function
induction
logarithms
algebra
functional equation

A function $ f: R \rightarrow R$,not constant with $ f(1)\equal{}e$.If $ f$ satisfy the condition $ f(x\plus{}y)\equal{}f(x)f(y)$ $ \forall x,y \in R$.Show that:
$ 1)$ $ f(x\minus{}y)\equal{}\frac{f(x)}{f(y)},\forall x,y \in R$

$ 2)$ $ f(kx)\equal{}f^k(x),\forall k \in Z$ and $ x\in R$

$ 3$ $ f(\frac{1}{n}x)\equal{}f^{\frac{1}{n}},\forall n \in N^*$

$ 4$ $ f(x)\equal{}e^x,\forall x\in Q$

[quote="Dimitris X"]A function $ f: R \rightarrow R$,not constant with $ f(1) \equal{} e$.If $ f$ satisfy the condition $ f(x \plus{} y) \equal{} f(x)f(y)$ $ \forall x,y \in R$.Show that:
$ 1)$ $ f(x \minus{} y) \equal{} \frac {f(x)}{f(y)},\forall x,y \in R$

$ 2)$ $ f(kx) \equal{} f^k(x),\forall k \in Z$ and $ x\in R$

$ 3$ $ f(\frac {1}{n}x) \equal{} f^{\frac {1}{n}},\forall n \in N^*$

$ 4$ $ f(x) \equal{} e^x,\forall x\in Q$[/quote]

Umm....... It's more like a solution than questions.

$ x \equal{} 1$ and $ y \equal{} 0$;

$ \Rightarrow \, f(1) \equal{} f(1).f(0)$

$ \Rightarrow\, f(0) \equal{} 1$.

$ y \equal{} \minus{} x \Rightarrow\, f(0) \equal{} f(x).f( \minus{} x)$

$ \Rightarrow\, 1 \equal{} f(x).f( \minus{} x)$

$ \Rightarrow\, f( \minus{} x) \equal{} \frac {1}{f(x)}$.

For 2):

By mathematical induction:

The basis step is

$ f(2x) \equal{} f(x \plus{} x) \equal{} f(x).f(x) \equal{} f^{2}(x)$

and the inductive step is straightforward.

P.S. I guessed someone would beat me to it.

I think that any function that satisfies the property $ f(x\plus{}y) \equal{} f(x).f(y)$ can be written as $ f(x) \equal{} a^x$

here, $ f(1) \equal{} e \ \Rightarrow \ a \equal{} e$

so, $ f(x) \equal{} e^x$ .

Rest of the proofs follow from the properties of exponential functions.

[hide]1. $ f(x \minus{} y \plus{} y) \equal{} f(x \minus{} y)f(y)$
         $ f(x \minus{} y) \equal{} \frac {f(x)}{f(y)}$

2. Induction using $ f(2x) \equal{} f(x)f(x)$

3. Using 2, let change $ x$ into $ \frac {x}{k}$

  $ f(x) \equal{} (f(\frac {x}{k}))^k$

  $ (f(x))^\frac {1}{k} \equal{} f(\frac {x}{k})$

4. Using 2 and 3 and $ f( \minus{} 1) \equal{} e^{ \minus{} 1}$[/hide]

[quote="stephencheng"][quote="Nora.91"]I think that any function that satisfies the property $ f(x \plus{} y) \equal{} f(x).f(y)$ can be written as $ f(x) \equal{} a^x$
[/quote]

Really? But it's true over rational numbers only.......

[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]

And the proof that it's true over rational numbers is $ 1) \to 2) \to 3) \to 4)$, i.e. you will do circular proof.[/quote]

But isn't Cauchy's functional equation : $ f(x\plus{}y) \equal{} f(x) \plus{} f(y) \ and \ not \ f(x).f(y)$    :?:

[url=http://en.wikipedia.org/wiki/Functional_equation]This page[/url] mentions that $ f(x\plus{}y) \equal{} f(x).f(y)$ is satisfied by[b] all exponential functions[/b].

_

[quote="Nora.91"][quote="stephencheng"][quote="Nora.91"]I think that any function that satisfies the property $ f(x \plus{} y) \equal{} f(x).f(y)$ can be written as $ f(x) \equal{} a^x$
[/quote]

Really? But it's true over rational numbers only.......

[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]

And the proof that it's true over rational numbers is $ 1) \to 2) \to 3) \to 4)$, i.e. you will do circular proof.[/quote]

But isn't Cauchy's functional equation : $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \ and \ not \ f(x).f(y)$    :?:

[url=http://en.wikipedia.org/wiki/Functional_equation]This page[/url] mentions that $ f(x \plus{} y) \equal{} f(x).f(y)$ is satisfied by[b] all exponential functions[/b].

_[/quote]
If $ f(x \plus{} y) \equal{} f(x).f(y) \Rightarrow f(x) \equal{} a^x\equal{}e^{cx}$ for $ f: R\rightarrow R$ and $ f$ is not a constant.

Then let $ g(x)\equal{}In(f(x))$ we get $ g(x\plus{}y)\equal{}g(x)\plus{}g(y) \Rightarrow g(x)\equal{}cx$ which is a wrong statement as in the link.

[quote="stephencheng"]
If $ f(x \plus{} y) \equal{} f(x).f(y) \Rightarrow f(x) \equal{} a^x \equal{} e^{cx}$ for $ f: R\rightarrow R$ and $ f$ is not a constant.

Then let $ g(x) \equal{} In(f(x))$ we get $ g(x \plus{} y) \equal{} g(x) \plus{} g(y) \Rightarrow g(x) \equal{} cx$ which is a wrong statement as in the link.[/quote]

Ahem... $ \text{In}(f(x))\equal{}\ln {x}$.

$ \ln{x\plus{}y}\neq\ln x \plus{} \ln y$.

$ \ln xy\equal{}\ln x \plus{}\ln y$.

[quote="hsiljak"][quote="stephencheng"]
If $ f(x \plus{} y) \equal{} f(x).f(y) \Rightarrow f(x) \equal{} a^x \equal{} e^{cx}$ for $ f: R\rightarrow R$ and $ f$ is not a constant.

Then let $ g(x) \equal{} In(f(x))$ we get $ g(x \plus{} y) \equal{} g(x) \plus{} g(y) \Rightarrow g(x) \equal{} cx$ which is a wrong statement as in the link.[/quote]

Ahem... $ \text{In}(f(x)) \equal{} \ln {x}$.

$ \ln{x \plus{} y}\neq\ln x \plus{} \ln y$.

$ \ln xy \equal{} \ln x \plus{} \ln y$.[/quote]

What do you mean?

We take log to $ f(x \plus{} y) \equal{} f(x).f(y) \Rightarrow f(x) \equal{} a^x \equal{} e^{cx}$

$ In(f(x\plus{}y))\equal{}In (f(x).f(y))\equal{}In(f(x))\plus{}In(f(y)) \Leftrightarrow g(x\plus{}y)\equal{}g(x)\plus{}g(y)$

Also $ f(x)\equal{}e^{cx} \Leftrightarrow g(x)\equal{}cx$ ,

So $ f(x \plus{} y) \equal{} f(x).f(y) \Rightarrow f(x) \equal{} a^x \equal{} e^{cx}$ is a statement equivalent to

$ g(x \plus{} y) \equal{} g(x) \plus{} g(y) \Rightarrow g(x) \equal{} cx$, isn't it?

1. Sorry, I thought you were using your $ In$ as notation for inverse function, and not the natural logarithm (please, use \ln in the future).

2. I obviously misunderstood the sentence

[quote]Then let $ g(x)\equal{}In(f(x))$ we get $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)\Rightarrow g(x)\equal{}cx$ which is a wrong statement as in the link.[/quote]

because I don't see anything wrong there, since log of exp is linear identity function.

[quote="hsiljak"]2. I obviously misunderstood the sentence

[quote]Then let $ g(x) \equal{} In(f(x))$ we get $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)\Rightarrow g(x) \equal{} cx$ which is a wrong statement as in the link.[/quote]

because I don't see anything wrong there, since log of exp is linear identity function.[/quote]

From this link
[url]http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation[/url]

There exists function $ f_1(x)$ which satisfies $ f_1(x\plus{}y)\equal{}f_1(x)\plus{}f_1(y)$ and is not of the form $ cx$

Define $ g(x)\equal{}e^{f_1(x)}$.

Then $ g(x)$ satisfies $ g(x\plus{}y)\equal{}g(x).g(y)$ and it's not of the form $ e^{cx}$

I am sorry for not presenting my idea well. :(

Now I get it  :wink:  You were showing the equivalence between Cauchy's equation and the Cauchy-type equation such as one given in the O.P.'s problem. Nice.


Find the area of the region enclosed by the line $3x+4y=12$ and the coordinate axes.

[hide]
The intercepts are 3 and 4.

$\text{Area}=\frac{1}{2}\cdot 3\cdot 4= \boxed{6}$. [/hide]

Using double intercept form:
$x/4+y/3=1$
you get the intercepts of $3 and 4$
Since it's a triangle
$3*4/2$
$\boxed{=6}$
hope I did the box correctly...


Se dau 3 cercuri de aceeasi raza 2 cate 2 secante, ce au in comun punctul M. Fie A;B;C celelalte puncte de intersectie, iar O' O" si O''' centele cercurilor.
Sa se demonstreze ca dreptele lui Euler ale triunghiurilor O'O"O''' si ABC coincid

In forma asta, pur si simplu nu e adevarat. Poate trebuie ca cercurile sa aiba raze egale?

Ai dreptate grobber cercurile sunt de aceeasi raza. imi cer scuze. Am facut modificarea  in enunt. Subiectul este dat la examenul de grad II in 1997

[hide="Simpatica problema, dar cam usurica."]Se arata ca punctul $M$ este simultan centrul cercului circumscris triunghiului $OO'O''$ si ortocentrul triunghiului $ABC$. Mai departe, se considera ortocentrul N al triunghiului $OO'O''$ si se arata usor ca N este si centrul cercului circumscris triunghiului $ABC$. De aici, concluzia este imediata.[/hide]
[hide="Observatie."]Enuntul problemei trebuia sa spuna ca centrele $O, O', O''$ ale cercurilor date nu trebuie sa fie situate in varfurile unui triunghi echilateral.[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

It's well-known that a group can't be the reunion of 2 subgroups different from itself. Here's a nice problem (even if it's kind of easy  :D ):

Prove that a field K (commutative or not) can't be the reunion of 3 subfields different from itself.

[i]Romanian NMO, final round, 1987, 12'th grade[/i]

Maybe you could use another problem I posted somewhere around here ("Proposed problems") and which you people disregarded as well... :D

Suppose it is possible to have K as the union of K1, K2, K3.
Then K1 not included in K2 U K3 and same for the others (otherwise K would be the union of only two subfields)

Take x in K1-(K2 U K3), y in K2-(K1 U K3)
x<>0, y<>0 () is in all subfields)
xy not in K1 or K2 so xy in K3
x*xy=x 2 y not in K1 or K3 so x2y in K2
y in K2 so x 2  in K2
x+y not in K1 or K2 so x+y in K3
x(x+y) not in K1 or K3 so x(x+y)=x 2 +xy in K2
x 2 in K2 so xy in K2
y in K2 so x in K2, false

I suppose this is all. Correct me if I'm wrong. :)

Good job  :)  blondu I moved ot to solved problems

Right!
Changed it now.
Thx for pointing it out. :)


If $ X$ and $ Y$ are iid r.v's, what can we say about $ E[X.1_A]$ and $ E[Y.1_A]$ for $ A\in \sigma(X,Y)$?

You cannot say anything deep if you don't make assumptions on your set $ A$
those expectations are not even equal ( take $ A\equal{}(X\in[a,b];Y\in[c,d])$ for suitable $ a,b,c,d$)

so what is your point ?

Can't we have any relationship between $ E[X.1_A]$ and $ E[Y.1_A]$ when $ X$ and $ Y$ are iid?

no, because there are a lot of events inside $ \sigma(X,Y)$ where $ X$ and $ Y$ can be very different.

I think, as siiz said, $ E[X.1_A] \equal{} E[Y.1_A]$ for events $ A \equal{} \{X \plus{} Y\le a\}$ and these events form a $ \pi$-system which generates the $ \sigma$- field $ \sigma(X\plus{}Y)$

oh! now you are talking about $ \sigma(X\plus{}Y)$, which is completely different of $ \sigma(X,Y)$ ..

[quote="limsup"]I think, as siiz said, $ E[X.1_A] \equal{} E[Y.1_A]$ for events $ A \equal{} \{X \plus{} Y\le a\}$ and these events form a $ \pi$-system which generates the $ \sigma$- field $ \sigma(X \plus{} Y)$[/quote]
Furthermore, we can easily show that
\[ E(X|\phi(X,Y)) \equal{} E(Y|\phi(X,Y)) \equal{} 0.5\phi(X,Y)
\]
for all symmetric Borel function $ \phi$ such that $ \phi(X,Y)$ is integrable. Hence $ E[X.1_A] \equal{} E[Y.1_A]$  for any $ A$ in sigma algebra generated by the family of $ \phi(X,Y)$, where $ \phi$ satisfies above requirements.

[quote="siiz"][quote="limsup"]I think, as siiz said, $ E[X.1_A] \equal{} E[Y.1_A]$ for events $ A \equal{} \{X \plus{} Y\le a\}$ and these events form a $ \pi$-system which generates the $ \sigma$- field $ \sigma(X \plus{} Y)$[/quote]
Furthermore, we can easily show that
\[ E(X|\phi(X,Y)) \equal{} E(Y|\phi(X,Y)) \equal{} 0.5\phi(X,Y)
\]
[/quote]
I am not completely sure about the last assertion. I think for example that
\[ E[X | X\plus{}Y] \equal{} E[X | 100(X\plus{}Y)] \equal{} \frac{X\plus{}Y}{2} \neq 50(X\plus{}Y)\]

You are right, as usual :D
Ok, let say it again:
\[ E(X|\phi(X,Y)) \equal{} E(Y|\phi(X,Y))
\]
for all symmetric Borel function $ \phi$. Hence $ E[X.1_A] \equal{} E[Y.1_A]$  for any $ A$ in sigma algebra generated by the family of $ \phi(X,Y)$, where $ \phi$ satisfies above requirements.
I have made so many mistakes recently, not only on this forum  :(


Find all $ x\in N$ for $ x^2\plus{}6x\plus{}2008$ is a perfect square number.

[quote="thanhnam2902"]Find all $ x\in N$ for $ x^2 \plus{} 6x \plus{} 2008$ is a perfect square number.[/quote]

$ y^2\equal{}x^2\plus{}6x\plus{}2008\equal{}(x\plus{}3)^2\plus{}1999$
$ \Leftrightarrow (y\plus{}x\plus{}3)(y\minus{}x\minus{}3)\equal{}1999$

As $ 1999$ is prime, $ \begin{cases}y\plus{}x\plus{}3\equal{}1999 \\ y\minus{}x\minus{}3\equal{}1 \end{cases}$

gives $ x\equal{}996$ as the only solution.


The players of a champion football team are to be given medals by their club managers. Including the reserves, the team of 15 players stand in a line along the pitch with one meter between them each. The manager will begin with any player and then continue presenting the medals to the others. If this procedure is to be completed using the shortest route, then beginning with player number 1 and continuing in line until player number 15 is the most appropriate. (In this case 14 meters would be covered). What will the total distance be if the same procedure is to be completed using the longest possible route for n players?

I have no idea if
[hide="this"]
the coach starts at position 1 and goes to 15, then 2, then 14, and so on, thus covering 14+13+12+...+1 meters or 105 meters
[/hide]
is right.

:rotfl: Wow--just wow. :o What a bad start. Problem fixed.

So is it right now?

I think it's right. At least I can't find a longer path... :P

Well taking a smaller example, with 1 2 3 4 5 players, lined up like that, then your path would be to go in this order: 15243 for a total distance of 4+3+2+1=10.

Consider this path instead: 25143 for a total distance of 3+4+3+1=11.

I don't know what the longest possible path is but I'll think about it some more, but I think by considering this smaller path, it shows that there is likely a longer path than the 1, 15, 2, 14, 3, 13, ... path.

He can wallk through path between men number
1,2 atmost 2 times as well as 14,15
2,3 atmost 4 times as well as 13,14
3,4 atmost 6 times as well as 12,13
4,5 atmost 8 times as well as 11,12
5,6 atmost 10 times as well as 10,11
6,7 atmost 12 times as well as 9,10
7,8 atmost 14 times as well as 8,9
which is 2(2+4+6+8+10+12+14) = 112 m because each path length is 1.
but he cannot walk 112 because of the beginning point. <= Sorry for unclear explanation. It's hard to expain.

So the he can walk at most
112 - 1 = 111 m

And the path 8,1,15,2,14,3,13,4,12,5,11,6,10,7,9 is 111 m


linear algebra
matrix
superior algebra
superior algebra unsolved

[quote="wiosna"]I have theorem:
Let $ A$ be an algebra with unity. If $ a$ contains matrix units $ e_{ij}, i,j \equal{} 1,2....,n$ then A is isomorfic $ M_{n}(e_{tt}Ae_{tt})$ for each $ t \equal{} 1,...,n.$

We can write $ a_{ij} \equal{} e_{ti}ae_{jt}$ and $ \phi : A \minus{} \minus{} > M_{n}(e_{tt}Ae_{tt})$ where $ \phi (a) \equal{} (a_{ij})$ I don't know why $ \phi$ is surjective.[/quote]

If $ b\equal{}(e_{tt}a_{ij}e_{tt}) \in M_n(e_{tt}Ae_{tt}),$ then let $ a\equal{}\sum_{i,j}e_{it}a_{ij}e_{tj}.$ It's easy to show that $ \phi(a)\equal{}b.$


When $ 0.\overline{36}$ is expressed as a common fraction, what is the sum of the numerator and denominator?

To find the fraction N that equals $ 0.\overline{36}$, you solve:

$ 100N \equal{} 36.\overline{36}$
$ 1 \quad N \equal{} 00.\overline{36}$.

(so that the repeating part will cancel out)

Subtracting the second equation from the first,
$ 99N \equal{} 36 \implies N \equal{} \frac{36}{99} \implies N\equal{} \frac{4}{11}$.
Thus, the sum of the numerator and denominator is $ 4 \plus{} 11 \equal{} 15$.


inequalities
absolute value
combinatorics proposed
combinatorics

Suppose $x_1,...,x_n$ are real numbers and $-1 \leq x_i \leq 1$ and $\sum x_i =0$
Prove there is a permutation $\sigma$ one the ${ 1,...,n \}}$ that for $1 \leq p \leq q \leq n$ we have
\[\mid x_{\sigma (p)}+...+x_{\sigma (q)} \mid \leq 2 - \frac 1n\]
Prove we can't use $2 - \frac 4n$ instead of $2- \frac 1n$
____________________________________
No pain, no gain

i'm sorry but I don't understant the pb ?? Could you reformulate it please ..

I think I have explained enough.
We have n real numbers between 1 and -1 with sum 0.
Prove we can put tem in a row that absolute value of sum of any block is not bigger than $2-\frac 1n$

Hey people solve this problem ;)  ;)  ;)

I'll reformulate it for you so pbornstein can solve it ;-)

[quote="Omid Hatami"]Consider $x_1,...,x_n \in \mathbb{R}$ and $-1 \leq x_i \leq 1$ and $\displaystyle \sum^n_{i=1} x_i = 0$

1. Show that there exists a permutation $\sigma$ of the numbers $1 \rightarrow n$, so that for each $p,q \in \mathbb{N}$ for which $1 \leq p \leq q \leq n$, we have $\mid x_{\sigma (p)}+...+x_{\sigma (q)} \mid \leq 2 - \frac 1n$

2. Prove that the above statement is false if we replace $2- \frac 1n$ by $2 - \frac 4n$.[/quote]

Well why didn't you understand what I said ?

Why me? It was Alekk asking  :(

Anyway, I think that this one has been proposed on ML before, maybe on the inequalities forum.
It also appear in the book from Andreescu and Feng so...

Pierre.

:P woops

sorry pierre, I saw the french flag without avatar and thought it was you  :blush:

should have known the post was to short  :lol:

Don't worry, Peter, even Valentin made this kind of mistake before :D (in the days when belenos had his French flag there for a time): [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=11271]click here[/url] ;) .

It passed 3 months and nobody has solution for this problem.I'll send my solution soon.


function
AMC
AIME
number theory
relatively prime
arithmetic sequence

This is a pretty fun problem.

A sequence is defined by $a_1=1995$, $a_{n+1}=\frac{a_n}{a_n+1}$ for all $n\geq 1$.  If $a_{1995}=\frac{m}{n}$, where $m$ and $n$ are relatively prime integers, find the remainder when $m+n$ is divided by $1000$.

Edit: This is my $29^2$ post!  :D

[quote="JesusFreak197"]This isn't from 1995 AIME? It looks exactly like an AIME problem, and $a_1=1995$... out of curiosity, what's the source?[/quote]Actually it's based off an old Math League question, I modified it some though, in addition to making it fit the AIME format.

Edit: Oh and $26$ is the correct answer.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

A matrix in M(n,C) such that

$tr(A)=0$
$tr(A^2)=0$
.............
$tr(A^{n-1})=0$
$tr(A^n)=k \neq 0$

prove that A is diagonalisable

Suppose the eigenvalues are not disinct:
There are $m_1$ eigenvalues equal to $a_1$,$m_2$ eigenvalues equal to $a_2$, and .., ,$m_r$ eigenvalues equal to $a_r$, with $m_1+..+m_r=n$, $r<n$, $m_i \geq 1$.
Consider the $r \times r$ Vandermonde $V(a_1,..,a_r)$ matrix whose $k$-th column is $(a_k,a_k^2,..,a_k^r)^t$. Then $V(m_1,..,m_r)=0$ and $V$ is invertible. So $m_1=0$, contradiction.

Unless I have made a silly mistake what do u think of this solution. From Newton's formulae it follows that the characteristic polynomial of A is x^n+d, where d is nonzero, and thus A has n distinct eigenvalues.


In how many ways can one arrange the numbers: 21,31,41,51,61,71 and 81 such that the sum of every four consecutive numbers is divisible by 3?


Χμ... Πριν λίγο είδα ότι η ΕΜΕ έβαλε τα θέματα και είδα των μικρών αρχικά...

Η ανισότητα είναι αρκετά απλή για 3ο θέμα ή μου φάνηκε εμένα ;  :huh:

Για $ a,b,c > 0$, ν.δ.ο.

$ \frac {(ab)^2}{a \plus{} b} \plus{} \frac {(bc)^2}{b \plus{} c} \plus{} \frac {(ca)^2}{c \plus{} a}\leq\frac {a^3 \plus{} b^3 \plus{} c^3}{2}$

Βάζω μια πρόχειρη λύση :

$ a \plus{} b\geq 2\sqrt {ab}$

Έτσι

$ RHS\leq\frac {(ab)^2}{2\sqrt {ab}} \plus{} \frac {(bc)^2}{2\sqrt {bc}} \plus{} \frac {(ca)^2}{2\sqrt {bc}} \equal{} \frac {ab\sqrt {ab} \plus{} bc\sqrt {bc} \plus{} ca\sqrt {ca}}{2}$

Όμως

$ \frac{a^3 \plus{} b^3 \plus{} c^3}{2} \equal{} \frac{(a\sqrt {a})^2 \plus{} (b\sqrt {b})^2 \plus{} (c\sqrt {c})^2}{2}$

και από την $ x^2 \plus{} y^2 \plus{} z^2\geq xy \plus{} yz \plus{} zx$

προκύπτει το ζητούμενο .

Χάνω κάπου ;

Στέλιος

Όχι σωστός είσαι. :wink:  Κι εγώ την είδα την ώρα του διαγωνισμού από ένα μικρό που καθόταν δίπλα και την έλυσα σε 2 λεπτά πριν ασχοληθώ (αποτυχώς) με τα θέματα των μεγάλων.  :rotfl: Ήταν όντως εύκολα τα θέματα των μικρών...

Εγω την ελυσα σε 1 ενα λεπτο  :lol:

Παντως την λυση εδωσε ο r_boris εδω [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1102051#1102051[/url]

Για εσας ειναι σιγουρα ενα πολυ ευκολο θεμα, εμενα ομως που τωρα αρχισα να ασχολουμε με εξωσχολικα μαθηματικα μου φανηκε κανονικο οπως και το 4ο θεμα των μικρων :huh: . Οριστε η λυση μου για το 3ο θεμα:

Η ανισωτητα $ xy(x \plus{} y) \leq x^3 \plus{} y^3 (1)$ ισχυει για καθε 2 πραγματικους αριθμους $ x, y$ αφου για $ (x \plus{} y)$ διαφορο του μηδεν ειναι ισοδυναμη με την $ (x \minus{} y)^2 \geq 0$, ενω για $ x \plus{} y \equal{} 0$ ισχυει η ισοτητα.

Ακομα απο την ανισωτητα αριθμιτικου - γεωμετρικου μεσου ευκολα προκειπτει οτι $ (x \plus{} y)^2 \geq 4xy (2)$, και για $ x, y$ θετικους εχουμε $ xy > 0$ και $ (x \plus{} y) > 0$,  αρα και $ \frac {xy}{x \plus{} y} > 0$, οποτε πολλαπλασιαζοντας και τα 2 μελη της (2) με $ \frac {xy}{x \plus{} y}$ λαμβανουμε
$ xy(x \plus{} y) \geq 4xy{\frac {xy}{x \plus{} y}} < \equal{} > xy(x \plus{} y) \geq \frac {4(xy)^2}{x \plus{} y} (3)$

Απο (1) και (3) προκειπτει οτι $ \frac {4(xy)^2}{x \plus{} y} \leq x^3 \plus{} y^3$ για καθε 2 θετικους αριθμους $ x,y$. Ετσι για τους θετικους αριθμους $ a, b, c$ εχουμε $ \frac {4(ab)^2}{a \plus{} b} \leq a^3 \plus{} b^3 (4) , \frac {4(ac)^2}{a \plus{} c} \leq a^3 \plus{} c^3 (5)$ , και $ \frac {4(bc)^2}{b \plus{} c} \leq b^3 \plus{} c^3 (6)$

Με προσθεση κατα μελη των $ (4), (5)$ και $ (6)$, λαμβανουμε:

$ \frac {4(ab)^2}{a \plus{} b} \plus{} \frac {4(ac)^2}{a \plus{} c} \plus{} \frac {4(bc)^2}{b \plus{} c} \leq 2(a^3 \plus{} b^3 \plus{} c^3) (7)$

και διαιρωντας και τα 2 μελη της $ (7)$ με το 4 εχουμε:

$ \frac {(ab)^2}{a \plus{} b} \plus{} \frac {(ac)^2}{a \plus{} c} \plus{} \frac {(bc)^2}{b \plus{} c} \leq \frac {a^3 \plus{} b^3 \plus{} c^3}{2}$

Ειναι σωστη η λυση η εχω κανει πουθενα κανενα λαθος;


i am trying to train my school for mathcounts since my teachers arnt doing much so im making a powerpoint and although i might not complete it this year ill definetly finish it over summer tpo have it ready for next year and if anyone reading this is from longfellow i will make sure u never get first again at mathcounts >:O (dont worry im not psychotic).
http://www.artofproblemsolving.com/Forum/weblog.php?w=1211

It's really hard to read your post if it's only one long, run-on sentence and every word is shortened.

What's the powerpoint about?

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779[/url].  If you insist on making more posts of this quality, I'll probably just start deleting them.

[quote]I want to help my school train for MathCounts. My teachers aren't doing too much, so I want to make a PowerPoint. I might not finish it this year, but I'll definitely finish it by the end of the summer and have it ready for next year.[/quote]

What do you think you'll accomplish by making a PowerPoint? I can see that you don't have much of a sense of direction because you did not tell us even vaguely what your PowerPoint will be about, or the goals of your PowerPoint, nor have you asked us about either of these.

[b]If you spend that same amount of time preparing for MathCounts yourself, you will do much better, thereby improving your team's total score. You will have a much more significant effect on your team's score by doing this than than by putting together a PowerPoint without a sense of direction.[/b]

Firstly, what do you intend to accomplish by posting this in this forum?

It looks like the PowerPoint presentation is supposed to be used for instruction, or something of an all-things-MATHCOUNTS reference/tutorial. I will assume this in the remainder of this post.

[quote="peregrinefalcon88"]i am trying to train my school for mathcounts[/quote]

It seems more time was put into the creating PowerPoint than into creating this topic. I hope your efforts won't be wasted... at the very least teaching others forces the teacher to have a thorough understanding, and can enhance a teacher's skill in a subject through new perspectives. However, there are quite a few things which should be taken into consideration. Creating a comprehensive curriculum for any subject is huge task, much bigger than I first thought (I once tried to do something similar). Then to teach it is yet another big endeavor - one that tends to fail when the targeted students lack motivation. People spend their entire lives doing this, mind you. Before proceeding, think of what you ultimately want to do with this, and if it is feasible.

People I didnt tell u what it was about so you'd be curious and go figure it out yourself. JBL i'm sorry about the post I didn't know it was wrong but I just wanted to see how I was doing on the powerpoint. TZF its nice to make a ppt because its relaxing and gives me something to present as opposed to standing there just talking and when I did the bases persentation it worked out very good. I like to make my powerpionts look fancy so i can have some fun while making them and people think they're really cool.

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding yours.[/color]

I aw your power point...but there are a lot of spelling mistakes... (sumation ?)

As I feared, the OP's powerpoint skills exceed his mathematical skills :(. For instance, slide 18 currently reads
[quote]
To find the sum of an arithmetic series such as you simply add the first and last terms and multiply by the number of terms there are. For example to sum 5, 10, 15, 20, 25, ……… 355 we have (5+355)(71)
[/quote]
in big black letters on the nice red background :P.

So, alas, I have to agree with TZF.

well i cant say that i really like making the powerpoint or that im good at grammar but i just make it to try and help out my school because im the only aopser and im trying to get others to become aopsers also and fedja thank you i see i forgot to divide by two in that part of the ppt (which is why i would like people to review it with me) but if we could all continue this thread on my blog comments i think it would be better because i dont want to keep returning to this forum thread after the mod said i wasnt supposed to post this here


algebra
polynomial
algorithm
calculus
superior algebra
superior algebra solved

Here are some problems from the GM 3/2000 :

1) Compute all the matrices A in M_2(Z) with the property that there exists n natural n>=2 so that:
A^(n+1)-nA=B.

where B=
2 2
2 2

2)Let f be in R[x] a polinomial (the polinomial is not constant). Prove the equivalence if the statements:
i) f doesn't have real roots;
ii)for any A in M_2(R), det(f(A))=0 iff f(A)=O2.

3)Compute the det:
D=
1/k       1/(k+1) ... 1/n
1/(k+1)   1/(k+2) ... 1/(n+1)
............................
1/n       1/(n+1) ... 1/(2n-k)
where k and N are natural, k<=n.

For the third:

Look at this determinant:
det(A) where A=(a_i,j) and a_i,j=1/(x_i+y_j). It is easier to compute this det. It can be computed using polynomial methods, but it does not have a nice form. It belongs to Cauchy. If you can't compute it look for the solution in Knuth's "The Art of Computer Programming - Fundamental Algorithms" (it has a lot of math in it).

For the first.

AB=BA. By writing A=
x y
z t
we get x=t and y=z, so A has the form
x y
y x
It is easy to see that A k has the form
x_k y_k
y_k x_k
and
x_(k+1)=x*x_k+y*y_k
y_(k+1)=y*x_k+x*y_k
Denote s_k=x_k+y_k and s=x+y.
Obviously s_k=s k.

Our problem turns to finding all x,y such that
x_(n+1)-nx=2
y_(n+1)-ny=2.
By adding these two we get s n+1-ns=4. In integers s and n with n \geq 2, this equation has the solutions: s=-1 n=3 and s=2 n=2.

If s=-1 then y=-1-x and x_(k+1)=x*x_k+(-1-x)*((-1) k-x_k)=(2x+1)*x_k+(-1) k+1+x.
Since n=3 we must have x_4=3x+2. But x_4=(2x+1)*x_3-1+x, so
(2x+1)*x_3=2x+3. This should be easy to solve.
Do smth similar 4 s=n=2.

for the third one
let A \in Mp(R).if A=(a_ij) with a_ij=1/(x_i+y_j)
we have that
det(A)=V(x1,x2,...,xp)*V(y1,y2,...,yp)/ \pi {i=1,j=1}(xi+yj), where V(x,y,z,...) is a van der Mont determinant...
which is not hard to proove... but it's a lot of calculus :D


let $P(x)$ be a polynomial with degree $3n$ such that:
$P(0)=P(3)=...=P(3n)=2$
$P(1)=P(4)=...=P(3n-2)=1$
$P(2)=P(5)=...=P(3n-1)=0$
and we know $P(3n+1)=730$
find all possible values of $n$
It is for years ago but i have got stuck with this. help is desparetly needed.

Please solve this

i have done what ever i can but i dont know, how on the earth can we find out the $n$?

Can someone find the solution for it in usa olympiad archives or something?

huh ?  :huh:
Lets take n=1 then P(3*1+1)=730
Now take n=2 then P(3*2-2)=1
But 1 =/= 730 ...

Am I missing something ?

You have shown that $n \neq 2$.

http://www.kalva.demon.co.uk/usa/usa84.html

I don't quite understand the problem  :D
(but to say true I don't give a d***  )


Engel mistakenly asserts the identity

$ \sum_{k \equal{} 1}^{\infty} \frac {1}{F_k} \equal{} 4 \minus{} \phi$

where $ F_n$ is the Fibonacci sequence and $ \phi$ is the golden ratio.  The value of this sum is in fact [url=http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html]not known[/url] (or at least not known in a form this neat).  However, can you figure out a reciprocal identity involving the Fibonacci numbers whose value is actually $ 4 \minus{} \phi$?

(I do not know the answer.  I also have no idea why such a blatant error is in Engel.)

Engel almost certainly means the so-called "Millin series":
[url]http://mathworld.wolfram.com/MillinSeries.html[/url]

While we are on the subject, here is another error I found.  On page 184, Engel asks to prove the inequality
\[ \sqrt {xy} < \frac {x \minus{} y}{\ln x \minus{} \ln y} < \frac {x \plus{} y}{2},
\]
where $ x > y > 0$.

In the solution on page 197, using a geometric argument, Engel derives the inequality
\[ \frac {2(x \minus{} y)}{x \plus{} y} < \ln x \minus{} \ln y,
\]
which gives the right inequality.  However, he also derives the inequality
\[ \frac {x \minus{} y}{\sqrt {xy}} < \ln x \minus{} \ln y,
\]
which is the opposite of what he wants!  So my challenge to anyone is to fix the argument.

I think this works to prove the left inequality . . .
[hide]
First, we find the minimum of

$ \sqrt{z}-\sqrt{\frac{1}{z}}-\ln(z)$.

Differentiating, we want

$ \frac{1}{2}z^{-1/2}+\frac{1}{2}z^{-3/2}-\frac{1}{z}=0$,

so we can multiply through by $ 2z^{3/2}$ to get

$ z-2\sqrt{z}+1=0$,

which clearly has a double root at $ 1$.  The original expression is equal to $ 0$ at $ z=1$, so $ 0$ is its minimum. We see that the expression is $ >0$ for $ z>1$.  Thus, substituting $ \frac{x}{y}$ for $ z$, we find that for $ x>y$,

$ ln(\frac{x}{y})<\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}$.

So

$ \ln x-\ln y< \frac{x-y}{\sqrt{xy}}$

as desired?
[/hide]


In 4 liter of water there are 200 gr $ NaCl$ .What is the molar concentration of $ NaCl$ in the water?

The molar concentration is the number of moles in 1 liter

Well $ \mbox{Molarity}\equal{}\frac{\mbox{mols}}{\mbox{Volume}}$ if I can recall correctly.  I don't have a periodic table in front of me so I don't know the molar mass of $ \mbox{NaCl}$ but once you know it, you just divide grams by molar mass to get mols and then divide by the 4 liters.

For this I'm only going to use $ 5$ significant figures to describe the numbers. According to my periodic table, $ \text{1 mol of NaCl \equal{} 58.442 g}$.
$ \text {200 g of NaCl}\times{\frac {\text{1 mol}}{\text{58.442 g of NaCl}}} \equal{} \text {3.4222 mols of NaCl}$
Even though this can be simplified earlier, we now divide $ \text{3.4222 mols of NaCl}$ by $ \text{4 liters}$, giving us the molar concentration of $ \text{0.85555 mols of NaCl per liter}$ of water.


probability geometry trigonometry rectangle inequalities LaTeX Support

The Mock AMC contest will take place on Friday, July 23rd, at 3:00 PM Eastern time. If you cannot make this date, the problems will still be left up in this forum, so you may take the contest whenever you want, albeit "unofficially". You can PM me your answers and I will send you back your score (or you can check your score on your own after consulting the answers), but unfortunately I will not be able to add it to the high score list, should it qualify.

The rules of this contest are the same as those of the AMC 10/12. The test contains 25 multiple-choice questions in approximate increasing order of difficulty. Each question has five answer choices, and only one of these is the correct answer. You will have 75 minutes to complete the test. Each correct answer is worth 6 points, each skipped question is worth 2.5 points, and each incorrect answer is worth 0 points. Therefore, it is [i]not[/i] to your advantage to guess on a question unless you have eliminated a significant number of answer choices. [url=http://www.collegeboard.com/counselors/hs/sat/registration/calc.html\]Any calculator acceptable for use on the SAT[/url] may be used for this contest. You may not use outside resources such as books, the internet, etc., but you may use graph paper, a ruler, compass, and a protractor.

At 3:00 PM Eastern time, I will post the problems in an announcement entitled "Problems of the Mock AMC Contest" in the AMC forum. The contest will end at precisely 4:15 PM Eastern time. In case your clock is a bit off, set a timer for 75 minutes and start it at the time that the questions are posted. Immediately after the contest is over, please PM me your answers (no solutions are necessary; all you need to enter on the PM is the letter corresponding your answer). The best way to do this would be to open a new PM before the contest and type in the following:

1.
2.
3.

...all the way to 25. As you work on the contest, enter your answers on the PM. That way, as soon as 75 minutes are up, you will have all your answers ready to be sent. I will allow a few minutes' buffer time for slow modems, etc., but not more than that, so please be prompt!

After I have graded the contest and sent back individual scores, I will post (in spoiler) the correct answers on the forum, and solutions may be discussed (but please do not discuss solutions in the thread with the problems, so as not to ruin the contest for the unofficial test takers).  I will also post a list of high scorers (all scorers of 100+ or the top 10 scorers, whichever would have more) on the forum.

If you have any questions, feel free to ask! Good luck to everyone!  ;)

I assume, that next to the 1, for example, there will be (a) or (b) ..... (e)?

JSRosen3 wrote:Here is a practice question:

A weighted coin has a [tex]\frac{2}{5}[/tex] chance of coming up heads and a [tex]\frac{3}{5}[/tex] chance of coming up tails. Given that the probability of getting exactly 4 heads out of 5 tosses is [tex]\frac{m}{n}[/tex], with m and n relatively prime positive integers, find [tex]m+n[/tex].

(A) 141 (B) 173 (C) 673 (D) 1202 (E) 1242

Spoiler:

[hide]

C.

2/5*2/5*2/5*2/5*3/5+2/5*2/5*2/5*3/5... there are 5 ways to order 2/5*2/5*2/5*2/5*3/5, so the probability is 5*2/5*2/5*2/5*2/5*3/5 = 48/625.  48 + 625 = 673.[/hide]

Approximately what number would that practice question be on the test?

Probably about #11 or #12. Hard to say for sure.

And yes, Rep, there will be plenty of harder questions.

Another few sample questions:

(approx. difficulty: #4) If [tex]2^{1000}-2^{999}-2^{998}+2^{997}=q*2^{997}[/tex], find the value of [tex]q[/tex].

(A) 1  (B) 2  (C) 3  (D) 4  (E) 5

(approx. difficulty: #13) Find the largest integer [tex]n[/tex] with [tex]0<n \leq 96[/tex] such that the triangle with sides [tex]n[/tex], [tex]2n-3[/tex], and [tex]3n-9[/tex] has integer area.

(A) 72  (B) 74  (C) 76  (D) 78  (E) 80

(approx. difficulty: #15) In quadrilateral [tex]ABCD[/tex], it is given that [tex] \angle A = 120 \degrees[/tex], angles [tex]B[/tex] and [tex]D[/tex] are right angles, [tex]AB=13[/tex], and [tex]AD=46[/tex]. Then [tex]AC[/tex] equals

(A) 60  (B) 62  (C) 64  (D) 65  (E) 72

(approx. difficulty: #18) A point (x,y) in the plane is called a [i]lattice point[/i] if both x and y are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to

(A) 4.0  (B) 4.2  (C) 4.5  (D) 5.0  (E) 5.6

Very nice questions!  I loved #3.  I used to hate geometry but now I love it!

3.  [hide]

Notice something very important: the quadrilateral is cyclic.  So I drew the quadrilateral in a cirlce.  Some observations:

If O is the center of the circle:

1.  AC = 2OA = 2OB = 2OC = 2OD

2.  Angle c = 60, so arc DAB is 120.  Thus angle DOB is 120

By the law of cosines, DB^2 = 2883.  Now since angld DAB = angle DOB and OD = OB = x, then 2883 = x^2 + x^2 - 2*x*x*cos 120 = 3x^2.  Solving for x gives x = 31.  So AC = 2*x = 62.[/hide]

Fun fun fun!

Here's number 1

[hide]C.  (This is going to be so embarassing if it's not C...)  Well I just changed the problem up a bit since 2^1000 is just too big.

Let's rewrite the problem like this:

2^5-2^4-2^3+2^2=q*2^2

32-16-8+4=q*4

12=4q

q=3

As for the geometry question, is there another way to solve the problem without identifying the quadrilateral as cyclic and using the law of cosines?  Surely the number 15 question could be solved in another method that does not involve using the law of cosines or cyclic quadrilaterals, both of which I am unfamiliar with.  Thanks!  (I spent 20 minutes trying to solve that geometry problem.  All I could do was extend some of the segments so we get a rectangle with a couple of 30-60-90 triangles, but that didn't seem to help)[/hide]

NightFlarer wrote:As for the geometry question, is there another way to solve the problem without identifying the quadrilateral as cyclic and using the law of cosines?  Surely the number 15 question could be solved in another method that does not involve using the law of cosines or cyclic quadrilaterals, both of which I am unfamiliar with.  Thanks!  (I spent 20 minutes trying to solve that geometry problem.  All I could do was extend some of the segments so we get a rectangle with a couple of 30-60-90 triangles, but that didn't seem to help)

Probably.  I'm probably just missing something stupidly obvious like I always do.

By the way:  #2 is [hide]B[/hide], but I'm trying to find a better method than the one I used.

EDIT:  Aha!  [hide]

2.  If a, b, and c are the sides of a triangle and s is the semiperimiter, then the area is sqrt[s(s-a)(s-b)(s-c)].  So the area of our triangle is

A = sqrt[(3n-6)(2n-6)(n-3)(3)]

   = sqrt[3*3*2*(n-2)(n-3)^2]

   = 3*(n-3)*sqrt[2]*sqrt[n-2]

Thus the problem reduces to:

Find the largest integer n<= 96 such that 2(n-2) is a perfect square.

144/2 +2= 74.  The next even perfect square is 196.  196/2 + 2 = 100, which is too big.  Thus our answer is 74.

[/hide]

Again, I probably did more than I needed to.

JSRosen3 wrote:Yes, when 75 minutes are up you should have the PM filled out with a, b, c, d, or e next to all the questions you wish to answer.

Here is a practice question:

A weighted coin has a [tex]\frac{2}{5}[/tex] chance of coming up heads and a [tex]\frac{3}{5}[/tex] chance of coming up tails. Given that the probability of getting exactly 4 heads out of 5 tosses is [tex]\frac{m}{n}[/tex], with m and n relatively prime positive integers, find [tex]m+n[/tex].

(A) 141 (B) 173 (C) 673 (D) 1202 (E) 1242

Easier way:

[[hide]spoiler]Use binomial probabilities. The probability of an event with probability x happening y times in z trials is yCz*x^y*(1-x)^(z-y). So, substituting, 4C5*(2/5)^4*(3/5)^1. It simplifies to 5*(16/625)*(3/5). This is nothing but 48/625. Since this is in the form of the fraction m/n, m=48, n=625, and n+m=673. Therefore the answer is (C).[/hide][/hide]

[quote="samus"]nr1337, how do you know that the quadrilateral is cyclic, and if so, what are some propertis you go from that observation?[/quote]

The quadrilateral is cyclic because B+D=A+C=90+90=120+60=180.  I listed the observations/properties in my solution.

[color=red]Or for the geometry problem, extend sides AB and CD to intersect at E to form [b]two[/b] 30-60-90 right triangles, one will be ADE and the other one is BCE.  From there it's easy.[/color]

Question:

Can we use the Windows calculator for the test? I don't think it has anything the TI-83 doesn't.

Is signing up for this contest necessary?  I just told some people about this contest and they might not get to it...

Also, will general statistics be posted? (overall mean, median score, etc.)

If your going to have statistics, and if there are enough people, is it possible to post high scorers by age group?

I doubt there would be enough people though.

For the question 4 (the one with lattice points),
I think the answer is (C)4.5.

Again, I spent less than 2 minute on this problem, so I think someone come up with better answer.

If somebody says I am right, then maybe I will explain how I got this answer

Answers to questions:

yif man12 is correct on #4--the answer is D, 5.0. The area is in fact exactly equal to 5.0.

You may use the windows calculator for the contest.

You do NOT need to be signed up to take the test. As long as I receive a PM with answers immediately after 75 minutes are up, I will score you in the official category.

As for statistics, average total score and average score per question will be posted, as well as high scores. I can't do a high score per age group because I don't know everyone's age. If you have any further requests for statistics, please let me know!

Also, as a note, NO DISCUSSION OF THE QUESTIONS IS ALLOWED UNTIL THE ANSWERS ARE POSTED.

Thanks! Good luck to everyone! Problems will be posted in 27 minutes.

Shoot, I just realized that I am in California.

It is almost 12:00 here.

[quote="Ultimatemathgeek"][quote="crayshot97"]Beta, your answer for the sample problem needs some refinements,

like you said, the mean of the series once a number is taken out is (n(n+1)-2k)/(2n-2), so k=(17n*2-1187n+1204)/34. However, if k is in the series, 1 :le: k :le: n. [color=red]Solving the 2 inequalities gives n :ge: 69, n :le: 70, or n=1. n=1 is obviously not feasible[/color], so n=69 or n=70. Of the 2, only 69 gives an integer value for k, so therefore k=7 and that is the only answer

You forgot the restrictions.[/quote]

What 2 inequalities did you use to get n :ge: 69, n :le: 70, or n=1?[/quote]

Well you basically have the 2 inequalities that k :ge: 1 and k :le: n. These two are both quadratic inequalities in n. The first gives you n :le: 1 and n :ge: 69. The second, where k :le: n gives 1 :le: k :le: 70. These is where you get n=69,70, or 1.

Any shot of us over-loading the AMC Forums today?  :P
Good luck everyone!

It's more AMC 12 level than AMC 11 level, I'd say. It turned out a little harder than I expected it would be. Don't worry--if most of the scores are low, I will lower the "AIME Qualifying Score" (which I put in quotes because there is no Mock AIME (unless an Advanced solver wants to make one (hint hint...(mmmm....nested parentheses are fun...)))) Don't be discouraged by too hard a test...if you work on the problems later, outside of the time limit, you may be able to solve a lot more than you could in the time limit.

When will the results be available (approximately)?

[quote="crayshot97"]

Well you basically have the 2 inequalities that k :ge: 1 and k :le: n. These two are both quadratic inequalities in n. The first gives you n :le: 1 and n :ge: 69. The second, where k :le: n gives 1 :le: k :le: 70. These is where you get n=69,70, or 1.[/quote]

Man, I am lost.  Am I stupid or what?  Or is there something obvious that I am missing?


modular arithmetic
number theory unsolved
number theory

Is it true that for k with k dividing $\varphi(p^{a})$
that $a^{k}\equiv 1 \mod{p^{a}}$  has exactly k solutions for p not being 2?????????

Are we to understand that [b]for each [/b]p>2 there exist exactly k integers a such that:

$k|\varphi(p^{a})$

and

$a^{k}\equiv 1 \mod{p^{a}}$

I apolgise if I have misunderstood the question, just wanted to make sure

Not $k$ integers, but $k$ different residues $b \mod p^{a}$ with $b^{k}\equiv 1 \mod p^{a}$ (he just used $a$ twice).
This follows from the existence of primitive roots $\mod p^{a}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=55473 ) for example.

Hi Fabian, nice to see you on ML!

Well, I assume that the $a$ in $p^{a}$ is another $a$ then the $a$ in $a^{k}$, otherwise the problem would not make much sense :D

Let's rephrase the problem:
[color=brown]Let $p$ be an odd prime and $a$ be a positive integer. Furthermore let $k$ be a divisor of $\varphi(p^{a})$. Then the congruence $x^{k}\equiv 1 \pmod{p^{a}}$ has exactly $k$ pairwise incongruent solutions in $x$.[/color]

Proof: Let $g$ be a primitive root modulo $p^{a}$. Then $x\equiv g^{y}$ for some $y \in \{1,\ldots,\varphi(p^{a})\}$.
We have
$x^{k}\equiv 1 \pmod{p^{a}}\Leftrightarrow g^{yk}\equiv g^{\varphi(p^{a})}\pmod{p^{a}}\Leftrightarrow yk\equiv \varphi(p^{a}) \pmod{\varphi(p^{a})}$.
Hence the number of solutions $x \mod p^{a}$ is equal to the number of solutions in $y\mod{\varphi(p^{a})}$. But $yk\equiv \varphi(p^{a}) \pmod{\varphi(p^{a})}$ has exactly $k$ solutions in $y$ since $gcd(k, \varphi(p^{a}))=k$ and $k\mid \varphi(p^{a})$ by definition.
$\Box$

Here we use the fact that $ax\equiv b \pmod{c}$ has solutions in $x$ iff $d=gcd(a,c)\mid b$ and that if it has solutions, the number of solutions is $d$.

Edit: ZetaX was faster.....once again :P

[quote="Fabian May"]danke recht schön, yimin.  ja, hab mich tatsächlich auf mathlinks verirrt.
also, bau das nämlich in mein spezialgebiet ein und bin auf den beweis nicht draufgekommen.
dann, gute nacht[/quote]

English here please. If not, use PM's.


This is from Singapore Math:

Peter and Paul each had an equal amount of money. Each day Peter spent 36 dollars
and Paul spent 48 dollars. When Paul used up all his money, Peter still had 240 dollars left.
How much money did each of them have at first?

I just stated that Peter saved 12 bucks everyday compared to Paul, and hence, Paul spent money for 20 days, and 20 x 48 is 960.  I was just wondering if I can only make this assumption/statement because they started off with the same amount.  Also, any other ways to solve?

Yes. That's correct. You can only make that assumption when you have the same starting amount.


I have this one math problem that my friend gave to me as a joke about an hour ago.

What is the gradient of the function y = [2x(x^2+3x-4)+4x-4]/[x-1] when x = 2?

So far I think I got this, but it's probably wrong.

[(4x^8+12x-16)+8-4]
[(520)+8-4]
[525] = y

I need someone to help me out with the answer, and the steps you took to get it.

[quote="LJ6580"]
What is the gradient of the function y = [2x(x^2+3x-4)+4x-4]/[x-1] when x = 2?

[(4x^8+12x-16)+8-4]
[(520)+8-4]
[525] = y
[/quote]
Your $ 4x^8$ term should be $ 4x^2$ instead.
Answer:
$ \frac{2x(x^2\plus{}3x\minus{}4)\plus{}4x\minus{}4}{x\minus{}1}\equal{} \frac{(2x)(x\plus{}4)(x\minus{}1)\plus{}4(x\minus{}1)}{x\minus{}1}\equal{} 2x(x\plus{}4)\plus{}4\\
2(2)(2\plus{}4)\plus{}4\equal{}4\cdot6\plus{}4\equal{}28$

How do I get the EXACT gradient, not the average?


function
limit
calculus
calculus computations

Can anyone help me set up this problem.

A pond initially contains 1,000,000 gal of water and an unknown amount of undesirable chemical. Water containing 0.01 gram of this chemical per gallon flows into the pond at a rate of 300 gal/hr. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond.

a). Write a differential equation for the amount of chemical in the pond at any time. I am confused with the units, including gal and grams. Am I supposed to work with grams or concentration?

Any help would be greatly appreciated. THx

My suggestion: make your main function, $y(t),$ be the mass of the the contaminant in the whole pond, measured in grams.

The concentration in the pond at any time is then $10^{-6}y(t),$ with the units of measurement there being grams per gallon. (Yes, that's a truly weird unit of concentration - but I've actually seen it. Just go along with it.)

Now, what happens during a small interval of time $\Delta t?$ (Use hours as the unit of measurement for $\Delta t.$)

The amount of the chemical flowing into the pond is $300\cdot 0.01\cdot \Delta t.$ The units of measurement there are

$\frac{\text{gal}}{\text{hr}}\cdot\frac{\text{gram}}{\text{gal} }\cdot \text{hr}=\text{gram}$

Now, what is the amount flowing out of the pond? The volume of water flowing out is $300\cdot\Delta t.$ The concetnration varies with time, but taking $\Delta t$ to be a very short interval, we approximate it as $10^{-6}\cdot y(t).$ Then the amount flowing out is approximately $10^{-6}\cdot y(t)\cdot 300\cdot\Delta t.$ Once again the units work out to be grams. So:

$y'(t)\cdot\Delta t\approx 300\cdot 0.01\cdot \Delta t-10^{-6}\cdot y(t)\cdot 300\cdot\Delta t$

$y'(t)=3-(3\times 10^{-4})y(t).$

I'll leave you to solve that equation yourself. It is both linear and separable and hence can be solved in more than one way.

Two questions:

1. What is the steady-state condition of the pond? In particular, what is $\lim_{t\to\infty}y(t)$ (or the concentration you could derive from that)? Hint: the "common sense" value is the correct value, and could be seen in this autonomous differential equation without bothering to find the solution.

2. What is the time scale? Here's an interesting alternative: imagine setting up a movable impermeable membrane between the inlet and the outlet so that the input solution never mixes with the original contents of the pond at all. It would take $\frac{10^6}{300}$ hours to completely replace the original contents of the pond. So that - about 333 hours - is some kind of fundamental time scale here. Now for my question: for the mixing process, what is the half-life? (Yes, that is a meaningful question, if you look at it correctly.)


geometry
trapezoid
circumcircle
geometry open

In trapezoid $ ABCD$, $ AD\parallel BC$. $ K$ on $ AB$ and $ L$ on $ CD$, so that $ \angle ABL \equal{} \angle CDK$. Prove that $ \angle ALB \equal{} \angle CKD$.

I've posted this problem in Pre-Olympiad forum, but since no one are keen to answer my problem, I decided to put it here

Thanks

I think that the problem is not true in general.

Because of $ \angle ABL \equal{} \angle CDK,$ we have that the circumcircles $ (O),\ (O'),$ of the triangles $ \bigtriangleup BKL,\ \bigtriangleup DKL$ respectively, are congruent.

It is easy to show that $ \angle ALB \equal{} \angle CKD,$ in the particular case when the vertices $ C,\ A,$ of the given trapezium $ ABCD,$ lie on the circles $ (O),\ (O'),$ respectively.

Kostas Vittas.


P(x) is a polinomyal with integers coefficients such that $P(n)>n$ ; $n \in Z$ , for each positive integer m, there's a term of the serie:

$P(1),P(P(1)),P(P(P(1))),...$

which is divisible by m.
Show that $P(x)= x+1$

$f(x)=P(x)-x$ is a polynomial of the same degree $d$ as $P(x)$.

Put $N=P(P(P(\dots(P(a))\dots)), m=f(N)$, where $P$ is applied $k$ times. Then $P(N)$ equals $N$ modulo $m$, so $P(P(n))$ equals $N$ modulo $m$, etc. Hence our sequence contains at most $k$ different residues modulo $m$. But $m>k$ for large $k$, the onliest exception is $P(x)=x+1$.

¿Puedes traducirlo al español???? >>>>>Esto para Mario-Peru
Yo no se mucho ingles y no entiendo muy bien la solucion :blush:

Por favor :oops:

$f(x)=P(x)-x$ es un polinomio del mismo grado $d$ que $P(x)$.

Ponemos $N=P(P(P(\dots(P(a))\dots)), m=f(N)$, donde $P$ es aplicado $k$ veces. Luego $P(N)$ es congruente con $N$ en módulo $m$, entonces $P(P(n))$ es congruente con $N$ en módulo $m$, etc. Entonces nuestra sucesión contiene como máximo $k$ residuos diferentes en módulo $m$. Pero $m>k$, para $k$ mayor, la única excepción es $P(x)=x+1$.

Gracias,no se mucho ingles ,recien estoy aprendiendo algo..... :blush:


Here is my solution to the second problem:

Verify that the number of such sequences is the number of paths from the leftmost $ 0$ to the rightmost $ 0$ in the following figure:

[code]
 0   0   0   0 . . . . .    0   0
   1   1   1  . . . . .   1   1
     2   2   2 . . . .  2   2
       3   3  . . .   3   3
         4   4. . . 4   4
           5  . . .   5
              . . .
                n     [/code]

where from each of the entries we may reach the next column either to the top-right or to the bottom-right entry. Here the $ i$th column contains all the possible values of $ a_i$.

Consider the following configuration of the above figure:

[code]
                    0
                 0  1
              0  1  2
                    .
                    .
      0  1  2 . . .n-2
   0  1  2  3 . . .n-1
0  1  2  3  4 . . . n
[/code]

where each of the numbers represents a lattice point in the region $ 0\le y\le x\le n$. Hence our problem is equivalent to finding the number of lattice paths from $ (0,0)$ to $ (n,n)$ which do not cross the line $ y \equal{} x$ and from a point we may go either up or right. This is the $ n$-th Catalan number $ \frac {1}{n \plus{} 1}\binom{2n}{n}$, which is well known. Q.E.D.

Yeah, that was all I got as well (with the same solution via the standard Catalan path count), though admittedely I was more focussed on other things (school exams and music stuff) the last couple of weeks.

Would the marks have come from showing it is equivalent to the standard catalan path, or would the marks be for the proof that they count to catalan numbers?

I messaged Valentin about this, and he assured me that just showing it was equivalent to the standard Catalan (going up in lattice points such that x>=y) and then quoting the formula was enough.


[quote="I Hate Watermelons"]Okay, so I got x = -17, which makes the statement true (at least it seems to me).

x - |x+4| = ±30
x+4 = -(x + 30)
x+4 = -x-30
x = -17

-----

x - |x+4| = ±30
x+4 = -(x - 30)
x+4 = -x+30
x = 13

13 does not make the statement true. :| Why is that? Can someone explain this please? Also, is -17 the only solution?[/quote]

$ |x|$----> if $ x\geq 0$---> $ |x|\equal{}x$
        ---->  if $ x<0$ ---> $ |x|\equal{}\minus{}x$

$ |x \minus{} |x\plus{}4\parallel{} \equal{} 30$

a) $ x\plus{}4\geq 0$-----> $ x\geq \minus{}4$
$ |x \minus{} (x\plus{}4)| \equal{} 30$---> $ |x \minus{} x\minus{}4| \equal{} 30$-----> $ |4| \equal{} 30$ ----> $ \phi$

$ x\plus{}4\leq 0$-----> $ x\leq \minus{}4$
$ |x \plus{} (x\plus{}4)| \equal{} 30$----> $ |2x\plus{}4| \equal{} 30$---> $ x\leq \minus{}4$---> $ \minus{}(2x\plus{}4)\equal{}30$----> $ \minus{}2x\minus{}4\equal{}30$----> $ x\equal{}\minus{}17$

yes, -17 is the only solution.

if x>=-4, the value of x-|x+4| is always -4, and absolute value of that is 4.

which doesnt equal 30.

thus, we know that 13, which is larger than -4, is not the solution.


algebra unsolved
algebra
Function equations

Note: Two problems below is not "unsolved problem"  :arrow:

[b]Problem 1[/b]: Find all $f: R\to R$ sastifying $f(x^{2}+f(y))=xf(x)+y$

[b]Problem 2[/b]: Find all $f: R\to R$ sastifying $f(xf(x)+f(y))=(f(x))^{2}+y$

Enjoy!

1)Firstly, by putting $x=0$ we get $f(f(y))=y$ and $f(x)=f(y)\Rightarrow f(f(y))=f(f(x))\Rightarrow x=y$.
This means $xf(x)=f(f(x))f(x) \rightarrow f(x^{2}+f(0))=f(f(x)^{2}+f(0))\Rightarrow$
$\Rightarrow x^{2}=f(x)^{2}\Rightarrow |x|=|f(x)|$.
So $\exists \sigma: \mathbb{R}\to \{1,\;-1\}\;/\;f(x)=\sigma(x)x$.  Notice in any case $f(0)=0$. From inyectivity and by letting $y=0$ respectively we get $\sigma(-x)=\sigma(x)$ and $\sigma(x)=\sigma(x^{2})$. Now suppose $\exists\; z>y\in \mathbb{R}^{+}\;/\;\sigma(z)\neq \sigma (y)$; then by letting $x=\sqrt{z-f(y)}$ we have
$xf(x)+y=\sigma(x)x^{2}+y =f(x^{2}+f(y))=f(z)=\sigma(z)z=$
$=\sigma(z)x^{2}+\sigma(y)\sigma(z)y=\sigma(z)x^{2}-y \Rightarrow 2y=x^{2}(\sigma (z)-\sigma(x^{2}))$;   here $y>0$ implies $2y=2x^{2}\Rightarrow y=z-f(y)$; now, since $z>y$ we have actually $2y=z$; however $\sigma (y^{2})=\sigma(y)\neq \sigma (2y)=\sigma (4y^{2})$ but since $(2x)^{2}>x^{2}>0$ repeating the reasonament we reach $2x^{2}=4x^{2}\rightarrow 2=4$, a contradiction. So $\forall x,y\in \mathbb{R}^{+}\;\sigma(x)=\sigma(y)$, which means either $\forall x\in \mathbb{R}\;f(x)=x$ either $\forall x\in \mathbb{R}f(x)=-x$.
2) It seems it is indeed almost the same as the previous one, once you have noticed biyectivity. If you let $A=f^{-1}(0)$, by letting $x=A$ you get $f(f(y))=f(Af(A)+f(y))=f(A)^{2}+y=y$. And then you can make directly, I think, the previous substitution $x$ by $f(x)$ or realize this is actually the same as equation 1) using the fact $f^{-1}(x)=f(x)$.
[size=59]But this is wrong.[/size]


The Fibonacci Series has the generating function

$\frac{x}{1-x-x^{2}}= \sum_{k=0}^{\infty}F_{k}x^{k}$

Plug in $x = \frac{1}{10^{k}}$. ;)


geometry
circumcircle
geometric transformation
reflection
angle bisector
perpendicular bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14044[/img]

 :(

$ AH$ is the A-altitude line, perpendicular to $ AH$ through $ T_a$ is the $ a \equiv BC$ sideline, cutting $ AH$ at the foot $ D$ of the A-altitude. $ O$ is the circumcenter, $ AO, AH$ are isogonals with respect to the angle $ \angle CAB$ $ \Longrightarrow$ $ AT_a$ bisects the angle $ \angle OAH.$ Let $ M_a$ be the midpoint of $ BC.$ Then $ OM_a \equal{} \frac{AH}{2}.$ Let $ K$ be the midpoint of $ AH.$ Reflect $ h_a \equiv AH \equiv AD$ in angle bisector $ AT_a$ into $ o_a.$ Reflect $ K$ in the perpendicular bisector of $ AD$ into $ K'.$ Parallel $ k'$ to $ a \equiv T_aD$ through $ K'$ cuts $ o_a$ at $ O.$ Circle $ (O)$ with radius $ OA$ cuts $ a \equiv T_aD$ at $ B, C.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14053[/img]

 :thumbup:


Find continuous$ F: \mathbb{R} \mapsto \mathbb{R}$ satisfying
for all$ x_1, x_2  \in \mathbb{Q}$,$ x_1 \minus{} x_2  \in \mathbb{Q}$ then $ f(x_1) \minus{} f(x_2) \in \mathbb{Q}$

Let $ g_a(x)\equal{}f(x\plus{}a)\minus{}f(x),a\in Q$. Because $ g_a(x)$ continiosly and $ g_a(x)\in Q \forall x$ we get $ g_a(x)\equiv b\in Q$
Therefore $ f(x\plus{}na)\equal{}f(x)\plus{}nb, f(x\minus{}na)\equal{}f(x)\minus{}nb$.
Let $ a_1,a_2,b_1,b_2\in Q$, suth that $ g_{a_i}\equiv b_i$.Exist $ n_i\in Z$, suth that $ n_1a_1\equal{}n_2a_2\to n_1b_1\equal{}n_2b_2$.
Therefore $ \frac{b_i}{a_i}\equiv c\in Q$ and $ f(x\plus{}a)\minus{}f(x)\equal{}ac, forall a\in Q,x\in R$. Therefore $ f(r)\equal{}f(0)\plus{}rc, \forall r\in Q$,
by continiosly $ f(x)\equal{}cx\plus{}d, c\in Q,d\in R$.

[quote="Rust"]Let $ g_a(x) \equal{} f(x \plus{} a) \minus{} f(x),a\in Q$. Because $ g_a(x)$ continiosly and $ g_a(x)\in Q \forall x$ we get $ g_a(x)\equiv b\in Q$
[/quote]
Sorry, Rust But I can't understand this line  :blush:  .Can you explain more?

$ g_a(x)$ is continiosly. If $ g_a(x_1)\equal{}b_1,g_a(x_2)\equal{}b_2>b_1$, then for any $ b\in (b_1,b_2)$ exist x, suth that $ g_a(x)\equal{}b$.
If $ g_a(x)$ is not constant, then we can chose $ b\in (b_1,b_2)$ - irrational. Therefore for any rational a $ g_a(x)\equiv b\in Q$.


function
algebra
domain
integration
complex analysis

Let $ c$ be a simple closed curve, and $ u$ be a function defined on a domain $ D$ disjoint from $ c$.  $ u$ is continuous on $ c$.
define a new function \[ h(z)\equal{}\int_c \frac{u(s)}{s\minus{}z}ds,\indent z\in D\]

Show that $ h$ is an analytic function.

We can see easily that $ h'(z)\equal{}\int_{C}\frac{u(z)}{(s\minus{}z)^2}ds$. So $ h$ is analytic.


If $ a_1,a_2,a_3$ are positive real numbers such that $ a_1\plus{}a_2\plus{}a_3\equal{}1$.
Prove that: $ \frac{a_1^2}{a_1\plus{}a_2}\plus{}\frac{a_2^2}{a_2\plus{}a_3}\plus{}\frac{a_3^2}{a_3\plus{}a_1} \ge \frac{1}{2}$

[quote="earldbest"]this is  directly from Cauchy-Schwarz.  :lol:
$ \frac {a_{1}^{2}}{a_{1} \plus{} a_{2}} \plus{} \frac {a_{2}^{2}}{a_{2} \plus{} a_{3}} \plus{} \frac {a_{3}^{2}}{a_{3} \plus{} a_{1}}\ge\frac {(a_{1} \plus{} a_{2} \plus{} a_{3})^2}{2(a_{1} \plus{} a_{2} \plus{} a_{3})}$[/quote]
- Can you explain me step by step,couse i`m having trouble to get it,thanks :)


topology
parameterization
analytic geometry
calculus
derivative
trigonometry
advanced fields

Can someone tell me how to embed the Klein bottle into R^4???
I know the intuition part...but I'm have no idea on how to figure out the parametrization of it in R^4.

Thanks alot

Represent the Klein bottle as $ \mathbb{R}^2/G$, where $ G$ is the group generated by $ (x,y)\mapsto (x\plus{}1,y)$ and $ (x,y)\mapsto (\minus{}x,y\plus{}1)$.

An embedding into $ \mathbb{R}^4$, represented as $ \mathbb{C}^2$:
$ f(x,y)\equal{}(e^{\pi iy}\sin(2\pi x),e^{2\pi iy}\cos(2\pi x))$

Symmetries of the first coordinate: $ (x,y)\mapsto (x\plus{}1,y),(x,y)\mapsto (\minus{}x,y\plus{}1)$, $ (x,y)\mapsto (\frac12\minus{}x,y)$. Symmetries of the second coordinate: $ (x,y)\mapsto (x\plus{}1,y)$, $ (x,y)\mapsto (\minus{}x,y\plus{}1)$, $ (x,y)\mapsto (\minus{}x,y)$.
The intersection is exactly the desired group.

We need both coordinates to avoid zero derivatives with respect to $ y$; $ \sin 2\pi x$ and $ \cos 2\pi x$ are never both zero.

The symmetries are the key to the construction; we need to build it with periodic functions.


mathfanatic wrote:1990 AHSME #30

If R_n = 1/2*(an + bn) where a = 3 + 2 :rt2:, b = 3 - 2 :rt2: , and n = 0, 1, 2, ...., then R_12345 is an integer.  Its units digit is

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

[hide]Yuck. This needs to be made more beautified.

a=3+2 :rt2:

a :^2: = 9 + 8 + 12 :rt2: = 17 + 12 :rt2:

a :^3: = 27 + 16 :rt2: + 72 + 54 :rt2: = 99 + 70 :rt2:

b=3-2 :rt2:

b :^2: = 9 + 8 - 12 :rt2: = 17 - 12 :rt2:

b :^3: = 27 - 54 :rt2: + 72 - 16 :rt2: = 99 - 70 :rt2:

So it's pretty obvious that the radicals cancel, and the integer terms are the same. And then we know that (1/2)(a^n + b^n) is the same as the integer part of a^n. (where I'm using "integer part" to mean the term without the radical)

n[tab]units digit of integer part of a^n

0[tab]1

1[tab]3

2[tab]7

3[tab]9

4[tab]7

5[tab]3

6[tab]1

7[tab]3

8[tab]7

9[tab]9

10[tab]7

11[tab]3

12[tab]1

Yeah, I used a calculator for some of that, but it's AMC so that's okay. It repeats in groups of 6. 12345 mod 6  :equiv: 3 so the answer is 9. E amirighthuhuhuhuh?[/hide]

It might help to note [hide]the recurrence relation Rn+2 = 6 Rn+1 - Rn.  That equation works because a and b are roots of the polynomial x2 - 6x + 1.[/hide]

My answer is 1. But, what was the time limit?


Let [tex]r_1,r_2,r_3,r_4[/tex] be the radii of four mutually externally
tangent circles.

Prove that [tex]\displaystyle\[\sum_{i=1}^4 \frac{2}{r_1^2} = \left( \sum_{i=1}^4 \frac{1}{r_i} \right)^2.\]
[/tex]

Does anyone have a solution?

[quote="WarpedKlown1335"](Descartes' circle theorem)
Let $r_1,r_2,r_3,r_4$ be the radii of four mutually externally
tangent circles. Prove that
[tex]\[
\sum_{i=1}^4 \frac{2}{r_1^2} = \left( \sum_{i=1}^4 \frac{1}{r_i} \right)^2.
\][/tex]

I know the theorem but I'm looking for a proof of it.[/quote]

You can find something [url=http://www.mathlinks.ro/viewtopic.php?t=352&highlight=descartes]here[/url].


geometry
geometric transformation
reflection
algebra
polynomial
number theory unsolved
number theory

Sorry this isn't really a contest problem per se, but this seemed like a good place to bring it up:

Let 0 < a, b < 1.  Evaluate this expression:

a + b*(a + b*(...))

where the ellipsis indicates (infinite) recursion.  The expression has a finite value e.g. when a = (1-r) and b = r^2 with 0 < r < 1, in which case you can think of it as an expression of the amount of light which escapes from a thin gap of air between two mirrors.  Things get even more interesting when you add more mirrors :-)

Well, assuming it converges (which is probably easy to prove with those bounds on a and b, but I'm not worried about that part), then we just have $a+bx = x$, so $x = \frac{a}{1-b}$ is the answer.

Er... right  :blush:

Here's the more general problem:

Say you shine a flashlight at a stack of n partial mirrors,  where each mirror reflects r of any light which falls on it, and lets 1-r through.  How much of the light from the flashlight makes it through all n mirrors?  Keep in mind that the mirrors reflect light in both directions.

It's pretty simple to solve for small n by mapping out the (2n-1) "states" of a light particle as it bounces around in the mirrors, e.g. for n = 2 you have:

0_right = (1-r)*1_right + r*0
1_right = (1-r)*2_right + r*2_left
1_left = (1-r)*0 + r*1_right
2_right = (1-r) + r*1_left
2_left = (1-r)*1_left + r*2_right

0_right is the state of a unit of light heading into the first mirror from the outside.  1_left is in between the first mirror and the second, heading back out the "wrong way", and 2_right is one step away from the other side of the mirrors.  Solving for 0_right gives you the answer after some hairy algebra, but I haven't found the general solution for any number of mirrors.

The idea for this problem came from an exercise in an optics textbook, which gave an answer of (1-r)^n.  That's actually the amount of light which is transmitted directly through all n mirrors, but it ignores the light which bounces around between the mirrors before escaping.

Hmm. I could solve "How much light gets through mirror number k" if there are an infinite number of mirrors by setting up a double recursion, but that doesn't quite work when there are a finite number as the recursion gets stuffed up by the end terms.. still thinking.

This is the answer I found empirically by looking at answers for N = 0,1,2,3,4,5:

\[ \frac{1-r}{(N-1)r+1} \]

No idea about any proof, though.

Ahha. Yep, that works, you can prove it by induction.

OK, It can be done inductively. Because no light is ever absorbed, the reflectivity of N mirrors with a reflectivity of r is $1 - \frac{1-r}{(N-1)r+1}$.

So these N mirrors are equivalent to [b]one[/b] mirror with a reflectivity of $1 - \frac{1-r}{(N-1)r+1}$. We can find the transmittivity of N+1 mirrors by solving a situation with two mirrors; one with reflectivity of r, and the other with reflectivity of $1 - \frac{1-r}{(N-1)r+1}$. The total transmittivity of the whole thing turns out to be $\frac{1-r}{Nr+1}$, completing the inductive step.

That's very neat.  Out of curiosity, how did you arrive at the answer?  I worked out the solution for a few n but always ended up with a quotient of polynomials which I couldn't simplify any further.

Well, I , uh, [size=75]used Maple[/size]. But it might have also helped that I actually worked with transmittivity instead of reflectivity, since that gives a numerator of just -t instead p-1. I'm guessing that'd make it a lot easier to find any common factors in the numerator and denominator.


geometry
trapezoid
calculus
integration
modular arithmetic

(1) A regular hexagon has side length 10.  What is the sum of the lengths of all the diagonals for the hexagon?  Express your answer in simplest radical form.

[hide="My Answer"]$ 60 \plus{} 72\sqrt{3}$[/hide]

[hide="The Answer"]Hm...I can't read it  :P (No, I'm not kidding.)[/hide]

(2)How many different arithmetic sequences are there with all the following properties: (1) The first term is 31, (2) The last term is 61, (3) The common difference is a whole number and (4) The total number of terms is at least 3?

[hide]7[/hide]

(3) An equilateral triangular region of side length 2 cm is cut form a corner of an equilateral triangular region with side length 6 cm.  What is the number of square centimeters in the area of the resulting trapezoid?

(4)Jill has two watches.  One gains 2 seconds per hour and the other loses 1 seconds every 2 hours.  If the clocks are set at the right time at 8:00 AM, at what time will the clocks be exactly one hour apart?

(5) the 9-digit number $ abbababa3$ is a multiple of 99 for some pair of digits $ a$ and $ b$.  What is $ b \minus{} a$

[hide="1"]
$ 3\cdot20 \plus{} 6\cdot10\sqrt3 \equal{} \boxed{60 \plus{} 60\sqrt3}$.

As a check, this encompasses all $ 3 \plus{} 6 \equal{} 9 \equal{} \dbinom62 \minus{} 6$ diagonals.
[/hide]
[hide="2"]
Our sequence is $ 31,31 \plus{} d,\ldots,61 \equal{} 31 \plus{} nd$.

Since $ n$ must be integral, $ n|(61 \minus{} 31 \equal{} 30)$ ($ 30 \equal{} 2\cdot3\cdot5$), which means that there are $ 2\cdot2\cdot2 \equal{} 8$ possible values for $ n$.

However, $ n \equal{} 1$ is invalid, since it produces the sequence $ 31,61$, with only two terms.

Our final answer is $ 8 \minus{} 1 \equal{} \boxed7$.
[/hide]
[hide="3"]
The area of an equilateral triangle of side $ s$ is $ \frac {s^2\sqrt3}4$, so the answer is

$ \frac {\sqrt3}4(6^2 \minus{} 2^2) \equal{} \boxed{8\sqrt3}$.
[/hide]
[hide="4"]
Every hours, the two clocks get $ 2 \plus{} \frac12 \equal{} \frac52$ seconds apart.

There are $ 60\cdot60 \equal{} 3600$ seconds in an hour, so it will take $ \frac {3600}{\frac52} \equal{} 1440$ hours.

$ 1440\equiv0\pmod{24}$ and $ 1440 \equal{} 60\cdot24$, so this will happen after $ 60$ days at $ 8: 00$ AM.
[/hide]
[hide="5"]
This number must be divisible by $ 9$ and $ 11$.

To be divisible by $ 11$, we must have

$ 3 \minus{} a \plus{} b \minus{} a \plus{} b \minus{} a \plus{} b \minus{} b \plus{} a \equal{} 3 \minus{} 2a \plus{} 2b\equiv0\pmod{11}$, or

$ 2b \minus{} 2a\equiv \minus{} 3\equiv8\pmod{11}\implies b \minus{} a\equiv4\pmod{11}$.

Since $ 0\leq a,b\leq9$, we can see that $ b \minus{} a \equal{} \boxed4$ is the only possible answer.
[/hide]


[quote="benjalizana"]Hello, im am writing to ask if there is a summation formula for
1^n +2^n +3^n + . . . + k^n =?
I do not know how to find it, i would appreciate any help that includes proof and formula. Thanks
Please reply as fast as possible.

Ben

PD: My email is benjalizana@gmail.com and msn benjamin_lizana@hotmail.com
 Aim's Screen Name: benjalizana[/quote]

I know the formulas for $n=1,2,3$:
$n = 1 \quad \frac{n(n+1)}{2}$

$n = 2 \quad \frac{n(n+1)(2n+1)}{6}$

$n = 3 \quad \left( \frac{n(n+1)}{2}\right)^{2}$

I don't know if someone could generalize this... They are all fairly simple to prove with induction. But, I don't know how to derive them.

[quote="benjalizana"]Hello, im am writing to ask if there is a summation formula for
1^n +2^n +3^n + . . . + k^n =?
I do not know how to find it, i would appreciate any help that includes proof and formula. Thanks
Please reply as fast as possible.

Ben

PD: My email is benjalizana@gmail.com and msn benjamin_lizana@hotmail.com
 Aim's Screen Name: benjalizana[/quote]

Bleh: http://en.wikipedia.org/wiki/Summation#Identities

See this pdf on the [url=http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf]sum of the $k^{th}$ powers[/url].  See page 45 to skip directly to your question.

yes i found it up to n=4 but cant find a relation :S Help Please

The formula was on the link I sent you...

[quote="binomial_4eva"]easy
$\sum_{i=1}^{n}n^{n}=\frac{n^{n}-1}{n-1}$ or $\frac{n^{r}-1}{r-1}$[/quote]

Isn't that for geometric series? That wouldn't work here...

right, the thing is its not n to the power of n, but n to any power different from n, lets say K.. The formula in that pdf.. or on wikipedia seems to be the answer, but i cant really understand the proof behind it; how and why does newtons binomial theory apply to this problem?? how does "Bk", the kth Bernoulli number relate to this problems solution..also in that pdf page 46 i do not understand what "fk(n + m)" Please help as fast as posible.

Thanks

[quote="vishalarul"][quote="binomial_4eva"]i think ur rite but no wait i think i am rite coz the formula for this sequence is $n^{n}$ isnt it ? so therefore its the sum of the geometric seq. $U_{n}=a(r)^{n-1}$ $s_{\infty}=\frac{1-r^{n}}{1-r}$ with $a=1$ $r=n-1$[/quote]

This:
$1^{1}, 2^{2}, 3^{3}, ..., N^{N}$
is [b]NOT[/b] a geometric series!

There is no common ratio for this. In a geometric series (strictly speaking, sequence), there is always a common ratio. For example,
$1, 2, 4, 8, 16, ...$
has a common ratio of $2$.
If you still don't believe me, look at this:
$2^{2}/1^{1}= 4$
$3^{3}/2^{2}= 27/4$
$4^{4}/3^{3}= 256/27$
Clearly, there is no common ratio.[/quote]
But still that isn't what the problem is asking for.
http://mathworld.wolfram.com/PowerSum.html

[quote="benjalizana"]..also in that pdf page 46 i do not understand what "fk(n + m)" Please help as fast as posible.[/quote]

The notation you're asking about is mentioned in the first couple of pages.  Our specific series is mentioned on page 7.  $f_{k}(n)$ denotes a sum of consecutive integers from $1$ to $n$, each of which are raised to the power $k$.  For example, $f_{2}(5) = 1^{2}+2^{2}+3^{2}+4^{2}+5^{2}$.

From Wikipedia:
$\sum_{i=0}^{n}i^{p}= \frac{(n+1)^{p+1}}{p+1}+\sum_{k=1}^{p}\frac{B_{k}}{p-k+1}{p\choose k}(n+1)^{p-k+1}$

    where Bk is the kth Bernoulli number.

Here's the exact article: http://en.wikipedia.org/wiki/Faulhaber%27s_formula


hey guys i'm sorry to bother you with this thing, but i just can't find a topic i've been looking for. it's the one where one posts a problem of an IMO prep level, and the one who solves it - posts the next.
thanks a lot!

You may refer to the following link. Hope that helps.

http://www.mathlinks.ro/viewtopic.php?t=288069

Regards,

Very funny.

He wants IMO problems, not MATHCOUNTS...
I don't know if it exists, but look at Olympiad section, you'll find many problems...


Could someone please tell when the JBMO 2007 will be hosting ?
Also there is a site for this ?Thank you

The jBMO 2007 is held in Shumen, Bulgaria, on 25-30 June (these days). The official site of the competion is [url]http://www.jbmo.eu/[/url].


algebra
polynomial
number theory proposed
number theory

Let$P(x)$  be a polynomial,such that :
 $P(x)\equiv a \mod(x-a)$
$P(x)\equiv b \mod(x-b)$
$P(x)\equiv c \mod(x-c)$
find$d$ such that : $P(x)\equiv d \mod(x-a)(x-b)(x-c)$

Isn't it obviously $x$?
Of course if $a,b,c$ are not pairwise unequal, you can not determine, but in the other case it is obviously $x$.


A man is standing on the platform of a train station. As soon as  a train pulls out of the station, he visualises a lady  aboard it waving. He runs over to the train, gets on it and kills the lady. Thereafter, he gets off the train. Then, he goes to the nearest polce station and confesses to murdering the lady.

However, he is able to  walk away from the police station without consequence.

How is this possible?

[hide]

If he only "visualises" the lady, then she isn't real and he's insane?

 [/hide]

i guess its pretty clear that the woman was most wanted. then, the man confesses and the police dont give him a reward for kiling her, but let the man walk away safe.

Hey people. you seem to misunderstand these riddles:
you shall not show how cool you are (by giving what you heard whereever else), but you should ask no-or-yes-questions which the author is supposed to answer.

[quote="K Sengupta"]The solution to the puzzle is as follows:

[hide]The lady was the man's wife. Years before she faked her own death and framed her husband. Her husband was convicted and served time for murder. In the country where this occurred, a person can not be prosecuted for the same crime twice. Since he had already served his time, he could not be sentenced again.[/hide][/quote]

K Sengupta, just a question..if that is the actual solution, how are we supposed to get that kinda far fethched answer from the puzzle and teh information that was provided to us?? just wondering, cuz don't riddles have to some contexual clues to obtain the answer? i don't mean to be disrespectful, i was just simply curious... :?:  :maybe:

As I already wanted to point out, this is a lateral (or however it's called in english), thus normally all are allowed to ask the person who gave the riddle questions; as a restriction, these questions have to have "yes" or "no" as only answers (possibly allowing finite casechecking, too, since that could always be realised by yes-no-questions).

That's a very good point. There should be some kind of rule saying if you already know the answer, don't spoil the problem for everyone else.

[hide]The lady is a terrorist/criminal/dangerous maniac!! lol :D [/hide]
or
[hide]the lady isn't real..she's just his imagination...and the police have him registered as a mental asylum patient[/hide]
dunno yaar...


[url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamoqual.shtml[/url]

I guess here is the preliminary list. A lot from my state this year, but it seems TJHSST and Vestivia Hills didn't get as many as they usually do... Exeter had like a freakin WALL of qualifiers  :P Stuy did pretty awesome as usual....

Good luck to all on USAMO!! :)

Thanks!!

Wow, I think I have met everyone from my state that made USAMO.

Wow, you're right, it's already up! Congratulations to all qualifiers!  Good luck on USAMO!

Summary:

11/12 Index = 204.0

Floor = 6

10 or below index (with floor) = 202.5

[b]NOTE[/b]: Some people who took the AIME II are not on the list YET but will qualify. That's why it is "preliminary."

Yay I'm on the list. Which means I made less than 3 bubbling mistakes on the AIME, if any.  :P
I recognize many names...lol. If I counted correctly, there are around 89 people from California who qualified. Yee.

Heh, to think that for the last few weeks I've been thinking that my 216 index was borderline . . .

Anyway, congratulations to everyone who qualified (including me :D )!

I think it's pretty awesome that the only states with more qualifiers than MA are CA (obviously), and NY, which only has like 3 more. Maybe there are other states that I misssed, but I'm pretty sure those are the only ones.

I should have qualified (AMC 10:126, AIME II:9), but I'm not on the list. This might be because on my AIME answer form, I put that my AMC 12 score was a 109.5.

Also, I know that the forms for my school have arrived, for I can see that one person from my school has already made USAMO.

Or it might be that

[quote="AMCDirector"]There are still a few (200 or less) AIME II answer forms to come in yet and be scored[/quote]

[quote="knexpert"]I should have qualified (AMC 10:126, AIME II:9), but I'm not on the list. This might be because on my AIME answer form, I put that my AMC 12 score was a 109.5.[/quote]

For you and anyone else who believes s/he may have qualified,

Please email the following information to amcinfo@maa.org after April 13th:

1) Your name.
2) Your school's name and CEEB number.  If you took the contests at more than one location, include both.
3) Your address.

As has been mentioned before the score you bubble on the answer form is just an initial identity check; it does not serve as the qualification score.

As E^(pi*i)=-1 says, we may not have received your school's answer forms, yet, and therefore you may appear later this week.

Elizabeth Claassen
AMC Senior Accounting Clerk
(and resident number cruncher)

Crap. My index=202, so I was two points off.

Actually, you were .5 points off.  Hope that makes you feel somewhat better, Eric. :)

Then again, it might not. :maybe:

But just feel good about the fact that you're still one of the most pwnzinating ARML kids in GA. :D

Hmm I wonder which state had the most qualifiers? I think it was CA...IL had a largeish number but not that many.

[quote="calc rulz"][b]NOTE[/b]: Some people who took the AIME II are not on the list YET but will qualify. That's why it is "preliminary."[/quote]

any description of those people?

just the 200 answer sheets who havn't been scored yet?

[quote="pythag011"]144/9. On to blue mop.

I need remedial additon.[/quote]
Whoa...6th grader aiming for MOP...I wish you luck.

Finally in, in my last chance.

(I'm the only one from IL outside Chicago, other than Alex Zhai)

I count as [i]at least [/i] half Illinois-ese, Aryth. So 1.5 Illinois freshmen this year =P.

[quote="Aryth"]haha, what a true illinoiser.
"in Chicago" meaning within two hours driving distance  :P
yay only freshman from il. no eighth graders this year. troubling.[/quote]

Eh, it's all the same up there when you live down near St. Louis like I do.

When do the official lists come out??? (if they change something)

When are you officially imformed through email?

[quote="firecricket91"]When are you officially imformed through email?[/quote]

You'll be emailed by tomorrow, according to the announcement.

[quote="gpl_coll4ever"]it's unfair!!!
if a junior took amc12 and he/her index was 203,he/she would not qualify to usamo! but the one who took amc10 (and his/her index was 203 too) is qualified !!!  :mad:[/quote]

That is false no matter how you look at it. A "junior" meaning 11th grade can not take the AMC 10. If you meant an underclassman (10th grade or below), then a 203 index qualified regardless of which test they took. It is unfair because it's easier to get a 203 with the AMC10 but not because of what you said.

[quote="gpl_coll4ever"]i am so sad.
i am in 10 grade , took amc12a(108) and 12b(118.5) at two different schools, and took aime at my high school where i got a lower score. so they added my aime score(9) to 108, and i am not qualified to usamo.[/quote]

You will still qualify.  The score reports sent to the schools will typically report scores from one school - not those taken at other schools.  The AMC folks have no easy way of knowing how to match your AIME score to the 12b score unless you reported the higher score on the AIME information form.  If you aren't on the qualifier list posted on the AMC website yet, call the AMC office and explain the situation.

i am very happy to hear that.
a teacher told me i had to try next year.
thank you very much!!!

gt59 wrote:
[quote] If you aren't on the qualifier list posted on the AMC website yet, call the AMC office and explain the situation.[/quote]

gpl_coll4ever (and others in the same circumstances):  Remember that the AMC has requested that you [b]WAIT until AFTER April 13 before you contact [/b]them to inquire about this.

And the AMC has requested that you [b]email [/b]them, NOT call them.

See excerpt from Dave Patrick's posting below:

[quote]Elizabeth Claassen of the AMC posts the following information for those people who think that they should be on the USAMO qualifiers list, but are not listed:
[quote]
For you and anyone else who believes s/he may have qualified,

Please email the following information to amcinfo@maa.org [b]after April 13th[/b]:

1) Your name.
2) Your school's name and CEEB number. If you took the contests at more than one location, include both.
3) Your address.[/quote]
I have added the emphasis on "after April 13th". They are still processing some scores and rechecking some others. Please let them finish this task before contacting them. There is still plenty of time to get properly registered for the USAMO if indeed you qualified but are not yet listed.[/quote]

Only freshman in Ohio...  :thumbup:  Hope to make it to Red MOP


Rumors say that MOP is expanding, who's got the details. Post 'em here please.

Thank you,
IW.OAV

My information is limited, but yes, the current plan is to expand MOP to about 60 people in 2004.

I believe 2002 was because of excessive sponsorship. I'm guessing that any further expansion will not be open to seniors, based upon the 2002 case and that under the current system the non-winners who attend are non-senior. Reason for this is probably because while it makes sense to accept all 12 potential IMO-quals for that year, any seniors who are not USAMO winners can't qualify for IMO the following year and thus it doesn't help the US team's future to accept nonwinner seniors, whereas students from lower grades can someday become a member of the team... thus it makes more sense to give them adequate training...

[quote="Ripper1"]I heard that MOP in 2002, the large one was for everyone who wasn't a senior and seniors who did very well.  Why is that the case? I am a senior and if the chances of making it to MOP were made a bit easier to achieve then that will motivate me to work harder.  If MOP is expanded, does it make it easier for anyone to get there or usually non-seniors? Ok thanks[/quote]

In 2002, the program was expanded.  But it didn't include [i]everyone[/i] who was a senior, only the top 120 or so, I believe.  In order to qualify for MOP, you had to be in the top 12 (senior or not) or you had to be a non-senior (preferably a young student) with a medium-rate score.

If MOP expands next year, it won't be to the same magnitude as in 2002.

In 2002, the top 12 were invited, and then every non-senior who took the USAMO was also invited, regardless of score. So some of the people at MOP 2002 got 0s on the USAMO.

oh sigh...if only they expanded that much this year

Oh...I didn't know that.  Oy, if only they'd do that again...

Hmm, I need to find a way to shave a couple years off my age.

How do you "shave" off a couple of years? :)
Maybe :evil: can help you... :)

So does anyone have more updated info about this MOP expansion? We're a ways into the school year, but the AMC website has not stated any expansion of the summer program.

I am a member of an advisory committee to the CAMC and while we all agreed that the expanded 2002 MOP was a good thing that we'd like to see repeated, this did not seem likely at our meeting last March.  As other people have indicated, the large 2002 MOP was due to very generous donations that year, primarily one large donation from Akamai if I remember correctly, and the current economic climate made us feel that such donations were unlikely any time soon.  For those who have heard "rumors" of an expanded MOP this year (Eve, IWoAV), where are you hearing them?  I'm very interested in learning more about this.  I agree with the general opinion that most of the expansion would be for non-seniors who might end up on future IMO teams.

I corresponded with Steve Dunbar, the AMC Director, about this around 3 months ago.  He indicated that they received more funding from Akamai & would be expanding MOP next summer by "about 30".  (TomC - I'm guessing this all came to fruition after that meeting in March.)  The selection criteria were not set at the time, but he suggested they would be included with the AMC-10/12 packets that your schools should receive this month.

Hi -- I'm here to answer a few questions and put to rest some rumors.

Yes, MOSP 2004 will have room for about 60 students, an expansion of "about 30" from our 2003 size of "about 30."

Richard is correct, this did come to fruition in about June-July of last summer, several months after the AMC Advisory Board meeting, and the provisional advice and input of the Advisory Board at that meeting was used in crafting the agreement during summer 2003.

Final decisions about the composition of the MOSP 2004, (i.e. who will be invited and under what criteria) have not been made at this moment, although a proposal has been circulated to our policy-proposing MAA Committee on the American Mathematics Competitions (CAMC) and the (financially advising and personnel managing) MAA Executive Committee.  The proposal should be discussed face-to-face at the Joint Mathematics Meetings in January and finalized shortly thereafter.  I will publicize the decisions through our website, and in the AIME-USAMO Teachers' Manual.

The reasons for inviting only non-seniors (except for the 12 USAMO Finalists) given by complex zeta, mathfanatic and J. Lin are essentially correct.

I would be interested in hearing ideas form this group about what should be the criteria for inviting "about 30" more students to MOSP 2004.  Remember that the decision making rests with several committees, and that not all your suggestions may be accommodated for practical and financial reasons which are beyond the scope of this forum!

Finally, please remember that our decision-making and planning cycle takes place on an (annual) timescale which is much longer than a volatile economic time scale (~3 months) which in turn is much longer than the information time scale of the Internet (~minutes).  That can be frustrating for everyone, including me!

Steve Dunbar, AMC Director

wow, thats cool, though im definitely not making this year... maybe in my junior year if it stays this size, i have a chance.

thats too bad for seniors... but i guess that they've had enough tries already :P

Maaaaaan, what a rip...

There goes my chance at MOP

Thanks a lot Jongmin Baek, pffft.

:P

What did Jongmin do? He's a good friend of mine, by the way.


Is it possible [b]not[/b] to qualify for the AIME with a 120. On the 2003 AMC10B, the AIME qualification score was [b]121+[/b], why is that? If I make a 120 in 2010, will I be guaranteed to move on to the AIME?

I depends on the how easy/hard the AMC was that year. But you can rest assured that the cutoff score won't deviate more than $ 5$ points from $ 120$.

Back in 2003, there was no rule that a 120 on the AMC 10 was guaranteed to make AIME.  Now there is such a rule.

If you take the AMC 10 practice test for that year, you may find that that test was relatively easy.  (Like I did. :))

This is a somewhat related thread:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321425

RaviB is correct, I checked the AMC 10s and from 2000-2003 the AIME qualification level from the AMC 10 was set at the top 1%.  After 4 years of experience the contest creation committee members felt that we had enough experience in setting the test that we could use a fixed level of 120 (or the top 1%, whichever is lower.)

If you have the 21st Century CD, look at the fine print on the covers of the AMC 10s for those years to see for yourself.

-- Steve Dunbar
MAA Director, American Mathematics Competitions


We have :

 $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{c+a} +1= \frac{a}{b+c}
+\frac{b}{a+c}+\frac{c}{a+b}$

So our inequality is equal to:

 $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \geq 3$

and it is easy from AM-GM.

here is a stronger form with condition $a+b+c \ge \frac{1}{5}$ with Perfect proof of darij .
 [url]http://www.mathlinks.ro/Forum/topic-51000.html[/url]
As the source you wrote :My Book ,quite enjoying yourself with these things ha ? :|  :|

\[\sum\frac{a^2+b}{b+c}=\sum\frac{a^2+ab+b(b+c)}{b+c}=\sum\frac{a(a+b)}{b+c}+\sum{a}=\\\sum\frac{(a+b)-(a+b)(b+c)}{b+c} +1=\sum\frac{a+b}{b+c} -(\sum{a+b}) +1=\\\sum\frac{a+b}{b+c}-1\geq 3-1=2\] where the last inequality holds by AM-GM

[quote="nttu"]This problem is in Old and New ineqs .
  The solution showed in that book is not nice (it's just my idea) . This one is nicer :
  $ \frac{a^2+b}{b+c}+a = \frac{a^2+b}{1-a}+a = \frac{a+b}{b+c} $
  It's enough to prove : $ \frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \geq 3 $ , which is true by AM-GM[/quote] This is exactly [i]xtar[/i]'s solution.  ;)

[quote="nttu"]This problem is in Old and New ineqs .
  What's the book??

because the ineq is homogenous we can give the problem without the condition $a+b+c=1$ Sure?

[quote="Makaveli"][quote="ALGEBRA"]let a,b,c>0 and a+b+c=1 prove that this inequality
$\frac{a^2+b}{b+c}\ + \frac{b^2+c}{c+a}\ + \frac{c^2+a}{a+b} \geq\ 2 $[/quote]
After taking common denominator, making cross product and cancelling common terms on LHS and RHS the inequality reduces to a^3+b^3+c^3>=a*(b^2)+b*(c^2)+c(a^2)
whic follows directly from rearrangement inequality..Namely, suppose a>=b>=c
(a^2)*a+(b^2)*b+(c^2)*c>=(a^2)8c+(b^2)*a+(c^2)*b[/quote]
  No, I think you can't suppose $ a \geq b \geq c $
 This ineq $ a^3+b^3+c^3 \geq ab^2+bc^2+ca^2 $ can be easily solved by AM-GM :
 $ a^3+b^3+c^3 \geq 3ab^2 $

I have a question . which site I can see the problem of Romanian mathmatic olympiad and the problem of Romanian olympiad  team?

well , you can find it in  "www.ajorza.org"

[quote="ALGEBRA"]let a,b,c>0 and a+b+c=1 prove that this inequality
$\frac{a^2+b}{b+c}\ + \frac{b^2+c}{c+a}\ + \frac{c^2+a}{a+b} \geq\ 2 $[/quote]
Expanding gives $\sum_{cyc}(a^4+a^3b-a^2b^2-a^2bc)\geq0,$ which is obviously true.

The book referred to earlier is either  [i]Old and New Inequalities [/i]by T. Andreescu, V. Cirtoaje, G. Dospinescu, and M. Lascu  or  [i]Old and New Inequalities, Volume 2[/i] by Vo Quoc Ba Can and Cosmin Pohoata.  Both are published by GIL Publishing House, Zalau, Romania (2004, 2008, respectively)

[quote="ALGEBRA"]let a,b,c>0 and a+b+c=1 prove that this inequality
$\frac{a^2+b}{b+c}\ + \frac{b^2+c}{c+a}\ + \frac{c^2+a}{a+b} \geq\ 2 $[/quote]

where did I go wrong here?

by cauchy, $\sum_{cyc} \dfrac{a^2+b}{b+c} = \sum_{cyc} \dfrac{a^2}{b+c} + \dfrac{b^2}{b^2+bc} \ge \dfrac{(a+b+c)^2}{2(a+b+c)} + \dfrac{(a+b+c)^2}{a^2 + b^2 + c^2 + ab + bc + ac} = \dfrac{1}{2} + \dfrac{1}{(a+b+c)^2 - ab - bc - ca}= \dfrac12 + \dfrac{1}{1-ab-bc-ca}$

What remains to be proven is that $ab + bc + ca \ge \dfrac13$ if $a+b+c = 1$ which is ... false? because by rearrangement, $a^2 + b^2 + c^2 \ge ab + bc + ca$ so $(a+b+c)^2 \ge 3ab + 3bc + 3ca$ or $\dfrac{1}{3} \ge ab + bc + ca$

Dear AkshajK, that is because i think the inequality $ \sum \frac{b}{b+c} \ge \frac{3}{2}$ isn't true for all positive numbers

[quote="AkshajK"][quote="ALGEBRA"]let a,b,c>0 and a+b+c=1 prove that this inequality
$\frac{a^2+b}{b+c}\ + \frac{b^2+c}{c+a}\ + \frac{c^2+a}{a+b} \geq\ 2 $[/quote]

where did I go wrong here?

by cauchy, $\sum_{cyc} \dfrac{a^2+b}{b+c} = \sum_{cyc} \dfrac{a^2}{b+c} + \dfrac{b^2}{b^2+bc} \ge \dfrac{(a+b+c)^2}{2(a+b+c)} + \dfrac{(a+b+c)^2}{a^2 + b^2 + c^2 + ab + bc + ac} = \dfrac{1}{2} + \dfrac{1}{(a+b+c)^2 - ab - bc - ca}= \dfrac12 + \dfrac{1}{1-ab-bc-ca}$

What remains to be proven is that $ab + bc + ca \ge \dfrac13$ if $a+b+c = 1$ which is ... false?[/quote]

[b]AkshajK[/b], because your estimations were weak.


combinatorial geometry
combinatorics proposed
combinatorics

Do there exist two disjoint infinite sets A and B of points in the plane such that: (1) no three points in the union of A and B are collinear; (2) the distance between any two points in the union of A and B is at least 1; (3) given any three points in B, there is a point of A in the triangle formed by the three points; (4) given any three points in A, there is a point of B in the triangle formed by the three points? A triangle means all points in or on the triangle (the vertices, all points on the sides of the triangle and all points in the interior of the triangle).

You simply need to find 5 points in A such that their convex hull does not contain any other points from A. Since A is discrete this is not difficult.
When you have this points  suppose the convex hull is 5-gon. Divide it into 3 disjoint triangles. You have point in B in each of this triangles. This gives you triangle formed by three points from B. So there is point in A in it. Contradiction.
It is the same when convex hull is quadrilateral or triangle.


4. Whats six times one-seventh?
This is the official number four in the chapters.
How many seconds did it take you to do that?

I disagree, I believe #3 on the Chapter Team Round was easier, I spent 2 minutes on that looking for pitfall, however, I just moved past this problem as there's no extra words that could've trapped you.

Link to 2006 Chapter Team Round:
http://www.mathcounts.org/webarticles/articlefiles/510-06CTeam.pdf

yes, i agree with Ignite168 on this one.  Our strategy was to assign everyone on our team 2 problems and do the last two together and i was assigned #3.   :D thats how i remember it

Yeah, that was one of them too, it was hard because it was so easy, I got confused for a minute.

I still have to say 6* 1/7 is easier, it tells you the answer... And less reading.

i say #4, 6*1/7 was easier. i have to be the slowest reader here and i hate "interpreting" graphs. i remember it. didn't take long.

In 2005 problem #1 on chapter was to find the distance between cities A and C given a mileage chart for cities A,B,C,D.

#1 at state was the same mileage chart except calculate the probability that two cities are more than 7000 mi apart.

I have to say that it is sadly pathetic...

[quote="kyyuanmathcount"]Yeah, I was appaled.

It took me 20 seconds because I was searching for some pitfall, I really didn't believe there was a problem THAT easy.[/quote]
Same here.  :D  :D :D

I didn't even think when I did that problem. :spider:  :icecream:

#3 of the team competition was a piece of cake! :icecream:

I read it 3 times to check for any tricks [hide=Will put something in here after we are allowed to discuss state][/hide]


[quote="mathservant"]f(xy/3) = f(x/3) + f(y)   then find  f(101)=?[/quote]
What is the domain of $f$?
If $f$ is defined on the whole set of reals, then let $x=0$. We get $f(0)=f(0)+f(y)\Rightarrow f(y)=0\forall y\in\mathbb{R}\Rightarrow f(101)=0$
If $f$ is only defined on positive reals, then by letting $x=3$, we get $f(y)=f(1)+f(y)\Rightarrow f(1)=0$. Substituting $a=\frac{x}{3}$ and $b=y$ in the given condition yields $f(ab)=f(a)+f(b)$. Since $f$ is continous at one point, it is continous everywhere, and so the solution to this Cauchy equation is $f(x)=c\log x$ where $c$ is a constant. Therefore $f(101)=c\log 101$. I think we don't have enough conditions to work out $c$ here... :(


ABC is an equilateral triangle with sides equal to 2 cm. Line BC is extended its own length to D, and E is the midpoint of Line AB. Line ED meets Line AC at F. Find the area of quadrilateral BEFC in square centimeters in simplest radical form.

[hide]Draw segment AD. We now have triangle ABD with medians AC and DE drawn. Now draw the thrid median from B to side AD at point Z. We have all medians drawn, and the triangle is divided into six triangles of equal area. So to find the area of BEFC, we need to find the area of two of the equal area triangles, or 2/3 of the area of ABC. \[\sqrt{3}\times \frac{2}{3}=\boxed{\frac{2\sqrt{3}}{3}}.\] [/hide]


function
Ross Mathematics Program
logarithms
calculus
derivative
inequalities

the followings are the standard way of proving [code]e^pi>pi^e[/code] :

1) By considering the maximality of the functions [code]x^(1/x)[/code] or [code]x/ln(x)[/code].
2) Another indirect proof given by Ross Honsberger in his book "Mathematical Morsels".

I am required at least 3 more independent proofs of the above inequality. Is there way to prove it directly from
[code]e^pi=1+pi+(pi^2/2!)+(pi^3/3!)+....[/code] ?

Can anyone help ? :!:  :(  :(  :?:

hello, let $ f(x)=\frac{x}{\ln(x)}$ then it is easy to show that $ f(x)$ has a global minimum ar $ x=e$, so we get since $ \pi>e$ implies ${ \frac{\pi}{\ln(\pi)}}>e=f(e)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, let $ f(x) = \frac {x}{\ln(x)}$ then it is easy to show that $ f(x)$ has a global minimum ar $ x = e$, so we get since $ \pi > e$ implies ${ \frac {\pi}{\ln(\pi)}} > e = f(e)$.
Sonnhard.[/quote]

I think that's the method Shanku is referring to in #1.

This is probably not a completely independent proof, but here goes:

We can prove the result that $ e^x > x^e$ for $ x > e$.  There exists $ k > 1$ such that $ x = ke$; we therefore must prove that $ e^{ke} = (e^e)^k > (ke)^e = (k^e)(e^e)$.  This is the same as requiring $ (e^e)^{k - 1} = (e^{k - 1})^e > k^e$; taking the ln of both sides, $ e^{k - 1} > k$.  Substituting $ g = k - 1$, we require $ e^g > g + 1$.  We know that $ e^0 = 0 + 1$, so by comparing derivatives, it is easy to prove that the inequality $ e^g > g + 1$ holds for $ g > 0$.  So $ e^x > x^e$ for $ x > e$, as needed.  Since $ \pi > e$ this concludes the proof.

Yes, Sonnhard's proof is the one I've already referred.
Thanx mqtrinh, though not entirely independent, I can consider your approach as a new proof because of the methodology. Pl keep me posted in case any other is found.

Nothing New Andrew.
I wonder whether these are the only ways to prove it.


let {a(n)} satisfies:
1.a(0)=a (a is an integer,a>=2)
2.if a(n) is a square number then a(n+1)=[a(n)]^(1/2)
3.if a(n) is not a square number then a(n+1)=a(n)+b.Here b is a odd prime.(a,b)=1

prove or disprove that we can find a number M such that for all n>M we have a(n+1)=a(n)+b

Oh,sorry.A mistake......
and what about "when a b are both odd primes"?
just a guessing of my friend.