numbers is constructed by starting with any 2 numbers, adding them to produce a third, adding the 2nd and 3rd to produce a 4th, adding the 3rd and 4th to produce a 5th, etc.  The longest possible "Fibonoid" chain of positive integers that ends with the number 2005 is produced.  Find the sum of the smallest two postive numbers in this chain.

are we allowed to use a calculator?
(this is because, i would strongly suspect a relation with phi)

Let $\{G_{n}\}$ be a sequence with the same recurrence formula as Fibonacci numbers, but with a different seed, and assume that this sequence contains a member $G_k = 2005$. Then

$G_k = G_{k-2} + G_{k-1} = 2005$
$G_{k-2} + (G_{k-2} + G_{k-3}) = G_{k-3} + 2G_{k-2} = 2005$
$G_{k-3} + 2(G_{k-3} + G_{k-4}) = 2G_{k-4} + 3G_{k-3} = 2005$
$2G_{k-4} + 3(G_{k-4} + G_{k-5}) = 3G_{k-5} + 5G_{k-4} = 2005$
$3G_{k-5} + 5(G_{k-5} + G_{k-6}) = 5G_{k-6} + 8G_{k-5} = 2005$
$5G_{k-6} + 8(G_{k-6} + G_{k-7}) = 8G_{k-7} + 13G_{k-6} = 2005$
$8G_{k-7} + 13(G_{k-7} + G_{k-8}) = 13G_{k-8} + 21G_{k-7} = 2005$

etc., until,

$F_{m-1} G_{k-(m-1)} + F_m G_{k-m} = 2005$

where $G_{k-(m-1)}, G_{k-m}$ are the 2 initial members (the seed) of the sequence $\{G_{n}\}$. Hence, it is necessary to solve the equation

$p F_{n-1} + q F_n = 2005$

or equivalently,

$r F_{n-1} + s F_n = 2005 \bmod F_{n+1}$

$r = p - \left[\frac{2005}{F_{n+1}}\right] \le 0, s = q - \left[\frac{2005}{F_{n+1}}\right] \ge 0$

for the largest possible $n$, where $F_n$ are Fibonacci numbers and $q > p > 0$ are positive integers, the initial members (the seed) of the desired sequence with the same recurrence formula as Fibonacci numbers.

I think that a solution of the above equation in integers always exists, because consecutive Fibonacci numbers are always coprime (not sure), but the problem requires the integers $q > p > 0$ to be positive.

$n = 1:$
$2005 \bmod F_2=  2005 \bmod 2 = 1,  r F_0 + s F_1 = r + s = 1$
$r = 0, s = 0, p = 0 + 1002 = 1002, q = 1 + 1002 = 1003$

$n = 2:$
$2005 \bmod F_3 =  2005 \bmod 3 = 1,  r F_1 + s F_2 = r + 2s = 1$
$r = -1, s = 1, p = -1 + 668 = 667, q = 1 + 668 = 669$

$n = 3:$
$2005 \bmod F_4 =  2005 \bmod 5 = 0,  r F_2 + s F_3 = 2r + 3s = 0$
$r = -3, s = 2, p = -3 + 401 = 398, q = 2 + 401 = 403$

$n = 4:$
$2005 \bmod F_5 =  2005 \bmod 8 = 5,  r F_3 + s F_4 = 3r + 5s = 5$
$r = -5, s = 4, p = -5 + 250 = 245, q = 4 + 250 = 254$

$n = 5:$
$2005 \bmod F_6 =  2005 \bmod 13 = 3,  r F_4 + s F_5 = 5r + 8s = 3$
$r = -1, s = 1, p = -1 + 154 = 153, q = 1 + 154 = 155$

$n = 6:$
$2005 \bmod F_7 =  2005 \bmod 21 = 10,  r F_5 + s F_6 = 8r + 13s = 10$
$r = -2, s = 2, p = -2 + 95 = 93, q = 2 + 95 = 97$

$n = 7:$
$2005 \bmod F_8 =  2005 \bmod 34 = 33,  r F_6 + s F_7 = 13r + 21s = 33$
$r = -12, s = 9, p = -12 + 58 = 46, q = 9 + 58 = 67$

$n = 8:$
$2005 \bmod F_9 =  2005 \bmod 55 = 25,  r F_7 + s F_8 = 21r + 34s = 25$
$r = -15, s = 10, p = -15 + 36 = 21, q = 10 + 36 = 46$

$n = 9:$
$2005 \bmod F_{10} =  2005 \bmod 89 = 47,  r F_8 + s F_9 = 34r + 55s = 47$
$r = -52, s = 33, p = -52 + 22 = -30, q = 33 + 22 = 55$

$n = 10:$
$2005 \bmod F_{11} =  2005 \bmod 144 = 133,  r F_9 + s F_{10} = 55r + 89s = 133$
$r = -17, s = 12, p = -17 + 13 = -4, q = 12 + 13 = 25$

$n = 11:$
$2005 \bmod F_{12} =  2005 \bmod 233 = 141,  r F_{10} + s F_{11} = 89r + 144s = 141$
$r = -123, s = 77, p = -123 + 8 = -115, q = 77 + 8 = 85$

$n = 12:$
$2005 \bmod F_{13} =  2005 \bmod 377 = 120,  r F_{11} + s F_{12} = 144r + 233s = 120$
$r = -38, s = 24, p = -38 + 5 = -33, q = 24 + 5 = 29$

$n = 13:$
$2005 \bmod F_{14} =  2005 \bmod 610 = 175,  r F_{12} + s F_{13} = 233r + 377s = 175$
$r = -318, s = 197, p = -318 + 3 = -315, q = 197 + 3 = 200$

$n = 14:$
$2005 \bmod F_{15} =  2005 \bmod 987 = 31,  r F_{13} + s F_{14} = 377r + 610s = 31$
$r = -97, s = 60, p = -97 + 2 = -95, q = 60 + 2 = 62$

and we can safely stop, because the seed $p$ is no longer positive.

Test:

$n = 1:\ 1002, 1003, 2005, ...$
$n = 2:\ 667, 669, 1336, 2005, ...$
$n = 3:\ 398, 403, 801, 1204, 2005, ...$
$n = 4:\ 245, 254, 499, 753, 1252, 2005, ...$
$n = 5:\ 153, 155, 308, 463, 771, 1234, 2005, ...$
$n = 6:\ 93, 97, 190, 287, 477, 764, 1241, 2005, ...$
$n = 7:\ 46, 67, 113, 180, 293, 473, 766, 1293, 2005, ...$
$n = 8:\ 21, 46, 67, 113, 180, 293, 473, 766, 1293, 2005, ...$

[quote="SJS"]This problem appeared on a 50-minute math contest with 19 other problems -- all of the rest were much less difficult.    Could one reasonably expect to tackle this problem and the others under the time limit?[/quote]

The idea of the solution is quick, but I doubt the sequences can be calculated without errors in such a short time with a calculator, unless you are the Rainman. I used MS Excel. Moreover, I took a second look at the negative seeds and I got this:

Test continues:
$n = 9:\ -30, 55, 25, 80, 105, 185, 290, 475, 765, 1240, 2005, ...$
$n = 10:\ -4, 25, 21, 46, 67, 113, 180, 293, 473, 766, 1239, 2005, ...$
$n = 11:\ -115, 85, -30, 55, 25, 80, 105, 185, 290, 475, 765, 1240, 2005, ...$
$n = 12:\ -33, 29, -4, 25, 21, 46, 67, 113, 180, 293, 473, 766, 1239, 2005, ...$
$n = 13:\ -315, 200, -115, 85, -30, 55, 25, 80, 105, 185, 290, 475, 765, 1240, 2005, ...$
$n = 14:\ -95, 62, -33, 29, -4, 25, 21, 46, 67, 113, 180, 293, 473, 766, 1239, 2005, ...$

Of course, the first few negative members of a sequence can be disregarded and the first 2 consecutive positive members used as a seed. The sequences seem to "alternate randomly" between 2 types - with the positive seed $p = 25, q = 21$ or with the positive seed $p = 55, q = 25$. Hence, the longest sequence does not have the seed $p = 21, q = 46$, as concluded previously, but $p = 25, q = 21$. The conditions of the problem do not require the sequence to be always increasing. Thus the answer to the problem question seems to be $p + q = 25 + 21 = 46$, rather than $p + q = 21 + 46 = 67$.

There may be an easier way.  You just use the fact that any "fibonoid" sequence has a_n=A*:phi:^n + B/( :phi:^n), where A and B are arbitrary constants.  This is from a well known fact about how to solve general linear recurrences using powers of the roots of polynomials (i.e. if A*a_n+B*a_n+...=0 then a_n = (arbitrary)*x_1^n+(arbitrary)*x_2^n+... where x_1, x_2, ... are the roots of the polynomial Ax^n+Bx^(n-1)+...).  I don't know where to go from there in the solution of this problem, but that method must work faster.

Yetti:

Consecutive Fibonacci numbers are always coprime, via the Euclidean algorithm.  This is because we have that:

$F_n=F_{n-1}+F_{n-2}$
$F_{n-1}=F_{n-2}+F_{n-3}$
$\ldots$
$F_3=F_2+F_1$
$F_1=1$
$\gcd(F_n,F_{n-1})=1$
$\Rightarrow \text{ consecutive fibonacci numbers are coprime}$

[quote="blahblahblah"]Consecutive Fibonacci numbers are always coprime, via the Euclidean algorithm.  ...[/quote]

Thanks. My math professors kept the number theory secret from me.