Let $ x_1, x_2,..., x_n$ be distinct real numbers. Find a polynomic function $ f_i$ of degree $ n\minus{}1$ such that $ f_i(x_i)\equal{}1$ and $ f_i(x_j)\equal{}0$ if $ i\neq{j}$.

[b]Follow-up:[/b]  Find a polynomial $ F$ of degree $ n\minus{}1$ such that $ F(x_i) \equal{} y_i$ for arbitrarily reals $ y_1, y_2, ... y_n$.


conics
hyperbola
geometry
function
calculus
calculus computations

EDIT: Fixed it!

I get the minimum area of the triangle as being $\frac{3\sqrt{3}}{2}$ and this occurs for the value $p=\sqrt{2}$, and these based on my calculation that the area of $\triangle PAB$ is given by \[Area(p)=\frac{1}{p^{2}}\left( p^{2}+1\right)^{\frac{3}{2}}\] which was based on the tangent points $A$ and $B$ being given by $\left( \pm\sqrt{\frac{1}{p^{2}}+1},-\frac{1}{p}\right)$, repectfully.

[quote="kunny"]your answer is incorrect.[/quote]

I dropped a square on $p$ early on... it's fixed now.

This time your answer is right. but you are to get minimum value.
You can use $A.M.-G.M.$

[size=117][b][u]Nice problem, Kunny !

[/u][/b][/size] [quote="kunny"][color=darkred]Let $p>0.$ Draw two tangent lines from the point $P(0,\ p)$ to the hyperbola $x^{2}-y^{2}=1.$ Let $A,\ B$ be the tangent points respectively.
Find the value of $p$ for which the area of the triangle $PAB$ is minimized.[/color][/quote]

[color=darkblue] Let $A(a,b)$ , $B(c,d)$ be two points of the hyperbola $\mathbb H$ so that $P\in AA\cap BB\ .$

The area $S$ of the triangle $APB$ is $S=\frac{1}{2}\cdot\mod|\begin{array}{ccc}0 & p & 1\\\ a & b & 1\\\ c & d & 1\end{array}|=\frac{1}{2}\cdot |(b-d)-a(p-d)+c(p-b)|\ .$

$%Error. "maltese" is a bad command.
\ \{\begin{array}{c}AA\ : \ ax-by=1\ ;\ P\in AA\Longrightarrow b=-\frac{1}{p}\\\ BB\ : \ cx-dy=1\ ;\ P\in BB\Longrightarrow d=-\frac{1}{p}\end{array}\Longrightarrow b=d=-\frac{1}{p}$ $\Longrightarrow$ $S=\frac{1}{2}(p+\frac{1}{p})|a-c|\ .$

$%Error. "maltese" is a bad command.
\ \{\begin{array}{c}A\in\mathbb H\Longrightarrow a^{2}-b^{2}=1\\\ B\in\mathbb H\Longrightarrow c^{2}-d^{2}=1\end{array}\ ,\ a\ne c\Longrightarrow a=\pm\frac{\sqrt{p^{2}+1}}{p}\ ,\ c=\mp\frac{\sqrt{p^{2}+1}}{p}$ $\Longrightarrow$ $\boxed{\ S(p)=\frac{(1+p^{2})\sqrt{1+p^{2}}}{p^{2}}\ }\ .$

Using the substitution $t=\sqrt{1+p^{2}}$ construct the function $f: (1,\infty )\rightarrow \mathbb R$ , $f(t)=\frac{t^{3}}{t^{2}-1}\ .$

We have $f'(t)\ .s.s.^{(*)}\ (t-\sqrt 3)\ ,$ i.e. $t_{\mathrm{min}}=\sqrt 3\ ,$ what means $p_{\mathrm{min}}=\sqrt 2$ and $S_{\mathrm{min}}=S(\sqrt 2)=\frac{3\sqrt 3}{2}\ .$

[b]Remark.[/b] Otherwise, using a geometrical interpretation, denote $A(a,-\sqrt{a^{2}-1})$ , $B(-a,-\sqrt{a^{2}-1}\right)\ ,$ where $a>1\ .$
Then the equation of the tangent $AA$ is $ax+y\sqrt{a^{2}-1}=1\ .$ From the condition $P\in AA$ obtain $\sqrt{a^{2}-1}=\frac{1}{p}\ ,$ i.e. $a=\frac{\sqrt{1+p^{2}}}{p}\ .$
The area $S$ of the triangle $APB$ is $S=\frac{1}{2}\cdot(p+\sqrt{a^{2}-1})\cdot 2a\ ,$ i.e. $S=(p+\frac{1}{p})\frac{\sqrt{1+p^{2}}}{p}$ a.s.o.
__________________
$^{(*)}\mathrm{.s.s.}\--\ \mathrm{same\ sign\ .}$[/color]


A bag of mixed peanuts and cashews contains two pounds of peanuts that cost $ \$3$ per pound and one pound of cashews that costs $ \$6$ per pound. How much should the bag of mixed peanuts and cashews cost?

2 pounds of peanuts at 3 dollars per pound is 6 dollars.
1 pound of cashews at 6 dollars per pound is 6 dollars.
Add them up to get 12 dollars.


I wonder how mathematicians (and anyone studying maths) use their draft (rough notebook)? :roll:
Could you tell me some of your experience in using your draft? :)

I don't understand your question.  Can you please rephrase it?

I mean when we solve some problems, we have to use drafts to write down what we are thinking. This makes us to think easier. :)
How do you use your drafts?
(I don't know whether or not I use the word "draft" exactly. I looked it up in Oxford Dictionary: "draft" means "a rough written version of sth that is not yet in its final form" )

[quote="mysmartmouth"]Do you mean scratch work/brainstorming?[/quote]
No, I don't. Draft = draft paper or paper for drafting :)

Oh.  OK.  I think we were thinking of different things...

I think that it changes from person to person i hardly write anything down when im solving a problem unless it is a proof.


When I save a tag like [code][hide=blah][/code] in my sig, it becomes [code][hide:ef4c9ca0fc=blah][/code] or something similar. The same happens with the quote tags, is there a way to fix this?

While the gibberish is always ef4c9ca0fc for me, it seems like it is different for different users, I saw someone else with the same thing happening in their sig but it was different letters and numbers.

This happened to me before in a different way.  I posted my name in LaTeX, kinda like this (except with dollar tags):

- \mathBB{S}\mathfrak{ean}

Resulting in:

- $%Error. "mathBB" is a bad command.
{S}\mathfrak{ean}$

However, it would add in random numbers like those above.


group theory
abstract algebra
superior algebra
superior algebra unsolved

For a prime number $ p$, let $ G_p\equal{}U_1 \cup U_2 \cup \cdots \cup U_n$ where $ n$ is all natural numbers, and $ U_k\equal{}$ {$ z \in C|{z}^{(p^n)}\equal{}1$}.
Prove that every proper subgroup of $ H$ is infinite when $ H\equal{}G_p$, where $ p$ is a prime number. ($ H$ is an infinite subgroup of $ C$)

Are you sure you meant union?  It looks like there are many cyclic finite proper subgroups.

I think you mean $ G_p: \equal{}\{z\in\mathbb{C}: z^{p^n}\equal{}1\text{ for some }n\in\mathbb{N}\}\le \mathbb{C}^\times$ for each prime $ p$
and $ H\le \mathbb{C}^\times$ is infinite with every proper sugroup finite $ \Leftrightarrow$ $ H\equal{}G_p$ for some prime $ p$.


linear algebra
matrix
superior algebra
superior algebra solved

Prove that if the trace of a square matrix $ A$ is $ 0$, then there exist two square matrices $ X$ and $ Y$ of the same size as $ A$ such that $ A\equal{}XY\minus{}YX$.

[quote="Valentin Vornicu"]Prove that if the trace of a square matrix $ A$ is $ 0$, then there exist two square matrices $ X$ and $ Y$ of the same size as $ A$ such that $ A \equal{} XY \minus{} YX$.[/quote]

Here the proof from the book
Roger A.Horn & Charles R.Johnson,
Topics in Matrix Analysis, Cambridge University Press

Chapter 4, Matrix equations and the Kronecker product
theorem 4.5.2 p
If trA=0, then the matrix equation XY-YX=A can be solved.To do so
we may replace A by any matrix similar to A because the property
of being a commutator is similarity invariant.
A is unitarily to a matrix whose diagonal entries are all equal to 1/n.trA
Let A=(a(i,j)) with a(1,1)=...=a(n,n)=0
Choose X=diag(x_1,...,x_n), x_i are pairwise distinct, with X fixed and A
given , XY-YX=A becomes a linear matrix equation in Y=(y(i,j))
XY-YX=((x_i-x_j)y(i,j))

y(i,j)= c(i,j)/(x_i-x_j) if i=/=j
y(i,j) arbitrary if i=j
gives a solution since x_i=/=x_j when i=/=j and a(i,i)=0

M.Omarjee

Nice solution.
Also much much shorter than the one I knew before.
Of course that solution was written in an elementary style for students whose knowledge on linear algebra raises to Hamilton-Cayley and not a bit further. For those students similarity in matrices is SF, so that solution had to prove everything.
The bottom line is that I couldn't understand that solution because it was to explicit and I too young, but I understand yours.

I think Valentin must now find a way to make this topic a [b]no sticky[/b], normal post.

Cool solution Moubi! This problem was proposed to the 11-graders at that contest I've mentioned b4, Gheorghe Titeica. I spent a lot of time on it in the 11'th grade (not in the contest; I tried to solve it for training), but got nowhere. When I eventually decided to look at bthe solution in a book, I got scared (it spanned over several pages), closed the book and ran away! :D  :D  :D

I know the feeling grobber. That was my first reaction to that proof too.

[quote="Moubinool"]
A is unitarily to a matrix whose diagonal entries are all equal to 1/n.trA
[/quote]

I am stupid. How do you prove it? I may prove it only for real case.


How many of the first 1000 positive integers can be expressed in the form
[2x] + [4x] + [6x] + [8x],
where x is a real number, and [z] denotes the greatest integer less than or equal to z?

Edit: oops your right, fixed

clearly $ x$ is positive. What we need to do is solve this problem graphically. The critical points are $ 1/8,2/8,3/8,..1$ and $ 1/6,2/6..1$ $ 1/4,2/4,..1$ and $ 1/2$ and $ 1$. Combining we get $ 1/8,1/6,1/4,1/3,3/8,1/2,5/8,2/3,6/8,5/6,7/8$ and $ 1$. At each critical point there will be an increase of atleast one and at special points like$ 1/4,3/4$ we will have increase of two and at $ 1/2,1$ we have an increase of $ four$. So the numbers that can be represented are $ 1,2,4,5,6,10,11,12,14,15,16,20$. So out of first $ 20$ we get $ 12$ so out of $ 1000$ we will get $ \boxed{600}$\\


geometry
3D geometry
algebra unsolved
algebra

$ w,x,y,z$ are positive rational numbers satisfying
$ w$ $ (\sqrt{y} \plus{} \sqrt{z})^3 \equal{} \sqrt{z}$
$ x$ $ (\sqrt{y} \plus{} \sqrt{z})^3 \equal{} \sqrt{y}$
then prove that
$ (a)$ $ 1\minus{}2xy\minus{}6xz$   is cube of a rational number.
$ (b)$ $ xy \plus{} wz \plus{} 3xz \plus{} 3wy \equal{} 1$.

[quote="rohitsingh0812"]$ w,x,y,z$ are positive rational numbers satisfying
$ w$ $ (\sqrt {y} \plus{} \sqrt {z})^3 \equal{} \sqrt {z}$
$ x$ $ (\sqrt {y} \plus{} \sqrt {z})^3 \equal{} \sqrt {y}$
then prove that
$ (a)$ $ 1 \minus{} 2xy \minus{} 6xz$   is cube of a rational number.
$ (b)$ $ xy \plus{} wz \plus{} 3xz \plus{} 3wy \equal{} 1$.[/quote]

It is very simple : you just have to compute the expressions.

==========$ (a)$============
$ 1 \minus{} 2xy \minus{} 6xz \equal{} 1\minus{}2\frac{y\sqrt{y}}{(\sqrt {y} \plus{} \sqrt {z})^3}\minus{}6\frac{z\sqrt{y}}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ 1 \minus{} 2xy \minus{} 6xz \equal{} \frac{(\sqrt {y} \plus{} \sqrt {z})^3\minus{}2y\sqrt{y}\minus{}6z\sqrt{y}}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ 1 \minus{} 2xy \minus{} 6xz \equal{} \frac{\minus{}y\sqrt{y}\plus{}3y\sqrt{z}\minus{}3\sqrt{y}z\plus{}z\sqrt{z}}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ 1 \minus{} 2xy \minus{} 6xz \equal{} \frac{(\sqrt {z} \minus{} \sqrt {y})^3}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ 1 \minus{} 2xy \minus{} 6xz \equal{} \frac{(\sqrt {\frac{z}{y}} \minus{} 1)^3}{(\sqrt {\frac{z}{y}} \plus{} 1)^3}$

$ 1 \minus{} 2xy \minus{} 6xz \equal{} \frac{(\frac{w}{x} \minus{} 1)^3}{(\frac{w}{x} \plus{} 1)^3}$

$ 1 \minus{} 2xy \minus{} 6xz \equal{} \left(\frac{w \minus{} x}{w\plus{}x}\right)^3$ and so is cube of a rational number.

==========$ (b)$============

$ xy \plus{} wz \plus{} 3xz \plus{} 3wy \equal{} \frac{y\sqrt{y}}{(\sqrt {y} \plus{} \sqrt {z})^3}$ $ \plus{}\frac{z\sqrt{z}}{(\sqrt {y} \plus{} \sqrt {z})^3}$ $ \plus{}\frac{3z\sqrt{y}}{(\sqrt {y} \plus{} \sqrt {z})^3}$ $ \plus{}\frac{3y\sqrt{z}}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ xy \plus{} wz \plus{} 3xz \plus{} 3wy \equal{} \frac{y\sqrt{y}\plus{}z\sqrt{z}\plus{}3z\sqrt{y}\plus{}3y\sqrt{z}}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ xy \plus{} wz \plus{} 3xz \plus{} 3wy \equal{} \frac{(\sqrt {y} \plus{} \sqrt {z})^3}{(\sqrt {y} \plus{} \sqrt {z})^3}$

$ xy \plus{} wz \plus{} 3xz \plus{} 3wy \equal{} 1$


Ross Mathematics Program
geometry
3D geometry
rotation
AMC
USA(J)MO
USAMO

This will be a hypothetical conversation between two or more persons. The scenario can be changed every 100 posts or so. The speech of the same person can not continue beyond one post, nor can two persons speak in the same post.

The first is between a mother (A) of a young chess enthusiast who has been left out of a tournament and two people (B and C) from the management.

errr...ok

A: why has my child been left out of the chess tournament!!!?!?!?!?!??! :mad:

B. We  are sorry madam, but we had no choice! Your son turned in a fake Maestro's Approval Ticket and we could not believe him when he said he had been framed!

(sorry, but maybe the conversation should include a mystery and a detective comes in or something)

C) Has your son considered doing mathematics...entering a mathematics competition instead?

A) No, of course not! Can't you imagine the headlines? World famous Opera singer's son enters mathematics competition! that's sooooooooo nerdy!

B) And chess isn't nerdy??? World famous Opera singer's son enters chess tournament!

C. Madam, please understand that police is on the way to arrest your son of being a fraud!

B. Madam, I told you! He submitted a fake Maestro's Approval Ticket! That is absolutey illegal!

A) does fraud mean something like...i don't know... against the law?

C. Yes, "c'est illegal" as they would say in France or any other Francophone country.

A: Wait let me get this clear. You mean my son is what you call a fraud?  And you also add that a fraud is someone who has done something against the law? So ading the two ... hang on, I am not good at these things, but even to me it seems that somehow you are trying to establish that the law is after my son!

B. I'm sorry, you're going to have to refer that to my lawyers.

A: So you are using this as an excuse to leave out my sweetie pie, are you?

lock this topic so nobody keeps this topic alive

it is a useless topic
but conversation II (new topic) should be open for everyone and since arnav and other spammers rnt here, they cant spam

stomps everyone and commits suicide so no one can talk.

Please everyone don't spam in this. If it is making an exit, let it be a graceful one. You already have the other topic. I don't want this to be locked. Just let it go down.

[quote="akalra1"]go spam on aim

chatting wit urself is really fun (ok not really)[/quote]

actually it is fun chatting with yourself.(especially if you've blocked yourself)

o wat is fun is blocking ur friend in the middle of the chat

nonono
blocking yourself not your friend.


Which point on the circle $ (x \minus{} 11)^2 \plus{} (y \minus{} 13)^2 \equal{} 116$ is farthest from the point $ (41, 25)$?


How many non-negative solutions exist for $1!a+2!b+3!c+\cdots+10!j = 20!$?

Observation:

If a person were looking at making equations by a "brute force" method, then there are (10 terms choose 1) plus (10 terms
choose 2) plus on up to (10 terms choose 10), which gives
(2^10) - 1 = 1023.  This means 1023 separate equations to solve, with varying numbers of variables and varying numbers of solutions for those equations.

For example, the first few equations could be:

1!a = 20!
2!b = 20!
3!c = 20!

And the last, 1023rd, equation would be the given equation in the original post.

HOWEVER, practically speaking, I would think the above procedure is not the way to go, because it is relatively very
time prohibitive for one person to work on alone, as it compares to other daily activities, in my opinion.

If you were going to use brute force, you might as well use the snazzy method, by which I mean the generating function

$P(x) = \frac{1}{(1-x^{1!})(1-x^{2!})...(1-x^{10!})}$

We want the $x^{20!}$ term.


calculus
integration
trigonometry
function
trig identities
Law of Cosines
calculus computations

$\int (a-\frac{b}{\cos x})(a^{2}+b^{2}-2ab \cos x)^\frac{1}{2}dx$

Mathematica could not do it
http://integrals.wolfram.com/index.jsp

Try it now, I made a change

${\int (a-b \cos x})(a^{2}+b^{2}-2ab \cos x)^\frac{1}{2}dx$

Mathematica managed to solve this one though but the answer is very complicated- got to be an easier way!!

Maybe you can tell us the whole problem? This looks to me like some triangle problem.

Also, that antiderivative is not an elementary function for most values of $a,b$. It's an elliptic integral (or a closely related object).

I managed to remove the square root sign, is it easier to do now?

$\int (\cos(\arctan \frac{b \sin x}{r-b\cos x}))(a^{2}+b^{2}-2ab \cos x)dx$

Yes this is a triangle problem The right hand bracket is the law of cosines

$(a^{2}+b^{2}-2ab \cos x)$

No. The square root comes back in when evaluating that $\cos\circ\arctan$.

I see
Is there some kind of a half angle indentity for $(a^{2}+b^{2}-2ab \cos x)^\frac{1}{2}$ ?

Here is the same problem but in a different form. Can you solve it this time?

$\int(a-b\cos x)((a-b\cos x)^{2}+(b\sin x)^{2})^\frac{1}{2}dx$

Not "tough". Impossible. Wasting time on an impossible problem is not something I like to do.

Wait

look at the second post

Mathematica did solve it, but the result was not what I call elegant.

So, what functions did it use? I think there were elliptic integral functions in there, which aren't elementary.


group theory
abstract algebra
superior algebra
superior algebra solved

$R$ is a commutative ring. $P$ is an ideal of $R$. For any ideals $\mathfrak a$ and $\mathfrak b$ of $R$, $\mathfrak{a}\mathfrak{b}\subset P$ implies that $\mathfrak{a}\subset P$ or $\mathfrak{b}\subset P$.

Prove that $P$ is a prime ideal. (If $ab\in P$ then $a\in P$ or $b\in P$)

hmm. We have
$Ra\subset P$ or $Rb\subset P$.

$R$ may not have identity  :?

Then?

Sorry, then. I assumed it had identity. Rings are usually assumed to have identity when we talk about prime ideals and such. I guess we're not done :).

[quote="grobber"]If $ab\in P$, apply the hypothesis to $\mathfrak a=aR,\mathfrak b=bR$.[/quote]

Let's try it for $I=aR+<a>$ and $J=bR+<b>$.

[quote="amfulger"]
Let's try it for $I=aR+<a>$ and $J=bR+<b>$.[/quote]
Is $<a>$ here an additive subgroup or the ideal generated by $a$?

I denote the ideal generated by an element a of a ring by $(a)$, not $<a>$. Why should it be the ideal generated by a?
I didn't expect you to ask this because you've seen me use this idea.


integration
real analysis
real analysis unsolved

Evaluate  the  sum

   \[ \sum \limits_{n=1}^{\infty} \frac{1}{(2n-1)(3n-1)}  \]

I found something like $ln\frac{4}{\sqrt27} + \frac{\pi}{2\sqrt3}$
[i hope there are no errors...:)]

Because $\frac{1}{(2k-1)(3k-1)} = \frac{2}{(2k-1)} - \frac{3}{(3k-1)}$ we consider sequences:
$a_n = \sum_{k = 1}^n \frac{1}{2k-1} - \frac{1}{2}ln n$
$b_n = \sum_{k = 1}^n \frac{1}{3k-1} - \frac{1}{3}ln n$
$c_n = \sum_{k = 1}^n \frac{1}{3k-2} - \frac{1}{3}ln n$
$x_n = \sum_{k = 1}^n \frac{1}{k} - ln n$

A classical result tells us $x_n \to  E \in (0, 1)$.
We have to compute $2a_n - 3b_n = 2a-3b$ if we assume $a_n \to  a, b_n \to  b$.

One can verify that $a_n = x_{2n} - \frac{1}{2}x_n + ln 2$, so $a = \frac{1}{2}E + ln 2$.
The same, $b_n + c_n = x_{3n} - \frac{1}{3}x_n + ln 3$, so $b + c = \frac{2}{3}E + ln 3$,
if $b_n \to  b, c_n \to  c$.
But it can be proved easily that $b_n$ is decreasing and being positive is convergent.
Because $b_n + c_n$ is convergent, $c_n$ is convergent.
To find b and c we observe that:
$c_n - b_n = \int_0^1 [1 + x + ... + x^{3n-3}]dx - \int_0^1 [x + x^4 + ... + x^{3n-2}]dx$
$c_n - b_n = \int_0^1 (1-x)(1 + x^3 + ... + x^{3n-3})dx$
$c_n - b_n = \int_0^1 (1-x)\frac{1-x^{3n}}{1-x^3}dx = \int_0^1 \frac{1-x^{3n}}{1+x+x^2}dx$
$c_n - b_n = \int_0^1 \frac{1}{1+x+x^2}dx - K_n$
But $K_n \leq \int_0^1 x^{3n}dx \to  0$.
It follows $c - b = \int_0^1 \frac{1}{1+x+x^2}dx = \frac{\pi}{3\sqrt3}$
Then $b = \frac{1}{3}E + \frac{1}{2}ln 3 - \frac{\pi}{6\sqrt3}$,
and $2a-3b=ln \frac{4}{\sqrt27} + \frac{\pi}{2\sqrt3}$


If 2^P-1 is a prime number then 2^(2^p-1)-1 is a prime? :)

Not true. $2^{13}-1$ is a prime while $2^{8191}-1$ is not. It is divisible by 210206826754181103207028761697008013415622289 according to [url=http://www.mersenne.org/ecm1.htm]this page[/url]. Some more factors can be found [url=http://mathworld.wolfram.com/DoubleMersenneNumber.html]here[/url]. Actually, I would prefer conjecture that the only primes $P=2^{n}-1$ where $2^{P}-1$ is prime too are $n\in\{2,3,5,7\}$.

Same for $p=17$ (needed over one hour on my PC...).

For p=13,17,19 \[2^{2^{p}-1}-1\] is not prime. I think, it is true only for primes p<10.

2^7-1=127=prime.Ibelieve that and 2^127-1is prime number so.....

Known that it is true for primes p=2,3,5,7, and it is not true for p=13,17,19 (Consider list of Mersene's primes). Unknown for p>=31.

Ι understand you Rust but  i believe that: 2^2-1=3,2^3-1=7,2^7-1=127,2^127-1=.......prime? so we earn many Merssene's primes.Can sb prove that the number 2^127-1 is prime?I will try to prove that it is a prime. :)

[quote="Αρχιμήδης 6"]Yes ZetaX.I see from mathworld that the number 2^127-1=x is a Merssene prime.I believe that and the number 2^x-1 is prime.[/quote]

My computer says it is false . :maybe:  the second you typed

How did you check $2^{2^{127}-1}-1$ for primality¿ The only way I can think of is that there is a rather small factor or something counterexample of whatever test you use (the number has $5.12175997193697 \cdot 10^{37}$ digits).

And even if the check is incorrect, I believe that your believe is wrong.

I don't know if this number is prime but i have seartch some primes (not all primes) and none of them devide this number.I believe that the sequence 2^..........2^(2^(2^(2^2-1)-1)-1)-1 gives many prime numbers.But i thought that there isn't any sequence whitch gives many prime number's isn't it?

I bet this sequence gives not more than some primes (it stoppes giving primes if one composite number is reached).
And Mersenne numbers never have small factors ;)

Hmm...maybe this number is the last Mersenne number that this sequence earn.I try to see only for small primes(2,3,5,7) and never devide the number.I bet that this number is an Mersenne prime number but it takes months to see this with PC'S.

Check divisors $\equiv 1 \mod 2^{127}-1$ only ;)
And not having this four divisors of you tells really nothing, and the amount of time is not months, it is aeons.

I believe that in aeons the computers will be faster.Also i believe that it is to stupid to try to see if this number is prime with this method.(only with use computers).We must find some tehnics. :wink:

We have the Lucas-Lehmer-test, which is quite fast in relation to most other known tests, and it is deterministic.

For current computers, testing it this way is out of reach. Quantum computers  would clearly help :D
But if it is not prime, we have a small (very small) chance to find a divisor even now.


inequalities search function calculus derivative limit

$ a,b,c\in \mathbb{R^\plus{}}$ and $ abc\equal{}1$ ,for $ k\in \mathbb{N}$ \ $ \left \{1\right\}$

Prove that

$ {a^k\over a\plus{}b} \plus{} {b^k\over b\plus{}c}\plus{}{c^k\over c\plus{}a}\ge {3\over2}$

 :roll:

[quote="Dimitris X"][quote="vinskman"]$ a,b,c\in \mathbb{R^ \plus{} }$ and $ abc \equal{} 1$ ,for $ k\in \mathbb{N}$ \ $ \left \{1\right\}$

Prove that

$ {a^k\over a \plus{} b} \plus{} {b^k\over b \plus{} c} \plus{} {c^k\over c \plus{} a}\ge {3\over2}$

 :roll:[/quote]
From the Andreescu inequality we have:
$ LHS \ge \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)}$
So we have to prove that:

$ \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)} \ge \frac {3}{2} \Leftrightarrow (a \plus{} b \plus{} c)^{k \minus{} 1} \ge 3^{k \minus{} 1}$

But from $ AM \minus{} GM$ the $ (a \plus{} b \plus{} c)^{k \minus{} 1} \ge (3 \sqrt [3]{abc})^{k \minus{} 1} \equal{} 3^{k \minus{} 1}$

Try the stronger now(generalization of IMO-1995):
For $ a,b,c > 0$ and $ abc \equal{} 1$ show that:
$ \frac {1}{a^n(b \plus{} c)} \plus{} \frac {1}{b^n(c \plus{} a)} \plus{} \frac {1}{c^n(a \plus{} b)} \ge \frac {3}{2}$ , for $ n \in N^*$ and $ n \ge 3$ :)[/quote]

Would you please post how to apply  the " Andreescu inequality " here? :oops:

[quote="PKMathew"]

Would you please post how to apply  the " Andreescu inequality " here? :oops:[/quote]
In the generalization you mean???

[quote="Dimitris X"][quote="vinskman"]$ a,b,c\in \mathbb{R^ \plus{} }$ and $ abc \equal{} 1$ ,for $ k\in \mathbb{N}$ \ $ \left \{1\right\}$

Prove that

$ {a^k\over a \plus{} b} \plus{} {b^k\over b \plus{} c} \plus{} {c^k\over c \plus{} a}\ge {3\over2}$

 :roll:[/quote]
From the Andreescu inequality we have:
$ LHS \ge \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)}$
So we have to prove that:

$ \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)} \ge \frac {3}{2} \Leftrightarrow (a \plus{} b \plus{} c)^{k \minus{} 1} \ge 3^{k \minus{} 1}$

But from $ AM \minus{} GM$ the $ (a \plus{} b \plus{} c)^{k \minus{} 1} \ge (3 \sqrt [3]{abc})^{k \minus{} 1} \equal{} 3^{k \minus{} 1}$

Try the stronger now(generalization of IMO-1995):
For $ a,b,c > 0$ and $ abc \equal{} 1$ show that:
$ \frac {1}{a^n(b \plus{} c)} \plus{} \frac {1}{b^n(c \plus{} a)} \plus{} \frac {1}{c^n(a \plus{} b)} \ge \frac {3}{2}$ , for $ n \in N^*$ and $ n \ge 3$ :)[/quote]

Would you please explain in details.
$ \boxed { \text{From the Andreescu inequality we have: }\\
LHS \ge \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)}}$

[quote="PKMathew"]
Would you please explain in details.
$ \boxed {\text{From the Andreescu inequality we have: } \\
LHS \ge \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)}}$[/quote]
For $ x_1,x_2...,x_n >0$ and $ a_1,a_2,...,a_n>0$ then:
$ \frac{x_1^k}{a_1}\plus{}\frac{x_2^k}{a_2}\plus{}...\plus{}\frac{x_n^k}{a_n} \ge \frac{(x_1\plus{}x_2\plus{}...\plus{}x_n)^k}{n^{k\minus{}2}(a_1\plus{}a_2\plus{}...\plus{}a_n)}$
So for $ n\equal{}3$ we extract the desired result...
Do you understand it now :?:

Try the generalization now :)

[quote="Dimitris X"][quote="PKMathew"]
Would you please explain in details.
$ \boxed {\text{From the Andreescu inequality we have: } \\
LHS \ge \frac {(a \plus{} b \plus{} c)^k}{3^{k \minus{} 2} \cdot 2 \cdot (a \plus{} b \plus{} c)}}$[/quote]
For $ x_1,x_2...,x_n > 0$ and $ a_1,a_2,...,a_n > 0$ then:
$ \frac {x_1^k}{a_1} \plus{} \frac {x_2^k}{a_2} \plus{} ... \plus{} \frac {x_n^k}{a_n} \ge \frac {(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^k}{n^{k \minus{} 2}(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)}$
So for $ n \equal{} 3$ we extract the desired result...
Do you understand it now :?:

Try the generalization now :)[/quote]

Thanks :lol:

But I am unfamiliar with Andreescu inequality or generalised Andreescu inequality , have any sources  showing  the proof  been  read ?

It's equivalent to Cauchy-Schwartz, so if you take CS inequality as a start, you can derive Andreescu lemma.

Here's a thread about it: http://www.mathlinks.ro/viewtopic.php?search_id=1512021779&t=263819

[quote="Dimitris X"][quote="PKMathew"]
Thanks :lol:

But I am unfamiliar with Andreescu inequality , have any sources  showing  the proof  been  read ?[/quote]
Take a look here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15125[/quote]

Thanks, I read Andreescu inequality/lemma.

 then how about generalised Andreescu inequality ?

I realized that this doesn't work but I typed alot so I don't want to delete it. It is hard to use Darji's principle (or Engel's) to prove this because of the n^{k-2}...

The sum of the minima of n functions is less or equal to the minimum of the sum of these functions.

[hide]

Consider the function
\[ f(x) \equal{} \frac {a x^{\alpha \plus{} \beta}}{\alpha \plus{} \beta} \minus{} \frac {bx^{\beta}}{\beta} \equal{} x^\beta\left(\frac {a x^{\alpha}}{\alpha \plus{} \beta} \minus{} \frac {b}{\beta}\right)
\]
where ($ a\ge 0$ ,$ b\ge 0$, $ \alpha,\beta\ge 1$). Taking the derivative,
\[ f'(x) \equal{} a x^{\alpha \plus{} \beta \minus{} 1} \minus{} b x^{\beta \minus{} 1} \equal{} (x^{\beta \minus{} 1})(a x^\alpha \minus{} b)
\]
Notice that $ f'(x)$ changes from negative to positive at $ (x_0)^\alpha \equal{} \frac {b}{a}$ so that is a local minimum. So the function value at that local minimum is
\[ f(x_0) \equal{} \left(\frac {b}{a}\right)^{\frac {\beta}{\alpha}}\left(\frac {a \frac {b}{a}}{\alpha \plus{} \beta} \minus{} \frac {b}{\beta}\right) \equal{} \minus{} \frac {1}{(\beta)(\alpha \plus{} \beta)}\cdot b\cdot \left(\frac {b}{a}\right)^{\frac {\beta}{\alpha}}
\]
which is less than zero so the other local min is not an absolute minimum.

$ f$ is continuous and differentiable (obviously) so $ f$ on any closed interval has a min and a max at either the endpoints or at a critical point. To extend this to a minimum over $ \mathbb{R}$, we need to have $ \lim_{x\rightarrow\pm\infty}f(x) \equal{} \infty$. So this only holds for $ \alpha \plus{} \beta$ is an even integer.

When all is said and done, we choose $ \alpha \equal{} 2$ and $ \beta \equal{} 2(k \minus{} 1)\ge 1$ (so $ k\ge \frac {3}{2}$).

So if we take $ f_i(x) \equal{} \frac {a_i x^{2k}}{2k} \minus{} \frac {b_i x^{2k \minus{} 2}}{2k \minus{} 2}$

Then
\[ \min(f_1(x)) \plus{} \min(f_2(x)) \plus{} ... \plus{} \min(f_n(x))\le \min (f_1(x) \plus{} f_2(x) \plus{} ... \plus{} f_n(x))
\]
Plugging in the minimum that we calculated before and rearranging gives us:
\[ \frac {b_1^k}{a_1^{k \minus{} 1}} \plus{} \frac {b_2^k}{a_2^{k \minus{} 1}} \plus{} ... \plus{} \frac {b_n^k}{a_n^{k \minus{} 1}}\ge \frac { (b_1 \plus{} b_2 \plus{} ... \plus{} b_n)^k}{(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^{k \minus{} 1}}
\]
This holds for $ k\ge 2$. As a comment, if we replace $ \alpha \equal{} 2m$ and $ \beta \equal{} 2m(k \minus{} 1)$ where $ m\in\mathbb{N}$, then $ \beta\ge 1\implies k\ge 1 \plus{} \frac {1}{2m}$ so by continuity, the inequality holds for $ k > 1$.

Equality occurs when $ \frac {b_i}{a_i} \equal{} C$ where $ C$ is a constant. [/hide]

Regarding that previous hidden content message:

As long as all of the parts of the answer
to the original question were meant to be
able to do in this "Intermediate Topics"
without using any calculus
(not that it couldn't be done WITH calculus),
then I'm fine with that.  Else move it to a
relatively advanced section.

As [b]PKMathew[/b] asked, what about the inequality $ \frac{x_{1}^{k}}{a_{1}}\plus{}\frac{x_{2}^{k}}{a_{2}}\plus{}...\plus{}\frac{x_{n}^{k}}{a_{n}}\ge\frac{(x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n})^{k}}{n^{k\minus{}2}(a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n})}$ ?
Could someone provide a proof for that?

@keyree 10, it is true by the Holder's Inequality.
$ \left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\frac {\sum x_{i}}{n^{k \minus{} 2}}$
$ \implies\sum\frac {x_{i}^{k}}{a_{i}}\geq\frac ({\sum x_{i})^{k}}{n^{k \minus{} 2}\sum a_{i}}$

[quote="Agr_94_Math"]@keyree 10, it is true by the Holder's Inequality.
$ \left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\frac {\sum x_{i}}{n^{k \minus{} 2}}$
$ \implies\sum\frac {x_{i}^{k}}{a_{i}}\geq\frac {\sum x_{i}^{k}}{n^{k \minus{} 2}\sum a_{i}}$[/quote]

Can you explain how $ {n^{k \minus{} 2}}$ comes when using the Holder's Inequality?

[quote="PKMathew"][quote="Agr_94_Math"]@keyree 10, it is true by the Holder's Inequality.
$ \left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\frac {\sum x_{i}}{n^{k \minus{} 2}}$
$ \implies\sum\frac {x_{i}^{k}}{a_{i}}\geq\frac {\sum x_{i}^{k}}{n^{k \minus{} 2}\sum a_{i}}$[/quote]Can you explain how $ {n^{k \minus{} 2}}$ comes when using the Holder's Inequality?[/quote]You might see it if you write it as \[ (1\plus{}1\plus{}\ldots \plus{} 1)(1\plus{}1\plus{}\ldots \plus{} 1)\cdots(1\plus{}1\plus{}\ldots \plus{} 1)\left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\sum x_{i}\]

[quote="aadil"]are the andreescu inequality and titus lemma same?[/quote]
Yes

[b]solution for the generalization i proposed in my first reply[/b]

Let $ a\equal{}\frac{1}{x},b\equal{}\frac{1}{y},c\equal{}\frac{1}{z}$
By the condition $ abc\equal{}1$, we obtain that: $ xyz\equal{}1$
Our ineq can be written as:
$ \sum\frac{1}{\frac{1}{x^n}(\frac{1}{y}\plus{}\frac{1}{z})}\equal{}\sum\frac{x^{n\minus{}1}}{y\plus{}z}$
From andreescu $ \sum\frac{x^{n\minus{}1}}{y\plus{}z} \ge \frac{(x\plus{}y\plus{}z)^{n\minus{}2}}{2}$
So we only have to prove that:
 $ (x\plus{}y\plus{}z)^{n\minus{}2} \ge 3$ which is straight application of $ AM\minus{}GM$ (notice that $ n \ge 3$) :wink:

[quote="minsoens"][quote="PKMathew"][quote="Agr_94_Math"]@keyree 10, it is true by the Holder's Inequality.
$ \left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\frac {\sum x_{i}}{n^{k \minus{} 2}}$
$ \implies\sum\frac {x_{i}^{k}}{a_{i}}\geq\frac {\sum x_{i}^{k}}{n^{k \minus{} 2}\sum a_{i}}$[/quote]Can you explain how $ {n^{k \minus{} 2}}$ comes when using the Holder's Inequality?[/quote]You might see it if you write it as
\[ (1 \plus{} 1 \plus{} \ldots \plus{} 1)(1 \plus{} 1 \plus{} \ldots \plus{} 1)\cdots(1 \plus{} 1 \plus{} \ldots \plus{} 1)\left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\sum x_{i}
\]
[/quote]

Is it a typo?
$ \boxed {(1 \plus{} 1 \plus{} \ldots \plus{} 1)(1 \plus{} 1 \plus{} \ldots \plus{} 1)\cdots(1 \plus{} 1 \plus{} \ldots \plus{} 1)\left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq\sum x_{i} }$

Should be
$ \boxed {(1 \plus{} 1 \plus{} \ldots \plus{} 1)(1 \plus{} 1 \plus{} \ldots \plus{} 1)\cdots(1 \plus{} 1 \plus{} \ldots \plus{} 1)\left(\sum\frac {x_{i}^{k}}{a_{i}}\right)(\sum a_{i})\geq (\sum x_{i})^k }$

Right? :)

[quote="aadil"]are the andreescu inequality and titus lemma same?[/quote]
You see, Titu Andresscu is a single person.  :rotfl:


calculus
derivative
function
calculus computations

I was given the graph the derivative, f'(x), and was asked to find the inflection points. Is there a way to tell by the graph without finding the 2nd derivative? I don't remember the equation, but the f'(x) graph looked like this...

http://img143.imageshack.us/img143/9231/35536700dx0.png

The inflection points of $ f$ are the places where $ f''$ changes sign. We have the graph of $ g\equal{}f'$; $ f''$ is the first derivative of $ g$. Can you see where the first derivative of a function $ g$ changes sign with its graph?


Hi. My problem is such, please solve.
Let ABCD be a quadratileral with a cycle inside (it's cyclic). And the center of the circle is A', we construct A'B'C'D' same with ABCD, then the two circle compose, and do the two quadratileral. So if the area(quadratileral)/area(circle)=a, then what is the section area(composed quadratilerals)/area(composed circles) equal to?

I doubt that anyone understands the problem statement... darij

so click on my signature "Tehranian" to see the recent state of this problem, please.


How about a suggestion.  A person doesn't die immediately, a person dies during the next player on the list's turn, except when there are 2 people.

I sign up, by the way.

[quote="Bugi"]Wait how would that change the game? So next one could challenge the loser of the last?[/quote]

It would simply add more suspense.

@ernie: Maybe it could be nice, but not this time (probably in next), ok?

So how would we play this??
(Organization)

Same like RR. Next player on the list challenges one from the remaining, gives him one pill, challenged player can accept or decline.

It's basically RR, just it's always one dies/one lives

[quote="Bugi"]Hello people!
Mewto first made this, but quickly died (viruses...). Now it's here!

Rules:
This is like Russian Roulette, just the guns are always 6 in one, 0 in another. Or, if we transfer to this game, you have 2 pills, one is poisonous, one isn't. You give to anyone challenged one of those, he accepts or declines, and you take the pills. One dies, one lives.

PM me with which pill is poison (A or B).

Signup in thread!

Signups:
1. ernie
2. EggyLv.999
3. spaceguy524
4. 15!!!
5. AwesomeToad[/quote]

I didn't sign up, just wondered how to play. Sorry for confusion :blush:

Then no. 5 is cf249 instead of AwesomeToad. If so little people join, IDK what to do...

I join! Sounds funilicious!
Signups:
1. ernie
2. EggyLv.999
3. spaceguy524
4. 15!!!
5. AwesomeToad
6.KING1
AwesomeToad..just stay :P

I offer cgyao15 Pill A (The poisonous one).

[hide="Match 4"]

KING1 was nervous... He mixed the pills! Is it A or B? Hm...

cgyao shouted: ''Come on, come on! You are deaaad!!!''
But KING just stood there, thinking... Which is the poisonous one? He asked the mod to say him again...
And I thought. Should I play a game on KING? Should I tell him the wrong pill? I said: Pill A is poisonous.

And KING offered A to cgyao.
''I take it!''
And who died? It's up to my choice about telling the truth :)

[hide="cgyao"]
died. I was good. But next time I maybe won't be
[color=red]cgyao15 dies[/color],[color=green] KING1 lives[/color]

Alive:

3. spaceguy524
5. cf249
6. KING1
7. Mewto55555

[color=blue]Mewto55555[/color], choose who to trick!

[/hide]
[/hide]

Blah no time for flavor I am bored

[hide="Match 5"]

[color=green]Mewto55555 lives[/color], [color=red]KING1 dies[/color]

[/hide]

Remaining contestants, please send in choices for Round 2. And let's do it!

Boo.  The whole point of the game is the flavor!  Why else would you sign up? :D

Ah... but that is the challenge. To survive non-flavor pshyocological battles.

Blah I'm too bored to write flavor... Probably I'll write some for R2... ernie you can give me some if you like, I'll write your name :P

I did.  I think Bugi just forgot about this.

Yeah I forgot about this... Blah let's play ernie's...
And Mewto, you DIDN'T send, space and cf did... Game over I win


Solve all solutions in integers of the equation

$ x^{3}+(x+1)^{3}+(x+2)^{3}+...+(x+7)^{3}= y^{3}$

Please see this book: [url]http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf[/url]

It is on page 50 (Hungary problem 5)


Αν f(x) είναι συνεχής, και η f'(x) υπάρχει όταν το χ≠ξ
και limf'(x)=κ όταν το χ τείνει στο ξ
Να δειχθεί ότι f'(ξ)=κ


I have some calculus problems where I'm asked to use a computer algebra system to find the
derivative of the function, then use the utility to graph the function and it's derivative.
What would be a good program to use to do this please?

[quote="AIME15"]You don't need to post this twice, even if somebody else tells you to.[/quote]

True, but I realised he was probably right :)


function
calculus
derivative
real analysis
real analysis unsolved

Let f(x,y,z) be a three variables class $ C^{1}$ function with f(0,0,0)=0 and $ D_{3}f(0,0,0) \neq 0$ ( derivative of f with respect to z at (0,0,0) is not 0).
Suppose w $ \in$ $ \mathbb{R}^{3}$ is perpendicular to gradf(0,0,0). Show that there exists g(t) a $ C^{1}$ space curve defined on $ ( \minus{} \delta, \delta)$ satisfying f(g(t))=0 and g'(0)=w.


invariant
combinatorics proposed
combinatorics

The integers $ 1,2,\dots,20$ are written on the blackboard. Consider the following operation as one step: [i]choose two integers $ a$ and $ b$ such that $ a\minus{}b \ge 2$ and replace them with $ a\minus{}1$ and $ b\plus{}1$[/i]. Please, determine the maximum number of steps that can be done.
[i]Yudi Satria, Jakarta[/i]

[hide="Invariant for the operation"]Notice that the sum of the two elements of an operation is invariant just because $ a\plus{}b\equal{}(a\minus{}1)\plus{}(b\plus{}1)$[/hide]
[hide="Restriction for the last configuration"]If there is no possible step then $ a\minus{}b\equal{}0$ or $ a\minus{}b\equal{}1$ for all pairs of $ a,b$ [/hide]
[hide="Invariant for the numbers on the blackboard"]but the sum of all elements keeps the same after each step and equal to $ 210$ [/hide]
[hide="Last Configuration"]then the last integers are $ (10,\cdots,10,11,\cdots,11)$ with ten $ 10$ and ten $ 11$.[/hide]

Is this really enough to answer the question as written?  Not all paths leading to the same configuration have the same length (for example, if we replace $ 20$ by $ 6$ we can get to the final configuration by

$ 123456 \rightarrow 223455 \rightarrow 233445 \rightarrow 333444$ ($ 3$ steps)

or

$ 123456 \rightarrow 124455 \rightarrow 224445 \rightarrow 233445 \rightarrow 333345 \rightarrow 333444$ ($ 5$ steps)

or possibly longer paths.  It's not clear from the final configuration what the longest possible path is, though you can get an upper bound by looking at the sums of the squares of the elements (replacing $ (a,b)$ by $ (a \minus{} 1, b \plus{} 1)$ reduces the sum of the squares by $ a^2 \plus{} b^2 \minus{} (a \minus{} 1)^2 \minus{} (b \plus{} 1)^2 \equal{} 2(a \minus{} b) \minus{} 2 \geq 2$).

It's easy to find an upper bound for the number of steps.

[hide="Invariant"]
Let the numbers on the blackboard be $ a_1,a_2,\cdots a_{20}.$
Let $ S\equal{}\sum_{i\equal{}1}^{20} a_i^2.$  At the beginning, $ S_0\equal{}\sum_{i\equal{}1}^{20} i^2\equal{}2870$.
As we showed above, the final setup must be 10,10,...,11,11,..., giving $ S_{final}\equal{}10(10^2\plus{}11^2)\equal{}2210$.

After performing the operation, $ S_n\equal{}S_{n\minus{}1}\minus{}a^2\minus{}b^2\plus{}(a\minus{}1)^2\plus{}(b\plus{}1)^2\equal{}S_{n\minus{}1}\minus{}2(a\minus{}b)\plus{}2\le S_{n\minus{}1}\minus{}2$.

Therefore since the invariant decreases by 4 each time, we can perform the operation at most $ \frac{S_{final}\minus{}S_0}{2}\equal{}330$ times.
[/hide]

Seems like a lot of trouble to find a construction though.

[i]Yes, the answer is $ 330$. Some contestants got $ 7$ by just showing the construction, with some explanation, which is actually not complete.[/i]

The construction is as follows:

(1) We start with the sequence 1-20. Operate on these pairs in order: (1,3) (2,4) (3,5) ... (18, 20). This transforms the sequence into 2-19+2,19 (that is the sequence from 2 to 19, and an extra 2 and 19). Essentially, what it does is to pull a "1" from the "20" and add it to the "1", leaving everything else intact.

(2) Now since we have 2-19+2,19, the sequence 2-19 is just similar to 1-20, and if we apply the same operations, we get 3-18+2,3,18,19. Do this iteratively and finally we get 10-11+2,3,...19, i.e. 10,11+2-19. Thus, starting from 1-20, we get 10,11+2-19 with all the above operations.

(3) Since 2-19 is similar to 1-20, if we apply all the above operations on it, we should get 10,10,11,11+3-18. Do this iteratively we will eventually get 10,10...11,11,...11.

Since each of the above operations involve x and x+2 for some x, it must be maximum in terms of the number of operations to get the last configuration.


Screenshots of my computer for sharing on AoPS


geometry
perimeter
inradius
circumcircle
trigonometry
trig identities
Law of Sines

[hide]Let $ r,s$ and $ A$ denote the inradius, semiperimeter and area of the equilateral triangle, respectively. It is well known that $ A=rs$. But $ s=\frac{3a}{2}$ and $ A=\frac{\sqrt 3}{4}a^{2}$. So $ \frac{\sqrt 3}{4}a^{2}=\frac{3}{2}ar\Longrightarrow a=2\sqrt 3r$.

Now as $ \pi r^{2}=48\pi=4^{2}\cdot 3\cdot \pi$ we get $ r= 4\sqrt 3$. Thus $ a=2\sqrt 3r=\boxed{24}$.[/hide]

[quote="xxxxx"]The area of a circle inscribed in an equilateral triangle is 48$ \pi$. The perimeter of this triangle is:

a. $ 72$
b. $ 48\sqrt{3}$
c. $ 36$
d. $ 24$[/quote]
the the area of the circle is $ 48\pi$ so $ 4\sqrt{3}$ is the raduis of the circle. since its inscribed, the  radius that intesects a base is perpendicular to it. therefore it coincides with the altitude. theres a formula or something that says the circumcenter of an equilateral triangle to the bisector of a base is 1/3 of the altitude or something. so thats what we have here. the altitude is therefore $ 12\sqrt{3}$ using special right triangles, we figure out the sidelength is 24 so the perimeter is 72.

abcadaea is right.. nayel found just the length of a side of the triangle.

I got $ 4\sqrt{3}$ in the same manner.
[hide="however"]These are legs to an isosceles tri. call it... $ ABC$. $ ABC$'s base($ b$) is half the length of the circumscribing triangle's... and one-sixth of it's perimeter($ p$).
[hide="We will use..."][b]The Law of Sines[/b]

$ \frac{4\sqrt{3}}{\sin 30^{\circ}}= \frac{b}{\sin 120^{\circ}}$

mentally, $ b = \frac{4(\sqrt{3})^{2}(2)}{2}=12 \Rightarrow b = \frac{p}{6}$
$ p = 72$[/hide]
[/hide]


Find an integer a such that (x-a)(x-10)+1 can be written as a product of (x+b)(x+c) of two factors with integers b and c.

[hide]

So you have a+10 = b+c

and bc = 10a+1

bc ends in 1, so it's either  9 9, 3 3 , 3 7, or 7 3.

You find that it's 99, so it's

(x-9):^2: = x:^2:-18x+81 = (x-8)(x-10) + 1

So a = 8

[/hide]


algebra polynomial calculus derivative function

Let $ f$ from R->R such that $ f(x)\equal{}\frac{x^{3}}{x^{2}\plus{}1}$  .Prove the existance of $ f^{\minus{}1}$ and find $ f^{\minus{}1}(x)$

[quote="silouan"]Let $ f$ from R->R such that $ f(x) \equal{}\frac{x^{3}}{x^{2}\plus{}1}$  .Prove the existance of $ f^{\minus{}1}$ and find $ f^{\minus{}1}(x)$[/quote]

I think $ f^{\minus{}1}$ is messy.

However, we can show that it exists. Since $ f(x)$ is continous, it is sufficient for it to be monotone. Show that $ f'(x)> 0$ and conclude that this is monotone (except at $ 0$ which isnt a problem).

[quote="sylow_theory"]
I think $ f^{\minus{}1}$ is messy. [/quote]

Yes, $ f^{\minus{}1}$ can be found by solving a degree_3 polynomial equation (with Cardan method, for example) but the result seems ugly and of a very poor interest.

Well there's still the first part of his question: prove that $ f^{\minus{}1}$ exists.

$ f(x) \equal{}\frac{x^{3}}{x^{2}\plus{}1}\equal{} x\left(1\minus{}\frac{1}{x^{2}\plus{}1}\right)$

$ f$ is defined and continuous for all real inputs.

Note that for positive $ x$, since $ x$ is strictly increasing and $ \left(1\minus{}\frac{1}{x^{2}\plus{}1}\right)$ is strictly increasing, $ f(x)$ is strictly increasing. Since $ f(x)$ is odd, $ f(x)$ is strictly increasing for negative $ x$, too. $ f(x)$ happens to have derivative 0 at $ x \equal{} 0$, but this is only at one point.

This means that $ f(x)$ is monotonically increasing, and is therefore a one-to-one function, which means that $ f^{\minus{}1}$ exists.


geometry
geometric transformation
rotation
complex analysis
complex analysis unsolved

Find a conformal mapping of the region outside the circle $ |z\minus{}1|\equal{}1$ and inside the circle $ |z|\equal{}2$ onto the unit disk $ |w|<1$.

Given a region ibounded by two circular arcs: (1) find their points of intersection; (2) send one of them to $ \infty$ and the other to $ 0$ - the region becomes a sector; (3) map the sector onto half-plane; (4) map the half-plane onto a disk.

Only in this case, we have only one point of intersection. If we map that to $ \infty,$ we'll get a strip, not a sector. The mapping $ z\mapsto\frac1{z\minus{}2}$ maps this region to the strip $ \minus{}\frac12<\text{Re}(z)<\minus{}\frac14.$

Thanks for the correction. I was too lazy to read the statement.  :|

Finishing the thought:

$ z\mapsto\frac1{z\minus{}2}$ sends our original set to the vertical strip named above.

$ z\mapsto e^{iz}$ sends that vertical strip to a sector.

$ z\mapsto e^{i\alpha}z^{\beta}$ for a properly chosen $ \beta$ sends that sector to a half plane (with the rotation $ e^{i\alpha}$ there so we can make it a familiar half plane.)

And, as mlok said, we can send the half plane to the disk.


My favorite store was having a 15% off everything sale. The jeans I wanted
were originally 49.99 dollars , but they were already marked down to 35.99 dollars
before the 15% discount. When I bought the jeans, what percent discount off
the original price did I pay? (Round to the nearest tenth of a percent.)

Hi;

You pay 30.5915 which is 85% of 35.99

Your total discount is 38.9%

Correcting slight error , the answer is .38804 rounding to the nearest tenth of a percent is .38.8 % not 38.9%, sorry!


Abel lemma: Suppose $ (x,y,z)$ is one solution of Fermat equation $ x^n\plus{}y^n\equal{}z^n$.
Under condition of $ (x,y)\equal{}1,(x,z)\equal{}1,(y,z)\equal{}1,n$ is odd prime, if $ (y,n)\equal{}1$, then we get $ y\equal{}pq,z\minus{}x\equal{}p^n,(p,q)\equal{}1$.

Since $ z \minus{} x|y^n,(y^n,y \plus{} 1) \equal{} 1$, hence $ (z \minus{} x,y \plus{} 1) \equal{} 1$.
So there exists $ k,k(y \plus{} 1) \equiv \minus{} x,\bmod z \minus{} x$.
Hence $ k^n(y \plus{} 1)^n \equiv \minus{} x^n,\bmod z \minus{} x$.
Since $ (x,z) \equal{} 1$, hence $ (x,z \minus{} x) \equal{} 1$.
Considering $ k(y \plus{} 1) \equal{} \minus{} x,\bmod z \minus{} x$, we get $ (k,z \minus{} x) \equal{} 1$.
Since$ z \equal{} z \plus{} ky \minus{} ky$, hence $ 0 \equal{} z^n \minus{} y^n \minus{} x^n \equal{} (z \plus{} ky \minus{} ky)^n \minus{} y^n \minus{} x^n \equal{} (z \plus{} ky)^n \plus{} C_n^1( \minus{} ky)(z \plus{} ky)^{n \minus{} 1} \plus{} A \plus{} C_n^{n \minus{} 1}( \minus{} ky)^{n \minus{} 1}(z \plus{} ky) \plus{} ( \minus{} ky)^n \minus{} y^n \minus{} x^n$

Since $ z + ky \equiv z - x - k,\bmod k(y + 1) + x$, hence
$ 0\equiv (z - x - k)^n + C_n^1( - ky)(z - x - k)^{n - 1} + B + C_n^{n - 1}( - ky)^{n - 1}(z - x - k) + ( - ky)^n - y^n - x^n,\bmod k(y + 1) + x$.
Considering $ k(y + 1) + x\equiv 0,\bmod z - x$,we get
$ 0\equiv (z - x - k)^n + C_n^1( - ky)(z - x - k)^{n - 1} + B + C_n^{n - 1}( - ky)^{n - 1}(z - x - k) + ( - ky)^n - y^n - x^n,\bmod z - x$.
Since $ z - x|y^n$, hence
$ 0\equiv ( - k)^n + C_n^1( - ky)( - k)^{n - 1} + C + C_n^{n - 1}( - ky)^{n - 1}( - k) + ( - ky)^n - x^n,\bmod z - x$.
Then we get $ 0\equiv ( - k)^n(1 - y)^n - x^n,\bmod z - x$.
Since $ 0\equiv k^n(y + 1)^n + x^n,\bmod z - x,(k,z - x) = 1,y = pq,z - x = p^n,(p,q) = 1,n$ is odd prime, hence $ p^3|2npq$.
Then we get $ z - x = p^n = 1$.

Suppose $ (a,b,c)$ is such positive solution of Fermat equation.
Since $ n$ is odd prime, hence $ a^n\plus{}b^n\equal{}c^n,(\minus{}a)^n\plus{}c^n\equal{}b^n$.
Suppose $ (n,b)\equal{}1$. Let $ z\equal{}c,y\equal{}b,x\equal{}a$, then we get $ c\minus{}a\equal{}1$.
If $ (n,a)\equal{}1$. Let $ z\equal{}c,y\equal{}a,x\equal{}b$, then we get $ c\minus{}b\equal{}1$.
Hence $ a\equal{}b\equal{}1$. This is contradictory.
If $ (n,c)\equal{}1$. Let $ z\equal{}b,y\equal{}c,x\equal{}\minus{}a$, then we get $ b\plus{}a\equal{}1$.
This is contradictory. T

Thank you for your view.

[quote="Taoli"]Since $ z + ky \equiv z - x - k,\bmod k(y + 1) + x$, hence
$ 0\equiv (z - x - k)^n + C_n^1( - ky)(z - x - k)^{n - 1} + B + C_n^{n - 1}( - ky)^{n - 1}(z - x - k) + ( - ky)^n - y^n - x^n,\bmod k(y + 1) + x$.
Considering $ k(y + 1) + x\equiv 0,\bmod z - x$,we get
$ 0\equiv (z - x - k)^n + C_n^1( - ky)(z - x - k)^{n - 1} + B + C_n^{n - 1}( - ky)^{n - 1}(z - x - k) + ( - ky)^n - y^n - x^n,\bmod z - x$.
Since $ z - x|y^n$, hence
$ 0\equiv ( - k)^n + C_n^1( - ky)( - k)^{n - 1} + C + C_n^{n - 1}( - ky)^{n - 1}( - k) + ( - ky)^n - x^n,\bmod z - x$.

Then we get $ 0\equiv ( - k)^n(1 - y)^n - x^n,\bmod z - x$.
[\quote]This is wrong.


probability
geometry
3D geometry
tetrahedron
combinatorics unsolved
combinatorics

It's been a while since the last post...

If you choose randomly 6 positive numbers (uniformly distributed) what is the probability of being the lenghts of a thetraedron?

Do you believe this one is unsolvable:  :maybe:

If you choose randomly 6 positive numbers (uniformly distributed in $[0,1]$) what is the probability of being the lenghts of a thetraedron?

Does anyone can fix it?!?

If six numbers are chosen at random, uniformly and independently, from the interval $[0,1]$, what is the probability that they are the lengths of the edges of a tetrahedron?

This is a nightmarish question for two reasons: first of all, it is not at all easy to decide whether 6 given numbers are edge lengths of a tetrahedron. Secondly, once one gets an analytic condition (which is possible but, as I said, very unpleasant and boring), there seems to be no way to evaluate the corresponding volume exactly. So, the answer is "Yes, nobody can solve this". I agree that it seems sort of strange that such a simply formulated problem has no reasonable solution but that's how life goes. If you really need it for something, we can think of the best way to evaluate the probability numerically. Otherwise, my advice would be just to forget it.

I conjecture that:

If you choose randomly $ 6$ positive numbers (uniformly distributed in $ [0,l]$ ) the probability of being the lenghts of some thetraedron is $ \frac{1}{3}$.

If you choose randomly $ C^2_n$ positive numbers (uniformly distributed in $ [0,l]$ ) the probability of being the lenghts of some $ n$-dimensional simplex is $ \frac{1}{n}$.

Any other comment is warmly welcomed.


Which state do you live in? Who's the tallest person in your family? Who's the tallest person yu've ever met?

state= alabama
talleast person= in my family, my dad who is 5'11 feet, if relatives count i have several cousins that are 6' up to 6'4
Tallest person I have met= 6'9 some guy on the airplane  :rotfl:


determine ALL x’s of:

 cos 2 (pi) (x − 1/4 )

such that cos x > 0.


topology
advanced fields
advanced fields theorems

Does anyone have recommendations for books on algebraic topology?

I'm reading Hatcher: Algebraic topology and consider it as a good book on the subject.
Since it's freely available on the net, you can take a look at it.


inequalities
geometry
angle bisector
geometry proposed

In $\triangle ABC$, $AN$ is the angle bisector of $\angle A$ where $N$ is on $BC$. The incirle of $\triangle ABC$ touches the sides $AB$ and $AC$ at $D$ and $E$. $S_a$ is the area of $ADNE$. $S_b$ and $S_c$ are derived in similar ways. If $S$ is the area of $\triangle ABC$, then prove that $ S \leq S_a + S_b + S_c \leq \frac {3S}2$

$\sum\frac{S_a}{S}=3-\sum\frac{(s-b)\frac{ac}{b+c}}{ac}=3-\sum\frac{s-b}{b+c}=3-\sum\frac{a}{b+c}$. Thus the inequality is equivalent to \[\frac{3}{2}\le\sum\frac{a}{b+c}<2\]which is well known


Is there an efficient way to solve the following problem? The answer is 43.

At Chicken Little's, you can order Chicken Fingers in boxes of 6, 9, or 20. If you order a box of 6 and a box of 9, you can get 15 fingers. What is the greatest number of fingers you cannot order?

Thanks

The formula is  ab-a-b for two numbers but I'm not so sure for three...

There is no known formula for three numbers. However, there is a formula for the largest not achievable with ab, bc, ca. a,b,c might have to be pairwise coprime, but I'm not sure about that. However, this problem isn't of that form. Anyway, with this problem, if you can show that you cannot get 43 but you can get 44, 45, 46, 47, 48, and 49, then you're done. (Do you see why?)

The following link seems to suggest there is a formula for 3 variables:[url]http://mathworld.wolfram.com/CoinProblem.html[/url].  I haven't read the papers referenced, however.  In any case, there are efficient algorithms known for 3 variables.

For a fixed number of variables, Ravi Kannan showed that there is a polynomial-time algorithm to solve the problem.  See the work on the Frobenius problem at  [url]http://www-2.cs.cmu.edu/~kannan/Papers/pubs.html[/url]

For the particular numbers here (6, 9, 20), these results are overkill, because as ComplexZeta noted, we can reason about this case directly.


% poznan1.tex
% j12-I-12
%

\input Jarnik.tex
\jdate      {10th April 2002}
\jyear      {12th}
\category   {I}
\problemno  {12}
\university {Adam Mickiewicz University in Pozna\'n}

\problem
Let $S(n)$ be the set of all permutations of the set $\{1,2,\ldots,n\}$.
For every $\pi\in S(n)$, put $A(\pi) = \{\eta \in S(n)\colon
\eta\circ\pi = \pi\circ\eta\}$.  Let $\Phi(\pi)$ be the number of all
elements of $A(\pi)$.  Find all values of\/~$\Phi$.
\endproblem

\solution
For any $a_1, a_2, \ldots, a_k \in \{1,2,\ldots,n\}$, $k \geq 2$, $a_i
\ne a_j$ for $i \ne j$, denote by $\pi_{a_1,a_2,\ldots,a_k}$ the
following permutation$
\pi_{a_1,a_2,\ldots,a_k}(x) = \cases {
	a_{ i+1  \bmod k} &
$if $x \in    \{a_1,a_2,\ldots,a_k\}$\cr
	x                 & if $x \notin \{a_1,a_2,\ldots,a_k\}$} \hbox{.}

Let $  \pi = \pi_{a_1^1, a_2^1, \ldots, a_{k_1}^1} \circ \pi_{a_1^2,
a_2^2, \ldots, a_{k_2}^2} \circ \cdots \circ \pi_{a_1^l, a_2^l, \ldots,
a_{k_l}^l} $ where $ \bigl \{a_1^i, a_2^i, \allowbreak\ldots,
a_{k_i}^i\bigl \} \cap \bigr \{a_1^j, a_2^j, \allowbreak\ldots,
a_{k_j}^j\bigl \} = \emptyset $ for every $i \ne j$.  Each permutation
has the form above.  Suppose that  there exists a $c \notin \bigl
\{a_1^i, a_2^i, \allowbreak\ldots, a_{k_i}^i\bigr \}$ such that $\eta(c)
= a_s^i$ for some $i$ and $s$.  Then
\]
\eqalignnn {
\pi \bigl (\eta(c)\bigr ) &= a_{ s+1  \bmod k_i}^i \cr
\noalign {\noindent and}
\eta\bigl (\pi(c)\bigr ) &= \cases {
	\eta(c) = a_s^i & if $c \notin \bigcup_{j\ne i}
		\bigl \{a_1^j,a_2^j,\ldots,a_{k_j}^j\bigr \}$\cr
	\eta(a_r^j)     & if $c = a_{ r-1  \bmod k_j}^j$} \hbox{.} }

Consequently, if $ \eta \circ \pi = \pi \circ \eta $, then $ \eta\Bigl
(\bigl \{a_1^i,a_2^i,\ldots,a_{k_i}^i\bigr \}\Bigr ) = \bigl \{a_1^i,
a_2^i, \allowbreak\ldots, a_{k_i}^i\bigr \} $ for every $ i=1, 2,
\allowbreak\ldots, l$.  If $\eta(a_s^i) = a_r^i$ for some $s,r \in \{1,
\allowbreak\ldots, k_i\}$, then $\pi\bigl (\eta(a_s^i)\bigr ) = a_{r+1
\bmod k_i}^i$ and $ \eta\bigl (\pi(a_s^i)\bigr ) = \eta\bigl (a_{s+1
\bmod k_i}^i\bigr )$.  Consequently, if $ \eta \circ \pi = \pi \circ
\eta$, then $ \eta\Bigl (\bigl \{a_1^i, a_2^i, \allowbreak\ldots,
a_{k_i}^i\bigr \}\Bigr ) = \bigl \{a_1^i,a_2^i,\ldots,a_{k_i}^i\bigr \}$
and $\eta(a_s^i)= a_{r_i+s \bmod k_i}^i$ for every $i = 1, 2,
\allowbreak\ldots, l$ and some $ r_i \in \{1, \allowbreak\ldots, k_i\}$.
It is clear that the conditions above are also sufficient.  Therefore
$\Phi(\pi) = k_1 \cdot k_2 \cdot \cdots \cdot k_l \bigl (n - (k_1 + k_2
+ \cdots + k_l)\bigr )!$.
\endsolution

\bye

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\documentclass[11pt,leqno]{article}
\usepackage{amsmath,amssymb,latexsym}
\usepackage{amscd}

\show \leqslant

  \topmargin -1.2in

\oddsidemargin 0in

\textwidth 16.5cm

\textheight 25cm

\pagestyle{empty}

\begin{document}
\vspace{1.0cm} {\bf  Problem } ( I category ) \newline Let $S(n)$
be the set of all permutations of the set $\{1,2,\dots,n\}$. For
every $\pi\in S(n)$ put $A(\pi)= \{\eta\in S(n):\eta\circ
\pi=\pi\circ \eta\}$.
 Let $\Phi(\pi)$ be the number of all elements of $A(\pi)$.
 Find all values of $\Phi$.
 \vspace{0.5cm}

{\bf Solution.} For any $a_1,a_2,\dots,a_k\in \{1,2,\dots,n\}$,
$k\geqslant 2$,
$a_i\ne a_j$ for $i\ne j$ denote by $\pi_{a_1,a_2,\dots,a_k}$
the following permutation
\]
\pi_{a_1,a_2,\dots,a_k}(x)=\left\{
\begin{array}{ll} a_{i+1\,mod\,k}  &
  \mbox{if $x\in\{a_1,a_2,\dots,a_k\}$} \\
x  &   \mbox{if $x \notin \{a_1,a_2,\dots,a_k\}$}.
\end{array}\right
\[
Let $\pi=\pi_{a_1^1,a_2^1,\dots,a_{k_1}^1}\circ
\pi_{a_1^2,a_2^2,\dots,a_{k_2}^2}
\circ\dots\circ \pi_{a_1^l,a_2^l,\dots,a_{k_l}^l}$ where
$\{a_1^i,a_2^i,\dots,a_{k_i}^i\}\cap \{a_1^j,a_2^j,\dots,a_{k_j}^j\}=
\emptyset$ for every $i\ne j$.
Each permutation has the form above.  Suppose that  there exists
$c\notin \{a_1^i,a_2^i,\dots,a_{k_i}^i\}$
such that
$\eta(c)=a_s^i$ for same $i$ and $s$. Then
\]
\pi(\eta(c))=a_{s+1\,mod\, k_i}^i
\[
 and
\]
 \eta(\pi(c))=\left\{ \begin{array}{ll}
\eta(c)=a_s^i & \mbox{if $ c \notin
\bigcup_{j\ne i}  \{a_1^j,a_2^j,\dots,a_{k_j}^j\} $ }\\
\eta(a_r^j) &  \mbox{if $c=a_{r-1\,mod \,k_j}^j $ }.
\end{array}\right
$$
Consequently,
if $\eta\circ\pi=\pi\circ \eta$,
then $\eta(\{a_1^i,a_2^i,\dots,a_{k_i}^i\})=\{a_1^i,a_2^i,\dots,a_{k_i}^i\}$
for every $i=1,2,\dots,l$.
If $\eta(a_s^i)=a_r^i$ for some $s,r\in \{1,\dots,k_i\}$, then
$\pi(\eta(a_s^i))=a_{r+1\,mod\, k_i}^i$ and
$\eta(\pi(a_s^i))=\eta(a_{s+1\,mod\, k_i}^i)$.
Consequently,
if $\eta\circ\pi=\pi\circ \eta$,
then $\eta(\{a_1^i,a_2^i,\dots,a_{k_i}^i\})=\{a_1^i,a_2^i,\dots,a_{k_i}^i\}$
 and
$\eta(a_s^i)=
a_{r_i+s\,mod\, k_i}^i$ for every $i=1,2,\dots,l$ and some
$r_i\in\{1,\dots, k_i\}$.   It is clear that the conditions above are also
sufficient. Therefore $\Phi(\pi)=k_1\cdot k_2\cdot\dots\cdot k_l
(n-(k_1+k_2+\dots+k_l))!$.

\end{document}

oh excuse me for this silly action ,I have a file in tex format and I couldnt read it so I post it here to read.
I do apologise. :blush:

You can always use the Preview button instead of Submit.  Or just download a free version of LaTeX.


Express $ 0.\overline{45} \plus{} 0.\overline{54} \plus{} 0.5 \plus{} 0.4$ as a common fraction.


[hide]
there are 10 ways to get even, each with a 1/16 probability
o,o,o,o
e,e,e,e
e,e,o,e
e,e,e,o
o,e,e,e
e,o,e,e
e,o,e,o
o,e,o,e
o,e,e,o
e,o,o,e
so, 5/8[/hide]

Is this the real answer because i got 9/16

All right, both have the same parities so $(\frac{3}{4})^{2}+(\frac{1}{4})^{2}=\frac{10}{16}=\frac{5}{8}=E$
I know this because I got it right.  :D

\documentclass{article}
\begin{document}

I guess the answer is $ \frac{9}{16}$

because that even case is $ \frac{3}{4}$ and odd case is $ \frac{1}{4}$.

therefore, $1-( \text{even -odd} ) \times 2
=1- ( \frac{1}{4} \times 2 frac{3}{4} )\times 2 = 9/16 $

\end{document}  :roll:

okay - quick LaTeX tip.  if you're using TeXnicCenter, you do everything with the begin document.  if you're posting stuff on AoPS, all you have to do is enclose the math stuff with dollar signs.  the text comes out normally without you doing anything

For example, $\$$ \sqrt{ab} $\$$ renders as

$\sqrt{ab}$ as desired.

this problem stumped me during the test but I understood it right after. It seemed tricky at first but as you proceed, it does seem to unfold to become an easy problem.

Sorry for reviving- I dont understand Xaero's solution with parities  :P

because of symmetry, the probability of $ ad$ being some parity is the same as the probability of $ bc$ being that parity (and clearly both need to be the same parity).

so probability of $ ad$ being even is $ 3/4$ and probability of it being odd is $ 1/4$, so then you get the result that xaero got.

how do we know the probabillity of ad being even is 3/4, and then why do we square??

Nevermind I understand:
3/4 * 3/4 chance of both even, 1/4 * 1/4 chance of both odd.
The reason I didnt get this on the test was because I didnt understand that "chosen independently" meant with replacement  :lol:

Given four distinct sets, each of which is made up of two possibilities(e or o),  there are 8 possibilities. The P(even)=P(odd) since there are the same number of evens and odds in a consecutive sequence starting at 0 and ending with an odd. therefore. E= E+E or O+O. which gives Given two integers, you can get an even sum in two ways e+e and o+o. Since there are two sets (ab and cd) , there are four possibilities for E+E. Odd can only occur in one way yielding 5/8

[quote=worthawholebean]Integers $ a$, $ b$, $ c$, and $ d$, not necessarily distinct, are chosen independently and at random from $ 0$ to $ 2007$, inclusive. What is the probability that $ ad \minus{} bc$ is even?

$ \textbf{(A)}\ \frac {3}{8}\qquad \textbf{(B)}\ \frac {7}{16}\qquad \textbf{(C)}\ \frac {1}{2}\qquad \textbf{(D)}\ \frac {9}{16}\qquad \textbf{(E)}\ \frac {5}{8}$[/quote]

When it says that $ad-bc$ is even, are they assuming that it could also be NEGATIVE? Because, based on the solutions, they consider the case when $a=0, d=0$ as a case when this works. This would not work if we were to only want $ad-bc$ to be a positive even integer. So am I right about $ad-bc$ being both positive AND negative?

Thank you!

Of course $ad-bc$ can be negative. $-2$ is an even integer, for example.


probability
ceiling function
counting
distinguishability

A pair of 8-sided dice have sides numbered 1 through 8. Each side
has the same probability (chance) of landing face up. What is the
probability that the product of the two numbers on the sides that
land face-up exceeds 36?

We can use a bit of casework.

First, we find $ \lceil 36 \div 8 \rceil$, to narrow our possibilities.  This is $ 5$.  So our combinations must have a number $ \ge 5$.

The possible ways:

(5,8)
(6,7)
(6,8)
(7,6)
(7,7)
(7,8)
(8,5)
(8,6)
(8,7)
(8,8)

There are $ 64$ ways for the dice to come up, so our answer is $ \frac{10}{64}$, or $ \boxed{\frac{5}{32}}$.

I made a kind of multiplication table.  8 rows and 8 columns.  Then count which ones are greater than 36.
A little more legwork, but it helped me to visualize.

I just want to make sure, are die distinguishable? Because I did this problem and it said that you have to find out how many was a certain number like this before and it said in the solution that you have to first find out how many ways could which die obtain the number. If you don't exactly get what I mean PM me.


my school is a secondary school but it doesnt offer amc 10 or 12 so wut do i do case i still want to take the test. :?:

You just asked this last week. We already answered it for you.


function
integration
trigonometry
calculus
limit
complex analysis

Hi,

Can you show that
$ \frac{1}{\Gamma(z)} \equal{} \int_{c\minus{} i\infty}^{c\plus{}i\infty} e^t t^{\minus{}z} dt \equal{} \frac{e^c c^{1\minus{}z}}{\pi} \int_0^{\frac{\pi}{2}} \cos^{z\minus{}2}\theta \cos\left({c\tan{\theta} \minus{}z\theta} \right) d\theta$

In the original presentation, we were to solve the Bromwich integral by a modified Hankel loop.

Cheers,

Nick

I think it should be:

\[ \frac{1}{\Gamma(z)}\equal{}\minus{}\frac{i}{2\pi}\int_{c\minus{}i\infty}^{c\plus{}i\infty}e^{t}t^{\minus{}z}dt\]

youriens,
You are absolutely right.

 I can solve this integral using a Henkel loop, I am just having trouble solving the Bromwich Integral, where this contour is defined in the usual way for Inverse Laplace transforms.

Then it just reduces to a lot of trigonometry: If $ t\equal{}c\plus{}iy\equal{}re^{i\theta}$, then for any $ c<\infty$, $ \displaystyle\lim_{r\to\infty} (\theta)\equal{}\pm \pi/2$.  So $ r\equal{}\sqrt{c^2\plus{}y^2}\equal{}\sqrt{c^2\plus{}c^2\tan^2 \theta}$ and therefore $ t\equal{}c\, \sec(\theta) e^{i\theta}$.  Now just plug everything in and run it from $ \minus{}\pi/2$ to $ \pi/2$.


Two player are playing the following game. Suppose we have an infinite chessboard. Player A puts on his turn 0 somewhere on the board and B puts 1. Winner is the player who has the first $2\times 2$ board full of his own numbers. Has someone a winning strategy?

Neither has a winning strategy. I will show first that B has no winning strategy. B can split the board into $2\times 1$ blocks like this

...
...aabbccddeeff...
...ghhiijjkkllm...
...nnooppqqrrss...
...tuuvvxxyyzzå...
...

We see that every $2\times 2$-block must contain some $2\times 1$-block B has thought. A can't win because he can put his number into the same $2\times 1$ block as A.

A can't win because he can play the same way as B on above and the first move of A can't make the situation worse than before the move.


Four cards are face down on a table.  One of them shows the sum of two numbers, another their product, a third the sum of their reciprocals, and the last one the sum of their squares.

These cards are labeled on the front what each number represents, but the backs are identical.  For example, if the numbers are 3 and 6, the front of the cards will read:

9: SUM
18: PRODUCT
1/2: SUM OF RECIPROCALS
45: SUM OF SQUARES

These numbers can be real or complex.

[b]1) What is the minimum number of cards you need to turn over to determine the two numbers?[/b]

Suppose you cut off the part of the card that told you what each number was.  So now the cards would read:

9
18
1/2
45

[b]2) Is it possible to determine what the two numbers are by looking at all four cards?  For example, let's say I gave you cards that read 0, 1, 2, and 2.  Is it possible to determine the numbers?[/b]

[b]3) Suppose you discarded the SUM OF SQUARES card.  Is it possible to look at the other three cards and determine the numbers?[/b]

The sum of recipricols is unnecessary. Just divide the sum by the product to get the recipricol.

I'm not too sure about the first problem, but the answer to the second one is solve the system of equations. Since (sum)/(product)=(sum of recipricols), we know that sum=2, and product=2, and sum of recipricols=1. Now it's easy to solve.

Follow the procedure in step 2 to get the answer for the third one.

if you know sum=a, product=b, why not just solve

$x^{2}-ax+b=0$?

[hide="1"]
I think it's [b]3[/b].
If:
$x+y=a , \ \ xy=b , \ \ \frac{1}{x}+\frac{1}{y}=c , \ \ \text{and}\ \ x^{2}+y^{2}=d$

Then $c=\frac{a}{b}$ and $d=a^{2}-2b$, But i cannot get a relationship between $c$ and $d$. If you pick any two other than $c,d$ then you can solve, but if you pick $c$ and $d$ you can try to solve, but end up getting two solutions. Thus you need $\boxed{3}$. I think.
[/hide]

2 is coming....

(edits: formatting)

[hide="About 1"]I know it's either 2 or 3.  If you can solve it knowing the sum of their squares and the sum of their reciprocals (and only those), it's 2.  Otherwise it's 3.[/hide]

[quote="Allan Z"]If you get $0$,$1$,$2$,$3$.

The attack point is $"0"$
If the sum is $0$, that means two numbers are $a$, and $-a$. It implies the sum of the reciprocals is also $0$. Therefore, $0$ can't be the sum nor sum of reciprocals.
If the sum of the squares is $0$, it implies two numbers are $0$, now everything is $0$ except the sum of reciprocals is undefined. Therefore, the sum of the squares can't be $0$.
If the product is $0$, that means at least one number is $0$, and the sum of the reciprocals is undefined, so the product can't be $0$ neither.

So the answer is no?[/quote]

The problem with this argument is that a set of cards [u]cannot[/u] have only one zero, without another zero or an undefined. (When dealing with only real numbers).

Put $x+y=a \ , \ \ xy=b , \ \ \frac{1}{x}+\frac{1}{y}=c \ , \ \text{ and }x^{2}+y^{2}=d$.

Then $a=0 \iff c=0 \ , \ \ b=0 \Longrightarrow c \text{ is undefined,}\ \ \text{and }\ d=0 \Longrightarrow a=0 , \ b=0 , \text{ and c is undefined}$.

This argument does not work, however, with complex numbers, ie $x=1 , \ y=i \ \Longrightarrow a=1+i \ , \ \ b=i \ , \ \ c=1-i \ , \ \ d=0$.

I think, with use of some inequalities and case work, that at least [i]some[/i] sets of cards can be solved.

Heres one (for which i have the solution): $-2, \ 0, \ 0, \ 1$.

Remember, complex numbers are allowed -- read [b]lingomaniac88[/b]'s first post again.


Ring Theory
superior algebra
superior algebra solved

Let $k$ be a field.
Prove that $k[x,y]/(x^3-xy)$ and $k[x,y]/(x^2-xy^2)$ are not isomorphic as $k$-algebras.

Let $R=k[x,y]/(x^3-xy)$ and $S=k[x,y]/(x^2-xy^2)$.

Assume $f: R\to S$ is an isomorphism.
The minimal primes of $R$ are $xR$ and $(x^2-y)R$.
The minimal primes of $S$ are $xS$ and $(x-y^2)S$.
$f$ takes minimal primes to minimal primes, so $f(\{xR,(x^2-y)R\})=\{xS,(x-y^2)S\}$.
We have $xR+(x^2-y)R=(x,x^2-y)R=(x,y)R$ which is a maximal ideal of $R$.
On the other hand, $xS+(x-y^2)S=(x,x-y^2)S=(x,y^2)S$ is not a maximal ideal of $S$.
If $f$ was an isomorphism then $f$ would take maximal ideals to maximal ideals, but instead $f((x,y)R)=f(xR+(x^2-y)R)=f(xR)+f((x^2-y)R)=xS+(x-y^2)S=(x,y^2)S$. We present ourselves with a contradiction.


floor function
number theory unsolved
number theory

Compare:  $ 1$  and $ \{(3 \plus{} \sqrt {5})^n\} \plus{} \{(3 \minus{} \sqrt {5})^n\}$,where$ n\in N$ and $ \{x\} \minus{}$ is fractional part of $ x$.

Easy to check $ x_n\equal{}(3\minus{}\sqrt{5})^n\plus{}(3\plus{}\sqrt{5})^n$ then $ x_n\in N$
So $ \{(3\plus{}\sqrt{5})^n\}\equal{}\{\minus{}(3\minus{}\sqrt{5})^n$
$ \Longrightarrow\{(3\plus{}\sqrt{5})^n\}\plus{}\{(3\minus{}\sqrt5)^n\}\equal{}1\minus{}\{(3\minus{}\sqrt{5})^n\}\plus{}\{(3\minus{}\sqrt{5})^n\}\equal{}1$

[quote="Ibrogimov"]Compare:  $ 1$  and $ \{(3 + \sqrt {5}^n\)\} + \{(3 - \sqrt {5})^n\}$,where$ n\in N$ and $ \{x\} -$ is fractional part of $ x$.[/quote]

Let $ a=(3+\sqrt{5})^n$ and $ b=(3-\sqrt{5})^n$

We have $ a>0$, $ b>0$ and $ a+b=2\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2k}3^{n-2k}5^{k}\in \mathbb{N}$

Since $ a>0$, $ b>0$ and $ a+b\in\mathbb{N}$, then either $ a\in\mathbb{N}$ and $ b\in\mathbb{N}$ and $ \{a\}+\{b\}=0$ , either $ \{a\}+\{b\}=1$

The case $ a\in\mathbb{N}$ is easy to eliminate (since $ a=u+v\sqrt{5}$ with $ v\neq 0$) and so

$ \{(3 + \sqrt {5})^n\)\} + \{(3 - \sqrt {5})^n\}=1$


My middle name has 6 letters, two of which are vowels. What is the probability that if the first two letters of my middle name are JO, my middle name is Joseph?

[quote="mathnerd314"]Well, there are only so many possible middle names that start with JO. But if you allow stuff like Joqbki (sorry if that's actually anyone's name :P) you get:

[hide]We want to find the number of ways you can make the last four letters be three consonants and one vowel. This is $5 \cdot 21^{3}\cdot 4! = 1,111,320$.

The probability is $\frac{1}{1,111,320}$[/hide][/quote]
[hide] Isn't it just $4\cdot5\cdot21^{3}$, becuase there are 4 ways: CCCV, CCVC, CVCC, VCCC, each with $5\cdot21^{3}$ possibilities. [/hide]

lotrgreengrapes, the c's in CCCV can be arranged differently if they are different letters, so you can get ghja, gjha, hgja, hjga, jhga, and jgha.

When you do $21^{3}$, you already include those different combinations of CCC...


geometry
geometric transformation
dilation
percent
3D geometry

The problem is attached as an image for reference.

First, I have looked at the Peter Atkins chapter on calculating pH for very dilute acids, but still have some questions.

1. How do we figure out that acetic acid can be the acid in question, and how do we know that the solution must be very dilute?

2. What is $ c$ and where does the equation $ c \equal{} [HA] \plus{} [A^{ \minus{} }] \equal{} [H^ \plus{} ]$ come from?

3. What is the motivation for the longest line of equations in the middle by the (2pts) in the picture?e

4. How exactly does their solution account for the condition that "the pH changed one unit after a tenfold dilution?"

5. What kind of book or website would be good to read about these kind of acid/base questions that require a bit more thinking than say AP chem-level acid base questions?

[quote="aufha"]2. What is $ c$ and where does the equation $ c \equal{} [HA] \plus{} [A^{ \minus{} }] \equal{} [H^ \plus{} ]$ come from?[/quote]2.The molar (total) concentration of the acid (c) that you can read in the label is the sum of the concentrations of the two forms of the acid in solution: the ionized form ($ A^ \minus{}$) and the undissociated form (HA).
$ c \equal{} [HA] \plus{} [A^{ \minus{} }]$
Then, the problem says that $ c\equal{}[H^\plus{}]$ too.
That's were the equation comes from.
[quote="aufha"]1. How do we figure out that acetic acid can be the acid in question, and how do we know that the solution must be very dilute?[/quote]In a solution of acid in water some of the hydrogen ions come from the acid and some from dissociation of water molecules. If the acid is a monoprotic weak acid (like acetic acid) it will be only partly dissociated and will contribute less than its total concentration to the concentration of hydrogen ions. There is a shortfall equal to $ [HA]$, the concentration of the undissociated acid. The shortfall needs to be made up by the dissociation of water in this problem. Water cannot provide a large concentration of hydrogen ions, so you need the concentration of total acid (c) to be very small, making the undissociated percent, and $ [HA]$ small enough. [quote="aufha"]3. What is the motivation for the longest line of equations in the middle by the (2pts) in the picture?[/quote]The longest line of equations in the middle is just a bit of algebra to get to an expression with only the concentration of hydrogen ion (What you want to solve for) as a variable. The first part of the long line of equations (from the left end to just before the second equal sign) is the equation for the dissociation constant for a weak acid. Then variables (concentrations) are replaced with equivalent expressions to get rid of all of them except the concentration of hydrogen ion.

I forgot to link to the original source: http://www.icho.hu/Files/theory_icho40_final_empty_lic.doc

2. Why is it that $ HA]\plus{}[A^{\minus{}}] \equal{} [H^{\plus{}}]$?

1. I checked up "shortfall," which I have not seen before and found "deficiency" as a definition. Replacing, I get "There is a deficiency equal to [HA], the concentration of the undissociated acid." What is there a deficiency of?

[quote="aufha"]I forgot to link to the original source: http://www.icho.hu/Files/theory_icho40_final_empty_lic.doc

2. Why is it that $ [HA] \plus{} [A^{ \minus{} }] \equal{} [H^{ \plus{} }]$?[/quote]I'll check your source and see what else I can understand of the problem.
$ c \equal{} [HA] \plus{} [A^{ \minus{} }]$ because the molecules of the acid are in one form or the other.
The problem says that $ c \equal{} [H^{ \plus{} }]$ was found by measuring pH.
Putting both together you have
$ c \equal{} [HA] \plus{} [A^{ \minus{} }] \equal{} [H^{ \plus{} }]$

I believe the other equation comes from considering that all the $ H^ \plus{}$ comes from only two sources: $ HA$ and water. Any $ H^ \plus{}$ coming from water left a $ OH^ \minus{}$ behind and any $ H^ \plus{}$ coming from $ HA$ left a loose $ A^ \minus{}$ behind. So $ [H^{ \plus{} }] \equal{} [A^\minus{}] \plus{} [OH^\minus{}]$

[quote="aufha"]1. I checked up "shortfall," which I have not seen before and found "deficiency" as a definition. Replacing, I get "There is a deficiency equal to [HA], the concentration of the undissociated acid." What is there a deficiency of?[/quote]I am sorry. I have been living in the US for too long and was trying to talk like an American-born. What I meant was this:
The problem says that the total concentration of acid (c) written in the label equals $ [H^ \plus{} ]$
$ c \equal{} [H^ \plus{} ]$
Not every molecule of $ HA$ forms a $ H^ \plus{}$ ion. Some just stay as $ HA$.
The concentration of $ H^ \plus{}$ ion coming from $ HA$ must be less than c, and the difference (shortfall, deficiency, missing amount) must be $ H^ \plus{}$ ions coming from water molecules.

2. In the below passage from the question, does "concentration" refer to the quantity $ [HA] \plus{} [A^{ \minus{} }]$?
[quote]The label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its "concentration" was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label. [/quote]
The reason I'm asking is because I usually see the concentration of a solution of a substance, say substance A, referred to as just [A].

1. Just to make sure that I am following the logic correctly, could you check if I am interpreting your answer correctly?

Are we using the fact that $ [H^ \plus{} ] > [A^ \minus{} ]$ (from $ [H^\plus{}] \equal{} [HA] \plus{} [A^{ \minus{} }]$) to justify that a significant amount of  $ [H^ \plus{} ]$ comes from water?

After that, are we using the fact that a significant amount of  $ [H^ \plus{} ]$ comes from water to justify why the acetic acid has to be very dilute?

[quote="aufha"]2. In the below passage from the question, does "concentration" refer to the quantity $ [HA] \plus{} [A^{ \minus{} }]$? [/quote]Yes. That's what you would put on a label: the amount of HA that was put into each liter of water, regardless of the fact that it will dissociate. A chemist using that reagent may call that [HA] because for most practical uses, the distinction does not matter. For this problem the distinction is very important.

[quote="aufha"]1. Just to make sure that I am following the logic correctly, could you check if I am interpreting your answer correctly?

Are we using the fact that $ [H^ \plus{} ] > [A^ \minus{} ]$ (from $ [H^ \plus{} ] \equal{} [HA] \plus{} [A^{ \minus{} }]$) to justify that a significant amount of  $ [H^ \plus{} ]$ comes from water?

After that, are we using the fact that a significant amount of  $ [H^ \plus{} ]$ comes from water to justify why the acetic acid has to be very dilute?[/quote]That's the correct interpretation, but the "significant amount of  $ [H^ \plus{} ]$" may not be very large, even when considered in comparison to the concentration (c) on the label.
Of course, if the concentration on the label was not already very small, you would know that it could not be acetic acid.

One more thing:

How do we get the last equation mentioned in the picture where they say we can find [H+] through iteration? How do we use the last one to find H+ exactly? Do we plug in 10^-7 for the [H+] on the right initially?

See pic

That looks right.
You can just solve for [H+] and use that to find the pH.
[quote="The problem"]
Starting with a neutral solution...
[/quote]

It appears that you do start off with the RHS [H+] as 10^-7

When we have a usual acid, we usually use k=[H+]/([HA]-[H+])
we iterate it just to make sure we have 5%
however, you can just iterate it to calculate the real value

[quote="aufha"]One more thing:

How do we get the last equation mentioned in the picture where they say we can find [H+] through iteration? [/quote]To get the expression just next to "(2 pt)" you do just a little algebra with the expression right before the equal sign, substituting $ K_w$ for $ [H^\plus{}][OH^\minus{}]$ twice.
In the numerator:
$ [H^\plus{}]([H^\plus{}](\minus{}[OH^\minus{}])\equal{}[H^\plus{}]^2\minus{}[H^\plus{}][OH^\minus{}]\equal{}[H^\plus{}]^2\minus{}K_w$
Now if you multiply numerator and denominator times $ [H^\plus{}]$, you get the final expression with $ [H^\plus{}][OH^\minus{}]$ in the denominator, and $ [H^\plus{}][OH^\minus{}]\equal{}K_w$.

From there they just solved for the $ [H^\plus{}]$ that is cubed, and got the last "full" equation with the cube root. Because $ [H^\plus{}]<<K$, they get the other "good approximation" cube root equation.

[quote="aufha"]How do we use the last one to find H+ exactly? Do we plug in 10^-7 for the [H+] on the right initially? [/quote]As you said, you plug values for the $ [H^\plus{}]$ on the right and calculate the $ [H^\plus{}]$ on the left repeatedly, getting closer approximations, until you are close enough and end up rounding the results to the same value.
For $ [H^\plus{}]\equal{}10^{\minus{}7}$, you get $ [H^\plus{}]\equal{}5.59*10^{\minus{}7}$.
For $ [H^\plus{}]\equal{}5.59*10^{\minus{}7}$, you get $ [H^\plus{}]\equal{}5.64*10^{\minus{}7}$.
For $ [H^\plus{}]\equal{}5.64*10^{\minus{}7}$, you get $ [H^\plus{}]\equal{}5.64*10^{\minus{}7}$.

Sorry to revive a month-old topic, but just to say, the iteration is sorta analogous to the divide-and-average method of finding square roots.

You have to keep doing it until you actually find the answer (to the number of decimal digits you want of course), because it would be a complete pain to actually solve that equation/you would need a calculator with a solver.


real analysis
induction
vector
combinatorial geometry
Functional Analysis
real analysis unsolved

Could anyone please post a proof of the Krein-(Millman theorem: any compact convex is the convex hull on its extremal points?  Thanks.

I can post the solution in french (since I know that you are living in Paris nowadays) but it is long to write down . If you prefer, you can find it in several books such as: the book on functional analysis by W. Rudin, the books by Bourbaki, the book "topologie" by G. Choquet, and probably the best reference is:
R. R. Phelps: lectures on choquet's theorem.

What do you prefer?

Michael

I can post the proof in Russian, since I am in Saint-Petersburg. I mean, this theorem and its proof is useful not only in Paris and not even only for French-speaking members.

Harazi, would you please specify: do you mean the finite dimensional, the Banach space, or the most general linear topological case?

I am sorry, Petrov,  I didn't want to give the impression that it is only for french members. I wanted to see a proof of a simple case of the theorem, since I am sure I cannot understand it. Thank you again.

The finite dimensional case should be a rather simple induction by dimension, I think :?.

[quote="Michael"]If you prefer, you can find it in several books such as: the book on functional analysis by W. Rudin, the books by Bourbaki, the book "topologie" by G. Choquet, and probably the best reference is:
R. R. Phelps: lectures on choquet's theorem.

What do you prefer?

Michael[/quote]

Thank you harazi for not responding :mad:  :mad:

I simply don't understand your attitude. Thank you for your indications, they were precious, but I didn't consider I should answer. What should I answer? They are all nice.  :? . Thanks again, if you wanted this.

I'll try to prove it for (real) normed vector spaces.

Suppose we know that $(A)$ every convex compact subset of a real normed vector space $\mathcal E$ has an extremal point (I'll try to prove this below). Let $\mathcal K$ be a convex compact in $\mathcal E$, and assume we can find a point $x$ of $\mathcal K$ outside the closed convex hull $\mathcal H$ of the extremal points of $\mathcal K$. Then by using [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=separating+hyperplane&t=69217]this problem[/url] (that's why I posted it: I didn't want to write the proof here, provided my proof is correct :)) for the particular case when the closed convex set is $\mathcal H$, and the compact convex set disjoint from it is $x$, we can find a linear functional $f$ such that $f(x)>f(y),\ \forall y\in\mathcal H\ (*)$.

Let $\mathcal T$ be the set of points of $\mathcal K$ where $f$ reaches its maximum $M$. This $\mathcal T$ is the intersection between the hyperplane $f^{-1}(M)$ and $\mathcal K$, so it can be regarded as a convex compact set in a real normed vector space $\mathcal F$ identified with $f^{-1}(M)$. Clearly, from $(*)$ we can see that $\mathcal T\cap\mathcal H=\emptyset$. By $(A),\ f^{-1}(M)\cap\mathcal K$ has an extremal point $x_0$. Now assume we have $x_0=tx_1+(1-t)x_2$ for some $t\in[0,1],\ x_1,x_2\in \mathcal K$. We have $M=f(x_0)=tf(x_1)+(1-t)f(x_2)$, which means that $f(x_1)=f(x_2)=M$, since $M=\max_{x\in\mathcal K}f(x)$. This, in turn, implies that $t=0$ or $1$, since $x_0$ is extremal in $f^{-1}(M)\cap\mathcal K$. We have thus found $x_0$ to be an extremal point of $\mathcal K$, outside $\mathcal H$. This is a contradiction.

All that's left now is to prove $(A)$ (this is actually the hard part of the proof anyway :)).

We construct a chain of compact subsets of $\mathcal K$ by transfinite induction, starting with $\mathcal K_0=\mathcal K$. Suppose $\alpha$ is an ordinal and we have constructed $\mathcal K_\beta,\ \beta<\alpha$. Notice that the family $(\mathcal K_\beta)_{\beta<\alpha}$ has non-empty intersection, since it’s a family of compact (and thus closed) sets in a compact topological space $\mathcal K$, every finite subfamily of which has a non-empty intersection. If $\bigcap_{\beta<\alpha}\mathcal K_\beta$ has only one point, we stop and declare $\alpha$ to be the smallest ordinal $\gamma$ for which $\mathcal K_\gamma$ is not defined. Otherwise, consider the set $\mathcal D_\alpha=\bigcap_{\beta<\alpha}\mathcal K_\beta$. It’s a compact convex subset of $\mathcal E$, so we can find a continuous linear functional $f_\alpha$ which is not constant on it, and this functional will reach its maximum $M_\alpha$ on $\mathcal D$. We take $\mathcal K_\alpha=f_\alpha^{-1}(M_\alpha)\cap\mathcal D_\alpha$.

Of course, all the ordinals we can reach in this manner have bounded cardinality, so there are ordinals we cannot reach. Let $\alpha$ be the first such ordinal. As we’ve seen above, we could define $\mathcal K_\alpha$ if the non-empty set $\bigcap_{\beta<\alpha}\mathcal K_\beta$ had more than one point, so this set must consist of one point $x_0\in\mathcal K$ only. We can prove now that $x_0$ is an extremal point of $\mathcal K$.

Suppose the contrary. We can then find points $x_1,x_2\in\mathcal K$ such that $x_0\in(x_1,x_2)$. Let $\gamma$ be the smallest ordinal which is the index of a $\mathcal K_\beta\ (\beta<\alpha)$ which contains at most one of $x_1$ and $x_2$. First we prove that $\mathcal K_\gamma$ actually contains none of $x_1,x_2$: if it contained $x_1$, say, then we would have $f_\gamma(x_1)=f_\gamma(x_0) =M_\gamma$, so this must be the value of $f_\gamma(x_2)$ as well. Since $x_2\in\bigcap_{\beta<\gamma}\mathcal K_\beta,\ \mathcal K_\gamma$ would contain $x_2$ as well. Now, this means that we have $f_\gamma(x_0)=M_\gamma>f_\gamma(x_1),f_\gamma(x_2)$, and this is impossible, since $x_0$ belongs to the open segment $(x_1,x_2)$.


I'm wondering... This years valedictorian at The Lincoln School was went to the IMO training camp, but her writing was really lacking.  (The salutatorian was the best English person).   I don't know, but my rationale is that if you're good at math, then you got all your sciences and math classes down with A's (Physics, Chemistry...)  However, with English. all you really have IS English.  Of course it does look good to have eloquent essays for history, science, and the like, but that only goes so far.   What do you guys think?

I think that you must be well rounded.  At least thats what everyone tells me.  If your good at only math then you have the science stuff kind of, but your science essays might be pretty bad.  I think you just have to work hard at everything.

I believe that your valedictorian was exceptional at mathematics, but that person must also be fairly decent at his/her other academic subjects. Valedictorian is based on grades/GPA, no? Such, as long as that person receives top marks in English, there's no need to be a very prolific writer.

What if the person in question has no desire to be a valedictorian?

There is no person in question.  Seriously, I was just wondering.

My school didn't [i]do[/i] a valedictorian. It would likely be controversial if they tried it, because there are kids ranging from special ed. classes, a career tech center, a strong arts and music department, and honors and AP classes, and nothing is weighted. So can you say to the kid who took all AP classes and had a 3.945 "you are not as intelligent as Susan, who took eight art, and four music credits and rounded out her schedule with the required minimum classes, but maintained a 4.0"? A small part of me says 'no'.

Danny, I am pretty sure you are wrong about this girl going to IMO training camp.  Seriously, a few years back this genius guy from my HS qualified for it, and then after that only 2 guys have qualified for it since then, and neither goes to Lincoln HS.

To answer your question - being "well rounded" is important in life.  You need to know how to write reports, communicate effectively, and read books or reports in order to survive in the professional/academic workplace.  However, you don't have to have a passion for both english and math related subjects.  Nobody says you have to go pick up Hamlet for fun reading on your vacation.  Also, about English only helping in English class - you have heard about the whole right brain, left brain theory?  So you are probably good at English/Languages/History/Art/Music or Math/Science/Technology/Computers (I think), so I think both math and english give skills to other subjects.

[quote="JS1527"]so I think both math and english give skills to other subjects.[/quote]
...and it's not exactly like your college math and science books will be written at a third grade reading level.

Julie - I don't know, but apparently she got a perfect score for some MIT scholarship search or something...I don't know, but there is a plaque there.  Hmm..my friend really has no idea what she's talking about.  :-P

English is important, especially for places like MIT because pretty much everyone who applies is good at math and science so they have to look at things like writing (at least, that's what they said when I visited).  Of the top people in my class, all are pretty good at both math and english-we get extra GPA points for AP classes and so anyone who wants to try for valedictorian has to be good enough to get into AP English, although some of my classmates in english were a bit...lacking, shall we say, in writing.  But the top kids are good essay-writers.

I find advanced math books a fine source of great literature.


The usamo teacher's manual says:
"Please leave a 1'' margin on all sides of the Answer Sheet and on any additional pages and write DARKLY and LEGIBLY, so your faxed copy can be easily read."
Does this mean student has to leave 1'' margin inside the box of the answer sheet as well? (Or is the margin already left by the box?) Also does this mean student can write on sheets other than the official answer sheet?

Thanks.

You can use other sheets of paper, and the given box on the official one already includes the 1-inch margin.

Generally, the proctor will print out several copies of the "official answer sheet" (5-10 usually suffice), so using something other than that is not necessary.  You won't have to fit your solutions onto one page  :D .

Hmm, I heard some USAMO masters use 10+ sheets per problem.

No one I've talked to does; granted, that's a relatively small universe of people. You should use as few as the problem requires - you need to have a good feeling for how long proofs should be. Too long and you can easily make a hidden mistake; too short and you leave things out.

It shouldn't take more than 5-7 pages. If it does, you likely a) have a really long-winded solution and have a higher chance of annoying the graders and getting points docked (not necessarily, but they will be angry at minor errors because they have to read 105971209751092 pages!), or b) have restarted some incorrect solutions (like I did last year on #4... I think I had to get more answer sheets printed XD)

but if you really need more sheets, you can work on other problems while they're printing. If you're crunched for time and need more sheets... just use blank paper? but label those sheets meticulously.

In case if we ever need a ton of sheets, is it allowed for us to print some before the test and bring them in?

(It's sorta inconvenient to have them printed during the test and teachers tend to underestimate the number of sheets needed...)

I would assume it would be okay to print sheets as long as you let the proctor inspect them beforehand.  If you need extra sheets though, it is easy to just take a blank page and a ruler and draw straight lines for a margin.  It is difficult to work for 4.5 hours without a break, and drawing lines with a ruler is a good way to use that time if you don't have enough pages.

If I remember correctly, the record length for a proof with a score of 7 on USAMO is 14 pages.  The record length for a proof with a score of 0 is 18 pages.  Both are from number 4 last year.  In general, if a proof is that long, it is most likely being very "talky"  and "hand-wavy" instead of being rigorous.  A little bit of time spent planning and clarifying a solution in your own mind can improve rigor and reduce the amount of time spent writing up solutions significantly.

[quote="Infinite_Epsilon"]The record length for a proof with a score of 0 is 18 pages. [/quote]

that would suck. ><

Where can a student find these "blank pages" to print out prior to test time? Can someone please tell me?

You can find them at the end of the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-USAMO-2009.pdf]Teacher's Manual[/url].


Posted a zillion times before, it's IMO 1995 problem 2, day 1  :wink:


http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=292149

Discuss here!


If for the angles $ \alpha , \beta$ and $ X$ the equality holds:
$ \sin^2 \alpha \plus{} \sin^2 \beta \plus{} \sin^2 X \equal{} 2$.
Prove that triangle has a right-angle,(is right-angled).

[quote="Obel1x"][color=darkblue]Let $ ABC$ be a triangle so that $ \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 B \equal{} 2$  .  Prove that the triangle is [b]right-angled[/b][/color] .[/quote]

I'll use the well-known relations: $ \sin^2 x \equal{} \frac {1 \minus{} \cos 2x}{2},\ \forall\ x\in\mathbb{R}\ \ (*)$ and, in a triangle $ \sum \cos 2A \equal{} \minus{} 1 \minus{} 4\prod \cos A\ (**)$ .

Therefore, $ \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C\stackrel{(*)}{ \equal{} }\frac 32 \minus{} \frac 12(\cos 2A \plus{} \cos 2B \plus{} \cos 2C)\stackrel{(**)}{ \equal{} }$ $ \frac 32 \minus{} \frac 12\left( \minus{} 1 \minus{} 4\prod\cos A\right) \equal{} 2 \plus{} 2\prod\cos A$ .

$ \Longrightarrow 2 \equal{} 2 \plus{} 2\prod\cos A \Longleftrightarrow \cos A\cos B\cos C \equal{} 0 \Longleftrightarrow$ $ 0\in\{\cos A,\cos B,\cos C\} \Longleftrightarrow 90^{\circ}\in\{A,B,C\}$ .

===============================================================================================

Now we can conclude that:   $ \boxed{\boxed{\triangle ABC\ \mbox{\small{is}}\ \begin{array}{ccc} \nearrow \mbox{\small acute \minus{} angled} & \Longleftrightarrow & \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C > 2\ \Longleftrightarrow\ p > 2R \plus{} r \\
 \\
\rightarrow \mbox{\small right \minus{} angled} & \Longleftrightarrow & \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C \equal{} 2\ \Longleftrightarrow\ p \equal{} 2R \plus{} r \\
 \\
\searrow \mbox{\small obtuse \minus{} angled} & \Longleftrightarrow & \sin^2 A \plus{} \sin^2 B \plus{} \sin^2 C < 2\ \Longleftrightarrow\ p < 2R \plus{} r \end{array} }}$ .


it's easy to see by the cyclic nature of the system that we must have $ x\equal{}y\equal{}z$. then

\[ x^3\equal{}x^2\plus{}x\plus{}\frac{1}{3}\\3x^3\equal{}3x^2\plus{}3x\plus{}1\\4x^3\equal{}x^3\plus{}3x^2\plus{}3x\plus{}1\\(\sqrt[3]{4}x)^3\equal{}(x\plus{}1)^3\\\sqrt[3]{4}x\equal{}x\plus{}1\\x\equal{}\frac{1}{\sqrt[3]{4}\minus{}1}\]

so the unique solution is $ x\equal{}y\equal{}z\equal{}\frac{1}{\sqrt[3]{4}\minus{}1}$.

find all complex root of system
$ \{x^2\equal{}y\plus{}3\\y^2\equal{}z\plus{}3\\z\equal{}x\plus{}3$

[quote="Nam Luu"][quote="pluristiq"]it's easy to see by the cyclic nature of the system that we must have $ x \equal{} y \equal{} z$. then
[/quote]
Why $ x \equal{} y \equal{} z$?
If $ x > y$ then $ y^{2} \plus{} y > z^{2} \plus{} z$ . How do you know $ y > z$?[/quote]

Because : $x^3=y^2+y+\frac{1}{3}>0 \Rightarrow x>0 $
Similarity, we have $y,z>0$
$ y^{2} \plus{} y > z^{2} \plus{} z\Rightarrow (y-z)(z+y+1)>0\Rightarrow y>z$

[b]HINT:[/b]
$
\mbox{If }f(x)=\sqrt[3]{x^2+x+\frac{1}{3}}\mbox{, then:}\\
\{\begin{array}{c}
x^3=y^2+y+\frac{1}{3}\\
y^3=z^2+z+\frac{1}{3}\\
z^3=x^2+x+\frac{1}{3}
\end{array}
\Leftrightarrow
\{\begin{array}{c}
x=\sqrt[3]{y^2+y+\frac{1}{3}}=f(y)\\
y=\sqrt[3]{z^2+z+\frac{1}{3}}=f(z)\\
z=\sqrt[3]{x^2+x+\frac{1}{3}}=f(x)
\end{array}
\Rightarrow \\
\Rightarrow x=f(y)=f[f(z)]=f\{f[f(x)]\}=(f\circ f\circ f)(x)=f^{(3)}(x)\Rightarrow\\
\Rightarrow x=\mbox{fixe point of function } f^{(3)}.
$

i figured out a way to solve that using Cardano's method , but after verifying i realised that it was not correct :(