[size=150]First day [/size]
31.05.2005.
Time allowed: 4h 30min

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39569]Problem 1[/url]
Let $\cos (\alpha)=\frac 35$. Proove that $\alpha$ is not $r\cdot \pi$ with $r$ rational.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39573]Problem 2[/url]
Let $xOy$ be convex angle and $M$ point inside it. Proove that there exists unique point $P$ in the plane of the angle, such that, for any line passing through $M$ and intersecting lines $xO$ and $yO$ (including extensions of rays $xO$ and $yO$) at points $X$ and $Y$, it holds that $\angle XPY \leq \frac \pi 2$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39570]Problem 3[/url]
Find all polynomails $P(x) \in \mathbb{R}[x]$ such that $P(x^2+1)=P(x)^2+1$ holds for all $x \in \mathbb{R}$

[size=150]Second day [/size]
01.06.2005.
Time allowed: 4h 30min

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39572]Problem 4[/url]
Let $T$ be centroid of triangle $ABC$. Proove that \[\frac 1{\sin \angle TAC} + \frac 1{\sin \angle TBC} \geq 4\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39571]Problem 5[/url]
Let $a,b,c$ be positive reals and $abc=1$. Proove that \[\frac a{a^2+2}+\frac b{b^2+2}+\frac c{c^2+2} \leq 1\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=247152]Problem 6[/url]
We are given $n \times n$ table and positive integer in each of $n^2$ squares such that any two of them wich are in the same row or column are different. For $n$ numbers (or unit squares) we say that they are in [i]general position[/i] if any two of them are in different coloumn and different row. We are given that sum of any $n$ numbers in general positions is same.
Proove that if numbers, whose product is the smallest of all products of $n$ numbers in general position, are sorted (by their values) at one diagonal, then the numbers whose product is the greatest of all $n$ numbers in general position are on the other diagonal of the square.

I am asking the question for Valentin this time: Did you take part in it and if so what did you do?  :lol: But I am Orlando so I say: Did you take part in it and did you have fun?  ;)

No, I finished my IMO contest time year ago, I am student now  :)
But Problem 6 was made and suggested by me, so I took part in some sense...


Let $\alpha$, $\beta$, and $\delta$ denote the angles of elevation of a tower measured at distances of $100$, $200$, and $300$ meters from the foot of the tower, respectively. Find the height of the tower if $\alpha + \beta + \delta = 90^{\circ}$

Ok, bare with me because i haven't been able to get latex to work, and i apologize for that. I hope this doesn't pain your eyes too much. Anyways, here is my solution:

[hide] first of all, since latex isn't working, here are my variables. a = alpha, b= beta, and c= delta, just so you aren't confused.

if you draw the diagram of the situation and are familiar with right triangle trigonometry, you will see that cos a = (h/100), cos b = (h/200), and cos c = (h/300). If you solve each term for h, you will see that h = 100 cos a = 200 cos b = 300 cos c.

Now. i will factor out 100, so this yields:

(h/100) = cos a = 2 cos b = 3 cos c. Let's ignore the (h/100) term and just worry about the angles.

if you take the inverse cosine of each term, you find that a = 2b  = 3c. We know that a + b + c = 90 degrees, so with our new knowledge, we can write all of those terms in terms of one angle. I will write them all in terms of a because that is how i solved this problem, but you could write them all in terms of b or c if you prefer.

so a + (a/2) + (a/3) = 90

if you find the denominator and simplify, you get that (11a/6) = 90. so 11a = 540. Thus a = (540/11) degrees. I will leave this as it is because i don't like messy decimals. We don't need to find the other angles. but i will just to show you that they meet the criteria a + b + c = 90.

using the relationships a = 2b = 3c, we find that b = (270/11) degrees and c = (180/11) degrees.

so a + b + c = (540 + 270 + 180)/11 = (990/11) = 90 degrees.

we can use any angle to solve for h. I will use a, just because that is what we have been working with so far.

cos a = (h/100), so 100 cos a = h.

h = 100 cos (540/11) = approximately 65.5 meters

[/hide]

Hmm... I don't think your answer works because trig functions and their inverses aren't proportional to the angles.

$\frac{A}{B} \neq \frac{cos^{-1} A}{cos^{-1} B}$

[b][u]Lemma.[/u][/b] $\boxed {x+y+z=90^{\circ}\Longrightarrow \tan x\tan y+\tan y\tan z+\tan z\tan x=1}\ .$

[b]Proof.[/b] $x+y+z=90^{\circ}$ $\Longrightarrow$ $x+y=90^{\circ}-z$ $\Longrightarrow$ $\tan (x+y)=\tan \left(90^{\circ}-z\right)$ $\Longrightarrow$

$\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{1}{\tan z}$ $\Longrightarrow$ $\tan x\tan y+\tan y\tan z+\tan z\tan x=1\ .$

[b][u]Proof of the proposed problem.

[/u][/b] $h=100\cdot k$ $\Longrightarrow$ $k=\tan \alpha =2\tan \beta =3\tan \delta$. From the previous lemma results :

$k^2\left(1\cdot \frac 12+\frac 12\cdot \frac 13+\frac 13\cdot 1\right)=1$ $\Longrightarrow$ $k=1$ $\Longrightarrow$ $h=100\ .$

[b]Remark.[/b] If $x+y+z=180^{\circ}$ then $\tan x+\tan y+\tan z=\tan x\cdot\tan y\cdot\tan z$
and $\alpha +\beta +\delta =180^{\circ}\Longrightarrow h=100\sqrt{11}\ .$


2000 can be factored into 10^3 * 2 = 2^4 * 5^3.

By divisor formula (a^x *b^y*c^z = n, and a b and c are prime factors, the num of divisors = (a+1)(b+1)(c+1)...)
So 5*4 = 20.

Brady im sure you are asking this because of that question in the Troy math meet.


[quote="linzk3"]The roots of the equation x^2 +mx + k = 0 are a and b.  Find a^2 +b^2 in terms of k and m.[/quote]

[hide="Method"]Since $a$ and $b$ are the roots,

$x^{2}+mx+k = (x-a)(x-b)$

Expanding,

$x^{2}+mx+k = x^{2}-(a+b)x+ab$

Now, note that this implies:

$a+b =-m$

$ab = k$

Try to prove this identity if it is not immediately obvious: $a^{2}+b^{2}= (a+b)^{2}-2ab$.

Using the identity,

$a^{2}+b^{2}= (-m)^{2}-2k = m^{2}-2k$[/hide]


Pokaz ze dla kazdego $n \in \mathbb{N}$
\[ \varphi ( x ) = n!  \]
ma rozwiazanie w liczbach calkowitych.

$\varphi \equiv \mbox{Euler Totient function}$.


Hello,
    I think it would be nice i we could discuss the Problems of IMO 2006 here in this forum as well, not necessarily the solutions,as there is a separate place for that, but other aspects such as Indian Performance and the mathematical content of the problems.
  To start off, Problem 1 is, as Arne has claimed, easy.Our team members must have cracked it.Problem 2 ,(I am working on,now) is quite interesting.I have tried some bijections and progressed to a certain extent, but get stuck at some points. Problem 3, on the outset, seems a long chase(anyway, I have not put in time on this) of algebraic manipulations.
  Rushil, What is your take on these problems?
Sathej

Curious to know how the team did this year. Any updates?

dang's result was really a surprise? what happened to him?vihang u must be knowing it.

[quote="shadysaysurspammed"]none of the aopsers are imoers right?(indians).]

I have a strong feeling imo was too hard for the indians this year..not expecting any silver medals from them as well.[/quote]

...

What do you mean too hard for Indians???

Hello,
    Anyway,in hindsight,IMO 2006 has proved to be too difficult to Indian contestants('Indians' abroad,well,no remark).The reasons for India's poor performances in IMOs(refreshingly IPhO,IChO,IBO are better) are obvious and manifold.Some of them have been outlined briefly in this forum earlier and I hope to write more on it soon.
Sathej

Hi!!!!!!!!!!!!!!
seems like a nice idea to discuss about the IMOs here..............but i feel sorry i don hav so much ideas(i kno the elementary ones though) and views on IMOs.........sombody help me out please.....thankyou


abstract algebra
superior algebra
superior algebra unsolved

Assume $ M,M',N,N'$ are $ A$-modules. Show that there exist a unique $ A$-linear map
$ f: \operatorname{Hom}_A(M,M')\otimes_A \operatorname{Hom}_A(N,N') \to \operatorname{Hom}_A(M\otimes_A N,M'\otimes_A N')$
such that $ f(u \otimes v) \equal{} u \otimes v$.

Define the map $ F: Hom_A(M,M')\times Hom_A(N,N')\to Hom_A(M\otimes_A N, M'\otimes_A N')$, $ F(u,v)\equal{}u\otimes v$ (this is well defined).
Prove now that $ F$ is $ A$-bilinear. From the universal property of the tensor product it follows that there exists a unique map $ f: Hom_A(M,M')\otimes_A Hom_A(N,N')\to Hom_A(M\otimes_A N, M'\otimes_A N')$ so that $ f(u\otimes v)\equal{}F(u,v)\equal{}u\otimes v$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A, B\in M_{n}(\mathbb{R})$ be diagonalisable such that $e^{A}=e^{B}$. Show that $A=B$.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=diagonalizable&t=19814]this topic[/url]. Scroll down to see grobber's solution.

a related usefull result, surely proved on the forum : $M \in M_{n}(\mathbb{C})$ is diagonlizable iff $e^{M}$ is diagonalizable.

Here's a proof of this latter problem:

The non-trivial part is proving that if $e^{M}$ is diagonalizable, so is $M$. Write $M$ as a sum of two commuting matrices $D,N$, with $D$ diagonalizable and $N$ nilpotent. We then have $e^{M}=e^{D}e^{N}$. $e^{D}$ is diagonalizable, $e^{N}$ is unipotent (i.e. $1+$a nilpotent matrix), and the two commute. This decomposition is unique (as was $M=N+D$), so, since $e^{M}$ is diagonalizable, we must have $e^{N}=1$. What we must prove has now been reduced to:

If $N$ is a nilpotent complex matrix such that $e^{N}=1$, then $N=0$.

Suppose $N\ne 0$. There is then a minimal $k\ge 2$ for which $N^{k}=0$. This means that $1=e^{N}=1+p(N)$, where $p$ is a polynomial of degree $k-1$. $p(N)=0$ contradicts the fact that the minimal polynomial of $N$ is $x^{k}$, and this finishes the proof.


algebra
polynomial
inequalities
inequalities unsolved

Actually wt is the difficulty of this mo?
it seems that it is easier than cmo... (china one of course)

Of course, it is only a regional MO, and taken part in by students of class 10.

Oh, I made a mistake, it should be class 10(高一). I'll correct it.
In russia, class 9 is the same as (高一), but in China, it is not.


inequalities
linear algebra
matrix
linear algebra unsolved

Can someone give a complete proof for the Sylvester inequality for matrices?

By the way here's an interresting application:

 If $A$ is a matrix of order $3$ such that $A^2=I_3$ then prove:

         $rank(I_3-A)+rank(I_3+A)=3$ ;)

I suppose you're referring to the inequality $r(AB)+n\ge r(A)+r(B)$ for $n\times n$ matrices, right? ($r(\cdot)$ is the rank)

We denote the linear transformations which have the matrices $A,B$ in the standard basis also by $A$ and $B$ (respectively). The inequality we want to prove translates to $\dim(\mbox{Im}\  B)-\dim(\mbox{Im}\  AB)\le n-\dim(\mbox{Im}\  A)=\dim(\ker\ A)$. Let $\tilde A$ be the linear transformation from $\mbox{Im}\ B$ to $\mbox{Im}\ AB$ induced by $A$. We have $\dim(\mbox{Im}\  B)-\dim(\mbox{Im}\ AB)=\dim(\ker\ \tilde A)$, which is $\le\dim(\ker\ A)$, since $\ker\ \tilde A$ is a subspace of $\ker A$.

Yes,grobber you're right. I haven't read all your post but on first look it is ok.It remains for me to read it carefully
and to check all details. thanks,anyway. ;)

maybe this is more elementary http://www.mathlinks.ro/Forum/viewtopic.php?p=461393#461393

[quote="spix"]This is also a nice way to prove Sylvester`s inequality:$rank(AB)\leq \min(rank(A),rank(B))$[/quote]
That's not Sylvester's inequality

Oups I confused them... I always take it as a double inequality $rank(A)+rank(B)-n\leq rank(AB)\leq \min(rank(A),rank(B))$


Is there a way to find a sum of the factors of a number?
If so, how?

Yes.

If a number $ n \equal{} p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{r}^{k_{r}}$,

then the number of factors of $ n$ is

$ \tau(n) \equal{} (k_{1} \plus{} 1)(k_{2} \plus{} 1)...(k_{r} \plus{} 1)$

and the sum of its factors is

$ \frac {p_{1}^{k_{1} \plus{} 1} \minus{} 1}{p_{1} \minus{} 1}\frac {p_{2}^{k_{2} \plus{} 1} \minus{} 1}{p_{2} \minus{} 1}...\frac {p_{r}^{k_{r} \plus{} 1} \minus{} 1}{p_{r} \minus{} 1}.$

This might be useful: http://sdp.mathforum.org/library/drmath/view/71550.html

what andrewtom said looked a bit confusing :P . Heres what he was basically saying:
Consider the number $ 600$. To find the number of factors of 600, prime factorize it ($ 2^35^23$) and then do the following:

Take the number that is the power ($ 2^{\boxed3}5^{\boxed2}3^{\boxed1}$), add one to each of the boxed numbers, and multiply them. In this case, the number of factors of $ 600$ is $ (3 \plus{} 1)(2 \plus{} 1)(1 \plus{} 1) \equal{} 24$

To find the sum of the factors, prime factor it. Note that each of the factors of 600 are the combinations of products of the various possible powers of it's prime factors:
$ 2^0$, $ 2^1$, $ 2^2$, $ 2^3$, $ 3^0$, $ 3^1$, $ 5^0$, $ 5^1$, $ 5^2$
Sum of all the factors (i.e. above combinations) is:
$ (1 \plus{} 2 \plus{} 4 \plus{} 8)(1 \plus{} 3)(1 \plus{} 5 \plus{} 25)$


When generating a PDF with TeXnicCenter it automatically puts a page number at the bottom of the page. Is it possible to get rid of the page number or instead of just having it say "1" is it possible to change it to "Page 1 of 7"

\pagestyle{empty} will remove page numbers. But to customise the footer properly you should use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancyhdr]fancyhdr package[/url]. Together with the lastpage package you can get Page 1 of 7 with [code]\cfoot{\thepage\ of \pageref{LastPage}}[/code]See the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]manual[/url] section 16.

[i][b]Please remove or resize the ridiculously large image in your signature.[/b][/i]


Let $f(x) = 4x - x^2$. Consider the sequence $x_1, x_2, x_3, ... $ where $x_i = f(x_{i-1})$ for $i>1$. Computer the number of values of $x_1$ such that $x_1, x_2, x_3$ are all distinct, but $x_i = x_3$ for all $i>3$.

Yea can anyone help?
I honestly don't even know what this problem means.

The problem is saying that ${x_n}$ is a recursive sequence defined by $x_n = f(x_{n-1})$.

[hide]
If we want $x_i = x_3$ for $i>3$, then we solve for $f(x)=x$ so we get the same value when plugging it in.

So $4x-x^2 = x \Rightarrow 3x-x^2 = x(3-x) = 0 \Rightarrow x = 0, 3$.

So we have $x_3 = 0, 3$ to satisfy that condition.

Then we construct the sequence backwards, using each value of $x_3$.

CASE 1: $x_3 = 0$

So $4x_2-x_2^2 = x_3 = 0 \Rightarrow x_2 = 0, 4$
Reconstruct $x_1$ from each of those...

Sub-Case 1: $x_2 = 0$
$4x_1-x_1^2 = x_2 = 0 \Rightarrow x_1 = 0, 4$

Sub-Case 2: $x_2 = 4$
$4x_1-x_1^2 = x_2 = 4 \Rightarrow x_1^2-4x_1+4 = (x_1-2)^2 = 0 \Rightarrow x_1 = 2$

CASE 2: $x_3 = 3$

$4x_2^2-x_2^2 = x_3 = 3 \Rightarrow x_2 = 1,3$

Sub-Case 1: $x_2 = 1$
$4x_1-x_1^2 = x_2 = 1 \Rightarrow x_1 = 2 \pm \sqrt{3}$

Sub-Case 2: $x_2 = 3$
$4x_1-x_1^2 = x_2 = 3 \Rightarrow x_1 = 1, 3$

Whew. There you go. So there are 7 distinct values of $x_1$.

[/hide]

[quote="gotztahbeazn"]hmm i get everything except for the very beginning when you say that you can set  $f(x) = x$
how do you know that??[/quote]

We want $x_3 = x_4 = x_5 = ...$ by the last condition, so $x_3$ must be some value such that $x_4 = f(x_3) = x_3$. So we solve the equation $f(x) = x$ to determine all such values of $x_3$.


If you roll a six sided die two times, what is the probablility the sum of the numbers rolled will be divisble by 2 or 3?

How do you solve this without writing out all the sums?

thanks

[hide]Let's make this a little easier and just find the dice that sum up to 5, 7, or 11
So we have (1, 4)(4, 1)(2, 3)(3, 2)(3, 4)(4, 3)(2, 5)(5, 2)(1, 6)(6, 1)(6, 5)(5, 6)

So it would be $ \frac{36 \minus{} 12}{36}$ or $ \frac{2}{3}$
[/hide]


The results of Part 2 of this years' MMPC have been released, for you mi people. As usual, the top 200 scores arent released (they will be rechecked, afterwards the scores of the bottom 100 will be realeased. the top 100 will be announced at awards banquet.)

the link is: [url]http://www.delta.edu/math/mmpc/[/url]


algebra
polynomial
function
calculus
derivative
limit
real analysis

In an older journal(Romany) between 2004-2006 I have found the next problem. I can not find the source , but in my notes I have it :

 Find all non constant ponymomial  finctions $ f$ , such that $ f(f'(x)) \equal{} f'(f(x))$ for all $ x$ .

 [i]I try again to solve the problem , but I fail. Probably I have forgotten a condition(integer coffiecents works for example?). Please , can someone help to put the least conditions ?As a solution I want th have $ f(x) \equal{} x$[/i] Regards - Babis

Iasu babis .Pos ise? :lol:
I think you mean polynomila instead  [quote="stergiu"].. ponymomial .. [/quote]. OK
answer is : it's true for polynomial $ f(x) = a_{0} + a_{1}x$ where $ a_{0} + a_{1}^{2} = a_{1}$
suppose $ n\geq 2$ and consider
$ f(x) = \sum_{0}^{n}a_{k}x^{k}$
and suppose $ a_{n}\neq 0$
and consider $ f(f'(x)) = f'(f(x))$ just take derivative and we have :
$ f'(f'(x))f''(x) = f''(f(x))f'(x)$
let's substitute instead $ f(x)$ this sum $ \sum_{0}^{n}a_{k}x^{k}$ and consider after calculation coeficient of maximal power $ x$ left and right side equation thus we have
$ .... + na_{n}(na_{n}x^{n - 1})^{n}n(n - 1)a_{n}x^{n - 2} = ... + n(n - 1)a_{n}(a_{n}x^{n})^{n - 2}na_{n}x^{n - 1}$(if $ n = 2$ then take $ x \to \infty$ and hence $ a_{n} = 0$)
if $ n > 2$ then
as we see power of $ x$ in left side  is equal to $ n^{2} - 2$  and in right side is equal to  $ n^{2} - n - 1$ also let's $ x \to \infty$ then coeficient of $ x^{n^{2} - 2}$ must be equal to zero , hence $ a_{n} = 0$, thus contradiction whith $ a_{n} \neq 0$ hence $ n < 2$ then take $ f(x) = a_{0} + a_{1}x$ and direct calculus show that
it's enough to say $ a_{0} + a_{1}^{2} = a_{1}$

[color=green] Unnecessary quote removed by moderator[/color]
Hallo!

 First thanks for the solution.It is nice and I will keep it in my archiv.Second sorry for ...killing the '' polyonymic function'' with my typos.The bell had ..dringngngngngng(ed) and I had to go in the classroom. I did not read the text again!
  Babis

Everything is fine except $ f(x)\equal{}\frac 12 x^2$ also solves the equation ;).

[quote="fedja"]Everything is fine except $ f(x) \equal{} \frac 12 x^2$ also solves the equation ;).[/quote]
 :lol:  yes. I have some simple mistake
I mean here
[quote="Extremal"]....$ .... \plus{} na_{n}(na_{n}x^{n \minus{} 1})^{n}n(n \minus{} 1)a_{n}x^{n \minus{} 2} \equal{} ... \plus{} n(n \minus{} 1)a_{n}(a_{n}x^{n})^{n \minus{} 2}na_{n}x^{n \minus{} 1}$([b]if $ n \equal{} 2$ then take $ x \to \infty$ and hence $ a_{n} \equal{} 0$[\b])
...[/quote] : in my proof case when $ n \equal{} 2$ let's consider again :
when $ n \equal{} 2$ idea about $ x\to \infty$ don't works because power of $ x$ in left and right side equation is same and equal to $ 1$ (for $ n > 2$ my proof works  :) ) hence we need to consider
$ f(x) \equal{} a_{0} \plus{} a_{1}x \plus{} a_{2}x^{2}$
after sabstitution and calculation we see that $ f(x) \equal{} a_{0} \plus{} \frac {1}{2}x^{2}$ works . Thanks fedja
or answer is:
$ f(x) \equal{} a_{0} \plus{} \frac {1}{2}x^{2}$ for all $ a_{o}\in R$
and
$ f(x) \equal{} b_{0} \plus{} b_{1}x$ with condition
$ b_{0} \plus{} b_{1}^{2} \equal{} b_{1}$

[quote="Extremal"]for $ n > 2$ my proof works[/quote]
Really?:roll: How about $ f(x)\equal{}\frac 19 x^3$ then? :P (actually, there are solutions in every degree...)

[quote="fedja"][quote="Extremal"]for $ n > 2$ my proof works[/quote]
Really?:roll: How about $ f(x) \equal{} \frac 19 x^3$ then? :P (actually, there are solutions in every degree...)[/quote]
really  :(  I have mistake in calculation instead
[quote="Extremal"]$ .... \plus{} na_{n}(na_{n}x^{n \minus{} 1})^{n}n(n \minus{} 1)a_{n}x^{n \minus{} 2} \equal{} ... \plus{} n(n \minus{} 1)a_{n}(a_{n}x^{n})^{n \minus{} 2}na_{n}x^{n \minus{} 1}$[/quote]
there must be :
$ .... \plus{} na_{n}(na_{n}x^{n \minus{} 1})^{n\minus{}1}n(n \minus{} 1)a_{n}x^{n \minus{} 2} \equal{} ... \plus{} n(n \minus{} 1)a_{n}(a_{n}x^{n})^{n \minus{} 2}na_{n}x^{n \minus{} 1}$hence
maximal powers is same and equal to $ n^{2}\minus{}n\minus{}1$ Hence I agree with fedja there is infinity many solution  :maybe: ..

Yes.By explict arguments we can prove the general form is $ n^{1\minus{}n}x^n$,when $ n>1$.

1) if we add condition that all coeficient is integer then my proof works. :blush:
really we get that :$ a_{n} \equal{} n^{1 \minus{} n}$ as it note  ffrog.
we have nice problem :
consider all polynomial $ f(x)$ satisfies condition $ f(f'(x)) \equal{} f'(f(x))$
if we take from each polynom of degree $ n$ coeficient of maximal power $ x$ say $ a_{n}$ only one times then
$ \lim _{n \to \infty}\sum_{j \equal{} 1}^{n}a_{j}$ exist and is irrational  :lol: .
for example :
suppose $ f(x) \equal{} \sum_{i \equal{} 1}^{n}b_{i}x^{i}$ satisfies equation :$ f(f'(x)) \equal{} f'(f(x))$ and  such as $ b_{n}\neq 0$ or polynom is degree $ n$ then we take $ b_{n}$
after we consider polynom of degree $ n \plus{} 1$ and choose $ b'_{n \plus{} 1}$
I note that for degree $ n$ we may have many polynomial satisfies condition,  from them we take any $ b_{n}\neq 0$
and e.t.c finally we consider sum of this coeficients and take $ n\to \infty$.
not a bad problem  :lol:


Represent the hypotenuse of a right triangle by $ c$ and the area by $ A$.  The atltidue on the hypotenuse is:

$ \textbf{(A)}\ \frac{A}{c} \qquad
\textbf{(B)}\ \frac{2A}{c} \qquad
\textbf{(C)}\ \frac{A}{2c} \qquad
\textbf{(D)}\ \frac{A^2}{c} \qquad
\textbf{(E)}\ \frac{A}{c^2}$

[quote="AIME15"]Represent the hypotenuse of a right triangle by $ c$ and the area by $ A$.  The atltidue on the hypotenuse is:

$ \textbf{(A)}\ \frac {A}{c} \qquad \textbf{(B)}\ \frac {2A}{c} \qquad \textbf{(C)}\ \frac {A}{2c} \qquad \textbf{(D)}\ \frac {A^2}{c} \qquad \textbf{(E)}\ \frac {A}{c^2}$[/quote]
[hide]Area of a triangle=base*height*1/2
$ A\equal{}\frac{ch}{2}$
$ 2A\equal{}ch$
$ \frac{2A}{c}\equal{}h$ so B.[/hide]

[quote=Archeon]Sorry for the bump, but there is a typo in the problem statement. It should say "The altitude on the hypotenuse is:"

(This correction was made for the sake of the contest collection)[/quote]

What is the source of the problem?


Euler
trigonometry
Olimpiada de matematicas

A pedido del publico,m otro problema de geo :D

Dado $\triangle ABC$ y un triangulo $DEF$ equilatero inscrito en el, demostrar que $DE \geq \frac{2\sqrt[]{2}[ABC]}{\sqrt[]{a^{2}+b^{2}+c^{2}+4\sqrt[]{3}[ABC]}}$.

$\bigstar$ Capitán Mandarina

[quote="conejita"]Creo que ya me estoy acercando a la solucion, te explico:

En la IMO  de 1961(creo) se demostro que a^2+b^2+c^2 >= 4(3^1/3)(ABC)
Entonces de ahi lo unico lo que tengo k demostrar se reduce a:
DE^2 >= (ABC)/(3^1/3)

Voy acercandome a la solucion o ya me desvie??[/quote]

Creo que deberias hacer algo escencial, cuando un triangulo equilatero inscrito en uno dado tienen perimetro minimo??

tarea jeje

Pues yo se que en un triangulo cualquiera, el triangulo inscrito de menor perimetro es el ortico (o sea, el formado por los pies de las alturas). A eso te refieres???

:blush: conejita lo que tu afirmas es correcto porque eso acabo de estudiarlo en el libro: MIRAR Y VER de Miguel Guzman, esta en el setimo ensayo titulado " El arte de mirar correctamente "  pag 49 jajajaja inclusive esta su demostración

Si gustas agregame conejita soy de Lima Perú

             edson_laura_galvez@hotmail.com       :)

Por tu propia seguridad, poné que tu dirección es abc, donde a=edson_laura_galvez, b=@ y c=hotmail.com (Hay muchos bots que lo único que buscan en Internet son direcciones de mail)

Ya que nadie lo resolvio lo are yo mismo

Empezaremos con algunos lemas y teoremas

Las coordenadas tripolares de un punto $P$ con respecto a $\triangle ABC$ es la terna $(x: y: z)$ tal que
$\frac{x}{AP}=\frac{y}{BP}=\frac{z}{CP}$.

[b]Lema 1. [/b](Euler) Dado $\triangle ABC$ y $P=(x: y: z)$ en coordenadas tripolares reales (las distancias reales), entonces:

$(i)$ $\sum(y^{2}+z^{2}-a^{2})^{2}x^{2}=4x^{2}y^{2}z^{2}+\prod(y^{2}+z^{2}-a^{2})$.
$(ii)$ $\sum(-a^{2}+b^{2}+c^{2})(y^{2}z^{2}+a^{2}x^{2})=a^{2}b^{2}c^{2}+(a^{2}x^{4}+b^{2}y^{4}+c^{2}z^{4})$

[b]Demostración. [/b]Demostraremos $(i)$ dejando $(ii)$ para uds. Por el teorema
de cosenos tenemos que $\cos\gamma=\frac{x^{2}+z^{2}-b^{2}}{2xz}$,
$\cos\alpha=\frac{y^{2}+z^{2}-a^{2}}{2yz}$ y
$\cos\beta=\frac{x^{2}+y^{2}-c^{2}}{2xy}$ pero de () tenemos que
$\cos^{2}\alpha+\cos^{2}\beta+cos^{2}\gamma=1+\cos\alpha \cos\beta \cos\gamma.$ Reemplazando se sigue que
$\sum(y^{2}+z^{2}-a^{2})^{2}x^{2}=4x^{2}y^{2}z^{2}+\prod(y^{2}+z^{2}-a^{2})$. $\blacksquare$

[b]Definición. [/b]Dado $\triangle ABC$, sean $V_{a}, V_{a}'$ los puntos de
interseccion de la $A-$ bisectriz interior y exterior
respectivamente con $\overline{BC}$. La circunferencia de diametro
$\overline{V_{a}V_{a}'}$ es la $A-$[i]circunferencia de Apolonio[/i] de
$\triangle ABC$. Analogamente para $B$ y $C$.

[b]Teorema 2. [/b]Las tres circunferencias de Apolonio tienen dos puntos en
comun.

[b]Demostración. [/b]Dado $\triangle ABC$, sean $J$ y $J'$ los puntos de
intersección de la $B-$circunferencia de Apolonio y la
$C-$circunferencia de Apolonio. De la definicion tenemos que
$\frac{CJ}{JA}=\frac{a}{c}$ y $\frac{AJ}{JB}=\frac{b}{a}$ y
multiplicando ambas expresiones tenemos que
$\frac{CJ}{JA}.\frac{AJ}{JB}=\frac{a}{c}.\frac{b}{a}$ con lo que
$\frac{CJ}{JB}=\frac{b}{c}$ y esto es, $J$ pertenece a la
$A-$circunferencia de Apolonio de $\triangle ABC$ y an\'alogamente
para $J'$. $\blacksquare$

Sean $J$ y $J'$ los puntos isodinamicos de $\triangle ABC$; esto es,
el punto de interseccion de las circunferencias de Apolonio de
$\triangle ABC$.

[b]Teorema 3. [/b]Los tri\'angulos pedal de $J$ y $J'$ son equil\'ateros.

Para demostrarlo usaremos el siguiente

[b]Lema 4.[/b] Las coordenadas tripolares de $J$ y $J'$ son iguales e
iguales a $(bc: ca: ab)$.

[b]Demostración. [/b]De la definición tenemos que
$\frac{BJ}{JC}=\frac{c}{b}$, sea $\lambda$ tal que
$BJ=\frac{ac}{\lambda}$ de ahi que $JC=\frac{ab}{\lambda}$. Ademas
tenemos que $\frac{AJ}{JC}=\frac{c}{a}$ con lo que
$JA=\frac{bc}{\lambda}$. Luego
$J=(\frac{bc}{\lambda}: \frac{ca}{\lambda}: \frac{ab}{\lambda})=(bc: ca: ab)$.
$\blacksquare$

Ahora pasaremos con la solucion del problema!
[b]
Solución. [/b]Dado $\triangle ABC$ sea $J_{1}J_{2}J_{3}$ el triangulo
pedal de $J$ son respecto a $\triangle ABC$. Tenemos que
$JJ_{2}=AJ.sen \angle JAJ_{2}$ y ademas $\frac{J_{2}J_{3}}{JJ_{2}}=\frac{sen \angle JJ_{3}J_{2}}{sen \angle J_{2}JJ_{3}}$. Como $\square JJ_{2}AJ_{3}$ es
c\'iclico tenemos que $m\angle JJ_{3}J_{2}=m\angle JAJ_{2}$ y $m\angle J_{2}JJ_{3}+m\angle J_{2}AJ_{3}=180º$ con lo que
$\frac{J_{2}J_{3}}{JJ_{2}}=\frac{sen \angle J_{2}AJ_{3}}{sen \angle JAJ_{2}}$,
es decir $J_{2}J_{3}=JJ_{2}.\frac{sen \angle J_{2}AJ_{3}}{sen \angle JAJ_{2}}$.
Por el lema 4 tenemos que $J=(bc: ca: ab)$ en coordenadas tripolares,
de ahi que si $AJ=\frac{bc}{\lambda}$ entonces $J_{2}J_{3}=AJ.sen A=\frac{bc}{\lambda}.sen A=\frac{4R^{2}}{\lambda}.sen A.sen B.sen C$.
Analogamente tenemos que $J_{3}J_{1}=\frac{4R^{2}}{\lambda}.sen A.sen B.sen C$ y $J_{1}J_{2}=\frac{4R^{2}}{\lambda}.sen A.sen B.sen C$ con lo
que $\triangle J_{1}J_{2}J_{3}$ es equil\'atero. $\blacksquare$

Ahora pasaremos a la solucion del problema:

Por el lema 1 tenemos: \[\sum (-a^{2}+b^{2}+c^{2})(\frac{c^{2}a^{2}}{\lambda^{2}}\frac{a^{2}b^{2}}{\lambda^{2}}+a^{2}\frac{b^{2}c^{2}}{\lambda^{2}})=a^{2}b^{2}c^{2}+(a^{2}\frac{b^{4}c^{4}}{\lambda^{4}}+b^{2}\frac{c^{4}a^{4}}{\lambda^{4}}+c^{2}\frac{a^{4}b^{4}}{\lambda^{4}}) .\] Separando t\'erminos y dividiendo por $a^{2}b^{2}c^{2}$ tenemos que \[\frac{1}{\lambda^{4}}\sum(-a^{2}+b^{2}+c^{2})a^{2}+\frac{1}{\lambda^{2}}(a^{2}+b^{2}+c^{2})=1+\frac{1}{\lambda^{4}}(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}).\] Esto es \[(2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4})+(a^{2}+b^{2}+c^{2})\lambda^{2}=\lambda^{4}+(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2})\] y reagrupando \[\lambda^{4}-(a^{2}+b^{2}+c^{2})\lambda^{2}+(a^{4}+b^{4}+c^{4}-b^{2}c^{2}-c^{2}a^{2}-a^{2}b^{2})=0\] con lo que \[2\lambda^{2}=(a^{2}+b^{2}+c^{2})+\sqrt[]{(a^{2}+b^{2}+c^{2})^{2}-4(a^{4}+b^{4}+c{}^{4}{}^{-}b^{2}c^{2}-c^{2}a^{2}-a^{2}b^{2})}\] y por el teorema de Heron
$2\lambda^{2}=(a^{2}+b^{2}+c^{2})+4\sqrt[]{3}[ABC]$ y de ahi que
$\sqrt[]{2}\lambda=\sqrt[]{(a^{2}+b^{2}+c^{2})+4\sqrt[]{3}[ABC]}$. Como
$J_{1}J_{2}=\frac{bc.sen A}{\lambda}=\frac{2[ABC]}{\lambda}$ se concluye
que
$\frac{2\sqrt[]{2}[ABC]}{\sqrt[]{a^{2}+b^{2}+c^{2}+4\sqrt[]{3}[ABC]}}$.
Sean $P_{1}\in \overline{BC}$, $P_{2}\in \overline{CA}$, $P_{3}\in \overline{AB}$ tal que $\triangle P_{1}P_{2}P_{3}$ es equil\'atero. La
semejanza espiral $\varphi$ tal que $\varphi(J_{i})=P_{i}$ para
$i=1,2,3$ con lo que se sigue que $P_{i}P_{i+1}\geq J_{i}J_{i+1}$. Por
lo tanto
$P_{i}P_{i+1}\geq\frac{2\sqrt[]{2}[ABC]}{\sqrt[]{a^{2}+b^{2}+c^{2}+4\sqrt[]{3}[ABC]}}$;
esto es el lado de todo triangulo equilatero inscrito en $\triangle ABC$ es mayor o igual que
$\frac{2\sqrt[]{2}[ABC]}{\sqrt[]{a^{2}+b^{2}+c^{2}+4\sqrt[]{3}[ABC]}}$.
$\blacksquare$

$\bigstar$ Capitán Mandarina!

[quote="conejita"]Ola hucht, sinceramente kreo k tu solucion no es nada bonita, muy tecnica y poco ingeniosa. Le geometria siempre es un poco de las 2, yo kreo k debe de aber alguna solucion mediante simples cosas euclidianas. Si de casualidad sabes alguna asi, agradeceria k la pongas, xq sinceramente no entendi naa de tu solucion, y el problema en si se me ace interesante.
Grax  :roll:[/quote]

Sorry pero la geometria euclideana esta bien como reseña istorica pero como herramienta en resolucion de problemas no sirve de mucho. Quiza tu puedas buscar una solucion con geometria euclideana yo no tengo mucho tiempo para perder con exigencias inutiles exquisiteses, resuelvo los problemas como puedo, quiza no sean "beliisimas" mis soluciones pero tambien puedo encontrar de esas que les gustan a ustedes con "congruencia" y teorema de tales y pitagoras, cosas poco utiles [url]http://www.personal.us.es/rbarroso/trianguloscabri/sol/sol19chav.pdf[url] - es mi solucion de un problema con geometria sintetica

Con todo, el mejor geometra que conozco (Darij Ginberg) tambien resuelve problemas asi, puedes verlo en donde demuestra dos teoremas de mi autoria usando miles de cosas [url]http://de.geocities.com/darij_grinberg/isogonal.pdf[/url]

Creo que tenés razón, ya con la desigualdad a demostrar se ve que la solución no va a ser muy hermosa porque los números y letras son "feos".
Además, mientras que resuelvas el problema está bien, las soluciones elegantes son más para problemas sencillo y con datos enteros o exactos

[quote="DarthTenebrus"]Yo creo que lo que dice hucth es correcto :

De hecho que para empezar en geometria los trazos y construcciones son buenos y muy bellos pero a medida que pasa el tiempo un joven olimpico no puede depender de herramienta tan simples para la solucion de problemas de geometria es decir . Debe aprender tambien las geometria euclidiana avanzada : Conjugacion armonica - Rectas Polares y la Inversion de Steiner . Tambien debe emplear para la maximizacion y minimizacion la tecnica de variacion parcial .

Pero eso no debe quedar alli tambien debe estar presente la geometria proyectiva y aun asi eso es poco .

En particular para mi mis mejores tecnicas en la geometria de esas que son detonates :

- Geometria vectiorial
-Geometria Analitica
-Geometria de los numeros complejos

Y Por ultimo porque no decirlo expresiones trigonometricas y todos los tipos de relacion de senos , cosenos y tangentes cuando los angulos suman 180 grados osea el triangulo .[/quote]

 :wink:  Yo tambien pienso lo mismo, mis tecnicas de resolucion las aprendi en el foro internacional, luego de mucho estudiar ahora puedo dominar:

Coordenadas baricentricas
Coordenadas Tripolares
Geometria con Numeros complejos
Semejanzas Espirales
Vectores
Relaciones metricas
Algunas congruencias
Proyectividades (entre ellas las colineaciones una colineacion particular es la polaridad)
Inversiones
Transformaciones Afines (Ahora estoy estudiando la Isogonologia)
Transformaciones asociadas al triangulo (Isogonales, Isotomicas, Isocriticas (es una que estoy inventando, mas informacion en el link de darij grinberg ya que los teoremas mios que el resolvio son parte de mi teoria  :D ))

$\bigstar$ Capitán Mandarina

Weno, weno, perdonen la ofensa, kiza me molesto un poco, que la estuve intentando muxo tiempo, y no me salio nada!!!! Ademas, yo todavia no domino muy bien otros temas, como son geometria vectorial , numeros complejos, y esos temas no euclidianos, k en realidad tienen razon, en muxisimos problemas son muy utiles!!

Disculpen el comentario!! :blush:  :blush:

NONO esperemos corregir algo la geometria sigue siendo euclideana lo nuevo entre comillas es las tecnicas de solucion a los problemas que por cierto no tiene mucho de nuevo .

YO PIENSO QUE LAS SOLUCIONES DE GEOMETRIA MEDIANTE TRAZOS QUE AHORRAN OPERACIONES ES LA + BELLA ESO ES GEOMETRIA PURA POR SU ELEGANCIA Y POR LO VISTOSA QUE ES!!

MIENTRAS QUE LA GEOMETRIA QUE NOTO EN TU SOLUCION ES MAS TIRANDO PARA EL ALGEBRA OSEA DEBERIAN LLAMARLA GEOMETRIA ALGEBRAICA O ALGO ASI ,YO COMPARTO TU APRECIACION CONEJITA

Y  A TI TE RESPETO HUCHT POR QUE TIENES HERRAMIENTAS DE SOBRA PARA ATACAR UN PROBLEMA MEN!!  oe pero mi problemas pues x  2x  7x  jejeje

Hola conejita, el problema que propuso hucht pertenece a una shortlist IMO 93, aqui te paso el link, el problema esta resuelto de una manera mas entendible, es el problema ISA 1, esta en la parte 2, solo usa un poco de trigo y opera de manera conveniente.

http://www.mathlinks.ro/Forum/viewtopic.php?t=15580

Aunque al igual que la mayoría creo que mientras más herramientas tengamos mayores son las posibilidades de resolver un problema.

PD. Edson ya coloque la solucion de tu problema x,2x,7x.

Es sabido que de todos los triángulos equiláteros circunscritos al triángulo $ ABC,$ aquel $ \triangle XYZ$ que es homotético al triángulo de Napoleón externo, es el que tiene mayor área (lado). Por tanto, este tiene lado doble del lado $ L$ del triangulo de Napoleón externo, A saber:
\[L \equal{} \sqrt {\frac {_1}{^6}(a^2 \plus{} b^2 \plus{} c^2) \plus{} \frac {2\sqrt {3}}{3}[ABC]}\]
Para cada triangulo equilátero $ \Delta$ circunscrito a $ \triangle ABC$ podemos asociar un triangulo $ \Delta'$  homotético a éste e inscrito a $ \triangle ABC.$ Así pues, aplicando Teorema de [i]Gergonne-Ann[/i] tenemos $ [ABC]^2 \equal{} [\Delta] \cdot [\Delta'].$ Entonces el área (lado) $ L'$ de $ \Delta'$ será mínimo si  el área (lado) de $ \Delta$ es máximo. En otras palabras $ [ABC]^2 \equal{} \frac {_3}{^{16}}(L')^2(4L^2)$

Sustituyendo el valor de $ L$ del lado del triángulo de Napoleón tenemos:

\[ [ABC]^2 \equal{} \frac {_1}{^8}(L')^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}[ABC]) \]
\[ L' \equal{} \frac {2\sqrt {2}[ABC]}{\sqrt {a^2 \plus{} b^2 \plus{} c^2 \plus{} 4\sqrt {3}[ABC]}}\]
Todo  equilátero $ \triangle DEF$ inscrito a $ \triangle ABC$ tiene pues lado mayor o igual que $ L'.$


This is probably a silly question, but it looks interesting for me. What condition must satisfy the sets $ A_1, A_2,... $ of positive integers such that there exists a function f from the positive integers to the same set such that $ f^k$ has the image $ A_k $ for all $k$. Here $ f^k$ is the k-th iterate of the function?

Sorry but I don't understand.
$f(f(x))$ can be written as both $f(f(x)$) and $f(y)$ for $y=f(x)$.
Hence it must lie in both $a_2$ and $A_1$. By the same idea
$A_{i+1}$ is a subset of $A_i$. Now this condition is sufficiemt but this looks too easy !

[quote="iura"]Now this condition is sufficiemt but this looks too easy ![/quote]
Wrong!
If $A_k=A_{k+1}$ and $A_{k+1}\neq A_{k+2}$ than such function doesn't exist.
(if $A_k$ has image $A_{k+1}=A_k$ of $f$ then $A_{k+1}$ has image $A_{k+2}=A_k$ of $f$)

I know I am idiot, but I simply do not understand from where it follows that the condition is also sufficient. It can' be hard, but...

And there is one more thing. You cannot construct such a function if $ A_1 $ is the set of integers greater than 1 and $ A_2 $ is the set of integers greater than 5. Thus, you see, there are quite enough conditions to be satisfied. I start to wonder if there are nice necessary and sufficient conditions.

Another observation. If $\exists k$, $ A_k $ is finite than $\forall i$, $ A_i $ is finite.

I want to add another condition:
$A_0=\{1,\cdots\}$ and $\|A_i-A_{i-1}\| \geq \|A_{i+1}-A_i\|$.
Now I think it's all OK

I think i found the answer for this problem. It comes immediately to me and it seems too easy to be ok! Anyway i will take the risk and i will post it ;)

Now, can someone post a (full) proof? :)

It is obviosly. If all terms $A_k$ are distinkt, it is easy construct f: $f(A_i)=A_{i+1}$.
If exist i,j: $A_i=A_j$, then $\{A_k | k\in N\}$ must be periodic. This conditions are nesessary and sufficiently.


search
algebra
polynomial
inequalities
triangle inequality
calculus
calculus computations

Hello all,

Can anyone explain what is a "M-adic" and a "M-ary"? I don't know how to search them in google. I got so many fail positive hits.

TIA

The second is the base-M equivalent of decimal or binary expansion.

The first I have usually seen as "p-adic", restricting p to be a prime. Represent the positive integers as polynomials $a_0+a_1p+a_2p^2+\dots+a_np^n$ with $0\le a_k<p$, then represent everything else as formal power series $a_{-n}p^{-n}+a_{-n+1}p^{-n+1}+\dots+a_0+a_1p+\dots$
When doing arithmetic, move your operations from the bottom up.

The $p$-adic numbers have a norm: $|a_{-n}p^{-n}+a_{-n+1}p^{-n+1}+\dots+a_0+a_1p+\dots|_p=p^n.$ Taking the distance between $a$ and $b$ as $|a-b|_p$, this satisfies the triangle inequality, and in fact the even stronger inequality that $|a-c|_p\le \max(|a-b|_p,|b-c|_p).$

We saw above how to express every rational number in $p$-adic form. The object called the $p$-adic numbers is the completion of the rationals with respect to the $p$-adic norm, just as the reals are the completion with respect to the usual norm.

thank you jmerry! It is p-adic and now I can search :)


A={factors of 361} B={perfect squares} C={odd #s greater than 2}

which # is in all 3 sets?

My answer[hide=":"]361[/hide]

their answer[hide=":"]9[/hide]

[hide="i dont get it"]how is it 9?

361= 19^2, so 9 cant be in that category[/hide]

It is a mistaken answer, there are several mistakes and the administartor(s) are trying to solve them all.
keep putting 9, you may be surprised at how quickly they can fiw these problems.
(by the way, the answer is 361)

[quote="stevenmeow"]
(by the way, the answer is 361)[/quote]

yea thats what i put....

ok i already reported it though

huh, is it a coincidence that $ 9$ is the answer if $ A$ was the set of factors of 360, which is only 1 less than 361?

That's what I always thought the intended problem was.

there is only one word on this problem i did yesterday, all it says it "1/2"!!!
LOL :rotfl:

@applepi

NOT the right place to put it

start a new thread

and thats not funny. it is a bug that probably should be reported

Oh, btw, I had this question but they fixed it into

Factors of 360  :thumbup:

@applepi: That has absolutely nothing to do with this, start a new

thread.


The scale of a backpacker's map is "1 cm on the map represents 24,000 cm on the ground." How many kilometers long is a hike that measures 10 cm on the map? Express your answer as a decimal to the nearest tenth.

The scale is $ 1\,\text{cm}: \equal{} 0.24\,\text{km}\implies10\,\text{cm}: \equal{} \boxed{2.4\,\text{km}}$. (Hopefully this is the right notation?)

Isn't it just $ 1\,\text{cm}: 0.24\,\text{km}\implies10\,\text{cm}: \boxed{2.4\,\text{km}}$ (no equals signs)


Spongebob's French!

(I'll give anyone a hundred bucks if they can figure out how the above comment is related to this thread)

stupid pi day i had to compose a humongous pi song for calculus. i rather celebrate Eienstein's Birthday March 14.

I know 20 digits, but only in binary.

11.001001000011111101

spikzz (u know who it is, since we are not suppose to use REAL names, said by tokenadult) said u stopped trying to memorize more of it, and said he is smart enough to not trying to memorize any cuz there is not use

yes... i stopped memorizing pi... cause all the math teachers got rid of thos long strips of pi on the walls...
and i now also think its a waste of time

I know up to 12, but that just happened naturally.  3.14 is enough for me.

memorizing pi is a good waste of time if you're really bored (like i am). i know.... hang on..... 244 digits. =]

yes, i happen to understand fully that there will be no use for pi except for 3.14 or 22/7 in finding a circle's area and circumference, but it's pretty fun, especially if you want to annoy people.

i wish pi day was on a weekday that way our teacher would bring us pie for class

In high school in my senior year, our principle got on the loudspeaker 1:59 and wished everyone a happy pi day.
http://www.wpdpi.com/mnemonics.shtml
that site has some pi mnemonics, the one on the bottom is something like 700 digits of pi that is contained in a poem based on Edgar Allan Poe's "The Raven"

wow.
oh, i get it! 3.14159...
im a bit slow...

[quote]The principal of your school knew it was pi day? That's incredible.[/quote]

we were impressed as well, but before becoming principal he used to be a math teacher, which probably explains it

interesting....... btw, thanks for that pi site, zscool!! it's really cool. i bookmarked it. =]

n u r gonna cum 2 school 2morrow with tenthousand digits memorized...

if you're talking to me........ then no. [i]unfortunately[/i], i can't memorize that many digits in one day. =[

Wouldn't  :pi: be great if it was rational? (Even if it is more than several trillion digits long, it'd be great if we found the end)

[quote="Tare"]Do we even know whether or not it's rational?[/quote]

textbooks tell us it's irrational...but heck...what do they know?

Well, I don't tend to trust textbooks, but I do trust Lindemann, who proved that :pi: is transcendental even though he wasn't as especially good mathematician (at least compared to Hermite, who proved that e is transcendental). I don't think the proof is that hard either.

transcendental=irrational=can not be shown as the ratio of two rational numbers I believe

No. A number is transcendental iff it is not the root of a polynomial with integer coefficients. A number that isn't transcendental is algebraic.

I wonder why more people don't use iffs in definitions. They seem to make more sense than just ifs.

Yall are talking but all I can see is blah blah blah blah. Blah blah blah blah. Blah blah blah. Quit argueing about definitions it's all the same blah blah blah. There is no one definition for anything in math. There is many many explinations and definitions for everything in math. So quit blah blah blahing about things that dont matter. Sorry Im mad at math.

Oooh... about 160 :)  I love Pi Day!  I recorded myself reciting the digits on my computer too! :):)  I can say 100 digits in 11 seconds, too ;-)

-- I am a pi nerd

You dun need to remember that many digits...just ask your calculator...

*gasps*
what happened? i thought this topic was deleted off the forum... this is scary

Some of you have way too much free time.
All I know is 3.141592653589 .
And I think that is enough.

teacher made us bring in pies!!!!that was cool

Stop reviving this darned topic! It's been over three months since Pi Day :-).

<puts on best Dogbert Impression> Bah!

I can understand how some people can choose to memorize enormous blocks of pi, but I doubt that that many would be able to begin reciting from a given starting point within what they've memorized.  I know this girl at my school who has memorized 287 digits.  I'm dying to ask her to start from digit 137 and see what happens.

Personally, I feel that memorizing the Periodic Table of Elements is somewhat more applicable in most situations.  I've gotten through the first 25 or so, but only in name.  Eventually, I hope to memorize their properties, such as various isotopes, radioactivity of said isotopes, atomic weight, etc.

to add on to what yasha said,
this topic really should get closed. pi day was more than 3.5 months ago! it's already summer! let's stop reviving these old topics guys. just try reading the sticky miths posted first.

Guys, maybe revive the thread again, it's now 4 days after pi day???

The last post was in 2004. Please do not revive dead threads.

get real. =p

but seriously, please do not revive old topics.

[size=85][color=#FF0000]mods, please lock[/color][/size]
[color=#00BB00][size=90]Locked. --PhireKaLk6781[/size][/color]


Prove that
\[ \ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} abc) \ge 4(a^2b \plus{} b^2c \plus{} c^2a)
\]
for all positive real numbers $ a, b, c$.

$ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} abc) \ge 2(a^3\plus{}b^3\plus{}c^3)\plus{}(a^2b\plus{}b^2c\plus{}c^2a)\plus{}(ab^2\plus{}bc^2\plus{}ca^2)$
$ \equal{}\sum (a^3\plus{}ab^2) \plus{} \sum a^2b \plus{}\frac{\sum(a^3\plus{}a^3\plus{}b^3)}{3}\ge 2\sum a^2b \plus{}\sum a^2b \plus{}\sum a^2b$
$ \equal{}4(a^2b\plus{}b^2c\plus{}c^2a)$

Very Nice and Neat proof, NextPeace!  :)
Here is another solution: (ofcourse uglier than yours)

we have:

$ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} abc) \minus{} 4(a^2b \plus{} b^2c \plus{} c^2a) \equal{}$

$ \equal{} \frac {(13a \plus{} 5b \minus{} 3c)(a \minus{} b)^2 \plus{} (13b \plus{} 5c \minus{} 3a)(b \minus{} c)^2 \plus{} (13c \plus{} 5a \minus{} 3b)(c \minus{} a)^2}{6}$

which is not hard to finish.


geometry
rhombus
trigonometry
circumcircle
ratio
perpendicular bisector
angle bisector

Let $M$ be the intersection of two diagonal, $AC$ and $BD$, of a rhombus $ABCD$, where angle $A<90^\circ$. Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$, provided that $O \neq M$. Construct a circle that has $O$ as centre and goes through $B$ and $D$. Let the intersections between the circle and $AB$ be $B$ and $X$. Let the intersections between the circle and $BC$ be $B$ and $Y$. Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$, respectively. Express $\frac{OQ}{OP}$ in terms of $t$.

Denote $\alpha = \angle DAB = \angle BCD,\ \beta = \angle ABC = \angle CDA,\ \alpha + \beta = 180^o$. The quadrilateral $XBYD$ is cyclic, hence the sums of the opposite angles are equal to $180^o$:

$\angle BYD = 180^o - \angle BXD$

$\angle XDY = 180^o - \angle XBY = 180^o - \beta = \alpha$

Thus the angle $\angle XDY \equiv \angle PDQ = \alpha$ is constant. Using the sine theorem for the triangles $\triangle BDX, \triangle BDY$,

$\frac{DX}{BC} = \frac{\sin{\widehat{XBD}}}{\sin{\widehat{BXD}}} = \frac{\sin{\frac \beta 2}}{\sin{\widehat{BXD}}}$

$\frac{DY}{BC} = \frac{\sin{\widehat{YBD}}}{\sin{\widehat{BYD}}} = \frac{\sin{\frac \beta 2}}{\sin{(180^o - \widehat{BXD})}} = \frac{\sin{\frac \beta 2}}{\sin{\widehat{BXD}}}$

Consequently, the segments $DX = DY$ are equal, which implies that the triangle $\triangle XYD$ is isosceles, the perpendicular bisector of the segment $XY$ passes through the vertex $D$ and it is identical with the bisector of the angle $\angle XDY \equiv \angle PDQ$. Since $(O)$ is the circumcircle of this isosceles triangle, the circumcenter $O$ lies on this perpendicular bisector and the angle bisector. Denote $\phi = \angle MDO$. Then

$MO = MD \tan \phi = MA \tan{\frac \alpha 2} \tan \phi$

$t = \frac{MA}{MO} = \frac{1}{\tan{\frac \alpha 2} \tan \phi} = \cot{\frac \alpha 2} \cot \phi$

$\frac{MD}{PD} = \cos{\left(\frac \alpha 2 - \phi\right)},\ \ \ \frac{MD}{QD} = \cos{\left(\frac \alpha 2 + \phi\right)}$

The angle bisector $DO$ divides the opposite side of the triangle $\triangle DPQ$ in the ratio of the adjacent sides $OP, OQ$. As a result,

$\frac{OQ}{OP} = \frac{QD}{PD} = \frac{\cos{\left(\frac \alpha 2 - \phi\right)}}{\cos{\left(\frac \alpha 2 + \phi\right)}} = \frac{\cos{\frac \alpha 2} \cos \phi + \sin{\frac \alpha 2} \sin \phi}{\cos{\frac \alpha 2} \cos \phi - \sin{\frac \alpha 2} \sin \phi} =$

$= \frac{\cot{\frac \alpha 2} \cot \phi + 1}{\cot{\frac \alpha 2} \cot \phi - 1} = \frac{t + 1}{t - 1}$

We have $\angle{AXD}=\angle{BYD}=\angle{BOA}$ hence $BOPX$ is cyclic.
Consider the inversion [b]I[/b] of pole $O$ and power $OB^2$. Then the circle circumscribed to $BOPX$ will transform to line $AB$. Hence $I(P)=A$, consequently $OP\cdot{OA}=OB^2$.
Similarly we obtain $OQ\cdot{OC}=OB^2$.
Hence $\frac{OQ}{OP}=\frac{OA}{OC}=\frac{MA+MO}{MA-MO}=\frac{\frac{MA}{MO}+1}{\frac{MA}{MO}-1}=\frac{t+1}{t-1}$


1.
$ \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{n^{n \plus{} 1} e}} {{\left( {n \plus{} 1} \right)^n }} \minus{} n} \right)$

2.
$ \lim_{n\to \infty}{n(e\minus{}(1\plus{}\frac{1}{n})^{n})}$

You're welcome.
We are glad to have helped you on your previous limit problem.
And we'll surely continue to solve all your problems, since you seem so happy when we do so.

You must not help me. But it your choice .

$ \lim_{n \rightarrow \plus{} \infty} n( e^{ 1\minus{}nln(1\plus{}\frac{1}{n})}\minus{}1 ) \equal{}\lim_{t \rightarrow 0^\plus{}} \frac{ e^{ 1\minus{}\frac{ln(1\plus{}t)}{t}}\minus{} 1}{t}$
$ \equal{}\lim_{t \rightarrow 0^\plus{}}      e^{ 1\minus{}\frac{ln(1\plus{}t)}{t}}.( \frac{ ln(1\plus{}t)}{t^2}\minus{}\frac{1}{t(t\plus{}1)})$
$ \equal{}\lim_{t \rightarrow 0^\plus{}}  \frac{ ln(1\plus{}t)}{t^2}\minus{}\frac{1}{t(t\plus{}1)}$
We have :
$ ln(1\plus{}t) \sim t\minus{}\frac{t^2}{2}$
$ \Rightarrow \lim_{t \rightarrow 0^\plus{}}  \frac{ ln(1\plus{}t)}{t^2}\minus{}\frac{1}{t(t\plus{}1)} \equal{} \Rightarrow \lim_{t \rightarrow 0^\plus{}} \frac{1}{t}\minus{}\frac{1}{2}\minus{}\frac{1}{t(t\plus{}1)}$
$ \equal{}\frac{1}{2}$

Have I gotten mistake somewhere?

hello, i have for the first $ \frac{1}{2}$ and for the second $ \frac{1}{2}e$
Sonnhard.


First  Day  (2006.1.12)

sorry.my  English is very  poor.

pro1.Let $a_1,a_2,...a_k$ be  real  numbers  and $a_1+a_2+...+a_k=0$,
Prove that $max(a_k^2)\leq [(a_1-a_2)^2+(a_2-a_3)^2+...+(a_{k-1}-a_k)^2]\frac{k}{3}$

pro2.For positive intergers $a_1,a_2 ,...,a_{2006}$ such that $\frac{a_1}{a_2},\frac{a_2}{a_3},...,\frac{a_{2005}}{a_{2006}}$ is distinct，Find the max amount of  the distinct positive intergers in the set$\{\frac{a_1}{a_2},\frac{a_2}{a_3},...,\frac{a_{2005}}{a_{2006}}\}$？

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=407830#p407830
Positive integers $k, m, n$ satisfy $mn=k^2+k+3$, prove that at least one of the equations $x^2+11y^2=4m$ and $x^2+11y^2=4n$ has one odd solution.

for problem 2, I don't understand the question. What if $\frac{a_i}{a_{i-1}}= 2^i$?

It is really strange that the first problem of CMO is not a geometry problem. So I guess tomorrow there will be one or two very difficult ones.  :)

[quote="zhaoli"][quote="wenxin"]First  Day  (2006.1.12)

pro2.For positive intergers $a_1,a_2 ,...,a_{2006}$ such that $\frac{a_1}{a_2},\frac{a_2}{a_3},...,\frac{a_{2005}}{a_{2006}}$ is distinct，Find the max amount of  the distinct positive intergers in the set$\{\frac{a_1}{a_2},\frac{a_2}{a_3},...,\frac{a_{2005}}{a_{2006}}\}$？[/quote]

The Chinese version says :

For positive integers $a_1,a_2 ,...,a_{2006}$ such that $\frac{a_1}{a_2},\frac{a_2}{a_3},...,\frac{a_{2005}}{a_{2006}}$ are pairwise distinct, find the minimum possible amount of  distinct positive integers in the set$\{a_1,a_2,...,a_{2006}\}$.[/quote]
are you sure?

4. In a right angled-triangle ABC, angle ACB = 90. Its incircle O meets BC, AC, AB at D,E,F respectively. AD cuts O at P. If angle BPC = 90, prove AE + AP = PD.

5. ${a_n}$:$a_1 = \frac{1}{2}$,$a_{k+1}=-a_k+\frac{1}{2-a_k}$ for $k = 1, 2,...$

Prove:$\left(\frac{n}{2(a_1+a_2+...+a_n)}-1\right)^n \leq \left(\frac{a_1+a_2+...+a_n}{n}\right)^n\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)...\left(\frac{1}{a_n}-1\right)$

6. Suppose X is a set with |X| = 56. Find the minimum value of n, so that for any 15 subsets of X, if the cardinality of the union of any 7 of them is greater or equal to n, then there exists 3 of them whose intersection is nonempty.

CMO 2006: (n家校本. Thanks: wenxin, Soarer, zhaoli and zhaobin)
Problem1: http://www.mathlinks.ro/Forum/viewtopic.php?p=407814#p407814
Real numbers $a_1,a_2\cdots a_k$ satisfy: $a_1+a_2+\cdots+a_k=0$. Prove:
\[ \max_{1 \le i \le k} (a_i^2)\leq \frac{k}{3}\cdot[(a_1-a_2)^2+(a_2-a_3)^2+\cdots+(a_{k-1}-a_k)^2] \]
Problem2:
For positive integers $a_1,a_2 ,...,a_{2006}$ such that $\frac{a_1}{a_2},\frac{a_2}{a_3},...,\frac{a_{2005}}{a_{2006}}$ are pairwise distinct, find the minimum possible amount of  distinct positive integers in the set$\{a_1,a_2,...,a_{2006}\}$.

Problem3: http://www.mathlinks.ro/Forum/viewtopic.php?p=407830#p407830
Positive integers $k, m, n$ satisfy $mn=k^2+k+3$, prove that at least one of the equations $x^2+11y^2=4m$ and $x^2+11y^2=4n$ has a solution $(x,y)$ with $x,y$ being odd.

Problem4: http://www.mathlinks.ro/Forum/viewtopic.php?p=408479#p408479
In a right angled-triangle $ABC$, $\angle{ACB} = 90^o$. Its incircle $O$ meets $BC$, $AC$, $AB$ at $D$,$E$,$F$ respectively. $AD$ cuts $O$ at $P$. If $\angle{BPC} = 90^o$, prove $AE + AP = PD$.

Problem5: http://www.mathlinks.ro/Forum/viewtopic.php?p=408392#p408392
$\{a_n\}$:$a_1 = \frac{1}{2}$,$a_{k+1}=-a_k+\frac{1}{2-a_k}$ for $k = 1, 2,...$.Prove:
$\left(\frac{n}{2(a_1+a_2+...+a_n)}-1\right)^n \leq \left(\frac{a_1+a_2+...+a_n}{n}\right)^n\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)...\left(\frac{1}{a_n}-1\right)$

Problem6:
Suppose X is a set with |X| = 56. Find the minimum value of n, so that for any 15 subsets of X, if the cardinality of the union of any 7 of them is greater or equal to n, then there exists 3 of them whose intersection is nonempt.

[quote="alexei"]Great thanks! Excuse me, do you know the cut-offs and results of the CMO2006? Where I can found this results?[/quote]
For the foreign teams participating in CMO,
Hong Kong has 2 silvers, 4 bronze
Russia (uncertain) has 1 silver, ? bronze

[quote="shobber"]How was mecrazywong, yptosi's score?[/quote]
For yptsoi, he didn't pariticipate in CMO (for some reason).
For mecrazywong, you'd better ask him yourself. It's not suitable to tell the others' scores here. :(

can I ask who is 方博成 and 钟安业 in mathlinks :)

They don't have a mathlinks account. :)
P.S. 方博成 should be 方博威. His name was wrong in all the publications of this year's CMO. (and this mistake has become a joke among us :D )

[quote="leepakhin"]They don't have a mathlinks account. :)[/quote]
Thanks. :P

[quote="shobber"]Gold: 84.
Silver: 57
Bronze: All under 57.[/quote]

Is 84  full score????

21 points for each problem, the full score is 21*6=126.


I though of this a while ago, and can think of no obvious method for solving it without a grapher or some other computer program. Solve:
                                                        $cos(x)=x$

You may  not meet such value of $x$ when you walk down  on the  street,  (or it may not have a nice symbol like $\pi$ but there is nothing wrong with graphing  ..The  answer  is around $0.7391 $ radians .

FWIW , More often than not, the answer to such problems do not have  "nice" solutions, but  in one sense it is not that different than getting values like  $\cos (\pi/2) = 0$ and for some $\pi /2 $ may not be as  "nice" a number as it can only be approximated  by a fraction...

Important part is to know  a method  where you can  get this  value (as accurate as you wished )  even without a calculator...

Im just wondering if theres a way to do it by hand, even a re-itteration or series or something similar... any ideas? Mcclaurin series?

Sure,

To find x such that cos(x) = x

compute cos(cos(cos(....cos(1))) .... ), the resulting sequence converges with error something like sqrt(3/n) (why?).

Interisting i must say... it seems to converge for most values of x anyway, due to the line bouncing off the graph $y=cos(x)$ and the line $y=x$? By the way, in $\sqrt{\frac{3}{n}}$ is $n$ the number of iterations?

I did a coursework on three methods -grin-


email Support MATHCOUNTS ARML AMC USA(J)MO USAMO

[quote="MCrawford"][quote="xxreddevilzxx"]Would the pensions be higher/lower? What are the working conditions???[/quote]

That depends on the school.

We have connections with a few schools, but overall our goal is more to match interested teachers with interested programs.  The details are not as much our concern, though I doubt they would be anything other than good circumstances.

However, the Art of Problem Solving Foundation is beginning to promote a support of high standard of benefits for teachers of high quality.  It might take a few years, but hopefully those efforts will be successful in some cases.[/quote]

I live in India.  Will I be able participate in the programme?

It's not a program.  We just try to connect capable and interested teachers will schools willing to develop high level math programs.  Unforutnately we have no contacts in India.

[quote="MCrawford"]Art of Problem Solving is happy to help connect teachers and potential teachers with schools are the country that are interested in building or already have top math programs and/or math teams in place.  Anyone interested in such a job can email us at classes@artofproblemsolving.com.

Additionally, schools and parent groups interesting in finding such teachers can contact us at that same email address.[/quote]

I am interested to teach. But is it OK for a foreigner??? I am now in a math graduate program in US and have some experience training the math olympiad team back in my country.

thanks

[quote="MCrawford"]If you are in the U.S. and are interested in pursuing a high school or middle school teaching position, you can contact us at the email address above.[/quote]

What if I dont want to pursue a high school teacher position, BUT I am interested in teaching math as an extracurricular activity??

thanks

[quote="MCrawford"]Then you will have to find a way to do that.  That's not really something we can help with from this side of the globe.  We could probably find programs to work with in the U.S., but we don't have any school or program network anywhere else.[/quote]

I think I didn't myself clear. I am a foreigner who now live and study math in US and I am interested in teaching math as an extracurricular activity. But I dont have any intention to become a higschool teacher. So is it possible for me to involve with this program?

Thanks

[quote="rrusczyk"]Where in the US are you?  We may be able to hook you up with a group or school.[/quote]

I will send the letter to the above address, to give detail about myself.

Thanks

Richard, In about three years I will be looking for someone to begin becoming involved with my programs and eventually take them over.  I am planning on retiring in about 7 years and I am afraid that if I do not find someone to take over my prgrams they will die a swift and sure death.  I have tried for several years to get someone  interested and they all get scared off or are not interested!! It will be like taking your finger out of a bucket of water.  After the ripples die you can't even tell that the finger was ever in the water.

I will have 40 years teaching, all in middle school.  I will be the one in the old folks home who has someone walk by their room and  hear someone teaching.  They stick in their head and I will be the only one in the room.

The students that I have had have benifited greatly but I would like the program to continue for the next generation of kids.

I'm very sad for your students, but it's good to hear you're already looking out for the longevity of the program even with though you'll still be there for 7 more years.

I have PM'ed you with a suggestion of someone to talk to, and I will continue to try to come up with others.

For those of you on the board considering teaching or in teaching and looking for an ideal opportunity to work with excellent students, I [b]strongly[/b] suggest you check this one out.  Jeff has built one of the most impressive programs in the country at [i]two[/i] schools.  You can read about it [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php]here[/url].  I taught high school for a half year right after college before I moved on to Shaw and ultimately, AoPS.  Had I been fortunate enough to end up in a situation like continuing a program like the one Jeff has built, it's possible I would never have left the classroom.

Hi, Richard,

I'm the math curriculum coordinator in Georgetown ISD (Texas) now (have coached math from elementary to high school, including MOEMS, MATHCOUNTS, AMC/Mandelbrot/high school competitions, asst. coach of Texas ARML team) - and we have a great opening at our high school for a math coach/teacher.  The class that I currently teach for gifted kids, Independent Study in Math/Topics in Math, is part of the assignment, and the job is specifically for someone who knows math competitions and can work with our Math Team and gifted math students.  We are interviewing now for Fall 2007.

We have some very bright students; 4 are going to ARML this May, and we have 3 feeder MATHCOUNTS schools.  In the past, this team has produced a MOSP-er (2001), quite a few Texas ARML team members, USAMO-ers, won the Rice University Math Tournament, etc.   Very interested population of families of these students.  Our new coach would be very much loved and appreciated by these kids and their families here in Georgetown, and they'll work really hard for him/her.  We have 3 feeder MATHCOUNTS middle schools, which have produced many State individuals and teams, with one National (2nd Nats Team 1998).

Georgetown is a beautiful town very close to Austin, Texas.  Parks, Lake, very pretty downtown area - friendly people and supportive, involved families.  Math is popular here!

I'm very worried that the pressure of filling this position might cause us to hire someone before we find the most qualified person.  If anyone is be interested, I need to hear from or about them ASAP!

The link to the job posting is:  http://georgetownisd.org/career/jobs/job.asp?jobpostingsid=1094
Applicants can apply online.   Be sure to read the job description and include your qualifications that match in the resume and the statement.

Thank you so much,
Tricia Rothenberg

Contact information for the position in Georgetown Texas is:

Tricia Rothenberg
Georgetown ISD
603 Lakeway Dr.
Georgetown, TX 78628
(512) 943-5000 X 7881
mathweb99@hotmail.com

(See previous message for more information about the position.)

Hi Everyone

    I am a math teacher intensely interested in advanced mathematics such as calculus and beyond. I went to Stuyvesant HS in New York when i was a teen (25 years ago!!!!!) and I would absolutely love to teach in a similar school (Stuy is a special school for gifted kids) Please write me back at amybethx@rocketmail.com if you can offer any help at all. Thank you in advance,

Amy

Is this sticky even active and are the programs still active? last post seems to be like June of 2008.

We do still hear from schools occasionally who seek teachers interested in developing their top students.

[quote]star99
In high school, I would like to help my middle school in maths, in tutoring or other after school activities. do you know of any thing out side the school in which I could partiipate in, or if i would be able to get service credit for these things?
[/quote]
sorry to reactivate this topic, but i had the exact question. Thanks.

I think the way to go about service credit is to negotiate with your school about what counts as credit.  We've had students do work at the San Diego Math Circle for credit, and I think others have coached MATHCOUNTS teams for credit.

I am interested in summer teaching involving math team coaching,etc. If anyone has any suggestions please email me at amybethx@rocketmail.com  Thanks!


Let $M$ be a point inside a $\triangle ABC$. The lines $AM,BM,CM$ intersect
$BC,CA,AB$ at $A1,B1,C1$, respectively. Assume that
$S_{MAC_1} + S_{MBA_1}+ S_{MCB_1} = S_{MA_1C} + S_{MB_1A} + S_{MC_1B}$.
Prove that one of the lines $AA_1,BB_1,CC_1$ is a median of the triangle
$ABC$.

Suppose w.l.o.g. that $[ABC]=1.$ For the interior point $M$ we have $[BMC]=x<1$, $[CMA]=y<1$, $[AMB]=z<1$ and $x+y+z=1$.
$[MBA_1]=\frac{xz}{y+z}$, $[MCA_1]=\frac{xy}{y+z}$ a.s.o. and you prove easily that $\boxed {x+y+z=1\ \ \wedge \ \ \sum_{cyc} \frac{xy}{y+z}=\sum_{cyc}\frac{xz}{y+z}\left(=\frac 12\right)\Longrightarrow (x-y)(y-z)(z-x)=0}\ .$


function
linear algebra
matrix
algebra open
algebra

I have a matrix that needs a certain values for every dimension. The values can be worked out trough functions.
The first set of values for dimension one is calculated by  $f_1(x)=x$
the second set of values for dimensions of two is calculated by the function $f_2(x)=\frac{x^{2}-x}{2}$
and the third is value is  $f_3(x)=\frac{x^{3}}{6}-\frac{x^2}{2}+\frac{x}{3}$

How do I get the function  $f_n(x)$
or is it not possible?

and maybe this is useful;

$x\in \mathbb{ N}$

 $n\leq x$

$f_x(x)=1$

if demanded I could try to get more info..but Im pretty much unsure if there is a solution...

I appologize if I put this into the wrong forum...

Correct me if I'm wrong, but it looks like $f_k (x) = \sum_{i=0}^{x-1} {i \choose k-1}$.

hey thats great, but how do you translate that... :? to ordinary functions??

heres the background on the problem. thanks


Consider a box of mass $m$ and dimensions $l*b*h$ that is moving on a rough horizontal floor having coefficient of kinetic friction ${\mu}_{k}$.A horizontal force $F$ acts at a distance d from the centre of this box.Then determine the conditions when -
(a)The box slides before toppling
(b)the box topples before slliding.

Can you be more specific about the point where the force acts?

assume that the area $l*b$ faces you .You may assume it even to be a two dimensional problem...I just need to understand what concept is involved in this.


Let $\left(I,r\right)$ be the incircle of a triangle $A_{1}A_{2}A_{3}$. For every $i\in\left\{1,2,3\right\}$, let $\left(O_{i},r_{i}\right)$ be the circle externally tangent to the circle $\left(I\right)$ and touching the sides $A_{i}A_{i+1}$ and $A_{i}A_{i+2}$ of triangle $A_{1}A_{2}A_{3}$ (hereby, indices are taken modulo $3$). Prove that

$r=\sqrt{r_{2}r_{3}}+\sqrt{r_{3}r_{1}}+\sqrt{r_{1}r_{2}}$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=55605

  darij


Is this the relevant forum to post? :huh:

I think he's trying to get people to go to his site so that people will click on the Google Ads.

[quote="mathstricks"]can you multiply 998 X 996 in 2 seconds

http://learnmathstricks.blogspot.com/

http://studymathematics.blogspot.com[/quote]

just about

1000x1000 = 1000000
                - 1996
               = 998004

i think


Show that for no positive integer $ n$ can both $ n\plus{}3$ and $ n^2\plus{}3n\plus{}3$ be perfect cubes.

[quote="Smartguy"]Show that for no positive integer $ n$ can both $ n \plus{} 3$ and $ n^2 \plus{} 3n \plus{} 3$ be perfect cubes.[/quote]

Assume that both of them are perfect cubes.
Let $ n\plus{}3\equal{}x^3$, then $ n^2 \plus{} 3n \plus{} 3\equal{}x^6\minus{}3x^3\plus{}3$

And as $ x \ge 2$, $ (x^2\minus{}1)^3<x^6\minus{}3x^3\plus{}3<(x^2)^3$.

So $ n^2 \plus{} 3n \plus{} 3$ cannot be a cube, contradiction.

We can write k, a perfect cube, as n+3. So k^2 would be n^2+6n+9 but we need n^2+3n+3. So when we subtract 6n+9-3n+3= 3n+6. Which is close to 3k. k^2-3k+3. So all we need to do is prove that k^2 -3k +3 can never be a perfect cube if k is also a perfect cube which I do not know how to do.

Sorry can be a perfect cube.

Can someone tell me if my solution is correct?

[hide="alternate solution"] We have $ n\plus{}3$ and $ n^2\plus{}3n\plus{}3$. If they are both perfect cubes, then their product will also be a perfect cube.

$ (n\plus{}3)(n^2\plus{}3n\plus{}3)\equal{}n^3\plus{}6n^2\plus{}12n\plus{}9$.

$ (n\plus{}2)^3\equal{}n^3\plus{}6n\plus{}12n\plus{}8$ and $ (n\plus{}3)^3\equal{}n^3\plus{}9n^2\plus{}27n\plus{}27$.

Since our expression is between two consecutive perfect cubes, it cannot be a perfect cube so one of $ n\plus{}3$, $ n^2\plus{}3n\plus{}3$ is not a perfect cube.[/hide]


geometry
AoPS Books
Science Olympiad
\/closed

America's true power is not gun and rocket, but culture and language. Any book written in English  should potentially have the whole world as market. In my professional field, books are so expensive, no company will consider publishing unless the market    is international. I am just wondering whether Mr. R ever    contemplate contacting book deals overseas to open a channel for AoPS to be known to foreign students. Taiwan, Singapore, Hong Kong, Japan, Korea , to name a few , will be good ones to start. Parents  in that area are known to pay anything to improve their kids' education. I have read some translated versions of japan's and Taiwan's math books.  I believe AoPS is still on top of thems, especislly with full solutions supplied.

Well, just as a comment, this website can be accessed from anywhere in the world...

I am talking about THE  AoPS  "books", the paper copy, not the website. By the way, any body has idea to popularize this web site overseas? I notice the overseas members are not as many as expected.

Um, maybe that's because most non-Singaporean Asians do not know much English by default. If we are going to export AoPS books we'll need a translator and the cost it takes to sell them might be over the profit. (unless there's volunteer translators) Either way since the U.S. is like the only developed country with mathematical education level this low I doubt the book'll do well overseas, but it could be worthing looking at.

Perhaps the more reasonable solution would be to make the books available for purchase (and readable unlike Amazon) online, however there's still some cons to that as well. Oh well.

We do sell the books overseas, but not in great volume.  We do have one company reselling the books in Singapore.  If you know of anyone interested in reselling in Japan or Taiwan, I would be happy to speak with them.

As for the site, if you have any recommendations on where to get cross-linked to draw more international students, please let me know.

Perhaps [url=http://www.showin-kids.com/]Showin-kids[/url] can be of some help...however they are completely Japanese and most does not know much English, however I do know there's approximately three who can write English so maybe they can help you.

Japan and Taiwan will be hard since there's so many educational books out...murph, but you could try.

Most CHinese people can read English. I was in china last summer and my cousin had her something like Middle School Final for Shanghai. One of the tests is english so chinese people know english pretty (at least shanghai).

You're kidding me, right? Some Chinese can't even read/write Chinese...

how come the books are not sold on Amazon and other similar sites?

tare china has a really good literacy rate in the cities, plus the whole country has a education thanks to the communist govenment goals

that's why they always score in the top 3 in the IMO.

china has the world's most genuine undisturbed culture, (we heard of greek and egypt being controlled by tons of country)

unlike usa we keep our ancient traditions

most chinese people strive to excellence, and are ambitious

plus china is in the top 3 in summer olympics and about top 5 in winter olympics

plus chinese people's average height is higher than all those other asian pepople (japan, vietnam, korea

) GO CHINA BASKETBALL ASIA CHAMPION I'm a yao ming, meng batter, and wang zhizhi fan.

plus china has a big population and big land, doesn't matter what you sell in china its going to sell

My cousin had some IMO books in China. in CHina they sell  of books called "subject" olympiads like science olympiad and math oylpiad. i think that is a competition there i'm not sure. She'll probably ace the AMC (or close to). Too bad the compeittion is too tough there.
It sucks that i can't read chinese. however we have enough math stuff here to go thourgh


[quote="P.H.Li"]a,b,c is positive real number such that
ab+bc+ca=3,prove or disprove :
3(a+b+c)+abc≧10[/quote]
It follows from the following inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128169

$ a,\ b,\ c\in{x^3 \minus{} (a \plus{} b \plus{} c)x^2 \plus{} 3x \minus{} abc \equal{} 0}$, let $ x \equal{} \sqrt{3}$, we have $ 3(a \plus{} b \plus{} c) \plus{} abc \equal{} 6\sqrt {3}\geq 10$.

[quote="kunny"]$ a,\ b,\ c\in{x^3 \minus{} (a \plus{} b \plus{} c)x^2 \plus{} 3x \minus{} abc \equal{} 0}$, let $ x \equal{} \sqrt {3}$, we have $ 3(a \plus{} b \plus{} c) \plus{} abc \equal{} 6\sqrt {3}\geq 10$.[/quote]
I am sorry, kunny. Think please about what you wrote. :wink:
Thank you!
Delete your last post and I'll delete the mine! :!:

Sorry, I have just noticed my mistake, but I can't delete my post.

[quote="P.H.Li"]a,b,c is positive real number such that
ab+bc+ca=3,prove or disprove :
3(a+b+c)+abc≧10[/quote]
Let $ b \equal{} min(a,b,c) \equal{}> b\le {1}$;
$ c \equal{} \frac{3\minus{}ab}{a\plus{}b}$; we must prove :    $ c \ge \frac{10\minus{}3a\minus{}3b}{3\plus{}ab} <\equal{}> (3\minus{}b^2)a^2\plus{}(6b\minus{}10)a\plus{}(3b^2\minus{}10b\plus{}9)\ge {0}$;
$ D \equal{} (b\minus{}1)^2(3b^2\minus{}4b\minus{}2)\le{0}$;


[quote="jagdish"][b](1) If f(0) =1,f(1) =2 and f(x) = 1/2{f(x+1)+f(x+2)}. Then find f(100) = [/b][/quote]

This is not a functional equation. This is just a sequence computation :

$ u_0\equal{}1$
$ u_1\equal{}2$
$ u_{n\plus{}2}\equal{}2u_n\minus{}u_{n\plus{}1}$

So $ u_n\equal{}\frac 43 \minus{}\frac 13(\minus{}2)^n$

So $ f(100)\equal{}u_{100}\equal{}\frac 43 \minus{}\frac 13(\minus{}2)^{100}$

[quote="jagdish"][b]Sir can you explain me

how can i get Un=4/3-1/3(-2)^-n from  Un+2 = 2Un-Un+1.[/b][/quote]

Hello,

This is a classical method to solve sequences like $ u_{n\plus{}2}\equal{}\alpha u_{n\plus{}1} \plus{} \beta u_{n}$ :
You look first for solutions in the form $ u_n\equal{}z^n$, which implies to solve $ z^2\equal{}\alpha z \plus{} \beta$

And then all the solutions are in the form $ az_1^n \plus{} b z_2^n$ with $ z_1$ and $ z_2$ solutions of $ z^2\equal{}\alpha z \plus{} \beta$

Here, the equation to solve is $ z^2\equal{}2\minus{}z$ whose roots are $ 1$ and $ \minus{}2$ and so $ u_n\equal{}a \plus{} b(\minus{}2)^n$
You just then have to compute $ a$ and $ b$ from $ u_0$ and $ u_1$ and you get the resuired result.

In this case, you can also use a direct method : $ u_{n\plus{}2}\minus{}u_{n\plus{}1}\equal{}(\minus{}2)(u_{n\plus{}1}\minus{}u_n)$ and so $ u_{n\plus{}2}\minus{}u_{n\plus{}1}\equal{}(\minus{}2)^{n\plus{}1}(u_1\minus{}u_0)$ ... and so on


simplify the product:

$ P=\cosh\cos2a\cos4a...\cos2^{n-1}a$

good luck!

sorry, I am Brazilian,
must simplify the P

I gave a hint. You can use $ \boxed{\sin 2\theta \equal{}2\sin \theta \cos \theta}$ repeatedly.

ahhhta, it was suspected that, but do not see the solution.

From quoting you first post, I can tell that you meant: $ P = \cos a\cos2a\cos4a...\cos2^{n - 1}a$. (By using "\cosa" with no space between "cos" and "a" the $ \text{\LaTeX}$ didn't show the $ \cos a$.

[hide]$ P = \cos a\cos2a\cos4a...\cos2^{n - 1}a$

Multiply both sides by $ \sin a$

$ P \sin a = \sin a\cos a\cos2a\cos4a...\cos2^{n - 1}a$

Now use $ \dfrac{1}{2}\sin(2\theta) = \sin(\theta)\cos(\theta)$ for $ \theta = a$:

$ P \sin a = (\sin a\cos a)\cos2a\cos4a...\cos2^{n - 1}a$

$ P \sin a = \dfrac{1}{2}\sin2a\cos2a\cos4a...\cos2^{n - 1}a$

Now use $ \dfrac{1}{2}\sin(2\theta) = \sin(\theta)\cos(\theta)$ for $ \theta = 2a$:

$ P \sin a = \dfrac{1}{2}(\sin2a\cos2a)\cos4a...\cos2^{n - 1}a$

$ P \sin a = \dfrac{1}{4}\sin4a\cos4a...\cos2^{n - 1}a$

Now use $ \dfrac{1}{2}\sin(2\theta) = \sin(\theta)\cos(\theta)$ for $ \theta = 4a$:

$ P \sin a = \dfrac{1}{4}(\sin4a\cos4a)...\cos2^{n - 1}a$

$ P \sin a = \dfrac{1}{8}\sin8a...\cos2^{n - 1}a$

Keep doing this until you get:
$ P \sin a = \dfrac{1}{2^n} \sin(2^n a)$

Thus, $ P = \dfrac{\sin(2^n a)}{2^n \sin a}$. [/hide]


Each of $ n$ cats has $ 2n$ fleas. If two cats (and their fleas) are removed, and three fleas are removed from each remaining cat, the total number of fleas remaining would be half the original total number of fleas. What is the value of $ n$?

Since there are $ n$ cats and $ 2n$ fleas per cats, there are $ 2n^{2}$ fleas in total.  If two cats are removed, then $ 4n$ fleas are removed.  Then we're also going to remove $ 3\times (n\minus{}2)\equal{}3n\minus{}6$ fleas from the remaining cats.  Hence there are now $ 2n^{2}\minus{}4n\minus{}3n\plus{}6\equal{}2n^{2}\minus{}7n\plus{}6\equal{}\frac{2n^{2}}{2}\equal{}n^{2}$.  Now just solve the quadratic.

how the heck
do u get the national sprint papers of 2008. Could any1 send me a link cause i can't find it anywhere. Is there something known as a mock trial just for state.

It's a robot.  It doesn't actually have access to these things.  I believe this appeared in FTW and someone couldn't solve it so they pressed a button and it was posted on the forum.  You should probably ask one of the site administrators.


modular arithmetic
number theory
relatively prime
number theory unsolved

[quote="pco"]Hello Rust
Nice demo, but you forgot at least $2^{t}-1=1$, $2^{t}+1=3^{y}7^{z}$ which gives $t=1$, $y=1$ and $z=0$ and $2^{2}=1+3^{1}7^{0}$

--
Patrick[/quote]
Yes, I forgot solution with z=0.
For first case better $7\not |2^{t}+1, \ \forall t\in Z_{+}$.

[hide]
If $y=0$, then $z=0\Rightarrow x=1,\ z=1\Rightarrow x=3$. For $x\geq 4$, it fails mod 16.

If $y\geq 1$, then $2^{x}\equiv 1\pmod{3}\Rightarrow x$ even, let $x=2m$ so that
\[(2^{m}+1)(2^{m}-1)=3^{y}\cdot 7^{z}\]
As $2^{m}+1$ and $2^{m}-1$ are relatively prime (clearly $m\neq 0$), one factor is a power of 3 and the other is a power of 7.

If $2^{m}-1=3^{y}$, then mod 8 shows $y\leq 1$, so $y=1,\ m=2$. But then $2^{m}+1$ is not power of 7, contradiction.

So $2^{m}-1=7^{z}$, mod 16 shows $z\leq 1$. Here we get $z=0,m=1,y=1$ and $z=1,m=3,y=2$.

The solutions are $(1,0,0),(2,1,0),(3,0,1),(6,2,1)$.
[/hide]


[hide]Let $(4+6+9)\cdots=P$
Notice that $4+6+9=2^{2}+2\cdot3+3^{2}$
So multiply by 3-2 on both sides.
LHS telescopes(is that the right word?) to
so
$P=3^{3^{2007}}-2^{3^{2007}}$
[/hide]
EDIT:fixed stupid mistake

oops I made a factoring mistake  :blush:
I should have multiplied by (2-3) on both sides right? um.. or by (3-2)
ok edited

Nice solutions :)  The generalized polynomial statement of this problem I found interesting, since it provides an immediate polynomial proof that every integer is expressible in base $3$ in exactly one way.  (The generalization to base $b$ is also immediate.)


hi!
i have just started to learn the java language.
can you give me the definition of the term ''class'' ?
what exactly we mean with this term i have not understand it yet :wacko:

From what I've gathered a class used to create objects in OOP. You can create objects using a constructor.

A class, in essence, contains the "blueprint" of an object - what it does, what it looks like, what it is.

To understand the point of the class, imagine someone who would want to create a game of Tetris.  This person would create various objects, such as an object that could display something to the screen, an object that could determine whether a line was made, an object that could recieve input from the keyboard, etc.  These objects, put together in the correct way, perhaps by another object, could produce the game of Tetris.

But in order to create all of these objects, the computer needs to know what the object looks like - and that is what a class is for.  By creating a Java class, you tell the computer what the object looks like, and what it should be able to do.

Of course, that's only one kind of class. An executable program with a main method is also a class, and there are classes which are really just bundles of static methods (such as the Math class, which has methods such as sqrt).

For the object sort, I'll use something I wrote recently as an example: a permutation class.

Each instance of this object contains some fields; in this case, an array of numbers to represent the values of the permutation at $ 1,2,\cdots,n$. You don't need to know that, because you're not allowed to look at or change those values directly. It's a private field, because I don't want someone coming in and setting a value so that $ f(1)\equal{}\minus{}33$ in this "permutation". Instead, all you need to know to use this object are some methods that make these objects or read information about them. For example, here's a snippet using my Permutation class:
[code]Permutation p = Permutation.random(12);
System.out.println(p);
System.out.println(p.cycleDecomposition());[/code]which results in output of [code]8 4 10 7 1 2 9 6 12 11 3 5
(1, 8, 6, 2, 4, 7, 9, 12, 5) (3, 10, 11)[/code]
So, what do each of those lines do? "random" is a static method in my Permutation class, which builds an array containing the numbers from 1 to n in random order, and then constructs a permutation from that array. It's not a constructor itself, but it calls one to make the actual object.
The third line calls the "cycleDecomposition" method, which looks at its object and returns a string representing the cycle decomposition.
The second line, printing the permutation p, actually calls the "toString" method of my Permutation object. If I hadn't written something there, it would have defaulted to printing the address of the object in memory, which isn't very useful.

To make this class, I had to write all of these methods and more. But then, from the outside, I can just make and use Permutation objects whenever I want to.

ok thanks!

i have understant that class has 2 meanings
1)class is a set of objects
2)practically a class is a subprogramme the we can call it when we need it in every programme
please correct me if i am wrong

Well, look at a class like this

It is the name of the main block of code that contains all the methods needed
to call a class, you have to import the package first
to use a method inside a class, you have to enter something like

myName.length();

as in you have to add parentheses behind the method. also, when you save the class, the file name should be that of the class


Three vertices of a square are $ A(2, 3), B(5, 1)$ and $ C(7, 4)$.

What are the coordinates of the fourth vertex $ D$?

What are the coordinates of the centre of the square?

[hide="One way to do it (there are more)"]Since the diagonals of the square bisect each other at the center of the square, we can find the midpoint of $ BC$, let it be $ O$, and set $ O$ as the midpoint of $ BD$:

The midpoint of a line is the average of its endpoints coordinates: Thus, the midpoint of $ AC$ is $ \left( \frac {2 \plus{} 7}{2}, \frac {3 \plus{} 4}{2} \right) \equal{} \boxed{(4.5,3.5)}$. This is the center of the square, $ O$.

Now, let the coordinates of $ D$ be $ (x,y)$. We must have $ \left( \frac {x \plus{} 5}{2},\frac {y \plus{} 1}{2} \right) \equal{} (4.5,3.5)$. Solving, we find $ (x,y) \equal{} \boxed{(4,6)}$, which are the coordinates of $ B$.
[/hide]


What number is one-fifth of one-fourth of one-third of one-half of 1200?

I think that FantasyLover's solution would be quicker. $ 1200*\frac{1}{5}*\frac{1}{4}*\frac{1}{3}*\frac{1}{2}=1200*\frac{1}{120}=1200*\frac{1}{5!}=10 $
I just realized that it was the same thing as $5$ factorial and I memmed 5 factorial.


$A_1, A_2, \ldots, A_{29}$ are $29$ different sequences of positive integers. For $1 \leq i < j \leq 29$ and any natural number $x,$ we define $N_i(x) =$ number of elements of the sequence $A_i$ which are less or equal to $x,$ and $N_{ij}(x) =$ number of elements of the intersection $A_i \cap A_j$ which are less than or equal to $x.$ It is given for all $1 \leq i \leq 29$ and every natural number $x,$ \[ N_i(x) \geq \frac{x}{e}, \] where $e = 2.71828 \ldots$ Prove that there exist at least one pair $i,j$ ($1 \leq i < j \leq 29$) such that \[ N_{ij}(1988) > 200. \]

Looks to me like the [url=https://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=404130]coat and patches problem[/url], but I haven't checked to see if the arithmetic works out quite so simply.

Indeed ! I am copying my proof from there, adapted to the problem at hand. Let $n=29$ (so $n=2m+1$, with $m=14$).

Denote by $\alpha_k$ the number of integers from $\{1,2,\ldots,x\}$ belonging to exactly $k$ of the sequences $A_1,A_2,\ldots,A_n$, for $1\leq k \leq n$. Then clearly
$(1) \ \ \sum_{k=1}^n \alpha_k \leq x$.
By the same token
$(2) \ \ \sum_{k=1}^n k\alpha_k = \sum_{k=1}^n N_k(x) \geq  \dfrac {x} {\textrm{e}}n$ and
$(3) \ \ \sum_{k=2}^n \binom {k} {2} \alpha_k = \sum_{1\leq i <j\leq n} N_{ij}(x)$.

Since $n$ is odd, take $n=2m+1$. Multiply relation (2) by $m$ and subtract from it relation (1) multiplied by $\binom {m+1} {2}$, and compare it with relation (3),  to get $\sum_{1\leq i <j\leq n} N_{ij}(x) \geq \dfrac {x} {\textrm{e}} m(2m+1) - \dfrac {m(m+1)} {2} x$ (where the sum in the LHS has $\binom {2m+1} {2}$ terms).

This means there exist $1\leq u < v \leq n$ such that $N_{uv}(x) \geq \dfrac {x} {\textrm{e}} - \dfrac {m+1} {2(2m+1)} x$. For our numerical values, and $x=1988$, it yields $N_{uv}(x) \geq 1988\left (\dfrac {1} {\textrm{e}} - \dfrac {15} {58}\right ) \approx 217.2 > 200$.


$ a+b > 2$ assume $ a$ and $ b \in \mathbf{Z}$. Thus, the minimum possible value for a and b is 1 and 2, respectively. Thus

$ 1^{4}+2^{4}> 2$
$ 16 > 2$

If we consider the function $ f(x) = x^{n}$, for any natural number $ n$, we have a function that increases over its entire domain. A similar increasing function would be $ g(x) = x^{n}+y^{n}$, which resembles $ a^{4}+b^{4}$. Thus for a>1 and b>2 there is no way the inequality won't be satisfied.

[quote="Elio (n)"]$ If a+b>2 demostrate$
$ that a^{4}+b^{4}>2$[/quote]
Proofs should go in a high school section.

[quote="Prime factorization"]...assume $ a$ and $ b \in \mathbf{Z}$... [/quote]

We can't assume that. And one of $ a,b$ could be less than $ 1$, i.e. $ a=\frac{4}{5}$.

[hide="Crude Solution"]
Working backwards:
\[ a^{4}+b^{4}> 2 \iff a^{4}> (2-b)^{4}> 2-b^{4}\iff b^{4}+12b^{2}+12 > 4b^{3}+16b \]
Now, I don't remember a lot of my inequalities, and I'm sure there are plenty that would make this one line. So, I'll go with this argument:

WLOG $ a \geq 1 \geq b$ with both equalities not at the same time. $ b \leq 0$ leads to $ a^{4}>2$, so we only consider positive $ b$ between $ 0$ and $ 1$.

Then, $ 4 \geq 4b^{3}$ and $ 8 \geq 8b$. So, it remains to prove that $ b^{4}+12b^{2}> 8b \iff 12b^{2}-8b >0 \iff b > \frac{2}{3}$.

Now, if we have $ b \leq \frac{2}{3}$ then $ a \geq \frac{4}{3}\implies a^{4}\geq \frac{256}{81}> 2$.

Thus, for all $ a,b$ we have $ a^{4}+b^{4}> 2$. [/hide]

[quote="Elio (n)"]$ If a+b>2 demostrate$
$ that a^{4}+b^{4}>2$[/quote]

[hide="Quick"] Say that $ a+b = 2+\epsilon > 2$.  Then

$ a^{4}+(2+\epsilon-a)^{4}\ge 2 \left( \frac{a+(2+\epsilon-a)}{2}\right)^{4}= 2 \left( 1+\frac{\epsilon}{2}\right)^{4}> 2$

By Power Mean. [/hide]

[quote="cincodemayo5590"]
\[ a^{4}+b^{4}> 2 \iff a^{4}> (2-b)^{4}> 2-b^{4}\]
[/quote]

$ a^{4}> (2-b)^{4}$ is more or less equivalent to the condition $ a+b > 2$, so this is actually the content of the problem.


How many $ 3\minus{}digit$ numbers are there such that $ a > b > c$ where $ a,b,c$ represent the digit in the hundreds , ten's and one's place respectively ?

NOTE: Plz dont count out the possible numbers, i would like the solution to be solved using combinatorial formulae. Thank You  :)

I am not sure,but it think that the answer is:
$ \binom{9}{2}\plus{}\binom{8}{2}\plus{}\dots\plus{}\binom{2}{2}$

[quote="JBL"]Yes, but it's just as easy to see directly that the answer is $ \binom{10}{3}$.  Problems of this difficulty really belong in the Intermediate forum.[/quote]
Could you plz explain how you got that ?

you can prove the well-known identity by the same way:
$ \binom{r}{r}\plus{}\binom{r\plus{}1}{r}\plus{}...\plus{}\binom{n}{r}\equal{}\binom{n\plus{}1}{r\plus{}1}.$


I've just created an [url=http://www.artofproblemsolving.com/Wiki/index.php/Successor_set]article[/url]. When I saw the preview, it was looking just as I wanted it to, but now it is not showing any of the $ \text{\LaTeX}$ part.

Also, how do you comment on the Talk part of the Wiki? :oops:

i can view that and i even edited that :)

Then there seems to be some problem with my server. I can't see $ \text{\LaTeX}$ on any 'article' page but it is visible on the 'history' and 'edit-preview' pages.

Try refreshing your cache (ctrl + f5) or logging out and logging back in.

You comment on the "talk section" by simply leaving your message and then signing it. To sign it, type for tildes: ~~~~ . The server will automatically replace that with your user name and the time and date. For organization, though, you may want to click on the "+" button next to the edit button, which will make a new section. Title the new section with a summary of your concern (similar to the topic title on the forums) and enter your message in to the textbox. Sign yourself with four tildes as in the other case.

This is a current problem, and so has been pinned.

You type the message in the BIG textbox.

See, when making a new topic, click the little "+" button in the upper right hand of the screen, and that'll give you the screen I was talking about. Then follow my instructions in the last post.

To leave a message after making a new topic, just edit the talk page like you would an article, leaving your message under your section and following my directions in the last post.

Yes, I figured that out. Anyway, thanks for the help.

Hi,

New here. I have been having fun with the AoPS wiki, but have found the exact same problem with the LaTeXing as described by shreyas_patankar. I have tried logging out and back in.

In fact, the LaTeX does render properly on some pages, just not all of them, which I think is kind of strange.

(I am using Internet Explorer 6, but am on a uni computer, so can't do too much about changing my settings etc.)

Any help would be much appreciated.

Thanks!

You [i]could[/i] try using a different browser like [url=http://www.mozilla.com/en-US/firefox/]Firefox[/url], though I'm not entirely sure if it'll take care of the LaTeX viewing problem. Well, LaTeX renders fine for me in Firefox.

But I do know that IE6 is _olddddd_ :) it'd be a good idea to use Firefox or at least upgrade to IE7.

EDIT: oops er I can't read - university computer, as in like a computer lab? eh nvm then :|

It's weird, usually university computers have Firefox installed on them (unless the computer is really really old and has almost no RAM).

Hi,

Thanks very much for the replies.

Since writing that post, I've started using Firefox to access the wiki, which has indeed fixed the problem :)

Thanks again!


In front of you there are three machines. One of them always tell the truth, one of them always lies, and one of them answers the xor of what the other two machines would answer*. Call these three machines the TRUTH, FALSITY, and XOR machines respectively.

Unfortunately, the very eccentric programmer who made the machines had them output their answers by using two different letters rather than "yes" and "no". You have no idea which letter corresponds to which answer, or even what two letters will be used. (all three machines use the same set of letters however).

The machines are, of course, omniscient and know everything.

You may ask three questions, each question being directed at a machine of your choice. You may not ask any questions that would lead to a dumb paradox, or else the machines will simply break. Determine the identity of each of the machines.

*In other words, if the other two machines would answer your question differently, the XOR machine answers "yes". Otherwise, it answers "no". For example, the XOR machine answers "does 2+2=4" with YES because the truth machine would answer YES and the falsity machine would answer NO. An example of a question that the XOR machine would answer NO would be, "Are you the TRUTH machine?"

I don't understand the xor machine, but here it goes:

Ask:
1)Does 2+2=4? The only one answering no is the FALSITY machine, put a purple sticky note on it to mark it.
2)Did you lie on the last question? The TRUTH and the FALSITY both say yes, and now you know the difference between the TRUTH and the FALSITY. The XOR says no, so you mark the XOR with a green sticky note.
3) Are you the TRUTH machine? All say yes, the only one without a note is the TRUTH machine.

TaDa!!!! (I don't understand the XOR, so this may be wrong.)

[quote="Brut3Forc3"]I think you can only ask each question to one machine.[/quote]

aw.... (I'm not good at reading directions)

First question: ask a machine: are you the truth machine? if it answer no, it is the XOR machine. otherwise, it is something else.

Second question: ask another machine: are you the truth machine? again, if it answers no, we have the XOR machine. Otherwise, the other one is the XOR machine.

Third question: Does 1+1=2? If it answer no, that is the false machine, if it answer yes, that is the truth machine. The other one is what is left.

You forgot that the answers are not "yes" or "no" ;)

Let yes be TRUE and no be FALSE.

Then everything works out.

You did not understand the problem ;)
Read the problem again.

[quote="pythag011"]First question: ask a machine: are you the truth machine? if it answer no, it is the XOR machine. otherwise, it is something else.[/quote]

The first machine answers "Q". Now what?

[quote="Ihatepie"]Ask the second machine.

  If the other machine answers the same thing, then the two machines you asked are the truth and false machines while the one you didn't ask is the xor. Then ask if 1+1=2 to one of the machines you have already asked, if he says q, then that is truth, if he says something else, he is false.

  If the other machine answers something different, then that means either the first or second
machines are the xor ones. So then ask the third machine[/quote]
But if the other machine answer "W", you don't know what do these answers mean, which one is YES and which one is NO? :help:

|?|?|?|y
(to machine 1: Are you the truth machine?)
(to machine 3: Are you the truth machine?)

|?|X|?| If same answer, record it (It signifies TRUE). Then to machine 1: Is 1+1=2? If same as last two answers it is a truth machine, else it is a falsity machine.

|?|?|?| If different answers, record them (say x,y), then to machine 2: If you were to be a machine of opposite parity (e.g. tell a lie if you are the truth, and vice-versa) then, is the machine to the left of you the XOR machine?

I've got most  of it but haven't finished it all up...

[quote="Ihatepie"]whatever the first two machines answer is yes, since both will say that they are the truth machine. Asking 1+1=2 shows which one of them is the truth and which one is false.[/quote]

This approach uses too many questions though.

Cryptic Hint: When you have figured out the solution, you'll get the information requested but you might not be able to figure out which answer is yes and which answer is No.


If $ S$ is a set of $ 10$ positive integers that less or equal than $ 53$, then prove that we can choose $ U$ and $ V$ as the subset of $ S$ such that $ |U| \equal{} |V| \equal{} 4$,$ U \cap V$ is an empty set, and the sum of all elements in $ U$ and in $ V$ are the same. Thanks in advance.

This is false. Let $ S\equal{}\{1,2,3,4,5,6,7,9,22,39\}$ be a set of 10 integers in the required range. There are no subsets $ U$, $ V$ of $ S$ such that $ |U| \equal{} |V| \equal{} 4$, $ U$ and $ V$ are disjoint and the sum of the elements of $ U$ and $ V$ are equal.

To see this, first suppose that 39 is a member of one of the sets, let's say $ 39 \in U$. Then the sum of elements of $ U$ is [b]at least[/b] $ 39\plus{}1\plus{}2\plus{}3 \equal{} 45$, while the sum of the elements of $ V$ is [b]at most[/b] $ 22\plus{}9\plus{}7\plus{}6 \equal{} 44$, contradiction. So 39 does not appear in either $ U$ or $ V$.

Now do the same with 22: since $ 22\plus{}1\plus{}2\plus{}3 > 9\plus{}7\plus{}6\plus{}5$ we once again get a contradiction, so $ 22$ is out as well.

Finally, $ \{1,2,3,4,5,6,7,9\}$ cannot be split into two disjoint subsets of equal sum (regardless of size) simply because the sum of all those numbers is odd.

So, you must have mis-stated the problem I'm afraid. Due to the fact that $ 53\plus{}52\plus{}51\plus{}50 \equal{} 206$ is almost $ 10 \choose {4}$, I think the problem should have been as stated but without the requirement that $ U \cap V$ is the empty set. That last requirement renders the problem incorrect, but without it it's a simple pigeonhole.

Can you check your source again?

By the way, you can keep the requirement $ U \cap V \equal{} \emptyset$ but drop the requirement on the size. You can then up the range from 53 to 100 and get an easy problem from IMO 1972 which also appears in Peter Winkler's book.