This is the thread that will serve a purpose as a board meeting. If you are not a board member, please refrain from posting here. You can pm the board members or ernie or myself.

The following are guaranteed board members (as of now)

Xoangieexo
Isabella2296
Vallon22
Superllamel
FantasyLover

Also Ernie and I as prez and VP can post here.

Our first discussion is with regards to the rating system. Please discuss which rating system for CTC you prefer, Ernie's, the ladder,or Elo (similar to FTW). Obviously specifics will be discussed later.

I for one support the Elo system due to its versatility, and you won't necessarily be dropped down a seed because of a lost match.

Yes, the ladder method is very restricting.  But no one like's my method. :(

You get a certain amount of points for winning.  You get more credit for the higher your opponent's rating is relative to yours.  You get more credit for winning 4-0 rather than winning 4-3.

ernie, ur type of rating system would be really complicated
since mewto has a website that can calculate rating based on elo, i think we should work with that

I agee with vallon. ALso because ernie is the only one who truly knows how his rating system works, if he cannot play AOPS for a period of time, we couldn't do a rating system.

Also, please post your preliminary votes (we can debate later, but if everyone is unanimous no point in discussion)

Ladder:0

Ernie's:0

Elo:7 (mewto55555, vallon22,isabella2296,fantasylover,superllamel)

I agree. The ELO Rating system has a calculator that I really like, with a weighing factor of 50. I think it works well.

So are your preliminary votes both for Elo?

i guess so
what's the rating we start with for elo
regular 1200, or highest rating or what?

As I said, specifics discussed later,but I would think 1200 would be good.

What do you mean by that? Everyone would start with a rating (presumably 1200) and then in first tournament be randomly seeded, and have elo results calculated from that.

u said tournies, does that mean regular games between two CTC members don't count towards your CTC rating?

No it does not, unless they wish to count it as such, in which case it must be a best 3 out of 5 cd series in which they report the results precisely.

Exactly. Where does rating come into this?

ok
i'm guessing most pplz want to keep the use of guest accounts in the tournies, so then we will
anyone with any questions, concerns, or points they want to bring up about this, plz pm me

It might be better if they posted in the Introducing...CTC! thread, because that way people won't argue/ask the same thing twice.

ok fine
idk where they put it
just bring it up

It's pretty obvious that ratings don't determine your true abilities. You have to take into account memming and only playing people way lower than you. For example, my rating is about 1860 higher than it should be. That said, I think guests should be allowed as long as it's confirmed that they are actually the person you are supposed to be playing.

Yeah, but there are guests who try to fake. (cough cough izzy.)

[quote="isabella2296"]Woah, ernie. I TOLD you this.

I was watching the chat as a guest and got bored. I didn't know you and RG were playing a tourney game, just that RG opened a cd, so I joined for fun.

I was not trying to fake you.[/quote]

Not theplace for personal arguments pm each other. (Dang,since when have I been advocating no spam in forums? :rotfl: )

Anyway, the players themselves are on honor system. super could be facing izzy and give me his pass so i can ownificate her. If you can't trustyour fellow geeks, who can you trust?

Actually, mewmew, I just always [b]let[/b] you win. :P

And sorry about the spam.

We'll just have to go on the honor system. I mean, when people sign up for my math tourneys, like the MO, they could easily ask other people for help, and I would probably never know, but I doubt people would do so.

[quote="vallon22"]that's impossible to impliment
how would you know whether a guest was your opponent or not (you wouldn't)
the only way to know that the guest is really your opponent is if he/she tells you, so we're on the honor system
there is no other way[/quote]
Sorry for posting here, but:
It's quite easy to tell, because they'll have to come in and view the game to tell you.  Especially on normal games: make it viewable with a pass.

Hmmm... for now, the person in the tourney could tell the other person what guest number they are, greatly reducing confusion.

I for one vote to approve isabella and ernie's winter break tourney.

lol. I don't think we need CTC's approval for that toruney. Oh and if its actually CTC and mewto still prez. I can't do that toruney. I dropped out :)


vector
linear algebra
matrix
linear algebra unsolved

Prove if  AB = BA  THEN A and B have same eigenvectors

this is certainly incorrect
take $A=0$
B=matrix(2,2,[0,0,0,1])

of course $AB=BA$ but every vector is an eigenvector of A, while that is not true for B

Sorry Man u are Wrong : My Proof
Let  be an eigenvector of the matrix A corresponding to the eigenvalue , i.e., it holds Let us multiply both sides  on the left by the matrix B. Due to the commutability of the matrices A and B, we get the chain:
Let x an eigenvector of A and k eigenvalue of A
Ax = k x =>AB x= B(Ax) = Bk x = k Bx then Bx is egenvector of A
Thus, if  x is an eigenvector of the matrix A corresponding to the eigenvalue k , thenBx   is also an eigenvector of the matrix A corresponding to the single eigenvalue k , then the vectors  x and Bx are collinear, i.e.,
then there is u such that Bx = ux then x is eigenvalue of B
Thus, the eigenvector  of the matrix A corresponding to the eigenvalue  is also the eigenvector of the matrix B corresponding to the eigenvalue
 :D

then the vectors x and Bx are collinear, i.e.,
then there is u such that Bx = ux then x is eigenvalue of B

what if the eigenspace is of dim > 1? I don't think your statement is true in general

I think you want $n$ distinct eigenvalues.  I also have a feeling that this problem has been posted before.

sparkytx , have you examined my counterexample carefully
I really think it is a good counterexample?
And I deliberately made the A so that your trick does not work
you would find that Bx and x are both eigenvectors of A with eigenvalue zero
but as you see here that does not imply collinearity


$ AE$ is the internal angle bisector in $ \triangle ABC$. $ AD$ is the internal angle bisector in $ \triangle ABE$. $ AF$ is the internal angle bisector in $ \triangle AEC$.
Prove that $ BD \cdot EF \cdot AC \equal{} CF \cdot DE \cdot AB$

By bisector theorem we have:

$ \left\|\begin{array}{c}\ \triangle ABE\ \implies\ \frac{AB}{AE}=\frac{BD}{DE}\\
\triangle AEC\ \implies\ \frac{AC}{AE}=\frac{FC}{EF}\end{array} \right\|\ \implies \frac{AB\cdot DE}{BD}=\frac{AC\cdot EF}{FC}\ \Longleftrightarrow$ $ \boxed{AB\cdot DE\cdot CF=AC\cdot EF\cdot BD}$

Also we can easily prove that $ \boxed{BD\cdot EF\cdot EC=CF\cdot DE\cdot BE}$ .


algebra
polynomial
inequalities
inequalities proposed

Prove that for all positive real numbers $x$ and $y$ the following inequality holds:

$x^3y^3 + x^3 + 3x^2y + 1 \geq x^3y^2 + x^2y^2 + x^3y + x^2 + xy + x$

and determine when equality holds.

Rushil, can you post your solution?
Anyway, here's mine:
Write the initial inequality as $x^2(xy+x+1)(y-1)^2+(x-1)^2+x(x-1)(y-1) \ge 0$
Since $x^2(xy+x+1)(y-1)^2+(x-1)^2 \ge 2x \sqrt{xy+x+1} |x-1||y-1| \ge 2x|x-1||y-1| \ge x|x-1||y-1|$, it's true.


A board with $6$ columns and $n\geq 2$ rows is filled with $0$s and $1$s in such a way that all the rows are different, and for every two rows $(a_{1},a_{2},...,a_{6}),(b_{1},b_{2},...,b_{6})$, there is a row $(a_{1}b_{1},a_{2}b_{2},...,a_{6}b_{6})$. Show that in each column, at least half the entries are $0$s.

Are you sure about that? It doesn't work for

1 1 1 1 1 1
1 0 0 0 0 0

or any board that doesn't have a row composed entirely of 0's.

I guess Cancer meant this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=60138

  Darij


algebra
polynomial
quadratics
complex numbers
Galois Theory
superior algebra
superior algebra unsolved

Find explicitly the $ 6$ roots of this polynomial:
$ 64x^6\minus{}160x^5\plus{}32x^4\plus{}136x^3\minus{}76x^2\plus{}4x\plus{}1$.

[quote="loup blanc"]Find explicitly the $ 6$ roots of this polynomial:
$ 64x^6 \minus{} 160x^5 \plus{} 32x^4 \plus{} 136x^3 \minus{} 76x^2 \plus{} 4x \plus{} 1$.[/quote]

The polynomial factors into quadratics over the cyclotomic field generated by a primitive 7th root of unity.  They split over the cyclotomic field generated by a 35th root of unity, z:

1/2*(-z^21 - z^20 - z^15 - z^14),
1/2*(z^21 - z^20 - z^15 + z^14 + 1),
1/2*(-z^23 - z^21 + z^20 - z^16 + z^15 - z^14 - z^9 + z^5 - z^2 + 1),
1/2*(-z^23 + z^21 + z^20 - z^16 + z^15 + z^14 - z^9 + z^5 - z^2 + 2),
1/2*(z^23 - z^21 + z^16 - z^14 + z^9 - z^5 + z^2),
1/2*(z^23 + z^21 + z^16 + z^14 + z^9 - z^5 + z^2 + 1)

OK JackSchmidt your solution is correct but you can do better:
1) Your $ z$ is algebraic of degree $ 24$. What is the degree of the algebraic extension from $ Q$ to the field of decomposition of our polynomial ?
2) The six roots are real and here you can give a real expression (that is without complex numbers) of these roots.
PS: What is the software you used ?

We finish the JackSchmidt's calculation.
Let $ p(x)\equal{}64x^{6}\minus{}160x^{5}\plus{}32x^{4}\plus{}136x^{3}\minus{}76x^{2}\plus{}4x\plus{}1$.
 $ f(u)\equal{}\minus{}1/2(u^{21}\plus{}u^{20}\plus{}u^{15}\plus{}u^{14})$ was considered by JackSchmidt with $ u$ a $ 35$th primitive root of the unity.
There exist $ 24$ primitive $ 35$th roots of unity: $ z_k\equal{}e^{2ki\pi/35}$ with $ gcd(k,35)\equal{}1$.
Let $ K$ be the field of decomposition of $ p$. According to Maple the $ p$-Galois group is $ C_6$. $ \mathbb{Q}\subset{K}\subset{\mathbb{Q}}(z_1)$, where the extensions are of degree $ 6$ and $ 4$.
The roots of $ p$ are the $ (f(z_k))_k$. Each root $ \alpha$ is reached $ 4$ times and $ K\equal{}\mathbb{Q}(\alpha)$.
The first $ 2$ roots given by JackSchmidt are generated by $ z_1^{20}\plus{}z_1^{15}$ and $ z_1^{21}\plus{}z_1^{14}$ and conversely these last values are algebraically generated by the first $ 2$ roots.
$ z_1^{20}\plus{}z_1^{15}\equal{}e^{8i\pi/7}\plus{}e^{6i\pi/7}\equal{}\minus{}2cos(\pi/7)$ and $ z_1^{21}\plus{}z_1^{14}\equal{}\minus{}2cos(\pi/5)$. Thus $ K\equal{}\mathbb{Q}(cos(\pi/7),cos(\pi/5))$. Moreover $ cos(\pi/7),cos(\pi/5)$ are algebraic over $ \mathbb {Q}$ with degree $ 3,2$. The conjugates of $ cos(\pi/7)$ are $ cos(3\pi/7),cos(5\pi/7)$ and the conjugate of $ cos(\pi/5)$ is $ cos(3\pi/5)$.
Conclusion: the $ 6$ roots of $ p$ are the $ f(z_k)$ for
$ k\equal{}1: cos(\pi/7)\plus{}cos(\pi/5)$
$ k\equal{}2: cos(5\pi/7)\plus{}cos(3\pi/5)$
$ k\equal{}3: cos(3\pi/7)\plus{}cos(3\pi/5)$
$ k\equal{}4: cos(3\pi/7)\plus{}cos(\pi/5)$
$ k\equal{}8: cos(\pi/7)\plus{}cos(3\pi/5)$
$ k\equal{}9: cos(5\pi/7)\plus{}cos(\pi/5)$.


function
inequalities
algebra proposed
algebra

Find the strictly increasing functions $f : \{1,2,\ldots,10\} \to \{ 1,2,\ldots,100 \}$ such that $x+y$ divides $x f(x) + y f(y)$ for all $x,y \in \{ 1,2,\ldots,10 \}$.

[i]Cristinel Mortici[/i]

Taking $y=x+1$, we obtain that $2x+1$ divides $xf(x)+(x+1)f(x+1)$. Since $2x+1$ also divides $xf(x)+(x+1)f(x)$ it follows that $2x+1$ divides $(x+1)(f(x+1)-f(x))$. But $\gcd (2x+1,x+1)=1$, therefore $2x+1$ divides $f(x+1)-f(x)$. This implies that $f(x+1)-f(x) \ge 2x+1$. Adding up these inequalities for $x=1,2,\ldots ,10$ yields $f(10)-f(1)\ge 99.$ Clearly, this is possible only if $f(10)=100, f(1)=1$ and all inequalities above are equalities. It is easy to check that these conditions imply $f(x)=x^2$ for all $x$.


You are given two circles of radius $r_{1}$ and $r_{2}$, with both external tanges drawn, and one internal tangent drawn. The internal tangent intersects the external tangents at $P$ and $Q$. What is the relationship between the length of $PQ$ and the lengths of the external tangents?

Let $r_{1}\geq r_{2}$ ,and assume that one of the external tangents touches circles ( with radius $r_{1}$ and $r_{2}$ with centers $O_{1}$ and $O_{2}$ respectively ) at $S_{1}$ and $S_{2}$ respectively and the other external tangent touches them at $T_{1}$ and $T_{2}$. to understand this solution, draw one of the internal tangents such that it intersect $S_{1}S_{2}$ at $Q$ and $T_{1}T_{2}$ at $P$ and also such that $QS_{1}\geq S_{2}Q$ and $T_{2}P \geq T_{1}P$. we'll prove that : $PQ = S_{1}S_{2}= T_{1}T_{2}$.
First of all I'll evaluate $QM$'s lenght in terms of $r_{1}$, $r_{2}$ and $\alpha$ ( $\alpha$ is the angle between $PQ$ and $O_{1}O_{2}$ ). It's easy to prove that $O_{1}O_{2}= \frac{r_{1}+r_{2}}{sin\alpha}$ and if $PQ$ is tangent to these circles ( with radius $r_{1}$ and $r_{2}$ ) at $N$ and $M$ respectively, then : $MN = \frac{r_{1}+r_{2}}{tan\alpha}$.
Now we know that $QS_{1}= QN = QM+\frac{r_{1}+r_{2}}{tan\alpha}$, and also $QS_{1}= S_{1}S_{2}-QS_{2}= \sqrt{(\frac{r_{1}+r_{2}}{sin\alpha})^{2}-(r_{1}-r_{2})^{2}}-QM$
As a consequence : $QM = \frac{\sqrt{(\frac{r_{1}+r_{2}}{sin\alpha})^{2}-(r_{1}-r_{2})^{2}}-\frac{r_{1}+r_{2}}{tan\alpha}}{2}$. Now in the same way we can prove that: $NP = \frac{\sqrt{(\frac{r_{1}+r_{2}}{sin\alpha})^{2}-(r_{1}-r_{2})^{2}}-\frac{r_{1}+r_{2}}{tan\alpha}}{2}$. So $NP = QM$.
Also we know that $S_{1}S_{2}= S_{1}Q+QS_{2}= QN+QM = QN+NP = PQ$  , and we're done.
( sorry for long long solution  :oops:  :D  :P )

that's fine. just what we needed. that's a tough problem.

Here's a significantly simpler solution. See the attached diagram.

By the tangent property:

$PA=PR$
$QD=QR$

Adding those two, we get

$PA+QD=PQ\qquad(1)$

Also

$PB=PS$
$QC=QS$

Adding those two, we get

$PB+QC=PQ\qquad(2)$

Adding $(1)$ and $(2)$ we get

$(PA+PB)+(QC+QD)=2PQ\implies AB+CD=2PQ$

But because of the symmetry, we have $AB=CD$, hence

$2AB=2PQ\implies PQ=AB$

Nice solution Farenhajt :D . I was also trying to find such a solution, but I think I made it much long...  :oops:

using fahrenhajt's diagram...

let $PR=x$, $RS=y$, and $SQ=z$
then $AP=x$, $PB=x+y$, $QC=z$, $QD=y+z$, then
$AB=2x+y$, and $CD=2z+y$, but $AB=CD$, so $x=z$

then $AB=x+y+x=x+y+z=PQ$ as desired


function
calculus
derivative
real analysis
real analysis unsolved

Define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = 1/(1+x^2)$.
Prove directly from the definition of uniform continuity that $f$ is uniformly continous on $\mathbb{R}$. Determine $\delta$ in terms of $\epsilon$ and show that your choice of $\delta$ actually works.

We could use the fact that this function has a bounded derivative, along with the mean value theorem, but what's the fun of that? Here's some pure algebra:

$f(x)-f(y)=\frac1{1+x^2}-\frac1{1+y^2}=\frac{y^2-x^2}{(1+x^2)(1+y^2)}$

$=\left( \frac{x+y}{(1+x^2)(1+y^2)}\right)(y-x)$

Now, $|2x|\le 1+x^2.$ (Easily proved directly; can be considered the AM-GM inequality.)

Thus $\frac{|x|}{(1+x^2)(1+y^2)}\le\frac{|x|}{1+x^2}\le\frac12.$

And similarly, $\frac{|y|}{(1+x^2)(1+y^2)}\le\frac12.$

Thus $\left| \frac{x+y}{(1+x^2)(1+y^2)}\right|\le\frac12+\frac12=1,$ and

$|f(x)-f(y)|\le|x-y|.$

Finally, given $\epsilon>0,$ if $|x-y|<\epsilon,$ then $|f(x)-f(y)|<\epsilon.$


There is an integer of each square of an $8x8$ chessboard. In one move, you may choose any $4x4$ or $3x3$ square and add 1 to each integer of the chosen square. Can you always get a table with each entry divisible by a) 2, b) 3?

If we increase all of a 3 by 3 square by 1, we add an odd number to either the 3rd row or the 6th row, leaving the sum of the other rows invariant mod 2. If we do the same for a 4 by 4 square, the sum of all numbers except those in the 3rd row and 6th row is still invariant. Thus, if the initial sum is odd, there will always be at least 1 odd term.

The argument for mod 3 is very similar.

hmm that is very similar to the solution given by Engel in his book, but what I don't understand is how does a sum of all the rows equal to $0 \bmod 2$ guarantee that each entry is an even number? especially if you omit the third and sixth rows from your sum? for example, couldn't you have like 62 even entries and 2 odd entries, yet the sum would still be even?

Yes this is from Engel.

I did not say that an even sum implies even terms, I simply said that an odd sum implies at least one odd term. If the sum of the terms is odd, it is impossible, if it is even, it may or may not be impossible.

Oh, but does that answer the original question? Because we showed that if the original sum is odd then we can't get a table with each entry divisible by $2$, but we haven't shown that we can get a table with each entry divisible by $2$... So even if the sum is even for all the entries this still doesn't mean each entry will be divisible by $2$ eventually

[quote="gotztahbeazn"]Oh, but does that answer the original question? [/quote]

[hide="answer"]The answer to the problem is "no".[/hide]

[hide="proof"]
Consider the mod 2-case:

1. (1+1=0) Since we work mod 2, using twice the same square is like doing nothing. Hence, a minimal method uses each square at most once.

2. (1+0=0+1) Addition being commutative, the order of the squares is irrelevant.

3. (1+1=0) If a given choice of 3x3 and 4x4 squares allows one to reduce the board to the blank board (i.e. with 0's), then the same choice applied to the blank board yields the board one started with.

4. There are 25 4x4 squares and 36 3x3 squares, which makes a total of 61 squares.

Thus the number of boards that can be constructed by the given method, starting from the blank board, is at most 2^61.

But there are 2^64 possible boards. With the condition that the sum of all numbers is even (i.e. 0 in our case), we still have 2^63 boards. QED

One can probably proove that only one eighth of all boards with even sum can be reduced to the blank board.

The same method will work in the mod 3 case as well.
[/hide]

---
zpz

zpz's proof is an extension of the problem, but mine is a sufficient answer to the original problem. The original question was, "is it always possible ..." and it is never possible when the sum of the all rows except the 3rd and 6th are odd.


analytic geometry
graphing lines
slope
calculus
derivative
conics
parabola

Ok ive been workin on this for since sunday and i am simply out of ideas
i know it may be super easy for some but if anybody could take the time to show me how to solve this, I would appreciate it greatly!!

$C_{1}: y=x^{2},\ C_{2}: y=-x^{2}+6x-5.$
Let $A(s,\ s^{2})$ be the tangent point on $C_{1}$ and $B(t,-t^{2}+6t-5)$ for $C_{2}.$ From $y=x^{2}\Longrightarrow y'=2x,\ y=-x^{2}+6x-5\Longrightarrow y'=-2x+6$ we have the tangent lines at these  points$y=2sx-s^{2}$ and $y=-2(t-3)x+t^{2}-5.$ The condition of coinciding these tangent lines gives .....

P.S.

We have at least another basic solutions without derivative.

sorry also i am a little confused how you found,
y=2sx-s^2 and y=-2(t-3)x+x^2 -5
and what they mean

thanks again for the help!

EDIT

ok so i think i found out, you got those numbers using the point slope form with the derivatives at x for the slope,....what is the next step!?

You seem to be not able to use the differentiation, :lol:

I will give you a solution.

Let $y=ax+b$ be the desired tangent line, we can write as follows. \[x^{2}-(ax+b)=(x-s)^{2}\]  \[-x^{2}+6x-5-(ax+b)=-(x-t)^{2}\] Since these are identical equation with respect to $x,$ we have .....

hmmm, lol perhaps this is too much for me haha, i did see how you got 2sx-s^2, but the -2(t-3)x.... i cannot get.
as for your last reply with the solutions, i am still lost as to where
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/a7719d550f083260f78bc1e8cb8caf94.gif[/img] came from.

sorry for not understanding  :blush:

Sorry You seemed to be able to use differentiation, :lol:

Set $x^{2}-(ax+b)=0,$ what does this root mean? :)

$x^{2}-(ax+b)=(x-s)^{2}=0\Longleftrightarrow x=s,$ which is the double root, that is to say the abcissa of the tangent point of the parabola $y=x^{2}$ and the tangent line $y=ax+b.$

ok,...i guess i am confused back when you went from
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/3bace7ca4e4761ae2af9ec63ce487af6.gif[/img] and [img]http://www.mathlinks.ro/Forum/latexrender/pictures/23fe31c2ea6cb3474b42736d18e72dae.gif[/img]
           to this....
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/a7719d550f083260f78bc1e8cb8caf94.gif[/img] and the -x2+6x...=-(x-t)^2

Generally the tangent line at the tangent point $(t,\ f(t))$ of $y=f(x)$ is $\boxed{y-f(t)=f'(t)(x-t)}.$

One line solution:

The tangent line $y=ax+b$ contacts with the palabolas simultaneously, thus  the following quadratic equations have double roots. \[x^{2}=ax+b.\]  \[-x^{2}+6x-5=ax+b.\] Set the discriminants$D_{1},\ D_{2}=0,$ we have $D_{1}=a^{2}+4b=0,\ D_{2}=(a-6)^{2}-4(b+5)=0.$

ahhhh,..sorry i see where i confused you,...I AM trying to solve this using differentiation.  sorry i was confused earlier

so using differentiation the equation for the tangent lines are [img]http://www.mathlinks.ro/Forum/latexrender/pictures/3bace7ca4e4761ae2af9ec63ce487af6.gif[/img] and [img]http://www.mathlinks.ro/Forum/latexrender/pictures/23fe31c2ea6cb3474b42736d18e72dae.gif[/img]????


That 'becomes grey' is for comments.  You can use these to leave notes in the file that won't come out when you typeset.


[color=red]ASKHSH 11[/color]

Estw oi diakekrimenoi migadikoi $z_{1},z_{2},z_{3},z_{4}$ me $|z_{1}|=|z_{2}|=|z_{3}|=|z_{4}|$. Ipothetoume oti iparxei pragmatikos arithmos $t\neq 1$ gia ton opoio isxiei

$|tz_{1}+z_{2}+z_{3}+z_{4}|=|z_{1}+tz_{2}+z_{3}+z_{4}|=|z_{1}+z_{2}+tz_{3}+z_{4}|$

Na apodeiksete oti oi eikones twn $z_{1},z_{2},z_{3},z_{4}$ sto migadiko epipedo sximatizoun orthogwnio parallilogrammo.

Alexandros

Θέτουμε $a=z_{1}+z_{2}+z_{3}+z_{4}$ και έχουμε

$|(t-1)z_{1}+a|=|(t-1)z_{2}+a| \Leftrightarrow z_{1}\overline{a}+\overline{z_{1}}a=z_{2}\overline{a}+\overline{z_{2}}a \Leftrightarrow z_{1}z_{2}\overline{a}=|z_{1}|a$ .

Η πρώτη ισοδυναμία προκύπτει υψώνοντας στο τετράγωνο και τα δύο μέλη της ισότητας και χρησιμοποιώντας την ιδιότητα $|z|^{2}=z\overline{z}$ για κάθε μιγαδικό $z$ και η δεύτερη πολλαπλασιάζοντας με $z_{1}z_{2}$ και τα δύο μέλη και χρησιμοποιώντας ότι $|z_{1}|=|z_{2}|$ .

Όμοια προκύπτει ότι $z_{1}z_{3}\overline{a}=|z_{1}|a$ και επειδή οι $z_{1},z_{2},z_{3}$ είναι διαφορετικοί μεταξύ τους θα ισχύει $a=0 \Leftrightarrow z_{1}+z_{2}+z_{3}+z_{4}=0$. Όμως οι εικόνες των $z_{1},z_{2},z_{3},z_{4}$ βρίσκονται στον ίδιο κύκλο κέντρου Ο(0,0) (αφού οι μιγαδικοί έχουν ίσα μέτρα), άρα στο μιγαδικό επίπεδο σχηματίζουν ορθογώνιο παραλληλόγραμμο.

Στέλιος

Etsi akrivws ksekinisa ki egw Stelio. Poli swsta! Stin askisi auti einai poli eukolo na xatheis stis prakseis an den ekmetaleuteis tin gewmetriki eukolia pou sou parexoun oi migadikoi.

Me tin isotita $|a-(1-t)z_{1}|=|a-(1-t)z_{2}|=|a-(1-t)z_{3}|$ simperainoume afenos (apo tin prwti isotita) oti o migadikos $a$ kineitai panw stin mesokatheti tou tmimatos pou prokiptei apo tis eikones twn migadikwn $(1-t)z_{1}$ kai $(1-t)z_{2}$ kai afeterou (apo tin deuteri isotita) oti o migadikos $a$ kineitai panw stin mesokatheti tou tmimatos pou prokiptei apo tis eikones twn migadikwn $(1-t)z_{2}$ kai $(1-t)z_{3}$.  Sinepws i eikona tou migadikou $a$ einai akrivws to kentro tou kiklou ston opoio kinountai oi $z_{1},z_{2},z_{3},z_{4}$ diladi i arxi twn aksonwn. Ara $a=0$.

Epeidi mporei na iparxei kapoios pou den katalavainei me profani tropo giati twra oi eikones twn $z_{1},z_{2},z_{3},z_{4}$ dimiourgoun orthogwnio grafw tin apodeiksi.

Exoume $z_{1}+z_{2}+z_{3}+z_{4}=0 \ \ (1)$ kai xwris vlavi tis genikotitas ipothetoume oti i thesi twn $z_{1},z_{2},z_{3},z_{4}$ panw ston kiklo einai me tin seira pou anaferontai (me tin thetiki fora,tin antistrofi tou rologiou)

 Tote oi $z_{1},z_{3}$ and $z_{2},z_{4}$  einai oi korifes twn diagwniwn tou tetrapleurou kai $|z_{1}+z_{3}|=|z_{2}+z_{4}|$ (Xrisimopoiwntas tin
($1$)). Omws genika ean $|z_{1}+z_{3}|=|z_{2}+z_{4}|$ and
$|z_{1}|=|z_{2}|=|z_{3}|=|z_{4}|$ tote einai eukolo na deiksoume (mia apli ipswsi sto tetragwno arkei) oti $|z_{1}-z_{3}|=|z_{2}-z_{4}| \ \ (2)$. Ap'tin alli meria oi sxeseis
$|z_{1}-z_{2}|=|z_{3}-z_{4}|$ (Apo tin $(1)$ pairnoume omoia $|z_{1}+z_{2}|=|z_{3}+z_{4}|$)  and
$|z_{1}-z_{4}|=|z_{2}-z_{3}|$ (Apo tin $(1)$ pairnoume omoia $|z_{1}+z_{2}|=|z_{3}+z_{4}|$) to opoio simainei oti to tetrapleuro einai parallilogrammo (afou oi apenanti pleures einai ises) to opoio simainei oti einai kai orthogwnio (dioti deiksame apo tin $(2)$ oti kai oi diagwnies einai ises)

Alexandros

[color=red]ASKHSH 12[/color]

$15$ Elefantes stekontai sti seira. Ta vari tous ekfrazontai se akeraio arithmo kilwn. To athroisma tou varous kathe elefanta - ektos apo ton teleutaio deksia - me to diplasio tou varous tou deksiou tou geitona, einai akrivws $15$ tonoi. Na vrethei to varos kathe elefanta.

Alexandros

μα τι το προκλιτικο εχει η εκφωνηση?αν δε θες ελεφαντες μπορουμε να βαλλουμε αλλα ζωα :P

Βάζω ένα πολύ ενδιαφέρον πρόβλημα .
[color=red]ΑΣΚΗΣΗ 13[/color]
Γράψτε το παρακάτω πολυώνυμο ως γινόμενο από μη παραγοντοποιήσιμα πολυώνυμα στο $Z[X]$
$f(X)=X^{2005}-2005X+2004$

[quote="RAKAR"]esto akolu8ia $a_{n}$ me$a_{1}=1$ kai
$a_{n}=n(a_{n-1}+1)$ gia ka8e $n\geq 2$
Na ipologisete to

$K=\prod(1+\frac{1}{a_{i}})$ (apo i=1 mexri apiro)

 :D[/quote]

[u][b]Λύση της άσκησης 14 [/b][/u]

Χρησιμοποιώντας την $a_{n}=n(a_{n-1}+1)$ παίρνουμε $\prod_{i=1}^{n}\frac{a_{i}+1}{a_{i}}= \frac{a_{n}+1}{n!}=b_{n}$ .

Επίσης $a_{n}=n(a_{n-1}+1) \Rightarrow \frac{a_{n}+1}{n!}= \frac{a_{n-1}+1}{(n-1)!}+\frac{1}{n!}\Rightarrow b_{n}=b_{n-1}+\frac{1}{n!}$

και η τελευταία μας δίνει $b_{n}=1+\sum_{i=1}^{n}\frac{1}{n!}$ άρα $\lim_{n \to \infty}b_{n}=e$ συνεπώς $K = e$ .

Καλές γιορτές σε όλους  :)

[u][color=red]ASKHSH 16[/color][/u]

Kalispera se olous kai Xronia Polla,

Me aformi tin askisi 13 tou Silouanou, kai ws genikeusi autou exw na valw mia askisi (den kserw an eixe kati analogo sto mialo tou o Silouanos otan tin egrapse). Pira aformi kai apo mia palaioteri aporia - eikasia (thimamai kala?) tou Silouanou gia to thema.

Na apodeixthei oti to poliwnimo $P(x)= x^{n}+2x^{n-1}+3x^{n-2}+\cdots+nx+(n+1)$ einai anagwgo sto $\mathbb{Z}[X]$ diladi den mporei na graftei san ginomeno dio (mi statherwn) poliwnimwn me akeraious sintelestes.

Alexandros

Y.G. Kanontas xrisi merikwn thewrimatwn tis thewrias Swmatwn (Thewria Galois) mporoume na vgaloume to apotelesma alla oxi me poli kompso tropo. Etsi to eixa antimetwpisei tote (giati den eimoun sigouros ean ipirxe stoixeiwdis lisi gi'auto) gi'auto to logo den eixa grapsei tin apopsi mou. Τwra pisteuw oti mporei na antimetwpisthei stoixeiwdws.

Αλέξανδρε, είχα διαβάσει κάπου ότι είναι ανοικτό πρόβλημα ακόμη το αν το πολυώνυμο $Q(x)=(n+1)x^{n}+nx^{n-1}+.....+2x+1$ είναι μη παραγοντοποιήσιμο στο $Z[x]$ για κάθε τιμή του φυσικού $n$. Αν λοιπόν θεωρήσουμε το πολυώνυμο $P(x)=x^{n}+2x^{n-1}+.....+nx+(n+1)$ προκύπτει ότι $Q(x)=x^{n}P\left(\frac{1}{x}\right)$. Συνεπώς το $Q(x)$ είναι μη παραγοντοποιήσιμο στο $Z[x]$ αν και μόνο αν το $P(x)$ είναι μη παραγοντοποιήσιμο στο $Z[x]$, άρα είναι και αυτό ανοικτό πρόβλημα. Μου φαίνεται δηλαδή περίεργο να υπάρχει στοιχειώδης λύση για αυτή τη γενική περίπτωση, εκτός βέβαια αν έχω κάπου λάθος στο συλλογισμό μου.

[quote="silouan"][color=red]ΑΣΚΗΣΗ 13[/color]
Γράψτε το παρακάτω πολυώνυμο ως γινόμενο από μη παραγοντοποιήσιμα πολυώνυμα στο $Z[X]$
$f(X)=X^{2005}-2005X+2004$[/quote]
Σιλουανέ έχεις λύση έστω και για αυτή την ειδική περίπτωση; Μοιάζει αρκετά δύσκολη άσκηση.

[quote="RAKAR"]Na diksete oti to polionimo

$P(x)=(x-a_{1})^{2}(x-a_{2})^{2}......(x-a_{n})^{2}+1$

opu $n\geq 1$ kai $a_{1},a_{2},...a_{n}$ ine diafori ana dio akerei ari8mi,

den analiete se ginomeno dio polionimon me akereus sintelestes.

 :D[/quote]

ASKHSH -15 LYSH
Dείτε τη λύση μου παρακάτω
http://www.mathlinks.ro/Forum/viewtopic.php?t=54591

Katarxin XRONIA POLLA kai KALH XRONIA me YGEIA SE OLOYS kai perissotera Mathimatika tin epomeni xronia! (Ante na kseperasoume kai ton Kanada, dioti tin Roumania se liges meres tha tin exoume afisei pisw).

[quote="stedes"]Αλέξανδρε, είχα διαβάσει κάπου ότι είναι ανοικτό πρόβλημα ακόμη το αν το πολυώνυμο $Q(x)=(n+1)x^{n}+nx^{n-1}+.....+2x+1$ είναι μη παραγοντοποιήσιμο στο $Z[x]$ για κάθε τιμή του φυσικού $n$. Αν λοιπόν θεωρήσουμε το πολυώνυμο $P(x)=x^{n}+2x^{n-1}+.....+nx+(n+1)$ προκύπτει ότι $Q(x)=x^{n}P\left(\frac{1}{x}\right)$. Συνεπώς το $Q(x)$ είναι μη παραγοντοποιήσιμο στο $Z[x]$ αν και μόνο αν το $P(x)$ είναι μη παραγοντοποιήσιμο στο $Z[x]$, άρα είναι και αυτό ανοικτό πρόβλημα. Μου φαίνεται δηλαδή περίεργο να υπάρχει στοιχειώδης λύση για αυτή τη γενική περίπτωση, εκτός βέβαια αν έχω κάπου λάθος στο συλλογισμό μου.
[/quote]

Exeis apolito dikio kai zitw siggnwmi gi'auto!! Eixa kati allo sto mialo mou to opoio nomiza oti genikeuotan se auto pou egrafa, alla apodeixthike teleiws lathos. Siggnwmi pou argisa kiolas na apantisw alla eleipa gia giortes!!

Alexandros

φίλε ράλλη είναι σίγουρο ότι έχει λύσεις γιατί εέλγξα την διακρινούσα στο τριώνυμο που εμφανίζεται (το πώς εσύ θα μου πείς) αλλά δεν έλεγξα τις λύσεις όπου πιθαν'ώς  να απορρίπτονται λόγω ριζικού

Aμβροσιε μηπως στο  β μελος της εξισωσης αντι για 15 ηθελες να γραψεις 15χ? (απλα μου παει καλυτερα).Παντος λυνεται ευκολα με 2 υψωσεις τετραγωνων αλλα η μεθοδος αυτη......... :(

φίλε μου κάνε πολλσμο με τησ υζυγη παράσταση κι προσέθεσε τις 2 ισότητεσ
κι θα πάρεις ένα ριζικό

:blush: Καλη σκεψη Αμβροσιε. :)

τι νόμιζες θα γέμιζα πόστς το φόρουμ χωρίς σοβαρό λόγο??
είπαμε καλό φόρουμ αλλα κι ουσιαστικό

Πολυ καλη σκεψη παντος Αμβροσιε. :wink:

ευχαριστώ
ειθε να αμειφθεί η σκέψη αυτή όπως πρέπει

ίδια ιδέα με την περσινή πρώτη άσκηση του Θαλή Β Λυκείου  :wink:

[color=red]ASKHSH 16[/color]

Enas mathitis Likeiou gia na lisei tin eksiswsi \[\sqrt{3x-2}-\sqrt{2x-3}=1 \ \ (1)\]
"ipswse" eis to tetragwno kai egrapse $3x-2-2x-3=1$ pou linontas tin kataligei sto $x=6$. Omws i teleutaia, einai lisi tis eksiswsis $(1)$.
Na apodeiksete oti iparxoun apeires tetoies tetrades $(a,b,c,d)$ gia tis opoies  i parapanw methodos "ipswsis eis to tetragwno" gia tin eksiswsi \[\sqrt{ax-b}-\sqrt{cx-d}=1\] odigei se swsti lisi.

Tha itan protimotero, ean dinate mia pliri lisi sto parapanw provlima kai oxi apla mia oikogeneia tetradwn pou ikanopoiei to provlima gia na fainetai kai o tropos lisis kai skepsis. Eimai sigouros oti de tha xreiastei na valw lisi se merikes imeres.

Alexandros

agapite cretanman,

mipos 8a mporouses na valeis ti lysi
giati prokeitai gia ena poly endiaferon provlima...kai arketa asyni8isto :wink:

pantos to siguro ine oti i tetrades $(a,b,c,d)=(a,-d-1,a,d)$ine lisis.. :P
[hide]cretanman.evgala mia sxesi pu prepi na isxii..ala iparxi ena <keno>....stis periptosis...
dialdi gia $a<>c$ an isxiun episis
$ad+a+bc\geq 0$
$cb+2cd+c-ad\geq 0$
tote $a=\frac{cb+c(d+1)^{2}}{d^{2}+d}$
ala an dn isxiun aytes i dio i sin8ikes?
exo provlima tote... :wallbash:
[/hide]
 :D

Mporei na min egina katanoitos prin legontas oti "tha itan protimoteri mia pliris lisi tou provlimatos". To "pliris lisi" pigaine sto na vreite me opoio tropo thelete kai me opoies sinthikes, mia oikogeneia apo apeires liseis! Den einai anagi na tis vreite OLES tis oikogeneies lisewn. Eksallou to provlima akrivws auto zitaei. Na apodeiksete oti iparxei apeiria lisewn. Etsi, mporeite na kanete  OPOIES paradoxes thelete, prokeimenou na parete mia oikogeneia lisewn kai na aplopoiisete to provlima.

Mono kai mono gia na iparxei o tropos skepsis eipa oti tha itan protimotero na grapsete pws vrikate tin lisi kai oxi na petaksete apla mia oikogeneia lisewn!

Egw exw sto mialo mou 2,3 oikogeneies lisewn. Se kamia periptwsi omws oles!

Alexandros


Δινεται ενας πρωτος $ p\geq2$.
Να βρεθει το υπολοιπο της διαιρεσης του $ \binom{2p}{p}$ με τον $ p$.

Ο συντελεστης ισουτε με:

$ \frac {2(p \plus{} 1)(p \plus{} 2)...(p \plus{} p \minus{} 1)}{(p \minus{} 1)!} \equiv k (modp) \Rightarrow 2(p \plus{} 1)(p \plus{} 2)...(p \plus{} p \minus{} 1) \equiv k(p \minus{} 1)! \equiv \minus{} k (modp)$

Ομως $ 2(p \plus{} 1)(p \plus{} 2)...(p \plus{} p \minus{} 1) \equiv 2(p \minus{} 1)! \equiv \minus{} 2 (modp)$

Αρα $ k \equiv 2 (modp)$ και το υπολοιπο ειναι $ 2$

ΥΓ: Γινεται και χωρις Wilson, αφου ο $ p$ ως πρωτος ειναι σχετικα πρωτος με τους: $ 1,2,...p \minus{} 1$, και επομενος ο $ (p \minus{} 1)!$ εχει αντιστροφο $ (modp)$


If $ x$, $ y$ and $ z$ are positive reals prove the inequality (sums are asssumed to be cyclic):
$ \sum {\frac{1}{2x^{2}+xy+xz}} \geq \sum {\frac{1}{x^{2}+xy+yz+zx}}$

[quote="delegat"]If $ x$, $ y$ and $ z$ are positive reals prove the inequality (sums are asssumed to be cyclic):
$ \sum {\frac {1}{2x^{2} + xy + xz}} \geq \sum {\frac {1}{x^{2} + xy + yz + zx}}$[/quote]
$ \sum_{cyc}\left( {\frac{1}{2x^{2}+xy+xz}}-{\frac{1}{x^{2}+xy+yz+zx}\right)=\sum_{cyc}\frac{(x-y)^2(2x+2y+z)}{2xy(x+y)(2x+y+z)(2y+x+z)}.}$


On a 8 X 12 grid of squares, a line is drawn from one corner to the opposite corner. How many squares does it pass through?

The ratio $ 8: 12 \equal{} 6: 9 \equal{} 4: 6 \equal{} 2: 3$
So the line passes through $ (8 \minus{} 1) \plus{} (12 \minus{} 1) \equal{} 18$ horizontal/vertical lines
Since two sets of lines are intersecting - at $ (2,3)$ and $ (4,6) and (6,9)$ - subtract $ 3$
So it passes through $ 15$ squares


Suppose $ a,c,d \in N$ and $ d|a^2b\plus{}c$ and $ d\geq a\plus{}c$
Prove that $ d\geq a\plus{}\sqrt[2b] {a}$

This problem was incorrectly stated,it should be:
Suppose that $ a,b,c,d$ are natural numbers such that $ d|a^{2b} \plus{} c$ and $ d\geq a \plus{} c$.
Prove that
\[ d\geq a \plus{} \sqrt [2b]{a}
\]
[b]Solution:[/b]
Suppose that $ d \equal{} a \plus{} k$,therefore:
\[ a \plus{} k|a^{2b} \plus{} c
\]
On the other hand $ a \plus{} k|a^{2b} \minus{} k^{2b}$ combining it with the given relation $ a \plus{} k|a^{2b} \plus{} c$,we conclude that
$ a \plus{} k|k^{2b} \plus{} c$.
Therefore,
$ \boxed{k^{2b} \plus{} c\geq a \plus{} k\geq a \plus{} c}$.
Finally
\[ k\geq\sqrt [2b]{a}
\]
But this means that $ d \equal{} a \plus{} k\geq a \plus{} \sqrt [2b]{a}$.


How many ways can a person fill bags of candy with 5 pieces of candy if his only choices are sour balls, jawbreakers, and black licorice?
A)5
B)6
C)7
D)8
E)9

Are you sure the answer is one of the choices. Because I did balls and urns at first, but when that didn't get the right answer, I actually started to list them. It's definitely more than 9.
If you did it with balls and urns:
[hide]
5 balls and 2 dividers, so $ \dbinom{7}{2}$ = 21?[/hide]

Are you sure that you are reading the problem right? I don't see how you need to use dividers in here. Or maybe you
're doing it some other way.

But, yeah, this is the correct problem. It is from the UIL "C" test.

[hide="Maybe?"]
I agree with mathcountsfreak;
SSSSJ
SSSJJ
SSJJJ
SJJJJ
JJJJJ
SSSSB
SSSBB
SSBBB
SBBBB
BBBBB

So trivially there are more than 10 (his solution is correct too under this interpretation). Unless the problem intended that each candy is picked at least once. Then each candy must be picked at least once so three of the candies are determined, and then of the remaining two there are $ 6$ possibilities (by simple casework $ {3\choose 1}\plus{}{3\choose 2}\equal{}6$ or by ball-and-urns: $ {2\plus{}2\choose 2} \equal{} 6$).
[/hide]

I also got an answer that wasn't out of the choices. If you pick 5 candies out of 3 types, then its the number of distinct arrangements of XXXXX||, where X is a candy chosen and the two |'s divide them into the three types. The number of distinct arrangements of this is $ \frac{7!}{5!2!} \equal{} 21$

@iamhe:  I looked at the UIL "C" test (this is problem #50), and you have the wrong answer choices.

The correct answer choices are:
A. 21
B. 30
C. 60
D. 125
E. 243

Looks like you accidentally used the answer choices from problem #49.


[quote="kimby_102"][quote="edyfox"]Show that the equation $5n^{2}=36a^{2}+18b^{2}+6c^{2}$ has no other integer solutions except $n=a=b=c=0$.[/quote]
[hide]
first note that $n$ has to be divisible by $6$.
so let $n=6\alpha$, for some integer $\alpha$.
Then our given equation becomes:
$5\cdot 6\alpha^{2}=6a^{2}+3b^{2}+c^{2}$.
then we observe that $c$ has to be divisible by $3$, so let $c=3\beta$.
thus we got:
$10\alpha^{2}=2a^{2}+b^{2}+3\beta^{2}$.
what should i do now?

[/hide][/quote]

I got to the exact same place and i stopped :\

and one more thing that $b$ and $\beta$ have the same parity.......

$b$ is divisible by $5$, but I can't get to a contradiction immediately, which annoys me.

I also got stuck. does someone have an idea?

$n^{2}=7a^{2}+3b^{2}+c^{2}+\frac{a^{2}+3b^{2}+c^{2}}{5}$

look at the square residues... $a,b,c$ all have to be even, or all odd

suppose they are all even...then substitute a new variable...and then we see n would be even also, then you cancel, but this can be repeated forever! so if they are even, then they are integers infinitely divisible by 0 giving $(n,a,b,c)=(0,0,0,0)$

but suppose that they are all odd...

stuck  :P


What is usually the score on the USAMO to make Red MOP?

Can eighth graders make it to RED MOP or do they have to qualify for black or blue MOP?

THANKS!! :spam:  :juggle:  :ddr:

1. depends on the year, last year it was 10, year before it was 9 but its usually around the 9 region.

2. no, 8th grader cannot qualify for red mop they have to qualify for black/blue MOP

To moderators and community in general: can we please please please have a big Q&A thread where these things can be answered? or at the very least a link to the AMC handbook. Please.

That's what I've been trying to do with the FAQ: Other Questions thread, but it hasn't worked much. Feel free to link to that.


[quote="jensen"]i want some easy problems in number theory.if you can and you have plz send me some easy problem,you'll help me alot . thanks . :D[/quote]Let the n-digit decimal number M:=1...1 (all digits equal 1) be such that  p:=n+1 > 5  is a prime. Then  p|M.  (Prove it).

Write $M = \sum_ {k=0}^{n-1} 10^k = \frac{10^n - 1}{9}$. Since $p = n+1$ is a prime $> 5$, by Fermat's little theorem $p \mid (10^n - 1)$. Hence by Euclid's lemma $p \mid M$, q.e.d.


Here is a problem that I have composed but not solved. If true, this would generalize the following original problem:

http://www.mathlinks.ro/viewtopic.php?p=20253#20253

I suspect that there is an easy counterexample, though. So, give it a try!

Let $A$ be a finite set of real numbers such that any $a \in A$ is expressible under the form $a=b+c$, where $b,c \in A$. Is it necessarily true that there exist pairwise distinct elements $a_1,\ldots,a_k \in A$ for which
\[
a_1 + \cdots + a_k = 0?
\]

[quote="vess"]Let $A$ be a finite set of real numbers such that any $a \in A$ is expressible under the form $a=b+c$, where $b,c \in A$. Is it necessarily true that there exist pairwise distinct elements $a_1,\ldots,a_k \in A$ for which
\[
a_1 + \cdots + a_k = 0?
\][/quote]
This statement is true. Consider a directed graph with vertices formed by elements of $A$ and directed edges $(a,b)$ and $(a,c)$ for every identity $a=b+c$ with $a,b,c\in A$. Then for every vertex there exists an outgoing edge, implying that there exists a directed cycle $(a_1,a_2,\dots,a_k,a_{k+1}=a_1)$.
Then for every $i=1,2,\dots,k$, we have $a_i = a_{i+1} + b_i$ for some element $b_i\in A$ and hence
\[\sum_{i=1}^k a_i = \sum_{i=1}^k ( a_{i+1} + b_i )\]
implying that
\[\sum_{i=1}^k  b_i = 0\]
as requested.

[This statement is true. Consider a directed graph with vertices formed by elements of $A$ and directed edges $(a,b)$ and $(a,c)$ for every identity $a=b+c$ with $a,b,c\in A$. Then for every vertex there exists an outgoing edge, implying that there exists a directed cycle $(a_1,a_2,\dots,a_k,a_{k+1}=a_1)$.
Then for every $i=1,2,\dots,k$, we have $a_i = a_{i+1} + b_i$ for some element $b_i\in A$ and hence
\[\sum_{i=1}^k a_i = \sum_{i=1}^k ( a_{i+1} + b_i )\]
implying that
\[\sum_{i=1}^k  b_i = 0\]
as requested.[/quote]

pairwise distinct elements?


For x,y,z positive reals such that x+y+z=1 prove that
\[  x^4+y^4+z^4 +\frac 1{27} \geq \frac 23 (xy+yz+zx)^2 . \]

\[ x^4+y^4+z^4+\frac 1{27} \geq \frac {(x+y+z)^4}{27}+\frac {1}{27} =\frac 2{27} \quad (1) \]
\[ \frac 2{27} \geq \frac 23 (xy+yz+zx)^2 \Leftrightarrow 1 \geq 9(xy+yz+zx)^2  \quad (2) \]
from (1) and (2) and $(x+y+z)^4 \geq 9(xy+yz+zx)^2$ which is obvious, the result follows.

[Admin edit: Made Latex Compliant]


calculus
derivative
analytic geometry
graphing lines
slope
function

not sure, but i think the calculus superbowl hello is talking about is run by this site, and info on it can be found here too.

http://www.calculus-help.com

Math is so much more enjoyable with functions of one variable compared to two variables.

[quote="JBL"]Well, I have no idea what the Calculus Superbowl is, but if you think about what a vertical tangent line means in terms of the derivative, you'll see that it's when the slope of the curve is infinite -- that is, the derivative should be undefined.  The equation of such a line is x = k, where k is just some real number.  Note that the derivative being undefined does not necessarily mean that there is a vertical tangent line -- consider the graph of y = |x| at x = 0 to see why -- but if there is a vertical tangent line, the derivative will not be defined.  An examle of this would be y = (x - 1)1/3 at the point x = 1.[/quote]

If you want a horizontal tangent, you could set the derivative equal to 0.

[quote="hello"]I was wondering whether anybody here is participating in the Calculus superbowl. Also can somebody help me by telling how to find an equations of a vertical tangent live when having the derivative.[/quote]

The calculus bowl is run by the calculus-help website. It is a 10 questions 3 hours test that is in a similar format to the free response ap problems, there are no calculators allowed and you have to work in groups of three. Also has anyone here done the physics bowl 40 multiple choices 45 minutes.

[quote="hello"]I was wondering whether anybody here is participating in the Calculus superbowl. Also can somebody help me by telling how to find an equations of a vertical tangent live when having the derivative.[/quote]

Take the reciprocal of the derivitive and set it equal to 0.

..... I don't get it, I just looked at the 2003 questions, and there is no possible way in all of hell that the top three students at any given school shouldn't get a perfect score on that test. Those questions are so easy... What's the catch, is it like 4 minutes per problem? Is it 30 minutes total for the test? I don't understand, I could do most of those problems in my head.

It is probably a not very high level competition and is similar to the level of the ap exam. This is irrelevent but can someone explain to me how to determine for solids of revolutions volume which function to subtract from which in the integrand. Thanks

[quote="hello"] Also has anyone here done the physics bowl 40 multiple choices 45 minutes.[/quote]

taking it for the first time this year  :)
but i am pretty familiar with it

[quote="hello"]This is irrelevent but can someone explain to me how to determine for solids of revolutions volume which function to subtract from which in the integrand. [/quote]

By graphing it you can easily see which to subtract. You usually subtract the bottom function from the top, unless the figure is backwards(this is kinda hard to explain and not too common anyway).

Also, have any tips on how you scored high on physicsbowl? Thanks

Um, can you pm me, there are some things I would rather just tell you. But I did like 10 practice contests. I would say it is just knowledge of key concepts and basic problems solving skills, in my opinion the toughest amc problem is much harder then a physics bowl problem.


$\frac{1}{cosx}+ctg3x=ctg\frac{3x}{2}$
solve for x?

[quote="midorya"]sorry for reviving thread
[hide="answer"]
x=p/8+pn/2
x=p/4+pn
[/hide][/quote]
Could you post your solution please?
[hide] $x=\frac{\pi}{8}+\frac{n\pi}{2}$, $x=\frac{\pi}{4}+n\pi$ [/hide]

[quote="Chocoboom"]What's $ctg$? cotangent?[/quote]
Yes. In some countries (including my own) tangent and cotanget are denoted as $tg$ and $ctg$.


I think this is middle school level. So others don't answer it. Its a really common question though.

An arab had 17 horses and 3 sons. After he died, his will was read out. It said "One-Half of my horses should goto the eldest son. One-Third of the Horses should goto the second son and One-Ninth should goto the youngest son". Ofcourse as you can see, 17 is not divisible by 2 or 3 or 9 so the sons started fighting and contemplating whether they should kill the horse(s). Then a wise old man came along (tut..tut.. what a stereotype) with his horse and donated it to the brothers. Now totally there were 18 horses.

The first son took half(9), the second son took 1/3rd (6) and finally(2). But what was surprising was that 9+6+2 = 17. Which means that the wise man's horse too was remaining  behind and happily he took it back home and all's well in the galaxy.

Nice story eh?? The question is.. What was the problem in the man's will? What was the fact that enabled it work with the wise man's horse but not with his own horses. Or simply said, give me the mathematical moral of the story. :D

[quote="Cyber-Shreyas"]9+6+3 = 17[/quote]
For a second there, I thought this might be interesting...

For further referrences on this exact topic refer: Math Adventures (or whatever) by Shairesh Shirali! There's an excellent account about this there  :)  including finding all genl solns for the sum

EDIT: As this topic says for middle school students i wont post more  :)

Fishy beat me...

The fractions didn't add up to one, but rather, 17/18.

wow......i tought it was the horse that are fake.......

[quote="Chronoz"]Yup! As simple as that.

The verdict is that the sons did not follow the advice of the father properly. Disobedient sons. Infact 9 is more than half of 17.... The 1st son got more than he should have. ;)[/quote]

You know, they could've just used the "Fight to the Death" option and avoided a math problem that they couldn't solve but a middle schooler could. I think the horses should be sold to get them in school.

huh?this is messed up but i kno that the first son got more than he should have also like the others said :D


Hi! I am someone who is preparing for the AMC 12 and AIME. I am going into Grade 10 and do not have a large amount of intuivity (certain questions which are all about intuitive thought I cannot solve; e.g. 2009 CMO #2). Can experience account for this (even at olympiad level) or will I always be at a disadvantage? Thanks guys for answering even though this is a weird question.

It's really a combination of both. People who've worked a long time have seen how different tools work in different occasions and are better fit to see where to use them. However, a lot of ingenuity does just come from natural talent.

I think you can actually improve your intuition by solving a difficult problem in many different ways. That should help build a different, more effective way of thinking.


How are the .gif files in AoPS's latexrender directory managed? I'm enrolled in an online homeschool, and I was thinking I could use the $ \text{\LaTeX}$ files rendered on AoPS to embed (with <img> tags) in the homework I submit electronically (my school doesn't support $ \text{\LaTeX}$.. what a pity). Files seem to be created in the latexrender directory even for post previews.. do those get deleted?

Hopefully this isn't against the TOS or anything.. I'll drop it if it is :P

I can't speak for AoPS policy so will suggest alternatives:
1. Go into the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url], create your gif then copy/save the picture to your computer and use it where you like;
or
2. use [url=http://www.forkosh.com/mathtex.html]mathTeX[/url] which allows you to create img tags on any computer which are converted to pictures. Thus <img src="http://www.forkosh.dreamhost.com/mathtex.cgi?c=\sqrt{a^2+b^2}" /> results in $ c\equal{}\sqrt{a^2\plus{}b^2}$. See [url=http://sixthform.info/steve/wordpress/?p=65]mathTeX helper[/url] for more details and a tool to make it simpler to use. LaTeX is not required on your computer or your school.

the mathTeX tool is very useful. Thanks for the heads-up!


Let A be the set of real numbers x such that 0<x<=1. For every x which is an element of A, let Ex be the set of real numbers y such that 0<y<x. Then prove :

the intersection of all sets Ex = empty

any ideas how to prove? and doesn't this contradict the Nested Interval Theorem?? :huh:

thx

You have $ E_x\equal{}(0,x).$

Prove that the intersection is empty? To prove that a set is empty, you must show that it has no elements. So, for any given $ z,$ is $ z\in\cap E_x?$

If $ z\le0,$ then $ z\not\in E_x$ for any $ x,$ so it's certainly not in the intersection.

If $ z>0,$ then we can find some $ x$ such that $ 0<x<z.$ $ x\equal{}\frac{z}{2}$ would do. Then for that particular $ x,$ $ z\not\in E_x$ and thus $ z$ is not in the intersection.

So there are no values of $ z$ that can be in the intersection and thus the intersection is empty.

As for the Nested Interval Theorem: that theorem states that a nested decreasing sequence of [b]closed[/b] intervals has a nonempty intersection. These intervals are not closed - indeed this example (well, something related to this example in which you replace the uncountably many $ x$ by some sequence, but that's not important) stands as a counterexample showing why we need the word "closed" in the hypotheses.

Clearly any member of $ \bigcap E_x$ must be greater than zero. However, if $ \epsilon > 0$ then we can pick $ 0 < x < \varepsilon$. Then $ \varepsilon\notin E_x$, so $ \bigcap E_x$ is empty.

The reason this doesn't contradict the nested intervals lemma is that these intervals are open, while the lemma applies to closed intervals.

Edit: Beaten to it.

ya I understand the point that for any z>0, there is always a x<z so z is not in the intersection. But when I think about it, isn't there always a z<x as well?

thx guys   :)


\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=142943
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=142944&sid=087f0cefd4ff39d30ead4270f4cff17a
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=142945&sid=087f0cefd4ff39d30ead4270f4cff17a
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=142946&sid=087f0cefd4ff39d30ead4270f4cff17a
\[\text{Day 2}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=142947&sid=087f0cefd4ff39d30ead4270f4cff17a
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=142950&sid=087f0cefd4ff39d30ead4270f4cff17a
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=142948&sid=087f0cefd4ff39d30ead4270f4cff17a
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=142951&sid=087f0cefd4ff39d30ead4270f4cff17a

Thanks to bilarev :)

Could you create a PDF file,if we ask you? :blush:

Thanks my friend :) !

By the way, I think you ought to become a moderator of this sub-forum,sooner or later :lol:.


I'm looking for an upper bound for the $n$-th highly composite number.
Can you help me ? :)

What do you mean by 'highly compostie?'  :|

Let be $d(n)$ the number of (positive) divisors of $n$.
Then a number $n$ is called highly composite iff $d(n) > d(k)$ for all $k<n$.
But I don't know any nontrivial upper bound... (a trivial one would be $2^n$)

Ok. I found it :)
Let $Q(x)$ be the number of higly composite number inferior to $x$.
Erdos proved that $(\mbox{ln } x)^a \leq Q(x) \leq (\mbox{ln } x)^b$ for two constants $a,b > 1$.
Putting $x=h_n$ where $h_n$ is the $n$-th highly composite number gives the result.
(since $Q(h_n)=n$)

@ZetaX: Can you give me the proof of the 'trivial' bound $2^n$ ?

When $n$ is a highly composite number, one of the numbers $n+1,n+2,...,2n$ will also be one since $d(2n) > d(n)$. Thus the result follows by induction.

Nice. Is this proof due to you or to folklore ?
(I might possibly use it in a publication so I want the right person to be acknowledged)

Well, the line above and the idea behind it is from me.
But I would be astonished if this wasn't known before.

Any ideas for an "elementary" (proved with elementary methods) lower bound ?
In particular I believe there exist an positive integer $c$ such that for any positive integer $n$, the number of highly composite number between $n$ and $2n$ is bounded by $c$...

just making sure you looked at wikipedia's entry, as they discuss some bounding for HCN.  Can certainly bound in terms of tau, the # of integer divisors


geometry
rectangle
combinatorics unsolved
combinatorics

You have many 1.1 squares. You may color their edges with one of four colors and glue them together along edges of the same color. Your aim is to get an m.n rectangle. For which m and n is this possible?

[quote="pleurestique"]the problem as stated is trivial: you can just color every edge of every square the same color, and then you can build any $m\times n$ rectangle. you might want to stipulate something more.[/quote]
To Slove!

[quote="pleurestique"]to solve? oh really, i didn't realize that's what you were driving at!![/quote]

I still don't get it :(

[quote="ZetaX"]What pleurestique said:
The problem is totally trivial because you made no restriction on the coloering.
Even if you want that any color occures exactly once at each square, it's still trivial.

Thus the problem is probably incomplete.[/quote]

Okay... I thought that Pleurestique realized what the problem meant...

no, i haven't realized what Huy Van meant his problem to mean. Huy Van needs to express himself more clearly.

I want to express myself more clearle but I don't know how to draw the picture. Who helps me?
I think we can glue together an m.n rectangle iff m and n have the same parity.


Solve the equation: \[\log_{2}\left(\sqrt x+\frac{3}{2}\right)+2^{x^{2}+\sqrt x-\frac{3}{4}}=2\]

there is no solution of this equation
let $\sqrt{x}+\frac{3}{2}= 2^{2^{y}}$
and $x^{2}+\sqrt{x}-\frac{3}{4}= z$
the given equation reduces to
$2^{y}+2^{z}= 2$
L.H.S must be the integer ad the only possible case when L.H.S gives integer is y=0,z=0 or y, z are positive integer or y,z = -1. Only y = 0,z=0 satisfies the given equation.
but by substituting this is the original equation. we get two different values of x hence there is no solution for the equation....


Asymptote
calculus
calculus computations

I scanned the question.. it can be found here
[url=http://img346.imageshack.us/img346/1217/picture8eq.jpg]img346.imageshack.us/img346/1217/picture8eq.jpg[/url]

I think the answers are I, III, IV, which would make it C.  Am i correct?


inequalities
number theory proposed
number theory

Prove that:
$ p_1p_2....p_n>p_{n\plus{}1}^2$
for $ n\geq 4$ where $ p_1\equal{}2,p_2\equal{}3,...$ is the increasing sequence of prime numbers.


Which of these consoles would you prefer based on games, function, and features, and why?

None, I don't play games, make that an option!

I have a wii. It is awesome. People say it is bad but they lie. PC works too.

The future of XBOX is the Xbox 720.

The future of PS is PS4.

I'd prefer a Wii! DK Barrel Blast rules!

someday, i really wish to have a wii, maybe for christmas, or something.

MAC FTW!!!

Well, It can have minesweeper, it's getting EA games, and it has some linux games that you can download(Like BZFlag and Armagetron).

[quote="Nerd_of_the_Ages"]PS3: WAYYY too expensive. Not too many extremely good games.

XBOX 360: Hmm, well, there's Halo....

PC: Yeah, there's some good games....

Wii: COMPLETELY backwards compatible with gamecube. So it can play Super Smash Brothers Melee. Therefore, Wii wins. QED[/quote]

That's a good point there...with some of the other consoles there aren't that many games that came out for them.

pc ftw
i love fps games and fps games are harder to aim on consoles with the joystick, and autoaim is for noobs.

[quote="perfect628"]*cough* everyone who doesny say wii shut up *cough*.[/quote]

wii?(why?)

Besides, what good is a Wii if you have a 4-year-old brother around 22-7? If he sees a game, he has to play it from sunrise to sunset each morning.

I've had that trouble with my computer once.

I like wii because its revolutionary, not controllers. Although pcs are very useful, internet, homework and guitar hero...
I still like wii because this is a video game place

[quote="pi guy"]I still like wii because this is a video game place[/quote]

With Lego Star Wars the Complete Saga, you can control the lightsaber, but it's just shaking it around, to tell you the truth.

legend of zelda:twilight princess
super mario galaxy
SSBB
best games ever
wii pwns

well.... SSBB isn't out yet..
except in Japan...

In SSBB, you can play as some freaking awesome characters, including:
Sonic(!)
Lucario(!!)
Wario(?)
Pit(!!!) and of course,
KING DEDEDE(!!!!!!1!!!!!!!) Which is why I want a Wii by March.

Don't post spoilers erabel. Some people want to be surprised. :wink:

I love to bump :cool:  :play_ball:  :gleam:

[quote=SansPapyrus683]where's the switch?[/quote]

this was made 10 years before the switch launched xD

[b]N I T N E N D O[/b] [color=#fff]a[/color] [b]W I I S T A T I O N 3 6 0[/b]

lol I'm creeped out about that the first posts are older than me myself...


limit
complex numbers
real analysis
real analysis unsolved

Is there a sequence indexed by the integers $z_n$ of non-zero complex numbers such that $z_{n+1}=z_n^2+z_n$ and $z_n$ tends to 0 for $n$ large and small.  :D

He means that $n\in\mathbb{Z}$ and that this is a double-ended sequence.

Harazi is asking for both $\lim_{n\to\infty}z_n=0$ and $\lim_{n\to -\infty}z_n=0.$

All I can say at first glance is that it's a non-trivial question.


Gah, this problem has been bothering me forever. Someone please solve it....

Let a string of 0's and 1's: a1,a2,a3,a4,a5,...an be replaced by |a1-a2|,|a2-a3|,...|an-a1|. For example, 00011 becomes 00101. Let n denote the length of the string, and let the period of the string of length-n "000000....0011" [edited, thanks Ravi B] be denoted c(n). For example, for n = 3, starting with 011, gives us 011 -> 101 -> 110 -> 011, a period of 3. Prove the following:

1. For n=2^r, we can always reach 00000000...00

2. Let n not equal to 2^r, and c(n) be the period. Prove c(2n) = 2c(n) and find all exceptions.

3. Prove that for odd n, 0000...0011 will always give a cycle.

4. Can we find any patterns in the function c(n)? Anything we can observe or prove?

[quote]Let n denote the length of the string, and let the period of the string n be denoted c(n). [/quote]
Is n the length of the string, or the string itself?

Fixed wording, thanks. n is the length of the string.

[quote]let the period of the string of length-n be denoted c(n)[/quote]
Okay, but now the quote above has me confused.  It says "the string of length-n", but there are many strings of length n.  Which one is used to define c(n)?  Is c(n) the maximum period of a string of length n?

I think the question usually lets the string of length n be a bunch of zeros and then two ones. i.e. 0000011 is the string of length 7.

[hide]I haven't had any time to actually work on it, but I found something very interesting... If we think about the sequence as 11000... instead of the reverse, then the first sequence (the one we started with) is {1C0, 1C1, 1C2,... 1C(n-1)}. Then the second sequence (after 1 iteration) is {2C0, 2C1, 2C2, 2C3,...2C(n-1)} mod 2. In general, the k'th sequence is {kC0+kC(n)+kC(2n)+..., kC(1)+kC(n+1)+kC(2n+1)+..., ........} mod 2 (this is easily proved by induction). Note that this immediately takes care of the first question asked, because when k = n (if n is a power of 2) we get all 0's (too lazy to show the proof for this). I don't know if that will get us anywhere on the other problems, but when i get more time I'll try to work on it.[/hide]

[hide]The process is just a function f(a_n-1, a_n)=(a_(n-1)+a_n)mod 2.

   a_1 mod2,  a_2 mod2,...  a_n mod 2

 (a_1+a_2)mod2, (a_2+a_3)mod2,...(a_n+a_1)mod2,

    (a_1+a_3)mod2, (a_2+a_4)mod2,...(a_n+a_2)mod2

   (a_1+a_2+a_3+a_4)mod2, (a_2+a_3+a_4+a_5)mod2,..

    (a_1+a_5)mod2, (a_2+a_6)mod2, ....

    ...................

     (a_1+a_2+...a_8)mod2,.......

    ...................

      (a_1+a_2+...a_(2^r))mod2,.... you probably do an induction, but I am not good at that

it will be 000000000000000000 when all the terms are same

n=2^r, we will be able to have all the terms to be(a_1+a_2+..a_n)mod2.[/hide]

This is the first time I post in Advance Topics, so it's probably wrong.

Yep. Thats my idea on the 2^r one as well.


AMC
USA(J)MO
USAMO
number theory solved
number theory

For a given positive integer n, prove that there are infinitely many positive integers m for which (m!)^(n-1)/((n!)^(m-1)) is an integer.

[i]USAMO 1985, training session, Bjorn Poonen[/i]

Hello, Orlando!

Let m=ns.

Suppose p is a prime number, p \leq n.
Let A=\sum [n/pk]. It is well known pA|n! and n!/pA+1 is not an integer.
Let B=\sum [(ns)!/pk], pB|(ns)!.

Now we are comparing (n-1)B and (ns-1)A, i.e. B and (ns-1+ns-2+...+1)A.
Suppose A=\sum {k=1,t}[n/pk]. We have simple estimation:
B \geq \sum {k=1,ts}[ns/pk] = \sum {k=1,t} + \sum {k=t+1,2t} +...+ \sum {k=ts-t+1,ts} \geq \sum {k=1,t}ns-1[n/pk] + \sum {k=1,t}ns-2[n/pk] +...+ \sum {k=1,t}[n/pk] {because n \geq pt and [na] \geq n[a]} = (ns-1+ns-2+...+1)A.


space pirates kidnap an earthling and hold him imprisoned on one of the planets of the solar system. with nothing else to do, the prisoner amuses himself by dropping his watch from eye level (170cm) to the floor. he observes that the watch takes 0.95 sec to fall. on what planet is he being held?

$1.7+\frac{1}{2}a\left(.95\right)^{2}=0\implies a\approx-3.77\,\textrm{m}/\textrm{s}^{2}$

According to Wikipedia, this would be Mercury.


It's supposed to be around late summer.  Unfortunately for me :(

I need Geometry and NT alcumus :P

Geometry is really good. That'll help me in next year Mathcounts!!!

We'll start working on algebra over the summer.


function
trigonometry
calculus
integration
limit
real analysis
real analysis theorems

Why is the function defined by f=${ \frac{1}{x}\cdot sin(\frac {1}{x^2})}, f(0) = 0$ not l-integrable on $ [ - 1,1]$? It should be $ < \infty$ a.e. -and therefore integrable, correct?

If this argument is incorrect, how do i pick my partition to show that it blows up?

Thanks in advance!

Los

g(x) = 1/x is finite a.e., so you claim it is integrable on [-1,1] ?

It's not (absolutely) integrable because it's like $ \frac1x$ near zero. $ \left|\sin\frac1{x^2}\right|\approx \frac2{\pi}$; it oscillates rapidly, but integrating something nice against it looks like integrating that nice function against a nonzero constant.

The improper integral $ \lim_{t\to 0^\plus{}}\int_t^1 \frac1x\sin\frac1{x^2}\,dx$ does exist; that's not the same thing.

Thanks for the quick answers!

A propos the first post, one can show that $ \frac {1}{x}\sin{\frac {1}{x^2}}$ is not measurable on $ [ \minus{} 1,1]$  by showing that $ 1/x$ is not measurable there. For instructional purposes I have attempted to show  that $ 1/x$ isnt measurable on [-1,1]..[b]can [/b]one use the following argument using the l. monotone convergence theorem (If a series  $ \sum_{k \equal{} 1}^\infty f_k(x)$ with $ f_k(x)\ge 0$ is measurable for every $ k\ge 1$ then the limit of the integral of $ f_k(x)$ is equal to the integral of the limit). Define

$ f_n(x) \equal{} \begin{cases} n, & \mbox{for } |x| \leq\frac {1}{n} \\
\frac {1}{x}, & \mbox{for } |x| > \frac {1}{n} \end{cases}$

This sequence is increasing in $ x$ and nonnegative. But here we have that

$ finite value \equal{} \lim_{n\to \infty }\int f_n \neq \int \lim_{n\to \infty } f_n \equal{} infinite value$.

Then $ f_n(x)$ is not measurable on [0,1], and so $ \frac {1}{x}$ is not measurable there.

Is that argument correct? It would be nice if it were, because would give a method for finding when functions are [i]not[/i] measurable, which would be great since finding if a fcn is measurable seems to be a lot of work.

Thanks in advance!

Los

Take care with your language! The function in question is most certainly measurable. What it's not is that it's not integrable.

As for your argument: that's really a monotone convergence theorem argument rather than a DCT argument.

$ \lim f_n(x)\equal{}f(x)$ [i]a.e.[/i] and for each $ x,$ $ f_n(x)$ is a nondecreasing sequence. Since each $ f_n\ge 0$ as well, the MCT applies. Hence by the MCT,

$ \int f\equal{}\int\lim f_n\equal{}\lim \int f_n\equal{}\infty.$

Note that $ \int_0^1\frac1x\,dx$ is clearly defined by the Lebesgue theory - but it's infinite. As a matter of convention, we say that it is not Lebesgue integrable.

$ g(x)\equal{}\frac1x\sin\left(\frac1{x^2}\right)$ isn't nonnegative, so when its integral turns out not to converge absolutely because both its positive and negative parts integrate to infinity, we don't define its Lebesgue integral at all.

Thanks again! I really like this method..it is an indirect sort of integrability criterion- you needn't know what the integral evaluates to here. I would like to read some more things like this if there is a good paper in the Monthly or nice book

Los

Its surprising to see that $ \int_{0}^{1}\frac{1}x\,dx\equal{}\infty$ but $ \int_{0}^{1}\frac{1}{\sqrt{x}}\,dx \equal{}2$, finite.

Thats because $ \int\frac{dx}{x^n }$ is convergent only for n < 1 .


limit
real analysis
real analysis unsolved

Hello,

Who can you proof that $\sum^{+\varpropto}_{n=1}\frac{1}{\sqrt{2n+1}}\leqslant \sum^{+\varpropto}_{n=1}\frac{1}{\sqrt{n}}$

Greets.

It is clear since
\[ \frac{1}{\sqrt{2n+1}}<\frac{1}{\sqrt{n}}  \]
for every $n$

but must not be $\frac{1}{n^p}$ where p must be greater than 1?

Greets.

I basically did not understand the proof. The logic is correct as compared by oldman, but both sums diverges. How can you possibly tell which one is bigger if both diverges, EVEN THOUGH each of the terms of one sum is larger than each of the terms of the second sum? Unless Bert has a 'special' meaning for \varpropto (which to me this is the proportionality symbol used in physics), I would assume that's the \infty symbol

Let $[r]$ be the hyperreal number $\{S_1, S_2, S_3, ...\}$, where $S_n$ is the $n^{th}$ partial sum of the first series. Let $[s]$ be another hyperreal number, defined similarly for the second series. Then $[r] \le [s]$ by oldman's argument. Note that $[r]$ and $[s]$ are greater than any real number.

10000th user: my point is that don't let the fact that the series are divergent suppress your creativity.  ;) There are other ways to think of this, too (that don't require something as exotic as hyperreals).

but we proof that the serie is not convergent? yes ?

Certainly, neither series is convergent.

but supose that and this is so that the sum of the second is intifine then it will posible taht the sum of the first will smaller maybe define?

If we want to prove that $\sum_{n=1}^\infty \frac1{\sqrt{2n+1}}$ diverges, finding a divergent series with larger terms tells us nothing. We need a smaller divergent series such as $\sum_{n=1}^\infty \frac1{2\sqrt{n}}$ to apply the comparison test usefully.

Here is a problem related to our theme from Tokyo University entrance exam 1990.

http://www.mathlinks.ro/Forum/viewtopic.php?t=19959&highlight=university+tokyo

thank you very well I see where I miss.

There are two things to know there is a different between proofing that $\sum^{\infty}_{n=1}\frac{1}{\sqrt{2n+1}}$  is smaller then $\sum^{\infty}_{n=1}\frac{1}{\sqrt{2n}}$

so that was my first question and oldman help me very well.

But ther is also an other thing to do that is proofing the divergention and this can't do by the first proof I need to find someting others.

So $L=\lim_{n\rightarrow \infty}|\frac{1}{\sqrt{2n+1}}|^2=\lim_{n\rightarrow \infty}\frac{1}{2n+1}=0$

normally we can say that is convergent but I now thet the root criterium is not enof.

Greets.

supose $\frac{1}{\sqrt{2x+1}}$ can I simple proof that $\frac{1}{2\sqrt{2x}}$ this is smaller?

if I multiply by $\frac{1}{2}$ then it wil be smaller but if I delete 1 then it wil be bigger but not so much who do I proof this?


calculus
integration
calculus computations

convergens becaese your integral is something like this integral
$ \int_{1}^{\plus{}oo}\frac{C}{x^{\frac{3}{2}}}dx$
its can be calculated

[quote="lord12"]Does the integral $ \int_{1}^{\infty}\frac{3\ dx}{2x\sqrt{4\plus{}5x}}$ converge or diverge? If it converges, what is its value?[/quote]

use the substitution $ 4\plus{}5x\equal{}t^{2}$ to get

$ \frac{3}{4}\int_{3}^{\infty}\;\left(\frac{1}{t\minus{}2}\minus{}\frac{1}{t\plus{}2}\right)\;dt$

[quote="lord12"]whats the actual value since i can't integrate it.[/quote]
Hint:
do next substitution:
$ 4\plus{}5x\equal{}y^{2}$
after you will get this integral
$ \int\frac{dy}{ay^{2}\plus{}b}$  which is standart integral


inequalities
geometry
circumcircle
inequalities proposed

Prove that in any trinagle $ABC$ the following inequality holds:

                           \[ \frac{1}{h_{a} r_{a}}+\frac{1}{h_{b} r_{b}}+\frac{1}{h_{c} r_{c}}\leq\frac{\sqrt{3}}{S}. \]

Assuming $h_a, r_a$ denote the altitude and exradii from $A$ and that $S$ equals area, we have

$r_a = \frac{S}{s-a}$ and $h_a = \frac{bc}{2R}$ where $R$ is the circumradius of triangle $ABC$.  The inequality hence becomes

$\sqrt{3} \geq \sum \frac{2R(s-a)}{bc} = \sum \frac{4Ra(s-a)}{2abc} = \sum \frac{a(s-a)}{2S}$

Converting this into $x = s-a, y=s-b, z=s-c$ yields

$\sqrt{3xyz(x+y+z)} \leq xy+yz+zx \iff \sum x^2(y-z)^2 \geq 0$


Aratati ca :log in baza xy din z+log in baza yz din x+log in baza xz din y>=3/2,unde x,y,z sunt numerele reale mai mari ca 1

$\log_{xy}z=\frac{\ln z}{\ln(xy)}=\frac{\ln z}{\ln x+\ln y}$.
notam $a=\ln x$, $a>0$, si celelalte, si inegalitatea devine :
$\sum \frac{a}{b+c}\geq \frac{3}{2}$,
o inegalitate foarte cunoscuta.

Ms .....era usoara......acum invat logaritmii:)


By starting with a million and alternatively dividing by 2 and multiplying by 5, Anisha created a sequence of integers that starts 1000000, 500000, 2500000, 1250000, and so on. What is the last integer in her sequence? Express your answer in the form $ a^b$, where $ a$ and $ b$ are positive integers and $ a$ is as small as possible.

$ 1000000\equal{}2^6 \times 5^6$, so we divide by $ 2$ $ 6$ times and multiply by $ 5$ $ 6$ times, so the answer is $ 5^{6\plus{}6}\equal{}\boxed{5^{12}}$.


Se tiene un triangulo ABC, siendo BC > AC > AB ; se traza AD con D en BC;  el <BAD = 2<DAC. Se toma un punto M en el lado AC tal que : <ADM = 90
Si: BD=2MC , AB=AD y AM-DC = 12
Calcule la Distancia de M a BC.
Alguna solucion :spider:

alguna idea para el dato de AM - DC = 12 :?:

sea $<BAD = 2<DAC=2x,$ trazamos $DT$ perpendicular a $AC$ ($T$ en $AC$) asi $2TD=BD,$ $DT^2=AT.TM$ y es facil notar que $m<DAC=m<TDM=m<MDC,$ asi $DM$ es la bisectriz de $<TDC,$ sea $Q$ ($Q$ en $BC$) tal que $DC$ es perpendicular a $MQ,$ asi $DT=DQ$ y $TM=MQ$ por lo tanto $DT/TM=DC/MC$ y de ahi que $AT=DC.$ Sea $P$ el punto medio de $BD,$ $tgx=PD/AP$ y $tg(2x)=(PD+AP)/AP$ y aplicnado la identidad del angulo doble tenemos que $x=45$, asi $TM=AT,$ asi, como $AM=12-AT$ se sigue que $TM=MQ=4.$

[quote="M4RI0"]pero se puede sin trigonometria :?:[/quote]

no lo se, a mi me parece que el problema es defectuoso

no defectuoso precisamente, me refiero a que estoy casi seguro de que la solucion a dicho problema dada en el libro donde fue propuesto es con congruencias, los problemas aqui en peru son antojadizos, ponen demasiadas condiciones para que puedan usar alguna propiedad de congruencias (ya que aqui, existe una seccion en los libros de casos especiales a los que se llaman "propiedades") y luego si tu lo examinas del todo llegas a conseguir todos los datos del triangulo (me refiero a que seguramente en la solucion oficial no llegaron a que los angulos eran de 45 y que muchos de los segmentos eran iguales) particularmente no me gustan esta clase de problemas  :|.


Why is the charge on a charged conducting sphere uniformly distributed throughout the volume of the sphere? Wouldn't the shell theorem mean that repulsive forces are minimized when the charge is distributed over the surface of the sphere?

On a conducting sphere, charge IS only on the surface of the sphere.

I must have misunderstood my physics teacher then. I guess that answers this question.


algebra
polynomial
inequalities
absolute value
algebra proposed

Some more (and partially related) problems ( first ones: http://www.mathlinks.ro/Forum/viewtopic.php?t=42609 ) on cyclotomic polynomials ($\Phi_n$ is the $n$-th cyclotomic polynomial again):

[b]a)[/b] find $\Phi_n(0)$, $\Phi_n '(0)$, $\Phi_n (1)$ and $\Phi_n (-1)$
[b]b)[/b] prove that $\Phi_n(x)$ is reciprocal for $n\neq 1$ (sorry if this one is to easy)
[b]c)[/b] let be $o(n)$ be the highest occuring absolute value of a coefficent in $\Phi_n (x)$. Prove that $o(n)$ is unbounded.

PS: it would be nice if some of you can give some more problems on this topics (to me or to all mathlinkers)  :)

There has been proved even more than c) : every integer can be coefficient of a cyclotomic polynomial.

To prove this i need the following:
Lemma:
For any positive integer $\ t\ge 3$ there exist primes$\ p_1< p_2<...<p_t$ such that $\ p_1+p_2>p_t$.
Proof: Fix $\ t \ge 3$. Suppose that for any set of primes $\ p_1< p_2<...<p_t$ we have $\ p_1+p_2 \le p_t$ implies that $\ 2p_1<p_t$. Now we have that between $\ 2^k$ and $\ 2^{k+1}$ there are less than t primes. Summing over $\ k \le n$ we have that there are at most kt primes less than $\ 2^k$ i.e. $\ \pi (k)<kt$ for every k.By Chebyshev Th. We have that $\ \pi(x)> \frac{cx}{lnx}$, where c is a positive constant. Hence $\ \frac{c2^k}{ln2^k}$ i.e. $\ c2^k<k^2t$ which is false for sufficiently large k.

Now let t > 3 be an odd integer. Select primes $\ p_1<p_2<...<p_t$ so that $\ p_1+p_2>p_t$ set $\ p=p_t$ consider the polynomial $\ \ phi_n (x)$ modulo $\ x^{p+1}$. For t odd we have:

${\ \phi_{p_1p_2..p_t}=\frac { (x^{p_1}-1)...(x^{p_t}-1) }{ x-1 } \frac {\prod(x^{p_ip_jp_k}-1)}{\prod(x^{p_ip_jp_kp_l}-1)})}...$
But $\ x^{p_ip_j}=0 ( mod x^{p+1}),x^{p_ip_jp_k}=0 (mod  x^{p+1})$. Hence
$\ \phi_{p_1p_2...p_t}=\frac{(1-x^{p_1})...(1-x^{p_t})}{x-1} (mod x^{p+1})$
$\ p_i+p_j>p_1+p_2>p_t$ impliy that $\ (1-x^{p_1})...(1-x^{p_t})= 1-x^{p_1}-...-x^{p_t} ( mod  x^{p+1})$
It is clear that $\ (1-x)^{-1}=1+x+x^2+...+x^p ( mod  x^{p+1})$. Hence $\ \phi_{p_1p_2...p_t}=(1+x+x^2+...+x^p)(1-x^{p_1}-...-x^{p_t}) ( mod x^{p+1})$. One can easily show that the coefficient of $\ x^p$ in $\ \phi_{p_1p_2...p_t}$ is $\ -t+1$ and the coefficient of $\ x^{p-2}$ is $\ -t+2$, for all $\ t>2$. Now we can extend the result to all set of integers using that $\ \phi_{2n}(x)= \phi_n(-x)$, and using the fact that the coefficients of $\ x^p$ and $\ x^{p-2}$ in the two polynomials differ only by sign.


let $x,y,z$, is integer $>0$.prove that if $(xy+1)(yz+1)(xz+1)$ is square number then $xy+1,yz+1,zx+1$ is square number ,too

Actually, I think the problem you meant is : http://www.mathlinks.ro/Forum/viewtopic.php?t=52039 .
It was posted several times.
All leading back to the same link : http://www-math.mit.edu/~kedlaya/papers/ .


AMC
AIME
probability
AMC 10
AMC 10 B
AMC 12 B
AMC 12 A

What is AMC-10 B? I thought there is only one test per year.  If I missed passing the AMC-10, do I get another chance this year, or do I have to wait till next year?

Request for moderator to delete post or at least the number of points. Please hold ALL discussion until Thursday.

You can take 10B if you're a sophomore or below.

I'd like to see this thread removed.

ln(dx/dy), no offense but here is what you said:

[quote="ln(dx/dy)"]This should be in the AMC section, but I shouldn't be talking. I've already had three things moved [/quote]

I should note that if your posts have been removed, you should not try to make another try anymore. ;)

To moderator, this quote has been taken from the post that was started by cj in here:

http://artofproblemsolving.com/Forum/topic-25390.html

Please delete both thread or lock it till Thursday evening.

Thank you

you don't "have" to lock the thread
merely delete the part about the score because he/she has a right to get an answer

How does the AMC 12-B compare in difficulty with the AMC 12-A, or are they very similar in difficulty? The AMC-12 B has a much higher average score, probably because the fewer people taking it are naturally much more qualified - and this is where the cutoff score of 100 is especially important.

well, I don't think that there should be too much difference between 12A and 12B (some questions are common to both).
for 10A and 10B... the qualifying for the AIME for 10A is generally less than that for B.... two or three points.

[quote]well, I don't think that there should be too much difference between 12A and 12B (some questions are common to both)[/quote]
There will be no questions in common.

[quote]for 10A and 10B... the qualifying for the AIME for 10A is generally less than that for B.... two or three points.[/quote]
Unfounded speculation. There is no preplanned difference between the two tests. Whatever the AMC may decide to do after the fact is, well, after the fact.

[quote]The AMC-12 B has a much higher average score, probably because the fewer people taking it are naturally much more qualified -[/quote]
Also unfounded. Some schools take only the A tests; some schools only the B tests. Any systematic difference in the qualifications of the students taking each would be an accident. Again, there's no preplanned or intentional difference in the difficulty of the two tests, but the AMC cannot prescreen them to determine the practical level of difficulty. You should assume that the probability of the B test being "easier" than the A test is 50%.


The roots of the equation $ ax^2 + bx + c = 0 $ will be reciprocal if:
\[ \text{(A)}\ a = b \qquad \text{(B)}\ a = bc \qquad \text{(C)}\ c = a \qquad \text{(D)}\ c = b \qquad \text{(E)}\ c = ab \]

[hide]
If the roots of the equation are reciprocal, their product is $ 1 $.
Since we know that the product of the roots is $ \frac{c}{a} $, we have $ \frac{c}{a}=1\Longleftrightarrow c=a $[/hide]

[hide]Let $ x1 $ and $ \frac{1}{x1} $ be the two roots of the equation.

By Vieta's formula:

$ x1\cdot\frac{1}{x1}=\frac{c}{a} $

$ 1=\frac{c}{a} $

$ c=a $

Answer: $ C $[/hide]


[quote="silouan"]Prove that  $(\left[ \sqrt{n}\right]+1)^{2}\geq n+1$[/quote]

It's very easy!
We have:  $(\left[ \sqrt{n}\right]+1)^{2}> (\sqrt{n}-1+1)^{2}> n\geq n+1$ because of $\left[ \sqrt{n}\right]+1\in\mathbb{Z}$

silouan, what is the significance of the polynomial in your signature?


induction
combinatorics proposed
combinatorics

The following problem by Peter Frankl has appeared in an old issue of the Amer. Math. Monthly:

[b]Problem. [/b] Let $S$ be a finite set, and let $\mathcal{P}=2^S$ be collection of all subsets of $S$. For $\mathcal{A} \subset \mathcal{P}$ and $\mathcal{B} \subset \mathcal{P}$, define $\mathcal{A} \star \mathcal{B}$ to be the subset of $\mathcal{P}$ consisting of all subsets $X \subset S$ such that $X \subset A \cup B$ for some $A \in \mathcal{A}, B \in \mathcal{B}$. If $|\mathcal{A}| + |\mathcal{B}| >  2^k$, prove that $|\mathcal{A} \star \mathcal{B}| \geq 2^k$.

Enjoy!   :)

--Vesselin

yesterday while i was generally looking at some older posts, this one caught my attention. I tried it once more and i think i have a solution.
Induct on $k$. For small $k$ it is trivial.(note however that we need the assumption that the families $\mathcal{A},\mathcal{B}$ are both nonempty).
Now let $x_0\in S$ be a fixed element and denote by $\mathcal{A}_0,\mathcal{B}_0$ the subfamilies of $\mathcal{A},\mathcal{B}$ resp. that consist of sets that contain the element $x_0$. Of course to have induction we must have $x_0$ such that both these families are nonempty. If such a choice for $x_0$ is impossible then it follows that for any $x_0$ contained in some $A\in\mathcal{A}, x_0\notin\displaystyle\bigcup_{B\in\mathcal{B}} B$. BUt then $A\cap B=\emptyset,\ \forall A\in\mathcal{A},B\in\mathcal{B}$ and then the conclusion is trivial.
Firstly note that $|\mathcal{A}_0|+|\mathcal{B}_0|>2^{k-1}$ or $|\mathcal{A}_1|+|\mathcal{B}_1|>2^{k-1}$. if the former holds, then by consider the families $\mathcal{A}_0',\mathcal{B}_0'$ by deleting the element $x_0$ from each of the sets of these families. By induction $|\mathcal{A}_0'\star\mathcal{B}_0'|\ge 2^{k-1}$. Now consider all the sets $X\cup\{x_0\},X\in \mathcal{A}_0'\star\mathcal{B}_0'$. Each of these sets doesn't occur in $\mathcal{A}_0'\star\mathcal{B}_0'$ and so these are $2^{k-1}$ new sets which however are elements of $\mathcal{A}\star\mathcal{B}$.
Suppose then that $|\mathcal{A}_1|+|\mathcal{B}_1|>2^{k-1}$(again if one of these families is empty then the problem is easily resolved by deleting the set of all points that occur in all the sets of either collection). By induction, $|\mathcal{A}_1\star\mathcal{B}_1|\ge 2^{k-1}$. Again note now $|\mathcal{A}_0|+|\mathcal{B}_1|>2^{k-1}$ or $|\mathcal{A}_1|+|\mathcal{B}_0|>2^{k-1}$. suppose the former holds. Again by induction,  $|\mathcal{A}_0'\star\mathcal{B}_1|\ge 2^{k-1}$, so that  all the sets $X\cup\{x_0\},X\in \mathcal{A}_0'\star\mathcal{B}_1$ again contribute new sets to the collection and which are elements of $\mathcal{A}\star\mathcal{B}$. This completes the induction and the proof.


PLAYER LIST:
[list]
miyomiyo
mnmath
Temperal
perfect628
woser456
1=2
Peter
Ars
Bery...
142857
xpmath
cheeseIsGood
lingomaniac88
zapi2007
erabel
math_explorer
azjps
bowei
shaggy75
LevatorNasolabialis
ASPEED
andersonw
alanchou
ZhangPeijin
kimmystar94
setheus_soul
sonickid01
jnub
cf249
[/list]

You should have all received role PMs. If not, send me a PM and I'll reply with your role PM.

Now, on to the story:

You all wake up, but something feels different. As you look around, you realize that you're surrounded not by your homes, but by white clouds. You quickly looks for any signs of your family, but it's useless. However, you see 28 other people standing doing the same thing. Finally, you realize that you and the 28 others are dead ghosts. Suddenly, you hear God talking to you.

[size=150]I have been informed by the devil that on this day, some of the dead will try to sabotage heaven. As such, I cannot allow you in until you have discovered and eradicated these evil doers. Thus is your task.[/size]

Day 1 begins immediately. Nights will last 48 hours. 15 to lynch. I guess I should note: You're all playing and allowed to post. The people who have been lynched, etc are called banished (to hell).

[size=150]GO[/size]

[b][u]The Rules[/u][/b]:[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Please be attentive and [b]unvote[/b], if necessary, before casting a new vote.  This is not [i]required[/i], but I’d appreciate it.
[*]Executions will require a simple majority of votes.  Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may [b]vote: no lynch[/b] – majority votes of this kind are necessary to end the day without a death.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so. - [b]If your role PM tells you the alignment of other players, you may privately talk to them unless otherwise stated.[/b]
[*]Once your death scene has been posted, you’re dead.  Stop typing.
[*]Don’t edit/delete previously submitted posts
[*]Don’t quote any PMs from me.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, tough!  No choice will be made.
[*][b]If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread.[/b]  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 [i]fun[/i].  Do your part.[/list]

So we [i][b]can[/b][/i] lynch ghosts. :)

setheus_soul is evil!

[b]Vote: setheus_soul[/b]

My role PM is the least informative one I have ever seen, which is not saying much. [b]Vote: No Banishment[/b]

Important Observation: Every single person here (excluding me) finds Berylliumnitride harder to memorize than LevatorNasolabialis. (No offense.)

This is my first Mafia game! Whee!

Eh, I don't have any knowledge of anyone just yet, so [b]Vote: No Banishment[/b], I guess.

Important reply to important observation: Berylliumnitride is a much easier name to remember than LevatorNasolabialis. I believe that an elemental compound is much easier to remember that what looks like a type of pasta. :P *shot*

Yeah, my role pm tells me basically nothing.
[b]vote:No Banishment[/b].

sonic: please prepend [b]Vote:[/b] to your votes in the future.

And yes, Berylliumnitride is much easier to remember than Levator... 1=2 was the one who added him to the player list and I was too lazy to change it ;)

Vote count:
setheus_soul (1): 1=2
No Banishment (3): math_explorer, sonickid01, xpmath

[b]Vote sonickid01[/b] for saying no banishment without a vote and calling LevatorNasolabialis a type of pasta.

btw, i can't remember both names.

[b]Vote:ASPEED[/b] for being the only person with more than 3 capital letters in their username.

BerylliumNitride is way easier to remember.

[quote="math_explorer"]Every single person here (excluding me) finds Berylliumnitride harder to memorize than LevatorNasolabialis. (No offense.)[/quote]

Not me!

Berylliumnitride
LevatorNasolabialis

I memorized Levator after Beryllium.

Since pasta is yummy, I want some. [b]Unvote, Vote: sonickid01[/b] for making me want pasta.

Oh wait, are we in heaven? *checks* Shouts: O MIGHTY LORD! ARE WE IN HEAVEN, OR A SUSPICION CHAMBER OF SOME SORT!
God: The latter.

O.K. sonickid01 made me hungry.

Theme good, but he could have spiced up the roles a bit.

P.S. Inactive:
miyomiyo
mnmath
Temperal
perfect628
woser456
Peter
Ars
Berylliumnitride
cheeseIsGood
lingomaniac88
zapi2007
erabel
azjps
bowei
shaggy75
LevatorNasolabialis
ASPEED
alanchou
ZhangPeijin
kimmystar94
setheus_soul
jnub
cf249

We must lynch them.

[b]Unvote, Vote: setheus_soul[/b]

Something is sketchy about him, yeah that's it, sketchy. MUAHAHAHAHA!

cf249 sounds like a virus. [b]vote: cf249[/b]

Already an inactivity list after just 3 hours?

Not much information yet, so [b]Vote:No Banishment[/b].

Yeah, sorry about the whole "vote:" thing. My bad. The post has been edited.

My first game and everyone already hates me. D:

Yeah, don't edit posts either...

[b]The Vote Count of Hating sonickid01[/b]
[list]
setheus_soul (1): 1=2
sonickid01 (1): 142857
ASPEED (1): andersonw
cf249 (1): Peter
No Banishment (4): math_explorer, sonickid01, xpmath, azjps
[/list]

[b]vote: miyomiyo[/b] You'll soon find out why.

[quote="zapi2007"]L+1 sounds like overkill
but that's cool[/quote]

And the game did end.  :)

So .. thoughts?

Edit: ASPEED killed perfect628 [mafia] and shaggy75 [mafia] killed Levator [mafia] .. nice .. and I was saved from a redirected mafia kill by a randomly redirected angel - what are the chances of that?  :lol:

THEY DIDN'T TAKE OVER HEAVEN YAY!!!

Wow I'm so proud of myself surviving until night 8.  :D

[quote="azjps"]
 .. and I was saved from a redirected mafia kill by a randomly redirected angel - what are the chances of that?  :lol:[/quote]

it's $ \frac1{(n \minus{} 1)^2}$, when n is the number of dead ghosts that night, wow

And also the angel only reduces the chance of banishment? [does that mean more randomness]

I was really confused on that night when the kills were kimmystar94 and 1=2 .. now I understand why. Hmm, and town won by a large margin; 7 survivors!

Okay explanation on the angels. There were two of them. If exactly one targeted a person, (s)he was immune from being banished. There was no randomness involved, but also there was a chance that both angels would target the same person (but this never happened - often only one angel saved anyone during the night)

Also interesting was that the sk kept targeting random people, while the mafia insisted on banishing zapi2007 (which was probably their downfall)

yeah, now that i see it, the mafia wasn't all that smart...
if they had been a little better at targeting, ceasing to target me after the first fail, they probably should have gone after someone else.
maybe, there only would have been 3 or so people left
it wouldn't have looked so ridiculous for the town

good job guys
(although i guess i did pretty good too  :)  )

intersting:
setheus_soul (mafia) BANISHES sonickid01 (mafia)
lingomaniac88 (angel) PROTECTS sonickid01 (mafia)

i'm also wondering why zapi was not killed. but they will know that someone is redirecting their kills, so keep on killing zapi is not a bad choice, unless the kill is redirected to themselves.

good job zapi, i've never seen a so successful inspector before.
also good job cheese for getting sonickid.

Hehe...the redirecting role was definitely fun, though I messed up the first few days.  I guess my role basically was used as a doctor, though I was afraid that a kill would be redirected towards me. :P

Anyways, awesome job, zapi.

I especially liked the night actions on the last day too:

shaggy75 (mafia) BANISHES zapi2007 REDIRECTED to azjps
Berylliumnitride (INCORRECT) TARGETS zapi2007
mnmath (angel) PROTECTS zapi2007 REDIRECTED to azjps

haha, I had a pretty pointless role - what does it mean that "cf249 DRAINS temperal"?

EDIT: oh, never mind. he just stole my pointless power. no wonder it didn't work!

[quote="sonickid01"]Wait, I drained people? What is that supposed to mean?
I was told I was a mafia roleblocker. I thought I was, but apparently not.[/quote]

What do you think drain means? It means you voided their night action...sheesh.

So if you drain a doctor, then he becomes townie?

from my understanding, i think it just negates a power for one turn

woohoo.

my team won even though i died. :D


The numbers 1 through 36 are arranged in a 6 x 6 square so that the sums of the numbers in each row, column, and diagonal are the same. What is that special sum?

[hide] The total sum of the numbers is $ \frac{(36)(37)}{2}$. Now, each of the six rows has the same sum, which is just $ \frac{1}{6}\cdot\frac{(36)(37)}{2}$. We can evaluate that expression to be 111. Notice that we might as well have used the six columns instead.[/hide]

These type of square are known as magic square.
if n is no.of rows then sum of each row or column or diagonal=n^2(n^2+1)/2n.

[quote="kabi"]These type of square are known as magic square.
if n is no.of rows then sum of each row or column or diagonal=n^2(n^2+1)/6.[/quote]

Hm...if we use that formula ($ \frac{n^2(n^2\plus{}1)}{6}$), then we get $ \frac{36(37)}{6}\equal{}222$, but if we add the highest 6 numbers: $ 36\plus{}35\plus{}34\plus{}33\plus{}32\plus{}31\equal{}201$, so it is impossible to get $ 222$.  I don't think your formula works.

[quote="007math"][quote="kabi"]These type of square are known as magic square.
if n is no.of rows then sum of each row or column or diagonal=n^2(n^2+1)/6.[/quote]

Hm...if we use that formula ($ \frac {n^2(n^2 \plus{} 1)}{6}$), then we get $ \frac {36(37)}{6} \equal{} 222$, but if we add the highest 6 numbers: $ 36 \plus{} 35 \plus{} 34 \plus{} 33 \plus{} 32 \plus{} 31 \equal{} 201$, so it is impossible to get $ 222$.  I don't think your formula works.[/quote]
Dude who said u 36  35  34  33  32  31 are going to come in a row or column or a diagonal.
Go through magic square.
Hopefully u'll get ur quires clear.

Anyway their is the derivation of that formula.
In a n*n square ,there are no. from 1,2,3,.....,n^2.
sum of all no. coming in this magic sq.=n^2(n^2+1)/2
sum of any row or column or diagonal=n^2(n^2+1)/2n=n(n^2+1)/2.

Actually, kabi, you're wrong. (You forgot to divide by 2).
Suppose there are $ n$ rows and columns, and the numbers $ 1,2,\cdots,n^2$ are placed in them to form a magic square.
Then the sum of all the numbers is $ \sum_{k\equal{}0}^{n^2}k\equal{}\frac{n^2(n^2\plus{}1)}{2}$.
Divide that sum by the number of rows $ n$ to get $ \frac{n^2(n^2\plus{}1)}{2n}\equal{}\frac{n(n^2\plus{}1)}{2}$.

Plugging in $ n\equal{}6$ this gives $ \frac{6(37)}{2}\equal{}111$ as expected.

007math's argument makes sense, actually, because the maximum sum of a row of a 6x6 magic square containing $ 1,2,\cdots,36$ must be $ 36\plus{}35\plus{}\cdots\plus{}31$.

I apologize for the mistake in writing that formula.I just forgot to write 2.
I have edited them correctly.
007math and grn_trtle sorry about the formula.
From now I will check my reply before posting it.


geometry
inequalities
trigonometry
function
geometry proposed

Let $ABC$ be an acute-angled triangle of area 1. Show that the triangle whose vertices are the feet of the perpendiculars from the centroid $G$ to
$AB$, $BC$, $CA$ has area between $\frac 4{27}$ and $\frac 14$.

Use the Euler's formula $S(G) = \frac{1}{4}(1-\frac{OG^2}{R^2})S$ and Leibnitz's theorem $R^2-OG^2 = \frac{GA^2+GB^2+GC^2}{3} = \frac{a^2+b^2+c^2}{9}$. Note that since $ABC$ is an acute triangle, we have $8R^2 \leq a^2+b^2+c^2 \leq 9R^2$.

[quote="msecco"]A better value for the left side of inequality is 2/9.[/quote]

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=287&t=348040[/url] :D

The right handside was proven already. For the left handside we get:
$ 3\sum a^2>16R^2\Leftrightarrow 3\sum(2R\sin\alpha)^2>16R^2\Leftrightarrow 3\sum(\sin\alpha)^2>4$.
Using the Cauchy-Swartz inequality we get that: $3\sum(\sin\alpha)^2\geq(\sum\sin\alpha)^2$ so it is enough to prove that: $\sum\sin\alpha >2$
But this follows from Karamata inequality aplied for the concave function $f(x)=\sin x$, and notice that for an acute angled triangle $(\alpha,\beta,\gamma)\succ (90^\circ,90^\circ,0)$, so $\sum\sin\alpha\geq\sin(90^\circ)+\sin(90^\circ)+\sin(0)=2$. But the equality can't hold, so $\sum\sin\alpha >2$

[quote=vlad1m1r]to prove that: $\sum\sin\alpha >2$[/quote]

why not use Jeason？


In a town, there are exactly 100 heterosexual couples and no one else.  Every wife is cheating on every husband.  Every husband knows that all other 99 wives are cheating and their husbands, but assumes his own wife is faithful.

The town tradition is that if you find out your wife is cheating, you are to short her the following noon in the town square.  However, the men never, ever talk to each other, and they never leave their homes, they never talk on the telephone--they don't communicate, it's not a trick question, so each man never finds out his own wife is cheating.  And life goes on....

Until a salesperson comes to town, and figures out what's going on.  He gathers every man in the town square, but the men do not communicate with each other at all.  When all the men arrive, the salesperson says "There is at least one wife cheating on her husband."  Then he leaves forever.

Prove that exactly 100 days later, every man drags his wife to the town square and shoots her.

HINT:  DO NOT LOOK AT THIS UNTIL YOU HAVE TRIED THE PUZZLE BECAUSE THIS MAKES IT PRETTY EASY: [hide]It can also be proven that if there are exactly n heterosexual couples in the town, the wives will die n days later, where n is a natural number.[/hide]

haha, i like this one,
 :)
but if the salesman says AT LEAST, would that make anydifference? because you already said, every man has faith about his wife, and since the men do not comunicate, then how can that help...?
I KNOW I'M MISSING SOMETHING.... :wow:

reminds me of certain gnome problems, lol.  I'll leave it to someone who doesnt know the answer.

[quote="jmadsen"][hide="answer"]Prove the general case that if there are n couples, the wives will be shot on the nth day. For n=1, this is trivial. If it is true for k couples, then the addition of one more faithful couple will not change the outcome, since everyone else knows they are faithful. Then, the shooting would still occur on the kth day. So, when the men see that the shooting does not occur on this day, they realize that their wives must be cheating, so all the wives are shot on the k+1th day. Use induction to prove this for all n.[/hide][/quote]I still don't believe it!!  :noo:

yes, it makes perfect sense...
all of the wives would die on the 100th day, because every husband thinks that his wife is not cheating, but after 100 days when no one gets shot, all of the husbands are going to shoot their wifes, because they didn't see any other person shoot their wife... :)

Of course, this implies that the husbands are intelligent. If they have never communicated ever, this doesn't seem to be a likely case.

and if they're that smart how come they think their own wife is faithful when they know that all 99 other wives are cheating?

There's just one problem, why [b]would[/b] they shoot their own wives if they believe their own is faithful, and according to the premises - each husband [b]knows[/b] that the other 99 are cheating.  Since this is the sole belief of every husband, even after the 100th day, they would not shoot their own wife, because in their minds, they are still knowing that there should have been at least 1 shooting, even if there wasn't.  Also that the "at least 1" doesn't prove anything, since they "know" the other 99 are cheating on their husbands - does not imply their own wife.

 Maybe a clarification on what's meant by "know" in this scenerio, and the town rules would help.

I think I'm with Ixeus on this one.  Unless it's made clearer what reasoning process all these people use, the question isn't complete -- after all, wouldn't every man just assume that the other men are simply to dumb to realize the obvious and then go on with his life?

why isn't bubala saying anything? i mean afterwards he proposed the problem so he must know the answer...

I think somebody said the answer already?


If $ p_{k}(n)$ is defined to be number of distinct partitions of $ n$ into parts no larger than $ k$, find:
$ \lim_{n \to \infty} \frac { p_{k}(n) }{ n^{k \minus{} 1} }$.
I think I've worked out that this limit is equal to $ \frac {1}{k!(k\minus{}1)!}$ .. I was wondering if someone could verify that.

That's correct. $ p_k(n)$ is equal to the number of nonnegative integer solutions of the equation $ x_1\plus{}2x_2\plus{}\ldots\plus{}kx_k\equal{}n$. It's well-known that for any $ a_1,\ldots,a_k\in\mathbb N$, such that $ \gcd(a_1,\ldots,a_k)\equal{}1$, the number of integer nonnegative solutions of the equation $ a_1x_1\plus{}\ldots\plus{}a_kx_k\equal{}n$ is $ \sim\frac{n^{k\minus{}1}}{a_1\ldots a_k(k\minus{}1)!}$ as $ n\to\infty$ ($ k$ fixed).


Let $ a,b,c\in[\frac{1}{3},3]$.Prove that:$ abc\plus{}3(a^2b\plus{}b^2c\plus{}c^2a)\ge2(ab^2\plus{}bc^2\plus{}ca^2)$

[quote="vo thanh van"]Let $ a,b,c\in[\frac {1}{3},3]$.Prove that:$ abc \plus{} 3(a^2b \plus{} b^2c \plus{} c^2a)\ge2(ab^2 \plus{} bc^2 \plus{} ca^2)$[/quote]
$ \sum_{cyc}(3b \minus{} 2c) > 0.$ Let $ 3b \minus{} 2c > 0.$
Hence, $ abc \plus{} 3(a^2b \plus{} b^2c \plus{} c^2a)\ge2(ab^2 \plus{} bc^2 \plus{} ca^2)\Leftrightarrow$
$ \Leftrightarrow(3b \minus{} 2c)a^2 \minus{} (2b^2 \minus{} bc \minus{} 3c^2)a \plus{} (3b \minus{} 2c)bc\geq0.$
Here we can assume $ 2b\minus{}3c\geq0.$ :wink:
 Id est, it remains to prove that
$ (2b^2 \minus{} bc \minus{} 3c^2)^2 \minus{} 4bc(3b \minus{} 2c)^2\leq0,$ which is equivalent to
$ (b \minus{} c)(b \minus{} 9c)(4b^2 \plus{} c^2)\leq0,$ which is true.